Reinforced concrete design 9780190269807, 9780190647049, 9780190269852, 0190269804

Citation preview

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REINFORCED CONCRETE DESIGN

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REINFORCED CONCRETE DESIGN CHU-​K IA WANG CHARLES G. SALMON JOSÉ A. PINCHEIRA GUSTAVO J. PARRA-​M ONTESINOS University of Wisconsin–​Madison EIGHTH EDITION

New York  Oxford OXFORD UNIVERSITY PRESS

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Oxford University Press is a department of the University of Oxford. It furthers the University’s objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and certain other countries. Published in the United States of America by Oxford University Press 198 Madison Avenue, New York, NY 10016, United States of America. © 2018 by Oxford University Press © 2007 by John Wiley & Sons, Inc. © 1997 by Addison Wesley Publishing Company For titles covered by Section 112 of the US Higher Education Opportunity Act, please visit www.oup.com/​us/​he for the latest information about pricing and alternate formats. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by license, or under terms agreed with the appropriate reproduction rights organization. Inquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above. You must not circulate this work in any other form and you must impose this same condition on any acquirer. Library of Congress Cataloging-​in-​Publication Data Names: Wang, Chu-Kia, 1917–author. Title: Reinforced Concrete Design / Chu-Kia Wang, Charles G. Salmon, José A. Pincheira,   Gustavo J. Parra-Montesinos. Description: New York: Oxford University Press, [2018] |   Includes bibliographical references and index. Identifiers: LCCN 2017000252 | ISBN 9780190269807 (hardcover) |   ISBN 9780190647049 (looseleaf) | ISBN 9780190269852 (eISBN) Subjects: LCSH: Reinforced concrete construction. Classification: LCC TA683.2 .W3 2018 | DDC 624.1/8341—dc23 LC record available at https://lccn.loc.gov/2017000252 987654321 Printed by Edwards Brothers Malloy, United States of America

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CONTENTS IN BRIEF Preface  xxi About the Authors  xxv Conversion Factors  xxvii

  1 INTRODUCTION, MATERIALS, AND PROPERTIES    2 DESIGN METHODS AND REQUIREMENTS 

1

31

  3 FLEXURAL BEHAVIOR AND STRENGTH OF BEAMS    4 T-​S ECTIONS IN BENDING 

43

94

  5 SHEAR STRENGTH AND DESIGN FOR SHEAR    6 DEVELOPMENT OF REINFORCEMENT 

110

176

  7 ANALYSIS OF CONTINUOUS BEAMS AND ONE-​WAY SLABS  239   8 DESIGN OF ONE-​WAY SLABS 

254

  9 DESIGN OF SLAB–​B EAM–​G IRDER AND JOIST FLOOR SYSTEMS  266 10 MEMBERS IN COMPRESSION AND BENDING 

321

11 MONOLITHIC BEAM-​C OLUMN CONNECTIONS  12 SERVICEABILITY 

403

13 SLENDERNESS EFFECTS ON COLUMNS 

463

380

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14 STRUT-​A ND-​T IE MODELS—​D EEP BEAMS, BRACKETS, AND CORBELS  528 15 STRUCTURAL WALLS 

569

16 DESIGN OF TWO-​WAY FLOOR SYSTEMS  17 YIELD LINE THEORY OF SLABS  18 TORSION 

622

720

748

19 FOOTINGS 

799

20 INTRODUCTION TO PRESTRESSED CONCRETE  21 COMPOSITE MEMBERS AND CONNECTIONS  Index  949

844

897

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CONTENTS Preface  xxi About the Authors  xxv Conversion Factors  xxvii

1 INTRODUCTION, MATERIALS, AND PROPERTIES  1.1

Reinforced Concrete Structures  1

1.2

Historical Background  2

1.3 Concrete 4 1.4 Cement 5 1.5 Aggregates 6 1.6 Admixtures 6 1.7

Compressive Strength  9

1.8

Tensile Strength  12

1.9

Biaxial and Triaxial Strength  14

1.10 Modulus of Elasticity  14 1.11 Creep and Shrinkage  16 1.12 Concrete Quality Control  18 1.13 Steel Reinforcement  19 1.14 Fiber-​Reinforced Concrete 

25

1.15 Units  26 Selected References  26

2 DESIGN METHODS AND REQUIREMENTS  2.1

Structural Design Process—​General  31

2.2

ACI Building Code  31

31

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2.3

Strength Design and Working Stress Methods  32

2.4

Working Stress Method  33

2.5

Strength Design Method  33

2.6

Safety Provisions—​General 

2.7

Safety Provisions—​ACI Code Load Factors and Strength Reduction Factors  36

2.8

Serviceability Provisions—​General 

2.9

Serviceability Provisions—​ACI Code  39

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2.10 Handbooks and Computer Software  39 2.11 Dimensions and Tolerances  40 2.12 Accuracy of Computations  41 Selected References  41

3 FLEXURAL BEHAVIOR AND STRENGTH OF BEAMS 

43

3.1

General Introduction  43

3.2

Flexural Behavior and Strength of Rectangular Sections  44

3.3

Whitney Rectangular Stress Distribution  47

3.4

Nominal Flexural Strength Mn—​Rectangular Sections Having Tension Reinforcement Only  48

3.5

Balanced Strain Condition  51

3.6

Tension-​and Compression-​Controlled Sections  52

3.7

Minimum Tension Reinforcement  58

3.8

Design of Rectangular Sections in Bending Having Tension Reinforcement Only  60

3.9

Practical Selection for Beam Sizes, Bar Sizes, and Bar Placement  64

3.10 Nominal Flexural Strength Mn of Rectangular Sections Having Both Tension and Compression Reinforcement  72 3.11 Design of Beams Having Both Tension and Compression Reinforcement  78 3.12 Nonrectangular Sections  84 3.13 Effect of As, As′ , b, d, fc′ , and fy on Flexural Behavior  86 Selected References  88 Problems  88

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4 T-​S ECTIONS IN BENDING 

94

4.1 General 94 4.2

Comparison of Rectangular and T-​Sections  95

4.3

Effective Flange Width  95

4.4

Nominal Moment Strength Mn of T-​Sections 

4.5

Design of T-​Sections in Bending  105

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Selected References  108 Problems  108

5 SHEAR STRENGTH AND DESIGN FOR SHEAR 

110

5.1 Introduction 110 5.2

Shear Stresses Based on Linear Elastic Behavior  111

5.3

Combined Normal and Shear Stresses  113

5.4

Behavior of Beams without Shear Reinforcement  114

5.5

Shear Strength of Beams without Shear Reinforcement—​ACI Approach  119

5.6

Function of Web Reinforcement  122

5.7

Truss Model for Reinforced Concrete Beams  125

5.8

Shear Strength of Beams with Shear Reinforcement—​ACI Approach  128

5.9

Deformed Steel Fibers as Shear Reinforcement  129

5.10 ACI Code Design Provisions for Shear  130 5.11 Critical Section for Nominal Shear Strength Calculation  135 5.12 Shear Strength of Beams—​Design Examples  136 5.13 Shear Strength of Members under Combined Bending and Axial Load  146 5.14 Deep Beams  151 5.15 Shear Friction  152 5.16 Brackets and Corbels  157 Selected References  168 Problems  172

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6 DEVELOPMENT OF REINFORCEMENT 

176

6.1 General 176 6.2

Development Length  177

6.3

Flexural Bond  179

6.4

Bond Failure Mechanisms  180

6.5

Flexural Strength Diagram—​Bar Bends and Cutoffs  182

6.6

Development Length for Tension Reinforcement—​ACI Code  185

6.7

Modification Factors ψt, ψe , ψs, and λ to the Bar Development Length Equations—​ACI Code  190

6.8

Development Length for Compression Reinforcement  194

6.9

Development Length for Bundled Bars  195

6.10 Development Length for a Tension Bar Terminating in a Standard Hook  195 6.11 Bar Cutoffs in Negative Moment Region of Continuous Beams  198 6.12 Bar Cutoffs in Positive Moment Region of Continuous Beams  201 6.13 Bar Cutoffs in Uniformly Loaded Cantilever Beams  202 6.14 Development of Positive Reinforcement at Simple Supports and at Points of Inflection  209 6.15 Development of Shear Reinforcement  211 6.16 Tension Lap Splices  213 6.17 Welded Splices and Mechanical Connections in Tension  215 6.18 Compression Lap Splices  216 6.19 End Bearing Connections, Welded Splices, and Mechanical Connections in Compression  217 6.20 Splices for Members under Compression and Bending  217 6.21 Design Examples  217 Selected References  234 Problems  236

7 ANALYSIS OF CONTINUOUS BEAMS AND ONE-​WAY SLABS  239 7.1 Introduction 239

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7.2

Analysis Methods under Gravity Loads  240

7.3

Arrangement of Live Load for Moment Envelope  241

7.4

ACI Code—​Arrangement of Live Load and Moment Coefficients  246

7.5

ACI Moment Diagrams  247

7.6

Shear Envelope for Design  250 Selected Reference  252 Problems  252

8 DESIGN OF ONE-​WAY SLABS 

254

8.1 Definition 254 8.2

Analysis Methods  254

8.3

Slab Design  255

8.4

Choice of Reinforcement  258

8.5

Bar Details  264 Selected References  265 Problems  265

9 DESIGN OF SLAB-​B EAM-​G IRDER AND JOIST FLOOR SYSTEMS  266 9.1 Introduction 266 9.2

Size of Beam Web  267

9.3

Continuous Frame Analysis for Beams  270

9.4

Choice of Longitudinal Reinforcement in Beams  274

9.5

Shear Reinforcement in Beams  285

9.6

Details of Bars in Beams  287

9.7

Size of Girder Web  294

9.8

Continuous Frame Analysis for Girders  297

9.9

Choice of Longitudinal Reinforcement in Girders  300

9.10 One-​Way Joist Floor Construction  306 9.11 Design of Joist Floors  307

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9.12 Redistribution of Moments—​Introduction to Limit or Plastic Analysis  312 Selected References  317 Problems  318

10 MEMBERS IN COMPRESSION AND BENDING 

321

10.1 Introduction 321 10.2

Types of Columns  322

10.3

Behavior of Columns under Pure Axial Load  322

10.4

Safety Provisions for Columns  325

10.5

Concentrically Loaded Short Columns  326

10.6

Strength Interaction Diagram  326

10.7

Slenderness Effects  328

10.8

Lateral Ties  329

10.9

Spiral Reinforcement and Longitudinal Bar Placement  330

10.10 Limits on Percentage of Longitudinal Reinforcement  332 10.11 Maximum Strength in Axial Compression—​ACI Code  333 10.12 Balanced Strain Condition  333 10.13 Nominal Strength of a Compression-​Controlled Rectangular Section  336 10.14 Nominal Strength of a Rectangular Section with Eccentricity e Greater than That at the Balanced Strain Condition  340 10.15 Design for Strength—​Region I, Minimum Eccentricity  342 10.16 Design for Strength—​Region II, Compression-​Controlled Sections (emin < e < eb )  345 10.17 Design for Strength—​Region III, Transition Zone and Tension-​Controlled Sections (e > eb )  351 10.18 Circular Sections Under Combined Compression and Bending  354 10.19 Combined Axial Tension and Bending  357 10.20 Combined Axial Force and Biaxial Bending  359 10.21 Design for Shear  368 Selected References  370 Problems  374

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11 MONOLITHIC BEAM-​C OLUMN CONNECTIONS  11.1 Introduction 380 11.2

Beam-​Column Joints Actions  381

11.3

Joint Transverse Reinforcement  383

11.4

Joint Shear Strength  387

11.5

Column-​to-​Beam Moment Strength Ratio  389

11.6

Anchorage of Reinforcement in the Joint Region  390

11.7

Transfer of Column Axial Forces through the Floor System  391

11.8 Examples 391 11.9

Additional Remarks  399 Selected References  399 Problems  401

12 SERVICEABILITY 

403

12.1 Introduction 403 12.2

Fundamental Assumptions  403

12.3

Modulus of Elasticity Ratio, n  404

12.4

Equilibrium Conditions  404

12.5

Method of Transformed Section  407

12.6

Deflections—​General  410

12.7

Deflections for Linear Elastic Members  411

12.8

Modulus of Elasticity  414

12.9

Effective Moment of Inertia  414

12.10 Instantaneous Deflections in Design  417 12.11 Creep Effect on Deflections under Sustained Load  428 12.12 Shrinkage Effect on Deflections under Sustained Load  431 12.13 Creep and Shrinkage Deflection—​ACI Code Method  435 12.14 Creep and Shrinkage Deflection—​Alternative Procedures  436 12.15 ACI Minimum Depth of Flexural Members  439

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12.16 Span-​to-​Depth Ratio to Account for Cracking and Sustained Load Effects  441 12.17 ACI Code Deflection Provisions—​Beam Examples  446 12.18 Crack Control for Beams and One-​Way Slabs  451 12.19 Side Face Crack Control for Large Beams  455 12.20 Control of Floor Vibrations—​General  456 Selected References  457 Problems  459

13 SLENDERNESS EFFECTS ON COLUMNS 

463

13.1 General 463 13.2

Buckling of Concentrically Loaded Columns  465

13.3

Effective Length Factor  468

13.4

Moment Magnification—​Members with Transverse Loads—​Without Joint Lateral Translation (i.e., No Sidesway)  470

13.5

Moment Magnification—​Members Subject to End Moments Only—​Without Joint Lateral Translation (i.e., No Sidesway)  472

13.6

Moment Magnification—Members with Sidesway—Unbraced (Sway) Frames  477

13.7

Interaction Diagrams—​Effect of Slenderness  479

13.8

ACI Code—​General  480

13.9

ACI Code—​Moment Magnifier Method for Columns in Nonsway Frames  482

13.10 ACI Code—​Moment Magnifier Method for Columns in Sway Frames  485 13.11 Alignment Charts for Effective Length Factor k  490 13.12 Second-​Order Analysis—​ACI Code  493 13.13 Minimum Eccentricity in Design  493 13.14 Biaxial Bending and Axial Compression  494 13.15 ACI Code—​Slenderness Ratio Limitations  494 13.16 Amplification of Moments in Beams  495 13.17 Examples  495 Selected References  523 Problems  526

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14 STRUT-​A ND-​T IE MODELS—​D EEP BEAMS, BRACKETS, AND CORBELS  528 14.1 Introduction 528 14.2

Deep Beams  542

14.3

Brackets and Corbels  559

14.4

Additional Remarks  565 Selected References  566 Problems  567

15 STRUCTURAL WALLS 

569

15.1 General 569 15.2

Minimum Wall Dimensions and Reinforcement Requirements—​ACI Code  569

15.3

Design of Nonbearing Walls  573

15.4

Design of Bearing Walls  573

15.5

Design of Shear Walls  576

15.6

Lateral Support of Longitudinal Reinforcement  596

15.7

Retaining Structures  597 Selected References  619 Problems  620

16 DESIGN OF TWO-​WAY FLOOR SYSTEMS 

622

16.1

General Description  622

16.2

General Design Concept of the ACI Code  624

16.3

Total Factored Static Moment  625

16.4

Ratio of Flexural Stiffnesses of Longitudinal Beam to Slab  633

16.5

Minimum Slab Thickness for Deflection Control  637

16.6

Nominal Requirements for Slab Thickness and Size of Edge Beams, Column Capital, and Drop Panel  639

16.7

Direct Design Method—​Limitations  644

16.8

Direct Design Method—​Longitudinal Distribution of Moments  645

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16.9

Direct Design Method—​Effect of Pattern Loadings on Positive Moment  647

16.10 Direct Design Method—​Procedure for Computation of Longitudinal Moments  647 16.11 Torsion Stiffness of the Transverse Elements  651 16.12 Transverse Distribution of Longitudinal Moment  656 16.13 Design of Slab Thickness and Reinforcement  662 16.14 Size Requirement for Beam (If Used) in Flexure and Shear  669 16.15 Shear Strength in Two-​Way Floor Systems  671 16.16 Shear Reinforcement in Flat Plate Floors  676 16.17 Direct Design Method—​Moments in Columns  686 16.18 Transfer of Moment and Shear at Junction of Slab and Column  687 16.19 Openings and Corner Connections in Flat Slabs  697 16.20 Equivalent Frame Method for Gravity Load Analysis  698 16.21 Equivalent Frame Models  710 16.22 Equivalent Frame Method for Lateral Load Analysis  711 Selected References  711 Problems  718

17 YIELD LINE THEORY OF SLABS 

720

17.1 Introduction 720 17.2

General Concept  720

17.3

Fundamental Assumptions  723

17.4

Methods of Analysis  724

17.5

Yield Line Analysis of One-​Way Slabs  725

17.6

Work Done by Yield Line Moments in Rigid Body Rotation of Slab Segment  728

17.7

Nodal Forces at Intersection of Yield Line with Free Edge  729

17.8

Nodal Forces at Intersection of Three Yield Lines  732

17.9

Yield Line Analysis of Rectangular Two-​Way Slabs  736

17.10 Corner Effects in Rectangular Slabs  742

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17.11 Application of Yield Line Analysis to Special Cases  743 Selected References  747 Problems  747

18 TORSION 

748

18.1 General 748 18.2

Torsional Stress in Homogeneous Sections  749

18.3

Torsional Stiffness of Homogeneous Sections  751

18.4

Effects of Torsional Stiffness on Compatibility Torsion  752

18.5

Torsional Moment Strength Tcr at Cracking  755

18.6

Strength of Rectangular Sections in Torsion—​Skew Bending Theory  757

18.7

Strength of Rectangular Sections in Torsion—​Space Truss Analogy  761

18.8

Strength of Sections in Combined Bending and Torsion  765

18.9

Strength of Sections in Combined Shear and Torsion  767

18.10 Strength Interaction Surface for Combined Bending, Shear, and Torsion  768 18.11 Torsional Strength of Concrete and Closed Transverse Reinforcement—​ACI Code  770 18.12 Combined Torsion with Shear or Bending—​ACI Code  772 18.13 Minimum Requirements for Torsional Reinforcement—​ACI Code  773 18.14 Examples  775 Selected References  791 Problems  796

19 FOOTINGS 

799

19.1

Purpose of Footings  799

19.2

Bearing Capacity of Soil  799

19.3

Types of Footings  800

19.4

Types of Failure  800

19.5

Shear Strength  802

19.6

Flexural Strength and Development of Reinforcement  803

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19.7

Proportioning Footing Areas for Equal Settlement  804

19.8

Investigation of Square Spread Footings  804

19.9

Design of Square Spread Footings  809

19.10 Design of Rectangular Footings  814 19.11 Design of Plain and Reinforced Concrete Wall Footings  818 19.12 Combined Footings  822 19.13 Design of Combined Footings  823 19.14 Pile Footings  841 Selected References  841 Problems  842

20 INTRODUCTION TO PRESTRESSED CONCRETE 

844

20.1 Introduction 844 20.2

Historical Background  844

20.3

Advantages and Disadvantages of Prestressed Concrete Construction  845

20.4

Pretensioned and Post-​tensioned Beam Behavior  846

20.5

Service Load Stresses on Flexural Members—​Tendons Having Varying Amounts of Eccentricity  849

20.6

Three Basic Concepts of Prestressed Concrete  853

20.7

Loss of Prestress  856

20.8

Nominal Strength Mn of Flexural Members  866

20.9

Cracking Moment  871

20.10 Shear Strength of Members without Shear Reinforcement  873 20.11 Shear Reinforcement for Prestressed Concrete Beams  881 20.12 Development of Reinforcement  883 20.13 Proportioning of Cross Sections for Flexure When No Tension is Permitted  885 20.14 Additional Topics  894 Selected References  894 Problems  895

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21 COMPOSITE MEMBERS AND CONNECTIONS  21.1 Introduction 897 21.2

Composite Action  897

21.3

Concrete Composite Flexural Members  901

21.4

Concrete-​Steel Composite Columns  916

21.5

Concrete-​Encased Steel Composite Columns  918

21.6

Concrete-​Filled Tube Columns  934

21.7

Moment Connections with Composite Columns  943 Selected References  944 Problems  947

Index  949

897

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PREFACE The eighth edition of this textbook has been substantially updated to incorporate the changes introduced by the publication of the 2014 American Concrete Institute (ACI) Building Code and Commentary for Structural Concrete, as well as to reflect changes in construction and design practices that have occurred in the last few years.

APPROACH This new edition follows the same philosophical approach that has gained wide acceptance of users since the first edition was published in 1965. Herein, as in past editions, consider­ able emphasis is placed on presenting to the student, as well as to the practicing engineer, the basic principles of reinforced concrete design and the concepts necessary to understand and properly apply the provisions of the ACI Building Code. Numerous examples are presented to illustrate the general approach to design and analysis. The material is incorporated into the chapters in a way that permits the reader to either study in detail the concepts in logical sequence or obtain a qualitative explanation and proceed directly to the design process using the ACI Code.

NEW TO THIS EDITION The eighth edition of this book incorporates the changes arising from the publication of the 2014 American Concrete Institute Building Code and Commentary (ACI 318-​14). While past editions of the ACI Code were largely structured around member actions (e.g., flexure, shear, and axial load), ACI 318-​14 is organized primarily by structural elements (e.g., beams, columns, walls). As a result, virtually all design provisions have changed in format and number, and are located under a new chapter designation in the  Code. Accordingly, all chapters and example problems have been revised to conform to the format and reorganization of the 2014 ACI Building Code (ACI 318-​14). In addition, content has been reorganized within existing chapters, moved to other chapters, or relocated as new, stand-​alone chapters for better continuity and presentation of the material. Main revisions, updates, and new material include the following. 1. A new chapter on Structural Walls (Chapter 15) has been added. This chapter includes the design of Non-​Bearing and Bearing Walls, as well as the design of Shear Walls. The design of Cantilever Retaining Walls (formerly Chapter 12) has been revised and is included at the end of the new Chapter 15. 2. The chapter on composite construction (Chapter 21) has been substantially revised and renamed “Composite Members and Connections” to better reflect its new scope. The first part covers the design of concrete-​concrete composite flexural members, including calculation of deflections for shored and unshored construction. In addition, the chapter now includes sections on Concrete-​Encased Steel Columns and Concrete-​Filled Tubes, along with a new section on Moment Connections between Composite Columns and Steel Beams.

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3. The material on the Strut and Tie Method and its application to the design of Deep Beams, Brackets, and Corbels (previously included in Chapter 5, Shear Strength) has been updated and is now presented as a stand-​alone, separate chapter (Chapter 14). 4. The material on Rectangular Sections in Bending under Service Load Conditions (formerly Chapter 4) and Deflections (formerly Chapter 14) was revised and combined into a single chapter dealing with Serviceability (new Chapter 12). 5. The design of T-​Sections in Bending (formerly Chapter 9) was relocated as Chapter 4, immediately after Chapter 3, Flexural Behavior and Strength of Beams, for better flow and continuity of the material on beam design. 6. The chapter on Slenderness Effects on Columns, now Chapter  13 (formerly Chapter 15) has been revised to include additional content and example problems for better understanding of the ACI Code procedures established to account for second-​ order effects in column design. 7. The alternative provisions of Appendix B and the alternative load factors of Appendix C included in past editions of the ACI Code have been removed from the Code. Accordingly, discussion and example problems corresponding to these appendices are not included in this edition of the textbook. 8. All sections have been revised to improve the flow and continuity of the material in accordance to the changes in ACI 318-​14. In addition to the content revisions indicated above, all the examples and the problems at the end of each chapter have been revised and updated to conform to the current ACI Code. Examples and problems have also been updated, and new examples have been added to reflect the strengths of the materials most commonly used in current practice. A  few examples, however, use less common values in order to emphasize specific aspects of the design process that students might otherwise overlook. To aid instructors, a solutions manual has been prepared for the end-​of-​chapter problems. Many problems are solved in Mathcad®, allowing alternative solutions to be easily arrived at by modifying a few parameters, either as suggested in this textbook or at the choice of the instructor.

COURSE SUGGESTIONS Depending on the proficiency required of the student, this book may provide material for two courses of three or four semester-​hours each. It is suggested that the beginning course in concrete structures for undergraduate students contain all or most of the material in Chapters 1 through 6, and Chapters 8 through 10. The second course may begin with Chapter 10, using that topic (members in compres­ sion and bending) to review many of the subjects in the first course, followed by Chapter 12 on serviceability, Chapter 13 on slenderness effects on columns, and Chapter 16 on two-​ way slab systems. In addition, one or two of the following may be included in a second course:  the remaining sections of Chapter  5 on shear strength affected by axial force; Chapter 15 on structural walls; Chapter 18 on torsion; Chapter 14 on strut-​and-​tie models, deep beams, brackets, and corbels; and Chapter 20 on prestressed concrete. Chapters on beam-​ column connections (Chapter  11), yield line theory of slabs (Chapter 17), footings (Chapter 19), and composite members and connections (Chapter 21) may serve as contents for a third course.

UNITS This edition continues the modest treatment of SI units used in previous editions. The 2014 ACI Code has an SI version (known as ACI 318-​14M), and the SI versions of the ACI Code equations appear in this book as footnote equations with the same equation number. According to the ACI Code, the designer must use in its entirety either the Inch-​Pound units version (ACI 318-​14) or the SI version (ACI 318-​14M), although the

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PREFACE

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Inch-​Pound units version is the official version of the Code. The authors believe that sufficient metrication should be included in a text on reinforced concrete to permit the reader to gain some familiarity with SI units, but suspect that too much would interfere with learning the basic concepts of concrete design; constant conversion back and forth between Inch-​Pound and SI units is more confusing than using either one exclusively. The text provides data on reinforcing bars in accordance with the American Society for Testing and Materials Inch-​Pound units, and also ASTM SI units (the “soft” conversion of the bar sizes and strengths approved in 1996). Some design tables are provided for bars and material strengths in SI units, a few numerical examples are given in SI units, and some problems at the ends of chapters are given with an SI alternate in parentheses at the statement concluding the problem. In all parts of this book that use metric units, force is expressed in the newton (N) or kilonewton (kN) unit. The SI unit of stress is the pascal (Pa), or newton per meter squared, which because of its typically large numerical value is usually expressed in megapascals (MPa): that is, 106 pascals. A few diagrams show, along the stress axis, the kilogram force per centimeter squared (kgf/​cm2) in addition to Inch-​Pound and SI units. For the convenience of the reader, some conversion factors for forces, stresses, uniform loading, and moments are provided on a separate page following this Preface. It is noted that throughout the textbook, conversion factors (for forces, stresses, and dimensions) used in example problems (when needed) are shown with a smaller font so as to not interfere with the values of the parameters actually involved in the calculations and to facilitate understanding of the problem solution.

ACKNOWLEDGMENTS The authors continue to be indebted to students, colleagues, and other users of the first seven editions of this book, who have suggested improvements of wording, identified errors, and recommended items for inclusion or omission. The authors are pleased to acknowledge the following reviewers, to whom they owe special thanks:  Mohammad Azarbayejani, University of Texas–​Pan American; Abdeldjelil Belarbi, University of Houston; Sergio F.  Breña, University of Massachusetts–​ Amherst; Norbert Delatte, Cleveland State University; Apostolos Fafitis, Arizona State University; Susan Faraji, University of Massachusetts Lowell; Catherine French, University of Minnesota; David Garber, Florida International University; Roberto Leon, Virginia Tech University; John B. Mander, Texas A&M University; Fatmir Menkulasi, Louisiana Tech University; Gregory K. Michaelson, Marshall University; Levon Minnetyan, Clarkson University; Ayman M. Okeil, Louisiana State University; Nima Rahbar, Worcester Polytechnic Institute; Michael Seek, Old Dominion University; Ahmed Senouci, University of Houston; Lisa Spainhour, FAMU–​ Florida State University; Andreas Stavridis, University at Buffalo; Jale Tezcan, Southern Illinois University–​Carbondale; Robin Tuchscherer, Northern Arizona University; Baolin Wan, Marquette University; Paul Ziehl, University of South Carolina. Their comments and suggestions have been carefully considered and the results of our review are reflected in this complete revision. Users of this eighth edition are urged to communicate with the authors regarding all aspects of this book, particularly on identification of errors and suggestions for improvement. We are indebted to late Professors Chu-​Kia (CK) Wang and Charles (Chuck) G. Salmon, who originated this textbook and entrusted us to carry on their legacy. Much of the new and expanded material presented in this eighth edition would not have been possible without their work in earlier editions of this book. Special thanks are due to the Higher Education Group, Oxford University Press—​in particular, Dan Kaveney, Executive Editor, Christine Mahon, Associate Editor, Claudia Dukeshire, Production Editor, Megan Carlson, Assistant Editor, and Nancy Blaine, former Senior Acquisitions Editor. We acknowledge the long-​time continuing patience and encouragement from our families and especially from our wives, Rebeca and Connie, throughout the preparation of this edition of the book. Nicole and Gabriel Parra, with their frequent smiles and unbounded

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love, were a continuous source of inspiration to their father. We also owe a special recognition to our parents, Paulina Peña, Hernán Pincheira, Gustavo Parra Pardi and Yolanda Montesinos Soteldo, who instilled in us from an early age the importance of learning, education, and hard work. To all of them we wholeheartedly dedicate this book. José A. Pincheira Gustavo J. Parra-​Montesinos

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ABOUT THE AUTHORS CHU-​KIA WANG* was Professor of Civil Engineering at the University of Wisconsin–​ Madison for more than 30 years. A devoted teacher throughout his career, he was the author or coauthor of many textbooks in the field of structural engineering as the outgrowth of lectures he prepared for his classes. The University of Wisconsin–​Madison recognized his contribution to the education of future engineers with the College of Engineering’s Benjamin Smith Reynolds Award for Excellence in Teaching. A fellow and lifetime member of the American Society of Civil Engineers, Professor Wang was also a member of the American Concrete Institute (ACI), the American Society for Engineering Education (ASCE), and other professional societies. CHARLES G. SALMON* was Professor Emeritus of Civil Engineering at the University of Wisconsin–​Madison. An accomplished author, educator, researcher, and professional structural engineer, Professor Salmon received numerous honors in recognition of his contributions to the field, including the Western Electric Award for excellence in teaching from the American Society for Engineering Education, the University of Wisconsin’s Emil H. Steiger Distinguished Teaching Award, and the American Concrete Institute’s Joe W. Kelly and Delmar L. Bloem Awards. He was a long-​time member of the ACI Building Code Committee for Structural Concrete (ACI 318), Committee 340 (Design Aids), and Committee 435 (Deflections of Concrete Structures). Professor Salmon was also an honorary member of ACI, an honorary member of the American Society of Civil Engineers; and a life member of the American Society for Engineering Education. *Deceased JOSÉ A.  PINCHEIRA is Associate Professor of Civil and Environmental Engineering at the University of Wisconsin–​Madison. His main research interests include the behavior and design of reinforced concrete structural systems subjected to earthquakes, as well as the seismic rehabilitation of concrete structures. Dr. Pincheira is a fellow of the American Concrete Institute, a member of subcommittee 318-​R (High Strength Steel Reinforcement) of the Building Code for Structural Concrete (ACI 318), and former member of subcommittee 318-​D (Flexure and Axial Loads). He is past chair and current member of ACI Committee 369, Seismic Repair and Rehabilitation of Concrete Buildings; member of ACI Committee 374, Performance-​Based Seismic Design of Concrete Buildings; and former member of the Committee on Seismic Rehabilitation of the American Society of Civil Engineers. Professor Pincheira has received several prestigious awards in recognition of his contributions to research and teaching, including the CAREER Award from the National Science Foundation; the James M.  Robbins Excellence in Teaching Award from Chi Epsilon, the Civil Engineering Honor Society; the Martin P. Korn Award from the Precast/​ Prestressed Concrete Institute; and the Wason Medal from the American Concrete Institute. GUSTAVO J.  PARRA-​MONTESINOS is the C.  K. Wang Professor of Structural Engineering in the Department of Civil and Environmental Engineering at the University of Wisconsin–​Madison. Professor Parra’s main research interests include the behavior and design of reinforced concrete, fiber-​reinforced concrete, and hybrid steel-​concrete structures. He is a member of the ACI Building Code Committee 318 and Chair of its

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subcommittee 318-​J (Joints and Connections). He is also Chair of ACI-​ASCE Committee 352 on Joints in Monolithic Reinforced Concrete Structures. In addition, he is a member of ACI-​ASCE Committee 335 on Composite and Hybrid Structures and of the Fédération Internationale du Béton (fib) Task Group T4.1 on fiber-​reinforced concrete. Professor Parra has received several prestigious awards for his contributions to research in the field of reinforced concrete, including the Wason Medal, the Chester Paul Siess Award, and the Charles S. Whitney Medal from the American Concrete Institute. In addition, he is a recipient of the Arthur J. Boase Award from the ACI Foundation, the Walter L. Huber Research Prize from the American Society of Civil Engineers, the Shah Family Innovation Prize from the Earthquake Engineering Research Institute, and the ACI Young Member Award for Professional Achievement. Professor Parra is also a fellow of the American Concrete Institute.

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CONVERSION FACTORS Some Conversion Factors, between Inch-Pound and SI Units, Useful in Reinforced Concrete Design

Forces

Stresses

Moments Uniform Loading

Density

To Convert kip force lb kN ksi psi MPa MPa ft-kip kN · m kip/ft kN/m kip/ft2 psf kN/m2 pcf

To kN N kip MPa (i.e., N/mm2) MPa ksi psi kN · m ft-kip kN/m kip/ft kN/m2 N/m2 kip/ft2 kg/m3

Multiply by 4.448 4.448 0.2248 6.895 0.006895 0.1450 145.0 1.356 0.7376 14.59 0.06852 47.88 47.88 0.02089 16.01846

Basis of Conversions: 1 in. = 25.4 mm; 1 lb force = 4.448 newtons. Basic SI units relating to structural design: Quantity length mass time

Unit meter kilogram second

Symbol m kg s

Derived SI units relating to structural design: Quantity force pressure, stress energy, or work

Unit newton pascal joule

Symbol N Pa J

Formula kg · m/s2 N/m2 N·m

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REINFORCED CONCRETE DESIGN

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CHAPTER 1 INTRODUCTION, MATERIALS, AND PROPERTIES

1.1 REINFORCED CONCRETE STRUCTURES The most common materials from which most structures are built are wood, steel, reinforced (including prestressed) concrete, and masonry. Lightweight materials such as aluminum and advanced composite materials, such as fiber-​reinforced polymers (FRP), are also becoming more common in use. Reinforced concrete is unique in that two materials, reinforcing steel and concrete, are used together; thus the principles governing structural design in reinforced concrete differ in many ways from those involving design in one material. Many structures are built of reinforced concrete:  buildings, bridges, viaducts, retaining walls, tunnels, tanks, conduits, and others. This text deals primarily with fundamental principles of behavior and design of reinforced concrete members subjected to axial force, bending moment, shear, torsion, or combinations of these. These principles are applicable to the design of any structure, as long as information is known about the variation of axial force, shear, moment, and so on, along the length of each member. Although analysis and design may be treated separately, they are inseparable in practice, especially in the case of reinforced concrete structures, which are usually statically indeterminate. In such cases, reasonable sizes of members are needed in the preliminary analysis that must precede the final design; so the final conciliation between analysis and design is largely a matter of trial, judgment, and experience. Reinforced concrete is a logical union of two materials: plain concrete, which possesses high compressive strength but little tensile strength, and steel, in the form of bars embedded in the concrete, which can provide the needed strength in tension and ductility to the member. For instance, the strength and deflection capacity of the beam shown in Fig. 1.1.1 are greatly increased by placing steel bars in the tension zone. Without steel reinforcement, the beam would undergo a brittle failure once the tensile stress at the bottom of the beam reached the tensile strength of the concrete. Adding sufficient longitudinal steel reinforcement in the tension zone, however, allows the beam to sustain additional load beyond the formation of a transverse (flexural) crack. As shown in Fig. 1.1.1, several flexural cracks will likely develop as the load is increased, providing some degree of warning prior to failure. Since reinforcement steel is capable of resisting compression as well as tension, it is also used to provide part of the carrying capacity in reinforced concrete columns, and frequently in the compression zone of beams to increase ductility and control deflections. Also, reinforcement is needed transversely to resist shear, to provide lateral support to longitudinal reinforcement, and to confine the concrete.

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Stratosphere Tower, Las Vegas; the tallest free-standing observation tower in the United States, 1149 feet high. A three-legged concrete tower is topped by a ring beam that supports the steel dome, completed in 1996 (Photo by C. G. Salmon).

Steel and concrete work readily in combination for several reasons: (1) bond (interaction between bars and surrounding hardened concrete) allows transfer of forces between the two materials; (2) proper concrete mixes provide adequate impermeability of the concrete against water intrusion and bar corrosion; and (3) sufficiently similar rates of thermal expansion—​that is, 0.0000055 to 0.0000075 for concrete and 0.0000065 for steel per degree Fahrenheit (ºF), or 0.000010 to 0.000013 for concrete and 0.000012 for steel per degree Celsius (ºC)—​introduce negligible forces between steel and concrete under atmos­ pheric changes of temperature.

1.2 HISTORICAL BACKGROUND Joseph Monier, the owner of an important nursery in Paris, is generally given the credit for making the first practical use of reinforced concrete. In 1867, Monier recognized many of its potential uses and successfully undertook to expand the application of the new method [1.1].1 Prior to his work, however, the method of reinforcing concrete with 1  Numbers in brackets refer to the Selected References at the end of the chapter.

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1.2  HISTORICAL BACKGROUND

P

Neutral axis

A

compression zone tension zone

Concrete

Steel bars

A Cracks

Steel bars

Section A–A

Figure 1.1.1  Use of steel bars as tension reinforcement in a reinforced concrete beam.

iron was known and in some cases was protected by patents. Ancient Grecian structures show that much earlier builders knew something about the reinforcing of stonework for added strength [1.2]. In the mid-​ 1800s, Lambot in France constructed and later exhibited at the Paris Exposition of 1854 a small boat, on which he received a patent in 1855. In Lambot’s patent was shown a reinforced concrete beam and a column reinforced with four round iron bars. Another Frenchman, François Coignet, published a book in 1861 describing many applications and uses of reinforced concrete. In 1854, W. B. Wilkinson of England took out a patent for a reinforced concrete floor. Monier acquired his first French patent in 1867 for iron-​reinforced concrete tubs. This was followed by his many other patents, such as for pipes and tanks in 1868, flat plates in 1869, bridges in 1873, and stairways in 1875. In 1880–​1881, Monier received German patents for innovations that included railroad ties, water feeding troughs, circular flower pots, flat plates, and irrigation channels. Monier’s iron reinforcement was made mainly to conform to the contour of the structural element and generally strengthen it. He apparently had no quantitative knowledge regarding its behavior or any method of making design calculations [1.1]. In the United States, the pioneering efforts were made by Thaddeus Hyatt, originally a lawyer, who conducted experiments on reinforced concrete beams in the 1850s. In a perfectly correct manner, the iron bars in Hyatt’s beams were located in the tension zone, bent up near the supports, and anchored in the compression zone. Additionally, transverse reinforcement (known as vertical stirrups) was used near the supports. However, Hyatt’s experiments were unknown until 1877 when he published his work privately. Built in 1870, the William Ward house in Port Chester, New York, is generally credited as the first cast-​in-​place reinforced concrete structure in the United States [1.3]. E. L. Ransome, head of the Concrete–​Steel Company of San Francisco, apparently used some form of reinforced concrete in the early 1870s. He continued to increase the application of wire rope and hoop iron to many structures and was the first to use and patent in 1884 the deformed (twisted) bar. M. K. Hurd has provided an interesting biographical sketch of Ernest L. Ransome [1.4]. In 1890, Ransome built the Leland Stanford Jr. Museum in San Francisco, a reinforced concrete building two stories high and 312 ft (95 m) long. Since that time, development of reinforced concrete in the United States was rapid. During the period 1891–​1894, various investigators in Europe published theories and test results; among them were Moeller (Germany), Wunsch (Hungary), Melan (Austria), Hennebique (France), and Emperger (Hungary), but practical use was less extensive than in the United States. Throughout the entire period 1850–​1900, relatively little was published, because the engineers working in the reinforced concrete field considered construction and computational methods to be trade secrets. One of the first publications that might be classified as a textbook was that of Considère in 1899. By the turn of the century, there was a multiplicity of systems and methods with little uniformity in design procedures, allowable stresses, and systems of reinforcing. In 1903, with the formation in the United States of a joint committee of representatives of all organizations interested in reinforced concrete, uniform application of knowledge to design was initiated. The development of standard specifications is discussed in Chapter 2. The earliest textbook in English was that of Turneaure and Maurer [1.5], published in 1907. In the first decade of the twentieth century, progress in reinforced concrete was

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rapid. Extensive testing to determine beam behavior, compressive strength of concrete, and modulus of elasticity was conducted by Arthur N. Talbot at the University of Illinois, by Frederick E. Turneaure and Morton O. Withey at the University of Wisconsin, and by Bach in Germany, among others. From about 1916 to the mid-​1930s, research centered on axially loaded column behavior. In the late 1930s and 1940s, eccentrically loaded columns, footings, and the ultimate strength of beams received special attention. Since the mid-​1950s, reinforced concrete design practice has made the transition from that based on elastic methods to one based on strength. Prestressed concrete (Chapter 20), wherein the steel reinforcement is stressed in tension and the concrete is in compression even before external loads are applied, has advanced from an experimental technique to a major structural composite material. There has been a transition from cast-​in-​place reinforced concrete to elements precast at a manufacturer’s plant and shipped to the job site for assembly. A summary of concrete building construction in the United States is given by Cohen [1.3]. Understanding of reinforced concrete behavior is still far from complete; building codes and specifications that give design procedures are continually changing to reflect latest knowledge.

1.3 CONCRETE Plain concrete is made by mixing cement, fine aggregate, coarse aggregate, water, and frequently admixtures (see Fig.  1.3.1). When reinforcing steel is placed in the forms and wet concrete mix is placed around it, the final solidified mass becomes reinforced concrete. The strength of concrete depends on many factors: notably the proportion of the ingredients and the conditions of temperature and moisture under which it is placed and cured. Subsequent sections contain brief discussions of the materials in and the properties of plain concrete. The treatment is intended to be only introductory; an interested reader should consult standard references devoted entirely to the subject of plain concrete [1.6–​1.8].

Figure 1.3.1  Cross section of concrete. Cement-​and-​water paste coats each aggregate particle and fills space between particles. (Photo by José A. Pincheira.)

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5

1.4 CEMENT Cement is a material that has adhesive and cohesive properties enabling it to bond mineral fragments into a solid mass. Although this definition can apply to many materials, the cements of interest for reinforced concrete construction are those that can set and harden in the presence of water—​the so-​called hydraulic cements. These consist primarily of silicates and aluminates of lime made from limestone and clay (or shale) which is ground, blended, fused in a kiln, and crushed to a powder. Such cements chemically combine with water (hydrate) to form a hardened mass. The usual hydraulic cement used for reinforced concrete is known as portland cement because of its resemblance when hardened to Portland stone found near Dorset, England. The name originated in a patent obtained by Joseph Aspdin of Leeds, England, in 1824. Concrete made with portland cement ordinarily requires several days to attain strength adequate to allow forms to be removed and construction and dead loads carried. The design or specified compressive strength of such concrete is typically assumed to be reached at about 28 days. This ordinary portland cement is identified by ASTM (American Society for Testing and Materials) C150/​C150M [1.9] as Type I. Other types of portland cement and their intended uses are given in Table 1.4.1. ACI Committee 225 provides a guide for selection and use of hydraulic cements [1.10]. There are also several categories of blended hydraulic cements (ASTM C595/​C595M [1.11]), such as portland blast-​furnace slag cement (Type IS), portland-​pozzolan cement (Type IP), portland-​limestone cement (Type IL), and ternary blended cement (Type IT). Ternary blended cements are defined in ASTM C595 as those “consisting of portland cement with either a combination of two different pozzolans, slag cement and a pozzolan, a pozzolan and a limestone, or a slag cement and a limestone.” Pozzolan is a finely divided siliceous or siliceous and aluminous material that possesses little or no inherent cementitious property; in the powdery form and in the presence of moisture, however, it will chemically react with calcium hydroxide at ordinary temperatures to form compounds possessing cementitious properties. Portland blast-​furnace slag cement has lower heat of hydration than ordinary Type I cement and is useful for mass concrete structures such as dams; and because of its high sulfate resistance, it is used in seawater construction. Portland-​pozzolan cement is a blended mixture of ordinary Type I  cement with pozzolan. Blended cements with pozzolan gain strength more slowly than cements without pozzolan; hence they produce less heat during hydration, and thus are widely used in mass concrete construction. Air-​entraining portland cement contains a chemical admixture finely ground with the cement to produce intentionally air bubbles on the order of 0.002 in. (0.05 mm) diameter uniformly distributed throughout the concrete. Such air entrainment will give the concrete improved durability against frost action, as well as better workability. Air-​entraining portland cement for Types I, II, and III, given in Table 1.4.1, is designated IA, IIA, or IIIA. Air-​entraining agents may also be added to the blended hydraulic cements in ASTM C595/​ C595M [1.11] at the time the concrete is mixed.

TABLE 1.4.1  TYPES OF PORTLAND CEMENTa Type Uses I Ordinary construction where special properties are not required II Ordinary construction when moderate sulfate resistance or moderate heat of hydration is desired III When high early strength is desired; has considerably higher heat of hydration than Type I cement IV When low heat of hydration is desired V When high sulfate resistance is desired a

According to ASTM C150/​C150M [1.9].

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1.5 AGGREGATES Since aggregate usually occupies about 75% of the total volume of concrete, its properties have a definite influence on the behavior of hardened concrete. Not only does the strength and stiffness of the aggregate affect the strength and stiffness of the concrete, its properties also greatly affect durability (resistance to deterioration under freeze-​thaw cycles). Since aggregate is less expensive than cement, it is logical to use the largest percentage feasible. In general, for maximum strength, durability, and best economy, the aggregate should be packed and cemented as densely as possible. Hence, aggregates are usually graded by size and a proper mix specifies percentages of both fine and coarse aggregates. Fine aggregate (sand) is any material passing through a No. 4 sieve2 [i.e., less than about 3 16 -​in. (5-​mm) diameter]. Coarse aggregate (gravel) is any material of larger size. The nominal maximum size of coarse aggregate permitted (ACI-​26.4.2.1)3 is governed by the clearances between sides of forms and between adjacent bars and may not exceed (a) 1 5 the narrowest dimension between sides of forms, nor (b)  1 3 the depth of slabs, nor (c) 3 4 the minimum clear spacing between individual reinforcing bars. Additional information concerning aggregate selection and use is to be found in a report of ACI Committee 221 [1.13]. Natural stone aggregates conforming to ASTM C33 [1.14] are used in the majority of concrete construction, giving a unit weight for such concrete of about 145 pcf (pounds per cubic foot) or 2320 kg/​m3 (kilograms per cubic meter). When steel reinforcement is added, the unit weight of normal-​weight reinforced concrete is taken for calculation purposes as 150 pcf or 2400 kg/​m3. Actual weights for concrete and steel are rarely, if ever, computed separately. For special purposes, lightweight or extra-​heavy aggregates are used. Structural lightweight concretes are usually made from aggregates conforming to ASTM C330 [1.15] which are produced artificially in a kiln, such as expanded clays and shales. The unit weight of such concretes typically ranges from 70 to 115 pcf (1120–​1840 kg/​m3) (see Fig. 1.5.1). Lightweight concretes ranging down to 30 pcf (480 kg/​m3), often known as cellular concretes, are also used for insulating purposes and for masonry units. When lightweight materials are used for both coarse and fine aggregates in structural concrete, it is termed all-​lightweight concrete. When only the coarse aggregate is of lightweight material but normal weight sand is used for the fine aggregate, it is said to be sand-​lightweight concrete. Often the term “sand replacement” is used in connection with lightweight concrete. This refers to replacing all or part of the lightweight aggregate fines with natural sand. Steiger [1.16] provides historical background for the use of lightweight aggregate concrete, Mackie [1.17] has discussed uses of lightweight concrete, and ACI Committee 213 has a guide for the use of structural lightweight aggregate concrete [1.18]. Heavyweight, high-​density concrete is used for shielding against gamma and X radiation in nuclear reactor containers and other structures [1.19]. Naturally occurring iron ores, titaniferous iron ores, “hydrous iron ores” (i.e., containing bound and adsorbed water), and barites are crushed to suitable size for use as aggregates. Heavyweight concretes typically weigh from 200 to 350 pcf (3200–​5600 kg/​m3).

1.6 ADMIXTURES In addition to cement, coarse and fine aggregates, and water, materials known as admixtures, mineral or chemical, are often added to the concrete mix immediately before or during the mixing. Admixtures are used to modify the properties of the concrete to make it better serve its intended use or for better economy.

2  4.75 mm according to ASTM Standard E11. 3  Numbers refer to sections in the “ACI Code,” officially 318-​14, Building Code Requirements for Structural Concrete [1.12].

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1.6 ADMIXTURES Low-density concrete

Moderate strength concrete

Structural concrete Expanded slag Sintering grate expanded shale, clay, or fly ash Rotary kiln expanded shale, clay, and slate

Scoria Pumice Perlite Vermiculite kgf/m3 pcf 20

400

600 40

800

1000

1200

1400

60 80 28-day air dry unit weight

1600 100

1800

kgf/m3 120 pcf

Figure 1.5.1  Approximate unit weight and use classification of lightweight aggregate concrete. (From Ref. 1.114.)

Mineral Admixtures Mineral admixtures are finely divided materials including pozzolans such as fly ash, cement slag, and silica fume. These admixtures are often used as cement replacement, but can also be used in addition to cement, as replacement of sand. In general, mineral admixtures reduce the heat of hydration and improve workability and durability. General information about mineral admixtures is available in [1.7] and [1.20]. Mielenz has given a history and background on mineral admixtures [1.21]. Fly ash Fly ash is a by-​product from the combustion of coal in power plants. Fly ash used in concrete shall meet ASTM C618 [1.22]. Since its cost is substantially lower than that of cement, it is typically used as cement replacement for economic reasons. The use of fly ash in concrete improves workability, reduces the heat of hydration, and increases durability. Albinger [1.23] provides general information on when to use and what to expect from fly ash concrete, and Ravina [1.24] discusses slump retention of fly ash concrete (see Section 1.7) with and without chemical admixtures. Cement Slag Blast-​furnace slag is a by-​product of iron production. This by-​product is first granulated and then ground to achieve particle sizes similar to those of cement. For use as mineral admixture in concrete, slag shall meet ASTM C989 [1.25]. Cement slag in concrete reduces the heat of hydration and provides increased durability by decreasing concrete permeability and increasing resistance to sulfate attacks. Although the strength gain of concrete with cement slag is lower in the first few days, compressive strength after that is typically greater than or comparable to that of concrete without slag cement. Additional information on the use of cement slag in concrete can be found in a report by ACI Committee 233 [1.26]. Silica Fume Silica fume is the finely divided solid-​microsilica material collected from the fumes of electric furnaces that produce ferrosilicon or silicon metal. When used in concrete, it shall conform to ASTM C1240 [1.27]. In addition to its use as a pozzolan, silica fume in the concrete mix produces a more impermeable concrete, able to resist chloride intrusion into concrete exposed to deicing chemicals. Silica fume is an important admixture in high-​performance

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concrete to achieve high strength and excellent durability. Further information about silica fume is available from Cohen, Olek, and Mather [1.28] and Durning and Hicks [1.29].

Chemical Admixtures A wide variety of chemical admixtures for concrete are available, most commonly air-​ entraining, water-​reducing, and set-​controlling admixtures. A history and background on chemical admixtures can be found in Durning and Hicks [1.30]. A report by ACI Committee 212 [1.31] provides an essential guidance for use of chemical admixtures. In this report, chemical admixtures are classified into categories that include air-​entraining admixtures, accelerating admixtures, water-​reducing and set-​retarding admixtures, admixtures for flowing concrete, and admixtures for self-​consolidating concrete. A brief discussion on these chemical admixtures follows. Air-​Entraining Admixtures These chemicals, meeting the requirements of ASTM C260 [1.32], can be added either to the hydraulic cement or as an admixture to the concrete mix. The chemical causes air in the form of minute bubbles (often 0.004 in. or 0.1 mm diameter or smaller) to be dispersed throughout the concrete mix, with the purpose of increasing workability and resistance to deterioration that results from both freeze-​thaw action and ice-​removal salts. Air-​entraining admixtures are probably the most widely used type of chemical admixture. In addition to resistance to freeze-​thaw cycles and to the corrosiveness of deicing chemicals, air entrainment improves plasticity and workability, permitting a reduction in water content. Uniformity of placement with little bleeding and segregation can be achieved. In addition, air-​entrained concrete is more watertight and increases resistance to sulfate action. For exposed concrete, the possible reduction in strength (approximately 5% for each percent of entrained air) is far less important than the improved durability in terms of resistance to freeze-​thaw action and deicing chemicals. Accelerating Admixtures Accelerating admixtures modify the properties of concrete, particularly in cold weather, to (a) accelerate the rate of early-​age strength development; (b) reduce the time required for proper curing and protection; and (c)  permit earlier start of finishing operations. Accelerators must not be used as antifreeze agents for concrete. Accelerators must meet the requirements of Type C or E in ASTM C494 [1.33]; calcium chloride, the best known and most common accelerator, must also meet the requirements of ASTM D98, Specifications for Calcium Chloride. Limits on water-​soluble chloride, however, are specified in ACI 318 (ACI-​19.3.2.1) to reduce potential for corrosion of reinforcement. Water-​Reducing and Set-​Retarding Admixtures Water-​reducing admixtures are used to reduce the amount of water required for a given slump or, when used without water reduction, to increase concrete workability. Some of these admixtures also increase the setting time for concrete. Water-​reducing and set-​ retarding admixtures must meet the requirements of ASTM C494 [1.33], where they are classified as water-​reducing admixtures (Type A); retarding admixtures (Type B); water-​ reducing and retarding admixtures (Type D); water-​reducing and accelerating admixtures (Type E); water-​reducing, high-​range admixtures (Type F); and water reducing, high-​range, and retarding admixtures (Type G). Water-​reducing, high-​range admixtures are sometimes referred to as “superplasticizers,” meaning that the quantity of mixing water is reduced by 12% or more. The last two classifications (Type F and Type G) are also covered by ASTM C1017 [1.34]. A report by ACI Committee 212 [1.31] lists seven general categories for materials used as water-​reducing admixtures, including lignosulfonic acids and their salts, hydroxylated carboxylic acids and their salts, carbohydrate-​based compounds and polysaccharides, and

9



1.7  COMPRESSIVE STRENGTH

9

polycarboxylates. Materials used for water-​reducing, high-​range admixtures, on the other hand, include sulfonated naphthalene condensates, sulfonated melamine condensates, modified lignosulfonates, and polycarboxylates. Ramezanianpour, Sivasundaram, and Malhotra [1.35] have discussed superplasticizers and their effect on strength properties of concrete. Set-​retarding admixtures are used primarily to offset the accelerating and damaging effect of high temperature, to keep concrete workable during placement, and to minimize form-​deflection cracks. A variety of water-​reducing admixtures also serve as set-​retarding admixtures. Admixtures for Flowing Concrete Flowing concrete is defined [1.34] as “concrete that is characterized as having a slump greater than 7 in. (190 mm) while maintaining a cohesive nature … .” These admixtures are classified by ASTM C1017 [1.34] into two types: Type I—​Plasticizing, and Type II—​ Plasticizing and Retarding. Flowing concrete is commonly used where high rates of casting are required or in highly reinforcement-​congested members [1.31]. Admixtures for Self-​Consolidating Concrete Self-​consolidating concrete is highly flowable concrete that requires no vibration. Given its high flowability, a flow slump, rather than slump, is measured in self-​consolidating concretes. In general, flow slumps between 22 and 30 in. (550 and 750 mm) are associated with these concretes. Two primary types of chemical admixtures are used in self-​consolidating concrete, high-​range, water-​reducing admixtures (typically polycarboxylate based) and viscosity-​modifying admixtures (polymer based or in the form of fine particles). High-​ range, water-​reducing admixtures are used to lower the yield stress of the material, while viscosity-​modifying admixtures serve to increase cohesion and plastic viscosity when a concrete with low yield stress and high plastic viscosity is desired [1.31]. More information about self-​consolidating concrete can be found in a 2007 ACI publication [1.36]. Besides the admixture categories listed above, ACI Committee 212 [1.31] lists the following categories for chemical admixtures: cold-​weather admixture systems, admixtures for very high-​early strength concrete, extended set-​control admixtures, shrinkage-​reducing admixtures, corrosion-​ inhibiting admixtures, lithium admixtures to reduce deleterious expansion from alkali-​silica reaction, permeability-​reducing admixtures, and miscellaneous admixtures. Miscellaneous admixtures are categorized as bonding admixtures; coloring admixtures; flocculating admixtures; fungicidal, germicidal, and insecticidal admixtures; rheology-​and viscosity-​modifying admixtures; and air-​detraining admixtures. Details are available in the ACI Committee 212 Report [1.31].

1.7 COMPRESSIVE STRENGTH The strength of concrete is primarily controlled by the proportioning of cement, coarse and fine aggregates, water, and various admixtures. In reinforced concrete design, “concrete strength” means uniaxial compressive strength measured by a compression test, typically of a standard test cylinder. The most important variable in determining concrete strength is the water to cement (w/​c) ratio, as shown in Fig. 1.7.1. The lower the water/​ cement ratio, the higher the compressive strength. This relationship has been recognized since the 1920s. In the past decades, with the increasing use of admixtures, many of which contain cemen­ titious materials, researchers have confirmed that any cementitious admixtures should be included with the cement in determining the proper mix to obtain a specified strength. This is recognized in ACI 318, where a water to cementitious ratio (w/​cm) shall be calculated, including the weight of fly ash and other pozzolans (ASTM C618 [1.22]), slag cement (ASTM C989 [1.25]), and silica fume (ASTM C1240 [1.27]). Popovics [1.37] and Popovics and

10

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500 For Type I portland cement

45

450

6000

4000

350 300

3000

kgf/cm2

5000

40

400

N on -a iren Ai tra r-e ine nt ra dc in on ed c re co te nc ret e

35 30

250

25

200

20

MPa

7000

Compressive strength, psi

10

15 4

5 6 7 Water–cement ratio, gallons per bag 0.4

0.5

0.6

8

0.7

Water–cement ratio, by weight

Figure 1.7.1  Effect of water-​to-​cement ratio on 28-​day compressive strength. Average values for concrete containing 1.5 to 2% trapped air for non-​air-​entrained concrete and no more than 5 to 6% air for air-​entrained concrete. (Curves drawn from data in Ref. 1.42, Table 6.3.4a.)

Popovics [1.38, 1.39] have reviewed the validity of strength based on the water to cementitious material ratio. A certain minimum amount of water is necessary for the proper chemical action in the hardening of concrete; extra water increases the workability (the ease of concrete flow) but reduces strength. A measure of the workability is obtained by a slump test. A truncated cone-​shaped metal mold 12 in. (300 mm) high is filled with fresh concrete, the mold is then lifted off, and a measurement is made of the distance to the top of the wet mass “slumps” from its position before the mold was removed. The smaller the slump, the stiffer and less workable the mix. In building construction, a 3-​to 4-​in. (75-​to 100-​mm) slump is common. Vibration of the concrete mix will greatly improve workability and even very stiff no-​slump concrete can be placed [1.40]. Proportioning of concrete mixes can be done in accordance with Design and Control of Concrete Mixtures [1.41], as well as ACI Standard 211.1 for normal-​weight, heavyweight, and mass concrete [1.42], ACI Standard 211.2 for structural lightweight concrete [1.43], and ACI Standard 211.3 for no-​slump concrete [1.40]. Strength of concrete in place in the structure is also greatly affected by quality control procedures for placement and inspection. Details regarding good practice are available in ACI Standard 304 [1.44] and in the ACI Manual of Concrete Inspection [1.45]. Durability, long recognized as an important quality of concrete, is related to the w/​cm ratio and compressive strength, among other factors. Durability requirements for concrete can be found in Chapter  19 of the ACI Code (Concrete:  Design and Durability Requirements). A source for obtaining durable concrete is the Committee 201 Guide to Durable Concrete [1.46]. The strength of concrete is denoted in the United States by fc′, which is the compressive strength in psi of test cylinders with diameter and height of either 6 in. (150 mm) and 12 in. (300 mm) or 4 in. (100 mm) and 8 in. (200 mm), typically measured at 28 days after casting. In many parts of the world, the standard test unit is the cube, frequently measuring 8 in. (200 mm) to a side. Since nearly all reinforced concrete behavior is related to the standard 28-​day compressive strength fc′, it is important to note that such strength depends on the size and shape of the test specimen and the manner of testing [1.47]. Properties such as tensile strength of concrete and size of contact area of the testing machine have more effect on cube strength than on cylinder strength. As an average, the 6 × 12 in. (150 × 300 mm) cylinder strength

1



11

1.7  COMPRESSIVE STRENGTH

is 80% of the 6-in. (150-​mm) cube strength and 83% of the 8-​in. (200-​mm) cube strength [1.48]. For lightweight concrete, cylinder strength and cube strength are nearly equal. Given the effect on cylinder compressive strength of numerous test variables (e.g., load rate, specimen dimensions, casting and curing conditions), it is clear that such strength will differ from the in-​place concrete strength in a structure. Results from a thorough investigation of in-​place versus molded cylinder concrete compressive strengths were reported by Bloem [1.49]. Compressive strengths obtained from tests of drilled cores were less than those of cylinders. Compared with strengths of field-​cured cylinders, the compressive strength of drilled cores averaged between 11 and 21% less, depending on curing conditions. An interesting discussion of the cylinder test is given by Shilstone [1.50], and Tait [1.51] has discussed the use of test results. When an assessment of the strength of in-​place concrete is desired, procedures ranging from tests of cylindrical cores cut from the structure to the use of nondestructive tests [1.52–​1.58] are available. ACI Committee 214 [1.59] has a recommended practice for evaluating strength test results from concrete cores.

Stress-​Strain Relationship The stress-​strain relationship for concrete depends on its strength, age at loading, rate of loading, aggregates and cement properties, and type and size of specimens [1.60, 1.61]. Typical curves for specimens (6 × 12 in. cylinders) loaded in compression at 28 days using normal testing speeds are shown in Fig. 1.7.2. The rate of applying strain during testing influences the shape of the stress-​strain curve, as shown in Fig. 1.7.3, particularly the portion after the maximum stress has been reached. Note from Fig. 1.7.2 that lower-​strength concrete has greater deformability (ductility) than higher-​strength concrete, as evidenced by the length and smaller slope of the descending portion of the curve after the maximum stress has been reached at a strain between 0.002 and 0.0025. Ultimate strain at crushing of concrete often varies from 0.003 to as high as 0.008.

0p

si

12

80

700

70

600

60

500

50

8000

8 7

6000

6

400

5000

5

4000

4

3000

3

kgf/cm2

9

MPa

12

10,000

fc’ =

10 Concrete compressive stress, fc, ksi

800

,00

11

40

300

30

200

20

100

10

2 1 0

0.001

0.002

0.003

0.004

0

Strain, in./in. (mm/mm)

Figure 1.7.2  Typical stress-​strain curves for concrete in compression under short-​time loading. (Curves represent a compromise adapted from curves and results given by Wang, Shah, and Naaman [1.61], Bertero [1.62], Naaman [1.63], and Nilson [1.64].)

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Ratio of concrete stress to cylinder strength

1.00

Strain rate 0.001 per 100 days 0.0

0.75

0.00

00

0.

00

0.50 0.25

01

0.

Cylinder strength fc’ = 3000 psi at 56 days

0.001

1

1

pe

rh

pe

rm

per

day

in

ou

r

ut

e

0.002 0.003 0.004 0.005 Concrete compressive strain, εc

0.006

0.007

Figure 1.7.3  Stress-​strain curves for various strain rates of concentric loading. (From Rüsch [1.65].)

In usual reinforced concrete design, specified concrete strengths fc′ of 3500 to 5000 psi (24–​35 MPa) are used for nonprestressed structures, and 5000 to 8000 psi (35–​56 MPa) are used for prestressed concrete. For special situations, particularly in columns of tall buildings, concretes ranging from 6000 to 14,000 psi (42–​97 MPa) have been used [1.66–​1.68]. On the Pacific First Center in Seattle, the specified strength was 14,000 psi (97 MPa) at 56  days [1.66]. The average strength obtained throughout the project was about 18,000 psi (124 MPa). Research continues on high-​strength concrete (often referred to as “high-​ performance” concrete) because in addition to high strength, the concrete must have other excellent characteristics [1.69, 1.70].

1.8 TENSILE STRENGTH The strength of concrete in tension is an important property that greatly affects the extent and size of cracking in structures. Tensile strength is usually determined by using the split-​ cylinder test in accordance with ASTM C496/​C496M [1.71]. In this test, the same size cylinder used for the compression test is placed on its side in the testing machine so that the compression load P is applied uniformly along the length of the cylinder in the direction of the diameter. The cylinder will split in half when the tensile strength is reached. The stress is computed by 2P/​[π(diameter)(length)] based on the theory of elasticity for a homogeneous material in a biaxial state of stress.4 Tensile strength is a more variable property than compressive strength and is about 10 to 15% of it. The split-​cylinder tensile strength fct has been found to be proportional to fc′ ,5 such that

fct = 6 fc′ to 7 fc′ psi for normal-weight conrete 6 fct = 5 fc′ to 6 fc′ psi for lightweight conrete 6

In the ACI Code, fct = 6.7 fc′ psi is assumed for normal-​weight concrete. The lower split-​ cylinder tensile strength exhibited by lightweight concrete compared to normal-​weight concrete is accounted for in the ACI Code through the use of a factor λ that multiples the

4  See, for example, S. Timoshenko and J. N. Goodier, Theory of Elasticity, 3rd ed., McGraw-​Hill, 1970, pp. 122–​123. 5  fc′ is in psi units; thus fc′ = 3000 psi, fc′ = 54.8 psi. When fc′ is in kilogram-​force per square centimeter (kgf/​cm2), the constant in front of fc′ is to be multiplied by 0.265; when fc′ is in newtons per square millimeter, that is, megapascals (MPa), the constant in front of fc′ is to be multiplied by 0.083. 6  In SI, with fc′ and fct in MPa,



fct = 0.5 fc′ to 0.6 fc′

for normal-weight conrete

fct = 0.4 fc′ to 0.5 fc′

for lightweight conrete

13



13

1.8  TENSILE STRENGTH

strength associated with normal-​weight concrete (see, e.g. Lightweight Concrete in Section 5.5). This factor λ may be determined from results of split-​cylinder tests as (ACI-​19.2.4.3)

λ=



fct 6.7 fcm

≤ 1.0

(1.8.1)



where fcm is the measured average compressive strength of the concrete. Alternatively to the use of results from split-cylinder tests, λ may be taken as 0.75 for all-​lightweight concrete i.e., fct = 5 fc′ and 0.85 for sand-​lightweight concrete i.e., fct = 5.7 fc′ (ACI-​19.2.4.2). For lightweight-​fine blend concrete, λ ranges between 0.75 and 0.85 depending on the absolute volume of normal-​weight fine aggregates as a fraction of the total absolute volume of fine aggregates. Similarly, for sand-​lightweight, course blend concrete, λ ranges between 0.85 and 1.0 based on the absolute volume of normal-​weight course aggregate as a fraction of the total absolute volume of course aggregate. Tensile strength in flexure, known as modulus of rupture, measured in accordance with ASTM C78 [1.72], is also important when considering cracking and deflection of beams. The modulus of rupture fr , computed from the flexure formula f = Mc / I , gives higher values for tensile strength than the split-​cylinder test, primarily because the concrete compressive stress distribution is not linear when tensile failure is imminent, as is assumed in the computation of the nominal Mc / I stress. It is generally accepted (ACI-​19.2.3.1) that an average value for the modulus of rupture fr may be taken as 7.5λ fc′ 0.62 λ fc′ MPa , where λ = 1.0 for normal-​weight concrete, while λ = 0.85 and 0.75 for sand-​lightweight concrete and all-​lightweight concrete, respectively, as discussed above. Because of the large variability in modulus of rupture, as shown in Fig. 1.8.1, the selection of the coefficient 7.5, or even the entire expression 7.5λ fc′, should be viewed as a practical choice for design purposes. One may note that neither the split-​cylinder nor the modulus of rupture tensile strength is correctly a measure of the strength under axial tension. However, axial tension is difficult to measure accurately and, when compared with the modulus of rupture or split-​cylinder strength, it does not give better correlation with tension-​related failure behavior such as inclined cracking from shear and torsion or splitting from interaction of reinforcing bars with surrounding concrete.

)

(

)

(

(

1000

10

20

f’c (MPa) 30 40

50

60

900

6.0

5.0

700 600

4.0

500

fr (MPa)

Modulus of rupture, fr (psi)

800

3.0

400

fr = 7.5 f’c (ACI-19.2.3.1)

300

2.0

200 1000

3000

5000

7000

9000

Concrete compressive strength, f’c (psi)

Figure 1.8.1  Comparison of test results for modulus of rupture of normal-​weight concrete with ACI Code expression. (Adapted from Mirza, Hatzinikolas, and MacGregor [1.73].)

)

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1.9 BIAXIAL AND TRIAXIAL STRENGTH Concrete is seldom subjected to only uniaxial compressive or tensile stress. For example, the presence of shear in flexural members generates biaxial stresses in the concrete and the restraint against lateral expansion provided by transverse reinforcement in members under axial compression leads to triaxial compressive stresses. Fig. 1.9.1 shows a biaxial stress interaction diagram from Ref. 1.74. Under biaxial compression, concrete compressive strength is greater than the uniaxial compressive strength, being approximately 30% higher on average for a 1:0.5 ratio of biaxial compressive stresses. Under combined tension and compression, the strength interaction is approximately linear with strengths lower than both the uniaxial compressive and tensile strength. Under equal biaxial tension, concrete strength is approximately equal to the uniaxial tensile strength. Compressive strength and deformation capacity are greatly increased by the presence of lateral compressive stresses. From tests of cylinders subjected to triaxial compression with the largest stress applied in the longitudinal direction, Richart, Brandtzaeg, and Brown [1.75] showed that concrete strength increases at a rate of approximately 4.1 times the magnitude of the lateral (confining) stress. Strain capacity also increased with an increase in lateral pressure, with strain at peak stress ranging between 0.5 and 7% for the range of lateral pressures considered (Fig. 1.9.2).

1.10 MODULUS OF ELASTICITY The modulus of elasticity of concrete varies, unlike that of steel, with strength. It also depends, though to a much lesser extent, on the age of the concrete, the properties of the aggregates and cement, the rate of loading, and the type and size of the specimen. Furthermore, since concrete exhibits some permanent set even under small loads, there are various definitions of the modulus of elasticity. Figure 1.10.1 represents a typical stress-​strain curve for concrete in compression. In the figure, the initial modulus (tangent at origin), the tangent modulus (at 0.5 fc′ ), and the secant modulus (also at 0.5 fc′ ) are noted. Usually the secant modulus at 25 to 50% of the compressive strength fc′ is considered to be the modulus of elasticity. For many years the modulus was approximated adequately as 1000 fc′ by the ACI Code; but with the rapidly increasing

fc (psi)

fc2/fc’

fc’ = 3660 psi

20,000

1.2 1.0

fc2

0.8

fℓ = 4090 psi

16,000

P

fc1

fℓ = 2010 psi

12,000

0.6

=

fℓ

f c2

f c1

8000

fℓ = 1090 psi

0.4

0.2

0.4

0.6

0.8

1.0

1.2

Figure 1.9.1  Strength of concrete under biaxial stress. (Adapted from Fig. 6 in Ref. 1.74.)

fc1/fc’

P

fℓ = 550 psi

4000

0.2

fℓ

0.01

0.02

0.03

0.04

0.05

0.06

εc

Figure 1.9.2  Stress-​strain response of concrete under triaxial compression. (Adapted from Fig. 23 in Ref. 1.75.)

15



15

1.10  MODULUS OF ELASTICITY

Concrete compressive stress

fc’

Tangent modulus at 0.5 fc’ Ultimate strain typically varies from 0.003 to 0.004

Initial modulus (tangent at origin) 0.5fc’

Secant modulus at 0.5 fc’

0

0.001

0.003

0.002

0.004

Concrete strain

Figure 1.10.1  Definitions of the modulus of elasticity for concrete in compression. Values of the modulus of elasticity for various concrete strengths appear in Table 1.10.1.

TABLE 1.10.1  VALUES OF MODULUS OF ELASTICITY (USING EC = 33 w1.5 c

f c′ FOR NORMAL-​WEIGHT CONCRETE WEIGHING 145 PCF) Inch-​Pound Units

SI Unitsb

fc′ (psi)

Ec (psi)

fc′ (MPa)

Ec† (MPa)

3000 3500 4000 4500 5000 6000 8000

3,150,000 3,400,000 3,640,000 3,860,000 4,070,000 4,460,000 5,150,000

21a 24 28 31 35 41 55

21,500 23,000 24,900 26,200 27,800 30,100 34,900

These metric values are rounded values approximating concrete strengths in Inch-​Pound units; actual equivalents for 3000, 3500, 4000, 4500, 5000, 6000, and 8000 psi are 20.7, 24.1, 27.6, 31.0, 34.5, 41.3, and 55.1 MPa, respectively. b Multiply MPa values by 10.2 to obtain kgf/​cm2. a

†

Using Ec = 4700

f c′ as per ACI 318-​14M.

use of lightweight concrete, the variable of density needed to be included. As a result of a statistical analysis of available data, the empirical formula given by ACI-​19.2.2.1 Ec = 33w1c .5 fc′



(1.10.1)7

was developed [1.76] for values of wc between 90 and 155 pcf. Equation (1.10.1) is representative of the secant modulus at a compressive stress of 0.45 fc′ . Reviews of the applicability of Eq. (1.10.1) have been made by Shih, Lee, and Chang [1.77] and also Oluokun, Burdette, and Deatherage [1.78]. For normal-​weight concrete weighing 145 pcf, Eq. (1.10.1) gives Ec = 57,600 fc′ . For normal-​weight concrete, ACI-​19.2.2.1 suggests Ec = 57, 000 fc′





7  For SI, with wc in kg/m3 and Ec and fc′ in MPa, Ec = 0.043w15 fc′ c 8  For SI, Ec and fc′ in MPa, and with Ec and fc′ in kgf /cm2.

(ACI 318-14M)

(1.10.2)8

(1.10.1) 

(ACI 318-14M)

(1.10.2) 

Ec = 15, 000 fc′ (approximate)

 

Ec = 4700 fc′

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1.11 CREEP AND SHRINKAGE Creep and shrinkage are time-​dependent deformations that, along with cracking, cause a great concern for the designer because of the inaccuracies and unknowns that surround them. Concrete may behave essentially elastic only under loads of short duration; and, because of additional deformation with time, the effective behavior is that of an inelastic material. Deflection after a long period of time is therefore difficult to predict, but its control is needed to assure serviceability during the life of the structure.

Creep Creep is the property of concrete (and other materials) by which it continues to deform with time under sustained loads at unit stresses within the accepted elastic range (say, below 0.5 fc′. This inelastic deformation increases at a decreasing rate during the time of loading, and its total magnitude may be several times as large as the short-​time elastic deformation. Frequently, creep is associated with shrinkage, since the two occur simultaneously and often provide the same net effect: increased deformation with time. As may be noted by the general relationship of deformation versus time in Fig. 1.11.1, the “true elastic strain” decreases, since the modulus of elasticity Ec is a function of concrete strength fc′, which increases with time. Although creep is separate from shrinkage, it is related to it. Detailed information is available for estimating creep [1.79, 1.80]. The internal mechanism of creep, or “plastic flow” as it is sometimes called, may be due to any one or a combination of the following: (1) crystalline flow in the aggregate and hardened cement paste, (2) plastic flow of the cement paste surrounding the aggregate, (3) closing of internal voids, and (4) the flow of water out of the cement gel due to external load and drying. Factors affecting the magnitude of creep are (1) the constituents—​such as the composition and fineness of the cement, the admixtures, and the size, grading, and mineral content of the aggregates; (2) proportions, such as water content and the water/cement ratio; (3) curing temperature and humidity; (4) relative humidity during period of use; (5) age at loading; (6) duration of loading; (7) magnitude of stress; (8) surface to volume ratio of the member; and (9) slump. Accurate prediction of creep is complicated because of the variables involved; however, a general prediction method developed by Branson [1.80] gives a standard creep coefficient equation (4 in. or less slump, 40% relative humidity, moist cured, and loading at 7 days or more), Ct =

creep strain initial elastic strain

t 0.60 Cu = 10 + t 0.60



(1.11.1)

shown in Fig.  1.11.2, where t is the duration of loading (days) and Cu is the ultimate creep coefficient. (Branson [1.80] suggests using an average of 2.35 for Cu under standard

Strain

Creep Shrinkage True elastic strain t0

Nominal elastic strain

Time

Figure 1.11.1  Change in strain of a loaded and drying specimen; t0 is the time at application of load.

17



17

1 . 1 1   C R E E P A N D S H R I N K AG E

creep strain Ct = elastic strain

conditions, but the range is shown to be from 1.3 to 4.15.) Correction factors are given for relative humidity, loading age, minimum thickness of member, slump, percent fines, and air content. For practical purposes, the only factors significant enough to require correction are humidity and age at loading. The effect of unloading may be seen from Fig. 1.11.3, where at a certain time t1 the load is removed. There is an immediate elastic recovery and a long-​time creep recovery, but a residual deformation remains. Creep of concrete will often cause an increase in the long-​term deflection of members. Unlike concrete, steel is not susceptible to creep. For this reason, steel reinforcement is often provided in the compression zone of beams to reduce their long-​term deflection.

Cu

Ct = 0.78Cu at 1 year

Ct = 0.90Cu at 5 years

Eq. (1.11.1)

100

200

300

400

500

600

Duration of loading, days

Deformation

Figure 1.11.2  Standard creep coefficient variation with duration of loading (for 4 in. or less slump, 40% relative humidity, moist cured, and loading at 7 days or more).

Elastic recovery Creep Creep recovery Elastic

t1 Time after load application

Figure 1.11.3  Typical relationship between creep and recovery with time.

Shrinkage Shrinkage, broadly defined, is the volume change during hardening and curing of the concrete. It is unrelated to load application. The main cause of shrinkage is the loss of water as the concrete dries and hardens. It is possible for concrete cured continuously under water to increase in volume; however, the usual concern is with a decrease in volume. A discussion of the mechanisms of shrinkage may be found in Ref. 1.7. In general, the same factors have been found to influence shrinkage strain as those that influence creep—​primarily those factors related to moisture loss. The Branson general prediction method [1.80] gives a standard shrinkage strain equation (for 4 in. or less slump, 40% ambient relative humidity, and minimum member dimension of 6 in. or less, after 7 days moist cured)

 t  ε sh =  (ε )  35 + t  sh u

(1.11.2)

shown in Fig.  1.11.4, where t is time (days) after moist curing, and (ε sh )u is the ultimate shrinkage strain. (Branson [1.80] suggests using 800 × 10 –6 for average conditions,

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Shrinkage strain, ε sh

(εsh)u

εsh = 0.91(εsh)u at 1 year

εsh = 0.72(εsh)u at 3 months

Eq. (1.11.2)

100

200 300 400 Time after moist curing, days

500

600

Figure 1.11.4  Standard shrinkage strain variation with time after 7 days of moist curing (for 4 in. or less slump, 40% ambient relative humidity, and minimum member dimension of 6 in. or less).

but the range is from approximately 400 to more than 1000 × 10 −6 .) Correction factors are given with the primary one relating to humidity H,

correction factor = 1.40 − 0.01H for 40% ≤ H ≤ 80% correction factor = 3.00 − 0.03H for 80% ≤ H ≤ 100%

Shrinkage, particularly when restrained unsymmetrically by reinforcement, causes deformations generally additive to those of creep. For proper serviceability, it is desirable to predict or compensate for shrinkage in the structure.

1.12 CONCRETE QUALITY CONTROL In reinforced concrete design, concrete sections are proportioned and reinforced using a specified compressive strength fc′. The strength fc′ for which each part of a structure has been designed should be clearly indicated on the design drawings. In the United States, as indicated in Section 1.7, fc′ is based on cylinder strength (6 × 12 or 4 × 8 in. cylinders). Because concrete is a material whose strength and other properties are not precisely predictable, test cylinders from a mix designed to provide, say, 4000 psi (roughly 28 MPa) concrete will show considerable variability. Therefore, mixes must be designed to provide an average compressive strength greater than the specified value  fc′. ACI-​26.4.3.1(b) allows concrete to be proportioned to achieve the specified compressive strength following ACI 301 [1.81], based on either field test data or trial mixes. When the ready-​mix plant or other concrete production facility has a field test record based on at least 15 consecutive strength tests, or two groups of consecutive strength tests with a total no less than 30 tests and at least 10 tests in a group, for materials and conditions similar to those expected, the standard deviation s can be computed based on those tests to establish how variable the concrete strength is. These records shall correspond to a concrete with compressive strength within 1000 psi (6.9 MPa) from the specified concrete strength and obtained within the past 12 months, from a period no less than 60 calendar days. When at least 30 consecutive strength tests are the basis for computing the standard deviation, s, ACI 301 indicates that the required average compressive strength fcr′ used for proportioning the mix must be taken as the larger of Eq. (1.12.1) and the appropriate one of either (1.12.2) or (1.12.3): 

fcr′ = fc′ + 1.34 s

(1.12.1)

fcr′ = fc′ + 2.33s − 500

(1.12.2)

and when fc′ ≤ 5000 psi,

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1.13  STEEL REINFORCEMENT

19

or when fc′ > 5000 psi,

fcr′ = 0.90 fc′ + 2.33s (1.12.3)

For example, if the designer has used a specified strength fc′ of 4000 psi, and the concrete producer has shown field test data with a standard deviation of 450 psi, the mix should be designed for an average strength of 4600 psi [i.e., the larger of 4000 + 1.34 ( 450 ) and 4000 + 2.33 ( 450 ) − 500 ]. When fewer than 30 tests are available, a modification factor will require using a higher value of  fcr′ . When data are not available to establish a standard deviation, the required average compressive strength fcr′ is calculated as follows when fc′ ≤ 3000 psi,

fcr′ = fc′ + 1000

when 3000 psi ≤ fc′ ≤ 5000 psi,

fcr′ = fc′ + 1200

when fc′ > 5000 psi,

fcr′ = 1.1 fc′ + 700

According to ACI-​26.12.3.1(b), concrete compressive strength is considered acceptable if both conditions below are satisfied. 1. Every arithmetic average of any three consecutive strength tests9 equals or exceeds  fc′. 2. No individual strength test falls below fc′ by more than 500 psi when fc′ is 5000 psi or less; or by more than 0.10 fc′ when fc′ exceeds 5000 psi. Statistical variations are to be expected, and strength tests failing to meet the above criteria will occur perhaps once in 100 tests even though all proper procedures have been followed. A discussion on the risks inherent in the consideration of limited test data is provided by Tait [1.82]. ACI-​26.12.4 specifies steps to be taken in case low-​strength test results are obtained. The foregoing discussion of concrete strength variation should merely give an awareness of the fact that concrete having a specified compressive strength fc′ cannot be expected to provide precisely known actual strength and other properties. Quality control in the broader sense for reinforced concrete construction is a subject of great importance, but generally lies outside the scope of this text. The ACI Committee 121 Report [1.83] and the papers by Tuthill [1.84], Mather [1.85], Newman [1.86], and Scanlon [1.87] provide an excellent overall treatment of this subject.

1.13 STEEL REINFORCEMENT Steel reinforcement may consist of bars, welded wire reinforcement, wires, or discrete fibers.

Deformed Bars For usual construction, bars (called deformed bars) having lugs or protrusions (deformations) are used (Fig. 1.13.1). Such deformations serve to deter slip of the bar relative to the concrete that surrounds it resulting from tension or compression in the bar. These 9  According to ACI 301-​4.2.2.8a, a strength test is the “average of at least two 6 × 12 in. cylinders or the average of at least three 4 × 8 in. cylinders made from the same concrete sample.”

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deformations can have different patterns depending on the bar producer (see Fig. 1.13.1), but they all have to meet minimum requirements of spacing, height, and gap according to ASTM standards. These deformed bars are available in the United States in sizes 3 8 to 2 1 4 in. (9.5–​57 mm) nominal diameter. Sizes of ASTM bars (in Inch-​Pound units) are indicated by numbers (see Table 1.13.1). For sizes #3 through #8, they are based on the number of eighths of an inch included in the nominal diameter of the bars. Bars designated #9 through #11 are round bars corresponding to the former 1 in. square, 1 1 8 in. square, and 1 1 4 in. square sizes, and bars designated #14 and #18 are round bars having cross-​sectional areas equal to those of 1 1 2 and 2 in. square sizes, respectively. The nominal diameter of a deformed bar is equivalent to the diameter of a plain bar having the same weight per foot as the deformed bar. For metric units, ASTM standards use “soft” conversion, as given in Table 1.13.2. Reinforcing bar steel in the United States is covered under ASTM designations as shown in Table 1.13.3 [1.88–​1.92]. The “Grade” of steel is the minimum specified yield stress10 expressed in ksi for Inch-​Pound reinforcing bar Grades 40, 50, 60, 75, 80, 100, and 120 and in MPa for SI reinforcing bar Grades 280, 350, 420, 520, 550, 690, and 830. Both Grades 40 and 60 exhibit the well-​defined yield point and elastic-​plastic strain behavior shown in Fig. 1.13.2(a). The overall relationships are shown in Fig. 1.13.2(b). Higher grade steels, on the other hand, often exhibit little or no yield plateau and lower ductility compared to Grades 40 and 60 steels. To ensure sufficient ductility for use in earthquake-​resistant structures as well as adequate weldability, ASTM A706/​A706M [1.91] has more restrictive mechanical and chemical properties than the other types of steel. Minimum elongation measured over a length of 8 in. (200 mm) ranges between 10 and 14% depending on the grade of steel and bar size. Also, restrictions apply to the ratio between tensile and actual yield strength, as well as the actual yield strength (see footnote d of Table 1.13.3). Deformation requirements for ASTM A615 and A615M steel [1.88] are lower than those for ASTM A706/​A706M steel, ranging between 6 and 9% for Grades 60, 75, and 80 steel. Higher deformation requirements apply to smaller bar sizes. No special requirements are specified for enhanced weldability. Axle and rail steel bars, both of which are rarely used now, are rerolled from old axles and rails and are generally less ductile than bars satisfying ASTM A615/​A615M. Grade 60 is the most widely used grade for reinforcing bars. However, higher-​strength steels are gaining popularity, since the ACI Code allows the use of a design yield strength of 80 ksi (550 MPa) for longitudinal reinforcement in structural members other than those of “special seismic systems” and of 100 ksi (690 MPa) for confinement reinforcement in earthquake-​resistant members and spirals in columns.

Figure 1.13.1  Deformed reinforcing bars. (Courtesy of Concrete Reinforcing Steel Institute.)

10  The term “yield stress” refers to either yield point, the well-​defined deviation from perfect elasticity, or yield strength, the value obtained by a 0.2% offset strain for material having no well-​defined yield point (ACI-​20.2.1.2).

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1.13  STEEL REINFORCEMENT

60

0

600 500

Smaller size bars, say 400 #8 and smaller 300

40 20

700

Grade 40 steels

Strain hardening begins at about 0.012 to 0.020

200

800

Tensile strength

100

700

Grade 60

600

80

500

Tensile strength

60

400

Grade 40

40

300 200

Yield stress Grade 40 20

100

0.002 0.006 0.010 0.014 0.004 0.008 0.012

Nominal stress, MPa

Strain hardening begins at strains as low as about 100 0.004 Larger bars, 80 Grade 60 say #11 and larger steels

120

800

Nominal stress, ksi

Nominal stress, ksi

120

Nominal stress, MPa



100

0

0.05 0.10 0.15 0.20 Strain

Strain (a) Enlarged portion in design

(b) Entire curve to rupture

Figure 1.13.2  Typical stress-​strain curves for Grade 60 and Grade 40 steel reinforcing bars in tension.

TABLE 1.13.1  ASTM STANDARD REINFORCING BAR DIMENSIONS AND WEIGHTS (BARS IN INCH-​POUND UNITS) Nominal Dimensions Diameter

Area

Weight

Bar Number

(in.)

(mm)

(sq in.)

(cm2)

(lb/​ft)

(kg/​m)

3 4 5 6 7 8 9 10 11 14 18

0.375 0.500 0.625 0.750 0.875 1.000 1.128 1.270 1.410 1.693 2.257

9.5 12.7 15.9 19.1 22.2 25.4 28.7 32.3 35.8 43.0 57.3

0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56 2.25 4.00

0.71 1.29 2.00 2.84 3.87 5.10 6.45 8.19 10.06 14.52 25.81

0.376 0.668 1.043 1.502 2.044 2.670 3.400 4.303 5.313 7.65 13.60

0.559 0.994 1.552 2.235 3.041 3.973 5.059 6.403 7.906 11.38 20.24

TABLE 1.13.2  1996 ASTM (“SOFT” METRIC) REINFORCING BAR DIMENSIONS AND WEIGHTS IN SI UNITS Metric Bar Number

Inch-​Pound Bar Number

Diameter (mm)

Mass (kg/​m)

Area (mm2)

10 13 16 19 22 25 29 32 36 43 57

3 4 5 6 7 8 9 10 11 14 18

9.5 12.7 15.9 19.1 22.2 25.4 28.7 32.3 35.8 43.0 57.3

0.560 0.994 1.552 2.235 3.042 3.973 5.060 6.404 7.907 11.38 20.24

71 129 199 284 387 510 645 819 1006 1452 2581

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TABLE 1.13.3  REINFORCING BAR STEELS Minimum Yield Stressa fy ASTM Designation A615/​A615Mb (Carbon steel)

A955/​A995M (Stainless steel) A996/​A996M (Rail steel,c axle steel)

A706/​A706Md (Low-​alloy steel) A1035/​A1035M (Low-​carbon, chromium steel)

Grade

Bar Sizes

ksi

40 60 75 80 280 420 520 550 60 75 420 520 40 50 60 280 350 420 60 80 420 550 100 120 690 830

#3–​#6 #3–​#18 #3–​#18 #3–​#18 10–​19 10–​57 10–​57 10-​57 #3–​#18 #3–​#18 10–​57 10–​57 #3–​#8 #3–​#8 #3–​#8 10–​25 10–​25 10–​25 #3–​#18 #3–​#18 10–​57 10–​57 #3–​#18 #3–​#18 10–​57 10–​57

40 60 75 80

MPa

Minimum Tensile Strength, fu ksi

MPa

70 90 100 105 280 420 520 550

60 75

500 620 690 725 90 100

420 520 40 50 60

620 690 70 80 90

280 350 420 60 80

500 550 620 80 100

420 550 100 120

550 690 150 150

690 830

1030 1030

The term “yield stress” refers to either yield point, the well-​defined deviation from perfect elasticity, or yield strength, the value obtained by a 0.2% offset strain for material having no well-​defined yield point (ACI-​20.2.1.2). a

b

Metric (SI) specification applies to bars designated numbers 10 through 57.

c

Although rail steel is no longer considered a practical source of bar reinforcement, its use is still permitted.

In addition to the yield and tensile strength limits given in the table, the tensile strength shall be at least 1.25 times the actual yield strength, where actual yield strength shall not exceed 78 and 98 ksi (540 and 675 MPa) for Grade 60 and Grade 80 steel, respectively. d

Wire Reinforcement Welded wire reinforcement is used in thin slabs, thin shells, thin webs of T-​beams, and other locations where available space would not permit the placement of deformed bars with proper cover and clearance. Welded wire reinforcement shall conform to either ASTM A1064 [1.93] for carbon steel or ASTM A1022 [1.94] for stainless steel. Welded wire reinforcement consists of cold-​worked wire, cold-​rolled or hot-​rolled from steel rod, in orthogonal patterns, square or rectangular, resistance welded at all intersections. The wires may be smooth or deformed. The wire is specified by the symbol W (for smooth wires) or D (for deformed wires), followed by a number representing the cross-​sectional area in hundredths of a square inch, varying from 1.5 to 45. On design drawings such reinforcement usually is indicated by the spacings of wires in the two orthogonal directions, followed by the type and wire sizes. Thus, 6 × 8—​W5 × W5 indicates welded wire reinforcement with 6-​ in. longitudinal wire spacing, 8-​ in. transverse wire spacing, and both sets of wires smooth and having a cross-​sectional area of 0.05 sq in. Unlike most hot-rolled steel bars, the wire used in welded wire reinforcement does not generally have a well-​defined yield point and is less ductile. Figure  1.13.3 shows

23



100

100

Tensile strength Nominal stress, ksi

Nominal stress, ksi

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1.13  STEEL REINFORCEMENT

80 60 40 20 1

2 3 Strain, %

4

Tensile strength

80 60 40 20

5

1

(a) Smooth

4

2 3 Strain, %

5

(b) Deformed

Figure 1.13.3  Typical stress-​strain curves for welded wire reinforcement.

typical stress-​strain curves for welded wire reinforcement. Additional information about welded wire reinforcement is available from the Wire Reinforcement Institute [1.95]. Wires in the form of individual wires conforming to either ASTM A1064/​A1064M or A1022/​A1022M can also be used as reinforcement for certain purposes. When deformed, wires can be used for confinement and as spiral reinforcement, among others. Plain wires, on the other hand, can be used only in the form of spirals.

Prestressing Reinforcement Wire reinforcement in the form of groups of wires forming strands or as individual wires are used for prestressing concrete. Wire and strands are available in great variety. The most prevalent strand is the 7-​wire strand, either stress relieved (i.e., normal relaxation) or low relaxation, conforming to ASTM A416 [1.96]. Low-​relaxation reinforcement is now regarded as the standard type. These strands have a center wire enclosed by six helically wound outside wires (see Fig.  1.13.4). Usual nominal diameters for 7-​wire strand are 1 4 , 3 8 , and 1 2 in. The minimum tensile strength for strands of Grade 250 is 250,000 psi (1725 MPa), and that of Grade 270 is 270,000 psi (1860 MPa); there is no well-​defined yield point. A  typical stress-​strain curve for stress-​relieved strands is shown in Fig. 1.13.5. ASTM A416 requires that the yield strength, measured at a 1% extension under load, should be at least 85% of the tensile strength for stress-​relieved strand and 90% of the tensile strength for low-​ relaxation strand. Typically, under service conditions, these prestressed strands have a stress of 150,000 to 160,000 psi (1030–​1100 MPa).

250

1724

fpu = 250 ksi

1000

100 500

0

1

2

3

4

5

Unit strain, %

Figure 1.13.4  Typical 7-​wire strand used in prestressed concrete construction (Photo by José A. Pincheira.)

Figure 1.13.5  Typical stress-strain curve for stress-relieved strands.

Unit stress, MPa

Unit stress, ksi

1500 200

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Additionally, for prestressing, uncoated stress-​relieved and low-​relaxation wire under ASTM A421 [1.97] and uncoated high-​strength steel bars under ASTM A722 [1.98] are used. The modulus of elasticity for all nonprestressed steel is permitted to be taken (ACI-​ 20.2.2.2) as 29,000,000 psi (200,000 MPa). For prestressing steel, the modulus of elasticity is lower and more variable; therefore it must be obtained from the manufacturer or from tests. A value of 27,000,000 psi (186,000 MPa) is often used for 7-​wire strands conforming to ASTM A416 [1.96].

Coated Reinforcement Corrosion of the steel reinforcement can occur when the structure is subjected to severe environmental conditions, such as in structures exposed to marine environments or bridge decks or parking garages subjected to deicing salts. The corrosion of a reinforcing bar embedded in concrete is a slow process that eventually will lead to cracking and spalling of the concrete cover. The repair of corrosion-​induced damage is often expensive and difficult. A proper concrete cover can effectively delay the corrosion of steel bars and it is generally agreed that larger covers will provide better protection. In severe environments, however, large concrete covers alone will not be effective. A common method to prevent or ameliorate corrosion of the reinforcement in concrete structures is the use of epoxy-​coated reinforcing bars. The surface of these bars is protected with a thin coat of epoxy (between 7 and 12 mils) to isolate the steel from the oxygen, moisture, and chlorides that will induce corrosion. Although the performance of epoxy-​coated bars has been the subject of some controversy in the past, many studies have shown that epoxy-​coated bars can effectively reduce corrosion of the reinforcement and extend the service life of concrete structures. Care must be exercised during transportation, handling, storage, and placing of the bars to prevent damage to the coating. Current practice requires that any damage to the coating (cracking, nicks, and cuts) be repaired before bar placement in the forms. The manufacturing requirements of these bars are presently covered by ASTM A775, Specification for Epoxy-​Coated Reinforcing Steel Bars [1.99], and ASTM A934, Specification for Epoxy-​Coated Prefabricated Steel Reinforcing Bars [1.100]. It must be noted that epoxy-​coating of bars will result in reduced slip resistance. The ACI Code contains specific provisions that account for the reduced ability of an epoxy-​coated bar to transfer the force in the reinforcement to the surrounding concrete. These provisions are discussed in detail in Chapter 6. Zinc-​coated (galvanized) bars are sometimes specified to reduce corrosion of steel reinforcement. Similar to epoxy-​coated bars, galvanized bars are protected with a thin layer of zinc on the surface. Although zinc coating will protect the steel bar, zinc will corrode in concrete [1.101]; and eventually, corrosion of the steel bar will also occur. While the use of galvanized bars can delay the onset of concrete spalling, it will not significantly extend service life in a severe chloride environment [1.102]. In this case, the use of zinc and epoxy dual-​coated bars would be more advantageous. A layer of zinc alloy is first applied, followed by a layer of epoxy. Requirements for the manufacture of galvanized bars are given in ASTM A767 Specification for Zinc-​Coated (Galvanized) Steel Bars for Concrete Reinforcement [1.102] and in ASTM A1055/​A1055M Standard Specification for Zinc and Epoxy Dual-​Coated Steel Reinforcing Bars [1.103] for zinc and epoxy dual-​coated bars.

Fiber Reinforcement Another type of steel reinforcement, permitted for use for the first time in the 2008 ACI Code, is deformed steel fibers. According to ACI-​26.4.1.5.1, deformed steel fibers, when used for shear resistance, shall conform to ASTM A820/​A820M [1.104] and have a length-​ to-​diameter ratio between 50 and 100. These fibers are typically 1 to 2.4 in. (25–​60 mm) in length with diameters between 0.015 and 0.04 in. (0.38 and 1 mm). Deformations are introduced to the fibers in order to increase bond with the surrounding concrete. Typical deformations in steel fibers include hooks, flat ends, and waves and twists along the fiber length. The tensile strength of wire used to manufacture deformed steel fibers typically ranges between 150 and 350 ksi (1030 and 2400 MPa).

25



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1.14  FIBER-REINFORCED CONCRETE

1.14 FIBER-​R EINFORCED CONCRETE Fiber-​reinforced concrete consists of concrete reinforced with short, typically randomly oriented, fibers. Fibers used in structural applications are often made of steel, with deformations to provide mechanical bond with the concrete (see section on fiber reinforcement above). Contrary to regular bar-​type reinforcement, however, fibers are generally expected to pull out rather than yield (except locally at or near fiber deformations). The behavior of fiber-​ reinforced concrete is thus controlled by the bond stress versus slip response of the fibers. Typically, steel fibers are used in dosages ranging between 0.5 and 1.5% by volume. For steel fibers, this corresponds to approximately 65 lb and 200 lb cu yard (40 and 120 kg/​m3). Fibers can be premixed with cement and aggregate or added after the concrete has been mixed. Fiber reinforcement provides the concrete with post-​cracking tensile resistance. Fiberreinforced concrete subjected to tension typically exhibits a softening response. In some cases, however, a pseudo-​strain-​hardening response in tension can be obtained, which is accompanied by multiple cracking and strains at peak stress typically greater than 0.5%. These strain-​hardening materials are often referred to as high-​performance fiber-reinforced concrete (HPFRC). Cracking strength (or first cracking strength in the case of HPFRC) is little sensitive to the presence of fibers, given the relatively small volume fractions used in practice. Fig.  1.14.1 shows qualitative tensile stress-​strain responses for plain, strain-​ softening, and strain-​hardening fiber-​reinforced concretes. Since there is no commonly accepted tensile test method for fiber-reinforced concrete, tensile behavior is generally evaluated indirectly, through a standard beam test. The most commonly used flexural test in the United States is that specified in ASTM C1609 [1.105], where a beam, typically 4 × 4 or 6 × 6 in. in cross section (100 × 100 or 150 × 150 mm) and with a span equal to three times its cross-​sectional side dimension, is tested under four-​ point bending up to a midspan deflection of 1150 times the span length. In Europe, flexural tests are often conducted on a notched beam, for which the crack mouth opening displacement is often measured to obtain an estimate of tensile stress versus crack width response. Fiber reinforcement also has an effect on concrete compressive behavior. While strength is little sensitive to the addition of fibers in typical amounts used in practice, ductility is increased by the presence of fibers [1.106], translating into a lower post-​peak stress decrease and higher strain capacity (Fig. 1.14.2). In the case of HPFRCs, a compressive stress-​strain response similar to that of well-​confined concrete can be obtained. The addition of fibers leads to a decrease in concrete workability, which should be taken into account when concrete mixes are selected. Typically, an increase in fiber volume fraction or fiber length to diameter ratio leads to a decrease in concrete workability. However, with the use of high-​range, water-​reducing admixtures and viscosity-​modifying agents,

ft fcr

Fiber-reinforced concrete (Strain-hardening)

fc/fc’ 1 Vf = fiber volume fraction Fiber-reinforced concrete

Fiber-reinforced concrete (Strain-softening) 0.5

εcr

Vf = 2% Vf = 1%

Plain concrete

Plain concrete εt

Figure 1.14.1  Qualitative tensile stress-​strain response of plain and steel fiber-​reinforced concretes.

1

2

3

4

5

εc/εo

Figure 1.14.2  Effect of fiber reinforcement on compressive post-​peak response of concrete. (Data from Ref. 1.106.)

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self-​consolidating concrete can be manufactured, even with steel fibers in a 1.5% volume fraction (200 cu yard or 120 kg/​m3) [1.107]. The enhancement of tension and compressive behavior of concrete with the addition of fibers makes the use of fiber-​reinforced concrete attractive in cases when additional shear resistance or enhanced bond between reinforcement and concrete is required, or where large amounts of confinement reinforcement are needed. Examples include the use of fibers as minimum shear reinforcement in beams and as punching shear reinforcement in slabs, confinement steel in beam-​column connections, and shear walls and coupling or link beams of earthquake-​resistant structures. Information about the effect of fiber reinforcement on shear resistance, bond between reinforcement and concrete, and confinement requirements can be found in References 1.108 through 1.112. Structural applications of fiber-​reinforced concrete in the United States have been limited, primarily because of lack of Code design provisions. It was not until 2008 that the ACI Code included provisions for the use of deformed steel fibers (as minimum shear reinforcement). Steel fibers have been used for a few years, however, in precast segmental linings and, more recently, in coupling or link beams of earthquake-​resistant coupled wall structures [1.112].

1.15 UNITS U.S. Customary units are the primary units used throughout this text. In some cases, values or equations are also expressed in SI units. Although not SI, the MKS (meter-​kilogram-​ second) system is used in most western hemisphere countries (the United States and Canada excepted), where instead of the kilogram (kg) as a unit of mass, as in SI, the kilogram (kgf) is used as a unit of force. Thus, curves involving stresses have sometimes three parallel axes: pounds per square inch (psi) or kips per square inch (ksi) for Inch-​Pound units, megapascals (MPa) for SI units, and kilogram-​force per square centimeter (kgf/​cm2) for the other metric (MKS) system. The authors believe that some familiarity with metric units is essential; on the other hand, overemphasis on units in design calculations and procedures detracts from concepts. In general, throughout the remaining chapters, design equations that involve units will have an SI version (usually ACI 318-​14M) given in a footnote. The reader interested in proper use of SI units should consult the IEEE/​ASTM American National Standard for Metric Practice [1.113].

SELECTED REFERENCES 1.1.

Hans Straub. A History of Civil Engineering. Cambridge, MA: MIT Press, 1964 (pp. 205–​215); also London: Leonard Hill Ltd., 1952. 1.2. R.  S. Kirby, Sidney Withington, A.  B. Darling, and F.  G. Kilgour. Engineering in History. New York: McGraw-​Hill, 1956. 1.3. PCA. “An Historical Look at Reinforced Concrete Structures Through the Eyes of Eli W. Cohen,” Engineered Concrete Structures, 8, 2 (August 1995). Skokie, IL: Portland Cement Association, 3 pp. 1.4. M.  K. Hurd. “Ernest L.  Ransome—​ Concrete Designer, Constructor, Inventor,” Concrete International, 18, May 1996, 50–​51. 1.5. Frederick E.  Turneaure and Edward R.  Maurer. Reinforced Concrete Design. New York: Wiley, 1907. 1.6. Adam M. Neville. Properties of Concrete (4 ed.). New York: John Wiley & Sons, 1996. 1.7. Sidney Mindess, J. Francis Young, and David Darwin. Concrete (2nd ed.) Upper Saddle River, NJ: Prentice-​Hall, 2003. 1.8. P.  Kumar Mehta. Concrete Structure, Properties, and Materials. Englewood Cliffs, NJ: Prentice-​Hall, 1986. 1.9. ASTM. Standard Specification for Portland Cement (C150/​C150M–​12). West Conshohocken, PA: ASTM International, 2012.

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1.10. ACI Committee 225. Guide to the Selection and Use of Hydraulic Cements (ACI 225R-​99 [Reapproved 2009]). Farmington Hills, MI: American Concrete Institute, 1999, 30 pp. Also ACI Journal, Proceedings, 82, November–​December 1985, 901–​928. 1.11. ASTM. Standard Specification for Blended Hydraulic Cements (C595/​C595M-​13). West Conshohocken, PA: ASTM International, 2013. 1.12. ACI. Building Code Requirements for Structural Concrete (ACI 318–​14) and Commentary (ACI 318R-​14). Farmington Hills, MI: American Concrete Institute, 2014. 1.13. ACI Committee 221. Guide for Use of Normal Weight and Heavyweight Aggregates in Concrete (ACI 221R-​ 96 [Reapproved  2001]). Farmington Hills, MI:  American Concrete Institute, 1996, 28 pp. 1.14. ASTM. Standard Specification for Concrete Aggregates (C33/​C33M-​13). West Conshohocken, PA: ASTM International, 2013. 1.15. ASTM. Standard Specification for Lightweight Aggregates for Structural Concrete (C330/​ C330M-​13). West Conshohocken, PA: ASTM International, 2013. 1.16. Richard W. Steiger. “Development of Lightweight Aggregate Concrete,” Concrete Construction, 30, June 1985, 519–​525. 1.17. George K. Mackie II. “Recent Uses of Structural Lightweight Concrete,” Concrete Construction, 30, June 1985, 497–​502. 1.18. ACI Committee 213. Guide for Structural Lightweight-​Aggregate Concrete (ACI 213R-​03). Farmington Hills, MI: American Concrete Institute, 2003, 38 pp. 1.19. ACI Committee 304. Heavyweight Concrete:  Measuring, Mixing, Transporting, and Placing (ACI 304.3R-​96 [Reapproved  2004]). Farmington Hills, MI:  American Concrete Institute, 1997, 8 pp. 1.20. V. M. Malhotra. “Use of Mineral Admixtures for Specialized Concretes,” Concrete International, 6, April 1984, 19–​24. 1.21. Richard C. Mielenz. “History of Chemical Admixtures for Concrete,” Concrete International, 6, April 1984, 40–​53. 1.22. ASTM. Standard Specification for Coal Fly Ash and Raw or Calcined Natural Pozzolan for Use as a Mineral Admixture in Concrete (C618-​12a). West Conshohocken, PA:  ASTM International, 2012. 1.23. John M.  Albinger, “Fly Ash for Strength and Economy,” Concrete International, 6, April 1984, 32–​34. 1.24. Dan Ravina. “Slump Retention of Fly Ash Concrete With and Without Chemical Admixtures,” Concrete International, 17, April 1995, 25–​29. 1.25. ASTM. Standard Specification for Slag Cement for Use in Concrete and Mortars (C989-​13). West Conshohocken, PA: ASTM International, 2013 1.26. ACI Committee 233. Slag Cement in Concrete and Mortar (ACI 213R-​03 [Reapproved 2011]). Farmington Hills, MI: American Concrete Institute, 2003, 18 pp. 1.27. ASTM. Standard Specification for Use of Silica Fume for Used in Cementitious Mixtures (C1240-​14). West Conshohocken, PA: ASTM International, 2014. 1.28. Menashi D. Cohen, Jan Olek, and Bryant Mather. “Silica Fume Improves Expansive-​Cement Concrete,” Concrete International, 13, March 1991, 31–​37. 1.29. T.  A. Durning and M.  C. Hicks. “Using Microsilica to Increase Concrete’s Resistance to Aggressive Chemicals,” Concrete International, 13, March 1991, 42–​48. 1.30. T. A. Durning and M. C. Hicks. Concrete Construction, 30, April 1985 (entire issue devoted to admixtures). 1.31. ACI Committee 212. Chemical Admixtures for Concrete (ACI 212.3R-​10). Farmington Hills, MI: American Concrete Institute, 2010, 61 pp. 1.32. ASTM. Standard Specification for Air-​Entraining Admixtures for Concrete (C260/​C260M-​10a). West Conshohocken, PA: ASTM International, 2010. 1.33. ASTM. Standard Specification for Chemical Admixtures for Concrete (C494-​99ae1). West Conshohocken, PA: ASTM International, 1999. 1.34. ASTM. Standard Specification for Chemical Admixtures for Use in Producing Flowing Concrete (C1017-​98). West Conshohocken, PA: ASTM International, 1998. 1.35. A. A. Ramezanianpour, V. Sivasundaram, and V. M. Malhotra. “Superplasticizers: Their Effect on the Strength Properties of Concrete,” Concrete International, 17, April 1995, 30–​35. 1.36. ACI Committee 237. Self-​ Consolidating Concrete (ACI 237R-​ 07). Farmington Hills, MI: American Concrete Institute, 2007, 30 pp. 1.37. Sandor Popovics. “Analysis of the Concrete Strength versus Water–​Cement Ratio Relationship,” ACI Materials Journal, 87, September–​October 1990, 517–​529. 1.38. Sandor Popovics and John S.  Popovics. “The Foundation of a Computer Program for the Advanced Utilization of w/​c and Air Content in Concrete Proportioning,” Concrete International, 16, December 1994, 21–​26.

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1.39. Sandor Popovics and John S.  Popovics. “Computerization of the Strength Versus w/​ c* Relationship,” Concrete International, 17, April 1995, 37–​40. 1.40. ACI Committee 211. Guide for Selecting Proportions for No-​Slump Concrete (ACI 211.3R-​02 [Reapproved 2009]). Farmington Hills, MI: American Concrete Institute, 2002, 26 pp.. 1.41. PCA. Design and Control of Concrete Mixtures (15th ed.). Skokie, IL:  Portland Cement Association, 2011. 1.42. ACI Committee 211. Standard Practice for Selecting Proportions for Normal, Heavyweight, and Mass Concrete (ACI 211.1-​91 [Reapproved 2009]). Farmington Hills, MI: American Concrete Institute, 1991, 38 pp. 1.43. ACI Committee 211. Standard Practice for Selecting Proportions for Structural Lightweight Concrete (ACI 211.2-​ 98 [Reapproved  2004]). Farmington Hills, MI:  American Concrete Institute, 1998, 20 pp. 1.44. ACI Committee 304. Guide for Measuring, Mixing, Transporting, and Placing Concrete (ACI 304R-​00 [Reapproved 2009]). Farmington Hills, MI: American Concrete Institute, 2000, 44 pp.. 1.45. ACI Committee 311. ACI Manual of Concrete Inspection (SP-​2), 10th ed. (ACI 311.1R-​07). Farmington Hills, MI: American Concrete Institute, 2007. 1.46. ACI Committee 201. Guide to Durable Concrete (ACI 201.2R-​ 008). Farmington Hills, MI: American Concrete Institute, 2008, 49 pp. 1.47. Walter H.  Price. “Factors Influencing Concrete Strength,” ACI Journal, Proceedings, 47, February 1951, 417–​432. 1.48. UNESCO. Reinforced Concrete:  An International Manual. London:  Butterworth, 1971 (pp. 19–​22). 1.49. Delmar L.  Bloem. “Concrete Strength in Structures,” ACI Journal, Proceedings, 65, March 1968, 176–​187. 1.50. James M.  Shilstone, Jr. “The Cylinder Test—​Reliable Informer or False Prophet,” Concrete International, 2, July 1980, 63–​68. 1.51. J. Bruce Tait “Making the Most of Concrete Strength Test Results,” ACI Journal, Proceedings, 83, May–​June 1986, 383–​388. 1.52. John R. Smith. “Estimating Later Age Strengths of Concrete,” ACI Journal, Proceedings, 81, November–​December 1984, 609–​612. 1.53. A. P. Keiller. “Assessing the Strength of In Situ Concrete,” Concrete International, 7, February 1985, 15–​21. 1.54. Robert S. Jenkins. “Nondestructive Testing—​An Evaluation Tool,” Concrete International, 7, February 1985, 22–​26. 1.55. William C. Stone and Bruce J. Giza. “The Effect of Geometry and Aggregate on the Reliability of the Pullout Test,” Concrete International, 7, February 1985, 27–​36. 1.56. Thomas J. Parsons and Tarun R. Naik. “Early Age Concrete Strength Determination by Maturity,” Concrete International, 7, February 1985, 37–​43. 1.57. Kal R.  Hindo and Wayne R.  Bergstrom. “Statistical Evaluation of the In-​Place Compressive Strength of Concrete,” Concrete International, 7, February 1985, 44–​48. 1.58. V. M. Malhotra. “Testing Early-​Age Strength of Concrete In-​Place,” Concrete International, 7, April 1985, 39–​41. 1.59. ACI Committee 214. Guide for Obtaining Cores and Interpreting Compressive Strength Results (ACI 214.4R-​10). Farmington Hills, MI: American Concrete Institute, 2010, 17 pp. 1.60. Hjalmar Granholm. A General Flexural Theory of Reinforced Concrete. New York: Wiley, 1965 (pp. 23–​36). 1.61. P. T. Wang, S. P. Shah, and A. E. Naaman. “Stress–​Strain Curves of Normal and Lightweight Concrete in Compression,” ACI Journal, Proceedings, 75, November 1975, 603–​611. 1.62. Vitelmo V. Bertero. “Inelastic Behavior of Structural Elements and Structures,” High Strength Concrete, Proceedings of a Workshop, University of Illinois at Chicago Circle, December 2–​4, 1979 (pp. 102–​103, 153). 1.63. Antoine E.  Naaman. Prestressed Concrete Analysis and Design Fundamentals. New York: McGraw-​Hill, 1982 (p. 54). 1.64. Arthur H. Nilson. Design of Prestressed Concrete. New York: Wiley, 1978 (p. 45). 1.65. Hubert Rüsch. “Researches Toward a General Flexural Theory for Structural Concrete,” ACI Journal, Proceedings, 57, July 1960, 1–​28. 1.66. Vaughan Randall and Kenneth Foot. “High-​Strength Concrete for Pacific First Center,” Concrete International, 11, April 1989, 14–​16. 1.67. Nathan L.  Howard and David M.  Leatham. “The Production and Delivery of High-​Strength Concrete,” Concrete International, 11, April 1989, 26–​30. 1.68. Gregory J.  Smith and Franz N.  Rad. “Economic Advantages of High-​Strength Concretes in Columns,” Concrete International, 11, April 1989, 37–​43.

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1.69. S. P. Shah, editor. High Strength Concrete, Proceedings of a Workshop, University of Illinois at Chicago Circle, December 2–​4, 1979 (Sponsored by National Science Foundation) (226 pp). 1.70. Anthony E. Fiorato. “PCA Research on High-​Strength Concrete,” Concrete International, 11, April 1989, 44–​50. 1.71. ASTM. Standard Test Method for Splitting Tensile Strength of Cylindrical Concrete Specimens (C496/​C496M-​11). West Conshohocken, PA: ASTM International, 2011. 1.72. ASTM. Standard Test Method for Flexural Strength of Concrete (Using Simple Beam with Third-​Point Loading) (C78/​C78M-​02). West Conshohocken, PA: ASTM International, 2010. 1.73. Sher Ali Mirza, Michael Hatzinikolas, and James G.  MacGregor. “Statistical Descriptions of Strength of Concrete,” Journal of the Structural Division, ASCE, 105, ST6 (June 1979), 1021–​ 1037. Disc. 106, ST7 (July 1980), 1659–​1660. 1.74. Helmut Kupfer, Hubert K. Hilsdorf, and Hubert Rusch. “Behavior of Concrete under Biaxial Stresses,” ACI Journal, Proceedings, 66, August 1969, 656–​666. 1.75. Frank E. Richart, Anton Brandtzaeg, and Rex L. Brown. A Study of the Failure of Concrete under Combined Compressive Stresses. Bulletin 185, Engineering Experiment Station, University of Illinois, Urbana, November 1928 (104 pp). 1.76. Adrian Pauw. “Static Modulus of Elasticity of Concrete as Affected by Density,” ACI Journal, Proceedings, 57, December 1960, 679–​687. 1.77. T. S. Shih, G. C. Lee, and K. C. Chang. “On Static Modulus of Elasticity of Normal-​Weight Concrete,” Journal of Structural Engineering, ASCE, 115, 10 (October 1989), 2579–​2587. 1.78. Francis A. Oluokun, Edwin G. Burdette, and J. Harold Deatherage, “Elastic Modulus, Poisson’s Ratio, and Compressive Strength Relationships at Early Ages,” ACI Materials Journal, 88, January–​February 1991, 3–​10. 1.79. ACI Committee 209. “Prediction of Creep, Shrinkage and Temperature Effects in Concrete Structures,” Designing for Creep and Shrinkage in Concrete Structures (SP-​76). Farmington Hills, MI: American Concrete Institute, 1982 (pp. 193–​300). 1.80. Dan E. Branson. Deformations of Concrete Structures. New York: McGraw-​Hill, 1977 (pp. 11–​ 27, 44–​55). 1.81. ACI Committee 301. Specifications for Structural Concrete (ACI 301-​10). Farmington Hills, MI: American Concrete Institute, 2010, 77 pp. 1.82. J. Bruce Tait. “Concrete Quality Assurance Based on Strength Tests,” Concrete International, 3, September 1981, 79–​87. 1.83. ACI Committee 121. Guide for Concrete Construction. Quality Systems in Conformance with ISO 9001 (ACI 121R-​08). Farmington Hills, MI: American Concrete Institute, 2008 (35 pp). 1.84. Lewis H.  Tuthill. “Obtaining Quality in Concrete Construction,” Concrete International, 8, March 1986, 24–​29. 1.85. Bryant Mather. “Selecting Relevant Levels of Quality,” Concrete International, 8, March 1986, 30–​36. 1.86. Ken Newman. “Common Quality in Concrete Construction,” Concrete International, 8, March 1986, 37–​49. 1.87. John M. Scanlon. “Quality Control During Hot and Cold Weather,” Concrete International, 1, September 1979, 58–​65. 1.88. ASTM. Standard Specification for Deformed and Plain Carbon-​ Steel Bars for Concrete Reinforcement (A615/​A615M-​13). West Conshohocken, PA: ASTM International, 2013. 1.89. ASTM. Standard Specification for Deformed and Plain Stainless-​Steel Bars for Concrete Reinforcement (A955/​A955M-​14). West Conshohocken, PA: ASTM International, 2014. 1.90. ASTM. Standard Specification for Rail-​Steel and Axle-​Steel Deformed Bars for Concrete Reinforcement (A996/​A996M-​14). West Conshohocken, PA: ASTM International, 2014. 1.91. ASTM. Standard Specification for Low-​Alloy Steel Deformed and Plain Bars for Concrete Reinforcement (A706/​A706M-​13). West Conshohocken, PA: ASTM International, 2013. 1.92. ASTM. Standard Specification for Deformed and Plain, Low-​Carbon, Chromium, Steel Bars for Concrete Reinforcement (ASTM A1035/​A1035M-​14). West Conshohocken, PA:  ASTM International, 2014. 1.93. ASTM. Standard Specification for Carbon-​ Steel Wire and Welded Wire Reinforcement, Plain and Deformed, for Concrete (A1064/​A1064M-​13). West Conshohocken, PA:  ASTM International, 2013. 1.94. ASTM. Standard Specification for Deformed and Plain Stainless Steel Wire and Welded Wire for Concrete Reinforcement (A1022/​ A1022M-​ 13). West Conshohocken, PA:  ASTM International, 2013. 1.95. WRI. Structural Welded Wire Reinforcement—​Manual of Standard Practice (8th ed.). Hartford, CT: Wire Reinforcement Institute (942 Main Street, Suite 300, Hartford, CT 06103), 2010. 1.96. ASTM. Standard Specification for Steel Strand, Uncoated Seven-​Wire for Prestressed Concrete (A416/​A416M-​12a). West Conshohocken, PA: ASTM International, 2012.

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 1.97. ASTM. Standard Specification for Uncoated Stress-​ Relieved Steel Wire for Prestressed Concrete (A421/​A421M-​10). West Conshohocken, PA: ASTM International, 2010.  1.98. ASTM. Standard Specification for Uncoated High-​Strength Steel Bar for Prestressing Concrete (A722/​A722M-​12). West Conshohocken, PA: ASTM International, 2012.  1.99. ASTM. Specification for Epoxy-​ Coated Steel Reinforcing Bars (A775/​A775M-​07b [Reapproved 2014]). West Conshohocken, PA: ASTM International, 2007. 1.100. ASTM. Specification for Epoxy-​Coated Prefabricated Steel Reinforcing Bars (A934/​A934M-​ 13) West Conshohocken, PA: ASTM International, 2013. 1.101. ACI Committee 222. Protection of Metals in Concrete Against Corrosion (ACI 222R-​01 [Reapproved 2010]). Farmington Hills, MI: American Concrete Institute, 2001, 41 pp. 1.102. ASTM. Specification for Zinc-​Coated (Galvanized) Steel Bars for Concrete Reinforcement (A767/​A767M-​09). West Conshohocken, PA: ASTM International, 2009. 1.103. ASTM. Standard Specification for Zinc and Epoxy Dual-​ Coated Steel Reinforcing Bars (A1055/​A1055M-​10e1) West Conshohocken, PA: ASTM International, 2010. 1.104. ASTM. Standard Specification for Steel Fibers for Fiber Reinforced Concrete (A820/​A820M-​ 11) West Conshohocken, PA: ASTM International, 2011. 1.105. ASTM. Standard Test Method for Flexural Performance of Fiber-​Reinforced Concrete (Using Beam with Third-​ Point Loading (C1609/​ C1609M-​ 12) West Conshohocken, PA:  ASTM International, 2012. 1.106. Duane E. Otter and Antoine E. Naaman. “Properties of Steel Fiber Reinforced Concrete under Cyclic Loading,” ACI Materials Journal, 85, July–​August 1988, 254–​261. 1.107. W.-​C. Liao, S.-​H. Chao, S.-​Y., Park, and A. E. Naaman. “Self-​Consolidating High-​Performance Fiber-​Reinforced Concrete (SCHPFRC)—​Preliminary Investigation.” Report No. UMCEE 06-​ 02. University of Michigan: Ann Arbor, 2006. 1.108. G.  J. Parra-​Montesinos. “Shear Strength of Beams with Deformed Steel Fibers,” Concrete International, 28, November–​December 2006, 57–​66. 1.109. M.-​ Y. Cheng and G.  J. Parra-​ Montesinos. “Evaluation of Steel Fiber Reinforcement for Punching Shear Resistance in Slab-​Column Connections. Part I:  Monotonically Increased Load,” ACI Structural Journal, 107, January–​February 2010, 101–​109. 1.110. S. H. Chao, A. E. Naaman, and G. J. Parra-​Montesinos. “Bond Behavior of Reinforcing Bars in Tensile Strain-​Hardening Fiber Reinforced Cement Composites,” ACI Structural Journal, 106, November–​December 2009, 897–​906. 1.111. G.  J. Parra-​Montesinos, S.  Peterfreund, and S.-​H. Chao. “Highly Damage Tolerant Beam-​ Column Joints through Use of High-​Performance Fiber-​Reinforced Cement Composites,” ACI Structural Journal, 102, May–​June 2005, 487–​495. 1.112. G. J. Parra-​Montesinos, J. K. Wight, C. Kopczynski, R. D. Lequesne, M. Setkit, A. Conforti, and J. Ferzli. “High-​Performance Fiber Reinforced Concrete Coupling Beams: From Research to Practice,” Proceedings of the 10th National Conference in Earthquake Engineering, Earthquake Engineering Research Institute, Anchorage, AK, 2014. 1.113. IEEE/​ASTM. American National Standard for Metric Practice (IEEE/​ASTM SI 10-​2010). West Conshohocken, PA: ASTM International, 2010. 1.114. ACI Committee 213. “Guide for Structural Lightweight Aggregate Concrete,” Concrete International, 1, February 1979, 33–​62.

CHAPTER 2 DESIGN METHODS AND REQUIREMENTS

2.1  STRUCTURAL DESIGN PROCESS—​G ENERAL The main objective of the structural design process is to ensure that the structure is able to withstand the load effects with an appropriate margin of safety against failure. In this context, failure is defined as that condition at which the structure ceases to fulfill its intended function. This condition may occur because the structure has achieved its maximum resistance and can no longer carry the imposed loads or, although it can still carry the loads, because it has undergone excessive deformations that have rendered the structure unusable. In other words, the structure must be designed so that it is both safe and functional (or serviceable). Strength and serviceability considerations such as a limit on deflection may be thought of as “limit states.” MacGregor [2.1] has provided an excellent treatment of limit states design as applied to reinforced concrete. Ensuring structural safety (i.e., protecting the life of the occupants) is a matter of public concern and is clearly the most important of these objectives. Nevertheless, serviceability can be as important in design to ensure that the structure is usable and can serve its intended purpose. To satisfy the structural safety requirement, structural components must be adequately proportioned to ensure that the structure is provided with enough strength to resist the effects of the anticipated loads. On the other hand, serviceability issues (such as control of deflections, vibrations, and cracking) are mostly a function of the stiffness of members. In this context, the proper design of a reinforced concrete structure consists in the judicious selection of member size and material strengths to provide the structure with enough strength and stiffness to ensure that both safety and serviceability are satisfied. Throughout the book, structural safety and serviceability are emphasized. However, other aspects, such as the proportioning of members for cost effectiveness and ease of construction, are also addressed. Maintenance and aesthetics are also considered where appropriate. In practice, the engineer must satisfy all these aspects while ensuring an economical design of the structure.

2.2  ACI BUILDING CODE When two different materials, such as steel and concrete, act together, the analysis for strength of a reinforced concrete member clearly must be, in part, empirical. These principles and methods are being constantly revised and improved as results of theoretical and experimental research accumulate. The American Concrete Institute, serving as a clearinghouse for these changes, issues building code requirements, the most recent of which is

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Lake Point Tower, Chicago; 70-story apartment building (Photo courtesy of Magnusson Klemencic Associates).

Building Code Requirements for Structural Concrete (ACI 318-​14), hereafter referred to as the ACI Code [1.12].1 The ACI Code is a Standard of the American Concrete Institute. To achieve legal status, it must be adopted by a governing body as a part of its general building code. The ACI Code is partly a specification-​type code, which gives acceptable design and construction methods in detail, and partly a performance code, which states desired results rather than details of how such results are to be obtained. A building code, legally adopted, is intended to prevent people from being harmed; therefore, it specifies minimum requirements to provide adequate safety and serviceability. It is important to realize that a building code is not a recommended practice, nor is it a design handbook, nor is it intended to replace engineering knowledge, judgment, or experience. It does not relieve the designer of the responsibility for having a safe, functional, and economical structure.

2.3  STRENGTH DESIGN AND WORKING STRESS METHODS Two philosophies have been used in the design of reinforced concrete structures in the past. The working stress method, the main approach used from the early 1900s until the early 1960s, focused on conditions at service loads (i.e., at the maximum loads under the

  The reader is advised to have the ACI Code as a ready reference while using this text.

1

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2.5  STRENGTH DESIGN METHOD

intended use of the structure). Today, with few exceptions, the strength design method is the prevalent procedure used in practice. The strength design method, which focuses on conditions at loads greater than service loads, when failure may be imminent, is deemed conceptually a more realistic approach to establish structural safety. Both approaches, however, provide approximately the same level of safety and serviceability.

2.4  WORKING STRESS METHOD In the working stress method, a structural element is so designed that the stresses resulting from the action of service loads (also called working loads) and computed by the mechanics of elastic members do not exceed some predesignated allowable values. Service load is the load (e.g., dead, live, snow, wind, earthquake) that is assumed actually to occur when the structure is in service. The working stress method may be expressed by the following: f ≤ fallow (2.4.1)

where

f = an elastic stress, such as by using the flexure formula f = Mc/​I for a beam, computed under service load and fallow = a limiting or allowable stress prescribed by a building code, for example, as a percentage of the compressive strength fc′ for concrete, or of the yield stress fy for the steel reinforcing bars. Some of the obstacles to the working stress method are as follows: 1. Since the limitation is on the stress under the total service load, there is no simple way to account for different degrees of uncertainty of various kinds of loads. Generally the dead load (gravity load due to weight of structural elements and permanent attachments) is known more accurately than the live load, which may have unknown and variable distribution. 2. Creep and shrinkage of concrete, which contribute major time-​dependent effects on a structure, are not easily accounted for by calculation of elastic stresses. 3. Concrete stress is not proportional to strain up to its crushing strength, so that the inherent safety provided is unknown when a percentage of fc′ is used as the allowable stress.

2.5  STRENGTH DESIGN METHOD In the strength design method (formerly called ultimate strength method), the service loads are increased by factors to obtain the load at which failure is considered to be “imminent.” This load is called the factored load or factored service load. The structure or structural element is then proportioned such that the strength is reached when the factored load is acting. The strength design method may be expressed by the following,

strength provided ≥ strength required ( to carry factored loads)

(2.5.1)

where the “strength provided” (such as moment strength) is computed in accordance with the provisions of a building code, and the “strength required” is that obtained by performing a structural analysis using factored loads. The “strength provided” is commonly referred to by practitioners as “ultimate strength.” The computation of this strength takes into account the nonlinear stress-strain behavior of concrete and reinforcing steel. However, it is a code-​defined value for strength and is not necessarily “ultimate” in the sense of being a value that cannot possibly be exceeded. The ACI Code uses a conservative definition of strength; thus the modifier “ultimate” is not entirely appropriate.

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When the strength design method is used, the comparison of provided strength with required strength (i.e., axial force, shear, or bending moment, caused by factored loads) does not imply that any material “yields” or “fails” under service load conditions. In fact, at service loads, the behavior of the structure is essentially elastic. The use of the term “imminent failure” under factored loads is only a mechanism for establishing adequate safety parameters. Comments on Strength Design Methods Historically, “ultimate” strength was the earliest approach to design, since the failure load could be measured by tests without knowledge of the magnitude or distribution of internal stresses. With the interest in and understanding of the elastic methods of analysis in the early 1900s, the elastic working stress method was adopted almost universally by codes as the best for design. As more detailed understanding of the actual behavior of reinforced concrete structures subjected to loads in excess of the service loads developed, adjustments in the theory and in the design procedures were made. The first modification of the elastic working stress method resulted from the study of axially loaded columns in the early 1930s. By 1940, the design of axially loaded columns was based on ultimate strength. Next, the working stress method was modified to account for creep of concrete in beams with compression steel and in eccentrically loaded columns. The early history of the ACI Code has been summarized by Kerekes and Reid [2.2]. An excellent discussion of what is involved in writing the ACI Code is given by Siess [2.3]. The 1956 ACI Code was the first that officially recognized and permitted the strength design method, the result of work by ACI-​ASCE Committee 327 [2.4]. The 1963 ACI Code treated the working stress method and the strength design method on an equal basis; but, actually, the major portion of the working stress method was based on strength. With the relegation of the working stress method to a small section referred to as the “alternate method,” the 1971 ACI Code entirely accepted the strength design method. Between 1971 and 1999, the ACI Code had the “alternate design method” in an appendix, but it was removed from the 2002 and subsequent editions of the ACI Code.

2.6  SAFETY PROVISIONS—​G ENERAL Structures and structural members must always be designed to carry some reserve load above what is expected under normal use. Such reserve capacity is provided to account for a variety of factors, which may be grouped into two general categories: factors relating to overload and factors relating to understrength (i.e., less strength than computed by acceptable calculating procedures). Overloads may arise from changing the use for which the structure was designed, from underestimation of the effects of loads by oversimplification in calculation procedures, and from effects of construction sequence and methods. Understrength may result from adverse variations in material strength, workmanship, dimensions, control, and degree of supervision, even though individually these items are within required tolerances. Conventionally, the term “safety factor” has been used in working stress design to designate the ratio between the yield stress (real, as for steel; nominally defined, as for concrete) and the allowable working stress. Such use has resulted in structures and structural elements having the same “safety factor” but considerably different variance in their strength to ser­ vice load ratio. Thus the term “safety factor” as conventionally applied has little meaning with respect to the prediction of strength. The variability in the ratio of the strength to service load in the working stress method was an important reason for the change to the strength design method. To distinguish between the term “safety factor,” as used in working stress design, and the ratio of strength to service load, the term “load factor” was adopted for the latter. The purpose of a safety provision is to limit the probability of failure and yet permit economical structures. Obviously, if cost is no object, it is easy to design a structure whose

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35

probability of failure is nil. To arrive properly at a suitable degree of safety, the relative importance of various items must be established. Some of those items are 1. 2. 3. 4. 5.

Seriousness of a failure, either to humans or goods. Reliability of workmanship and inspection. Expectation of overload and to what magnitude. Importance of the member in the structure. Chance of warning prior to a failure.

By assigning percentages to those five items and evaluating the circumstances for any given situation, proper values for overload factors U and strength reduction factors φ can be established. The background and practicalities leading to the present strength design procedure are discussed later in this section. The ACI Code strength design method has traditionally divided the safety provisions into two parts; U factors to account for the probability of overload, and φ factors to account for probability of understrength. The requirement for strength design may be expressed as follows:

design strength ≥ required strength

(2.6.1)

φ Sn ≥ Su

(2.6.2)

or

where Sn represents the code-​defined strength of the member computed by using standard assumptions and specified material strengths; it is referred to as the “nominal” strength. The term φ Sn is called the “design strength.” The “required strength,” Su, corresponds to the member actions (or load effects) computed from a structural analysis using the U factors. In terms of member actions (or load effects), Eq. (2.6.2) may be written as follows:

φ Pn ≥ Pu

(2.6.3)



φ M n ≥ M u

(2.6.4)



φ Vn ≥ Vu

(2.6.5)



φ Tn ≥ Tu (2.6.6)

where Pn, Mn, Vn, and Tn are the “nominal” strengths in axial compression (or tension), bending moment, shear, and torsion, respectively. Similarly, Pu, Mu, Vu, and Tu are the member actions (or load effects) computed under factored loads in axial compression (or tension), bending moment, shear, and torsion, respectively. For many years the ACI Code used U and φ factors that resulted from combined experience and historical precedent to arrive at the appropriate numerical values. Over time, attention focused on using the theory of probability as a basis for a design code, thus providing a more rational basis for the components comprising the U and φ factors. A series of papers [2.5–​2.9] gives an excellent overview of this approach, and an interesting collection of discussion and opinion is also available [2.8]. One U factor combination that has been used for many years is that for gravity load:

U = 1.2 D +1.6 L

(2.6.7)

where D and L are the service dead load and live load (or dead load and live load effects, such as axial force, bending moment, or shear). In this load combination, the service live load is assigned a larger load factor than that of the service dead load to account for the greater uncertainty that exists in computing the value and distribution of the actual service live load. In other words, it is implied that dead loads can be computed with greater certainty than live loads.

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The previous example provides some rationale for the use of different load factors with loads of different types (live loads, snow loads, wind loads, etc.), but it gives no insight into the appropriateness of the assigned values—that is, whether the load factors provide for adequate safety. Benjamin and Lind [2.6] have stated the following five safety conditions, which form the basis for the current practice in structural analysis and design: 1. The probability of a real loading in excess of the nominal service load D + L must be satisfactorily small. 2. The probability of a real loading in excess of the factored loading, say U = 1.2D + 1.6L, must be very small or near to zero during the life of the structure. 3. The probability of unsatisfactory performance at the factored load U must be satisfactorily small. 4. The probability of unsatisfactory performance under a load test must be very small or near to zero. 5. The probability of unsatisfactory performance at the service load D + L must be practically zero. It is noted that conditions 1 and 2 relate to the overload factors U along with analysis methods, whereas conditions 3, 4, and 5 relate to the factor φ for understrength as well as the methods for computing strength. The probability of overload is certainly independent of the material (i.e., concrete, steel, wood, or masonry) supporting the load. Detailed discussions of the implications of using design reliability and probability-​based load criteria are given by Ellingwood [2.10, 2.11], MacGregor, Mirza, and Ellingwood [2.12], and MacGregor [2.13]. The subject is treated in general by Galambos, Ellingwood, MacGregor, and Cornell [2.14, 2.15]. Corotis [2.16] and Israel, Ellingwood, and Corotis [2.17] reflect thinking regarding unified load criteria for all materials. Today, the load factors and load combinations used in design are based on probabilistic analysis and survey data elicited from design experts and are specified by the ASCE/​SEI 7-​ 10 Standard, Minimum Design Loads for Buildings and Other Structures [2.18]. Developed for use with design specifications for traditional structural materials, including structural concrete, steel, masonry, and wood, the ASCE/​SEI 7 standard forms the basis for the load factors and load combinations specified by model codes such as the International Building Code [2.19]. Accordingly, the 2014 ACI Code uses the ASCE/​SEI 7-​10 [2.18] load factors and load combinations U along with appropriate φ factors in the main body of the Code.

2.7 SAFETY PROVISIONS—​A CI CODE LOAD FACTORS AND STRENGTH REDUCTION FACTORS Load Factors and Load Combinations U The basic load factors and load combinations U as given by ACI-​5.3.1 are U = 1.4 D

(2.7.1)

U = 1.2 D + 1.6 L + 0.5 ( Lr or S or R ) (2.7.2) U = 1.2 D + 1.6 ( Lr or S or R ) + (1.0 L or 0.5W ) (2.7.3) U = 1.2 D + 1.0W + 1.0 L + 0.5 ( Lr or S or R ) (2.7.4) U = 1.2 D + 1.0 E + 1.0 L + 0.2 S (2.7.5) U = 0.9 D + 1.0W

(2.7.6)

U = 0.9 D + 1.0 E

(2.7.7)

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37

where D is dead load; L is live load; Lr is roof live load; S is snow load; R is rain load; W is wind load; and E is the earthquake-​induced load. The load factor on the live load L in Eqs. (2.7.3), (2.7.4), and (2.7.5) is permitted to be reduced to 0.5 except for garages, areas occupied as places of public assembly, and all areas where the live load L is greater than 100 psf. The load factors in Eqs. (2.7.1) through (2.7.7) assume that D, L, Lr, S, and R are all unfactored, service-​level loads. However, the wind load factor in Eqs. (2.7.3), (2.7.4) and (2.7.6) assumes that the wind load W is based on strength-​level loads, computed in accordance with the provisions of ASCE/​SEI 7-​10 [2.18]. If the wind load W is based on service-​ level loads, then 0.8W must be used in place of 0.5W in Eq. (2.7.3), and 1.6W instead of 1.0W in Eqs. (2.7.4) and (2.7.6). Similarly, the earthquake load factor of 1.0 in Eqs. (2.7.5) and (2.7.7) assumes that the earthquake-​induced load E is based on strength-​level seismic forces. If the earthquake-​induced load E is computed using service-​level seismic forces, a higher load factor on the earthquake-​induced load E must be used. If present, loads due to weights and pressures of fluids, F, loads due to lateral earth pressure, H, and structural actions (or load effects) due to the restraint of volume change caused by temperature changes, by creep and shrinkage, and by differential settlement, T, must also be included in the basic load combinations given above. Appropriate load factors for these and other loads are given in ACI-​5.3.6 through ACI-​5.3.12. It is noted that in applying the load combinations, the effect of one or more loads not acting simultaneously also must be investigated. Also note that loads must always be taken to cause the more severe effect and that their effect can be either additive or subtractive. Comments on Load Factors and Load Combinations The load factors and combinations are based on the notion that in addition to the dead load (considered to be permanent), one of the variable loads takes its maximum value, while the rest take an arbitrary value less than the maximum. Consider, for example, the load combination given by Eq. (2.7.2). In this load combination, the live load, L, is considered to take its maximum value, with a load factor of 1.6, while Lr, S, or R takes an arbitrary value less than the maximum, and thus is assigned a lower load factor. In contrast, in the load combination given by Eq. (2.7.3), Lr, S, or R is assumed to be acting at its maximum value, with a load factor of 1.6, while the live load, L, (or the wind load, W) takes an arbitrary value less than the maximum, and thus has a lower load factor. The term 1.2D is common to all load combinations except in Eqs. (2.7.1), (2.7.6) and (2.7.7). The load combination in Eq. (2.7.1) considers the dead load only with a load factor of 1.4 and represents an extreme case. The load combinations in Eqs. (2.7.6) and (2.7.7) have a reduced load factor of 0.9 on the dead load, D, because they are intended to capture the cases in which structural actions (or load effects) from wind, W, or earthquake, E, counteract those from gravity loads. A common example is a roof girder, where the effects from wind loads may cause uplift, thereby counteracting the actions from the girder self-​weight.

Strength Reduction Factors φ The factors φ for understrength are called strength reduction factors according to Chapter 21 of the ACI Code. These factors account for the probability of understrength due to varying material properties, varying dimensions and tolerances, imperfections, and different failure modes. All other things being equal, brittle failures are assigned lower strength reduction factors to safeguard the structure against failures that may be catastrophic, or that may occur with little or no warning. Conversely, noncatastrophic, ductile failure modes, where member strength can be reached with ample warning of failure, are assigned larger strength reduction factors.

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The φ factors in ACI-​21.2.1 and ACI-​21.2.2 are as follows:

φ Factor 1. Flexure (with or without axial force)a Tension-​controlled sections Compression-​controlled sections Spirally reinforced Other 2. Shear and torsion 3. Bearing on concrete 4. Post-​tensioned anchorage zones 5. Struts, ties, nodal zones, and bearing areas in strut-​and-​tie models 6. Brackets and corbels 7. Plain concrete elements

0.90 0.75 0.65 0.75 0.65 0.85 0.75 0.75 0.60

For combined axial load and flexure, both axial load and bending moment are subject to the same φ factor, which may be variable as discussed in Chapter 10. a

2.8  SERVICEABILITY PROVISIONS—​G ENERAL As noted earlier in this chapter, the structure must be designed so that it protects the life of the occupants (i.e., it is safe), but also so that it can serve its intended purpose throughout its service life (i.e., it must provide for serviceability). Some serviceability issues that can be as important as strength in the design of reinforced concrete structures are excessive deflection, detrimental cracking, excessive amplitude or undesirable frequency of vibration, and excessive noise transmission. Any one, or a combination, of the strength and serviceability factors may provide a criterion for the limit of structural usefulness. For example, serviceability can govern the design of beams and slabs in long span floors, where the minimum depth of these members is often dictated by deflection or vibration limits, rather than by strength considerations. While serviceability issues (e.g., excessive deflection) do not generally pose a threat to the safety of the occupants, they can have severe economic consequences. Thus, the designer must adequately address them in design. The design requirement for serviceability may be expressed as follows:

allowable limit ≥ performance under service loads

(2.8.1)

Sallowable ≥ Sservice_loads

(2.8.2)

or

where Sallowable represents the specified limit not to be exceeded to ensure adequate behavior and functionality of the structure during its service life. The term Sservice_​loads is the calculated performance of the structural member or system under service-​level loads. In terms of the service conditions that are commonly considered in the design of reinforced concrete structures, Eq. (2.8.2) may be expressed, for example, as follows:

∆ allowable ≥ ∆ service_loads

(2.8.3)



wallowable ≥ wservice_loads

(2.8.4)



Fallowable ≥ Fservice_loads

(2.8.5)

where Δallowable represents the allowable limit on the deflection of a member or on the lateral drift of the structural system; wallowable is the maximum permitted crack width; and Fallowable is the limit on the frequency of vibration of the floor system. Similarly, Δservice_​loads, wservice_​loads,

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39

and Fservice_​loads are the deflection, crack width, and frequency of vibration of the member or structural system, respectively, computed from a structural analysis under unfactored, service loads.

2.9  SERVICEABILITY PROVISIONS—​A CI CODE Serviceability limits depend primarily on the intended function and occupancy of the structure, and on the type of structural system. It is thus difficult to specify allowable serviceability limits that apply to all types of structures. Chapter  24 of the ACI Code provides minimum requirements for serviceability in reinforced concrete structures. For example, maximum allowable deflections [i.e., Δallowable in Eq. (2.8.3)] in roofs and floor systems are given in Table 24.2.2 of the ACI Code. This table provides limits on the immediate and long-​term deflections for elements that support or are attached to non-​structural elements, as well as for elements that do not support or are not attached to structural elements. These limits, though largely based on engineering judgment and years of experience, are not necessarily adequate in every case. Depending on the functional needs of the structure, tighter, more restrictive deflection limits may be required. It must be emphasized that the performance of the structural member or system (e.g., Δservice_​loads) is computed using service-​level, unfactored loads:  that is, using the loads expected from normal use. Service-​level load combinations are not provided in the main provisions of the ACI Code, though some guidance and recommendations are given in the ACI  Code Commentary. In addition, the commentary to Appendix C of ASCE/​SEI 7-​10 [2.18] provides recommendations for load combinations to be used in serviceability checks. Note that the wind load W used in the load combinations of the safety provisions (Section 2.7) is based on strength-​level loads and should not be used for checking serviceability; its use for evaluating serviceability would be too conservative. The selection of the appropriate wind load for evaluating the performance of the structure under service conditions is, in part, a matter of engineering judgment. The commentary to Appendix C of ASCE/​SEI 7-​10 [2.18] provides guidance for the calculation of the wind load, Wa, based on service-​level wind speeds that are appropriate for use in checking serviceability requirements. Design for serviceability (control of deflections, cracking, and vibrations) in accor­ dance with Chapter  24 of the ACI Code is treated in detail later in Chapter  12 of this textbook. However, basic design considerations and recommended practices for adequate serviceability are discussed throughout the text, together with strength requirements where appropriate.

2.10  HANDBOOKS AND COMPUTER SOFTWARE This textbook does not devote much space to the use of design aids. Once concepts and principles are thoroughly understood, the use of curves and tables can greatly speed up design. Several handbooks are in common use: The Reinforced Concrete Design Manual [2.20], published by ACI; CRSI Design Handbook 2008 [2.21], published by the Concrete Reinforcing Steel Institute; and PCI Design Handbook [2.22], published by the Prestressed Concrete Institute. These publications contain useful tables and charts that can speed up design for the experienced designer. The importance of correct and clear detailing work cannot be overemphasized. For this the reader is referred to ACI Detailing Manual (SP-​66) [2.23], which contains typical detailing of steel reinforcement, engineering and placing drawings, and other reference data regarding materials and sizes. Today, computer software is used for a considerable portion of the design computations. The designer, especially the inexperienced engineer, should be wary of using commercial software (for analysis or design, or for both) to do computations that the designer/​engineer does not thoroughly understand and would be unable to formulate in the absence of a computer program.

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2.11  DIMENSIONS AND TOLERANCES Although the designer may tend to think of dimensions, clearances, and bar locations as exact, practical considerations require that there be accepted tolerances. These tolerances are the permissible variations from dimensions given on drawings. Overall dimensions of reinforced concrete members are usually specified by the engineer in whole inches for beams, columns, and walls, sometimes half-​inches for thin slabs, and often 3-​in. increments for more massive elements such as plan dimensions for footings. Formwork for the placing of these members must be built carefully so that it does not deform excessively under the action of workmen, construction machinery loads, and wet concrete [2.24]. Based on accepted practice, ACI Committee 117 [2.25] has provided recommended tolerances for concrete construction and materials. Examples of accepted tolerances for variation in cross-​sectional dimensions of columns and beams and in the thickness of walls are + 1 2 in. and  –​3 8 in. when the specified dimension is greater than 12 in. but not exceeding 3 ft [2.25]. For concrete footings, accepted variations in plan dimensions are +2 in. and  –​1 2 in. [2.25], whereas the thickness has an accepted tolerance of  –​5% of the specified thickness [2.25]. The strength reduction factor φ is intended to account for (among other reasons) a case in which several acceptable tolerances might adversely combine to reduce the strength from that computed using specified dimensions. Reinforcing bars are normally specified in 3-​in. length increments. Placement tolerances for reinforcement are given in the ACI Code [ACI-​26.6.2.1(a)] in terms of the effective depth d (distance from compression face of concrete to center of tension steel), and in terms of the specified concrete cover in flexural members, walls, and compression members, as follows: Effective Depth, d

Tolerances On Effective Depth d

(in.)

(mm)

(in.)

(mm)

On Specified Concrete Cover (in.)

(mm)

Smaller of (–1/3) × specified cover or d≤8

200a

± 3 8

±10a

– ​3 8

–​10a

d>8

200a

± 1 2

±13a

– ​1 2

–​13a

These values are from ACI 318-​14M, and are not hard conversions from ACI 318-​14.

a

Since the effective depth and the clear concrete cover are both components of total depth, the tolerances on those dimensions are directly related. When the tolerances on bar placement and cover accumulate, the overall dimension tolerance may be exceeded; thus field adjustment may have to be made [2.25]. This may be particularly important for very thin sections such as in prestressed, precast, and shell structures. For location of bars along the longitudinal dimension, and of bar bends, the tolerance is ±2 in. (±50 mm) except at discontinuous ends of brackets and corbels, where the tolerance shall be ± 1 2 in. (±13 mm), and ±1 in. (±25 mm) at the discontinuous ends of other members [ACI-​26.6.2.1(b)].

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SELECTED REFERENCES

41

2.12  ACCURACY OF COMPUTATIONS When one understands that variations in material strength, for both steel and concrete, and in dimensions are inevitable (and acceptable), it becomes clear that design calculations for reinforced concrete structures do not require a high degree of precision. The designer should place highest priority on determining proper location and length of steel reinforcement to carry the tension forces. Failures, when they occur, generally result from gross underestimation of tensile forces or failure to identify ways in which the structure or element will behave under loads. Failures are rarely the result of carrying too few significant figures in the design computations. However, significant figures may be lost in arithmetic operations, and gross errors sometimes result from sloppiness. Results indicated on the display of a calculator or computer, no matter how many digits, rarely indicate a precision of more than two or three significant figures. Recording of results on computation sheets should not exceed four significant figures, primarily for systematic control and checking of computations. While the extra digits recorded may not at first seem harmful, the resulting many-​digit values make difficult the scanning of computations to detect gross errors (in addition to being a waste of time).

SELECTED REFERENCES 2.1. J. G. MacGregor. “Safety and Limit States Design for Reinforced Concrete,” Canadian Journal of Civil Engineering, 3, December 1976, 484–​513. 2.2. Frank Kerekes and Harold B. Reid Jr. “Fifty Years of Development in Building Code Requirements for Reinforced Concrete,” ACI Journal, Proceedings, 50, February 1954, 441–​472. 2.3. Chester P. Siess. “Writing the Code—​More Than 40 Years on Committee 318” [interview conducted by Nancy L. Gavlin], Concrete International, 20, November 1998, 37–​42. 2.4. ACI–​ASCE Committee 327. “Ultimate Strength Design,” ACI Journal, Proceedings, 52, January 1956, 505–​524. 2.5. Robert G. Sexsmith and Mark F. Nelson. “Limitations in Application of Probabilistic Concepts,” ACI Journal, Proceedings, 66, October 1969, 823–​828. 2.6. Jack R. Benjamin and N. C. Lind. “A Probabilistic Basis for a Deterministic Code,” ACI Journal, Proceedings, 66, November 1969, 857–​865. 2.7. C.  Allin Cornell. “A Probability-​ Based Structural Code,” ACI Journal, Proceedings, 66, December 1969, 974–​985. 2.8. R. C. Reese, D. E. Allen, C. A. Cornell, Luis Esteva, R. N. White, R. G. Sexsmith, and George Winter. “Probabilistic Approaches to Structural Safety,” ACI Journal, Proceedings, 73, January 1976, 37–​49. 2.9. Haresh C. Shah and Robert G. Sexsmith. “A Probabilistic Basis for the ACI Code,” ACI Journal, Proceedings, 74, December 1977, 610–​611. 2.10. Bruce Ellingwood. “Reliability of Current Reinforced Concrete Designs,” Journal of the Structural Division, ASCE, 105, ST4 (April 1979), 699–​712. 2.11. Bruce Ellingwood. “Reliability Based Criteria for Reinforced Concrete Design,” Journal of the Structural Division, ASCE, 105, ST4 (April 1979), 713–​727. 2.12. J.  G. MacGregor, S.  A. Mirza, and B.  Ellingwood. “Statistical Analysis of Resistance of Reinforced and Prestressed Concrete Members,” ACI Journal, Proceedings, 80, May–​June 1983, 167–​176. 2.13. James G.  MacGregor. “Load and Resistance Factors for Concrete Design,” ACI Journal, Proceedings, 80, July–​August 1983, 279–​287. 2.14. Theodore V.  Galambos, Bruce Ellingwood, James G.  MacGregor, and C.  Allin Cornell. “Probability Based Load Criteria:  Assessment of Current Design Practice,” Journal of the Structural Division, ASCE, 108, ST5 (May 1982), 959–​977. 2.15. Bruce Ellingwood, James G.  MacGregor, Theodore V.  Galambos, and C.  Allin Cornell. “Probability Based Load Criteria:  Load Factors and Load Combinations,” Journal of the Structural Division, ASCE, 108, ST5 (May 1982), 978–​997. 2.16. Ross B.  Corotis. “Probability-​ Based Design Codes,” Concrete International, 7, April 1985, 42–​49.

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2.17. Morris Israel, Bruce Ellingwood, and Ross Corotis, “Reliability-​Based Code Formulations for Reinforced Concrete Buildings,” Journal of Structural Engineering, ASCE, 113, 10 (October 1987), 2235–​2252. 2.18. ASCE. American Society of Civil Engineers, Minimum Design Loads for Buildings and Other Structures (ASCE/​SEI 7-​10). Reston, VA: American Society of Civil Engineers, 2010. 2.19. International Code Council. International Building Code. Country Club Hills, IL: International Code Council, Inc., 2012. 2.20. Ronald Janowiak, Michael Kreger, and Antonio Nanni, Editors. The Reinforced Concrete Design Manual [SP-​17(11)1)]. American Concrete Institute, Farmington Hills, MI, 2012, 338 pp. 2.21. CRSI Design Handbook 2008 (10th ed.), Concrete Reinforcing Steel Institute, Schaumberg, IL, 2008. 2.22. PCI Design Handbook—​Precast and Prestressed Concrete (7th ed.), Prestressed Concrete Institute, Chicago, 2010, 804 pp. 2.23. ACI Committee 315. ACI Detailing Manual–​2004 [SP-​66(04)]. American Concrete Institute, Farmington Hills, MI, 2004, 212 pp. 2.24. ACI Committee 347. Guide to Formwork for Concrete. American Concrete Institute, Farmington Hills, MI, 2014, 36 pp. 2.25. ACI Committee 117. Specification for Tolerances for Concrete Construction and Materials (ACI 117-​10) and Commentary, (ACI 117-​10). Farmington Hills, MI: American Concrete Institute, 2010, 78 pp. 2.26. CRSI. “Construction Tolerance Conflicts in Reinforced Concrete,” Engineering Data Report Number 40. Schaumberg, IL: Concrete Reinforcing Steel Institute, 1995. 2.27. P. R. Morgan, T. E. Ng, N. H. M. Smith, and G. D. Base. “How Accurately Can Reinforcing Steel Be Placed? Field Tolerance Measurement Compared to Codes,” Concrete International, 4, October 1982, 54–​65.

CHAPTER 3 FLEXURAL BEHAVIOR AND STRENGTH OF BEAMS

3.1 GENERAL INTRODUCTION Reinforced concrete beams are primarily subjected to a combination of flexure and shear. Axial forces, if present, are generally negligible. If properly designed, a reinforced concrete beam should be able to undergo large, noticeable deformations prior to failure, providing ample warning to occupants about the risk of a potential failure. It is thus of utmost importance to have a thorough understanding of the behavior of beams and the factors that control their flexural response. The material presented in this chapter is intended to serve this purpose.

University of Wisconsin Stadium, Madison; rectangular tapered beams, cantilevers, and rigid frame (Photo by C. G. Salmon). 

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C H A P T E R   3     F L E X U R A L B E H A V I O R A N D S T R E N G T H O F   B E A M S

3.2 FLEXURAL BEHAVIOR AND STRENGTH OF RECTANGULAR SECTIONS Consider in Fig. 3.2.1 a simply supported reinforced concrete beam with two concentrated loads on top and supported at the bottom. Under such loading and support conditions, flexure-​induced stresses will cause compression at the top and tension at the bottom of the beam. Concrete, which is strong in compression but weak in tension, resists the stresses in the compression zone, while steel reinforcing bars are placed in the bottom of the beam to resist tension once flexural cracking has occurred. As the applied load is gradually increased from zero to failure of the beam (ultimate condition), a well-​designed beam may be expected to behave in the following manner: 1. Initially, when the applied load is low, the stress and strain distribution is essentially linear over the depth of the section. The tensile stresses in the concrete are low enough that the entire cross section remains uncracked. The flexural strain and stress distribution is as shown in Fig. 3.2.1(a). 2. On increasing the applied load, the tensile stress at the bottom of the beam reaches the tensile strength of the concrete, at which point a transverse (flexural) crack forms. Right after cracking, the neutral axis moves upward and a sudden increase in tensile stress in the steel reinforcement occurs. Just below the neutral axis, tensile stresses remain below the tensile strength of the concrete, leaving a small depth of the beam uncracked. These tensile stresses in the concrete, however, offer only a small contribution to flexural strength. At this stage, concrete behavior in compression is essentially linear. 3. As load is further increased, the neutral axis moves farther upward and the steel reinforcement reaches its yield strength [Fig. 3.2.1(b)]. After yielding, flexural cracks widen as the beam deflection increases. The concrete stress distribution in the compression zone becomes highly nonlinear [see Fig. 3.2.1(c)]. It is possible for the steel reinforcement to develop large tensile strains and reach strain hardening. A relatively modest increase in load beyond first yield is to be expected. Beam deflections, however, are large at the point of maximum load, with strains in the steel several times the yield strain. 4. Failure occurs when the concrete strain at the top of the beam reaches the crushing strain of concrete. Prior to failure, a small decrease in load is expected as the concrete becomes highly nonlinear and the neutral axis depth starts to increase, leading to a decrease in the lever arm between the internal compressive and tensile forces.

Basis of Flexural Strength Consider the rectangular beam cross section shown in Fig. 3.2.2(a). Such a beam is assumed to reach its flexural strength when the extreme concrete fiber in compression reaches the crushing strain of the concrete εcu. The assumed strain and stress distributions for computing flexural strength are shown in Fig. 3.2.2. Two main assumptions have been made in drawing this figure: 1. It has been assumed that plane sections have remained plane after bending up to failure of the beam. This traditional beam theory assumption, strictly valid for elastic, homogeneous beams, has been verified by tests and found to be a good assumption for reinforced concrete beams loaded up to failure [3.1]. This assumption permits the use of a linear strain distribution over the beam depth [see Fig. 3.2.2(b)]. 2. The steel reinforcing bars are assumed to be perfectly bonded to the surrounding concrete such that there is no slip of bars relative to concrete. This assumption implies that the change in strain in the steel reinforcement and that in the concrete surrounding the steel are the same. While locally at a crack this assumption is not true, the average strain measured over several cracks indicates the concrete and the steel bars work together reasonably well.

45



3. 2  F L E XU RAL B E H AV I O R A N D S T R E N G T H O F R E C TA N G U L A R S E C T I O N S

b

h

45

fc

εc d εs

fs (a)

εc

fc

εs = εy

fs = fy (b)

εc

εs > εy

fs ≥ fy (c)

Figure 3.2.1  Strain and stress distributions in a reinforced concrete beam under increasing load. 

From Fig. 3.2.2 it may also be noted that tensile stresses in the concrete are ignored, as their contribution to flexural strength is negligible. This implies that the portion of the beam cross section below the neutral axis does not affect the flexural strength. Thus, the important depth dimension for computing strength is the effective depth d rather than the overall depth h. The effective depth is defined as the distance from the extreme fiber in compression to the centroid of the tension steel area. When the tension steel comprises bars in several layers, the effective depth d is taken as the distance from the extreme compression fiber to the centroid of the combined area, with all bars assumed to have the same strain for calcula­ tion of tensile stress and force in the reinforcement. Although the compressive stress distribution in a beam may be expected to have the same general shape as that obtained from a test cylinder (or cube) specimen [1.60, 3.1, 3.2], the strain at crushing and the peak compressive stress will not be necessarily the same. A cylinder or a cube specimen is subjected to a uniform strain distribution during a standard test, while the compression zone of a beam has a strain gradient [Fig. 3.2.2(b)]. Also, the in-​place concrete compressive strength will likely differ from that obtained using a cylinder compressive test (see Section 1.7). The exact shape of the stress distribution in the compres­ sion zone of a beam cannot thus be determined, but it can be expressed in terms of three coefficients ( k1 , k2 , k3 ) that define the magnitude and position of the internal compression force resultant, as shown in Fig. 3.2.2(c). The k3 coefficient represents the ratio between the peak stress in the compression zone of a beam to the cylinder compressive strength fc′. The coefficient k1 represents the ratio between the average stress and the peak stress k3 fc′. The k2 factor, on the other hand, defines the location of the resultant compressive force C from the extreme compressive fiber, as a fraction of the neutral axis depth c. According to Hognestad [3.1, 3.2], this approach was first proposed by Stüssi in 1932. Further developments occurred during the 1950s, led by Hognestad himself and others [3.1–​3.6]. The stress-​strain response of the reinforcing steel is typically assumed to be the same as that obtained from a tension test. Thus, if the strain in the steel is known, the tensile stress and corresponding tensile force can be calculated. In many cases, such as in the calculation

46

46

C H A P T E R   3     F L E X U R A L B E H A V I O R A N D S T R E N G T H O F   B E A M S k3 f’c

b

εcu = crushing strain

MR c

N.A.

C = k1k3 f’c bc

d As

k2 c

T = Asfy (for εs ≥ εy) εs

d = effective depth; distance from compression face to centroid of tension steel (a) Singly reinforced beam

(b) Strain condition when MR is reached

(c) Stress condition when MR is reached

Figure 3.2.2  Strain and stress conditions when flexural strength is reached. 

of the nominal flexural strength (see Section 3.4), any increase in stress beyond yield due to strain hardening is ignored. When concrete crushing occurs (commonly a sudden occurrence), the strain in the tension steel may be either larger or smaller than the yield strain ε y. If the steel area As relative to the area of the beam cross section is low enough, the steel will yield prior to crushing of the concrete; the result will be a ductile failure mode in which there is large deformation. On the other hand, the reinforcing steel, when used in large quantities, would remain elastic at the time of crushing of the concrete, causing a brittle or sudden mode of failure. Sections in which the steel yields prior to the concrete reaching the strain εcu are said to be underreinforced. Otherwise, the section is overreinforced. The ACI Code has provisions that limit the strain in the tension steel, with the intent of ensuring a ductile mode of failure (see Section 3.6). The internal compressive force C [Fig. 3.2.2(c)] is the resultant of the compressive stresses acting on the compression concrete area, which may be considered a “stress solid” volume as

C = favg cb = k1 k3 fc′ cb

(3.2.1)

Assuming that the steel yields prior to crushing of the concrete, the tensile force T is

T = As f y

(3.2.2)

Equilibrium requires C = T , from which

c=

As f y k1 k3 fc′ b

(3.2.3)

The flexural strength, M R , may then be expressed as

M R = T (arm ) = T ( d – k2 c ) = As f y ( d – k2 c )

(3.2.4)

Substituting Eq. (3.2.3) for c into Eq. (3.2.4) gives

 k As f y  M R = As f y  d − 2 k1 k3 fc′ b  

(3.2.5)

One may note that if the flexural strength MR is the quantity of interest, it is readily obtainable from Eq. (3.2.5) if the quantity k2 / ( k1 k3 ) is known. It is thus not necessary to have values for k1, k2, or k3 individually if the value for the combined term is known. Experimental results [1.59, 3.3] have established values for the combined term, as well as the individual k values, with some of the results shown in Fig. 3.2.3. From this figure, k2 / ( k1 k3 ) ranges from about 0.55 to 0.63. It should be mentioned that these values, which were experimentally determined when crushing of the concrete occurred at the compression face, necessarily involved variation in the crushing strain εcu for the various tests.

47



47

3 . 3   W H I T N E Y R E C TA N G U L A R S T R E S S D I S T R I B U T I O N fc’, MPa 20

1.0

25

30

Coefficients

k1

k2/(k1k3)

0.5 k2

2500

3000

4000

3500

4500

5000

f’c , psi 250

200

300

350

f’c , kgf/cm2

Figure 3.2.3  Values of coefficients that define compression zone in a flexural member. (Adapted from Hognestad, Hanson, and McHenry [3.3].) 

3.3 WHITNEY RECTANGULAR STRESS DISTRIBUTION Flexural strength based on the approximately parabolic stress distribution of Fig. 3.2.2(c) may be computed by using Eq. (3.2.5) with given values of k2 /(k1 k3 ). However, it is desirable for the designer to have a simple method in which basic static equilibrium is used. Although the use of a stress block in lieu of a more sophisticated inelastic stress distribution has been proposed since the early 1910s [3.1], the most widely used stress block is that developed by Whitney in the 1930s [3.7, 3.9]. Whitney realized that it was not necessary to know the real stress distribution in the compression zone of a beam to accurately predict its flexural strength. He further added [3.7]: “Whatever it actually is, it must have an average intensity, fc, and an effective depth, a.” Assuming a peak concrete compressive stress equal to fc′ (i.e., k3 = 1.0), Whitney proposed the use of an average stress of 0.85 fc′ over a depth a [Fig. 3.3.1(c)].

0.85 f’c

f’c

b

a 2

k2 c

c d

h

d–

N.A. As

C

a = β1 c

C

T

Mn

a 2

T

d = effective depth; distance from compression face of concrete to centroid of tension steel (a) Beam

(b) Assumed stress distribution

Figure 3.3.1  Definition of Whitney rectangular stress distribution.

(c) Whitney rectangular stress block

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A relationship between the depth of the stress block a and the neutral axis depth c was proposed by Mattock, Kriz, and Hognestad [3.6], where a = β1c, determined so that a /2 = k2 c. For fc′ ≤ 4000 psi, β1 = 0.85, decreasing at a rate of 0.05 for each 1000 psi of fc′ in excess of 4000 psi.1 The value of β1 is not to be taken less than 0.65. The ACI Code explicitly accepts the Whitney rectangular stress block (ACI-​22.2.2.4.1) in combination with the β1 factor proposed by Mattock, Kriz, and Hognestad (ACI-​22.2.2.4.3) for the determination of nominal flexural strength (see Section 3.4).

3.4 NOMINAL FLEXURAL STRENGTH M n —​ RECTANGULAR SECTIONS HAVING TENSION REINFORCEMENT ONLY The quantities defining a rectangular section with tension reinforcement only are b, d, and As [Fig. 3.3.1(a)]. Such a section is said to be singly reinforced. The steel area As is, of course, furnished by the combined area of an actual number of reinforcing bars. Protective concrete cover is necessary around the bars not only to make the steel and concrete act together but also very importantly, to provide protection against corrosion and fire. Minimum cover requirements are generally prescribed by code (see ACI-​20.6.1). For computation of nominal flexural strength Mn, the following assumptions (ACI-​22.2) are made: 1. The strength of members shall be based on satisfying the applicable conditions of equilibrium and compatibility of strains. 2. Strain in the steel reinforcement and in the concrete shall be assumed directly proportional to the distance from the neutral axis (except for deep members covered under ACI-​9.9). 3. The maximum usable strain εcu at the extreme concrete compressive fiber shall be assumed equal to 0.003 (Fig. 3.4.1.). 4. The tensile strength of the concrete is to be neglected (except for certain prestressed concrete conditions).

Steel reinforcement

fy

εy = 0.00207 for fy = 60 ksi

Stress

Concrete εy = fy /Es εcu = ACI Code-defined crushing strain

f’c 1

2

3

4

5

6

7

8

9

10

× 10–3 Strain

Figure 3.4.1  Stress-​strain curves for concrete and steel reinforcement, and definitions of εy and εcu.  1  For SI, ACI 318-​14M gives β1 = 0.85 for fc′ ≤ 28 MPa and reduces by 0.05 for each 7 MPa of fc′ in excess of 28 MPa, but not less than 0.65.

49



3.4  NOMINAL FLEXURAL STRENGTH

49

5. The stress-​strain relationship for nonprestressed steel shall be assumed to be elastic–​ perfectly plastic (i.e., strain hardening is ignored) (Fig. 3.4.1). The modulus of elasticity of such steel reinforcement may be taken as 29,000,000 psi (200,000 MPa or 2,040,000 kgf/​cm2). 6. For practical purposes, the relationship between the concrete compressive stress distribution and the concrete strain when nominal strength is reached may be taken as an equivalent rectangular stress distribution (ACI-​22.2.2.4), wherein a concrete stress intensity of 0.85 fc′  is assumed to be uniformly distributed over an equivalent compression zone bounded by the edges of the cross section and a straight line located parallel to the neutral axis at a distance a = β1c from the fiber of maximum compressive strain. The distance c from the fiber of maximum strain to the neutral axis is measured in a direction perpendicular to that axis. The value of β1 is given by the following equations: For  fc′ ≤ 4000 psi,

β1 = 0.85

(3.4.1)

 f ′ − 4000  ≥ 0.65 β1 = 0.85 − 0.05  c  1000 

(3.4.2)

For  fc′ > 4000 psi,

It should be noted that assumption (6) describes the Whitney rectangular compressive stress distribution with the provisions for determination of neutral axis depth by Mattock, Kriz, and Hognestad [3.6] (see Section 3.3). However, other shapes of stress solids, such as the trapezoid and the parabola, have been used [3.10] and are acceptable for use according to ACI-​22.2.2.3. The nominal flexural strength M n, using the equivalent rectangular stress block, is obtained from Fig. 3.3.1(c) as follows: 

C = 0.85 fc′ ba

(3.4.3)

T = As f y

(3.4.4)

where the use of f y assumes that the steel yields prior to crushing of the concrete (required for beams, as discussed in Section 3.6). Equilibrium requires that C = T ; thus

a=

As f y 0.85 fc′ b

(3.4.5)

Taking moments about the compressive force resultant, C

a a   M n = T  d −  = As f y  d −     2 2

(3.4.6)

which, on substituting Eq. (3.4.5) into Eq. (3.4.6), gives As f y   M n = As f y  d − 0.59 fc′ b  

(3.4.7)

Note that 0.59 corresponds to k2 /( k1 k3 ) of Eq. (3.2.5). One may also note that β1 is needed only to establish the neutral axis location for determining the steel strain. As long as the steel strain exceeds ε y , as shown in Fig. 3.2.2, the nominal flexural strength is not affected by the value of  β1 .

50

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C H A P T E R   3     F L E X U R A L B E H A V I O R A N D S T R E N G T H O F   B E A M S

EXAMPLE 3.4.1 Determine the nominal flexural strength Mn of the rectangular section shown in Fig. 3.4.2. given fc′ = 5000 psi, f y = 65, 000 psi, b = 14 in., d = 21.5 in., and As = 4–#8 bars. a = 0.80c εcu = 0.003

14”

0.85f’c

a 2 C

4.32”

a 2

d–

21.5” 17.18” As = 3.16 sq in.

T

εs (a) Cross section

(b) Assumed strain and stress distribution

Figure 3.4.2  Singly reinforced beam of Example 3.4.1. 

SOLUTION Assume the steel has already yielded when the nominal strength is reached. From Fig. 3.4.2 the internal forces are C = 0.85 fc′ ba = 0.85(5)(14)a = 59.5a T = As f y = 3.16(65) = 205 kips



For equilibrium, C = T ; therefore a=

205 = 3.45 in. 59.5

β1 = 0.80 for fc′ = 5000 psi



The neutral axis position is c=

a 3.45 = = 4.32 in. β1 0.80

The strain in the tension steel when the compressive strain 0.003 is reached at the extreme concrete fiber is, by straight-​line proportion,

εs = εy =

d −c 17.18 (0.003) = (0.003) = 0.012 c 4.32 fy Es

= 0.0022

When the nominal flexural strength is reached, ε s is five times ε y , which means that large deformation has occurred before the crushing of concrete. The assumption that the steel yields has been validated. The nominal flexural strength is a a   M n = C  d −  or T  d −    2 2  

       = 205 ( 21.5 − 1.73)

1 12

= 339 ft-kips

51



51

3.5  BALANCED STRAIN CONDITION

3.5 BALANCED STRAIN CONDITION At the balanced strain condition [Fig. 3.5.1(b)], the maximum strain at the extreme concrete compressive fiber just reaches the crushing strain εcu when the tension steel reaches a strain εy. As mentioned earlier, the ACI Code requires the use of ε cu = 0.003 ( ACI − 22.2.2.1). The balanced strain condition will exist for an amount of tension steel Asb such that the neutral axis depth will be cb, as shown in Fig. 3.5.1(b). If the actual As were greater than Asb , equilibrium of internal forces ( C = T ) would require an increase in the depth a of the compression stress block (and thereby would also make c exceed cb), so that the strain εs would be less than εy when the extreme fiber in compression reaches ε cu (= 0.003 according to the ACI Code). The failure of this beam will be sudden when the concrete reaches the strain 0.003, with little deformation (steel does not yield) to warn of impending failure. On the other hand, when the actual As is less than Asb , the tensile force decreases so that internal force equilibrium leads to a reduction in the depth a of the compression stress block (and thereby makes c less than cb), giving a strain εs greater than ε y . In this case, with the steel having yielded and if As is substantially less than Asb , the beam will have noticeable deflection prior to the concrete reaching the crushing strain of 0.003. Therefore, the relative amount of tension steel with respect to that in the balanced strain condition can be used to determine whether the failure is ductile (gradual, giving warning) or brittle (sudden, without warning).

Reinforcement Ratio at Balanced Strain Condition for Rectangular Beam Having Tension Reinforcement Only The symbol ρ , the tension reinforcement ratio (often called tension reinforcement percentage), may be conveniently used to represent the relative amount of tension reinforcement in a beam. Thus, using the dimensions of Fig. 3.5.1,

ρ=



As bd

(3.5.1)

The reinforcement ratio in the balanced strain condition, ρb , may be obtained by applying the equilibrium and compatibility conditions. From the linear strain condition, Fig. 3.5.1(b), cb ε cu 0.003 87, 000 = = = d ε cu + ε y 0.003 + f y / 29, 000, 000 87, 000 + f y



The compressive force Cb is Cb = 0.85 fc′ bab = 0.85 fc′ bβ1cb

  The tensile force Tb  is

Tb = Asb f y = ρb bdf y

 

0.85f’c

b

εcu = 0.003 cb

ab = β1cb

Cb

d

Asb = ρbbd Tb = Asbfy

εs = εy = fy /Es (a) Cross section

(b) Strain diagram

Figure 3.5.1  Balanced strain condition.

(c) Internal forces

(3.5.2)

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Equating Cb to Tb  gives 0.85 fc′ bβ1cb = ρb dbf y

ρb =



0.85 fc′  cb  β1    d fy

(3.5.3)

which on substitution of Eq. (3.5.2) gives

ρb =



0.85 fc′  87, 000  β1   fy  87, 000 + f y 

2

(3.5.4)

where the stresses f y and fc′ are in psi. It may be noted that ρb depends only on the material properties and is independent of the beam size. In other words, given fc′ and f y , the balanced reinforcement ratio ρb can be readily determined. Values of ρb for typical concrete and steel yield strengths are shown in Table 3.6.1. It can be seen that ρb increases with an increase in fc′, while it decreases with an increase in f y . For the common case of fc′ = 4000 psi and f y = 60 ksi, ρb = 2.85%, increasing to 3.77% for fc′= 6000 psi . When the yield strength is increased to 80 ksi while the same concrete strength is maintained, ρb becomes 66% of that for fy = 60 ksi. On the other hand, if Grade 40 steel is used, ρb increases by about 70% compared to ρb for fy = 60 ksi.

3.6 TENSION-​ AND COMPRESSION-​C ONTROLLED SECTIONS The ACI Code classifies reinforced concrete sections as either tension-​ or compression-​ controlled (ACI-​21.2.2) depending on the magnitude of the net strain in the reinforcement closest to the tension face, ε t , as the concrete reaches the assumed crushing strain ε cu of 0.003 as follows (see Fig. 3.6.1): Tension-​controlled sections:       εt ≥ 0.005 Compression-​controlled sections:   εt ≤ εy For sections having more than one layer of tension steel, the strain limits must be checked for the layer of reinforcement closest to the tension face of the member (at a distance dt from the extreme compressive fiber; see Fig. 3.6.1). For a typical Grade 60 reinforcing bar,

εy =

 

fy Es

=

60 = 0.0021 29, 000

Thus, the ACI Code allows the compression-​controlled strain limit εy to be taken as 0.002 for Grade 60 reinforcement. Sections with tension steel strains between the compression-​ and tension-​controlled strain limits are considered to be in a transition region.

Minimum Net Tension Steel Strain εt, min To have reasonable assurance of a ductile mode of failure, ACI-​9.3.3.1 requires that the net tension steel strain εt be equal to or greater than 0.004 in beams subjected to a factored axial load less than 0.1Ag fc′ (where Ag is the gross area of the cross section). For Grade 60 steel reinforcing bars, this requirement corresponds to about twice the yield strain ε y . 2 For SI,



ρb =

0.85 fc′  600  β1   fy  600 + f y 

with fc′ and fy in MPa, and β1 as given in footnote 1 in Section 3.3.

(3.5.4) 

53



53

3.6  TENSION- AND COMPRESSION-CONTROLLED SECTIONS

The writers, however, recommend that beam sections be designed to be tension controlled (i.e., ε t ≥ 0.005), to achieve a more ductile design with ample warning prior to failure. As discussed in Example 3.6.1, while the area of steel corresponding to the tension-​control limit is slightly less than that for ε t = 0.004, the design flexural strength will be approximately the same, the beam with larger steel area being less ductile. When reinforcement with yield strength higher than 60 ksi is used (e.g., Grade 80 steel), the tension-​controlled strain limit may need to be increased above 0.005 because the amount of plastic deformation in the steel reinforcement when ε cu = 0.003 will be less. For example, for Grade 80 steel, the plastic deformation for a net tensile strain of 0.005 would be only 1.8ε y compared to 2.5ε y for Grade 60 steel. In such cases, to ensure adequate inelastic deformation prior to failure, it may be desirable to design for a higher net tensile strain.

Reinforcement Ratio Corresponding to Strain Limit for Tension-​Controlled Sections ρtc Similar to the concept of balanced reinforcement ratio, there is a unique amount of reinforcement that will cause the tension steel to reach the net tensile strain ε t of 0.005 for tension-​controlled sections just as the extreme concrete fiber in compression reaches ε cu of 0.003. For sections with one layer of tension reinforcement, dt is equal to d and, using the procedure discussed in Section 3.5, the following expression for the reinforcement ratio ρ can be obtained

ρ (εt ) =



0.003 + ε y 0.003 + ε t

ρb

(3.6.1)

For the strain limit corresponding to tension-​controlled sections, Eq. (3.6.1) becomes

ρtc = ρ ( ε t = 0.005) =



0.85β1 fc′ 0.003 β f′ = 0.319 1 c fy 0.008 fy

(3.6.2)

This value represents the maximum reinforcement ratio that ensures a tension-​controlled section design. In other words, beam sections reinforced with a reinforcement ratio ρ ≤ ρtc will develop a net tensile strain ε t greater than or equal to 0.005. It should be noted that because Eq. (3.6.2) was derived assuming all tension reinforcement in a single layer (i.e., dt = d), the reinforcement ratio obtained from this equation will lead to a strain ε t slightly greater than 0.005 for sections with more than one layer of tension reinforcement. Further, it should be kept in mind that Eq. (3.6.2) applies only to rectangular sections or nonrectangular sections with a rectangular compression zone (see Section 4.4).

b

εcu = 0.003 c

dt = distance from extreme fiber in compression to centroid of the layer of reinforcement closest to the tension face.

dt

εt ≥ 0.005 εt ≤ εy

Tension controlled Compression controlled

Figure 3.6.1  Definition of tension-​and compression-​controlled sections (ACI-​21.2.2).

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TABLE 3.6.1  REINFORCEMENT RATIO ρ AT VARIOUS STRAIN CONDITIONS FOR SINGLY REINFORCED RECTANGULAR BEAMS fc′ = fy

3000 psi

4000 psi

5000 psi

6000 psi

8000 psi

β1 = 0.85

β1 = 0.85

β1 = 0.80

β1 = 0.75

β1 = 0.65

40,000 psi

ρb = ρtc = ρ (ε t = 0.005) = ρ (ε t = 0.0075) = ρ (ε t = 0.01) =

0.0371 0.0203 0.0155 0.0125

0.0495 0.0271 0.0206 0.0167

0.0582 0.0319 0.0243 0.0196

0.0655 0.0359 0.0273 0.0221

0.0757 0.0414 0.0316 0.0255

60,000 psi

ρb = ρtc = ρ (ε t = 0.005) = ρ (ε t = 0.0075) = ρ (ε t = 0.01) =

0.0214 0.0135 0.0103 0.0083

0.0285 0.0181 0.0138 0.0111

0.0335 0.0212 0.0162 0.0131

0.0377 0.0239 0.0182 0.0147

0.0436 0.0276 0.0210 0.0170

80,000 psi

ρb = ρtc = ρ (ε t = 0.005) = ρ (ε t = 0.0075) = ρ (ε t = 0.01) =

0.0141 0.0102 0.0077 0.0063

0.0188 0.0135 0.0103 0.0083

0.0221 0.0159 0.0121 0.0098

0.0249 0.0179 0.0137 0.0110

0.0288 0.0207 0.0158 0.0128

Values of ρ corresponding to net tensile strains ε t of 0.005, 0.0075, and 0.01 for typical concrete and steel yield strengths are shown in Table 3.6.1. As discussed earlier, it is desirable to use a higher net tensile strain limit when reinforcing steel with yield strength higher than 60 ksi is used.

Strength Reduction Factors φ A tension-​controlled section will exhibit a ductile failure mode with visible flexural cracks and deflection. For these types of members, the strength reduction factor φ is 0.90 (see Section 2.7). In contrast, compression-​controlled sections are expected to fail suddenly, with little or no warning of impending failure. This lack of ductility is accounted for by using a lower strength reduction factor. The ACI Code specifies two strength reduction factors for compression-​controlled sections depending on the type of transverse reinforcement used. Sections containing spirals as transverse reinforcement (spirally reinforced) have a φ factor of 0.75, while those with stirrups or ties have a φ factor of 0.65. In practice, spiral reinforcement is used almost exclusively in columns. Because spirals provide better confinement to the concrete (see Chapter 10), a larger φ factor is assigned to spirally reinforced sections. For sections that fall in the transition region between a compression-​controlled and a tension-​controlled section, the φ factor must be computed by using linear interpolation, as illustrated in Fig. 3.6.2. The linear equations in terms of the extreme tensile strain εt are as follows: For sections with stirrups or ties,

φ = 0.65 + 0.25

(ε

t

− εy

)

(0.005 − ε ) y

≤ 0.90



(3.6.3)

5



55

3.6  TENSION- AND COMPRESSION-CONTROLLED SECTIONS

For spirally reinforced sections,

φ = 0.75 + 0.15



(ε

t

− εy

)

(0.005 − ε )

≤ 0.90

(3.6.4)



y

where ε y is the yield strain of the tension reinforcement layer closest to the tension face, which for Grade 60 steel can be taken as 0.002. The φ equations can also be expressed in terms of the ratio c / dt , where c is the neutral axis depth measured from the compression face of the member and dt is the distance from the extreme compressive fiber to the layer of reinforcement closest to the tension face. When Grade 60 bars are used for the longitudinal reinforcement, the strength reduction factor φ can be obtained as follows: For sections with stirrups or ties,  1 5 φ = 0.65 + 0.25  −  ≤ 0.90  c / dt 3 



(3.6.5)

For spirally reinforced sections, (3.6.6)

 1 5 φ = 0.75 + 0.15  −  ≤ 0.90  c /d t 3 



It should be noted that beam sections designed with a reinforcement ratio ρ less than or equal to ρtc will be tension controlled and thus, they will be assigned a strength reduction factor φ = 0.90. εt, min = 0.004 for beams with factored axial load less than 0.1Agf’c (ACI-9.3.3.1)

φ 0.90 0.75 0.65 Compression controlled

ρ ≤ ρtc (Eq. 3.6.2)

Spiral Other Transition εt = εy

Tension controlled εt = 0.005

Figure 3.6.2  Variation of φ with net tensile strain εt. (Adapted from ACI-​R21.2.2.)

EXAMPLE 3.6.1 The beams shown in Fig. 3.6.3 have the same cross-​sectional dimensions but have been reinforced with different amounts of tension steel. Both have two layers of reinforcement, but the beam of Fig. 3.6.3(a) has 3–#8 bars for the upper layer and 3–#9 bars for the bottom layer. The beam of Fig. 3.6.3(b) has 3–#8 bars in both layers. Compute the nominal flexural strength, the corresponding φ factor, and the design flexural strength φMn. Use fc′ = 4500 psi and  f y = 60, 000 ksi. (Continued)

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Example 3.6.1 (Continued) 0.85fc’

14” εcu = 0.003 c = 7.29” dt = 17.5”

a/2 C = 322 K

a = 6.02”

d = 16.40” 3 – #8 εs

2.5”

T=C

εt = 0.0042 3 – #9

(a) 0.85f’c

14”

εcu = 0.003 c = 6.44” a = 5.31”

dt = 17.5”

a/2 C = 284 K

d = 16.25” 3 – #8 εs

2.5”

T=C

εt = 0.0052 3 – #8 (b)

Figure 3.6.3  Beam sections and strain and stress diagrams at nominal condition for Example 3.6.1.

SOLUTION (a) Referring to Fig. 3.6.3(a), the effective depth d, which is measured to the centroid of the reinforcement, is calculated as  

d=

3 (1.0 )(17.5) + 3 ( 0.79)(15) 3 (1.0 ) + 3 ( 0.79)

= 16.40 in.

Because there is more than one layer of tension steel, the distance dt is calculated to the layer of steel closest to the extreme tension face. In this case, dt = 17.5 in. Assuming the steel yields before the concrete reaches the strain ε cu, the internal tensile and compressive forces, as well as the depth of the stress block a, are calculated as  

T = As f y = 3 ( 0.79) + 3 (1.0 ) (60 ) = 322 kips = C

 

a=

322 C = = 6.02 in. 0.85 fc′ b 0.85 ( 4.5)(14 )

For = 4500, β1 = 0.825. Thus, the neutral axis depth c is  

c=

6.02 a = = 7.29 in. β1 0.825

Once the depth of the neutral axis c is known, the strain at the level of the centroid of the steel and at the layer of steel closest to the extreme tension face can be readily calculated:   

εs =

0.003 0.003 (d − c) = (16.40 − 7.29) = 0.0037 c 7.29

 

εt =

0.003 0.003 dt − c ) = (17.50 − 7.29) = 0.0042 ( c 7.29

(Continued)

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3.6  TENSION- AND COMPRESSION-CONTROLLED SECTIONS

Example 3.6.1 (Continued) The strain at the level of the centroid of the steel reinforcement εs is greater than the yield strain for Grade 60 steel (0.0021). Thus, the assumption that the steel yields prior to the concrete reaching the strain εcu was correct. While the strain at the centroid of the steel reinforcement εs is less than 0.004, the strain at the layer of steel closest to the extreme tension face (ε t = 0.0042) is greater than the minimum strain ε t,min of 0.004 required by ACI-​9.3.3.1. Because εt is less than 0.005, the section is in the transition region for the purpose of determining the strength reduction factor φ. Using Eq. (3.6.3), a value of φ = 0.83 is obtained. The nominal flexural strength of the section M n and the design flexural strength φ M n can now be calculated as  

a a 6.02  1    M n = T  d −  = C  d −  = 322  16.40 −  = 359 ft-kiips    2 2 2  12   φ M n = 0.83 (359) = 300 ft-kips

(b) The analysis of the section shown in Fig. 3.6.3(b) follows the same procedure as for the section in Fig. 3.6.3(a). Since the area of steel in the layer closest to the extreme tension fiber is smaller, the effective depth d needs to be calculated again. The depth dt , however, remains the same as the location of this layer of steel has not changed. Thus,  

d=

3 ( 0.79)(17.5) + 3 ( 0.79)(15) 6 ( 0.79)

= 16.25 in.

Since the area of steel has been reduced compared to the section analyzed in part (a), it is already known that the steel will yield prior to the concrete reaching the strain ε cu. The internal tensile and compressive force resultants, as well as the depth of the stress block a and of the neutral axis c, are then calculated as follows:  T = As f y = 6 ( 0.79) (60 ) = 284 kips = C

   

a=

 

284 C = = 5.31 in. 0.85 fc′ b 0.85 ( 4.5)(14 ) c=

a = 6.44 in. β1

With the neutral axis depth c known, the strain at the level of the centroid of the steel (ε s) and at the layer of steel closest to the extreme tension face (ε t ) can be readily calculated:   

εs =

0.003 0.003 (d − c) = (16.25 − 6.44) = 0.0047 6.44 c

 

εt =

0.003 0.003 ( dt − c ) = 6.44 (17.50 − 6.44) = 0.0052 c

In this case, the section is tension controlled (ε t > 0.005 ) and φ = 0.9 [compared with φ = 0.83 in part (a)]. The nominal and design flexural strengths are then calculated as  

a a 5.31 1    M n = T  d −  = C  d −  = 284  16.25 −  = 322 ft-kiips    2 2 2  12

 

φ M n = 0.90 (322 ) = 290 ft-kips (Continued)

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Example 3.6.1 (Continued) On comparing the design flexural strength φ M n for both sections, it can be seen that while the area of steel in the beam of part (b) was 88% of that used in part (a), the design flexural strengths of the sections differed by only 3%. The result is due to the lower φ factor assigned to beam sections in the transition region, such as the beam of part (a), compared to that assigned to beam sections that are tension controlled. From an economical and performance viewpoint, it is thus better to design beam sections to be tension controlled (ε t ≥ 0.005) rather than in the transition region (0.004 ≤ ε t < 0.005 ).

3.7 MINIMUM TENSION REINFORCEMENT When steel reinforcement in a flexural member is only a small amount because the factored moment Mu is low, the beam may perform uncracked at service load. The computation of nominal moment strength M n, in accordance with Section 3.4, assumes the tension concrete to be cracked. Thus, the computed nominal strength M n for a section having a small amount of reinforcement could be less than the strength of the same section of plain uncracked concrete (i.e., no reinforcement). Since a ductile failure mode is desired, the lowest amount of steel permitted should be at least the amount that would equal the strength of an unreinforced section. The desired relationship then becomes

strength of reinforced  strength of plain  concrete beam, φ M  ≥ concrete beam, M  n  cr   

(3.7.1)

The flexural strength of a plain concrete beam, known as the cracking moment M cr , is achieved when the extreme fiber in tension reaches the modulus of rupture fr (see Section 1.8). For normal-​weight concrete, ACI-​19.2.3.1 uses fr = 7.5 fc′



(3.7.2)3

As discussed in Section 1.8, this modulus of rupture is meant to approximately represent an average value from test results. Modulus of rupture, however, is highly variable, as shown in Fig. 1.8.1. Assuming plain (nonreinforced) concrete as an elastic homogeneous material, the flexure formula gives Mcr as M cr = fr



Ig yt

= fr S t



(3.7.3)

where Ig = moment of inertia of the gross concrete cross section yt = distance from the neutral axis to the extreme fiber in tension St = elastic section modulus with respect to extreme fiber in tension; Cbwh2/​6 C = coefficient to account for flanges of T-​sections; C = 1.0 for rectangular section bw = width of section; width of web for T-​section h = overall depth of the section On expanding, Eq. (3.7.3) becomes M cr = 7.5 fc′ C





3  For SI, with fr and fc′ in MPa,

fr = 0.62 fc′

bw h 2 6

(3.7.4)

(3.7.2) 

59



59

3.7  MINIMUM TENSION REINFORCEMENT

For a reinforced concrete section, using Eq. (3.4.6), a  φ M n = φ As f y  d −   2



(3.7.5)

Substituting Eqs. (3.7.4) and (3.7.5) into Eq. (3.7.1) gives

7.5 fc′ Cbω h 2 a  φ As f y  d −  ≥ M cr =  6 2

(3.7.6)

Estimating a/​2 as 0.05d, and using φ = 0.9 for flexure, Eq. (3.7.6) gives



 7.5 fc′   h  2  C  As ,min =      bω d  f y   d   5.13 

(3.7.7)

Rectangular Sections—For rectangular sections, C = 1.0 and h/​d varies from about 1.05 to 1.2. For such sections, Eq. (3.7.7) becomes

As ,min =

1.6 fc′ fy

bω d to As ,min =

2.1 fc′ fy

bω d



(3.7.8)

T-​Sections Having Slab in Compression—For this case, C will vary from about 1.3 to 1.6 for practical ranges of ratios of flange thickness to overall depth and of flange width to web width. Taking C = 1.5 along with h/​d from 1.05 to 1.2, Eq. (3.7.7) becomes

As ,min =

2.4 fc′ fy

bω d to As ,min =

3.2 fc′ fy

bω d



(3.7.9)

The minimum area of tension reinforcement required in the ACI Code is (ACI-​9.6.1.2) As ,min =



3 fc′ fy

bw d ≥ 200

bw d fy

(3.7.10)4

Note that 200bw d / f y in Eq. (3.7.10) will govern for concrete strengths fc′ less than 4444 psi. For Grade 60 steel and fc′ ≤ 4444 psi, As ,min = 0.0033bw d . For higher concrete strengths, As,min increases, being As ,min = 0.0045bw d for fc′ = 8000 psi. This minimum area of tension reinforcement is required “at every section where tensile reinforcement is required by analysis …” (ACI-​9.6.1.1), unless the area of reinforcement provided is at least one-​third greater than the area required by analysis (ACI-​9.6.1.3). T-​Sections Having Slab in Tension—For this case, C will vary from about 3.0 to 4.0 for a practical range of ratios of flange thickness to overall depth and of flange width to web width. Taking C = 3.5 along with h/​d from 1.05 to 1.2, Eq. (3.7.7) becomes

As ,min =

5.6 fc′ fy

bω d to As ,min =

7.4 fc′ fy

bω d



(3.7.11)

ACI-​9.6.1.2 states that minimum reinforcement “For a statically determinate beam with a flange in tension, … ” is to be computed using Eq. (3.7.10), with bw replaced by either 2bw or the width of the flange, whichever is smaller.

4  For SI, with fy and fc′ in MPa, and bw and d in mm,



As ,min =

fc′ 4 fy

bw d

(3.7.10) 

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3.8 DESIGN OF RECTANGULAR SECTIONS IN BENDING HAVING TENSION REINFORCEMENT ONLY In the design of rectangular sections in bending with tension reinforcement only, the problem is often to determine b, d, and As , from the required value of M n = Mu / φ, and the given material properties. The two conditions of equilibrium are C = T

(3.8.1a)

a a   Mn = T  d −  = C  d −    2 2

(3.8.1b)

and

Since there are three unknowns but only two conditions, there are many possible solutions. If the reinforcement ratio ρ is preset, then, from Eq. (3.8.1a),  

C = 0.85 fc′ ba = ρ bd f y = T



 fy  d a = ρ  0.85 fc′

(3.8.2)

Substituting Eq. (3.8.2) into Eq. (3.8.1b),



 ρ  fy   M n = ρ bd fy  d −  d 2  0.85 fc′  

(3.8.3) 2

A strength coefficient of resistance Rn is obtained by dividing Eq. (3.8.3) by bd  and letting m=



fy 0.85 fc′

(3.8.4a)

Thus

Rn =

Mn  1  = ρ f y  1 − ρ m  2  bd 2

(3.8.4b)

The relationship between ρ and Rn for various values of fc′ and fy is shown in Fig. 3.8.1. In some situations the values of b and d may be preset, which is equivalent to having Rn preset; then ρ may be determined by solving the quadratic equation Eq. (3.8.4b). Thus  

 1  Rn = ρ f y  1 − ρ m  2 

from which



ρ=

2 mRn  1 1 − 1 −  m fy 

(3.8.5)

The procedure (neglecting certain practical decisions for now) to be used in the strength design of rectangular sections with tension reinforcement only involves the following steps. 1. Assume a value of ρ equal to or less than the reinforcement ratio corresponding to the tension control limit, ρtc [see Eq. (3.6.2), repeated here, or Table 3.6.1], but greater

61



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3 . 8   D E S I G N O F R E C TA N G U L A R S E C T I O N S

than the minimum of ACI-​9.6.1.2 (Eq. 3.7.10). As the section would then be tension controlled, the minimum strain limit ε t = 0.004 of ACI-​9.3.3.1 will be satisfied.

ρ ( ε t = 0.005) = ρtc = 0.319



β1 fc′ f y [3.6.2]

where for f c′ ≤ 4000 psi β1 = 0.85

 f c′ − 4000  ≥ 0.65  for f c′ > 4000 psi β1 = 0.85 − 0.05   1000  2. If b and d are unknown, determine the required bd 2  from

required bd 2 =

 

required M n Mu / φ = Rn Rn

in which  1  Rn = ρ f y  1 − ρ m  2 

  with

m=

 

fy 0.85 fc′

As the section is being designed with ρ ≤ ρtc , φ = 0.9.

f’c = 8000 fy = 80,000

1400

f’c = 6000 fy = 60,000

1200 f’c = 4000 fy = 80,000 f’c = 4000 fy = 60,000

100

9

90

8

6 5

600

4

400

Mn = Rn bd 2 200 Min ρ controls 0

0

0.005

0.01

0.025 0.015 0.02 Reinforcement Ratio, ρ

0.03

110

10

7

800

120

11

f’c = 6000 fy = 80,000

1000

12

Coefficient of Resistance, Rn, MPa

1600

Coefficient of Resistance, Rn, psi

f’c = 8000 fy = 60,000

80 70 60 50 40

3

30

2

20

1

10

0 0.035

Coefficient of Resistance, Rn, kgf/cm2

1800

0

Figure 3.8.1  Strength curves (Rn vs ρ) for singly reinforced rectangular sections. Upper

limit of curves is at ρmax, that is, reinforcement ratio corresponding to ε t  = 0.004. 

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3. Choose a suitable set of values of b and d so that the bd 2 provided is approximately equal to the bd 2 required. (Note: Actually, d is not chosen; rather, the overall depth h is chosen and d is computed from h while maintaining the desired minimum protective cover.) 4. Determine the revised value of ρ after computing Rn = M n / bd 2 for the selected section using one of the following methods: a. By formula (most exact),

ρ=

 

2 mRn  1 1 − 1 −  m fy 

b. By curves (Fig. 3.8.1). c. By approximate proportion (on the safe side if revised Rn is smaller than original  Rn), the revised ρ is

ρ ≈ (original ρ)

 

(revised Rn ) (original R n )

It may be noted from Fig. 3.8.1 that the relationship between Rn and ρ is approximately linear at low values of ρ , even though the actual equation is a quadratic one. 5. Compute As  from  

As = (revised ρ) (actual bd )

6. Select reinforcement and check the strength of the section to be certain that  

φ M n ≥ Mu

EXAMPLE 3.8.1 Determine a set of values b, d, and As that will carry a factored bending moment Mu of 430 ft-​kips. Use fc′ = 5000 psi and  f y = 60, 000 psi. SOLUTION (a) Establish the limits within which the reinforcement ratio ρ can be chosen. From Eq. (3.6.2), the reinforcement ratio corresponding to the tension control limit is  

ρtc = 0.319

β1 fc′ fy

where β1 = 0.80 for fc′ = 5000 psi.  Thus  

ρtc = 0.319

(0.80)(5) = 0.0212 60

From Eq. (3.7.10) (ACI-​9.6.1.2), the minimum percentage area of tension reinforce­ment is  As ,min =  ρmin =

3 fc′ fy

bw d

3 fc′ fy

As 3 5000 is greater than 200 psi, ρmin = 0.00354 .

=

3 5000 = 0.00354 60, 000 (Continued)

63



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3 . 8   D E S I G N O F R E C TA N G U L A R S E C T I O N S

Example 3.8.1 (Continued) (b) Choose reinforcement ratio ρ, determine corresponding required beam size, and select actual beam size. Arbitrarily assume that ρ = 0.016 , which is within the prescribed limits of ρmin and ρtc . Also note that for the chosen value of ρ = 0.016 , ε t = 0.0075 (see table 3.6.1.). Thus,  m =

 

fy 0.85 fc′

=

60 = 14.12 0.85 ( 5)

 1   1  Rn = ρ f y  1 − ρ m = 0.016 (60, 000 ) 1 − ( 0.016 )(14.12 )  2   2  = 852 psi As the selected ρ ≤ ρtc , the section will be tension controlled and φ = 0.90.

 

 required M n =

Mu 430 = = 478 ft-kips φ 0.9

required bd 2 =

required M n 478 (12, 000 ) = = 6733 sq in. Rn 852

Try b = 16 in., d =  6733 / 16  = 20.5 in. The effective depth d is usually 2 3 8 to 2 5 8 in. less than overall beam depth h if one layer of steel can be used. Assuming that d is 2.5 in. less than the overall beam depth, h = 23 in. is obtained. However, even numbers are usually preferred for overall beam depths; thus, h = 24 in. is selected with d = 21.5 in. and b = 16 in. (c) Compute required As and select actual steel bars, using actual beam size. The effective depth exceeds that corresponding to ρ = 0.016 . Thus, the required Rn and required ρ must be revised.  

required Rn =

 ρ =

 =  

required M n 478 (12, 000 ) = = 775 psi 2 b d2 16 ( 21.5) 2 mRn  1 1 − 1 −  m fy  2 (14.12 )( 775)  1  1 − 1 −  = 0.0144 14.12  60, 000 

As = ρ bd = 0.0144 (16 )( 21.5) = 4.95 sq in.

Select 5–#9 bars, all in one layer, As = 5 (1.0 ) = 5.0 sq in., which is only slightly larger than the required As. Other bar combinations are possible, but they may not fit in one layer or give a total area much larger than that required. The practical selection of bars is discussed in Section 3.9. (d) Check design. A review of the correctness of the above computations in which formulas are used should be made using the basic statics shown in Fig. 3.8.2.    

 

T = As f y = 5 (60 ) = 300 kips a=



300 T or C = = 4.41 in. 0.85 fc′ b 0.85 ( 5)(16 )

a 4.41 1   M n = (T or C )  d −  = 300  21.5 −  = 482 ft-kips   2  12 2

(Continued)

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Example 3.8.1 (Continued) Since the chosen reinforcement ratio was less than ρtc , the section will be tension controlled (i.e., ε t ≥ 0.005). For illustration purposes, however, the strain εt will be calculated. From Fig. 3.8.2, the depth of the neutral axis is  

c=

a 4.41 = = 5.52 in. β1 0.80

and the strain in the tension steel is  

εs = εt =

d −c 21.5 − 5.52 (0.003) = (0.003) = 0.0087 > 0.005 c 5.52

as was anticipated. (e) Final decision and design sketch. Every design requires a clear decision as its conclusion, along with an appropriate design sketch. Use 5–#9 bars in the 16 × 24 in. (d = 21.5 in.) beam, as shown in Fig. 3.8.2. 16”

0.85f’c εcu = 0.003 c = 5.52” 4.41”

C = 300 K

19.29”

24” 21.5” 5 – #9 εs = εt = 0.0087

T = 300 K

Figure 3.8.2  Section for Example 3.8.1. 

3.9 PRACTICAL SELECTION FOR BEAM SIZES, BAR SIZES, AND BAR PLACEMENT In the preceding section, the procedure and example for the design of rectangular sections in bending with tension reinforcement only were based on the assumption that the factored moment Mu was known. This is rarely the case, however, because the factored moment must include the effect of the weight of the beam itself, which has not yet been designed. In reality, then, the dead weight of the beam must be assumed at the outset; a trial beam size is then obtained and may be readjusted if its effect on the factored moment significantly differs from the assumed value. For typical design situations, a preliminary estimate for the self-​weight of the beam could be obtained by assuming that the beam depth is 10 to 20% greater than the minimum depth for which deflections are assumed to be satisfactory (ACI-​ 9.3.1.1) and a beam depth-​to-​width ratio of 1.5. These minimum depths are L /​16 for simply supported spans, L /​18.5 for spans with one end continuous, L /​21 for spans with both ends continuous, and L /​8 for cantilever beams, where L is the span length. Calculation and control of deflections in flexural members is covered in Chapter 12. The choice of the steel reinforcement ratio ρ is very much dependent on the desired level of ductility and limitation on the deflection of the beam. The selection of high reinforcement ratios, while allowing the use of smaller beam sections, may lead to limited ductility and unacceptable deflections. Years of experience with the working stress method showed that deflection problems were rarely encountered with beams having a steel reinforcement ratio ρ not more than one-​half the maximum value permitted in the 2002 and earlier editions of the ACI Code (i.e., 0.75 ρb). For Grade 40, 60, and 80 steel, this corresponds

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3.9  PRACTICAL SELECTION FOR BEAM SIZES

approximately to 0.7ρtc , 0.6ρtc and 0.5ρtc , respectively, where ρtc is the reinforcement ratio at the tension control limit [see Eq. (3.6.2)]. In general, a tension reinforcement ratio for singly reinforced sections between 1.0 and 1.5% represents a good preliminary choice to ensure adequate ductility and reduce the likelihood of encountering deflection problems. Abendroth and Salmon [3.11] have presented a sensitivity study to show how the variables in simply supported beam design affect the economics of the design. TABLE 3.9.1  TOTAL AREAS FOR VARIOUS NUMBERS OF REINFORCING BARS

Bar Size #3 #4 #5 #6 #7 #8 #9 #10 #11 #14a #18a a

Nominal Diameter (in.) 0.375 0.500 0.625 0.750 0.875 1.000 1.128 1.270 1.410 1.693 2.257

Number of Bars Weight (lb/​ft) 0.376 0.668 1.043 1.502 2.044 2.670 3.400 4.303 5.313 7.65 13.60

1

2

3

4

5

6

7

8

9

10

0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56 2.25 4.00

0.22 0.40 0.62 0.88 1.20 1.58 2.00 2.54 3.12 4.50 8.00

0.33 0.60 0.93 1.32 1.80 2.37 3.00 3.81 4.68 6.75 12.00

0.44 0.80 1.24 1.76 2.40 3.16 4.00 5.08 6.24 9.00 16.00

0.55 1.00 1.55 2.20 3.00 3.95 5.00 6.35 7.80 11.25 20.00

0.66 1.20 1.86 2.64 3.60 4.74 6.00 7.62 9.36 13.50 24.00

0.77 1.40 2.17 3.08 4.20 5.53 7.00 8.89 10.92 15.75 28.00

0.88 1.60 2.48 3.52 4.80 6.32 8.00 10.16 12.48 18.00 32.00

0.99 1.80 2.79 3.96 5.40 7.11 9.00 11.43 14.04 20.25 36.00

1.10 2.00 3.10 4.40 6.00 7.90 10.00 12.70 15.60 22.50 40.00

#14 and #18 bars are used primarily as column reinforcement and are rarely used in beams.

TABLE 3.9.2  MINIMUM BEAM WIDTH (INCHES) ACCORDING TO THE ACI CODEa Number of Bars in Single Layer of Reinforcement Add for Each Added Bar

Size of Bars

2

3

4

5

6

7

8

#4 #5 #6 #7 #8 #9 #10 #11

6.8 6.9 7.0 7.2 7.3 7.6 7.8 8.1

8.3 8.5 8.8 9.0 9.3 9.8 10.4 10.9

9.8 10.2 10.5 10.9 11.3 12.2 12.9 13.8

11.3 11.8 12.3 12.8 13.3 14.3 15.5 16.6

12.8 13.4 14.0 14.7 15.3 16.6 18.0 19.4

14.3 15.0 15.8 16.5 17.3 18.8 20.5 22.2

15.8 16.7 17.5 18.4 19.3 21.1 23.1 25.0

1.50 1.63 1.75 1.88 2.00 2.26 2.54 2.82

#14 #18

9.1 10.8

12.5 15.3

15.9 19.8

19.3 24.3

22.6 28.8

26.0 33.3

29.4 37.9

3.40 4.51

Table shows minmium beam widths when #3 stirrups are used for bars #11 and smaller, and #4 stirrups are used for bars #14 and #18. For additional bars, add dimmension in last column for each added bar. For bars of different size, determine from table the beam width for smaller size bars and then add last column figure for each larger bar used. aAssumes maximum aggregate size does not exceed three-​fourths of the clear space between bars (ACI-​25.2.1). 1 A = 1 -​in. clear cover to stirrup 2 B = stirrup bar diameter

B A

C

D

C = For #11 and smaller bars, use twice the diameter of #3 stirrups (i.e., C = 0.75 in.). For #14 and #18 bars, use C = 0.5db D = clear distance between bars  = db,1 in. or 4 the maximum 3 aggregate size, whichever is greater (db = diameter of largest adjacent longitudinal bar)

Diameter of corner bar is assumed to be located to intersect the horizontal tangent to stirrup bend

6

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For the selection of an actual number of bars to meet a total steel area requirement, it is desirable to tabulate the combined area of several bars at a time. Table 3.9.1 gives bar areas for up to 10 bars of different sizes. Note that once the area of one bar, say of #8, has been set at 0.79 sq in., the area of 10 bars becomes 7.90, not 7.85; this practice has been carried on by tradition. For placement of bars in reinforced concrete members, the ACI Code specifies a minimum concrete cover for the reinforcement based on member and exposure types (ACI-​ 20.6.1.3). For cast-​in-​place beams not exposed to weather or in contact with ground, the minimum specified cover for the longitudinal reinforcement and stirrups is 1 1 2 in. Since beams are nearly always provided with stirrups surrounding the longitudinal reinforcement, the minimum clear cover to the stirrups will govern. In addition to minimum cover requirements, ACI-​25.2.1 specifies the clearance needed between bars in a single layer to permit proper concrete placement around them. This clearance for bars in a layer is 1 in., or the nominal diameter of the largest bar, or 4 3 the maximum aggregate size, whichever is greater. When two or more layers of bars are required, the minimum clearance between layers is 1 in. (ACI-​25.2.2). The bars in different layers must be located directly above one another and not staggered (ACI-​25.2.2). Table 3.9.2 gives minimum beam widths for various numbers of equal-​sized bars, computed in the manner described above, but assuming that aggregate size does not control minimum clear distance between bars. Developing the tensile force in the reinforcing steel requires the bars to be bonded with the surrounding concrete. Larger than minimum spacing and clear cover of bars will improve the bond between the bars and the concrete. Since this chapter has treated only the selection of cross section and bars thus far, suffice it to say that shorter bar lengths can be used when 2db clear lateral spacing (minimum width in Table 3.9.3) and 3db clear lateral spacing (minimum width in Table 3.9.4) can be provided. Bars are supported from the bottom of the forms, and layers of bars are separated by bar supports of various types; these are known as bolsters and chairs, some of which are shown in Table 3.9.5. Bar supports may be made of concrete, metal, or other approved materials. Usually they are standard factory-​made wire bar supports. These remain in place after the concrete is cast and must have special rust protection on the portions nearest the face of the concrete. TABLE 3.9.3  MINIMUM BEAM WIDTH (INCHES) TO SATISFY 2 BAR DIAMETERS CLEAR SPACING Number of Bars in Single Layer Bar Size

2

3

4

5

6

7

8

#4 #5 #6 #7 #8 #9 #10 #11

6.8 7.1 7.5 7.9 8.3 8.6 9.1 9.5

8.3 9.0 9.8 10.5 11.3 12.0 12.9 13.7

9.8 10.9 12.0 13.1 14.3 15.4 16.7 17.9

11.3 12.8 14.3 15.8 17.3 18.8 20.5 22.2

12.8 14.6 16.5 18.4 20.3 22.2 24.3 26.4

14.3 16.5 18.8 21.0 23.3 25.6 28.1 30.6

15.8 18.4 21.0 23.6 26.3 28.9 31.9 34.9

#14 #18

10.8 13.0

15.9 19.8

20.9 26.6

26.0 33.3

31.1 40.1

36.2 46.9

41.2 53.7

Table assumptions: a. Side cover 1.5 in. each side. b. #3 stirrups for bars #11 and smaller. c. #4 stirrups for bars #14 and #18. d. Since stirrups are bent around 4 stirrup bar diameters, the distance from centroid of bar nearest side face of beam to inside face of #3 stirrup is taken as 0.75 in. for bars #11 and smaller; and equal to the bar radius for #14 and #18 bars.

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3.9  PRACTICAL SELECTION FOR BEAM SIZES

TABLE 3.9.4  MINIMUM BEAM WIDTH (INCHES) TO SATISFY 3 BAR DIAMETERS CLEAR SPACING Number of Bars in Single Layer Bar Size

2

3

4

5

6

7

8

#4 #5 #6 #7 #8 #9 #10 #11

7.3 7.8 8.3 8.8 9.3 9.8 10.3 10.9

9.3 10.3 11.3 12.3 13.3 14.3 15.4 16.5

11.3 12.8 14.3 15.8 17.3 18.8 20.5 22.2

13.3 15.3 17.3 19.3 21.3 23.3 25.6 27.8

15.3 17.8 20.3 22.8 25.3 27.8 30.7 33.5

17.3 20.3 23.3 26.3 29.3 32.3 35.7 39.1

19.3 22.8 26.3 29.8 33.3 36.8 40.8 44.7

#14 #18

12.5 15.3

19.2 24.3

26.0 33.3

32.8 42.4

39.6 51.4

46.3 60.4

53.1 69.5

Table assumptions: a. Side cover 1.5 in. each side. b. #3 stirrups for bars #11 and smaller. c. #4 stirrups for bars #14 and #18. d. Since stirrups are bent around 4 stirrup bar diameters, the distance from centroid of bar nearest side face of beam to inside face of #3 stirrup is taken as 0.75 in. for bars #11 and smaller; and equal to the bar radius for #14 and #18 bars.

TABLE 3.9.5  STANDARD TYPES AND SIZES OF BAR SUPPORTS (ADAPTED FROM REF. 2.23) Symbol

Bar Support Illustration

Type of Support

Standard Sizes

Slab bolster

, 1,1 1 2, and 2 in. heights in 5-​ft and 10-​ft lengths

Slab bolster upper

Same as SB

Beam bolster

1, 1 1 2, 2, over 2 to 5 in. heights in increments of 1 4 in. in lengths of 5 ft

Beam bolster upper

Same as BB

BC

Individual bar chair

3

HC

Individual high chair

2 to 15 in. heights in increments of 1 4 in.

CHC

Continuous high chair

Same as HC in 5-​ft and 10-​ft lengths

Continuous high chair upper

Same as CHC

SB 5”

SBU

3

4

5”

BB 2 21 ”

BBU 2 21 ”

4

, 1, 1 1 2, and 1 3 4, in. heights

8”

CHCU 8”

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The following guidelines are suggested to assist the designer further in making choices for beam sizes, bar sizes, and bar placement. These suggestions may be regarded as accepted practice; they are not ACI Code requirements. Undoubtedly, situations will arise in which the experienced designer will, for good and proper reasons, make a selection not conforming to the guidelines. For Beam Sizes 1. Use whole inches for overall dimensions; slabs, however, may be in 1 2-​in. increments. 2. Beam stem widths are most often in multiples of 2 or 3 in., such as 9, 10, 12, 14, 15, 16, and 18 in. 3. Minimum specified clear cover is measured from outside the stirrup or tie to the face of the concrete. (Thus beam effective depth d has rarely, if ever, a dimension to the whole inch.) 4. An economical rectangular beam proportion is one in which the overall depth-​to-​ width ratio is between about 1.5 to 2.0. 5. For T-​shaped beams, the flange thickness typically represents about 20% of overall depth (see Chapter 4 for treatment of T-​shaped sections). For Reinforcing Bars 6. Maintain bar symmetry about the centroidal axis, which lies at right angles to the bending axis (i.e., symmetry about the vertical axis in usual situations). 7. Use at least two bars wherever flexural reinforcement is required. Use bars #11 and smaller for usual sized beams. The use of a larger number of bars of smaller diameter leads to improved bond and better crack control than can be achieved with fewer bars of larger diameter. 8. Use no more than two bar sizes and no more than two standard sizes apart for steel in one face at a given location in the span (e.g., #7 and #9 bars may be acceptable, but #9 and #4 bars would not). 9. Place all bars in one layer if practicable. Try to select bar size so that no less than two and no more than five or six bars are put in one layer. 10. Follow requirements of ACI-​25.2.1 and 25.2.2 for clear distance between bars and between layers, and for arrangement between layers. 11. When different sizes of bars are used in several layers at a location, place the largest bars in the layer nearest the face of the beam.

EXAMPLE 3.9.1 Select an economical rectangular beam size and select bars using the ACI Code requirements. The beam is a simply supported span of 40 ft; it is to carry a live load of 1.38 kips/​ft and a dead load of 1 kip/​ft (not including beam weight). Without actually checking deflection, use a reinforcement ratio ρ such that excessive deflection is unlikely. Use fc′ = 4000 psi, and fy = 60,000 psi. SOLUTION (a) Decide on a reinforcement ratio ρ to use. To have reasonable expectation that deflection will not be excessive, choose ρ at about 0.6ρtc = 0.0108 as recommended. Use ρ = 0.011. (b) Determine the desired Rn (corresponding to the desired ρ) using Eq. (3.8.4).  

m=

fy 0.85 fc′

=

60 = 17.65 0.85(4)

1 Rn = ρ f y (1 − ρm) 2

= 0.011(60,000)[1 –​0.5(0.011)(17.65)] = 596 psi (Continued)

69



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3.9  PRACTICAL SELECTION FOR BEAM SIZES

Example 3.9.1 (Continued) (c) Determine required Mn using ACI-​5.3.1 and a φ factor of 0.9. M u = 1.2 M D + 1.6 M L 1.38 ( 40 )



2

 

ML =

8

= 276 ft-kips

As the beam span is 40 ft long and simply supported, the minimum required depth in ACI-​9.3.1.1 to avoid the need for deflection check would be L /​16 = 40(12)/​16 = 30 in. For the purpose of estimating the beam weight, assume a depth of 32 in. and a width of 22 in. (approximately two-​thirds of beam depth). This leads to a beam weight estimated at 0.733 kip/​ft, based on a unit weight of reinforced concrete of 0.15 kips/​cu ft. MD =

(1.0 + 0.733)( 40)2 8

= 347 ft-kips

thus  

M u = 1.2 (347) + 1.6 ( 276 ) = 416 + 442 = 858 ft-kips  required M n =

Mu = 953 ft-kips φ

(d) Determine required bd2 from desired Rn. Use required Mn = 953 ft-​kips, required bd 2 =

 

M n 953 (12, 000 ) = = 19,190 cu in. Rn 596

(e) Establish beam size. Select width b and determine the corresponding required value for effective depth d. Make a table of possibilities. Chosen b

Required d

12 15 18 20

40.0 35.8 32.7 ← try 31.0

Selecting the 18-​in. width will give a beam whose overall depth is between 1 1 2 and 2 times its width (suggested guideline). Verify the assumed weight by determining overall depth. Assuming that the bars to be selected will fit in one layer, the minimum overall depth may be computed (see Fig. 3.9.1). Assume #8 bars for the longitudinal reinforcement and #3 bars for stirrups.  

 

h = d + 1 12 in. cover + 83 in. diameter stirrup + bar radius, say 12 in.



  = d + (2 83 to 2 12 in.)   = 32.7 + 2.5 = 35.2 in. The overall depth would be in whole inches; a depth of either 36 or 34 in. may thus be selected (even numbers are often preferred for overall beam depths). Since the guideline value of Rn = 596 psi for ρ = 0.011 is not a rigorous requirement, the overall depth selected could be somewhat less or somewhat more than the computed requirement and still obtain a desired dimension. In this case, the overall depth (h) of 34 in. is selected (Continued)

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Example 3.9.1 (Continued)

Small bars for holding stirrup in place d

h

Bar radius

Stirrup, usually

3„ 1 to „ diam. 8 2

1 min. 2 (ACl-20.6.1.3.1) Clear cover, 1

Figure 3.9.1  Quantities added to effective depth d to get overall depth h for beams having one layer of tension steel. 

which, as it is less than the required overall depth for the target reinforcement ratio, will lead to a slightly larger reinforcement ratio than originally assumed. The stirrup is reinforcement to provide shear strength for the beam and should always be allowed for at this stage of the design. The actual size and spacing of stirrups is determined after the cross section and tension bars have been selected, as explained in Chapter 5. Try h = 34 in. (f) Check weight, revise Mu, and select reinforcement.  w = revised M D  

18 (34 ) 144

(0.15) = 0.637 kip / ft

2 1.0 + 0.637)( 40 ) ( =

8

= 328 ft-kips

revised Mu = 1.2 (328) + 1.6 ( 276 ) = 835 ft-kips  revised required M n = 835 = 927 ft-kips 0.90

Compute the value of d from the overall dimension h, actual d = h – ( ≈ 2.5 in.) for one layer of bars = 34 – 2.5 = 31.5 in. When the overall depth is increased or decreased, the clear cover distance (ACI-​ 20.6.1.3.1) is the one that is held constant; thus the effective depth will shift.  

required Rn =

M n 927 (12, 000 ) = = 623 psi bd 2 18 (31.5)2

The steel requirement may be determined from Eq. (3.8.5) or from the curves of Fig. 3.8.1, or approximated by straight-​line proportion. Using the latter and knowing that   R = 596 psi for ρ = 0.011 n

find approximate ρ for Rn = 623 psi,  623  approximate ρ = 0.011  = 0.0115    596  [using Eq. (3.8.5) would have given about the same value] approximate As = ρ bd = 0.0115 (18)(31.5) = 6.52 sq in.

(Continued)

71



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3.9  PRACTICAL SELECTION FOR BEAM SIZES

Example 3.9.1 (Continued) Select 4–#10 and 2–#9 bars, As = 7.08 sq in. (Table 3.9.1). (Note that 3–#10 and 3–#9 having As = 6.81 sq in. provide a smaller As , but the bars cannot be placed in one layer and still have symmetry about the vertical beam axis, and 7–#9 will not fit in one layer.) Check whether 4–#10 and 2–#9 will fit into an 18-​in. width in one layer. 18 − 2(1.5) − 2(0.375) − 4(1.27) − 2(1.128) approx. clear spacing between bars =      5 = 1.38 in. > [1.27in. = diameter of # 10] OK Subtracted from the overall width are the combined values of the minimum clear cover on both sides (3.0 in.), one stirrup diameter on both sides (0.75 in.), 4–#10 bar diameters (5.08 in.), and 2–#9 bar diameters (2.26 in.). The result is divided by the number of spaces between bars, and this is the approximate clearance that must exceed the diameter of the larger bar (ACI-​25.2.1). Note that the above clearance computation is approximate because it assumes the #3 stirrup may be bent tightly around the corner longitudinal bar. ACI-​25.3.2 requires the inside diameter of bends for stirrups to be not less than four stirrup bar diameters for #5 stirrups and smaller; thus for #3 stirrups the actual curve of the stirrup at the corner has a radius of 3 4 in., which is larger than the longitudinal bar radius for #11 bars and smaller. Table  3.9.2 is based on the conservative assumption that the diameter of the corner longitudinal bar is located to intersect the horizontal tangent to the stirrup bend (see sketch accompanying Table 3.9.2). Using Table 3.9.2, the minimum required width for 4–#10 and 2–#9 is  

min b = 7.6 + 4 ( 2.54 ) = 17.76 in.



The 7.6 in. is the minimum width for 2–#9 and the 2.54 in. applies to each of the additional #10 bars (bar diameter plus a spacing of one bar diameter). (g) Check strength and provide design sketch. Using computed d = 31.5 in., C = 0.85 fc′ ba = 0.85(4)18a = 61.2 a T = As f y = 7.08(60) = 425 kips a=

 

425 = 6.95 in. 61.2



a 1  M n = As f y  d −  = 425(31.5 − 3.47) = 994 ft-kips  12 2 The net strain in the tension steel is

εs =  

d − a / β1 d −c (0.003) = (0.003) c a / β1 =

31.5 − 6.95 / 0.85 (0.003) = 0.0086 > 0.005 6.95 / 0.85

Thus, the section is tension controlled and the φ factor is 0.90, as assumed.  

     [φ M n = 0.90(994) = 895 ft-kips] > [ Mu = 835 ft-kips] OK

Use 18 × 34 beam with 4–#10 and 2–#9 bars, as shown in Fig. 3.9.2. (Continued)

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Example 3.9.1 (Continued) a 2 = 3.47”

0.85f’c

C = 61.2a

#3 stirrup (assumed)

a = 6.95” d = 31.5”

34” #10 #9

1 12 ”

T = 425k

18”

Figure 3.9.2  Design for Example 3.9.1. 

3.10 NOMINAL FLEXURAL STRENGTH M n OF RECTANGULAR SECTIONS HAVING BOTH TENSION AND COMPRESSION REINFORCEMENT Rectangular sections having both tension and compression reinforcement are also called “doubly reinforced” sections. Because the compressive strength of concrete is high, the need for compression reinforcement to obtain adequate strength is not great. In beams where compression reinforcement might be used to reduce the size of the cross section, deflections may be excessive and it may be difficult to place all the tension reinforcement within the width of the beam, even if two or more layers of bars are used. In addition, the shear stress will become high, possibly calling for a large amount of shear reinforcement. In fact, the usual reason for the use of compression reinforcement is for deflection control (to reduce the creep- and-shrinkage-induced deflection). Another reason to provide compression reinforcement is to increase the ductility of the beam when, for example, the tension steel yields but the strain εt is less than the minimum required by the ACI Code (ε t,min = 0.004). Adding compression steel will raise the neutral axis and decrease the depth of the compression stress block, resulting in an increase in the strain in the tension reinforcement. Similarly, compression reinforcement can be used to achieve a tension-​controlled section design when the required amount of tension steel in a singly reinforced beam would lead to a compression-​controlled section design. An area of compression steel equal to one-​quarter that of the tension steel can be considered a reasonable amount for the purpose of increasing ductility and controlling deflections. For members subjected to earthquake effects, where large inelastic deformations may be expected, the use of areas of compression steel greater than one-​half that of the tension steel is common. The nominal moment strength Mn of a doubly reinforced section such as that shown in Fig. 3.10.1 involves using variables b, d, d′, As, As′, fc′, and fy. The computation is similar to that for the singly reinforced beam except the compressive force C consists of two parts, one in the concrete and the other in the steel. The compression steel at nominal strength M n of the beam may or may not be at yield depending on the position of the neutral axis. The stress in the compression steel used in the nominal strength computation must be compatible with the strain diagram at crushing of concrete. In Fig. 3.10.1, the area of the concrete under compression in a beam with compression steel is given by  

Aconc = ab − As′

and the corresponding compression resultant equals 0.85 fc′Aconc. The point of action of this resultant requires, however, the calculation of the centroid of the shaded area in

73



ε cu = 0.003

Compression face d’

As’ h d

c

εs’

0.85f’c

0.85f’c d’ Cs = As’ (fs’ – 0.85fc’) a/2 Cc = 0.85fc’ ab

a

εs

b

(a) Cross section

(b) Strain diagram

Cs

Cc =

Aconc As

Tension face

73

3.10  NOMINAL FLEXURAL STRENGTH

(d – a–2)

T = As fy

Mnc

+

(d – d’)

(As – Asc)fy

Asc fy

(c) Stress diagram

(d) Moment carried as a singly reinforced beam, Mnc

Mns

(e) Moment carried by the compression steel, Mns

ρ’ =

As’ bd = Compression steel reinforcement ratio

f’s = Compression steel stress = Es ε’s ≤ fy

ρ=

As = Tension steel reinforcement ratio bd

Asc = Part of the tension steel to balance Cc

Figure 3.10.1  Strain and stress condition at nominal flexural strength for doubly reinforced beam. 

Fig. 3.10.1(a), which is often cumbersome. This calculation, however, can be avoided by treating the concrete compression zone as rectangular and adjusting the stress in the compression steel, as illustrated next. Equilibrium requires that  

C = T

or  0.85 fc′Aconc + As′ fs′ = As f y  0.85 fc′ ( ab − As′ ) + As′ fs′ = As f y  

0.85 fc′ ab + As′ ( fs′− 0.85 fc′ ) = As f y

or Cc + C s = T

  where

Cc = 0.85 fc′ ab

  and

Cs = As′ ( fs′− 0.85 fc′ )

 

The line of action of Cc is at a /2, while that of Cs is at d ′ from the extreme fiber in compression; both distances are known or easily determined. Nominal flexural strength is then calculated with respect to the centroid of the tension steel (Fig. 3.10.1)  

a  M n = M nc + M ns = Cc  d −  + Cs ( d − d ′ )  2

It is emphasized that the actual stress in the compression steel is fs′ and that the term ( fs′− 0.85 fc′) is fictitious but mathematically convenient. The components Cc and Cs are hereinafter computed as shown above [see Fig. 3.10.1(c)]. Note that the use of this fictitious stress is valid as long as a ≥ d ′. When the depth of the stress block a is sufficiently small that a < d ′ but c ≥ d ′, the compression steel would fall outside the compression stress block and Cs would equal As′ fs′ . In other cases, however, c < d ′ (i.e., “compression” steel is in tension). In such situations, the area of steel As′ can be ignored and the section treated as a singly reinforced section.

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The stress in the compression steel fs′, and thus the resultant force Cs , will depend on the depth of the neutral axis, which is not known. If both the tension and compression steel have the same yield strength, the compression steel will yield if the following inequality is satisfied f y ( As − As′ )

 

0.85 fc′ bβ1

≥ γ d′



where γ is equal to 1.85 and 3.22 for Grade 40 and Grade 60 steel, respectively. Equilibrium requires Cc + C s = T

 

For the case of compression steel that has yielded,

(

)

0.85 fc′ ab + As′ f y − 0.85 fc′ = As f y

 

(

 

)

As f y − As′ f y − 0.85 fc′

a=

0.85 fc′ b



On the other hand, when the strain in the compression steel is less than the yield strain (i.e., ε s′ < ε y ), Cc + C s = T

  or

0.85 fc′ bβ1c + As′ [ fs′− 0.85 fc′] = As f y

  where

fs′ = Es ε s′ < f y

  and

 c − d′ ε s′ = 0.003   c 

 

Replacing ε s′ and fs′, and rearranging terms leads to the following quadratic equation for the neutral axis depth, c,

0.85 fc′ bβ1c 2 +  As′ ( 0.003Es − 0.85 fc′ ) − As f y  c − 0.003Es As′d ′ = 0

(3.10.1)

Once c is known, the following procedure can be used to calculate the nominal flexural strength.  a = β1c  c − d′   fs′ = Es ( 0.003)  c 

 or  

 Cs = As′ ( fs′− 0.85 fc′)

for a ≥ d ′

Cs = As′ fs′

for a < d ′ and  c ≥ d ′

a  M n = Cc  d −  + C s ( d − d ′ )  2

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3.10  NOMINAL FLEXURAL STRENGTH

EXAMPLE 3.10.1 Determine the nominal moment strength Mn of the rectangular section shown in Fig.  3.10.2, given fc′  =  5000 psi, fy  =  60,000 psi, b  =  14 in., d  =  26 in., d′  =  3 in., As′  = 2–#8 and As = 8–#10 bars. Use 1-​in. clear spacing between layers. 3”

a = 8.77”

εcu = 0.003

A’s = 2 – #8

26”

εs2 = 0.0038

A’s = 8 – #10

0.85f’c = 4.25 ksi Cs = 88k Cc = 522k

εs‘ = 0.00218

c = 10.96”

14”

1”

T = 610k

εs1 = 0.0044

Figure 3.10.2  Strain and stress condition, and internal force resultants for section of Example 3.10.1. 

SOLUTION (a) Determine the neutral axis distance c at nominal strength. From Fig. 3.10.2, use the requirement of equilibrium (i.e., C = T). Compression steel will yield if f y ( As − As′ )

 

0.85 fc′ bβ1

≥ γ d ′

where γ is equal to 3.22 for Grade 60 steel. Thus  

60 (10.16 − 1.58)

0.85 ( 5)(14 )( 0.80 )

= 10.82 ≥ γ d ′ = 3.22 (3) = 9.66

Therefore, compression steel yields. T = f y As = 60(10.16) = 610 kips Cc = 0.85 fc′ ba = 0.85(5)(14)a = 59.5a Cs = ( fs′− 0.85 fc′ ) As′ = (60 − 4.25)1.58 = 88 kips  

C=T a=

T − Cs 610 − 88 = 8.77 in. = 59.5 59.5

c=

a 8.77 = = 10.96 in. β1 0.80



(Continued)

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Example 3.10.1 (Continued) (b) Determine whether the amount of tension steel complies with the minimum net tension steel strain per ACI-​9.3.3.1. From the strain diagram, the strain in the bottom layer of reinforcement is 1 1.27   − 10.96  26 + +  2 2 ε s1 = 0.003 = 0.0044 > 0.004    OK 10.96

 

Similarly, the strain in the second layer of reinforcement is 1 1.27   − 10.96  26 − −  2 2 ε s2 = 0.003 = 0.0038 > ε y     OK 10.96

 

Thus, both layers of steel have yielded. (c) Compute Mn  

M n = Cc ( d − a / 2 ) + C s ( d − d ′ ) = 522[26 − 8.77 / 2]

1 12

+ 88(26 − 3)

1 12

= 940 + 169 = 1109 ft-kips

Noted that while ε t = 0.0044 > 0.004 , the section is in the transition region. In this case, it is advisable to increase the area of compression steel so that the section is tension controlled (i.e., ε t ≥ 0.005).

EXAMPLE 3.10.2 Repeat the solution for Example 3.10.1, except that As = 4–#11 bars instead of As = 8–#10 bars. Also, determine the percentage increase in the strength (and the net tensile strain) of the beam over the same beam without compression steel. SOLUTION (a) Check if the compression steel yields at nominal strength. Compression steel will yield if f y ( As − As′ )

 

0.85 fc′ bβ1

≥ γ d ′

where γ is equal to 3.22 for Grade 60 steel. Thus  

60 (6.24 − 1.58)

0.85 ( 5)(14 )( 0.80 )

= 5.87 < γ d ′ = 3.22 (3) = 9.66

Compression steel does not yield. (Continued)

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3.10  NOMINAL FLEXURAL STRENGTH

Example 3.10.2 (Continued) 14”

3”

εcu = 0.003

Cs = 66.9k

a = 5.17”

c = 6.46”

A’s = 2 – #8

0.85f’c = 4.25 ksi

εs’ =

Cc = 308k

εcu (c – 3) c

26”

As = 4 – #11

T = 374k εs

Figure 3.10.3  Strain and stress condition, and force resultants for section of Example 3.10.2. 

(b) Determine actual c location. Because compression steel does not  yield, solve Eq. (3.10.1). 4.25 (14 ) 0.80c 2 + 1.58 (0.003 ( 29, 000 ) − 4.25) − 6.24 (60 ) c

− 0.003 ( 29, 000 )(1.58) 3 = 0



c = 6.46 in. a = β1c = 0.80 (6.46 ) = 5.17 in.

Then   Cc = 4.25 (14 )( 5.17) = 308 kips  6.46 − 3.00   ε s′ =  med)  0.003 = 0.00161 < ε y (as previously confirm  6.46   Cs = 29, 000 ( 0.00161) − 4.25 (1.58) = 66.9 kips     Cc + Cs = 308 + 66.9 = 375 kips   T = As f y = 6.24 (60 ) = 374 kips M n = 308 26 – 0.5 ( 5.17)

 

1 12

+ 66.9

M n = 600 + 128 = 728 ft-kiips

(26 – 3) 12

(Check)

1



Compute the net tensile strain and check ACI-​9.3.3.1,  

εt = εs =

d −c 26 − 6.46 (0.003) = (0.003) = 0.0091 > 0.004       OK c 6.46 (Continued)

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Example 3.10.2 (Continued) (c) Determine Mn if the compression steel had not existed.   C = 0.85 fc′ba = 4.25 (14 ) a   T = 6.24 (60) = 374 kips  a =

374 = 6.29 in. 4.25 (14 )

1  M n = 374 26 – 0.5 (6.29) = 713 ft-kips 12

Compute the net tensile strain and check ACI-​9.3.3.1, 6.29 = 7.86 . 0.80           26 − 7.86 OK εt = εs = (0.003) = 0.0069 > 0.004 7.86 c=

Thus, the addition of the compression steel to the singly reinforced beam increased the nominal flexural strength only 2.1%. The net strain in the tension steel, however, increased 31% when compression steel with an area about 25% that of the tension steel was added, which shows the effectiveness of compression steel to increase the ductility of the beam at nominal strength.

3.11 DESIGN OF BEAMS HAVING BOTH TENSION AND COMPRESSION REINFORCEMENT As discussed above, the main reason for using compression reinforcement in a beam, in addition to reducing long-​term deflections caused by creep and shrinkage, is to increase ductility. For singly reinforced sections in which the amount of tension reinforcement required is such that the strain in the layer of extreme tension reinforcement is less than permitted (ε t < 0.004 ) or desired, the use of compression steel may provide an acceptable solution without the need for increasing section dimensions. It should be noted that the addition of compression steel will lead to only a slight increase in flexural strength (see Example 3.10.2 and Section 3.13). However, it allows an increase in area of tension steel beyond that permitted for a singly reinforced beam, with the associated increase in flexural strength. Having decided that compression steel is to be used, be it required for ductility or desirable for deflection control, the designer now needs to select the appropriate tension steel As and compression steel As′. For cases in which the flexural strength of a singly reinforced section with the maximum permitted amount of steel is less than the required strength, the minimum area of compression steel can be calculated based on the additional area of tension steel needed to achieve the required flexural strength. It is thus useful to calculate the flexural strength of the section as the summation of the strength of a singly reinforced section with an area of tension reinforcement Asc [see Fig.  3.10.1(d)] and the moment

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3.11  DESIGN OF BEAMS

corresponding to a couple consisting of the compression force Cs and a tension force of equal magnitude [Fig. 3.10.1(e)] as follows: M n = M nc + M ns

  or  

a a   M n = Cc  d −  + Cs ( d − d ′ ) = Asc f y  d −  + ( As − Asc ) f y ( d − d ′ )   2 2

The required area of tension steel As and the minimum area of compression steel As,min ′ needed for a tension-​controlled section can be calculated as follows Asc =

 

Cc 0.85 fc′ ab = fy fy

The neutral axis depth c for the tensile strain that defines the limit of tension-​controlled sections (i.e., ε t = 0.005) is (see Fig. 3.6.1)  0.003  c= d  0.008  t

  Thus

 0.003  a = β1  d  0.008  t

  and  

M a  M n = Asc f y  d −  + ( As − Asc ) f y ( d − d ′ ) ≥ u  2 φ

where φ = 0.90 because the section is designed to be tension controlled. Solving for As

 

Mu a  − Asc f y  d −   φ 2 As = + Asc fy (d − d ′ )

The minimum area of compression steel required such that the section is tension controlled is then As′,min =

 

( As − Asc ) fy fs′



where  

fs′ = 0.003

(c − d ′ ) E c

s

= 0.003

( a / β1 − d ′ ) E a / β1

s

≤ fy



Typically, the minimum amount of compression steel required for the section to be tension controlled should not be greater than about one quarter of As . This is because larger areas would indicate that the section dimensions are too small, which would lead to reinforcement congestion, excessive deflections, or both.

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It should be noted that the required area of tension and compression steel can be calculated for any target strain in the extreme layer of tension steel εt by following the same procedure outlined above, the only difference being in the calculated value of the stress block depth a, which will decrease as the target tensile strain εt is increased (see Example 3.11.2). The procedure for designing a doubly reinforced section is illustrated in Examples 3.11.1 and 3.11.2.

EXAMPLE 3.11.1 Determine the steel areas As and As′ required to carry a service live load moment of 414 ft-​kips and a service dead load moment of 234 ft-​kips. It is desired that the section be tension controlled. Use the ACI code and b = 14 in., d = 26 in., d′ = 3 in., fc′ = 5000 psi, and f y = 60, 000 psi, as shown in Fig. 3.11.1(a). εcu = 0.003

d = 26”

c = 9.75”

d’ = 3”

4.25 ksi

a = 7.80”

14”

C

T εs = εtc = 0.005 (b) Singly reinforced section with εs = εtc = 0.005

(a) Beam size

εcu = 0.003 Cs Cc

εs’

4.25 ksi

a = 4.80”

4.25 ksi c = 6.00”

εs’

a = 7.80”

c = 9.75”

εcu = 0.003

Cc Cs

T

T

εs = εtc = 0.01

εs = εtc = 0.005 (c) Doubly reinforced section with εs = εtc = 0.005

(d) Doubly reinforced section with εs = εtc = 0.01

Figure 3.11.1  Section, strain and stress condition, and force resultants for Examples 3.11.1 and 3.11.2. 

SOLUTION (a) Determine the required nominal strength, using ACI-​5.3 for calculating Mu and 21.2.2 for φ factors.  

M u = 1.2 ( 234 ) + 1.6 ( 414 ) = 281 + 662 = 943 ft-kips



Assuming a φ factor of 0.90 (i.e., tension-​controlled section),  

required M n =

Mu 943 = = 1048 ft-kips φ 0.90

(Continued)

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3.11  DESIGN OF BEAMS

Example 3.11.1 (Continued) (b) Determine the nominal flexural strength and maximum reinforcement allowed for a tension-​controlled singly reinforced section. The location of the neutral axis for this condition may be obtained as in Fig. 3.11.1(b);  c =

0.003 0.003 dt = 26 = 9.75in. 0.008 0.008

  a = β1C = 0.80 ( 9.75) = 7.80 in.  

C = 0.85 fc′ ab = 0.85 ( 5)( 7.80 )(14 ) = 464 kips

Thus, the maximum As in a tension-​controlled, singly reinforced section is 464/​60  = 7.74 sq in. and the corresponding nominal moment is 7.80  1  M n = 464  26 −  = 855 ft-kips  2  12

 

The required Mn exceeds the maximum strength obtainable without compression steel for a tension-​controlled section. In this case, therefore, compression steel is needed to increase the amount of tension steel enough to achieve the required strength while maintaining εt ≥ 0.005. (c) Determine the minimum compression reinforcement required. Maintain c at εt = 0.005, which is 9.75 in. [see Fig. 3.11.1(c)]. Let    

M nc = 855ft-kips

[from part(b)]

M ns = M n − M nc = 1048 − 855 = 193ft-kips  required Cs =

193 (12 ) 26 − 3



= 101 kips

Will compression steel yield when c = 9.75 in.?  ε s′ = 0.003

(c − d ′ ) = 0.003 (9.75 − 3) = 0.0021 > ε 9.75

c

y

Thus, compression steel will yield.

(

)

 Cs = As′ f y − 0.85 fc′ = 101 kips  required As′ =

101 = 1.81 sq in. 60 − 4.25

  T = Cc + Cs = 464 + 101 = 565 kips  required As =

565 = 9.42 sq in. 60

This amount of tension steel corresponds exactly to the tension-​controlled limit for the beam with the compression reinforcement area calculated. In this example the remaining steps of selecting actual bars and making a final check of strength are not shown.

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EXAMPLE 3.11.2 Redesign the section of Example 3.11.1 so that strain in the extreme layer of tension reinforcement ε t is approximately 0.01. From Example 3.11.1, Mu = 943 ft-kips. SOLUTION (a) Determine M nc corresponding to locating the neutral axis such that ε t = 0.01. Start by referring to Fig. 3.11.1(d); the location of the neutral axis at this strain condition may be obtained as  c =

ε cu 0.003 26 = 6.00 in. dt = ε cu + 0.01 0.003 + 0.01

Once the neutral axis location is known, the depth of the stress block a, internal force Cc, and Mnc can be determined as follows   a = β1c = 0.80 (6.00 ) = 4.80 in.  

C = 0.85 fc′ ab = 0.85 ( 5)( 4.80 )(14 ) = 286 kips   C = T1 = 286 kips  M nc = 286  26 − 4.80  1 = 562 ft-kips  2  12

Let Asc be the part of tension steel to match the concrete compressive force; then   Asc = T1 = 286 = 4.76 sq in. fy 60 (b) Determine steel requirements for both faces of the beam. Because the section is tension-​controlled, φ  =  0.90 and the required nominal moment strength M n is 943 / 0.90 = 1048 ft-kips. Thus,  

M ns = 1048 − 562 = 486 ft-kips  Cs = T2 =

 

486 (12 )

ε s′ = 0.003

26 − 3

= 254 kips

(c − d ′ ) = 0.003 (6.00 − 3.0) = 0.0015 < ε c

6.00

y



Compression steel does not yield. If c = 6.00 in. is still to be maintained, then fs′ = 0.0015(29,000 ksi) = 43.5 ksi so that the stress used is consistent with the strain on the compression steel.  

As′ =

T2 254 = = 6.46 sq in. fs′− 0.85 fc′ 43.5 − 4.25

  Ass = As − Asc = T2 = 254 = 4.23 sq in. fy 60  

As = Asc + Ass = 4.76 + 4.23 = 8.99 sq in. (Continued)

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3.11  DESIGN OF BEAMS

Example 3.11.2 (Continued) Select 4–#10 and 4–#9 in two layers for tension steel (As = 9.08 sq in.), and 4–#11 as compression reinforcement ( As′ = 6.24 sq in.). The 4–#11 is less than the amount required, but it is the maximum steel that will fit into one layer. (c) Check the strength of the section. The above solution of As = 8.99 and As′ = 6.46 sq in. is the correct one for ε t = 0.01. When As′ = 6.24 sq in. is used instead, both c and As will have to change. A trial-​and-​error procedure is presented below to determine the stress fs′ acting in the compression steel, an alternative to that used in Example 3.10.2, where a quadratic equation was solved for the neutral axis location c. Since a smaller amount of compression steel is used than that required for ε t = 0.01, the actual neutral axis will be lower if the section is to carry the same bending moment. Also, the stress fs′ must exceed 43.5 ksi, as it was calculated for c = 6.00 in. (c for ε t = 0.01). Estimate the compression steel stress to be 45 ksi.   Cc = 0.85 fc′ ba = 0.85 ( 5)(14 ) a = 59.5a   Cc = As′ ( fs′ − 0.85 fc′ ) = 6.24 ( 45 − 4.25) = 254 kips   T = As f y = 9.08 (60 ) = 545 kips   C c + Cs = T  a =

545 − 254 4.89 = 4.89 in.; c = = 6.11 in. 59.5 0.80

6.11 − 3.00 (0.003) = 0.00153 6.11   fs′ = 0.00153(29, 000) = 44.3 ksi  ε s′ =

Since this does not agree with the 45 ksi assumed, make a new assumption, say,    fs′ = 44.5 ksi : revised Cs = 6.24(44.5 –​ 4.25) = 251 kips  a = 545 − 251 = 4.94 in. 59.5  c = 4.94 = 6.18 in. 0.80  

ε s′ = 0.003

(6.18 − 3.0) = 0.00154 6.18

fs′ = 0.00154(29,000) = 44.8 ksi Repeated trials may be made until computed fs′ agrees as closely as desired with the assumed value. In this case, assume that the present agreement is close enough.    

Cc = 59.5a = 59.5 ( 4.94 ) = 294 kips Cs = 251 kips M n = 294[26 − 0.5(4.94)]

1 12

+ 251(26 − 3)

1 12

= 576 + 481 = 1057 ft-kips

φ Mn = 0.90(1057) = 951 ft-​kips ≥ Mu = 943 ft-​ kips          OK (Continued)

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Example 3.11.2 (Continued) It may be observed that when there is a large amount of compression steel that does not yield, varying the amount of compression steel has essentially the effect of changing only the proportions in M nc and M ns. The choice of 4–#10 and 4–#9 (tension steel) and 4–#11 (compression steel) is acceptable.

3.12 NONRECTANGULAR SECTIONS When nonrectangular beams are used, the shape of the compression zone dictates whether the formulas developed in this chapter are applicable. (T-​sections are treated in Chapter 4.). The concept of using the Whitney rectangular compressive stress distribution, however, may be used in accordance with ACI-​22.2.2.4.1 for any shape of the compression zone in the cross section. Determination of stress block depth a and neutral axis depth c is performed as for rectangular sections by using equilibrium of forces (i.e., C = T ). Since the shape of the compression zone is no longer rectangular, the resultant concrete compressive force, which is needed for calculation of moment strength, will not be located at a/2 from the extreme compressive fiber. Since ε t or c /dt is used to determine the appropriate φ factor to use, the design procedures presented earlier are identical for all shapes and all kinds of reinforcements, including compression steel.

EXAMPLE 3.12.1 For the trapezoidal beam section shown in Fig. 3.12.1, determine the required tension reinforcement to carry a factored bending moment Mu of 155 ft-​kips. Use fc′ = 4500 psi and f y = 60, 000 psi. 0.85f’c

c = 5.16”

d = 17.5”

20”

8”

εcu = 0.003

a = 4.25”

C = 144 K

15.31” 4 – #7 εs = εt = 0.0072

T = 144 K

16”

Figure 3.12.1  Section, strain and stress condition, and force resultants for Example 3.12.1. 

SOLUTION (a) Determine the required nominal moment strength assuming the section is tension controlled.  

required M n =

Mu 155 = = 172 ft-kips φ 0.90 (Continued)

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3 . 1 2   N O N R E C TA N G U L A R S E C T I O N S

Example 3.12.1 (Continued) (b) Determine the area of tension steel. As the compression zone is not rectangular, it is more convenient to first estimate the area of tension steel based on an estimate of the moment arm. Given that the section is narrowest at the level of the extreme compressive fiber, a moment arm of 0.80d should be a conservative assumption.  

M n = T (arm ) = As f y ( 0.80 d ) = As (60 )( 0.80 )(17.5)

  As ≥

172 (12 )

(60)(0.80)(17.5)

1 12

≥ 172 ft-kips

= 2.46 sq in.

Calculate depth of stress block a using equilibrium of forces. The area on which the compression stress block acts is taken as the summation of a rectangular area equal to 8a, and the summation of the area of two triangles, equal to a2/​5.  

T = As f y = 2.46 (60 ) = 148 kips

 

  a2  a2  C = T = 148 kips = 0.85 fc′ 8a +  = 0.85 ( 4.5) 8a +  5 5  

Solving for a gives  

a = 4.35 in.

for which the resultant compressive force is located at 2.25 in. from the extreme compressive fiber. The corresponding moment arm is then  

arm = d − 2.25 in. = 15.25 in. = 0.87d



As the arm is nearly 10% greater than the originally assumed value of 0.80d, recalculate the required area of steel.  

As ≥

172 (12 )

(60)(15.25)

= 2.26 sq in.

Select 4–#7 bars (As = 2.4 sq in.) and calculate nominal flexural strength. From Table 3.9.2, a minimum width of 10.9 in. is required to satisfy bar spacing requirements, and this is less than the section width at the level of the tension reinforcement (15 in.).  

T = As f y = 2.4 (60 ) = 144 kips

 

  a2  a2  C = T = 144 kips = 0.85 fc′ 8a +  = 0.85 ( 4.5) 8a +  5 5  



Solving for a gives  

a = 4.25 in. (Continued)

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Example 3.12.1 (Continued) For this value of a, the resultant compressive force is located at 2.20 in. from the extreme compressive fiber, which leads to a moment arm of 15.31 in. Thus M n = T (arm ) = 144 (15.31)

 

1 12

= 184 ft-kips > 172 ft-kips

(c)  Verify that the section is tension controlled. For fc′ = 4500 psi, β1 = 0.825. Thus  

c=

4.25 a = = 5.16 in. β1 0.825

and  

ε t = ε s = 0.003

( d − c ) = 0.003 (17.5 − 5.16) = 0.0072 > 0.005 5.16

c

Section is tension controlled. (d) Determine minimum area of tension reinforcement. Given that the width of the web varies and is maximum in the extreme tension fiber, the use of bw = 16 in. is recommended. From Eq. 3.7.10 (ACI-​9.6.1.2), the minimum percentage area of tension reinforcement is   As ,min =

3 fc′ fy

bw d =

3 4500 (16)(17.5) = 0.94 sq in. 60, 000

 As 3 4500 is greater than 200 psi, As,min = 0.94 sq in. < 2.4 sq in. (e) Final decision and design sketch. Every design requires a clear decision as its conclusion, along with an appropriate design sketch. Use 4–#7 bars with d = 17.5 in. as shown in Fig. 3.12.1.

3.13 EFFECT OF A S , As′ , b, d, fc′ AND f y ON FLEXURAL BEHAVIOR The flexural behavior of reinforced concrete sections, and the effect of variables such as percentage of tension and compression steel, concrete compressive strength, and yield strength of reinforcement can be better understood through the use of a moment-​curvature analysis. In such analysis, a measure of section deformation is selected, typically strain in the extreme compressive fiber, εc, and the corresponding moment and curvature (ϕ = εc /c) are calculated. This process is repeated for concrete compressive strains ranging from those corresponding to elastic behavior up to the concrete crushing strain. In a moment-​curvature analysis, it is assumed that plane sections prior to loading remain plane after loading (i.e., linear strain distribution over section depth) and that the steel reinforcement is perfectly bonded to the concrete. The relationship between compressive strain and stress in the compression zone of a beam may be expected to have the same general shape as that obtained from a test cylinder. A  parabolic shape for the ascending portion of the stress-​strain response is typically assumed, with a linear post-​peak descending branch up to the concrete compressive strain capacity ε cu. The peak concrete compressive stress is then assumed equal to k3 fc′ (see Fig. 3.2.2). For beams, however, k3 is typically taken as 1.0. The stress-​strain behavior for the steel reinforcement is assumed to be the same as that obtained from a standard tensile test. Thus, increases in stress beyond the yield strength due to strain hardening of the steel are often considered.

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3 . 1 3   E F F E C T O F F L E X U R A L B E H AV I O R

To illustrate the effect of design variables on flexural behavior, Fig. 3.13.1 plots results from a series of moment-​curvature analyses. To facilitate comparison, the moments and curvatures were normalized by the maximum moment and maximum curvature of the singly reinforced rectangular section shown in the figure (Control). Also, the stress-​strain response of the steel was assumed to be bilinear, with a constant stress f y beyond the yield strain ε y (i.e., no strain hardening). Failure was assumed when the strain in the extreme compressive fiber reached 0.004. The moment-​curvature responses shown in Fig. 3.13.1 are characterized by three main regions: uncracked, linear-​elastic behavior; cracked, elastic behavior; and flexural yielding. The maximum curvature for all the sections shown in Fig. 3.13.1 was governed in each case by the assumed compressive strain capacity of concrete of 0.004. The singly reinforced control section, with a reinforcement ratio of approximately 1%, exhibited an ultimate curvature approximately 6 times that at first yield of the longitudinal steel. Depending on the changes made to the section dimensions and reinforcement, changes in strength and/​or curvature capacity occurred with respect to that of the control section. A 50% increase in beam width led to only a little increase in flexural strength. This can be explained by noting that while the depth of the compressive zone decreased, the change in distance between the resultant tensile and compressive forces was small. Deformation capacity, however, increased because for the same maximum strain in the extreme compressive fiber, the reduction in neutral axis depth led to larger strains in the tension steel, and thus to a higher curvature. The same situation occurs with the addition of compression steel ( As′ = 0.25 As ) or with an increase in concrete compressive strength [  fc′ = 2( fc′ ) control].   A 25% increase in yield strength of steel led to a proportional increase in flexural strength. However, a larger neutral axis depth is needed to balance the larger tensile force in the reinforcement, leading to a decrease in curvature for a given maximum compressive strain, and thus to a decrease in ductility. Increasing the section effective depth by 50% also led to a similar increase in flexural strength. No change in curvature capacity occurred, however, because the depth of the neutral axis depth remained unchanged since the tensile force remained the same (no strain hardening considered). The strain in the tension steel, however, increased in comparison to the control section, as the distance from the tension steel to the neutral axis increased.

1.6

d = 1.5dcontrol

14”

1.4

f’c = 2(f’c)control

fy = 1.25(fy)control

1.2 M/(Mmax)control

d = 21.5”

1 b = 1.5bcontrol

0.8

Control

4#8 2.5”

As’ = 0.25As

0.6

fc’ fy As ρ

0.4 0.2 0

= 4000 psi = 60,000 psi = 3.16 in2 = 0.0105

Mmax = 310 ft-kips ϕmax = 9.9 × 10–4 rad/in

0

0.5

1

ϕ/(ϕmax)control

1.5

Figure 3.13.1  Effect of As, As′, b, d, fc′, and fy on moment-​curvature response. 

2

8

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The responses shown in Fig. 3.13.1 can be summarized as follows: 1. Flexural capacity is rather insensitive to changes in section width, concrete compressive strength, and addition of compression steel. 2. Maximum curvature increases with an increase in beam width, concrete compressive strength, and the addition of compression steel. 3. An increase in yield strength of steel leads to an increase in strength but a reduction in curvature capacity. 4. An increase in effective depth leads to an increase in strength but no or little change in maximum curvature.

SELECTED REFERENCES   3.1. Eivind Hognestad. A Study of Combined Bending and Axial Load in Reinforced Concrete Members. Bulletin Series No. 399. Engineering Experiment Station, University of Illinois, November 1951.   3.2. Eivind Hognestad. “Fundamental Concepts in Ultimate Load Design of Reinforced Concrete Members,” ACI Journal, Proceedings, 48, June 1952, 809–​828.  3.3. Eivind Hognestad, N.  W. Hanson, and Douglas McHenry. “Concrete Stress Distribution in Ultimate Strength Design,” ACI Journal, Proceedings, 52, December 1955, 455–​479.  3.4. Jack R.  Janney, Eivind Hognestad, and Douglas McHenry. “Ultimate Flexural Strength of Prestressed and Conventionally Reinforced Concrete Beams,” ACI Journal, Proceedings, 52, February 1956, 601–​620.   3.5. Eivind Hognestad. “Confirmation of Inelastic Stress Distribution in Concrete,” Journal of the Structural Division, ASCE, 83, Paper No. 1189, ST2, March 1957.  3.6. Alan H.  Mattock, Ladislav B.  Kriz, and Eivind Hognestad. “Rectangular Concrete Stress Distribution in Ultimate Strength Design,” ACI Journal, Proceedings, 57, February 1961, 875–​928.   3.7. Charles S.  Whitney. “Design of Reinforced Concrete Members under Flexure or Combined Flexure and Direct Compression,” ACI Journal, Proceedings, 33, 1937, 483–​498.   3.8. Charles S. Whitney. “Plastic Theory of Reinforced Concrete Design,” Transactions ASCE, 107, 1942, 251–​326.   3.9. C. S.Whitney and Edward Cohen. “Guide for Ultimate Strength Design of Reinforced Concrete,” ACI Journal, Proceedings, 53, November 1956, 455–​475. 3.10. ACI-​ASCE Joint Committee. “Report on ASCE–​ACI Joint Committee on Ultimate-​Strength Design,” ASCE, Proceedings, 81, Paper No. 809. October 1955. See also ACI Journal, Proceedings, 52, January 1956, 505–​524. 3.11. Robert E. Abendroth and Charles G. Salmon. “Sensitivity Study of Reinforced Concrete Simple Beams,” ACI Journal, Proceedings, 83, September–​October 1986, 764–​771.

PROBLEMS

All problems* (except Problems 3.1 and 3.2), are to be done in accordance with the ACI Code; all loads given are service loads, unless otherwise indicated. Wherever possible, basic principles are to be used for solutions, avoiding the direct use of formulas. All design problems require a clear statement of the final choice at the end of the calculations, along with a design sketch drawn to scale.

Review Problems in Mechanics of Materials 3.1 An elastic homogeneous beam of material capable of carrying both tension and compression, has the dimensions shown in the figure for Problem 3.1; it is simply supported over a span of 20 ft (6 m) center-​to-​center

of supports. The beam carries a uniformly distributed load of 2 klf (kips per linear ft) (30 kN/​m) in addition to a concentrated load of 10 kips (45 kN) located at 5 ft (1.5 m) from the right end of the span.

*  Most problems may be solved as problems stated in Inch-​Pound units, or as problems in SI units using quantities in parentheses. The metric conversions are approximate to avoid implying higher precision for the given information in metric units than that for the Inch-​Pound units.

89



Case

h

b1

89

PROBLEMS

t1

tw

b2

Reinforced Concrete Problems

t2

(Dimensions in Inches) 1 2 3 4 5 6 7 8 9

30 20 20 20 24 28 30 30 20

12 40 40 60 50 40 30 10 10

2 3 4 4 4 6 4 3 4

1 4 4 12 14 14 12 2 20

12 40 20 12 14 14 12 10 20

2 3 4 0 0 0 0 3 0

(Dimensions in millimeters) 10 11

750 500

300 1500

50 100

25 300

300 300

50 0

b1

t1 h

tw

3.3 For the case (or cases) assigned, do the following. (a)  Compute the nominal moment strength Mn of the cross section shown in the figure for Problem 3.3 using basic statics (i.e., no formula) with the Whitney rectangular stress block and the internal couple. Metric bar dimensions for cases 6 and 7 are in Table 1.13.2. Case Beam- Beam Bars Effective Width Depth b d

fs′

(Dimensions in inches)

(Stresses, psi)

1 2 3

19.5 19.5 19.5

12 12 10

3–#7 3500 60,000 Rectangular 3–#10 3500 60,000 Rectangular 6–#7 3500 40,000 Rectangular (Two layers of three)

4

14.7

10

5

36.25

18

5–#9 3500 40,000 Rectangular (3–#9 outside layer and 2–#9 inside layer) 8–#11 4000 60,000 Rectangular (Two layers of four)

(Dimensions in Millimeters)

b2

Shape

fy

(Stresses, MPa)

6 7

495 495

300 300

3–#19M 25 3–#32M 25

400 Rectangular 400 Rectangular



(b)  Compute the service moment capacity (M D + M L ) using the factored gravity dead load plus live load combination U of ACI-​ 5.3.1 along with the φ factors of ACI-​21.2 (assume M L = 1.5 M D ).

t2

Problem 3.1 

(a) Compute independently the maximum flexural tensile and compressive stresses on the section shown in the accompanying figure (use the case assigned by the instructor), using basic statics involving internal forces and internal couple, that is, C = T and M = (C or T) times (moment arm between points of action of C and T). (b) Check the correctness of part (a) by computing both extreme fiber stresses using the flexure formula, f = Mc/​I. 3.2 The cross section of Problem 3.1 (use the case assigned by instructor) is used for an elastic homogeneous beam on a simply supported span of 25 ft (7.5 m); the beam carries a uniformly distributed load of 1 kip/​ft (15 kN/​m) plus a concentrated load of 9 kips (40 kN) at 8 ft (2.4 m) from the left end of the span. Compute the stresses as required by parts (a) and (b) of Problem 3.1.

d

As b

Problem 3.3

90

90

C H A P T E R   3     F L E X U R A L B E H A V I O R A N D S T R E N G T H O F   B E A M S

(4 – #32M) 4 – #10 14” (350 mm)

Problem 3.4   

14” (350 mm)

3.4 For the rectangular section shown in the figure for Problem 3.4, prove by calculating strains in the tension steel whether there is yield of the tension steel when the nominal strength M n is reached (i.e., when the extreme fiber in compression reaches a strain of 0.003). Use basic statics with the Whitney rectangular stress block and the internal couple (i.e., no formulas). Is the beam section tension controlled? Use fc′ = 4000 psi and fy = 58,000 psi ( fc′ = 30 MPa and f y = 400 MPa).

3.6 For the case assigned in Problem 3.5, design the largest practical size beam using the limits of ACI-​9.6.1.2. This time, assume that the given dead load does not include the beam weight; and make the practical design decisions, including the choice of overall dimensions, selection of the bars, and checking of the strength. The beam is to be designed using tension steel only (i.e., no compression steel). Use the U factors of ACI-​ 5.3.1 along with the φ factors of ACI-​21.2. 3.7 For the case assigned in Problem 3.5, design the smallest practical size for a tension-​controlled section. All other requirements are as in Problem 3.6. 3.8 Design a rectangular cross section having tension steel only of a size that would obviate the need to check deflections. The beam is simply supported and is to carry given uniformly distributed dead and live loads, in addition to the weight of the beam. Make all the practical decisions for size of cross section, select the reinforcement, and check the strength using statics and basic principles. Do the case assigned by the instructor.

3.5 For a rectangular section, determine the required fy wD wL Span Approx fc′ effective size (width b rounded to whole inches Case (psi) (psi) (kips/​ft) (kips/​ft) (ft) d/​b and effective depth d) and theoretical steel area 1 3500 40,000 1.0 2.0 30 1.5–​2.0 As to carry uniformly distributed dead and live 2 3000 60,000 0.8 1.8 30 1.5–​2.0 loads on a simply supported span as given. Do 3 3500 60,000 1.3 2.5 30 1.5–​2.0 the case assigned by the instructor. Do not make 4 4000 60,000 1.2 2.5 30 1.5–​2.0 the practical decisions of choosing actual over5 4000 60,000 0.8 1.5 30 1.5–​2.0 all beam dimensions and selecting bars. Assume 6 4000 60,000 1.2 2.2 32 1.5–​2.0 the dead load includes the beam weight. Use 7 4000 60,000 1.6 2.5 28 1.5–​2.0 basic principles (i.e., no formulas) to obtain 8 4000 60,000 1.8 2.7 26 1.5–​2.0 minimum (ACI-​ 9.6.1.2) and maximum (ACI-​ 9 3000 60,000 0.8 2.8 24 1.5–​2.0 9.3.3.1) reinforcement ratio limits, as well as the maximum reinforcement ratio corresponding to a tension-​controlled section (verify the latter by comparing with Table 3.6.1). Derive the formula 3.9 For the case in Problem 3.8 assigned by the for Rn in terms of ρ and then use it to get the instructor, design the smallest practical size for limiting values. a tension-​ controlled rectangular cross section (a) Obtain the largest effective size permitted having tension steel only. The beam is simply by ACI-​9.6.1.2. supported and is to carry given uniformly dis (b) Obtain the smallest effective size permitted tributed dead and live loads, in addition to the by ACI-​9.3.3.1. weight of the beam. Make all the practical decisions for size of cross section, select the reinforcement, and use statics and basic principles to check the strength. fy wD wL Span Approx fc′ 3.10 Design a rectangular beam to carry a live load of Case (psi) (psi) (kips/​ft) (kips/​ft) (ft) d/​b 1.5 kips/​ft and a dead load of 1.5 kips/​ft, in addition to the beam weight, for a simple span of 32 1 3000 60,000 1.3 1.0 24 1.75 ft. Use a steel ratio ρ corresponding to a net ten2 4000 60,000 1.3 1.0 24 2.0 sile strain ε t of approximately 0.0075, and also 3 4000 60,000 1.3 1.0 24 1.25 satisfy ACI-​Table 9.3.1.1. Use fc′ = 4000 psi and 4 5000 60,000 1.0 1.6 24 1.5 f y = 60, 000 psi. (Live load  =  22 kN/​m; dead 5 5000 60,000 0.8 1.2 18 1.5 load  =  22 kN/​m; span  =  9.8 m; fc′ = 30 MPa; 6 3000 40,000 0.8 1.4 18 2.0 f y = 400 MPa.)

91



91

PROBLEMS

3.11 Select an economical reinforcement for a beam 12 in. wide by 22 in. deep overall to carry dead load and live load moments of 15 and 60 ft-​kips, respectively. Use fc′ = 5000 psi, f y = 60, 000 psi. (Width  =  300  mm; depth  =  560  mm; dead load moment  =  20 kN·m; live load moment  = 80 kN·m; fc′ = 35 MPa; f y = 400 MPa.) 3.12 Select economical reinforcement for a beam 20 in. wide by 40 in. overall depth to carry a live load moment of 500 ft-​kips and a dead load moment (including beam weight) of 300 ft-​kips. Use a singly reinforced section. Though a check of deflection cannot be made with the given information, would you expect such a check to show that deflection is excessive? Explain your answer. Use fc′ = 4000 psi and f y = 60, 000 psi. (Beam size  =  500  mm × 1000   mm; live load moment = 700 kN·m; dead load moment = 400 kN·m; fc′ = 30 MPa; f y = 400 MPa.) 3.13 Assuming no deflection limitation, and without using compression steel, select a rectangular section having the smallest practical size permitted by ACI-​9.3.3.1; use the same size for the entire length (main span plus cantilever) of beam, as shown in the figure for Problem 3.13. The loads given are in addition to the beam weight. Select reinforcement for both the positive and negative moment regions. Use the case assigned by the instructor. For SI, use the reinforcing bars from Table  1.13.2. (Note: Live load is always to be applied in the locations and distributions that will cause the most severe effects; spans may be fully loaded, partially loaded, or unloaded, as necessary to obtain maximum effects.)

Main span

fc′

3500 3500 3500 4000 4000 4000 4000 4000 4000

40,000 40,000 60,000 60,000 60,000 60,000 60,000 60,000 60,000

Case (psi) 1 2 3 4 5 6 7 8 9 10 11

Main Cantilever wD wL Span Span (kips/​ft) (kips/​ft) (ft) (ft) 1.0 1.25 1.0 1.0 0.8 1.2 0.8 1.2 1.2

Main span

Cantilever spans

Problem 3.14   

fc′

fy (psi)

3500 3500 3500 4000 4000 4000 4000 4000 3000

60,000 60,000 60,000 60,000 60,000 60,000 60,000 60,000 60,000

Case (psi) 1 2 3 4 5 6 7 8 9

Main Cantilever wD wL Span Span (kips/​ft) (kips/​ft) (ft) (ft) 1.0 1.0 1.3 1.2 0.8 1.2 1.6 1.8 0.8

1.5 1.8 2.5 2.5 1.5 2.2 2.5 2.7 2.8

30 28 28 26 32 32 24 24 24

10 10 9 9 8 12 8 6 8

Cantilever span

Problem 3.13   

fy (psi)

3.14 For the double overhanging cantilever beam shown, design the smallest practical rectangular cross section (without compression steel) within the ρ limits of ACI-​9.3.3.1. The same overall size is to be used for the entire beam length (including cantilevers). The given dead load does not include the beam weight. Select reinforcement for both positive moment and negative moment regions. Do the case assigned by the instructor. Assume there is no deflection limit, and refer to the note at the end of Problem 3.13.

1.5 2.0 1.5 2.0 1.5 2.2 1.5 2.2 2.5

(MPa) (MPa) (kN/​m) (kN/​m) 25 300 15 20 25 300 17 30

20 24 16 18 20 20 22 22 24

8 8 10 8 8 8 10 10 10

(m) 6 7.3

(m) 2.5 2.4

3.15 For the loading and beam span conditions of Problem 3.14, design the beam as part of a floor system supporting nonstructural elements not likely to be damaged by large deflections. (Hint: Refer to ACI-​Table  9.3.1.1 and Problem 3.8.) Do the case assigned by the instructor. 3.16 Use principles of statics with internal couple to compute the nominal strength M n for the assigned case of a beam having compression steel. As a normal part of the procedure, verify by means of basic principles whether compression steel has yielded when nominal strength is reached; if it does not yield, use a compression steel stress proportional to the strain in the compression steel. Verify that tension steel does not exceed the maximum permitted by ACI-​ 9.3.3.1. Determine the appropriate φ factor to use according to ACI-​21.2.

92

92

C H A P T E R   3     F L E X U R A L B E H A V I O R A N D S T R E N G T H O F   B E A M S

Case

fc′ (psi)

fy (psi)

b (in.)

d (in.)

d′ (in.)

1 2 3 4 5 6 7 8 9

4000 3000 4000 5000 3500 3500 3500 3500 4000

40,000 60,000 60,000 60,000 60,000 40,000 60,000 60,000 60,000

18 16 16 16 12 12 12 14 20

36.1 27.3 27.3 27.3 21.5 20.5 17.6 19.5 28.4

2.5 10–#11 6–#7 3.0 4–#11 3–#11 3.0 4–#11 3–#11 3.0 4–#11 3–#11 2.5 3–#10 2–#10 2.44 3–#10, 3–#9 3–#9 2.38 3–#9 3–#8 2.5 2–#11, 2–#9 2–#10 2.38 6–#10, 2–#9 2–#8

10 11

As Bars

(MPa) (MPa) (mm) (mm) (mm) Bars   30   300 460 917 64 10-​36Ma   30   400 400 693 76   4-​36M

As′ Bars

Compression face

b d’ A’s

d

As

Bars 5-​25M 3-​35M

Tension face

Use reinforcing bars from Table 1.13.2.

a

Problem 3.16   

3.17 A rectangular section with b = 14 in . and effective depth d = 21.5 in. has 4–#10 as tension reinforcement and 2–#10 as compression reinforcement centered 2.5 in. from the face of the beam. Determine the strength M n for this section. How much can the strength be increased by adding tension steel only when the case is governed by ACI-​9.3.3.1? At the point where compression steel is needed to further increase strength by adding tension steel, what ratio of As′ to As would require to be added? Use fc′= 5000 psi and f y = 60, 000 psi (b = 350 mm; d = 546 mm; tension steel, 4–#32M; compression steel, 2–#32M centered 63.5 mm from face; fc′ = 35 MPa; f y = 400 MPa). 3.18 For the cross section obtained for the assigned data case of Problem 3.13, redesign as a beam having compression steel such that the net reinforcement ratio (ρ – ρ′) is about that corresponding to a net tensile strain ε t = 0.0075 for a singly reinforced beam. 3.19 For the conditions of Problem 3.14, Case 1, use a rectangular cross section 14 × 22 in. overall, and control creep and shrinkage deflection by designing compression steel such that As′ is about 0.5As. 3.20 For the beam shown in the figure for Problem 3.20, it is desired to utilize 3–#7 bars as compression reinforcement. The factored moment Mu to be carried is 210 ft-​kips; fc′ = 4500 psi, and f y = 60, 000 psi. Determine the tension steel required, select the bars, and check the section strength.

Compression face

3 – #7 #3 22”

1 12 ” clear cover all sides

As = ? 10”

Problem 3.20   

NONRECTANGULAR SECTION PROBLEMS 3.21–​3.22  For the beam cross sections shown in the figures for Problems 3.21 and 3.22, assuming fc′ = 3000 psi, f y = 60, 000 psi, and the clear cover from the bottom (tension) face to the bars = 2 in., perform the following: (a)  Compare the given tension reinforcement with the maximum permitted by ACI-​9.3.3.1. (b)  Using basic principles with the Whitney rectangular stress distribution, compute the nominal strength M n for the cross section. (c) Neglecting the reinforcement and assuming that the concrete is a homogeneous elastic material, compute the cracking moment M cr when the extreme fiber in tension reaches the modulus of rupture value given by ACI (i.e., 7.5 fc′ for normal-​weight concrete).

93



(d) Using the result in part (c), determine the reinforcement ratio As ,min / bw d that makes φ M n for a reinforced concrete beam (ignore the given bars for this part) equal to M cr . Compare with ACI-​9.6.1.2.

4”

6”



(e)  Assuming that compression steel of an amount equal to the given tension steel were to be used and located with 2-​in. clear cover to the top (compression) face of the beam, determine the maximum amount of tension steel that ACI-​9.3.3.1 would permit.

4”

8” 8” 8” 20”



93

PROBLEMS

5 – #6 24”

16” 3 – #8

14”

Problem 3.21   

Problem 3.22 

CHAPTER 4 T-​S ECTIONS IN BENDING

4.1 GENERAL Chapter 3 covered the analysis and design of isolated rectangular beams and briefly discussed nonrectangular sections (Section 3.12). In most building frame structures, however, beams are built monolithically with the slab, leading to a T-​shaped beam section; hence the name T-​beams. Beams with a T shape may also be individually built as such, as is commonly done in precast construction, where the flange may later serve as support for a slab topping or floor finishing. For the purpose of analysis and design, each beam cast monolithically with the slab is assumed to include a portion of the slab projecting as a flange from each side of the stem. For negative bending moment the flange is on the tension side of the neutral axis, thus the T-​section is in effect a rectangular section with a compression area of width equal to the width of the stem or web. For positive bending moment the flange does provide considerably more compression area than in the negative bending moment portion of the span. Analysis and design of T-​sections subjected to bending are discussed in sections 4.2 through 4.5.

City Hall, Boston. (Photo by C. G. Salmon).

95



95

4.3  EFFECTIVE FLANGE WIDTH

b

b

N.A.

N.A. d

d N.A.

bw (a)

(b)

Figure 4.2.1  Two equivalent sections in bending.

Figure 4.2.2  T-​section under positive bending.

4.2 COMPARISON OF RECTANGULAR AND T-​S ECTIONS A comparison of the two sections shown in Fig. 4.2.1 indicates that the flexural strength of a rectangular section is the same as that of a T-​section as long as both sections have compression zones of equal width and depth, and the same steel area at the same effective depth. Thus, as far as bending is concerned, any T-​section with a rectangular compression area, such as shown in Fig. 4.2.1(b), may be regarded as a rectangular section regardless of the shape of the tension zone. When the compression zone of a T-​section subjected to positive bending extends below the flange, as shown in Fig. 4.2.2, computation of its flexural strength requires different treatment than for a rectangular section; this is because the shape of the compression zone will no longer be rectangular.

4.3 EFFECTIVE FLANGE WIDTH Very wide beams do not conform in behavior to the assumption of the elementary theory of bending. In ordinary theory, bending stresses are assumed not to vary across the beam width. Simple theory, therefore, would dictate a constant stress at, say, the extreme fiber over the entire flange width of a T-​section, no matter how great the overhang from the stem. In reality, the stress, based on the plate theory, decreases the more distant a point is from the stem of the beam. Thus, for a flange of infinite width, the compressive stress in the flange varies as shown in Fig. 4.3.1. Theoretical investigations for an infinitely long continuous beam on equidistant supports, with an infinitely large flange width and a thickness that is small in comparison to the beam depth, have led to the determination of an effective flange width bE over which the compressive stress may be considered constant. The total compressive force carried by the equivalent system is the same as that carried by the actual system. For such assumptions, the equivalent width of overhang λ depends only on the type of loading and the span length of the beam. The theory, first developed by T. von Kármán, along with certain results, is summarized in Timoshenko and Goodier [4.1] and Girkmann [4.2]. Whereas the aforementioned theory gives the equivalent width bE for an infinite flange width as a function of the span length L, in practical situations other variables are important as well. These variables are the spacing of the beams, the width of the stem or web of the beam bw, and the relative thickness of the slab with respect to the total beam depth t/​h. To illustrate the effect of loading, several cases (from Girkmann [4.2]) showing the variation of effective flange projection for a flange of zero thickness between beams (t/​h = 0) are given in Fig. 4.3.2. In the case of a floor slab built monolithically with floor beams, there is transverse bending in the slab between beams, which also tends to reduce the effectiveness of the slab in carrying compression at points remote from the beam web. Thus, there is a valid reason for using a conservatively low effective flange width. The effective section of a T-​beam in a floor system is shown shaded in Fig. 4.3.3.

96

96

C hapter   4     T- S ections in B ending

λ = equivalent width for uniform stress and same compressive force as actual stress distribution

bE

Actual extreme fiber compressive stress fc for infinitely wide flange λ

λ

t h

bW

Figure 4.3.1  Actual and equivalent stress distribution over flange width. w sin πx/L

x

w L

L bw +0.363L

0.191L

bw

bw +0.194L

0.191L

w L/2

bw

P

L

L

bw +0.152L bw +0.36L

bw

bw +0.550L

bw

bw +0.764L

Max moment 3L/8

bw +0.196L

Figure 4.3.2  Equivalent flange widths for beams with infinitely wide flanges and a rib cross-​ sectional area of 0.1 tL. (Adapted from Girkmann [4.2].)

The effective width bE of the flange is an essential factor in the neutral axis location. Available information indicates that as the strain εc at the extreme compressive fiber increases toward its ultimate value εcu, the width of the effective compression flange increases [4.4]. Therefore, it is safe to use the smaller effective widths applicable under service load conditions based on linear elastic behavior.

97



4 . 4   N O M I N A L M O M E N T S T R E N G T H M n O F T-​S E C T I O N S

97

bE

t h

2bo bw

Figure 4.3.3  Effective section of a T-​beam with the slab in compression in a beam-​slab floor system.

The ACI Code (ACI-​6.3.2.1) prescribes a limit on the effective flange width bE of inter­ ior T-​sections to the smallest of the following: 1.  bE = bw + Ln /4 2.  bE = bw + 16t  3.  bE = center-to-center spacing of beams

(4.3.1a) (4.3.1b) (4.3.1c)

where Ln is the clear span length of the beam, and bw and t are as shown in Fig. 4.3.3. For exterior T-​sections (slab on only one side of web), ACI-​6.3.2.1 prescribes for effective width bE the smallest of the following: 1.  bE = bw + Ln /12 2.  bE = bw + 6t  3.  bE = bw + 1 / 2 ( clear distance to next beam )

(4.3.2a) (4.3.2b) (4.3.2c)

For isolated T-​sections, ACI-​6.3.2.2 requires

bE ≤ 4bw

(4.3.3a)

1 bw 2

(4.3.3b)

and

t≥

Since the true effective width depends greatly on the type of loading and on the relationships (Fig. 4.3.3) t/​h, L /​bw, and L /​b0, the ACI criteria are very much simplified. They seem properly adequate for certain loading cases and are unduly conservative for others. Detailed procedures for determining the values of effective flange width are presented in [4.3, 4.4].

4.4 NOMINAL MOMENT STRENGTH M n OF T-​S ECTIONS In computing the nominal moment strength Mn of a T-​section under positive bending, the neutral axis location determines whether the compression zone is T-​shaped or rectangular. Since the option of using Whitney rectangular stress distribution (Section 3.3) is also available for nonrectangular sections, use of the rectangular stress distribution would mean that as long as its depth a does not exceed the flange thickness t, the moment strength Mn would be the same as for a rectangular section having b = bE. Strictly speaking, rectangular section behavior should be based on the depth of the neutral axis c and not of Whitney’s stress block. However, the use of a = t as the boundary between rectangular section and T-​section behavior leads to adequate results in terms of flexural strength. Two cases are possible for computation of the positive nominal moment strength Mn of a T-​section: (1) the depth a of the rectangular stress distribution is equal to or less than t, and (2) the depth a is greater than t. As for most cases a ≤ t, it is convenient to first calculate the depth of the stress block assuming that Case 1 governs.

98

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C hapter   4     T- S ections in B ending

Case 1: Rectangular section behavior, a ≤ t [Fig. 4.4.1(a)] For this case, a is calculated taking b = bE . Thus, the analysis is exactly as presented for rectangular sections in Chapter 3, using bE for b. The following discussion is limited to singly reinforced T-​sections. In general, the effect of compression steel reinforcement, if provided, can be neglected because it will have little effect on moment strength. Further, εt >> 0.005 in typical singly reinforced T-​sections with rectangular section behavior (φ = 0.90), and thus there is no need to account for the beneficial effect of compression steel on section ductility in determining the strength reduction factor φ. If compression steel must be included in the calculation of moment capacity, it can be done following the same procedure outlined in Section 3.10. The effect of compression reinforcement on the flexural strength and normal strain distribution in T-​sections is illustrated in Example 4.4.4. The depth of the stress block a is then calculated accordingly and compared with the flange thickness t as follows:  a=



As f y 0.85 fc′ bE

≤t

(4.4.1a)



obtained from equating C to T, where 

C = 0.85 fc′ bE a

(4.4.1b)



T = As f y

(4.4.1c)

bE

c = a/β1

εc = 0.003

0.85fc’

C

d

Moment arm = (d – 21 a )

N.A.

a≤t

t

As εs ≥ εy

bw

(a) a ≤ t

bE 1 2

A2

t

C1

c

d

N.A.

C2

Moment arm = (d – 21 t)

A1

Moment arm < (d – 21 t)

A2

εc = 0.003

a>t

1 2

T = Asfy

As bw

εs ≥ εy

T = C1 + C2 = Asfy

(b) a > t

Figure 4.4.1  Strain and stress conditions, and force resultants of T-​sections in bending: a) a ≤ t and b) a > t.

9



99

4 . 4   N O M I N A L M O M E N T S T R E N G T H M n O F T-​S E C T I O N S

Case 2: a > t [Fig. 4.4.1(b)] For this case, the area on which the uniform stress 0.85 fc′ acts is T-​shaped. Except for cases in which compression steel is needed for ductility so that the section is tension controlled (not common), its presence can be neglected in the calculation of moment strength as in Case 1. The discussion that follows deals with singly reinforced T-​sections. Calculation of nominal moment strength in T-​sections with compression reinforcement and a  >  t is illustrated in Example 4.4.5. It is desirable to separate the total compressive force into forces C1 and C2, with C1 resulting from the stress on area A1 and C2 on area A2. The moment arm for C2 is equal to d –​ t/​2, but that of C1 is less than d  –​ t/​2. Thus  a t   M n = C1  d −  + C2  d −    2 2



(4.4.2a)

in which 



C1 = 0.85 fc′ bw a

(4.4.2b)

C2 = 0.85 fc′ (bE − bw )t

(4.4.2c)

and since  C1 + C2 = As f y = T

then 

a=



T − C2 0.85 fc′ bw

(4.4.2d)

The tensile force T and the tension steel area As may also be separated into T1 and T2 and As1 and As2, respectively.

EXAMPLE 4.4.1 Determine the nominal moment strength Mn within the span of a floor beam (Fig. 4.4.2) whose projection below a 4 1 2 -​in. slab is 13 × 24 in. (effective depth is 25 in. for two layers of steel). Tension reinforcement is 8–​#8 bars. The span length of the beam, measured to the center of the supporting columns, is 24 ft, and the beams are centered 13 ft apart. Supporting columns have a square cross section with 28-​in. side dimension. Use fc′ = 4500 psi and fy = 60,000 psi. 4 21 ”

Effective width bE = 78”

0.85fc’

c

1 2

a C

a 25” 24” 8 – #8

As = 6.32 sq in. T

13”

Figure 4.4.2  T-​section for Example 4.4.1.

(Continued)

10

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C hapter   4     T- S ections in B ending

Example 4.4.1 (Continued) SOLUTION (a) Determine the effective flange width. Following ACI-​6.3.2.1, the effective flange width bE from Eqs. (4.3.1) is the smallest of 13 + ( 24 )(12 ) − 28 / 4 = 78 in., 13 + 16 ( 4.5) = 85 in ., or (13)(12 ) = 156 in. Thus bE = 78 in. (b) Assume rectangular section behavior (a ≤ t) and yielding of the reinforcement.  As f y (6.32)(60) a= = = 1.27 in. ≤ t = 4.5 in. 0.85 fc′ bE (0.85)(4.5)(78) Assumption of rectangular section behavior is correct. For such a small value of a, it is clear that εs > εy. (c) Treat as a rectangular section with a = 1.27 in. a moment arm = d − = 25 − 0.64 = 24.36 in. 2 M n = (6.32)(60)(24.36)



1 12

= 770 ft-kips

EXAMPLE 4.4.2 Determine the nominal moment strength Mn of the isolated T-​section shown in Fig. 4.4.3 when As = 12.48 sq in. (8–​#11). Use fc′ = 4000 psi and  f y = 60, 000 psi. εc = 0.003

30” 1 2

A2

A1

1 2

A2

7”

0.85fc’

a

c

3 21 ”

1 2

a

C2 C1

N.A. 36”

As εs ≥ εy

T

14”

Figure 4.4.3  T-​section for Examples 4.4.2 and 4.5.1.

SOLUTION (a) Check requirements for isolated T-​sections. According to ACI-​6.3.2.2, bE cannot exceed 4bw = 56 in., and t must be at least bw  /​  2 = 7 in. Thus, flange thickness is satisfactory and bE = 30 in. is less than 4bw. (b) Calculate stress block depth a assuming rectangular section behavior. As f y (12.48)(60) a= = = 7.34 in. > t = 7 in. . f b ( 0 .85)(4)(30) 0 85 c′ E Assumption of rectangular section behavior is not correct. Section behaves like a T-​section. (c) Treat T-​section by the two-​couple method (Fig. 4.4.3). T = As f y = 12.48(60) = 749 kips C = C1 + C2 = 0.85 fc′ A1 + 0.85 fc′ A2

(Continued)

10



101

4 . 4   N O M I N A L M O M E N T S T R E N G T H M n O F T-​S E C T I O N S

Example 4.4.2 (Continued) 749 = 3.4(14a ) + 3.4(16)(7) a = 7.73 in. C1 = 3.4(14)(7.73) = 368 kips C2 = 3.4(16)(7.0) = 381 kips M n = C1 [36 − 0.5(7.73)]

1 12

+ C2 (36 − 3.5)

1 12

M n = 985 + 1032 = 2017 ft-kips

EXAMPLE 4.4.3 For the T-​section shown in Fig. 4.4.4, determine the design strength φMn in accordance with the ACI Code. The tension steel is 9–​#11 (As = 14.04 sq in.) with three bars in each of three layers. Assume #4 stirrups are used, and there is one inch between layers. Use fc′ = 4000 psi and fy = 60,000 psi. 30” c = 11.40”

0.003 7”

d

40”

dt

9 – #11 εt = 0.0068

14”

Figure 4.4.4  T-​section and strain distribution for Example 4.4.3.

SOLUTION (a) Determine the effective depth d and calculate the stress block depth a, assuming rectangular section behavior. For a clear cover of 1.5 in. and a #4 stirrup,

d = 40 – 1.5 ( i.e., cover ) – 0.5 ( i.e., stirrup ) – 1.5 (1.41) ( i.e., 1.5 bar diameters) – 1 ( i.e., between layers) = 34.9 in.

The depth of the compressive stress block is 

a=

As f y 0.85 fc′ bE

=

(14.04)(60) = 8.26 in. > t = 7 in. (0.85)(4)(30)

Thus, the compression zone is T-​shaped and the neutral axis is in the web. The internal forces are  T = As f y = 14.04(60) = 842 kips C1 = 0.85 fc′ bw a = 0.85(4)(14)a = 47.6 a

C2 = 0.85 fc′ (bE − bw)t = 0.85(4)(16)7 = 381 kips C = C1 + C2 = T

(Continued)

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C hapter   4     T- S ections in B ending

Example 4.4.3 (Continued)

a=

842 − 381 = 9.69 in. > t 47.6

c=

a = 11.40 in. β1

(b) Compute the nominal moment strength Mn. Using the two-​couple method, a t   M n = C1  d −  + C2  d −    2 2 9.69  1  1 M n = 47.6(9.69)  34.9 −  12 + 381(34.9 − 3.5) 12  2



M n = 1155 + 997 = 2152 ft-kips (c) Compute the net tensile strain εt at the extreme tension steel. The distance dt to the extreme tension steel is dt = d + 1.41 / 2 + 1 + 1.41 / 2 = 34.9 + 2.41 = 37.3 in.



ε t = 0.003

dt − c 37.3 − 11.40 = 0.003 = 0.0068 c 11.40

Because εt exceeds the limit 0.005 for tension-​controlled sections (see Fig. 3.6.2), the appropriate φ factor is 0.90 (ACI-​21.2.2). Thus,

φ M n = 0.90 ( 2152 ) = 1937 ft-kips



EXAMPLE 4.4.4 Determine the design moment strength φMn in accordance with the ACI Code of the same section used in Example 4.4.3 but with 5–​#8 bars (  fy = 60 ksi) as compression reinforcement (Figure 4.4.5). 30” 5 – #8

0.003 c = 7.29”

d’ = 2.5”

7”

d

dt

40” 9 – #11

εt = 0.0123

14”

Figure 4.4.5  T-​section and strain distribution for Example 4.4.4.

SOLUTION (a) Determine depth of stress block and internal forces. Assuming the compressive steel yields and rectangular section behavior at nominal strength condition,

a=

As f y − As′ ( f y − 0.85 fc′ ) 0.85 fc′ bE

=

(14.04)(60) − 3.95[60 − 0.85(4))] = 6.07 in. < t = 7 in. (0.85)(4)(30) (Continued)

103



4 . 4   N O M I N A L M O M E N T S T R E N G T H M n O F T-​S E C T I O N S

103

Example 4.4.4 (Continued) Note that the stress in the compression steel has been adjusted to account for the area of displaced concrete in compression. Verify that for the calculated stress block depth, the compression steel reinforcement yields as assumed. c=



ε s′ = 0.003

a = 7.14 in. β1

c − d′ 7.14 − 2.5 = 0.003 = 0.00195 < ε y = 0.00207 c 7.14

Compression reinforcement does not yield. Rather than developing a closed-​ form solution, it is easier to assume a value for the depth of the stress block and verify that equilibrium is satisfied. Since the compression steel is near yield, assume only a small increment in the depth of the stress block a. Thus, taking a = 6.20 in., c=

a = 7.29 in. β1

T = As f y = 14.04(60) = 842 kips Cc = 0.85 fc′ bE a = 0.85(4)(30)(6.20) = 632 kips

ε s′ = 0.003

c − d′ 7.29 − 2.5 = 0.003 = 0.00197 c 7.29

Cs = As′ (ε s′ Es − 0.85 fc′ ) = 3.95 [(0.00197)29000 − 0.85(4)] = 212 kips C = Cc + Cs = 844 kips ≈ T By adding compression reinforcement with an area approximately equal to one-​third that of the tension steel, the section now exhibits a rectangular section behavior (Case 1, a  ≤  t). (b) Compute the nominal moment strength Mn. Taking moments with respect to the centroid of the tension steel, a  M n = Cc  d −  + Cs (d − d ′ )  2  1 6.20   M n = 632  34.9 − + 212(34.9 − 2.5)   2    12



M n = 1676 + 573 = 2249 ft-kips This nominal moment strength is only 5% greater than that calculated without compression reinforcement in Example 4.4.3. (c) Compute the net tensile strain εt at the extreme tension steel.

ε t = 0.003

dt − c 37.3 − 7.29 = 0.0123 > 0.005 = 0.003 c 7.29

Thus, φ = 0.90 and

φ M n = 0.90 (2249) = 2024 ft-kips

Note that the net tensile strain εt nearly doubled compared to that of the same section without compression steel. Therefore, the addition of compression steel, though having little effect on flexural strength, did have a significant effect on section ductility.

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C hapter   4     T- S ections in B ending

EXAMPLE 4.4.5 Determine the design moment strength φMn in accordance with the ACI Code of the same section used in Example 4.4.4 but using a flange thickness of 5 in. (Figure 4.4.6). 30”

0.003 c = 8.58”

d’ = 2.5”

5” 5 – #8

d

40”

dt

9 – #11

εt = 0.010

14”

Figure 4.4.6  T-​section and strain distribution for Example 4.4.5.

SOLUTION (a) Determine depth of stress block and internal forces. Assuming rectangular section behavior and yielding of the compression steel, the depth of the compressive stress block is



a=

As f y − As′ ( f y − 0.85 fc′ ) 0.85 fc′ bE

=

(14.04)(60) − 3.95[60 − 0.85(4))] = 6.07 in. > t = 5 in. (0.85)(4)(30)

Section falls within Case 2 (i.e., T-​section behavior). The internal forces, assuming yielding of compression steel, are T = As f y = 14.04(60) = 842 kips C1 = 0.85 fc′ bw a = 0.85(4)(14)a = 47.6 a C2 = 0.85 fc′ (bE − bw )t = 0.85(4)(16)5 = 272 kips

Cs = As′ ( f y − 0.85 fc′ ) = 3.95 [60 − 0.85(4)] = 224 kips C = C1 + C2 + Cs = T a=

842 − 272 − 224 = 7.29 in. 47.6

C1 = 0.85 fc′ bw a = 0.85(4)(14)(7.29) = 347 kips c=

a = 8.58 in. β1

Verify that the compression steel reinforcement yields as assumed.

ε s′ = 0.003

c − d′ 8.58 − 2.5 = 0.003 = 0.00212 > ε y = 0.00207 c 8.58

Compression reinforcement yields as assumed. (Continued)

105



105

4 . 5   D E S I G N O F T- S E C T I O N S I N B E N D I N G

Example 4.4.5 (Continued) (b) Compute the nominal moment strength Mn. Taking moments with respect to the centroid of the tension steel, a t   M n = C1  d −  + C2  d −  + Cs (d − d ′ )   2 2



 1 7.29  5.00    M n = 347  34.9 −  + 272  34.9 −  + 224(34.9 − 2.5) 12  2 2   M n = 904 + 734 + 605 = 2243 ft-kips

(c) Compute the net tensile strain εt at the extreme tension steel.

ε t = 0.003

dt − c 37.3 − 8.58 = 0.010 > 0.005 = 0.003 c 8.58

Thus, φ = 0.90 and 

φ M n = 0.90(2243) = 2019 ft-kips

4.5 DESIGN OF T-​S ECTIONS IN BENDING The design of T-​shaped isolated beams involves the dimensions of the flange and web and the area of tension steel, or a total of five unknowns—​one more if compression steel is used. Thus, there are many possible solutions to the problem. The more common T-​sections are those in the region of positive bending moment in continuous monolithic slab-​beam-​girder systems. The size of the available flange is therefore known after the slab thickness has been selected, and only the web size needs to be designed. The selection of web size for such beams is treated in Chapter 9, and it is often based on negative moment requirements at the supports, where the web will be in compression, or on shear strength requirements. Once the overall dimensions have been established, the design problem is to determine the amount of positive moment reinforcement. Thus, it is important first to ascertain the location of the neutral axis associated with the positive moment. This may be done by first computing the moment strength at which the depth a of the rectangular compressive stress block is equal to the flange thickness t. If the depth a is less than t, then the compression zone is rectangular and the beam is designed as a rectangular section. If a is greater than t, then the compression zone is T-​shaped and the two-​couple method can be used.

EXAMPLE 4.5.1 Determine the amount of tension steel required in the T-​section of Fig. 4.4.3 to carry a dead load moment of 560 ft-​kips and a live load moment of 700 ft-​kips, using fc′ = 4000 psi and  fy = 60,000 psi. SOLUTION (a) Determine the factored moment  Mu. 

Mu = 1.2(560) + 1.6(700) = 1792 ft-kips (Continued)

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C hapter   4     T- S ections in B ending

Example 4.5.1 (Continued) Assuming  φ = 0.90,  required M n =



Mu 1792 = = 1991 ft-kips φ 0.90

(b) Determine whether the depth a of the rectangular stress distribution will be greater than t = 7 in. For a = t,  C = 0.85 fc′ bE t = 0.85(4)(30)(7) = 714 kips

t 7 1   M n = C  d −  = 714  36 −  = 1934 ft-kips    2 2  12

Since the required Mn exceeds 1934 ft-​kips, the actual a must exceed t. (c) Use the two-​couple method (Fig. 4.4.3) to obtain As. a t   M n = 0.85 fc′ A1  d −  + 0.85 fc′ A2  d −     2 2

a  1991(12) = 3.4(14 a )  36 −  + 3.4(16)(7)(36 − 3.5)  2 a 2 − 72 a = −484 a = 7.50 in. > t C1 = 0.85 fc′ bw a = 0.85(4)(14)(7.50) = 357 kips As1 =

T1 357 = = 5.95 sq in. fy 60

C2 = 0.85 fc′ (bE − bw) t = 0.85(4)(16)(7) = 381 kips As 2 =

T2 381 = = 6.35 sq in. fy 60

As = As1 + As 2 = 5.95 + 6.35 = 12.30 sq in. Check minimum tension reinforcement requirement. Since in this case 3 fc′ < 200 psi, the minimum required amount of tension reinforcement is (see Section 3.7) 

As ,min =

200 200 bw d = (14)(36) = 1.68 sq in. < 12.30 sq in. fy 60, 000

Try 8–​#11 bars in two layers (As = 12.48 sq in.). From Table 3.9.2, the minimum beam width required for 4–​#11 is 13.8 in., which does not exceed the 14 in. available, and is acceptable. Assuming 1-​in. clear cover between layers, the distance dt to the extreme tension steel is  dt = d + 0.5 + 1.41 / 2 = 36 + 1.21 = 37.2 in.

ε t = 0.003

dt − a / β1 37.2 − 7.50 / 0.85 = 0.003 = 0.0096 a / β1 7.50 / 0.85

Since εt exceeds the limit of 0.005 for tension-​controlled sections (see Fig. 3.6.2), then φ = 0.90 as assumed. Use 8–​#11 bars in two layers.

107



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4 . 5   D E S I G N O F T- S E C T I O N S I N B E N D I N G

EXAMPLE 4.5.2 Design the steel reinforcement for the section of Fig. 4.4.2 to carry a factored moment Mu of 735 ft-​kips. Use fc′ = 3000 psi and  f y = 50, 000 psi. SOLUTION (a) Determine whether the depth a of the rectangular stress distribution will be greater than t. For a = t, C = 0.85 fc′ bE t = 0.85(3)(78)(4.5) = 895 kips 4.5  1 t   M n = C  d −  = 895  25 −  = 1697 ft-kips   2 2  12

required M n =

M n 735 = = 817 ft-kips 0.90 φ



Since the required Mn is less than the amount necessary to cause a = t, the section can be designed as a rectangular section. (b) Design as a rectangular section. The computation of the coefficient of resistance Rn = Mu  /​(φ bE  d 2) and the solving for ρ may be made exactly as described in Chapter 3 using Eq. (3.8.5). However, because of the wide flange (bE = 78 in.) and the likelihood of a relatively small value for a, a trial procedure may be preferred. First use 0.9d as a conservative (i.e., low) trial value for the moment arm (d –​ a/​2); then

required As =

required M n 817(12) = = 8.71 sq in. ( d − a / 2) f y 0.9(25)50

because in this case 3 fc′ < 200 psi, the minimum required tension reinforcement is (see Section 3.7) 200 200 bw d = (13)(25) = 1.08 sq in. < 8.71 sq in. As ,min = fy 60, 000 Try 4–​#9 and 4–​#10, As  =  9.08 sq in. (minimum width = 12.9 in. from Table  3.9.2). Check:  C = 0.85 fc′ bE a = 0.85(3)(78)a = 198.9a T = As f y = 9.08(50) = 454 kips 454 = 2.28 in. 198.9 arm = 25 − 0.5(2.28) = 23.86 in. (i.e., 0.95d ) a=

A second trial gives 

required As =

817(12) = 8.22 sq in. 23.86(50)

Revise to 4–​#8 and 4–​#10, As = 8.24 sq in. Check:  C = 198.9a (as before) T = 8.24(50) = 412 kips 412 = 2.07 in. 198.9 arm = 25 − 0.5(2.07) = 23.96 in. ≈ 23.86 in. used a=



(Continued)

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C hapter   4     T- S ections in B ending

Example 4.5.2 (Continued) No further saving can be made. Using a 1-​in. clearance between layers,  dt = d + 0.5 + 1.27 / 2 = 25 + 1.14 = 26.1 in. d − a / β1 26.1 − 2.07 / 0.85 = 0.029 ε t = 0.003 t = 0.003 a / β1 2.07 / 0.85



Since εt  > 0.005 for tension-​controlled sections, then φ = 0.90 as assumed; thus, M n = 412(23.96)



1 12

= 823 ft-kips



[φ M n = 0.90(823) = 740 ft-kips ] > [ Mu = 735 ft-kips]      OK

Use 4–​#8 and 4–​#10. Note that the strain in the extreme tension steel (2.9%) is large compared to that in a typical rectangular section. This result is usual for floor T-​beams and is due to the wide flange, which requires only a small depth of the compression zone to balance the tensile force in the steel.

SELECTED REFERENCES 4.1

S. Timoshenko and J. N. Goodier. Theory of Elasticity (3rd ed.). New York: McGraw-​Hill, 1970 (pp. 262–​268). 4.2 Karl Girkmann. Flachentragwerke (3rd ed.). Vienna: Springer-​Verlag, 1954 (pp. 116–​123). 4.3 Franco Levi. “Work of the European Concrete Committee,” ACI Journal, Proceedings, 57, March 1961, 1049–​1054. 4 .4 Gottfried Brendel. “Strength of the Compression Slab of T-​Beams Subject to Simple Bending,” ACI Journal, Proceedings, 61, January 1964, 57–​76. 4.5 Antoine E.  Naaman. “Rectangular Stress Block and T-​Section Behavior,” PCI Journal, 47, September–​October 2002, 107–​112.

PROBLEMS All problems* are to be done according to the ACI Code. All loads given are service loads unless otherwise indicated. 4.1 (a) Determine the nominal moment strength Mn for the beam cross section shown in the figure for Problem 4.1. Use fc′ = 4000 psi and f y = 60, 000 psi. ( fc′ = 30 MPa; f y = 400 MPa; slab t = 125 mm; beam h = 900 mm; bw = 380 mm.)

(b) Determine the maximum tension reinforcement As permitted for this beam by ACI-​ 9.3.3.1. What is the maximum As permitted if the section is to be tension controlled? (c)  Determine and compare the design moment strengths φ M n corresponding to the two areas of tension steel determined in part (b). 4.2 (a)  Design the reinforcement for the beam shown in the figure for Problem 4.2, if the

* Problems may be solved as problems stated in Inch-​ Pound units, or as problems in SI units using the quantities in paren­ theses. The SI values are approximate conversions to avoid implying higher precision for given information in metric units than for Inch-​Pound units.

109



109

PROBLEMS

5” d

36”

15”

Inch-Pound As = 9.08 sq in. (4 – #9, 4 – #10, two layers) Span = 30 ft

4” (100 mm)

SI 4 – #30M 4 – #35M

8’ – 0” (2.4 m)

Problem 4.1 

dead load moment is 90 ft-​kips and the live Use fc′ = 5000 psi and f y = 60, 000 psi. load moment is 140 ft-​kips. (M D = 122 kN ⋅ m ; M L = 190 kN ⋅ m ; fc′ = (b) Determine the maximum tension reinforce35 MPa; f y = 400 MPa.) ment As permitted for this beam by ACI-​ 4.3 Repeat Problem 4.2 for a dead load moment 9.3.3.1 and for a tension-​controlled section. of 65 ft-​ kips and a live load moment of 100 ft-​ kips. Use fc′ = 3500 psi and f y = 4” (100 mm) 48” (1.2 m) 60,000 psi. (M D = 88 kN ⋅ m; M L = 135 kN ⋅ m ; fc′ = 24 MPa; f y = 400 MPa.) 24” (610 mm) As

15” (380 mm)

Problem 4.2 

CHAPTER 5 SHEAR STRENGTH AND DESIGN FOR SHEAR

5.1 INTRODUCTION This chapter treats the shear strength of nonprestressed one-​way flexural members. Effects of axial compression and tension are included in Section 5.13. Deep beams behave differently than slender beams, and so their design is usually performed using strut-​and-​tie models, as covered in Chapter 14. Consideration is also given to the shear-​ friction concept (Section 5.15) and its application to the design of corbels and brackets (Section 5.16). Shear strength of prestressed concrete members is treated in Chapter 20. Shear effects arising from torsion and its interaction with shear due to bending are treated in Chapter 18. For the simple beam shown in Fig. 5.1.1, the bending moment M at section A-​A causes compressive stresses in the concrete above the neutral axis and tensile stresses in the reinforcement (and in the concrete below the neutral axis if it has not yet been cracked). To satisfy vertical force equilibrium, the resultant of the vertical shear stresses across the section must be equal to the shear force V. In an element located at the neutral axis, there is a state of pure shear as shown in Fig. 5.1.2, which gives rise to a principal tensile stress

Shear failure of reinforced concrete beam without web reinforcement. (Photo by Gustavo J. Parra-​Montesinos).

1



5 . 2   S hear S tresses B ased on   L inear E lastic B eha v ior

111

A

A N.A.

M V

A A

Figure 5.1.1  Shear force and bending moment at a section in a simply supported beam. v ft (max) = v v

v

v 45° v

v

Figure 5.1.2  State of pure shear stress (i.e., no tensile or compressive stresses on the faces of the element).

equal to the shear stress, and acting on a 45° plane from the longitudinal beam axis. This diagonal tension will cause diagonal cracking, which can lead to premature failure of the beam (i.e., prior to development of the full flexural strength with ample deformation). Thus the failures in beams commonly referred to as “shear failures” are often actually diagonal tension failures caused by the formation of inclined cracks. Among the earliest to recognize this phenomenon were Emil Mörsch [5.1] in Germany and Arthur Talbot in the United States [5.2] in the early 1900s. Bresler and MacGregor [5.3] have presented an excellent systematic treatment of the various situations in which shear-​related cracks develop. Their work was expanded by ACI-​ASCE Committee 426 [5.4]. Collins [5.5], Collins and Mitchell [5.6], Marti [5.7], Vecchio and Collins [5.8, 5.10], Schlaich, Schäfer, and Jennewein [5.9], Ramirez and Breen [5.11], Al-​Nahlawi and Wight [5.12], Collins, Mitchell, Adebar, and Vecchio [5.13], and Tureyen and Frosch [5.14], among others, have presented alternative models for estimating the shear strength of beams. A good review of additional approaches for the shear design of reinforced concrete members has been published by ASCE-​ACI Committee 445 [5.15]. This chapter begins with the concepts of horizontal and vertical shear, and the resulting principal tensile stress, so that the reader may gain insight into the potential direction of inclined cracks. Then, the variables affecting shear strength, leading up to the current ACI Code provisions, are presented.

5.2 SHEAR STRESSES BASED ON LINEAR ELASTIC BEHAVIOR A simply supported, uniformly loaded rectangular beam is shown in Fig. 5.2.1. Consider the free body of the elemental block abcd, as shown in Fig. 5.2.1(b), where kd is the elastic neutral axis distance. Horizontal force equilibrium requires

vy b dz = C2 − C1

(5.2.1)

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C hapter   5     S hear S trength and D esign for   S hear 2

1 a c e g 1

w

b d f

N.A.

h dz

b

2

(a)

w fc1 C1 y

fc1y

d

y

vb dz

a

b

c

d vy b dz

fc2

e

C2 kd

fc2y

f

T1

T2 g

h

N.A. (b)

(c)

Figure 5.2.1  Horizontal shear stress in a beam.

in which vy is the unit horizontal shear stress on a plane at a distance y from the neutral axis. Under service loads, the flexural stress distribution in the compression region may be assumed to vary linearly, thus 1 ( fc1 + fc1 y )b(kd − y) 2 y f c1 y = f c1 kd   y  2  1 1  y C1 = fc1  1 +  (kd − y)bb = fc1bkd 1 −    2 2  kd    kd   C1 =



(5.2.2)

Dividing the bending moment M1 on section 1-​1 by the arm of the internal couple gives the full internal compressive force on section 1-​1, or M1 1 = fc1b(kd ) arm 2 Substitution of fc1 from the above expression into Eq. (5.2.2) gives

C1 =

M1  y2  1 −  arm  (kd )2 

(5.2.3)

C2 =

M2  y2   1 − arm  (kd )2 

(5.2.4)

Similarly,

Substituting Eqs. (5.2.3) and (5.2.4) into Eq. (5.2.1) gives y2   M − M1  1  1− vy =  2    dz  b (arm)  (kd )2 

y2  V  = 1 −   b (arm)  (kd )2 

(5.2.5)

Equation (5.2.5) is valid from y = 0 at the neutral axis to the extreme concrete compression fiber, where y = kd . Proceeding from the extreme compressive fiber to the neutral axis,

13



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5 . 3   C o m bined N or m al and S hear S tresses

the differential horizontal force C2 – C1 increases to a maximum. Thus at the neutral axis

( y = 0), Eq. (5.2.5) gives the maximum shear stress v=



V b(arm)

(5.2.6)

The horizontal shear stress v of Eq. (5.2.6) is also the vertical shear stress, because shear stresses on two perpendicular planes must be equal. For a homogeneous, linear elastic rectangular beam of overall depth h, the arm is equal to (2 / 3) h, which leads to a maximum shear stress at the location of the neutral axis equal to 1.5 times the average shear stress V / ( bh ). For strength design, the shear force V is the factored shear force and the distribution of flexural stress is no longer linear in the compression region. The shear stress v of Eq. (5.2.6) is then not the actual shear stress. Distribution of shear stresses over the depth of a cracked section is difficult to estimate. However, the relative magnitude obtained from Eq. (5.2.6) is still a measure of the potential for inclined cracking. In the ACI Code, the denominator in Eq. (5.2.6) is thus replaced by b times the full value of the effective depth d.

5.3 COMBINED NORMAL AND SHEAR STRESSES If at a certain point below the neutral axis in a homogeneous beam the tensile stress is ft and the shear stress is v [Fig. 5.3.1(a)], the principal tensile stress ft ( max ) is given by 2

1 1  ft (max) = ft +  ft  + v 2 2  2



(5.3.1)

The derivation of Eq. (5.3.1) is available in most textbooks on mechanics of materials, but because of its importance, it is shown here, as well. Using equilibrium of the forces acting on the free body in the directions of ft ′ and v ′ shown in Fig. 5.3.1(b) and calling the width of the beam b,  bdz   bdz   bdz  cos α + v(bdz ) cos α + v  sin α ft ′  = ft   sin α   tan α   tan α   bdz   bdz   bdz  v′  sin α + v(bdz )sin α − v  cos α = ft   sin α   tan α   tan α 

from which

1 ft (1 + cos 2α ) + v sin 2α 2 1 v ′ = ft sin 2α − v cos 2α 2 ft ′ =



dz v

v

ft

ft

ft v

v

dz tan α

v

ft

v’

α

v

αmax

ft’

v

(a)

(b)

Figure 5.3.1  Stress condition on an infinitesimal element block.

αmax

ft (max)

dz sin α

v

v’ = 0

(c)

14

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ft (max)

Figure 5.3.2  Directions of potential cracks in a simply supported beam under uniform loading.

The value of the principal angle α max (i.e., the angle α that makes ft ′ maximum and at the same time makes v′ zero) may be found either by differentiating the expression for ft ′ with respect to α or by setting the expression for v′ to zero. This leads to the following expression, for which the principal angle can be determined if the shear stress v and normal stress ft are known.



tan 2α max =

v 1 f 2 t

(5.3.2)

Equation (5.3.1) may then be obtained by substituting Eq. (5.3.2) into the expression for ft ′. The principal tensile stress ft ( max ), which is at an angle α max with the beam axis, is at least as large as either ft or v. It is nearly equal to the longitudinal tensile stress ft if the shear stress v is small (i.e., near the extreme tension fiber of a beam) and its direction is nearly horizontal. It is nearly equal to the shear stress v if the longitudinal tensile stress ft is small, and its direction is nearly at 45° with the beam axis. Since concrete is weak in tension, these principal tensile stresses are undoubtedly correlated to inclined cracking, as shown in Fig. 5.3.2.

5.4 BEHAVIOR OF BEAMS WITHOUT SHEAR REINFORCEMENT As discussed in Section 5.3, diagonal tension caused by shear stress on a beam may result in the formation of inclined cracks. Typically the propagation and width of inclined cracks are controlled by using transverse reinforcement (known as “shear reinforcement”) in the form of closed or U-​shaped stirrups; normally the stirrups are placed perpendicular to the beam longitudinal axis to enclose the main longitudinal reinforcement along the faces of the beam (see Fig. 5.6.1 in Section 5.6). Before discussing the shear behavior and strength of beams with shear reinforcement, however, it is necessary to have a clear understanding of shear behavior of beams without transverse reinforcement. Inclined cracking in the webs of reinforced or prestressed concrete beams may develop either in the absence of flexural cracks or as an extension of a previously developed flexural crack. An inclined crack occurring in a beam with no cracks due to flexure is known as a web-​shear crack, as shown in Fig. 5.4.1(a). An inclined crack originating at the top of and becoming an extension to a previously existing flexural crack is known as a flexure-​shear crack, as shown in Fig. 5.4.1(b). In this case, the critical flexural crack is referred to as the “initiating crack.” Flexure-​shear cracks are the usual type found in both reinforced and prestressed concrete. Web-​shear cracks are relatively rare in nonprestressed beams. These cracks occur in thin-​webbed I-​shaped beams having relatively large flanges, common only in prestressed concrete construction. This is discussed further in Chapter 20, which is devoted entirely to prestressed concrete. Web-​shear cracks may also occur near the inflection points or bar cutoff points on continuous reinforced concrete beams subjected to axial tension [5.15, 5.16].

15



5 . 4   B eha v ior of   B ea m s W itho u t S hear R einforce m ent

115

A

A (a) Web-shear crack

Section A–A

Flexure-shear crack

Initiating crack Secondary crack (b) Flexure-shear crack

Figure 5.4.1  Types of inclined cracks. (From Bresler and MacGregor [5.3].) Va = aggregate interlock (interface shear)

Arm, ha

C Vcz = shear resistance of uncracked concrete

T Vd = dowel force w z

Figure 5.4.2  Shear resistance components after formation of inclined crack.

Shear Resisting Mechanisms The transfer of shear in reinforced concrete members without shear reinforcement occurs by a combination of the following mechanisms [5.4], as shown in Fig. 5.4.2: 1. Shear resistance of the uncracked concrete in the compression zone, Vcz . 2. Aggregate interlock (or interface shear transfer) force Va, tangentially along a crack [5.18–​5.20], and similar to a frictional force due to irregular interlocking of the aggregates along the rough concrete surfaces on each side of the crack. 3. Dowel action, Vd, the resistance of the longitudinal reinforcement to a transverse force [5.21–​5.26]. 4. Arch action [see later: Fig. 5.4.5(a)] on relatively deep beams. The ability of a beam to carry additional load after an inclined crack has formed depends on whether the portion of shear formerly carried by uncracked concrete can be redistributed across the inclined crack and the uncracked compression zone. Mechanisms 1 through 4 listed above all participate in the redistribution, the success of which determines the shear strength and the degree of seriousness of the crack formation. For rectangular beams without shear reinforcement, it is reported [5.4] that after an inclined crack has formed, the proportion of the shear transferred by the various mechanisms is roughly as follows: 15 to 25% by dowel action, 20 to 40% by the uncracked concrete compression zone, and 30 to 50% by aggregate interlock or interface shear transfer. As diagonal cracks increase in length and width, however, the relative contributions of these mechanisms will change. For example, the contribution of aggregate interlock to

16

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C hapter   5     S hear S trength and D esign for   S hear

the beam shear strength will decrease as diagonal cracks widen with increasing load. This reduction in shear resistance contributed by aggregate interlock (and also by dowel action) with increasing deformation may lead to a shear failure even after the beam has reached its flexural strength, as progressive flexural yielding causes cracks to widen. It is therefore critical that a beam be capable not only of reaching its flexural strength, but also of exhibiting substantial flexural deformations prior to the development of a shear failure should the beam ultimately fail in shear. Inclined cracks begin and grow depending on the relative magnitudes of shear stress v and flexural stress ft , as shown in Section 5.3. These controlling stresses may be expressed as v = k1

V bd

(5.4.1a)

ft = k2

M bd 2

(5.4.1b)



where k1 and k2 are proportionality constants. The discussion in Section 5.2 may serve to justify the shear stress as proportional to V / bd ; in the presentation of Chapters 3 and 4, the flexural capacity M was related to bd 2 through the use of a coefficient of resistance R, which has the same units as the flexural stress. From Section 5.3 one may note that the principal tensile stress is a function of the ratio ft / v. Another expression for ft /v may be obtained from Eqs. (5.4.1); thus ft k2 M M = = k3 v k1 Vd Vd



(5.4.2)

For a simply supported beam symmetrically loaded with two equal concentrated loads (see Fig. 5.4.3), the ratio M / V may be thought of as the distance a over which the shear is constant. This distance a is known as the shear span. For the general case where the shear is continually varying, the “shear span” may be expressed as a=



M V

(5.4.3)

which varies along the length of the beam. Using Eq. (5.4.3) in Eq. (5.4.2), the ratio ft / v becomes ft  a = k3    d v



(5.4.4)

The ratio of shear span to effective depth a / d has also been shown experimentally to be a highly influential factor in establishing shear strength [5.3, 5.4, 5,17, 5.27]. When factors

a

P

P

a

V = +P

V = –P

M = Va

Figure 5.4.3  Definition of shear span a.

17



5 . 4   B eha v ior of   B ea m s W itho u t S hear R einforce m ent

117

other than a / d are kept constant, shear capacity tends to decrease with an increase in a / d ratio. The variation in shear capacity, expressed in terms of the failure moment Va, with a / d ratio, may be illustrated by Fig. 5.4.4 using the results for rectangular beams. Member depth has also been found to significantly affect shear strength of flexural members without shear reinforcement. Results from tests of large flexural members [5.28–​5.30] have shown that an increase in member depth results in a decrease in shear stress at failure. This phenomenon has often been attributed to an increase in crack spacing with member depth, which results in wider cracks and reduced aggregate interlock [5.10, 5.29, 5.30]. Thus, this so-​called shear size effect has been linked not only to member depth, but also to aggregate size. Shear size effect has also been reported to be more pronounced in members constructed with high-​strength concrete [5.29]. With the use of shear reinforcement, however, shear size effect is greatly diminished or even eliminated, as transverse steel contributes to a better cracking distribution and controls the growth of diagonal cracks.

Failure Modes From Fig. 5.4.4, four general categories of failure may be established: (1) deep beams with a /d < 1; (2) short beams with a /d ratios from 1 to about 2 1 2 , in which the shear strength exceeds the inclined cracking capacity; (3) usual beams of intermediate length having a /d ratios from about 2 1 2 to 6, in which the shear strength equals the inclined cracking strength; and (4) long beams with a /d ratios greater than 6, whose flexural strength is typically less than their shear strength. Deep Beams, a/d ≤ 1 For a deep beam, shear stress has the predominant effect. After inclined cracking has occured, a simply supported beam such as that shown in Fig. 5.4.5(a) tends to behave like a tied arch wherein the load is carried by direct compression extending around the shaded area and by tension in the longitudinal steel. Once the shear-​related crack develops, the beam transforms quickly into a tied ​arch which exhibits considerable reserve capacity. Possible modes of failure are indicated in Fig. 5.4.5(b) [5.3]; they are (1) an anchorage failure (i.e., pullout of the tension reinforcement has occurred at the support) (2) a crushing failure at the reactions; (3) a “flexural failure” arising from either crushing of concrete near the top of the arch or yielding of the tension reinforcement; and (4) failure of the arch rib due to an eccentricity of the arch thrust, resulting in either a tension crack

Compression arch

Flexural strength

Shear-compression strength Failure moment = Va

Inclined cracking strength, Vc

Tension tie (a) Arch action

Deep beams

Shear-tension and shear-compression failures

Flexural failures Diagonal tension failures

0

1

2

3

4

5

6

3

4 1

2

5 3

1. Anchorage failure 2. Bearing failure 3. Flexural failure 4. & 5. Arch-rib failure

7

a/d

Figure 5.4.4  Variation in shear strength with a/​d for rectangular beams. (Adapted from Bresler and MacGregor [5.3].)

(b) Types of failure

Figure 5.4.5  Modes of failure in deep beams, a/d ≤ 1.0. (Adapted from Bresler and MacGregor [5.3].)

18

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over the support at region 4 of Fig. 5.4.5(b) or crushing of concrete on the underside of the rib at region 5. Short Beams, 1 < a/​d ≤ 2 12 Like deep beams, short beams have a shear strength that exceeds the inclined cracking strength. After an inclined crack develops, the crack extends further into the compression zone as the load increases. It may also propagate as a secondary crack toward the tension reinforcement and then progress horizontally along that reinforcement. Failure eventually results, either by (1)  an anchorage failure of the tension reinforcement, called a “shear-​ tension” failure [Fig. 5.4.6(a)]; or (2) a crushing failure in the concrete near the compression face, called a “shear-​compression” failure [Fig. 5.4.6(b)]. Intermediate Length Beams, 2 12 < a/​d ≤ 6 For intermediate length beams, vertical flexural cracks are the first to form, followed by the inclined flexure-​shear cracks. At the beginning, several flexural cracks tend to incline, creating beam segments between cracks. Kani [5.31] proposed a “comb” analogy, where the “teeth” of the comb are the concrete pieces in between flexural cracks (Fig. 5.4.7). He proposed that, as a result of the increasing number of flexural cracks, when the root of the “tooth” is so reduced in size that it becomes unable to carry the moment arising from ΔT, it breaks to form the inclined flexure-​shear crack. At the sudden occurrence of the inclined crack, the beam is not able to redistribute the load, as is the case in beams with smaller a/​d ratio. In other words, the formation of the inclined crack represents the shear strength of beams in this category, for which the term “diagonal tension failure” has been given [5.3]. This is the most common category for beam design. Moreover, for beams in this category, a shear failure may develop before or after the beam has reached its flexural strength. As discussed above, widening of cracks caused by flexural yielding may lead to a shear failure soon after flexural yielding begins due to a decrease in the shear resistance provided by aggregate interlock and dowel action. Long Beams, a /​d > 6 The failure of long beams starts with yielding of the tension reinforcement and ends by crushing of the concrete at the section of maximum bending moment. In addition to the nearly vertical flexural cracks at the section of maximum bending moment, prior to failure, slightly inclined (from the vertical) cracks may be present between the support and the section of maximum bending moment. Nevertheless, the strength of the beam is entirely dependent on the magnitude of the maximum bending moment, as the shear force required for the beam to reach its flexural strength is less than its shear capacity. In summary, shear tends to cause inclined cracks. When no such cracks form before the nominal flexural strength is reached, the effect of shear is often negligible, although it is possible for a beam to fail in shear after the initiation of flexural yielding, since aggregate interlock and dowel action decrease with increasing deformation. Beams may fail on formation of an inclined crack, as for the so-​called diagonal tension failures when a/​d is between about 2 1 2 and 6, or there may be considerable reserve strength, as is the case for shorter beams. Experimental results also indicate a reduction in shear strength with increased member depth. For beams with reserve capacity to maintain a state of equilibrium, forces must be redistributed after an inclined crack has formed. For the design of all but deep beams, however, the shear strength of beams without shear reinforcement is assumed to be reached when the inclined crack forms.

19



5.5  STRENGTH OF BEAMS WITHOUT SHEAR REINFORCEMENT

119

Crushing of concrete

Loss of bond due to crack (a) Shear-tension failure

(b) Shear-compression failure

Figure 5.4.6  Typical shear failures in short beams, a / ​d = 1 to 2 1 2 . (Adapted from Bresler and MacGregor [5.3].)

Tooth T

Figure 5.4.7  “Diagonal tension failure” or “tooth-​cracking failure” as suggested by Kani [5.31] on intermediate length beams; a/​d about 2 1 2  to 6.

5.5 SHEAR STRENGTH OF BEAMS WITHOUT SHEAR REINFORCEMENT—​A CI APPROACH Except for deep beams (see Section 5.14), the strength at which an inclined crack forms [usually a flexure-​shear crack as in Fig. 5.4.1(b)] is taken to be the shear strength of a beam without shear reinforcement, according to the intent of the ACI Code. After establishing on a rational basis the variables involved, the relationship between them was statistically determined from test results [5.17]. It is assumed that the strength is reached when the principal tensile stress in Eq. (5.3.1) reaches the tensile strength of concrete, which is assumed to be proportional to fc′ . Although the exact distributions of the flexural and shear stresses in a cross section are not known, it may be assumed that flexural tensile stress ft varies as Ec / Es times the tensile stress in the reinforcement and that v varies as the average shear stress. Assume that Ec is also proportional to fc′ . As already shown by Eq. (5.4.1a), the average shear stress when the inclined crack forms may be written v = k1



V bd

(5.5.1)

The stress in the steel is proportional to M /( As d ), and the tensile stress ft in the concrete then becomes

ft ∝

M fc′ Ec f s EM M  fc′  ∝ c ∝ ∝ 2  Es Es dAs Es dAs bd  ρEs 

The above expression for ft may be written as

ft =

k4  fc′  M   Es  ρ  bd 2

(5.5.2)

in which k4 is a dimensionless constant and Es has a known value. Also, the tensile strength of concrete may be represented by

ft (max) = k5 fc′

(5.5.3)

120

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C hapter   5     S hear S trength and D esign for   S hear

Substituting Eqs. (5.5.1) to (5.5.3) into the principal stress equation, Eq. (5.3.1),



2    1 k M fc′  V  1 k4 M fc′ 2  4 k + + fc′ =   1  bd  2 Es Vd ρ  2 Es Vd ρ   

k5

or V

bd fc′

= k5

2    1 k4 M fc′ +  1 k4 M fc′  + k 2    1   2 E Vd ρ 2 Es Vd ρ   s  

(5.5.4)

In Eq. (5.5.4) the variables are observed to be V / (bd fc′) and M fc′ / ( Es ρVd ). It may be noted that these two variables are nondimensional quantities because fc′ has units of force per unit area. In the statistical study, the shear force V was defined as that causing the critical inclined crack and M as the corresponding moment at the location of diagonal cracking which, for shear spans longer than 2d, was taken at a distance d from the section of maximum moment. On the basis of 440 tests [5.17], as shown in Fig. 5.5.1, the relationship between these two variables, on using the numerical value of Es , was obtained as follows: V

bd fc′

= 1.9 + 2500

ρ Vd M fc′

≤ 3.5

(5.5.5)

Note that in Fig. 5.5.1, V on the left-hand side is taken as the nominal shear strength, while V and M on the right-hand side are taken as the applied shear and corresponding moment at the critical section. Equation (5.5.5) is generally considered [5.4, 5.28, 5.31] to be acceptable for predicting the flexure-​shear cracking load, particularly for M / (Vd ) (i.e., shear span/​depth) ratios of about 2 1 2 to 6, with considerable conservatism for lower M / (Vd ) values. Thus the favorable factors for the shear strength of beams without shear reinforcement are a high percentage ρ of longitudinal reinforcement and a high ratio of Vd to M, that is, a low a / d ratio. Since 1963, the ACI Code has accepted the relationship of Eq. (5.5.5) as the shear (inclined cracking) strength of beams without shear reinforcement. Thus defining Vc as the nominal strength of such beams, using the web width1 bw for b and adding a multiplier λ to fc′ to account for the use of lightweight concrete (see Section 1.8), Eq. (5.5.5) becomes

 ρ V d Vc =  1.9λ fc′ + 2500 w u  bw d ≤ 3.5λ fc′bw d Mu  

(5.5.6)2

which is given in Table  22.5.5.1 of the ACI Code. For normal weight concrete, λ = 1.0. Values for λ when lightweight concrete is used are discussed shortly in this section (see subsection on lightweight concrete). Note that in Eq. (5.5.6), the factored shear force Vu and the factored moment Mu acting concurrently with Vu at the section under consideration are used. Also note that the reinforcement ratio ρw = As / ( bw d ) is used in the ACI Code formula, where bw is the web width for a T-​section rather than the flange width. The ACI Code defines bw as “web width.” For tapered webs, however, such a definition is unclear. In general, when the web is subject to flexural tension, the “average web width” should be used as bw in Eq. (5.5.6). On the other hand, when the web is subject to flexural compression (as for negative moment regions) the use of average web width may be unsafe [5.32]. For such negative moment regions, the use of a value lower than the average, perhaps the minimum, web width, is more appropriate. The value of Vu d / Mu in Eq. (5.5.6) shall not be taken greater than 1.0 except when axial compression is present [see Eq. (5.13.2), where a modified moment replaces Mu ], which has the effect of limiting Vc at and near the points of inflection. 1  The term “web” refers to the width dimension of a beam at its neutral axis. For T-​shaped beams the flange width b is distinguished from the stem, or web, width bw. 2  For SI, ACI 318-​14M, with fc′ in MPa, gives  ρ V d Vc =  0.16 λ fc′ + 17 w u  bw d ≤ 0.29λ fc′ bw d (5.5.6)  M   u

12



5.5  STRENGTH OF BEAMS WITHOUT SHEAR REINFORCEMENT

121

6.0 Vn = Vc 5.0

V bwd fc’

4.0 3.0

Vn =2 bw d fc’

2.0

Vn ρ Vu d = 1.9 + 2500 ≤ 3.5 bw d fc’ Mu fc’

1.0

Inverse scale 0.2

0.4

0.8

0.6

(

ρ Vu d 1000 Mu fc’

1.0

1.5 2.0

5

)

Figure 5.5.1  Comparison of shear stresses calculated using Eq. (5.5.5), with test results (shown as dots) for beams without shear reinforcement, that is, the nominal shear strength Vn = Vc. (Adapted from ACI-​ASCE Committee 326 Report [5.17].)

In lieu of Eq. (5.5.6), the ACI Code (ACI-​22.5.5.1) allows the use of a simplified expression, as follows:

Vc = 2 λ fc′ bw d

(5.5.7)

Note that except for a few experimental results, this equation represents a lower bound of the test data shown in Fig. 5.5.1. More recent work [5.29, 5.30], however, has shown that large beams without web reinforcement may fail at shear stresses significantly lower than those obtained using either Eq. (5.5.6) or (5.5.7).

Continuous Beams The application of Eq. (5.5.6) for continuous beams has been subject to question [5.4]. The​ strut action on a continuous beam is shown in Fig. 5.5.2. The analogy in Fig. 5.4.3 that M / V equals the shear span a implies that at the point of zero moment there is a support to accommodate a diagonal strut [see Fig. 5.4.5(a)]. For a continuous beam as in Fig. 5.5.2, the distances a1 and a2 are analogous to a in Fig. 5.4.3; however, there is no support to resist a ​strut reaction at the inflection point. ACI-​ASCE Committee 426 [5.16] recommended that Eq. (5.5.6) [ACI Table 22.5.5.1] no longer be used in such situations; instead Vc should be taken as 2λ  fc′ bwd [Eq. (5.5.7)].

Lightweight Concrete It has been shown [5.33–​5.35] that the same general relationships as those for normal-​ weight concrete are valid for lightweight concrete when a multiplier λ , typically less than 1.0, is applied to fc′ [see Eq. (5.5.6)]. This multiplier accounts for the fact that for lightweight concrete, the tensile strength fct based on the split-​cylinder test (see Section 1.8) provides a better correlation with inclined cracking strength than does fc′ . Tests have shown [5.33] that inclined cracking strength for lightweight concrete varies from about 60 to 100% of the values for normal-​weight concrete of the same nominal compressive strength, depending on the particular aggregates used. Studies [5.35] have shown that using 0.75 to 0.85 of the tensile strength–​related term fc′ for normal-​weight concrete in shear strength equations is a reasonable and generally conservative approach. Thus, alternatively to the use of results from split-cylinder tests, λ may be taken as 0.75 for all-​ lightweight concrete and 0.85 for sand-​lightweight concrete (ACI-​19.2.4.2) (see Section).

12

122

C hapter   5     S hear S trength and D esign for   S hear

t

stru

a1 = Mneg /V

a2 = Mpos /V Mpos +

–

Point of inflection

Mneg

Figure 5.5.2  Crack pattern, diagonal strut, and bending moment diagram near support on a continuous beam.

High-​Strength Concrete In response to the increasing use of concrete having fc′ greater than 8000 psi, studies [5.36–​5.43] have indicated that shear strength does not continue to increase in proportion to fc′ . Though a definitive relationship for the shear strength of beams using high-​ strength concrete has not been established, the ACI Code Committee (ACI-​22.5.3.1) limits fc′ to 100 psi in shear strength calculations of members with fc′ above 10,000 psi. However, values of fc′ greater than 100 psi are permitted in computing Vc as per ACI-​ 22.5.3.2 when the area Av of shear reinforcement (see Section 5.8) satisfies the minimum shear (and torsion per ACI-​9.6.4.2; see Chapter 18) reinforcement required in ACI-​9.6.3.3 [see Section 5.10, Eq. (5.10.8)].

Size Effect As mentioned in Section 5.4, shear stress at failure has been reported to decrease with an increase in member depth and a decrease in aggregate size. This size effect, however, is not accounted for in Eqs. (5.5.6) and (5.5.7). Research [5.29, 5.30] has shown that the use of these equations may lead to unconservative strength predictions in large flexural members without shear reinforcement. It is thus recommended that at least a minimum amount of shear reinforcement (see Section 5.10) always be used in members with depths greater than 24 in.

5.6 FUNCTION OF WEB REINFORCEMENT The types of shear reinforcement recognized for reinforced nonprestressed concrete by the ACI Code (ACI-​22.5.10.5.1) are (1) vertical stirrups (Fig. 5.6.1), ties or hoops perpendicular to the longitudinal axis of the member; (2) welded wire fabric (see Section 1.13) with wires located perpendicular to the axis of the member; and (3) spiral reinforcement. Inclined stirrups, where the plane of the stirrup makes an angle of 45° or more with the longitudinal reinforcement, are also permitted (ACI-​22.5.10.5.2), as well as longitudinal tension bars bent toward the compression zone of the member so that the axis of the bent portion makes an angle of 30° or more with the axis of the longitudinal portion (22.5.10.6.1). Vertical stirrups, ties, or hoops (as opposed to inclined stirrups and bent bars) are used almost exclusively for shear reinforcement in current design practice. The most generally accepted model for the behavior of reinforced concrete beams containing shear reinforcement is the truss model, originated by Ritter [5.44] and Mörsch [5.1]. The current thinking related to the truss model is well presented by Schlaich, Schäfer, and Jennewein [5.9]. In a simply supported steel truss, as shown in Fig. 5.6.2(b), the upper and

123



123

5 . 6   F u nction of   W eb R einforce m ent A

Stirrup

Vertical stirrups

Section A–A

A

Single-leg stirrup

U stirrup with 90° hooks

U stirrup with 135° hooks

Closed stirrup with 90° hooks

Closed stirrup with 135° hooks

Figure 5.6.1  Vertical stirrups as shear reinforcement.

(a) Reinforced concrete beam with vertical shear reinforcement

(b) A steel truss Concrete

Web reinforcement

(c) Truss action in a reinforced concrete beam

Figure 5.6.2  Truss analogy.

lower chords are in compression and tension, respectively, and the diagonal and vertical members, called web members, are in compression and tension. In the reinforced concrete beam of Fig. 5.6.2(c), the concrete performs the task of carrying the compressive forces while steel reinforcement is used to resist tensile forces. Since the vertical transverse reinforcement (stirrups) acts in a reinforced concrete beam like the tension web members of a steel truss, the term “web reinforcement” is often used for shear reinforcement. The shear reinforcement wraps around the longitudinal tension reinforcement and must be anchored in the compression zone, usually by hooking it around the top longitudinal bars. (These longitudinal bars either are compression reinforcement or are provided solely to hold in place and anchor the shear reinforcement.) While shear reinforcement provides shear strength, its contribution to the strength occurs only after inclined cracks form. Prior to the formation of inclined cracks, the concrete performs the task of carrying the shear. However, shear reinforcement is necessary to allow a redistribution of internal forces across any inclined crack that may form, and thus prevent a sudden failure upon formation of the crack.

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C hapter   5     S hear S trength and D esign for   S hear

stirrups

Va = aggregate interlock (interface shear)

s

Fv dowel force = Vd

Fv

Fv

Arm, ha

C Vcz = shear resistance of uncracked concrete

T Fv = Avfv w

z

Figure 5.6.3  Shear resisted by stirrups crossed by an inclined crack. Splitting along longitudinal bars Internal resisting shear

124

Vcz = shear carried by concrete

Loss of interface shear transfer

Vd = dowel action

Va = aggregate interlock (or interface shear transfer) Vs = shear carried by web reinforcement Vc

Vs Vs

Stirrups yield Failure Applied shear

Vc Inclined crack forms Flexural cracking occurs

Figure 5.6.4  Distribution of internal shears in beams with shear reinforcement. (Adapted from ACI-​ASCE Committee 426 Report [5.4].)

The role of web reinforcement in resisting shear is also illustrated in Fig. 5.6.3, which is the same as Fig. 5.4.2, except for the presence of stirrups. In addition to the contributions to shear strength assigned to the concrete (i.e., Vcz , Va, and Vd), the stirrups crossing the inclined crack contribute a resistance Vs equal to the number of stirrups intersected by the crack (w / s ) times the force in each stirrup (Fv = Av fv , where Av is the cross-​sectional area of all legs in each stirrup and fv is the stress in the stirrup). Therefore, shear reinforcement has several functions that combine to allow a redistribution of the internal forces upon the formation of an inclined crack. The primary functions [5.4] (see Figs. 5.6.3 and 5.6.4) are (1) to carry part of the shear Vs ; (2) to restrict the growth of the inclined crack and thus help maintain aggregate interlock (also known as interface shear transfer) strength Va; and (3) to tie the longitudinal bars in place and thereby increase their strength Vd to resist transverse forces (known as dowel action). In addition, dowel action on the stirrups may transfer a small force across a crack, and the confining action of the stirrups on the compression concrete may slightly increase its strength. If the amount of shear reinforcement is too little, it will yield immediately at the formation of an inclined crack and the beam will fail without warning. If, on the other hand, the amount of shear reinforcement is too large, a web-​crushing failure will occur prior to yielding of the transverse reinforcement. The ideal amount of shear reinforcement should be such that the shear reinforcement and the compression zone of the beam each continues to carry increasing shear after the formation of the inclined crack, allowing the shear reinforcement to yield and providing some warning prior to shear failure.

125



5 . 7   T r u ss Model for   R einforced C oncrete   B ea m s

125

5.7 TRUSS MODEL FOR REINFORCED CONCRETE BEAMS The reinforced concrete beam shown in Fig. 5.7.1(a) contains vertical stirrups spaced at a constant spacing s. The behavior of a cracked reinforced concrete beam can be represented, in its simplest form, by the truss model shown in Fig. 5.7.1(b). In this model, the diagonals, such as gd, carry compressive forces and are represented by concrete struts. The top chord, such as cd, would carry the internal flexural compressive force. The bottom chord (i.e., the main tension reinforcement) would carry the internal flexural tensile force. The vertical members, such as gc, represent the stirrups and carry tensile forces. Rather than representing a single stirrup, each vertical member represents stirrups distributed over the tributary width of the vertical member [i.e., w in Fig. 5.7.1(b)]. Note that the contributions to shear strength (see Fig. 5.6.4) from Vcz , Va and from dowel action in the longitudinal reinforcement Vd are ignored in the truss model. In other words, the shear resistance is assumed to be provided entirely by the stirrups. This drawback in the use of truss models for determining shear strength of beams had already been recognized in the 1920s by Slater, Lord, and Zipprodt [5.45] and Richart [5.46], who proposed the addition of a shear contribution as a function of the concrete compressive strength fc′ to the strength obtained through the use of a truss model. Figure 5.7.1(c) shows a refined truss model for the same beam, except that the vertical member spacing has been reduced to s. In this case, each vertical member represents a single stirrup. In this model, it is noted that the diagonals, such as fc, f ′c ′, and f ′′c ′′, do not have to extend from the top of one vertical member to the bottom of the next vertical member. This idealization stems from the observation that, in reality, the shear is resisted by a continuous field of diagonal compressive stresses, often called a compression field, rather than by discrete diagonal struts. The angle of the compression field depends primarily on the size and spacing of the stirrups, amount of longitudinal reinforcement, and axial load, but it is often taken between 30 and 45°. In the vicinity of concentrated loads [under the concentrated load and at supports in Fig.  5.7.1(c)], the traditional beam theory, which assumes that plane sections remain plane, is not applicable, and the shear transfer mechanism cannot be assumed to be a continuous compression field. In such regions, the load will spread or “fan out” to more than one vertical tension member to equilibrate the force. In the model, “fanning out” of the concentrated load at midspan is represented by diagonals gd, jd, kd, and id. Similarly, diagonals eb′, eb, and ec′ represent the “fanning out” of the reaction force at the support. Once a truss model has been conceived, the truss member forces can be obtained from statics. The member forces in the diagonals, verticals, and top and bottom chords are then checked against prescribed limits. For example, the compressive stresses in the concrete struts (top chord and diagonals in Fig. 5.7.1) are limited by an effective concrete stress fc which is less than fc′. The value of fc depends on the stress field in the region where the member is located. For instance, a diagonal concrete strut that crosses a stirrup in a region with shear cracks has a lower strength than a concrete strut in the compression zone of the beam. The strength of the vertical members and bottom chord is usually taken as the area of the reinforcement times its yield strength. It is noted that truss models can be used to design not only the shear reinforcement, but also the longitudinal reinforcement. Marti [5.47] provides several practical examples of the use of truss models in the design of reinforced concrete beams. Strut-​and-​tie models, which are in effect truss models, and their application to the design of structural members, are discussed in Chapter 14. The truss analogy has served as the basis for the development of other approaches, such as the variable-​angle truss models of Marti [5.47] and Nielsen [5.48], the compression field theory of Mitchell and Collins [5.49], the modified compression field theory of Vecchio and Collins [5.8], and the softened truss model of Hsu [5.50]. The zone in the vicinity of a concentrated load, or at an abrupt discontinuity in the member (such as an abrupt change in member depth) is called a D-​region, where D stands for discontinuity, disturbance, or detail [5.9]. In such regions, the plane section theory is not

126

126

C hapter   5     S hear S trength and D esign for   S hear

P

Stirrups

Main tension steel P 2

s

P 2

(a) Reinforced concrete beam P a

c

b

1

d

arm α e

P 2

f

g

h

1

j

P 2

6 @ w = 6w (w = panel width) (b) Truss model for concrete beam P a

P 2

b’

e

f’

b

f

c’

f”

s

c

c”

g

j

d

h

k

P 2

i

(c) Concrete beam showing possible compression fields Requirement for main tension reinforcement with stirrups spaced at s

s n PL 4(arm)

o

m (d) Internal tensile or compressive force from bending moment, M/arm

Figure 5.7.1  Truss model for reinforced concrete beam.

applicable, and the true forces are not those obtained by shear and moment diagrams and first-​order elastic beam theory, particularly when the sections are cracked. In general, the reinforcement design of the D-​region has relied on “past experience” or “good practice.” However, truss models or strut-​and-​tie models have become accepted tools for the design of D-​regions. In the regions of a beam where the assumption of plane sections remaining plane after loading (the Bernoulli hypothesis) is appropriate, such as panel bc of Fig.  5.7.1(b), and

127



127

5 . 7   T r u ss Model for   R einforced C oncrete   B ea m s

therefore design for the forces at each section along the member is appropriate, the term “B-​region” is used (where B stands for beam or Bernoulli). The truss model also shows that the use of forces from the homogeneous beam bending moment diagram, such as mno in Fig. 5.7.1(d), to design the main tension reinforcement is not correct. Rather, since the shear reinforcement is provided at discrete intervals (stirrups at spacing s), the tensile force T in the main reinforcement obtained from a truss model corresponds to the stepped diagram of Fig. 5.7.1(d), with the tensile force being constant over the panel distance (i.e., the tensile force changes only at the location of a vertical member). Since the real situation is not a pin-​jointed truss, and the angle of the compression field may vary, the requirement for the main tension steel would actually be somewhere between the M/​arm [based on ordinary bending moment diagram) and the stepped diagram for a truss, as shown in Fig. 5.7.1(d)]. Shown in Fig. 5.7.2 is a segment along a beam, together with the internal forces acting on that segment. At the section x, the forces are Ccx (concrete compression chord), Cwx (the diagonal “strut”), and Tsx (the tensile force carried by the main tension steel). Equilibrium in the vertical direction requires that the shear force Vx be resisted by the vertical component of the diagonal strut. Thus Vx = Cwx sin θ

(5.7.1)

The horizontal forces Tsx and Ccx can be calculated as

Tsx =

Mx (a − z ) + Vx arm arm

(5.7.2)

Ccx =

Mx Vz − x arm arm

(5.7.3)

where z is the distance measured from the vertical member of the truss to section x (see Fig. 5.7.2). The important observation is that the truss model recognizes that the force in the tension steel is often larger than that obtained from the bending moment diagram. In general, this effect has been accounted for in ACI Code design by extending the longitudinal bars a distance d beyond the point where they are no longer required. This is discussed further in Chapter 6, which is devoted to determining bar lengths and the distance along the span that bars must extend. However, the truss model also shows that the shear reinforcement behavior and the forces in longitudinal bars cannot be entirely separated. The truss model also illustrates the importance of reinforcement at the “joints” of the truss. The behavior of the stirrups, their proper detailing, and their effectiveness have been studied by Hsiung and Frantz [5.51], Johnson and Ramirez [5.52], Anderson and Ramirez [5.53], Mphonde [5.54], Mphonde and Frantz [5.55], and Belarbi and Hsu [5.56].

(arm) cos θ Ccx1

Mx

Mx1

Ccx Cwx

Vx1 arm

Cwx1 θ

θ Vx

Tsx

Tsx1 a = (arm) cot θ

a x1

x

x1

x z

Figure 5.7.2  Beam actions and internal forces in truss model for equilibrium of beam element.

128

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C hapter   5     S hear S trength and D esign for   S hear

5.8 SHEAR STRENGTH OF BEAMS WITH SHEAR REINFORCEMENT—​A CI APPROACH The traditional ACI approach to design for shear strength is to consider the total nominal shear strength Vn as the sum of two parts, Vn = Vc + Vs (5.8.1)



in which Vn is the nominal shear strength; Vc is the shear strength of the beam attributable to the concrete, and Vs is the shear strength attributable to the shear reinforcement (see Figs. 5.6.3 and 5.6.4, and Section 5.5). Note that regardless of whether shear reinforcement is provided, the ACI Code uses the same expression for Vc . As discussed earlier, Vc is intended to represent the shear corresponding to inclined cracking in members without shear reinforcement. In the case of members with shear reinforcement, however, Vc is meant to account for the contribution to shear strength of the concrete compression zone, aggregate interlock, and dowel action once diagonal cracking has occurred. For simplicity, these two quantities are assumed to be equal. An expression for Vs may be developed from the truss analogy. Consider the truss model shown in Fig. 5.8.1, where the shear reinforcement (stirrups) is assumed to be inclined at an angle α with respect to the horizontal and spaced at a distance s. The angle θ of the diagonal struts is assumed as shown. Equilibrium of joint A of the truss shows that the vertical component of the tensile force developed in the stirrups must balance the shear force on the cross section. Assuming that the web reinforcement yields prior to crushing of the concrete (i.e., fv = f yt in Fig. 5.8.1, where f yt is the yield strength of the shear reinforcement), its contribution to shear strength can be obtained as Vs = Av f yt sin α



(5.8.2)

where Av is the area of the shear reinforcement within a distance s. From the geometry of the truss, s = arm(cot θ + cot α )

(5.8.3)

The force per unit length in a stirrup is Av f yt s

=

Vs sin α (arm)(cot θ + cot α )

(5.8.4)

Thus Vs =



Av f yt (arm)sin α (cot θ + cot α ) s



(5.8.5)

Since the early twentieth century, the angle θ of the diagonal struts has typically been assumed to be 45°, though in reality it will vary depending on the amount and spacing of

C Avfv α s

A

Vs arm α

θ s

Figure 5.8.1  Truss model for a beam with inclined stirrups.

T

129



5 . 9   D efor m ed S teel F ibers as   S hear R einforce m ent

129

the stirrups, the amount of longitudinal steel, and the axial load, if any. Assuming a 45° angle for the diagonal struts, Eq. (5.8.5) becomes

Vs =

Av f yt d (sin α + cos α ) s



(5.8.6)

where the arm of the internal couple has been approximated as the effective depth d. For a beam containing stirrups perpendicular to its longitudinal axis, that is, α = 90°, Vs =

Av f yt d s



(5.8.7)

Equation (5.8.7) can also be obtained by summing the forces in the stirrups crossed by a diagonal crack (Fig. 5.6.3) whose projection along the longitudinal axis of the beam (w in Fig. 5.6.3) is equal to d. The force in each stirrup is equal to Av f yt , while the number of stirrups crossed by a crack within a projection along the beam axis of d is equal to d / s.

5.9 DEFORMED STEEL FIBERS AS SHEAR REINFORCEMENT Steel reinforcement in the form of randomly oriented deformed fibers (see Section 1.14) has been shown to be effective in increasing shear strength of reinforced concrete beams [5.62]. Fibers contribute to shear resistance by transferring tensile stresses across diagonal cracks and restraining their widening, which in turn increases aggregate interlock. The use of deformed steel fibers as shear reinforcement was recognized in the ACI code for the first time in the 2008 edition. Although no shear strength equation was provided then or in the 2014 ACI Code, the contribution of fibers is implicitly recognized by allowing their use in lieu of minimum stirrup reinforcement in normal-​weight concrete beams with depths not greater than 24 in., fc′ ≤ 6000 psi, and Vu ≤ φ 2 fc′ bw d (ACI 9.6.3.1), where φ is the strength reduction factor for shear, equal to 0.75. Deformed steel fibers shall comply with ASTM A820 [1.104] and have a length-​ to-​diameter ratio between 50 and 100 (ACI- 26.4.1.5.1). Because many types of deformed steel fibers can be used in various dosages, performance criteria are provided in the ACI Code, based on the use of a third-​point load beam test as defined in ASTM 1609 [1.105]. To be considered as minimum shear reinforcement, fiber-​reinforced concrete must satisfy the following [ACI-​26.12.5.1 and ACI-​26.4.2.2(d)]: •​ Residual flexural strength at a midspan deflection of 1/​300 of the span length of at least 90% of the first peak strength. •​ Residual flexural strength at a midspan deflection of 1/​150 of the span length of at least 75% of the first peak strength. •​  A minimum fiber content of 100 lb/​cu yard of concrete. Residual flexural strength is determined assuming linear elastic, uncracked behavior (i.e., using an elastic modulus of bh 2 / 6). First peak flexural strength is not to be taken less than 7.5 fc′. The minimum required fiber content was determined based on the test results of fiber-reinforced concrete beams, which indicated a minimum shear failure stress of 3.6 fc′ in beams with at least 100 lb/​ cu yard (fiber volume fraction of 0.75%) of deformed fibers [5.62]. Similarly, the performance criteria based on ASTM 1609 tests were determined from performance of materials similar to those used in experimental investigations.

130

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C hapter   5     S hear S trength and D esign for   S hear

Shear failure of fiber-reinforced concrete beam without stirrups. (Photo by Gustavo J. Parra-Montesinos.)

5.10 ACI CODE DESIGN PROVISIONS FOR SHEAR In the ACI design method for shear, it is required that

φVn ≥ Vu (5.10.1)



where Vu is the factored shear force and φ Vn is the design strength in shear. The strength reduction factor φ is 0.75 for shear (ACI Table 21.2.1). The nominal shear strength Vn is Vn = Vc + Vs (5.10.2)



where Vc and Vs are the portions of the shear strength attributable to the concrete (see Section 5.5) and to the shear reinforcement (see Section 5.8), respectively.

Strength Vc Attributable to Concrete The development of the detailed equation, Eq. (5.5.6), for Vc was shown in Section 5.5 (see also Fig. 5.5.1). Since that equation is not easy to use as a design equation, and because of the wide scatter of test results, ACI-​22.5.5.1 permits using either of the following: (1)  For the simplified method,

Vc = 2 λ fc′ bw d

(5.10.3)3

where bw is the web width, d is the effective depth, and λ is 0.75 for all-​lightweight concrete and 0.85 for sand-​lightweight concrete (ACI-​19.2.4.2) (see Section 5.5.5, subsection on lightweight concrete and Section 1.8 for more information). From Fig. 5.5.1 the value obtained using Eq. (5.10.3) appears to be conservative; however, studies [5.3, 5.59–​5.61] have shown otherwise when ρw is below about 0.012. For values of ρw (that is, As / bw d ) lower than 0.012, the following, more conservative expression, is suggested [5.59], where the λ factor has been added to account for the use of lightweight concrete:



Vc = (0.8 + 100ρw )λ fc′ bw d

3  For SI, ACI 318-​14M, with fc′ in MPa, gives

Vc = 0.17λ fc′ bw d

(5.10.4)

(5.10.3) 

13



5 . 1 0   A C I C ode D esign P ro v isions for   S hear

131

(2)  For the more detailed method,



 ρ V d Vc =  1.9λ fc′ + 2500 w u  bw d ≤ 3.5λ fc′ bw d (5.10.5)4 Mu  

Equation (5.10.5) is identical to Eq. (5.5.6). The value of Vu d / ​Mu may not exceed 1.0, and Mu is the factored moment occurring simultaneously with the Vu for which shear strength is being provided. ACI–​ASCE Committee 426 recommended [5.16] against further use of Eq. (5.10.5) (see Section 5.5, subsection on continuous beams). The primary practical use of that equation is and has been to justify slightly larger stirrup spacings in high shear regions where spacings using Eq. (5.10.3) are small (say, less than 3 in.).

Strength Vs Attributable to Shear Reinforcement The contribution of shear reinforcement, as developed in Section 5.8, is (ACI-​22.5.10.5.4) Vs =



Av f yt d s

(sin α + cos α )

(5.10.6)

where fyt is the yield strength of the shear reinforcement. When vertical stirrups (i.e., stirrups perpendicular to the axis of the member) are used (α = 90°) (ACI-​22.5.10.5.3), Vs =



Av f yt d s



(5.10.7)

Lower and Upper Limits for Amount of Shear Reinforcement As noted in Section 5.6, the amount of shear reinforcement should be neither too low nor too high in order to prevent a sudden failure if diagonal cracking develops and to ensure yielding of the transverse steel (i.e., fv = f yt) when the failure strength in shear is reached. When shear reinforcement is used, a minimum area Av is required for nonprestressed beams (ACI-​9.6.3.3), slabs (ACI-​7.6.3.3), and columns (ACI-​10.6.2.2), equal to

min Av = 0.75 fc′

bw s 50bw s ≥ f yt f yt

(5.10.8)5

in which bw is the member web width and s is the spacing of the shear reinforcement in inches, f yt is the yield strength of the shear reinforcement in psi, and fc′ is the specified compressive strength of the concrete in psi. From Eq. (5.8.7) and using the lower limit of Eq. (5.10.8), this minimum amount corresponds to



Vs =

Av f yt d s

=

f yt d  bw s   50  = 50bw d s  f yt 

(5.10.9)6

4  For SI, ACI 318-​14M, with fc′ in MPa, gives



 ρ V d Vc =  0.16 λ fc′ + 17 w u  bw d ≤ 0.29λ fc′ bw d Mu  

(5.10.5) 

5  For SI, ACI 318-​14M, for Av in mm2, bw and s in mm, f y and fc′ in MPa, ACI-​9.6.3.3 gives



min Av = 0.062 fc′

bw s 0.35bw s ≥ f yt f yt

6  For SI, ACI 318-​14M, for bw and d in mm and Vs in newtons (N), gives Vs = (0.35MPa ) bw d

(5.10.8)  (5.10.9) 

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or in terms of nominal unit stress on area bw d , vs =



Vs b d = 50 w = 50 psi bw d bw d

(5.10.10)

To ensure that the amount of shear reinforcement is not too high, the designer will usually keep Vs below the following range: Vs ≤ 6 fc′ bw d to 8 fc′ bw d



ACI-​22.5.1.2 gives the upper limit for Vs as 8 fc′ bw d . Although this requirement is imposed on the shear reinforcement, it is intended to limit the maximum stress in the diagonal concrete struts, which may cause a sudden and premature crushing of the web, as well as to limit the width of diagonal cracks under service loads.

Design Categories and Requirements For design, the shear envelope for Vu is the starting point. From a practical point of view it is better to plot the Vu diagram rather than the “required Vn” diagram (equal to Vu / φ ). The basic diagrams used should be those of Mu and Vu owing to factored load; the φ  factor, which is different for moment and for shear, should not be included in the load-​related diagrams. The design for shear may be separated into the following categories: 1. Vu ≤ 0.5φ Vc (5.10.11) In this category, no shear reinforcement is required (ACI-​9.6.3.1). However, because of the potential for large members to exhibit shear strengths lower than those obtained from ACI Code strength equations (see Section 5.5), it is recommended that members with depths of 24 in. or greater be provided with at least minimum shear reinforcement satisfying Eq. 5.10.8 (ACI-​9.6.3.3). 2. 0.5 φ Vc ≤ Vu ≤ φ Vc (5.10.12) Minimum shear reinforcement is required in beams (ACI-​9.6.3.1) except for (a)  beams where the total depth does not exceed 10 in. (b) beams with depth less than the greater of 2 1 2 times the flange thickness for T-​shaped sections, or one-​half of the web width, but in all cases beam depth shall not exceed 24 in. (c) normal-​weight fiber-reinforced concrete beams with depth no greater than 24 in., fc′ ≤ 6000 psi, and Vu ≤ φ 2 fc′ bw d . The fiber-reinforced concrete used shall satisfy the performance criteria in ACI-​26.12.5.1 and contain at least 100 lb/​cu yard of deformed steel fibers (see Section 5.9). (d)  one-​way joist systems. (e)  one-​way slabs and footings (ACI-​7.6.3.1). For this category, the shear reinforcement must satisfy ACI-9.6.3.3 and ACI-​9.7.6.2.2, as follows:  required φVs = min φVs = φ 0.75 fc′ bw d ≥ φ (50)bw d



(5.10.13)7

and

maximum spacing s ≤

d ≤ 24 in. 2

7  For SI, ACI 318-​14M, for bw and d in mm and Vs in N, ACI-​9.7.6.2.2 gives



(

)

(

min φ Vs = φ 0.062 fc′ bw d ≥ φ 0.35 fc′

)

(5.10.14)

(5.10.13) 

13



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5 . 1 0   A C I C ode D esign P ro v isions for   S hear

The requirement of minimum shear reinforcement may be waived if test results show that the required flexural and shear strengths can be developed (ACI-​9.6.3.2). 3. φ Vc < Vu ≤ [φ Vc + min φ Vs ](5.10.15) For all flexural members, including those exempted from shear reinforcement in Category 2, shear reinforcement must be provided satisfying Eqs. (5.10.13) and (5.10.14). 4. [φ Vc + min φ Vs ] < Vu ≤ [φ Vc + φ (4 fc′ ) bw d ](5.10.16)8 For this category, the computed shear reinforcement requirement will exceed the min φVs requirement, and the shear reinforcement must satisfy ACI Code Eq. (22.5.10.1), ACI-​ 22.5.10.5.3 and ACI-​9.7.6.2.2, as follows:  required φ Vs = Vu − φ Vc

φ Av f yt d

provided φVs =

s

maximum s =

(for α = 90° )

d ≤ 24 in. 2

(5.10.17) (5.10.18) (5.10.19)

Note that in terms of nominal stress, vs = Vs / ( bw d ) = 4 fc′ psi is the maximum vs for which the d/​2 maximum spacing limit applies. 5. [φ Vc + φ (4 fc′ )bw d ] < Vu ≤ [φ Vc + φ (8 fc′ )bw d ](5.10.20)9 The difference between Categories 4 and 5 is that for all regions of a beam where the nomi­ nal stress vs to be taken by shear reinforcement is between 4 fc′ and 8 fc′ the maximum shear reinforcement spacing s may not exceed d /4 nor 12 in. These reduced spacing limits are intended to provide adequate diagonal cracking control at high shear stresses, which helps improve aggregate interlock and dowel action. The shear reinforcement provided in this category must satisfy ACI Code Formula (22.5.10.1), ACI-​22.5.10.5.3 and ACI-​ 9.7.6.2.2, as follows required φ Vs = Vu − φ Vc

provided φ Vs = maximum spacing s ≤

φ Av f yt d s

(for α = 90°)

d ≤ 12 in. 4

(5.10.21) (5.10.22) (5.10.23)

In addition, factored shear Vu may not exceed the upper limit of Eq. (5.10.20) according to ACI-​22.5.1.2. The ACI Code provisions for shear strength of beams described in this section are summarized in Table 5.10.1.

8  For SI, ACI 318-​14M gives in place of 4 fc′ psi ,

fc′ / 3 when fc′ is in MPa.

9  For SI, ACI 318-​14M gives in place of 4 fc′ and 8 fc′ psi,

fc′ /3 and 2 fc′ /3 , respectively, when fc′ is in MPa.

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TABLE 5.10.1  SHEAR STRENGTH OF MEMBERS UNDER BENDING ONLY—​ACI CODE Strength Design (φ = 0.75)

Item 1

2 3

φ Vn ≥ Vu Maximum Vu at a distance d from face of support in usual situations (three exceptions; see Section 5.11) Vn = Vc + Vs fc′ ≤ 100 psi unless Av ≥ 0.75 fc′

4

Code 9.5.1.1, 9.4.3.2

Formula (22.5.1.1) 22.5.3.1

bw s 50bw s ≥ f yt f yt

Simplified method: Vc = 2 λ fc′ bw d

Formula (22.5.5.1)

More detailed method:  ρ V d Vc =  1.9λ fc′ + 2500 w u  bw d ≤ 3.5λ fc′ bw d Mu   Vu d/​Mu not to exceed unity

Table 22.5.5.1

Allow 10% increase for one-​way joists Lightweight concrete factor λ when fct is specified:

9.8.1.5

λ=

fct 6.7 fcm

≤ 1.0

Lightweight concrete factor λ when fct is not specified: Use λ between 0.75 and 0.85 depending on the absolute volume of normal-​weight fine aggregates as a fraction of the total absolute volume of fine aggregates. For sand-​lightweight, coarse blend concrete, λ ranges between 0.85 and 1.0 based on the absolute volume of normal-​weight coarse aggregate as a fraction of the total absolute volume of coarse aggregate. 5

6

Vs =

Av f yt d (sin α + cos α ) s

19.2.4.3 19.2.4.2

Formula (22.5.10.5.4)

Vs = Av f y sinα ≤ 3 fc′ bw d    (single bar or single group of bent-up bars)

22.5.10.6.2

f yt not to exceed 60,000 psi, except for welded deformed wire reinforcement not to exceed 80,000 psi

Table 20.2.2.4a

Vs not to exceed 8 fc′ bw d

22.5.1.2

For 0.5 φ Vc < Vu ≤ φ Vc min φ Vs = 0.75 φ fc′ bw d ≥ φ 50 bw d 9.6.3.1, 9.6.3.3 Av f yt Av f yt d use s = ≤   (max s ≤ ≤ 24 in.) 50bw 0.75 fc′ bw 2 except for slabs; footings; joists; small beams shallower than 10 in., 2 1 2 times flange thickness, or bw / ​2, but with depth no greater than 24 in.; steel fiber-reinforced, normal-weight concrete beams with depth no greater than 24 in., fc′ ≤ 6000 psi, and Vu ≤ φ 2 fc′ bw d (Continued)

135



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5 . 1 1   N O M I NA L S H E A R S T R E N G T H CA L C U L AT I O N

TABLE 5.10.1  (Continued) Strength Design (φ = 0.75)

Item 7

Code

For φ Vc < Vu ≤ [φ Vc + minφ Vs ]

9.6.3.1, 9.6.3.3

min φ Vs = φ 0.75 fc′ bw d ≥ φ 50 bw d use s = 8

Av f yt 50bw

≤

Av f yt

d      max s ≤ ≤ 24 in.   2 0.75 fc′ bw

(

)

For [ φ Vc + min φ Vs ] < Vu ≤ φ Vc + φ 4 fc′ bw d    Design shear reinforcement. For α = 90°, Vs =

Av f yt d

s d max s = ≤ 24 in. 2 9

≥

Vu − Vc φ

For [φVc + φ (4  fc′ bw d)] < Vu ≤  [φVc + φ (8  fc′ bw d)] Design shear reinforcement. For α = 90°, Vs = max s =

Av f yt d s

≥

Formula (22.5.10.1), Formula (22.5.10.5.3), 9.7.6.2.2

Formula (22.5.10.1), Formula (22.5.10.5.3), 9.7.6.2.2

Vu − Vc φ

d ≤ 12 in. 4

5.11 CRITICAL SECTION FOR NOMINAL SHEAR STRENGTH CALCULATION In experimental work the critical section for computing the nominal shear strength was the location of the first inclined crack. Since most testing was made on simply supported beams under simple loading arrangements, it was difficult to extend such results to generalized loadings on continuous structures. For nonprestressed beams subjected to distributed loading near their top surface and supported at the bottom such that the support reaction in the direction of shear introduces compression, ACI-​9.4.3.2 permits taking the first critical section at a distance d from the face of support. Thus the Code recognizes that the portion of the distributed loading applied over a distance d from the face of the support will transfer directly to the support, without requiring additional shear reinforcement. Shear reinforcement must be provided, however, between the face of support and the distance d therefrom, using the same requirements as at the critical section. The critical section must be taken at the face of support when one of the following occurs: 1. Factored shear Vu does gradually decrease from the face of support, but the support is itself a beam or girder and therefore does not introduce compression into the end region of the member (see Fereig and Smith [5.58]). 2. Loads are not applied at or near the top of beam. 3. A concentrated load occurs between the face of support and a section at a distance d therefrom. 4. Any loading that may cause a potential inclined crack to occur at the face of support or extend into instead of away from the support (see Fig. 5.11.1).

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Figure 5.11.1  Inclined crack when the reaction induces tension in the member. Critical section is at the face of support.

5.12 SHEAR STRENGTH OF BEAMS—​D ESIGN EXAMPLES Three examples are presented to illustrate the basic procedure of designing vertical stirrups. In the first example, the complete design procedure for flexure and shear is shown for a simply supported beam using the simplified procedure. Some of the more practical aspects are discussed in the second example. The third example illustrates the use of metric units. Design of shear reinforcement in the continuous spans of a slab-​beam-​girder floor system is treated in Chapter 9.

EXAMPLE 5.12.1 For the given beam [Fig. 5.12.1], first determine the maximum uniform dead and live loads under service condition permitted by the ACI Code. Then use those maximum service loads to design the shear reinforcement using vertical stirrups and the simplified procedure using constant Vc . Assume that the service live load to dead load ratio is 1.5, fc′ = 4000 psi (normal-​weight concrete), and f yt = 60, 000 psi. SOLUTION  (a) Compute the nominal moment strength M n of the given section. Assume the compression steel yields at nominal strength M n. Then from statics, Cc = 0.85 fc′ ba = 0.85(4)(14)a = 47.6 a Cs = ( f y − 0.85 fc′ ) As′ = (60 − 3.4)2.40 = 136 kips T = f y As = 60(8.0) = 480 kips Cc + C s = T

a=

(480 − 136) = 7.23 in.; 47.6

c=



a 7.23 = = 8.51 in. β1 0.85

Check to see whether compression steel yields; compute strain ε s′ at the location of the compression steel,

ε s′ =

c − d′ 8.51 − 2.50 ε cu = 0.003 = 0.00212 c 8.51

The compression steel yields as assumed because

fy   60 [ ε s′ = 0.00212] >  ε y = = = 0.00207 E 29 000 , s  

(Continued)

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Example 5.12.1 (Continued) 4 – #7 throughout 8 – #9 throughout 23’–0” clear 24’–0” center to center 2.5”

As’ = 2.40 sq in.

εcu = 0.003 εs’

4 – #7 d = 22.5” dt

26”

c = 8.51 in.

8 – #9 As = 8.00 sq in.

bw = 14”

εs

(b) Strain diagram at Mn

(a) Cross section

Figure 5.12.1  Beam for Example 5.12.1.

Since there is no axial load, check that the net tensile strain is at least 0.004 (ACI-​ 9.3.3.1). To obtain dt to the layer of steel reinforcement closest to the tension face, add 0.5 in. (assuming 1 in. between layers) plus the bar radius (0.564 in. for a #9 bar) to the effective depth d. Thus,

dt = d + 0.5 + 0.564 = 22.5 + 0.5 + 0.564 = 23.6 in.

The strain εt at the extreme tension steel is 

ε t = 0.003

dt − c 23.6 − 8.51 = 0.0053 > 0.004     OK = 0.003 c 8.51

Since the tensile strain exceeds 0.005, the section is tension controlled (ACI-​21.2.2), and the appropriate φ factor is 0.90. The nominal moment strength M n is M n = Cc ( d − a / 2 ) + C s ( d − d ′ ) = 47.6(7.23)(22.5 − 7.23 / 2)

1 12

+ 136(22.5 − 2.5)

1 12

= 542 + 227 = 769 ft--kips



(b) Compute the maximum permitted service loads wD and wL using the U and φ factors of ACI-​5.3 and ACI-​21.2.2. 1 wu L2 = φ M n 8 8φ M n 8(0.90)769 = = 9.61 kips/ft wu = (24)2 L2 Mu =

wL = 1.5 wD wu = 1.2 wD + 1.6 (1.5wD )

9.61 = 2.67 kips//ft 1.2 + 2.4 service live load wL = 4.00 kips/ft

service dead load wD =

(Continued)

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Example 5.12.1 (Continued) (c) Design of shear reinforcement using the simplified method with a constant value for Vc . The factored shear to be designed for must be the maximum that may possibly act at each point along the span; that is, an envelope of maximum shear is needed. A  knowledge of influence lines tells the designer that for bending moment on a simply supported span, the maximum value occurs at every point along the span when the full span is loaded with live load. However, for shear, the maximum shear occurs with partial span loading for every point along the span except at the supports. Unless the designer can justify other treatment, the live load should always be treated as variable position loading acting wherever it may cause the greatest effect, whereas the dead load would be a fixed position loading. For most ordinary situations an approximate shear envelope may be acceptable, using a straight-​line relationship between the maximum shear at the support and the maximum shear at midspan. Such a procedure will always be conservative.

Vu = 1.2VD + 1.6 VL’ kips

CL of support True envelope for maximum Vu 115.2

Straightline approximation for maximum shear envelope Vu due to factored loads

+

19.2

Shear diagram for Vu with full dead plus live load on entire span

19.2

–

Midspan 24’–0”

115.2

Figure 5.12.2  Factored shear Vu diagram for Example 5.12.1.

For this beam (see Fig. 5.12.2), the maximum factored shear at the centerline of support is

Vu =

wu L [2.67(1.2) + 4.00 (1.6)] 24 = = 115.2 kips 2 2

The maximum factored shear possible at midspan occurs with live load on half the span, Vu =

wL L 4.00(1.6)(24) = = 19.2 kips 8 8

The dead load shear (with full span loaded) is zero at midspan. The critical section for determining the closest stirrup spacing may be taken, according to ACI-​9.4.3.2, at a distance d from the face of support; in this case, d = 22.5 in. and the support width is 12 in., making the critical section (22.5 + 6 = 28.5 in.) 2.38 ft from the center of the support. By linear interpolation, Vu at d from the face of support is  115.2 − 19.2  Vu = 115.2 −   2.38 = 96.2 kips  12 The design requirement between the face of support and the critical section is considered to be constant; in this case, 96.2 kips. (Continued)

139



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Example 5.12.1 (Continued) The factored shear Vu diagram for the left half of this symmetrical structure is shown enlarged in Fig.  5.12.3. The design may be done primarily on this factored shear Vu diagram. The design shear strength attributable to the concrete is

(

φ Vc = φ 2 λ fc′ bw d

)

  φ Vc = 0.75[2(1) 4000 ](14)(22.5)  1  = 29.99 kips  1000 



The difference (the portion cross-​hatched in Fig. 5.12.3) between Vu and φ Vc must be provided for by shear reinforcement. For most beams (ACI-​9.6.3.1) shear reinforcement may not be terminated when Vu equals φ Vc, but rather must be continued until Vu = 0.5φ Vc (Fig. 5.12.3). The distance over which shear reinforcement is required from the face of support is in this case the entire span because 0.5φ Vc never exceeds Vu .

12”

11 @ 3”

7 @ 4”

5 @ 10”

4 @ 6”

# 3 stirrups

3” 12’ –0“ Face of support

Vu = 1.2VD + 1.6 VL, kips

115.2k

Critical section 96.2k

φVn = 85.6k for s = 4” s = 5” s = 6” s = 8” s = 10” s = 11.25”

d 2.38’

29.9k 19.2k

φVc

CL of support

4”

5”

0.5φVc

6”

8” 10” 11.25”

15.0k

Midspan

Figure 5.12.3  Design of shear reinforcement in Example 5.12.1 using constant φ Vc.

Examination of Fig. 5.12.3 shows that Vu exceeds φ Vc over most of the span, putting most of the design in Categories 3 through 5 of Section 5.10. At the critical section, required φ Vs = Vu − φ Vc = 96.2 − 29.9 = 66.3 kips Check limits on φ Vs ,  1  min φ Vs (ACI − 9.6.3.3) = 0.75(0.75 4000 )14(22.5)  = 11.2 kips  1000  but not less than

0.75(50)14(22.5)

1 1000

= 11.8 kips (Continued)

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Example 5.12.1 (Continued) Thus, the 11.8 kip limit controls.



d    1  max φ Vs  for s = ≤ 24 in. = 0.75(4 4000 )14(22.5)     1000  2 = 59.8 kips (ACI-9.7.6.2.2)



d    1  max φ Vs  for s = ≤ 12 in. = 0.75(8 4000 )14(22.5)     1000  4 = 119.5 kips (ACI-9.7.6.2 .2)

Since the required φ Vs equals 66.3 kips at the first critical section (which is the maximum for the beam) and lies between 59.8 and 119.5 kips, design for the portion of the beam between the face of support and the critical section falls into Category 5 (max s = d /4). Using #3 vertical U stirrups, in which two stirrup bar areas comprise Av in Eq. (5.10.18), s (at critical section) =

φ Av f yt d φ Vs

=

0.75(2)(0.11)(60)(22.5) = 3.4 in. 66.3

This means that #3 stirrups may not be spaced farther apart than 3.4 in. for the region from the face of support to the critical section at a distance d from the face of support. The determination of max s at the critical section usually controls the size of stirrups to be used. If the spacing for #3 stirrups were required to be too small, say less than about 3 in., the stirrup bar size would be increased to #4, or occasionally #5. In this case, a spacing of 3 in. is considered acceptable. As the Vu requirement decreases toward midspan, the stirrup spacing can be increased. A table of potentially acceptable spacings (s vs. φ Vs ) should be made, as in the first two columns of Table 5.12.1. Values are chosen until the spacing reaches the maximum permitted. Since φ Vs does not exceed 4 fc′ bw d beyond a distance of 3.19 ft (38.25 in.) from the center of the support, the maximum spacing beyond this section is the lesser of d /2 (11.25 in. for this beam) or 24 in. (ACI-​9.7.6.2.2). In the region where a minimum area TABLE 5.12.1  SPACING AND STRENGTH RELATIONSHIPS FOR VERTICAL STIRRUPS IN EXAMPLE 5.12.1

s 3.4 in. 4 5 6 8 10 11.25 max

φ Vs =

φ Av f yt d s (1000)

=

222.8 s

66.3 kips (max) 55.7 44.6 37.1 27.9 22.3 19.8

z = Distance from face of support to intersection of φ Vn (i.e., φ Vc + φ Vs ) with Vu 66.3 − φ Vs (144 − 28.5) = 22.5 + 96.2 − 19.2 0 to 22.5 in. 38 55 66 80 89 92–​138 in. (Continued)

14



5 . 1 2   S hear S trength of   B ea m s — D esign E xa m ples

141

Example 5.12.1 (Continued) of shear reinforcement is required (ACI-​9.6.3.3), the spacing required for #3 stirrups is the smaller of

s=

0.22(60, 000) 0.75 4000 (14)

= 19.9 in.

or

s=

0.22(60, 000) = 18.9 in. 50(14)

Therefore, the maximum spacing of 11.25 in. based on beam depth controls. In Table 5.12.1, the d /2 spacing limit results in a shear strength of 49.7 kips (φ Vc + φ Vs) at a distance z  =  92 in. from the face of the support. This means that between z = 92 in. and midspan, the spacing of the stirrups cannot be increased even though Vu continues to decrease. Thus, a stirrup spacing of 11.25 in. (or less) should be used for this region of the span until stirrups are no longer required. As noted earlier, however, Vu is always larger than φ Vc /2 and, therefore, stirrups are required over the entire span in this case. To determine the set of spacings to actually be used, the designer may prefer to draw on the Vu diagram the lines representing the φ Vn provided by different spacings, as shown in Fig. 5.12.3, and then scale from the diagram the locations z where the various spacings are permissible (marked along the baseline of Fig. 5.12.3); or, alternatively, the locations z may be computed as in the third column of Table 5.12.1. Note that in this column, the quantity (144 –​28.5)/​(96.2 –​19.2) is the distance in which the value of Vu drops 1 kip. Whether the location z is computed or scaled, a table of φ Vs versus s should be used. The limiting spacings of 3.4 and 11.25 have already been explained. The intermediate spacings of 4, 5, 6, 8, and 10 in. are those chosen by the designer as practical possibilities. Since the spacing of stirrups cannot be varied continuously, they are varied by using discrete increments. One conservative policy is to use a spacing, say, 6 in., only beyond the theoretical point at which a 6-​in. spacing may be used (in this case, 66 in. or more from the face of support). In a less conservative manner, the next larger spacing may be used somewhat before the point at which this spacing may be used. With this in mind, the set of spacings to be used is 3 in. | 11 @ 3 in. | 7 @ 4 in. | 4 @ 6 in. | 5 @ 10 in. | = 138 in. 3

36

64

88

138 (midspan)

To make uniform spacing at midspan, the first stirrup is placed at 3 in. from the support, more than the usual half-​space. Many designers prefer to place the first stirrup a full space from the face of support for closely spaced stirrups. Note that in the adopted set of spacings, 4-​in. spacing is used after z = 36 in. (z = 38 in. required); 6-​in. spacing is used after z = 64 in. (z = 66 in. required); and 10-​in. spacing is used after z = 88 in. (z = 89 in. required).

EXAMPLE 5.12.2 Design the locations of #3 vertical U stirrups to be used in the beam of Fig.  5.12.4. The beam is to carry service dead and live loads of 5.2 and 6.0 kips/​ft, respectively. Use fc′ = 3500 psi (normal-​weight concrete) and f y = f yt = 60, 000 psi. Use the alternate approach of plotting vn (vn is the nominal shear stress, which is the required nominal shear strengthVu / φ divided by bw d ) diagram and then providing the vc attributable to the concrete and vs attributable to shear reinforcement. (Continued)

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Example 5.12.2 (Continued) SOLUTION (a) The factored shear force Vu and required nominal shear stress vn diagrams (envelope of maximum values for different loading conditions) for design are as shown in Fig. 5.12.4. The effect of partial span live load on shear is approximated by passing a straight line through the maximum shear values at the support and at midspan. 1.2 wD = 1.2(5.2) = 6.24 kips/ ft 1.6 w L = 1.6(6.0) = 9.60 kips/ft At the center of the support, Vu =

1 (6.24 + 9.60)(12) = 95 kips 2

vn =

Vu 95, 000 = = 490 psi φ bw d 0.75(12)21.5

Setting vn = vu / ​φ

At midspan, 1 Vu = (9.60)(12) = 14.4 kips 8 Vu 14, 000 vn = = = 74.4 psi φ bw d 0.75(12)(21.5) Note that the shear at midspan due to dead load is zero; but positive live load shear at midspan is largest when only the right half of the span is loaded. (b) Simplified method. vc = 2 λ fc′ = 2(1) 3500 = 118 psi The required nominal shear stress at the distance d from the face of the support is (6 + 21.5) (490 − 74.4) = 331 psi 72 required vs = vn − vc = 3311 − 118 = 213 psi vn (at d ) = 490 −



4 fc′ = 237 psi (ACI − 9.7.6.2.2) Where vs > 4 fc′ , the maximum spacing of vertical stirrups may not exceed d /4; otherwise, d /2 (ACI-​9.7.6.2.2). In this case, the spacing limitation d /2 applies wherever stirrups are needed. Note that this simple check on whether required vs exceeds 4 fc′ often gives advantages over using the criterion shown in item 9 of Table 5.10.1. Try #3 vertical U stirrups.



Av f yt = 2(0.11)(60, 000) = vs bw s 13, 200 vs s = = 1100 psi. in. 12

which for the vs of 213 psi at the critical section permits

max s (at critical section) =

1100 = 5.2 in. 213 (Continued)

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5 . 1 2   S hear S trength of   B ea m s — D esign E xa m ples

Example 5.12.2 (Continued)

d = 21.5”

4 – #9 2”

5 @ 5” 12”

3 @ 9”

2 @ 6”

#3 U stirrups

12”

12’ –0” center to center of supports Symmetrical about CL of beam

(a) The beam

95 Vu shear force, kips

12”

Critical section Straightline approximation True maximum shear envelope for uniform live load

d = 21.5”

14.4

Face of support (b) Shear force diagram

490 Shaded portion indicates stress to be carried by 331 s = 6” shear reinforcement s = 8” s = 10” Required vs

d = 21.5” Vu φbwd shear stress, psi

74.4 vc/2 vc = 118 psi carried by concrete (c) Required shear stress vn diagram 12 s stirrup spacing, inches 2”

8 4

6”

Max. s (d/2) = 10.75”

8” 10”

Face of support Permitted 5 @ 5 = 2’ –1”

Provided 2 @ 6”

3 @ 9”

#3 stirrups

(d) Stirrup requirement diagram

Figure 5.12.4  Stirrup placement for Example 5.12.2.

The maximum spacing for shear reinforcement to provide for a minimum vs of 50 psi or 0.75 fc′, whichever is greater, is 1100 = 24.8 in. . 0 75 3500 1100 d  but not less than = 22 in. >  = 10.75 in. 50 2  max s (for min Av ) =



Thus the spacing cannot be more than 10.75 in. even when vs gets small. If the above computation had given a value less than d /2 instead of 22 in., then that lesser value would be the maximum s for the region wherever stirrups are required (vn > vc /2). (Continued)

14

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Example 5.12.2 (Continued) The placement of stirrups will be done by scaling from the diagram of Fig. 5.12.4(c). To facilitate this, Table 5.12.2 is computed using spacings considered desirable by the designer. The shear stress vs values are plotted as horizontal lines marked s = 6, 8, and 10 in. on Fig. 5.12.4(c); then their intersections with the required shear stress vn line are projected downward to the base line. Stirrups are then laid out with a scale beginning a distance s /2 from the face of support. (Some designers place the first stirrup a full space from the face of support for small spacings.) The same spacing necessary at the critical section is specified by ACI-​9.4.3.2 to be used between the face of support and the critical section. Thus one may start at s /2 from the face of the support with a 5-​in. spacing until within s /2 of the capacity line for the next desired spacing, which in this case is slightly beyond the vertical line projected from s = 6 in. The ACI Code requires shear reinforcement until the stress vn ≤ vc / 2. In this case, the entire beam must be provided with stirrups, because the smallest vn ( 74.4 psi ) is larger than vc /2 (59 psi). To illustrate clearly what has been done, a diagram showing permitted (actual spacings must be below permitted line) and provided stirrup spacings is given in Fig. 5.12.4(d). In this case, some shifting of spacing has been made to avoid leaving a small fragmental space at midspan. One more space at 5 in. was used than necessary and the three spaces adjacent to midspan were reduced to 9 in. from the permitted 10 in. in order to eliminate a small space near midspan. The final design details are shown in Fig. 5.12.4(a). TABLE 5.12.2  SPACING OF VERTICAL STIRRUPS (EXAMPLE 5.12.2)

s 5.2 6 8 10 max 10.75

vs =

1100 s

vu /φ = vn = vc + vs

213 183 138 110 102

331 301 256 228 220

EXAMPLE 5.12.3 Determine the vertical stirrup requirement for the beam of Fig.  5.12.5. Use #10M bars (see Table  1.13.2) for U stirrups, fc′ = 25 MPa (normal-​weight concrete), and f y = f yt = 300 MPa. The service live load is 30 kN/​m and the service dead load is 43 kN/​m (including beam weight). SOLUTION  (a) Determine maximum factored shear Vu envelope.

wu = 1.2(43) + 1.6(30) = 99.6 kN/ m (Continued)

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Example 5.12.3 (Continued) For maximum shear at d (480 mm) from face of support, place live load on remainder (3.2 − 0.15 − 0.48 = 2.57 m) of the span.

Vu at critical section =1.2(43)(1.6 − 0.15 − 0.48)+1.6(30)

(2.57)2 2(3.2)

= 50. 0 + 49. 6 = 99.6 kN 1 Vu at midspan = (1.6)(30)(3.2) = 19.2 kN 8 The straight-line approximation maximum shear envelope is given as Fig.  5.12.5(c). (Note that, in the previous example, the straightline approximation was made between centerline of support and midspan. The present approximation is closer to the “exact” maximum shear curve.) (b) Determine stirrup spacing. If the factored shear Vu diagram is used, vc =

fc′ 6

(from ACI 318 − 14M with fc′ in MPa)

= 25 / 6 = 0.833 MPa

φ Vc = φ vc bw d = 0.75(0.833)(280)(480)

1 1000

= 84.0 kN

At the critical section, required φ Vs = Vu − φ Vc = 99.6 − 84.0 = 15.6 kN The limiting φ Vs = φ ( fc′ / 3)bw d for d /2 stirrup spacing limitation can be conveniently obtained from 2φVc ; thus

limiting φ Vs = 2φ Vc = 2 (84.0 ) = 168 kN > [required φ Vs = 15.6 kN ]

Thus maximum spacing cannot exceed d /2. For #10M U stirrups, Av = 2 ( 71) = 142 mm 2 . The spacing requirement for strength is

s=

φ Av f yt d φ Vs

=

(0.75)(142)(300)480 φ Vs

1 1000

=

15, 228 φ Vs (kN)

The stirrup spacing requirements can be summarized as follows: 1. Maximum spacing for strength requirement at the critical section,  

max s =

15, 228 = 976 mm 15.6

max s =

d 480 = = 240 mm 2 2

2. Maximum spacing d /2,  

Controls (Continued)

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Example 5.12.3 (Continued) 280 mm Support width = 300 mm #10M stirrup

3.2 m

d = 480 mm

(a)

4 – #19M

40 mm cover

Support width = 300 mm

(b)

Face of support

Factored shear Vu, kN

Stirrups required = 1.17 m

120 mm

99.6 84.0 Midspan

d φVc

42.0 φVc 2

4 @ 240 mm

19.2 #10M U stirrups

(c)

Figure 5.12.5  Beam and stirrup design for Example 5.12.3.

3. Maximum spacing for minimum shear reinforcement (ACI-​9.6.3.3),

)

(

min φ Vs = φ ( 0.35 MPa ) bw d > φ 0.062 fc′ bw d (ACI 318-14M) = 0.75 (0.35)(280)(480) max s =

1 1000

= 35.3 kN

15, 228 d = 431 mm > 35.3 2



4. Conclusion. Use 240-​mm spacing for the portion where stirrups are required (i.e., from face of support to location where Vu = 0.5φ Vc). The final stirrup arrangement is shown in Fig. 5.12.5(c).

5.13 SHEAR STRENGTH OF MEMBERS UNDER COMBINED BENDING AND AXIAL LOAD The presence of an axial compressive load on a reinforced concrete flexural member decreases the longitudinal tensile stress and the resulting tendency for inclined cracking. Conversely, the addition of an axial tensile load increases the longitudinal tensile stress and the tendency for inclined cracking. Thus, for the same bending moment, the shear strength of a member is increased by the addition of an axial compressive load and decreased by an

147



147

5.13  SHEAR STRENGTH OF MEMBERS

axial tensile load. Some experimental work is available on shear strength in the presence of axial load [5.4, 5.63–​5.67].

Axial Compression Since inclined cracking is dependent on the combination of tensile or compressive stress due to flexure and shear stress, as discussed in Sections 5.3 through 5.5, the addition of axial compression tends to delay the formation and opening of inclined cracks and limit their propagation due to an increase in the size of the compression zone. When the simplified method is used with axial compression, a linear increase with axial compression is given by ACI-​22.5.6.1:  Nu  Vc = 2  1 +  λ fc′bw d  2000 Ag 



(5.13.1)10

Note that Nu is the factored axial compressive load (positive quantity), Ag is the gross area of the concrete cross section, and N u / Ag is expressed in psi. A more detailed equation for Vc is also permitted by 22.5.6.1,   Vu d Nu   Vc =  1.9λ fc′ + 2500ρw bw d ≤ 3.5λ fc′bw d 1 +  ( 4h − d ) 500 Ag   Mu − N u   8 This expression, however, is not applicable when Mu − N u

(5.13.2)11

( 4h − d ) ≤ 0. 8

( 4h − d ) in lieu of Mu in Eq. (5.10.5) may 8 be seen by referring to Fig. 5.13.1, which shows a free body of a short length of beam dz. Moment equilibrium about point A, the line of action of the internal compressive force C, gives The rationale for the use of the term Mu − N u

a   h a T  d −  = Mu − N u  −    2 2 2



(5.13.3)

When the moment arm (d – a /2) is approximated as 7d /8, thereby making a /2 = d /8, the ( 4h − d ) right side of Eq. (5.13.3) becomes Mu − N u , and the left side represents an equiva8 lent moment. This equivalent moment procedure is to be used in the more detailed method for combined axial compression and bending only. Experience [5.63] has shown the method to be unsafe for axial tension. The upper limit for the nominal shear strength Vc of members without shear reinforcement subject to bending only is given in Eq. (5.10.5) as 3.5λ fc′ bw d . In the presence of Nu axial compression, however, this upper limit is increased by a factor 1 + , whose 500 Ag rationale is discussed next. 10  For SI, ACI 318-​14M, for Nu /​Ag and fc′ in MPa, ACI-​22.5.6.1 gives



 N  Vc = 0.17  1 + u  λ fc′ bw d Ag  14 

(5.13.1) 

11  For SI, ACI 318-​14M, with Nu /​Ag and fc′ in MPa, gives



    Vu d Nu Vc  0.16λ f c′ + 17ρw  bw d ≤ 0.29λ f c′ bw d 1 + h d 4 − 3.5 Ag ( )   M u − Nu   8

(5.13.2)

 

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a 2 h 2 h

A

Mu

C d– a 2

Nu

T dz

Figure 5.13.1  Actions and internal forces in a member under combined axial compression and bending moment. Nu/Ag, MPa 5

2.5

0

6

(d)

Curve (a): ACI Table 22.5.6.1(a) Eq. (5.13.2) [Mu/(Vud) = 2, ρw = 0.03, and fc’ = 2500 psi] Curve (b): ACI Table 22.5.6.1(a) Eq. (5.13.2) [Mu/(Vud) = 5, ρw = 0.005, and fc’ = 5000 psi] Curve (c): ACI Eq. (22.5.6.1) Eq. (5.13.1) Curve (d): ACI Table 22.5.6.1(b) Eq. (5.13.5) Curve (e): ACI Eq. 22.5.7.1 Eq. (5.13.6) Vc = Vu d = 0.9h

–2.5

Vc bwd

5

fc’

(a) 4 (c)

3 2

(b) 1 Compression 1000

(e)

Tension

500

0

–500

Nu/Ag, psi

Figure 5.13.2  Effect of axial load on ACI-calculated inclined cracking shear stress. Lightweight concrete factor λ = 1. (Adapted from MacGregor and Hanson [5.63])

The formula for the principal tensile stress ft ( max ) in terms of the tensile stress ft and the shear stress v was derived in Section 5.3 as ft (max) =

1 1 ft +  2 2

2

 ft  + v 2 

Solving this equation for v, v = ft (max) 1 −

ft ft (max)

(5.13.4)

It may be seen from Eq. (5.13.4) that the shear strength of a member under bending only becomes a constant if ft is zero, that is, if the bending moment approaches zero. Empirically, this constant is the upper limit 3.5λ fc′. For a member under an axial compressive load N u without bending moment, ft is a constant that is equal to – N u / Ag . Substituting this value of ft in Eq. (5.13.4) and using an average value of 500 psi for ft ( max ) inside the square root, the upper limit of strength Vc in members under combined bending and axial compression becomes

Vc (upper limit) = vbw d = 3.5λ fc′ bw d 1 +

where N u / Ag is to be expressed in psi (see Fig. 5.13.2).

Nu 500 Ag

(5.13.5)

149



149

5.13  SHEAR STRENGTH OF MEMBERS

Axial Tension When a flexural member is subject to axial tension, a simple linear reduction for Vc is specified in ACI-​22.5.7.1 as follows:  Nu  Vc = 2  1 +  λ fc′ bw d  500 Ag 



(5.13.6)12

Note that N u is negative for tension. It is recommended, however, that Vc = 0 for members in which the axial tensile load is considered significant. Figure 5.13.2 shows the relationship between ACI formulas and experimental results. The cross-​hatched portion on the compression side represents a reasonable range when the more detailed procedure is used. Parts of the ACI Code are summarized in Table 5.13.1 so that the shear strength of members under bending only may be compared with that of members under combined bending and axial load.

TABLE 5.13.1  EFFECT OF AXIAL LOAD ON THE SHEAR STRENGTH OF MEMBERS WITHOUT SHEAR REINFORCEMENT—​ACI CODE

Bending only

Simplified Method

More Detailed Method

Formula (22.5.5.1)

Table 22.5.5.1

Vc = 2 λ fc′bw d

 ρ V d Vc =  1.9λ fc′ + 2500 w u  bw d ≤ 3.5λ fc′ bw d Mu   Vu d / Mu not to exceed unity

Bending and axial compression

22.5.6.1

Table 22.5.6.1

 Nu  Vc = 2 1 +  λ fc′ bw d  2000 Ag 

  Vu d   Vc =  1.9λ fc′ + 2500ρw b d ( 4h − d )  w   Mu − N u   8 ≤ 3.5λ fc′ bw d 1 +

Nu is positive for compression and Nu / ​Ag is in psi Bending and axial tension

Although not required, it is recommended to take Vc = 0 if axial tension is considered significant.

Nu 500 Ag

Nu is positive for compression and Nu / ​Ag is in psi 22.5.7.1  Nu  Vc = 2  1 +  λ fc′bw d  500 Ag  N u is negative for tension and N u / Ag is in psi

12  For SI, ACI 318-​14M, for N u / Ag and fc′ in MPa, ACI-​22.5.7.1 gives



 Nu  Vc = 0.17  1 +  λ fc′ bw d 3.5 Ag  

(5.13.6) 

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C hapter   5     S hear S trength and D esign for   S hear

Example 5.13.1 Show the effect of axial load on the ACI shear strength Vc for the beam of Example 5.12.1 (Fig. 5.12.1) when it contains no shear reinforcement. Compute Vc for the critical section at d from the face of support. As in Example 5.12.1, use normal-​weight concrete with fc′ = 4000 psi, b = 14 in, d = 22.5 in, h = 26 in., and 8–​#9 bars for the tension steel. SOLUTION  (a) Axial compression. Using the simplified method, Eq. (5.13.1) or ACI-​22.5.6.1  Nu  Vc = 2  1 +  λ fc′ bw d  2000 Ag   Vc Nu  = 2 1 + λ fc′ bw d  2000 Ag 



The right side of the above equation is 2 for N u / Ag = 0; it varies linearly to 2.8 for N u / Ag = 800 psi. This linear expression is plotted on the left side of the vertical axis in Fig. 5.13.3. Using the more detailed equation, Eq. (5.13.2) or ACI Table 22.5.6.1,

ρw =

Vc

fc′ bw d

= 1.9(1.0) +

As 8(1.00) = = 0.0254 bw d 14(22.5) Vu d Nu ≤ 3.5(1.0) 1 + [ 4(26) − 22.5] 500 Ag Mu − N u 8(12, 000)

2500(0.0254) 4000

where Mu is in ft-​kips and N u is in lb. vc Vud vc = 1.9 fc’ + 2500ρw Mm 4h–d Mm = Mu – Nu 8

(

)

fc’ 6.0 5.0 4.0

vc = 3.5 fc’

1+ 0.002Nu /Ag

3.0

vc = 2(1+ 0.0005Nu/Ag)

2.0

vc = 2(1+ 0.002Nu /Ag)

fc’

fc’

1.0 800

600

400

200

Compression (Nu = +), psi

100 200 300 400 500 Tension (Nu = –), psi

Figure 5.13.3  Shear strength variation with axial load: Example 5.13.1, with fc′ = 4000 psi (normal-​weight concrete), ρw = 0.0254, Mu = 247 ft-​kips, Vu = 96.2 kips, d = 22.5 in., h = 26 in., and Ag = 364 sq in. Lightweight concrete factor λ = 1 (not shown in equations). (Note that vc = Vc  /​  bwd.)

(Continued)

15



151

5 . 1 4   D eep   B ea m s

Example 5.13.1 (Continued) At the critical section, considering live load applied to the entire span, Vu = 92.5 kips and Mu =



wu 9.61 (2.38)(24 − 2.38) = (2.38)21.62 = 247 ft-kips 2 2

Thus, Vc

fc′bw d

= 1.9 +

Nu 92.5 (1.88) ≤ 3.5(1.0) 1 + 247 − 0.00085 N u 500 Ag

where Ag = (14 )( 26 ) = 364 sq in. Note that Vu and Mu at the critical section must correspond to the same loading condition, rather than the maximum value for each. Values for the detailed formula as well as the upper limit equation, Eq. (5.13.5) [ACI Table 22.5.6.1], are tabulated in Table 5.13.2. The results are shown in Fig. 5.13.3. (b) Axial tension. The simple linear expression, Eq. (5.13.6) [ACI 22.5.7.1], is plotted on the right side of Fig. 5.13.3. TABLE 5.13.2  SHEAR STRENGTH WITH AXIAL COMPRESSION— ​EXAMPLE 5.13.1 N u / Ag (psi)

Mm (ft-​kips)a

1.88Vu / M m

231 216 185 123 61 0

0.75 0.81 0.94 1.41 2.83 —​

 50 100 200 400 600 800

Vc /

(

fc′ bw d 2.65 2.71 2.84 3.31 4.73 —​

)

3.5 1 + 0.002 N u / Ag  3.5(1.05) = 3.67 3.5(1.095) = 3.83 3.5(1.183) = 4.14 3.5(1.342) = 4.70 3.5(1.484) = 5.19 3.5(1.612) = 5.64

a M = M − N (4 h − d ) . m u u 8

5.14 DEEP BEAMS Deep beams are flexural members in which the assumption of plane sections remaining plane after loading is not appropriate because of significant nonlinearity in the longitudinal strain distribution over the member depth. In the ACI Code (ACI-​9.9.1.1), deep beams are defined as members loaded on one face and supported on the opposite face such as to allow the development of diagonal struts. For a member to qualify as a deep beam in the ACI Code, it must also have a clear ratio of span to overall depth not exceeding 4.0 and/​or be subjected to concentrated loads within twice the member depth from the face of the support. Since shear resistance in deep beams is primarily provided through diagonal compression between the loads and supports, the procedures developed previously are not applicable for deep beams. The design of deep beams, both for shear and flexure, is typically performed using the strut-​and-​tie method, as permitted in ACI-​9.9.1.3. Design of deep beams is discussed in Chapter 14.

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5.15 SHEAR FRICTION Even though uncracked concrete is relatively strong in shear, and shear-​related cracks in usual beams are inclined cracks (diagonal tension cracks), such shear-​related cracks become more vertical as the member becomes deeper in comparison to the shear span, as discussed in Section 5.4. The ACI Code design procedures for beams as discussed in Sections 5.5 through 5.13 are intended either to prevent inclined cracking (diagonal ten­ sion cracking) or to allow a redistribution of stresses and an increase in load beyond that at diagonal cracking through the use of web reinforcement. For situations in which a crack may form and slippage along that crack interface might occur if no steel reinforcement crosses the crack, and the usual design procedures for shear reinforcement are inappropriate (such as for a / ​d less than about 1.0), the shear-​friction concept of shear transfer represents a valid tool for design. The use of the shear-​friction concept is appropriate for providing a shear transfer mechanism in such cases as the following: 1. At the interface between concretes cast at different times. 2. At the junction of a corbel (bracket) with a column, such as shown in Fig. 5.15.1(a). 3. At the junction of elements in precast concrete construction, such as the bearing detail shown in Fig. 5.15.1(b). 4. At an interface between steel and concrete, such as the steel bracket attachment to a concrete column shown in Fig. 5.15.1(c). The crack for which shear-​friction reinforcement is required may not have been caused by shear. It could, for instance, have been caused by shrinkage. However, once the crack has occurred (from whatever source), a shear transfer mechanism must be provided. The design approach is to assume that a crack will occur and then provide reinforcement across the assumed crack to resist relative displacement along the crack. Consider a block of concrete as in Fig. 5.15.2(a) acted on by collinear shear forces V such that a failure plane would form along plane a-​a. Since the crack along a-​a will tend to be rough, the sliding motion will produce a separation, as in Fig. 5.15.2(b). One might imagine slippage along a sawtooth that forces the crack to open; as it opens, the reinforcement is

An =

Nuc φfy a

An + part of Avf

Vu

Bars welded to angle Nuc Assumed crack

Remainder of Avf

Assumed crack and shear plane

Avf Bars welded to angle Vu

(a) Corbel (see Sec. 5.16 for details)

(b) Bearing region of beam Vu

Embedded studs, Avf

Assumed crack

Studs welded to plate, steel angles bracket welded to face plate (c) Column face plate

Figure 5.15.1  Examples of applications of shear-​friction concept.

153



153

5 . 1 5   S hear F riction

V

V

Section a–a with exaggerated roughness

a

Reinforcement in which tension is induced

a

V

V (a)

(b)

Figure 5.15.2  Idealization of the shear-​friction concept.

put in tension, with a resulting compression or clamping force on the concrete. A frictional force is then developed equal to the compression in the concrete (or the tension in the bars) times the coefficient of friction. If one may assume that the separation and the slip are sufficient to load the steel reinforcement to its yield stress, the shearing resistance then equals the frictional force; thus (ACI-​22.9.4.2) states Vn = Avf f y µ



(5.15.1)

where Avf is the area of reinforcement extending across the potential crack at 90° to it, and µ is the coefficient of friction between materials along the potential crack. This concept of shear friction has been verified experimentally [5.68–​5.81]. ACI-​22.9.1.1 states that shear-​friction provisions apply “where it is appropriate to consider shear transfer across any given plane, such as an existing or potential crack, an interface between dissimilar materials, or an interface between two concretes cast at different times.” If the shear-​friction reinforcement is inclined at an angle to the assumed crack, such that the shear force produces tension in the shear-​friction reinforcement, as shown in Fig. 5.15.3, the shear strength Vn becomes (ACI-​22.9.4.3) Vn = Avf f y ( µ sin α f + cos α f )



(5.15.2)

where α f is the angle between the shear-​friction reinforcement and the shear plane. This equation does not apply if the shear-​friction reinforcement is oriented such that it is subjected to compression. T sin αf T

T cos αf

Force (and its horizontal and vertical components) that develops in shearfriction reinforcement

C = T sin αf Compressive force acting to keep crack from opening

Assumed crack

Applied shear

µC

Shear-friction reinforcement, Avf αf T

Figure 5.15.3  Action of shear-​friction reinforcement when inclined to shear plane.

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The logic of Eq. (5.15.2) maybe observed from Fig. 5.15.3, where the shear strength provided along the potential crack consists of two parts: the vertical component Tcos α f of the force in the reinforcement Avf and the frictional force µC, which is the same as µT sin α f . Thus, Vn = T cos α f + µC

(5.15.3)

One should note that the component of the tensile force in the shear-​friction reinforcement normal to the potential crack causes a compression at the crack interface, giving rise to the friction force µC. The required nominal shear-​friction strength Vn is Vu / φ, in which case Eq. (5.15.1) becomes required A vf =

Vu φ fy µ

(5.15.4)

and when α f is less than 90°, the required Avf from Eq. (5.15.2) is

required Avf =

Vu φ f y (µ sin α f + cos α f )

(5.15.5)

Note that Eq. (5.15.5) becomes Eq. (5.15.4) when α f = 90°. Just as for regular stirrups, f y may not be taken greater than 60,000 psi. The strength reduction factor φ is taken to be 0.75 for shear in Eqs. (5.15.4) and (5.15.5). The coefficient of friction µ is to be taken (ACI Table 22.9.4.2) as follows: 1. Concrete cast monolithically. µ = 1.4λ 2. Concrete placed against hardened concrete with surface intentionally roughened to a full amplitude of approximately 1 4 in. µ = 1.0λ 3. Concrete placed against hardened concrete not intentionally roughened. µ = 0.6λ 4. Concrete placed against as-​rolled structural steel where shear transfer is achieved through headed studs or welded bars or wires, with as-​rolled steel “clean and free of paint.” µ = 0.7λ In the above expressions for µ, the multiplier λ shall be 1.0 for normal-​weight concrete, 0.85 for “sand-​lightweight” concrete, and 0.75 for “all-​lightweight” concrete (ACI-​19.2.4.2). For other lightweight concretes, λ varies depending on the volumetric ratios of lightweight and normal-​weight aggregate, but is not to exceed 0.85. When normal-​weight concrete is cast against hardened concrete intentionally roughened to a full amplitude of approximately 1 4 in., the maximum nominal shear force that can be transferred may not exceed the least of [ACI Table 22.9.4.4(a)–​(c)]: a) 0.2 fc′Ac b) (480 psi +0.08 fc′) Ac c) (1600 psi) Ac where Ac is the area of concrete section over which shear transfer is achieved (sq in.). For all other cases, the nominal shear that can be transferred may not exceed 0.2 fc′Ac nor (800 psi)Ac [ACI Table 22.9.4.4(d) and (e)]. When concretes with different compressive strengths are used, the lesser fc′ value should be used. Since the steel Avf across the potential crack as determined by Eqs. (5.15.4) and (5.15.5) is only that necessary to provide the clamping action that produces friction, any external direct tension on the assumed crack must be provided for by additional reinforcement (ACI-​22.9.4.6). On the other hand, any permanent net compression may be added to the clamping force provided by the reinforcement (ACI-​22.9.4.5). To ensure development of the required clamping force, shear-​friction reinforcement should be properly anchored on both sides of the shear plane to develop its yield strength

15



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5 . 1 5   S hear F riction

(ACI-​22.9.5.1). Shear-​friction reinforcement should also be uniformly distributed along the shear plane unless a moment is present, in which case it is recommended that shear-​ friction reinforcement be placed primarily on the tension side (ACI-​R22.9.5.1). The application of the shear-​friction provisions to brackets and corbels appears in Section 5.16, which is devoted entirely to that topic. For guidance in the application of shear friction to bearings and other special situations commonly encountered in precast concrete construction, the designer should consult the PCI publication Design and Typical Details of Connections for Precast and Prestressed Concrete [5.82], as well as PCI Design Handbook [2.22]. The following example presents an application for cases other than brackets and corbels.

EXAMPLE 5.15.1 Design the reinforcement needed at the bearing region of a precast beam 14 in. wide by 28 in. deep supported on a 4-​in. bearing pad. The factored shear Vu is 95 kips. The horizontal force resulting from restraint of volume change movements due to creep, shrinkage, and temperature effects is 0.3 of the factored shear Vu . Grade 60 steel is to be used for reinforcement. A rolled structural steel angle is used for confinement across the width of the beam at the support. SOLUTION  (a) Identify the potential crack location. One should assume that a crack will form in the most undesirable manner. According to the PCI Design Handbook [2.22], a vertical crack is likely to develop [i.e., crack angle θ ≈ 0° in Fig. 5.15.4(a)]. The crack may intersect the bottom of the beam immediately adjacent to the bearing pad, which in this example is taken as 4 in. from the end of the beam. (b) Determine the shear-​friction reinforcement Avf required. Presumably, it would be appropriate to resolve the Vu and N uc in Fig. 5.15.4(a) into components parallel and perpendicular to the potential crack when θ is assumed other than zero. Even in such a case, it will be simpler and more practical to assume that all of Vu will act parallel to the crack. Thus, using Eq. (5.15.4),

required Avf =

Vu 95 = = 1.51 sq in. φ f y µ 0.75(60)1.4

(c) Determine the additional reinforcement An to provide for the net tension across the potential crack. It will be conservative not to use the sum of components Vu and N uc perpendicular to any assumed nonvertical crack, but rather to use N uc as if for a vertical crack. According to ACI-​22.9.4.6, required An =

N uc 0.3(95) = = 0.63 sq in. φ f y 0.75(60)

Note that N uc was given as 0.3 of Vu , presumably as the result of an analysis for volume change effects. It is recommended [2.22, 5.79] (and required by ACI-​16.5.3.5 in the case of brackets and corbels) that unless all tensile force N uc can be eliminated by appropriate design, the value of N uc should not be taken less than 0.2Vu . The φ factor of 0.75 for shear is considered appropriate for the above calculation, even though 0.90 for axial tension is indicated by ACI-​21.2.2. For brackets and corbels, in the similar situation of reinforcement for the tensile force N uc , ACI-​21.2.1 specifies taking φ as 0.75. (d) Total reinforcement to restrain primary crack [Fig. 5.15.4(a)]. The total reinforcement required is As = Avf + An = 1.51 + 0.63 = 2.14 sq in. (Continued)

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Example 5.15.1 (Continued) Possible crack

Asv θ ≈ 0°

Avf + An 15°

Nuc

Nuc Main longitudinal reinforcement

Vu

Vu (b) Secondary possible crack and reinforcement Asv to provide shear transfer

(a) Primary possible crack and reinforcement to provide shear transfer

Figure 5.15.4  Shear-​friction concept applied to bearing region of a beam.

Use 5–​#6, As = 2.20 sq in. Distribute as shown in Fig. 5.15.5, place at the recommended [5.82] maximum angle of 15° with the horizontal, weld to the steel angle on one end, and embed the other end into the beam to develop the tensile strength of the #6 bars beyond the potential crack. (Development length requirements are treated in Chapter 6.) (e) Reinforcement for the potential secondary horizontal crack that may form as shown in Fig. 5.15.4(b). If a vertical crack begins near the corner region where the main shear-​friction reinforcement terminates, then either with or without the tensile force N uc acting, there would be a potential horizontal crack owing to the tensile force developed in the main shear-​friction reinforcement. The maximum shear that could act along such a failure plane would be the horizontal shear-​friction force arising from the tensile capacity of the main shear-​friction reinforcement. Thus the required vertical stirrup shear-​friction reinforcement Asv [Fig. 5.15.4(b)] is tensile capacity of shear - friction reinforcement µ fy 2.20(60) = = 1.57 sq in. 1.4(60)

required Asv =

Use 4–​#4 U stirrups. Asv = 4 ( 0.4 ) = 1.60 sq in.

4 – #4 U stirrups @ 4“ spacing

1 – #4 closed stirrup 2 – #3 U stirrups

Main reinforcement

28”

15°

Nuc

2’ – 0” Vu

3 – #6 2 – #6

Figure 5.15.5  Final design for Example 5.15.1.

(Continued)

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Example 5.15.1 (Continued) (f) Additional confinement reinforcement. A conservative approach, recommended by Mast [5.69], is to provide reinforcement to prevent splitting in the vertical plane of the beam equal to 25% of the support reaction. This confinement reinforcement is divided equally into horizontal Ach and vertical Acv portions. Thus, Ach = Acv =



Vu 95 = = 0.20 sq in . 8 f y 8(60)

Use 1-​#4 vertical closed stirrup and 2-​#3 U stirrups horizontal. The final design is shown in Fig. 5.15.5.

5.16 BRACKETS AND CORBELS Brackets and corbels projecting from the faces of columns are widely used in precast concrete construction to support beams and girders, as shown in Fig. 5.16.1. It is inappropriate to design brackets and corbels as cantilever beams using the customary beam provisions for shear as described in Sections 5.5 through 5.8 and 5.10 through 5.13. As discussed in Section 5.4 and shown in Fig. 5.4.4, when a / d is less than about 1.0, deep beam theory, rather than simple flexural theory, should apply. Brackets and corbels, furthermore, differ from deep beams in that design calculations for horizontal forces must also be made. Because the beams are attached to the bracket, the restraint on the beams due to creep, shrinkage, and temperature deformations give rise to horizontal forces in the bracket or corbel (N uc in Fig. 5.16.1). Typically in the past, reinforcement for brackets or corbels consisted of several bars across the width of the bracket, bent as shown in Fig.  5.16.2(a). When minimum bend radii are considered, the actual arrangement is as in Fig.  5.16.2(b), where the dashed line indicates a potential failure surface. When the outer face is too shallow, the critical inclined crack will form in the location shown in Fig. 5.16.2(c). When the bracket is deep enough, the crack will tend to extend back into the column [Fig. 5.16.2(d)], with the portion between the crack and the sloping face acting as a compression element. If the strut can be developed, the bracket will have reserve capacity after the crack forms; if the strut cannot develop [as in Fig. 5.16.2(c)], failure will be instantaneous upon formation of the crack. Vu a

Asc = ρbd

Nuc h

d

Ah

Figure 5.16.1  Bracket or corbel.

N

V

V

V

N

N

V N

Potential failure surface (a) As shown on drawing

(b) As bent

(c) Cracking in corbel with too shallow an outer face

Figure 5.16.2  Corbel details and possible failure modes.

(d) Cracking in corbel with outer face of sufficient depth

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Research by Mattock et al. [5.74, 5.75] has shown that the shear-​friction concept can be applied to bracket (corbel) design for a /d as high as 1.0. The design recommendations of Mattock [5.83], the suggestions of ACI-​ASCE Committee 426 [5.16], and the further discussions of MacGregor and Hawkins [5.87] are the basis for the traditional provisions of ACI-​16.5. Additional discussions of the subject are provided by Shaikh [5.84] and Solanki and Sabnis [5.85]. Starting with the 2002 edition, the ACI Code allows the design of brackets and corbels using strut-​and-​tie models. These provisions are applicable for shear span-​to-​depth ratios a /d less than 2; thus, they can be used for larger a /d ratios than the traditional provisions, which are limited to a /d less than 1. A design example of a corbel using a strut-​and-​tie model is given in Chapter 14.

Basic Equilibrium Equations Using the shear-​friction concept and referring to Fig. 5.16.3, the strengths in shear Vn and in tension N nc are related to the internal forces such that statics is satisfied. From vertical force equilibrium,

Vn = µC (5.16.1)

From horizontal force equilibrium,

Nnc = T –​ C (5.16.2)

and from moment equilibrium, taken about point A, a  a    Vn a + N nc  h − 1  = T  d − 1     2 2 



(5.16.3)

Substituting Eq. (5.16.1) for C into Eq. (5.16.2), and taking T = Asc f y , N nc = Asc f y −



Vn µ

or Asc =



Vn N nc + fy µ fy

Vn a T = Ascfy

Nnc Asc

d– h

d

a1

a1 2

h–

µC

C a1 2

0.85fc’

a1 2

A

Forces contributed by horizontal stirrups or ties are neglected

Figure 5.16.3  Equilibrium of forces acting on a bracket or corbel.

(5.16.4)

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5 . 1 6   B rac k ets and C orbels

159

Substitution of Asc f y for T in Eq. (5.16.3) gives

a  a    Vn a + N nc (h − d ) + N nc  d − 1  = Asc f y  d − 1    2 2 

and solving for As  gives Asc =



Vn a + N nc (h − d ) N nc + a1 fy  fy  d −    2

(5.16.5)

Equations (5.16.4) and (5.16.5) give the formulas for Asc to provide the required strengths Vn and N nc. For design, the factored loads Vu and N uc divided by φ are the required strengths Vn and N nc, respectively. Thus, Eq. (5.16.4) becomes required Asc =

Vu N + uc f fy µ f fy

(5.16.6)

and Eq. (5.16.5) becomes



required Asc =

Vu a + N uc (h − d ) N uc + a1  φ fy  φ fy  d −    2

(5.16.7)

Note that Nuc /(​φ fy) is the reinforcement An required for axial tension (using the symbol An in Example 5.15.1) and Vu /(​φ fy µ) is the shear-​friction reinforcement Avf given by Eq. (5.15.4). Furthermore, observe that if the numerator of the first term in Eq. (5.16.7) is treated as an “equivalent” moment, the first term would represent the reinforcement A f required for a beam, corresponding to As of Eq. (3.4.6). To summarize the steel area requirements for brackets and corbels, the following may be stated: required Asc = Avf + An (5.16.8) or required Asc = Af + An (5.16.9) in which



Avf =

Vu φ fy µ

(5.16.10)

An =

N uc φ f y

(5.16.11)

Af =

equivalent Mu φ f y (d − a1 / 2 )

(5.16.12)

and equivalent Mu = Vu a + Nuc(h –​ d)(5.16.13)

Minimum Horizontal Stirrups In addition to the steel Asc in Fig. 5.16.3, stirrups in the horizontal plane are needed across the vertical potential crack to prevent a premature failure. These must be closed stirrups or ties, or hoops, having a total area Ah. It was conservative to neglect this steel Ah in the

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development of Eqs. (5.16.8) and (5.16.9) because Ah could have been deducted from the right side of those equations. Thus Eq. (5.16.8) could become required Asc = Av f + An –​ Ah (5.16.14) Tests [5.76] on brackets (corbels) indicate that minimum Ah for the horizontal stirrups must be

min Ah ≥

1 Af 2 (5.16.15)

min Ah ≥

1 Av f 3 (5.16.16)

and

ACI Code Provisions ACI-​16.5.5.1 requires the area of primary tension reinforcement Asc to be the larger of the following: required Asc = A f + An (5.16.17)



required Asc =

2 Avf + An 3

(5.16.18)

but not less than



 f ′ Asc = 0.04  c  bw d  fy 

(5.16.19)

The minimum Asc in Eq. (5.16.19) is intended to prevent a sudden failure should cracking occur. According to ACI-​16.5.5.2, closed stirrups or ties parallel to Asc must also be used, having a total area Ah not less than the following:

required Ah ≥ 0.5 ( Asc − An ) (5.16.20)

The area Ah is to be uniformly distributed within two-​thirds of the effective depth from Asc (ACI-​16.5.6.6). Note that the ACI requirements expressed by Eqs. (5.16.17), (5.16.18), and (5.16.20) automatically satisfy the basic equilibrium requirements of Eqs. (5.16.9) and (5.16.14), as well as the minimum Ah requirements of Eqs. (5.16.15) and (5.16.16). That Eq. (5.16.16) is always satisfied can be proved by substituting Eq. (5.16.18) into Eq. (5.16.20), or



2  required Ah ≥ 0.5  A ν f + An − An  3  1 required Ah ≥ A ν f 3

Equation (5.16.18) satisfies Eq. (5.16.14) in recognition that Ah is at least

1

3

Av f .

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5 . 1 6   B rac k ets and C orbels

161

In addition to Eqs. (5.16.17) to (5.17.20), ACI Code limitations and requirements in the design of brackets and corbels, including those described in the introductory material, are as follows. 1. Shear span-​to-​depth ratio a /d may not exceed 1.0 (ACI-​16.5.1.1). 2. Factored tensile force N uc may not exceed factored shear Vu (ACI-​16.5.1.1). 3. Factored tensile force N uc may not be taken less than 0.2Vu unless special precautions are taken to avoid tensile forces (ACI-​16.5.3.5). 4. Critical section is at face of support, where the effective depth d is to be measured (ACI-​16.5.2.1), as shown in Fig. 5.16.4. The effective depth d may not be more than twice the depth d1 at the outer edge of the bearing area (ACI-​16.5.2.2). 5. The strength reduction factor φ is to be taken as 0.75 for all calculations relating to the design of brackets and corbels (ACI-​21.2.1). 6. The maximum strength Vn (or Vu / φ) for which brackets and corbels may be designed using normal-​weight concrete is (ACI-​16.5.2.4) max Vn = 0.2 fc′ bw d ≤ (480 psi + 0.08 fc′ ) bw d ≤ (1600 psi)bw d (5.16.21)



according to ACI-​11.9.3.2.1. For “all-​lightweight” or “sand-​lightweight” concrete, the maximum (ACI-​16.5.2.5) is (280 psi) a a   max Vn =  0.2 − 0.07  fc′ bw d ≤  800 psi −  bw d   d d



(5.16.22)

7. Primary reinforcement Asc at front face must be anchored (a) by a structural weld to a transverse bar of at least equal size to develop a force of Asc f y or (b) by bending Asc bars back to form a horizontal loop or (c) by some means of positive anchorage (ACI-​16.5.6.3). 8. Bearing area must not project beyond straight portion of Asc bars, nor beyond the interior face of the transverse anchor bar if one is provided (ACI-​16.5.2.3).

Additional Recommendations for Detailing Kriz and Raths [5.86] provide several recommendations for detailing, as shown in Fig. 5.16.5. Detail A in that figure is essentially that of ACI-​16.5.6.3. Alternatively, a confinement angle as in Fig. 5.16.6 can be used, to which the main tension bars are welded at the underside. The use of the confinement angle is one of the recommendations in PCI Design Handbook [2.22]. Kriz and Raths also recommended that the outer edge of a bearing plate resting on a corbel should be placed not closer than 2 in. from the outer edge of the Outer edge of the bearing area

Vu a

Nuc

Asc

ρ=

Asc

bd b = corbel width

d1

d ≤ 2d1

Figure 5.16.4  Effective depth of bracket or corbel.

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Note: Distance x should be large enough to prevent contact between outer corbel edge and beam due to possible rotations

min

Elastomeric pad 5 8

2” min

x

min

Asc

Weld must develop the yield strength of Asc Detail A

Restrained precast beam

Unrestrained precast beam

x

1 ” 2

Same bar size

Steel PL

Beam reinforcement not shown

2” min ++++++++ ++++++++++++

5 min 8”

see A

s

5 min 8”

see A

s

3 4 ” min

h/2 min

s

3 4 ” min

h/2 min

s

d

d

h

h

4” min

4” min (a) Corbels subject to vertical load only

(b) Corbels subject to vertical load and restrained creep and shrinkage force. Steel PL ‘s welded or not welded.

Figure 5.16.5  Recommended corbel details. (From Kriz and Raths [5.86].) Face of column Steel angle say 6 × 4 ×

1 2

Vu Asc Main tension steel welded to angle along contact between bars and inside of angle

Figure 5.16.6  Anchorage of main steel provided by welding to a confinement angle.

corbel. PCI Design Handbook recommends a 1-​in. minimum setback when no confinement angle is used, but does not indicate a setback when a confinement angle is used. Further, for good practice, when corbels are designed to resist horizontal forces, steel bearing plates welded to the tension reinforcement should be used to transfer the horizontal forces directly to the tension reinforcement. Details and design aids for brackets and corbels are given in the PCI Handbook [2.22].

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EXAMPLE 5.16.1 Design a typical interior bracket that projects from a 14-​in. square tied column. It must support a dead load reaction of 26 kips and a live load reaction of 51 kips, resulting from gravity loads. Assume that suitable bearings are provided for the supported prestressed concrete girder to avoid the development of horizontal restraint forces. The tolerance gap between the beam end and column face is 1 in. Use fc′= 5000 psi, f y = 60, 000 psi, and the ACI Code. SOLUTION  (a) Factored loads.

Vu = 1.2 VD +1.6 VL = 1.2 ( 26) +1.6 (51) = 113 kips

(b) Preliminary bracket size. The shear span a is dependent on the bearing length required to support the reaction on the concrete. ACI-​22.8.3.2 gives nominal bearing strength as 0.85 fc′A1, where A1 is the bearing area, so that Vu = φ (0.85 fc′ ) A1

using φ = 0.65 (ACI-​21.2.1),

bearing plate width =

113, 000 = 2.9 in. 0.65(0.85)(5000)14

Use 3 in. for bearing plate width. Allowing the tolerance gap of 1-​in. clear between face of column and beam for possible overrun in beam length and also because the beam might be 1 in. too short, the shear span is

1 a = 2 + (bearing plate width) = 2 + 1.5 = 3.5 in. 2

(c) Determine depth of bracket. Based on the maximum strength Vn [Eq. (5.16.21), repeated below] permitted by ACI-​16.5.2.4,     max Vn = 0.2 fc′ bw d ≤ (480 psi + 0.08 fc′ )bw d ≤ (1600 psi)bw d

[5.16.21]

Since 0.2 fc′ = 1000 psi, max vn = Vn / ( bw d ) = 480 + 0.08 ( 5000 ) = 880 psi ; then

min d =

Vu 113, 000 = = 12.2 in. φ bw (max vn ) 0.75(14)(880)

Select overall h = 15 in., d ≈ 13.5 in. (allowing 1-​in. cover). Check:

a 3.5 = = 0.26 < 1.0  (ACI-​16.5.1.1) d 13.5

(d) Determine the shear-​friction reinforcement Av f . Using Eq. (5.16.10) according to ACI-​16.5.4.4,

Aν f =

Vu 113 = = 1.79 sq in . φ f y µ 0.75(60)1.4

using µ = 1.4 for monolithically cast concrete.

(Continued)

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Example 5.16.1 (Continued) (e) Determine main tension reinforcement Asc. First calculate the requirement A f for flexure (ACI 16.5.3.1). Mu = Vu a + Nuc(h –​ d) = Vu a = 113 (3.5) required Rn =



1 12

= 33.0 ft-kips

Mn 33.0(12,000) = = 207 psi 2 φ bd 0.75(14)(13.5)2

From Eq. (3.8.5), required ρ = 0.0035



From ACI-​16.5.5.1, the main steel Asc requirement is the largest of Eqs. (5.16.17) through (5.16.19), as follows:

Asc = A f + An ( zero in this example ) = ( 0.0035)(14 )(13.5) = 0.66 sq in.

or  2  2 Asc =   Avf + An ( zero in this example ) =   (1.79) = 1.119 sq in.  3  3   

Controls

or

Asc = 0.04

fc′  5 bw d = 0.04   (14)(13.5) = 0.63 sq in.  60  fy

Use 4–​#5 (As = 1.24 sq in.). (f) Design closed stirrups or ties. In accordance with ACI-​16.5.5.2, Eq. (5.16.20) requires

required Ah = 0.5( Asc − An ) = 0.5 (1.19) = 0.60 sq in.

Note that in this example, with small a / d = 0.28, the flexure requirement A f does not affect the design; the shear-​friction requirement Avf for Vu controls, with Avf = Asc + Ah [Eq. (5.16.14)]. Use 3–​#3 closed stirrups [Ah = 2(3)0.11 = 0.66 sq in.]. The stirrups must be placed within the upper two-​thirds of the effective depth (ACI-​16.5.6.6). (g) Final design. Referring to Fig. 5.16.7, the overall depth at the face of column is

h = 13.5 +1( cover ) + 0.3125 ( bar radius ) = 14.8 in.

Use h = 15 in. The 1-​in. cover is used here but it could be as little as the 5 8 -​in. bar diameter for precast columns (ACI Table 20.6.1.3.3), but would presumably have to be 1 1 2 in. for cast-​in-​place members (ACI Table 20.6.1.3.1). At the outer edge of the bearing area, the effective depth d1 must be at least half of that used at the face of the column (see Fig. 5.16.4). In this case, making the outer face 8 in. will satisfy the minimum required d1 of 13.7/​2. Another important aspect of the bracket design is the provision of adequate anchorage into the column so that the tensile force Asc f y is available at the face of the column. Development of reinforcement is treated in Chapter 6. (Continued)

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Example 5.16.1 (Continued) a = 3.5” #5 bars welded to underside of steel angle

Available embedment length = 14 – 1.5 – 0.375 = 12.1” (see Chapter 6)

3”

1

1 2 ” clear cover 4 – #5

8” 15” #3 closed stirrups @ 3” spacing #4 7” 14” square column

Figure 5.16.7  Final design for Example 5.16.1.

EXAMPLE 5.16.2 Design a bracket that is to support gravity dead and live loads of 15 and 25 kips, respectively. The vertical reaction is 10 in. from the face of a 14-​in. square column. The bracket is also subjected to a horizontal reaction of 9.5 kips due to creep and shrinkage of a restrained beam. Use fc′ = 5000 psi, f y = 40, 000 psi, and the ACI Code. SOLUTION  (a) Factored loads.

Vu = 1.2 (15) +1.6 ( 25) = 18 + 40 = 58 kips

ACI-​16.5.3.4 states that N uc is to be regarded as live load when it results from creep, shrinkage, or temperature change. Thus N uc = 1.6 (9.5) = 15.2 kips



Also, Nuc must be greater than or equal to 0.2Vu (ACI-​16.5.3.5). N uc 15.2 = = 0.26 > 0.20 min OK             Vu            58 (b) Depth of bracket for shear.

max Vn = 0.2 fc′ bw d ≤ (480 psi + 0.08 fc′ )bw d ≤ (1600 psi)bw d

[5.16.21]

Since 0.2 fc′ = 1000 psi, max vn = Vn / ( bw d ) = 880 psi; then

min d =

Vu 58, 000 = = 6.3 in. φ bw (880) 0.75(14)(880) (Continued)

16

166

C hapter   5     S hear S trength and D esign for   S hear

Example 5.16.2 (Continued) Since this is very small, perhaps the flexure requirement will require a larger d (this is a good possibility because the load on the bracket is 10 in. from the face of column). (c) Depth of bracket for flexure. M u = Vu a + N uc (h − d ) = 58 (10 ) + 15.2(h − d ) Estimating (h – d) at 2 in., Mu = 58 (10 ) +15.2 ( 2 ) = 610 in. -kips Using the minimum reinforcement ratio, fc′  5 = 0.04   = 0.005  40  fy

min ρ = 0.04

which corresponds to minimum Rn = 193 psi (see Section 3.8),

required d =

Mu 610, 000 = = 17.3 in. φ Rn b 0.75(195)14

For a reinforcement ratio of 3%, maximum Rn = 1031 psi, which gives required d =

610, 000 = 7.5 in. 0.75(1031)14

(d) Select bracket depth. Since the provisions of ACI-​16.5 for bracket and corbel design apply only when a / d does not exceed 1.0, min d = a = 10 in. Try a bracket somewhat deeper, say 15 in. overall. This would make d ≈ 13.5 in . (e) Determine the shear-​friction reinforcement Avf . Using Eq. (5.16.10) according to ACI-​16.5.4.4, required A ν f =

Vu 58 = = 1.38 sq in. φ f y µ 0.75(40)1.4

where µ = 1.4 for monolithic concrete. (f) Determine the flexure reinforcement A f . required Rn =

Mu φ bd 2

where (ACI-​16.5.3.1) Mu = Vu a + N uc (h − d ) = 58 (10 ) +15.2 (1.5) = 603 in.-kips required Rn =

603, 000 = 315 psi 0.75(14)(13.5)2

required ρ = 0.0082 [from Eq. (3.8.5) or Fig. 3.8.1] required Af = 0.0082(14)13.5 = 1.55 sq in.

(Continued)

167



167

5 . 1 6   B rac k ets and C orbels

Example 5.16.2 (Continued) (g) Determine additional reinforcement An for axial tension. In accordance with ACI-​ 16.5.4.3, using Eq. (5.16.11), required An =

N uc 15.2 = = 0.51 sq in. φ f y 0.75(40)

(h) Total main tension reinforcement Asc. From ACI-​16.5.5.1, the main steel Asc requirement is the largest of Eqs. (5.16.17) through (5.16.19), as follows: required Asc = Af + An = 1.55 + 0.51 = 2.06 sq in.

Controls

or required Asc =

2 2 A ν f + An = (1.38) + 0.51 = 1.43 sq in. 3 3

or Asc = 0.04

fc′  5 bw d = 0.04   (14)(13.5) = 0.63 sq in.  60  fy

The required Asc = 2.06 sq in. Use 5–​#6 for main tension steel, Asc = 2.20 sq in. (i) Determine stirrup (or tie) requirements. According to ACI-​16.5.5.2, required Ah = 0.5( Asc − An ) = 0.5 ( 2.06 − 0.51) = 0.78 sq in.



Use 3–​#4 closed stirrups, Ah = 0.40 (3) = 1.20 sq in., the spacing of which should be (2/​3)[13.5/​3] = 3.0 in. (ACI-​16.5.6.6). Use 3-​in. spacing.

14” square column

3”

Steel plate welded to main steel

2”

10” Welded

7 21”

5 – #6

#7 welded

15” #4 ties @ 3” spacing 13 21

#4

Figure 5.16.8  Final design for Example 5.16.2.

(Continued)

168

168

C hapter   5     S hear S trength and D esign for   S hear

Example 5.16.2 (Continued) (j) Overall bracket dimensions. Assuming that a 1-​in.-​thick bearing plate is to be welded to the main tension reinforcement, the overall depth is h = bearing plate + bar radius + effective depth d = 1 + 0.375 + 13.5 = 14.9 in., say 15 in. bearing plate length = =

Vu φ 0.85 fc′ (column width) 58, 000 = 1.50 in. 0.65(0.85)5000(14)

Use a 3-​in. plate length as the practical minimum. length of bracket projection = 2 in. + 12 bearing plate + shear span, a

= 2 + 1.5 + 10 = 13.5 in

depth of outer face of bracket =

1 2

overall depth = 7 12  in.

Final design is shown in Fig. 5.16.8.

SELECTED REFERENCES General   5.1. Emil Mörsch. Concrete-​ Steel Construction (Der Eisenbetonbau), The Engineering News Publishing Company, New York, 1909, 368 pp.  5.2. Arthur N.  Talbot. “Tests of Reinforced Concrete Beams:  Resistance to Web Stresses,” Engineering Experiment Station, University of Illinois, January 1909, 85 pp.   5.3. Boris Bresler and James G. MacGregor. “Review of Concrete Beams Failing in Shear,” Journal of the Structural Division, ASCE, 93, ST1 (February 1967), 343–​372.   5.4. ACI-​ASCE Committee 426. “The Shear Strength of Reinforced Concrete Members—​Chapters 1 to 4,” Journal of the Structural Division, ASCE, 99, ST6 (June 1973), 1091–​1187.   5.5. Michael P.  Collins. “Towards a Rational Theory for RC Members in Shear,” Journal of the Structural Division, ASCE, 104, ST4 (April 1978), 649–​666.   5.6. Michael P.  Collins and Denis Mitchell. “Shear and Torsion Design of Prestressed and Non-​ Prestressed Concrete Beams,” PCI Journal, 25, September–​October 1980, 32–​100. Disc. 26, November–​December 1981, 96–​118.   5.7. Peter Marti. “Basic Tools of Reinforced Concrete Beam Design,” ACI Journal, Proceedings, 82, January–​February 1985, 46–​56.  5.8. Frank J.  Vecchio and Michael P.  Collins. “The Modified Compression-​ Field Theory for Reinforced Concrete Elements Subjected to Shear,” ACI Journal, Proceedings, 83, March–​April 1986, 219–​231. Disc. 84, January–​February 1987, 87–​90.   5.9. Jörg Schlaich, Kurt Schäfer, and Mattias Jennewein. “Toward a Consistent Design of Structural Concrete,” PCI Journal, 32, May–​June 1987, 74–​150. 5.10. Frank J.  Vecchio and Michael P.  Collins. “Predicting the Response of Reinforced Concrete Beams Subjected to Shear Using Modified Compression Field Theory,” ACI Structural Journal, 85, May–​June 1988, 258–​268. 5.11. Julio A.  Ramirez and John E.  Breen. “Evaluation of a Modified Truss-​Model Approach for Beams in Shear,” ACI Structural Journal, 88, September–​October 1991, 562–​571. 5.12. Khaled A. Al-​Nahlawi and James K. Wight. “Beam Analysis Using Concrete Tensile Strength in Truss Models,” ACI Structural Journal, 89, May–​June 1992, 284–​289. 5.13. Michael P.  Collins, Denis Mitchell, Perry Adebar, and Frank J.  Vecchio. “A General Shear Design Method,” ACI Structural Journal, 93, January–​February 1996, 36–​45.

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5.14. A. Koray Tureyen and Robert J. Frosch. “Concrete Shear Strength: Another Perspective,” ACI Structural Journal, 100, September–​October 2003, 609–​615. 5.15. ASCE-​ACI Committee 445. “Recent Approaches to Shear Design of Structural Concrete,” Journal of Structural Engineering, ASCE, 124, 12 (December 1998), 1375–​1417. 5.16. ACI–​ASCE Committee 426. Suggested Revisions to Shear Provisions for Building Codes. Detroit: American Concrete Institute, 1979, 82 pp. 5.17. ACI-​ASCE Committee 326. “Shear and Diagonal Tension,” ACI Journal, Proceedings, 59, January, February, and March 1962, 1–​30, 277–​344, and 352–​396.

Aggregate Interlock 5.18. T.  Paulay and P.  J. Loeber. “Shear Transfer by Aggregate Interlock,” Shear in Reinforced Concrete, Vol. 1 (SP-​42). Detroit: American Concrete Institute, 1974 (pp. 503–​537). 5.19. Theodossius P. Tassios and Elizabeth N. Vintzeleou, “Concrete-​to-​Concrete Friction,” Journal of Structural Engineering, ASCE, 113, 4 (April 1987), 832–​849. 5.20. Sandro Dei Poli, Pietro G. Gambarova, and Cengiz Karakoc. “Aggregate Interlock Role in R.C. Thin-​Webbed Beams in Shear,” Journal of Structural Engineering, ASCE, 113, 1 (January 1987), 1–​19.

Dowel Action 5.21. David W. Johnston and Paul Zia. “Analysis of Dowel Action,” Journal of the Structural Division, ASCE, 97, ST5 (May 1971), 1611–​1630. 5.22. Rafael Jimenez, Richard N. White, and Peter Gergely. “Cyclic Shear and Dowel Action Models in R/​C,” Journal of the Structural Division, ASCE, 108, ST5 (May 1982), 1106–​1123. 5.23. Parviz Soroushian, Kienuwa Obaseki, Maximo C.  Rojas, and Jongsung Sim. “Analysis of Dowel Bars Acting Against Concrete Core,” ACI Journal, Proceedings, 83, July–​August 1986, 642–​649. 5.24. E.  N. Vintzeleou and T.  P. Tassios. “Behavior of Dowels under Cyclic Deformations,” ACI Structural Journal, 84, January–​February 1987, 18–​30. 5.25. P.  Soroushian, K.  Obaseki, M.  Rojas, and H.  S. Najm. “Behavior of Bars in Dowel Action Against Concrete Cover,” ACI Structural Journal, 84, March–​April 1987, 170–​176. 5.26. Elizabeth N. Vintzeleou and Theodossius P. Tassios. “Eccentric Dowels Loaded Against Core of Concrete Sections,” Journal of Structural Engineering, ASCE, 116, 10 (October 1990), 2621–​2633.

Beams Without Shear Reinforcement 5.27. JoDean Morrow and I. M. Viest. “Shear Strength of Reinforced Concrete Frame Members without Web Reinforcement,” ACI Journal, Proceedings, 53, March 1957, 833–​869. 5.28. G.  N. J.  Kani, “How Safe Are Our Large Reinforced Concrete Beams?” ACI Journal, Proceedings, 64, March 1967, 128–​141. 5.29. Michael P. Collins and Daniel Kuchma. “How Safe Are Our Large, Lightly Reinforced Concrete Beams, Slabs, and Footings?” ACI Structural Journal, 96, July–​August 1999, 482–​491. 5.30. Edward G.  Sherwood, Evan C.  Bentz, and Michael P.  Collins. “Effect of Aggregate Size on Beam-​Shear Strength of Thick Slabs,” ACI Structural Journal, 104, March-​April 2007, 180–​190. 5.31. G. N. J. Kani. “The Riddle of Shear Failure and Its Solution,” ACI Journal, Proceedings, 61, April 1964, 441–​467.

Lightweight Aggregate 5.32. John M.  Hanson. Square Openings in Webs of Continuous Joists (RD001.01D). Skokie, IL: Portland Cement Association, 1969. 5.33. J.  A. Hanson. “Tensile Strength and Diagonal Tension Resistance of Structural Lightweight Concrete, ACI Journal, Proceedings, 58, July 1961, 1–​40. 5.34. E.  Hognestad, R.  C. Elstner, and J.  A. Hanson. “Shear Strength of Reinforced Structural Lightweight Aggregate Concrete Slabs,” ACI Journal, Proceedings, 61, June 1964, 643–​656. 5.35. Don L. Ivey and Eugene Buth. “Shear Capacity of Lightweight Concrete Beams,” ACI Journal, Proceedings, 64, October 1967, 634–​643.

170

170

C hapter   5     S hear S trength and D esign for   S hear

High-​Strength Concrete 5.36. Andrew G. Mphonde and Gregory C. Frantz. “Shear Tests of High-​and Low-​Strength Concrete Beams without Stirrups,” ACI Journal, Proceedings, 81, July–​August 1984, 350–​357. 5.37. Ashraf H.  Elzanaty, Arthur H.  Nilson, and Floyd O.  Slate. “Shear Capacity of Reinforced Concrete Beams Using High-​Strength Concrete,” ACI Journal, Proceedings, 83, March–​April 1986, 290–​296. 5.38. Shuaib H. Ahmad, A. R. Khaloo, and A. Poveda. “Shear Capacity of Reinforced High-​Strength Concrete Beams,” ACI Journal, Proceedings, 83, March–​April 1986, 297–​305. 5.39. Ashraf H.  Elzanaty, Arthur H.  Nilson, and Floyd O.  Slate. “Shear Capacity of Prestressed Concrete Beams Using High-​Strength Concrete,” ACI Journal, Proceedings, 83, May–​June 1986, 359–​368. 5.40. Shuaib H.  Ahmad and D.  M. Lue. “Flexure-​Shear Interaction of Reinforced High-​Strength Concrete Beams.” ACI Structural Journal, 84, July–​August 1987, 330–​341. Disc. 85, May–​June 1988, 354–​358. 5.41. M.  Keith Kaufman and Julio A.  Ramirez. “Re-​evaluation of the Ultimate Shear Behavior of High-​Strength Concrete Pre-​stressed I-​Beams,” ACI Structural Journal, 85, May–​June 1988, 295–​303. 5.42. Miguel A. Salandra and Shuaib H. Ahmad. “Shear Capacity of Reinforced Lightweight High-​ Strength Concrete Beams,” ACI Structural Journal, 86, November–​December 1989, 697–​704. 5.43. John J. Roller and Henry G. Russell. “Shear Strength of High-​Strength Concrete Beams with Web Reinforcement,” ACI Structural Journal, 87, March–​April 1990, 191–​198.

Truss Models and Shear Reinforcement 5.44. W.  Ritter. “Die Bauweise Hennebique,” Schweizerische Bauzeitung, 33, February 1899, pp. 59-​61. 5.45. Willis A.  Slater, Arthur R.  Lord, and Roy R.  Zipprodt. “Shear Tests of Reinforced Concrete Beams,” Technologic Papers of the Bureau of Standards, No. 314, 20, April 1926, pp. 389-​495. 5.46. Frank E. Richart. “An Investigation of Web Stresses in Reinforced Concrete Beams,” Bulletin No. 166, University of Illinois Engineering Experiment Station, June 1927, 105 pp. 5.47. Peter Marti. “Truss Models in Detailing,” Concrete International, 7, 12 (December 1985), 66–​73. 5.48. M.  P. Nielsen. Limit Analysis and Concrete Plasticity (2nd ed.). Boca Raton, FL:  CRC Press, 1998. 5.49. Denis Mitchell and Michael P.  Collins. “Diagonal Compression Field Theory—​A Rational Model for Structural Concrete in Pure Torsion,” ACI Journal, 71, August 1974, 396–​408. 5.50. Thomas T.C. Hsu. “Softened Truss Model Theory for Shear and Torsion,” ACI Structural Journal, 85, November–​December 1988, 624–​635. 5.51. Wayne Hsiung and Gregory C. Frantz. “Transverse Stirrup Spacing in R/​C Beams,” Journal of Structural Engineering, ASCE, 111, 2 (February 1985), 353–​362. Disc. 113, 1 (January 1987), 174–​177. 5.52. Mark K. Johnson and Julio A. Ramirez. “Minimum Shear Reinforcement in Beams with Higher Strength Concrete,” ACI Structural Journal, 86, July–​August 1989, 376–​382. Disc. 87, May–​ June 1989, 362–​364. 5.53. Neal S. Anderson and Julio A. Ramirez. “Detailing of Stirrup Reinforcement,” ACI Structural Journal, 86, September–​October 1989, 507–​515. 5.54. Andrew G. Mphonde. “Use of Stirrup Effectiveness in Shear Design of Concrete Beams,” ACI Structural Journal, 86, September–​October 1989, 541–​545. 5.55. Andrew G. Mphonde and Gregory C. Frantz. “Shear Tests of High and Low-​Strength Concrete Beams with Stirrups,” High-​Strength Concrete (SP-​87). Detroit: American Concrete Institute, 1985 (pp. 179–​196). 5.56. Abdeldjelil Belarbi and Thomas T. C. Hsu. “Stirrup Stresses in Reinforced Concrete Beams,” ACI Structural Journal, 87, September–​October 1990, 530–​538. 5.57. CSA. Design of Concrete Structures for Buildings (CAN3-​A23.3-​04). Rexdale, Ontario: Canadian Standards Association (178 Rexdale Blvd., Rexdale, Ontario, Canada M9W IR3), 2004. 5.58. S. M. Fereig and K. N. Smith. “Indirect Loading on Beams with Short Shear Span,” ACI Journal, Proceedings, 74, May 1977, 220–​222. 5.59. K. S. Rajagopalan and P. M. Ferguson. “Exploratory Shear Tests Emphasizing Percentage of Longitudinal Steel,” ACI Journal, Proceedings, 65, August 1968, 634–​638. Disc. 66, 150–​154. 5.60. Barrington deV. Batchelor and Mankit Kwun. “Shear in RC Beams without Web Reinforcement,” Journal of the Structural Division, ASCE, 107, ST5 (May 1981), 907–​921. 5.61. Michael N. Palaskas, Emmanuel K. Attiogbe, and David Darwin. “Shear Strength of Lightly Reinforced T-​Beams,” ACI Journal, Proceedings, 78, November–​December 1981, 447–​455.

17



S elected R eferences

171

Fiber Reinforced Concrete 5.62. Gustavo J. Parra-​Montesinos. “Shear Strength of Beams with Deformed Steel Fibers,” Concrete International, 28, November 2006, 57-​66.

Axial Load Effect 5.63. James G.  MacGregor and John M.Hanson. “Proposed Changes in Shear Provisions for Reinforced and Prestressed Concrete Beams,” ACI Journal, Proceedings, 66, April 1969, 276–​288. Disc. 849–​851. 5.64. Alan H. Mattock. “Diagonal Tension Cracking in Concrete Beams with Axial Forces,” Journal of the Structural Division. ASCE, 95, ST9 (September 1969), 1887–​1900. 5.65. Munther J.  Haddadin, Sheu-​ Tien Hong, and Alan M.  Mattock. “Stirrup Effectiveness in Reinforced Concrete Beams with Axial Force,” Journal of the Structural Division, ASCE, 97, ST9 (September 1971), 2277–​2297. 5.66. Shrinivas B. Bhide and Michael P. Collins. “Influence of Axial Tension on the Shear Capacity of Reinforced Concrete Members,” ACI Structural Journal, 86, September–​October 1989, 570–​581. 5.67. Alan H. Mattock and Zuhua Wang. “Shear Strength of Reinforced Concrete Members Subject to High Axial Compressive Stress,” ACI Journal, Proceedings, 81, May–​June 1984, 287–​298.

Shear Transfer; Shear Friction; Corbels 5.68. Philip W. Birkeland and Halvard W. Birkeland. “Connections in Precast Concrete Construction,” ACI Journal, Proceedings, 63, March 1966, 345–​368. 5.69. R. F. Mast. “Auxiliary Reinforcement in Precast Concrete Connections,” Journal of the Structural Division, ASCE, 94, ST6 (June 1968), 1485–​1504. 5.70. J. A. Hofbeck, I. O. Ibrahim, and Alan H. Mattock. “Shear Transfer in Reinforced Concrete,” ACI Journal, Proceedings, 66, February 1969, 119–​128. Disc. 66, August 1969, 678–​680. 5.71. A. H. Mattock and N. M. Hawkins. “Research on Shear Transfer in Reinforced Concrete,” PCI Journal, 17, March–​April 1972, 55–​75. 5.72. Bjorn R. Hermansen and John Cowan. “Modified Shear-​Friction Theory for Bracket Design,” ACI Journal, Proceedings, 71, February 1974, 55–​60. 5.73. A.  H. Mattock. Disc. of “Modified Shear-​Friction Theory for Bracket Design,” by B.  R. Hermansen and J. Cowan, ACI Journal, Proceedings, 71, August 1974, 421–​423. 5.74. A.  H. Mattock, “Shear Transfer in Concrete Having Reinforcement at an Angle to the Shear Plane,” Shear in Reinforced Concrete, Vol. 1 (SP-​42). Detroit:  American Concrete Institute, 1974 (pp. 17–​42). 5.75. Alan H.  Mattock, L.  Johal, and H.  C. Chow. “Shear Transfer in Reinforced Concrete with Moment or Tension Acting Across the Shear Plane,” PCI Journal, 20, July–​August 1975, 76–​93. 5.76. Alan H.  Mattock, W.  K. Li, and T.  C. Wang. “Shear Transfer in Lightweight Reinforced Concrete,” PCI Journal, 21, January–​February 1976, 20–​39. 5.77. Joost Walraven, Jerome Frenay, and Arjan Pruijssers. “Influence of Concrete Strength and Load History on the Shear Friction Capacity of Concrete Members,” PCI Journal, 32, January–​ February 1987, 66–​84. 5.78. Thomas T. C. Hsu, S. T. Mau, and Bin Chen. “Theory of Shear Transfer Strength of Reinforced Concrete,” ACI Structural Journal, 84, March–​April 1987, 149–​160. 5.79. Robert A. Bass, Ramon L. Carrasquillo, and James O. Jirsa. “Shear Transfer Across New and Existing Concrete Interfaces,” ACI Structural Journal, 86, July–​August 1989, 383–​393. 5.80. G. Annamalal and Robert C. Brown, Jr. “Shear Transfer Behavior of Post-​Tensioned Grouted Shear-​Key Connections in Precast Concrete-​Framed Structures,” ACI Structural Journal, 87, January–​February 1990, 53–​59. 5.81. Nijad I. Fattuhi. “Reinforced Corbels Made with High-​Strength Concrete and Various Secondary Reinforcements,” ACI Structural Journal, 91, July–​August 1994, 376–​383. Disc. 92, May–​June 1995, 386–​387. 5.82. PCI Committee on Connection Details. Design and Typical Details of Connections for Precast and Prestressed Concrete (2nd ed). Chicago: Prestressed Concrete Institute, 1988. 5.83. Alan H. Mattock, “Design Proposals for Reinforced Concrete Corbels,” PCI Journal, 21, May–​ June 1976, 18–​42. Disc., 22, March–​April 1977, 90–​109. 5.84. A Fattah Shaikh, “Proposed Revisions to Shear-​Friction Provisions,” PCI Journal, 23, March–​ April 1978, 12–​21.

172

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C hapter   5     S hear S trength and D esign for   S hear

5.85. Himat Solanki and Gajanan M.  Sabnis. “Reinforced Concrete Corbels—​ Simplified,” ACI Structural Journal, 84, September–​October 1987, 428–​432. 5.86. L. B. Kriz and C. H. Raths, “Connections in Precast Concrete Structures—​Strength of Corbels,” PCI Journal, 10, 1 (February 1965), 16–​47. 5.87. J.  G. MacGregor and N.  M. Hawkins. “Suggested Revisions to ACI Building Code Clauses Dealing with Shear Friction and Shear in Deep Beams and Corbels,” ACI Journal, Proceedings, 74, November 1977, 537–​545. Disc., 75, May 1978, 221–​224.

PROBLEMS All problems13 are to be worked in accordance with the ACI Code unless otherwise indicated, and all stated loads are service loads. Use the “correct” shear envelope (i.e., live load must be appropriately applied such as to cause maximum effect). All shear diagrams are to be drawn to scale in terms of Vu directly below the diagram of the beam, also drawn to scale. Use the Vu diagram for design by scaling values from it; also scale from it the locations where stirrup spacings are permitted. Computation of Vu values and locations of stirrup spacings may be made to confirm scaled information. Unless otherwise indicated, use the load factors U of ACI-​5.3.1 and the φ factors of ACI-​21.2. 5.1 The simply supported beam of 16-​ft span, shown in the figure for Problem 5.1, is to carry a uniform dead load of 1.6 kips/​ft (including beam weight) and a uniform live load of 2.6 kips/​ft. Use fc′ = 3500 psi and f yt = 60, 000 psi. (a) Determine the adequacy of the #3 U stirrups that are spaced at 8 in. Use the simplified method with constant Vc . (b) Draw superimposed on the factored shear Vu diagram the diagram of strength provided φVn. From the comparison of φVn with Vu, determine if the stirrups are adequate for the entire beam. If not, make recommendations to obtain a beam having adequate shear strength. 5.2 The beam in the figure for Problem 5.2 carries a uniform live load of 3.6 kips/​ft in addition to its own weight. Assume a support width of 12 in., and use fc′ = 4500 psi and f yt = 60, 000 psi. (a) Draw the maximum factored shear Vu envelope. Calculate the value at the critical section, at the 1 4 point, and at midspan; then

draw a smooth curve through these points for design use. (b) Draw the curve for required stirrup spacing using the simplified method for calculating Vc (ACI ​ Formula 22.5.5.1); show on the same diagram the spacings provided. (c) Use the simplified method of constant Vc (ACI F ​ ormula 22.5.5.1) to evaluate whether the spacings used satisfy the ACI Code. (d) Repeat part (b) using the more detailed procedure of ACI Table 22.5.5.1; show also the spacings provided. (e) Evaluate whether the spacings used are satisfactory according to the analysis of part (d). 5.3 The beam in the figure for Problem 5.3 is to carry 1.6 kips/​ft live load and 0.90 kip/​ft dead load (including beam weight). Using fc′ = 3000 psi and f yt = 40, 000 psi, investigate the beam for stirrup adequacy according to the simplified method using constant Vc . If design is not adequate, indicate what revision is necessary.

18.5” 12”

4”

3 @ 9” 4 @ 8” # 3U stirrups

5 – #8

5 – #8

14”

Sym. abt CL

8’ – 0”

Problem 5.1 

13  Some problems may be solved as problems stated in Inch-​Pound units, or as problems in SI units using quantities in parentheses. To avoid implying higher precision for the given information in SI units than that given for the Inch-​Pound units, the metric conversions are approximate.

173



173

P roble m s

25”

5 – #9 6 @ 10” = 5’–0” 8”

Same stirrups as at other support

#3 U stirrups

5 @ 6” = 2’–6”

12”

20’–0”

Problem 5.2  bE = 60”

Symmetrical about CL span 4”

d = 18”

12” supports 9”

9 @ 6”

3 – #7

22”

#3 U stirrups

3 – #9 12”

22’–0”

Problem 5.3 

d = 19.5” 12”

12” 9’–0” 58k

Factored shear Vu

CL of span

18k

Problem 5.4 

5.4 For the portion of the continuous beam shown 5.5 For the beam shown in the figure for Problem in the figure for Problem 5.4, with the given 5.5 and the case assigned by the instructor, use portion of the factored shear Vu envelope, the simplified method of ACI ​Formula 22.5.5.1 determine the spacings to be used for #3 U to design vertical U stirrups (i.e., specify their stirrups. Dimension and show the stirrups on size, dimension their locations, and show them the given portion of the beam. Use the simplion a side view of the beam). Use whole inches fied method of constant Vc , with fc′ = 3500 psi for spacings. The beam is simply supported and and f yt = 60, 000 psi. (Beam: b = 300 mm; has a support width of 18 in.; and the given dead d  =  530  mm; support width  =  300  mm; half-​ load does not include the beam weight. Choose span = 2.7 m; Vu at support = 260 kN; Vu at midthe stirrup size to ensure that the spacing will not span = 80 kN; use #10M stirrups; fc′ = 25 MPa; be closer than 3 in. Explicitly state the length of f yt = 400 MPa.) the beam over which stirrups are required.

Case Span (ft)

bw (in.)

d (in.)

As (sq in.)

1

20

18

36.3

2 3 4

26 30 28

18 16 18

29.6 24.6 29.6

5-​#10 5-​#11 6-​#9 9-​#9 6-​#9

Dead Load (kips/​ft)

Live Load (kips/​ft)

fc′ (psi)

fyt (psi)

6

10

4000

40,000

1.6 1.2 1.6

2.7 2.5 2.5

4000 4000 4000

60,000 60,000 60,000

174

174

C hapter   5     S hear S trength and D esign for   S hear

d

h

As bw

18”

Symmetrical about CL of span

L/2

Problem 5.5 

5.6 For the simply supported beam shown in the figure for Problem 5.6 having a span of 32 ft, with uniform dead load of 2.3 kips/​ft (including beam weight) and live load of 3.7 kips/​ft, design #4 vertical U stirrups using whole inches for spacings. Show the stirrups on a side view of the beam and dimension their locations. Use fc′ = 3750 psi and f yt = 50, 000 psi, and the simplified method with constant Vc . The support width is 12 in.

d = 24.5”

15”

Problem 5.6 

5.7 A  reinforced concrete simply supported beam (b = 250 mm, d = 410 mm) must carry on a span of 5.5 m a single concentrated moving load of 45 kN plus a uniform dead load of 30 kN/​m (including beam weight). Design and detail the stirrup spacing (use 20-​ mm increments) for #10M vertical U stirrups; support width is 300 mm; fc′ = 25 MPa; f yt = 400 MPa. Apply the simplified method using a constant value for Vc. See SI footnote to Eq. (5.10.3). 5.8 The beam shown in the figure for Problem 5.8 is to carry dead load of 1.4 kips/​ft (including beam weight) and live load of 1.6 kips/​ft. Use fc′ = 3500 psi, f yt = 40, 000 psi, and neglect any

compression steel effect. Use a “correct” factored shear Vu envelope. Using the simplified procedure of constant Vc , design and detail the spacings for #3 U stirrups for the beam. Use whole inches for spacings.   5.9 Repeat Problem 5.8, except use an 18-​ft main span and a 7-​ft cantilever, with dead load of 1.5 kips/​ft (including beam weight) and live load of 1.7 kips/​ft. All other dimensions and reinforcing bars are the same as in Problem 5.8. 5.10 The beam shown in the figure for Problem 5.10 is to carry a dead load of 55 kN/​m (including beam weight) and a live load of 72 kN/​m. Use fc′ = 25 MPa and f yt = 400 MPa. Using the simplified procedure of constant Vc, design and detail on the beam 20-​mm increment spacings of U stirrups for the beam. Choose a stirrup size to ensure that the 20-​mm increment spacing will not be closer than 80 mm. 5.11 For a simply supported beam of 16-​ft span, having support widths of 12 in., design and detail on the beam U stirrups (use no spacing less than 3 in.). The dead load is 1.6 kips/​ft (including weight of beam) and the live load is 3.0 kips/​ft. Use fc′ = 3000 psi and f yt = 60, 000 psi. Use the ACI Code simplified procedure with constant Vc . Assume bw = 14 in. and d = 21.5 in. 5.12 A  reinforced concrete simply supported beam of span 5.5 m carries a concentrated dead load of 23 kN at 1.8 m from the left support and a uniform dead load of 90 kN/​m. The width of support is 300 mm. The rectangular beam has a 300-​mm width and a 650-​mm effective depth d.

9’–0” 2 – #8 4’–6” 2 – #9

4 – #9 12” supports

2 – #8

4 – #8

24” 12”

2’–6” 20’–0” 18’–0”

Problems 5.8 and 5.9 

4’–0”

8’–0” 7’–0”

Prob. 5.8 Prob. 5.9

175



175

P roble m s

2.5 m

3 – #45M

300 mm

660 mm Supports

4 – #30M 5.6 m

2.8 m

360 mm

Problem 5.10 

Use fc′ = 30 MPa and f yt = 400 MPa. Design and specify by dimensioning the spacings to be used for #10M U stirrups. Use the simplified method of constant Vc (ACI-​Formula 22.5.5.1). 5.13 For a rectangular beam of 14 in. width and effective depth 22.5 in. with fc′ = 4000 psi and f yt = 60, 000 psi, determine the maximum factored shear Vu for this beam for the following conditions: (a) When no stirrups are to be used. (b) When minimum amount of shear reinforcement (#3 U stirrups) is used according to ACI-​9.6.3.3; specify the spacing to be used. (c)  When maximum amount of shear reinforcement is used (#4 U stirrups) according to ACI-​22.5.1.2; specify the spacing to be used. 5.14 Completely design and detail the stirrups for the beam of Problem 5.3 (ignore the spacings given), using the more detailed ρVd / M method of ACI Table 22.5.5.1. 5.15 For the case assigned by the instructor, repeat the requirements of Problem 5.5, except use the more detailed ρVd / M method of ACI Table 22.5.5.1. 5.16 Repeat the requirements of Problem 5.8, except use the more detailed ρVd / M method of ACI Table 22.5.5.1. 5.17 If a factored axial compression N uc of 140 kips is acting additionally on the beam of Example  5.13.1, determine the number of stirrups that may be eliminated by taking the compression into account when computing Vc by the more detailed procedure involving ρVd / M . (Note: This compressive force is approximately 0.1 fc′Ag and might reasonably be neglected in designing the section for flexure.) 5.18 Reinvestigate the shear reinforcement for the beam of Problem 5.1 if an axial tensile force of 35 kips live load is acting. Redesign stirrups for the beam using ACI Formula 22.5.7.1). 5.19 Redesign the stirrups for the beam case of Problem 5.5 assigned by the instructor if an

axial compressive force of 70 kips dead load and 120 kips live load is acting on the beam. (a) Use simplified method using ACI Formula (22.5.6.1). (b)  Use more detailed ρVd / M method (ACI Table 22.5.6.1). 5.20 Design the details of the bearing shoe on a prestressed girder of 12 in. width. Assume a vertical crack forms at the support as shown in Fig. 5.15.4(a). The reaction is 35 kips dead load and 40 kips live load. The girder concrete has fc′ = 6000 psi. Assume no horizontal restraint is developed. 5.21 Design a bracket (corbel) that projects from one side of a 16 × 16 column to support a verti­ cal load of 35 kips dead load and 65 kips live load. Assume that suitable bearings are provided to prevent any horizontal restraint. The reaction is located 5 in. from the column face. Use fc′ = 5000 psi and f y = 60, 000 psi . (Column size = 400 × 400 mm; dead load = 160 kN; live load = 290 kN; reaction 130 mm from column face; fc′= 35 MPa; f y = 400 MPa.) 5.22 Redesign the bracket (corbel) of Problem 5.21 if the reaction is 3.5 in. (90 mm) from the column face. 5.23 Design for the conditions of Problem 5.21 except take the reaction location 9 in. (230 mm) from the column face. 5.24 Repeat Problem 5.21 if the reaction is from a restrained beam that induces a horizontal tension equal to 50% of the total gravity reaction. 5.25 Redesign the bracket (corbel) of Problem 5.24 if the reaction is 3.5 in. (90  mm) from the column face. 5.26 Repeat Problem 5.23 if the reaction is from a restrained beam that induces a horizontal tension equal to 40% of the total gravity reaction. 5.27 Redesign the bracket (corbel) of Example 5.16.1 considering that the supported prestressed girder is welded to the bracket. Creep, shrinkage, and temperature effects on the restrained girder induce a horizontal force of 50 kips (unfactored) on the bracket.

CHAPTER 6

DEVELOPMENT OF REINFORCEMENT

6.1 GENERAL A basic requirement in reinforced concrete construction is the adequate and reliable transfer of the force in the reinforcement to the surrounding concrete. Consider the bar embedded in a concrete block a length L, with an applied tensile force as shown in Fig. 6.1.1. As the bar is loaded, the tensile force T must be balanced by stresses acting on the bar surface over the embedded length L. These interacting stresses between the bar and the surrounding concrete have been traditionally referred to as bond stresses. If the embedded length of the bar, L, is too short, the bond stresses may not be able to balance the applied bar force T. Consequently, the bar will begin to slip and eventually pull out or split the concrete. The length of embedment necessary to transfer the full bar force into the concrete is called development length. In design, the development length is usually calculated to develop the specified yield strength of the bar. In members subjected to seismic actions, however, a greater development length may be required.

Concrete cantilever beams, Denver (Photo by C. G. Salmon).

17



6.2 DEVELOPMENT LENGTH

177

T

Steel bar

L

Concrete

Figure 6.1.1  Bond stresses between a bar and surrounding concrete along the bar length.

The sources of bond stresses include chemical adhesion, friction, and bearing against the concrete of the raised ribs, or “lugs,” of the deformed bars. Chemical adhesion between the steel reinforcing bars and the concrete does not offer great resistance. For smooth bars (i.e., without ribs), very long embedment lengths would be required to develop the bar yield strength before breaking the adhesion between the bar surface and the surrounding concrete. Friction can provide added resistance, but it is not always a reliable transfer mechanism. As the bar is stretched in tension, its diameter will be slightly reduced owing to Poisson’s effect, but enough to gradually detach it from the surrounding concrete. The higher the bar force, the greater the reduction in the bar diameter and the smaller the friction resistance. For these reasons, smooth bars (i.e., without ribs) are not used in practice (except for confinement of the concrete). Instead, only deformed bars with specially designed rib patterns, rib angles, and rib gap (see Section 1.13) are specified. The bond transfer mechanism associated with bearing of the bar ribs against the surrounding concrete, as well as the concepts relating to the transfer of force between reinforcement and the surrounding concrete over the development length, with or without additional mechanical end anchorage, are presented in the next several sections. The mechanics of the behavior has been explained by ACI Committee 408 [6.1], Lutz and Gergely [6.2], Orangun, Jirsa, and Breen [6.3, 6.4], Jirsa, Lutz, and Gergely [6.5], Yankelevsky [6.6], and Kemp [6.7].

6.2 DEVELOPMENT LENGTH Design of longitudinal and shear reinforcement to accommodate the moment and shear at sections along a beam has been treated in Chapters 3, 4, and 5. To resist bending moment, an area of longitudinal steel is provided to carry the tensile force. However, the bars must be embedded in the concrete sufficiently to allow the tensile force to develop. If there is inadequate development length, the bars will either pull out or split the surrounding concrete. The flexural strength of a beam is, therefore, a three-​dimensional relationship involving not only the cross-​sectional properties at a location along the span, but also the embedment lengths of the steel bars in both directions therefrom. Consider a uniformly loaded cantilever beam as shown in Fig. 6.2.1(a), which has been properly proportioned so that at nominal strength Mn the steel force is As fy. To illustrate the concept of development length, assume the tension reinforcement consists of a single bar of diameter db. Consider the free-​body diagram of the bar segment A-​B, as shown in Fig. 6.2.1(b). The tensile force at B, which is f y (πdb2 / 4) , must be transmitted to the concrete by the interaction between the bar and the surrounding concrete over the development

178

178

C hapter   6     D evelopment of R einforcement

length L1 = AB. If us, the failure stress against slippage acting over the nominal surface area πdb L1, is assumed to be constant over L1, then equilibrium requires that at a slippage failure us π db L1 = f y π



db2 4

or L1 =

fy 4us

db

(6.2.1)



On the other hand, if ub is the failure stress against splitting and Abr is the average bearing area per unit length (also assumed constant over L1), then, at a splitting failure ub Abr L1 = f y π



db2 4

or L1 =

fy Abr ub

π

db2 4

(6.2.2)

In other words, the bar must be provided with a development length at least equal to or greater than that given by Eq. (6.2.1) (for the bar to reach yield prior to pullout) or by Eq. (6.2.2) (to prevent a splitting failure). The same situation exists in free body B-​C, as shown in Fig. 6.2.1(c). Thus the maximum tensile force at B has to develop by embedment in both directions from B: that is, both the A-​B and B-​C distances. Where space limitations prevent providing the proper amount of straight embedment, such bars may be terminated by standard hooks (as defined in ACI-​25.3). A standard hook is permitted to be considered as contributing to an equivalent development length by mechanical action (ACI-​25.4.3), thus reducing the total embedment length required. Section 6.10 treats the subject of providing development length with standard hooks. Adequate development length must be provided for a reinforcing bar in compression as well as in tension.

w A

B

C

L2

L1

(a)

u

L1 T=

A

B

πdb2 f 4 y

u

T

L2

C

B

(b)

Figure 6.2.1  Development of reinforcement.

(c)

179



179

6.3 FLEXURAL BOND

6.3 FLEXURAL BOND As bending moment varies along a span, the tensile force in the steel also varies; this induces flexural bond, that is, the longitudinal interaction between the bars and the surrounding concrete. High localized stress, either at the bar surface or at the bearing of the steel reinforcement lugs against the concrete, exists at locations along the span where the rate of change of tensile force in the bars is high. Consider a segment D - D ′ of the reinforcing bar in the cantilever beam used in Section 6.2. As shown by the free body of D - D ′ in Fig. 6.3.1(b), TD is slightly greater than TD′. The bending moment MD equals the internal force C or T times the moment arm between them; thus, TD =



MD arm

and

TD ′ =

MD′ arm (6.3.1)

Also, from horizontal force equilibrium, us π db dz + ub Abr dz = TD – TD ′ (6.3.2)



in which us is the localized surface stress over the nominal contact area between the steel bar and the concrete, db is the diameter of the single bar, and ub is the localized bearing stress over the area Abr per unit length between the lugs and the concrete. Whether the action represented by the left side of Eq. (6.3.2) is more “pullout” or “splitting,” the magnitude depends on (TD – TD ′ ) / dz , which from Eq. (6.3.1) is dM / (arm ) dz  = V / (arm ). Thus, this localized interaction (i.e., flexural bond) between bar and surrounding concrete is proportional in magnitude to the shear. Even though high localized surface stress on bars resulting from the rate of change of moment may seem to be important for proper design, research and experience in practice have shown that prevention of bar “pullout” failure or “splitting” failure by having adequate development length of embedment is a sufficient criterion. Several cases may be identified where the local situation has little significance in design, provided the bars have adequate development length. 1. In a region of low bending moment and the concrete is uncracked: thus the change in force, say (TD – TD ′ ) in Fig. 6.3.1, in the bars is overestimated because concrete actually carries part of the tensile force. 2. At a point where high bending moment (and low shear) exists and therefore where a flexural crack will likely occur, the low rate of change of moment (i.e., shear) would indicate low (TD − TD ′ ) . Adjacent to the flexural crack (see Fig. 6.3.2), however, the

D B

A

D’ C

dz

(a) D

u

D’

TD

TD’ dz (b)

Figure 6.3.1  Flexural bond in a tension bar.

180

180

C hapter   6     D evelopment of R einforcement

Flexural cracks

Calculated fs (at crack)

fs between cracks

Near max ultimate surface stress 0

0

Figure 6.3.2  Probable surface stress between cracks when beam shear is zero.

surface stress is likely to be high because the concrete shares in carrying the tension, whereas at the crack, the concrete flexural tension is zero [6.12, 6.49]. 3. In the vicinity of a shear-​related inclined crack, such as that of Fig. 5.4.2, not only does high surface stress exist adjacent to the crack but, in addition, the tensile force T computed at z from the support actually acts much closer to the support (refer to the truss model discussion in Section 5.7 and Fig. 5.7.1). 4. At locations where some bars are terminated in a tension zone, the abrupt change in the distribution of the total tensile force among the bars gives rise to very high surface stresses. Thus, the local effects related to the rate of change in moment are not directly correlated with the development-​length-​related strength of the member. When the bars are properly anchored—​that is, when they have adequate development length provided and continue to carry their required tensile force—​the localized stress condition is not of concern.

6.4 BOND FAILURE MECHANISMS The term “bond failure” has been given to the mechanism by which failure occurs when the provided development length is inadequate. Years ago, when plain bars (relatively smooth bars without lug deformations) were used, slip resistance (“bond”) was thought of as adhesion between concrete paste and the surface of the bar. Yet even with low tensile stress in the reinforcement, there was sufficient slip immediately adjacent to a flexural crack in the concrete to break the adhesion, leaving only friction to resist bar movement relative to the surrounding concrete over the slip length. Shrinkage can also cause frictional drag against the bars. Typically, a hot-​rolled plain bar may pull loose by longitudinal splitting when the adhesion and friction resistances are high, or just pull out, leaving a cylindrical hole when adhesion and friction resistances are low. Deformed bars were created to change the behavior pattern so that there would be less reliance on friction and adhesion (though these still exist) and more reliance on the bearing of the lugs against the concrete. These bearing forces act at an angle to the axis of the bar, which cause longitudinal and radial outward components against the concrete, as shown in Fig. 6.4.1. The radial component causes circumferential tensile stresses [Fig. 6.4.1(d)] and will tend to crack (split) the concrete around the bar. When inadequate development length is provided, deformed bars in normal-​weight concrete usually give rise to a splitting mode of failure (i.e., “bond failure”) [6.1, 6.5, 6.7]. A  splitting failure occurs when the wedging action of the steel lugs on a deformed bar

18



181

6.4  BOND FAILURE MECHANISMS Bearing and friction forces on bar

Adhesion and friction forces along the surface of the bar (a) On bar

(b) On concrete

(c) Components

radial

(d) Radial component tends to split surrounding concrete

Figure 6.4.1  Forces between bar and surrounding concrete. Final splitting failure

Whole layer suddenly splits after initial horizontal splits at sides

First splitting (a)

(b)

(c)

Figure 6.4.2  Splitting cracks and ultimate splitting failure modes. (From ACI Committee 408 [6.1].)

causes cracks in the surrounding concrete parallel to the bar. These cracks occur between the bar and the nearest concrete face, as shown in Fig. 6.4.2(a) and 6.4.2(b), or between bars when bars are closely spaced, as in Fig. 6.4.2(c). When small-​size bars are used with large cover, the lugs may crush the concrete by bearing and result in a pullout failure without splitting the concrete. This nonsplitting failure has also been reported for larger bars in structural lightweight concrete [6.1]. Although splitting is the usual failure mode, an initial splitting crack on one face of a beam is not considered failure. The distress sign indicating failure is progressive splitting. The presence of transverse reinforcement (stirrups, ties, or spirals) will provide confinement and restrain the growth of splitting cracks, thereby delaying failure (defined as an increase in deformation that results in no increase in resistance) until several splitting cracks have formed. Originally, development length requirements were based on pullout tests [6.8] of plain bars, followed by pullout tests [6.9–​6.15] of deformed bars, including the related load-​slip data. These tests often consisted of a concrete block with an embedded bar. In such tests, the surrounding concrete provides ample confinement, and a pullout rather than a splitting failure occurs. Thus, the results of pullout tests were not representative of splitting failures commonly observed in beams.

182

182

C hapter   6     D evelopment of R einforcement

s Cs

db

Cs Cs

db

Cs

Cs1

Cs2

db Cs2

db

Cs1

Failure plane Cb

Cb Cb > Cs, ...C = Cs

... C = Cb

Cs1 > Cb Cs2 > Cb

(a)

Cylinder of concrete tributary to bar

(b)

Figure 6.4.3  Concrete cylinder hypothesis for splitting failure. (From Orangun, Jirsa, and Breen [6.3].)

The splitting failure mode has been studied in detail by Orangun, Jirsa, and Breen [6.3, 6.4] and many others [6.7, 6.16–​6.24]. The studies of Orangun, Jirsa, and Breen [6.3] and Untrauer and Warren [6.16] have hypothesized that the action of splitting arises from a stress condition analogous to a concrete cylinder surrounding a reinforcing bar and acted on by the outward radial components [Fig.  6.4.1(d)] of the bearing forces from the bar. The cylinder would have an inner diameter equal to the bar diameter db and a thickness C equal to the smaller of Cb, the clear bottom cover, or Cs, half of the clear spacing to the next adjacent bar (see Fig. 6.4.3). The tensile strength of this concrete cylinder determines the resistance against splitting. If Cs < Cb , a side-​split type of failure occurs [Fig. 6.4.2(c)]. When Cs > Cb , longitudinal cracks through the bottom cover form first [first splitting cracks in Fig. 6.4.2(a) and 6.4.2(b)]. If Cs is only nominally greater than Cb, the secondary splitting will be side splitting along the plane of the bars. If Cs is significantly greater than Cb, the secondary splitting will also be through the sides [Fig. 6.4.2(a)] or through the bottom cover to create a V-​notch failure [Fig. 6.4.2(b)]. The current ACI Code provisions for development length of straight bars are based on a proposal from a 1979 ACI Committee 408 report [6.25] that recognized the cylinder hypothesis for splitting failure. Although the development length rules in the ACI Code have changed in format over the years, the basis of the provisions has remained unchanged since 1979.

6.5 FLEXURAL STRENGTH DIAGRAM—​B AR BENDS AND CUTOFFS As stated in Section 6.2, the flexural strength of a beam at any section along its length is a function of its cross section and the actual embedment length of its reinforcement. The concept of a diagram showing this three-​dimensional relationship can be a valuable aid in determining cutoff or bend points of longitudinal reinforcement. It may be recalled from Chapter 3 that in terms of the cross section, the nominal flexural strength for a singly reinforced rectangular beam may be expressed as:



a  M n = As f y  d –   2

[3.4.6]

Equation (3.4.6) assumes that the steel reinforcement comprising As is adequately embedded in each direction by the required development length Ld from the section where Mn is computed such that the stress fy is reached.

183



183

6 . 5   F L E X U R A L S T R E N G T H D I AG R A M

► EXAMPLE 6.5.1 For the beam of Fig. 6.5.1, compute and draw qualitatively the flexural strength diagram. SOLUTION  The flexural strength in each region is represented by the horizontal portions of the diagram in Fig. 6.5.1. In this example, there are five bars of one size in section C–​C; thus the flexural strength provided by each bar is in this case approximately one-​fifth of the total capacity. Actually, the sections with four and two bars will have a little more than four-​fifths and two-​fifths, respectively, of the total strength of the section containing five bars, owing to the slight increase in moment arm when the number of bars in the section decreases. At point a, the location where the fifth bar terminates, this bar has zero embedment length to the left and thus has zero capacity. Proceeding to the right from point a, the bar may be counted on to carry a tensile force proportional to its embedment from point a up to the development length Ld. Thus, in Fig. 6.5.1, point b represents the point where the fifth bar is fully developed through the distance Ld and can therefore carry its full tensile capacity. The other cutoff points are treated in the same way. A

B

2 bars

Point a

4 bars

A

C

5 bars

B

C

Note: Showing the tip of a bar bent up is a scheme used throughout this text to show the bar termination. The bars are actually straight and lie in a common plane. CL of span

Section A–A

Section B–B

Section C–C Ld

Ld

b

a

Mn of 5 bars

Ld

Mn of 4 bars

Mn of 2 bars

Figure 6.5.1  Flexural strength diagram.

184

184

C hapter   6     D evelopment of R einforcement

► EXAMPLE 6.5.2 Demonstrate qualitatively the practical use of the diagram for the design strength, φ M n , for verification of the locations of cutoff or bend points in a design. Assume that the main cross section with five equal-​sized bars provides exactly the required strength at midspan for this simply supported beam with uniform load, as shown in Fig. 6.5.2. SOLUTION  (a) Compute the design strength, φ M n , for each potential bar grouping that may be used; in the present case, for five bars, four bars, and two bars. (b) Decide which bars must extend entirely across the span and into the support. In beams, ACI-​9.7.3.8.1 states, “At simple supports, at least one-​third of the maximum positive moment reinforcement shall extend along the beam bottom into the support at least 6 in. …” In this example, two bars should extend into the support. (c) Decide on the order of cutting or bending the remaining bars. The least amount of longitudinal reinforcement will be obtained when the resulting φ M n diagram is closest to the factored moment Mu diagram. With that thought in mind, and proceeding from maximum moment region to the support, cut off one bar as soon as permissible. (d) Cutoff restrictions. Point A of Fig. 6.5.2 is the theoretical location where the design strength, φ M n , of the remaining four bars is adequate. To provide for a safety factor against shifting of the moment Mu diagram (especially in continuous spans) and to provide partially for the complexity arising from a potential diagonal crack, the ACI Code Uniformly distributed load

d 2 bars

4 bars

5 bars

Cross section at midspan

Center of support

CL of span Ld center bar

Ld interior bars

φMn of 5 bars

φMn diagram Ld corner bars

φMn of 4 bars

B A

d or 12 diam

Theoretical cutoff point

Factored moment diagram, Mu C Theoretical cutoff point

φMn of 2 bars

d or 12 diam Mu is zero at centerline of simple support At end of bar: no capacity; develops full strength by “bond” at length Ld

Figure 6.5.2  Factored moment Mu and design flexural strength φ Mn, diagrams for determining bar cutoffs.

(Continued)

185



6.6  DEVELOPMENT LENGTH FOR TENSION REINFORCEMENT

185

Example 6.5.2 (Continued) states that there must be an extension beyond the point where a bar theoretically may be terminated, or it may be bent into the compression face. Accordingly, ACI-​9.7.3.3 requires that the reinforcement be extended beyond the point at which it is no longer required to resist flexure for a distance equal to the effective depth of the member or 12 bar diameters, whichever is greater, except at supports of simply supported spans and at free ends of cantilevers. (e) Once cutoff or bend points have been located, a check is made by drawing the φ M n diagram to ensure no encroachment on the factored moment Mu diagram. (f) Other restrictions. Since points B and C of Fig. 6.5.2 are bar terminations in a tension zone, the stress concentrations described in Section 6.3 are present (Fig. 6.3.2), effectively reducing the shear strength of the beam [6.26, 6.27]. Thus, for cutoffs to be acceptable, one of the three special conditions of ACI-​9.7.3.5 must be satisfied. These conditions are discussed in Section 6.11. If these bars were bent up and anchored in the compression zone, no further investigation would be necessary; this, however, is seldom done in today’s practice.

6.6 DEVELOPMENT LENGTH FOR TENSION REINFORCEMENT—​A CI CODE The term “development length” was defined in Section 6.2 as the length of embedment needed to develop the yield stress in the reinforcement. As described in Section 6.4, the development length requirement is primarily a function of the splitting resistance of the concrete surrounding the bars rather than a frictional-​adhesional pullout resistance. The splitting resistance is roughly proportional to the bar area, indicated by Eq. (6.2.2), whereas the pullout resistance is roughly proportional to the bar diameter, indicated by Eq. (6.2.1). The 2014 ACI Code provisions are based on the basic relationship developed by Orangun, Jirsa, and Breen [6.3, 6.4] and by a later study by Sozen and Moehle [6.28]. The general equation for computing the development length of deformed bars is given in ACI-​ 25.4.2.3 as ACI Formula (25.4.2.3a),    ψ t ψ e ψ s  3 fy Ld =  d  40 λ fc′  cb + K tr   b   d     b



(6.6.1)1

where Ld = development length db = nominal diameter of bar or wire cb = cover or spacing dimension = the smaller of (1) distance from center of bar or wire being developed to the nearest concrete surface, and (2) one-​half the center-​to-​center spacing of bars or wires being developed

1  For SI, with fy and fc′ in MPa,



    fy ψt ψeψs   Ld = d  1.1λ fc′  cb + K tr   b   d     b

(6.6.1)

 

186

186

C hapter   6     D evelopment of R einforcement

Ktr = transverse reinforcement index defined as follows: K tr =



40 Atr sn

(6.6.2)2

where Atr = total cross-​sectional area of all transverse reinforcement which is within the spacing s that crosses the potential plane of splitting through the reinforcement being developed s = maximum center-​to-​center spacing of transverse reinforcement within development length Ld n = number of bars being developed along the plane of splitting The transverse reinforcement index, Ktr, reflects the improvement and increase in strength obtained from the confinement provided by the transverse steel. It is noted, however, that ACI-​25.4.2.3 allows the use of K tr = 0 even if transverse reinforcement is present. This is conservative. The modification factors ψ t , ψ e , ψ s , and λ are discussed in detail later in Section 6.7. In the use of Eq. (6.6.1), the cover and transverse reinforcement terms cannot be taken greater than 2.5; thus,  cb + K tr   d  ≤ 2.5 b



(6.6.3)

The upper limit on this parameter represents a pullout failure and indicates that beyond a certain point, increasing the amount of cover or the amount of transverse reinforcement is no longer effective at increasing the bond stress at failure. Indeed, providing a very large cover or extremely heavy transverse reinforcement will result in pullout rather than a bond-​ splitting failure. In computing the transverse reinforcement index, Ktr, it is important to identify the potential plane of splitting and the total area of reinforcement crossing the potential splitting plane, Atr. To be fully effective, the transverse reinforcement must be adjacent to the bar being developed and must cross the splitting plane on the outside of the bar [6.5]. In Fig. 6.6.1, examples showing the potential splitting planes and the effectiveness of the transverse reinforcement in beams are illustrated. In this figure, it is assumed that all shown bars are being developed at the same location. If one-​half the center-​to-​center spacing of the bars is smaller than the distance from the center of the bar to the nearest concrete surface, then cracking between bars with a horizontal splitting plane can be expected to occur as shown in Fig 6.6.1(a). For the bar arrangement and transverse reinforcement shown in Fig. 6.6.1(a), the stirrup is effective only for the outer bars. In such a case, the designer could choose to calculate a different Ld for the inner and outer bars, ignore the effect of the transverse reinforcement, or include the effect of the transverse reinforcement as an average over the bars being developed. In the last case, the total area of reinforcement crossing the splitting plane would be computed as Atr = 2 Aleg ( Aleg = area of one leg of the stirrup) for the four bars being developed and thus, n = 4 in Eq. (6.6.2). If, on the other hand, the bottom cover controls, vertical splitting would be expected to occur, as shown in Fig. 6.6.1(b). In this case, the stirrup crosses each splitting crack and the transverse reinforcement can be considered to be effective for all the bars. Thus, the total area of transverse reinforcement may be computed as Atr = 3 Aleg with n = 3 in Eq. (6.6.2). For the bar arrangement shown in Fig. 6.6.1(c), the transverse reinforcement is effective for only four of the seven bars being developed and thus, Atr = 4 Aleg and n = 7 in Eq. (6.6.2). 2  For SI,



K tr =

40 Atr sn

(6.6.2) 

187



187

6.6  DEVELOPMENT LENGTH FOR TENSION REINFORCEMENT

Atr = 2Aleg n =4

Atr = 3Aleg n =3

Splitting

Splitting plane

(a) Atr is effective only for two outer bars

(b) Atr is effective for all bars

Atr = 4Aleg n =7

Atr = 2Aleg n =2 Splitting

Splitting plane

(c) Atr is effective only for four bars

(d) Atr is effective for all bars

Figure 6.6.1  Transverse reinforcement effectiveness for various bar arrangements and splitting planes when all bars shown are developed at the same location. (Adapted from Ref. 6.5.)

For other cases, the designer must exercise judgment in identifying the splitting planes and in computing the contribution of the transverse reinforcement. For example, if the side and bottom cover are the same (a common case in practice), then vertical or side ​splitting, or both vertical and side splitting, can occur in practice as shown in Fig. 6.6.1(d). In fact, diagonal splitting cracks may also occur, as shown in Fig. 6.4.2. For the purpose of computing the contribution of the transverse reinforcement, the vertical and side splitting cracks occurring around the same bar are considered as a single splitting plane. Thus, for the example shown in Fig.  6.6.1(d), the transverse reinforcement is considered effective for both bars, with Atr = 2 Aleg and n = 2 in Eq. (6.6.2). In any case, it is always conservative to ignore the contribution of the transverse reinforcement, although in some cases it may be too conservative and ignoring it will result in development lengths much greater than those required.

Simplified Equations of ACI-​25.4.2.2 The use of Eq. (6.6.1) is a rather involved way of determining development length Ld. For many practical situations, ACI-​25.4.2.2 provides simplified equations that can be also used. These simplified equations are divided into two categories, as follows. Category A Either one of the following two conditions will satisfy this most favorable situation: 1. (a) clear lateral spacing between bars at least db, and (b) clear cover at least db, and (c) minimum Code-​specified stirrups or ties along the development length or 2. (a) clear lateral spacing between bars at least 2db, and (b) clear cover not less than db.

18

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C hapter   6     D evelopment of R einforcement

Simplification of Eq. (6.6.1) after substituting ( cb + K tr ) / db = 1.5 and a bar size modification factor ψ s = 0.8 for #6 and smaller bars (see Section 6.7) becomes: for #6 and smaller bars:  fy ψ t ψ e  Ld =   db  25λ fc′ 

(6.6.4)3

 fy ψ t ψ e  Ld =   db  20 λ fc′ 

(6.6.5)4

for #7 and larger bars:

Category B Anything not in Category A is in Category B. Simplification of Eq. (6.6.1) after substituting (cb + K tr ) / db = 1.0 and a bar size modification factor ψ s = 0.8 for #6 and smaller bars (see Section 6.7) becomes for #6 and smaller bars:  3 fy ψ t ψ e  Ld =   db (6.6.6)5  50 λ fc′  for #7 and larger bars:

 3 fy ψ t ψ e  Ld =   db  40 λ fc′ 

(6.6.7)6

It is noted that no strength reduction factor, φ, is used in the development length equations. An allowance for strength reduction has already been included in both the general and simplified equations (Eqs. 6.6.1, 6.6.4–​6.6.7) for computing development lengths.

Useful Tables Development length for common values of fc′ and Grade 60 reinforcement are shown in Tables 6.6.1 and 6.6.2 for all bar sizes using the simplified equations of the ACI Code [i.e., Eqs. (6.6.4)–​(6.6.7)]. Note the descriptive heading on the top of these tables. Also note that









3  For SI, with fy and fc′ in MPa, for #20 (see Table 1.13.2) and smaller:   fy ψ t ψ e  Ld =   db  2.1λ fc′  4  For SI, with fy and fc′ in MPa, for #22 (see Table 1.13.2) and larger:   fy ψ t ψ e  Ld =   db  1.7λ fc′  5  For SI, with fy and fc′ in MPa, for #20 (see Table 1.13.2) and smaller:   fy ψ t ψ e  Ld =   db  1.4 λ fc′  6  For SI, with fy and fc′ in MPa, for #22 (see Table 1.13.2) and larger:   fy ψ t ψ e  Ld =   db  1.1λ fc′ 

(6.6.4) 

(6.6.5) 

(6.6.6) 

(6.6.7) 

189



6.6  DEVELOPMENT LENGTH FOR TENSION REINFORCEMENT

189

the modification factors ψ t, ψ e, and λ have all been assumed equal to 1.0 (see Section 6.7 for the definition and values of these factors). These tables show that for Grade 60 reinforcement and normal-​weight concrete with fc′ = 4000 psi, the simplified equations require a development length of about 40 bar diameters for the smaller bar sizes and about 50 bar diameters for the larger bars.

Practical Application of the Simplified Development Length Equations The practicality for applying the simplified equations in ordinary reinforced concrete construction is that most beams will contain at least Code-​specified minimum stirrups (thereby satisfying Category A, item 1c); in addition, clear spacing must satisfy the larger of the bar diameter db , 1 in. or 4/​3 of the maximum aggregate size (ACI-​25.2.1), and cover must satisfy the minimum specified in ACI-​20.6.1.3.1 in any case. Using the minimum 1.5 in. of cover on beams will commonly provide the Category A  minimum of db . For slab-like elements without shear reinforcement, clear spacing will usually satisfy the Category A, item 2a, minimum of 2db; and 3/4-​in. minimum cover required by ACI-​20.6.1.3.1 will commonly provide at least the db cover required for Category A, item 2b. Regarding bar spacing for beams, minimum width tables for satisfying the minimum of ACI-​25.2.1 (db ≥ 1 in.), 2db, and 3db are given in Section 3.9 as Tables 3.9.2, 3.9.3, and 3.9.4, respectively. All bar arrangements in beams must satisfy Table 3.9.2; Category A, item 2, beams must satisfy Table 3.9.3; and good practice should provide 3db in accordance with Table 3.9.4. It must be mentioned that the simplified equations, while practical, generally result in conservative values of the required development length. Where congestion of reinforcement is anticipated, it is preferable to use the general equation, Eq. (6.6.1).

TABLE 6.6.1  DEVELOPMENT LENGTH Ld IN INCHES FOR GRADE 60 BARS, CATEGORY A,* EQS. (6.6.4) AND (6.6.5) WITH ψt , ψe , AND λ = 1.0 fc′ (psi) Bar

3000

4000

5000

#3 #4 #5 #6 #7 #8 #9 #10 #11 #14 #18

16.4 21.9 27.4 32.9 47.9 54.8 61.8 69.6 77.2 92.7 124

14.2 19.0 23.7 28.5 41.5 47.4 53.5 60.2 66.9 80.3 107

12.7 17.0 21.2 25.5 37.1 42.4 47.9 53.9 59.8 71.8 95.8

*(a) Clear spacing and clear cover ≥ db and minimum stirrups, or (b) Clear spacing ≥ 2db and clear cover ≥ db.

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TABLE 6.6.2  DEVELOPMENT LENGTH Ld IN INCHES FOR GRADE 60 BARS, CATEGORY B,* EQS. (6.6.6) AND (6.6.7) WITH ψt , ψe , AND λ = 1.0 fc′ (psi) Bar

3000

4000

5000

#3 #4 #5 #6 #7 #8 #9 #10 #11 #14 #18

24.6 32.9 41.1 49.3 71.9 82.2 92.7 104 116 139 185

21.3 28.5 35.6 42.7 62.3 71.2 80.3 90.4 100 120 161

19.1 25.5 31.8 38.2 55.7 63.6 71.8 80.8 89.7 108 144

* Everything not in Category A.

6.7 MODIFICATION FACTORS ψ t , ψ e , ψ s , AND λ TO THE BAR DEVELOPMENT LENGTH EQUATIONS—​A CI CODE The symbols ψ t , ψ e , ψ s , and λ in Eq. (6.6.1) are modification factors to the required bar development length per ACI-​25.4.2.4 that consider the main parameters known to affect bond strength for various bar surface conditions that may be commonly encountered in practice.

Casting Position, ψt Horizontal bars placed so that more than 12 in. (300 mm) of fresh concrete is cast in the member below the development length or splice region are often referred to as top-​cast bars. Research has shown that top-​cast bars have lower bond strengths owing to increased settlement and bleed of water at the bar as the depth of concrete below the bar increases [6.1, 6.11, 6.29]. For this reason, whenever the depth of concrete mixture placed in one lift under the horizontal bar exceeds 12 in., the ACI Code requires a 30% increase in development length (i.e., a modification factor ψ t = 1.3). Otherwise, ψ t = 1.0.

Epoxy-​Coated Bars, ψe Epoxy coating of reinforcing bars reduces adhesion and lowers the coefficient of friction between the bar surface and the surrounding concrete. As a result, bond strength of epoxy-​ coated reinforcement is reduced (Cleary and Ramirez [6.30], Choi, Hadje-​Ghaffari, Darwin, and McCabe [6.31], Hamad, Jirsa, and D’Abreu de Paulo [6.32], Hadje-​Ghaffari, Choi, Darwin, and McCabe [6.33], Bartoletti and Jirsa [6.34], Treece and Jirsa [6.50], Johnston and Zia [6.51], and Mathey and Clifton [6.52]). The effect of epoxy coating on bond strength is reflected in ACI-​25.4.2.4 through three different modification factors ψ e, depending on the type of coating, concrete cover, and clear spacing of the bars. The values are 1. Epoxy-​coated, or zinc and epoxy dual-​coated bars: With clear cover less than 3db , OR with clear spacing less than 6 db ψ e = 1.5 All other conditions ψ e = 1.2 2. Uncoated or zinc-​coated (galvanized) bars ψ e = 1.0

19



6 . 7   M O D I F I CAT I O N FAC TO R S

191

When the epoxy-​coated bars are in the unfavorable category of ψ e = 1.5 and the bars are top-​cast bars (ψ t = 1.3), ACI-​25.4.2.4 states that the product ψ t ψ e need not exceed 1.7 (it would otherwise be 1.95).

Bar Size, ψs This factor reflects the results of tests that suggested that shorter development lengths would be required for smaller diameter bars. Accordingly, the modification factor ψ s = 0.8 for #6 and smaller bars. For #7 and larger bars, ψ s = 1.0. This factor was obtained based on limited test data. Subsequent data and a reevaluation of the influence of bar size on bond strength by ACI Committee 408 [6.53] suggests that no reduction factor should be used. While the subject is still being debated, the latest report of ACI Committee 408 [6.53] does not support the use of this factor.

Lightweight Aggregate Concrete, λ There is increased potential for a pullout failure mode in lightweight aggregate concrete. Thus, a modification factor λ = 0.75 is to be used for lightweight aggregates of all types. Alternatively, a greater λ factor is permitted based on the composition of the aggregate in the concrete mixture per ACI Table 19.2.4.2 or when the splitting tensile strength of the lightweight concrete is specified per ACI-​19.2.4.3 (see Section 1.8). For normal-​weight concrete, λ = 1.0.

Excess Reinforcement, αexs When steel reinforcement in excess of that required has been provided, the designer is permitted to reduce the development length by the ratio (required As / ​provided As) in limited situations (ACI-​25.4.10.1). However, the development length is not permitted to be reduced by excess reinforcement in the following situations (ACI-​25.4.10.2): a) when anchorage to develop fy is specifically required. For example, the factor does not apply for the development of the temperature and shrinkage reinforcement required by ACI-​24.4.3.4 and of the reinforcement required by the structural integrity provisions of ACI-​8.7.4.2, 8.8.1.6, 9.7.7 and 9.8.1.6. b) at noncontinuous supports. c) where bars are required to be continuous. d) for headed and mechanically anchored deformed reinforcement. e) when the flexural member is part of the seismic force resisting system in structures assigned to Seismic Design Categories D, E, or F in accordance to ASCE 7 [2.18].

Minimum Development Length After all modifications factors have been applied, the resulting development length Ld cannot be less than 12 in. (300 mm) in accordance with ACI-​25.4.2.1(b).

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Summary The modification factors for both the general equation, Eq. (6.6.1), and the simplified equations, Eqs. (6.6.4) through (6.6.7), are summarized  below. For conditions not listed, the value of the modification factor is 1. Condition

Modification Factors ψ t , ψ e , ψ s, and λ

Casting positiona

ψ t = 1.3

Epoxy-​coated or zinc and epoxy dual-​coated barsa

ψ e = 1.5 for bars with clear cover  0.005 a / β1 7.92 / 0.85

Thus, the section is tension controlled and φ = 0.90:

φ M n = 0.90 (647) = 582 ft-kips ≈ Mu = 590 ft-kips       OK (b) Determine the theoretical cutoff point for the 2–​#8 bars. The remaining design strength φ Mn with 3–​#10 bars is C = 40.8a T = 3.81(60) = 229 kips a=

229 = 5.61 in. 40.8

φ M n = 0.90(229)[28 − 0.5(5.61]

1 12

= 433 ft-kips (Continued)

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205

Example 6.13.1 (Continued) Plot φ M n on the factored moment Mu diagram and locate the theoretical cutoff point A. Extend to the right 12 bar diameters (of the #8 bars that are to be cut) or the effective depth of the member, whichever is greater, to arrive at point B.

d = 28 in. ( 2.33 ft ) > [12 db = 12 (1.0 ) = 12 in.]

(c) Use the simplified equations to determine the development length Ld for #8 bars. Can Category A, the more favorable one, be used? Check the clear spacing of bars. Assuming the bars, though unequal in size, are uniformly spaced, the clear spacing between them is

clear spacing =

16 − 2 (1.5) − 2 (0.5) − 3(1.27) − 2 (1.0) = 1.55 in. 4

Since only the 2–​#8 bars are being developed, and the 3–​#10 are presumed to continue beyond the #8 cutoff location, it is the spacing between the two #8 bars that determines the category. The possible failure modes to consider are splitting from a #8 bar to the side or top face of the member and splitting between the two #8 bars. The ACI Code rules consider a bar (or bars) as essentially inert when it is not being developed within the development region of other bars. Thus, when the #10 bars of this example have a development length from their termination near the free end of the cantilever that is less than the distance to the #8 bar cut, the #10 bars are considered to have no influence on Ld for the #8 bars. That is, in this case the full spacing between the #8 bars, 2(1.55) + 1.27 diam of #10 = 4.37 in. The writers believe it appropriate in this case to consider the spacing of the #8 to be 4.37 in. for the purpose of satisfying a Category A requirement, assuming Ld for the #10 bars does not overlap the Ld for the #8 bars. Even if the concrete width between #8 bars were taken as 2(1.55) = 3.10 in., it would still exceed the 2db for the #8 bars to satisfy category A, item 2(a), given in Section 6.6, as well as item 2(b), because cover to the top face of the beam is 2 in., which exceeds db needed for that item. Thus, Category A applies! Using simplified Eq. (6.6.5) for #7 and larger bars,  fy ψ t ψ e  Ld =   db  20 λ fc′ 



[6.6.5]

The casting position factor ψ t = 1.3 applies, while both the epoxy-​coated bar factor ψ e and the lightweight aggregate concrete factor λ are 1.0. Thus, Eq. (6.6.5) gives

ψψ  60, 000 ψ t ψ e  Ld =  1.0 = 54.8 t e   20 λ 3000  λ

Ld = 54.8

ψt ψe (1.3)(1.0) = 54.8 = 71.2 in. λ 1.0

The 54.8 in. can be verified from Table 6.6.1. Thus,

Ld ( for # 8) = 71.2 in. ( 5.9 ft )

(d) Use the general equation, Eq. (6.6.1), to determine the development length Ld for #8 bars. That equation is



   ψ t ψ e ψ s  3 fy Ld =  d  40 λ fc′  cb + K tr   b   d     b

[6.6.1]

(Continued)

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Example 6.13.1 (Continued) The cover or spacing dimension cb is the smaller of (1) the distance from the center of the bar being developed to the nearest concrete surface and (2) one-​half the center-​ to-​center spacing of the bars being developed. The distance cb is the smaller of the following two values: top cover = 1.5 (i.e., clear) + 0.5 (i.e.,stirrup) + 0.5 ( i.e., bar radius) = 2.5 in. one-​half center-​to-​center spacing = 4.37 [see part (c)]/​2 = 2.19 in. Thus, one-​half the center-​to-​center spacing governs and cb = 2.19 in. This result indicates a potential failure mode that involves horizontal splitting. In this case, the transverse reinforcement is considered to be effective only for the two #10 corner bars [see Fig. 6.6.1(a)], which are not being developed at this location. Whether the transverse reinforcement will provide some confinement to the concrete and help improve the bond strength of the center bars is a matter of opinion. It is clear, however, that if horizontal splitting occurs, the stirrups will not cross the horizontal splitting plane adjacent to the #8 bars being developed. Therefore, the contribution of the transverse reinforcement can be conservatively ignored in this case (i.e., Ktr = 0). Evaluating Eq. (6.6.3),

  cb + K tr 2.19 + 0 = = 2.19 < 2.5 max  1.0   db

Thus, ( cb + K tr ) / db = 2.19. Evaluate Eq. (6.6.1),



   ψ t ψ e ψ s   3 60, 000 1.3(1.0)1.0  3 fy 1.0 = 48.8 in. (4.1 ft ) Ld =  db =   40 λ fc′  cb + K tr    40 1.00 3000 2.19    d     b

This value is significantly lower than the 71.2 in. obtained from the simplified equation in part (c). Use Ld = 48.8 in. for the design strength diagram in Fig. 6.13.2. Since point B, the proposed cutoff point, lies only about 3.5 ft from the support, the #8 bars would not have full capacity at the support. Therefore, extend the proposed cutoff to point C, which is located at Ld (for #8) = 4.1 ft from the support. (e) Check ACI-​9.7.3.5 for cutting bars at point C in the tension zone. The shear strength, including contribution of stirrups, is first computed. Using the simplified method of constant Vc with λ = 1.0,

Vc = 2 λ fc′ bw d = 2(1) 3000 (16)(28)

1 1000

= 49.1 kips

For the 14-​in. spaced #4 stirrups in the vicinity of the potential cut point C,

Vs =

Av f yt d s

=

2(0.20)(60)28 = 48.0 kips 14

The shear strength φ Vn at point C is

φ Vn = φ (Vc + Vs ) = 0.75(49.1 + 48.0) = 72.8 kips

Vu 71.9 = = 99% > 75% φ Vn 72.8



NG (Continued)

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207

Example 6.13.1 (Continued) Even when only 50% of the design strength φ M n is used by Mu, Vu /(φVn) cannot exceed 75% [see Condition 3, preceding Eqs. (6.11.3) and (6.11.4)]. Try using one more stirrup at 8-​in. spacing to cover the potential cut at point C, and see whether or not Condition 1, Eq. (6.11.1), is satisfied.  14  Vs = 48.0   = 84.0 kips  8 Vu 71.9 = 72% = φ Vn 0.75(49.1 + 84)



This is close to satisfying the two-​thirds limit of ACI-​9.7.3.5(a) [Eq. (6.11.1)]. Extending the #8 bars to point C′, 4.5 ft from the face of support, would result in Vu 64.5 = = 65% φ Vn 0.75(49.1 + 84)



OK

(f) Check whether the continuing #10 bars have adequate development length to the right of point C′. The clear spacing between the continuing three #10 bars is

clear spacing =

16 − 2(1.5) − 2(0.5) − 3(1.27) = 4.1 in. 2

which exceeds the 2db = 2.54 in. required for Category A, item 2(a). Top cover of 2.64 in. [i.e.,(1.5 + 0.5 + 1.27 / 2 ) = 2.64 in.] to the center of the #10 bars exceeds the db requirement of Category A, item 2(b). Thus, the simplified equation, Eq. (6.6.5) for #7 and larger bars, can be used



 fy ψ t ψ e  ψψ  60, 000 ψ t ψ e  Ld =  1.27 = 69.6 t s  db =    20 λ 3000  λ  20 λ fc′  Ld = 69.6

ψt ψe 1.3(1.0) = 69.6 = 90.5 in. λ 1.0

where a casting position factor ψ t = 1.3 has been used. Both the epoxy-​coated bar factor ψ e and the lightweight aggregate concrete factor λ are 1.0. Calculate the development length Ld based on the general equation, Eq. (6.6.1). The distance cb is the smaller of the following two values: top and side cover = 1.5 (i.e., clear) + 0.5 (i.e., stirrup) + 0.635 (i.e., bar radius) = 2.6 in. 4.1 one-​half center-​to-​center spacing =  + 0.635 (i.e., bar radius) = 2.7 in. 2 Although the cover parameter for top and side cover splitting with cb = 2.6 in. would govern, the difference between the two computed values is too small to predict the splitting failure plane with certainty. Here, the parameter will be computed assuming that both top and side cover splitting and horizontal splitting may occur. If top and side cover splitting is assumed, then the transverse reinforcement may be assumed to be effective for all three bars being developed [as in Fig. 6.6.1(b) and 6.6.1(d)]. Thus, the total area of transverse reinforcement is Atr = 3 Aleg for three bars being developed: that is, n = 3. Using the given 14-​in. spacing near the free end of the cantilever for computation, Eq. (6.6.2) gives

K tr =

40 Atr 40(3)(0.20) = = 0.57 14(3) sn

(Continued)

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Example 6.13.1 (Continued) If horizontal splitting is assumed [as in Fig. 6.6.1(a)], then the transverse reinforcement is effective for only two of the three bars being developed. Thus Atr = 2 A leg and n = 3. Eq. (6.6.2) gives then K tr =



40 Atr 40(2)(0.20) = = 0.38 sn 14(3)

which is more conservative. Use K tr = 0.38. Evaluating Eq. (6.6.3),

  cb + K tr 2.7 + 0.38 = = 2.43 < 2.5 max  1.27   db

Thus, ( cb + K tr ) / db = 2.43. Evaluate Eq. (6.6.1),    ψ t ψ e ψ s  3 fy Ld =  d  40 λ fc′  cb + K tr   b   d     b

ψψ ψ  3 60, 000 ψ t ψ e ψ s  1.27 = 42.9 t e s =  40 λ 3000 2.43  λ = 42.9

(1.3)(1.0)(1.0) 1.0

= 55.8 in. (4.65 ft ) This embedment of 4.65 ft measured from the end of straight #10 bars would overlap the development length region of the #8 bars, possibly requiring longer development length Ld for the #8 bars because the center-​to-​center spacing then would be the reduced value based on five bars in the 16-​in. width. The #10 bars would satisfy literally the statement of ACI-​ 9.7.3.4, which requires “Continuing flexural tension reinforcement shall have an embedment length at least Ld beyond the point where bent or terminated tension reinforcement is no longer required to resist flexure.” In other words, the distance from point A to the free end of the cantilever must be at least Ld (for #10 bars). The authors believe in a somewhat more conservative approach, requiring the design strength φ Mn diagram to have an offset from the factored moment Mu diagram, except at or near a simple support or the free end of a cantilever, equal to 12 bar diameters or the effective depth d, whichever is greater. In this case, try standard 90º hooks [see Fig. 6.10.1(b)] on the ends of the #10 bars. Since the beam has the usual 1.5-​in. clear cover and #4 stirrups, the cover to the hooked bars is 2 in., which is less than the 2 1 2 in. required by ACI-​25.4.3.3; thus, the special provisions of that code section must be satisfied. The development length Ldh for the #10 hooked bar is given by Eq. (6.10.1) and Table 6.10.2. Thus, for a #10 hooked bar,

Ldh = 27.8 in.

which exceeds the minimum 8db or 6 in., whichever is greater (ACI-​25.4.3.1). The Ldh of 27.8 in. is dimensioned from the outside face of the tail of the hook, as shown in Fig. 6.13.2. In accordance with ACI-​25.4.3.3, stirrups spaced at not more than 3db (3.81 in.) must be provided along the 27.8 in. of development distance. Also, the first stirrup must be provided within 2db (2.54 in.) of the outside bend, as shown in Fig. 6.13.2. (Continued)

209



6.14  DEVELOPMENT OF POSITIVE REINFORCEMENT

209

Example 6.13.1 (Continued) (g) Design strength φ M n diagram. The full design strength, φ M n , for the beam with 3–​#10 hooked bars will be available at 27.8 in. from the outside of the hook on the end of the beam. Assuming 1.5-​in. cover, full capacity is available at 29.3 in. (2.44 ft) from end of beam (point D). The dashed line in Fig. 6.13.2 goes from zero strength at the end of the hook to full strength φ M n = 433 ft-kips at 2.44 ft from end of beam; however, the drawing is not intended to imply that the hooked bar develops its strength linearly, since that is highly improbable. (h) Final decision. Cut 2–​#8 bars at 4 ft-​6 in. from the support; use 90° standard hooks on the 3–​#10 bars; use #4 U stirrups @ 3.5-​in. spacing as confinement over the Ldh distance, as shown in Fig. 6.13.2 by the dashed stirrups. The use of #10 bars in this cantilever beam is not a practical design but serves to illustrate the need for extending the cut location from B to C and then C′, and the need for and treatment of hooked bars.

6.14 DEVELOPMENT OF POSITIVE REINFORCEMENT AT SIMPLE SUPPORTS AND AT POINTS OF INFLECTION The concept of requiring the development of reinforcement on both sides of a section where the bars are to be fully stressed may also be applied to the continuation of positive moment tension reinforcement beyond either the centerline of a simple support or a point of inflection.

Simple Supports Referring to Fig. 6.14.1, consider point A on the factored moment Mu curve near a simple support, where the factored moment Mu equals the design strength φ M n of the bars continu­ ing into the support. The area of the shaded portion of the shear diagram equals the change in moment between the center of support and point A; thus the ordinate on the factored moment diagram at A is

Vavg (say, 0.9Vu ) x = φ M n

Thus, the distance x between the point A and the centerline of support in Fig.  6.14.1 is approximately equal to

x=

Mn Vu

Therefore, the available embedment length of the bars continuing into the support must equal or exceed Ld, or

La + x ≥ Ld

or

La +

Mn ≥ Ld Vu

(6.14.1)

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where Mn = nominal flexural strength at the section with all reinforcement assumed to be stressed at fy Vu = factored shear at the support, and La =  straight embedment length beyond the centerline of support to the end of the bars Equation (6.14.1) is given in ACI-​9.7.3.8.3(b) for beams [and in ACI 7.7.3.8.3(b) for one-​ way slabs]; it applies when the end of the reinforcement is not confined by a compressive reaction. In recognition of the fact that bars extending into a simple support have less tendency to cause splitting when confined by a compressive reaction [6.42], the ACI Code allows a 30% increase in the value of x or M n /Vu in such cases [ACI-​9.7.3.8.3(a) for beams and ACI-​7.7.3.8.3(a) for one-​way slabs], or (ACI-9.7.3.9.3 or ACI- 7.7.3.8.3) La + 1.3



Mn ≥ Ld Vu

(6.14.2)

Equations (6.14.1) or (6.14.2) need not be satisfied “If reinforcement terminates beyond the centerline of supports by a standard hook, or a mechanical anchorage at least equivalent to a standard hook. …” (ACI 9.7.3.8.3 and 7.7.3.8.3)

Inflection Points Since an inflection point is a point of zero moment located away from a support (refer to Fig. 6.14.2), bars in that region are not confined by a compressive reaction; therefore the 1.30 factor is interpreted as not to apply. In this case, the embedment length that must exceed the required development length Ld may be stated as  actual La , but  M available emdedment length =  not exceeding the  + n ≥ Ld Vu  larger of 12db or d 



Use 1.3 Mn/Vu when reaction causes compression around bars (i.e., at usual simple supports) La

Mn/Vu

ACI–9.7.3.8.3 (beams) ACI–7.7.3.8.3 (one-way slabs)

≥ Ld

Factored moment diagram, Mu A

Design strength of bars at support φMn

Vu Vavg

x

Figure 6.14.1  Development of reinforcement at a simple support.

[6.14.1]

21



211

6.15  DEVELOPMENT OF SHEAR REINFORCEMENT

For computation, La may not be taken greater than 12db or d

Actual La

Usable La

Mn/Vu Must exceed Ld

Available anchorage

Factored moment diagram, Mu

CL support

A φMn

Based on bars which extend past inflection point

Inflection point

Vu

Figure 6.14.2  Development of reinforcement at an inflection point.

where Mn and Vu refer to the nominal flexural strength and the factored shear at the point of inflection. The limitation of the usable La to 12 bar diameters or the effective depth has been applied because, according to ACI Commentary R9.7.3.8.3, there is no experimental evidence to show that longer extensions will improve bar anchorage. Additional development of reinforcement at the face of support is required by ACI-​ 9.7.3.8.2 when the beam is part of the primary lateral load resisting system.

6.15 DEVELOPMENT OF SHEAR REINFORCEMENT Reinforcement in the web of a beam, whether it be for shear or for torsion (see Chapter 18), must be properly anchored so that its full tensile capacity is available at or near the middepth of a beam. For proper function, the shear reinforcement must “extend as close to the compression and tension surfaces of the member as cover requirements and proximity of other reinforcement permits and shall be anchored at both ends” (ACI-​25.7.1.1). It is especially important to extend the stirrups as close to the compression face as possible because diagonal cracks may extend deeply into the compression zone when the nominal strength of the member is approached. The ends of single leg, simple U, or multiple U stirrups of deformed bars or wires must be anchored by means of a standard stirrup hook around longitudinal reinforcement, as shown in Fig. 6.15.1. Such stirrups may be inclined, but the angle between the stirrups and the longitudinal bars must be at least 45° in accordance with ACI-​22.5.10.5.2. The requirements for standard hooks for stirrups and ties are somewhat relaxed in comparison to the standard hooks used for the main tension reinforcement bars shown in Fig. 6.10.1. Also, hooked stirrups and hooked ties are not permitted for bars larger than #8. As noted in Chapter 5, stirrups and ties are usually #3 or #4 bars, occasionally #5 or #6, and rarely, if ever, larger than #8, so the ACI Code limit on size of hooked stirrups should rarely apply. Note also that the 180° hook is not typically used for stirrups and ties. The requirements for standard hooks at ends of ties and stirrups are shown in Fig. 6.15.2.

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C hapter   6     D evelopment of R einforcement

Standard stirrup hook (see Fig. 6.15.2) Longitudinal bar required As close as feasible, ACI–25.7.1.1

As close as feasible, ACI–25.7.1.1 0.014db

fyt

λ fc’ minimum

h 2

(b) For #6, #7, #8 stirrups with fyt > 40,000 psi (ACI–25.7.1.3b)

(a) For #5 bar and D31 wire, and smaller, or for #6, #7, and #8 bar with fyt ≤ 40,000 psi (ACI–25.7.1.3a)

Figure 6.15.1  Development of deformed bar or deformed wire stirrups.

6db ≥ 3 in.

135°

12db 6db ≥ 3 in.

(a) 90° hook (for #5 bars and smaller)

(b) 90° hook (for #6, #7, and #8 bars)

(c) 135° hook (for #8 bars and smaller)

Figure 6.15.2  Standard hooks for stirrups and ties (ACI-​25.3.2). As close as feasible, ACI–25.7.1.1 d 4

2”

max

d 4

max

d 4

max

d 2” min

8 wire diameter bend (min) (a)

(b)

Figure 6.15.3  Development of welded plain wire reinforcement stirrups (ACI-​25.7.1.4).

When closed stirrups are desired, one practical procedure is to use a pair of U stirrups without hooks, placed to form a closed unit. If this is done, ACI-​25.7.1.7 requires laps of 1.3Ld for proper splicing. When members are at least 18 in. deep and the yield strength of the stirrup does not exceed 9 kips per leg, splices are adequate if the legs extend the full available depth of the member. For U stirrups of welded plain wire reinforcement, anchorage may be accomplished (ACI-​25.7.1.4) by using either “(a) Two longitudinal wires spaced at a 2-​in. spacing along the member at the top of the U” or “(b) One longitudinal wire located not more than d/​4 from the compression face and a second wire closer to the compression face and spaced not less than 2 in. from the first wire. The second wire shall be permitted to be located on the stirrup leg beyond a bend, or on a bend with an inside diameter of bend of at least 8db.” These provisions are illustrated in Fig. 6.15.3. For single-​leg stirrups of welded plain or deformed wire reinforcement, ACI-​25.7.1.5 includes the recommendations [6.43] of the PCI/​WRI Ad Hoc Committee on Welded Wire

213



See ACI–25.7.1.1

d 2 d

213

6.16  TENSION LAP SPLICES

Two horizontal wires top and bottom

2” minimum Plain or deformed vertical wires

Main reinforcement “as close to tension surface as cover requirements and proximity of other reinforcement will permit” (ACI–25.7.1.1)

Greater of d/4 or 2”

Greater of d/4 or 2”

2” minimum Outer wire not above lowest main reinforcement

Figure 6.15.4  Development of single-​leg plain or deformed welded wire shear reinforcement (ACI-​25.7.1.5).

Fabric for Shear Reinforcement. Single-​leg stirrups are practical and common in precast prestressed concrete T-​sections. The provisions of ACI-​25.7.1.5 are illustrated in Fig. 6.15.4. For guidance in placing stirrups and ties to obtain proper anchorage, reference should be made to the (2004) ACI Detailing Manual [2.23].

6.16 TENSION LAP SPLICES Wherever bar lengths required in a structure exceed the length available or exceed the length that may be economically shipped or erected, splices are necessary. Splicing may be accomplished by simple lapping of bars side by side, either in contact or separated, over a specified length, referred to as the splice length (see Fig. 6.16.1). Alternatively, welded splices or mechanical connectors (though generally more expensive) may be used to provide continuity of the reinforcement (see Section 6.17). In the lapped splice of Fig. 6.16.1, the total force in the tension reinforcement must be transferred from bars a to bars b over the splice length, Lst. Test data suggest that the force transfer mechanism for lapped splices is similar to that for single bars [6.3, 6.4], where the forces from one bar are mainly transferred to the concrete by mechanical locking of the lugs to the surrounding concrete, and then from the concrete to the adjacent bar. Furthermore, the concrete cylinder hypothesis for splitting failure of single bars (see Fig. 6.4.3) may be extended for a lap splice where the cylinders of concrete tributary to each bar interact to form an oval ring, as shown in Fig. 6.16.2. Analysis of the data on splice and development length tests suggest that the overlap distance required in tension lap splices, Lst, should be equal to the development length Ld of the bar. However, because stress concentrations near the splice tend to produce splitting at early stages of loading, a splice length greater than the development length Ld is required unless special precautions are taken [6.3–​6.5, 6.23, 6.25, 6.28, 6.44–​6.48]. The ACI Code defines two classes of tension lap splices, Class A or Class B, depending on (1) the percentage of bars being spliced within the required lap distance and (2) the stress level in the unspliced bars at the splice location (ACI-​25.5.2.1). As shown in Table 6.16.1, Class A and Class B are defined to have overlap distances of 1.0Ld and 1.3Ld, respectively, but not less than 12 in., where Ld is the development length of the bar calculated in accordance with Section 6.6. These provisions apply equally to deformed bar or deformed wire splices, but tension lap splices are not permitted for bars larger than #11 (ACI-​25.5.1.1). The ratio (As required)/​(As provided) column in Table  6.16.1 refers to the percentage of available capacity that is utilized. The ratio may also be considered as the percent of fy

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Figure 6.16.1  Bar arrangement and force transfer in a typical lap splice.

S’ 2

2db

S’

2db

S’ 2

Cs = S’ 2

2db

S’ = 2Cs

2db

S’ 2

Failure plane Cb

Cb

Cb > Cs

Cs > Cb Cs >> Cb

C = Cs = S’ 2

C = Cb

Failure patterns as for single bars

Side split failure

At failure (CS >> Cb) V-notch failure

Just before failure

At failure (CS > Cb) Face-and-side split failure

Figure 6.16.2  Concrete cylinder hypothesis and split failure patterns for lapped spliced bars. (From Orangun, Jirsa, and Breen [6.3].)

215



6.17  WELDED SPLICES AND MECHANICAL CONNECTIONS

215

TABLE 6.16.1  TENSION LAP SPLICES (ACI-​25.5.2.1)  As required   A provided  s over splice length

a

Maximum Percent of As Spliced

Splice Class

Required Lap, Lsta

Notes

≤ 0.5

 50 100

A B

1.0Ld 1.3Ld

Desirable OK

> 0.5

100

B

1.3Ld

Avoid more than 0.5

12 in. minimum.

to which the bars are stressed. When the factored moment Mu is only 50% of the design strength φ M n, the ratio would be considered 0.5. In general, temperature and shrinkage reinforcement should be considered as fully stressed for the purpose of designing splices. The modification factors described in Section 6.7 to account for the location of the reinforcement ψ t , epoxy coating ψ e, bar size ψ s, and lightweight concrete λ, also apply to splices, except that Ld cannot be reduced if reinforcement in excess of that required by analysis is provided (ACI-​25.5.1.4). A beam with splices should be as ductile as one without splices. In general, splices should be located away from points of maximum tensile stress, and splicing should be staggered along the length of the bars. In other words, all bars should not be spliced at one location. The ACI Code provisions are intended to prevent splice failure when the full nominal strength in flexure is reached at the spliced location. Requirements for minimum clear spacing of contact splices (ACI-​25.5.1.2) are to ensure that an adequate amount of surrounding concrete is provided to resist splitting. In noncontact lap splices, however, the individual bars should not be spaced transversely too far apart; otherwise, an unreinforced section is created. Noncontact splices in flexural members may not be farther apart than one-​fifth of the required lap length nor 6 in. (ACI-​25.5.1.3). The lap lengths prescribed in Table  6.16.1 (also stated earlier:  Section 6.9) shall be increased 20% for a three-​bar bundle and 33% for a four-​bar bundle (ACI-​25.6.1.5). For determining the appropriate modification factors, a unit of bundled bars is to be treated as a single bar of a diameter derived from the equivalent total area. As stated in ACI-25.6.1.7, lap splices of individual bars within a bundle must be staggered so that they do not overlap.

6.17 WELDED SPLICES AND MECHANICAL CONNECTIONS IN TENSION A welded tension butt splice or mechanical connection is used when large tensile forces are to be transmitted across a splice or large bars need to be spliced and lapping may be impractical or prohibited. Bars larger than #11 may not be lap spliced to carry tension (ACI-​25.5.1.1). Tension tie members may not be lap spliced either (ACI-​25.5.7.4). A tension tie is a member (a)  carrying a tensile force large enough to cause tension over the entire section, (b) having a stress level in the reinforcement high enough to require every bar to be fully effective, and (c) having limited concrete cover on its sides (ACI-​R25.5.7.4). In the case of tension tie members, welded or mechanical splices in adjacent bars shall be staggered at least 30 inches. Minimum requirements for mechanical and welded splices are provided in ACI-​25.5.7. These splices are required to develop in tension at least 1.25fy (ACI-​25.5.7.1). According to ACI-​R25.5.7.1, “the 25 percent increase above the specified yield strength was selected as both an adequate minimum for safety and a practicable maximum for economy.” It is noted that mechanical and welded splices with strengths less than 1.25fy (allowed in the 2011 and previous editions of the ACI Code) are no longer permitted in the 2014 ACI Code. The welded splice is intended primarily for relatively large bars (#6 and larger) in main members (ACI-​R25.5.7.1). The tensile capacity required is intended to ensure sound

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full penetration welds—​that is, to produce splices capable of developing 1.25As fy of the spliced bars.

6.18 COMPRESSION LAP SPLICES Whereas bars larger than #11 acting in tension are not permitted to be lap spliced, #14 and #18 bars in compression may be lap spliced to #11 and smaller bars (ACI-​25.5.5.3). The minimum overlap in compression lap splices when fc′ is not less than 3000 psi is the following (ACI-​25.5.5.1): for f y ≤ 60, 000 psi,  Lsc = 0.0005 f y db

(

or 12 in. ( whichever is larger )

)

for f y > 60, 000 psi,  Lsc = 0.0009 f y − 24 db

(6.18.1)9

or 12 in. ( whichever is larger ) (6.18.2)9

When fc′ is less than 3000 psi, the lap length is to be increased by one-​third. Also, when bars of two different sizes are lapped (ACI-​25.5.5.4), the lap splice length is to be the larger of (1) the compression splice length of the smaller bar or (2) the compression development length Ld of the larger bar (see Section 6.8). For compression members whose main steel is surrounded by closed ties throughout the lap length, the required lap may be taken at 0.83 of that otherwise required by Eqs. (6.18.1) and (6.18.2), but not less than 12 in. (ACI-​10.7.5.2.1). A minimum percentage of column tie area is also required by ACI-​10.7.5.2.1. Column ties are discussed in Chapter 10 (Section 10.8). For members whose main steel is surrounded by a closely wound spiral, the required lap may be taken at 0.75 of that otherwise required, but not less than 12 in. (ACI-​10.7.5.2.1). Spiral reinforcement is discussed in Chapter 10 (Section 10.9). The number of bar diameters required for the overlap in compression lap splices is summarized in Table 6.18.1.

TABLE 6.18.1  BAR DIAMETERS REQUIRED FOR COMPRESSION LAP SPLICES FOR fc′ ≥ 3000 PSI (ACI-​25.5.5.1 AND ACI-​10.7.5.2.1) Yield Stress fy (ksi)

40 50 60 75 80

Bar Diametersa

Spiral Column

Tied Column

15 18.75 22.5 32.6 36.0

16.6 20.75 24.9 36.2 39.9

Others 20 25 30 43.5 48.0

When computing splice length, the minimum to be used is 12 in.

a

9  Eqs (6.18.1) and (6.18.2), for SI, Lsc and db in mm, fc′ and fy in MPa, give, respectively, For f y ≤ 420 MPa,



Lsc = 0.071 f y db

or 300 mm

For f y > 420 MPa, Lsc = (0.13 f y − 24) db

or 300 mm

(6.18.1)  (6.18.2) 

217



6.21  DESIGN EXAMPLES

217

6.19 END BEARING CONNECTIONS, WELDED SPLICES, AND MECHANICAL CONNECTIONS IN COMPRESSION End bearing connections are allowed for compression only, wherein the load in the bars is transmitted by bearing of square cut ends held in concentric contact by a suitable device. According to ACI-​25.5.6.3, bar ends must terminate in flat surfaces within 1.5° of a right angle to the axis of the bars and must be fitted within 3° of full bearing after assembly. End bearing splices are only permitted when the member contains closed ties, closed stirrups, spirals, or hoops (ACI-​25.5.6.2). When welded splices or mechanical connections are used in compression, the requirements are the same as for tension splices—​that is, they must be capable of developing 125% of the yield strength of the bars (ACI-​25.5.7.1).

6.20 SPLICES FOR MEMBERS UNDER COMPRESSION AND BENDING For compression members, there are three categories of special splice provisions (ACI-​10.7.5.1).

Compressive Bar Stress Due to Factored Loads When the factored load causes compression in the bars, any type of splice may be used, including lap splices according to ACI-​25.5.5.1 and 25.5.5.3, mechanical and welded splices according to ACI-​25.5.7, and an “end bearing splice” according to ACI-​25.5.6 and 10.7.5.3.1. For lap splices, the lap length must satisfy Eq. (6.18.1) or (6.18.2) and cannot be less than 12 in.

Tension Bar Stress Not Exceeding 0.5fy Due to Factored Loads When moderate tension bar stresses under factored loads exist but do not exceed 0.5fy, lap splices are permitted according to ACI-​10.7.5.2.2. Alternatively, mechanical or welded splices are permitted according to the requirements of ACI-​25.5.7. When lap splices are used (for tension not exceeding 0.5fy), Class B tension lap splices must be used when more than 50% of the bars are spliced at a section. When fewer than 50% of the bars are spliced at a section and the splices are staggered by at least Ld, Class A splices may be used (ACI-​10.7.5.2.2).

Tension Bar Stress Greater than 0.5fy due to Factored Loads These are tension splices. When lap splices are used, they must be Class B splices according to ACI-​10.7.5.2.2. As an alternative to lap splices, mechanical or welded splices are permitted according to ACI-​25.5.7.

6.21 DESIGN EXAMPLES Two complete examples on the design of reinforced concrete flexural members are presented for the purpose of showing the design for flexure, shear, and development of reinforcement, all in the same beam.

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EXAMPLE 6.21.1 Design the simply supported beam shown in Fig. 6.21.1(a). The dead load is 1.08 kips/​ft, not including the weight of the beam. The live load consists of a concentrated load of 13.8 kips at midspan. Use normal-​weight concrete with fc′ = 3000 psi and fy = fyt = 40,000 psi steel reinforcement. SOLUTION  (a) Design a cross section for flexure. Assume that a rectangular section is desired with tension reinforcement only. A small beam is desired, having a reinforcement ratio ρ somewhat less than that corresponding to a net tensile strain ε t = 0.005. This will satisfy the minimum net tensile strain limit of 0.004 (ACI-​9.3.3.1). From basic principles as illustrated in Section 3.6, or the value from Table 3.6.1,

ρtc = ρ (ε t = 0.005) = 0.0203



Select ρ = 0.02 , and using Eq. (3.8.4b) or strain compatibility and equilibrium, as shown in Section 3.8, obtain Rn = 675 psi. (b) Compute the factored loads using ACI-​5.3. Assume that the weight of the beam is 0.2 kip/​ft,

wu = 1.2 D +1.6 L = 1.2 (1.08 + 0.2 ) +1.6 ( 0 ) = 1.54 kips/ft ( dead load ) Wu = 1.2 ( 0 ) +1.6 (13.8) = 22.1 kips ( live load ) Mu =

1 1 (1.54)(20)2 + ( 22.1) (20) = 77 + 111 = 188 ft-kips 8 4

Assuming φ = 0.90 for a tension-​controlled section, Mu 188 = = 209 ft-kips φ 0.90

required Mu =

(c) Determine beam size. For the selected Rn ≈ 675 psi, the corresponding bd 2 is

required bd 2 =

Mn 209 (12,000) = = 3716 cu in. 675 chosen Rn

If b = 12 in., d=



3716 = 17.6 in. 12

1 required h = d + approx 2 in. for one layer of bars 2 = 17.6 + 2.5 = 20.1 in. 12(20) weight of beam = (0.15) = 0.25 kip/ft 144

Use 12 × 20 cross section. (Continued)

219



219

6.21  DESIGN EXAMPLES

Example 6.21.1 (Continued) 1’ – 0”

#3 to support stirrups

4’ – 6”

12”

20” 3 12 ”

#10 #11

13 spaces @ 8 12 ”

Symmetrical about CL

#3 U stirrups 10’– 0” (a)

φMn

Ld = 2.6’

d = 1.46’

191’k

Ld = 3.1’ for #11 B

145’k

A Factored moment, Mu

φMn with 2 – #11

#10 bar cut here

5’ (b) Face of support φVn = 30.9 k 27.1k

d

φVs = 13.6k for s = 8 12 ” Max required φVs = 6.7 k Factored shear, Vu

19.1k

11.1k φVc = 17.3k

0.5φVc = 8.7 k (c)

Figure 6.21.1  Simply supported beam of Example 6.21.1.

(d) Correct the loads for beam weight and select reinforcement. revised wu = 1.2(1.08 + 0.25) = 1.60 kips/ft (dead load)



1 1 revised Mu = (1.60)(20)2 + (22.1)(20) = 80 + 111 = 191 ft-kips 8 4 Mu 191 required Mu = = = 212 ft-kips φ 0.90 M 212(12,000) = 692 psi required Rn = n2 = bd 12(17.5)2

(Continued)

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C hapter   6     D evelopment of R einforcement

Example 6.21.1 (Continued) The required steel reinforcement ratio ρ may next be computed from Eq. (3.8.5), or approximated by straight-​line proportion,

 692  As ≈ 0.020(12)(17.5)  = 4.3 sq in.  675 

Note that when revised Rn is larger, as in this case, the approximation using straight-​line proportion is slightly nonconservative (i.e., it underestimates the ρ value). Try 2–​#11 and 1–​#10 bars (As = 4.39 sq in.). Assuming #3 U stirrups, the minimum width of beam to accommodate these bars is 10.8 in. (see Table 3.9.2), which is less than the width of beam being used. [Note that the entry of Table 3.9.2 must be with the two corner bars (#11); then add for the one smaller bar (#10), and then add the difference in diameters between the larger and smaller bars, giving 10.64 in. from the table plus 0.14 in., making a total of 10.78 in. (say, 10.8 in.). Alternatively, enter Table 3.9.2 with 3–#11 and subtract the difference in bar diameters.] (e) Check the strength of the section. Even when tables and design approaches should give a result satisfying the ACI Code if no errors have been made reading tables and curves, a check of strength should always be made to verify that no mistakes have been made. C = 0.85 fc′ ba = 0.85 (3)12 a = 30.6 a T = As f y = 4.39 ( 40 ) = 176 kips a = 176 / 30.6 = 5.75 in. a  M n = (C or T )  d –   2 = 176 17.5 − 0.5 ( 5.75) = 215 ft-kips 12 1



(f) Compute the net tensile strain εt at the extreme tension steel,

β1 = 0.85;

c = a/ β1 = 5.75 / 0.85 = 6.76 in.

For this beam having one layer of steel, dt = d = 17.5 in.,

ε t = 0.003

dt − c 17.5 − 6.76 = 0.003 = 0.00477 c 6.76

Since ε t is less than the limit 0.005 of tension-​controlled sections (see Fig. 3.6.2), the appropriate φ factor is slightly less than 0.90 (ACI-​21.2.2). For sections with stirrups, the linear relationship is given by Eq. (3.6.3),

φ = 0.65 + 0.25

(ε t − ε y ) (0.005 − ε y )

≤ 0.9



where

εy =

fy Es

=

40, 000 = 0.00138 29, 000, 000

Thus,

φ = 0.65 + 0.25

(0.0047 − 0.00138) = 0.88 (0.005 − 0.00138)

[φ M n = 0.88(215) = 189 ft-kips] ≈ [ Mu = 191 ft-kips]

The section is acceptable.

(Continued)

21



221

6.21  DESIGN EXAMPLES

Example 6.21.1 (Continued) (g) Design of shear reinforcement; simplified method with constant Vc. Vu (at centerline of support) = 1.60 (10) + 11.1 = 27.1 kips Vu (at d from face of support) = 27.1 − 1.60 (6 + 17.5)

φ Vc = φ (2 fc′ bw d ) = 0.75(2 3000 )(12)(17.5)

1 12

1 1000

= 24.0 kips = 17.3 kips

required φ Vs = Vu − φ Vc = 24.0 − 17.3 = 6.7 kips



min φ Vs = φ 0.75 fc′ bw d ≥ φ (50)bw d

min φ Vs = 0.75[0.75 3000 (12)17.5]

1 1000

≥ 0.75[50(12)17.5]

1 1000

min φ Vs = 6. 5 kips ≤ 7.9 kips thus, the 7.9-​kip limit controls. Since required φ Vs = 6. 5 kips is less than min φ Vs = 7.9 kips to satisfy the requirement of ACI-​9.6.3.3, use min φ Vs = 7.9 kips to determine the stirrup spacing at the criti­ cal section. For #3 U stirrups, using Eq. (5.10.7),

φ Vs =

φ Av f yt d s



Thus,

required s =

φ Av f yt d φ Vs

=

0.75(0.22)(40)17.5 = 14.6 in. 7.9

However, the stirrup spacing may not exceed d / 2 = 8.75 in. Try #3 stirrups @ 8 1 2 -​in. spacing. Stirrups at this spacing must be used until Vu ≤ 0.5φ Vc = 17.3 / 2 = 8.7 kips, which for this beam [see Fig.  6.21.1(c)] means the entire span. (h) Make the preliminary selection of the cutoff point for 1–​#10. The remaining design strength with 2–​#11 bars is C = 30.6 a T = 2(1.56)40 = 125 kips

a=

125 = 4.08 in. 30.6



Compute the net tensile strain εt,

ε t = 0.003

17.5 − 4.08 / 0.85 = 0.00794 4.08 / 0.85

Since εt exceeds 0.005, the section is tension controlled and φ = 0.90.

φ M n = 0.90(125) [17.5 − 0.5(4.08)]

1 12

= 145 ft-kips (Continued)

2

222

C hapter   6     D evelopment of R einforcement

Example 6.21.1 (Continued) The value of 145 ft-​kips is plotted on the factored Mu diagram to locate the theoretical cutoff point A in Fig. 6.21.1(b). This point is located 6.73 ft from the center of the support. The actual cutoff location (point B) is found by extending from point A toward the support a distance of 12 bar diameters or the effective depth of the member, whichever is greater (ACI-​9.7.3.3).

 1.27  12 db = 12  = 1.27 ft  12  d=

17.5 12

= 1.46 ft

(Controls)

Therefore, the actual cutoff at point B would be located at (6.73 − 1.46 ) = 5.27 ft from the center of the support. Make the proposed cut at 5 ft from the support center line. This cutoff point will be acceptable only if the shear does not exceed two-​thirds of the shear strength at point B [ACI-​9.7.3.5(a)]. An alternative is to provide extra stirrups in accordance with ACI-​9.7.3.5(c). (i) Check cutoff point for satisfying the shear requirement of ACI-​9.7.3.5(a) for cutting bars in the tension zone. The shear strength provided by #3 stirrups at 8 1 2 -​in. spacing is

φ Vn = φ Vc + φ Vs = 17.3 +

0.75(0.22)(40)(17.5) = 30.9 kips 8.5

Vu at propo osed cutoff point B = 19.1kips



Vn 19.1 = = 62% < 67% φ Vn 30.9

OK

Use 1–​#10, 10′–​0 ″ long, placed symmetrically about midspan. (j) Use the simplified equations to determine the development length Ld for a #10 bar that is to be terminated at point B. Since the #11 bars are (presumably) not being developed over the portion of the beam where the #10 will be developed, the #11 bars do not influence the determination of Category A or B. Since the #10 has nearly 5-​in. cover to the side faces of the beam and has cover to the bottom face of 1 7 8 in. (with #3 stirrups), both items 1 and 2 of Category A are satisfied. Using simplified Eq. (6.6.5) for #7 and larger bars,

 fy ψ t ψ e  Ld =   db  20 λ fc′ 

[6.6.5]

The modification factors ψ t , ψ e , λ are all 1.0 for this beam. Thus, Eq. (6.6.5) gives

 40, 000(1.0)1.0  Ld (for # 10) =  1.27 = 46.4 in. (3..9 ft)  20(1.0) 3000 

(k) Use the general equation, Eq. (6.6.1), to determine the development length Ld for #10 bars. That equation is



   ψ t ψ e ψ s  3 fy Ld =  d  40 λ fc′  cb + K tr   b   d     b

[6.6.1]

(Continued)

23



223

6.21  DESIGN EXAMPLES

Example 6.21.1 (Continued) The distance cb is the smaller of (a)  bottom or side cover and (b)  one-​half center-​ to-​center spacing; in this case,

cb = bottom cover = 1.5 (i.e.,clear) + 0.375 (i.e.,stirrup) + 0.635 (i.e., bar radius) = 2.51 in.

In other words, the potential splitting plane is expected to be vertical between the center bar and the bottom face of the beam, similar to that shown for the center bar in Fig. 6.6.1(b). To compute Ktr, use the #3 stirrups with spacing s = d / ​2 = 8.5 in. along the development length. The number n of bars being developed is 1. The area of transverse reinforcement that crosses the potential splitting plane within the spacing s is Atr = 1(0.11) sq in. Thus, evaluating Eq. (6.6.2) gives K tr =



40 Atr 40(1)(0.11)  40  =   = 0.345 sn 8.5(1)  60 

where the ratio 40/​60 has been applied to account for the yield strength of 40,000 psi. The 2008 and earlier editions of the ACI Code included the yield strength of the transverse reinforcement in the calculation of Ktr. The coefficient in the current equation for Ktr (i.e., 40) includes, in effect, the yield strength of the transverse reinforcement, and, for Grade 60 reinforcement, replaces the term f yt / 1500 ( = 60, 000 /1500 = 40 ) in the preceding expression for Ktr. Whether Ktr should be adjusted for the Grade of steel is a matter of opinion. According to ACI Commentary R25.4.2.3, the yield strength of transverse reinforcement was removed from the expression for Ktr “because tests demonstrate that transverse reinforcement rarely yields during a bond failure.” Because a Grade lower than 60 ksi steel is used in this example, the authors have adjusted the value of Ktr for Grade 40. This will result in a lower contribution of the transverse reinforcement, Ktr, a result that is conservative. Evaluating Eq. (6.6.3),

  cb + K tr 2.51 + 0.345 = = 2.25 < 2.5 max  1.27   db

Thus, ( cb + K tr ) / db = 2.25. Note that whenever the cover and transverse reinforcement term (cb + Ktr)/​db is greater than 1.5, the general equation will give a smaller Ld than the simplified equations. Evaluate Eq. (6.6.1),

 3 40, 000 1.0(1.0)1.0  Ld (for # 10) =  1.27 = 30.9 in. (2.6 ft) 2.25   40 (1.0) 3000

which is 33% less than Ld computed with the simplified equation. (l) Use the general equation, Eq. (6.6.1), to determine the development length Ld for the #11 bars that extend into the support. A check of Table 3.9.4 giving minimum beam width such that clear spacing of 2–​#11 bars will equal 3db, shows a beam 10.9 in. wide or more will qualify. Since this is less than the 12 in. used, the #11 bar spacing exceeds 3db, meaning that cover, rather than bar spacing, will control the dimension cb. Thus,

cb = bottom or side cover = 1.5 (i.e.,clear) + 0.375 (i.e.,stirrup) + 0.705 (i.e., bar radius) = 2.58 in. (Continued)

24

224

C hapter   6     D evelopment of R einforcement

Example 6.21.1 (Continued) The potential plane of splitting is between the bars and the bottom and/​or side face of the beam [see Fig. 6.6.1(d)]. For the stirrups in the development region, use s = d / 2 = 8.5 in . spacing for computation of Ktr. The number n of bars being developed is 2, and Atr = 2(0.11) sq in. Thus, evaluation of Eq. (6.6.2) gives K tr =



40 Atr 40(2)(0.11)  40  =   = 0.345 sn 8.5(2)  60 

Evaluating Eq. (6.6.3),

  cb + K tr 2.58 + 0.345 = = 2.07 < 2.5 max  1.41   db

Thus, ( cb + Ktr ) / db = 2.1. Evaluate Eq. (6.6.1),

 3 40, 000 1.0(1.0)1.0  Ld (for # 11) =   1.41 = 36.8 in. (3.1 ft) 2.1  40 (1.0) 3000

(m)  Check development length requirement at the support (Section 6.14). According to ACI-​9.7.3.8.3(a), it is required that 1.30



Mn + La ≥ Ld Vu

M n for 2 – #11 bar =

145 = 161 ft-kips 0.9

Vu = factored shear at centerline of support = 27.1 kips



La = embedment length beyond the center of support; assume zero here 1.30

161(12) = 92.7 in.> [ Ld (# 11) = 36.8 in.] 27.1

OK

Actually, the design strength φ M n diagram provides this same check but more conser­ vatively (i.e., without the 1.30 factor), since the horizontal distance from the center of support to point A exceeds Ld. (n) Design sketch. The final conclusions are presented in Fig. 6.21.1(a). The crack control provisions of ACI-​24.3 need to be checked. If deflection control is important to prevent damage to partitions or other construction, deflections must be checked according to ACI-​24.2.2. Computation for crack control and deflections is treated in Chapter 12.

EXAMPLE 6.21.2 Design the overhanging beam shown in Fig. 6.21.2. The superimposed service uniform dead load is 4.0 kips/​ft. Use fc′ = 4000 psi (normal-weight concrete), f y = f yt = 60, 000 psi. (Live load is not used in this example because its purpose is to show how flexure, shear, and development of reinforcement are to be considered together in one simple case.) (Continued)

25



225

6.21  DESIGN EXAMPLES

Example 6.21.2 (Continued) SOLUTION  (a) Design for flexure. Since tension reinforcement will be required in the top of the overhang, it may be desirable to run some bars straight across the top of the entire beam. This would also help reduce creep and shrinkage deflection under sustained load. To have reasonable expectation that deflection will not be excessive, choose ρ at about 0.6ρtc = 0.011 as recommended in Section 3.9. Use ρ = 0.011 . For the selected ρ, one can use Eq. (3.8.4b) to obtain Rn = 596 psi. Alternatively, choose Rn ≈ 600 psi from Fig. 3.8.1. (b) Compute the factored loads and the corresponding shear and moment diagrams. Estimate the beam weight as 0.3 kip/​ft (roughly 2 sq ft, say 12 × 24 in.),

wu = 1.4 D = 1.4 ( 4.0 + 0.3) = 6.02 kips / ft ( dead load )

The factored shear at the left support is

Vu =

6.02(18) 0.5(6.02)(6.5)2 − = 47.1 kips 2 18

max(+ ) Mu =

(47.1)2 = 184 ft-kips 2(6.02)

max( −) Mu =

1 (6.02)(6.5)2 = 127 ft-kips 2

The shear and moment diagrams for factored loads are shown in Fig. 6.21.2. (c) Determine beam size. For the chosen Rn ≈ 600 psi, the corresponding bd 2 is

required bd 2 =

Mu 184(12,000) = = 4090 cu in. φ (chosen Rn ) 0.90(600)

where φ = 0.90 is assumed. Thus, b (in.)

d (in.)

h (in.)

12 14

18.5 17.5

21 20

Assuming one layer of tension steel, 2.5 in. has been added to d to give the overall dimension h. When checking strength, compute d, starting from overall h. Use 14 × 20 cross section. (d) Select the reinforcement. Correct the beam weight and update the factored loads, shears, and moments, if needed.

revised weight of beam =

14(20) 144

(0.15) = 0.29 kip/ft

The revised factored load wu is 6.01 kips/​ft, which is approximately the same as that initially assumed. The factored shear and moment diagrams remain unchanged (Fig. 6.21.2). The corrected effective depth d, assuming one layer of steel, is

revised d = h − cover − stirrup diam − bar radius d = 20 − 1.5 − 0.375 ( #3 est.) − 0.5 ( est.) = 17.6 in. (Continued)

26

226

C hapter   6     D evelopment of R einforcement

Example 6.21.2 (Continued) At section A-​A of Fig. 6.21.2, the revised required Rn then becomes required Rn =



Mn 184(12, 000) = = 566 psi 2 0.90(14)(17.6)2 φ bd

The required steel reinforcement ratio ρ may next be computed from Eq. (3.8.5), obtained from the Fig. 3.8.1 curves, or approximated by straight-​line proportion,  566  As ≈ 0.011(14)(17.6)  = 2.6 sq in.  600 



Use 2–​#9 and 1–​#7 (As = 2.60 sq in.) in the midspan region of the 18-​ft span. wD = 4.0 kips/ft (not including beam weight)

18’– 0

(wu = 6.01 kips/ft with corrected beam weight)

6’– 6”

Width of supports = 18” 1

4’– 0”

A

72 ”

B

4 – #6 × 10’– 6” 20”

A

5’ 1

14” B

2 – #9 × 18’– 7 2 ”

Vu, kips

47.1k

4”

1 – #7 × 13’

d

39.1k

33.8k 23.4k φVc

d 23.6k

#3 stirrups 4” 14 @ 7” = 8’– 2”

0.5φVc

0.5φVc 0.5φVc

7 @ 8”

φVc

#3 stirrups

5 @ 8”

#3 stirrups

4”

23.4k 7.82’ Ld (#9) = 2.7’

47.8k Ld (#7) = 2.1’

d

Ld (#9) = 2.7’

187’k 184 Mu, ft-kips

61.2k

B

A Moment capacity diagram, φMn Min 12db or d

2 – #6 φMn = 67’k

Factored moment, D Mu

1.46’ (ACI–9.7.3.8.4)

Tensile capacity 2 – #9 φMn = 147’k

Min 12db or d = 17.6”

C Ld = 2.3’

Ld = 2.5’

127’k 131’k

Ld = 2.3’

18” support width

Figure 6.21.2  Overhanging beam for Example 6.21.2.

(Continued)

27



227

6.21  DESIGN EXAMPLES

Example 6.21.2 (Continued) At section B-​B, obtain by proportion

required As = 2.6

− Mu 127 = 2.6 = 1.8 sq in. + Mu 184

Use 4–​#6 (As = 1.76 sq in.) in the negative moment region. (e) Check the strength of the section. At section A-​A,

C = 0.85 fc′ ba = 0.85(4)14 a = 47.6 a T = As f y = 2.60(60) = 156 kips a = 156 / 47.6 = 3.28 in. M n = (C or T )(d − a /2) = 156[17.6 − 0.5(3.28)]

1 12

= 207 ft-kips



Compute the net tensile strain ε t ,

ε t = 0.003

17.6 − 3.28 / 0.85 = 0.0107 3.28 / 0.85

This far exceeds the limit of 0.005 for tension-​controlled sections; therefore, the appropriate φ factor is 0.90.

φ M n = 0.90 ( 207) = 187 ft-kips > [ Mu = 184 ft-kips]

OK

At section B-​B, C = 0.85 fc′ ba = 0.85(4)14 a = 47.6 a T = As f y = 1.76(60) = 106 kips a = 106 / 47.6 = 2.22 in.

M n = 106 [17.6 − 0.5(2.22)] = 146 ft-kips 1 12



Compute the net tensile strain ε t ,

ε t = 0.003

17.6 − 2.22 / 0.85 = 0.0172 2.22 / 0.85

Since εt exceeds 0.005, the section is tension ​controlled and φ = 0.90.

φ M n = 0.90 (146 ) = 131 ft-kips > [ Mu = 127 ft-kips]

OK

The reader is reminded that for a given Rn, the required ρ expression is a quadratic function; however, for practical use it may be approximated to be linear (see Fig. 3.8.1). A check of strength should always be made. The arrangement of main reinforcement finally selected is shown in Fig. 6.21.2. Note that in the design of the flexural reinforcement for the maximum negative moment, the center of support value of 127 ft-​kips was used instead of the lesser value at the face of support, as could have been used in accordance with the discussion of the negative moment on continuous beams (Section 6.11). On this statically determinate beam, the authors prefer a more conservative approach of using the center of support value. For beams on wider supports, corrections to the face of support would be appropriate. (Continued)

28

228

C hapter   6     D evelopment of R einforcement

Example 6.21.2 (Continued) (f) Shear reinforcement. At d from the face of right support on the 18-​ft span, Vu = 61.2 − 6.01(17.6 + 9)

1 12

= 47.8 kips

φ Vc = φ (2 λ fc′) bwd 1 = 0.75 2(1.0) 4000  (14)(17.6) = 23.4 kips 1000 where λ = 1.0 for normal-weight concrete

required φ Vs = Vu − φ Vc = 47.8 − 23. 4 = 24. 4 kips  max permitted φ Vs  for s = 

d  = 2φ Vc = 46.8 kips > [ required φ Vs = 24..2 kips] OK 2

Note that maximum permitted φ Vs calculated is based on the nominal stress vs = 4 fc′ at the maximum spacing of d/​2 (ACI-​9.7.6.2.2), which gives maximum φ Vs = 2φ Vc when the simplified Vc expression is used. For minimum percentage of stirrups (ACI-​9.6.3.3),

(

)

(

min φ Vs = φ 0.75 fc′ bw d = 0.75 0.75 4000 (14)(17.6)

)

1 1000

= 8.8 kips

but not less than

φ (50bw d ) = 0.75[50(14)(17.6)]

1 1000

= 9.2 kips



Thus, the 9.2-​kip limit controls. For strength, using #3 U stirrups, φ Av f yt d 0.75(0.22)(60)(17.6) 174 φ Vs = = = s s s or

s=

174 = 7.2 in. 24.2

Thus, the strength requirement permits stirrups no farther apart than 7.2 in. and is more restrictive than the d/​2 limit of 8.8 in. Use #3 U stirrups at 7-​in. spacing, as dimensioned on the shear diagram in Fig. 6.21.2. (Ordinarily the stirrups should be dimensioned on the design sketch, i.e., the side view of the beam in Fig. 6.21.2.) For the cantilever portion and to the right of the left support of the beam, it can be verified that only minimum stirrups at d / 2 = 8.5 in. maximum spacing are needed. Thus, Use #3 U stirrups at 8-​in. spacing in these regions, as shown in Fig. 6.21.2. (g) Development length using the simplified method of ACI-​25.4.2.2. Because the clear cover for bars being developed is at least db, clear lateral spacing is at least db (see Table 3.9.2), and at least minimum stirrups are used, Category A, item 1 is satisfied. Use simplified equations, Eqs. (6.6.4) and (6.6.5). From Table 6.6.1, the following Ld values are obtained when the modifications ψ t , ψ e , and λ = 1.0:

Ld ( #6 ) = 28.5 in. ( 2.4 ft )



Ld ( #7) = 41.5 in. (3.5 ft )



Ld ( #9) = 53.5 in. ( 4.5 ft )

(Continued)

29



6.21  DESIGN EXAMPLES

229

Example 6.21.2 (Continued) For the #6 bars in the negative moment zone, a casting position factor ψt of 1.3 must be used. Thus, for the #6 bars,

Ld ( #6 ) = 28.5 (1.3) = 37.1in. (3.1ft)

(h) Development length using the general equation, Eq. (6.6.1).



   ψ t ψ e ψ s  3 fy Ld =  d  40 λ fc′  cb + K tr   b   d     b

[6.6.1]

The distance cb is the smaller of (a) top or side cover, and (b) one-​half center-to-​center spacing. For the #9 corner bars: bottom and side cover = 1.5 + 0.375 + 1.128 / 2 = 2.44 in.

Controls

Assuming that the #7 bar will not be developed at the same location, one half of the bar spacing dimension is

1 1  1.128   spacing = 14 − 2(1.5) − 2(0.375) − 2  = 4.6 in.  2   2 2

Thus, the potential plane of splitting for these corner bars is between the bars and the bottom and/​or side face of the beam [see Fig. 6.6.1(d)]. To the right of the left support, stirrup spacing is 8 in. Thus, Ktr is computed by using Eq. (6.6.2) as follows:

K tr =

40 Atr 40(2)(0.11) = = 0.55 sn 8(2)

To the left of the right support, stirrup spacing is 7 in. Thus, 8 K tr = 0.55 = 0.63 7 For the #7 center bar: Assuming that the #9 corner bars will not be developed at the same location, then vertical splitting and thus the bottom cover dimension governs for the #7 bar: bottom cover = 1.5 + 0.375 + 0.875/​2 = 2.3 in. (vertical splitting governs) Toward the left support, stirrup spacing is 8 in; thus,

K tr =

40 Atr 40(1)(0.11) = = 0.55 sn 8(1)

Toward the right support, stirrup spacing is 7 in; thus, 8 K tr = 0.55 = 0.63 7 For the #6 bars (negative moment region): cover = 1.5 + 0.375 + 0.75/2 = 2.3 in. 1 1  0.75   1 = 1.6 in. (horizontal splitting governs) spacing = 14 − 2(1.5) − 2(0.375) − 2  2  2   3 2 (Continued)

230

230

C hapter   6     D evelopment of R einforcement

Example 6.21.2 (Continued) To the left of the right support, stirrup spacing is 7 in. Since horizontal splitting governs, the stirrups are effective only for the two corner bars out of the four bars being developed. Thus, 40 Atr 40(2)(0.11) = = 0.31 sn 7(4)

K tr =



For the development of the #6 bars near the free end of the cantilever K tr = 0. Compute (cb + K tr ) / db for the #9, #7, and #6 bars of this design.



For #9 bars

 cb + K tr 2.44 + 0.55  = = 2.65 > 2.5 max  1.128  db 

For #7 bars

 cb + K tr 2.30 + 0.55  = = 3.26  > 2.5 max  0.875  db 

In the calculations shown above, the smaller value of Ktr (i.e., larger stirrup spacing) was used for the #9 and #7 bars. Use of the larger Ktr value will result in an even larger value of (cb + K tr ) / db and thus the 2.5 limit would still control. To the left of the right support

For #6 bars

 cb + K tr 1.60 + 0.31  = = 2.55 > 2.5 max  0.75  db 

For the development of the #6 bars near the free end of the cantilever K tr = 0; thus,

For #6 bars

 cb + K tr 1.60 + 0  = = 2.13 < 2.5 max  0.75  db 

Substituting into Eq. (6.6.1) gives

 3 60, 000 1.0(1.0)1.0  Ld (# 9) =   1.128 = 32.1 in. (2.7 ft) 2.5  40 (1.0 4000 )



 3 60, 000 1.0(1.0)1.0  Ld (# 7) =   0.875 = 24.9 in. (2.1ft) 2.5  40 (1.0 4000 )

At the 18-​ft span side of the right support, using ψ t = 1.3 for top bars,

 3 60, 000 1.3(1.0)1.0  Ld (# 6) =   0.75 = 27.7 in. (22.3ft) 2.5  40 (1.0 4000 )

At the free end of the cantilever, using ψ t = 1.3 for top bars,

 3 60, 000 1.3(1.0)1.0  0.75 = 32.5 in. (2.7 ft) Ld (# 6) =   40 (1.0 4000 ) 2.13 

Although ACI-​25.4.2.4 permits a bar size factor ψ s = 0.8 to be used for #6 and smaller bars, it was not applied in the above calculations. In this example, however, the authors have adopted the recommendations of ACI Committee 408 [6.53] which do not support the use of this factor (see Section 6.7). The above computation of development length Ld has assumed #3 stirrups at either 7-​ or 8-​in. spacing for the high-shear ​regions near the supports. The 8-​in. spacing is very (Continued)

231



231

6.21  DESIGN EXAMPLES

Example 6.21.2 (Continued) near the maximum permitted of d/​2 (8.75 in.). Good practice will put stirrups in the entire beam. Note again that because the cover and transverse reinforcement factor, (cb + K tr ) / db , is larger than 1.5 in. in all Ld calculations, the simplified equations give much larger Ld values [part (g)]. (i) Examine the feasibil­ity of cutting the #7 bar in the positive moment zone. The remaining 2–​#9 would provide the following strength: C = 47.6 a

T = 2 (1.0 ) 60 = 120 kips

a = 2.52 in.

2.52  1  φ M n = 0.90 (120 )  17.6 −  = 147 ft-kips  2  12

Based on the design strength alone, the #7 bar could be cut at points A and B of Fig. 6.21.2; however, ACI-​9.7.3.5 must also be satisfied to cut bars in the tension zone. This is examined in part (j). For the #7 bar, the bar arrangement must be established before development length can be accurately determined. If the #7 bar is cut as shown in Fig. 6.21.2, then its development and the #9 bars development are independent. However, if all three bars (2–​#9 and 1–​#7) are run into the support, then the center-​to-​center spacing between bars might control, requiring increased development length for both the #7 and the #9 bars. At the point in a design when the final check is made, of which the design strength diagram is a part, the bar arrangement is known and from that the development lengths can be computed. The shorter development lengths computed using the general equation, Eq. (6.6.1), were used in drawing the design strength diagram of Fig. 6.21.1. (j) Check ACI-​9.7.3.5 for cutting the #7 bar at points A and B in the tension zone (see Section 6.11). i. Since φ Mn for the continuing #9 bars is not twice the factored moment Mu at the cutoff point, ACI-​9.7.3.5(b) cannot be satisfied. ii. Check if ACI-​9.7.3.5(a) can be satisfied for #3 U stirrups at 8 in. at A and 7 in. at B as provided. At point A The distance from the center of the left support to point A is

 1 (4.2 ft − d ) =  4.2 − (17.6)  = 2.7 ft 12  

Make the cut location 2 ft 8 in. from the left support. Then,

Vu (at A) = 47.1 − 2.66 (6.01) = 31.1 kips

φ Vs (at A) =

φ Av f yt d s

=

0.75(0.22)(60)17.6 = 21.8 kips 8

The percent stressed at A is then,

Vu 31.1 = = 69% > 67% max φ (Vc + Vs ) 23.4 + 21.8

NG

The percent stressed is very close to the maximum permitted; thus, a possible strategy to satisfy this requirement would be to reduce stirrup spacing to, say, 7 in. In such a (Continued)

23

232

C hapter   6     D evelopment of R einforcement

Example 6.21.2 (Continued) case, the percent stressed at A would become 64%. Alternatively, check ACI-​9.7.3.5, as shown below. At point B The distance from the center of the right support to point B is

 1 (6.6 ft − d ) = 6.6 − (17.6)  = 5.13 ft 12  

Make the cut location 5 ft from the right support. Then, Vu (at B ) = 61.2 − 5.0 (6.01) = 31.2 kips

φ Vs (at B) =

φ Av f yt d s

=

0.75(0.22)(60)17.6 = 24.9 kips 7

The percent stressed at B is then

Vu 31.2 = = 65% < 67% max φ (Vc + Vs ) 23.4 + 24.9

OK

iii.  Check if ACI-​9.7.3.5(c) can be satisfied for #3 U stirrups at 8 in. from A to ( 3 4  )d to the right of A, as provided. Vu (at A) = 31.1 kips



The required area of stirrups spaced at 8 in. at this location is

Av =

(31.1 − 23.4) 8 = 0.08 sq in. 0.75 60(17.6)

The area of stirrups in excess of that required at this location is

excess Av = 0.22 − 0. 08 = 0. 14 sq in.

which is greater than the minimum required,

60 bw s / f yt = 60 (14 )(8) / 60, 000 = 0.11 sq in.

Therefore, the #7 bar could be cut at the proposed point A. Since cutoff point A is close to the left support (2 ft 8 in.) and only one bar is being cut off, it may be just as cost effective to simply extend the bar into the support. The latter approach is adopted in this design example. Cut 1–​#7 bar only at point B and extend the bar into the left support (length = 13 ft). (k) Determine cut of 2–​#6 bars in the negative moment zone. In trying to cut the 2–​#6 bars from the negative moment region near section B-​B, the bars must be extended into the span to the left of the support farther than 12 bar diameters or the effective depth d, whichever is greater, beyond the point where they are no longer needed (i.e., where φ M n (2 − # 6) = Mu ). Since the potential cut at point C (at Ld from the face of support) is not in a tension zone, no further investigation of this cutoff location is required. On determining that the remaining 2–​#6 bars can be cut off only a short distance farther into the span at point D, beyond the point of inflection per ACI-​9.7.3.8.4, the decision is made to cut all 4–​#6 bars at point D, that is, at 1.46 + 2.40 = 3.86 ft, say, at 4 ft from the center of the support. If for some reason, such as the desire to (Continued)

23



233

6.21  DESIGN EXAMPLES

Example 6.21.2 (Continued)

have compression steel for deflection control, two of the #6 bars could be extended across the 18-​ft span instead of being cut at D; then the other 2–​#6 would be cut at point C. On the right side of the support in the negative moment region, any potential cut location would be so near the end of the cantilever that it was decided to run all 4–​#6 bars to the end (say, 1.5-​in. clear).

(l) Check ACI-​9.7.3.8.3 for development of positive moment reinforcement at the left simple support and at the point of inflection near the right support (see Section 6.14). At the inflection point, ACI-​9.7.3.8.3(b) requires that

Mn + La ≥ Ld Vu

Assuming that the 1–​#7 bar will be terminated (a conservative assumption for this computation), only the 2–​#9 bars contribute to Mn. The actual La from the inflection point to the end of the bars exceeds d (17.6 in.); thus, use La = 17.6 in.

  147(12)  0.90(47.1) + 17.6 = 59.2 in. > [ Ld = 32.1 in.]  

OK

For the simple support, ACI-​9.7.3.8.3(a) requires

1.30 M n + La ≥ Ld Vu

In this case, La = 7.5 in. since the bars extend this amount beyond the center of the support. Assuming that the #7 bar does not extend into the support (a conservative assumption for this computation),

 1.30(147)12  0.90(47.1) + 7.5 = 61.6 in. > [ Ld = 32.1 in.]  

OK

(m) Design strength diagram. Figure 6.21.2 compares the factored moment Mu diagram and the design strength φ M n diagram provided by the selected reinforcing bars. In computing the design strength, the steel in the compression side of the beam has been neglected because it would have a negligible effect and because it was not required for strength. At the left end of the beam, the #7 and the 2–​#9 bars have been extended as far as is feasible into the support. In drawing the design strength diagram for the left end of the beam, it was assumed that the design strength will vary linearly with the development length of the #9 bars, although the #7 bar requires a shorter development length. This is conservative. (n) Deflection. The reinforcement ratio ρ being used was expected to control the deflections. Nevertheless, the deflection must be investigated if damage to partitions or other construction is of concern. (o) Design sketch. The final arrangement of longitudinal steel and stirrups is shown in Fig. 6.21.2. The stirrup locations are dimensioned on the Vu diagram. Omitted from the elevation view of the beam are the nominal-​sized (say, #3 or #4) longitudinal bars arbitrarily added for the stirrups to wrap around wherever the stirrups would otherwise have no support to hold them in vertical position. These pairs of bars would be located in this beam at both top and bottom faces of the beam where no longitudinal bars are shown in the figure. Cross-​sectional views showing bars, omitted here, are always part of the design sketch.

234

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SELECTED REFERENCES  6.1. ACI Committee 408. “Bond Stress—​The State of the Art,” ACI Journal, Proceedings, 63, November 1966, 1161–​1190. Disc., 63, 1569–​1570.   6.2. LeRoy A. Lutz and Peter Gergely. “Mechanics of Bond and Slip of Deformed Bars in Concrete,” ACI Journal, Proceedings, 64, November 1967, 711–​721. Disc., 65, 412–​414.   6.3. C. O. Orangun, J. O. Jirsa, and J. E. Breen. The Strength of Anchored Bars: A Reevaluation of Test Data on Development Length and Splices, Research Report 154–​3F, Center for Highway Research, University of Texas, Austin, January 1975.   6.4. C.  O. Orangun, J.  O. Jirsa, and J.  E. Breen. “A Reevaluation of Test Data on Development Length and Splices,” ACI Journal, Proceedings, 74, March 1977, 114–​122. Disc., 74, September 1977, 470–​475.  6.5. James O.  Jirsa, LeRoy A.  Lutz, and Peter Gergely. “Rationale for Suggested Development, Splice, and Standard Hook Provisions for Deformed Bars in Tension,” Concrete International, 1, July 1979, 47–​61.   6.6. D. Z. Yankelevsky. “Bond Action Between Concrete and a Deformed Bar—​A New Model,” ACI Journal, Proceedings, 82, March–​April 1985, 154–​161.   6.7. Emory L. Kemp. “Bond in Reinforced Concrete: Behavior and Design Criteria,” ACI Journal, Proceedings, 83, January–​February 1986, 50–​57.   6.8. Herbert J. Gilkey, Stephen J. Chamberlain, and Robert W. Beal. Bond Between Concrete and Steel. Iowa Engineering Experiment Station Bulletin No. 147, Iowa State College, 1940.   6.9. Arthur P. Clark. “Bond of Concrete Reinforcing Bars,” ACI Journal, Proceedings, 46, November 1949, 161–​184. 6.10. Phil M.  Ferguson and J.  Neils Thompson. “Development Length for Large High Strength Reinforcing Bars,” ACI Journal, Proceedings, 62, January 1965, 71–​93. Disc., 62, 1153–​1156. 6.11. Phil M.  Ferguson, John E.  Breen, and J.  Neils Thompson. “Pullout Tests on High Strength Reinforcing Bars,” ACI Journal, Proceedings, 62, August 1965, 933–​950. 6.12. Ervin S. Perry and J. Neils Thompson. “Bond Stress Distribution on Reinforcing Steel in Beams and Pullout Specimens,” ACI Journal, Proceedings, 63, August 1966, 865–​875. 6.13. E. L. Kemp, F. S. Brezny, J. A. Unterspan. “Effect of Rust and Scale on the Bond Characteristics of Deformed Reinforcing Bars,” ACI Journal, Proceedings, 65, September 1968, 743–​756. Disc., 66, 224–​226. 6.14. Saeed M.  Mirza and Jules Houde. “Study of Bond Stress–​Slip Relationships in Reinforced Concrete,” ACI Journal, Proceedings, 76, January 1979, 19–​46. 6.15. S. Soretz and H. Holzenbein. “Influence of Rib Dimensions of Reinforcing Bars on Bond and Bendability,” ACI Journal, Proceedings, 76, January 1979, 111–​125. 6.16. Raymond E. Untrauer and George E. Warren. “Stress Development of Tension Steel in Beams,” ACI Journal, Proceedings, 74, August 1977, 368–​372. 6.17. E. L. Kemp and W. J. Wilhelm. “Investigation of the Parameters Influencing Bond Cracking,” ACI Journal, Proceedings, 76, January 1979, 47–​72. 6.18. Shiro Morita and Tetsuzo Kaku. “Splitting Bond Failures of Large Deformed Reinforcing Bars,” ACI Journal, Proceedings, 76, January 1979, 93–​110. 6.19. R. Jimenez, R. N. White, and P. Gergely. “Bond and Dowel Capacities of Reinforced Concrete,” ACI Journal, Proceedings, 76, January 1979, 73–​92. 6.20. S.  Ali Mirza. “Bond Strength Statistics of Flexural Reinforcement in Concrete Beams,” ACI Structural Journal, 84, September–​October 1987, 383–​391. 6.21. J. P. Moehle, J. W. Wallace, and S.-​J. Hwang. “Anchorage Lengths for Straight Bars in Tension,” ACI Structural Journal, 88, September–​October 1991, 531–​537. 6.22. David Darwin, Steven L. McCabe, Emmanuel K. Idun, and Steven P. Schoenekase. “Development Length Criteria: Bars Not Confined by Transverse Reinforcement,” ACI Structural Journal, 89, November–​December 1992, 709–​720. Disc., 90, September–​October 1993, 581–​583. 6.23. LeRoy A.  Lutz, S.  Ali Mirza, and Narendra K.  Gosain. “Changes to and Applications of Development and Lap Splice Length Provisions for Bars in Tension (ACI 318–​89),” ACI Structural Journal, 90, July–​August 1993, 393–​406. 6.24. Shyh-​Jiann Hwang, Yih-​Ren Leu, and Han-​Lin Hwang. “Tensile Bond Strengths of Deformed Bars of High-​Strength Concrete,” ACI Structural Journal, 93, January–​February 1996, 11–​20. 6.25. ACI Committee 408. “Suggested Development, Splice, and Standard Hook Provisions for Deformed Bars in Tension,” Concrete International, 1, July 1979, 44–​46. 6.26. Phil M. Ferguson and Farid N. Matloob. “Effect of Bar Cutoff on Bond and Shear Strength of Reinforced Concrete Beams,” ACI Journal, Proceedings, 56, July 1959, 5–​23. 6.27. Anthony M. Kao and Raymond E. Untrauer. “Shear Strength of Reinforced Concrete Beams Terminated in Tension Zones,” ACI Journal, Proceedings, 72, December 1975, 720–​722.

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SELECTED REFERENCES

235

6.28. Mete A. Sozen and Jack P. Moehle. A Study of Experimental Data on Development and Lap-​ Splice Lengths for Deformed Reinforcing Bars in Concrete. Report to The Portland Cement Association, Skokie, IL, and The Concrete Reinforcing Steel Institute, Schaumburg, IL, August 1990. 6.29. Paul R.  Jeanty, Denis Mitchell, and M.  Saeed Mirza. “Investigation of ‘Top Bar’ Effects in Beams,” ACI Structural Journal, 85, May–​June 1988, 251–​257. 6.30. D.  B. Cleary and J.  A. Ramirez. “Bond Strength of Epoxy-​ Coated Reinforcement,” ACI Materials Journal, 88, March–​April 1991, 146–​149. 6.31. Oan Chul Choi, Hossain Hadje-​Ghaffari, David Darwin, and Steven L.  McCabe. “Bond of Epoxy-​Coated Reinforcement: Bar Parameters,” ACI Materials Journal, 88, March–​April 1991, 207–​217. 6.32. Bilal S.  Hamad, James O.  Jirsa, and Natalie I.  D’Abreu de Paulo. “Anchorage Strength of Epoxy-​Coated Hooked Bars,” ACI Structural Journal, 90, March–​April 1993, 210–​217. 6.33. Hossain Hadje-​Ghaffari, Oan Chul Choi, David Darwin, and Steven L.  McCabe. “Bond of Epoxy-​ Coated Reinforcement:  Cover, Casting Position, Slump, and Consolidation,” ACI Structural Journal, 91, January–​February 1994, 59–​68. 6.34. Stacy J. Bartoletti and James O. Jirsa. “Effects of Epoxy Coating on Anchorage and Development of Welded Wire Fabric,” ACI Structural Journal, 92, November–​December 1995, 757–​764. 6.35. LeRoy A. Lutz. “Crack Control Factor for Bundled Bars and for Bars of Different Sizes,” ACI Journal, Proceedings, 71, January 1974, 9–​10. 6.36. N.  W. Hanson and Hans Reiffenstuhl. “Concrete Beams and Columns with Bundled Reinforcement,” Journal of the Structural Division, ASCE, 84, ST6 (October 1958), 1–​23. 6.37. Frank D. Steiner. “Suggested Applications for Bundled Bars,” ACI Journal, Proceedings, 64, April 1967, 213–​214. 6.38. John A. Hribar and Raymond C. Vasko. “End Anchorage of High Strength Steel Reinforcing Bars,” ACI Journal, Proceedings, 66, November 1969, 875–​883. Disc., 67, 423–​424. 6.39. John Minor and James O. Jirsa. “Behavior of Bent Bar Anchorages,” ACI Journal, Proceedings, 72, April 1975, 141–​149. 6.40. José L. G. Marques and James O. Jirsa, “A Study of Hooked Bar Anchorages in Beam–​Column Joints,” ACI Journal, Proceedings, 72, May 1975, 198–​209. 6.41. Robert L. Pinc, Michael D. Watkins, and James O. Jirsa. Strength of Hooked Bar Anchorages in Beam–​Column Joints. CESRL Report No. 77-​3, Department of Civil Engineering, University of Texas, Austin, November 1977. 6.42. Raymond E. Untrauer and Robert L. Henry. “Influence of Normal Pressure on Bond Strength,” ACI Journal, Proceedings, 62, May 1965, 577–​586. 6.43. Joint PCI/​WRI Ad Hoc Committee on Welded Wire Fabric for Shear Reinforcement, PCI Technical Activities Committee (Leslie D. Martin, Chairman). “Welded Wire Fabric for Shear Reinforcement,” PCI Journal, 25, July–​August 1980, 32–​36. 6.44. Phil M. Ferguson and John E. Breen. “Lapped Splices for High Strength Reinforcing Bars,” ACI Journal, Proceedings, 62, September 1965, 1063–​1078. 6.45. John P.  Lloyd and Clyde E.  Kessler. “Splices and Anchorages in One-​Way Slabs Reinforced with Deformed Wire Fabric,” ACI Journal, Proceedings, 67, August 1970, 636–​642. 6.46. M. A. Thompson, J. O. Jirsa, J. E. Breen, and D. F. Meinheit. “Behavior of Multiple Lap Splices in Wide Sections,” ACI Journal, Proceedings, 76, February 1979, 227–​248. 6.47. Telvin Rezansoff, Jim A.  Zacaruk, and Rob Topping. “Tensile Lap Splices in Reinforced Concrete Beams under Inelastic Cyclic Loading,” ACI Structural Journal, 85, January–​February 1988, 46–​52. 6.48. Tel Rezansoff, Adeniyi Akanni, and Bruce Sparling. “Tensile Lap Splices under Static Loading:  A  Review of the Proposed ACI 318 Code Provisions,” ACI Structural Journal, 90, July–​August 1993, 374–​384. 6.49. T. D. Mylrea. “Bond and Anchorage,” ACI Journal Proceedings, 44, March 1948, 521–​552. 6.50. R. A. Treece and J. O. Jirsa. “Bond Strength of Epoxy-​Coated Reinforcing Bars,” ACI Materials Journal, 86, March–​April 1989, 167–​174. 6.51. D. W. Johnston and P. Zia. Bond Characteristics of Epoxy-​Coated Reinforcing Bars. Report No. FHWA/​NC/​82-​002, Department of Civil Engineering, North Carolina State University, Raleigh, August 1982. 6.52. R. G. Mathey and J. R. Clifton. “Bond of Coated Reinforcing Bars in Concrete,” Journal of the Structural Division, ASCE, 102, January 1976, 215–​228. 6.53. ACI Committee 408. “Bond and Development of Straight Reinforcing Bars in Tension,” ACI 408R-​03. American Concrete Institute, Farmington Hills, MI, 2003, 49 pp.

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PROBLEMS All problems are to be done in accordance with the ACI Code, and all loads given are service loads, unless otherwise indicated. Assume that all beams have at least minimum cover satisfying ACI-​20.6.1.3.1, as well as minimum stirrups according to ACI-​9.6.3.3 and ACI-​9.7.6.2.2, unless otherwise indicated. All design strength φ M n diagrams used in these problems must be drawn to scale directly below a side view of the beam drawn to the same longitudinal scale on the same page. Design strength φ M n diagrams must have critical numerical values stated on the diagram, and horizontal distances to the critical points must be dimensioned with numerical values. Use the load factors and load combinations per ACI-​5.3 and the φ factors of ACI-​21.2.1. 6.1 Draw the free-​body diagram for the 3-​in. slice shown cross-​hatched on the beam given in the figure for Problem 6.1. Assuming linear elastic behavior (see Sections 12.2–​12.5), compute and show on the diagram values for the internal forces (tension, compression, and shear) on each side of the slice. (a) Compute the average flexural bond stress on the bars over the 3-​in. slice. (b) Compute the average flexural bond stress on the bars over the distance from section A to the left end of the beam. (c) What is the average anchorage bond stress resisting the tensile force at section A? (d) Explain what happens if the flexural bond stress in part (a) is so high that slippage occurs over the 3-​in. slice. What determines the adequacy of the beam? Assume that loading is within the usual service range. 6.2 For the simply supported beam shown in the figure for Problem 6.2, draw the design strength A

B

(φ M n ) diagram. Assume that the cutoff location 3 ft from the support satisfies the requirements of ACI-​9.7.3.5 and ACI-​9.7.3.8.3. What is the maximum uniformly distributed service load that the beam may be permitted to carry (assume 50% live load and 50% dead load)? Use fc′ = 3500 psi and f y = 40, 000 psi. 6.3 The beam of the figure for Problem 6.3 has 3–​#7 bars as the reinforcement at section A-​A with a 1–​#7 bar cut at 4 ft-​0 in. from the face of the support. Use fc′ = 4000 psi and f y = f yt = 60, 000 psi to investigate the adequacy for development of reinforcement if the beam is subjected to a uniform live load of 2.02 kips/​ft and uniform dead load of 1.86 kips/​ft (including beam weight). Compare factored moment Mu to the design strength φ M n by superimposing both diagrams. Consider all factors involved, including ACI-​ 9.7.3.5 for cutting bars in the tension zone. Assume that #3 U stirrups at 8-​in. spacing are used in the vicinity of the cutoff point.

5 kips/ft (total dead load)

n=8 17.5”

3”

3 – #9 2’– 0”

3”

12” 10’– 0”

Problem 6.1  A

2 #8 1 #7

1’– 0”

2 #8 cut here

3’– 0”

A

14”

1 #7

21.5” 24”

4 #8 Section A–A

16’– 0” c–c of supports Symmetrical about midspan

Problem 6.2 

237



237

P roblems

A

3 bars

3 – #7

1 bar

20’ 17.5”

4’– 0” 2’– 0”

6’– 0”

12” Section A–A

A

Problem 6.3  2 – #8B 2 – #9C

Bars C Bar A

Bars B

1 – #6A 20”

3’– 0”

3’– 0”

3’– 0”

2’– 0”

8’– 0” #3 stirrup

14”

Problem 6.4  A 6’– 0”

3 – #9 × 30’– 0” 2 – #9 × 20’– 9”

12”

A

B

1–#9 × 11’– 0” 2–#8

4’– 0”

4’– 3” 2 – #9 × 10’– 3”

20’– 0” A

4’– 0”

25 21 ”

12” 10’– 0”

14” Section A–A

Problem 6.5 

6.4 If the beam in the figure for Problem 6.4 is to carry uniformly distributed loads of 4.25 kips/​ft live load and 2.21 kips/​ft dead load (including beam weight), determine the adequacy of the bar cutoffs and the development of reinforcement. Stirrups are #3 at 6-​in. spacing where bars A are cut, and #3 at 8-​in. spacing where bars B are cut. As part of the solution, compare the factored moment Mu with the design strength φ M n by superimposing the diagrams. Use fc′ = 5000 psi and f y = f yt = 60, 000 psi. 6.5 For the beam shown in the figure for Problem 6.5, use fc′ = 3000 psi and f y = f yt = 40, 000 psi. (a)  Neglecting any compression reinforcement effect, plot the design strength φ M n diagram (positive moment over the 20-​ft span and negative moment over the support). (b) Check the development of reinforcement at the simply supported end and at the point of inflection closest to the right support on

the 20-​ft span. The loads are 1.63 kips/​ft dead load (including beam weight) and 2.98 kips/​ft live load. 6.6 For the beam of Problem 6.5, determine the acceptability of the cut locations at points A and B near the right support. (a) Assume that #3 U stirrups are spaced at 10 in. in the vicinity of the cut locations. (b) Assume that #3 U stirrups are spaced at 12 in. in the vicinity of the cut locations. 6.7 A  beam 14 in. wide by 24 in. deep (effective depth = 21.5 in.) is used as the section for a 20-​ft simply supported span having an 8-​ft cantilever at one end. The positive moment reinforcement is 4–​#9 bars, and the negative moment reinforcement is 4–​#8 bars. The loading to be carried is a live load of 2.01 kips/​ft and a dead load of 1.86 kips/​ft (including beam weight). Determine the lengths of bars (3-​in. increments) if two of the four bars in both the positive and the negative moment regions

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C hapter   6     D evelopment of R einforcement



are to be terminated as soon as practicable. The remaining bars are to be extended as required by the ACI Code. Width of supports is 15 in., and #3 U stirrups spaced at 10 in. are used in any potential cutoff region. Verify your design by showing for one set of axes the factored moment Mu envelope and the provided design strength φ M n diagram. Use fc′ = 3500 psi and f y = f yt = 40, 000 psi. 6.8 In the figure for Problem 6.8, a cantilever slab (e.g., for a retaining wall) varies in thickness from 24 in. at its supported end to 12 in. at its free end, with 2 1 2 -​in. clear cover over the reinforcement. Assuming that #7 bars at 9-​in. spacing are effective at the supported end, at what distance from the supported end may every other #7 bar be cut off? No shear reinforcement is used. Verify your result by showing the provided design strength φ M n diagram superimposed on the factored moment Mu diagram. Use fc′ = 4000 psi and f y = 60, 000 psi. Assume that the entire loading is earth pressure and neglect the beam weight.

Problems Using Concepts of Chapters 1 Through 6 For all designs, use at least one bar cutoff, even if such cutoff seems impractical. Use 0.6 ρtc as a reinforcement guideline where ρtc is the reinforcement ratio corresponding to εt = 0.005 (tension-control limit). 6.9 Design, including a design sketch, a rectangular cantilever beam 14 ft long to carry a live load of 2.7 kips/​ft. The beam size may not exceed 15 in. wide and 24 in. deep. Use fc′ = 4000 psi  and f y = f yt = 60, 000 psi. 6.10 The floor system shown in the figure for Problem 6.10 consists of 6-​in. precast slab sections of 10-​ft span. The live load is 66 psf, and the maximum depth available is 30 in. from the top of the floor slab. Completely design, as a simply supported beam, the beam indicated on the figure. Use whole-​inch increments for beam depth and width. Use fc′ = 4000 psi and f y = f yt = 60, 000 psi.

0.7 k/ft 0.3 k/ft

12”

24” 13’– 0”

Problem 6.8 

3 @ 10’– 0” = 30’– 0”

Live load = 50 psf

6” precast slab Beam to be designed

10’– 0” Typical section through 30’– 0” span beams 30’– 0”

Problem 6.10 

CHAPTER 7 ANALYSIS OF CONTINUOUS BEAMS AND ONE-​WAY SLABS

7.1 INTRODUCTION Reinforced concrete building systems commonly consist of floor slabs, beams, girders, and columns cast to form a continuous, monolithic structure. Consider the plan of a typical slab-​beam-​girder floor construction. In Fig. 7.1.1, section A-​A through the slab shows that the slab is supported on 10 beams. Intermediate beams such as B1, B2, and B3 are supported on the girders, whereas beams on the column lines such as B4, B5, and B6 are built integrally with and supported directly by the columns. Girders such as G1, G2, and G3 are also supported directly by the columns.

Rigid-​frame bridge piers. (Photo by C. G. Salmon).

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G6

G2

G3

B6

B3

B3

G4

G1

A B5

B2

B2

A

B4

B1

B1

G1

G4

G5

G6

Section A–A

B1 G4

B2 G1

B3 G1

G4

Beams monolithic with girders

B4

B5

B6

Beams monolithic with columns

G1

G2

G3

Girders monolithic with columns

Figure 7.1.1  Slab-​beam-​girder floor system.

7.2 ANALYSIS METHODS UNDER GRAVITY LOADS An elastic analysis of the structural system under gravity loads shown in Fig. 7.1.1, which may consist of, say, 10 or 15 stories or more, may be done by building a three-​dimensional model of the entire structure with the aid of standard structural analysis software. Alternatively, experience has shown that under gravity loads, analysis of the floor system may be simplified without great loss of accuracy by treating each floor level as a subassembly consisting of the members in the level being considered (slab, beams, and girders) and the columns above and below that level. Furthermore, for the purpose of gravity load analyses, the far ends of adjacent columns may be assumed to be fixed (ACI-​6.3.1.2), as shown in Fig. 7.2.1. A further simplification can be made for regular building frames, by subdividing the structure into two-​dimensional frames in the longitudinal and transverse direction of the building (e.g., along B4, B5, and B6, and along G1, G2, and G3 in Fig. 7.1.1), and analyzing each frame under its corresponding tributary load. Intermediate beams, such as B1, B2, and B3, may be modeled as a continuous beam supported on girders G1 and G4. In this case, it may be necessary to account for the torsional rigidity of the supporting girders depending on the magnitude of the flexural stiffness of the beam relative to the torsional stiffness of the girders. This is discussed later in Chapter 9 (Section 9.3). Regardless of the chosen analysis approach, it cannot be emphasized enough that the structural engineer must fully understand the capabilities and limitations of the software, as well as the effect of the modeling assumptions. In addition, the engineer should be competent in the use of traditional methods of analysis to be able to correctly interpret and

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241

Figure 7.2.1  Approximate deflected shape of a slab-​girder-​beam floor subassembly under a uniformly distributed load. Far ends of columns (above and below the floor) are assumed to be fixed.

corroborate the computer output. Paramount to obtaining a realistic distribution of member actions is the appropriate selection of the relative stiffness of the members; in particular, the correct inclusion of the effects of cracking on member stiffness. Guidance for the appropriate computation of member stiffness is provided later (see Sections 16.20 and 16.21).

7.3 ARRANGEMENT OF LIVE LOAD FOR MOMENT ENVELOPE The live load positions that cause the largest bending moments in slabs, beams, and girders are discussed in this section. To be able to establish the loading conditions appropriate for obtaining the bending moment and shear envelopes, the reader should review the subject of influence lines. Bending moments to be used in the design of columns are treated separately later. Consider the continuous beam ABCDEFGH with its adjacent columns shown in Fig. 7.3.1(a). The influence lines for bending moment at any point in the central portion of span CD and at a section an infinitesimal distance to the left of support D are shown in Fig. 7.3.1(b) and 7.3.1(c), respectively. From these influence lines the following cases for uniform live load are indicated: 1. For maximum positive moment within a span, load that span and all other alternate spans. 2. For maximum negative moment within a span, load the two spans adjacent to that span and all other alternate spans (all the spans not loaded in 1). 3. For maximum negative moment at a support, load the two spans adjacent to that support and all other alternate spans. 4. For maximum positive moment at a support, load the two spans beyond each of the two spans adjacent to that support and all other alternate spans (all the spans not loaded in 3). Note that loading cases 1 and 2 in the preceding list are complementary; that is, their combination results in all spans being loaded. Loading cases 3 and 4 are also complementary. It may be noted further that loading cases 1 and 3 are primary; that is, they result in moments of the same sign as those due to dead load only. Loading cases 2 and 4, on the other hand, are secondary; they result in moments opposite in sign to those due to dead load. When the secondary live load moment is numerically larger than the dead load moment, there is moment reversal.

24

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A

B

C

D

E

F

G

H

(a)

+

+ –

–

+

+ –

(b) Influence line for moment at section in central portion of span CD

–

+

+ –

–

+ –

(c) Influence line for moment at section infinitesimally to left of support D

–

+

+ –

–

+

–

(d) Influence line for moment at section between the fixed point and the support

+

–

+

–

(e) Influence line for moment at the fixed point near right end of span

Figure 7.3.1  Influence lines for continuous spans.

The loading cases mentioned thus far involve loading entire spans; that is, no partial loading of a span. These full span loading cases correctly give maximum and minimum bending moment in the midspan region (roughly the middle 50–​60%) of a span as well as at the support. The correct maximum (or minimum) values are not obtained, however, for approximately 20 to 25% of the span nearest the supports. A qualitative examination of Fig. 7.3.1(b) and 7.3.1(c) shows that if an influence line were drawn for a series of specific points along span CD between the midspan and the support, the peak ordinate in the positive portion of such influence line would get smaller and smaller. Also, the slope of the influence line at point C goes upward to the right for the midspan influence line but downward to the right for the support D influence line. As successive influence lines are drawn for the points along the span from midspan to the support, there must be some location for which the slope of the influence line at C is horizontal. Such point is called a fixed point, giving an influence line illustrated by Fig. 7.3.1(e). Any loading on spans to the left of the span under study will cause no moment at this fixed point. The fixed point is the closest location to the support for which full span loadings give the correct maximum or minimum bending moments. The influence line for a section between a fixed point and support is as shown by Fig. 7.3.1(d), indicating partial loading for the span in question to obtain maximum or minimum bending moment. For practical reasons, partial span loading for maximum or minimum bending moments is rare in the design of building frames—the effect on the design is too small to justify the effort. For the design of long spans, such as in highway bridges, partial span loading, as indicated by influence lines, for locations between the fixed point and the support would usually be considered. The application to a six-​span continuous beam with upper and lower columns is illustrated in the following example.

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EXAMPLE 7.3.1 Determine the maximum and minimum moments at the middle and the ends of each span in a beam-​column frame with six equal spans as shown in Fig. 7.3.2. Assume that the uniform live load wL is twice the uniform dead load wD, and that the flexural stiffness (EI/​L) of the columns is twice the stiffness of the beams. Express all moments in terms of wL2 in which w = wD + wL and L is the span length. 2 A

1

2

2 B

1

2 C

2

2

1

2 D

Relative stiffness

2

1

2 E 2

1

2 F 2

1

2 G 2

Figure 7.3.2  Six-​span continuous frame of Example 7.3.1 with relative stiffnesses.

SOLUTION All moments will be expressed numerically in terms of wL2 × 10 – 4. Seven loading conditions, as shown in Fig. 7.3.3, need to be investigated. The frame can be analyzed by means of any classical method of structural analysis (e.g., the moment distribution method) or with the aid of standard structural analysis software. The moments at the ends of the beams in each span for each of the seven loading conditions shown in Fig. 7.3.3 are summarized in Table 7.3.1. In this table, the designer’s sign convention for bending moment, in which a positive moment causes compression on the top side of the beam, is used. The controlling values of moments at the left and right ends of each span, M L and M R , for the various critical conditions are taken from Table 7.3.1 and entered in Table 7.3.2. LC1

DL + LL

DL

Max positive moment in spans AB, CD, and EF

LC2 Max negative moment at B

LC3 Min negative moment at B LC4 Max negative moment at C

LC5 Min negative moment at C

LC6 Max negative moment at D

LC7 Min negative moment at D

Figure 7.3.3  Loading conditions for six-​span continuous frame.

(Continued)

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Example 7.3.1 (Continued) TABLE 7.3.1  BEAM END MOMENTS FOR EXAMPLE 7.3.1 Joint

A

Member

LC1 LC2 LC3 LC4 LC5 LC6 LC7

B

C

AB

D

BC

E

CD

F

DE

G

EF

FG

ML

MR

ML

MR

ML

MR

ML

MR

ML

MR

ML

MR

–​712 –​668 –​227 –​187 –​709 –​712 –​184

–​804 –​911 –​293 –​390 –​813 –​805 –​398

–​340 –​888 –​241 –​745 –​385 –​343 –​787

–​331 –​729 –​374 –​878 –​225 –​323 –​780

–​777 –​325 –​785 –​882 –​228 –​731 –​378

–​777 –​337 –​774 –​732 –​379 –​883 –​228

–​334 –​778 –​333 –​325 –​786 –​883 –​228

–​332 –​777 –​332 –​336 –​774 –​731 –​378

–​772 –​331 –​772 –​772 –​330 –​323 –​780

–​790 –​340 –​790 –​790 –​340 –​343 –​787

–​399 –​804 –​399 –​399 –​804 –​805 –​398

–​184 –​712 –​184 –​184 –​712 –​712 –​184

Note: Values of moments in wL2 × 10–​4 using designer’s sign convention.

The moment at the midspan M s may be determined by superposition of the effect of end moments with that of the simply supported beam moment due to transverse loading,

1 M s = M 0 − ( M L + M R ) 2

where M 0 is the moment at the midspan for a simply supported beam. In the above equation, M 0, ML, and M R are taken as positive values. When the beam end moments are not equal, the maximum moment in the span does not occur at midspan, but its value is close to that at midspan. Because of symmetry, the controlling values obtained for a given span would also occur in the symmetric span when the frame is loaded with the mirror image of the load condition being considered. For example, the controlling values obtained for maximum positive moment in spans AB, CD, and EF (lines 1, 2, and 3 in Table 7.3.2), would also occur in spans FG, DE, and BC, respectively, if the frame were to be loaded with the mirror image of load condition LC1. Similarly, the controlling values obtained for minimum positive or maximum negative moment at midspan in spans BC, DE, and FG (lines 4, 5, and 6 in Table 7.3.2), would also occur in spans EF, CD, and AB, respectively. For the purpose of illustration, the moment diagram for the maximum and minimum positive moments at midspan, using the results of the first loading condition, LC1, in Fig. 7.3.3, and its mirror image is shown in Fig. 7.3.4. First, the simple beam moment diagrams for the total load and for dead load only are drawn to scale in Fig. 7.3.4(a). Next, the end moments for each span are taken from Table 7.3.2 and plotted in Fig. 7.3.4(b). Because of symmetry, only the moment diagrams for the first three spans are shown. The final moment diagrams in Fig. 7.3.4(b) are drawn by rotating the baselines for zero end moments in Fig. 7.3.4(a) to those connecting the end moments in Fig. 7.3.4(b). In this example, the dead load has been applied on all the spans in each of the seven loading conditions of Fig. 7.3.3. This has been done because an important purpose of the example is to justify the use of approximate moment coefficients as discussed in the next section. An alternative approach would be to use eight loading conditions: one condition for dead load only plus the same seven live load conditions but without the dead load applied simultaneously. Separation of the effects of dead and live load is desirable, because if a loading combination that includes wind-​or earthquake-​induced loads must be considered, different live load factors must be used according to ACI-​5.3.1 (see Section 2.7). (Continued)

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245

Example 7.3.1 (Continued) TABLE 7.3.2  SUMMARY OF RESULTS—CONTROLLING VALUES OF MOMENTS AT BEAM ENDS FROM TABLE 7.3.1 Line Number

Span

ML

MR

Value of ML and 1 MR from Load M s = M 0 − 2 � ( M L + M R ) Condition No.

1 2 3

For maximum positive moment at midspan

AB CD EF

–​712 –​777 –​772

–​804 –​777 –​790

LC1 LC1 LC1

+492 +473 [M0 = +1250] +469

4 5 6

For minimum positive or maximum negative moment at midspan

BC DE FG

–​340 –​334 –​399

–​331 –​332 –​184

LC1 LC1 LC1

+81 +84   [M0 = +417] +125

7 8 9 10 11 12

For maximum negative moment

A of AB B of AB B of BC C of BC C of CD D of CD

–​712 (–​668) –​888 (–​745) –​882 (–​731)

(–​804) –​911 (–​729) –​878 (–​732) –​883

LC1 LC2 LC2 LC4 LC4 LC6

13 14 15 16 17 18

For minimum negative or maximum positive moment

A of AB B of AB B of BC C of BC C of CD D of CD

–​184

LC7 LC3 LC3 LC5 LC5 LC7

–​293 –​241 –​225 –​228 –​228

Note: Values of moments are shown in wL2 × 10–​4 using designer’s sign convention. Numbers in parentheses are to be used in shear envelope computations (see Section 7.6).

1250 wL2 (10–4)

417 wL2 (10–4)

w=

wD + wL

1250 wL2 (10–4)

wD

=

1

1250 wL2 (10–4)

417 wL2 (10–4)

417 wL2 (10–4)

w

3

(a) Simple beam moment diagrams 469

492 A

473

B

C

125

184

399 712

D

81

804

340 790

84

331

334

332

772 777

777

(b) Moment diagrams for maximum and minimum positive midspan moments

Figure 7.3.4  Moment diagrams for the first three spans under LC1 of Example 7.3.1.

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7.4 ACI CODE—​A RRANGEMENT OF LIVE LOAD AND MOMENT COEFFICIENTS For the design of floor or roof systems under gravity loads, ACI-​6.4.1 and ACI-​6.3.1.2 permit the assumption that the live load is applied only to the level under consideration, with its upper and lower columns fixed at their far ends, as discussed earlier in Section 7.2. Furthermore, for one-​way slabs and beams, ACI-​6.4.2 allows computation of the (a) maximum positive moment near midspan by positioning the live load on the span under consideration and on alternate spans (LC1 in Fig. 7.3.3) and (b) maximum negative moment at a support by positioning the live load on adjacent spans only (LC2, LC4, and LC6 in Fig. 7.3.3). Alternatively, ACI-​6.5 allows an approximate method for calculating the required factored moments (and shears—see Section 7.6) for nonprestressed continuous beams and one-​way slabs that satisfy all of the following (ACI-​6.5.1): (a) Members are prismatic. (b) Loads are uniformly distributed. (c) The live load does not exceed three times the dead load. (d) There are at least two spans. (e) The longer of two adjacent spans does not exceed the shorter by more than 20%. Table 7.3.2 shows that for the six-​span frame of Fig. 7.3.2, which would meet the requirements of ACI-​6.5.1, critical values of moments may vary within the following limits: Exterior span: Exterior end Midspan Interior end First interior span: Exterior end Midspan Interior end Second interior span: Exterior end Midspan Interior end

–​0.0184wL2 and –​0.0712wL2 +0.0125wL2 and +0.0492wL2 –​0.0293wL2 and –​0.0911wL2 –​0.0241wL2 and –​0.0888wL2 +0.0081wL2 and +0.0469wL2 –​0.0225wL2 and –​0.0878wL2 –​0.0228wL2 and –​0.0882wL2 +0.0084wL2 and +0.0473wL2 –​0.0228wL2 and –​0.0883wL2

Similar values may be worked out for other values of the relative stiffness between the columns and the beams and of the wL / wD ratio. It may be observed that the maximum posi­ tive moments in the first and second interior spans are similar, that the maximum positive moment in the exterior span is higher than that in the interior spans, that the maximum negative moment at the interior end of the exterior span has the largest numerical value, and that the maximum negative moments at both ends of all interior spans are about equal. Based on these and similar analyses, ACI-​6.5.2 provides approximate moment coefficients for nonprestressed continuous beams and one-​way slabs, as shown in Table 7.4.1. A  comparison of these coefficients with the largest possible theoretical values [7.1] is shown in Table 7.4.1. Certainly, the largest possible theoretical values will be for the case of wL / wD = 3, which is the limit set forth in ACI-​6.5.1. In this instance, secondary live load moments with signs opposite to that of dead load occur infrequently; if they do occur, their values are small. Thus, as long as the ratio of live load to dead load does not exceed 3 and span lengths do not differ considerably, the ACI moment coefficients will be reasonably close to the theoretical values and, in general, conservative. Note that the ACI moment coefficients are given in terms of wL2n, where Ln is the clear span for positive moment and the average of the two adjacent clear spans for negative moment—​negative moments being those at the face of supports, not at the centerline of support. On the other hand, the theoretical coefficients are in terms of wL2 , in which L is

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TABLE 7.4.1  COMPARISON OF ACI MOMENT COEFFICIENTS WITH LARGEST THEORETICAL COEFFICIENTS [7.1] Largest Theoretical Coefficients

Location of Section Positive moment at End spans If discontinuous end is unrestrained If discontinuous end is integral with the support Interior spans: Negative moment at Exterior face of first interior support: Two spans More than two spans Other faces of interior supports: Face of all supports for: (a) slabs with spans not exceeding 10 ft, or (b) beams and girders where (ΣKcol)/​Kbeam exceeds 8 at each end of the spana

ACI  

Value

Number of Spans

(ΣKcol)/​Kbma

wL/​wD

+

1 11

+0.094

3

0

3

+

1 14

+0.073

3

0.5

3

+

1 16

+0.063

4 or more

0.5

3

–​0.111

2

0.5

3

–​0.107

4 or more

0.5

3

–​0.092

4 or more

2

3

–​0.083

any number

∞

any ratio

–​0.036 –​0.050

4 or more 4 or more

0.5 1

3 3

–​0.064

4 or more

2

3

 

1  9 1 − 10 1 − 11 −

−

1 12

Negative moment at Interior faces of exterior supports for members built integrally with their supports: Where the support is a spandrel beam or girder Where the support is a column: a

1 24 1 − 16

−

K = EI/L

the distance between centerlines of supports, and coefficients for negative moments refer to those at the centerlines of support. Although span lengths between centerlines of supports are always used in elastic analysis, the ACI Code states that for beams (ACI-​9.4.2.1) and one-​way slabs (ACI-​7.4.2.1) built integrally with supports, moments computed at faces of supports may be used for design.

7.5 ACI MOMENT DIAGRAMS In designing any span in a multispan continuous frame subjected to live load with the moment coefficients, two primary sets of shear and moment diagrams are inherently being assumed. In the general case, one will result from the loading position that causes

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maximum positive moment within the span, and the other will result from assuming that the maximum negative moments occur simultaneously at both ends. Actually, the loading position that causes maximum negative moment at one end is different from that which causes maximum negative moment at the other end. However, by assuming that both maximum negative end moments occur simultaneously, a critical curve having greater magnitude than either of the two actual curves is obtained. The ACI moment coefficients (ACI-​6.5.2) shown in Table 7.4.1 are the common values from the two primary conditions as described in the preceding paragraph. No secondary moment coefficients are suggested by the ACI Code, the reason being that as long as the design ratio of live to dead load is limited to 3, moment reversal will be unlikely to occur; that is, there can be only positive moment in the midspan region and only negative moment in the support region. Figures 7.5.1 through 7.5.4 show the two primary sets of shear and moment diagrams to be used in the design of continuous spans in accordance with the ACI moment coefficients. These diagrams are applicable to the actual clear span, which is also used to compute the positive moment. For negative moment, Ln is the average of adjacent clear spans (ACI-​6.5.2). The reader should utilize the fundamentals of shear and moment diagrams to verify the numerical ordinates on these diagrams. For instance, in case of maximum positive moment in Fig.  7.5.2(a), the distance x from the left support to the point of zero shear may be

0.0736 wu Ln2

wu

0.4264 wu Ln

0.5736 wu Ln

0.4000 wu Ln Ln

0.5736 Ln

+0.4264 wu Ln

0.4000 Ln

0.6000 Ln

+0.4000 wu Ln

–0.6000 wu Ln

–0.5736 wu Ln 1 + 11 wu Ln2 (given)

0.4264 Ln

(given)

0.6000 wu Ln

Ln 0.4264 Ln

1 w L 2 10 u n

wu

+0.0800 wu Ln2

0.4264 Ln

0.4000 Ln

0.4000 Ln

–0.0736 wu Ln2 1 – 10 wu Ln2 (given)

(a) Maximum in the positive zone

(b) Maximum in the negative zone

Figure 7.5.1  Exterior span with discontinuous end unrestrained.

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determined from the relationship that the change of moment between any two sections is equal to the area of the shear diagram between these two sections. Thus wu x 2  1 1 =  +  wu L2n  2 24 14 

from which

x = 0.4756 Ln



Also, the distance x between the section of maximum positive moment to the point of zero moment is wu x 2 1 = wu L2n 2 14

from which

x = 0.3780 Ln



Any time a designer uses moment coefficients for determining the factored moments, as permitted by the approximate method of ACI-​6.5, the moment diagrams that correspond to 1 w L 2 (given) 24 u n

0.0661 wu Ln2

wu

0.4756 wu Ln

0.5244 wu Ln

0.4756 Ln

Ln

0.5244 Ln

+0.4756 wu Ln

1 w L 2 (given) 24 u n

1 w L 2 (given) 10 u n

wu

0.4417 wu Ln

0.5583 wu Ln

0.4417 Ln

Ln

0.5583 Ln

+0.4417 wu Ln

–0.5244 wu Ln 1 + w L 2 (given) 14 u n

–0.5583 wu Ln +0.0558 wu Ln2

0.3780 Ln 0.3780 Ln 1 2 – w L (given) 24 u n

0.3341 Ln 0.3341 Ln 1 – w L 2 (given) 24 u n

–0.0661 wu Ln2 – (a) Maximum in the positive zone

1 w L 2 (given) 10 u n

(b) Maximum in the negative zone

Figure 7.5.2  Exterior span with exterior support built integrally with spandrel beam or girder.

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wu

0.5175 wu Ln

0.0450 wu Ln2

1 w L 2 (given) 16 u n

0.4825 wu Ln

0.4825 wu Ln

0.5375 wu Ln

Ln 0.5175 Ln

1 w L 2 (given) 10 u n

wu

Ln 0.4825 Ln

0.4625 Ln

+0.5175 wu Ln

0.5375 Ln

+0.4625 wu Ln

–0.5375 wu Ln

–0.4825 wu Ln +0.0445 wu Ln2

+ 1 wu Ln2 (given) 14

0.2983 Ln 0.2983 Ln 0.3780 Ln

0.3780 Ln

– 1 wu Ln2 (given) 16

–0.0450 wu Ln2 1 2 – w L (given) 16 u n (a) Maximum in the positive zone

– 1 wu Ln2 (given) 10 (b) Maximum in the negative zone

Figure 7.5.3  Exterior span with exterior support built integrally with column.

such coefficients should be used when establishing bar bend or cutoff locations. The use of moment coefficients implies a statically compatible moment diagram.

7.6 SHEAR ENVELOPE FOR DESIGN Inasmuch as the design of shear reinforcement depends on the variation of shear forces along the span, it is necessary to compute the maximum factored shear force due to the combination of dead and live loads at all sections. When the live load consists of important concentrated loads, such as in the design of highway bridge spans, accurate calculations of such maximum shears at all sections must be performed. The proper position of live load for maximum shear at a section can be determined by examining the influence line for shear at that section along the span. For instance, the influence lines for end shear at C of span CD and for the shear at midspan of CD in the frame of Fig. 7.6.1(a) are shown in Fig. 7.6.1(b) and 7.6.1(c), respectively. Figure 7.6.1(b) shows that the position of uniform live load for maximum end shear at C is identical to that for maximum negative moment at C [see Fig. 7.3.3 (LC4)]. Likewise, the position of uniform live load for maximum end shear at D is identical with that for maximum negative moment at D. From Fig. 7.6.1(c), it is apparent that partial span loading of uniform live load is indicated to give maximum shear at any point within the span. The partial span loading to obtain the correct maximum shear at locations within the span has already been used in the shear strength design examples of Section 5.12 for those statically determinate situations.

251



251

7.6  SHEAR ENVELOPE FOR DESIGN

0.0625 wu Ln2

0.0625 wu Ln2

wu

0.5000 wu Ln

0.5000 wu Ln

1 w L 2 (given) 11 u n

0.5000 wu Ln

0.5000 wu Ln

Ln 0.5000 Ln

1 w L 2 (given) 11 u n

wu

Ln 0.5000 Ln

0.5000 Ln

+0.5000 wu Ln

0.5000 Ln

+0.5000 wu Ln

–0.5000 wu Ln +

–0.5000 wu Ln

1 wu Ln2 (given) 16 +0.0341 wu Ln2

0.3535 Ln 0.3535 Ln –0.0625 wu Ln2

0.2611 Ln 0.2611 Ln

–0.0625 wu Ln2 – 1 wu Ln2 (given) 11

(a) Maximum in the positive zone

– 1 wu Ln2 (given) 11

(b) Maximum in the negative zone

Figure 7.5.4  Interior span.

A

B

D

C

E

G

F

(a)

+ –

+

–

+

–

(b) Influence line for maximum end shear at C of span CD

–

+

+ –

–

+

(c) Influence line for maximum shear at midspan of span CD

Figure 7.6.1  Influence lines for shear in a continuous frame.

–

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C H A P T E R   7     A N A L Y S I S O F C O N T I N U O U S B E A M S A N D O N E - W A Y   S L A B S

In buildings of usual types of construction, spans, and story heights, wherein the idealized rigid frame, such as shown in Fig. 7.6.1(a), is taken into consideration, the use of partial span loading of uniform live load is commonly ignored, although theoretically it is necessary for the computation of maximum shear at any section within the span. When partial span loading is not considered necessary, the maximum shears at the ends can be used alone to establish an approximate shear envelope. Note that the loading condition for maximum shear at one end is different from that for maximum shear at the other end. These two critical shear diagrams for each span, when a continuity analysis is performed, may be easily obtained by using the values of the negative end moments (including those in parenthesis) contained in lines 7 through 12 of Table 7.3.2. When the ACI moment coefficients are used, it is generally assumed that the shear diagrams accompanying the critical moment diagrams as shown in Figs. 7.5.1 through 7.5.4 may be used in the design. Alternatively, for beams and one-way slabs satisfying ACI6.5.1 [see (a) through (e) in section 7.4], ACI Table 6.5.4 allows the use of a shear force of 1.15wu Ln /2 at the exterior face of the first interior support. A shear force of wuLn /2 is permitted at the face of all other supports.

SELECTED REFERENCE 7.1 A.  J. Boase and J.  T. Howell. “Design Coefficients for Building Frames,” ACI Journal, Proceedings, 36, September 1939, 21–​36.

PROBLEMS 7.1 Compute and draw to scale the bending moment and shear diagrams for the loading conditions 1 through 7 of Fig.  7.3.3; that is, verify the results given in Tables  7.3.1 by using any structural analysis method assigned by the instructor. 7.2 Compute and draw to scale the envelope of bending moments (diagram showing range over which bending moment may vary) due to factored loads for the beams of the frame of the figure for Problems 7.2 and 7.3 using any structural analysis method assigned by the instructor. The uniform factored dead load is 1 kip/​ft, and the uniform factored live load is 2 kips/​ft.

7.3 For the beams of the frame in the figure, compute and draw the bending moment envelope using the coefficients of ACI-​6.5.2. (If Problem 7.2 has also been solved, compare the moments by giving the percentage difference in the maximum values obtained by coefficients as compared with the more exact values of a structural analysis.) 7.4 Consider an equal-​span, uniform-​section continuous beam over many supports. Compute and show diagrams for dead load coefficients of wL2 for moments at critical locations in the exterior and first interior spans. Could the coefficients for the first interior span be applied appropriately to the other interior spans? Recommend dead load coefficients for equal spans. Symmetrical about CL

Relative stiffness, I/L 2

2 5

A

2

2

16’– 0”

Problems 7.2 and 7.3 

2

2 4

B

5

C

2

18’– 0”

2 6

D

2

16’– 0”

E

2

14’– 0”

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PROBLEMS

7.5 Repeat Problem 7.4 for the case of alternate spans that are 20% longer than the others (1.2L), taking the exterior span as a short one (L). 7.6 For an equal-​span, uniform-​section continuous beam over many supports, compute and show diagrams for live load coefficients in terms of wL2 for maximum negative moments at (a) the first interior support (b) the second interior support (c) a typical interior support Recommend live load coefficients.

7.7 For an equal-​span, uniform-​section continuous beam over many supports, compute and show diagrams for live load coefficients in terms of wL2 for the maximum positive moment in (a) the exterior span (b) the first interior span (c) a typical interior span Recommend live load coefficients.

CHAPTER 8 DESIGN OF ONE-​WAY SLABS

8.1 DEFINITION One of the most common types of floor construction is the slab-​beam-​girder system, as briefly described in Section 7.1. The slab panel, bounded on its two long sides by the beams and on its two short sides by the girders, is usually at least twice as long as it is wide. In such a condition, the dead and live load acting on the slab area may be considered as being entirely supported in the short direction by the beams (see Section A-​A in Fig. 7.1.1), hence the term “one-​way slab.” Two-​way floor systems, with or without beams on column lines, are treated in Chapter 16, and ribbed-​joist floor construction is described in Sections 9.10 and 9.11. The determination of an optimum floor framing plan—​that is, the spacing of columns, beams, and girders—​depends on both the functional and the structural requirements. In most cases preliminary calculations are necessary for several different layouts and, after comparison, the most suitable and economical plan is chosen.

8.2 ANALYSIS METHODS Because the loading on a one-​way slab is nearly all transferred in the short direction, such a slab continuous over several supports may be treated as a beam. Because sufficiently accurate results are obtained, ACI-​6.5.1 permits the use of moment and shear coefficients in the case of two or more approximately equal spans (the larger of two adjacent spans not exceeding the shorter by more than 20%) with loads uniformly distributed, where the unit live load does not exceed three times the unit dead load. These coefficients are in terms of clear span Ln and the values given are for critical locations, that is, faces of support for shears and negative moments and midspan regions for positive moments. When the conditions of ACI-​6.5.1 are not satisfied, a structural analysis, typically a first-​ order elastic analysis, is required (see Section 7.2). Various live load arrangements according to ACI-​6.4 should be investigated in order to maximize design negative and positive moments and shear forces (see Section 7.3). ACI-​6.4.2 permits the determination of maximum negative moments at supports by applying the live load on adjacent spans, while maximum positive moment near midspan is determined by applying live load on alternate spans. For one-​way slabs built integrally with the supports and with clear spans not exceeding 10 ft, ACI-​6.6.2.3 permits the use of a model consisting of a continuous flexural member on knife-​edge supports with span lengths equal to the clear span lengths of the slab. As such a model would lead to zero moment at the end supports, the application of a moment equal to wu Ln 2 / 24 is recommended, where wu is the factored distributed load and Ln is the clear span length. For clear span lengths greater than 10 ft, no explicit guidelines for the use of a

25



8 . 3   S lab   D esign

255

Escala Apartments, Seattle, WA (photo courtesy of Cary Kopczynski & Associates).

simplified model are provided in the ACI Code. The writers believe that in lieu of a three-​ dimensional analysis of the entire floor system, a simplified model consisting of a continuous beam on knife-​edge supports can still be used. Rather than using the clear span length, however, a span length equal to the distance between centerlines of supporting beams is recommended. Such a model has been found to be adequate for determination of negative moments in beams supported by girders [8.1]. Rigid zones may be used to model the intersection between the slab and supporting beams, which will facilitate the determination of slab design moment and shear at the faces of the supporting beams. As in the simplified model allowed in ACI-​6.6.2.3, the application of a moment equal to wu L2n / 24 at exterior supports is recommended.

8.3 SLAB DESIGN In designing a one-​way slab, a typical imaginary strip 12 in. wide is usually considered. The continuous slab may then be designed as a continuous beam having a known width of 12 in.; the slab thickness, however, is unknown. The thickness of the slab depends on the deflection, bending, and shear requirements. Deflection requirements are imposed to prevent excessive deformations that might adversely affect the serviceability of the structure. According to ACI Table 7.3.1.1, one-​way slabs not supporting or not attached to partitions or construction likely to be damaged by large deflections must have at least a minimum slab thickness (for Grade 60 steel) of L  /​20, L  /​24, L  /​28, or L /​10 depending on whether L is the length of a simply supported, a one-​ end-​continuous, a both-​ends-​continuous, or a cantilever span. Thinner slabs are permitted, however, provided deflections are calculated and determined to be less than the limits in ACI Table 24.2.2. If the slab supports or is attached to construction likely to be damaged by large deflections, deflections must also be computed and shown to satisfy the limits of ACI Table 24.2.2. An extended treatment of deflections can be found in Chapter 12.

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TABLE 8.3.1  AVERAGE AREA PER FOOT OF WIDTH PROVIDED BY VARIOUS BAR SPACINGS Bar Size Nominal Number Diameter (in.)

3 4 5 6 7 8

0.375 0.500 0.625 0.750 0.875 1.000

Spacing of Bars in Inches

4

5

6

7

8

9

10

12

14

16

18

0.33 0.59 0.92 1.33 1.80 2.36

0.27 0.47 0.74 1.06 1.44 1.88

0.22 0.39 0.61 0.88 1.20 1.57

0.19 0.34 0.53 0.76 1.03 1.35

0.17 0.29 0.46 0.66 0.90 1.18

0.15 0.26 0.41 0.59 0.80 1.05

0.13 0.24 0.37 0.53 0.72 0.94

0.11 0.20 0.31 0.44 0.60 0.79

0.095 0.17 0.26 0.38 0.52 0.67

0.83 0.15 0.23 0.33 0.45 0.59

0.074 0.13 0.20 0.29 0.40 0.52

In most cases slab thickness is controlled by deflection requirements rather than flexural strength requirements. Minimum thickness required to resist the factored moment Mu , however, can be checked in accordance with the principles of Chapter 3. In this case, the desired coefficient of resistance Rn can be computed for the required effective depth (and slab thickness) determined using M n = Mu / φ = Rn bd 2 . In equal continuous spans, the negative moment at the exterior face of the first interior support is the largest; therefore, this negative moment is used to determine minimum slab thickness based on bending requirements. Table 8.3.1 lists reinforcement areas per foot of width for various bar sizes and spacings, which can be used once the required area of flexural reinforcement per 12-​in.-​wide strip has been determined. The shear requirement does not usually control, but it must be checked. Because of practical space limitations, shear reinforcement is not used in typical slabs; thus the governing factored shear Vu , which in equal continuous spans occurs at the exterior face (at a distance d therefrom) of the first interior support, must be kept below φ Vc of ACI-​22.5.5.1. According to ACI Table 20.6.1.3.1, the concrete protective covering for reinforcement (#11 and smaller) in slabs shall be not less than 3 4 in. at surfaces not exposed directly to the ground or to the weather. Also, when the top of a monolithic slab is the wearing surface and when unusual wear is expected as in buildings of the warehouse or industrial class, it has been customary to use an additional depth of 1 2 in. of concrete protective covering over that required by the design of the member. The ACI Code does not require extra slab thickness for wearing surface, but ACI-​7.3.1.2 permits, at the discretion of the designer, a monolithic floor finish to be considered as part of the structural member. For nonstructural purposes, any concrete floor finish may be considered as part of the required cover (ACI-​20.6.1.2).

EXAMPLE 8.3.1 Establish the thickness of the floor slab shown in the second-​floor framing plan of Fig. 8.3.1 (see Fig. 8.4.2 for a section through slabs 2S1-​2S2-​2S3) for a superimposed dead load of 20 psf and a service live load of 100 psf. Use fc′ = 4000 psi, f y = 60,000 psi, and the ACI Code. Assume an exterior staircase so that no openings are to be made in the slabs. (Continued)

257



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8 . 3   S lab   D esign

Example 8.3.1 (Continued) SOLUTION Because live load does not appear to exceed three times the dead load, the design will be done in accordance with the ACI moment and shear coefficients (see Sections 7.4, 7.5 and 7.6). (a) Minimum thickness based on deflection control requirements. For spans with one end and both ends continuous, the respective minimum thicknesses h from ACI Table 7.3.1.1 are L /​24 and L  /​28. L 13(12) = = 6.5 in. 24 24

min h =

L 13(12) min h = = = 5.6 in. 28 28

(for 2S1) (for 2S 2 and 2S 3)

Assume a 6 1 2 -​in.-​thick slab.1 The weight of the slab is (6.5/​12)(0.15) = 0.081 kips/​sq ​ft.

2S3

2S3

2S3

2G1

2G2

2G2

2S2

2S1

2G1

2S2

2S3

2S1

2S2

2S3

2B2

A 2S1

2S3

2S2

2S1

2B1

A

2S3

2S3

2S2

2S1

3 @ 26’ = 78’

2S2

2B1

2S1

4 @ 39’ = 156’

Figure 8.3.1  Second-​floor framing plan (for Section A-​A, see Fig. 8.4.2).

(b) Flexural strength requirements. Assume width of supporting beams to be 14 in. w = (0.081 + 0.02) = 0.101kips/ft /ft of width D

wL = 0.1 kips/ft /ft of width wu = 1.2(0.101) + 1.6(0.1) = 0.281 kips/ft /ft of width clear span = 13 −

14 12

= 11.83 ft

From Table 7.4.1, the negative moment at the exterior face of the first interior support is Mu =

1 (0.281)(11.83)2 = 3.94 ft-kips /ft of width 10

Strictly speaking, the slab thickness must be checked against bending requirements. Unless unusually large loads are applied to the slab, however, the minimum thickness required for deflection control will be adequate. For a tension-controlled section,

Mn ≥

Mu 3.94 = = 4.38 ft-kips /ft of width φ 0.90 (Continued)

1  Note that a thinner slab may be permitted by ACI-7.3.1.1 if it can be shown that the deflection limits of ACI Table 24.2.2 are satisfied.

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Example 8.3.1 (Continued) Assuming #4 bars are used (db = 0.5 in.) and using a minimum cover of 3 4 in. (surface not exposed directly to the ground or the weather), the effective depth of the slab is d = h−



db − 0.75 in. = 5.5 in. 2

Required area of tension steel. Following Eq.(3.8.4b), Eq. (3.8.4a) and Eq. (3.8.5) Rn = m=



ρ=

Mu 4.38(12, 000) = = 145 psi 2 φ bd 12(5.5)2 fy 0.85 fc′

=

60, 000 = 17.65 0.85(4000)

2 mRn  As 1  2(17.65)(145)  1 1− 1− = 0.0025 = 1 − 1 − =  60, 000  bd m  f y  17.65  As = ρ bd = 0.0025(12)(5.5) = 0.16 sq in. /ft of width

(c) Shear strength requirement.

max Vu = 1.15

wu Ln (0.281)(11.83) = 1.15 = 1.92 kips /ft of width 2 2

The design shear strength φ Vc for a member without shear reinforcement is



(

)

φ Vc = φ 2 λ fc′ bd = 0.75(2(1.0) 4000 )(12)(5.5)

1 1000

= 6.26 kips / ft off width > 1.92 kips / ft of width



The slab is acceptable for a member without stirrups. Note that the shear force at a distance d from the face of support could have been used, in which case the factored shear would have been even less than 1.92 kips/​ft of width.

8.4 CHOICE OF REINFORCEMENT Flexural Reinforcement The choice of flexural reinforcement depends primarily on the steel area and secondarily on development length requirements. The steel areas required at the principal sections, namely, those at the middle and at the ends of each span, are first computed. The calculated areas of flexural reinforcement cannot be less than the minimum specified in ACI ​Table 7.6.1.1 as follows: For f y < 60, 000 psi, As ,min = 0.0020bh (8.4.1a) For

f y ≥ 60, 000 psi, As ,min =

0.0018(60, 000) bh ≥ 0.0014bh fy

(8.4.1b)

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8 . 4   C hoice of   R einforcement

259

Once a tentative choice of reinforcement has been made based on flexural strength requirements, development length requirements (Chapter  6) are examined (see Fig.  8.4.1 for a possible arrangement of reinforcement). Critical sections for development of reinforcement are those where maximum stress in the reinforcement occurs, or where bent or terminated reinforcement is no longer required to resist flexure (ACI-​7.7.3.2). Reinforcement should be extended at least the greater of d or 12 db beyond the point at which it is no longer required to resist flexure except at simple supports or free ends of cantilevers (ACI-​7.7.3.3). Also, continuing reinforcement must be extended at least Ld beyond the point at which the bent or terminated reinforcement is no longer required (ACI-​7.7.3.4). Further, tension reinforcement should not be terminated in a tensile zone. ACI-​7.7.3.5, however, allows tension reinforcement to be terminated in the tensile zone of a slab without shear reinforcement when Vu < ( 2 3 ) φ Vc , or when continuing reinforcement, consisting of #11 bars or smaller, is at least twice the required area of tension steel and Vu ≤ ( 3 4 ) φ Vc . At simple supports, at least one-​third of the maximum positive moment reinforcement should be extended into the support (ACI-​7.7.3.8.1). For straight bars (i.e., bars not terminated by a standard hook or mechanical anchorage), the diameter of this reinforcement must be such that (ACI-​7.7.3.8.3)

Ld ≤ α M n /Vu + La (8.4.2)

where α = 1.3 if the reinforcement is confined by a compressive reaction force or α = 1.0 otherwise, and La is the embedment length beyond the center of the support. In Eq. (8.4.2), M n is to be calculated assuming all tension reinforcement yields. This same requirement for bar diameter applies to positive moment reinforcement at inflection points, where La in Eq. (8.4.2) is the embedment length beyond the point of inflection, but not to exceed the greater of d and 12 db. In this case, α = 1.0, since there is no compressive reaction force that could confine the bar being developed. For continuous supports, no less than one-​fourth of the maximum positive moment reinforcement must be extended at least 6 in. into the support (ACI-​7.7.3.8.2). At least one-​third of the negative moment reinforcement at supports must be extended at least the greatest of d, 12 db, and Ln /16 beyond the point of inflection (ACI-​7.7.3.8.4). Spacing s between flexural bars in slabs shall not exceed three times the slab thickness nor more than 18 in. (ACI-​7.7.2.3). Bar spacing, however, must also satisfy the maximum spacing required for crack control, as discussed in Section 12.18.

Shrinkage and Temperature Reinforcement Reinforcement perpendicular to the main flexural reinforcement (see Fig.  8.4.1), in accordance with ACI-​7.6.4, is required in one-​way slabs to resist stresses and control cracking due to shrinkage and temperature changes. ACI-24.4.3.2 requires the same minimum amount of shrinkage and temperature reinforcement as that for flexure [Eqs. (8.4.1)] spaced no farther apart than the lesser of 5 times the slab thickness or 18 in. (ACI-​ 24.4.3.3). This reinforcement must be capable of developing its tensile yield strength (ACI-​24.4.3.4). Note that the Code-​required amount of shrinkage and temperature reinforcement is a minimum; larger amounts of reinforcement may be needed in certain cases to ensure adequate control of cracking caused by stress-​induced shrinkage and changes in temperature [8.2]. Shrinkage and temperature reinforcement may be placed near the top or the bottom face of the slab, or distributed on both faces, the latter option being more appropriate for thicker slabs. This may be provided in the form of deformed bars, welded wire reinforcement or welded deformed bar mats.

260

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C hapter   8     D esign of O ne - W ay   S labs Top flexural bars

Bottom flexural bars

Shrinkage & temperature reinforcement

Figure 8.4.1  One possible arrangement of slab reinforcement.

Example 8.4.1 Choose the arrangement of reinforcement for the 6 1 2 -​in. floor slab 2S1, 2S2, and 2S3, of Example 8.3.1. SOLUTION (a) Steel area requirements. The ACI moment coefficient, the bending moment, the steel area required, and the tentative choice of reinforcement at the critical sections are shown in lines 1 to 6 of Table 8.4.1. The arrangement of reinforcement is shown in Fig. 8.4.2. The flexural reinforcement arrangement must satisfy the minimum reinforcement and detailing requirements of ACI-​7.6.1 and ACI-​7.7. Also, reinforcement perpendicular to the flexural bars must satisfy the shrinkage and temperature reinforcement requirement of ACI-​7.6.4, which refers to ACI-​24.4. For Grade 60 steel, the minimum required flexural reinforcement ratio based on the gross concrete area is 0.0018. Thus

As,min = 0.0018 (12 )(6.5) = 0.14 sq in./ft of width



For #4 bars, a spacing of 16 in. would correspond to an area of 0.15 sq in. per foot of width (Table 8.3.1), which satisfies the minimum reinforcement requirement. Maximum spacing is given by three times the slab thickness or 18 in.; thus

max s = 3 (6.5) = 19.5 in. or 18 in.



Use s = 16 in. The selection of the longitudinal steel should begin at the typical interior support. In this case, #4 bars at 16-​in. spacing ( As = 0.15 sq in.) are chosen for all interior supports except the first. At the first interior support, #4 bars at 14 in. spacing are chosen because of a slight increase in the design moment. At the middle of the first and typical interior spans, as well as the exterior support, minimum reinforcement requirements control. At these locations, the required area is furnished by #4 bars at 16 in. spacing. Also, it can be easily verified that the tensile strain in the reinforcement far exceeds the value of 0.005 in all sections. Thus, all sections are tension controlled and the corresponding φ factor is 0.90 as initially assumed. (b) Development length requirements. To confirm the choice of longitudinal reinforcement made on the basis of the required areas only, it is necessary to review the development length requirements, as covered earlier in this section. The requirements of ACI-​7.7.3.8.3 must be checked at the exterior support (point 1 of Fig. 8.4.2) and at inflection points 2 to 6. Also, positive reinforcement must extend at least 6 in. into interior supports (ACI-​7.7.3.8.2). In addition, embedment equal to the required development length must be provided in both directions from the maximum moment points at the faces of supports (points 7–​12). Finally, embedment of the bars at the top of the slab beyond extreme points of inflection (points 13–​18) must satisfy ACI-​7.7.3.8.4. (Continued)

#4@16

0.14b

0.0018 0.17

b

a

#4@14

0.0025

145

–​3.94

+2.81 103

1 – ​ 10

Support

1 +  14

Middle

Assuming a linear relationship between ρ and Rn (see Example 8.4.1 for ρ = 0.0025 and Rn = 145 psi). Minimum per ACI Table 7.6.1.1 controls [As,min = 0.0018(12)(6.5) = 0.14 sq in.].

#4@16

6. Provided As

0.0010 0.14b

line(3) × 0.0025 145

5. Required As = line (4) × 12(5.5)

4. Requireda ρ ≈

60

–​1.64

2. Mu = line(1) × 0.281(11.83)2

line(2) × 12,000 0.9(12)(5.5) 2

1 – ​ 24

1. ACI moment coefficient

3. Required Rn (psi) =

Support

Line Number

0.15

0.0023

132

–​3.58

1 – ​ 11

Support

#4@16

0.14b

0.0016

91

+2.46

1 +  16

Middle

2S2

0.15 #4@16

0.0023

132

–​3.58

1 – ​ 11

Support

0.15

0.0023

132

–​3.58

1 – ​ 11

Support

#4@16

0.14b

0.0016

91

+2.46

1 +  16

Middle

2S3

(Continued)

#4@16

0.15

0.0023

132

–​3.58

1 – ​ 11

Support



2S1

TABLE 8.4.1  REINFORCEMENT FOR ONE-​WAY SLAB OF EXAMPLE 8.4.1, USING ACI MOMENT COEFFICIENTS

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Example 8.4.1 (Continued) 8” 2’– 0”

3’ – 7”

#4 @ 16

3’ – 7”

3’ – 7”

#4 @ 14

7 1 13

14 2

8

#4 @ 16

9

3’ – 7”

#4 @ 16

#4 @ 16 3

16

15

4

10

11

5 17

18

6

12

#4 @ 16

#4 @ 16

Std. hook

3’ – 7”

#4 @ 16 (S&T) 2S3

2S2

2S1

Figure 8.4.2  Reinforcement in slab of Example 8.4.1.

Positive moment reinforcement. From the ACI shear and moment diagrams in Figs. 7.5.2 and 7.5.4, the shears at inflection points on the typical 1-​ft width of slab are

      V1 = V2 = 0.3780 wu Ln = 0.3780 ( 0.281)(11.83) = 1.26 kips



V3 = V4 = V5 = V6 = 0.3535wu Ln = 0.3535 ( 0.281)(11.83) = 1.18 kips

In computing the development length Ld for the bottom #4 bars that extend past the inflection points into the supports, the cb / db value used in Eq. (6.6.1) is based on cover, which is clearly smaller than one-​half the center-​to-​center bar spacing. Thus,

  cb 0.75 (i.e.,clear cover) + 0.25 (i.e., bar radius / 2) 1.00 = = 2.00  < 2.5 max  = 0.50 0.50   db

Because there are no stirrups, K tr = 0. Thus, Eq. (6.6.1) gives    f ψ t ψ e ψ s  3 y Ld =  d  40 λ fc′  cb + ktr   b   d     b

[6.61]

 3 60, 000 1.0(1.0)(0.8)  Ld =   0.5 = 14.2 in.  40 1.0 4000 2.0

In the above equation, the reinforcement size factor ψ s = 0.8 for bars #6 and smaller. Note again that the Ld computed above using the general equation has to be smaller than the Category A simplified equation [Eq. (6.6.4)] value of 19.0 in. from Table 6.6.1, where ( cb + K tr ) / db is taken as 1.5. The requirement of ACI-​7.7.3.8.3 at inflection points (such as point 2) is Mn + La ≥ Ld Vu

For #4 @ 16 in.,

C = 0.85(4)12 a = 40. 8 a T = 0.20(60) = 12.0 kips a=



12.0 = 0.29 in. 40.8

0.29  1  M n = 12.0  5.5 −  = 5.35 ft-kips / ft  2  12 Vu = 1.26 kips (computed above)

(Continued)

263



263

8 . 4   C hoice of   R einforcement

Example 8.4.1 (Continued) La = 12 db (max) = 6.0 in.

5.35(12) + 6.0 = 57.0 in. > Ld = 14.2 in. 1.26

OK

This calculation applies identically to other inflection points (3–​6). At the exterior end, which might be considered a simple support, the requirement of ACI-​7.7.3.8.3

1.30

Mn + La ≥ Ld Vu

is satisfied by inspection. This reinforcement is extended 8 in. into the exterior support. Because the bottom reinforcement is continuous over interior supports (Fig. 8.4.2), there is no need to check the minimum embedment length of 6.0 in. Negative moment reinforcement. The distance from the face of support to the cutoff location (beyond point 14, for instance) must be (1) at least Ld , and (2) adequate to satisfy ACI-​7.7.3.8.4. For the #4 bars,        Ld = 14.2 in. (computed above) To satisfy ACI-​7.7.3.8.4, the moment diagram of Fig. 7.5.2(b) corresponding to the ACI coefficients may be used to locate the point of inflection. The length required from the exterior face of the first interior support to the cutoff location is

(0.5583 − 0.3341)(11.83) +

11.83 = 3.39 ft 16

which compares favorably with the commonly used value of 0.3 of the clear span (3.55 ft), as suggested in the 2004 ACI Detailing Manual [2.23]. For all other interior supports [Fig. 7.5.4(b)],

(0.5 − 0.2611)(11.83) +

11.83 = 3.57 ft 16

Note that the second term in the two expressions above corresponds to Ln /16, which is greater than d and 12db. Both requirements above can be satisfied by extending the top slab bars 3 ft 7 in. from the face of each interior support. Note that while slightly different lengths could have been used, such a small difference in length at the first interior support versus the other interior supports would be impractical. For the top reinforcement at the exterior support, the minimum extension from the face of the support is (Fig. 7.5.2b)

(0.4417 − 0.3341)(11.83) +

11.83 = 2.0 ft > Ld 16

This reinforcement is anchored into the exterior support through a standard 180° hook. (c) Shrinkage and temperature reinforcement Minimum amount of shrinkage and temperature reinforcement is (ACI Table 24.4.3.2) 0.0018bh = 0.0018(12)(6.5) = 0.14 sq in. at a spacing not greater than 5 times the slab thickness (32.5 in.) and 18 in. Thus, provide #4 bars at 16 in. spacing perpendicular to the slab flexural reinforcement. Because the slab is only 6.5 in. thick, there is no need to distribute this reinforcement near the top and bottom faces of the slab. In this case, the shrinkage and temperature reinforcement is provided at the bottom of the slab, directly on top of the bottom flexural reinforcement (labeled as S&T in Fig. 8.4.2).

264

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C hapter   8     D esign of O ne - W ay   S labs

8.5 BAR DETAILS Consistent with the shear and moment diagrams given in Chapter 7 for the ACI Code coefficients, acceptable standard bar bend distances, extensions, and anchorage lengths have been developed. For instance, typical bar details in end and interior spans of one-​way slabs as adapted from the 2008 design handbook of the Concrete Reinforcing Steel Institute [2.21] are reproduced in Fig. 8.5.1. Although typical bar details are of importance in office practice, they must be used with discretion and care to ensure that they conform to the prevailing shear and moment diagrams. ACI standard hook (tilt from vertical if necessary to maintain ¾” clearance) CL

Size and spacing as tabulated 0.25Ln

Symmetrical about CL Temperature – shrinkage bars

¾” Clear

X

¾” Clear

1½” cover minimum

X

X

X = Bar spacing

Extend all bottom Bars into support

Slab thickness

Ln = Clearspan

Single span, simply supported ACI standard hook (tilt from vertical if necessary to maintain ¾” clearance) Size and spacing As tabulated 0.25Ln 0.3Ln or 0.3Ln1 ¾” Clear Greater

0”

1” cover minimum

Slab thickness

¾” Clear 0.3Ln or 0.3Ln1

0.125Ln

¾” Clear

Extend all bottom Bars into support Ln = Clearspan

Temperature – shrinkage Bars

Greater

X

6”*

X

X

X = Bar spacing

Ln1

End span, simply supported CL

¾” Clear 0.3Ln or 0.3Ln1 Greater

0.3Ln or 0.3Ln1 Greater

Symmetrical about CL Temperature – shrinkage Bars

0”

0.125Ln Ln1

6”*

¾” Clear

Ln = Clearspan

X

X

X

X = Bar spacing Slab thickness

Interior span, continuous *min. 6”, unless otherwise specified by the engineer

Figure 8.5.1  Typical details for one-​way solid slabs. (Adapted from CRSI Design Handbook 2008 [2.21].)

265



265

P roblems

SELECTED REFERENCES 8.1. Phil M. Ferguson. “Analysis of Three-​Dimensional Beam-​and-​Girder Framing,” Proceedings, Journal of the American Concrete Institute, 47, September 1950, 61–​72. 8.2. R. Ian Gilbert. “Shrinkage Cracking in Fully Restrained Concrete Members.” ACI Structural Journal, 89, March–​April, 1992, 141–​149.

PROBLEMS All problems are to be worked in accordance with the 2014 ACI Code. All stated loads are service loads unless otherwise indicated. 8.1 Design for a warehouse a continuous one-​way slab supported on beams 12 ft on centers as shown in the figure for Problem 8.1. Assume that beam stems are 12 in. wide. The dead load is 25 psf in addition to the slab weight, and the live load is 200 psf. Use fc′ = 3000 psi and f y = 60, 000 psi. Use ACI coefficients if permissible. (For an SI problem, use 3.7-​m spans, 400-​mm beam widths; superimposed dead load = 1.2 kN/​m; live load = 9.6 kN/​m; fc′ = 25 MPa; f y = 400 MPa.) 8.2 Repeat Problem 8.1 using a live load of 250 psf with fc′ = 4000 psi and f y = 60, 000 psi.

Spandrel-type end

12’– 0”

12’– 0”

12’– 0”

Problems 8.1 through 8.4  Cantilever end

8’– 0”

Problem 8.5 

16’– 0”

16’– 0”

(For SI:  live load = 12.0 kN/​m; fc′ = 30 MPa; f y = 400 MPa.) 8.3 Repeat Problem 8.1 using a live load of 300 psf with fc′ = 4000 psi and  f y = 60, 000 psi. 8.4 Repeat Problem 8.1 using a live load of 350 psf with fc′ = 4000 psi and  f y = 60, 000 psi. 8.5 Design a one-​way slab for the conditions shown in the figure for Problem 8.5. Assume that beam stems are 12 in. wide. The live load is 175 psf and the slab will not be the final wearing surface. Use fc′ = 4000 psi and fy = 60,000 psi.

CHAPTER 9 DESIGN OF SLAB-​B EAM-​G IRDER AND JOIST FLOOR SYSTEMS 9.1 INTRODUCTION The design of rectangular and T-​sections was treated in Chapters  3 and 4, respectively; shear strength along with stirrup design in Chapter  5; development of reinforcement in Chapter 6; and continuity analysis in Chapter 7. The slab-​beam-​girder type of floor construction has been described generally; Chapter 8 illustrates the design of one-​way slabs. This chapter presents complete designs of a typical floor beam and girder in a monolithic slab-​beam-​girder system. Primarily, what has been developed in the preceding chapters is applied. Thus the reader may consider the material of this chapter as an integrated review of the subjects in the aforementioned chapters. Also included in this chapter is the design of one-​way concrete joist floors. On continu­ ous spans, the concrete joists should be regarded as having rectangular sections near the supports and T-​sections in the positive moment region within the span. In some instances, precast slabs, either conventionally reinforced or prestressed, are placed on a monolithic beam-​and-​girder framing system. T-​sections are not involved in such systems because the slab rests on, but does not act with, the beams and girders. The procedures discussed in this chapter are also generally applicable to this simpler system of rectangular beams.

Noho III office building, North Hollywood, CA (Photo courtesy of Cary Kopczynski & Co).

267



9.2  SIZE OF BEAM WEB

267

Reinforced concrete building design may be a complicated venture, involving irregular floor plans and intricate structural framing. The simple floor framing plan used in the examples does not necessarily reflect a typical design practice, but it does serve to illustrate the basic essentials of design.

9.2 SIZE OF BEAM WEB The size of the beam web for continuous T-​shaped sections is usually controlled by the flexural and shear strength requirements at the exterior face of the first interior support. For typical conditions of equal spans and uniform dead and live loads, a negative moment of 110  wL2n, and a shear of 0.60wLn (see Fig. 7.5.1) or 1.15wLn /​2 (see ACI-​6.5.4) may be used in estimating the size of the beam. The following discussion sets out the detailed considerations involved in selecting the beam cross section based on the critical bending moment and shear.

Negative Moment Requirement The section of the beam resisting the negative bending moment at the face of the support behaves as a rectangular section, even in T-​section construction, because compression occurs in the lower part of the section (i.e., in the lower portion of the web). The tension reinforcement is most often provided by straight bars extending across the top of the beam, as required for negative moment (bent bars, though still permitted by the ACI Code, are rarely if ever used today). In accordance with ACI-​9.7.3.8.2, at least one-​fourth of the maximum positive moment reinforcement must continue along the bottom of the beam into the support. When this reinforcement is continuous or developed into the support, it may be utilized as compression reinforcement for increased ductility and deflection control. The guideline reinforcement ratio ρ to be used for deflection control of singly reinforced rectangular beams has been suggested (in Section 3.9) to be about 0.6ρtc for Grade 60 steel. For the negative moment requirement on continuous T-​sections, a higher value for ρ should be acceptable without contributing to higher deflection, because the gross moment of inertia of a typically proportioned T-​section is roughly twice that of its rectangular portion. Thus it may be reasonable to design the negative moment region of a T-​shaped section with about twice the percentage of reinforcement that would be used for deflection control on a completely rectangular beam (or more if compression steel is utilized).

Positive Moment Requirement In the positive moment region, the flange of the T-​section is in compression. In practice, the effective flange width is often large and thus the depth a of the Whitney rectangular stress distribution will rarely extend below the bottom of the flange. On the tension side, the amount of steel required will be inversely proportional to the depth of the section. For sizing the beam web, the only consideration utilizing the positive moment is to establish the width required to maintain adequate clearances for a given number of bars. This should be examined because a somewhat deeper or shallower beam might still permit the required steel to fit into one or two layers, as the case may be.

Shear Strength Requirement The designer may wish to establish the beam web size to limit the maximum nominal shear stress. This may be desirable for economical stirrup size and spacing. The ranges of five categories of design for shear reinforcement are given in Section 5.10. The upper limit of Category 4 (i.e., Vs = 4 fc′bw d ) serves as a practical guideline for ordinary design, typically permitting stirrup spacing from 3 in. to a maximum of d/​2 or 24 in.

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C hapter   9     S lab - B eam - G irder A N D J O I S T F L O O R S Y S T E M S

When Vc is taken as 2 λ fc′ bw d according to the simplified procedure, the total Vn at the upper limit in Category 4 is Vn = Vc + Vs = 2 λ fc′ bw d + 4 fc′ bw d



= (2 λ + 4) fc′ bw d

(9.2.1)

or, for a given factored shear Vu, the bwd required to satisfy this guideline is

required bw d =

required Vn (2 λ + 4) fc′

=

Vu

(9.2.2)

φ (2 λ + 4) fc′

Except for unusual conditions involving short spans, heavy concentrated loads, heavily doubly reinforced sections, or combinations of these, the maximum nominal shear stress Vu /(φ bw d ) should be less than 6 fc′ for reasonable stirrup size and spacing.

EXAMPLE 9.2.1 Use the ACI Code provisions to establish the preliminary size for the continuous floor beams 2B1-​2B2-​2B1 supported by girders as shown in the floor framing plan of Fig. 8.3.1. Assume a slab thickness of 5.5 in. Use fc′ = 4000 psi and f y = 60, 000 psi. SOLUTION (a) Negative moment requirement. Beam sections located at the face of the support are designed as rectangular sections with the compression zone located at the lower part of the section. These sections may be sized for the maximum reinforcement ratio corresponding to tension-​controlled sections because deflections will be governed primarily by the stiffness (moment of inertia) of the T-​sections (with the flange in compression) within the span. For fc′ = 4000 psi and f y = 60, 000 psi, the maximum reinforcement ratio recommended in Section 3.6 may be computed from Eq. (3.6.2) or from Table 3.6.1,

recommended ρmax = ρtc = 0.0181

Estimating the weight of the stem (portion of the web below the slab) at 0.3 kips/​ft (2 sq ft of area), and using the load factors of ACI-​5.3.1,

Factored dead load = 1.2 5.5 ( 0.15 /12 )(13) + 0.3 = 1.43 kips/ft Factored live load = 1.6 ( 0.10 )(13) = 2.08 kips/ft



Using Eqs. (3.8.4a) and (3.84b), the Rn corresponding to ρ = 0.0181 is m=

fy 0.85 fc′

= 17.65

1 Rn = ρ f y (1 − ρ m) = 913 psi 2



Assume width of supporting girders to be 18 in., then the clear span Ln is

Ln = 26 – 1.5 = 24.5 ft

(Continued)

269



269

9.2  SIZE OF BEAM WEB

Example 9.2.1 (Continued) Using the moment coefficient from ACI-​6.5.2 (see Table 7.4.1) and φ = 0.90 (tension-​ controlled section), 1 (1.43 + 2.08)(24.5)2 = 211 ft-kips 10 M 211(12, 000) required bd 2 = u = = 3081 cu in. φ Ru 0.90(913) max Mu =



If b = 13 in., required d =



3081 = 15.4 in. 13

The steel area required would then be As = 0.0181(13)(15.4 ) = 3.62 sq in.



In the negative moment region the flange at the top (slab) is available so that the steel does not have to fit within the web width. The shear requirement and the steel requirement for positive moment should be examined before a decision is made. (b) Shear requirement. Using the approximate shear coefficient from ACI-​6.5.4,

w L   3.51(24.5)  max Vu = 1.15  u n  = 1.15   = 49.4 kips   2  2

If it is desired that the nominal shear stress υn = Vu /(φ bw d ) does not exceed 6 fc′ = 379 psi, required bw d =



49, 400 = 174 sq in. 0.75(379)

For b = 13 in., then d = 13.4 in. (Actually the maximum shear may be taken at a distance d from the face of the support which would result in a smaller required beam depth.) (c) Positive moment requirement. The effective flange width bE is the smallest of the following (see Section 4.3): 24.5(12) = bw + 73.5 in. 4 2. bw + 16t = bw + 16(5.5) = bw + 88 in. 1.  bw + Ln / 4 = bw +

Controls!

3. center-​to-​center spacing = 13(12) = 156 in. For bw = b = 13 in., the effective width will be 86.5 in., or about 6.7 times wider than the web width, bw. Try d = 16 in. and estimate the moment arm at 0.95d ≈ 15 in. (This is a reasonable assumption for T-​beams with the flange in compression.) Using the moment coefficients from ACI-​6.5.2 for an end span with discontinuous end integral with the support and assuming φ = 0.90 (tension-​controlled section), Mu = required As =

1 3.51 wL2n = (24.5)2 = 150 ft-kips 14 14 Mu 150(12) ≈ = 2.22 sq in. φ f y (arm ) 0.90(60)(≈ 15)

(Continued)

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C hapter   9     S lab - B eam - G irder A N D J O I S T F L O O R S Y S T E M S

Example 9.2.1 (Continued) This amount of steel is satisfied with 3 #8 bars (As = 2.37 sq in.), for example, which easily fit into a 13-​in.-​wide beam in one layer (see Table 3.9.2). (d) Minimum depth. According to ACI Table 9.3.1.1, the minimum depth, h, of a beam with one end continuous (beams 2B1 are more restrictive than the center beam, 2B2) cannot be less than

min h =

L 26(12) = = 16.9 in. 18.5 18.5

unless the deflection limits given in ACI-​9.3.2 are satisfied. The designer must also keep in mind that if excessive deflection may cause damage, the deflection must be computed in accordance with ACI-​24.2 and must satisfy ACI Table 24.2.2, even if the limits of ACI Table 9.3.1.1 are satisfied. The above check is only a minimum requirement, where deflection is not likely to cause damage to nonstructural elements. (e) Choice of size. Based on the calculations in parts (a) through (c), a beam with a 13-​in.-​wide web and an effective depth, d, of about 16 in. would be adequate. Allowing 2.5 in. for the clear cover and stirrups, this would result in a minimum overall depth, h, of 18.5 in., or 19 in. Frequently the size is chosen larger than required. Sometimes this occurs because the designer wants a large number of beams to have the same external dimensions for economy of forming, or perhaps because ductwork or pipes are to pass through the beams, necessitating larger sizes. Use bw = 13 in., h = 22.5 in. (which gives d ≈ 20 in.). It is common to make the stem portion below the flange a whole-inch increment, such as 17 in. for this case. The arbitrary size selected is larger than necessary; any size at least equal to that indicated by steps (a) through (d) would serve to illustrate the design procedure.

9.3 CONTINUOUS FRAME ANALYSIS FOR BEAMS The shear and bending moment diagrams to be used in the design of the floor beams in the preceding example may be obtained by using the ACI moment coefficients (ACI-​6.5.2). However, because those coefficients are more suitable for use in frames involving, say, more than four continuous spans, an elastic analysis will be used to illustrate how to obtain the shear and bending moment envelopes for beams 2B1-​2B2-​2B1. Some designers would assume the girders to provide only vertical support to floor beams resting on them; thus a pure continuous beam analysis would be made. The girders, however, have torsional stiffness that acts to restrain the end rotation of the beams over these support girders. The ACI Code permits torsional effects to be ignored whenever the factored torsional moment, Tu, is less than the design threshold torsion, φ Tth (ACI-​9.5.4.1; see Chapter 18). In the case of a beam supported with relatively large and stiff girders [see Fig. 9.3.1(a)], the torsional stiffness provided by the girder may reasonably be taken as twice the flexural stiffness of the beam. An equivalent beam and column frame model for this beam-​girder system is shown in Fig. 9.3.1(b). For beams that frame into columns, however, the combination of the bending stiffness of the columns plus the torsional stiffness of the girders results in a relative end restraint greater than in the aforementioned case. To design such beams, the sizes of the members may be estimated and the (∑Kcol /​∑Kbeam) ratio computed therefrom. With regard to a T-​section, opinions as to how its stiffness in a continuous beam should be computed differ considerably. A T-​section certainly provides more stiffness in the positive moment region where the flange is in compression than it does as a rectangular section in the negative moment region. For gravity load analyses, it has been common practice to use the gross moment of inertia, neglecting reinforcement, in computing the flexural stiffness of such elements. For T-​sections in the positive moment region, the gross section usually includes the effective flange width.

271



271

9 . 3   C O N T I N U O U S F R A M E A N A LY S I S F O R   B E A M S A

B

C

D

2B2

2B1 26’–0”

2B2

26’–0”

26’–0”

(a) Actual continuous beam supported by girders 1 2

1 2

1 2

Relative stiffness

2

(b) Frame model to approximate floor beams supported on girders

Figure 9.3.1  Actual continuous beam and equivalent beam and column model.

One of the accepted methods [9.1] has been to use a T-​section moment of inertia equal   to 2  1 bw h3  , which is equivalent to using an effective flange width of about 6 times the  12  web width. Since the true stiffness is that of a span with variable moment of inertia along its length with the T-​section stiffness over, say, the middle half of the span and the rectangular section stiffness over the end quarters, the equivalent system is obtained approximately by using the T-​section with a flange width only twice the web width over the entire span. The ACI Code provides some guidance for calculating section properties to be used in the analyses under factored lateral loads (ACI-​6.6.3), but it is silent for the analysis under gravity loads. In ACI-​R6.6.3.1.1, however, it is stated that the moment of inertia of T-​beams should be based on the effective flange width defined in ACI-​6.3.2.1 or ACI-​6.3.2.2 (see Section  1  4.3), or as 2  bw h3  as discussed above. Examination of analyses using various stiffness  12  ratios will show that a fairly wide variation in stiffness ratio may be accommodated with relatively small changes in bending moments. In the following example, only the analysis of the floor beams supported on girders is shown, using the analytical frame model in Fig. 9.3.1(b). An overestimation of the beam stiffness will give a conservative design for the beam; but such an assumption should be reevaluated if any columns are to be designed.

EXAMPLE 9.3.1 Using an elastic statically indeterminate analysis, obtain the shear and moment diagrams to be used in compiling the moment and shear envelopes for the design of beams 2B1 and 2B2 in Example 9.2.1. Because the girders are not expected to be significantly larger and stiffer than the beams, assume the torsional stiffness of the girders to be the same as the flexural stiffness of the beams. SOLUTION Using the beam dimensions chosen in part (e) of Example 9.2.1, the actual weight of the portion of the web below the slab is 0.15(13/12)(17/12) = 0.23 k/ft, say 0.25 k/ft. The revised factored load then becomes Factored dead load = 1.2[5.5(0.15)(13/12) + 0.25] = 1.37 kips/ft Loading conditions 1 through 5 as shown in Fig. 9.3.2 are established by visualizing the influence lines for positive moment in the midspan region of each span and for negative moment at each support (see Section 7.3). The structural analysis may be performed by any method of statically indeterminate analysis—​for example, the moment distribution method—​or by using any standard structural analysis software package. Since the specific (Continued)

27

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C hapter   9     S lab - B eam - G irder A N D J O I S T F L O O R S Y S T E M S

Example 9.3.1 (Continued) WD

1. A

B

C

D

WL 2.

WL 3.

WL 4.

WL 5.

Figure 9.3.2  Loading conditions for Example 9.3.1.

structural analysis technique is outside the scope of this text, only the end moments resulting from the analysis for each of the five loading cases are given in Table 9.3.1. The primary shear and moment diagrams for maximum positive and negative moments for beams 2B1 and 2B2 are given individually in Figs. 9.3.3 and 9.3.4, respectively. The secondary shear and moment diagrams are shown in Figs. 9.3.5 and 9.3.6, respectively. The primary and secondary moment diagrams are drawn separately for clarity in the example. Ordinarily they are not drawn separately but, instead, are superimposed to become the moment envelope as shown later (see Fig. 9.6.2). a

TABLE 9.3.1  END MOMENTS (ft-​kips) FOR THE FIVE LOADING CONDITIONS OF FIG. 9.3.2 Joint

A

Member

AB

BA

BC

CB

CD

Case 1, M = Case 2, M =

–​27.1 –​54.2

+94.8 +95.2

–​82.1 –​35.6

+82.1 +35.6

–​94.8 –​95.2

+27.1 +54.2

Case 3, M = Case 4, M =

+13.1 –​39.0

+48.8 –​89.1 +153.2 –​143.9

+89.1 +69.8

–​48.8 –​39.6

–​13.1 –​11.1

Case 5, M =

–​2.0

+54.9 –​104.4

+52.2

B

–​9.2

C

+19.2

D DC Dead load only Live load spans 1 and 3 Live load span 2 Live load spans 1 and 2 Live load span 3

Clockwise moments acting on member ends are taken positive.

a

(Continued)

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9 . 3   C O N T I N U O U S F R A M E A N A LY S I S F O R   B E A M S

Example 9.3.1 (Continued) 81.3

190.0

3.45 k/ft

A

66.1

B

Ln = 24.5’

A

Vu , kips

40.7

B

Ln = 24.5’ 37.9

14.20’

15.00’ 11.00’

11.80’ 49.0

158.8 2.20’ Mu , ft-kips

248.0

3.45 k/ft

4.60’

+ 1 9.60’

81.3

1.91’

9.08’

66.1

1.45’

5.94’

+

–

9.60’

x1

2 – #7 2 – #5

2

51.8

142.1

–

9.08’

2 – #8

190.0

x2

2 – #7 248.0

x3 (b) Maximum in negative region at B (loading conditions 1 + 4)

(a) Maximum in positive region (loading conditions 1 + 2)

Figure 9.3.3  Primary shear and moment diagrams for 2B1. 171.2

171.2

3.45 k/ft

B

C

Ln = 24.5’

B

C

Ln = 24.5’

13’

12.17’

13’

13.83’

4.61’

44.9

121.0

2 – #7

42.0 103.8

4.61’

+

6.07’

2 – #4 Mu , ft-kips

151.9

3.45 k/ft

47.7

Vu , kips

44.9

226.0

–

3

8.39’

8.39’

4

–

–

2 – #8

7.76’

2 – #7

171.2 x4

x4

(a) Maximum in positive region (loading conditions 1 + 3)

171.2

4.42’

+ 7.76’

–

151.9

226.0

x5

(b) Maximum in negative region at B (loading conditions 1 + 4)

Figure 9.3.4  Primary shear and moment diagrams for 2B2.

(Continued)

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Example 9.3.1 (Continued) 14.0

143.6

1.37 k/ft

A

B

26’

12.8 Vu, kips

9.3’

Mu, ft-kips

8.18’

14.6’ 11.4’

20.0 59.7 +

–

8.18’

B

26’

2.05’

7.77’

+

14.0

A

22.8 45.8

86.6

1.37 k/ft

15.6

16.7’

1.17’

29.1

–

29.1

9.32’

9.32’

86.6

143.6 (b) Minimum in negative region at B (loading conditions 1 + 5)

(a) Minimum in positive region (loading conditions 1 + 3)

Figure 9.3.5  Secondary shear and moment diagrams for 2B1. 117.7

B

C

26’

17.8

17.8

–

137.0

1.37 k/ft

B

C

26’

15.05’ 10.95’

5.65’ –

2.07 117.7

62.9

15.0

13.0’ 13.0’

Mu , ft-kips

Vu , kips

117.7

1.37 k/ft

20.7 19.2

5.3’

5.3’

–

62.9 117.7

(a) Minimum in positive region (loading conditions 1 + 2)

137.0 (b) Minimum in negative region at B (loading conditions 1 + 5)

Figure 9.3.6  Secondary shear and moment diagrams for 2B2.

9.4 CHOICE OF LONGITUDINAL REINFORCEMENT IN BEAMS It cannot be overemphasized that the choice of longitudinal reinforcement depends on both the steel area and development length (or anchorage) requirements. The design of the main reinforcement in floor beams 2B1-​2B2-​2B1 is shown in the following example.

275



9.4  CHOICE OF LONGITUDINAL REINFORCEMENT IN BEAMS

275

EXAMPLE 9.4.1 Choose the arrangement of main reinforcement in the floor beams 2B1-​2B2-​2B1 of Example 9.3.1. SOLUTION (a) Flexural strength requirements. The critical moment at the face of the support, or at any point along the beam length, can be obtained from the diagrams shown in Figs. 9.3.3 through 9.3.6. Alternatively, a good estimate for preliminary design may 1 be obtained by subtracting ∆M = [V × (width of support )] from the moment at the 3 center of support. The shear V for this calculation is taken as that at the face of support. Start with A-​A, C-​C, and D-​D (Fig. 9.4.1), Section A-​A (rectangular section): Mu at face = 81.3 −



38.1(1.5) = 81.3 − 19.1 = 62.2 ft-kips 3

Assuming φ = 0.90, required Rn =



Mu 62.2(12, 000) = = 159 psi 2 φ bd 0.90(13)(20)2

Using Eq. (3.8.5), or Fig. 3.8.1, required ρ = m=



required ρ =

2 mRn  1 1 − 1 −  m fy  fy 0.85 fc′

=

60 = 17.65 0.85(4)



1  2(17.65)(159)  1− 1− = 0.0027  17.65  60, 000 

The minimum reinforcement for a statically indeterminate T-​beam section with the flange in tension is given by As ,min =



3 fc′ fy

bw d [3.7.10]*

but not less than 200bw d /f y . Thus,   As,min =

3 4000 (13)20 = 0.82 60, 000

or

200(13)20 = 0.87; 0.87 sq in. Controls! 60, 000

The required area based on the applied loads is

required As = 0.0027 (13) 20 = 0.70 sq in. <  As ,min = 0.87 sq in.

Therefore, use As = As,min = 0.87 sq in. (Continued)

*  Equation numbers in brackets indicate material from another section or chapter.

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Example 9.4.1 (Continued) A 4

B 1

F

8

7

A

B

C

10 2

F

D

5

6

C

E

11 9

3

D

24’–6”

E 12’–3” 2B2

2B1

Symmetrical about CL

2 – #8 (A = 2.78) s 2 – #7

2 – #7 (As = 1.20) 2.73

1.04

2.49

1.78

1.36 2 – #7 (A = 1.60) s 2 – #4

2 – #7 (A = 1.82) s 2 – #5

Figure 9.4.1  Longitudinal reinforcement areas required (sq in.) and bars selected for beams 2B1 and 2B2 in Example 9.4.1.

Section C-​C (rectangular section):



49.2(1.5) = 248.0 − 24.6 = 223 ft-kips 3 223(12, 000) = 572 psi required Rn = 0.90(13)(20)2 Mu at face = 248.0 −

From Fig. 3.8.1, or Eq. (3.8.5), required ρ = 0.0105,

required As = 0.0105(13)20 = 2.73 sq in. >  As ,min = 0.87 sq in.



Section D-​D (rectangular section):



45.1(1.5) = 226.0 − 22.6 = 203 ft-kips 3 203(12, 000) = 521 psi required Rn = 0.90(13)(20)2 Mu at face = 226.0 −

Computing the required steel area As in proportion to Rn,

required As = 2.73(521 / 572) = 2.49 sq in. >  As ,min = 0.87 sq in.

For sections B-​B and E-​E (Fig. 9.4.1): The effective flange width is the smaller of: 24.5(12) = 13 + 73.5 = 86.5 in 4 2. bw + 16t = bw + 16(5.5) = 13 + 88 = 101 in. 3. center-​to-​center spacing = 13(12) = 156 in. 1. bw + Ln / 4 = bw +

Controls!

Thus, bE = 86.5 in. For T-​sections having the flange in compression, the minimum reinforcement is given by

As ,min =

3 fc′ fy

bw d

[3.7.10] (Continued)

27



9.4  CHOICE OF LONGITUDINAL REINFORCEMENT IN BEAMS

277

Example 9.4.1 (Continued) but not less than 200bw d /​fy. In Eq. (3.7.10), bw is the width of the web. Thus, As,min = 0.87 sq in. as computed earlier. Section B-​B (T-​section with flange in compression): Estimate moment arm = 0.9d = 18 in.

required As =

Mu 158.8(12) = = 1.96 sq in. φ f y (arm) 0.90(60)(≈ 18)

Check: C = 0.85 fc′ bE a = 0.85(4)(86.5)a = 294 a T = As f y = 1.96(60) = 118 kips 118 = 0.40 in. 294 0.40 arm = 20 − = 19.8 in. 2 a=



revised required As =



158.8(12) = 1.78 sq in. > [As ,min = 0.87 sq in.] 0.90(60)19.8

OK

Section E-​E (T-​section with flange in compression):



required As =

Mu 121.0(12) = φ f y (arm) 0.90(60)(≈ 19.8)

= 1.36 sq in. > [As ,min

= 0.87 sq in.]

OK

Check: C = 294 a T = As f y = 1.36(60) = 82 kips

82 = 0.28 in. 294 0.28 arm = 20 − = 19.9 in. 2 a=

OK

On the basis of the areas required at sections A-​A, B-​B, C-​C, D-​D, and E-​E, the arrangement of main reinforcement shown in Figs. 9.4.1 and 9.4.2 is tentatively chosen. Note that the 2–​#5 bottom bars in beam 2B1 and the 2–​#4 bottom bars in beam 2B2 furnish the minimum one-​fourth of the positive moment reinforcement that must be extended at least 6 in. into the support in each span (ACI-​9.7.3.8.2). Also note that the two 2–#8 top bars in section E-E are needed to satisfy negative moment strength requirements at midspan as discussed later, in part (h) of Example 9.6.1. Confirming the choice of main reinforcement calls for checking the design strength φ Mn at the critical sections. The effect of any compression reinforcement will be small and is thus neglected. Negative moment at section A-​A: 2–​#7 top bars (As = 1.20 sq in. > required As = 0.87 sq in.)

C = 0.85(4)(13)a = 44.2 a 72.0 a= = 1.63 in. 44.2

T = 2(0.60)60 = 72.0 kips 1.63 c= = 1.92 in. 0.85

(Continued)

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Example 9.4.1 (Continued) 1

2 – #8

5 2” 2 – #7 17” 2 – #5

2 – #3 Stirrup supports 2 – #5

13”

Section B–B

2 – #8

2 – #5

2 – #4 2 – #5 (C–C) 2 – #4 (D–D)

2 – #7

Section A–A

2 – #8

2 – #7

Sections C–C and D–D

2 – #7 Section E–E

2 – #7 Section F–F

Figure 9.4.2  Typical sections for 2B1 and 2B2 in Example 9.4.1. Stirrups not shown for clarity.

The net tensile strain in the extreme tension steel is

εt =



d −c 20 − 1.92 (0.003) = (0.003) = 0.028 > 0.005 c 1.92

Thus, φ = 0.90

φ M n = 0.90(72.0)(20 − 0.82)

1 12

= 104 ft-kips [> 62.2 ft-kips]

OK

Positive moment at sections B-​B and F-​F: 2–​#5 and 2–​#7 bottom bars (As = 1.82 sq in. > required As = 1.78 sq in.) C = 0.85(4)(86.5)a = 294a  T = 1.82(60) = 109 kips a = 0.37 in.        c = 0.44 in.

εt = 0.133 > 0.005

φ M n = 0.90(109)(20 − 0.19 )

1 12

= 162 ft-kips [> 158.8 ft-kips]

OK

Negative moment at sections C-​C and D-​D: 2–​#8 and 2–​#7 top bars [As = 2.78 sq in. > required As = max. of (2.73, 2.49) sq in.] C = 44.2a   T = 2(0.79 + 0.6)60 = 167 kips a = 3.77 in.   c = 4.44 in.

εt = 0.0105 > 0.005

φ M n = 0.90(167)(20 − 1.89)

1 12

= 227 ft-kips [> (223 and 203) ft-kkips]

OK

Positive moment at section E-​E: 2–​#4 and 2–​#7 bottom bars (As = 1.60 sq in. > required As = 1.36 sq in.) C = 294a   T = 2(0.20 + 0.60)60 = 96 kips a = 0.33 in.   c = 0.38 in.

εt = 0.153 > 0.005

φ M n = 0.90(96)(20 − 0.17)

1 12

= 143 ft-kips[> 121 ft-kips]

OK

(Continued)

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9.4  CHOICE OF LONGITUDINAL REINFORCEMENT IN BEAMS

279

Example 9.4.1 (Continued) In addition to checking the design strength at the above critical sections, the design strength of the cross sections at locations where some of the longitudinal reinforcement is to be terminated must be computed. The design strength at these locations is required for later computation of the appropriate bar cutoff locations to ensure that the φ Mn diagram does not encroach the moment envelope, as will be illustrated later in Example 9.6.1. Positive moment near exterior support, after termination of 2–#7 bottom bars: 2–#5 (As = 0.62 sq in.) It is noted that at this location the provided amount of reinforcement is less than As, min (= 0.87 sq in.). According to ACI-​9.6.1.3, As, min need not be satisfied when As provided is at least one-​third greater than that required by analysis. Assume that the 2–#7 bars will be cut at a distance such that ACI-​9.6.1.3 will be satisfied with 2–#5 bars. Thus, C = 294a   T = 0.62(60) = 37.2 kips a = 0.13 in.    c = 0.15 in.

εt = 0.40 >>> 0.005

φ M n = 0.90( 37.2)(20 − 0. 065)

1 12

= 56 ft-kips

Note that the computed net tensile strain is much greater than the elongation at fracture for a typical Grade 60 reinforcing bar (see Fig. 1.13.2), which implies that the tension steel will not only yield but will also strain harden and likely fracture before concrete crushing. This behavior is due to the small amount of flexural reinforcement provided at this location and the wide flange (effective width), which results in a very small neutral axis depth (note that even when As, min is provided, the net tensile strains are also computed to exceed the expected elongations at fracture). It will be shown later that the design strength is much greater than the required strength and that ACI-​9.6.1.3 is satisfied by a large margin and thus As, min need not be provided at this location. Negative moment near sections C-​C and D-​D, after termination of 2–#7 top bars: 2–​#8 (As = 1.58 sq in.) C = 44.2a     T = 2 (0.79) 60 = 94.8 kips a = 2.14 in.   c = 2.52 in.

εt = 0.021 > 0.005

φ M n = 0.90(94.8)(20 − 1.07)

1 12

= 135 ft-kips

Positive moment near section D-​D, after termination of 2–#7 bottom bars: 2–​#4 (As = 0.40 sq in.) C = 294a    T = 2 (0.20)60 = 24 kips a = 0.082 in.   c = 0.10 in.

εt = 0.60 >>> 0.005

φ M n = 0.90(24)(20 − 0.041)

1 12

= 36 ft-kips

These moment capacities are compared later with a moment envelope (see Fig. 9.6.2), which shows that flexural strength requirements are met with the chosen reinforcement at all critical sections. (Continued)

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Example 9.4.1 (Continued) (b) Development lengths using the simplified equations, Eqs. (6.6.4) through (6.6.7); Tables 6.6.1 and 6.6.2 can be used. The development lengths are Bars

Location

Category A

Category B

#4 #5 #7

3 (bottom) 1 and 2 (bottom) 4, 5, and 6 (top) 7, 8, and 9 (bottom) 5, 6, 10, and 11 (top)

19.0 in. 23.7 in. 54.0 in. 41.5 in. 61.6 in.

28.5 in. 35.6 in. 81.0 in. 62.3 in. 92.6 in.

#8

where a modification factor for the casting position, ψt = 1.3, is applied for the bars located at the top of the beams. In this example, nearly all the bars meet Category A, and thus, the values shown in the table could be used for determining bar cutoff locations. As noted in Chapter 6, however, the simplified equations often result in development lengths greater than those required by the general equation, Eq. (6.6.1). Although the use of Eq. (6.6.1) requires additional calculations and effort, it often yields shorter development lengths, which could result in important cost savings due to the use of less material, less congestion of reinforcement, and easier fabrication and erection. (c) Development lengths using the general equation, Eq. (6.6.1) i. Development lengths for the #5 bottom bars at locations 1 and 2 (Fig. 9.4.1). The proposed layout of the reinforcement at these locations is shown as section A-​A and section C-​C in Fig.  9.4.2. Here the 2–​#5 extend into the support after the 2–​#7 bars are cut off. Assume that the #5 bars are fully developed between the supports into the span before the cutoff location of the #7 bars. Thus, the 2–​#5 are treated as if alone in a 13-​in.-​wide stem enclosed within at least minimum #3 U stirrups at these locations. The general equation, Eq. (6.6.1), is reproduced below



   f ψ t ψ e ψ s  3 y Ld =  d [6.6.1]  40 λ fc′  cb + K tr   b   d     b

but no less than 12 in. The cover or spacing dimension cb is the smaller of (1) distance from center of bar being developed to nearest concrete surface, and (2) half the center-​to-​center spacing of bars being developed. Thus, the distance cb will be the smaller of the following two values: top and side cover = 1.5 (i.e., clear) + 0.375 (i.e., stirrup) + 0.313 (i.e., bar radius) = 2.19 in.      Controls! 1 1 c-c spacing = [13 − 2(1.5) − 2(0.375) − (0.625)] = 4.3 in. 2 2 If there were no stirrups (which in fact there are), Ktr = 0 and (cb + Ktr) /​db = (2.19 + 0)/​0.625 = 3.5 > 2.5 max Thus,  3 60, 000 1.0(1.0)1.0  Ld =   0.625 = 17.8 in. (1.5 ft)  40 1.0 4000 2.5 where the modification factors ψt, ψe, and λ are all equal to 1.0. Although a bar size factor of ψs = 0.8 is permitted by the ACI Code, it was not applied here, in conformity to the recommendations of ACI Committee 408 [6.53], as discussed in Section 6.7. This is conservative. (Continued)

281



9.4  CHOICE OF LONGITUDINAL REINFORCEMENT IN BEAMS

281

Example 9.4.1 (Continued) ii. Development length for the #4 bottom bars at location 3 extending into the interior support (Fig. 9.4.1). The proposed layout of the reinforcement is shown as section D-​D in Fig. 9.4.2. Assuming that the 2–​#4 are fully developed before the #7 bars are cut off, they are then treated as if alone in a 13-​in.-​wide stem enclosed within at least minimum #3 U stirrups. The distance cb is the smaller of the following two values: top and side cover = 1.5 (i.e., clear) + 0.375 (i.e., stirrup) + 0.25 (i.e., bar radius) = 2.13 in.      Controls! 1 1 c-c spacing = [13 − 2(1.5) − 2(0.375) − (0.50)] = 4.38 in. 2 2 If there were no stirrups (Ktr = 0), then (cb + Ktr)/​db = (2.13 + 0)/​0.5 = 4.3 > 2.5 max The general equation using the maximum (cb + Ktr)/​db of 2.5 gives  3 60, 000 1.0 (1.0 )1.0  Ld =   0.5 = 14.2 in. (1.2 ft ) 2.5  40 1.0 4000 where no bar size reduction factor was applied (i.e., ψs = 1.0). iii. Development length for the #8 top bars extending from the interior support into beams 2B1 and 2B2 (near locations 10 and 11 in Fig. 9.4.1). Although the #8 bars are within the flange, they will be treated as if alone in a 13-​in.-​wide stem enclosed within at least minimum #3 U stirrups (see sections E-​E and F-F in Fig. 9.4.2). The distance cb is the smaller of the following two values: top and side cover = 1.5 (i.e., clear) + 0.375 (i.e., stirrup) + 0.50 (i.e., bar radius) = 2.38 in.      Controls! 1 1 c-c spacing = [13 − 2(1.5) − 2(0.375) − (1.00)] = 4.13 in. 2 2 Compute cb + K tr 2.38 + 0 = = 2.38 (without stirrup contribution) < 2.5 max db 1.00 The general equation, taking ψt = 1.3 for the casting position (i.e., more than 12 in. of fresh concrete cast below the bar), gives  3 60, 000 1.3(1.0)1.0  1.00 = 38.9 in. (3.2 ft) Ld =   40 1.0 4000 2.38  iv. Development length for the 2–​#7 top bars at location 4 (Fig. 9.4.1) extending into the exterior support. The 2–​#7 are treated as if alone enclosed within at least minimum #3 U stirrups in a 13-​in.-​wide stem. The distance cb is the smaller of the following two values (section A-​A of Fig. 9.4.2): top and side cover = 1.5 (i.e., clear) + 0.375 (i.e., stirrup) + 0.438 (i.e., bar radius) = 2.31 in.      Controls! 1 1 c-c spacing = [13 − 2(1.5) − 2(0.375) − (0.875)] = 4.19 in. 2 2 (Continued)

28

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Example 9.4.1 (Continued) Compute cb + K tr 2.31 + 0 = = 2.64 (without stirrup contribution) > 2.5 max db 0.875 The general equation taking ψt = 1.3 for the casting position (i.e., more than 12 in. of fresh concrete cast below the bar) gives  3 60, 000 1.3(1.0)1.0  Ld =   0.875 = 32.4 in. (2.7 ft)  40 1.0 4000 2.5 To develop properly the 2–​#7 bars into the support a straight embedment of 32.4 in. is required, which exceeds the width of the support (girder) of 18 in. Thus, use a standard 90° hook. For a #7 standard hook, the development length Ldh, according to Eq. (6.10.1) (ACI-​25.4.3.1), is  fy ψ e ψ c ψ r  Ldh =   db [6.10.1]  50 λ fc′ 



but not less than 8db = 8(0.875) = 7 in. nor less than 6 in. Assume concrete cover of at least 2 1 2 in. normal to the plane of the hook, and assume that cover on the bar extension beyond the 90° hook is at least 2 in., which would satisfy Condition 1 of Table 6.10.1. Thus, ψc = 0.7 and Ldh becomes

 60, 000(1.0)(0.7)1.0  Ldh =  0.875 = 11.6 in. > 6 in.  50(1.0) 4000 

where ψr has been assumed to be 1.0. The available Ldh (see Fig. 6.10.3) is available Ldh = 18(support width ) − 2 (cover on tail) = 16 in. > 11.6 in.

OK

OK

Thus, the 2–​#7 bars are assumed to be fully effective at point 4 of Fig. 9.4.1 when 90° hooks are used. Since a girder frames in at each side of beam 2B1, there will automatically be at least 2 1 2 -​in. side cover on the hooks. v. Development length for the 2–​ #7 bottom bars near locations 7, 8, and 9 (Fig. 9.4.1). Assume that the 2–​#7 are cut as soon as permitted, and that the 2–​#5 (or 2–#4) are extended into the support. The proposed layout of the reinforcement for section B-​B of beam 2B1 is shown in Fig. 9.4.2. Assuming the 2–​#5 (or 2–#4) are fully developed before the 2–​#7 are cut off, then the calculation of the clear spacing is based on the 2–​#7 bars only. Since the 2–​#7 bars are placed toward the center of the cross section with the 2–​#5 (or 2–#4) bars in the corners, side cover will not govern. Thus, the distance cb is the smaller of the bottom cover or half the center-​to-​center spacing between the 2–​#7 bars as follows: bottom cover = 1.5 (i.e., clear) + 0.375 (i.e., stirrup) + 0.438 (i.e., bar radius) = 2.31 in. The clear spacing between bars may be obtained from the geometry of the cross section with the aid of Table 3.9.3. From this table, the minimum width for a beam with 2–​#5 bars and 2db clear spacing is 7.1 in. For a beam with 2–​#5 and 2–​#7, the clear spacing between bars may be computed as follows: clear spacing =

1 {13 − [7.1 − 2(0.625)] − 2(0.875)} = 1.8 in. or 2.11db 3 (Continued)

283



9.4  CHOICE OF LONGITUDINAL REINFORCEMENT IN BEAMS

283

Example 9.4.1 (Continued) Therefore, the center-​to-​center spacing of the 2–​#7 bars is

1 1 c-c spacing = [1.8 + (0.875)] = 1.34 in. 2 2

Governs!

Assuming no contribution from stirrups, Ktr = 0 and (cb + Ktr)/​db = (1.34 + 0)/​0.875 = 1.53 < 2.5 max Thus,  3 60, 000 1.0(1.0)1.0  0.875 = 40.7 in. (3.4 Ld =  4 ft)  40 1.0 4000 1.53  Alternatively, if minimum #3 U stirrups spaced at d/​2 = 10 in. are assumed to restrain the opening of the splitting plane, Ktr will be greater than zero. Because the potential splitting plane is horizontal and two of the four bars are being developed, K tr =

40 Atr 40(2)(0.11) = = 0.44 sn 10(2)

and (cb + Ktr)/​db = (1.34 + 0.44)/​0.875 = 2.03 < 2.5 max Thus, the development length could be reduced to Ld = 40.7

1.53 = 30.7 in. (2.6 ft) 2.03

At location 9 (see section E-​E in Fig. 9.4.2) where 2–#4 bars instead of 2–#5 bars are used, it is reasonable and practical to assume that the center-​to-​center spacing of the #7 bars at the bottom of beam 2B2 is the same as that computed for locations 7 and 8 above. Thus, use the same development length at all three locations. vi. Development length for the 2–​#7 top bars near locations 5 and 6 (Fig. 9.4.1). Assume that the 2–​#7 are cut as soon as permitted, and that the 2–​#8 are extended into the spans of beams 2B1 and 2B2. The proposed layout of the reinforcement for sections C-​C of beam 2B1 and D-​D of beam 2B2 is shown in Fig. 9.4.2. Assuming that the 2–​#8 are fully developed before the 2–​#7 are cut off, the calculation of the clear spacing is based on the 2–​#7 bars only. Since the 2–​#7 bars are placed toward the center of the cross section with the 2–​#8 bars in the corners, side cover will not govern. Thus, the distance cb is the smaller of the top cover or one-​half the center-​to-​center spacing between the 2–​#7 bars as follows: top cover = 1.5 (i.e.,clear ) + 0.375 (i.e.,stirrup) + 0.438 (i.e., bar radius) = 2.31 in. The clear spacing between bars may be obtained from the geometry of the cross section with the aid of Table 3.9.3. From this table, the minimum width for a beam with 2–​#8 bars and 2db clear spacing is 8.3 in. For a beam with 2–​#8 and 2–​#7 bars, the clear spacing between bars may be computed as follows: clear spacing =

1 {13 − [8.3 − 2(1.00)] − 2(0.875)} = 1.65 in. or 1.9db 3

Therefore, one-​half of the center-​to-​center spacing of the 2–​#7 bars is

1 1 c-c spacing = [1.65 + (0.875)] = 1.26 in. 2 2

governs! (Continued)

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Example 9.4.1 (Continued) Assuming no contribution from stirrups, Ktr = 0 and (cb + Ktr)/​db = (1.26 + 0)/​0.875 = 1.44 < 2.5 max Thus,  3 60, 000 1.3(1.0)1.0  0.875 = 56.2 in. (4.7 Ld =  7 ft)  40 1.0 4000 1.44  where ψt = 1.3 is applied to account for the casting position of the bar. Alternatively, if minimum #3 U stirrups spaced at d/​2 = 10 in. are assumed to restrain the opening of the splitting plane, Ktr is computed under the assumption that the potential splitting plane is horizontal and that two of the four bars are being developed as follows: K tr =

40 Atr 40(2)(0.11) = = 0.44 sn 10(2)

and (cb + K tr ) / db = (1.26 + 0.44) / 0.875 = 1.94 < 2.5 max Thus, the development length could be reduced to Ld = 56.2

1.44 = 41.7 in. (3.5 ft) 1.94

Based on the general equation, Eq. (6.6.1), the development lengths at the various locations are Bars #4 #5 #7 #8

Location 3 (bottom) 1 and 2 (bottom) 4 (top) 5 and 6 (top) 7, 8, and 9 (bottom) 5, 6, 10, and 11 (top)

Ld (in.) 14.2 17.8 32.4 41.7 30.7 38.9

A comparison of these values with those computed using the simplified equations (Categories A and B) in part (b) reveals that the simplified equations can be very conservative. For example, for the #8 bars, the simplified equations (Category A) require a development length that is about 60% greater than that required by the general equation. Also note that for the #4 and #5 bars, a bar size factor ψs = 1.0 was used in the general equation, while the simplified equations already include ψs = 0.80. (d) Crack control requirements. At this point it is useful to check the distribution and spacing between bars for crack control in accordance with the serviceability requirements of ACI-24.3. For the purposes of this example it will be assumed that the provided reinforcement satisfies the requirements for crack control. Crack control provisions are presented in Chapter 12.

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9.5  SHEAR REINFORCEMENT IN BEAMS

9.5 SHEAR REINFORCEMENT IN BEAMS The design of shear reinforcement in beams 2B1-​2B2-​2B1 is shown in the following example.

EXAMPLE 9.5.1 Design the shear reinforcement in the floor beams 2B1-​2B2-​2B1 of Example 9.4.1. Use fc′ = 4000 psi and f y = f yt = 60, 000 psi. SOLUTION The factored shear Vu envelope for which shear reinforcement will be provided is taken from Figs. 9.3.3 and 9.3.4 and shown in Fig. 9.5.1. For these beams, the maximum factored shear may be taken at a distance equal to the effective depth from the face of support. Let V1, V2, and V3 be the maximum factored shear in regions 1, 2, and 3 as shown in Fig. 9.5.1; then using wu = 3.45 kips/​ft, 26’–0”

13’–0” 15.00’

40.7

V1 = 32.4 Region 1

47.7 d 1.67’

12.58’ or 151.0”

Symmetrical about CL V3 = 39.4 Region 3

x2 = V, kips

φVc = 24.7

43.0”

φVc/2

x1 = 26.8”

65.1”

φVc/2

φVc/2

43.0”

x3 = 51.1”

9.38’

0.75’

or 112.6”

d 1.67’

d 1.67’

Region 2 V2 = 43.4

11.80’

43.0” 11.41” or 136.9”

13.83’

51.8

2B1

2B2

Figure 9.5.1  Factored shear Vu envelope used for 2B1 and 2B2 in Example 9.5.1.

V1 = 3.45(9.38) = 32.4 kips V2 = 3.45(12.58) = 43.4 kips V3 = 3.45(11.41) = 39.4 kips



The concrete contribution to shear strength, φVc, is

(

)

1 φ Vc = φ 2λ fc′ bw d = 0.75 2(1.0) 4000  (13)(20) 1000 = 24.7 kips

Thus, the distance from the location where V1, V2, and V3 occur to the location where Vu = φVc is (see Fig. 9.5.1):  32.4 − 24.7  x1 = 112.6   = 26.8 in.  32.4 

 43.4 − 24.7  x2 = 151.0   = 65.1 in.  43.4   39.4 − 24.7  x3 = 136.9   = 51.1 in.  39.4 

(Continued)

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Example 9.5.1 (Continued) The regions spanning from the face of the support to the location where Vu = φVc  /​2, which represent the portions of the beam where shear reinforcement is required, are drawn from Fig. 9.5.1 as regions 1, 2, and 3 in Fig. 9.5.2. Assume #3 vertical U stirrups. From Fig. 9.5.2 the maximum required φVs anywhere on the beam is (43.4 –​24.7) = 18.7 kips. As shown below, the largest required φVs does not exceed that based on a nominal stress vs of 4 fc′,

limit φ Vs = φ (4 fc′bw d ) = 2(φVc ) = 2(24.7) = 49.4 kips



max required φ Vs = 18.7 kips < limit φVs Thus, the upper limit on stirrup spacing (Eq. 5.10.19) is d/​2 = 10 in. (< 24 in.) For a minimum percentage of shear reinforcement (Eq. 5.10.13), min φVs = φ 0.75 fc′ bw d ≥ φ (50)bw d = 0.75(0.75 4000 )(13)(20)

≥ 0.75(50)(13)(20)

1 1000

[5.10.13]

1 1000

= 9.2 kips

= 9.8 kips

Thus, the 9.8-kip limit controls. Face of support Face of support

Required φVs = Vu – φVc

13 @ 10”, 1 @ 5” Vu – φV c , kips

7.7

65.1” 20”

14.7

128.1”

26.8”

20” 20”

90”

82.6”

18.7

1 @ 5”, 9 @ 10”

51.1”

1 @ 5”, 8 @ 10”

Required φVs = Vu – φVc (a) Region 1

(b) Region 2

(c) Region 3

Figure 9.5.2  Portions of factored shear Vu in excess of φVc (i.e., required φVs diagram) for beams 2B1 and 2B2 in Example 9.5.1.

The strength requirement to provide for the shear represented by the shaded areas of Fig. 9.5.2 is Vs = or s=

Av f y d s

φ Av f y d φVs

=

0.75(0.22)(60)20 198 kips = in. φVs φVs

For region 1, a spacing of d/​2 governs because the maximum required φVs = 7.7 kips is less than the 9.8 kips required for minimum stirrups, and φVs = 9.8 kips permits spacing of 20.2 in., which exceeds d/​2. (Continued)

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Example 9.5.1 (Continued) Use 1 @ 5 in. and 9 @ 10 in. (95 in. from face of support). For region 2, the required spacing for the maximum required φVs of 18.7 is 198/​18.7 = 10.6 in., which is larger than the maximum permitted spacing of d/​2 = 10 in. Use 1 @ 5 in. and 13 @ 10 in. For region 3, to satisfy the strength requirement at the critical section,

max s =

198 = 13.5 in. 14.7

Thus, d/​2 = 10 in. controls. Use 1 @ 5 in. and 8 @ 10 in. As a practical matter, many designers would place stirrups by scaling from the factored shear envelope (in terms of force or unit stress), rather than accurately compute distances as illustrated here. Further, this example uses the simplified procedure of ACI-​ 22.5.5.1 for the value of Vc. It is noted that stirrup spacing will be modified on both beams near the interior support based on development length requirements, as shown in the next section.

9.6 DETAILS OF BARS IN BEAMS For typical conditions of equal spans and uniform load where moment and shear coefficients of ACI-​6.5.2 are used, standard bar details such as provided in the ACI Detailing Manual—​2004 [2.23] may be used. When the moment and shear envelopes are available, cutoff locations and embedment lengths into the supports should be determined therefrom, as illustrated shortly in Example 9.6.1.

Structural Integrity Reinforcement In addition to satisfying the basic structural safety requirement [design strength ≥ required strength: (Eq. 2.6.1)], the ACI Code has reinforcement and detailing requirements aimed at enhancing the overall integrity (i.e., redundancy and ductility) of the structure, in the event of severe damage to a major supporting element. Such requirements are often referred to as structural integrity requirements. For beams located along the perimeter of the structure, a minimum amount of structural integrity reinforcement must be provided at supports as follows (ACI-​9.7.7.1 and ACI-9.7.7.4): (a) At least one-​quarter of the maximum positive moment reinforcement, but not less than two bars, must be continuous. (b) At least one-​sixth of the negative moment reinforcement at the support, but not less than two bars, must be continuous. (c) At noncontinuous supports, structural integrity reinforcement must be anchored at the face of the support to develop fy. This longitudinal integrity reinforcement must be enclosed by closed stirrups in accordance with ACI-​25.7.1.6 (see Fig. 9.6.1) or hoops along the clear span of the perimeter beams. For nonperimeter beams, the longitudinal reinforcement must be enclosed by closed stirrups in accordance with ACI-​25.7.1.6 or by hoops along the clear span of the beam (ACI-​9.7.7.2). Alternatively, at least one-​quarter of the maximum positive moment reinforcement, but not less than two bars, must be continuous and, at noncontinuous supports,

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135-degree hooks (a)

U stirrup with 135-degree hooks (b)

Figure 9.6.1  Details of closed stirrups used for enclosing the longitudinal structural integrity reinforcement (ACI-​25.7.1.6) (a) Single piece with ends terminating with 135° standard hooks and (b) two-​piece stirrup made of a U stirrup with 135° hooks and a cap tie.

be anchored to develop fy at the face of the support, as for perimeter beams. This requirement will usually require a standard hook or a headed bar. For beams or girders supported on columns or walls, the structural integrity reinforcement is required to pass through the region bounded by the column (or wall) reinforcement (ACI-​9.7.7.3). Also, the structural integrity reinforcement may be spliced if necessary. In such a case, splices must satisfy the requirements of ACI-​9.7.7.5 and 9.7.7.6.

EXAMPLE 9.6.1 Determine the lengths of the main reinforcing bars in 2B1 and 2B2 of Example 9.4.1. SOLUTION (a) In determining bar cutoff locations, it is necessary first to locate the theoretical points where the bars are no longer required. Lines representing full-​strength φMn of the various bar combinations are computed and compared with the factored moment Mu, as shown in Fig. 9.6.2. (b) Extension of positive moment reinforcement into supports. To satisfy ACI-​ 9.7.3.8.2, the #5 straight bars in 2B1 must extend at least 6 in. into the supports. At the first interior support, the 6-​in. minimum embedment is sufficient, since there is no computed tensile requirement for these bars unless the bars are (1) to be used as compression reinforcement or (2) are required for “structural integrity” under ACI-​ 9.7.7. In this design, beam 2B1 is not a perimeter beam and the bottom bars are not utilized as compression reinforcement. While #3 U stirrups will be provided, they must be closed by using, for example, the details shown in Fig. 9.6.1. In such a case, an extension greater than the 6-​in. minimum is not required. However, since the extra cost of providing more ductility and tying the structure together is minimal, the writers recommend that, in general, the positive moment bars required to extend into the support be capable of developing fy at the face of support, a property that often requires a Class A splice when the bars are not continuous through the support. At the exterior support, the 6-​in. embedment will not be adequate if the support is a column and the flexural member is part of the primary lateral load resisting system. ACI-​ 9.7.3.8.2 requires development of the full yield stress in tension at the face of support for such cases. In this example, the supporting member is a girder and is not a part of (Continued)

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Example 9.6.1 (Continued) the primary lateral load resisting system; thus a 6-​in. embedment is sufficient, provided that closed stirrups are used along the clear span. Similarly, the 2–​#4 bars in beam 2B2 are extended into the supports 6 in. The total length of the #4 and #5 straight bars is [24 ft 6 in. + 2(6 in.)] = 25 ft 6 in. (Bar lengths are usually specified in 3-​in. increments.) (c) Theoretical cutoff locations. In establishing the cutoff for the #7 bottom bars near sections A-​A and C-​C in beam 2B1, and near section D-​D in beam 2B2, the distances x1, x2, and x4, as marked in Fig. 9.6.2 [also shown as x1, x2 in Fig. 9.3.3(a) and as x4 in Fig. 9.3.4(a)] must be computed. Thus, referring to Fig. 9.3.3(a),

3.45(11.05 − x1 )2 3.45(13.45 − x2 )2 = = 159 − 56 = 103 ft-kips 2 2

which gives:  x1 = 3.31ft;

and

x2 = 5.72 ft.

And, from Fig. 9.3.4(a), 3.45(12.25 − x4 )2 = 121 − 36 = 85 ft-kips 2 x4 = 5.23 ft



In accordance with ACI-​9.7.3.3, the reinforcement must be extended a distance of 12 bar diameters or the effective depth of the member, whichever is larger, beyond the point at which it is no longer needed to resist flexure. In this case, the effective depth of the member, 1.67 ft, controls for all bar sizes. Thus, the cutoff locations (from the face of the support) for the #7 bottom bars could be: cutoff based on x1 = 3.31 − 1.67 = 1.64 ft

cutoff based on x2 = 5.72 − 1.67 = 4.05 ft cutoff based on x4 = 5.23 − 1.67 = 3.56 ft

Similarly, to determine the cutoff locations for the #7 top bars near section C-​C in beam 2B1 and near section D-​D in beam 2B2, the distances x3 and x5, as marked in Fig.  9.6.2 [x3 in Fig.  9.3.3(b) and x5 in Fig.  9.3.4(b)] are computed as follows. From Fig. 9.3.3(b), 3.45(14.25 − x3 )2 = 142 + 135= 277 ft-kips 2 x3 = 1.58 ft



cutoff based on x3 = 1.58 + 1.67 = 3.25 ft And, from Fig. 9.3.4(b),



3.45(13.08 − x5 )2 = 104 + 135 = 239 ft-kips 2 x5 = 1.31 ft cutoff based onn x5 = 1.31 + 1.67 = 2.98 ft

The distances x1 through x5 with the effective depth d = 1.67 ft either added or subtracted locate correctly the theoretical cutoff points. However, additional requirements for bars terminated in a tension zone may require an extension beyond the calculated cutoff locations above. These requirements are checked in the next section. (Continued)

(Continued)

Figure 9.6.2  Moment envelope, φMn diagram, and bar arrangement for beams 2B1 and 2B2 of Example 9.2.1, as treated in Example 9.6.2 (all moments Mu and φMn are given in ft-​kips). Stirrups are not shown for clarity; LL = live load.

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Example 9.6.1 (Continued) (d) Check cutoff location for the 2–​#7 bottom bars in beam 2B1. The proposed cutoff locations at x1 –​ d (1.64 ft) and at x2 –​ d (4.05 ft) are both in a tension zone [see Figs. 9.3.3(a) and 9.6.2]. Therefore, ACI-​9.7.3.5 must be satisfied for the cutoff to be acceptable at these locations. From Fig. 9.6.2, it is clear that the continuing bars (2–​#5) provide more than double the area required for flexure at the potential cutoff point. Thus, it is necessary to check only whether the factored shear Vu exceeds three-​fourths of the shear strength φVn [Eq. (6.11.3)]. At x1 –​ d (= 1.64 ft) [see Fig. 9.3.3(a)]: Vu = wu (11.80 − 0.75 − x1 + d )



Vu = 3.45(11.80 − 0.75 − 1.64) = 32.5 kips

It is required that



Vu ≤ [0.75φVn = 0.75φ (Vc + Vs )] [6.11.3]

The strength provided, including stirrups, is

φ Vc + φ Vs = 24.7 +

198 = 44.5 kips 10

Vu 32.5 = 73% < 75% permitted = φVn 44.5

OK

Had the requirement not been satisfied, the stirrup spacing could have been reduced so that Vs would increase. Alternatively, the 2–​#7 bars may be extended to the inflection point, where they will no longer be in the tension zone. There is little material saved by cutting so close to the support, but the case serves to illustrate the procedure. At x2 –​ d (= 4.05 ft) [see Fig. 9.3.3(a)]: Vu = wu (14.20 − 0.75 − x2 + d ) Vu = 3.45(14.20 − 0.75 − 4.05) = 32.4 kips Since the strength provided by the stirrups is the same as that at x1 –​ d, it is clear that the requirement is also satisfied at this location. (e) Check cutoff for 2–​#7 bottom bars in beam 2B2. At the proposed location, x4 –​ d (3.56 ft), the bars terminate in the compression zone [see Figs. 9.3.4(a) and 9.6.2]. Therefore, the proposed locations is acceptable. (f) Check cutoff location for the 2–​#7 top bars in section C-​C of beam 2B1. The proposed cutoff locations at x3 + d (3.25 ft) is a tension zone (see Fig. 9.6.2). Check whether the continuing bars (2–​#8) provide at least double the area required for flexure at the potential cutoff point (Fig. 9.6.2). At the cutoff location, it is required that

φ M n (2 −# 8) ≥ 2 Mu at (x3 + d )



135 ft-kips ≥ 2(67. 4) = 134.8 ft-kips

OK

Therefore, it is necessary only to check whether the factored shear Vu exceeds three-​ fourths of the shear strength φVn (Eq. 6.11.3). At x3 + d (= 3.25 ft) [see Fig. 9.3.3(b)]:

Vu = wu (15.00 − 0.75 − x3 − d ) Vu = 3.45(15.00 − 0.75 − 3.25) = 38.0 kips

(Continued)

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Example 9.6.1 (Continued) Thus, Vu 38.0 = 85% > 75% permitted = φ Vn 44.5



NG

In this case, two possible solutions are as follows: (a) extend the 2–​#7 bars to a point where the required shear, Vu, is equal to or less than 0.75(44.5) = 33.4 kips (i.e., at 4.57 ft from the face of the support), or (b) reduce the stirrup spacing so that Vs increases in this region. In this example, alternative (b) is chosen to illustrate the procedure. Try reducing stirrup spacing to 6 in.

φ Vc + φ Vs = 24.7 +

198 = 57.7 kips 6

Vu 38.0 = = 66% < 75% permitted φ Vn 57.7



OK

Thus, make the cutoff at x3 + d = 3.25 ft, but use a 6-​in. stirrup spacing between the face of the support and the cutoff location. It is good practice to extend stirrups beyond the exact cutoff location. (g) Check cutoff location for the 2–​#7 top bars in section D-​D of beam 2B2. The proposed cutoff locations at x5 + d (2.98 ft) is in a tension zone (see Fig. 9.6.2). Check whether the continuing bars (2–​#8) provide at least double the area required for flexure at the potential cutoff point. At the cutoff location,

φ M n (2 −# 8) = 135 ft-kips < 2 Mu at (x 5 + d ) = 2(72. 2) = 144. 4 ft-kips

NG

Therefore, the cut cannot be made at this location unless the factored shear Vu at the cutoff point does not exceed two-​thirds of the design shear strength φ Vn [(ACI-​9.7.3.5(a)]. At x5 + d (= 2.98 ft) (see Fig. 9.3.4(b)]:

Vu = wu (13.83 − 0.75 − x5 − d ) Vu = 3.45(13.83 − 0.75 − 2.98) = 34.8 kips



It is required that 2 2  Vu ≤  φ Vn = φ ( Vc + Vs ) 3 3 

Vu 34.8 = 78% > 67% permitted = φ Vn 44.5

[6.11.1] NG

Again, two possible solutions are suggested: (a) extend the 2–​#7 bars to a point where the required shear, Vu, is equal to or less than 0.67(44.5) = 29.8 kips or (b) reduce the stirrup spacing so that Vs increases in this region. Using a stirrup spacing of 6 in.,

φ Vc + φ Vs = 24.7 +

198 = 57.7 kips 6

Vu 34.8 = = 60% < 67% permitted φ Vn 57.7

OK

Thus, make the cutoff at x5 + d = 2.98 ft, but use a 6-​in. stirrup spacing between the face of the support and the cutoff location (good practice is to extend stirrups at 6 in. beyond the exact cutoff location). (Continued)

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Example 9.6.1 (Continued) (h) Inflection point extensions. Top bars (negative moment) ACI-​9.7.3.8.4 (see Section 6.11) requires that at least one-​third of the negative moment reinforcement at a support have an extension beyond the point of inflection a distance equal to the largest of one-​sixteenth of the clear span, the effective depth of the member, or 12 bar diameters. In this case, the effective depth of the member, 1.67 ft, controls. In members 2B1 and 2B2, the continuing bars (2–​#8) represent more than one-​third of the maximum negative moment reinforcement at the interior support (sections C-​C and D-​D). From Fig. 9.6.2 and Fig. 9.3.5(a), the cutoff point for the #8 bars in member 2B1 may be made at 7.77 (inflection point) + 1.67 = 9.44 ft from the face of the support. Make the cutoff at 9 ft 6 in. from the face of the support (see Fig. 9.6.2). In member 2B2, a negative moment of 2.1 ft-​kips is required at midspan [Fig. 9.3.6(a)]; make the #8 bars continuous in this span. At the end support, all of the negative moment reinforcement (2–​#7) is to be extended into the span. From Fig. 9.3.3(a), this reinforcement may be terminated at (2.20 –​0.75 + 1.67) = 3.12 ft. Extend the bars into the span a distance of 3 ft 3 in. from the face of the support. Bottom bars (positive moment) At the inflection points (see Section 6.14), ACI-​9.7.3.8.3(b) requires that Mn + La ≥ Ld Vu

At location 1 in Fig. 9.4.1

For 2–​#5, extending 6 in. beyond face of support [actual length from inflection point (see Fig. 9.3.3) = 2.20 –​0.25 = 1.95 ft], 56 = 62 ft-kips 0.90 Vu = 40.7 − 3.45(2.20) = 33.1 kips

Mn =

La = d = 1.67 ft < actual length 1.95 ft





Mn 62 + La = + 1.67 = 3.5 ft > Ld (# 5) = 1.5 ft. Vu 33.1

OK

At location 2 (Fig. 9.4.1) For 2–​#5, since Mn and Vu are identical to their values at location 1, this check is made by inspection. At location 3 (Fig. 9.4.1) For 2–​#4, 36 = 40 ft-kips 0.90 Vu = 44.9 − 3.45(4.61) = 29.0 kips

Mn =



Mn 40 + 1.67 = 3.05 ft > Ld (# 4) = 1.2 ft. + La = Vu 29.0

OK

(i) The summary of bar arrangement, lengths, dimensions, and design strength provided was shown in Fig. 9.6.2. Bar cutoff locations have been rounded off so that bar lengths are specified in 3-​in. increments. It is noted that the 2–​#7 top bars are (Continued)

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Example 9.6.1 (Continued)

not fully developed at the face of the interior support. However, the design strength exceeds the required flexural strength at that location, as it does everywhere else in the spans for both beams. Furthermore, it can be shown that the development length for these bars could be reduced to 3 ft (instead of 3.5 ft) when the reduction in stirrup spacing to 6 in., as computed in parts (f) and (g), is accounted for in the general equation for computing development length in part (c), item vi, of Example 9.4.1. Regardless of whether the reduction in stirrup spacing is included in the calculations, the bar arrangement and cutoff locations are acceptable. The cross sections for the final choice at each designated section of Fig.  9.6.2 were shown in Fig. 9.4.2.

(j) Stirrup development. For stirrup supports, 2–​#3 bars are provided in the central part of beam 2B1. The dimensions of the stirrups are shown in Fig. 9.6.3. Closed stirrups with ends terminating in 135-​degree hooks must be provided over the entire clear span in accordance with the structural integrity requirements (ACI-​9.7.7.2 and ACI-​25.7.1.6). To facilitate fabrication and erection of the steel reinforcement cage, the stirrup end closest to the slab may be terminated in a 90-​degree hook (Fig. 9.6.1) because potential spalling may be assumed to be restrained by the slab [ACI-​25.7.1.6(b)]. As an alternative to providing closed stirrups over the entire span, continuing reinforcement in the bottom of the beam could be made continuous over the interior supports (ACI-​ 9.7.7.2) and anchored to develop fy at the noncontinuous end support (ACI-​9.7.7.4). 10”

This end could terminate in a 90-degree hook because 19 12” spalling is restrained by the slab

3 8”

Figure 9.6.3  Stirrup details in beams 2B1 and 2B2.

9.7 SIZE OF GIRDER WEB For the purpose of determining the size of the girder web, the maximum negative moment at the exterior face of the first interior support may be used, taken as 0.8 of the maximum positive moment in a simple span with span length equal to the clear span of the girder and with loadings identical to those on the girder. The maximum shear at the same location may be taken as 1.20 times the reaction to the simple span described above. Note that because concentrated loads are involved, the coefficients of ACI-​6.5.2 cannot be used in lieu of an elastic analysis. The concentrated loads on the girder may be taken as half of the total factored dead and live loads on the clear span of the beams on both sides of the girder. The dead and live uniform load on the girder will be the weight of the concrete and the live load on the floor, respectively, both within the width of the girder. This loading transfer is a sufficiently good approximation, although it is probable that the slab weight and floor load on a narrow strip parallel and close to either edge of the girder would act on the girder as uniform load instead of being carried by the one-​way slab to the adjacent beams and then to the girder as concentrated loads.

295



295

9.7  SIZE OF GIRDER WEB

EXAMPLE 9.7.1 Design the floor girders 2G1-​2G2-​2G2-​2G1 as shown in the floor framing plan of Fig. 8.3.1. Assume a slab thickness of 5.5 in. Work through the choice of the size of the girder web in this example according to the ACI Code ( fc′ = 4000 psi; fy = 60,000 psi). SOLUTION The concentrated reactions from the beams using the 24.5-​ft clear span are dead load = 1.2(24.50)[(0.069)(13) + 0.25] = 33.7 kips

live load = 1.6(24.50)(0.100)(13) = 51.0 kips total load = 33.7 + 51.0 = 84.7 kipss



Assuming an 18 × 35 in. web, the uniform loads on the girder are uniform dead load = 1.2 [18(40.5)(0.150)]

1 144

= 0.91 kips/ft

uniform live load = 1.6(0.100)(1.5) = 0.24 kip/ ft uniform total load = 0.91 + 0.24 = 1.15 kips/ft



Maximum positive moment on a simple span (see Fig. 9.7.1) is

1 Mu = 84.7(12.25) + (1.15)(37.5)2 = 1240 ft-kips 8

The estimated maximum negative moment in the girder at the exterior face of the first interior support is 0.8(1240) = 992 ft-​kips. From Fig. 9.7.1, estimate

dneg = 40.50 − 4.75 = 35.75 in. dpos = 40.50 − 3.50 = 37 in.



The negative moment requirement is

required Rn =

Mu 992 (12, 000) = = 575 psi 2 φ bd 0.90(18)(35.75)2

From Fig. 3.8.1, ρ ≈ 0.011 which is less than ρtc = 0.0181 (see Table 3.6.1). Thus, the section is tension controlled and required As for (–​M) ≈ 0.011(18)35.75 = 7.08 sq in. For the positive moment requirement, using the coefficients of ACI-​6.5.2 to obtain the approximate proportion between positive and negative moments, 10 10 ( − M ) = (992) = 709 ft-kips 14 14 Mu 709(12) = = 4.77 sq in. required As ≈ φ f y (arm) 0.90(60)(≈ 33)

estimated (+ M ) ≈

This may well fit into one layer (5–​#9 require a beam width of 14.3 in. according to Table 3.9.2). For the shear requirement, max Vu ≈ 1.20 [84.7 + 1.15(18.75)] = 128 kips

vu =

128, 000 = 265 psi = 4.2 fc′ 0.75(18)(35.75) (Continued)

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Example 9.7.1 (Continued) The stem could be made smaller; with this relatively large depth, deflection is unlikely to be excessive. The 18 × 35 stem appears to be somewhat large, though it could certainly be used. In this case, because of the large depth, reduce the stem size to 18 × 29. Estimated effective depths become dneg = 29.75 in.



dpos = 31 in.

18 × 18 column

P

12’–3”

13’–0”

P



12’–3”

18 × 18 column

w

Clear span = 37’– 6” 202 ft-kips Simple beam moments 1035 ft-kips

3 14 ” clear (to allow placement of slab and beam reinforcement)

4.75”

5 12 ”

Measured to top of girder steel 34.5” (40.5” initially assumed) 2” clear

18”

3.5”

Figure 9.7.1  Loading information for floor girders 2G1 and 2G2 in Example 9.7.1.

revised girder weight =1.2 [18(34.5)(0.15)]

1 144

= 0.78 kip/ft

wu = 0.78 + 0.24 = 1.02 kips/ft 1 revised Mu (simple beaam) = 1038 + (1.02)(37.5)2 = 1217 ft-kips 8 required Rn =

vn at support = Use 18 × 29 stem section.

0.8(1217)(12, 000) = 814 psi < 913 psi 0.90(18)(29.75)2 (max. for tensioncontrolled sections) 124, 600 = 310 psi = 4.9 fc′ 0.75(18)(29.75)

OK

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9 . 8   C O N T I N U O U S F R A M E A N A LY S I S F O R G I R D E R S

297

9.8 CONTINUOUS FRAME ANALYSIS FOR GIRDERS Although floor girders subjected to large concentrated loads are structural members of common occurrence, the ACI Code makes no mention of moment coefficients for these cases. In the following example, elastic statically indeterminate structural analysis is used for 2G1-​2G2-​2G2-​2G1 (Figure 8.3.1).

EXAMPLE 9.8.1 By the use of an elastic statically indeterminate structural analysis, determine the shear and moment diagrams to be used in the design of girders 2G1 and 2G2 in Example 9.7.1 ( fc′ = 4000 psi; fy = 60,000 psi). SOLUTION As discussed in Section 9.3, there are various ideas regarding what constitutes the correct stiffness for continuous T-​sections. In accordance with the thought that the true stiffness is that of a span with a variable cross section, the effective flange width for an equivalent uniform moment of inertia section may be assumed to be twice the web width (Fig. 9.8.1). The centroid of the gross area of the T-​section is at y=



18(29)14.5 − 36(5.5)(2.75) 7025 = = 9.76 in. 18(29) + 36(5.5) 720

The gross moment of inertia Ig is 36(5.5)3 18(29)3 + − 720(9.76)2 = 79, 700 in.4 3 3 79, 700 = = 2040 in.4 /ft 39

Ig = K girder

If the size of the upper and lower columns were 18 × 18 in. and the column height were 15 ft, K col =



18(18)3 15

1 12

= 583 in.4 /ft

5 12 ”

36”

y

29”

18”

Figure 9.8.1  T-​section dimensions for stiffness computation in Example 9.8.1.

(Continued)

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Example 9.8.1 (Continued)

2G1

2G2

2G2

2G1

1st span

2nd span

3rd span

4th span

Figure 9.8.2  Equivalent frame model for 2G1-​2G2-​2G2-​2G1.

Analysis of the frame shown in Fig. 9.8.2 yields the girder end moments shown in Table 9.8.1 under the various loading conditions considered. The controlling moment envelopes for spans 2G1 and 2G2 are given, respectively, in Figs. 9.8.3 and 9.8.4. In practice, this composite diagram should be made instead of the individual moment and shear diagrams for each loading case.

a

TABLE 9.8.1  END MOMENTS OF GIRDERS 2G1-​2G2-​2G2-​2G1 IN EXAMPLE 9.8.1 Joint

A

Member

AB

B BA

C BC

CB

D CD

E

DC

DE

ED

+438

–​467

+153

+361

–​213

–​45

+186

–​362

+228

+346

–​205

–​43

+279

–​165

–​35

+253

–​402

+220

+171

–​353

+230

Dead Load Only M=

–​153

+467

–​438

+365

–​365

Live Load Only Spans 1 and 3 M=

–​230

+353

–​171

+137

–​306

Live Load Only Spans 1, 2, and 4 M=

–​177

+606

–​599

+230

–​91

Live Load Only Span 3 M=

–​8

–​40

+67

+214

–​353

Live Load Only Spans 2 and 3 M=

+35

+165

–​279

+566

–​566

Live Load Only Spans 1 and 4 M=

–​220

+402

–​253

–​123

+123

Live Load Only Spans 2 and 4 M= a

+45

+213

–​361

+306

–​137

Clockwise moments acting on member ends are taken as positive.

(Continued)

29



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9 . 8   C O N T I N U O U S F R A M E A N A LY S I S F O R G I R D E R S

Example 9.8.1 (Continued) A

B

3 – #11

C 3 – #11

3 – #9

2 – #10

2 – #9

3 – #10

2 – #10

B

A

C 39’–0”

742

LL spans 1 and 3

598

18” column

18” column

6.8’

3.0’ LL spans 2 and 4

A

B

LL span 3

108 11.0’

3.7’

383

427 LL spans 1, 2, and 4

680 820 1073

Figure 9.8.3  Moment envelope and longitudinal bar arrangement for girder 2G1 (span 1). D

E

3 – #11 3 – #11

2 – #9

4 – #9

4 – #9

F

4 – #9 st

2 – #9

4 – #9 E

F

D 39’–0”

557

514 18” column 7.2’

18” column 6.4’

LL spans 2 and 4

LL spans 1 and 4

LL spans 1 and 3 B

371 609 799

LL span 3 14.6’ LL spans 1, 2, and 4

11.6’ LL spans 2 and 3

C 242 502 671 931

1037

Figure 9.8.4  Moment envelope and longitudinal bar arrangement for girder 2G2 (span 2).

(Continued)

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Example 9.8.1 (Continued) 13’–0”

13’–0”

92.4

CL of support

93.2

79.9

Spans 1 and 3 with LL

9”

13’–0”

CL of support

Face of support A

B 12.4

25.7

9”

(a) Girder 2G1 Face of support

Spans 1, 2, and 4 with LL 115.7

114.9

110.2 102.5

122.7

18.0

123.5

4.7

B

C 12.1 Spans 2 and 3 with LL

(b) Girder 2G2

96.6 109.1

109.9

Figure 9.8.5  Shear envelopes for girders 2G1 and 2G2 (full-​span loadings only).

For investigating development length requirements at points of inflection, for checking cutoff acceptability in the tension zone, and for designing stirrups, an approximate shear envelope (only full-​span loadings) is used, as given in Fig. 9.8.5.

9.9 CHOICE OF LONGITUDINAL REINFORCEMENT IN GIRDERS As with the design of the floor beams, the choice of longitudinal reinforcement in girders depends on both the steel area and the development length requirements as dictated by the moment and shear envelopes such as shown in Figs. 9.8.3 through 9.8.5. Attention is called to ACI-​7.5.2.3, which is applicable when the main reinforcement in the slab is parallel to the longitudinal axis of the girder. For this situation, reinforcement perpendicular to the longitudinal axis of the beam must be provided in the top of the slab to carry the load on the portion of the slab acting effectively as the flange of the girder. The overhanging flange is assumed to act as a cantilever. The spacing of the bars may not exceed 5 times the thickness of the flange or, in any case, 18 in. (ACI-​7.7.2.4).

301



9.9  CHOICE OF LONGITUDINAL REINFORCEMENT IN GIRDERS

301

EXAMPLE 9.9.1 Choose the arrangement of the main reinforcement in the floor girders 2G1-​2G2-​2G2-​ 2G1 of Example 9.7.1. SOLUTION (a) Sections (rectangular sections) A-​A, C-​C, D-​D, and F-​F (Figs. 9.8.3, 9.8.4, and 9.9.1). Section A-​A: The moment at the face of the supports may be obtained directly from the bending moment diagrams or for preliminary design by subtracting ΔM = 1 3 V × (width of support) from the moment at the center of support. Using this approximation: 92.4(1.5) = 383 − 46 = 337 ft-kips 3 Mu 337(12, 000) 265 required Rn = = = psi φ bd 2 0.90(18)(30.7)2 Mu at face = 383 −



where d is taken as 30.7 in. (see Fig. 9.9.1). Either from Eq. (3.8.5) or from Fig. 3.8.1,

required ρ ≈ 0.0045

required As = 0.0045(18)30.7 = 2.49 sq in.

Sections C-​C and D-​D: The larger moment is at section C-​C, 122.7(1.5) = 1012 ft-kips 3 1012(12, 000) required Rn = = 885 psi 0.90(18)(29.1)2 Mu at face = 1073 −



From Fig. 3.8.1, ρ ≈ 0.0175, required As ≈ 0.0175(18)(29.1) = 9.2 sq in. Section F-​F:



109.1(1.5) = 931 − 55 = 876 ft-kips 3 876(12, 000) = 741 psi required Rn = 0.90(18)(29.6)2 Mu at face = 931 −

Using straight-​line approximation,

 741 required As ≈ 0.0175  (18)29.6 = 7.9 sq in.  873 

(b) Sections (T-​sections) B-​B and E-​E (Figs. 9.8.3, 9.8.4, and 9.9.1): Effective flange width bE is the smallest of: Ln (39 − 1.5)(12) = 18 + = 131 in. 4 4 2. bw + 16t = 18 + 16(5.5) = 106 in. 1. bw +

Controls

3. center-​to-​center spacing = 26(12) = 312 in. It is likely that the depth a of the rectangular stress distribution will be less than the flange thickness t. Estimate a = 2 in. (about one-​half flange thickness). (Continued)

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Example 9.9.1 (Continued) 3 14 ” clear

2 – #9 2 – #10

2.6” (approx)

18” (typical)

d = 29.1” (approx)

6 – #11 d = 30.6” (approx)

d = 30.7” (approx)

3 – #9

2 – #10 (Section C–C) 2 – #9 (Section D–D)

3 – #10

2’ clear Section A–A

Sections C–C and D–D

Section B–B

4 – #9

4 – #9

d = 29.6” (approx)

29” (typical)

d = 31.9” (approx)

5 12 ”

4 – #9

2 – #9 2.6 (approx)

Section E–E

Section F–F

Figure 9.9.1  Typical sections for 2G1 and 2G2 (refer to Figs. 9.8.3 and 9.8.4) for Example 9.9.1.

Section B-​B: estimated d = 30.6 in. required As =



Mu 742(12) = = 5.6 sq in. φ f y (arm) 0.90(60)(30.6 − 1)

Check: C = 0.85(4)(106)a = 360.4 a T = 5.6(60) = 336 kips; a = 0.93 in. 742(12) revised reequired As = = 5.5 sq in. 0.90(60)(30.6 − 0.47)



Section E-​E: estimated d = 31.9 in. required As =



557(12) = 4.0 sq in. 0.90(60)(31.9 − 1)

Check:



C = 360.4 a;

T = 4.0(60) = 240 kips

a = 0.67 in



557(12) revised required As = = 3.9 sq in. 0.90(60)(31.9 − 0.33) (Continued)

30



303

9.9  CHOICE OF LONGITUDINAL REINFORCEMENT IN GIRDERS

Example 9.9.1 (Continued) (c) Confirmation of the tentative arrangement of main reinforcement as summarized in Fig. 9.9.2 awaits the check of development of reinforcement, crack control, and deflection. Note in Fig. 9.9.2 that the requirements indicated in the compression zone at B and C are merely the minimums of one-​fourth of the positive moment steel required to satisfy ACI-​9.7.3.8.2. (As = 3.00) 3 – #9

(As = 8.00) 4 – #9 4 – #9

(As = 9.36) 6 – #11

2.5

A

2G1

9.2

5.5

1.4

NA NA B

3 – #10 2 – #9 (As = 5.81)

2G2

7.9

3.9

1.0

4 – #9 (As = 4.00)

C

Figure 9.9.2  Longitudinal reinforcement areas (sq in.) required and bars selected for girders 2G1 and 2G2 for Example 9.9.1.

The details of determining bar lengths are not shown; for the general arrangement, see Figs. 9.8.3 and 9.8.4. (d) Check design strength at critical sections based on the selected arrangement of steel reinforcement. (The effect of compression reinforcement is ignored.) At section A-​A, 3–​#9: C = 0.85(4)(18)a = 61.2 a

T = 3.0(60) = 180 kips

a = 2.94 in.

c = 3.46 in.

d−x 30.7 − 3.46 (0.003) = (0.003) = 0.0236 > 0.005 x 3.46 φ = 0.90

ε1 =



  1 φ M n = 0.90(180)(30.7 − 1.47) 12 = 395 ft-kips  > [ M u = 337 ft-kips]  

OK

At section C-​C, 6–​#11 (two layers): C = 61.2 a a = 9.18 in. dt = 34.5 − 3.25 −

T = 9.36(60) = 561.6 kips c =10.80 in. 1.41 = 30.5 in. (distance to the layer of reinforcement 2 closest to the tension face)

30.5 − 10.8 (0.003) = 0.0055 > 0.005; φ = 0.90 10.8   1 OK φ M n = 0.90(561.6)(29.1 − 4.6) 12 = 1032 ft-kips > [ Mu = 1012 ft-kips]   Similar calculations can be made for the rest of the sections. The results are summarized in Table 9.9.1. Note that all sections satisfy minimum reinforcement requirements. Also, they are all tension controlled, and thus φ = 0.9. εt =

(e) Check the development length requirement at the positive moment inflection points. From analysis of the girders under the various loading conditions considered (Example 9.8.1), the factored shear values at the inflection points are as follows: For span 1,

Vu (near left end ) = 89 kips (live load on spans 1 and 3)

Vu (near right end) = 114 kips (live load on spans 1, 2 and 4)

(Continued)

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Example 9.9.1 (Continued) TABLE 9.9.1  CALCULATED FLEXURAL STRENGTH FOR THE GIRDERS OF EXAMPLE 9.9.1 Section

As (sq in.)

d (in.)

εt

φMn (ft-​kips)

Mu (ft-​kips)

A-​A C-​C (D-​D) F-​F B-​B E-​E

3–​#9 (3.00) 6–​#11 (9.36) 8–​#9 (8.00) 2–​#9, 3–​#10 (5.81) 4–​#9 (4.00)

30.7 29.1 29.6 30.6 31.9

0.0236 0.0055 0.0066 0.0774 0.1184

395 1032 924 787 568

337 1012 876 742 557

OK OK OK OK OK

For span 2,

Vu (near left end ) = 106 kips (live load on spans 1,2 and 4) Vu (near right end) = 101 kips (live load on spans 2 and 3)



At the span 1 inflection points, 2–​#10 bars continue along the bottom of the girder. For 2–#10 bars. C = 0.85(4)(106)a = 360.4 a T = 2(1.27)(60) = 152.4 kips

a = 0.42 in. M n = 152.4 [31.9 − 0.5(0.42)]

1 12

= 402 ft-kips

The development length Ld for the 2–​#10 in the bottom of span 2G1 depends on the clear spacing of those bars. From Table 3.9.3, in order to satisfy 2db clear spacing, the minimum width is 9.1 in. Thus, Category A, item 2 of the simplified development length equations is satisfied. From Table 6.6.1,

Ld (for # 10) = 60.2 in.(5.0 ft )

The critical location is at the right inflection point, where maximum Vu occurs (i.e., Vu = 114 kips). Thus,

  Mn 402 + La = + 2.66 = 6.19 ft  > Ld = 5.0 ft  114   Vu

OK

where La = effective depth, d, governs for this situation (see Section 6.14). Use of the general equation, Eq. (6.6.1), shows the value of (cb + Ktr)/​db to be larger than 1.5, thus resulting in a shorter requirement for Ld. In this case, the double-check is not needed because the Ld requirement has been satisfied by the simplified equation. In span 2, 2–​#9 continue past the inflection point and into the support. For 2–​#9 bars, C = 0.85(4)(106)a = 360.4 a T = 2(1.00)(60) = 120 kips

a = 0.33 in. M n = 120 [31.9 − 0.5(0.33)]

1 12

= 317 ft-kips (Continued)

305



9.9  CHOICE OF LONGITUDINAL REINFORCEMENT IN GIRDERS

305

Example 9.9.1 (Continued) The development length Ld for the 2–​#9 in the bottom of span 2G1 near the left support depends on the clear spacing of those bars. From Table 3.9.3, in order to satisfy 2db clear spacing, the minimum width is 8.6 in. Thus, Category A, item 2 of the simplified development length equations is satisfied. From Table 6.6.1,

Ld (for # 9) = 53.5 in. (4.5 ft )

The critical location is at the left inflection point where Vu = 106 kips. Thus,  Mn  317 + La = + 2.66 = 5.7 ft  > Ld = 4.5 ft  106  Vu 



OK

(f) The required steel reinforcement in the top flange, perpendicular to the longitudinal axis of the girder (ACI-​7.5.2.3), may be computed approximately as follows (Fig. 9.9.3): 1 [0.069(1.2) + 0.10(1.6)] (3.67)2 = 1.64 ft-kips /ft 2 1.64(12, 000) required Rn = = 111 psi 0.9(12)(4.06)2 Mu =

Main slab steel, #4 typical

6” or 8”

wD = 69 psf; wL = 100 psf

44”

bE

18” Girder

d = 5.5” –0.75” cover –0.50” bar diam (#4) –0.19” bar rad (#3) 4.06”

44” #3 @ 11”

(Girder steel not shown)

Figure 9.9.3  Steel reinforcement across the top of girder for Example 9.9.1.

Observing Fig. 3.8.1, this Rn indicates ρ ≈ 0.0025, which is less than the minimum ρ for beams. Since this reinforcement is for a slab of uniform thickness, ACI-​7.6.1.1 applies. The minimum amount of reinforcement for Grade 60 deformed bars is 0.0018 bh. Thus, the steel required is the larger of the following: or

As = 0.0025 bd = 0.0025(12)4.06 = 0.12 sq in. As = 0.0018 bh = 0.0018(12)5.5 = 0.12 sq in.



The maximum spacing of this reinforcement is the lesser of 3 times the slab thickness (i.e., 3 × 5.5 = 16.5 in.) or 18 in. (ACI-​7.7.2.3). Use #3 @ 11-​in. spacing. (As = 0.12 sq in./​ft) (g) The girder longitudinal bar arrangement is shown in Figs. 9.8.3 and 9.8.4 and cross sections were shown in Fig. 9.9.1. The shear reinforcement and dimensions of longitudinal reinforcement are not shown, because little new information will be involved in their presentation.

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9.10 ONE-​W AY JOIST FLOOR CONSTRUCTION One-​way concrete joist floor construction (Fig. 9.10.1), sometimes called ribbed-​slab construction, consists of regularly spaced ribs monolithically built with a top floor slab and arranged to span in one direction. Such a system may also be designed as a two-​way system (waffle slab) according to the procedures for two-​way floor systems treated in Chapters 16 and 17. The dimensions of the one-​way joist system are usually such that only temperature and shrinkage reinforcement is required in the slab. The slab is usually in the range of 2- to 4-​in. thick but may occasionally be as much as 6 in. The ribs (joists) of at least 4-in. width are usually tapered [Fig. 9.10.1(b)] and are designed so that the clear spacing between adjacent ribs does not exceed 30 in. During construction, removable and reusable form fillers are inserted into spaces between the joists. Such fillers may be standard-​sized steel “pans” in 20-​or 30-​in. widths and depths of 6, 8, 10, 12, 14, 16, and 20 in. Sometimes form fillers are made from hardboard, fiberboard, glass-​reinforced plastic, or corrugated cardboard. Occasionally, permanent fillers are used consisting of lightweight or normal-​weight concrete blocks or clay tile blocks [see Fig. 9.10.1(c)]. To limit deflection on floor joist construction, the minimum depth requirements of ACI-​ 9.3.1.1 may be applied. Whenever excessive deflection may cause cracking or other adverse effects, deflections must be computed even if ACI-​9.3.1.1 has been satisfied. Concrete joist floor construction is referred to in ACI-​9.8. Some of the requirements are as follows. 1. The joists shall not be farther apart than 30 in. face to face. The ribs shall be not less than 4 in. wide and of a depth not more than 3.5 times the minimum width of the rib. 2. The vertical shells of permanent fillers in contact with the joists are permitted to be included in strength calculations involving shear or negative bending moment provided the filler material has a unit compressive strength at least equal to that of the concrete in the joists. In this case, the minimum slab thickness is 1 1 2 in. or 112 of the clear distance between joists, whichever is larger. 3. When removable forms or fillers having less compressive strength than required in item 2 are used, the thickness of the concrete slab shall not be less than 112 of the clear distance between ribs, nor less than 2 in. For more information on concrete joist construction, see Chapter  10 of Manual of Standard Practice [9.1], Chapter 4 of Rice, Hoffman, Gustafson, and Gouwens [9.2], and Chapter 8 of CRSI Design Handbook—​2008 [2.21].

12 1 20” or 30” (b)

12” or 16” (a)

Figure 9.10.1  Concrete joist floor construction.

(c)

307



307

9.11  DESIGN OF JOIST FLOORS

9.11 DESIGN OF JOIST FLOORS The design of concrete joist floors involves the slab, the joists, and the girders. Generally, temperature and shrinkage reinforcement is placed at right angles to the joists, and the concrete slab is treated as if it were of plain concrete. The short clear span between joists may be considered as being fixed at both ends. The joist itself may be designed as a floor beam having a rectangular section in the region of negative bending and a T-​section in the region of positive bending. The critical design moment curves for each span may either follow the ACI moment coefficients or be determined by a continuity analysis. Largely because of the interaction of the slab with the closely spaced joists, ACI-​9.8.1.5 permits the shear strength Vc provided by the concrete to be 10% higher than for regular beams. The girder is designed as a floor girder, but the load from the joists may be considered as being uniformly distributed along the span. In the case of joist floors over removable steel pans, tapered end forms are available that increase the effective joist width 2 in. on each side for 20-​in.-wide forms and 2 1 2 ​in. on each side for 30-​in.-​wide forms over a distance of 3 ft from the end. This increased width may be necessary to take the large shear or negative bending moment near the end of the span.

EXAMPLE 9.11.1 Design the typical interior span of a concrete joist floor, using 30-​in.-​wide forms, for a center-​to-​center span of 26 ft and clear span of 24 ft 6 in. The live load is 80 psf, and the additional superimposed dead load is 10 psf. Use normal-weight concrete with fc′ = 3000 psi and fy = 60,000 psi. Use the moment and shear coefficients of ACI-​6.5.2. SOLUTION (a) Slab design. Assume a 3-​in. slab for weight estimation purposes. The factored load is

wu = 1.2(3.0)(0.15)

1 12

+ 1.2(0.010) + 1.6(0.080) = 0.185 ksf

Assuming that the slab is fixed at its junction with the joist,

Mu ≈

1 1 (0.185)(30)2 = 0.096 ft-kip/ft 144 12

The nominal flexural strength Mn of a plain concrete member in accordance with ACI-​14.5.2.1 is M n = 5λ fc′Sm (9.11.1)



where Sm is the elastic section modulus. Using a strength reduction factor φ of 0.60 for plain concrete (ACI-​21.2.1), the design strength, φ Mn, of the slab is  bh 2  φ M n = 0.60(5)(1.0) 3000   6 



Setting φ Mn = Mu and b = 12 in. for a typical 1-​ft strip, the required slab thickness, h, is found.

h=

0.096(12, 000) = 1.87 in., say, 2 in. 1 (0.60) 5(1.0) 3000 12 6

(

)

(Continued)

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Example 9.11.1 (Continued) Thus, a 3-​in. slab will be thick enough; perhaps 2 1 2 in. would be sufficient. Use h = 3 in. Shrinkage and temperature reinforcement must be provided per ACI-​9.8.1.7 in an amount equal to (ACI-​24.4.3.2) As = 0.0018(12)(3.0) = 0.065 sq in./ft Use welded wire reinforcement, 4 × 12—​W2.5 × W1.4 (As = 0.075 sq in.). The selection is made from Table 9.11.1; As in direction perpendicular to joists is 0.075 sq in./​ft, and As in direction parallel to joists is 0.014 sq in./​ft. The “4 × 12” means 4-​in. spacing of the wires running perpendicular to the joist and 12-​in. spacing of the wires running parallel to the joist. TABLE 9.11.1  COMMON WELDED WIRE FABRIC FOR TEMPERATURE AND SHRINKAGE REINFORCEMENT Designation Spacing (in.) of Longitudinala and Transverse Wires

Wire Size Designation Longitudinal × Transverse

As Longitudinal Direction (sq in.)

As Transverse Direction (sq in.)

4 × 12 4 × 12 4 × 12 4 × 12 4 × 12

W1.4 × W1.4 W2 × W1.4 W2.5 × W1.4 W3 × W2 W3.5 × W2

0.042 0.060 0.075 0.090 0.105

0.014 0.014 0.014 0.020 0.020

a

Longitudinal refers to the slab span, which is perpendicular to the joist span.

(b) Joist design. The overall depth of the joist floor must satisfy the minimum requirement of ACI-​9.3.1.1 unless deflections are computed; thus, for an interior span min h =



L 26(12) = = 14.9 in. 21 21

If excessive deflection may cause damage to nonstructural elements, deflection must be computed in all cases. Assuming that deflection must later be computed, it will be prudent to make joists deeper than indicated by ACI-​9.3.1.1. Assume use of joists 12 in. deep below the bottom of the slab with a width of 5 in. at the bottom (i.e., average 6 in.). The weight of the joist will then be

wD = [(3.0 + 12)6 + 30(3.0)](0.150)

1 144

= 0.19 kip/ft

The factored load is

wu = 1.2(0.19) + 1.2(0.010)(35)

1 12

+ 1.6(0.080)(35)

1 12

= 0.636 kip//ft

The maximum factored negative moment is

Mu =

1 (0.636)(24.5)2 = 34.7 ft-kips 11

(Continued)

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9.11  DESIGN OF JOIST FLOORS

Example 9.11.1 (Continued) Using the minimum cover of 3 4 in. (ACI-​20.6.1.3.2) and assuming #5 bars for the main steel, d = 15 – 0.75 – 0.31 = 13.94 in. The required Rn then becomes required Rn =



Mu 34.7(12, 000) = = 476 psi 2 φ bd 0.90(5)(13.94)2

Using Eq. (3.8.5),

required ρ = 0.009 required As (over support) = 0.009(5)13.94 = 0.63 sq in.



The minimum reinforcement for the negative moment region of a continuous T-​section in a statically indeterminate member where the flange would be in tension is given by As ,min =



3 fc′ fy

bw d

[3.7.10]

but not less than 200bw d / f y. Thus,

As,min =



but [Vu = 7.05 kips]  5.0 

OK

The maximum factored positive moment near midspan is

Mu =

1 (0.636)(24.5)2 = 23.9 ft-kips 16

(Continued)

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Example 9.11.1 (Continued)

Joist width

2” for 20” forms 2 12 ” for 30” forms

Usually 20” or 30” pans Girder

3’–0”

Figure 9.11.1  Plan view for example Example 9.11.1 showing ends of tapered pan joists.

For the T-​section, assume that the depth of the rectangular stress distribution falls within the flange; estimate arm to be ≈ 0.95d =0.95(13.94) = 13.2 in. Then, by trial, required As ≈



Mu 23.9(12) = = 0.40 sq in. φ f y (arm ) 0.90(60)(13.2)

Determine moment arm more accurately by using bE as 35 in. C = 0.85 fc′ ba = 0.85(3)(35)a = 89.3a T = As f y = 0.40(60) = 24.0 kips a = 24.0/89.3 =0.269 in.





23.9(12) revised required As = = 0.39 sq in. 0.90(60)(13.94 − 0.13) Try 2–​#4 bottom bars and 2–​#5 top bars (see Fig. 9.11.2), thus provided As over support = 0.62 sq in. (2 − # 5)



provided As at midspan = 0.40 sq in. (2 − # 4) A

Distribution rib



2 – #5

2 – #4

1 – #4

A 24’–6” 3”

12

12”

1 5”

2 – #4

30” Section A–A

Figure 9.11.2  Concrete joist floor using removable pans for Example 9.11.1.

(Continued)

31



311

9.11  DESIGN OF JOIST FLOORS

Example 9.11.1 (Continued) To confirm the choice of main reinforcement, the design strength φ Mn must be checked at midspan and at the support. At the support (neglecting the increased width due to tapered ends), C = 0.85(3)(5)(a ) = 12.75a

T = 0.62(60) = 37.2 kips 2.92 = 3.44 in. 0.85

a=

37.2 = 2.92 in. 12.75

εt =

d −c 13.94 − 3.44 (0.003) = (0.003) = 0.0092 > 0.005 c 3.44

c =



Thus, φ = 0.90

φ M n = 0.90(37.2) [13.94 − 0.5(2.92)]

1 12

= 34.8 ft-kips ≈ [ Mu = 34.7 ft-kips]

OK

At midspan, C = 89.3a



T = 0.40(60) = 24.0 kips

a=

24.0 = 0.27 in. 89.3

εt =

13.94 − 0.32 (0.003) = 0.128 > 0.005 0.32

c=

0.27 = 0.32 in. 0.85

φ M n = 0.90(24.0) [13.94 − 0.5(0.27)]



1 12

= 24.8 ft-kips >[Mu = 23.9 ft-kips]

OK

Use 2–​#4 bottom bars and 2–​#5 top bars (c) Development length requirements. Consistent with a loading for the maximum positive moment in an interior span, the inflection point is at 0.354Ln from centerline of span [see Fig. 7.5.4(a)]. Thus the development length requirement at inflection points (ACI-​9.7.3.8.3) must be checked. Only one #4 bar will be extended beyond the inflection point and into the support at least 6 in. C = 0.85 fc′ bE a = 0.85(3)(35)a = 89.3a T = 0.20(60) = 12.0 kips a = 0.13 in.

M n = 12.0 [13.94 − 0.5(0.13)]

1 12

= 13.9 ft-kips



Vu = 0.636(0.354)(24.5) = 5.5 kips La = 13.9 in. (larger of 12db or effective depth d) The equivalent embedment length provided is

Mn 13.9(12) + La = + 13.9 = 30.3 + 13.9 = 44.2 in. Vu 5.5 (Continued)

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Example 9.11.1 (Continued) The development length Ld for the #4 that extends beyond its cutoff location depends entirely on the amount of cover, because there are no other bars in the development length and no stirrups are used. The bottom cover is the minimum 3 4 in., which exceeds db but is less than 2db for the #4 bar; thus Category A in the simplified method is not satisfied. From Table 6.6.2 for Category B,

Ld (for # 4) = 32.9 in.(2.7 ft)



which exceeds the 12-​in. minimum. Thus, the check of ACI-​9.7.3.8.3 is completed by showing that 44.2 in., the equivalent embedment provided, exceeds Ld of 32.9 in. and thus is satisfactory. Frequently a transverse distribution rib (Fig.  9.11.2) is used, having a 4-​in. minimum width and containing at least one #4 bar both top and bottom. Such a rib would be located at the third points of the span for spans greater than 30 ft. For spans between 20 and 30 ft, one rib should be used near midspan; for spans less than 20 ft, the rib may be omitted [2.21].

9.12 REDISTRIBUTION OF MOMENTS—​ INTRODUCTION TO LIMIT OR PLASTIC ANALYSIS Methods of proportioning beams for flexure, shear, and bar development requirements have been discussed in Chapters 3, 4, 5, and 6 and further illustrated in Sections 9.1 through 9.11. When the inelastic behavior of concrete and steel has been accounted for in the design of a cross section based on its strength, it may seem somewhat illogical to have used an elastic analysis to determine the design moments and shears. However, because the evaluation of the true ultimate strength, or limit strength, of an entire structure requires a difficult and elaborate inelastic analysis, the procedures presented in the preceding sections and chapters are used. These are both conservative and safe. The concept of redistribution of moments introduced into the ACI Code in 1963 was the result of considerable research [9.3–​9.9] into the limit behavior of the entire structure (primarily continuous beams) beyond the elastic range to the point where the collapse load is reached. The use of plastic theory requires that the sections involved actually behave plastically. Figure 9.12.1(a) shows the ideal relationship of moment, M, to curvature, φ, where there is a perfectly elastic portion and an ideal plastic portion. Such an ideal relationship is often referred to as elastic−perfectly plastic relation. In reality, reinforced concrete may exhibit an M-​φ relation as shown in Fig.  9.12.1(b) and Fig. 3.13.1, which may often be approximated by such an ideal behavior. Consider the statically determinate, simply supported beam of Fig. 9.12.2 with a concentrated load at midspan. The limit load on such a system is reached when the extreme fiber of concrete in compression reaches the crushing strain εcu (taken by ACI as 0.003) after the longitudinal reinforcement has yielded. In limit analysis, the nominal moment strength, Mn, is taken as Mp, the plastic moment in the ideal system. When Mp is reached, the section at midspan begins to deform (rotate) into the plastic range without further increase in the applied moment (load). In other words, after reaching Mp, the cross section at midspan behaves as a real hinge (with no further moment resistance) with plastic deformations, thus it is referred to as a plastic hinge. For such simply supported beams, development of a plastic hinge at midspan causes the beam to become a mechanism with no further load-​carrying capacity. Therefore, the load that causes Mp to develop at one location along the span represents the maximum load that can be applied to the beam: that is, the limit of the system for such an ideal section behavior. It will be shown, however, that for statically indeterminate systems,

31



313

9.12  REDISTRIBUTION OF MOMENTS

Mn

Plastic

Elastic Mp = plastic moment

ϕy

Moment M

Moment M

My

Mn

ϕu

(

Curvature ϕ i.e.,

d 2y dx2

ϕy

)

ϕu

(

Curvature ϕ i.e.,

(a) Ideal elastic-plastic

d 2y dx2

)

(b) Reinforced concrete

Figure 9.12.1  Moment curvature relationships. Wu

θup Plastic hinge

Figure 9.12.2  Limit condition for simply supported beam.

the development of a plastic hinge at one location does not necessarily correspond to the maximum load that can be carried by the structure. Though typically shown as a single point, the plastic hinge is actually a region of concentrated inelastic deformations (curvatures) that spread out from the point of maximum moment a certain length, often called the plastic hinge length, lp. This length depends on the details of the longitudinal and transverse reinforcement, the shear span, among others, but it has been found to vary in practice between 0.5d and 2.5d on each side of the critical section [9.4, 9.8, 9.12]. For practical purposes, the beam may be treated as two rigid elements connected by a plastic hinge having a maximum concentrated rotation angle, or plastic hinge rotation, θup, at ultimate ( just before collapse) as shown in Fig. 9.12.2. This plastic hinge rotation may be approximately computed by assuming that the plastic curvature is constant over the length lp on each side of the point of maximum moment, in which case

θup = (ϕu − ϕ y )2l p



(9.12.1)

This maximum rotation that can be developed by the plastic hinge is often referred to as plastic rotation capacity. Most reinforced concrete beams have a reasonable amount of plastic rotation capacity since they are usually designed with a net reinforcement ratio ρ or ρ –​ρ′ of 50% or less of the balanced amount ρb. Consider the statically indeterminate, fixed-​end beam of Fig. 9.12.3(a) subjected to a uniformly distributed load, along with its corresponding moment diagram computed from an elastic analysis. As the load is applied to the beam, the sections with the largest moments (i.e., at the fixed ends) achieve Mp first. Thus, at the ends [see Fig. 9.12.3(b)],

wy L2 12

= M p and wy =

12 M p

L2

where wy is the load carried when Mp is just reached at an end, accompanied by a curvature ϕy [see Fig. 9.12.1(a)]. Upon the slightest increase in the applied load, the ends of the beam

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would behave as plastic hinges and would rotate but would carry no additional moment. The moments at all other sections along the beam span, however, remain within the elastic range, below Mp. In particular, the section at midspan has a moment of only Mp /​2. Therefore, as the load is increased above wy, the beam will behave as a simply supported beam with plastic hinges at the ends. In carrying this load, the beam is as stable as a simple beam and is capable of carrying load above wy. Increasing the load will cause the moments to increase everywhere along the beam span except at the ends, with maximum increase in moment occurring at midspan. It can be shown that the limit condition for this beam requires three plastic hinges (two at the ends and one at midspan), as shown in Fig. 9.12.3. Also shown in the figure is the corresponding moment diagram at the limit (collapse) condition. Statics may be used in computing the collapse load, for example, by drawing a free-​body diagram of the beam between the end and the plastic hinge at midspan and taking moments about the beam ends: Mp + Mp −

wu L L = 0 2 4 2M p = wu =

wu L2 8 16 M p L2

Thus the beam may carry 33% additional load after the plastic moment has been reached at the fixed ends, provided the rotation of the plastic hinges at the ends not exceed the rotation capacity (θup) prior to developing the plastic hinge at midspan (i.e., prior to reaching the load wu). In other words, rotation capacity permitting, the positive and negative w

wL2 24 + wL2 12

–

–

wL2 12

(a) wy Mp

Mp (b) wu

Mp

Mp

Mp

Plastic hinges

Mp

Mp

Mp (c)

Figure 9.12.3  Limit condition for a fixed-​end beam.

315



9.12  REDISTRIBUTION OF MOMENTS

εc

ϕu Strain occurring during redistribution of moment

Total steel strain

315

Ultimate concrete strain, εcu = 0.003

ϕy Steel strain at first yield, εy

Figure 9.12.4  Strain diagram for reinforced concrete beams in which steel reaches εy before maximum concrete strain εc = εcu.

moments under any particular loading condition tend to equalize (assuming, of course, that the strength of the section at the two regions is the same). For a thorough treatment of limit analysis the reader is referred to the work of Baker [9.8]. Reinforced concrete beam sections with a low net percentage of reinforcement ρ –​ ρ′ (i.e., tension-​controlled sections) will develop yielding of the tension steel reinforcement while the maximum concrete strain is still of low magnitude (see Fig. 9.12.4). Therefore, sufficient reserve curvature φu –​ φy, or plastic hinge rotation capacity, θup, will be available, allowing redistribution of moments to occur before the ultimate concrete strain is reached.

Redistribution of Negative Moments under ACI-​6.6.5 As noted earlier, redistribution of moments can occur only when there is enough reserve curvature after first yield of the tension steel. For this condition to exist, the tension steel must develop strains well above the yield strain before the concrete is crushed in the compression zone (see Fig. 9.12.4). Accordingly, redistribution of negative moments is permitted only when εt is equal to or greater than 0.0075 at the section at which moment is reduced [ACI-​ 6.6.5.1(b)]. Also, there is a limit to the amount of moment redistribution that can be allowed, irrespective of the net tensile strain at the cross section. ACI-​6.6.5.3 states that the percentage permitted for redistribution is 1000εt percent, with a maximum of 20%. The strain εt is the net tensile strain in the extreme tension steel at nominal strength. An easy way to remember the idea is to recall that a 10% redistribution limit corresponds to a value εt = 0.01. The limits of applicability of this provision may be summarized as follows. 1. Application is limited to continuous flexural members (i.e., statically indeterminate systems), ACI-​6.6.51(a). 2. No more than 20% reduction of moments for any given loading arrangement may be applied [ACI-​6.6.5.3]. 3. Bending moments used in such an adjustment must be obtained by an elastic analysis (ACI-​6.6.5.1); moments from use of coefficients or other approximate methods may not be adjusted (ACI-​6.5.3). 4. Adjustment, when permitted, is made for each given loading condition. The envelope of the adjusted diagrams from all loading conditions is then used to proportion the members (ACI-6.6.5.4 and ACI-6.6.5.5). In approaching a design, it may be advantageous to apply ACI-​6.6.5 directly. It is likely, however, that more frequent use will occur when the design is in progress and the designer realizes that the conditions of the redistribution provision are met and savings appear to be possible. The following example illustrates an application of this provision.

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EXAMPLE 9.12.1 Show the effects of redistribution on the moments obtained by elastic analysis for the floor beams 2B1 and 2B2 in Example 9.3.1. Use the provisions of ACI-​6.6.5. SOLUTION The beams, along with the critical moment diagrams, are shown in Fig.  9.12.5. The elastic end moments Mu are those computed under factored loads and tabulated in Table 9.3.1 for each loading condition. The elastic moment diagrams are shown in Figs. 9.3.3 through 9.3.6. (a) Investigate negative moment region at B to determine the maximum percent of moment that can be redistributed. The net tensile strain at B (sections C-​C and D-​D) A

B

C

26’–0’

26’–0’

3.45 k/ft (factored load)

1.37 k/ft (factored load) Spans 1 and 3 with LL

159 ft-kips

Symmetrical about CL of structure

144 From elastic analysis

2 14

81 90 190 210

118 130

118 130

(a) 3.45 k/ft 148

Spans 1 and 2 with LL

142

105 104

From elastic analysis

66 73

From elastic analysis 224 248

152 168

205 226 (b) 3.45 k/ft

1.37 k/ft Span 2 with LL 46 40

121 103

From elastic analysis

14 15 144 159

171 189

171 189

(c)

Figure 9.12.5  Redistribution of elastically computed moments for Example 9.12.1 according to ACI-​6.6.5.

(Continued)

317



SELECTED REFERENCES

317

Example 9.12.1 (Continued)

was computed earlier in part (a) of Example 9.4.1 as 0.0105, which is greater than 0.0075. Thus, moment redistribution is permissible at B. percent adjustment permitted = 1000 εt = 1000(0.0105) = 10.5%

Since the steel in the compression zone was not fully developed at the faces of the support (see Fig. 9.6.2), it may not be counted as compression steel. It might well be economical to develop the capacity of the steel in the compression zone at the support. This would increase the ductility in that region and allow a higher percentage of moment redistribution. (b) Make adjustments to elastic moments. The adjustment of the negative moments may be either an increase or a decrease as long as the positive moments are also adjusted to satisfy static equilibrium. Examine first the loading for maximum positive moment in span AB, Fig. 9.12.5(a). Increasing the negative moments by 10.5% reduces the maximum positive moment for this loading from 159 to 144 ft-​kips. The 10.5% adjustment for Case (b) is made by reducing the negative moment at B and increasing those at A and C, thus minimizing the effect on the positive moments. In Case (c), the negative moments are increased 10.5%, thus reducing the positive moment in span 2 from 121 to 103 ft-​kips. Note that, after these adjustments, the increased negative moment at B for Case (a) is still less than the controlling negative moment of 224 ft-kips occurring under Case (b). The increased positive moment in span 1 for Case (b) is only 3% greater than the reduced controlling positive moment of 144 ft-kips from Case (a). Similarly, the increased positive moment in span 2 under Case (b) is only 2% greater than the reduced positive moment of 103 ft-kips under Case (c). The envelope of adjusted moments would then be used to design the sections. The net effect on the envelope is a reduction for both positive and negative moments. This is not actually a reduction in the safety factor below that used for a simply supported beam; it is a reduction in the excess strength that the system has, by virtue of its continuity, one span with another. A simple application of limit design concepts to practical design has been presented by Furlong [9.10]. Further discussion of Furlong’s proposal has been presented [9.11] as a proposed alternate to the use of ACI-​6.6.5. It must be noted, however, that such factors as shear, development of reinforcement, and deflection may still control the design; thus, limit analysis for flexure may offer little, if any, economic advantage.

SELECTED REFERENCES 9.1. 9.2. 9.3. 9.4. 9.5. 9.6. 9.7. 9.8.

CRSI. Manual of Standard Practice (27th ed.). Schaumburg, IL:  Concrete Reinforcing Steel Institute, 2001. Paul F.  Rice, Edward S.  Hoffman, David P.  Gustafson, and Albert G.  Gouwens. Structural Design Guide to the ACI Building Code (3rd ed.). New York: Van Nostrand Reinhold, 1985. A. H. Mattock. “Limit Design for Structural Concrete,” Journal of the Research and Development Laboratories, Portland Cement Association, 1, May 1959, 14–​24. W. G. Corley. “Rotational Capacity of Reinforced Concrete Beams,” Journal of the Structural Division, ASCE, 92, ST5 (October 1966), 121–​146. (Also PCA Development Department Bulletin D108.) ACI-​ASCE Committee 428. “Progress Report on Code Clauses for ‘Limit Design,’” ACI Journal, Proceedings, 65, September 1968, 713–​720. Disc. 66, 221–​223. M. Z. Cohn. “Limit Design for Reinforced Concrete Structures: An Annotated Bibliography,” ACI Bibliography No. 8. Detroit: American Concrete Institute, 1970. Harold W.  Conner, Paul H.  Kaar, and W.  Gene Corley. “Moment Redistribution in Precast Concrete Frame,” Journal of the Structural Division, ASCE, 96, ST3 (March 1970), 637–​661. A.  L. L.  Baker. Limit State Design of Reinforced Concrete. London:  Cement and Concrete Association, 1971.

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9.9. M.  Z. Cohn. “Inelasticity of Reinforced Concrete and Structural Standards,” Journal of the Structural Division, ASCE, 105, ST11 (November 1979), 2221–​2241. 9.10. Richard W. Furlong. “Design of Concrete Frames by Assigned Limit Moments,” ACI Journal, Proceedings, 67, April 1970, 341–​353. 9.11. Richard W. Furlong and Carlos Rezende. “Alternate to ACI Analysis Coefficients,” Journal of the Structural Division, ASCE, 105, ST11 (November 1979), 2203–​2220. 9.12. Robert Park and Thomas Paulay. Reinforced Concrete Structures. New  York:  John Wiley & Sons, 1975, 769 pp.

PROBLEMS All problems are to be worked in accordance with the provisions of ACI Code, and all stated loads are service loads, unless otherwise indicated. A  design sketch showing all design decisions is required for each problem. For verifying bar lengths, draw the φ Mn diagram directly below a side view of the beam showing the bars, as in Fig. 9.6.2.

CONTINUOUS BEAM PROBLEMS 9.1 Design a rectangular beam continuous over three spans as shown in the figure for Problem 9.1. The live load is 2.75 kips/​ft, and the dead load is 1.0 kip/​ft in addition to the beam weight. The 9.2 floor is to be a prefabricated system. Assume the supports to be 15 in. wide. Use fc′ = 4000 psi and fy = fyt = 60,000 psi, and do not apply ACI-​ 6.6.5 for moment redistribution. (a) Determine the moment envelope using fac- 9.3 tored loads. (b) Determine the bar cutoff locations, directly from the moment envelope. Rectangular beam

15’– 0”

21’– 0”

15’– 0”

Problem 9.1 

2

2

14’–0”

8

A

14’–0”

(c) Use only full-​span loadings for computing the shear envelope and use U stirrups of #3 size if possible. Redesign the beam of Problem 9.1 as a monolithic T-​section floor system. The beams are spaced 8 ft on centers and the slab is 6 in. thick. The 1.0-​kip/​ft dead load includes the slab but not the beam stem. Design the beam ABC of the frame shown in the figure for Problem 9.3 in which the relative stiffnesses EI/​L are given by the encircled numbers. The beams are T-​sections having a 6-​in. slab. The dead load is 0.40 kip/​ft (not including beam stem or slab) and the live load is 3.75 kips/​ft. Assume the columns to be 15 in. square. Use fc′ = 4000 psi and fy = fyt = 60,000 psi. 9.4 Design the transverse beam indicated for the floor plan given in the figure for Problem 9.4. Assume that a warehouse live load of 375 psf is to be used. Assume also a 5-​in. slab placed monolithically with beams and girders; the

2

Problem 9.3 

6

B

2

30’–0”

2

Transverse spacing of frames = 12 ft

40’–0”

C

2

319

Longitudinal girder

7’–0” 7’–0” 7’–0”

21’–10”

21’–0”

21’–0” 85’–8”

10”

CL of bearing

Transverse beam

15” 21’–10” Section at wall

Problem 9.4 

21’–10”

61’–8”

21’–0” Transverse beam

9.5 Redesign the transverse beams of Problem 9.4, except use spans 22-​20-​22 instead of the original spans 21-​18-​21, and use fc′ = 4000 psi and fy = fyt = 60,000 psi. All other details are the same as in Problem 9.4.

21’–10”

Suggested additional problems (no solutions manual)

18’–0”



9.6 Redesign the transverse beams of Problem 9.4, except use spans 24-​21-​24 instead of original spans 21-​18-​21, and use fc′ = 4000 psi and fy = 60,000 psi. All other details are the same as in Problem 9.4. 9.7 Design the four-​span longitudinal girder indicated for the floor system of Problem 9.4. In lieu of the more accurate loadings from the results of Problem 9.4, use concentrated dead and live loads of 15 and 60 kips, respectively, coming to the girder from each side, plus the weight of the girder. Assume that the columns are 18 in. square and 15 ft high and that the beam receives equivalent restraint from monolithic attachment to the 15-​ in. reinforced concrete exterior wall. Assume also that the columns are fixed at the far ends. Use fc′ = 3500 psi and fy = 60,000 psi. 9.8 Design the continuous 26-​ft-span floor beams along a column line in Fig. 8.3.1. Assume that the floor slab is 5 in. thick. Assume the columns are 15 in. square and 10 ft long. Use fc′ = 4000 psi,

18’–0”



width of support at longitudinal girders is 18 in., and only a nominal minimum of moment restraint is provided by the exterior wall support (i.e., assume hinge for elastic analysis). Use fc′ = 3500 psi and fy = fyt = 60,000 psi. For a comparison of the effect of considering torsional stiffness of longitudinal girders, divide the class into three parts, each using one of the following assumptions: (a) zero torsional stiffness of the two longitudinal girders (b) torsional stiffness equal to 25% of the bending stiffness of the 21-​ft span beam (c) torsional stiffness equal to 50% of the bending stiffness of the 21-​ft span beam

21’–0”



319

PROBLEMS

21’–10”



320

320

C hapter   9     S lab - B eam - G irder A N D J O I S T F L O O R S Y S T E M S

fy = fyt = 60,000 psi, and a 100 psf live load. Make a preliminary design without structural analysis; then, using the preliminary size, make the structural analysis to obtain the factored moment and factored shear envelopes for the final design.

CONCRETE JOIST PROBLEMS 9.9 Design a concrete joist, using 30-​ in.-​ wide removable pans, for a typical interior span of 28 ft center to center of supporting girders. Assume a support width of 18 in. Use a live load of 100 psf, fc′ = 4000 psi, and fy = 60,000 psi. 9.10 Determine the service live load capacity for a single-​span joist of 20-​ft clear span, using a

2 1 2 -​in. slab, 20-​in.-​wide × 8-​in.-​deep forms, and 4-​in.-​wide joists with 2–​#5 bars. No taper is used. Use fc′ = 3000 psi and fy = 60,000 psi. 9.11 Design an end-​span joist for a continuous system to carry a live load of 225 psf, using 20-​in.-​ wide removable pans, for a clear span of 18 ft. Allow an extra 1 2 in. of thickness for dead load purposes only, since the concrete slab is to serve as the final wearing surface. Use fc′ = 4500 psi and fy = 60,000 psi.

MOMENT REDISTRIBUTION PROBLEM 9.12 Redesign the beam of Problem 9.1, taking into account permissible moment redistribution. Compare with work on Problem 9.1.

CHAPTER 10 MEMBERS IN COMPRESSION AND BENDING

10.1 INTRODUCTION Columns subjected to pure axial load rarely, if ever, exist. All columns are subjected to some bending moment, which may be caused by (1) end restraint arising from the monolithic placement of floor beams and columns; (2) accidental eccentricity from imperfect alignment and variable materials; (3) asymmetrical floor loads; (4) eccentric loads such as crane loads in industrial buildings, and (5) lateral loading such as from wind or induced by an earthquake. Concrete construction is usually monolithic; thus reinforced concrete frames and arches (Fig. 10.1.1) are common and advantageous. All sections in the two structures shown in Fig.  10.1.1 are subjected to combined bending and axial load. The vertical members in Fig. 10.1.1(a) and sections near the supports in Fig. 10.1.1(b) may be subjected to a high

Reinforced concrete tied columns under construction (Courtesy of Cary Kopczynski & Co.).

32

322

C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G

(a)

(b)

Figure 10.1.1  Reinforced concrete rigid frame and arch.

ratio of axial force to bending moment, while the horizontal member in Fig. 10.1.1(a) and sections near the crown in Fig. 10.1.1(b) may be subjected to a low ratio of axial force to bending moment. This chapter deals first with columns having minimal bending moment, commonly called axially loaded members. The effect of medium and large amounts of bending in columns is considered later in this chapter. Only the basic strength of short compression members is considered, however. Slenderness effects in columns are treated in Chapter 13.

10.2 TYPES OF COLUMNS Reinforced concrete columns are principally of two types, classified according to the manner in which the longitudinal reinforcing bars are laterally supported. A tied column, usually of square or rectangular shape, is one in which the longitudinal reinforcing bars are held in position by separate lateral ties, as shown in Fig. 10.2.1(a) and photo on page 321. A spirally reinforced column, usually of circular or square shape, is one in which the longitudinal reinforcing bars are arranged in a circle and wrapped by a continuous closely spaced spiral, as shown in Fig. 10.2.1(b). For a column to be treated as a spirally reinforced column, however, the spiral reinforcement must satisfy the requirements of ACI-​25.7.3 (Section 10.9). Otherwise, the column is treated as a tied column. A composite column is one that uses a structural steel shape, pipe, or tubing, with or without additional longitudinal bars. One common composite column arrangement may contain a structural steel shape completely encased in concrete, which is further reinforced with both longitudinal and lateral reinforcement (spiral or ties), as shown in Fig. 10.2.1(c). In a second kind of composite column, the steel may encase a concrete core, which may or may not contain longitudinal reinforcing bars, as shown in Fig. 10.2.1(d). Composite columns are treated in Chapter 21.

10.3 BEHAVIOR OF COLUMNS UNDER PURE AXIAL LOAD Many tests were made in the early 1900s on reinforced concrete columns under axial loads [10.1–​10.5], but loading was generally of short duration. As early as 1911, however, Withey [10.5], at the University of Wisconsin, observed that as load increased beyond the service load range, a transfer of load from concrete to steel took place. In the early 1930s, ACI Committee 105 reported [10.6] on the results of 564 column tests, primarily at Lehigh University and the University of Illinois, where attention was given to column size, quality of concrete, quality and amount of longitudinal and lateral reinforcement, rate of application of load, and shrinkage and creep under sustained loads. By 1940, code design procedures for axially loaded columns were based on the ultimate strength results of the aforementioned extensive investigation. Joint ACI-​ASCE Committee 441 has provided an excellent annotated bibliography of the early studies on reinforced concrete columns [10.7].

32



1 0 . 3   B E H AV I O R O F C O L U M N S U N D E R P U R E A X I A L   L OA D Ties

323

Steel tubing

Spiral

Concrete filled Spiral Pipe

(a) Tied column

(b) Spirally reinforced column

(c) Composite column (spiral bound encasement around structural steel core)

(d) Composite column (steel encased concrete core)

Figure 10.2.1  Types of columns.

When concrete and steel act together in compression, the proportion of loading carried by each changes continuously during the time the load is acting. Initially, the stress in the steel is Es /​Ec times the stress in the concrete, according to the elastic theory. As the time-​ dependent effects of creep and shrinkage occur, the steel gradually carries relatively more load than its elastic share. Creep and shrinkage deformations were introduced in Section 1.11, and are treated in detail with regard to deflections in Chapter 12. Members that are subjected to axial compression, either alone or in combination with bending, frequently have a substantial portion of the total load sustained. Consequently, the transfer of load to the steel from the concrete due to time-​dependent deformation is more pronounced in these members than in beams. The axial load versus axial deformation behavior of spirally reinforced columns is significantly different from that of tied columns. Spirally reinforced columns exhibit a marked “yielding,” followed by considerable axial deformation before failure, as shown in Fig. 10.3.1. On reaching the first peak point, the shell spalls off (Fig. 10.3.2) and the spiral begins to restrain lateral expansion, confining the concrete in the core. As discussed in Section 1.9, concrete compressive strength increases with lateral confinement. Thus, as the column core expands and the spiral reinforcement provides confinement, the compressive strength of the column core increases. If spiral reinforcement is provided in sufficient amount, the peak load resisted by the confined column core will be greater than the peak load prior to spalling of the column shell. The first peak has sometimes been referred to as the “yield” strength of a spirally reinforced column, while the second peak represents the ultimate strength. Typical appearance of spirally reinforced columns under concentric loading at failure is shown in Figure 10.3.2. Tied columns, particularly those with rectilinear ties, do not exhibit the deformation capacity of spirally reinforced columns. As shown in Fig. 10.3.1, the axial load versus axial deformation response shows only one peak, which corresponds to the point at which the shell begins to spall off and buckling of longitudinal reinforcement between the ties initiates (Fig. 10.3.3). Thus, there is no “yield” point in tied columns; the first peak represents the ultimate strength.

324

C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G

Based on the discussion above, the ultimate strength of a spirally reinforced column (i.e., after spalling of the column shell) may be expressed as follows [10.6]:

Pult = k3 fc′( Ach − Ast ) + Ast f y + ks ρs f yt Ach

(10.3.1)

where k3 = coefficient (taken equal to 0.85) to account for the difference between concrete strength in the column and that in a test cylinder fc′ = standard 28-​day cylinder strength Ach = area of concrete core, taken to the outside edges of the spiral reinforcement (see later: Fig. 10.9.1) Ast = total area of longitudinal reinforcement fy = yield stress of longitudinal reinforcement ρs = ratio of volume of spiral reinforcement to volume of concrete core calculated as (see Fig. 10.9.1)

ρs =



4 Asp Dc s

(10.3.2)

ks = constant that varies typically from 1.5 to 2.5 Asp = cross-​sectional area of spiral reinforcement (see Fig. 10.9.1) Dc = diameter of concrete core (see Fig. 10.9.1) s = spiral spacing or pitch (see Fig. 10.9.1) fyt = yield stress of spiral steel Note that the third term in Eq. (10.3.1) represents the contribution from confinement of the concrete core by the spiral reinforcement after spalling of the concrete shell. The “yield ” strength of a spirally reinforced column (first peak in Fig. 10.3.1) and ultimate strength of a tied column, on the other hand, can be expressed as

Pult = k3 fc′( Ag − Ast ) + Ast f y (10.3.3)

where Ag is the gross cross-​sectional area of the column. Studies on ductility of spirally reinforced column have been made by Priestley, Park, and Potangaroa [10.8], Ahmad and Shah [10.9], and Martinez, Nilson, and Slate [10.10]. Studies on the behavior of tied columns by Sheikh and Uzumeri [10.11], Scott, Park, and Priestley [10.12], Park, Priestley, and Gill [10.13], Fafitis and Shah [10.14], Moehle and Cavanagh [10.15], Sheikh and Yeh [10.16, 10.20, 10.21], Özcebe and Saatcioglu [10.17], Yong, Nour, and Nawy [10.18], and Abdel-​Fattah and Ahmad [10.19] show how the deformation capacity of tied columns varies with tie and longitudinal bar arrangements. Razvi and Saatcioglu [10.22] have studied the effects of confinement using welded wire reinforcement. Much of this work has been concerned with ductility under cyclic loading, such as for earthquake-​resistant design. As will be shown later, though columns of both types have approximately the same strength, a higher factor of safety should be provided for the tied column than for the spirally reinforced column because of the sudden failure and lack of toughness (energy absorption) exhibited by tied columns. Spirally reinforced column shows ability to deform “Yield” point (spirally reinforced prior to failure column shell spalls off)

Load

324

Tied column fails suddenly Tied and spirally reinforced columns

Deformation (unit shortening)

Figure 10.3.1  Typical load-​deformation curves for tied and spirally reinforced columns.

325



1 0 . 4   S A F E T Y P R OV I S I O N S F O R C O L U M N S

325

Figure 10.3.2  Failure of spirally reinforced columns under concentric axial force. (Courtesy of Portland Cement Association.)

Figure 10.3.3  Failure of tied columns under concentric axial force. (Courtesy of Portland Cement Association.)

10.4 SAFETY PROVISIONS FOR COLUMNS The load factors used in the design of columns are the same as for any other type of member. The design of columns in regions of little or no seismic risk is often governed by load combinations that include wind, whereas beams are usually governed by the gravity-​load combinations. As discussed in Section 2.7, for gravity loads,



U = 1.4 D

[2.7.1]

U = 1.2 D + 1.6 L + 0.5( Lr or S or R)

[2.7.2]

For loadings including wind,

U = 1.2 D + 1.6( Lr or S or R ) + (1.0 L or 0.5W )

[2.7.3]

U = 1.2 D + 1.0W + 1.0 L + 0.5( Lr or S or R )

[2.7.4]

U = 0.9 D + 1.0W

[2.7.6]

The other load combinations are not repeated from Section 2.7. The strength reduction factor φ for compression-​controlled columns (ACI-​21.2.2) is 0.75 for spirally reinforced columns and 0.65 for tied columns.

326

326

C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G

The strength reduction factor φ, which is lower for tied columns than for spirally reinforced columns, reflects the lower toughness exhibited by tied columns. The difference in the behavior of tied and spirally reinforced axially loaded columns is further accounted for by the use of a different maximum nominal compression capacity, Pn(max), as discussed in Section 10.11. For combined compression and bending, the φ value is permitted to be variable and may increase to 0.90 for tension-​controlled columns. This variation of φ is treated later (see Section 10.17).

10.5 CONCENTRICALLY LOADED SHORT COLUMNS According to the ACI Code, the strength for a concentrically loaded short column, Po, can be calculated based on Eq. (10.3.3) as follows (ACI-​22.4.2.2):

Po = 0.85 fc′( Ag − Ast ) + f y Ast

(10.5.1)

where Ag is the gross column area and Ast is the total area of longitudinal reinforcement. For the rectangular column shown in Fig. 10.5.1, Ag = bh and Ast = (A1 + A2). As discussed in Section 10.3, the 0.85 factor in Eq. (10.5.1) accounts for the difference between the compressive strength of concrete in a column and that of a concrete cylinder. This factor, thus, is not the same factor used in Whitney’s stress block. Po may also be expressed as

Po = Ag 0.85 fc′(1 − ρg ) + f y ρg 

(10.5.2)

where ρg = Ast /​Ag. When the terms including ρg are combined, Eq. (10.5.2) becomes

Po = Ag 0.85 fc′ + ρg ( f y − 0.85 fc′ )

(10.5.3)

Figure 10.5.1  Internal force resultants in a concentrically loaded column.

10.6 STRENGTH INTERACTION DIAGRAM When axial compression combined with bending moment acts on a member having low slenderness ratio (unbraced length Lu to radius of gyration r) such that column buckling is not a potential mode of failure, the strength of a member is governed by the material strength (corresponding to yielding in a homogeneous elastic material) of the cross section. ACI-​22.2.2.1 states that the nominal strength of this so-​called short column subjected to combined axial load and bending is achieved when the extreme concrete compression fiber reaches the strain εcu = 0.003.

327



327

1 0 . 6   S T R E N G T H I N T E R AC T I O N D I AG R A M

For a given cross s​ ection there is an infinite number of strength combinations at which Pn and Mn act together. These strength combinations lie on a curve, as shown in Fig. 10.6.1, which is called the strength interaction diagram or P–​M interaction diagram. Depending on the ratio of Mn to Pn (see Fig. 10.6.1), the strain diagram will exhibit two distinct categories. There may be (1) compression over most or all of the section such that the compressive strain in the concrete reaches 0.003 before the tension steel reaches the yield strain εy = fy /​Es or (2) tension in a large portion of the section such that the strain εs in the tension steel exceeds the yield strain εy when the compressive strain in the concrete reaches 0.003. The balanced strain condition in combined bending and axial load is represented by the point Pn = Pb and Mn = Mb on this diagram. The balanced strain condition corresponds to the “compression control” limit (see Section 3.6 for a discussion on compression- and tension-​controlled sections). Thus, member sections with a resultant axial force higher than that corresponding to the compression control limit are said to be compression controlled. In this case, the strength reduction factor φ is 0.75 for spirally reinforced columns and 0.65 for tied columns (ACI-​21.2.2). The region below the compression control limit is divided into two parts. A  section having a combination of Pn and Mn such that the strain in the extreme layer of tension reinforcement ε t ≥ 0.005 , is said to be tension controlled; the corresponding strength reduction factor φ is 0.90 (ACI-​21.2.2). This would be the case of a member subjected to pure flexure (i.e., no axial load applied), or to flexure and low levels of axial load. Sections such that f y / Es < ε t < 0.005 are said to be in the transition zone of the P–M diagram for which the strength reduction factor φ varies linearly with εt between 0.90 and 0.75 for spirally reinforced columns and between 0.90 and 0.65 for tied columns. This maximum strength interaction relationship for short rectangular members has been verified by research [10.23–​10.28]. Nonprismatic members have also been studied [10.29, 10.30], as well as members subject to nonproportionally varying axial load [10.31]. The major emphasis in this chapter is the analysis and design of sections whose nominal strength (Pn and Mn) lies at various points on the interaction diagram (Fig. 10.6.1). In studying Fig. 10.6.1, the reader may note that radial lines from the origin (Pn = 0, Mn = 0) represent constant ratios of Mn to Pn; that is, they represent eccentricities e of the load Pn from the plastic centroid, which was defined in Fig. 10.5.1. Under usual conditions of symmetrical reinforcement, the plastic centroid coincides with the centroid of the gross section. That e is equal to M/​P may be observed from Fig. 10.6.2, where it is shown that a column subject to an eccentric load is statically equivalent to a member under the combined action of an axial load and a bending moment. Thus, in Fig. 10.6.1 the vertical axis represents e = 0 and the horizontal axis represents e = ∞. This concept of replacing axial load and bending moment by a single eccentric load provides the basis for the practical approach to analysis and design computations in reinforced concrete. Pn - axis

Axis of bending

Po

e=

Mn Pn

εt < εy

0.003

e=0 Compression-controlled section Pb

Balanced strain condition Transition zone

e

= eb

Tension-controlled section

e=∞ M0

Mb

Mn - axis

0.003 fy εt = εy = E s 0.003 εt = 0.005

Mn, bending moment

Figure 10.6.1  Typical Pn–​Mn interaction diagram for axial compression and bending moment about one axis. Transition zone is where εy < εt 6”

8 bar ≤ 6”

> 6”

12 bar 6” max 6” max

16 bar

6” max

Figure 10.8.1  Common column tie arrangements. (Adapted from Ref. 2.23.)

3. Rectilinear ties shall be so arranged that every corner and alternate longitudinal bar shall have lateral support provided by the corner of a tie having an included angle of not more than 135° and no bar shall be farther than 6 in. clear on either side from such a laterally supported bar. 4. Where the bars are located around the periphery of a circle, a complete circular tie may be used. The ACI detailing manual [2.23] suggests tie arrangements for various numbers of bars, some common ones being shown in Fig. 10.8.1. In regions where longitudinal bars are spliced or offset bent, ACI Commentary R10.7.6.1.5 says that “it is prudent to provide a set of ties at each end of lap spliced bars, above and below end-​bearing splices, and at minimum spacings immediately below sloping regions of offset bent bars.” In addition, ACI-​10.7.6.1.5 waives the lateral reinforcement requirements where “tests and structural analyses demonstrate adequate strength and feasibility of construction.”

10.9 SPIRAL REINFORCEMENT AND LONGITUDINAL BAR PLACEMENT The spiral provides the column with the ability to absorb considerable deformation prior to failure [10.8]. This toughness is the principal gain that is achieved by the use of spirally

31



331

10.9  SPIRAL REINFORCEMENT

reinforced columns. Today’s knowledge of spiral behavior is based on the column research of the early 1930s [10.6]. Although the spiral does actually contribute strength to the column through confinement (as early as 1903, Considère [10.1, 10.2] indicated that the spiral was 2.4 times as effective as longitudinal reinforcement in providing column capacity), the conservative policy of ACI specifications since about 1940 has been to provide spiral reinforcement sufficient to increase the capacity of the core by an amount equal to the capacity of the shell, thus maintaining the column axial load capacity when the shell spalls off. Using the third term of Eq. (10.3.1) with an average ks of 2, the axial strength P contributed by the spiral reinforcement is P = 2ρs f yt Ach



(10.9.1)

As discussed in Section 10.3, this strength increase due to the spiral reinforcement was proposed by ACI Committee 105. Information on the derivation of this term, based on compressive strength exhibited by concrete under triaxial compressive stress, has also been presented by Huang [10.41]. Equating Eq. (10.9.1) to the strength of the shell and taking the concrete strength of the shell as about 90% of that in the core, or 0.75 fc′ , 2.0 f yt ρs Ach = 0.75 fc′ ( Ag − Ach )

from which

 Ag  f′ ρs = 0.375  − 1 c  Ach  f yt



(10.9.2)

Using an additional factor of safety of 1.20 to assure that the strength provided by the spiral through confinement exceeds the shell strength, Eq. (10.9.2) becomes  Ag  f′ ρs = 0.45  − 1 c  Ach  f yt



(10.9.3)

which is specified in ACI-​25.7.3.3. It is noted that the yield strength of the spiral, fyt, may not exceed 100,000 psi (see ACI-​Table 20.2.2.4a). A review of the spiral steel requirement has been made by Gamble [10.42]. According to ACI-​25.7.3.1, the clear spacing between spirals must be at least the greater of 1 in. and 4 3 the maximum aggregate size, but shall not exceed 3 in. Also, the spiral in cast-​in-​place construction shall have a diameter not less than 3 8 in. (ACI-​25.7.3.2).

db Ach =

πDc2 4

πh2 Ag = 4 ρs =

4Asp Dcs

Dc h

Figure 10.9.1  Spirally reinforced column.

Asp = area of spiral s

32

332

C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G Lower bar

1 12 diam or 1 12 ” min

Upper bar

Upper bar Lower bar 1 12 diam or 1 12 ” min

Figure 10.9.2  Bar arrangement in spirally reinforced columns.

Anchorage of spiral reinforcement shall be provided by 1 1 2 extra turns of spiral bar or wire at each end of the spiral unit (ACI-​25.7.3.4). Splices, when necessary, shall be either welded or mechanical splices conforming to ACI-​25.5.7, or lap splices in accordance with ACI-​25.7.3.6 for yield strengths not exceeding 60,000 psi. For uncoated deformed bars or deformed wire, the minimum splice length shall be 48 spiral bar or wire diameters but not less than 12 in. (ACI-​25.7.3.6). The spiral reinforcement is to be protected by the usual 1 1 2 in. minimum clear cover required by ACI-​20.6.1.3 for nonprestressed beams and columns. In spirally reinforced columns with large amounts of longitudinal steel, it sometimes becomes necessary at splice locations to lap bars (see ACI-​10.7.5 for lap splices of longitudinal bars in columns) inside the main circle of bars. This is done to maintain a minimum clear distance between individual longitudinal bars not less than the greatest of 1 1 2 times the nominal bar diameter, 1 1 2 in., and 4 3 times the maximum aggregate size (ACI-​25.2.3). The preferred and alternate bar arrangements are shown in Fig. 10.9.2. Bars may also be butt spliced by welding or mechanical connections (ACI-​10.7.5.1.1), thus permitting more utilization of available space. In a large column, an inner core of bars wrapped with a spiral or by ties may also be used.

10.10 LIMITS ON PERCENTAGE OF LONGITUDINAL REINFORCEMENT The percentage of total longitudinal reinforcement area Ast in terms of the gross cross-​ sectional area Ag must be between 1 and 8% (ACI-​10.6.1.1): that is, ρg = Ast /​Ag between 0.01 and 0.08. However, ACI-​10.3.1.2 permits basing the percentage on a reduced concrete area Ag when the gross concrete area is in excess of that needed for load considerations; but in no case may the reduced concrete area be less than half the actual concrete area (i.e., reinforcement ratio ρg may not be less than 0.005 based on the gross area provided). The primary purpose of these provisions for minimum steel is to prevent the failure mode from becoming that of a plain concrete column, which might be more disastrous than the sudden failure of tied columns previously described; in addition, these provisions are intended to avoid yielding of longitudinal steel under service loads due to load transfer from the concrete to the steel caused by creep. The upper limit on the amount of longitudinal reinforcement is a practical one in that if proper clearances are maintained between bars, little more than ρg = 0.08 can be put into the section. Thus the maximum ρg is, in a way, a double check on the minimum spacing restrictions of ACI-​25.2.3. The need for amounts of longitudinal steel close to this upper limit indicates that a small cross-​sectional area is being provided relative to the design loads, which could lead to problems associated with column stability and excessive shear stresses. A review of the longitudinal steel limits has been made by Lin and Furlong [10.43]. In order “to permit the use of reinforced concrete columns with small cross sections in lightly loaded structures, such as low-​rise residential and light office buildings,” ACI makes no restrictions on column dimensions (ACI Commentary-​R10.3.1). In such cases, “there is a greater need for careful workmanship, and shrinkage stresses have increased significance.”

3



10.12  BALANCED STRAIN CONDITION Pn Po

333

Pn(max) = 0.80Po for tied columns = 0.85Po for spirally reinforced columns

Pn(max)

e= 0

e = emin

Mn

e= ∞

Figure 10.11.1  Maximum nominal axial strength.

10.11 MAXIMUM STRENGTH IN AXIAL COMPRESSION—​A CI CODE Since a truly concentrically loaded column is rare, if not nonexistent, some minimal eccentricity should be provided for. Accidental eccentricity may result from end conditions, inaccuracy of manufacture, or variation in materials even when the load is theoretically concentric. Therefore, to account for a minimum eccentricity, the ACI Code prescribes (ACI-​22.4.2.1) that the maximum axial load nominal strength Pn(max) may not exceed 0.80Po for tied columns and 0.85Po for spirally reinforced columns (Fig. 10.11.1), with Po given by Eq. (10.5.1) or (10.5.2).

10.12 BALANCED STRAIN CONDITION The balanced strain condition, or compression control limit, represents the dividing point between the “compression-​controlled” and the “transition zone” regions of the strength interaction diagram (Fig.  10.6.1). Defined in the same manner as in Chapter  3, it is the simultaneous occurrence of a compressive strain of 0.003 in the extreme fiber of concrete and the tensile yield strain εy  =  fy /​Es on the layer of steel reinforcement closest to the tension face. Referring to the rectangular section in Fig. 10.12.1, the balanced strain condition gives cb 0.003 = d f y /Es + 0.003

cb =

0.003 87, 000 d d= f y + 87, 000 f y /[29(106 )] + 0.003

(10.12.1)

Force equilibrium requires

Pb = Cc + Cs − T

(10.12.2)

The compressive strength of concrete in a column subjected to pure axial load is k3 fc′ (see Eq. (10.3.3) where k3 is taken equal to 0.85. Under combined bending and axial load, the stress intensity of Whitney’s stress block should be taken as k3 (0.85 fc′ ). It has become common practice, however, to use a stress block intensity in columns of 0.85 fc′ , as in beams (i.e., k3 = 1.0). Such a difference in stress block intensity will not have significant impact in the calculated strength of a column below the balanced point; it may, however, lead to appreciable differences as the axial load approaches the pure axial load strength of the column. For

34

334

C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G e’ d”

Plastic centroid

N.A.

e = eb

As’

As

Pn = Pb

b

εy =

fy Es

cb

εs’

εcu = 0.003

d’

d h

Pn = Pb

Actual stress distribution

Average = 0.85k3fc’ T = Asfy

Cc

Cs

a = β1cb

Figure 10.12.1  Balanced strain condition—​rectangular section.

the balanced strain condition, this difference may be neglected and thus, k3 will be taken equal to 1.0. Based on this, the resultant concrete compressive force Cc in Eq. (10.12.2) can be expressed as

Cc = k3 (0.85 fc′ ab) = 1.0(0.85 fc′ ab) = 1.0(0.85 fc′β1cb b)

(10.12.3)

and the tension force T = As f y



(10.12.4)

If, moreover, compression steel yields at the balanced strain condition, Cs = As′ ( f y − 0.85 fc′ )

(10.12.5)

Pb = 0.85 fc′β1cb b + As′ ( f y − 0.85 fc′ ) − As f y

(10.12.6)

Thus Eq. (10.12.2) becomes

The eccentricity eb is measured from the plastic centroid, which was defined in Fig. 10.5.1. For symmetrical sections, the plastic centroid is at the middepth of the section. Moment equilibrium of the forces in Fig. 10.12.1 is satisfied by taking moments about any point such as the plastic centroid,

a   M b = Pb eb = Cc  d − − d ′′ + Cs (d − d ′ − d ′′ ) + Td ′′   2

(10.12.7)

35



335

10.12  BALANCED STRAIN CONDITION

EXAMPLE 10.12.1 Determine the axial force Pb and eccentricity eb corresponding to the balanced strain condition (compression control limit) for the section of Fig. 10.12.2. Use fc′ = 3000 psi, k3 = 1.0,  fy = 50,000 psi, and the ACI Code.

Figure 10.12.2  Balanced strain condition for Example 10.12.1.

SOLUTION (a) Locate the neutral axis for the balanced strain condition. cb =



0.003(21.6) = 13.72 in. 0.00172 + 0.003

ab = β1cb = 0.85(13.72) = 11.67 in. The value β1 is to be taken at 0.85 for fc′ ≤ 4000 psi (ACI-​22.2.2.4.3). (b) Compute the forces Cc, Cs, and T. Cc = 0.85(3.0)(11.67)(15) = 446 kips T = 50.0(2.37) = 119 kips

fy  13.72 − 2.4  ε s′ = 0.003  = 0.00248 > compression steel yiields   13.72  Es Cs = (2.37)[50.0 − 0.85(3.0)] = 119 − 6 = 113 kips

(c) Compute Pb and eb.

Pb = Cc + Cs − T = 446 + 113 − 119 = 440 kips

(Continued)

36

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C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G

Example 10.12.1 (Continued) For moment equilibrium about the plastic centroid, Pb eb = [ 446(12 − 11.67 / 2) + 113(12 − 2.4) + 119(12 − 2.4)]

1 12

= 229 + 90 + 95 = 414 ft-kips 414(12) eb = = 11.3 in. 440



On the given section, if Pn > 440 kips (or e < 11.3 in.), the section is referred to as compression controlled; if Pn < 440 kips (or e > 11.3 in.), the section may be in the “transition zone” (see Fig. 10.6.1) or, when e is large enough, the section would be tension controlled.

10.13 NOMINAL STRENGTH OF A COMPRESSION-​ CONTROLLED RECTANGULAR SECTION When the nominal compression strength Pn exceeds the balanced nominal strength Pb, or when the eccentricity e is less than the balanced value eb, or when εt at the extreme layer of steel at the face opposite the maximum compression face is less than the tensile yield strength εy, the section is “compression controlled.” The tensile force T [see Fig. 10.13.1(c)] will then be based on a tensile strain less than εy (see Fig. 10.6.1) and may actually be a compressive force if the eccentricity is small enough. The nominal strength Pn for a given eccentricity e < eb may be obtained by considering the strain variation (or neutral axis depth c) as the unknown and using the principles of statics. This is the most rational approach. As an alternative to the direct solution for the neutral axis as shown in the following examples, Reed [10.44] has presented an iterative procedure for analysis.

EXAMPLE 10.13.1 Determine the nominal compressive strength Pn for the section of Fig. 10.13.1 for an eccentricity e = 8 in. Use fc′ = 3000 psi, k3 = 1.0, fy = 50,000 psi, and the ACI Code.

Figure 10.13.1  Strain distribution and internal force resultants in a compression-​controlled section for Example 10.13.1.

(Continued)

37



10.13  COMPRESSION-C ​ O N T R O L L E D R E C TA N G U L A R S E C T I O N

337

Example 10.13.1 (Continued) SOLUTION (a) Determine whether the given eccentricity e is larger or smaller than eb. Since the balanced strain condition was computed in Example 10.12.1 as

Pb = 440 kips,

eb = 11.3 in



it is known that the section is compression controlled for e < 11.3 in. For e = 8 in., however, the position of the neutral axis c is not known. (b) Determine the location of the neutral axis. Since the actual c for e = 8 in. should exceed the value of cb = 13.72 in. and the value of ε s′ exceeds εy at the balanced strain condition, it is certain that ε s′ > ε y (see strain diagram of Fig. 10.12.2). Thus, referring to Fig. 10.13.1 for the forces and taking k3 = 1.0, Cs = As′ ( f y − 0.85 fc′ ) = 2.37(50.0 − 2.55) = 113 kips Cc = 0.85 fc′ b ( 0.85c ) = 2.55 (15)( 0.85c ) = 32.5c

Because ε s < ε y

T = As fs = As ( Es ε s ) = 2.37

(29, 000)(0.003)(21.60 − c) 4450 − 206c = c c

Taking moments about Pn,  0.85c  4450 − 206c (21.6 − 4.0) 0 = 113(4.0 − 2.40) − 32.5c  − 4.0 +  2  c

0 = c 3 − 9.41c 2 + 249.6c − 5673



c = 16.0 in. (c) Compute internal forces and strength Pn. Cs = 113 kips Cc = 32.5(16.0) = 520 kips

T=

4450 − 206(16.0) = 72.11 kips 16.0

Pn = 520 + 113 − 72.1 = 561 kips (d) Check by a moment equation about the plastic centroid.

561(8) ≈ 113(9.6) + 72.1(9.6) + 520(12 − 6.80) 4487 ≈ 4481



OK

EXAMPLE 10.13.2 Repeat the calculation of the nominal strength Pn for the section of Fig. 10.13.1, except consider that 1–​#8 is added in each of the 24-​in. faces of the member, at the middepth 12 in. from the compression face of the member (see Fig. 10.13.2). The eccentricity of Pn is 8 in., fc′ = 3000 psi, and fy = 50,000 psi. Use k3 = 1.0. (Continued)

38

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C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G

Example 10.13.2 (Continued) SOLUTION (a) Estimate whether the given eccentricity will cause the section to be compression controlled, in the transition zone, or tension controlled based on Fig. 10.6.1. Since the extra 2–​#8 bars are at the center of the section, they are unlikely to have an important effect on eb. Assume that the section will be compression controlled, since eb for Example 10.13.1 exceeds the 8-​in. eccentricity. Note that even if a wrong assumption is made, the neutral axis location thus obtained will reveal whether ε s′ < ε y or ε s > ε y , in which case the revised expression for Cs or T will have to be used in a new solution. (b) Estimate for each layer of reinforcement whether it is in compression or tension, and whether the strain at the layer will be greater or smaller than εy = 0.00172. The assumptions are as follows. For layer 1, 3–​#8 at 2.4 in. from the compression face; bars are in compression and are assumed to yield Cs1 = As′1 ( f y − 0.85 fc′ ) = 2.37(50 − 2.55) = 113 kips For layer 2, 2–​#8 at 12 in. from the compression face; bars are assumed to be in compression within the compressive stress block and assumed not to yield: 0.003 (c − 12.0) c  0.003  (c − 12.0)29, 000 − 0.85 fc′ Cs 2 = As′2  c  

ε s′2 =

 87  = 1.58  (c − 12.0) − 2.55 c  1650 = 137.5 − − 4.03 c 1650 = 133.5 − c





For layer 3, 3–​#8 at 21.6 in. from the compression face; bars assumed to be in tension and assumed not to yield  0.003  T = As  (21.6 − c)29, 000   c 



 4454  87 − 206.2 = 2.37  (21.6 − c) = c c 



(c) Determine the compressive force Cc as a function of the neutral axis depth c. From Example 10.13.1, Cc = 32.5c (d) Compute the neutral axis distance c. Taking moments about Pn,  0.85c  0 = 113(4.0 − 2.4) − 32.5c  − 4.0  2 

−

133.5c − 1650 4454 − 206.2c (8.0) + (21.6 − 4.0) c c

0 = c 3 − 9.41c 2 + 327.1c − 6629 c = 15.63 in.

a = 0.85(15.63)=13.3 in. (Continued)

39



10.13  COMPRESSION-C ​ O N T R O L L E D R E C TA N G U L A R S E C T I O N

339

Example 10.13.2 (Continued)

Figure 10.13.2  Analysis of a section containing an intermediate layer of steel for Example 10.13.2.

(e) Verify assumptions (see Fig. 10.13.2). For layer 1, 3–​#8 at 2.4 in. from the compression face,

0.003 (15.63 − 2.40) = 0.00254 > ε y 15.63

ε s′1 =

(as assumed )

For layer 2, 2–​#8 at 12 in. from the compression face,



ε s′2 =

0.003 (15.63 − 12.0) = 0.00070 < ε y 15.63

fs′2 = ε s′2 Es = 0.00070(29, 000) = 20.3 ksi

(as assumed )



These bars are in compression as assumed, and the depth a is greater than 12 in. Thus, the correction for displaced concrete (i.e., the −0.85 fc′ in the Cs2 force above) was appropriate. For layer 3, 3–​#8 at 21.60 in. from the compression face,

εs =

0.003 (21.60 − 15.63) = 0.00115 < ε y 15.63

(as assumed )

These bars are in tension as assumed. (Continued)

340

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C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G

Example 10.13.2 (Continued) (f) Compute internal forces and strength Pn. Cs1 = 113 kips Cs 2 = As′2 ( fs′2 − 0.85 fc′ ) = 1.58(20.3 − 2.55) = 28.0 kips

T = As ε s Es = 2.37(0.00115)29, 000 = 79.0 kips



Cc = 32.5c = 32.5(15.663) = 508 kips Pn = 113 + 28.0 − 79.0 + 508 = 570 kips (g) Check by a moment equation about the plastic centroid.

570(8) = 113(9.6) + 28.0(0) + 79.0(9.6) + 508(12 − 6.64) 4556 ≈ 4561

OK



Note that the additional two bars placed at the plastic centroidal axis increase Pn only from 561 kips (Example 10.13.1) to 570 kips (this example) for the same eccentricity of 8 in.

10.14 NOMINAL STRENGTH OF A RECTANGULAR SECTION WITH ECCENTRICITY e GREATER THAN THAT AT THE BALANCED STRAIN CONDITION When the eccentricity e is greater than the balanced value eb, or when the nominal compression strength Pn is less than the balanced nominal strength Pb, the net tensile strain εt at the extreme layer of steel at the face opposite the maximum compression face is greater than εy = fy /​Es. In such a case, the section behaves closer to a beam than a column (see Fig. 10.6.1). A rational approach to determine any point in the Pn–​Mn diagram below the balanced point corresponding to an eccentricity e is to designate the actual neutral distance c as unknown and apply statics. Alternatively, a value for c < cb may be first selected and the corresponding axial force, moment and eccentricity of the axial force determined through statics.

EXAMPLE 10.14.1 Determine the nominal compressive strength Pn for the member shown in Fig. 10.14.1 for an eccentricity e = 20 in., using fc′ = 3000 psi, fy = 50,000 psi, and the ACI Code. SOLUTION From the result of Example 10.12.1 it is known that e = 20 in. exceeds the eccentricity at the balanced strain condition eb = 11.3 in.; therefore the strain εs on the tension steel exceeds εy. It is assumed (initially) that the strain ε s′ on the compression steel is at least equal to the yield strain εy, although the validity of the assumption must be verified before the solution is accepted. Referring to Fig. 10.14.1 and using k3 = 1.0, the forces T, Cc, and Cs are T = As f y = 3(0.79)(50.0) = 119 kips

Cc = 0.85 fc′ab = 0.85(3.0)(0.85)(c)(15) = 32.5c



Cs = As′ ( f y − 0.85 fc′ ) = 3(0.79)(50 − 2.55) = 113 kips (Continued)

341



1 0 . 1 4   A   R E C TA N G U L A R S E C T I O N W I T H E C C E N T R I C I T Y e > e b

341

Example 10.14.1 (Continued)

Figure 10.14.1  Nominal strength Pn when e > eb for Example 10.14.1.

Force equilibrium requires Pn = Cc + Cs − T = 32.5c + 113 − 119 = 32.5c − 6.0 Taking moments arbitrarily about the tension steel, moment equilibrium gives

d − d′ a   Pn  e +  = Cc  d −  + Cs (d − d ′ )  2 2



(32.5c − 6.0)(20 + 9.6) = 32.5c(21.6 − 0.425c) + 113(19.2) c 2 + 18.82c − 169.9 = 0 c = 6.69 in. Therefore Cc = 32.5(6.69) = 217 kips



Pn = 217 − 6.0 = 211 kips



Verifying the correctness of the strain condition on the compression steel,



c − d′  6.69 − 2.40  = 0.003   = 0.00192  c 6.69  50 εy = = 0.00172 < 0.00192 29, 000

ε s′ = ε cu

OK

The nominal moment capacity Mn is then, M n = Pn e = 211(20)

1 12

= 350 ft-kips (Continued)

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Example 10.14.1 (Continued) The complete strength interaction diagram for the section is given in Fig. 10.14.2.

Figure 10.14.2  Strength interaction diagram for the section of Fig. 10.14.1.

10.15 DESIGN FOR STRENGTH—​R EGION I, MINIMUM ECCENTRICITY The approach to the design of members subjected to combined axial force and bending in accordance with the strength design method of the ACI Code may be divided into three categories (see Fig. 10.15.1): (a) design in Region I for a member having small or negligible bending moment (i.e., maximum permitted axial strength Pn(max) governs); (b) design for Region II in which a section is compression controlled but the strength Pn is less than maximum permitted strength of 0.80Po (tied columns) or 0.85Po (spirally reinforced columns); and (c) design for Region III in which e > eb or εt > εy. Designs in Region I, although permitted by the ACI Code, are nevertheless discouraged, since members subjected to compressive forces close to their pure axial load capacity possess low deformation capacity, and thus may fail without warning. Design in Region I occurs under the following conditions. 1. For braced or nonsway members (i.e., k ≤ 1) having low slenderness ratio kLu /​r such that the effects of slenderness may be neglected according to ACI-​6.2.5 (see also Section 10.7): (a) The member is subject to axial compression where the bending moment is considered negligible and is not computed. (b) The bending moment on the member is computed, but the corresponding eccentricity e = Mu /​Pu is less than emin (Fig. 10.15.1) corresponding to the maximum axial capacity. 2. For braced or nonsway members, where the effects of slenderness must be considered, computation of bending moment is required, with the result magnified by the factor δns in accordance with ACI-​6.6.4 (see Chapter 13) or obtained directly from a second-​ order analysis. When the computed eccentricity is less than emin (Fig. 10.15.1), design is in Region I.

34



343

10.15  DESIGN FOR STRENGTH—R ​ EGION I

Figure 10.15.1  Design categories for strength of a section under combined axial compression and bending moment.

An unbraced or sway member (i.e., k > 1.0) will rarely if ever be a design in Region I; details of the design procedure for unbraced compression members are to be found in Chapter 13. As all sections designed in Region I are compression controlled, a strength reduction factor φ = 0.65 (tied columns) or φ = 0.75 (spirally reinforced columns) must be used in design.

EXAMPLE 10.15.1 Design an axially loaded, spirally reinforced circular column for a gravity dead load of 345 kips and a live load of 405 kips using approximately 3.5% longitudinal reinforcement ratio. The column is of average height, and it will be assumed slenderness effects can be neglected. Use fc′ = 4000 psi, fy = 60,000 psi, and the ACI Code. SOLUTION (a) Determine the required nominal strength Pn.

Pu = 1.2(345) + 1.6(405) = 414 + 648 = 1062 kips required Pn =

Pu 1062 = = 1416 kips φ 0.75

(b) Determine the column size. ACI-​22.4.2.1 gives the maximum nominal strength in axial compression for spirally reinforced columns as

Pn(max) = 0.85Po



where Po is given by Eqs. (10.5.1) or (10.5.3). Thus Pn (max) = 0.85 Ag 0.85 fc′ + ρg ( f y − 0.85 fc′ ) 1416 = 0.85 Ag [0.85(4) + 0.035(60 − 3.40)] required Ag =

1416 = 310 sq in., 4.57

Try h = 20 in. with Ag = 314 sq in.



h(diameter) = 19.8 in. (Continued)

34

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Example 10.15.1 (Continued) (c) Determine reinforcement. Solve the Pn(max) equation for ρg. 1416 = 0.85(314)[3.40 + ρg (60 − 3.40)] required ρg = 0.0337





required Ast = ρg Ag = 0.0337(314) = 10.57 sq in. Use 20-​in. diameter column with 7–​#11 bars (As = 10.92 sq in.). No further check is required because no moment has been computed and the maximum nominal axial strength given by ACI-​22.4.2.1 governs in this case. (d) Design the spiral reinforcement. Using Eq. (10.9.3), which is ACI Formula (25.7.3.3),  Ag  f′ ρs = 0.45  − 1 c  Ach  f yt



Using 1.5-​in. clear cover to the spiral Ach =

π(20 − 3)2 = 227 sq in. 4

 314  4.0 − 1 ρs = 0.45  = 0.0115  227  60.0

Applying Eq. (10.3.2) gives

smax =

4 Asp

ρs Dc

=

4 Asp 0.0115(17)



which gives the data in Table 10.15.1. TABLE 10.15.1a  SPACING OF SPIRAL REINFORCEMENT FOR EXAMPLE 10.15.1

a

Bar

Asp (sq in.)

smax (in.)

#3 #4

0.11 0.20

2.25 4.10

Maximum Clear Spacing (in.) 1.87 3.59

Limitations (ACI-​25.7.3.1): 1.  Clear spacing 1 in. (assumed to be also greater than 4 3 times the maximum aggregate size).

Use #3 spiral at 2-​in. spacing.

345



345

10.16  DESIGN FOR STRENGTH—R ​ EGION II

10.16 DESIGN FOR STRENGTH—​R EGION II, COMPRESSION-​C ONTROLLED SECTIONS (e min < e  437 ft-kkips The calculated nominal moment capacity is approximately 25% greater than the required capacity, and thus a reduction in the amount of longitudinal steel is possible. Try 8–​#8 bars ( ρg = 0.016 > 0.01). For a design with 8–​#8 bars at Pn = 868 kips, c = 12.6 in., and Mn = 452 ft-​kips > 438 ft-​kips. Use 20-​in. square column with 8–​#8 bars, equally distributed in each face (Fig. 10.16.2). (g) Select lateral ties. Applying the provisions of ACI-​25.7.2, try #3 ties. Spacing limitations: 1.  Least lateral dimension = 20 in. 2.  16 longitudinal bar diameters = 16(1.0) = 16 in.  3 3.  48 tie bar diameters = 48   = 18 in.  8 Use #3 ties, 1 closed tie and a crosstie in each direction per set, at 16-​in. spacing. The final cross section is shown in Fig. 10.16.2. #3 crossties @ 16” Ast = 8 – #8 #3 ties @ 16”

20”

1.5”

20”

Figure 10.16.2  Section for Example 10.16.1.

Practical Design Approach In practice, compression member design is rarely done in the detailed manner illustrated throughout this chapter. Instead, designers usually use interaction diagram charts in nondimensional format or computer software. Various design aids are available, such as ACI-​SP-​17 (The Reinforced Concrete Design Handbook) [2.20], (which contains nondimensional interaction diagrams; the very useful ACI predecessor document SP-​7, by Everard and Cohen [10.46]; the CRSI Design Handbook 2008 [2.21]; and for SI, the Canadian Metric Design Handbook [10.47]. Typical interaction charts ( fc′ = 4000 psi, fy = 60,000 psi, and γ = 0.6 and 0.8) from Ref. 2.20 are given as Fig. 10.16.3. For L-​shaped columns, Marin [10.48] has provided design aids. Circular columns are treated in Section 10.18 of this book. Today, however, most column designs are performed using computer software based on a sectional analysis.

348

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Figure 10.16.3  Strength interaction diagrams for uniaxial bending and compression, symmetrical reinforcement,  fc′ = 4 ksi, fy = 60 ksi, γ = 0.60 and 0.80. (From ACI SP-​17 [2.20]).

349



349

10.16  DESIGN FOR STRENGTH—R ​ EGION II

EXAMPLE 10.16.2 Design a square tied column containing about 2% longitudinal reinforcement to carry a dead load axial compression of 770 kN and a bending moment of 68 kN·m, and a live load axial compression of 503 kN and bending moment of 33 kN·m. Use fc′ = 30 MPa, fy = 400 MPa, and the ACI Code. Assume the longitudinal reinforcement is distributed in equal amounts on two opposite faces. SOLUTION (a) Calculate factored axial force and moment and compute the eccentricity Pu = 1.2(770) + 1.6(503) = 1729 kN Mu = 1.2(68) + 1.6(33) = 134 kN ⋅ m



e=

134(1000) = 77.5 mm 1729

(b) Estimate e/​h and use the strength interaction charts in Fig.  10.16.3, which are for φ = 1.0. Note that the chart values of fc′ = 4000 psi and fy = 60,000 psi correspond closely to the given metric data. Alternatively, the Canadian Metric Design Handbook [10.47] can be used. Try the chart (Fig. 10.16.3) for γ = 0.8. Try e/​h = 0.2 (radial line from origin to vertical axis 1.0 intersected with horizontal axis 0.2) and use ρg = 0.02. From the chart, obtain Pn ≈ 0.70 fc′ Ag



Note that for e/​h  =  0.2, Pn > Pb and thus the section is compression controlled. Use φ = 0.65.



Pu 1729(1000) Pn φ 0.65 required Ag ≈ = = = 126, 700 mm 2 0.70(30) 0.70 fc′ 0.70(30)

Try a section 360 mm square (Ag = 129,600 mm2). Check assumed value for γ. For a cover to the center of the longitudinal bar equal to 65 mm,

γ=



360 − 2(65) = 0.64 < 0.8 360

Since the assumed value for γ  is not conservative (bars are farther from the column faces than assumed), either use chart for γ = 0.6 or interpolate between two charts. Compute chart horizontal axis value ( Pn / fc′Ag ) ( e / h ),

Pu /φ fc′ Ag

 e  1729(1000) / 0.65  77.5    =   = 0.684(0.215) = 0.147 h 30(129, 600)  360 

Enter Fig. 10.16.3 (interpolating for γ) with ( Pn / fc′Ag ) ( e / h ) = 0.147 and Pn / fc′Ag = 0.684; find ρg ≈ 0.023.

required Ast = 0.023(129, 600) = 2980 mm 2



Try 6–​#25M bars from Table 1.13.2, Ast = 3060 mm2. (Continued)

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Example 10.16.2 (Continued) (c) Check by statics. Based on the e/​h ratio and the curves of Fig. 10.16.3, the section is expected to be compression controlled. Referring to Fig. 10.16.4, Cc = 0.85 fc′ ba = 0.85(30)(360)a

1 1000

= 9.18a

a = β1c = 0.85c (see footnnote, Section 3.3) Cc = 7.80c

Assume compression steel yields: Cs = 3(510) [ 400 − 0.85(30)]



1 1000

= 573 kN

Taking Es = 200,000 MPa and d = 296 mm,  296 − c  1 T = 3(510)  (0.003)(200, 000) 1000  c 



271, 730 − 918c T= c



360 mm

#10M

6 – #25M 360 mm

40-mm clear cover

0.003 c = 277 mm 77.5 mm Pn

0.85fc’ 64 mm

T 235 mm

Cc Cs

Figure 10.16.4  Section, strain distribution, and internal force resultants for Example 10.16.2.

(Continued)

351



351

10.17  DESIGN FOR STRENGTH—R ​ EGION III

Example 10.16.2 (Continued) Taking moments about Pn gives T (296 − 102.5) + Cs (102.5 − 64) + Cc (102.5 − 0.425c) = 0

271, 730 − 918c (193.5) + 573(38.5) + 7.80c(102.5 − 0.425c) = 0 c c 3 − 241c 2 + 46, 930c − 15, 861,160 = 0 c = 277 mm

Check strains:

εy =

fy Es

=

400 = 0.00200 200, 000

At Cs,

 277 − 64  ε s′ = 0.003  = 0.00231 > ε y (compression steel yieldds)  277 

At T,

 296 − 277  ε s = 0.003  = 0.00021 < ε y  277 

This steel is in tension but does not yield; the section is compression controlled, as assumed. Compute forces: Cc = 7.80c = 7.80(277) = 2161 kN C s = 573 kN

T = 3(510)(0.00021)(200) = −64 kN



Pn = 2670 kN [φPn = 0.65(2670) = 1736 kN] > [ Pu = 1729 kN required]

OK

Compute nominal moment capacity by taking moments with respect to the plastic centroid: M n = [2161[180 − 0.5(235)] + 573(116) + 64(116)

1 1000

= 208 kN ⋅ m

φ M n = (0.65)208 = 135 kN ⋅ m > Mu = 134 kN ⋅ m



OK

Use 360 mm square column with 6–​#25M bars, as shown in Fig. 10.16.4.

10.17 DESIGN FOR STRENGTH—​R EGION III, TRANSITION ZONE AND TENSION-​ CONTROLLED SECTIONS (e > e b ) The “transition zone” (see Fig. 10.6.1) corresponds to the region where εt is greater than εy (e > eb), but less than 0.005. When εt exceeds 0.005, the section is tension controlled and φ = 0.90.

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The transition zone provides the variation in strength reduction factor φ from that corresponding to compression-​controlled sections to the φ factor for tension-​controlled sections, in accordance with ACI-​21.2.2. The φ factors of 0.65 and 0.75 for tied and spirally reinforced columns, respectively, are to be used when a section is compression controlled (i.e., when εt < εy). In the transition zone (Fig. 10.6.1), where the extreme tensile strain εt is between εy and 0.005, the φ factor is to be obtained by linear interpolation using either Eq. (3.6.3) or (3.6.4), as shown in Fig. 3.6.2. The linear equations in terms of the extreme tensile strain εt for Grade 60 bars are 1. For tied sections:  250  φ = 0.65 + (ε t − 0.002)  ≤ 0.90  3 



(10.17.1)

2. For spirally reinforced sections:

φ = 0.75 + ( ε t − 0.002 ) 50 ≤ 0.90



(10.17.2)

The φ equations can also be expressed in terms of the ratio c/​dt, where c is the neutral axis depth measured from the compression face of the member and dt is the distance from the extreme compressive fiber to the layer of reinforcement closest to the tension face. 1. For tied sections:  1 5 φ = 0.65 + 0.25  −  ≤ 0.90 / c d 3  t 



(10.17.3)

2. For spirally reinforced sections:



 1 5 φ = 0.75 + 0.15  −  ≤ 0.90  c /d t 3 

(10.17.4)

EXAMPLE 10.17.1 Design a rectangular tied column, not over 14 in. wide and with about 3% reinforcement, equally distributed on all four faces, to carry a service dead load of PD = 65 kips and MD = 144 ft-​kips and a service live load of PL = 49 kips and ML = 115 ft-​kips. Use fc′ = 4500 psi, fy = 60,000 psi, and the ACI Code. SOLUTION (a) Required nominal strengths and eccentricity. Pu = 1.2(65) + 1.6(49) = 156 kips

Mu = 1.2(144) + 1.6(115) = 357 ft -kips 357(12) e= = 27.5 in. 156 (Continued)

35



353

10.17  DESIGN FOR STRENGTH—R ​ EGION III

Example 10.17.1 (Continued) Assume a square section as a first approximation. Using Fig. 10.16.3 with γ = 0.8 ( fc′ is not the same as that given in the figure but is close enough; also, for designs below the balanced point, such a small difference in concrete compressive strength will have a negligible effect on column design) and e/​h = 27.5/​14 ≈ 2 (radial line from origin to vertical axis 0.2 intersected with horizontal axis 0.4) shows that for ρg = 0.03, Pn is well below Pb and probably εt > 0.005. Assuming φ = 0.85, required Pn = required M n =

156 = 184 kips 0.85 357 = 423 ft-kips 0.85

(b) Check assumed column depth. For e/​h = 2 and ρ = 0.03, the values of Kn and Rn in Fig. 10.16.3 are approximately 0.1 and 0.2, respectively. The required column area would thus be Ag ≥



Pn 184 = = 410 sq in. fc′ K n 4.5(0.10)

which requires a column depth h ≥ 29.3 in. The required column depth is much greater than the initially assumed value of 14 in. However, as the section depth increases, the normalized eccentricity e/​h decreases, leading to an increase in Kn. Assuming h = 20 in., e/​h ≈ 1.5. From Fig. 10.16.3 and using ρ = 0.03, the values of Kn and Rn are approximately 0.14 and 0.21, respectively. The minimum required column cross sectional area Ag = 292 sq in., which corresponds to h ≥ 20.9 in. Try a section 14 in. wide and 20 in. depth. For a reinforcement ratio ρg = 0.03 (note that the actual required reinforcement ratio should be slightly larger than 0.03 because the section depth selected is slightly smaller than the minimum required for that reinforcement ratio),

Ast ≥ ρg Ag = 0.03(14)(20) = 8.4 sq in.

Select 8–​#10 bars, Ast = 10.16 sq in., as shown in Fig. 10.17.1. (d) Check section strength using statics (Fig. 10.17.1). Calculate the neutral axis depth c corresponding to Pn = 184 kips. Assume first c = 7.5 in.

a = β1c = 0.825(7.5) = 6.19 in.

Strains and stresses in the reinforcing steel are



ε s1 = 0.0019,

fs1 = 55.1 ksi (compression)

ε s 2 = −0.0010,

fs 2 = −29.0 ksi (tension)

ε s 3 = −0.0039,

fs 3 = −60.0 ksi (tension)



Compression force in concrete: Cc = 0.85 fc′ ba = 0.85(4.5)(14)(6.19) = 331 kips Net axial force: Pn = Cc + Ast1 ( fs1 − 0.85 fc′ ) + Ast 2 fs 2 + Ast 3 fs 3

Pn = 331 + 3.81[ 55.1 − 0.85(4.5)] + 2.54( −29.0) + 3.81( −60) = 224 kips



(Continued)

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Example 10.17.1 (Continued) The calculated Pn is approximately 20% greater than the required Pn = 184 kips. The neutral axis depth c will thus be less than 7.5 in. For c = 7.1 in.,

a = β1c = 0.825(7.1) = 5.86 in.

Strains and stresses in the reinforcing steel are



ε s1 = 0.0018,

fs1 = 53.3 ksi (compression)

ε s 2 = 0.0012,

fs 2 = −35.5 ksi (tension)

ε s 3 = 0.0043,

fs 3 = −60.0 ksi (tension)



Compression force in concrete:

Cc = 0.85 fc′ ba = 0.85(4.5)(14)(5.86) = 314 kips



Net axial force:

Pn = Cc + Ast1 ( fs1 − 0.85 fc′ ) + Ast 2 fs 2 + Ast 3 fs 3 Pn = 314 + 3.81[53.3 − 0.885(4.5)] + 2.54( −35.5) + 3.81( −60) = 183 kips ≈ 184 kips



As εt = 0.0043, φ = 0.84 (Eq. 10.17.1) ≈ 0.85, as originally assumed. Calculate moment strength. Taking moments about column plastic centroid,  h a h  h  h  M n = Cc  −  + Ast1 ( fs1 − 0.85 fc′ )  − d1  + Ast 2 fs 2  − d2  Ast 3 fs 3  − d3   2 2 2  2  2   1 5.86   M n = 314  10 − 2  + 3.81 [53.3 − 0.85(4.5) ] (10 − 2.75) + 0 + 3.81( −60)(10 − 17 ..25) 12 = 437 ft-kkips > 423 ft-kips

Figure 10.17.1  Section, strain distribution, and internal force resultants and statics for Example 10.17.1.

10.18 CIRCULAR SECTIONS UNDER COMBINED COMPRESSION AND BENDING The concepts presented for the calculation of the strength Pn (acting at an eccentricity e from the plastic centroid) for a rectangular section are equally applicable to a circular section. The rectangular stress distribution may be applied to the concrete area under compression according to ACI-​22.2.2.3 and 22.2.2.4, though the use of statics requires knowing the area and centroid of area for circular segments. This information may be computed from formulas or by the use of coefficients for circular sections, as in Fig. 10.18.1. For a value on the strength interaction diagram for a tension-​controlled section, where the portion of the circular section within the rectangular stress distribution may be small, the relationship between the centroid of a circular segment and its area given by Fig. 10.18.2 may be useful.

35



355

10.18  CIRCULAR SECTIONS

When the steel reinforcement is in a circular arrangement, each bar may be located and treated in the same manner as a layer of steel in a rectangular section. The analysis would require first an assumption regarding whether the bar (or bars) will yield when εcu = 0.003, as for Example 10.13.2. When about eight or more bars are used, the analysis may treat the bars as a steel tube. Design aids for circular columns, both tied and spirally reinforced, are available from the Portland Cement Association [10.49], the American Concrete Institute [2.20], and the Concrete Reinforcing Steel Institute [2.21]. Mekonnen [10.50] has provided strength equations for circular beam-​columns. The CRSI has presented the equations to use for the interaction diagram using a programmable calculator [10.51], and several computer programs are available to compute interaction diagrams based on a sectional analysis. y

h cos θ 2 dx = h dθ sin θ 2 (x, y)

dθ α

h sin θ 2

θ

x

ηh 2 A of segment = h 2

Q0 of segment =

h3 4

4 I0 of segment = h 8

α

0

sin2 θ dθ = h2 α

0

0

h α – sin α cos α 4

sin2 θ cos θ dθ = h3

α

sin2 θ cos2 θ dθ = h4

sin3 α 12 4α – sin4 α 256

0.60

0.085

0.027

0.50

0.080

0.026

0.40

0.075

0.025

0.30

0.070

0.024

0.20

0.065

0.023

0.10

0.060

0.022

η = 0.3 0.4 0.5 0.6 0.7 A = (coefficient) h2

η = 0.3 0.4 0.5 0.6 0.7 Q0 = (coefficient) h3

η = 0.3 0.4 0.5 0.6 0.7 I0 = (coefficient) h4

Figure 10.18.1  Properties of circular segments.

Figure 10.18.2  Centroid of circular segment (measured from center of circle) versus its area.

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EXAMPLE 10.18.1 For the circular column shown in Fig.  10.18.3, calculate the axial force and bending moment corresponding to the balanced condition (i.e., compression control limit). Use fc′ = 5000 psi,  fy = 60,000 psi, and the ACI Code.

Figure 10.18.3  Cross section, strains, and internal force resultants in column of Example 10.18.1.

SOLUTION Referring to Fig. 10.18.3, the depth of the neutral axis corresponding to the balanced condition or compression control limit can be calculated as  1.27    20 −  1.5 + 0.5 +      d5 2   = 10.28 in. cb = 0.003   = 0.003   0.00207 + 0.003   ε y + 0.003     



and the depth of the compression stress block, a, is a = β1cb = 0.80(10.28) = 8.23 in. which corresponds to an angle α  =  79.8° in Fig.  10.18.1 and 10.18.3. Referring to Fig. 10.18.1, the resultant compressive force in the concrete can be calculated as Cc = 0.85 fc′ Acomp = 0.85(5)(121.8) = 518 kips



which is located at yc = 5.22 in. from the center of the column. The tensile strains in the longitudinal steel can be calculated based on the position of the reinforcement with respect to the neutral axis depth (third column of Table 10.18.1). Once these strains have been calculated, the stresses and corresponding forces in the reinforcement can be determined, as shown in Table 10.18.1. TABLE 10.18.1  Layer i

Asti (in.2)

ysi + (cb −h/​2) (in.)

εsi

fsi (ksi)

Fsi (kips)

1 2 3 4 5

1.27 2.54 2.54 2.54 1.27

7.66 5.50 0.28 –​4.93 –​7.09

0.00223 0.00160 0.00008 –​0.00144 –​0.00207

60–​4.25a 46.5–​4.25a 2.40 –​41.7 –​60

70.8 107 6.09 –​106 –​76.2

a

Stress is adjusted to account for displaced concrete. Tensile strains, stresses, and forces are negative.

(Continued)

357



357

10.19  COMBINED AXIAL TENSION AND BENDING

Example 10.18.1 (Continued) The axial force corresponding to the balanced condition or compression control limit can then be determined as

i=5

Pn = Cc + ∑ Fsi = 518 + 2.1 = 520 kips i =1

The corresponding nominal flexural strength can be calculated by taking moments about the plastic centroid of the column (i.e., column center).

h  i=5 M n = Cc  − yc  + ∑ Fsi ysi = 225 + 183 = 408 ft-kips 2  i =1

10.19 COMBINED AXIAL TENSION AND BENDING In the relatively uncommon situation of axial tension in combination with bending moment, the strength interaction diagram can be considered to extend on the negative Pn side of the axis, as shown in Fig. 10.19.1. The strength analysis for tension values of Pn is similar to that used for compressive load in the region for a tension-​controlled section (Fig. 10.6.1). When the eccentric load is tensile, the neutral axis distance c will be smaller than it is under pure bending (M0). When axial tension and bending moment exist, the usual tendency will be to proportion the section in a manner similar to the design as a beam. The subject of axial tension combined with bending moment has been discussed by Harris [10.52], Moreadith [10.53], and Villalta, Carreira, and Erler [10.54]. For sections subjected to pure axial tension force, the nominal axial tensile strength of the section is taken as the yield strength of the longitudinal reinforcement (ACI-​22.4.3.1). Examples 10.19.1 and 10.19.2 illustrate the calculation of points on the tension side of the interaction diagram (Fig. 10.19.1).

EXAMPLE 10.19.1 Determine the maximum value for the tensile force Pn when no bending moment is acting on the section shown in Fig. 10.19.1.

Figure 10.19.1  Strength interaction diagram showing both axial compression and axial tension regions.

(Continued)

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Example 10.19.1 (Continued) SOLUTION Since the concrete will crack before the steel yields, only the steel participates in carrying axial tension. Thus, according to ACI-​22.4.3.1,

Pn = Ast f y = 6(0.79)50 = 237 kips

This is plotted on Fig. 10.19.1.

EXAMPLE 10.19.2 Determine the axial tensile strength Pn on the section of Fig. 10.19.1 when the eccentricity is 20 in. SOLUTION Figure 10.19.2 shows that the eccentric tensile force Pn must be acting on the tension side of the plastic centroid. In this case, the neutral axis distance c is smaller than its value for pure bending. For bending alone, Cc = 0.85 fc′ ba = 0.85(3)(15)a = 38.3a



T = As f y = 3(0.79)50 = 118.5 kips



Assuming the compression steel yields, Cs = 3(0.79)(50 − 2.55) = 112.5 kips

C=T 118.5 − 112.5 a= = 0.16 in.;; 38.3

0.16 c= = 0.19 1

Design is therefore not adequate according to the load contour method.

(Continued)

367



10.20  COMBINED AXIAL FORCE AND BIAXIAL BENDING

367

Example 10.20.2 (Continued) 2500

Pn (kips)

2000 1500 Pnx = 925 kips

1000 500

ey =

0

100

in.

6.1

Mox

200

300

400

500

600

Mnx (ft-kips) 2500

Pn (kips)

2000 1500 Pny = 1075 kips

1000

.1

e

500

0

100

=4 x

in.

200

Moy 300

400

500

600

Mny (ft-kips)

Figure 10.20.9  Determination of Pnx and Pny for section of Example 10.20.2.

(d) A more exact analysis of this section may be made by following the procedure in Example 10.20.1. For a nominal axial force Pn = Pu /​φ = 440/​0.65 = 677 kips and a moment ratio Mnx /​Mny = 1.5, the following values are obtained: c = 14.55 in.

α = 47.3°

M nx = 324 ft-kips M ny = 217 ft-kips

For this case, the strain in the extreme tension bar εt = 0.0015, which is less than the yield strain εy. Thus, φ = 0.65. Even if εt had been slightly larger than εy, the use of φ corresponding to the compression control limit would still be recommended, since using a larger φ value based on only one bar reaching a strain greater than εy would be questionable. The design moments for bending about the x-​ and y-​axes are then

φ M nx = 0.65(324) = 211 ft-kips < 225 ft-kips φ M ny = 0.65(217) = 141 ftt-kips < 150 ft-kips

The section is therefore inadequate, as already demonstrated through the use of the reciprocal load method.

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10.21 DESIGN FOR SHEAR Design of columns for shear follows the procedures outlined in Chapter 5. Because columns are often subjected to significant levels of axial compression, the shear strength contribution attributed to the concrete, Vc, is typically calculated using Eq. (5.13.1), given in ACI-​22.5.6.1, which is repeated below for convenience.



 Nu  Vc = 2  1 +  λ fc′bw d  2000 Ag 

[5.13.1]

For members with axial tension, Eq. (5.13.6) may be used. The authors recommend, however, that the concrete contribution to shear strength be ignored (i.e., Vc = 0) when there is significant axial tension. In applying Eq. (5.13.1) for columns, a question may arise as to the value of the effective depth d to be used in the calculations. Because the amount of steel in tension depends on the eccentricity (e = Mu  /​Pu) being considered, different values of d may be computed. For example, at very low eccentricities, it is possible that no reinforcement is in tension (i.e., all reinforcement is in compression), whereas at high eccentricities several layers of reinforcement might be in tension. For circular sections of diameter D, ACI-22.5.2.2 allows the use of the same shear strength equations used for rectangular sections, with d = 0.8D and bw = D. However, the ACI Code is silent on the calculation of d for rectangular columns. In this case, the authors recommend taking d = 0.8h, similar to the case of circular columns. Once Vc has been calculated, the amount of transverse reinforcement required for shear resistance is obtained by Vs ≥



Vu − Vc φ

(10.21.1)

where (ACI-​22.5.10.5.3 and ACI-22.5.1.2)

Vs =

Av f yt d s

≤ 8 fc′ bw d

[5.10.7]

In Eq. (5.10.7), Av is the area of shear reinforcement within a spacing s, fyt is the yield strength of the transverse reinforcement, and bw is the web width (for solid rectangular sections, bw is equal to the section dimension perpendicular to the direction of shear). However, wherever Vu > 0.5φ Vc , the area of shear reinforcement Av must be at least (ACI-​10.6.2),

min Av = 0.75 fc′

bw s 50bw s ≥ f yt f yt

[5.10.8]

Maximum spacing of shear reinforcement depends on the shear force required to be resisted by the transverse reinforcement as follows,



For Vs ≤ 4 fc′ bw d ,

s ≤ d / 2 ≤ 24 in.

For Vs > 4 fc′ bw d ,

s ≤ d / 4 ≤ 12 in.

In addition to these requirements, transverse reinforcement in tied columns must satisfy the requirements for lateral support discussed in Section 10.8. For spirally reinforced columns, spiral reinforcement must satisfy the minimum volumetric ratio and spacing requirements discussed in Section 10.9.

369



369

10.21  DESIGN FOR SHEAR

EXAMPLE 10.21.1 The column section shown in Fig. 10.21.1(a) is subjected to a factored shear force Vu = 75 kips parallel to the long direction and a factored axial compressive force Pu = 420 kips. Design the transverse reinforcement such that shear and lateral support requirements are satisfied. Use fc′ = 4000 psi (normal weight) and fyt = 60 ksi. #3 ties @ 8 in.

10 – #7 bars 2.5” 6.5” 18” 6.5” 2.5” 2.5” 5.5” 6” 5.5” 2.5”

2.5” 5.5” 6” 5.5” 2.5”

22”

22”

(a) Column section

(b) Layout of lateral ties

Figure 10.21.1  Cross section and layout of lateral ties for column in Example 10.21.1.

SOLUTION (a) Determine the shear strength attributed to the concrete (Vc) by using Eq. (5.13.1) and the required contribution from shear reinforcement. Taking d = 0.8h = 17.6 in.,  Nu  Vc = 2  1 +  λ fc′ bw d  2000 Ag  420, 000   1 = 2 1 +  (1.0) 4000 (18)(17.6) 1000  2000(18)(22) 

= 61.3 kips V 75 1 = 160 kips Vs ≥ u − Vc = − 61.3 = 38.7 kips < 8 4000 (18)(17.6) 1000 φ 0.75



(b) Determine maximum spacing of shear reinforcement.

Vs = 38.7 kips < 4 fc′bw d = 4 4000 (18)(17.6)

1 1000

= 80.1 kips

Thus, s ≤ d/​2 = 17.6/​2 = 8.8 in. ≤ 24 in. Select s = 8 in. (c) Verify that transverse reinforcement spacing satisfies requirement for lateral bar support. Assume #3 ties. From Section 10.8, maximum tie spacing shall not exceed the smallest of 16 times the longitudinal bar diameter [16( 7 8 ) = 14 in.)], 48 times the tie diameter [48( 3 8 ) = 18 in.], or the minimum cross-​sectional dimension (18 in.). The selected spacing s  =  8 in. is therefore adequate for lateral bar support. (d) Determine required area of shear reinforcement. From Eqs. (10.21.1) and (5.10.7),

Av ≥

Vs s (38.7)(8) = = 0.29 sq in. f yt d (60)(17.6) (Continued)

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Example 10.21.1 (Continued) Select #3 ties with three legs parallel to the long direction [Av = 3(0.11) = 0.33 sq in.] by using a closed tie and a crosstie as shown in Fig. 10.21.1(b). The crosstie is anchored by a 135° hook on one end and a 90° hook on the other end. The closed tie, on the other hand, may be anchored by 90° hooks. In this example, however, 135° hooks are used [Fig. 10.21.1(b)]. (e) Verify that the selected reinforcement satisfies minimum shear reinforcement requirement. From Eq. (5.10.8) and noting that 0.75 fc′ < 50 psi for  fc′ = 4000 psi,

min Av =

50bw s 50(18)(8) = = 0.12 sq in. < 0.33sq in. f yt 60, 000

OK

(f) Provide additional ties (if needed) to satisfy requirements for lateral support of longitudinal bars. As discussed in Section 10.8, ties shall be arranged such that every corner and alternate longitudinal bar is laterally supported by the corner of a tie having an included angle of not more than 135°. In addition, no longitudinal bar shall be farther than 6 in. clear on either side from such a laterally supported bar. This will require additional ties parallel to the short direction of the column, to support the two intermediate longitudinal bars along each long face of the column. A closed tie enclosing the four intermediate longitudinal bars, as shown in Fig. 10.21.1(b), satisfies this requirement. Maximum spacing for these ties would be 14 in., assuming that shear reinforcement requirements in that direction do not control. For practical purposes, however, it is convenient to use the same spacing for all ties. Thus, select a spacing of 8 in. Final design for column transverse reinforcement is shown in Fig. 10.21.1(b).

SELECTED REFERENCES Historical 10.1 A.  Considère. “Compressive Resistance of Concrete Steel and Hooped Concrete, Part I,” Engineering Record, December 20, 1902, 581–​583; Part II, December 27, 1902, 605–​606. 10.2 A. Considère. “Concrete-​Steel and Hooped Concrete.” Reinforced Concrete, 1903, p. 119. 10.3 A. N. Talbot. “Tests of Concrete and Reinforced Concrete Columns,” Bulletin, No. 10, 1906, and No. 20, 1907, University of Illinois, Urbana. 10.4 M. O. Withey. “Tests of Plain and Reinforced Concrete Columns,” Engineering Record, July 1909, 41. 10.5 M. O. Withey. “Tests on Reinforced Concrete Columns,” Bulletin, No. 300, 1910, and No. 466, 1911, University of Wisconsin, Madison. 10.6 ACI Committee 105. “Reinforced Concrete Column Investigation,” ACI Journal, Proceedings, 26, April 1930, 601–​612; 27, February 1931, 675–​676; 28, November 1931, 157–​158; 29, September 1932, 53–​56: 29, February 1933, 275–​284; 30, September–​October 1933, 78–​90; 30, November–​December 1933, 153–​156. 10.7 Joint ACI-​ASCE Committee 441. Reinforced Concrete Columns (annotated bibliography), ACI Bibliography No. 5. Detroit: American Concrete Institute, 1965.

Confinement, Spirals, and Ties 10.8 M.  J. N.  Priestley, R.  Park, and R.  T. Potangaroa. “Ductility of Spirally-​Confined Concrete Columns,” Journal of the Structural Division, ASCE, 107, ST1 (January 1981), 181–​202. 10.9 S.  H. Ahmad and S.  P. Shah. “Stress–​ Strain Curves of Concrete Confined by Spiral Reinforcement.” ACI Journal, Proceedings, 79, November–​December 1982, 484–​490.

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SELECTED REFERENCES

371

10.10 Salvador Martinez, Arthur H. Nilson, and Floyd O. Slate. “Spirally Reinforced High-​Strength Concrete Columns,” ACI Journal, Proceedings, 81, September–​October, 1984, 431–​442. 10.11 Shamim A. Sheikh and S. M. Uzumeri. “Strength and Ductility of Tied Concrete Columns,” Journal of the Structural Division, ASCE, 106, ST5 (May 1980), 1079–​1102. 10.12 B. D. Scott, R. Park, and M. J. N. Priestley. “Stress-​Strain Behavior of Concrete Confined by Overlapping Hoops at Low and High Strain Rates,” ACI Journal, Proceedings, 79, January–​ February 1982, 13–​27. 10.13 Robert Park, M. J. Nigel Priestley, and Wayne D. Gill. “Ductility of Square-​Confined Concrete Columns,” Journal of the Structural Division, ASCE, 108, ST4 (April 1982), 929–​951. 10.14 Apostolos Fafitis and Surendra P. Shah. “Predictions of Ultimate Behavior of Confined Columns Subjected to Large Deformations,” ACI Journal, Proceedings, 82, July–​August 1985, 423–​433. 10.15 Jack P. Moehle and Terry Cavanagh. “Confinement Effectiveness of Crossties in RC,” Journal of Structural Engineering, ASCE, 111, 10 (October 1985), 2105–​2120. 10.16 Shamim A. Sheikh and C. C. Yeh. “Flexural Behavior of Confined Concrete Columns,” ACI Journal, Proceedings, 83, May–​June 1986, 389–​404. 10.17 Güney Özcebe and Murat Saatcioglu. “Confinement of Concrete Columns for Seismic Loading,” ACI Structural Journal, 84, July–​August 1987, 308–​315. 10.18 Yook-​Kong Yong, Malakah G. Nour, and Edward G. Nawy. “Behavior of Laterally Confined High-​Strength Concrete under Axial Loads.” Journal of Structural Engineering, ASCE, 114, 2 (February 1988), 332–​351. 10.19 Hisham Adbel-​Fattah and Shuaib H.  Ahmad. “Behavior of Hoop-​Confined High-​Strength Concrete under Axial and Shear Loads,” ACI Structural Journal, 86, November–​December 1989, 652–​659. 10.20 Shamim A.  Sheikh and Ching-​Chung Yeh. “Tied Concrete Columns under Axial Load and Flexure,” Journal of Structural Engineering, ASCE, 116, 10 (October 1990), 2780–​2800. 10.21 Shamim A.  Sheikh and C.  C. Yeh. “Concrete Strength in Tied Columns,” ACI Structural Journal, 87, July–​August 1990, 379–​385. 10.22 Salim R.  Razvi and Murat Saatcioglu. “Confinement of Reinforced Concrete Columns with Welded Wire Fabric,” ACI Structural Journal, 86, September–​October 1989, 615–​623.

Interaction Diagrams 10.23 E.  Hognestad. “A Study of Combined Bending and Axial Load in Reinforced Concrete Members,” Bulletin, No. 399, November 1951, Engineering Experiment Station, University of Illinois, Urbana (117 references). 10.24 ACI-​ASCE Committee 327. “Report on Ultimate Strength Design,” Proceedings ASCE, 81, October 1955, Paper No. 809. See also ACI Journal, Proceedings, 52, January 1956, 505–​524. 10.25 A. H. Mattock, L. B. Kriz, and Eivind Hognestad. “Rectangular Concrete Stress Distribution in Ultimate Strength Design,” ACI Journal, Proceedings, 57, February 1961, 875–​928. 10.26 E.  O. Pfrang, C.  P. Siess, and M.  A. Sozen. “Load–​Moment–​Curvature Characteristics of Reinforced Concrete Cross Sections,” ACI Journal, Proceedings, 61, July 1964, 763–​778. Disc., 62, 3, 1673–​1683. 10.27 Parviz Soroushian and Kienuwa Obaseki. “Strain Rate-​ Dependent Interaction Diagrams for Reinforced Concrete Sections,” ACI Journal, Proceedings, 83, January–​February 1986, 108–​116. 10.28 Shamim A. Sheikh and C. C. Yeh. “Analytical Moment–​Curvature Relations for Tied Concrete Columns,” Journal of Structural Engineering, ASCE, 118, 2 (February 1992), 529–​544. 10.29 Michael C. Head and J. Dario Aristizabal-​Ochoa. “Analysis of Prismatic and Linearly Tapered Reinforced Concrete Columns,” Journal of Structural Engineering, ASCE, 113, 3 (March 1987), 575–​589. 10.30 Abi O. Aghayere. “Design of Nonprismatic Concrete Columns by the Degree of Fixity Method,” ACI Structural Journal, 87, July–​August 1990, 473–​478. 10.31 M.  Ala Saadeghvaziri and Douglas A.  Foutch. “Behavior of RC Columns under Nonproportionally Varying Axial Load,” Journal of Structural Engineering, ASCE, 116, 7 (July 1990), 1835–​1856. 10.32 James G.  MacGregor, John E.  Breen, and Edward O.  Pfrang. “Design of Slender Concrete Columns,” ACI Journal, Proceedings, 67, January 1970, 6–​28.

372

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10.33 F. E. Richart, J. O. Draffin, T. A. Olson, and R. H. Heitman. “The Effect of Eccentric Loading, Protective Shells, Slenderness Ratios, and Other Variables in Reinforced Concrete Columns,” Bulletin No. 368, Engineering Experiment Station, University of Illinois, Urbana, 1947, 130 pp. 10.34 B.  Bresler and P.  H. Gilbert. “Tie Requirements for Reinforced Concrete Columns,” ACI Journal, Proceedings, 58, November 1961, 555–​570; Disc., 58, 897–​907. 10.35 James F.  Pfister. “Influence of Ties on the Behavior of Reinforced Concrete Columns,” ACI Journal, Proceedings, 61, May 1964, 521–​537. 10.36 Edwin G.  Burdette and Hubert K.  Hilsdorf. “Behavior of Laterally Reinforced Concrete Columns,” Journal of the Structural Division, ASCE, 97, ST2 (February 1971), 587–​602. 10.37 S. T. Mau. “Effect of Tie Spacing on Inelastic Buckling of Reinforcing Bars,” ACI Structural Journal, 87, November–​December 1990, 671–​677. 10.38 Koji Sakai and Shamim A.  Sheikh. “What Do We Know about Confinement in Reinforced Concrete Columns? (A Critical Review of Previous Work and Code Provisions),” ACI Structural Journal, 86, March–​April 1989, 192–​207. 10.39 Grant T. Halvorsen and Craig A. Carinci. “Tie Requirements for Prestressed Concrete Columns,” PCI Journal, 22, July–​August 1987, 46–​79. 10.40 J. S. Ford, D. C. Chang, and J. E. Breen. “Behavior of Concrete Columns Under Controlled Lateral Deformation,” ACI Journal, Proceedings, 78, January–​February 1981, 3–​20. 10.41 Ti Huang. “On the Formula for Spiral Reinforcement,” ACI Journal, Proceedings, 61, March 1964, 351–​353. Disc., 61, 9 1241–​1248. 10.42 William L. Gamble. “Re-​examination of Spiral Steel Requirements,” Concrete International, 8, November 1986, 33–​34. 10.43 Chien-​Hung Lin and Richard W. Furlong. “Longitudinal Steel Limits for Concrete Columns,” ACI Structural Journal, 92, May–​June 1995, 282–​287.

Design 10.44 Paul W. Reed. “Simplified Analysis for Thrust and Moment of Concrete Sections,” ACI Journal, Proceedings, 77, May–​June 1980, 195–​200. 10.45 Curtis J. Young. “Direct Selection of Concrete Dimensions in Columns,” Concrete International, 3, October 1981, 27–​31. 10.46 N. J. Everard and Edward Cohen. Ultimate Strength Design of Reinforced Concrete Columns (SP-​7). Detroit: American Concrete Institute, 1964, 182 pp. 10.47 Metric Design Handbook for Reinforced Concrete Elements in Accordance with the Strength Design Methods of CSA Standard CAN3-​A23.3-​M77. (Edited by Murat Saatcioglu). Ottawa, Ontario, Canada: Canadian Portland Cement Association, 1980. 10.48 Joaquin Marin. “Design Aids for L-​Shaped Reinforced Concrete Columns,” ACI Journal, Proceedings, 76, November 1979, 1197–​1216. 10.49 PCA. Ultimate Load Tables for Circular Columns. Chicago: Portland Cement Association, 1960. 10.50 Bekele Mekonnen. “Reinforced Concrete Column Design Equations,” ACI Journal, Proceedings, 81, May–​June 1984, 242–​250. 10.51 CRSI. Interaction Diagrams for Uniaxial Bending Moment and Axial Force on Reinforced Concrete Circular Sections Using a Programmable Calculator, Structural Bulletin No. 12. Schaumburg, IL: Concrete Reinforcing Steel Institute, June 1985.

Axial Tension and Bending 10.52 E. C. Harris. “Design of Members Subject to Combined Bending and Tension,” ACI Journal, Proceedings, 72, September 1975, 491–​495. 10.53 F. L. Moreadith. “Design of Reinforced Concrete for Combined Bending and Tension,” ACI Journal, Proceedings, 75, June 1978, 251–​255, Disc., 75, December 1978, 721–​723. 10.54 Fernando Villalta, Domingo J. Carreira, and Bryan Erler. Discussion of “Design of Reinforced Concrete for Combined Bending and Tension,” ACI Journal, Proceedings, 75, December 1978, 721–​723.

37



SELECTED REFERENCES

373

Biaxial Bending 10.55 Boris Bresler. “Design Criteria for Reinforced Columns under Axial Load and Biaxial Bending,” ACI Journal, Proceedings, 57, November 1960, 481–​490. Disc., 1621–​1638. 10.56 Richard W. Furlong. “Ultimate Strength of Square Columns under Biaxially Eccentric Loads,” ACI Journal, Proceedings, 57, March 1961, 1129–​1140. 10.57 F. N. Pannell. “Failure Surfaces for Members in Compression and Biaxial Bending,” ACI Journal, Proceedings, 60, January 1963, 129–​140. 10.58 Alfred L.  Parme, Jose M.  Nieves, and Albert Gouwens. “Capacity of Reinforced Rectangular Columns Subject to Biaxial Bending,” ACI Journal, Proceedings, 63, September 1966, 911–​923. 10.59 Donald C.  Weber. “Ultimate Strength Design Charts for Columns with Biaxial Bending,” ACI Journal, Proceedings, 63, November 1966, 1205–​1320. Disc., 64, 6, Part 2, Vol.64 No.6, Part 2. 1583–​1586. 10.60 Anis Farah and M.  W. Huggins. “Analysis of Reinforced Concrete Columns Subjected to Longitudinal Load and Biaxial Bending,” ACI Journal, Proceedings, 66, July 1969, 569–​575. 10.61 J. C. Smith. “Biaxially Loaded Concrete Interaction Curve,” Computers and Structures, 3 (1973), 1461–​1464. 10.62 K. N. Smith and W. H. Nelles. “Columns Subjected to Biaxial Bending—​Preliminary Selection of Reinforcing,” ACI Journal, Proceedings, 71, August 1974, 411–​413. 10.63 Peter D. Heimdahl and Albert C. Bianchini. “Ultimate Strength of Biaxially Eccentrically Loaded Concrete Columns Reinforced with High Strength Steel,” Reinforced Concrete Columns (SP-​50). Detroit: American Concrete Institute, 1975 (pp. 93–​117). 10.64 S.  I. Abdel-​ Sayed and N.  J. Gardner. “Design of Symmetric Square Slender Reinforced Concrete Columns under Biaxially Eccentric Loads,” Reinforced Concrete Columns (SP-​50). Detroit: American Concrete Institute, 1975 (pp. 149–​164). 10.65 A. K. Basu and P. Suryanarayana. “Analysis of Restrained Reinforced Concrete Columns under Biaxial Bending,” Reinforced Concrete Columns (SP-​50). Detroit: American Concrete Institute, 1975 (pp. 211–​232). 10.66 Albert J.  Gouwens. “Biaxial Bending Simplified,” Reinforced Concrete Columns (SP-​50). Detroit: American Concrete Institute, 1975 (pp. 233–​261). 10.67 W.  F. Chen and M.  T. Shoraka. “Tangent Stiffness Method for Biaxial Bending of Reinforced Concrete Columns,” IABSE Publications, International Association for Bridge and Structural Engineering, Part 35-​I, 1975, 23–​44. 10.68 Richard W.  Furlong. “Concrete Columns under Biaxially Eccentric Thrust,” ACI Journal, Proceedings, 76, October 1979, 1093–​1118. 10.69 M. Pinto de Magalhaes. “Biaxially Loaded Concrete Sections,” Journal of the Structural Division, ASCE, 105, ST12 (December 1979), 2639–​2656. 10.70 L. N. Ramamurthy and T. A. Hafeez Khan. “L-​Shaped Column Design for Biaxial Eccentricity,” Journal of Structural Engineering, ASCE, 109, 8 (August 1983), 1903–​1917. 10.71 Michael A. Taylor. “Direct Biaxial Design of Columns,” Journal of Structural Engineering, ASCE, 111, 1 (January 1985), 158–​173. 10.72 Cheng-​Tzu Thomas Hsu. “Biaxially Loaded L-​Shaped Reinforced Concrete Columns,” Journal of Structural Engineering, ASCE, 111, 12 (December 1985), 2576–​2595. 10.73 Cheng-​Tzu Thomas Hsu. “Reinforced Concrete Members Subject to Combined Biaxial Bending and Tension,” ACI Structural Journal, 83, January–​February 1986, 137–​144. 10.74 David A.  Ross and J.  Richard Yen. “Interactive Design of Reinforced Concrete Columns with Biaxial Bending,” ACI Structural Journal, 83, November–​December 1986, 988–​993. 10.75 Cheng-​Tzu Thomas Hsu. “Channel-​Shaped Reinforced Concrete Compression Members under Biaxial Bending,” ACI Structural Journal, 84, May–​June 1987, 201–​211. 10.76 Troels Brøndum-​ Nielsen. “Concrete Sections under Biaxial Bending,” Journal of Structural Engineering, ASCE, 113, 10 (October 1987), 2137–​2144. 10.77 Cheng-​Tzu Thomas Hsu. “Analysis and Design of Square and Rectangular Columns by Equation of Failure Surface,” ACI Structural Journal, 85, March–​April 1988, 167–​179. 10.78 Issam E. Harik and Hans Gesund. “Flowcharts for Biaxial Bending in R/​C Tied Columns,” Journal of Structural Engineering, ASCE, 114, 6 (June 1988), 1230–​1249; Disc., 116, 2 (February 1990), 556–​557. 10.79 Chu-​Kia Wang. “Solving the Biaxial Bending Problem in Reinforced Concrete by a Three-​Level Iteration Procedure,” Microcomputers in Civil Engineering, 3 (1988), 311–​320. 10.80 F. A. Zahn, R. Park, and M. J. N. Priestley. “Strength and Ductility of Square Reinforced Concrete Column Sections Subjected to Biaxial Bending,” ACI Structural Journal, 86, March–​April 1989, 123–​131. 10.81 Cheng-​Tzu Thomas Hsu. “T-​Shaped Reinforced Concrete Members under Biaxial Bending and Axial Compression,” ACI Structural Journal, 86, July–​August 1989, 460–​468.

374

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PROBLEMS All problems* are to be worked in accordance with the ACI Code unless otherwise indicated, and all stated loads are service loads. Note that eccentricities ex and ey are measured along the x-​ and y-​axes, respectively (see Fig. 10.20.1). For all design problems a design sketch (to scale) of the cross section is required: show section dimensions, location and size of bars, and tie or spiral size and spacing.

Problems on Rectangular Sections Subject to Uniaxial Bending x 3” 4” 20” (500 mm)

10.1 Calculate the nominal strength Pn for the section of the figure for Problem 10.1 for an eccentricity ey (measured along the y-​axis) = 0.1h = 1.8 in. Use fc′ = 3000 psi and fy = 40,000 psi. Use basic principles of statics to obtain solution, considering the effect of compression concrete displaced by steel; compare the result with the maximum Pn given by ACI-​22.4.2 (e = 46 mm; fc′ = 20 MPa; fy = 300 MPa).

y

6” 10 – #11 (10 – #36M)

4” 3” #3 @ 18 (#10M)

3” 6 – #9

y 18” (460 mm)

(6 – #29M)

3”

12” (310 mm) 18” (460 mm)

Problems 10.1 and 10.2 

3”

(75 mm)

9” (234 mm)

9” (234 mm)

3” (66 mm)

24” (600 mm)

Problems 10.3, 10.4, 10.5, and 10.6 

bending with respect to the x-axis (h = 24 in.). To obtain points for the diagram, compute Pn for the following cases in addition to eb (see Problem 10.3): (a) e = 0 (P0)

(d) e = 0.7h

(b) e = 0.1h

(e) e = h

(c) e = 0.3h

(f) e = ∞ (M0)

10.5 Same as Problem 10.3 except use fc′ = 4000 psi 10.2 Same as Problem 10.1 except fc′ = 5000 psi and and fy  =  60,000 psi. ( fc′ = 3000 MPa;  fy = 60,000 psi. ( fc′ = 35 MPa; f y = 400 MPa.) f y = 400 MPa.) 10.3 Compute the nominal strength Pn  =  Pb for the 10.6 Same as Problem 10.4 except use fc′ = 4000 psi balanced strain condition for bending about the and fy = 60,000 psi. (For eb, see Problem 10.5.) strong axis (measured as ey) for the column of 10.7 For the section of the figure for Problem 10.7, the figure for Problem 10.3. Compute also the use statics to compute and plot the strength eccentricity eb. Use basic statics, including the interaction diagram of Pn–​Mn for bending about effect of the compression concrete displaced the x-​axis. Compute the balanced condition in by steel, fc′ = 3000 psi and fy  =  40,000  psi.  addition to those points indicated for Problem ( fc′ = 20 MPa; f y = 300 MPa.) 10.4. Use fc′ = 3000 psi and fy  =  60,000 psi. 10.4 Using basic statics and taking into account con( fc′ = 20 MPa; f y = 400 MPa.) crete displaced by steel in compression, calculate and plot the Pn–​Mn strength interaction 10.8 Repeat Problem 10.7 except consider bending about the y-​axis. diagram for the section of Problem 10.3. Take *  Many problems may be solved either as problems stated in Inch-​Pound units or as problems in SI units using quantities in parentheses at the end of the statement. The SI conversions are approximate to avoid implying higher precision for the given information in metric units than that for the Inch-​Pound units.

375



375

PROBLEMS

10.9 For the section of Problem 10.7, compute the 10.11 For the section of the figure for Problem nominal strength Pn for an eccentricity ey of 5 10.11, compute the nominal strength Pn for in. with respect to the x-​axis. Use fc′ = 5000 psi the eccentricity given below (as assigned by and fy = 60,000 psi. (e = 125 mm; fc′ = 35 MPa;  the instructor) with respect to the x-​axis. Use f y = 400 MPa.) fc′ = 3000 psi and fy = 40,000 psi. ( fc′ = 20 MPa; fy = 300 MPa.) #3 ties @ 14” x (#10M ties) 2.5” (66 mm)

1

1 2 ” clear cover (40 mm) y

14” (350 mm)

4.5” 2.5”

2.5”

6.5” 6.5” (164 mm) 18” (460 mm)

(e) e = 30 in. (750 mm)

(b) e  = 8 in. (200 mm)

(f) e = 40 in. (1000 mm)

(c) e  = 12 in. (300 mm) (g) e = 200 in. (5 m)

4.5” (109 mm) 4 – #10 (4 – #32M)

(a) e = 6 in. (150 mm)

(d) e  = 20 in. (500 mm) (h) e = 2000 in. (50 m) 10.12 For the section of the figure for Problem 10.12, compute the nominal strength Pn for the eccentricity given below (as assigned by the instructor) with respect to the x-​axis. Use fc′ = 3000 psi and fy = 40,000 psi.

2.5”(66 mm)

Problems 10.7, 10.8, 10.9, 10.10, 10.39, and 10.40 

(a) e = 2 in. 10.10 For the section of Problem 10.7, compute the nominal strength Pn. Repeat Problem 10.9, except use an eccentricity ey of 22 in. with respect to the x-​axis. (ey = 560 mm.)

(e) e = 40 in.

(b) e = 3.6 in.  (f) e = 70 in. (c) e  = 10 in.

(g) e = 200 in.

(d) e = 15 in.

(h) e = 1000 in.

x 3”

(66 mm)

(117 mm) 4 12 ”

16 – #10 bars (16 – #32M bars)

1 4 2”

24” (600 mm)

y

1

4 2” 1

4 2” 3” (66 mm)

3”

7” (184 mm)

10”

10” (250 mm) 40” (1000 mm)

7”

3”

Problem 10.11 and 10.46  x 3” 1 42”

14 – #9

1

24”

42”

y

1

42” 1

42” 3” 1

22 ”

10”

11” 36”

Problem 10.12 

10”

1

22 ”

376

376

C H A P T E R   1 0     M E M B E R S I N C O M P R E S S I O N A N D B E N D I N G

10.13 For the section of the figure for Problem 10.13, use basic statics to compute and plot the strength interaction diagram of Pn–​Mn for bending about the x-​axis. Compute the balanced condition, in addition to enough other points to plot the curve. (Hint: By successively setting the neutral axis distance c at specific values, the points can be computed without solving a cubic equation.) Use fc′ = 4000 psi and fy  =  60,000 psi. ( fc′ = 30 MPa; fy = 400 MPa.)

10.15 For the case assigned by the instructor, design the smallest (in whole inches) square tied column having not more than 4% reinforcement to carry an axial compression (i.e., M  =  0 or neglected). Compute the strength Pn when e = 0.10h. Use fc′ = 3500 psi and fy  =  60,000 psi. ( fc′ = 25 MPa; f y = 400 MPa.) PD (dead load)

PL (live load)

(a) 100 kips (450 kN)   80 kips   (350 kN) (b) 200 kips (900 kN) 160 kips   (700 kN)

12 – #9 (12 – #29M)

(c) 350 kips (1500 kN) 150 kips   (700 kN)

x

(d) 400 kips (1800 kN) 320 kips (1400 kN) 12” (300 mm)

(e) 600 kips (2700 kN) 480 kips (2100 kN)

1

22 ”

3”

3” 14”

3”

1

22 ”

10.16 Repeat Problem 10.15, except use fc′ = 4000 psi and fy  =  60,000 psi. ( fc′ = 30 MPa and fy = 400 MPa.) 10.17 Repeat Problem 10.15, except use fc′ = 3500 psi and fy  =  40,000 psi. ( fc′ = 25 MPa and fy = 300 MPa.) 10.18 For the case assigned by the instructor, design a square tied column having symmetrical reinforcement of about 3 1 2 % to carry service loads as given. Use basic statics to check final answer. Use whole inches with fc′ = 4000 psi and fy = 50,000 psi.

(64 mm)

(75 mm)

(350 mm)

Problem 10.13 and 10.45 

  PD   PL   MD   ML (dead load) (live load) (dead load) (live load)

10.14 Using basic principles, compute the nominal strength Pn for an eccentricity ey (as assigned by the instructor) on the column of the figure for Problem 10.14. Use fc′ = 3500 psi and fy = 40,000 psi. ( fc′ = 25 MPa; fy = 300 MPa.)

(a) 100 kips   90 kips   40 ft-​kips

30 ft-​kips

(b) 200 kips 175 kips   75 ft-​kips

65 ft-​kips

(c) 300 kips 270 kips 120 ft-​kips 100 ft-​kips

(a) e = 3 in. (75 mm)

(f) e = 30 in. (750 mm)

(d) 300 kips 270 kips 250 ft-​kips 200 ft-​kips

(b) e = 6 in. (150 mm)

(g) e = 60 in. (1.5 m)

(e) 300 kips 270 kips 500 ft-​kips 400 ft-​kips

 (c) e = 9 in. (220 mm)

(h) e = 150 in. (3.8 m)

(f) 600 kips 550 kips 125 ft-​kips 100 ft-​kips

(d) e = 15 in. (380 mm) (i) e = 300 in. (7.5 m)

(g) 600 kips 550 kips 250 ft-​kips 200 ft-​kips

 (e) e = 20 in. (500 mm) (j) e = 2000 in. (50 m)

(h) 600 kips 550 kips 500 ft-​kips 400 ft-​kips εy

2 – #11 4 – #7 (2 – #36M) (4 – #22M)

16” (420 mm)

2.70” (68 mm)

12.30” (312 mm)

12.50” (318 mm)

30” (760 mm)

Problem 10.14 

#4 ties (#13M) 1 12

cover ” (40 mm)

2.5” (62 mm)

Pn

y

37



377

PROBLEMS

10.19 Redesign the column of Problem 10.18 using a rectangular section not over 15 in. wide instead of a square section. If the depth h would be greater than 30 in., use h/​b  =  2.0 and exceed the 15 in. limit. 10.20 Design a square tied column with symmetrical reinforcement of about 4% to carry a dead load of P = 225 kips and M = 220 ft-​kips and a live load of P = 200 kips and M = 190 ft-​kips. Use basic statics to check final answer. Use whole inches (20 mm for SI) for size with fc′ = 4000 psi and fy  =  40,000 psi. (DL:  P  =  1000 kN, M = 300 kN·m; LL: P = 900 kN, M = 260 kN·m; fc′ = 30 MPa ; fy = 300 MPa.) 10.21 Redesign the column of Problem 10.20 using a rectangular section not over 16 in. (400 mm) wide instead of a square section. 10.22 Design a square tied column with about 4% reinforcement to carry a dead load of P = 250 kips and M = 90 ft-​kips, and a live load of P = 185 kips and M  =  70 ft-​kips. Use whole inches (20 mm for SI) for size with fc′ = 4000 psi and fy = 60,000 psi. (DL: P = 1100 kN, M = 120 kN·m; LL: P = 820 kN, M = 95 kN·m; fc′ = 30 MPa; fy = 400 MPa.) 10.23 Design a square tied column with symmetrical reinforcement of about 2% to carry a dead load of P = 25 kips and M = 125 ft-​kips and a live load of P = 10 kips and M = 50 ft-​kips. Use whole inches (20 mm for SI) for size with fc′ = 4000 psi and fy  =  60,000 psi. (DL:  P  =  110 kN, M = 170 kN·m; LL: P = 45 kN, M = 68 kN·m; fc′ = 30 MPa; fy = 400 MPa.) 10.24 Redesign the column of Problem 10.23 as a rectangular column having a depth-​width ratio of between 1.5 and 2.0 along with unsymmetrical reinforcement with respect to the bending axis. 10.25 Design a square tied column with symmetrical reinforcement of about 2 1 2 % to carry a dead load of P  =  75 kips and M  =  125 ft-​kips and a live load of P = 30 kips and M = 50 ft-​kips. Use whole inches (20 mm for SI) for size with fc′ = 4000 psi and fy = 60,000 psi. (DL: P = 340 kN, M = 170 kN·m; LL: P = 140 kN, M = 70 kN·m; fc′ = 30 MPa; fy = 400 MPa.) 10.26 Redesign the column of Problem 10.25 using a rectangular member but still using symmetrical reinforcement.

Problems on Circular Sections 10.27 Determine the strength Pn = Pb and the eccentricity eb for the balanced strain condition on the section of the figure for Problem 10.27 by using the method of statics and the rectangular stress

Ds = 14.87” 1

1 2 ” cover

10 – #9

#4 spiral

36º

2.30” 6.01” 7.44” 20” d = 17.44” c = 11.08” εy = 0.00172

0.003

eb

Pb

a = 9.41” 0.85fc’ = 3.4 ksi T1 T2

T3

Cs3 C Cs1 s2 x Cc

Problem 10.27, 10.28, and 10.31 

distribution as for beams. Use fc′ = 4000 psi and fy = 50,000 psi. 10.28 Determine the nominal compressive strength Pn for the section of Problem 10.27 for an eccentricity e = 5 in. by the direct application of statics. Use fc′ = 4000 psi and fy = 50,000 psi. 10.29 Compute the nominal strength Pn of the spirally reinforced column of the figure for Problem 10.29 when the loading has an eccentricity e  =  0.05h  =  0.9 in. Use fc′ = 3000 psi and fy = 40,000 psi. Use basic statics with the circular section, including the effect of compression concrete displaced by steel. Compare with maximum Pn obtained from ACI-​22.4.2. (e = 23 mm; fc′ = 20 MPa; fy = 300 MPa.) 10.30 Same as Problem 10.29, except use fc′ = 5000 psi and fy = 60,000 psi. ( fc′ = 35 MPa; fy = 400 MPa.) 10.31 Use a direct application of statics to determine the nominal compressive strength Pn for the

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1

1 2 ” cover (40 mm)

6 – #8 (6 – #25M)

#3 spiral (#10M)

18” (460 mm)

Problem 10.29 and 10.30 

section of Problem 10.27 for an eccentricity e = 20 in. Use fc′ = 4000 psi and fy = 50,000 psi. 10.32 For the column of the figure for Problem 10.32, determine the nominal strength Pn at an eccentricity of 21 in. Use fc′ = 4000 psi and fy  =  50,000 psi. Apply the basic princi1

1 2 ” cover (40 mm)

16 – #14 (16 – #43M)

#5 spiral (#16M)

28” (720 mm)

Problem 10.32 and 10.47 

ples of statics. (e  =  530  mm; fc′ = 30 MPa; fy = 350 MPa.) 10.33 Same as Problem 10.15, except use a spirally reinforced circular column. 10.34 Redesign the column of Problem 10.18 as a spirally reinforced circular column. 10.35 Redesign the column of Problem 10.20 as a spirally reinforced circular column. 10.36 Design a circular spirally reinforced column for the conditions given in Problem 10.22. Make statics check using the designed circular section.

10.37 Redesign the column of Problem 10.25 as a spirally reinforced circular column.

Problems on Biaxial Bending and Compression 10.38 Compute the nominal strength Pn for the column of Problem 10.3 under an eccentricity ey of 8 in. about the strong axis and an eccentricity ex of 5 in. about the weak axis. Compute strength using (a) the Bresler reciprocal load method and (b) statics and compatibility. (Hint: for part (a), use the interaction curve of Problem 10.4 and calculate a similar interaction curve for the weak axis.) Use fc′ = 3000 psi and fy = 40,000 psi. (ey = 200 mm; ex = 130 mm; fc′ = 20 MPa ; fy = 300 MPa.) 10.39 Determine the adequacy of the rectangular tied column section of Problem 10.7 when subjected to an axial load of 90 kips, applied with an eccentricity ey of 5 in. with respect to the x-​axis and with an eccentricity ex of 3 in. with respect to the y-​axis. Use fc′ = 3000 psi and fy = 60,000 psi (use information developed in Problems 10.7 and 10.8). Compare results using (a) the Bresler reciprocal load method and (b) statics and compatibility. 10.40 Repeat Problem 10.39, except the axial load is 45 kips applied with eccentricities of 10 in. with respect to the x-​axis and 6 in. with respect to the y-​axis. 10.41 For the case assigned by the instructor, compute the nominal strength Pn for biaxial bending and compression for the section of Problem 10.7. Use both the Bresler reciprocal load method and statics and compatibility. (If available, the strength interaction diagrams from Problems 10.7 and 10.8 may be used.) Use fc′ = 3000 psi and fy = 60,000 psi. (a) ey = 1.8 in.

ex = 1.4 in.

(b) ey = 5.4 in.

ex = 1.4 in.

(c) ey = 15 in.

ex = 2 in.

(d) ey = 30 in.

ex = 3 in.

(e) ey = 40 in.

ex = 6 in.

(f) ey  = 40 in.

ex = 20 in.

10.42 Design a square tied column with about 4% reinforcement uniformly distributed around its sides. The loads are dead load:  P  =  225 kips, Mx  =  236 ft-​kips, My  =  108 ft-​kips, and

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PROBLEMS

live load:  P  =  200 kips, Mx  =  204 ft-​kips, My  =  64 ft-​kips. Select sizes in whole inches that are multiples of 2, using fc′ = 4000 psi and fy = 40,000 psi. 10.43 Repeat Problem 10.42, except use fy = 60,000 psi. 10.44 Repeat Problem 10.42, except reduce moments Mx about the x-​axis to 215 ft-​kips dead load and 162 ft-​kips live load, and increase fy to 60,000 psi.

Problems on Shear 10.45 For the section of the figure for Problem 10.13, design the shear reinforcement for Vu = 42 kips parallel to the long direction of the column.

379

Verify that lateral support requirements for longitudinal bars are satisfied. Use fyt = 60 ksi. 10.46 For the section of the figure for Problem 10.11, design the shear reinforcement for Vu  =  215 kips parallel to the short direction of the column. Verify that lateral support requirements for longitudinal bars are satisfied. There is no need to use the same transverse reinforcement layout shown in the figure. Use fyt = 60 ksi. 10.47 For the section of the figure for Problem 10.32, determine the required spiral pitch if the section is subjected to a shear force Vu = 165 kips. Verify that the final spiral pitch selected satisfies the requirements for spiral reinforcement in the ACI Code. Use fyt = 50 ksi.

CHAPTER 11 MONOLITHIC BEAM-​C OLUMN CONNECTIONS

11.1 INTRODUCTION Chapters 3 through 10 emphasized the design of flexural members (beams) and members subjected to combined axial force and bending (columns) for bending, shear, and development of reinforcement. Some attention has been given to development of reinforcement at exterior supports, including the use of hooks (Section 6.10). Often in design, not enough attention is given to the details of connections—​how the forces in beams and columns interact and get transmitted through the joint—​that is, the column region over the depth of the deepest beam framing into the column. A connection, on the other hand, consists of the joint and adjacent beam, slab, and column regions. The ACI Code (Chapter 15) provides little guidance specifically directed to detailing of beam-​column joints in regions of no or low seismicity, for which design is typically governed by gravity and wind loads. On the other hand, provisions for shear strength and detailing of joints in special moment frames, for which large shear reversals are expected under a design-​level earthquake, are provided in Section 18.8 of the Code. The state of the art regarding the design of beam-​column connections is summarized by ACI-​ASCE Committee 352, Joints and Connections in Monolithic Reinforced Concrete Structures [11.1]. That report contains detailed provisions for the design of two classes of beam-​column connections: •  Type 1 connections, primarily for members designed to satisfy ACI Code strength requirements, where no significant inelastic deformations are expected. •  Type 2 connections, usually for members subject to earthquake loading, where there is need for sustained strength under deformation reversals into the inelastic range. In general, Type 1 connections require only nominal ductility, typical of connections in a continuous moment-​resisting frame structure designed to resist gravity and wind loads. The primary source of what follows is from the 2002 ACI-​ASCE Committee 352 report [11.1]. These recommendations, which apply to beam-​column connections of normalweight concrete with strengths up to 15,000 psi, include connections where the beam width is larger than the column depth, often called wide-​beam connections. The recommendations also apply to eccentric connections (i.e., where the beam centerline does not pass through the column centroid) provided that all beam reinforcement is anchored or passes through the column core. Requirements for transverse reinforcement in Type 2 beam-​column connections discussed in Section 11.3 are those in Section 18.8 of the 2014 ACI Code; these have been

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11.2  BEAM-COLUMN JOINT ACTIONS

381

Beam-​column connection reinforcement. (Photo courtesy of Cary Kopczynski & Co.)

modified with respect to previous editions of the code and may in certain situations be more stringent than those in the ACI-​ASCE Committee 352 recommendations. The following discussion of some general concepts incorporates three examples. A more detailed treatment of the subject is outside the scope of this book. Also, some of the more readily available references are listed at the end of this chapter.

11.2 BEAM-​C OLUMN JOINT ACTIONS Just like the members themselves, beam-​column joints need to be designed for all actions that may act on them: axial load, bending moment, shear, and torsion, caused by external loads and other actions, such as ground motions, wind, creep, shrinkage, temperature, or settlement of supports. Assuming that the members have themselves been properly designed, the critical factor in joint design is the transmission into and through the joint of the forces that are present at the ends of the members. Figure 11.2.1 shows an interior connection with beams framing into the joint from all sides of a column (the slab is not shown for clarity). Figure 11.2.2 shows a beam-​column connection region of a frame subjected to lateral forces such that beams framing from opposite sides of the column are subjected to negative bending and positive bending (right and left beam in Fig. 11.2.2, respectively). The scenario in which both beams are subjected to negative bending, as is the case in a frame subjected to gravity loads only, is of less interest because little or no shear is introduced into the joint. Referring to Fig. 11.2.2, the forces T1 and C1 are the tension and compression resultants for negative bending in the beam framing into the joint from the right side; the forces C2 and T2 represent positive bending in the beam framing in from the left side; the forces Vcol represent the shears in the column just above and below the joint. The horizontal shear within the joint that potentially may cause the diagonal crack shown may be expressed as

Vu ( joint ) = T1 + C2 − Vcol = T1 + T2 − Vcol

(11.2.1)

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or Vu ( joint ) = α f y Ast + α f y Asb − Vcol



(11.2.2)

For Type 1 joints, ACI-​ASCE Committee 352 [11.1] recommends that joint shear demand be computed based on the flexural strength of the beams or the columns, whichever applies, without strength reduction factors [11.1, Sections 3.1 and 3.3.4.]. For Type 2 joints, on the other hand, flexural yielding is expected to occur primarily in the beams, and thus joint shear demand is determined under the assumption that the beams reach their flexural strength. Thus, T1 and T2 in Eq. (11.2.1) are computed as α fy Ast and α fy Asb rather than the forces calculated from the bending moment determined by structural analysis at that section. The parameter α is a stress multiplier based on the degree of inelastic deformation expected as follows:



α ≥ 1.0

for Type 1 connections

α ≥ 1.25

for Type 2 connections

A value of 1.0 is permitted for Type 1 connections because only limited inelastic deformation is expected in the adjacent members. For Type 2 connections, however, a minimum value of 1.25 is recommended because of the significant inelastic deformations expected in the members framing into this type of joint. The value of α = 1.25 is intended to account for the actual yield stress of the bar (typically 10–​25% higher than the nominal value) and for possible strain hardening of the bars at large plastic rotations [11.1, Section 3.3.4]. Note that Ast in Eq. (11.2.2) should include the beam reinforcement plus any effective flange reinforcement. For Type 1 connections, the effective flange width should be taken as that prescribed for flanges in tension in accordance with ACI-​24.3.4 [11.1, Section 3.3.1]. For Type 2 joints, the effective flange width is based on that prescribed for flanges in compression in ACI-​6.3.2.1. However, additional restrictions exist depending on the classification of the connection—​interior, exterior, or corner connection [11.1, Section 3.3.2].

Pu

Vuy

Mux Vux

Muy

Mux Vuy Vux

Muy

Pu

Figure 11.2.1  Actions on members framing into a joint.

38



383

11.3  JOINT TRANSVERSE REINFORCEMENT

Column steel

Column steel Ast Vcol

T1 = Ast αfy

s/2

C 2 = T2 s

Vu (joint)

Beam steel

s T2 = Asbαfy Shear crack

C 1 = T1 Vcol

α ≥ 1.0 for Type 1 joints α ≥ 1.25 for Type 2 joints (a)

Asb

Ash

s/2 Ties at less than s/2 from resultant T or C force in beam are probably not effective (b)

Figure 11.2.2  Horizontal shear and area of transverse reinforcement in a beam-​column joint (column transverse steel not shown for clarity).

11.3 JOINT TRANSVERSE REINFORCEMENT The primary roles of transverse reinforcement in beam-​column joints are to provide confinement to the concrete, to serve as tie reinforcement for joint shear resisted through a truss mechanism (see Section 11.4), and to provide lateral support to longitudinal column bars. Below is a summary of transverse reinforcement requirements for Type 1 and Type 2 joints. For Type 1 joints, these requirements are primarily based on the ACI-​ASCE Committee 352 recommendations [11.1], supplemented with requirements in Chapter 15 of the ACI Code. For Type 2 joints, transverse reinforcement requirements are primarily those of Section 18.8 of the ACI Code, which have changed with regard to the 2002 ACI Code, to which the latest ACI-​ASCE Committee 352 recommendations refer.

Type 1 Joints For a Type 1 joint, the transverse reinforcement (ties or spiral) may be omitted when “adequate” lateral confinement is provided by beams framing into the joint. ACI-​ASCE Committee 352 [11.1, Section 4.2.1.4] has defined confinement as shown diagrammatically in Fig. 11.3.1. The joint region within the depth of the shallowest member framing into a column may be considered adequately confined when beams frame in from all four sides and each beam has a width b at least three-​fourths the column width and no more than 4 in. of column width is exposed on each side of the beam. Where beams frame in from only two opposite sides, and their widths cover at least three-​ fourths of the column width at those sides, and not more than 4 in. of column width is exposed on each side of the beams, confinement occurs only at those faces. In this case, transverse reinforcement (Fig. 11.3.2) is needed through the joint only in the direction parallel to the two sides of the joint into which the beams frame. This provision has no effect when confinement is from a spiral or circular hoop, since to provide confinement the spiral or circular hoop must extend through the joint, thus providing full confinement (all directions) to the joint in any case. When ties must be provided over the joint depth, at least two layers of transverse reinforcement should be used between the top and bottom levels of beam reinforcement in the deepest member framing into the joint; the center-​to-​center spacing of tie or circular, continuously wound bar or wire,1 sh, should not exceed 12 in. [11.1, Section 4.2.1.3]. When the 1  ACI-​ 25.7.3 has specific requirements in order for a circular, continuously wound bar or wire to be classified as a spiral. Either a spiral or a circular continuous wound bar or wire not satisfying ACI-​25.7.3 may be used in joints. The latter, however, must be treated as a circular hoop.

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C hapter   1 1     M onolithic B eam - C olumn C onnections

z

hc y

bc

≥ 34 hc

≥ 34 bc Beam B2

Joint (shaded) Beam B1

b2 b1

x ≤ 4” ≤ 4”

Shallowest member

Note: Not more than 4 in. of column face width exposed at any edge

Figure 11.3.1  Confinement at a joint by members framing into all four sides of the column. Column

Required transverse reinforcement (cross-ties shown)

≤ 4”

Required transverse reinforcement must be used through core at spacing along column axis not to exceed 12 in. or one-half the depth of shallowest beam for Type 1 joint.

Figure 11.3.2  Confinement by members framing into two opposite faces of the column only.

joint is part of the nonseismic lateral load resisting system, the spacing should not exceed 6 in. In no case, however, should the spacing exceed one-​half the depth of the shallowest beam framing into the column (ACI-​15.4.2.2). Also, ACI-​15.4.2 requires a minimum area of tie reinforcement within a tie spacing sh as follows:

Ash = 0.75 fc′

bc sh bs ≥ (50 psi) c h f yt f yt

(11.3.1)

with sh not greater than one-half the depth of the shallowest beam framing into the column (ACI-15.4.2.2).

385



385

11.3  JOINT TRANSVERSE REINFORCEMENT

A Free horizontal face

Hooked bars

Vertical transverse reinforcement

≥Ld

Vertical transverse reinforcement Section A–A

Elevation

A

Figure 11.3.3  Transverse reinforcement details in joints with discontinuous columns. (Adapted from Fig. 4.2 of ACI-​ASCE Committee 352 [11.1].)

When required, the joint transverse reinforcement should satisfy the provisions of the ACI Code for tied or spirally reinforced columns (ACI-​10.7.6; see Sections 10.8 and 10.9) in addition to the joint confinement recommendations presented above. For Type 1 joints with a free horizontal face at the discontinuous end of a column (such as at the roof or at a mezzanine level) and with discontinuous beam reinforcement near the free horizontal face (knee joints), vertical transverse reinforcement is required through the full height of the joint. At least two layers of vertical transverse reinforcement should be provided between the outermost longitudinal column bars at a spacing not exceeding 12 in., or 6 in. when the joint is part of the nonseismic lateral load resisting system [11.1, Section 4.2.1.5]. An example of the transverse reinforcement details at a discontinuous end of a column is shown in Fig. 11.3.3.

Type 2 Joints For Type 2 joints, a minimum amount of transverse reinforcement is required regardless of whether “adequate” confinement is provided by beams framing into the joint. The transverse reinforcement requirements below are those given for earthquake-resistant structures in ACI-​18.8.3.1. For columns confined by spirals or circular hoops with factored axial load Pu ≤ 0.3 Ag fc′ and fc′ ≤ 10, 000 psi, the volumetric ratio of transverse reinforcement ρs must be at least the greater of



 Ag  f′ ρs = 0.45  − 1 c A  ch  f yt

(11.3.2)

and

ρs = 0.12



fc′ f yt

(11.3.3)

For columns with factored axial load Pu > 0.3 Ag fc′ or fc′ > 10, 000 psi, the volumetric ratio ρs of spiral or circular hoop reinforcement must be at least the greatest obtained from Eqs. (11.3.2), (11.3.3), and (11.3.4).

 P  ρs = 0.35k f  u   f yt Ach 

(11.3.4)

where

kf =

fc′ + 0.6 ≥ 1.0 25, 000

(11.3.5)

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and Ag = gross cross-sectional area of compression member Ach = area of rectangular core measured out-to-out of transversse reinforcement f yt = yield stress of transverse reinforcement, but not to be taken greater than 100,000 psi When rectilinear hoops and cross-​ties are used as transverse reinforcement in accor­ dance with ACI-​18.7.5.2, the total required cross-​sectional area of hoops and cross-​ties within a spacing sh, Ash (see Fig. 11.3.4), for columns with factored axial load Pu ≤ 0.3 Ag fc′ and  fc′ ≤ 10, 000 psi, is Ash = 0.3sh bc



 fc′  Ag − 1  f yt  Ach 

(11.3.6)

but not less than Ash = 0.09 sh bc



fc′ f yt

(11.3.7)

where bc is the cross-​sectional dimension of the column core measured to the outside edges of confining reinforcement, perpendicular to the transverse reinforcement area Ash being calculated. For columns with factored axial load Pu > 0.3 Ag fc′ or fc′ > 10, 000 psi, area of transverse reinforcement must be at least the largest from Eqs. (11.3.6), (11.3.7), and (11.3.8)  P  Ash = 0.2 k f kn  u  sh bc  f yt Ach 



(11.3.8)

where kf is computed from Eq. (11.3.5) and kn =



nl nl − 2

(11.3.9)

where nl is the number of longitudinal bars laterally supported by the corner of the hoop or by a seismic hook. In ACI 318, a seismic hook is defined as a hook with a bend no less than 135° and an extension of six bar diameters, but not less than 3 in. Transverse reinforcement must be arranged such that longitudinal column bars laterally supported by the corner of a hoop leg or a cross-​tie are no farther apart than a distance hx = 14 in. around the perimeter of the column, except for columns with Pu > 0.3 Ag fc′ or fc′ > 10, 000 psi, where each longitudinal bar must be laterally supported by the corner of a hoop or by a seismic hook and spaced no farther apart than hx = 8 in. The center-​to-​center spacing sh of the transverse reinforcement should not exceed the following: a. For joints of special moment frames (ACI-​18.7.5.3): 1 4 of the minimum column dimension, 6 times the diameter of the smallest longitudinal bar, or so where

  14 − hx   4 in. ≤  so = 4 +  ≤ 6 in.  3   

(11.3.10)

and hx is taken as the largest distance between laterally supported longitudinal bars around the column perimeter.

387



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11.4  JOINT SHEAR STRENGTH Direction of shear being considered

#3 sh

bc

dh

Ash = 4(0.11) = 0.44 sq in. bc = outside-to-outside of hoops dh = outside-to-outside of hoops Ach = bcdh

Figure 11.3.4  Example of rectilinear hoop reinforcement in height sh of joint core.

b. For joints that connect members that are not part of the primary seismic-​force-​resisting system, but such members are designed to sustain deformations in the inelastic range due to lateral displacement compatibility [11.1, Section 4.2.2.4], center-​to-​center spacing should not exceed 1 3 of the minimum column dimension, or 12 in. The amount of transverse reinforcement may be reduced by one-​half and maximum spacing increased to 6 in. when the joint is completely confined on all sides by beams whose width is at least three-​fourths the column width (ACI-​18.8.3.2). The ACI Committee 352 report [11.1] contains additional recommendations for beam bars closest to the free horizontal face in Type 2 joints with a discontinuous column. These recommendations are based on experimental tests on knee joints [11.47, 11.50, 11.60] and are similar to those described for Type 1 joints. Because of the inelastic behavior of Type 2 joints, the requirements are, however, more stringent [11.1, Section 4.2.2.8].

11.4 JOINT SHEAR STRENGTH In a well-​designed beam-​column joint, shear resistance is assumed to be provided by a diagonal strut mechanism and a truss mechanism (Fig. 11.4.1). The diagonal strut is activated by the compression zones of the beams and columns framing into the joint. The truss mechanism, on the other hand, relies on force transfer to the joint through bond between the longitudinal beam and column bars and the surrounding concrete. Transverse reinforcement thus enhances joint shear strength by providing confinement to the diagonal strut, as well as resisting horizontal tensile forces as part of the truss mechanism. As in any reinforced concrete design, design for shear strength must satisfy

φ Vn ≥ Vu

(11.4.1)

where Vu is calculated according to Eq. (11.2.2) and φ = 0.75. It is not possible to separate the contributions to joint shear strength from the strut and truss mechanisms. Thus, the approach taken by ACI Committee 352 (and ACI Committee

38

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Strut

(a)

(b)

Figure 11.4.1  Shear-​resisting mechanisms in a beam-​column joint: (a) diagonal strut (a) and (b) truss.

318)  is to define nominal joint shear strength Vn as a function of Section 4.3.1],

Vn = γ fc′ b j h

fc′ as follows [11.1, (11.4.2)

where bj = effective joint width (see Fig. 11.4.2)  bbeam + bcol ,  2  mh  = smallest of bbeam + ∑ , 2  b  col

or

The term mh / ​2 should not be larger than “the extension of the column beyond the edge of the beam.” Also, the summation term applies on “each side of the joint where the edge of the column extends beyond the edge of the beam.” bbeam = design beam width in the direction of loading. When there is a beam on only one face of the column in the direction of the load, bbeam is the width of that beam. When there are beams on opposite faces of the column, bbeam is the average of the widths of those beams. bcol = column width perpendicular to the direction of the load being considered h = depth of column in the direction of the load being considered m = coefficient that accounts for the joint eccentricity = 0.3 when the eccentricity between the beam centerline and the column centroid exceeds bcol  /​ 8 = 0.5 for all other cases γ = constant that depends on joint type and classification, as shown in Table 11.4.1.

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1 1 . 5   C O L U M N - TO - B E A M M O M E N T S T R E N G T H   R AT I O

bj ≤ bj

h

bbeam + bcol , 2 mh , or bbeam + Σ 2 bcol bj

Column

bbeam

Column

Direction of loading

h

e

bcol

bbeam bcol e = Eccentricity between column centroid and beam centerline

Plan views

Figure 11.4.2  Determination of effective joint width bj.

TABLE 11.4.1  VALUES OF JOINT SHEAR CONSTANT γ (FROM ACI COMMITTEE 352 [11.1]) Joint Classification Joint Type

Confineda on All Four Vertical Faces

Confineda on Three or Other Two Opposite Vertical Cases Faces

1

With a continuous column With a discontinuous column

24 20

20 15

15 12

2

With a continuous column With a discontinuous column

20 15

15 12

12 8

a

By beams with width at least ¾ the width of the column face into which they frame.

11.5 COLUMN-​T O-​B EAM MOMENT STRENGTH RATIO The design of Type 2 connections that are part of the seismic-​force-​resisting system requires that the summation of the nominal moment strengths of the columns framing into a joint be at least 1.2 times the summation of the nominal moment strengths of the beams, calculated at the faces of the joint [11.1, Section 4.4.2]. This minimum strength ratio, although too low to prevent the occurrence of column hinging, is considered acceptable to prevent the formation of a story mechanism. It is not applicable to roof connections, however, as either column or beam hinging is acceptable in this type of connections. The column-​ to-​beam moment strength ratio has also been linked to joint performance under shear reversals [11.9], joint behavior improving with an increase in moment strength ratio. No minimum column-​to-​beam moment strength ratios are specified for Type 1 joints, since column hinging is possible in frame design for which Type 1 joints apply.

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11.6 ANCHORAGE OF REINFORCEMENT IN THE JOINT REGION Adequate anchorage of beam and column longitudinal reinforcement in the joint region is critical because the members must be able to develop the required strength and transfer forces into and through the joint without unacceptable reinforcement slip. For Type 1 joints, the ACI Committee 352 recommendations include provisions for anchorage of beam and column longitudinal reinforcement terminating at a joint (e.g., exterior and roof connections). The critical section for development of beam longitudinal bars is taken at the face of the column. For longitudinal column bars, on the other hand, the critical section is taken at the outside edge of the bottom longitudinal reinforcement passing through (e.g., T-​joints) or anchored into (e.g., knee joints) the joint. When bars are terminated by a standard 90° hook, the development length is determined as [11.1, Section 4.5.2.3],2



fy ψ c ψ r

Ldh =

50 fc′

db

(11.6.1)

where ψc is a cover factor equal to 0.7 for #11 bars and smaller with side cover not less than 2.5 in. and cover beyond the hook tail greater than or equal to 2 in.; otherwise, ψc = 1.0. The confinement factor ψr is equal to 1.0 unless the hook is enclosed by horizontal or vertical ties along development length at a spacing not greater than 3db of the bar being terminated, in which case ψr = 0.8. For Type 2 joints, because there is potential for loss of concrete cover due to inelastic deformation reversals, the critical section for development of reinforcement is to be taken at the outside edge of the column core. In this case, the development length for longitudinal bars terminating at the joint is calculated as [11.1, Section 4.5.2.4],



Ldh =

α fy ψ r 75 fc′

db



(11.6.2)

where α ≥ 1.25. In all cases, Ldh must not be taken less than 6 in. or 8db, whichever is greater. Also, the outside edge of the hook tail should be within 2 in. from the farthest edge of the core, with the hook oriented toward the inside of the joint such that a diagonal strut can be properly developed (Fig. 11.6.1). For beams with more than one layer of flexural reinforcement, the hook tails of each additional layer should be located within 3db of the adjacent hook tail. To prevent excessive slip of longitudinal beam and column bars passing continuously through Type 2 joints, the following ratios between member depth and longitudinal bar diameter must be satisfied:  fy  ≥ 20   60, 000 

(11.6.3)

h(column)  fy  ≥ 20    db ( beam )  60, 000 

(11.6.4)

h(beam)

db ( column )

but no less than 20, where h refers to member depth and db to longitudinal bar diameter.

2  This equation is the same as Eq. (6.10.1) when applied to uncoated reinforcement embedded in normal-weight concrete. The lightweight concrete factor λ is not included in the ACI Committee 352 recommendations, which apply to normal-​weight concrete. The cover and confinement factors, ψc and ψr, respectively, although not explicitly included in the ACI Committee 352 document, are used here for consistency with expressions used in ACI 318-​14 and Chapter 6 of this book.

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11.8 EXAMPLES ≤ 2 in. Critical section (Type 1) Critical section (Type 2)

Joint ties

Figure 11.6.1  Anchorage details of flexural reinforcement with 90° hook.

The ACI Committee 352 recommendations also include provisions for the use of headed reinforcement in lieu of reinforcement terminating in a standard hook for both Type 1 and Type 2 joints. The reader is referred to these recommendations for further information.

11.7 TRANSFER OF COLUMN AXIAL FORCES THROUGH THE FLOOR SYSTEM The use of concretes with different compressive strengths for the columns and floor system (slab and beams) is not uncommon. When the specified compressive strength of the concrete used in the columns is greater than 1.4 times that of the concrete used in the floor system, the ACI Code requires that transmission of column axial load through the floor system be investigated (ACI-​15.3). For cases in which the design axial strength of the column joint calculated using fc′ of the floor system is less than the factored axial force, one of the following measurements should be taken to increase the axial strength of the connection, i. Concrete with the same specified compressive strength as that of the column should be provided within the floor depth and extended beyond the column faces at least 2 ft. ii. Addition of vertical dowels enclosed by spirals, designed to provide required additional axial strength. For beam-​column connections confined on all four sides by beams of approximately the same depth framing into the column and where the specified strength of the column concrete does not exceed 2.5 times that of the floor concrete, fc′ used in the calculation of axial strength can be taken as 75% that of the column plus 35% that of the floor system.

11.8 EXAMPLES Three examples are presented to illustrate the recommendations as they apply to Type 1 joints used in ordinary building construction. The requirements for Type 2 joints, where ductility and dissipation of energy through inelastic deformations are required, are more stringent. For examples illustrating the design of Type 2 joints, the reader is referred to the ACI Committee 352 recommendations [11.1].

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EXAMPLE 11.8.1 Design the exterior beam–​column joint shown in Fig. 11.8.1 for the case of the main beam subjected to negative bending. The joint is to be a Type 1 joint, and strength is the primary criterion. Use fc′ = 6000 psi, fy = fyt = 60,000 psi, and a factored axial load in the column Pu = 350 kips. Assume the story heights below and above the connection are equal to 11 ft and 10 ft, respectively. 20”

20” 8 – #9

Column

Main beam Spandrel beam 24” 4 – #10 20”

3 – #9 2 – #9

2 – #8 15”

10”

Figure 11.8.1  Design Example 11.8.1.

SOLUTION (a) Evaluate anchorage of #10 bars in the main beam. From Eq. (11.6.1), the required development length for #10 bars terminating with a standard hook in a Type 1 joint is

Ldh =

fy ψ c ψ r 50 fc′

db =

60, 000(0.7)(1.0) 50 6000

(1.27) = 13.8 in.

where ψc is taken as 0.7 given that the minimum side cover is satisfied (column is 5 in. wider than beam, which ensures a minimum side cover greater than 2.5 in.) and assuming a minimum of 2 in. of cover beyond the hook. The required development length is clearly less than the available length for anchorage. Also, the hooks for these bars should be located within 2 in. of the furthest edge of the core (i.e., core side closest to exterior face of column). (b) Examine the shear on the column to be transmitted through the joint. The moment on the columns may be assumed to be zero at midheight (see Fig. 11.8.2). In this case, with 12-​ft and 10-​ft column lengths and assuming that the beam reaches its flexural strength prior to the columns reaching their strength,

Vcol

(h1 + h2 ) = M pr ,b 2

where Mpr,b is the probable flexural strength of the beam. (Note that the moment caused by the beam shear with respect to the column centroid has been neglected in the determination of column shear.) (Continued)

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11.8 EXAMPLES

Example 11.8.1 (Continued) Vcol h1 = 5’ 2

Midheight of column assumed inflection 20” point Mpr,b

h2 = 6’ 2

Vcol = 45.1k T = 305k C = 305k 45.1k

Vcol

Figure 11.8.2  Column shear at joint in Example 11.8.1.

Because this is a Type 1 connection, Mpr,b may be taken as the nominal flexural strength using α = 1.0 (see Fig. 11.2.2). Thus, ignoring the bottom compression steel, C = 0.85 fc′ ba = 0.85(6)(15a ) = 76.5a T = α Ast f y = 1.0(4)(1.27)(60) = 305 kips





a = 3.99 in. Taking d = 21.5 in.,



a 3.99  1   M pr ,b = T  d −  = 305  21.5 −  = 496 ft-kips    2  12 2 3.99 21.5 − d −c 0.75 (0.003) = 0.0091 (0.003) = εt = εs = 3.99 c 0.75

Thus, the tension reinforcement yields and the section is tension controlled.

Vcol = M pr ,b

2 496 = = 45.1 kips (h1 + h2 ) 11

From Eq. (11.2.2), the joint horizontal shear is then

joint Vu = 305 − 45.1 = 260 kips

The nominal strength of the joint is given by Eq. (11.4.2). The design beam width bbeam in the direction of loading is 15 in. The effective joint width bj is the smallest of bbeam + bcol 15 + 20 = = 17.5 in. 2 2 mh (ii) bbeam + ∑ 2 (i)

Governs!

where mh / ​2 cannot exceed the extension of the column beyond the edge of the beam on each side. In the direction of loading being considered, the beam centerline and the column centroid intersect. Thus, the coefficient m is equal to 0.5. Accordingly,

mh 0.5(20) = = 5 in. 2 2 (Continued)

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Example 11.8.1 (Continued) The extension of the column beyond the edge of the beam is (20 –​15) /​ 2 = 2.5 in. Thus, mh /​2 = 2.5 in. and bbeam + ∑



mh = 15 + 2(2.5) = 20 in. 2

(iii) bcol = 20 in. Therefore, bj = 17.5 in. According to ACI-​ASCE Committee 352 [11.1, Section 4.3.2], a joint side is effectively confined when the beam covers at least three-​fourths of the width of the column, and the total depth of such a member is not less than three-​fourths of the total depth of the deepest member framing into the joint. Here the spandrel beams have a width of 10 in. and do not cover three-​fourths of the column width of 20 in. Because the joint is not confined on at least two opposite faces, the value γ for this connection, according to Table 11.4.1, is 15. The nominal shear strength is then (Eq. 11.4.2) and

Vn = γ fc′ b j h = 15 6000 (17.5)(20)

1 1000

= 407 kips

φ Vn = 0.75(407) = 305 kips > Vu = 260 kips



The joint shear strength is therefore adequate. (c) Horizontal column ties through the joint. For the purpose of establishing transverse reinforcement requirements, a core is not considered confined unless beams frame in from all four sides and cover at least three-​fourths of the column width, and not more than 4 in. of column width is exposed on each side of the beam (see Section 11.3). In this example the joint is not confined. Thus, according to the ACI Committee 352 recommendations, at least two layers of transverse reinforcement should be provided between the top and bottom longitudinal reinforcement of the deepest beam framing into the joint, not spaced farther than 12 in. Also, ACI-​15.4.2.2 specifies a maximum spacing of one-​half the depth of the shallowest beam framing into the column (10 in. in this case). For unconfined joints, transverse reinforcement should also satisfy the requirements for columns in ACI-​10.7.6.1.2 which, for the case of tied columns, refers to ACI-​25.7.2. For #10 longitudinal column bars or smaller, ACI-​25.7.2.2 requires the use of tie bars no smaller than #3. Maximum spacing for a column with #3 ties is limited to the smallest of (ACI-​25.7.2.1) 16 longitudinal bar diameters = 16(1.128) = 18 in. 48 tie diameters = 48(0.375) = 18 in. least cross-sectional dimension of column = 20 in. The above would not control for this joint, since 18 in. is approximately equal to the clear distance between the top and bottom layers of beam longitudinal reinforcement and at least two layers should be provided. Use #3 hoops @ 6 in. spacing, as detailed in Fig. 11.8.3. (Continued)

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11.8 EXAMPLES

Example 11.8.1 (Continued) For each layer of transverse reinforcement, provided area of transverse reinforcement must satisfy (ACI-​15.4.2)    Ash,min = 0.75 fc′

bcol sh (20)(6) = 0.75 6000 = 0.12 sq in. < 2(0.11 sq in.) f yt 60, 000

OK

Note that because fc′ = 6000 psi, 0.75 fc′ > 50 psi. (d) Transmission of axial load through the connection. Because concrete with the same compressive strength is used in both the columns and the beams, transmission of axial force through the connection is not a concern.

4 – #10

6 in. 6 in.

#3 joint hoops

6 in.

2 – #9

6 in.

Figure 11.8.3  Joint reinforcement details for Example 11.8.1.

EXAMPLE 11.8.2 Compute the effective joint width bj for the joint of Example 11.8.1 for loading in the direction parallel to the spandrel beams. SOLUTION In this direction, the centerline of the spandrel beams does not pass through the column centroid (see Fig. 11.8.1). However, all beam bars pass through the column core, and thus the recommendations of the ACI-​ASCE Committee 352 report apply [11.1, Section 2.2.2]. The effective joint width bj is the smallest of (i)

bbeam + bcol 2

(ii) bbeam + ∑

=

10 + 20 = 15 in. 2

mh 2

(Continued)

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Example 11.8.2 (Continued) The eccentricity between the column centroid and the centerline of the spandrel beams is (20 –​ 10) / ​2 = 5 in., which is greater than bcol / ​8 = 2.5 in. Therefore, the coefficient m is 0.3. Accordingly, mh 0.3(20) = = 3 in. ≤ [10 in.(extension of the column beyond thee edge 2 2 of the beam)]

and

bbeam + ∑



mh = 10 + 3 = 13 in. 2

Governs!

where ∑ mh/2 applies only to the side of the joint where the edge of the column extends beyond the edge of the beam; the other side of the beam is flush with the column face. (iii)  bcol = 20 in. Therefore, bj = 13 in.

EXAMPLE 11.8.3

11’–0 (assumed distance between points of inflection)

For the interior joint shown schematically in Fig. 11.8.4, check the adequacy of the joint shear strength and determine the transverse reinforcement required if the joint is Type 1 and the structure is subjected to non-seismic lateral loading in both directions. Use fc′ = 6000 psi, fy = 60,000 psi, and a factored axial load in the column Pu = 570 kips. Beam, 15 × 20, 4 – #9 top 4 – #8 bottom

Beam, 16 × 24 4 – #9 top 4 – #8 bottom Column, 20 × 20, 8 – #9 10 ft long above beam 12 ft long below beam

Figure 11.8.4  Joint for Example 11.8.3.

Figure 11.8.5  Forces in direction of 16 × 24 beams in Example 11.8.3.

(Continued)

397



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11.8 EXAMPLES

Example 11.8.3 (Continued) SOLUTION (a) Development of reinforcement. All longitudinal steel is to be extended through the joint. (b) Shear in direction of 16 × 24 beams. Due to lateral loading (from left to right) on the structure, the moments [Fig. 11.8.5(a)] M pr− ,b and M pr+ ,b have the same rotational direction and give the highest shear through the joint. Again, it is assumed that the probable moments equal the nominal strengths Mn for both positive and negative bending. If, however, the shear transferred into the joint is governed by the nominal moment capacity of the columns, the beam moments will be less than their nominal capacities. This is checked after calculation of probable moment capacities in the beams. The probable negative moment strength, M pr− ,b , based on tension in the 4–​#9 bars and neglecting compression reinforcement, is C = 0.85 fc′ ba = 0.85(6)(16)a = 81.6 a T = α Ast f y = 1.0(4)(1.00)(60) = 240 kips



a = 2.94 in. Taking d = 21.5 in., a 2.94  1   M pr− ,b = T  d −  = 240  21.5 −  = 400 ft-kips   2  12 2

d −c (0.003) = εt = εs = c

2.94 0.75 (0.003) = 0.0135 2.94 0.75

21.5 −

Thus the tension reinforcement yields and the section is tension controlled. For bending in the positive direction (tension in the 4–​#8 bars), neglecting compression reinforcement, T = α Ast f y = 1.0(4)(0.79)(60) = 190 kips



a = 2.32 in.



Taking d = 21.5 in.,

a 2.32  1   M pr+ ,b = T  d −  = 190  21.5 −  = 322 ft-kips   2 2  12

Because the area of positive (bottom) reinforcement is less than that of the negative (top) reinforcement, the tension reinforcement also yields and the section is tension controlled. The nominal moment capacity of the column for a factored axial force of 570 kips is 580 ft-​kips. Because the summation of the nominal moment capacities of the columns framing into the joint (1140 ft-​kips) is much greater than the summation of the moment capacities of the beams (722 ft-​kips), joint shear is governed by the moment capacity of the beams, as initially assumed. Thus,

(

Vcol = M pr− ,b + M pr+ ,b

) (h +2 h ) = (40011+ 322) = 65.6 kips 1

2

The joint horizontal shear is then joint Vu = 240 + 190 − 65.6 = 364 kips (Continued)

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Example 11.8.3 (Continued) The nominal strength of the joint is given by Eq. (11.4.2). The design beam width bbeam in the direction of loading is the average of the widths of the two beams (i.e., 16 in., since both are 16 × 24). The effective joint width bj is the smallest of bbeam + bcol 16 + 20 = = 18 in. 2 2 mh ,  where m = 0.5 (no joint eccentricity). Thus, (ii) bbeam + ∑ 2

Governs!

(i)

mh 0.5(20)  (20 − 16)  = = 5 in. >  = 2 in. (column extension beyond edge) 2 2 2   Thus, mh / ​2 = 2 in., and bbeam + ∑

mh = 16 + 2(2) = 20 in. 2

(iii) bcol = 20 in. Thus, bj = 18 in. Because there are beams on all four sides of the column whose widths (16 in. or 15 in.) cover at least three-​fourths of the width of the column (15 in.) and because their total depth (20 or 24 in.) is not less than three-​fourths of the total depth of the deepest member framing into the joint (18 in.), the joint is considered to be effectively confined in all four vertical faces. Thus, the value of γ is 24 (see Table 11.4.1). The nominal shear strength is (Eq. 11.4.2) and

Vn = γ fc′ b j h = 24 6000 (18)(20)

1 1000

= 669 kips

φ Vn = 0.75(669) = 502 kips > Vu = 364 kips



OK

The joint shear strength is therefore adequate. (c) Shear in the direction of the 15 × 20 beams. If the lateral loading on the structure requires moments to be transfered in this direction, similar to the force system in Fig. 11.5.5, the process used in step (b) would be repeated for this direction. The same top and bottom reinforcement is used in these beams as in the 16 × 24 beams, which means that the shear force introduced into the joint will be close to that calculated for the other direction (there will be a slight difference in column shear force because the moment capacities of the beams will be smaller due to their shallower depths). The joint shear strength will also remain approximately the same, since the effective joint width for this direction is 17.5 in., which is almost the same as that in the perpendicular direction (18 in.). Given the large difference between design joint shear strength and the joint shear demand calculated in part (b), joint shear strength in this direction should also be adequate. (d) Horizontal column ties through the joint. As discussed in Section 11.3, a core is considered confined when beams frame in from all four sides and cover at least three-​ fourths of the column width and no more than 4 in. of column width is exposed on each side of the beam. In this case, the joint is confined. Thus, column ties need not extend through the joint. (e) Transmission of axial load through the connection. Because concrete with the same compressive strength is used in both the columns and the beams, transmission of axial force through the connection is not a concern.

39



SELECTED REFERENCES

399

11.9 ADDITIONAL REMARKS The factors relating to the design of joints intended to resist primarily static loads within the elastic range (Type 1) have been discussed and treated using three examples. For the design of Type 2 joints requiring greater ductility, the reader is referred to the Committee 352 report [11.1], as well as ACI-​18.8. For clarity, the slab reinforcement was omitted from the calculations of the tensile resultant at the top of the beams in the examples. As recommended by the ACI-​ASCE Committee 352 report, slab reinforcement should be included in the calculations. This is particularly important for thick, heavily reinforced slabs, since this factor may result in an important increase in joint shear demand. Beam-​to-​column connections resisting seismic loads have been given considerable attention by several researchers, with the subject reviewed by Meinheit and Jirsa [11.10]. The behavior of beam-​to-​column connections in precast concrete has been studied by Pillai and Kirk [11.41] and others [11.15, 11.17, 11.25]. A comprehensive list of references on the subject of behavior and design of beam-​column joints is provided in the References section.

SELECTED REFERENCES  11.1. ACI-​ ASCE Committee 352. “Recommendations for Design of Beam–​ Column Joints in Monolithic Reinforced Concrete Structures” (ACI 352R-​02). Farmington Hills, MI: American Concrete Institute, 2002, 37 pp.   11.2. Norman W. Hanson and Harold W. Conner. “Seismic Resistance of Reinforced Concrete Beam-​ Column Joints,” Journal of the Structural Division, ASCE, 93, ST 5 (October 1967), 533–​560.   11.3. Norman W. Hanson. “Seismic Resistance of Concrete Frames with Grade 60 Reinforcement,” Journal of the Structural Division, ASCE, 97, ST 6 (June 1971), 1685–​1700.   11.4. L. M. Megget and R. Park. “Reinforced Concrete Exterior Beam-​Column Joints Under Seismic Loading,” New Zealand Engineering (Wellington), 26, 11 (November 1971), 341–​353.  11.5. S.  M. Uzumeri and M.  Seckin, “Behavior of Reinforced Concrete Beam-​Column Joints Subjected to Slow Load Reversals,” Report 74 05, Department of Civil Engineering, University of Toronto, Ontario, Canada, March 1974.   11.6. John Minor and James O. Jirsa. “Behavior of Bent Bar Anchorages,” ACI Journal, Proceedings, 72, April 1975, 141–​149.   11.7. James K. Wight and Mete A. Sozen. “Strength Decay of Reinforced Concrete Columns under Shear Reversals,” Journal of the Structural Division, ASCE, 101, ST5 (May 1975), 1053–​1065.   11.8. José G. L. Marques and James O. Jirsa. “A Study of Hooked Bar Anchorages in Beam–​Column Joints,” ACI Journal, Proceedings, 72, May 1975, 198–​209.   11.9. Duane L. N. Lee, Robert D. Hanson, and James K. Wight. “RC Beam-​Column Joints under Large Load Reversals,” Journal of the Structural Division, ASCE 103, 12 (December 1977), 2337–​2350. 11.10. Donald F. Meinheit and James O. Jirsa. “Shear Strength of R/​C Beam–​Column Connections,” Journal of the Structural Division, ASCE, 107, ST11 (November 1981), 2227–​2244. 11.11. Erik Skettrup, Jørgen Strabo, Niels Houmark Anderson, and Troels Brøndum-​ Nielson. “Concrete Frame Corners,” ACI Journal, Proceedings, 81, November–​ December 1984, 587–​593. 11.12. M. R.Ehsani and J. K. Wight. “Effect of Transverse Beams and Slab on Behavior of Reinforced Concrete Beam-​to-​Column Connections,” ACI Journal, Proceedings, 82, March–​April 1985, 188–​195. 11.13. Ahmad J. Durrani and James K. Wight. “Behavior of Interior Beam-​to-​Column Connections Under Earthquake-​Type Loading,” ACI Journal, Proceedings, 82, May–​June 1985, 343–​349. 11.14. M. R. Ehsani and J. K. Wight. “Exterior Reinforced Concrete Beam-​to-​Column Connections Subjected to Earthquake-​Type Loading,” ACI Journal, Proceedings, 82, July–​August 1985, 492–​499. 11.15. Prabhakara Bhatt and D. W. Kirk. “Tests on an Improved Beam Column Connection for Precast Concrete,” ACI Journal, Proceedings, 82, November–​December 1985, 834–​843. 11.16. T.  Ueda, I.  Lin, and N.  M. Hawkins. “Beam Bar Anchorage in Exterior Column–​Beam Connections,” ACI Journal, Proceedings, 83, May–​June 1986, 412–​422. 11.17. Charles W. Dolan, John F. Stanton, and Richard G. Anderson. “Moment Resistant Connections and Simple Connections,” PCI Journal, 32, March–​April 1987, 62–​74.

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11.18. Ahmad J.  Durrani and James K.  Wight. “Earthquake Resistance of Reinforced Concrete Interior Connections Including a Floor Slab,” ACI Structural Journal, 84, September–​October 1987, 400–​406. 11.19. Daniel P.  Abrams. “Scale Relations for Reinforced Concrete Beam–​Column Joints,” ACI Structural Journal, 84, November–​December 1987, 502–​512. 11.20. S.  E. El-​Metwally and W.  F. Chen. “Moment–​Rotation Modeling of Reinforced Concrete Beam–​Column Connections,” ACI Structural Journal, 85, July–​August 1988, 384–​394. 11.21. Parviz Soroushian and Ki-​Bong Choi. “Local Bond of Deformed Bars with Different Diameters in Confined Concrete,” ACI Structural Journal, 86, March–​April 1989, 217–​222. 11.22. Olga Velez Ammerman and Catherine Wolfgram French. “R/​ C Beam–​ Column–​ Slab Subassemblages Subjected to Lateral Loads,”Journal of Structural Engineering, ASCE, 115, 6 (June 1989), 1289–​1308. 11.23. Roberto T.  Leon. “Interior Joints with Variable Anchorage Lengths,” Journal of Structural Engineering, ASCE, 115, 9 (September 1989), 2261–​2275. 11.24. Thomas Paulay. “Equilibrium Criteria for Reinforced Concrete Beam–​Column Joints,” ACI Structural Journal, 86, November–​December 1989, 635–​643. 11.25. Catherine Wolfgram French, Michael Hafner, and Viswanath Jayashankar. “Connections between Precast Elements—​ Failure within Connection Region,” Journal of Structural Engineering, ASCE, 115, 12 (December 1989), 3171–​3192. 11.26. Roberto T. Leon. “Shear Strength and Hysteretic Behavior of Interior Beam–​Column Joints,” ACI Structural Journal, 87, January–​February 1990, 3–​11. 11.27. S.  J. Pantazopoulou and J.  P. Moehle. “Truss Model for 3-​D Behavior of R.  C. Exterior Connections,” Journal of Structural Engineering, ASCE, 116, 2 (February 1990), 298–​315. 11.28. M. R. Ehsani and J. K. Wight. “Confinement Steel Requirements for Connections in Ductile Frames,” Journal of Structural Engineering, ASCE, 116, 3 (March 1990), 751–​767. 11.29. M. Seckin and H. C. Fu. “Beam–​Column Connections in Precast Reinforced Concrete,” ACI Structural Journal, 87, May–​June 1990, 252–​261. 11.30. Philip K.  C. Wong, M.  J. N.  Priestley, and R.  Park. “Seismic Resistance of Frames with Vertically Distributed Longitudinal Reinforcement in Beams,” ACI Structural Journal, 87, July–​August 1990, 488–​498. 11.31. Parviz Soroushian and Ki-​Bong Choi. “Analytical Evaluation of Straight Bar Anchorage Design in Exterior Joints,” ACI Structural Journal, 88, March–​April 1991, 161–​168. 11.32. Parviz Soroushian, Ki-​Bong Choi, Gill-​Hyun Park, and Farhang Aslani. “Bond of Deformed Bars to Concrete: Effects of Confinement and Strength of Concrete,” ACI Structural Journal, 88, May–​June 1991, 227–​232. 11.33. Ian N.  Robertson and Ahmad J.  Durrani. “Gravity Load Effect on Seismic Behavior,” ACI Structural Journal, 88, May–​June 1991, 255–​267. 11.34. Mohammad R. Ehsani and Fadel Alameddine. “Design Recommendations for Type 2 High-​ Strength Reinforced Concrete Connections,” ACI Structural Journal, 88, May–​June 1991, 277–​291. 11.35. Fariborz Barzegar, Rainer Echle, and Mehrdad Foroozesh. “Moment Transfer and Slab Effective Widths in Laterally Loaded Edge Connections,” ACI Structural Journal, 88, September–​ October 1991, 615–​623. 11.36. A. G. Tsonos, I. A. Tegos, and G. Gr. Penelis. “Seismic Resistance of Type 2 Exterior Beam–​ Column Joints Reinforced with Inclined Bars,” ACI Structural Journal, 89, January–​February 1992, 3–​12. 11.37. Gilson N.  Guimaraes, Michael E.  Kreger, and James O.  Jirsa. “Evaluation of Joint-​Shear Provisions for Interior Beam–​Column–​Slab Connections. Using High-​Strength Materials,” ACI Structural Journal, 89, January–​February 1992, 89–​98. 11.38. A.  H. Mattock and J.  F. Shen. “Joints between Reinforced Concrete Members of Similar Depth,” ACI Structural Journal, 89, May–​June 1992, 290–​295. 11.39. Stavroula Pantazopoulou and John Bonacci. “Consideration of Questions about Beam–​Column Joints,” ACI Structural Journal, 89, January–​February 1992, 27–​36. 11.40. Richard H. Scott. “Intrinsic Mechanisms in Reinforced Concrete Beam–​Column Connection Behavior,” ACI Structural Journal, 93, May–​June 1996, 336–​346. 11.41. S.  U. Pillai and D.  W. Kirk. “Ductile Beam–​Column Connection in Precast Concrete,” ACI Journal, Proceedings, 78, November–​December 1981, 480–​487. 11.42. B.  Abdel-​Fattah and James K.  Wight. “Study of Moving Beam Plastic Hinging Zones for Earthquake-​Resistant Design of R/​C Buildings,” ACI Structural Journal, 84, January–​February 1987, 31–​39. 11.43. Sergio M. Alcocer. “R/​C Frame Connections Rehabilitated by Jacketing,” Journal of Structural Engineering, ASCE, 119, 5 (May 1993), 1413–​1431.

401



PROBLEMS

401

11.44. Sergio M. Alcocer and James O. Jirsa. “Strength of Reinforced Concrete Frame Connections Rehabilitated by Jacketing,” ACI Structural Journal, 90, May–​June 1993, 249–​261. 11.45. O. V. Ammerman and Catherine Wolfgram-​French. “R/​C Beam–​Column–​Slab Subassemblages Subjected to Lateral Loads,” Journal of Structural Engineering, 115, 6 (June 1989), 1298–​1308. 11.46. P. C. Cheung, Thomas Paulay, and Robert Park. “Mechanisms of Slab Contributions in Beam–​ Column Subassemblages,” Design of Beam–​Column Joints for Seismic Resistance (SP-​123). Farmington Hills, MI: American Concrete Institute, 1991 (pp. 259–​289). 11.47. P.  A. Cote and John W.  Wallace. “A Study of RC Knee-​Joints Subjected to Cyclic Lateral Loading,” Report No. CU/​CEE-​94/​04, Department of Civil and Environmental Engineering, Clarkson University, Potsdam, NY, 1994. 11.48. Catherine Wolfgram-​French and Jack P. Moehle. “Effect of Floor Slab on Behavior of Slab-​ Beam-​Column Connections,” Design of Beam–​Column Joints for Seismic Resistance (SP-​123). Farmington Hills, MI: American Concrete Institute, 1991 (pp. 225–​258). 11.49. Roberto T. Leon and James O. Jirsa. “Bi-​directional Loading of RC Beam–​Column Joints,” Earthquake Spectra, 2, 3 (1986), 537–​564. 11.50. S. Mazzoni, Jack P. Moehle, and C. R. Thewalt. “Cyclic Response of RC Beam-​Column Knee Joints:  Test and Retrofit,” Report No. UCB/​EERC-​91/​14, Earthquake Engineering Research Center, University of California, Berkeley, October 1991, 24 pp. 11.51. Stavroula Pantazopoulou, Jack P. Moehle, and B. M. Shahrooz. “Simple Analytical Model for T-​Beam in Flexure,” Journal of Structural Engineering, 114, 7 (July 1988), 1507–​1523. 11.52. Robert Park, M.  J. N.  Priestley, and W.  D. Gill. “Ductility of Square-​Confined Concrete Columns,” Journal of the Structural Division, ASCE, 108, ST4 (April 1982), 929–​950. 11.53. C. G. Quintero-​Febres and J. K. Wight. “Investigation of the Seismic Behavior of RC Interior Wide Beam-​Column Connections,” Report No. UMCEE 97-​15, Department of Civil and Environmental Engineering, University of Michigan, Ann Arbor, September 1997, 292 pp. 11.54. G. S. Raffaelle and James K. Wight. “R/​C Eccentric Beam–​Column Connections Subjected to Earthquake-​Type Loading,” Report No. UMCEE 92-​18, Department of Civil and Environmental Engineering, University of Michigan, Ann Arbor, 1992, 234 pp. 11.55. E.  I. Saqan and Michael Kreger. “Evaluation of U.S. Shear Strength Provisions for Design of Beam–​ Column Connections Constructed with High-​ Strength Concrete,” High-​Strength Concrete in Seismic Regions (SP-​176). Farmington Hills, MI:  American Concrete Institute, 1998 (pp. 311–​328). 11.56. S.  Sugano, T.  Nagashima, H.  Kimura, and A.  Ichikawa. “Behavior of Beam–​Column Joints Using High-​Strength Materials,” Design of Beam–​Column Joints for Seismic Resistance (SP-​ 123). Farmington Hills, MI: American Concrete Institute, 1991 (pp. 359–​378). 11.57. John W. Wallace, S. W. McConnell, P. Gupta, and P. A. Cote. “Use of Headed Reinforcement in Beam–​Column Joints Subjected to Earthquake Loads,” ACI Structural Journal, 95, September–​ October 1998, 590–​606. 11.58. Catherine Wolfgram-​French and A. Boroojerdi. “Contribution of R/​C Floor Slab in Resisting Lateral Loads,” Journal of Structural Engineering, 115, 1 (January 1989), 1–​18. 11.59. H.  E. Zerbe and A.  J. Durrani. “Seismic Response of Connections in Two-​Bay R/​C Frame Subassemblies,” Journal of Structural Engineering, 115, 11 (November 1989), 2829–​2844. 11.60. S. W. McConnell and John W. Wallace. “Behavior of Reinforced Concrete Beam-​Column Knee Joints Subjected to Reversed Cyclic Loading,” Report No. CU/​CEE-​95/​07, Department of Civil and Environmental Engineering, Clarkson University, Potsdam, NY, June 1995.

PROBLEMS All problems are to be worked using the ACI Code and the recommendations of ACI-​ASCE Committee 352 [11.1]. In all cases, assume that joint shear is governed by the flexural capacity of the beams. 11.1 Check the anchorage of bars and determine the transverse reinforcement for the exterior beam-​column joint of the figure for Problem 11.1. Assume that lateral loading provides the critical condition at the joint; assume that columns are bent in double curvature with an inflection point at midheight. Assume a

Type 1 joint and use fc′ = 4000 psi and fy = fyt = 60,000 psi. Show detail of transverse reinforcement in joint. 11.2 Repeat Problem 11.1, except consider the column to be 18 in. square, the spandrel 14 × 24 with 3–​#9 bars (top and bottom), and the beam 18 × 28 with 4–​#10 bars (top and bottom).

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402

C hapter   1 1     M onolithic B eam - C olumn C onnections

Column 24 × 24

Column 11’ height

11’

Beam, 21 × 28, 4 – #11 bars top 4 – #9 bars bottom Spandrel, 18 × 30, 3 – #10 bars top 3 – #9 bars bottom

Problem 11.1 

11.3 Repeat Problem 11.1, except consider the joint to be an interior one with the 21 × 28 beam on both sides of the column. The spandrel then becomes an interior beam. Assume that all bars are continuous through the column. The critical loading condition for the joint is with clockwise moments from the 21 × 28 beam acting on the joint at both sides of the column.

CHAPTER 12

SERVICEABILITY

12.1 INTRODUCTION As discussed in Chapter 2, a reinforced concrete structure must be designed not only to protect the life of the occupants, but also to ensure that it serves its intended purpose throughout its service life:  that is, it must be functional or serviceable. The first requirement is satisfied by meeting the safety provisions of the ACI Code (see Sections 2.6 and 2.7). The second requirement, hereafter referred to as serviceability, will be satisfied by meeting the ACI Code serviceability provisions, which may be expressed as given earlier.

Sallowable ≥ Sservice_loads [2.8.2]

(see Sections 2.8 and 2.9). In this equation, Sallowable represents the specified limit not to be exceeded in order to ensure satisfactory behavior and functionality of the structure during its service life, while the term Sservice_​loads is a calculated performance quantity of the structural member under service-​level loads. Note that the allowable serviceability limit, Sallowable, for satisfactory performance of a given member or structural system is not unique and will depend on the use of the structure. For example, a floor system supporting highly sensitive instrumentation must be designed for stricter deflection and vibration limits than one supporting partitions in a standard residential building. Common serviceability issues that affect the design of reinforced concrete members are excessive deflection, detrimental cracking, excessive amplitude or undesirable frequency of vibration, and excessive noise transmission. While serviceability issues do not generally pose a threat to the safety of the occupants, they can have severe economic consequences and thus, the structural engineer must adequately address them in design. This chapter discusses in detail design for the two most common serviceability concerns: control of deflections and control of crack widths. Vibration and noise control, which may also be important, lie outside the scope of this book and are not treated here in detail. Nonetheless, a brief discussion of current design criteria and some guidance for the control of floor vibrations is provided in Section 12.20.

12.2 FUNDAMENTAL ASSUMPTIONS Consider the flexural behavior of the reinforced concrete beam section under increasing applied loads described in Section 3.2. When designed to satisfy the strength requirements of the ACI Code, the strain and stress distributions in the concrete and the steel reinforcement at ultimate (i.e., under factored loads) will be as shown in Fig. 3.2.1(c). This means that under service-​level loads, the strains and stresses in the concrete and the steel

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reinforcement will be much smaller than those at the ultimate condition. Indeed, the behavior of the concrete in the compression zone is expected to be essentially linear and the steel strains less than the yield strain. For the purpose of computing the behavior of reinforced concrete members under service loads, it will be assumed that the stresses in the concrete in compression and the steel reinforcement are proportional to the strains:  that is, their behavior is linear elastic. The assumption that stresses are proportional to strains for the concrete in compression is reasonably correct for stresses below about one-​half fc′, the 28-​day compressive strength. This will typically be the case for beams with low and moderate amounts of tension reinforcement. The assumptions that (1)  plane sections remain plane, (2) concrete does not carry tension (it cracks under tension), and (3) no slip occurs between the steel bars and the surrounding concrete during the development of the tensile forces in the bars also hold true under service load levels.

12.3 MODULUS OF ELASTICITY RATIO, n As discussed in Chapter  1, the stress-​strain relation for the steel reinforcement is linear below the yield stress, but that of concrete is only approximately linear, even at or below the service load stresses. The modulus of elasticity of steel, Es, varies little with its strength, whereas that of concrete varies with its density and strength. In ACI-​20.2.2.2, Es is taken to be 29,000,000 psi (200,000 MPa); ACI-​19.2.2.1 gives Ec as 33w1c .5 fc′, in which wc is the unit weight of concrete in lb/cu ft. Table 1.10.1 gives values for the modulus of elasticity Ec for normal-​weight concrete of various strengths. It will be shown that the ratio between Es and Ec, or the modular ratio, n = Es /​Ec, is often used for computing stresses under service loads rather than the actual values of Ec or Es. These stresses are called service stresses (or working stresses, in reference to the working stress method used in the past). It is suggested that the values in Table 12.3.1 be used for normal-​weight concrete.

12.4 EQUILIBRIUM CONDITIONS Two equilibrium conditions apply to a section subjected to bending only: (1) the resultant internal compressive force must be equal to the resultant internal tensile force; and (2) the moment of the internal couple, composed of the resultant compressive and tensile forces, must be equal to the applied bending moment on the cross section. These two equilibrium conditions must hold true under any applied loads (e.g., under service-​level loads and at the ultimate condition). Under service loads, however, it is assumed that the stress distribution across the depth is linear and proportional to strains as discussed in Section 12.2.

TABLE 12.3.1  PRACTICAL VALUES FOR MODULAR RATIO n FOR NORMAL-​WEIGHT CONCRETE Inch-​Pounds

SI

fc′ (psi)

n

3000 3500 4000 4500 5000 6000

9 8.5 8 7.5 7 6.5

a

a

fc′ (MPa)

For practical use, multiply MPa by 10 to obtain value in kgf/​cm2.

20 25 30 35 40

n 9 8 7.5 7 6.5

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12.4  EQUILIBRIUM CONDITIONS

The resultant compressive force may be entirely from concrete stresses, or it may be from a combination of the stresses in concrete and those in the compression reinforcement. The resultant tensile force, of course, is provided entirely by the tension reinforcement, since the contribution of the concrete in tension is neglected.

EXAMPLE 12.4.1 Using the equilibrium conditions, determine the stresses under service loads in the tension steel reinforcement and on the extreme compression fiber of concrete for the beam of Example 3.9.1. The beam is simply supported with a span of 40 ft and was designed to carry a live load of 1.38 kips/​ft and a dead load of 1 kip/​ft (not including beam weight), with fc′ = 4000 psi, and fy = 60,000 psi. The final design called for an 18 × 34 in. cross section reinforced with 2–​#9 and 4–​#10 bars as tension steel (see Fig. 3.9.2). Ignore the presence of the top bars in the compression zone. SOLUTION (a) Compute the applied moment under service-level loads. From part (a) of the solution to Example 3.9.1, the moment under service live loads, ML = 276 ft-​kips, and from part (f), the moment due to service dead loads (including self-​weight), MD  =  328 ft-​kips. Therefore, the applied moment under service-​level loads is

Mservice = M D + M L = 328 + 276 = 604 ft-kips b = 18”

fc = Ec εc

εc

c/3 C

c d = 31.5”

NA 31.5 – 4 #10

c 3

No tensile stresses except in steel reinforcement εs

2 #9 (a) Section

(b) Strains

fs = Es εs

T

(c) Stresses

Figure 12.4.1  Assumed strain and stress distributions for the beam of Example 12.4.1 under service loads.

(b) Compute stresses under service loads. The assumptions of linearly varying strains and stress proportional to strain are shown in Fig. 12.4.1(b) and 12.4.1 (c). The first step in the solution is to locate the neutral axis (N.A.). The internal compressive force is obtained by integration of the stress times the area on which it acts. This is equivalent to computing the volume of the stress solid. Thus, the internal compressive force in the concrete is, C=



1 fc bc = 9.0 fc c 2

The internal tensile force is

T = fS As = fs [ 2(1.00) + 4(1.27)] = 7.08 fs (Continued)

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Example 12.4.1 (Continued) Equating C to T gives fs 9.0c = fc 7.08



The ratio of fs to fc may also be obtained by using the linear strain relationship and taking stress proportional to strain, as shown in Fig. 12.4.1,

ε s 31.5 − c = c εc



fs Es ε s ε = = n s fc Ec ε c εc

where n is the modular ratio defined in Section 12.3. From Table 12.3.1, choose n = 8 for fc′ = 4000 psi. Therefore, fs ε  31.5 − c  = n s = 8   fc εc c 



Note that the actual values of Es and Ec are not needed. Equating the two expressions for fs /​fc, 9.0c 8(31.5 − c) = 7.08 c 9c 2 = 56.64(31.5 − c)



c 2 + 6.29c = 198.24 c = 11.28 in. Note that other than the ratio of the moduli of elasticity, only the properties of the section (depth, width, and steel area) affect the position of the neutral axis. The loading does not affect the neutral axis location. The moment arm of the internal couple, or the distance from the centroid of the compressive solid to the centroid of the tension steel, is

arm = 31.5 −

c 11.28 = 31.5 − = 27.74 in. 3 3

Then, using the second equilibrium condition,

Mservice = 604 ft-kips = 7248 in.-kips = (C or T ) × arm C=T =



Mservice 7248 = = 261.3 kips arm 27.74

The stresses are determined from the expressions for C and T, fc =

C 261, 300 = = 2574 psi [or ≈ 0.64 fc′] 9c 9(11.28)

261, 300 T = 36, 907 psi [or ≈ 0.62 f y ] fs = = 7.08 7.08 (Continued)

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12.5  METHOD OF TRANSFORMED SECTION

Example 12.4.1 (Continued) The stresses in the tension steel reinforcement are less than the yield strength, and therefore the assumption that the behavior of the reinforcement is linear elastic is correct. As noted earlier, behavior of the concrete is nonlinear even at low stress levels, but it can be assumed to behave linearly up to stresses of approximately of 0.5 fc′ . In this example, the maximum stress is greater than this value, and therefore it may be expected that concrete behavior near the top of the beam will be nonlinear. In reference to Fig. 1.10.1, it may be seen, however, that while behavior of the concrete at 0.64 fc′ is nonlinear, the assumption of linear elastic behavior is still reasonable. In effect, it may be shown that an analysis of the beam cross section using a nonlinear stress-​strain distribution for the concrete will yield results similar to those obtained with the linear elastic assumption.

12.5 METHOD OF TRANSFORMED SECTION In the method of the transformed section, the cross section composed of steel reinforcement and concrete is transformed into an equivalent isotropic, homogeneous section of either concrete or steel alone. Most commonly, the steel is transformed into an equivalent area (i.e., imaginary area) of concrete where the transformed section is treated as a cross section made entirely of concrete. In the transformation, two conditions must be satisfied. Consider in Fig. 12.5.1(a) a reinforced concrete section in a tension zone where the concrete is assumed to be cracked. Let As and At be the areas of the actual steel reinforcement and the transformed, equivalent concrete, respectively. Similarly, let fs and ft be the tensile stresses in the actual steel and the equivalent concrete, respectively. First, the equilibrium condition requires that the total tensile force be the same in the actual and the transformed section, or As fs = At ft



(12.5.1)

where the contribution of the cracked concrete is ignored. Second, compatibility of deformations requires that the strain be the same, or fs f = t E s Ec



a

(12.5.2)

As nAs

b Actual section

Transformed section

(a) Actual and transformed area of steel in a tension zone (cracked concrete) As’ a

(n–1)As’

b Actual section Transformed section (b) Actual and transformed area of steel in a compression zone

Figure 12.5.1  Transformed areas of steel (a) in a tension zone and (b) in a compression zone.

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Utilizing the modular ratio n = Es /​Ec and combining Eqs. (12.5.1) and (12.5.2), the equivalent area of concrete (transformed area of steel) is

At = nAs

(12.5.3)

fs n

(12.5.4)

and the corresponding stress is ft =



Thus, the equivalent concrete area At is n times the actual steel area in tension, and the tensile stress ft (i.e., imaginary stress) of the equivalent area of concrete is 1/​n times the actual tensile steel stress. Similar expressions may be obtained for a cross section in the compression zone where the steel is in compression and the concrete is subjected to a stress fc. As before, let As′ and At be the areas of the actual steel reinforcement and the transformed, equivalent concrete, respectively [Fig. 12.5.1(b)]. Similarly, let fs′ and ft be the compressive stresses in the actual steel and the equivalent concrete, respectively. Equilibrium requires that the total compressive force be the same in the actual and the transformed section, or

As′ fs′+ Aconc fc = At ft + abfc

where Aconc = ab − As′ Therefore,



As′ fs′+ (ab − As′ ) fc = At ft + abfc  f′  As′  s − 1 fc = At ft f 

(12.5.5)

c

Compatibility of deformations require that the strain be the same in the concrete and the steel in the actual section and in the equivalent area of concrete in the transformed section, or

fs′ f f = t = c E s Ec Ec

(12.5.2′)

Rearranging and substituting Eq. (12.5.2′) in Eq. (12.5.5),



E  As′  s − 1 = At  Ec 

or

As′ (n − 1) = At

(12.5.6)

and

ft =

fs′ n

(12.5.7)

In other words, the transformed, equivalent concrete area At is (n–​1) times the actual steel area in compression, and the compressive stress ft (i.e., imaginary stress) is 1/​n times the actual compressive steel stress. By using Eqs. (12.5.3), (12.5.4), (12.5.6), and (12.5.7), a reinforced concrete section may then be treated as a section of one material (concrete). These equations are very useful whenever elastic properties are needed—​for example, to compute deflections under service-​level loads.

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409

EXAMPLE 12.5.1 Use the transformed section method to locate the elastic neutral axis and to compute the stresses in the tension steel reinforcement and in the extreme fiber in compression in the concrete for the beam of Example 12.4.1 under service loads. SOLUTION (a) Locate the elastic neutral axis using transformed area [Fig.  12.5.2(b)]. For fc′ = 4000 psi, the modular ratio n may be taken as 8 (see Table 12.3.1). The area 18.0c above the neutral axis is in compression, whereas the equivalent tension area nAs = 8(7.08) = 56.64 sq in. is assumed to be concentrated at the centroid of the steel reinforcement at a distance (31.50 –​c) below the neutral axis. Equating the first moments of the compression and tension areas about the neutral or centroidal axis of the transformed section,

1 (18.0)c 2 = 56.64 (31.50 − c) 2 c = 11.28 in.

Figure 12.5.2  Transformed section for the beam of Example 12.5.1.

(d) Use the transformed section method to find stresses. The transformed cracked section moment of inertia about the neutral axis (centroid of the transformed section) is

1 I cr = (18.0)(11.28)3 + 56.64(20.22)2 3 4 = 8611 + 23157 ≈ 31, 770 in.

The applied moment under service loads was computed as 604 ft-​kips [part (a) of Example 12.4.1]. By the flexure formula, the stress in the extreme fiber in compression in the concrete at the top is

fc =

604 (12, 000 ) (11.28) = 2573 psi 31, 770

and in the equivalent area of concrete (transformed area of steel),

ft =

604 (12, 000 ) ( 20.22) = 4613 psi 31, 770

The actual stress in the steel, fs, is found from Eq. (12.5.4) as fs = nft = 8(4613) = 36, 904 psi The values computed for fc and fs are the same as those computed in Example 12.4.1, as was expected.

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EXAMPLE 12.5.2 Compute the moment of inertia Icr of the transformed cracked section to be used in deflection calculation for the doubly reinforced section shown in Fig.  12.5.3, using fc′ = 4000 psi (n = 8). SOLUTION (a) Locate the neutral axis. The steel is transformed into equivalent concrete by using At = nAs for the steel in tension [Eq. (12.5.3)] and At = (n − 1) As′ for the compression steel [Eq.(12.5.6)]. Referring to Fig. 12.5.3, and equating the first moments of the compression and tension area,

13c 2 + 10.5(c − 2.5) = 56.64(22 − c) 2

Solving for c,

c = 9.76 in.



(b) Determine the transformed cracked section moment of inertia.

1 I cr = (13)(9.76)3 + 10.5(9.76 − 2.5)2 + 56.64(22 − 9.76)2 3 = 4030 + 550 + 8480 = 13,100 in.4 (n – 1)As’ = 10.5 sq in.

2 – #6 2 – #5 As’ = 1.50 sq in.

22”

13”

c = 9.76”

2.5”

4 – #10 2 – #9 As = 7.08 sq in.

nAs = 56.64 sq in.

Figure 12.5.3  Doubly reinforced section of Example 12.5.2.

12.6 DEFLECTIONS—​G ENERAL Over the years, the use of higher strength concrete has allowed a more efficient design of reinforced concrete members, which in turn has resulted in smaller, shallower sections. These smaller sections, while still providing the required strength, have less stiffness and consequently exhibit larger deflections. In fact, control of deflections rather than strength is often the governing design criterion for many structures built today. As noted earlier, the permissible deflection is governed by the serviceability requirements for the structure, such as the amount of deformation that can be tolerated by the interacting components of the structure. Excessive deflection of the member may not, in itself, be detrimental, but the effect on nonstructural (and structural components) that are supported by the deflecting member frequently determines the acceptable amount of deflection. Both the short-​term (instantaneous or immediate) and the long-​term (time-​dependent) effects must be considered. The acceptable deflection depends on many factors, among which are the intended use or function of the building (warehouse, school, factory, residence, etc.), the types of finish (presence of brittle tiles, plastered ceilings, brick veneer, etc.), the type and arrangement of partitions, the sensitivity of equipment to deflection, and

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411

the magnitude and duration of the transient loads (i.e., live load, snow load, or rain loads). Note that even in situations in which excessive deflections are not likely to cause damage, large noticeable deflections could be psychologically disturbing to humans and thus should be avoided. The ACI Code provides guidance for computing the minimum member depth to avoid excessive deflections, and also provides limits on the maximum permitted deflections for common situations. For nonprestressed members “not supporting or attached to partitions or other construction likely to be damaged by large deflections,” ACI-​Table 9.3.1.1 provides the minimum depth required for beams (see Section 12.15). Similar requirements for the minimum depth of one-​way slabs are given in ACI-​Table 7.3.1.1. When these minimum depth requirements are not satisfied or if excessive deflection may cause a problem, the calculated immediate and time-​dependent deflections must satisfy the limits of ACI-​24.2.2. Accordingly, the maximum acceptable immediate deflections due to live load, for flat roofs or floors that do not support and are not attached to nonstructural elements such as plastered ceilings or frangible partitions likely to be damaged by large deflections, are prescribed by ACI Table 24.2.2 to be (a) for flat roofs, L /​180, and (b) for floors, L /​360. Further, in recognition of the increase of deflection with time, ACI Table 24.2.2 also prescribes limits for the sum of the time-​dependent deflection due to all sustained loads plus any immediate deflection due to any additional live load. The two stated limits for this deflection combination are (a) L /​480 when nonstructural elements are likely to be damaged, and (b) L /​240 when no damage to nonstructural elements is likely. Limitations on deflection are somewhat arbitrary. Historically, L /​360 has been the accepted limit to prevent the cracking of plastered ceilings. Other limits should be considered as guidelines, with the designer having the responsibility for evaluating the possible adverse effects of excessive deflection in any given situation. Recommendations for allowable deflections for a great variety of situations are provided by ACI Committee 435 [12.1–​12.6]. The general concepts dealt with in this chapter are applicable to both one-​way (beams and slabs) and two-​way systems. Specific examples are applied only to one-​way systems. The reader is referred to Section 16.20 for a brief discussion of two-​way system deflections, along with specific references.

12.7 DEFLECTIONS FOR LINEAR ELASTIC MEMBERS Various methods are available in structural analysis for computing deflections of members with uniform or variable moment of inertia in statically determinate and indeterminate structures. In general, the maximum deflection in an elastic member may be expressed as

∆ max = β a

ML2 EI c

(12.7.1)

where M = a reference value of bending moment such as the maximum positive value L = span length E = modulus of elasticity Ic = moment of inertia of member cross section βa = a coefficient that depends on the degree of fixity at supports, the variation in moment of inertia along the span, and the distribution of loading; such as 5 48 for simply supported uniformly loaded beam; 1 4 for uniformly loaded cantilever The deflection coefficient βa for simple cases may be found in handbooks, such as the ACI Design Manual [2.20], or Handbook of Concrete Engineering [12.7]. The following example uses one elastic method, the conjugate beam method, to derive one of the most useful deflection expressions.

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EXAMPLE 12.7.1 Using the conjugate beam method, derive the general expression for the elastic midspan deflection for a uniformly loaded span with unequal end moments, as shown in Fig. 12.7.1. SOLUTION In the conjugate beam method, the deflection at a given point equals the bending moment at that point for a beam loaded with the M/​EI diagram. Thus the system of Fig. 12.7.1 may be regarded as being composed of three separate conjugate beams, as shown in Fig. 12.7.2. w Ma

Mb L Ms Mo =

+

wL2 8

–

– Constant I

Ma

Mb L/2

L/2

Figure 12.7.1  Typical bending moment diagram for a uniformly loaded span of a continuous beam. Mo El + Simply supported uniform load L/2

L (a)

Ma El End moment at left end of span

–

(b) Mb El –

End moment at right end of span

(c)

Figure 12.7.2  Component conjugate beams.

(Continued)

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Example 12.7.1 (Continued) The midspan deflections, or the bending moments on the three conjugate beams, are uniform load, ∆ s = left end moment, ∆ a =

2  L   M 0   L 3  L   5 M 0 L2 −   =   3  2   EI   2 8  2   48 EI 1  − Ma   L   L  1  − Ma   L   1   L    −     3  EI   2   2  2  2 EI   2   3   2 

− M a L2 = 16 EI



− M b L2 16 EI

right ennd moment, ∆ b = The total midspan deflection Δm is

∆m = ∆s + ∆a + ∆b

=

5 M 0 L2 M a L2 M b L2 − − 48 EI 16 EI 16 EI

=

L2 [5M0 − 3( M a + Mb )] 48 EI

(12.7.2)

The net positive midspan moment Ms is 1 M s = M 0 − ( M a + M b ) (12.7.3) 2



Then, on substituting Eq. (12.7.3) in Eq. (12.7.2), one obtains ∆m =

L2 48 EI

5L2 = 48 EI

5   5 M s + 2 ( M a + M b ) − 3( M a + M b )   1    M s − 10 ( M a + M b )  

(12.7.4)

Equation (12.7.4) may be used with satisfactory results for practically all prismatic (i.e., constant EI) beams, even though the absolute maximum deflection will be obtained only when the loading is uniform and the end moments are equal.

Continuous Beams Having Variable Flexural Rigidity When significant changes of gross cross section along the span length are involved, such changes in cross-​sectional properties should be included when statically indeterminate analyses are performed. When deflections are desired for such cases, the best approach is to account explicitly for the variable flexural rigidity EI, either mathematically in the component conjugate beams (see Fig. 12.7.2), or by the use of numerical integration. Section 12.9 treats this subject further in a discussion of what value of the moment of inertia should be used for reinforced concrete sections.

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12.8 MODULUS OF ELASTICITY As discussed in Section 1.10, variance in interpretation is encountered in references to the modulus of elasticity of concrete. Ordinary beam theory presumes the modulus in tension to be the same as that in compression for a homogeneous material. In a reinforced concrete section, creep affects the apparent modulus primarily in the compression zone. Even the modulus under short-​duration loading, measured as the secant modulus, is considerably more variant than the compressive strength, fc′. The secant modulus in tension is essentially the same as that in compression when the stress magnitude is low, but the modulus reduces markedly as the stress nears the cracking level. Also, in regions of high bending moment the concrete will be cracked in the tension zone, while at low moment sections the concrete may be uncracked. In other words, the actual modulus of elasticity in both tension and compression varies not only with the magnitude of stress from top to bottom at a section, but also along the span. Furthermore, creep and shrinkage over a period of time exert an effect equivalent to reducing the modulus in compression and will generally magnify deflections by a factor of 2 to 3. Referring to Fig. 1.11.1, one may note that the true elastic strain decreases slightly with time, indicating that the modulus of elasticity increases. The modulus increases because it is dependent on fc′, which increases with age. In design practice, the apparent elastic strain used is often computed with a code-​specified Ec, presumably corresponding to the 28-​day age at loading. Beyond 28 days, however, the increase in modulus of elasticity is relatively slight.

12.9 EFFECTIVE MOMENT OF INERTIA In addition to the modulus of elasticity, the flexural rigidity includes the cross-​sectional property, moment of inertia. Even for sections whose external dimensions are constant, the moment of inertia varies considerably along a span. Consider a span with a T-​shaped section in a continuous beam or frame, as shown in Fig. 12.9.1. At or near the support in such a continuous structure, the concrete slab is cracked, so that the effective section is section A-​A; while in the positive moment zone it is the stem that is cracked, and the effective section is section B-​B. Furthermore, near points of inflection the entire section may be uncracked and fully effective, as shown in section C-​C. The problem is further compounded because the amounts of reinforcement at the top and bottom of the beam vary along the span. A

C

B

A

w

A

C

B

w

A

Section A–A

Section B–B

Section C–C

Figure 12.9.1  Effective moment of inertia for continuous T-​shaped sections.

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In an elastic continuity analysis as discussed in Chapters 7 and 9, only relative stiffness values are required; in deflection computations, however, the absolute magnitudes for Ec and I must first be determined or assumed. Moreover, the amount of deflection under each increment of load is not constant. The flexural rigidity Ec I is greater at a low load level, since the full uncracked section provides the greatest effective moment of inertia. As the load increases, cross sections will gradually crack along the span, thereby reducing the flexural rigidity Ec I. The subject of effective moment of inertia for computing deflections due to short-​term loading was studied by Yu and Winter [12.8], and by Branson [12.9, 12.10]. It is known that the flexural rigidity Ec I varies with the magnitude of bending moment in the general manner shown in Fig. 12.9.2. Of course, the moment of inertia Icr of the transformed cracked section (see Section 12.5) increases roughly proportional with an increase in the percentage of reinforcement. Sections with higher percentages of reinforcement exhibit less change in rigidity under increasing load than those with low percentages of reinforcement. Duan, Wang, and Chen [12.11] reviewed and proposed an alternative expression for flexural rigidity. For loads below the cracking load (see Fig. 12.9.3), deflections may be based on the gross concrete section, with generally a small difference arising from whether the transformed area of reinforcement is also included. However, as the load increases above the cracking load, the moment of inertia approaches that of the cracked transformed section, although it may be greater between cracks. Generally, as shown by the typical load-​deflection curve in Fig. 12.9.3, the use of gross section underestimates the deflection, and the use of transformed cracked section overestimates the deflection. However, the degree of accuracy is affected by the magnitude of the service load compared to the load that causes cracking. Great accuracy in deflection computations is rarely justified in design because (a) it is difficult to estimate the actual concrete properties, in particular, its time-​dependent properties associated with creep and shrinkage; (b) deflection limits are rather arbitrary; and (c) except for dead loads, the magnitude and duration of actual loads are very difficult to predict with accuracy.

EcI based on gross section plus transformed area of reinforcement

EcI

EcI based on cracked transformed section 0.2 Mu

Mu

Figure 12.9.2  Typical variation of flexural rigidity with applied bending moment.

Service load

Load

Computed deflection based on transformed cracked section Computed deflections using gross I Actual deflection Cracking load

Deflection

Figure 12.9.3  Typical load-​deflection curve for reinforced concrete beams under service loads.

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ACI Effective Moment of Inertia Ie To obtain an estimate of the flexural rigidity of the member, the ACI Code uses the expression developed by Branson [12.9] given as ACI Formula (24.2.3.5a),



3   M 3  M cr  I g + 1 −  cr   I cr ≤ I g Ie =   M max    M max  

(12.9.1)

where Mcr = fr Ig /​yt = cracking moment [ACI Formula (24.2.3.5b)] Mmax = maximum service load moment acting at the condition under which deflection is computed Ig = moment of inertia of gross uncracked concrete section about the centroidal axis, neglecting reinforcement Icr = moment of inertia of transformed cracked section (see Section 12.5) fr = modulus of rupture of concrete (see Section 1.8), taken by ACI Code as 7.5λ fc′; in general, may be taken as 0.65 wc fc′, where wc is the unit weight of concrete in lb/cu ft [12.1] yt = distance from neutral axis to extreme fiber of concrete in tension Equation (12.9.1) provides a smooth transition between the moment of inertia Icr of the transformed cracked section and the moment of inertia Ig of the gross uncracked concrete section. The effective moment of inertia, Ie, is intended to be calculated at the location of maximum moment as a single value for the entire span in the case of simply supported beams, or as a single value between points of inflection in continuous beams. If one wishes to recognize the continuous variation of the moment of inertia along the span, Branson proposed using the fourth power instead of the third power in Eq. (12.9.1). In this case, Mcr and Mmax are the cracking moment and the applied moment, respectively, at each section along the span. The span may be broken into segments, with each segment having a different moment of inertia and numerical integration used to compute the deflection. Equation (12.9.1) was developed from a statistical study of 54 test specimens that had Mmax  /​Mcr values from 2.2 to about 4 and Ig /​Icr values from 1.3 to 3.5. The study included simple-​span rectangular beams [12.12], T-​beams [12.8], and two-​span continuous rectangular beams [12.13]. Branson has provided an excellent summary [12.10, 12.14, 12.15] of background material relating to Eq. (12.9.1). Lutz [12.16] has provided an ingenious set of diagrams to allow graphical evaluation of Eq. (12.9.1), and Shahrooz [12.17] has provided a simplified method for computing Ie. The recommendation of ACI Committee 435 to compute Ig neglecting reinforcement was made for practical simplicity rather than for any improvement in accuracy. Al-​Shaikh and Al-​Zaid [12.18] have studied the effect of the reinforcement ratio on Ie and suggested a modification to the power of 3 in Eq. (12.9.1) based on the reinforcement ratio ρ. The effective moment of inertia approach is also applicable to prestressed concrete (see Chapter  20); Branson and Trost [12.19] have presented unified procedures using the Ie method for partially cracked members, whether prestressed or nonprestressed.

Single Value of Effective Moment of Inertia for Practical Use As an approximation, a single value of effective moment of inertia is suggested for practical use. Various methods have been suggested depending on the support conditions and the expected cracking on the member along the span. 1. Midspan value alone. Recognized in ACI-​24.2.3.7 for prismatic beams and one-​way slabs, this assumption is

I e = I m (12.9.2)

where Im is the effective moment of inertia at midspan for prismatic simply supported and continuous spans, and at the support section for cantilevers. This is the simplest method, and

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when the inflection point is between about 0.2L and the support, the results are within ±20% of those obtained using variable I, as long as 0.33 ≤ α ≤ 1.0 [where α = (Im at midspan) /​(Ie at end)]. When α lies between 0.50 and 1.0, the results are within ±5% of those obtained using variable I. Zuraski, Salmon, and Shaikh [12.20] have suggested that this method is satisfactory for ordinary design situations, and it is endorsed by ACI Committee 435 [12.1]. 2. Simple average. For continuous members, ACI-​24.2.3.6 permits the moment of inertia to be taken as the average of the values for the critical positive and negative moment sections. This average is to be taken as



1 ( I e1 + I e 2 ) + I m average I e = 2 2

(12.9.3)

where Ie1 and Ie2 are the effective moments of inertia at the two ends of the span. The use of both Ie1 and Ie2 is appropriate only when there are end moments at both ends. For spans having one end continuous, use

average I e = 0.75I m + 0.25I e1 (12.9.4)

3. Weighted average. In this method, an adjusted I is obtained by weighting the moments of inertia in accordance with the magnitudes of the end moments [12.3]. The following weighted average expression has been recommended by ACI Committee 435 [12.1, 12.3] as giving a somewhat better result than use of the simple average. For spans with both ends continuous,

average I e = 0.70 I m + 0.15[ I e1 + I e 2 ]

(12.9.5)

For spans with one end continuous,

average I e = 0.85I m + 0.15I e1

(12.9.6)

For uniform loading on continuous spans, Eq. (12.9.5) is slightly more accurate (say, on the order of 5%) than using only the midspan value, but for concentrated loads it is less accurate [12.20]. It may be emphasized that when an average value is used as permitted by ACI-​24.2.3.6, this should be done in accordance with Eq. (12.9.3), rather than taking the sum of Im, Ie1, and Ie2 and dividing by 3 [12.3]. For a single heavy concentrated load, averaging reduces accuracy [12.20]; the midspan value from Eq. (12.9.2) should be used in such cases.

12.10 INSTANTANEOUS DEFLECTIONS IN DESIGN Throughout the history of reinforced concrete construction, computation of immediate or short-​term deflection has usually involved using either transformed cracked section or gross uncracked section. In either case, Eq. (12.7.1) is suitable after it has been slightly rewritten using Ec and Ie:

 ML2  ∆ = βa   Ec I e 

where βa = coefficient based on load and support conditions Ie = effective moment of inertia Ec = modulus of elasticity of concrete

(12.10.1)

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For immediate deflection and also generally for long-​term deflection under sustained loads, ACI-​24.2.3.4 permits the modulus of elasticity of concrete to be taken in accordance with ACI-​19.2.2 as

Ec = 33w1c,5 fc′

for concrete having a unit weight between 90 and 160 pcf. For normal-​weight concrete,

Ec = 57, 000 fc′

The generally accepted effective moment of inertia for use in Eq. (12.10.1) is Eq. (12.9.1), using a single value in accordance with Eqs. (12.9.2) through (12.9.6). To compute deflection at different load levels, such as dead load or dead load plus live load, the effective moment of inertia Ie should be computed by means of Eq. (12.9.1) for the total load level in each case. The incremental deflection, such as for live load only, is then computed as the difference between the deflections due to dead plus live load and dead load only. It should be assumed that the live load cannot act in the absence of dead load. Computation of the live load deflection as ∆D+L –​∆D gives the live load deflection occurring during the first application of live load. Figure  12.10.1 shows the typical idealized relationship between moment and deflection. For repeated loadings, the upper envelope is nearly the same as the single-​loading curve for both reinforced and prestressed concrete members, even though creep and cracking effects cause increasing residual deflections [12.2, 12.22]. Thus it seems reasonable to compute short-​term deflections using Ie as described above and the residual deflection separately, as discussed in Sections 12.13 and 12.14 on creep and shrinkage. Figure  12.10.2 compares the measured short-​term deflections with the deflections computed by Eq. (12.9.1) [12.6]. A study by ACI Committee 435 [12.4] indicates that by using the Code criteria for deflection for simply supported beams under controlled laboratory conditions, “there is approximately a 90% chance that the deflections of a particular beam will be within the range of 20% less-than to 30% more-than the calculated value.” Examples 12.10.1 and 12.10.2 demonstrate the computation of immediate deflections.

Figure 12.10.1  Typical idealized moment deflection diagram for short-​term loading. (Adapted from Refs. 12.2, 12.14, and 12.21.)

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20

%



Measured deflections

20

%

2.0”

1.0”

Branson (ACI–since 1971)

Computed deflections 0

1.0”

2.0”

Simply supported rectangular beams Simply supported T beams 2 span continuous rectangular beams

Figure 12.10.2  Comparison of computed and measured short-​term deflections (From Reference 12.6.).

EXAMPLE 12.10.1 Investigate the immediate, short-​term loading deflection for the simply supported beam of Fig. 12.10.3 over a span of 40 ft. Assume that the member has been designed to satisfy the strength requirements of the ACI Code using fc′ = 4000 psi (normal weight) and fy = 60,000 psi. SOLUTION According to ACI Table 9.3.1.1 the minimum required depth for the beam is computed as L /​16 = 40(12)/​16 = 30 in., which is greater than the beam depth of 24 in. Therefore, the deflection limits of ACI Table 24.2.2 must be satisfied regardless of whether excessive deflection is of concern. (a) Dead load short-​term deflection. The gross moment of inertia is Ig = M max

1 (18)(24)3 = 20, 700 in.4 12

0.45(40)2 = = 90 ft-kips 8



Using a modulus of elasticity ratio n = 8 (see Table 12.3.1), the neutral axis position for the transformed cracked section shown in Fig. 12.10.3(b) is 18c 2 = 57.3(20.7 − c) 2

c 2 + 6.37c = 131.8 c = 8.73 in.



1 I cr = (18)(8.73)3 + 57.3(20.7 − 8.73)2 = 12, 200 in.4 3 (Continued)

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Example 12.10.1 (Continued) The effective moment of inertia Ie is dependent on the bending moment Mcr that causes cracking at the extreme tension fiber. For normal-​weight concrete, the modulus of concrete is computed as (ACI-​19.2.3)

fr = 7.5λ fc′ = 7.5(1.0) 4000 = 474 psi



The cracking moment is computed from ACI Formula (24.2.3.5b)

M cr =

fr I g yt

=

0.474(20, 700)  1   12  = 68 ft-kips 12

Note that yt is the distance h/​2 for the 24-​in.-​deep beam when the gross section is used and the reinforcement is neglected. 3



 M  M cr 68 = (dead load only) = 0.756;  cr  = 0.431 M max 90  M max 

From ACI Formula (24.2.3.5a), Eq. (12.9.1), the effective moment of inertia is 3   M 3  M  I e =  cr  I g + 1 −  cr   I cr  M max    M max  



= 0.431(20, 7000) + 0.569(12, 200) = 15, 860 in.4 Ec = 33wc1.5 fc′ = 33(145)1.5 4000 = 3.64 × 106 psi (∆ i )D =



5wL4 5(0.45)(40)4 (12)3 = = 0.45 in. 384 EI 384(3.64)(103 )15, 860

c = 8.73”

Figure 12.10.3  Beam for Example 12.10.1

(Continued)

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Example 12.10.1 (Continued) This deflection may not be harmful because it may be accommodated by creating an upward deflection or camber during construction. Even if camber is not used, such dead load instantaneous deflection will not affect plastered ceilings or other items that are put into place after the immediate dead load deflection has taken place. Concern is primarily with the instantaneous deflection from live load and the long-​term creep and shrinkage deflection from sustained loads. (b) Dead load plus live load short-​term deflection. The maximum service load moment at this load level is M max =

25(40) + 90 = 340 ft-kips 4

(for dead load plus live load)

M cr 68 = (dead load + live load) = 0.20; M max 340

3

 M cr   M  = 0.008 max

I e = (0.008)(20, 700) + (0.992)(12, 200) = 12, 270 in.4 At this higher load level, Ie is only slightly larger than Icr. Using Ie = Icr = 12,200 in.4,



( ∆ i )beam weight =

5(0.45)(40)4(12)3 = 0.58 in. 384(3640)(12, 200)

( ∆ i )conc load =

25(40)3(12)3 = 1.30 in. 48(3640)(12, 200)

( ∆ i ) D + L = 0.58 + 1.30 = 1.88 in. (c) Live load short-​term deflection. Consistent logic dictates that live load deflection be obtained indirectly as

( ∆ i )L = ( ∆ i )D + L − ( ∆ i )D

= 1.88 − 0.45 = 1.43 in.

It is assumed that live load cannot act in the absence of dead load. Thus if the effective moment of inertia when dead load alone is acting is considerably different from that when dead load plus live load is acting, the live load deflection is properly obtained only by subtracting ΔD from ΔD+L. Even if the service live load has been preceded by a construction load of equal magnitude, the procedure is proper for repeated loadings, as discussed in the literature [12.14, 12.19, 12.21]. From a practical viewpoint, Ie = Icr may be used whenever (Mcr /​Mmax)3 is less than about 0.1. Assuming this to be a floor beam that is not supporting partitions, the acceptable immediate live load deflection from ACI Table 24.2.2 is

allowable (∆ i )L =

L 40(12) = = 1.33 in. < 1.43 in. 360 360

NG

If plastered ceilings or frangible partitions are to be supported, the long-​term deflection due to creep and shrinkage must be added to that due to live loads; the accept­ able limit for such deflection is L /​480 (ACI Table 24.2.2). One might expect that the deflection of this beam would be excessive since the reinforcement ratio ρ is 0.0192. This amount, corresponds to 1.06ρtc, which greatly exceeds the guideline value of about 0.6ρtc, or 0.375ρb, suggested in Chapter  3 as the maximum reinforcement ratio for deflection control.

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EXAMPLE 12.10.2 Compute the immediate deflection of the continuous beam of Fig.  12.10.4. Use fc′ = 3000 psi (normal weight), fy = 60,000 psi, and the ACI Code. SOLUTION This continuous girder supports two smaller beams that frame to it and some uniform loading that comes directly to it. ACI-​24.2.3.6 indicates that Ie may be computed as an average value for the critical positive and negative moment regions. For prismatic members, the effective moment of inertia Ie at the positive moment region may be used instead of the average value. In fact, for loading that is largely concentrated load, as in the situation for this example, ACI Committee 435 recommends [12.1, 12.2] against using an average value. The results will be compared using the various assumptions for the single value of Ie, as discussed earlier (see Section 12.9). (a) Section at the left support. For the gross section, neglecting reinforcement (see Fig. 12.10.5), c=

36(18)18 + 90(4.5)38.25 27,160 = = 25.79 in. 36(18) + 90(4.5) 1053

1 1 I g = (18) (25.79)3 + (14.71)3  + (72)(4.5)3 + (72)(4.5)(12.46)2 3 12

= 173, 000 in.4



If the reinforcement is included, c would have been 25.29 in. and Ig 200,000 in.4 It is generally accepted that Ig for use in ACI Formula (24.2.3.5a) should not include steel reinforcement. For the transformed cracked section [see Fig. 12.10.6(a)],  c 18c   + 40.6(c − 2.6) = 40.9(36.7 − c)  2 c 2 + 9.05c = 178.2 n. c = 9.56 in





1 I cr = (18)(9.56)3 + 40.6(9.56 − 2.6)2 + 40.9(36.7 − 9.56)2 3 = 37, 200 in.4

It is noted that a modulus of elasticity ratio n = 9 (see Table 12.3.1) is used to transform the area of steel into an equivalent area of concrete. Compute the cracking moment for the beam with tension on the flange. fr = 7.5λ fc′ = 7.5(1.0) 3000 = 411 psi M cr =

fr I g yt

=

0.411(173, 000)  1   12  = 402 ft-kips 14.71

M cr 402 (dead load) >1; = M max 89.6

use I e = I g = 173, 000 in.4

M cr 402 = (dead load +live load) >1;; M max 200

use I e = I g = 173, 000 in.4



(Continued)

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Example 12.10.2 (Continued) (b) Midspan section [Fig. 12.10.6(b)]. Determine whether the neutral axis is located in the flange by taking moments about the bottom of the flange, 90(4.5)(2.25) < 86.6(32.3) Thus, the neutral axis is in the stem. Locate the neutral axis, including the effect of compression in the stem.  1 90(4.5)(c − 2.25) + 18(c − 4.5)2   = 86.6(36.8 − c)  2

c 2 + 45.6c = 435



c = 8.09 in.

Figure 12.10.4  Data for Example 12.10.2.

(Continued)

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Example 12.10.2 (Continued) bE = 90”

4 12 ”

36” c = 25.79”

18”

Figure 12.10.5  Cross section of beam for Example 12.10.2.

Figure 12.10.6  Transformed cracked sections for Example 12.10.2.



1 1 (72)(4.5)3 + 72(4.5)(5.84)2 + (18)(8.09)3 + 86.6(36.8 − 8.09)2 12 3 = 86, 200 in.4

I cr =

The cracking moment for the beam with tension in the stem is M cr =

fr I g yt

=

0.411(173, 000)  1   12  = 229.5 ft-kips 25.79

M cr 229.5 = (dead load) >1; M max 229

use I e = I g = 173, 000 in.4 3



 M  M cr 229.5 = (dead load + live load) = 0.45;  cr  = 0.091 M max 510  M max 



3   M 3  M  I e =  cr  I g + 1 −  cr   I cr  M max    M max  

= 0.091(173, 0000) + 0.909(86, 200) = 94, 000 in.4 (Continued)

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Example 12.10.2 (Continued) (c) Section at the right support [Fig. 12.10.6(c)]. Use transformed cracked section and locate the neutral axis,  c 18c   + 40.6(c − 2.6) = 122.6(35.7 − c)  2 c 2 + 18.13c = 498





c = 15.02 in. 1 I cr = (18)(15.02)3 + 122.6(35.7 − 15.02)2 + 40.6(15.02 − 2.60)2 3 = 79, 000 in.4 M cr = 402 ft-kips (same as left support)

M cr 402 (dead load) >1; = M max 377

use I e = I g = 173, 000 in.4



3

 M  M cr 402 = (dead load + live load) = 0.68;  cr  = 0.314 M max 591  M max  I e = 0.314(1773, 000) + 0.686(79, 000) = 109, 000 in.4 (d) Summary of values for Ie. The values of effective moment of inertia are

Left end Midspan Right end

For DL

For DL + LL

Ie = 173,000 in.4 Ie = 173,000 in.4 Ie = 173,000 in.4

Ie = 173,000 in.4 Ie =  94,000 in.4 Ie = 109,000 in.4

Having obtained the above values, usual practice is to use a single adjusted value of Ie as discussed in Section 12.9. (e) Compute a single adjusted value for Ie using the various procedures of Section 12.9. 1. Midspan value:

I e = 173, 000 in.4 (for DL only) I e = 94, 000 in.4 (for DL + LL)



2. Simple average:



I e = 173, 000 in.4 (for DL only)  1  173, 000 + 109, 000  I e =   + 94, 000  2  2  = 118, 000 in.4 (for DL + LL)

(Continued)

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Example 12.10.2 (Continued) 3. Weighted average: I e = 0.70 I m + 0.15( I e1 + I e 2 ) I e = 173, 000 in.4 (for DL only)



I e = 0.70(94, 000) + 0.15(173, 000 + 109, 000)



= 108,100 in.4 (for DL + LL) The task of determining the immediate deflection is actually one of analyzing a continuous beam with variable moment of inertia. Because the most “exact” computations at best give deflections within probably ±20%, procedures more complex than those illustrated here are not justified. (f) Immediate dead load deflection. Ie for dead load only is 173,000 in.4 regardless of whether midspan, simple average, or weighted average is used. M a = 89.6 ft-kips



w = 0.76 kip/ft

M b = 377 ft-kips P = 24.48 kips

Note that the span is taken as that measured between the centerlines of supports, and the end moments are those computed for the same locations. Equally acceptable results are obtained by using the clear span and the face-​of-​support moments. Referring to Fig. 12.10.4, the total midspan deflection is ∆=

106 (3.64 P + 1.333wL − 0.164 M a − 0.164 M b ) Ec I e

Ec = 33wc1.5 fc′ = 57, 000 3000 = 3.15 × 106 psi

(∆ i ) D = =

1 [3.64(24.48) + 1.333(0.76)(39) − 0.164(89.6 + 377.0)] 3.15(173) 89.1 + 39.5 − 76.5 52.2 = = 0.10 inn. 545 545

As previously mentioned, the immediate dead load deflection will usually cause no difficulty, since most of it may be compensated for by the construction. However, it is used as a basis for determining the long-​term creep and shrinkage deflection, which is discussed in the next section. (g) Immediate live load deflection using the midspan value of Ie. The midspan value is 94,000 in.4 as computed in part (e). ( ∆ i ) D = 0.10 in. [see part (f)] M a = 200 ft-kips

M b = 591 ft-kips

(dead load + live load)

w = 0.91 kip/ft

P = 56.33 kips

(dead load + live load)

( ∆ i )D + L = =

1 [3.64(56.33) + 1.333(0.91)(39) − 0.164(200 + 591))] 3.15(94) 205.0 + 47.3 − 129.7 122.6 = = 0.41 in. 296 296

(∆ i )L = 0.41 − 0.10 = 0.31 in.

(Continued)

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Example 12.10.2 (Continued) (h) Immediate live load deflection using the simple average value of Ie. ( ∆ i ) D = 0.10 in.

[see part(f)]

simple average value of I e = 118, 000 in.4 [see part (e )]

 94, 000  ( ∆ i ) D + L = 0.41  = 0.33 in.  118, 000 



( ∆ i )L = ( ∆ i ) D + L − ( ∆ i ) D = 0.33 − 0.10 = 0.23 in. (i) Immediate live load deflection using the weighted average value of Ie. ( ∆ i ) D = 0.10 in.

[see part(f)]

weighted average value of I e = 108,100 in. [see part (e )] 4



 94, 000  ( ∆ i ) D + L = 0.41  = 0.36 in.  108,100  ( ∆ i )L = 0.36 − 0.10 = 0.26 in.

(j) Conclusion. The immediate live load deflection is computed to be 0.31, 0.23, and 0.26 in., respectively, depending on whether the midspan, simple average, or weighted average value of Ie is used for dead and live load deflection. The use of midspan Ie seems appropriate; for this example, one can conclude that the dead load deflection is about 0.1 in. and the live load deflection is about 0.3 in., using one significant figure. If this is a floor that does not support frangible partitions, the limiting permissible deflection is L /​360, or

L 39(12) = = 1.3 in. > 0.3 in (computed) 360 360

OK

Because the beam of this example is continuous and has different reinforcement amounts at the supports and at midspan, it is difficult to correlate the provided amounts of reinforcement with the calculated deflections. Furthermore, the beam behaves like a T-​beam in the positive moment region and as a rectangular cross section in the negative moment regions. It is noted, however, that Ie is dominated largely by the midspan value. Since the reinforcement ratio of about 0.2 ρtc or 0.14ρb for the positive moment region was below the guideline value suggested in Chapter 3 for deflection control; thus excessive deflection was not expected.

Recommended Values for Maximum Reinforcement Ratio ρ for Deflection Control For beams or one-​way slabs, ACI Committee 435 [12.1, 12.2] has recommended the following values of the maximum reinforcement ratio ρ to be used in the positive moment zone for deflection control: 1 For members of normal-​weight concrete not supporting or not attached to nonstructural elements likely to be damaged by large deflections, Rectangular cross section T-​beams or box cross section

0.35ρb 0.40ρb

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2. For members of normal-​weight concrete supporting or attached to nonstructural elements likely to be damaged by large deflections, Rectangular cross section T-​beams or box cross section

0.25ρb 0.30ρb

3. For members of lightweight concrete, use 0.05ρb less than that indicated in items 1 and 2. Additional guidance on deflection control may be found in Chapter 5 of the ACI Committee 435 report [12.1].

12.11 CREEP EFFECT ON DEFLECTIONS UNDER SUSTAINED LOAD The total long-​term deflection consists of the instantaneous elastic deflection plus the contributions from creep and shrinkage. Creep is inelastic deformation with time under sustained loads at unit stresses within the accepted elastic range (say, below 0.5 fc′ ), as shown in Fig. 12.11.1. This inelastic deformation increases at a decreasing rate during the time of loading. Factors that affect the magnitude of creep deformation [12.23, 12.24] are (1)  the constituents—​such as the composition and fineness of the cement, the admixtures, and the size, grading, and mineral content of the aggregates; (2) proportions, such as water content and water-​cement ratio; (3) curing temperature and humidity; (4) relative humidity during storage; (5)  size of the concrete member, particularly the thickness and the volume-​to-​ surface ratio; (6) age at loading; (7) duration of loading; and (8) magnitude of stress. Since, as seen from Fig. 12.11.1, the result of creep is an increase in strain with constant stress, one of the ways of accounting for it is by the use of a modified modulus of elasticity Ect. An alternative, and generally preferred, procedure is to apply a multiplier Ct to the short-term deflection ∆i. To understand qualitatively the effect of creep on beam deformation, consider the strain distributions before and after creep has occurred in the singly reinforced concrete cross section of Fig. 12.11.2. Over time, the strains in the concrete will increase due to creep and will cause the strain distribution to be as shown in Fig. 12.11.2(b). It is noted that the strain at the tension steel is essentially unchanged because the concrete carries little if any tension and because ordinary deformed steel reinforcement exhibits little creep. Since the neutral axis increases, three observations may be made: (1) the concrete stress reduces at the compression face (i.e., same compressive force acting and ccp exceeds ci); (2) the curvature of the cross section can increase significantly; and (3) the increase in compressive strain is much greater than the increase in curvature ϕ. This increase in curvature results in increased deflection over time.

Figure 12.11.1  Typical concrete stress-​strain curves for instantaneous and long-​term loading.

429



429

12.11  CREEP EFFECT ON DEFLECTIONS εi

εi

ci

α

ci

ccp

εcp

Creep effect (shaded)

ϕi

ϕi + ϕcp

As εs

εs

(a) Instantaneous strain due to initial loading

Little change due to creep

(b) Strain after creep has occurred

Figure 12.11.2  Creep effect on beam curvature.

For the purpose of computing deflections, it is frequently desirable to use a creep coefficient Ct defined as the ratio of creep strain to elastic strain, Ct =



ε cp εi

(12.11.1)

and the deflection resulting from creep is generally taken as ∆ cp = Ct ( ∆ i ) D



(12.11.2)

where (∆i)D is the instantaneous, immediate deflection due to all sustained loads. ACI Committee 209 has recommended [12.23] the hyperbolic-​type equation of Branson et al. [12.14, 12.23] for the creep coefficient, as follows:  t 0.60  Cu Ct =   10 + t 0.60 



(12.11.3)

where Ct = ratio of creep strain to elastic strain at any time t after a basic curing period t = time in days after loading Cu = ultimate creep coefficient; recommended average value is 2.35 for 40% humidity The general relationship of Ct /​Cu is shown in Fig. 1.11.2, Equation (12.11.3) applies to the standard condition of 40% ambient relative humidity, 4 in. (100 mm) or less slump, average thickness of member of 6 in. (150 mm), and loading age of 7 days for moist-​cured concrete or 1 to 3 days for steam-​cured concrete. For other conditions, the standard condition value is to be multiplied by the following correction factors (CF): 1. Age at loading. For moist-​cured concrete,

(CF )a = 1.25t a−0.118

(12.11.4a)

(CF )a = 1.13t a−0.095

(12.11.4b)

For steam-​cured concrete,

In Eqs. (12.11.4), ta is the age at loading, in days, after the initial period of curing. Several useful values of these equations appear in Table 12.11.1.

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TABLE 12.11.1  CREEP CORRECTION FACTOR (CF)a FOR AGE AT LOADING, EQS. (12.11.4) Correction Factor, (CF)a ta, Age in Days after Initial Curing Period

Moist Cured for 7 Days Initial Curing Period

10 20 30 60 90

Steam Cured for 1–​3 Days Initial Curing Period

0.95 0.87 0.83 0.77 0.74

0.90 0.85 0.82 0.76 0.74

2. Humidity. For H ≥ 40%, (CF )h = 1.27 − 0.0067 H



(12.11.5)

where H is the ambient relative humidity in percent. Values for this correction factor appear in Table 12.11.2.

TABLE 12.11.2  CREEP CORRECTION FACTOR (CF)h FOR HUMIDITY, EQ. (12.11.5) Ambient Relative Humidity, H (%) 40 or less 50 60 70 80 90 100

Correction Factor, (CF)h 1.00 0.94 0.87 0.80 0.73 0.67 0.60

3. Average thickness of member. Where the average thickness of the member in inches exceeds 6 in. (150  mm), a correction factor (reduction factor) may be applied. However, for most design purposes such a correction may be neglected. For members whose average thickness greatly exceeds 12 in. (300 mm), Meyers and Branson [12.25] provide a chart that may be used to correct for the effect of average thickness. 4. Other correction factors. Additional correction factors are available [12.14, 12.25] to account for slump greater than 4 in., cement content, percent of fine aggregate, and air content. However, these tend either to be small or to offset one another and may generally be neglected.

Compression Steel Effect on Creep The presence of compression steel decreases the deformation due to creep (and shrinkage as discussed in the next section). Evaluation of the effect of compression steel has been reported by Washa and Fluck [12.13], Yu and Winter [12.8], and Hollington [12.26], and a

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multiplier factor has been given by Branson [12.27] and recommended by Committee 435 [12.1, 12.2], as follows: kr =



0.85 1 + 50ρ′ (12.11.6)

where ρ′ is As′ / bd , the compression steel reinforcement ratio. Thus, the creep deflection ∆cp would become ∆ cp = kr Ct ( ∆ i ) D



(12.11.7)

instead of Eq. (12.11.2), when compression steel is present. Paulson, Nilson, and Hover [12.28] have provided an evaluation of Eq. (12.11.6) as it applies to high-​strength concrete beams; these authors also recommended modifications.

12.12 SHRINKAGE EFFECT ON DEFLECTIONS UNDER SUSTAINED LOAD Shrinkage of concrete in beams may have a similar effect on the deflection as creep. Shrinkage of an isolated plain concrete member would merely shorten it without causing curvature. When steel reinforcement is added, however, bond between concrete and steel restrains the shrinkage. Thus, a singly reinforced beam, having its shrinkage restrained at the reinforced face and unrestrained at the unreinforced face, may have considerable curvature. Generally, it is difficult to separate the effects of creep and shrinkage. Shrinkage is more pronounced than creep during the first few months. Typically, 90% of the shrinkage will have occurred by the end of 1 year, whereas 90% of the creep will not have occurred until the end of 5 years. A number of investigators have studied shrinkage effects separately from those of creep [12.14, 12.23–​12.25, 12.29–​12.31]. If the free shrinkage strain is known, shrinkage curvature ϕsh can be determined as a function of shrinkage strain. Such curvature will be dependent on the relative amounts of compression and tension steel just as creep is so affected. Finally, the shrinkage deflection will involve the geometry of the support system. Shrinkage deflection ∆sh may be expressed as [12.23] ∆ sh = α1ϕ sh L2



(12.12.1)

where α1 is a factor relating to the geometry of the support system and may be taken as the following:

α1 = 0.50



for cantilever beams

= 0.125

for simply supported beamss

= 0.086

for beams continuous at one end only for beams continuous at both ends

= 0.063

and L is the span length of the beam.

Shrinkage Strain, εsh ACI Committee 209 has recommended [12.23] that the following expressions by Branson et al. [12.14] be used for shrinkage strain εsh: For any time t after age 7 days for moist-​cured concrete,

ε sh =

t (ε sh )u 35 + t

(12.12.2a)

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For any time t after age 1 to 3 days for steam-​cured concrete,

ε sh =



t (ε sh )u 55 + t

(12.12.2b)

where εsh = shrinkage strain at any time t after initial curing t = time in days after initial curing (εsh)u = ultimate shrinkage strain; average value suggested is 800 × 10–​6 for 40% humidity Equation (12.12.2a) is shown graphically in Fig.  1.11.4. For conditions other than 40% ambient relative humidity, the standard condition value of Eqs. (12.12.2) is to be multiplied by the following correction factor (CF):

(CF )h = 1.40 − 0.010 H ,

40 ≤ H ≤ 80%

(12.12.3)

(CF )h = 3.00 − 0.030 H ,

H ≥ 80%

(12.12.4)

where H is the relative humidity in percent. Values for (CF)h appear in Table 12.12.1.

TABLE 12.12.1  SHRINKAGE CORRECTION FACTOR (CF)h FOR HUMIDITY, EQS. (12.12.3) AND (12.12.4) Ambient Relative Humidity, H (%)

Correction Factor, (CF)h

40 or less 50 60 70 80 90 100

1.00 0.90 0.80 0.70 0.60 0.30 0

Other correction factors may normally be neglected. Should such factors be desired, corrections for average thickness other than 6 in., slump greater than 4 in., cement content, percentage of fines, and air content are available [12.14, pp. 45–​47].

Shrinkage Curvature, φsh Several investigators [12.9, 12.23, 12.31] have developed expressions for curvature due to warping that arises from nonuniform shrinkage. Reinforcement of different amounts in the two faces of a beam is the principal cause of shrinkage warping. Miller [12.31] established the following relationship for singly reinforced beams. Referring to Fig. 12.12.1, by straight-​line proportion,



ϕ sh =

ε sh − ε s ε sh = d d

 εs   1 − ε  sh

(12.12.5)

where εs is the compressive strain induced in the steel from shrinkage; εsh is the free shrinkage strain at the unreinforced face. Miller empirically established values for εs /​εsh as a function of the percentage of reinforcement ρ.

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433

εsh

h

εs

d

ϕsh

Figure 12.12.1  Shrinkage strain related to beam curvature for a singly reinforced beam. (After Miller [12.31].)

Branson [12.9] modified Miller’s equation and empirically extended the results to give equations including the effects of compression steel.



ϕ sh = 0.7 ϕ sh =

ε sh  ρ − ρ′  (ρ − ρ′ )1 3  h  ρ 

12

ε sh h

for (ρ − ρ′ ) ≤ 3%

(12.12.6)

for (ρ − ρ′ ) > 3%

(12.12.7)

Note that ρ and ρ′ are in percent, 100(As or As′ )/​(bd). Equations (12.12.6) and (12.12.7) are recommended [12.2, 12.14] as the most appropriate relationships.

Geometry of Warping from Shrinkage The well-​known moment area theorems for beam deflections may be used to establish the factor α1 in Eq. (12.12.1) for the four typical cases. Since the quantity M/​(EI) is in fact the curvature due to bending moment, the φsh diagrams in Fig. 12.12.2 may be regarded as the equivalent M/​(EI) diagrams. In the derivations below, it is assumed that beam sections are singly reinforced and that the same reinforcement is used for positive and negative bending. For the cantilever beam [Fig. 12.12.2(a)],

∆ sh = BB ′ = moment of ϕ sh diagram between A and B about B  L = (ϕ sh L )   = 0.50ϕ sh L2  2 For the simply supported beam [Fig. 12.12.2(b)],



θ A = area of ϕ sh diagram between A and C  L = ϕ sh    2 ∆ sh = CC ′ = CC1 − C1C ′  L = θ A   − (moment of ϕ sh diagram between A and C about C)  2  L  L  L  L 2 = ϕ sh     − ϕ sh     = 0.125 ϕ sh L  2  2  2  4

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Figure 12.12.2  Geometry of warping due to shrinkage.

For the beam fixed at one end only [Fig. 12.12.2(c)],



BB′ = moment of ϕ sh diagram between A and B about B  x12  L − x1   + ϕ =0 = −ϕ sh ( L − x1 )  x1 + sh   2   2 

from which x1 =



2L 2

For the tangent at C′ to be horizontal, the distances AD and DC must be equal; thus  2 AD = DC = L − x1 =  1 − L 2   ∆ sh = CC ′ = moment of ϕ sh diagram between A and C about C = moment of a couple = ϕ sh ( L − x1 )2 2



1  1   2  = ϕ sh L2  1 − 2 +  = ϕ sh L2  1 −  2   2 = 0.086ϕ sh L2



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For the beam fixed at both ends [Fig. 12.12.2(d)], if the slope is to be horizontal at midspan for symmetry,

x1 =

L 4

Then ∆ sh = CC ′ = moment of ϕ sh diagram between A and C about C

 L = moment of a couple =ϕ sh    4 2 = 0.063ϕ sh L

2



Compression Steel Effect on Combined Shrinkage and Creep Whenever shrinkage is to be included in combination with creep in a deflection computation, a multiplier similar to that of Eq. (12.11.6) may be applied to the short-​term deflection [12.23]. The following multiplier kr has been recommended by ACI Committee 435 [12.1, 12.2] as most appropriate to account for the compression steel effect on the long-​term sustained load deflection ∆cp+sh,

kr =

1 1 + 50ρ′

(12.12.8)

where ρ′ = As′ / bd. Branson [12.27] and Shaikh [12.32] have discussed the use of this expression to account for the compression steel effect. Equation (12.12.8) has been used in the ACI Code since 1983 as a part of the combined multiplier λ∆ = kr ξ, which combines the compression steel effect kr and the time-​dependent effect ξ.

12.13 CREEP AND SHRINKAGE DEFLECTION—​A CI CODE METHOD In the ACI Code method, the creep and shrinkage deflection due to sustained load is obtained by multiplying the immediate, short-​ term deflection by a factor λ∆ (ACI-​24.2.4.1.1). Thus, where

∆ cp + sh = kr ξ( ∆ i ) D = λ ∆ ( ∆ i ) D

λ ∆ = kr ξ =

ξ 1 + 50ρ′

(12.13.1) (12.13.2)

and (∆i)D is the instantaneous deflection due to all sustained loads (usually dead load). The value of ξ is permitted by ACI-​24.2.4.1.3 to be taken in accordance with the duration of the sustained load as follows: 5 years or more 1 year 6 months 3 months

2.0 1.4 1.2 1.0

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EXAMPLE 12.13.1 For the beam of Example 12.10.1 (see Fig. 12.10.3), determine the creep and shrinkage deflection according to the ACI Code. Assume that only the dead load is sustained. SOLUTION First it is necessary to compute the immediate, short-​term deflection due to all sustained loads, in this case the dead load. From Example 12.10.1, part (a) of the solution,



( ∆ i ) D = 0.45 in.

Since no compression steel is used, Eq. (12.13.2) gives for 5 years or more load duration,

λ∆ =



ξ = 2.0 1 + 50ρ′

Then from Eq. (12.13.1),

∆ cp + sh = λ ∆ ( ∆ i ) D = 2.0(0.45) = 0.90 in.



If part of the live load were considered as sustained (as, e.g., in certain types of equipment whose placement or installation is not expected to change for a period of 5 years or more), it would be necessary to compute ∆i for the dead load plus the sustained live load. Under the ACI Code, an additional effective moment of inertia Ie would be computed using Mcr /​Mmax, where Mmax is due to dead load plus sustained live load.

12.14 CREEP AND SHRINKAGE DEFLECTION—​ ALTERNATIVE PROCEDURES Separate Creep and Shrinkage Multiplier Procedure A procedure for computing the deflections due to creep and shrinkage separately was recommended by ACI Committee 435 [12.1, 12.2], based on the work of Branson [12.9], as modified by improvement in the prediction of creep and shrinkage [12.23]. Thus

∆ cp + sh = ∆ cp + ∆ sh

(12.14.1)

where, using an equation given earlier

∆ sh = α1ϕ sh L2

[12.12.1]

∆ cp = kr Ct ( ∆ i ) D

(12.14.2)

and

For evaluation of Eqs. (12.12.1) and (12.14.2),

α1 = constant [see (Eq. 12.12.1)] φsh = shrinkage curvature [Eqs. (12.12.6) and (12.12.7)] L = span length kr = compression steel factor [Eq. (12.11.6)] Ct = creep coefficient, using Eq. (12.11.3) with correction factors of Eqs. (12.11.4) and (12.11.5), or per Ref. 12.23, “For average conditions, ultimate Ct = 1.6 may be used.” (∆i)D = immediate deflection due to all sustained loads.

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Combined Creep and Shrinkage Multiplier Procedure This method is similar to the ACI Code method, except the time-​dependent factor ξ can be more accurately evaluated [12.14, 12.23, 12.26]. It may be stated as ∆ cp + sh = kr ξ( ∆ i ) D



(12.14.3)

where kr =1/ (1+ 50ρ′ )(same as ACI )

(12.14.4)

ξ =  time-​ dependent coefficient (creep plus shrinkage), which may be taken from Table 12.14.1

TABLE 12.14.1  TIME-​DEPENDENT COEFFICIENT ξ INCLUDING BOTH CREEP AND SHRINKAGE EFFECTS, FOR BOTH NORMAL-​WEIGHT AND LIGHTWEIGHT CONCRETE MEMBERS OF COMMON TYPES, SIZES, AND a COMPOSITION (FROM BRANSON [12.14], P. 278). Average Relative Humidity and Age (days) When Loaded 100% Concrete Strength ≤7d fc′ at 28 Days

70%

50%

14d

≥28d

≤7d

14d

≥ 28d

≤7d

14d

≥ 28d

2500–​4000 psi (17–​28 MPa)

2.0

1.5

1.0

3.0

2.0

1.5

4.0

3.0

2.0

>4000 psi (28 MPa)

1.5

1.0

0.7

2.5

1.8

1.2

3.5

2.5

1.5

It is suggested that the following percentages of the values in the table be used for sustained loads that are maintained for the periods indicated: a

25% for 1 month or less 50% for 3 months 75% for 1 year 100% for 5 years or more The 50% values may normally be used for average relative humidities lower than 50%, which might be the case in heated buildings, for example.

EXAMPLE 12.14.1 For the beam of Example 12.10.1 (see Fig. 12.10.3), determine the ultimate (i.e., 5-year duration of load) creep and shrinkage deflection using the alternative methods discussed in Section 12.14. Assume that only the dead load is sustained, the ambient relative humidity is 70%, and the age at loading is 20 days after the initial moist-​curing period. SOLUTION It is noted that ACI-​24.2.4.1.1 permits computation of long-​term deflection by a “more comprehensive analysis,” which could include either of these alternative methods. (a) Separate creep and shrinkage multiplier procedure.

∆ cp + sh = ∆ cp + ∆ sh

(Continued)

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Example 12.14.1 (Continued) Using Eq. (12.14.2), ∆ cp = kr Ct ( ∆ i ) D

where, kr =

0.85 = 0.85 1 + 50ρ′

for ρ′ = 0

and from Eq. (12.11.3), for t = 5(365) days,  t 0.60  Ct =  Cu = 0.90Cu  10 + t 0.60  which for Cu = 2.35 as recommended for average conditions gives the basic value of Ct as Ct = 2.12 Adjusting for 70% humidity, (CF)h = 0.80 from Eq. (12.11.5) or Table 12.11.2, and for 20-​day age of loading after initial moist-​curing period, (CF)a = 0.88 from Eq. (12.11.4a). Thus the adjusted Ct is Ct = 2.12(0.80)(0.88) = 1.49 From Example 12.10.1 using the ACI Code method, ( ∆ i ) D = 0.45 in. Then

∆ cp = kr Ct ( ∆ i ) D = 0.85(1.49)0.45 = 0.57 in.

For shrinkage, from Eq. (12.12.1), ∆ sh = α1ϕ sh L2 where α1 = 0.125 for simply supported beams. For this beam, since ρ = 1.92% and ρ′ = 0, Eq. (12.12.6) gives ε  ϕ sh = 0.7  sh  3 ρ  h  Using ε sh from Eq. (12.12.2a),

ε sh =

t (ε sh )u 35 + t

which for t = 5(365) days is, for average conditions,

ε sh ≈ ( ε sh )u = 800 × 10 −6 Adjusting for 70% humidity, (CF)h = 0.70, from Eq. (12.12.3) or Table  12.12.1, the adjusted εsh is

ε sh = (800 × 10 −6 )0.70 = 560 × 10 −6 Then,  560 × 10 −6  3 ϕ sh = 0.7  1.92 = 20.3 × 10 −6 rad/in.  24   ∆ sh = α1ϕ sh L2 = 0.125(20.3 × 10 −6 )(480)2 = 0.58 in. ∆ cp + ∆ sh = 0.57 + 0.58 = 1.15 in. (Continued)

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Example 12.14.1 (Continued) (b) Combined creep and shrinkage multiplier procedure. Using Eq. (12.14.3), ∆ cp + sh = kr ξ( ∆ i ) D kr =



1 = 1.0 1 + 50ρ′



ξ = value from Table 12.14.1 ≈ 1.88 Note that age at loading is after initial curing period.

( ∆ i ) D = 0.45 in. ∆ cp + sh

(from Example 12.10.1) = 1.0(1.8)0.45 = 0.81 in.

A comparison of computation methods for creep and shrinkage deflection may be obtained from the following summary. Method

∆sh+cp

1. ACI, using λ∆ = ξ/​(1 + 50ρ′) = 2.0

0.90 in. (Example 12.13.1)

2. Separate creep and shrinkage per Eqs. (12.12.1), (12.14.1) and (12.14.2).

1.15 in. [Example 12.14.1, part (a)]

3. Combined creep and shrinkage using Eq. (12.14.3) with krξ = 1.8

0.81 in. [Example 12.14.1, part (b)]

12.15 ACI MINIMUM DEPTH OF FLEXURAL MEMBERS The minimum depths specified in ACI-​9.3.1.1 for beams and in ACI-​7.3.1.1 for one-​way slabs must be satisfied unless computation of deflection indicates that a lesser thickness can be used without adverse effects. The minimum depth (thickness) values apply to members where large deflection is not likely to damage partitions, ceilings, or other frangible attachments. When large deflection may cause such damage, calculated deflections must satisfy the limits given in ACI-​Table 24.2.2, regardless of whether the minimum thickness requirement is satisfied. The minimum depths prescribed by any such table are arbitrary and not always conservative. For this reason, deflections should be computed even when the minimum depth requirements of ACI-​9.3.1.1 and ACI-​7.3.1.1 are satisfied. The logic behind the limitation on span-​depth ratio given in the ACI Code as an attempt to control deflection is explained in the following. The deflection at midspan of a simply supported beam is

∆=

5wL4 384 EI

(12.15.1)

where w is the service uniformly distributed load. The maximum bending moment is

M=

wL2 fI = 8 y

(12.15.2)

where f is the service load stress and y is the distance from the neutral axis to the extreme fiber, where f is computed.

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Substituting Eq. (12.15.2) in Eq. (12.15.1) gives

∆=



5L2 f 48 E y

(12.15.3)

Assuming a cracked section under service load conditions, fs f f = = c y n(d − c) c



(12.15.4)

where n is the modular ratio and d is the effective depth. Assuming that the tension steel is stressed at, say, fs = 24,000 psi and that (d − c) ≈ 0.6h. 40, 000 Ec E f 24, 000 ≈ = = c y ( Es /Ec )(0.6h) hEs 725h



(12.15.5)

using Es = 29 × 106 psi. Substituting Eq. (12.15.5) in Eq. (12.15.3) and noting that E = Ec, ∆ 5  1 L =   L 48  725  h min h =



1  L  L 6960  ∆ 

(12.15.6)

Equation (12.15.6) represents an approximate relationship between depth, span, and span-​to-​deflection ratio for a fully stressed section under short-​term loading. If the member is under a reduced stress, the depth may be decreased proportionally to give the same short-​term deflection. To account for the sustained load creep and shrinkage deflection, the depth must be increased. Table 12.15.1 shows the evaluation of Eq. (12.15.6) to give the minimum depth required for various deflection limitations under fully and partially stressed conditions. The last two columns in Table 12.15.1 assume that total deflection including creep and shrinkage effects is twice the immediate deflection. Although the assumed stress under service loads in the tension steel is fs = 24,000 psi, in practice, the tension steel stress under service loads is usually higher; in such a case, the limits shown in Table 12.5.1 would be nonconservative. Regardless of the assumed values, any selection of limiting values for minimum depth from Table  12.15.1 can be considered only a crude attempt to control deflection. Table 12.15.2 shows the minimum depth requirements for beams and one-​way slabs given in ACI Table 9.3.1.1 and 7.3.1.1, respectively. These values correspond to a compromise between the relative conservative recommendations of ACI Committee 435 and the values that practicing engineers believe to be suitable on the basis of experience. When large deflections may cause cracking of partitions and other frangible attachments, the total deflection that occurs after installation of such elements is limited to L /​480 TABLE 12.15.1  MINIMUM DEPTH h FOR VARIOUS EQUIVALENT IMMEDIATE DEFLECTIONS AND PERCENTAGE STRESSEDa Percent Stressed 100 67 60 50 a

∆ = L /​300

∆ = L /​360

L /​23 L /​35 L /​39 L /​46

Assumed fs = 24,000 psi at 100% stressed.

L /​19 L /​29 L /​32 L /​39

∆ = L /​480 L /​15 L /​22 L /​24 L /​29

2∆ = L /​300 L /​12 L /​17 L /​19 L /​23

2∆ = L /​360 L /​9.7 L /​15 L /​16 L /​19

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TABLE 12.15.2  MINIMUM DEPTH h FOR BEAMS AND ONE-​WAY SLABS, FOR MEMBERS NOT SUPPORTING OR ATTACHED TO PARTITIONS OR OTHER CONSTRUCTION LIKELY TO BE DAMAGED BY LARGE a DEFLECTIONS Type of Member

Simple Support

One End Continuous

Both Ends Continuous

Cantilever

Beams (ACI Table 9.3.1.1)

fy = 60 ksi fy = 40 ksi

L /​16 L /​20

L /​18.5 L /​23

L /​21 L /​26

L /​8 L /​10

One-​way slabs (solid) (ACI Table 7.3.1.1)

fy = 60 ksi fy = 40 ksi

L /​20 L /​25

L /​24 L /​30

L /​28 L /​35

L /​10 L /​12.5

For structural lightweight concrete having weights wc from 90 to 115 pcf, multiply table values by 1.65 –​0.005wc but not less than 1.09. (60 ksi = 420 MPa; 40 ksi = 280 MPa, approximately.) a

(ACI-​24.2.2). This shows that the minimum depths of ACI Table 9.3.1.1 and 7.3.1.1 are likely to be too small; hence those tables do not apply for such cases. In general, minimum depth as a proportion of span is an inadequate criterion for controlling deflection; computation of deflection should be made whenever deflection is of concern.

12.16 SPAN-​T O-​D EPTH RATIO TO ACCOUNT FOR CRACKING AND SUSTAINED LOAD EFFECTS The general development, presented here, which is similar to that of Branson [12.33], illustrates the effects of the many variables on the span-​to-​depth ratio. Grossman [12.34] has also provided a procedure for determining the minimum thickness that would approximately satisfy any deflection limitation given by ACI Table 24.2.2. The procedure proposed by Grossman uses an approximated effective moment of inertia and eliminates the need to compute the moment of inertia of the cracked section, Icr. The development presented in this section, which perhaps is less practical than the method suggested by Grossman, is intended to illustrate how the variables interrelate and to show why it is impossible to have a simple table of minimum thicknesses. Along with Branson’s discussion [12.35] and Grossman’s closure [12.34], the work by Grossman [12.34] provides further insight and a useful discourse on the subject of deflection control. Rangan [12.36] has also presented a minimum thickness approach similar to that presented in the following. The short-​term deflection ∆i may be expressed, according to Eq. (12.10.1), as  M L2  ∆ i = β a  max   Ec I e 



(12.16.1)

where M max = maximum moment at the stage for which deflection is caalculated I e = Eq.(12.9.1)[which is ACI code Formula (24.2.3.5a)] = ( M cr /M max )3 I g + 1 − ( M cr /M max )3  I cr ≤ I g M cr = cracking moment = fr I g /yt

fr = modulus of rupture = 0.65 wc fc′ (ACI uses 7.5λ fc′) y t = distance from neutral axis to extreme fiber in tenssion



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Multiplying Eq. (12.16.1) by fr Ig /​(yt Mcr), which is equal to unity, gives  M L2  fr I g ∆ i = β a  max   Ec I e  yt M cr



(12.16.2)

Solving Eq. (12.16.2) for L /​yt gives L ∆ i  Ec   M cr  I e = yt L  fr β a   M max  I g



(12.16.3)

Since both Ec and fr are proportional to  fc′ , let

βw =



33w1c .5 fc′ Ec = = 50.8wc fr 0.65wc0.5 fc′

(12.16.4)

For normal-​weight concrete,

β w = 50.8wc = 50.8(145) = 7370



Conversion from normal-​weight concrete to lightweight concrete may be made by multiplying L /​yt by the ratio of the unit weight of lightweight concrete to 145 pcf. Letting γ = Ie /​Ig and βw = Ec /​fr in Eq. (12.16.3), L ∆ i  β w   M cr  = γ yt L  β a   M max 



(12.16.5)

where



γ=

3 3 I e  M cr    M cr   I cr   = + 1 − ≤1 I g  M max    M max   I g  

(12.16.6)

The ratio of the moment of inertia of the cracked section to that of the gross section may be computed for various shapes of beams using, for example, the transformed section method (see Section 12.5). As an example, for a singly reinforced rectangular beam, assuming d = 0.9h, I cr bc 3/ 3 + nAs (d − c)2 = Ig bh3 /12 2  ( c / d )3 c   = 8.75  + nρ  1 −    d    3



(12.16.7)

where c /d = (ρn)2 + 2ρn − ρn



(12.16.8)

Charts are given by Lutz [12.16] to obtain Ig and Icr for T-​sections. For dead load deflection, Mmax = MD, which makes γ = γD; Eq. (12.16.5) then becomes (∆ i )D βa  M D   L   1  = β w  M cr   yt   γ D  L

(12.16.9)

(∆ i )D + L βa  M D + L   L   1  = β w  M cr   yt   γ D + L  L

(12.16.10)

For dead load plus live load,

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As discussed in Section 12.10, since live load cannot act in the absence of dead load, the live load deflection must be obtained indirectly, ( ∆ i )L ( ∆ i )D + L ( ∆ i )D = − L L L



(12.16.11)

which, using Eqs. (12.16.9) and (12.16.10), and letting CL = ML /​MD, gives

 ( ∆ i )L β a  L   M D + L   1 1 = −   L β w  yt   M cr   γ D + L γ D (1 + CL ) 

(12.16.12)

When excessive deflection may cause damage to partitions and other nonstructural construction, it is the sum of deflections due to live load plus creep and shrinkage that is of concern. The instantaneous (short-​term) dead load deflection will have occurred when forms are removed and before any breakable attachments are put in place. Thus using Eq. (12.13.1), and the ACI time-​dependent multiplier λ∆, ∆ cp + sh = λ ∆ ( ∆ i )D



Finally, the deflection to be controlled to minimize possible damage is ( ∆ i )L ∆ cp + sh β a  L   M D + L   1 (λ ∆ − 1)  + = +       L L β w  yt   M cr   γ D + L γ D (1 + CL )    

=

β a  L   M D + L  1 + CL + (λ ∆ − 1)( γ D + L / γ D )  1   β w  yt   M cr   1+ + CL  γ D+L

(12.16.13)

Solving for L /​yt gives  L  CL + 1  ∆  β   M   =    w   cr    γ D+L  y    L  β a   M D + L   CL + 1 + (λ ∆ − 1)( γ D + L / γ D )  t limit for L + cp + sh    

(12.16.14)

For instantaneous dead load plus live load, Eq. (12.16.10) gives  L  y  t



limit for D+L

 ∆  β   M  =    w   cr  γ D + L  L   βa   M D + L 

(12.16.15)

As shown below, if the span-​to-​depth ratio limit is available for short-​term deflection under dead load plus live load, such as from Table 12.15.1, the effect of creep and shrinkage can be obtained by the use of a multiplier. Comparing Eqs. (12.16.14) and (12.16.15), in which ∆ /​L may be taken as a stated limit,



 L  y  t

L + cp + sh

 L =   yt  D + L

  C L +1   + + − C 1 ( λ 1 )( γ / γ ) ∆ D+L D   L

(12.16.16)

For the situation in which the Ie under dead load only is approximately the same as Ie under dead plus live load, γD+L ≈ γD, Eq. (12.16.16) becomes



 L  y  t

L + cp + sh

 L  CL + 1  =   yt  D + L  CL + λ ∆ 

(12.16.17)

Charts are available [12.33] for the L /​h ratio for short-​term effects of dead load plus live load, Eq. (12.16.10). The chart for singly reinforced rectangular beams ( yt = 0.5h) is given in Fig. 12.16.1.

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EXAMPLE 12.16.1 Determine the depth of beam required for the loading conditions of Example 12.10.1 (Figure 12.10.3) if the sum of the immediate live load plus creep and shrinkage deflection must not exceed L /​480. Assume only the dead load is sustained with a sustained load factor λ∆ = 2 as given by ACI-​24.2.4.1. SOLUTION (a) Use ACI Table 9.3.1.1, or Table 12.15.2. L = 16 h This considers average conditions and includes some effect of sustained load deflection. min h =

480 = 30 in. 16

130 120

Boundary conditions Simple – use curve values directly Cantilever – multiply curve values by 0.417 One end continuous (hinged–fixed) – multiply curve values by 2.41 Both ends continuous (fixed–fixed) – multiply curve values by 3.33

110 100 h 90 Rectangular beams singly reinforced

80 70 L h

60 50 40 30 20 10 0

0

Curve No.

fc’ (psi)

ρ (%)

1 2 3 4 5 6 7 8 9 10 11 12

3000 4000 5000 3000 4000 5000 3000 4000 5000 3000 4000 5000

3.0 3.0 3.0 2.0 2.0 2.0 1.0 1.0 1.0 0.5 0.5 0.5 1

1 2 3 4 65 78 9 10 11 12

Mmax

2

3

Mcr

Figure 12.16.1  Curves of L /​h versus Mmax /​Mcr for the conditions Δ = L /​360, normal-​weight concrete and uniformly distributed short-​term loading, and for different boundary conditions, steel percentages, and concrete strengths. (From Branson [12.33].)

(Continued)

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1 2 . 1 6  S P A N -​T O -​D E P T H R A T I O

Example 12.16.1 (Continued) (b) Use more accurate procedure with Fig. 12.16.1.

ρ = 0.0192 From Example 12.10.1, part (a) M cr = and

fr I g yt

= 68 ft-kips

M max 250 + 90 = = 5 > 3; use 3 for this example. M cr 68

Compute CL M 250 = 2.78 CL = L = M 90 D    



From Fig. 12.16.1 for Mmax /​Mcr = 3, fc′ = 4000 psi, and ρ = 0.0192, find ∆ 1  L = 23, for =   h D+L L 360  L  360  = 23  = 17.3,    480  h D+L

for

1 ∆ = L 480

 C +1   L  2.78 + 1  = 17.3  = 17.3  L = 13.7      2.78 + 2  h L + cp + sh  CL + λ ∆  min h =

480 = 35 in. 13.7

which is more severe than the minimum depth required by ACI Table 9.3.1.1. (c) Required depth and adequacy of beam in Example 12.10.1. For the given beam with h = 24 in.,

( ∆ i )L = 1.43 in. (Example 12.10.1) ∆ cp + sh = 0.90 in. (Example 12.13.1)



where



allowable ∆ =

L 480 = = 1 in. 480 480



calculated ∆ = 1.43 + 0.90 = 2.33 in. which is clearly inadequate, as was expected based on the results obtained in parts (a) and (b). Increasing the beam depth will increase the moment of inertia, but it will also add weight to the beam, which is sustained and will offset some of the reduction in deflection that would result due to an increase in depth. Because an important part of the total deflection is due to the sustained load, a more effective strategy is to add some compression steel to reduce the long-term deflection in addition to increasing beam depth.

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12.17 ACI CODE DEFLECTION PROVISIONS—​B EAM EXAMPLES The ACI Code provisions (ACI-​24.2) regarding deflection computations under service loads may be summarized as follows: 1. For members not supporting or not attached to elements likely to be damaged by large deflections that satisfy the minimum member depth requirements (ACI Table 9.3.1.1 for beams or ACI Table 7.3.1.1 for one-​way slabs) are “… considered to satisfy the requirements of the Code…” (ACI-​R24.2). For members that do not meet the mini­ mum depth requirements, both the immediate and the time-​dependent deflections must be computed and must satisfy the limits of ACI Table 24.2.2. 2. For members supporting or attached to elements likely to be damaged by large deflections, deflections must be computed and must satisfy the limits of ACI Table 24.2.2 regardless of whether they meet the minimum depth requirements. As noted in Section 12.15, the use of a minimum depth as a proportion of span is an inadequate criterion for controlling deflections. The use of a minimum depth (ACI Table 9.3.1.1 for beams and 7.3.1.1 for one-​way slabs) does not ensure that excessive deflections will not occur. Therefore, computation of deflection should always be made whenever deflection is of concern (as is often the case in practice), even for members under Category 1 above.

EXAMPLE 12.17.1 Investigate the deflection for the beam of Fig. 12.17.1 used on a simple span of 25 ft. The maximum bending moments under service load are 158 ft-​kips dead load and 105 ft-​kips live load. Assume that all loading is uniformly distributed and that none of the live load is sustained. The beam supports partitions and other construction likely to be damaged by large deflections. Use fc′ = 4000 psi (normal weight), fy = 60,000 psi, and the ACI Code. b = 14”

c = 8.31” 21.5”

h = 25” 13.19” 2 – #8

3 – #9

Figure 12.17.1  Beam cross section for Example 12.17.1.

SOLUTION (a) Because the beam supports elements likely to be damaged by large deflections, the deflection limits of ACI Table 24.2.2 must be satisfied even if the minimum depth requirement of ACI Table 9.3.1.1 is satisfied. Nonetheless, it is instructive to compute the minimum required depth from ACI Table 9.3.1.1 (or from Table 12.15.2 above for fy = 60 ksi) and compare it with the provided beam depth. Thus,

min h =

L 25(12) = = 19 in.< 25 in. provided 16 16

(Continued)

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447

Example 12.17.1 (Continued) Since the minimum depth requirement of ACI Table 9.3.1.1 is satisfied, one might expect that deflections for a member not supporting partitions likely to be damaged would be acceptable. (b) Examine reinforcement ratio ρ

ρ=



As 4.58 = = 0.015 bd 14(21.5)

This exceeds the 0.25ρb = 0.007 limit recommended by ACI Committee 435 [12.1] for members supporting nonstructural elements likely to be damaged by large deflections (see Section 12.10). From this check, one might expect deflection to be a problem. (c) Determine the moment of inertia Ig for the gross uncracked section without steel and Icr for the cracked transformed section. For the gross uncracked section,

Ig =

1 (14)(25)3 = 18, 200 in.4 12

For the cracked section, locate the neutral axis under service loads. Using a modular ratio n = 8,



14c 2 = 4.58(8)(21.5 − c) 2 c = 8.31 in.



1 I cr = (14)(8.31)3 + 4.58(8)(13.19)2 = 9050 in.4 3 (d) Determine the effective moment of inertia Ie (ACI Formula 24.2.3.5a). The cracking moment is

M cr =

fr I g yt

=

7.5 4000 (18, 200) = 57.6 ft-kips 12.5(12, 000)

For dead load deflection, M cr 57.6 = = 0.365; M max 158

3

 M cr   M  = 0.05 max

3   M 3  M  I e =  cr  I g + 1 −  cr   I cr  M max    M max  



= 0.05(18, 200) + 0.95(9050) = 9500 in.4 For dead load plus live load deflection,



M cr 57.6 = = 0.22 M max 263

3

 M cr   M  = 0.01 max

I e ≈ I cr = 9050 in.4 (e) Compute immediate deflections.

Ec = 57, 000 fc′ = 57, 000 4000 = 3.6 × 106 psi

(Continued)

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Example 12.17.1 (Continued) For dead load,

(∆ i )D =

5wL4 5 ML2 5(158)(12)(300)2 = = = 0.52 in. 384 EI 48 EI 48(3.6)(103 )(9500)

For dead load plus live load, (∆ i )D + L =



5(263)(12)(300)2 = 0.91 in. 48(3.6)(103 )(9050)

Then the immediate live load deflection is

( ∆ i )L = ( ∆ i ) D + L − ( ∆ i ) D = 0.91 − 0.52 = 0.39 in.

(f) Compute creep and shrinkage deflection. From ACI-​24.2.4.1, the multiplier is

λ ∆ = kr ξ =

2.0 = 2.0 1 + 50ρ′

for sustained load at 5 years or more. ∆ cp + sh = λ ∆ ( ∆ i ) D = 2.0(0.52) = 1.04 in. (g) Check limitation of ACI Table 24.2.2. For roof or floor construction supporting or attached to nonstructural elements likely to be damaged by large deflection, ( ∆ i )L + ∆ cp + sh ≤



L 480

This limit is for the deflection that is estimated to occur after the nonstructural elements have been put in place. Whatever portion, if any, of the live load or creep and shrinkage deflection that has occurred prior to the placement of nonstructural elements may be excluded from the L /​480 limitation. Further, if adequate measures are taken to prevent damage to supported or attached elements, the L /​480 limit may be exceeded (footnote, ACI Table 24.2.2). For this example,

300  L  [0.39 + 1.04 = 1.43 or 1.4 in.] >  = = 0.63 in.  480 480 

which is not acceptable.

EXAMPLE 12.17.2 Investigate the deflection for the one-​way continuous slab shown in Fig. 12.17.2. The slab is 5 in. thick. Assume that 60% of the 100 psf live load is sustained. Use fc′ = 3000 psi (normal weight) and fy = 60,000 psi. Assume the slab supports non-structural elements not likely to be damaged by large deflections. SOLUTION Equation (12.7.4) may be used to compute maximum deflection:

∆m =

5L2 48 EI

1    M s − 10 ( M a + M b ) [12.7.4]   (Continued)

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449

Example 12.17.2 (Continued) Center-​to-​center of supports will be used as the span L. However, when coefficients are used to compute moments as discussed in Chapter 7 using clear span Ln, the Ln should be used in computing deflections [12.2]. This mathematical model incorporates a conservative assumption of zero moment at the exterior support. Though Eq. (12.7.4) provides midspan deflection, maximum deflection occurs between the location of maximum moment and midspan. It is reasonable and practical to use maximum moment for Ms in that equation.

Ec = 57, 000 fc′ = 57, 000 3000 = 3.12 × 106 psi



(a) Determine Ig and Icr using a 1-​ft width of section. 1 (12)(5.0)3 = 125 in.4 12 For the positive moment region, d = 3.94 in. Using n = 9,



Ig =

12c 2 = 9(0.37)(3.94 − c) 2 c = 1.23 in.



1 I cr = (12)(1.23)3 + 9(0.37)(3.94 − 1.23)2 = 31.9 in.4 3 (b) Determine effective moment of inertia Ie (ACI-​24.2.3.5). Note that for prismatic one-​way slabs, ACI-​24.2.3.7 permits Ie to be computed at midspan for simple and continuous spans. fr = 7.5 fc′ = 7.5 3000 = 411 psi

M cr =

fr I g yt

=

0.411(125) = 1.71 ft-kips 2.50(12)

3   M 3  M  I e =  cr  I g + 1 −  cr   I cr  M max    M max  

Figure 12.17.2  End-​span details for continuous slab of Example 12.17.2.

(Continued)

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Example 12.17.2 (Continued) In the positive moment region, for dead load

M cr M cr 1.71 = = > 1; M D M max 0.83

I e = I g = 125 in.4

and for the dead load plus live load,



M cr M cr 1.71 = = = 0.8; M D + L M max 2.14

3

 M cr   M  = 0.51 max

I e = 0.51(125) + 0.49(31.9) = 79.4 in.4 (c) Immediate live load deflection. ( ∆ i )L = ( ∆ i )D + L − ( ∆ i )D (∆ i )D + L =

5(12)2 144 1.73    2.14 −  (12) = 0.21 in. 10  48(3.12)(103 )79.4 

5(12)2 144 0.67   (∆ i )D =  0.83 −  (12) = 0.05 in. 3 10  48(3.12)(10 )125



( ∆ i )L = 0.21 − 0.05 = 0.16 in. Check short-term deflection limit from ACI Table 24.2.2 for immediate live load deflection, L 12(12) allowable ∆ = = = 0.40 in. > 0.16 in. OK 360 360 (d) Consider the effect of sustained live load that contributes to the creep and shrinkage deflection. The immediate deflection due to all sustained loads is required as the base value on which to apply the time-​dependent multiplier. For the positive moment region, using sustained load moment of 0.83 + 0.6(1.31) = 1.62 ft-​kips/​ft, M cr 1.71 = > 1.0 M max 1.62



This result suggests that one could use Ie = Ig for computing the immediate deflection due to all sustained loads. However, because the presence at any point in time of the design live load will cause cracking in the positive moment region, it would be nonconservative to use Ie = Ig, as upon removal of the transient live load, immediate deflections would be those of a cracked slab. The actual stiffness of the cracked slab is unknown, but it seems reasonable to assume that it would be similar to that corresponding to Ie computed under dead load plus live load, i.e., Ie = 79.4 in4. (e) Creep and shrinkage deflection. The immediate deflection due to sustained  loads may be computed as,

( ∆ i ) D + sustL =

5(12)2 144 1.31   1.62 −  (12) = 0.16 in. 3 10  48(3.12)(10 )(79.4)

where the sustained moment at the support is 0.67 + 0.6(1.06) = 1.31 ft-​kips/​ft. Considering 5 years or more load duration, and with no compression steel, λ∆= 2.0

∆ cp + sh = λ ∆ ( ∆ i ) D + sustL = 2.0(0.16) = 0.32 in.

(Continued)

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Example 12.17.2 (Continued) (f) Check long-term deflection limit of L /​240 in ACI Table 24.2.2,

( ∆ i )*L + ∆ cp + sh ≤

L 240

where ( ∆ i )*L represents the immediate deflection due to any additional (not sustained) live load (see Section 12.6). At this stage, the slab is assumed to have been previously cracked by the presence of live load and thus ( ∆ i )*L will be computed using Ie = 79.4 in4 calculated in part (b). Thus,  0.4 L  ( ∆ i )*L = ( ∆ i ) D + L   D + L   0.4(0.10)  = 0.21   0.163 



= 0.05 inn. and





( ∆ i )*L + ∆ cp + sh = 0.05 + 0.32 = 0.37 in. allowable ∆ =

L 12(12) = = 0.60 in. > 0.37 in. 240 240

OK

The decision regarding whether part of the live load is sustained should be based on a consideration of the actual loading and its duration. For instance, a floor system supporting library stacks as the live load might well be considered to have a significant part of the live load treated as being sustained. In most situations it is acceptable to consider that only the dead load is sustained and thereby affects the magnitude of creep and shrinkage deflection. Although the deflection calculations have been illustrated throughout this chapter without permitting any shortcuts on steps in the formal procedure, deflection computations should be made with practicality in mind. The effective moment of inertia Ie theoretically should be computed at each total load level for which deflection is of concern—usually dead load, dead load plus live load, and dead load plus sustained live load. To obtain an average, as is encouraged by the ACI Code, Ie should be computed at each end and at the midspan region. However, when the true accuracy obtainable from a deflection computation is recognized, the designer should use the Ie equation only when the result will be significantly different from using either Icr or Ig. For most situations, furthermore, only the midspan Ie need be computed; an average does not measurably improve accuracy. The authors believe these practical suggestions satisfy the spirit of the ACI Code. For additional practical deflection examples based on the ACI Code and ACI Committee 435 recommendations, including composite beams and two-​way systems, see the chapter by Branson in Handbook of Concrete Engineering [12.2], the PCA Notes [12.37], and the CRSI Design Handbook [2.21]. Also, see Example  21.3.1 in Chapter  21 for a complete design example, including deflections, of a composite beam.

12.18 CRACK CONTROL FOR BEAMS AND ONE-​W AY SLABS Cracking in concrete is generally the result of the following actions [12.38–​12.40]: (1) volumetric change, including that due to drying shrinkage, creep under sustained load, thermal stresses, and chemical incompatibility of concrete components; (2) internal or external direct stress due to continuity, reversible load, long-​term deflection, camber in prestressed concrete, or differential movement in structures; and (3) flexural stress due to bending.

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Visible cracking is generally initiated by either internal microcracking (volumetric change would usually include this type) or flexural microcracks. Flexural microcracks are surface cracks that are not visible except by careful close investigation and are generally initiated by flexural stress. Once flexural microcracks have formed, a slight increase in load causes these cracks to open up suddenly to measurable widths. Under service loads, wide cracks could form, which might be detrimental to steel reinforcement under a corrosive environment. Factors such as humidity, salt air, and alternate wetting and drying or freezing and thawing may accelerate corrosion of the steel reinforcement and contribute to concrete deterioration in the vicinity of large-​width cracks. Increased cover provides thicker protection but may result in wider cracks at the beam face, influencing corrosion [12.41, 12.42]. Today, epoxy-​coated reinforcing bars are widely used to prevent and ameliorate corrosion of the steel reinforcement in concrete structures. However, even when corrosion is not of concern, wide cracks may be unsightly and contribute to doubt about structural safety. Although cracking cannot be eliminated, it is generally more desirable to have many fine hairline cracks than a few wide cracks. Thus, crack control is a matter of controlling the distribution and size of cracks rather than eliminating them. Excellent sources of information on control of cracking resulting from flexure and other causes are two ACI Committee 224 reports [12.38, 12.39]. To control cracking, it is better to use several smaller bars at moderate spacing than larger bars at large spacing. The objective is, therefore, one of distributing the reinforcement in the tension zone; hence ACI-​24.3, entitled “Distribution of flexural reinforcement …,” contains the crack control provisions for beams and one-​way slabs. Control of cracking is particularly important when reinforcement with a yield stress in excess of 40,000 psi is used, as in most cases, or when tension reinforcement ratios exceed about 0.375ρb. A good bar arrangement in the cross section will usually lead to adequate crack control. Entirely satisfactory structures have been built, particularly in Europe, using design yield stresses exceeding 80,000 psi, which is the current specified limit in ACI Table 2​ 0.2.2.4a for the main reinforcement in non-seismic-force-resisting systems. Many studies [12.43–​12.52] have verified the generally accepted belief that crack width is proportional to steel stress. Two other significant variables have been found to be the thickness of concrete cover and the area of concrete surrounding each individual reinforcing bar in the zone of maximum tension. Since even in a controlled laboratory environment crack widths are highly variable and difficult to predict accurately, they may be expected to vary widely within a given structural member. Rather than specifying limits on the maximum allowable crack width, the ACI Code requires a proper arrangement and spacing of the reinforcing bars that will usually lead to adequate crack control. Accordingly, the maximum permitted center-​to-​center spacing, s, of the nonprestressed reinforcement closest to the tension face is given in ACI Table 24.3.2 as:  40, 000  s = 15  − 2.5cc  fs 



1

(12.18.1)

but not greater than 12(40,000/​fs), where fs in psi is the stress in the reinforcement closest to the tension face at service loads computed based on the unfactored moment (ACI-​24.3.2.1). Recognizing that additional elastic analysis under service-level loads would be required to determine fs for use in Eq. (12.18.1), ACI-​24.3.2.1 permits taking fs as 2 3 of the specified yield stress fy. For Grade 60 reinforcement, this approach is generally conservative because the actual service load stress in the bar is usually less than ( 2 3 ) fy. The clear cover, cc, to be used in Eq. (12.18.1) is measured from the nearest surface in tension to the surface of the flexural tension reinforcement.

  For SI, ACI 318-​14M,

1



 280   280  s = 380  − 2.5cc but not greater than 300   fs   fs  where s and cc are in mm and fs is in MPa.

(12.18.1) 

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For two-​way slabs, the foregoing spacing limit formula does not apply. Other recommendations are given in the ACI Committee 224 Report [12.38]. Chapter 16 of this text also contains a brief treatment on the subject. When structures are subject to very aggressive exposure or designed to be watertight, the provisions of ACI-​24.3.2 are not sufficient. Special investigations and precautions are required for such structures. For guidance in the design of sanitary structures, the reader is referred to the work of ACI Committee 350 [12.51].

EXAMPLE 12.18.1 Check the crack control provisions of the ACI Code for the beam cross section of Example 12.4.1 (Fig. 12.4.1). The selected beam has b = 18 in., h = 34 in., 2–​#9 bars and 4–​#10 bars in one layer. Use #3 stirrups, 1.5-​in. clear cover at bottom. SOLUTION Compute cc,

cc =1.5 in. ( clear cover to bottom ) + 0.375 (stirrup ) =1.875 in.

From Example 12.4.1, part (b), the calculated stress in the tension steel under service loads is fs = 37,000 psi, which is somewhat less than ( 2 3 ) fy (i.e., fs = 40,000 psi, as permitted by ACI-​24.3.2.1.). The maximum, center-​ to-​ center spacing of the reinforcement, according to Eq. (12.18.1), is

 40, 000   40, 000  s = 15  − 2.5(1.875) = 11.5 in. − 2.5cc = 15    37, 000   fs 

but not greater than 12(40,000/​fs) = 12(40,000/​37,000) = 13 in. The section of Fig. 12.4.1 clearly meets the 11.5-​in. maximum spacing.

EXAMPLE 12.18.2 Investigate crack control at the maximum positive moment region for the beam of Example 6.21.2. SOLUTION As shown in Fig. 6.21.2, the beam is reinforced with one #7 and 2–​#9 bars in the positive moment region. Using

fs =

2 2 f y = 60, 000 = 40, 000 psi 3 3

as permitted by ACI-​24.3.2.1, and assuming a 1.5-​in. clear cover,

cc = 1.5(cover) + 0.375(stirrup) = 1.875 in.



then, according to Eq. (12.18.1)

 40, 000  s = 15  − 2.5(1.875) = 10.3 in  40, 000 

which is not greater than 12(40,000/​40,000) or 12 in.

(Continued)

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Example 12.18.2 (Continued) Using #3 stirrups and 1.5-​in. clear side cover, the center-​to-​center spacing between the bars is   sprovided =

14(beam width) − 2[1.5(cover ) + 0.375(stirrup)] 1.128(db# 9 ) − = 4.6 in. OK 2 2

which is less than the limit of 10.3 in. The above calculations are somewhat conservative. First, the calculation for sprovided does not account for the #3 bend radius, which would result in an even smaller bar spacing. Also, the actual service load stress in the bar is usually less than the 2 3 fy. A lesser value could be computed if needed to satisfy the crack control limitation.

EXAMPLE 12.18.3 Investigate the crack control requirements for the floor girders 2G1-​2G2-​2G2-​2G1 of Example 9.9.1. SOLUTION Using Eq. (12.18.1), the maximum allowed spacing of the reinforcement closest to the tension face is  40, 000  s = 15  − 2.5cc (12.18.1)  fs 



but not greater than 12(40,000/​fs). With reference to Fig. 9.9.1, the most critical location will occur for the reinforcement that is farthest from the concrete surface. In this case, this will occur for the negative moment reinforcement, where cc = 3.25 in. Using

fs =

2 2 f y = 60, 000 = 40, 000 psi 3 3

the maximum allowed spacing is

 40, 000  s = 15  − 2.5(3.25) = 6.9 in.  40, 000 

which is not greater than 12(40,000/​40,000): that is, 12 in. The provided center-​to-​center spacing between the bars at section A-​A (largest spacing between bars) is

sprovided =

18 − 2(1.5 + 0.375) 1.128(db# 9 ) − = 6.6 in. 2 2

OK

As noted before, this check is somewhat conservative because the bend radius of stirrups has not been included in the computation of sprovided and because the actual service load stress in the bar is usually less than ( 2 3 ) fy. For flanges of T-beams in tension, the ACI Code requires that part of the flexural tension reinforcement be distributed over the effective flange width bE, but not wider than Ln /10. If the effective flange width bE exceeds Ln /10, additional longitudinal reinforcement must be provided in the outer portions of the flange (ACI-24.3.4). This requirement (Continued)

45



12.19  SIDE FACE CRACK CONTROL FOR LARGE BEAMS

455

Example 12.18.3 (Continued) is intended to control the crack widths in the slab near the web and in the outer regions of the flange when the negative moment beam reinforcement is concentrated within or near the web. In this example, the flanges of girders are provided with the slab reinforcement which, if spaced in accordance to Eq. (12.18.1), should provide adequate control of cracking (see Fig. 9.9.3). In heavily reinforced beams, such as sections C-C, D-D, and F-F of Fig. 9.9.1, the top reinforcement should be provided in a single layer across the flange width.

12.19 SIDE FACE CRACK CONTROL FOR LARGE BEAMS On deep beams, the maximum crack width may occur along the side faces between the neutral axis and the main tension reinforcement [12.53–​12.56]. In such cases, the main reinforcement provides a restraining effect on the opening of cracks near the extreme tension face. Thus, the maximum width occurs somewhere between the neutral axis, where there should be no flexural cracking, and the main tension reinforcement, where the crack opening is restrained. Unacceptably wide side face cracks, sometimes three times as wide as the crack width at the level of the main tension reinforcement, have been observed [12.54, 12.55]. Braam [12.56] has also treated crack width control in deep beams. Therefore, ACI-​9.7.2.3 has provisions requiring side face (skin) reinforcement when the depth h of a beam or joist exceeds 36 in. Such skin reinforcement must be distributed along both side faces of the member for a distance h/​2 from the tension face (Fig. 12.19.1). The spacing s of the skin reinforcement is computed using ACI Table 24.3.2, Eq. (12.18.1) for deformed bars, where cc is the clear cover from the skin reinforcement to the side face. According to ACI-​R9.7.2.3, the area of the skin reinforcement is not specified because “research has indicated that the spacing rather than the bar size is of primary importance.” Bar sizes of #3 to #5 with a minimum area of 0.1 sq in. per foot of depth are typically provided. It is permitted to include the skin reinforcement in strength computations when a strain compatibility analysis is made.

Ask h > 36 in. s

h 2

As

Figure 12.19.1  Side face (skin) reinforcement for deep beams.

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EXAMPLE 12.19.1 Design the “skin” reinforcement according to the ACI Code for a rectangular beam 18 in. wide by 48 in. deep, having 10–​#9 as tension reinforcement (Fig.  12.19.2); fc′ = 4000 psi and fy = 60,000 psi. SOLUTION (a) Establish whether skin reinforcement is needed. Since the overall depth exceeds 36 in., longitudinal skin reinforcement is needed according to ACI-​9.7.2.3. (b) Determine the spacing to be used for the skin reinforcement. Try #3 bars and conservatively assume fs as 2 3 of the specified yield strength. Using a side cover cc of 2 in., the maximum spacing of the skin reinforcement is  40, 000   40, 000  s = 15  − 2.5cc = 15  − 2.5(2) = 10.0 in.   40, 000   fs 



but not greater than 12(40,000/​fs)  =  12(40,000/​40,000)  =  12 in. Considering that the center of the top layer of longitudinal reinforcement is located at about 4.2 in. from the bottom of the beam, use two layers of #3 bars spaced at 10 in. to satisfy the spacing and distribution requirements of ACI-​9.7.2.3. Details are shown in Fig. 12.19.2.

#4 Stirrup

48”

d = 44.9” 10”

10 – #9

#3 @ 10” 18”

Figure 12.19.2  Side face (skin) longitudinal reinforcement, for Example 12.19.1.

12.20 CONTROL OF FLOOR VIBRATIONS—​G ENERAL The tendency to use smaller, more efficient structural members has resulted in floor systems that are more flexible, and thus, susceptible to undesirable vibrations under the normal use of the structure. Excessive vibrations may, for example, impair proper functioning of sensitive equipment or of high-​precision manufacturing machines. Even when equipment malfunction is not of concern, floor vibrations may disrupt the tasks being performed and result in loss of productivity, or may simply cause discomfort to the occupants. Human response to floor vibrations is quite complex, and it depends on many factors (e.g., the source, type and magnitude of the vibration, the damping of the floor system, and the activities being performed by the occupants). To date, there is no unique criterion for establishing an acceptable floor vibration level. However, two key parameters in evaluating the serviceability of floor systems subject to vibrations are the peak acceleration and the vibration frequency of the motion. In office and

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residential construction, it has been found that peak accelerations for human comfort should not exceed about 0.5% of the acceleration of gravity, g [12.57, 12.58]. On the other hand, participants in some activities (e.g., weight lifting, dancing) can tolerate accelerations of about 5% of g [12.57]. The maximum accelerations that will develop on a floor system will depend on the source and nature of the motion but, in particular, on the relationship between the frequency of vibration of the source and the natural frequency of vibration of the floor system. For rhythmic excitation (e.g., aerobic dancing), peak accelerations can be largely amplified owing to the phenomenon referred to as resonance. Resonance occurs when the vibration frequency of the source is close to or coincides with the natural frequency of the structural member. In a floor system, the vibration frequency of the source, fsource, is dictated by the nature of the motion (e.g., dancing). On the other hand, the natural frequency of the floor system, fn, depends on the mass and stiffness of its members. Therefore, current design practice for controlling floor vibrations is generally based on “fine-​tuning” the stiffness of the floor to avoid resonance—​ that is, so that the natural frequency of the floor, fn, does not coincide with the frequency of the source, fsource. Increasing the amount of damping will help reduce floor vibrations. However, the practice of adding damping to a floor system has not always been successful [12.57], and it requires the installation of tuned dampers, which may be costly. The current ACI Code does not contain specific provisions for the control of vibration of reinforced concrete structures. However, a good summary of the background, acceptance criteria, and design examples can be found in Refs. 12.59 and 12.60.

SELECTED REFERENCES 12.1. ACI Committee 435. “Control of Deflections in Concrete Structures,” Report ACI 435R-​ 95 (Reapproved 2000; Appendix B added 2003). Farmington Hills, MI:  American Concrete Institute, 2003, 89 pp. 12.2. ACI Committee 435. “Proposed Revisions By Committee 435 to ACI Building Code and Commentary Provisions on Deflections” ACI Journal, Proceedings, 75, June 1978, 229–​238. 12.3. ACI Committee 435, Subcommittee 7.  “Deflections of Continuous Beams,” ACI Journal, Proceedings, 70, December 1973, 781–​787. 12.4. ACI Committee 435. “Variability of Deflections of Simply Supported Reinforced Concrete Beams,” ACI Journal, Proceedings, 69, January 1972, 29–​35. 12.5. ACI Committee 435, Subcommittee 1. “Allowable Deflections,” ACI Journal, Proceedings, 65, June 1968, 433–​444. Disc., 65, 1037–​1038. 12.6. ACI Committee 435. “Deflections of Reinforced Concrete Flexural Members,” ACI Journal, Proceedings, 63, June 1966, 637–​674. 12.7. Mark Fintel (Editor). Handbook of Concrete Engineering. New York: Van Nostrand Reinhold, 1985 (892 pp.) (see Chapter 2, “Deflections,” p. 53). 12.8. Wei-​Wen Yu and George Winter. “Instantaneous and Long-​Time Deflections of Reinforced Concrete Beams under Working Loads,” ACI Journal, Proceedings, 57, July 1960, 29–​50. Disc., 57 1165–​1171. 12.9. Dan E. Branson. “Instantaneous and Time-​Dependent Deflections of Simple and Continuous Reinforced Concrete Beams,” Part 1, Report No. 7. Alabama Highway Research Report, Bureau of Public Roads, August 1963 (1965) (pp. 1–​78). 12.10. Dan E.  Branson. Discussion of “Variability of Deflections of Simply Supported Reinforced Concrete Beams,” by ACI Committee 435, ACI Journal, Proceedings, 69, July 1972, 449–​451. 12.11. Lian Duan, Fu-​Ming Wang, and Wai-​Fah Chen. “Flexural Rigidity of Reinforced Concrete Members,” ACI Structural Journal, 86, July–​August 1989, 419–​427. Disc., 87, May–​June 1990, 364–​365. 12.12. G. W. Washa and P. G. Fluck. “Effect of Compressive Reinforcement on the Plastic Flow of Reinforced Concrete Beams,” ACI Journal, Proceedings, 49, October 1952, 89–​108. 12.13. G.  W. Washa and P.  G. Fluck. “Plastic Flow (Creep) of Reinforced Concrete Continuous Beams,” ACI Journal, Proceedings, 52, January 1956, 549–​561. 12.14. Dan E. Branson. Deformation of Concrete Structures. New York: McGraw-​Hill, 1977. 12.15. Dan E. Branson. Discussion of “Proposed Revision of ACI 318-​63 Building Code Requirements for Reinforced Concrete,” ACI Journal, Proceedings, 67, September 1970, 692–​693. 12.16. LeRoy A. Lutz. “Graphical Evaluation of the Effective Moment of Inertia for Deflection,” ACI Journal, Proceedings, 70, March 1973, 207–​213. Disc., 70, September 1973, 662–​663.

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12.17. Bahram M.  Shahrooz. “A Simplified Method for Computing Effective Moment of Inertia,” Concrete International, 14, January 1992, 38–​40. 12.18. Abdulrahman H.  Al-​Shaikh and Rajeh Z.  Al-​Zaid. “Effect of Reinforcement Ratio on the Effective Moment of Inertia of Reinforced Concrete Beams,” ACI Structural Journal, 90, March–​April 1993, 144–​149. 12.19. Dan E.  Branson and Heinrich Trost. “Unified Procedures for Predicting the Deflection and Centroidal Axis Location of Partially Cracked Nonprestressed and Prestressed Concrete Members,” ACI Journal, Proceedings, 79, March–​April 1982, 119–​130. 12.20. P. D. Zuraski, C. G. Salmon, and A. Fattah Shaikh. “Calculation of Instantaneous Deflections for Continuous Reinforced Concrete Beams,” Deflections of Concrete Structures (SP-​43). Detroit: American Concrete Institute, 1974 (pp. 315–​331). 12.21. N. H. Burns and C. P. Siess. “Repeated and Reverse Loading in Reinforced Concrete,” Journal of the Structural Division, ASCE, 92, ST5 (October 1966), 65–​78. 12.22. Koladi M.  Kripanarayanan and Dan E.  Branson. “Short-​Time Deflections of Beams Under Single and Repeated Load Cycles,” ACI Journal, Proceedings, 69, February 1972, 110–​117. 12.23. ACI Committee 209. “Prediction of Creep, Shrinkage, and Temperature Effects in Concrete Structures,” Designing for Effects of Creep, Shrinkage, and Temperature in Concrete Structures (SP-​27). Detroit: American Concrete Institute, 1971 (pp. 51–​93). 12.24. ACI Committee 209. “Effects of Concrete Constituents, Environment, and Stress on the Creep and Shrinkage of Concrete,” Designing for Effects of Creep, Shrinkage, and Temperature in Concrete Structures (SP-​27). Detroit: American Concrete Institute, 1971 (pp. 1–​42). 12.25. B. L. Meyers and D. E. Branson. “Design Aid for Predicting Creep and Shrinkage Properties of Concrete,” ACI Journal, Proceedings, 69, September 1972, 551–​555. 12.26. M. R. Hollington. A Series of Long-​Term Tests to Investigate the Deflection of a Representative Precast Concrete Floor Component, Technical Report TRA 442. London: Cement and Concrete Association, April 1970, 43 pp. 12.27. Dan E.  Branson. “Compression Steel Effect on Long-​ Time Deflection,” ACI Journal, Proceedings, 68, August 1971, 555–​559. 12.28. Kent A. Paulson, Arthur H. Nilson, and Kenneth C. Hover. “Long-​Term Deflection of High-​ Strength Concrete Beams,” ACI Materials Journal, 88, March–​April 1991, 197–​206. 12.29. T. C. Hansen and A. H. Mattock. “Influence of Size and Shape of Member on Shrinkage and Creep of Concrete,” ACI Journal, Proceedings, 63, February 1966, 267–​289. 12.30. Hans Gesund. “Shrinkage and Creep Influence on Deflections and Moments of Reinforced Concrete Beams,” ACI Journal, Proceedings, 59, May 1962, 689–​704. 12.31. Alfred L.  Miller. “Warping of Reinforced Concrete Due to Shrinkage,” ACI Journal, Proceedings, 54, May 1958, 939–​950. 12.32. A. F. Shaikh. Discussion of “Proposed Revision of ACI 318–​63 Building Code Requirements for Reinforced Concrete,” ACI Journal, Proceedings, 67, September 1970, 722–​723. 12.33. Dan E. Branson. “Design Procedures for Computing Deflections,” ACI Journal, Proceedings, 65, September 1968, 730–​742. 12.34. Jacob S. Grossman. “Simplified Computations for Effective Moment of Inertia Ie and Minimum Thickness to Avoid Deflection Computations,” ACI Journal, Proceedings, 78, November–​ December 1981, 423–​439. Disc., 79, September–​October 1982, 413–​419. 12.35. Dan E. Branson. Discussion of “Simplified Computations for Effective Moment of Inertia Ie and Minimum Thickness to Avoid Deflection Computations,” ACI Journal, Proceedings, 78, November–​December 1981, 423–​439, ACI Journal, Proceedings, 79, September–​October 1982, 413–​414. 12.36. B.  Vijaya Rangan. “Control of Beam Deflections by Allowable Span–​Depth Ratios,” ACI Journal, Proceedings, 79, September–​October 1982, 372–​377. 12.37. PCA. Notes on ACI 318–​05 Building Code Requirements for Structural Concrete and Commentary. Skokie, IL: Portland Cement Association, 2005. 12.38. ACI Committee 224. “Control of Cracking in Concrete Structures,” Report ACI 224R-​01. Farmington Hills, MI: American Concrete Institute, 2001, 49 pp. 12.39. ACI Committee 224. “Causes, Evaluation, and Repair of Cracks in Concrete Structures,” Report ACI 224.1R-​07. Farmington Hills, MI: American Concrete Institute, 2007, 26 pp. 12.40. Edward G.  Nawy. “Crack Control in Reinforced Concrete Structures,” ACI Journal, Proceedings, 65, October 1968, 825-​836. Disc. 66, 308–​311. 12.41. David Darwin, David G.  Manning, Eivind Hognestad, Andrew W.  Beeby, Paul F.  Rice, and Abdul Q. Ghowrwal. “Debate: Crack Width, Cover, and Corrosion,” Concrete International, 7, May 1985, 20–​35. 12.42. A. W. Beeby. “Cracking, Cover, and Corrosion of Reinforcement,” Concrete International, 5, February 1983, 35–​40.

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PROBLEMS

12.43. Bengt B. Broms and LeRoy A. Lutz. “Effects of Arrangement of Reinforcement on Crack Width and Spacing of Reinforced Concrete Members,” ACI Journal, Proceedings, 62, November 1965, 1395–​1410. Disc. 62, 1807–​1812. 12.44. Peter Gergely and LeRoy A.  Lutz. “Maximum Crack Width in Reinforced Concrete Flexural Members,” Causes, Mechanism, and Control of Cracking in Concrete (SP-​20). Detroit: American Concrete Institute, 1968 (pp. 87–​117). 12.45. A. W. Beeby. “The Prediction of Crack Widths in Hardened Concrete,” The Structural Engineer, 57A, 1, January 1979, 9–​17. 12.46. R.  J. Frosch. “Another Look at Cracking and Crack Control in Reinforced Concrete,” ACI Structural Journal, 96, May-​June 1999, 437–​442. 12.47. LeRoy A. Lutz, Nand K. Sharma, and Peter Gergely. “Increase in Crack Width in Reinforced Concrete Beams Under Sustained Loading,” ACI Journal, Proceedings, 64, September 1967, 538–​546. 12.48. A.  W. Beeby. “The Prediction and Control of Flexural Cracking in Reinforced Concrete Members,” Cracking, Deflection, and Ultimate Load of Concrete Slab Systems (SP-​30). Detroit: American Concrete Institute, 1971 (pp. 55–​75). 12.49. John P.  Lloyd, Hassen M.  Rejali, and Clyde E.  Kesler. “Crack Control in One-​Way Slabs Reinforced with Deformed Wire Fabric,” ACI Journal, Proceedings, 66, May 1969, 366–​376. 12.50. A. W. Beeby. “An Investigation of Cracking in Slabs Spanning One Way,” Technical Report No. TRA 433. London: Cement and Concrete Association, April 1970, 32 pp. 12.51. ACI Committee 350. “Concrete Sanitary Engineering Structures,” ACI Journal, Proceedings, 80, November–​December 1983, 467–​486. 12.52. LeRoy A. Lutz. “Crack Control Factor for Bundled Bars and for Bars of Different Sizes,” ACI Journal, Proceedings, 71, January 1974, 9–​10. 12.53. Gregory C.  Frantz and John E.  Breen. “Cracking on the Side Faces of Large Reinforced Concrete Beams,” ACI Journal, Proceedings, 77, September–​October 1980, 307–​313. 12.54. Gregory C.  Frantz and John E.  Breen. “Design Proposal for Side Face Crack Control Reinforcement for Large Reinforced Concrete Beams,” Concrete International, 2, October 1980, 29–​34. 12.55. Perry Adebar and Joost van Leeuwen. “Side-​Face Reinforcement for Flexural and Diagonal Cracking in Large Concrete Beams,” ACI Structural Journal, 96, September–​October 1999, 693–​704. 12.56. C.  R. Braam. “Control of Crack Width in Deep Reinforced Concrete Beams,” Heron, 35 (1990), 4 (published jointly by Delft University of Technology, Delft, The Netherlands, and TNO—​Institute for Building Materials and Structures, Rijswijk, The Netherlands). 12.57. David E. Allen. “Building Vibrations from Human Activities,” Concrete International, 12, June 1990, 66–​73. 12.58. “Evaluation of Human Exposure to Whole-​Body Vibration, Part 2: Vibration in Buildings (1 Hz to 80 Hz),” (ISO 2631-​2). International Standards Organization, Geneva, 2003. 12.59. David A. Fanella and Mike Mota. “Design Guide for Vibrations of Reinforced Concrete Floor Systems,” 1 ed. Schaumburg, IL: Concrete Reinforcing Steel Institute, 2014. 12.60. Applied Technology Council (ATC). ATC Design Guide 1—​Minimizing Floor Vibration. ATC, Redwood City, CA, 1999.

PROBLEMS All problems are to be done in accordance with the ACI Code unless otherwise indicated. 12.1 For the case assigned by the instructor, calculate the stress fs at the centroid of the steel and fc(max) at the extreme compression fiber of the concrete due to application of the given service load bending moment Mw.

(a) using internal equilibrium conditions (see Example 12.4.1) (b)  The flexure formula method with transformed section (see Example 12.5.1)

Case Es /​Ec

Mw d (in.) b (in.) Tension Bars (ft-​kips)

1 2 3 4

10 9 9 8

85 52 92 400

19.5 19.5 19.5 32.7

12 12 12 16

5 6

8 8

400 110

27.5 18.5

22 12

3–​#9 3–​#7 3–​#10 7–​#9 (2 layers; 5 and 2 bars) 7–​#10 2–​#8, 2–​#10

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C H A P T E R   1 2     S erviceability

12.2 Compute the immediate deflections due to dead load and live load on the beam of the figure for Problem 12.2. The span of the uniformly loaded, simply supported beam is 30 ft, and the maximum service load moments are 80 ft-​kips for dead load and 95 ft-​kips for live load. Use fc′ = 3500 psi (n = 8.5) and fy = 60,000 psi. 12”

12.4 Compute the immediate deflections due to dead load and live load on the beam of the figure for Problem 12.4. The uniformly loaded, simply supported beam on a span of 30 ft must resist maximum service load moments of 20 ft-​kips dead load and 14 ft-​kips live load. Use fc′ = 3000 psi (n = 9) and fy = 40,000 psi. 12.5 Repeat Problem 12.4, except the service load moments are 29 ft-​kips dead load and 19 ft-​kips live load, and fy = 60,000 psi. 10”

d = 19.5”

22”

3 – #10

13”

15”

Problems 12.2, 12.8, and 12.9 

2 – #8

12.3 Compute the immediate deflections due to dead load and live load on the beam of the figure for Problem 12.3. The uniformly loaded, simply supported beam on a span of 29 ft must resist maximum service load moments of 300 ft-​kips dead load and 500 ft-​ kips live load. Use fc′ = 4000 psi (n = 8) and fy = 40,000 psi. 18”

d = 36.25”

40”

2 – #9 8 – #11

Problems 12.3, 12.10, and 12.11  10”

Problems 12.4, 12.5, 12.12, and 12.13 

12.6 Investigate the acceptability of the beam of the figure for Problem 12.6 for immediate live load deflection if the limitation is L/​360. Use fc′ = 3500 psi (n = 8.5) and fy = 40,000 psi. 12.7 Investigate the acceptability of the beam of the figure for Problem 12.7 for immediate live load deflection if the limitation is L/​360. Use fc′ = 3000 psi (n = 9) and fy = 60,000 psi. 12.8 Investigate the acceptability of the beam of Problem 12.2 if the limit for live load plus creep and shrinkage deflection is L/​360. Consider that none of the live load is sustained. Use data computed in Problem 12.2 if that problem was previously assigned. 12.9 Repeat Problem 12.8, except use the separate creep and shrinkage multiplier procedure instead of the ACI Code method. Assume that humidity is 80%, age at loading is 28 days after initial curing period, and duration of sustained load is 5 years. B

10”

A w

6 – #7

d = 15.6” 5 – #9

A

19”

d = 15.9”

20’– 0” +

Section A–A

Problems 12.6 and 12.14 

Section B–B

M

42 ft-kips (DL) 75 ft-kips (LL)

B 38.4 ft-kips (DL) 0 ft-kips (LL)

Service load moments

461



461

PROBLEMS

A

B

C

w 7 – #9 each layer 40’–0”

630 ft-kips (DL) 750 ft-kips (LL)

A

382 ft-kips (DL) 309 ft-kips (LL)

B

+ Service load moments

45”

8 – #11 9 – #10 4 – #9

24” Section B–B

6 12 ”

d = 35.1” 40”

C

40” d = 36.25”

405 ft-kips (DL) 505 ft-kips – (LL)

24” Section A–A

6 12 ”

4 – #11 and 3 – #9 each layer

40” d = 36”

24” Section C–C

Problems 12.7 and 12.15 

12.10 Investigate the acceptability of the beam of Problem 12.3 if the beam supports partitions and other nonstructural construction likely to be damaged by large deflections. Consider that 20% of the live load is sustained. 12.11 Repeat Problem 12.10, except use the separate creep and shrinkage multiplier procedure instead of the ACI Code method. The deflection limit is the same as for ACI Code, however. Assume that humidity is 90%, age at loading is 60 days after initial curing period, and duration of sustained load is 5 years. 12.12 Investigate the acceptability of the beam of Problem 12.4 if the beam supports partitions, and so on, which limit the maximum deflection due to live load plus creep and shrinkage to L/​360. Assume that none of the live load is sustained, the relative humidity is 50%, age at loading is 20 days after initial curing, and sustained load will be in place for 1 year.

(a)  Use ACI Code method. (b) Use separate creep and shrinkage multiplier procedure. (c) Use combined creep and shrinkage multiplier procedure.

12.13 Investigate the acceptability of the beam of Problem 12.5 if the beam supports partitions,

and so on, which limit the maximum deflection due to live load plus creep and shrinkage to L/​480. Assume that none of the live load is sustained, the relative humidity is 90%, age at loading is 30  days after initial curing period, and duration of sustained load is 5  years or more.

(a)  Use ACI Code method. (b) Use separate creep and shrinkage multiplier procedure.

12.14 Investigate the acceptability of the beam of Problem 12.6 if the live load plus creep and shrinkage deflection is limited to L/​250 and 10% of the live load is considered sustained. Assume that relative humidity is 50%, age at loading is 10  days after initial curing period, and duration of sustained loading is 1 year.

(a)  Use ACI Code method. (b) Use separate creep and shrinkage multiplier procedure.

12.15 Investigate the acceptability of the beam of Problem 12.7 if the beam supports partitions and other nonstructural construction likely to be damaged by large deflections. Assume that 10% of the live load is sustained.

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C H A P T E R   1 2     S erviceability

12.16 Investigate the acceptability of an interior 45-​ft span (see the figure for Problem 12.16) of a continuous T-​section with regard to deflection. The service load moments at midspan are 40 ft-​kips dead load and 100 ft-​kips live load; at both supports these moments are 50 ft-​kips dead load and 114 ft-​kips live load. Assume that the L/​480 limit of ACI Table 24.2.2 applies and that none of the live load should be considered as sustained. Use fc′ = 3500 psi (n = 8.5) and fy = 60,000 psi.

27”

2 12 ”

6 – #8

18”

12.17 Check the ACI crack control provisions for the tension steel detail of the figure for Problem 12.17. Assume 1.5-​in. clear cover and that #4 stirrups are used. 12.18 Check the ACI crack control provisions for the tension steel detail of the figure for Problem 12.18. Assume 1.5-​in. clear cover and that #4 stirrups are used.

27”

4”

4” 2 12 ”

3 – #8

5 – #8

2”

12”

12”

End sections Regions of negative moment

Center section Regions of positive moment

2 12 ”

Problem 12.16  2 12 ”

2 12 ”

10 – #11 4 – #11 16”

Problem 12.17 

18”

Problem 12.18 

40”

30”

d = 27.3”

d = 36.1”

6 – #7

3 – #11

CHAPTER 13 SLENDERNESS EFFECTS ON COLUMNS

13.1 GENERAL In the basic treatment of compression members in Chapter  10, it was assumed that the effects of buckling and lateral deflection on strength were small enough to be neglected. Short compression members—​that is, those having a low slenderness ratio L  /​r (L = column height and r = radius of gyration = I / A )—​experience a material failure (crushing of concrete) prior to reaching a buckling mode of failure. Further, the lateral deflections of short compression members subjected to bending moments are small, thus contributing little secondary bending moment PΔ as shown in Fig. 13.1.1. It is these buckling and deflection effects that reduce the strength of a compression member below the value computed according to the principles of Chapter 10. Over time, the use of higher-​strength steel and concrete has led to the design of more slender members. A stocky member having an L /​r less than 20 will essentially exhibit a material failure whose strength may be computed by using the procedures discussed in Chapter 10. In contrast, a member having L /​r greater than about 70 will have a considerable reduction in strength, not only due to the increased likelihood of buckling but because of an amplification of the bending moment due to second-​order effects. To permit the greatest flexibility in structural design, specifications must provide for adequate determination of strength with any slenderness ratio. Thus the provisions of the ACI Code take into account the slenderness effects on long compression members. To compute the radius of gyration, r, ACI-​6.2.5.1(a) permits the designer to use r=



Ig Ag

where Ig and Ag are the gross moment of inertia and cross-​sectional area of the column, respectively. Alternatively, r may be computed for a rectangular column having b as the width and h as the depth as [ACI-​6.2.5.1(b)]



r=

Ig Ag

=

1 3 bh 12 = 0.288h ≈ 0.30h bh

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C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

Long columns in Denver. (Photo by C. G. Salmon.) W P

P

L Mi = primary bending moment

Figure 13.1.1  Primary and secondary moment for beam-​columns.

and for a circular column having h as the diameter as [ACI-​6.2.5.1(c)]



r=

Ig Ag

=

π h 4 ( 4) = 0.25h 64 π h 2

These values actually should be slightly larger owing to the effect of the reinforcement. In the ACI Code, the evaluation of the effect of slenderness may be approximated by using the moment magnifier method, whereby the sum of the primary and secondary moments (Fig. 13.1.1) is treated as being equal to the product of the primary moment and a magnification factor δ. The general idea relating to this approach is derivable from the differential equation of a beam-​column. In the next several sections, the general concepts relating to the effect of slenderness on the strength of compression members are presented.

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1 3 . 2   B U C K L I N G O F C O N C E N T R I C A L LY L OA D E D C O L U M N S

465

13.2 BUCKLING OF CONCENTRICALLY LOADED COLUMNS Over 250 years ago Leonhard Euler derived the well-​known Euler formula [13.1] for concentrically loaded columns stressed below the proportional limit. Engesser [13.2] in 1889 proposed the tangent modulus modification of the Euler formula. In 1910, von Kármán [13.3] performed a series of careful tests verifying Engesser’s assumptions. The tangent modulus formula has now been accepted as representing the lower bound for buckling strength of concentrically loaded columns, Pc =



π 2 Et I (kLu )2

(13.2.1)

where Pc Et I kLu

= buckling load = tangent modulus of elasticity of concrete at the buckling load = effective moment of inertia of the section = equivalent pin-end length (Lu = actual unbraced length)

Although von Kármán was the first to use a rational analytical method for the inelastic buckling of slender columns, his work did not consider reinforced concrete. The fact that concentrically loaded columns rarely, if ever, exist in reinforced concrete structures led investigators [13.4–​13.22] to focus attention on the interaction of long columns with beams in frame structures, resulting in the more rational provisions for slenderness effects on compression members beginning with the 1971 ACI Code. The basic design limitation of a maximum axial strength equal to 80 or 85% of the concentric capacity Po (see Fig. 10.11.1) means, of course, that from a practical viewpoint the concentrically loaded column is considered not to exist. However, to help the reader understand the effect of the slenderness ratio on the behavior of beam-​columns over the entire range from Pn = Po with Mn = 0 to Pn = 0 with Mn = Mo (see Fig. 10.6.1), the limiting case of the concentrically loaded column is considered first. To apply Eq. (13.2.1), a realistic expression for Et of concrete must be used. Since buckling may occur at practically any value of concrete strain, it is necessary to know as accurately as possible the stresses at all strain levels. The idealized stress-​strain diagram for steel was shown in Chapter 3 (Fig. 3.4.1), where the modulus of elasticity is taken at 29,000,000 psi. A realistic stress-​strain diagram for the concrete in the compression zone is that of Hognestad [3.1], shown in Fig. 13.2.1, in which the initial modulus of elasticity for concrete is taken as

Ec = 1, 800, 000 + 500 fc′′ psi

(13.2.2)

where fc′′= 0.85 fc′. Such a stress-​strain relationship may be used along with three assumptions: concrete resists no tensile stress, linear strain variation exists across the section, and the deflected shape is part of a sine wave. On the basis of these assumptions, evaluation of Eq. (13.2.1) gives the typical column strength curves for concentric loading, such as those of Fig. 13.2.2. A study of Fig. 13.2.2 shows that in the curves for fy = 40,000 psi there occurs a flat leveling off (such as portion BC of Fig. 13.2.2) of the curve, indicating yielding of the steel with a sudden drop in Es from 29 × 106 psi to zero. In such cases where εy < ε0, the concrete may still increase the capacity (such as portion AB in Fig. 13.2.2) with an increased strain up to ε0. For fy = 50,000 psi and fc′ = 4000 psi, the strains εy = 0.00173 and ε0 = 0.00194 are nearly equal, which means that little increase above the plateau of yielding in the steel can occur. The effect of creep on long-​time loading may be noted by the cross-​hatched region.

46

466

C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

[ ( )]

ε ε fc = fc′′ 2 εc – ε c 0 0

2

Stress fc

0.15fc′′

fc′′ = 0.85 fc’ Ec ε0 = 2fc′′/Ec

εcu = 0.0038

Strain εc

Figure 13.2.1  Hognestad’s stress-​strain diagram for flexure [3.1].

EXAMPLE 13.2.1 Calculate the ordinate and abscissa of points A, B, and C of Fig. 13.2.2. Use fc′ = 4000 psi, fy = 40,000 psi, and a steel area As = 0.02bh (0.01bh in each opposite face located at 0.45h from the center). Use the stress-​strain diagrams of Figs. 13.2.1 and 3.4.1. SOLUTION The basic quantities of the concrete stress-​strain diagram are computed first: fc′′ = 0.85 fc′ = 0.85(4) = 3.4 ksi Ec = 1800 + 500 fc′′ = 1800 + 500(3.4) = 3500 ksi

ε0 = εy =

2 fc′′  3.4  = 2 = 0.00194  3500  Ec fy Es

=



40 = 0.00138 29, 000

(a) Point A, maximum strength of section according to the principles of Chapter  10 (upper limit, εc = ε0 > εy),

Pn = 0.85 fc′ bh + As f y



without correcting for the displaced concrete. Pn = 0.85 fc′ bh + 0.02bh(40)

 0.02(40)  = 0.85 fc′ bh 1 + 0.85(4)  



Pn = 1.235 (ordinate of point A) 0.85 fc′ bh (Continued)

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1 3 . 2   B U C K L I N G O F C O N C E N T R I C A L LY L OA D E D C O L U M N S

467

Example 13.2.1 (Continued) (b) Point B, εc = εy = 0.00138 < ε0, Es = 0.   ε   ε 2 fc = fc′′ 2  c  −  c     ε 0   ε 0     1.38   1.38  2  = 3.4 2   −     1.94   1.94   = 3.4[2(0.712) − 0.508] = 3.12 ksi Pn = fc bh + As f y = 3.12bh + 0.02bh(40) = 3.92bh



Pn 3.92 = = 1.153 (ordinate of point B) 0.85 fc′ bh 3.4

0.45h 1.6 fy = 50 ksi fy = 40 ksi

1.4

fy = 50 ksi ρ = 0.04

0.45h

b

A 1.2

C

Pn /(0.85fc’bh)

B

h

1.0 fy = 50 ksi ρ = 0.01

ρ = 0.02

0.5 fy = 50 ksi ρ = 0.02 long-time loading

140 kL Slenderness ratio, u r 70

210

Figure 13.2.2  Strength curves for reinforced concrete ( fc′ = 4000 psi) concentrically loaded pin-​end columns. (Adapted from Ref. 13.23.)

(Continued)

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C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

Example 13.2.1 (Continued) Applying Eq. (13.2.1) with I of the uncracked transformed section, Et =

 2 2ε  Eε dfc = fc′′  − 2c  = Ec − c c dεc ε0  ε0 ε0 

 1.38  = 3500  1 − = 1008 ksi  1.94  I=



E  bh3 + 0.02bh  s  (0.45h)2 12  Et 

1  0  2 3 = bh3  + 0.02   (0.45)  = 0.0833bh  12 1008   (kLu )2 =

π 2 Et I Pc

(kLu )2 =

π 2 (1008)(0.0833bh3 ) = 212h 2 3.92bh

kLu = 14.55 h Where the first equation for (kLu )2 is a restatement of Eq. (13.2.1). For a rectangular section, r = h / 12,

kLu = 14.55 12 = 50.3 (abscissa of point B) r

(c) Point C, εc at an infinitesimal amount less than εy; E = 29,000 ksi. 1   29, 000  (0.45)2  I = bh3  + 0.02    1008  12  = bh3 (0.0833 + 0.1167) = 0.20bh3



(kLu )2 =

π 2 (1008)(0.20bh3 ) = 507h 2 3.92bh



kLu = 22.5 h kLu = 22.5 12 = 78 r

(abscissa of point C )

13.3 EFFECTIVE LENGTH FACTOR For conditions other than pin ends, where the factor k in Eq. (13.2.1) is 1.0, the equivalent pin-​end length (also called effective length) factor k must be determined for various rotational and translational end restraint conditions. Where translation at both ends is adequately prevented, the distance between points of inflection is shown in Fig. 13.3.1. For all such cases the equivalent pin-​end length is less than or equal to the actual unbraced length (i.e., k ≤ 1.0).

469



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13.3  EFFECTIVE LENGTH FACTOR

Figure 13.3.1  Equivalent pin-​end (i.e., effective) lengths; no joint lateral translation.

P

P

P

Lu

Lu

Lu

kLu = Lu P

(a) End rotation fully restrained

kLu = 2Lu P

(b) One end rotation fully restrained, other end unrestrained

kLu > 2Lu Partial restraint P

(c) One end rotation partially restrained, other end unrestrained

Figure 13.3.2  Equivalent pin-​end (i.e., effective) lengths; joint lateral translation possible.

If sidesway or joint lateral translation is possible, as in the case of an unbraced frame, the equivalent pin-​end length is greater than or equal to the actual unbraced length (i.e., k ≥ 1.0), as shown in Fig. 13.3.2. Reinforced concrete columns are in general part of a larger frame, and thus it is necessary to understand the concepts of the braced frame (where joint lateral translation may be assumed to be prevented by rigid bracing, shear walls, or attachment to an adjoining structure) and the unbraced frame (where stability is dependent on the stiffness of the beams and columns that constitute the frame). As shown in Fig. 13.3.3(a) and 13.3.3(c), the effective length kLu for cases where joint lateral translation is prevented may never exceed the actual length Lu. In an unbraced frame [Fig. 13.3.3(b) and 13.3.3(d)] instability results in a sidesway type of buckling with the effective length kLu always exceeding the actual length Lu. Values for effective length factor k are discussed in Section 13.11.

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C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

P

P

P kLu 2

Lu

0.7Lu < kLu < Lu

(a) Braced frame, hinged base

P

P kLu > 2Lu

Lu

(b) Unbraced frame, hinged base

0.5Lu < kLu < 0.7Lu P

P

P

Lu

Lu

Lu < kLu < 2Lu (c) Braced frame, fixed base (d) Unbraced frame, fixed base

Figure 13.3.3  Equivalent pin-​end (i.e., effective) lengths for columns in frames.

13.4 MOMENT MAGNIFICATION—​M EMBERS WITH TRANSVERSE LOADS—​W ITHOUT JOINT LATERAL TRANSLATION (i.e., NO SIDESWAY) As stated previously, nearly all compression members are simultaneously subjected to some bending moment that causes deflections. A deflected compression member with no sidesway (i.e., with no joint lateral translation) will be further subjected to a secondary bending moment PΔ, as shown earlier (see Fig. 13.1.1). One may consider this as a magnification of the primary or first-​order bending moment. An approximate determination of the magnifying effect may be made by considering the member as finally achieving a deflection Δmax, which is composed of the deflection ∆0 due to the primary bending moment Mi and the additional deflection ∆1 due to the secondary moment from axial compression (see Fig. 13.4.1). For a simply supported member, it may be assumed that the secondary bending moment takes the shape of a sine curve (very nearly exact for members with no end restraint and whose primary bending moment and deflection are both maximum at midspan). The midspan deflection ∆1 equals the moment of the M/​(EI) diagram (for secondary bending moment) between the support and midspan taken about the support. Thus

∆1 =

P PL2  L 2  L ( ∆ 0 + ∆1 )     = ( ∆ 0 + ∆1 ) 2  2 π  π EI π EI

(13.4.1)

from which

 PL2 / (π 2 E I )   α  ∆1 = ∆ 0   = ∆ 0   2 2 1− α 1 − PL / (π EI ) 

(13.4.2)

where α = PL2/​(π 2EI). Since Δmax is the sum of Δ0 and Δ1,

∆  α  ∆ max = ∆ 0 + ∆1 = ∆ 0 + ∆ 0  = 0  1− α 1− α

(13.4.3)

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1 3 . 4   M E M B E R S W I T H T R A N S V E R S E L O A D S — N O S I D E S WAY

Figure 13.4.1  Primary and secondary bending moments for a simply supported member with transverse loads.

The maximum bending moment, including the effect of axial load, becomes

M max = M m + P ∆ max = M m +

P∆0 1− α

(13.4.4)

On substituting the expression for Δmax of Eq. (13.4.3) and substituting P = απ 2EI/​L2, Eq. (13.4.4) becomes

 C  M max = M m  m  = M m δ 1− α

(13.4.5)

where

δ=

Cm = magnification factor 1− α

(13.4.6)

 π 2 EI ∆ 0  Cm = 1 +  − 1 α 2  Mm L 

(13.4.7)

and



Thus, for common cases of single curvature deflection, the magnification factor to be applied to the primary bending moment is Cm  /​(1 –​ α). For Cases 1 to 7 shown in Table 13.4.1, rigorous solution of the differential equation may be obtained (see Section 13.5). The approxi­mate values of Cm for positive moment shown in this table for Cases 1 to 3, computed by using Eq. (13.4.7), are in good agreement with those from theoretical solutions. Derivations for approximate values of Cm for negative moment are presented by Iwankiw [13.24]. Positive moments for the restrained end cases (4–​7) are given to correlate with theoretical solutions [13.24]. Inherent in the theoretical solutions for these restrained cases is the assumption that the slope at a restrained support is zero, the practicality of which can be argued. One may note that for all cases this Cm value will be close to 1.0, because in actual concrete structures α (i.e., the ratio between the applied axial load and the column buckling strength) rarely exceeds about 0.3. The approximate treatment of slenderness in ACI-​6.6.4.5.3(b) conservatively requires that Cm be taken as 1.0 for all cases with transverse loading between supports.

472

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C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

TABLE 13.4.1  SUGGESTED VALUES OF Cm FOR COMMON SITUATIONS WITH NO JOINT TRANSLATION Case

M

1

2

3

4

5

6

M

P

w

P

L/2

P

P

Q

P

w

P

P

w

P

P

L/2

P

Q

L/2

7

P

Q

P

Cm (Positive Moment)

Cm (Negative Moment)

1 + 0.2α

–​

Mm +

1.0

–​

Mm +

1 –​ 0.2α

–​

1 –​ 0.3α

1 –​ 0.4α

1 –​ 0.2α

1 –​ 0.4α

P

1.0

1 –​ 0.3α

P

1 –​ 0.2α

1 –​ 0.2α

Primary Bending Moment

Mm +

Mm +

–

Mm +

+

8

MA

P MA = MB

MB

Eq. (13.4.7)

–

Mm –

–

Mm +

q P

–

not available

+

–

Mm –

13.5 MOMENT MAGNIFICATION—​M EMBERS SUBJECT TO END MOMENTS ONLY—​W ITHOUT JOINT LATERAL TRANSLATION (i.e., NO SIDESWAY) Consider the general case shown in Fig. 13.5.1 wherein the end moments M1 and M2 constitute the primary bending moment Mi, which is a function of z. The sum of primary and secondary moments causes the member to have a deflection y, and thus the secondary moment is equal to P y. Stating the total moment Mz at the section z of Fig. 13.5.1 gives

M z = Mi + Py = − EI

d2 y dz 2

(13.5.1)

for members with constant EI, and dividing by EI gives

M d2 y P + y= − i 2 EI EI dz

(13.5.2)

473



1 3 . 5   M E M B E R S W I T H E N D M O M E N T S O N LY — N O S I D E S WAY

M1

473

M2

P

P

y

z

z y M1

M2 Py

Primary moment, Mi M2 > M1

Secondary moment, Py

Figure 13.5.1  Primary and secondary moments for a member subject to end moments only and no joint lateral translation.

For design purposes, the general expression for moment Mz is of greater importance than the deflection y. Differentiating Eq. (13.5.2) twice gives d4 y P d2 y 1 d 2 Mi + =− 4 2 EI dz 2 EI dz dz



(13.5.3)

From Eq. (13.5.1), M d2 y =− z 2 EI dz



and

2 d4 y 1 d Mz = − EI dz 2 dz 4

Substitution into Eq. (13.5.3) gives −



2 1 d Mz 1 d 2 Mi P  Mz  + − =−   2 EI dz EI  EI  EI dz 2

Simplifying and letting λ2 = P/​EI, d 2 Mz



dz

2

+ λ2 Mz =

d 2 Mi dz 2

(13.5.4)

which is of the same form as the deflection differential equation, Eq. (13.5.2). The homogeneous solution for Eq. (13.5.4) is M z = A sin λ z + B cos λ z



(13.5.5)

To this must be added the particular solution that will satisfy the right-​hand side of the differential equation. In the special case of unequal end moments acting without transverse loading,  M − M1  Mi = M1 +  2  z  L

Since

d 2 Mi = 0 dz 2

(13.5.6)

47

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C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

Eq. (13.5.4) becomes a homogeneous equation, in which case Eq. (13.5.5) represents the entire solution. To determine the maximum moment,

dM z = 0 = Aλ cos λ z − Bλ sin λ z dz

(13.5.7)

or tan λ z =



A B

(13.5.8)

At maximum Mz,

sin λ z =

A A +B 2

2

and cos λ z =

B A + B2 2

(13.5.9)

Substitution of Eq. (13.5.9) in Eq. (13.5.5) gives M max =

=

A2 A2 + B 2

B2

+

A2 + B 2

A2 + B 2

(13.5.10)

Now the constants A and B are evaluated by applying the boundary conditions to Eq. (13.5.5). The conditions are (1) at z = 0,

M z = M1 ∴ B = M1

(2) at z = L,

M z = M2 M 2 = A sin λ L + M1 cos λ L



∴A=

M 2 − M1 cos λ L sin λ L



so that

 M − M1 cos λ L  M2 =  2  sin λ z + M1 cos λ z  sin λ L

(13.5.11)

and 2

 M − M1 cos λ L  2 M max =  2  + M1  sin λ L

= M2

1 − 2( M1 /M 2 ) cos λ L + ( M1 /M 2 )2 sin 2 λ L

(13.5.12)

For the general case of a member subject to end moments, the maximum moment may be either (1) the larger end moment M2 at the braced (supported) location, [Fig. 13.5.2(a)] or (2) the magnified moment given by Eq. (13.5.12) that occurs within the span [Fig. 13.5.2(b)], depending on the ratio M1/​M2 and the value of α, where α is the ratio of P to the critical load π 2EI/​L2, making λ L = π α . Note that in Eq. (13.5.12) the bending moment ratio M1/​M2 is positive when the member is bent in single curvature, as shown in Fig. 13.5.1, and negative when the member is bent in double curvature.

475



1 3 . 5   M E M B E R S W I T H E N D M O M E N T S O N LY — N O S I D E S WAY

475

To compute the required strength of a member, one needs to know whether the maximum moment occurs at a location between the supports, and if so, the correct location. The concept of equivalent uniform moment [Fig. 13.5.2(c)] can eliminate the need for such information. Thus, for the case with unequal end moments, use of the equivalent moment assumes that Mmax is at midspan. To establish the equivalent moment, let M1 = M2 = Mequiv in Eq. (13.5.12), 2(1 − cos λ L ) sin 2 λ L

M max = M equiv



(13.5.13)

Equating Eq. (13.5.12) and Eq. (13.5.13) gives M equiv = M 2



( M1 / M 2 )2 − 2( M1 / M 2 ) cos λ L + 1 2(1 − cos λ L )

(13.5.14)

According to the procedure of Section 13.4, the approximate expression for maximum moment is  C  M max = M m δ = M m  m  1− α



(13.5.15)

For the case of uniform moment (M1 = M2 = Mequiv), ∆0 =

M equiv L2 8 EI

M m = M equiv

Mz < M2 for all values of α M2

(a) Max moment at end Zero slope

Mmax > M2

M2

M1 (b) Max moment not at end Mmax

Mequiv

Mequiv (c) Equivalent uniform moment with max magnified moment at midspan

Figure 13.5.2  Combined primary and secondary bending moment diagrams for members having end moments without transverse loading.

476

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C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

Replacing these expressions into Eq. (13.4.7),  π 2 EI  M equiv L2  Cm = 1 +   2  − 1 α ≈ 1  L  8 EIM equiv 

Thus

 1  M max = M equiv   1 − α 



(13.5.16)

Substitution of Eq. (13.5.14) into Eq. (13.5.16) gives  C  M max = M 2  m  1− α



(13.5.17)

in which



Cm =

( M1 /M 2 )2 − 2( M1 /M 2 ) cos λ L + 1 2(1 − cos λ L )

(13.5.18)

Comparing Eq. (13.5.17) with Eq. (13.5.16), Cm M2 may be considered to be the equivalent uniform moment along the span. Equation (13.5.18) assumes that the strength of the member is limited by excessive deflection in the plane of bending. Also, it does not fully cover the double-​curvature cases where M1/​M2 lies between –​0.5 and –​1.0. The actual failure mode of members bent in double curvature with such bending moment ratios is generally one of “unwinding” from double to single curvature in a sudden type of buckling. Massonnet [13.25] and the AISC specification for steel buildings1 have suggested approximate expressions for Cm to be used for design in place of Eq. (13.5.18). The ACI Code has adopted the long-​used expression from AISC to compute Cm [ACI-​6.6.4.5.3(a)]:

Cm = 0.6 − 0.4

M1ns M 2 ns

(13.5.19)

where the additional subscript ns denotes that these moments are those acting on a nonsway compression member. Note that the M1ns /​M2ns ratio in Eq. (13.5.19) is positive when the member is bent in double curvature, and negative when the member is bent is single curvature. Figure 13.5.3 compares Eq. (13.5.18) with the recommendations of Massonnet [13.25] and the ACI expression. The reader should note that for a given value of α, the curves plotted using Eq. (13.5.18) are terminated when the moment M2 at the end of the member exceeds the magnified moment. The straight line adopted by ACI-​6.6.4.5.3(a) falls near the upper limit for Cm at any given bending moment ratio, and thus it provides a realistic and simple approximation. Also, note that for members bent in double curvature, Cm will vary between 0.2 and 0.6, while for members bent in single curvature, Cm will vary between 0.6 and 1.0.

1 See Manual of Steel Construction (14 Ed.). Chicago: American Institute of Steel Construction, 2011.

47



477

1 3 . 6   M O M E N T M AG N I F I C AT I O N — M E M B E R S W I T H S I D E S WAY

P

Cm =

Equiv uniform moment Max moment M2 at end

1.0

α=

0.1

Cm = 0.6–0.4

M1

M1ns [Eq. (13.5.19); M2ns ACl-6.6.4.5.3(a)]

M2 > M1

α = 0.2

0.8

M2

α = 0.3 α = 0.4

0.6 Cm = 0.4

0.2

1.0

P

α = 0.5 M 2 M 2 0.3 M1 + 0.4 M1 + 0.3 2 2 α = 0.6 (Massonnet–Ref. 13.25) α= α= 2P 2 0.7 Curves plotted π El/L α= using Eq. (13.5.18) 0 α = .8 0.9

0.75

( )

( )

0.5 0.25 0.50 0 0.25 End moment ratio M1 / M2 or M1ns / M2ns

0.75

1.0

Double curvature

Single curvature

Notes: · M1 /M2 is positive for single curvature in Eq. (13.5.18) and in Massonnet’s equation [Ref. 13.25] · M1ns /M2ns is positive for double curvature in Eq. (13.5.19) [ACI-6.6.4.5.3(a)]

Figure 13.5.3  Comparison of theoretical Cm with design recommendations for members subject to end moments only, without joint lateral translation.

13.6 MOMENT MAGNIFICATION—MEMBERS WITH SIDESWAY—​U NBRACED (SWAY) FRAMES In a braced (nonsway) frame, relatively stiff diagonal bracing or shear walls restrain lateral movement, so that any end lateral displacement (or sidesway) of the member can be assumed to be small. Moment magnification in columns of braced (nonsway) frames is thus due to deflections between the ends of the member and may be computed using the procedures described in Sections 13.4 and 13.5. In an unbraced (sway) frame, however, the frame must rely on the stiffness of the beams and columns to limit the lateral translation [see earlier: Figs. 13.3.2 and 13.3.3(b) and 13.3.3(d)] and thus, a relatively larger sidesway deflection ∆ will be expected. Consequently, secondary moments will develop in columns of unbraced frames as a result of both member end lateral translation and deflections occurring between member ends. To visualise the basic concepts of secondary moments that develop in an unbraced (sway) frame, refer to Fig. 13.6.1(a), where the moments M1s and M2s (the subscript “s” denotes “sway”) are the moments resulting from a first-​order analysis. From equilibrium,

M1s + M 2 s = Vu Lc

(13.6.1)

In the model of Fig. 13.6.1(a), acted on by Vu, the top of the member will exhibit a first-​ order lateral translation ∆0 with respect of the bottom. This means that the gravity load, ΣPu, is acting with an eccentricity ∆0, which, in turn, will increase the primary or first-​order overturning moment Vu Lc by the amount ΣPu∆0. This additional moment will further increase the relative lateral deflection, which upon equilibrium of the structure in the final displaced

478

478

C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

∑Pu

∑Pu M1s 1

Vu

1’

Lc

Vu

2’

2

M2s ∑Pu

∑Pu

(a) First-order analysis: For equilibrium, M1s + M2s = VuLc ∑Pu

∑Pu δsM1s Vu

1

1’

Lc

Vu

2

2’

δsM2s ∑Pu

∑Pu

Figure 13.6.1  First-​and second-​order end moments and sidesway deflection of columns in one story of a multistory unbraced frame.

position will be ∆2u (larger than ∆0), as shown in Fig. 13.6.1(b). The total moments at the column ends (primary plus secondary) then may be expressed as

δ s ( M1s + M 2 s ) = Vu Lc + ∑ Pu ∆ 2 u

(13.6.2)

As shown by the right-​hand side of Eq. (13.6.2), the total moment can be thought of as the sum of the primary plus the secondary part, or as on the left, it can be thought of as the primary moment magnified by the factor δs. Second order effects due to lateral translation (sidesway) are usually more significant under the action of lateral loads, but they can also occur under gravity loads alone. Indeed, when the structural configuration or the loading condition is asymmetric, an unbraced frame will exhibit sidesway under gravity loads. Furthermore, even if the frame and loading condition were intended to be perfectly symmetric, randomly occurring construction inaccuracies can cause columns to be “out of plumb” (i.e., the centroid of the top of the column is not directly over the centroid of the bottom of the column) which will induce an unintended eccentricity of the gravity loads (see study by MacGregor [13.28]). As noted earlier, the lateral displacement in a braced system is usually small; thus the secondary bending moments from any small amount of sidesway (the P-​∆ effect) may ordinarily be neglected. In unbraced frames, however, the relatively larger sidesway deflection usually cannot be ignored and must be accounted for in design. It is possible, however, for a braced frame to undergo large sidesway deflection (e.g., if bracing is too flexible) or, conversely, for an unbraced frame to develop very small sidesway deflection (e.g., if beams and columns are very stiff). In other words, whether the P-​∆ effects due to sidesway can or

479



1 3 . 7   I N T E R AC T I O N D I AG R A M S — E F F E C T O F S L E N D E R N E S S

479

cannot be neglected is a function of the amount of sidesway expected to occur under the load combination being considered, regardless of whether the system is physically provided with lateral bracing. Prior to 1995, the ACI Code Commentary indicated that the sidesway effect could be neglected except in cases of asymmetrical gravity loading or structural configuration where “appreciable” sidesway might occur. The ACI Code Commentary further indicated that “appreciable” sway would be considered to exist when the calculated sway ∆ from a first-​order frame analysis divided by the clear height Lu exceeded 1/​1500. While this recommendation no longer appears in the ACI Code Commentary, it does provide a sense of the amount of sway under gravity load that could be permitted without incurring significant P-​∆ effects. The 2014 ACI Code requires that columns in frames be classified as either nonsway or sway based on the expected amount lateral deflection or the percent increase of first-​order moments, as described later in Section 13.8. For a more detailed treatment of braced and unbraced elastic frames, the reader is referred to Refs. 13.26 and 13.27. Also, MacGregor and Hage [13.29] and MacGregor [13.30] have provided an excellent summary of the various methods and the basic concepts regarding the design of slender columns.

13.7 INTERACTION DIAGRAMS—​E FFECT OF SLENDERNESS To understand the ACI Code procedure and its approximations as discussed in later sections of this chapter, the general approach is described for determining a point on the Pn–​Mn interaction diagram for kL/​r not equal to zero. Chapter 10 treated the basic strength of a section with zero kL/​r, giving the interaction diagram shown in Fig. 10.6.1, where the curve corresponds to the one designated kL/​r = 0 in Fig. 13.7.1. Points A and B represent combinations of Pn and Mn with the neutral axes (N.A.) located at cA and cB, respectively, from the extreme compression fiber whose crushing strain is taken as 0.003, as prescribed by the ACI Code. The curve labeled kL /​r ≈ 60 in Fig. 13.7.1 represents a typical strength interaction curve that includes slenderness effects. Pfrang and Siess [13.6], Pfrang [13.12], and MacGregor, Breen, and Pfrang [13.32] have provided excellent discussions of the slenderness effects on interaction diagrams. A slender column may fail in one of two ways: (1) by material failure, that is, by reaching a combined Pn–​Mn as computed by the methods of Chapter 10; or (2) by instability when an infinitesi­ mal increase in axial load results in additional deflection such that equilibrium cannot be achieved (i.e., buckling or instability failure). Referring to Fig.13.7.2(a), consider a member with an increasing axial load P acting at an initial eccentricity e. A “short” column (kL /​r = 0) will follow the path of the straight line to point C at P = P1 and can continue to carry an additional axial load and moment up to point Pshort that lies on the interaction diagram for the member, that is, where material failure

P0

cA kL ≈ 0 r kL ≈ 60 r A

εc = 0.003 EIA

N.A. cB

Pn

εc = 0.003 B

N.A.

EIB

M0 Mn

Figure 13.7.1  Slenderness effects on column strength interaction diagram.

480

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C H A P T E R   1 3     S L E N D E R N E S S E F F E C T S O N C O L U M N S

Pn Mm = P1e

P0

Pshort kL = 0 r

Pn P1

Primary eccentricity line

Material failure

C D

E

Plong

kL ≈ 100 r

Mn

M0

(a) Slender column behavior

D of Fig. 13.7.2(a)

P1

kL = 0 r

C

Instability failure

100

70 35

130

M0

Mr

(b) Interaction diagrams

Figure 13.7.2  Slender column interaction diagrams. (Adapted from Ref. 13.32.)

would occur. Consider a “long” column with large slenderness, say kL /​r = 100. In such a case, the member will follow the path to point D on the interaction diagram, where the material failure is reached at an axial load Plong much less than Pshort. This behavior occurs because lateral deflection of the column will in effect increase the axial load eccentricity and thereby increase the applied moment to Mmax = Mm + P1Δmax as shown in Fig. 13.7.2(a). Note that both the “short” and “long” columns would exhibit a material failure in this instance. It is also possible, however, that the column could fail by instability. In such a case, the column would follow the path (dashed) up to point E in figure 13.7.2(a). That is, it would be unable to reach the material strength interaction diagram. It is possible then to construct an interaction diagram where a material failure occurring at D on the kL /​r = 0 curve due to P1 plus a magnified moment Mmδ (equal to Mm + P1Δmax) is plotted for the particular loading arrangement as point C on the primary moment radial line for kL = 100, as shown in Fig. 13.7.2(b). Whatever the primary moment loading arrangement—​such as equal end eccentricities of axial load, unequal end eccentricities of axial load, or lateral transverse loading—​the deflection of the member will differ, and so the secondary bending moment will differ. This means that different types of primary moments will cause a member of kL /​r = 100 to follow a curved path to intersect the kL /​r = 0 material strength interaction diagram at different locations such as point D. The radial line through point C, however, is uniquely a function of the primary moment, which is the same for all slenderness ratios. The development of a correct interaction diagram for members of large slenderness, such as Fig. 13.7.2(b), would require an elaborate analysis for each structure, taking into account such factors as the following: (1) a realistic moment-​curvature relationship, (2) the time-​dependent and cracking effects on deflections, and (3) influence of axial load on the flexural stiffness of members. Mockry and Darwin [13.33] have made such an analysis and prepared a practical set of design charts. Parme [13.34] has treated the subject in a similar manner, and the Portland Cement Association has published design aids [13.35]. In practice, however, rather than constructing interaction diagrams for various kL /r values [Fig. 13.7.2(b)] the design of slender columns is done using the interaction diagram for “short” columns (kL /r = 0) with the magnified moment. Instability or buckling failures [point E in Fig. 13.7.2(a)] are not permitted.

13.8 ACI CODE—​G ENERAL Unless slenderness effects can be neglected (see Section 13.15), the design of columns in both sway and nonsway frames in accordance with the ACI Code is to be done (a) by an approximate moment magnifier method or (b) by conducting a second-​order analysis from which the magnified moments are obtained directly (see Section 13.12).

481



481

13.8  ACI CODE—GENERAL

Section Properties In the moment magnifier method, a first-​order analysis of the structure is used to compute the first-​order or primary bending moments. The moments thus computed are then multiplied by a moment amplification factor to obtain the total moments, including second order effects. Ideally, the structure should be analyzed by using section properties that take into account the influence of axial loads, the presence of cracked regions along the length of the member, and the effects of the duration of the loads. However, the intricacies and the computational effort involved in estimating the appropriate modeling parameters make such an approach often impractical for use in everyday design practice. For this reason, alternative approaches based on simplified modeling assumptions are often preferred. For a first-​order, elastic analysis under factored loads, the ACI Code permits the use of the properties listed in Table 13.8.1. TABLE 13.8.1  MODULUS OF ELASTICITY, MOMENT OF INERTIA, AND CROSS-​SECTIONAL AREA FOR ELASTIC ANALYSIS AT FACTORED LOADS (a) Modulus of elasticity

Ec from ACI-​19.2.2

(b) Moments of inertia [ACI-​Table 6.6.3.1.1(a)] Beams Columns Walls—​uncracked Walls—​cracked Flat plates and flat slabs

0.35Ig 0.70Ig 0.70Ig 0.35Ig 0.25Ig

(c) Area [ACI Table 6.6.3.1.1(a)]

1.0Ag

The moment-​ of-​ inertia values of Table  13.8.1 are based on those suggested by MacGregor  and Hage [13.29] and include the effects of cracking and axial load on the member stiffness. They also include a stiffness reduction factor, φK, as described next. Consistent with the ultimate strength design approach, in which a strength reduction factor is used, member stiffness should be also multiplied by a stiffness reduction factor, φK, to account for the variability that exists in the member stiffness. Studies have shown that the strength reduction factor φ and the stiffness reduction factor φK are not the same [13.60]. For example, for an isolated column, φK was found to be about 0.75 whether the column was tied or spirally reinforced. On the other hand, the column deflections in a multistory frame will depend on the average concrete strength of all the columns in the frame. The average concrete strength, and consequently, the average modulus of elasticity of the columns in a frame, will be higher than those in a single understrength column (ACI-​ R6.6.4.5.2). Based on these considerations, the moments of inertia given in Table 13.8.1 were obtained by using a stiffness reduction factor φK of 0.875 (larger than that of an isolated column) to account for the variability of member stiffness. Alternative moment of inertia expressions, more refined than those given in Table 13.8.1 [ACI Table 6.6.3.1.1(a)], are given in Table 13.8.2 [ACI Table 6.6.3.1.1.(b)]. These expressions are based on the study by Khuntia and Ghosh [13.61], and require knowledge of the axial force level, eccentricity, and reinforcing ratio. For this reason, these equations are more appropriate for a more refined analysis rather than a preliminary or first estimate of slenderness effects. To account for the effects of load duration, the moments of inertia must be modified when sustained loads act, as discussed later (see Sections 13.9 and 13.10).

Nonsway or Sway Frames To compute the appropriate moment magnification by means of the moment magnifier method, columns in frames must be classified as either nonsway or sway. According to

482

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TABLE 13.8.2  ALTERNATIVE MOMENTS OF INERTIA FOR ELASTIC ANALYSIS [ACI TABLE 6.6.3.1.1(b)] Member

Minimum

I

Maximum

Columns and walls

0.35Ig

 Ast   Mu P − 0.5 u  I g  0.80 + 25   1 − Ag   Pu h Po  

0.875Ig

Beams, flat plates, and flat slabs

0.25Ig

b   (0.10 + 25ρ)  1.2 − 0.2 w  I g  d

0.5Ig

Po is the axial compressive strength of the member at zero eccentricity.

ACI-​6.6.4.3, it is permitted to consider a frame as nonsway if one of the following conditions is met. 1. The increase in column end moments due to second-​order effects does not exceed 5% of the first-​order end moments. 2. The story stability index, Q, is less than or equal to 0.05, where Q=



∑ Pu ∆ 0 Vu Lc

(13.8.1)

and ΣPu = total vertical factored load in the story under consideration ∆0 = relative lateral deflection between the top and bottom of the story in question due to Vu, computed using a first-​order elastic frame analysis Vu = factored shear in the story in question Lc = length of the compression member under consideration, measured from center to center of the joints in the frame Alternatively, if bracing elements resisting lateral movement of a story have a stiffness of at least 12 times the gross lateral stiffness of the columns in the direction considered, ACI6.2.5 permits the columns to be considered as braced against sidesway (i.e., nonsway). Note that even if a column is part of a braced frame it may need to be treated as part of a sway frame. For example, if the bracing elements (e.g., walls) are too flexible and relative lateral end displacement cannot be ignored.

13.9 ACI CODE—​M OMENT MAGNIFIER METHOD FOR COLUMNS IN NONSWAY FRAMES It was shown in Sections 13.4 and 13.5 that the maximum second-​order moment in an elastic member of a nonsway frame may be computed from Eq. (13.4.4), above, as follows:

M max = M m + P ∆ max = M m +

P∆0 1− α

[13.4.4]

where α = P/​Pc and Pc = π 2EI/​L2. Furthermore, Eq. (13.4.5) for members with transverse loading, and Eq. (13.5.15) for members with end moments only, indicate that the maximum second-​order moment may also be expressed as the maximum primary moment Mm times a magnification factor,

 C  M max =  m  M m 1− α

[13.4.5]

483



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483

or M max = δ ns M m



(13.9.1)

which is the same as ACI Formula (6.6.4.5.1)2, where

δ ns =



Cm 1 − α

(13.9.2)

in which the subscript ns denotes nonsway. Several studies have shown [13.18, 13.32, 13.40] that this procedure is acceptable for reinforced concrete members. The design strength of columns in nonsway frames is thus computed to satisfy the following equations:

φPn ≥ Pu

and

φ M n ≥ δ ns M m

(13.9.3)

where design of the member is based on the factored axial load Pu combined with the corresponding factored moment, Mm,2 computed from an elastic, first-​order analysis, and magnified by the factor δns given by ACI-​6.6.4.5.2 as



δ ns =

Cm ≥ 1.0 Pu 1− 0.75Pc

(13.9.4)

The primary moment, Mm, in Eqs. (13.9.1) and (13.9.3) corresponds to the largest moment occurring in the member and, in general, may occur at either end, or along the length of the member. Equation (13.9.4) is basically the same as Eq. (13.9.2), where the critical buckling load, Pc, has been multiplied by a stiffness reduction factor, φK, to account for the variability in member stiffness. A single value φK = 0.75 is used for both tied and spirally reinforced isolated columns as suggested by Mirza et al. [13.60]. The design strengths φPn, φMn are obtained from the interaction diagram for short columns (kL /r = 0) as discussed in Section 13.7.

Factor Cm For members with transverse loading, the quantity Cm may be computed from Eq. (13.4.7) for simply supported members or from Table 13.4.1 for other support conditions. Since, however, Cm will usually vary between 0.9 and 1.0 in practice, there is little advantage to using the values given in Table 13.4.1. Accordingly, the ACI Code requires Cm to be taken as 1.0 for members with transverse loading. For members with end moments only, the ACI Code uses the approximate expression given by Eq. (13.5.19) for Cm. In summary, for columns in nonsway frames, Cm is computed under ACI-​6.6.4.5.3 as follows. 1. Members subject to transverse loading,

Cm = 1.0

(13.9.5)

2. Members subject to end moments only,

Cm = 0.6 − 0.4

M1ns M 2 ns

(13.9.6)

where M1ns and M2ns are the first-​order end moments. Note that M2ns is taken as the larger of the end moments, and that the ratio M1ns/​M2ns is negative when the member is bent in single curvature.

2  Note that the ACI Code uses M2 instead of Mm in Eq. (13.9.1) [ACI Formula (6.6.4.5.1)]. Mm in Eq. (13.9.1) is the maximum moment occurring in the member and, in general, may occur at either end or, if there is transverse loading, within the span. Here Mm is used instead to distinguish it from the larger end moment M2, in members with end moments only (see Section 13.5).

48

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Stiffness Parameter EI The other quantity required for evaluating the moment magnifier δns is the critical buckling load, which in accordance with ACI-​6.6.4.4.2 is computed as

Pc =

π 2 ( EI )eff (knonsway Lu )2

(13.9.7)

where Lu is the unsupported length of the member. The current ACI Code does not provide a definition for Lu as in past editions, but it should be taken as the clear length of the column. The parameter knonsway corresponds to the effective length factor k (see Section 13.3) and is computed as described later (see Section 13.11) for nonsway (braced) frames. To account for the effects of cracking, load duration, and nonlinearity of the concrete stress-​strain curve, an effective stiffness parameter, (EI )eff, is used in Eq. (13.9.7). MacGregor, Breen, and Pfrang [13.32] proposed the use of the larger of two simple expressions. These appear in ACI-​6.6.4.4.4 as Formulas (6.6.4.4.4a) and (6.6.4.4.4b), respectively,

( EI )eff =

0.4 Ec I g

(13.9.8)

(1 + β dns )

or

( EI )eff =

0.2 Ec I g + Es I se (1 + β dns )



(13.9.9)

where Ec = concrete modulus of elasticity = 57, 000 fc′ for normal-​weight concrete (ACI-​19.2.2.1) Ig = gross moment of inertia of concrete section, neglecting reinforcement Ise = moment of inertia of reinforcement about the concrete section centroid βdns = ratio between the maximum factored axial load that is considered sustained and the maximum factored axial load associated with the same load combination (usually the ratio between the factored axial dead load and the total factored axial load) Alternatively, the stiffness parameter, (EI )eff, may be computed by using ACI Formula (6.6.4.4.4c) as follows

( EI )eff =

Ec I (1 + β dns )

(13.9.10)

where I is computed in accordance with Table 13.8.2 [ACI Table 6.6.3.1.1(b)]. Under sustained loads, creep effects will increase the long-​term deflections and, as a result, they will also increase the second-​order moments. To account for these effects, the value of EI is divided by (1 + βdns) in Eqs. (13.9.8), (13.9.9), and (13.9.10). The relative accuracy of Eqs. (13.9.8) and (13.9.9) as a function of the axial load ratio, P/​Pc, for the case of no sustained load (βdns = 0) is shown in Fig. 13.9.1 [13.32]. In the study of Eqs. (13.9.8) and (13.9.9), EI values for about 100 cases were estimated by means of theoretical load-​moment-​curvature diagrams for columns of various dimensions, strengths, and steel percentages. Effective EI values were also computed for the University of Texas frame tests [13.5, 13.8, 13.9], and for a series of computer simulations. It may be seen that Eqs. (13.9.8) and (13.9.9) provide a lower-​bound estimate of EI for most axial load ratios, except at levels exceeding P/​Pc ratios of about 0.8. Such high P/​Pc ratios are rarely if ever encountered in practice. Equation (13.9.8) is easier to use because it does not require knowledge of the amount of reinforcement (which is not known in the initial stages of design); however, it greatly underestimates the contribution of the reinforcement for heavily reinforced columns.

485

EI from (13.9.8)

5

EcIg 2.5 ρ = 0.08

4 3 2 1 0

ρ = 0.01 0.6

0.7

P Pc

0.8

(a) Eq. (13.9.8)

0.9

EI = Theoretical EI

EI = Theoretical EI

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EI from Eq. (13.9.9)



2

ρ = 0.01

1

0

EcIg + EsIse 5

ρ = 0.08

0.6

0.7

P Pc

0.8

0.9

(b) Eq. (13.9.9)

Figure 13.9.1  Comparison of Eqs. (13.9.8) and (13.9.9) for EI for short-​duration loading (βdns = 0) with EI values from moment-​curvature diagrams. (Adapted from Ref. 13.32.)

Alternative expressions for EI have also been proposed by MacGregor, Oelhafen, and Hage [13.41], Medland and Taylor [13.42], and Zeng, Duan, Wang, and Chen [13.43]. As a result of these studies it seemed appropriate to divide only the concrete term in Eq. (13.9.9) by the factor (1 + βdns). However, when this is done, the number of cases in which the ratio of theoretical EI to the formula-​computed EI is less than 1.0 becomes 3 times more than otherwise, particularly when the reinforcement ratios are low. Thus, ACI Committee 318 concluded that the (1 + βdns) factor should be retained as the divisor for both the steel and concrete terms. Studies of flexural stiffness expressions for EI include those of Diaz and Roesset [13.44] and Mirza [13.45] for rectangular sections, and those of Sigmon and Ahmad [13.46] and Ehsani and Alameddine [13.47] for circular sections. The sustained load effect on slender columns has also been studied by Goyal and Jackson [13.48], Drysdale and Huggins [13.49], and Rangan [13.50].

13.10 ACI CODE—​M OMENT MAGNIFIER METHOD FOR COLUMNS IN SWAY FRAMES In this method, second-order effects are computed by combining the magnified moments due to the deflections that occur along the length of the member (i.e., nonsway) with those due to lateral translation at the member ends (i.e., sway). This is an approximate method and can be used only in frames with loads causing deflections in the plane of the frame. If significant torsion in plan exists, this procedure may significantly underestimate the magnified moments. In such a case, a three-​dimensional, second-​order analysis must be used (ACI-​R6.6.4.6.1). The moment magnifier method for sway frames consists of computing the first-​order nonsway moments, Mns, and the first-​order sway moments, Ms, separately. The moments thus computed are then amplified by a moment magnifier to find the second-​order moments. This approach of separating the nonsway and sway portions of the frame action is based, in part, on the recommendations of Ford, Chang, and Breen [13.51]. While refinements to this procedure have occurred over the years, the basic approach remains the same.

Computation of Nonsway Moments According to the ACI Code, the nonsway moments, Mns, are to be computed from a first-​ order, elastic frame analysis under loads that cause no appreciable sidesway. Traditionally, this has been interpreted as an analysis of the frame under gravity loads alone. Clearly, however, unless the layout of the frame, support conditions, and load distribution are perfectly symmetric, gravity loads will induce sidesway of the frame. The amount of appreciable sidesway is not explicitly defined in the current ACI Code. Earlier editions of the Code (ACI 318-​92 Commentary) suggested that lateral deflections due to vertical loading

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would be considered appreciable for a deflection ratio Δ /​Lu greater than 1/​1500. In other words, if Δ /​Lu is less than or equal to 1/​1500, the effect of the sway component due to gravity loads can be neglected and the corresponding moments can be considered as nonsway moments. If that condition is not satisfied, the sway component due to vertical loads cannot be ignored and must be considered. An alternative approach to computing the nonsway moments, Mns, that does not depend on whether lateral deflections under gravity loads are appreciable, is presented next.

(a) Frame subjected to factored vertical and lateral loads under the load combination being considered HR

H3

H2

H1

(b) Model for computing first-order nonsway moments HR

H3

H2

H1

(c) Model for computing first-order sway moments

Figure 13.10.1  Computation of nonsway and sway moments

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487

Consider the four-​story, two-​bay frame of Fig. 13.10.1(a) subjected to factored vertical and lateral loads under a given load combination. In this approach, the nonsway moments are computed by analyzing the frame under conditions that prevent lateral translation in each floor (i.e., a nonsway condition is enforced). In this step, all the factored loads (i.e., vertical and lateral loads) for the load combination being considered are applied, as shown in Fig. 13.10.1(b). The bending moments thus obtained will then correspond to the first-​order, nonsway moments, Mns. Analysis of the frame under these load and support conditions will generate horizontal reactions at the lateral supports from (a) the applied lateral loads and (b) any sway component due to the applied vertical loads. These reactions are then used to compute the sway moments, as described next. If no lateral loads exist under the load combination being considered, any sway of the frame caused by gravity loads will result in nonzero horizontal reactions. Thus, whether the amount of sway due to gravity loads is appreciable is irrelevant because any sway effects due to gravity loads are included in the horizontal reactions and the resulting sway moments when this approach is used.

Computation of Sway Moments To obtain the first-​order sway moments, Ms, a separate frame analysis is then conducted by allowing lateral translation and by applying only the horizontal reactions obtained from the nonsway analysis as lateral loads in the opposite direction, as shown in Fig. 13.10.1(c). The actual loads acting on the frame are not applied in this analysis. The idea is that any tendency to sway caused by the actual loads (vertical and lateral) is already captured by the horizontal reactions computed from the nonsway analysis with lateral translations prevented.

Nonsway Moment Magnification Factor, δns Consider the simple frame of Fig. 13.10.2(a) subjected to factored vertical and lateral loads. The first-​order moment diagrams obtained from the nonsway and sway analyses are also shown in the figure. For the loading and support conditions shown, the column has no transverse loading along its length and is bent in double curvature. This is the most common situation found in practice (the special case of columns subjected to transverse loading or in single curvature will be discussed later). The nonsway moment magnification, δns, may be computed by using Eq. (13.9.4). Since the column in Fig. 13.10.2 has no transverse loading, it can be treated as a member subjected to end moments only, and thus, Cm may be computed from Eq. (13.9.6). For a member bent in double curvature, the M1ns /​M2ns ratio is always positive, and thus Cm will be at most 0.6 when M1ns /​M2ns = 0 (see Fig. 13.5.3). On the other hand, the ratio of Pu  /​(0.75Pc) in Eq. (13.9.4) will rarely, if ever, exceed 0.4 in practice. As a result, the nonsway moment magnification for a column bent in double curvature will generally be at most

0.6 δ ns < ∼ 1 − 0.4 = 1.0

(13.10.1)

This means that although the first-​order moments are amplified along the length of the member, they are smaller than at the ends, as shown in Fig. 13.10.2(b). In other words, the largest nonsway moment for a column bent in double curvature, including second-​order effects, will most commonly occur at the member ends and will be equal to the first-​order moment M2ns [M2 in Fig. 13.5.2(a)]. This result is typical of most columns found in practice. For this reason, for columns in sway frames, the ACI Code requires the nonsway moments, including second order effects, to be taken as the first-​order, factored moments at the member ends, M1ns or M2ns. In other words, the ACI Code assumes δns = 1.0. If the column has transverse loading along its length or if it is bent in single curvature with end moments only, the maximum second-​order moment can occur along the member length and can be larger than that at the ends. This situation is discussed later when the second-​order nonsway and sway moments are combined.

48

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P

H

Mtop ns

H First order

(a)

Second order

bot Mns

(b)

top + Mns δs Mstop

δs Mstop

R Mstop

First order

R

First order

Moment magnifier method “True” second order

Msbot

Moment magnifier method

bot + Mns δs Msbot

δs Msbot

(d)

(c)

Figure 13.10.2  Computation of second-​order moments using the moment magnifier method: (a) simple frame subjected to gravity and lateral loads; (b) first-​and second-​order nonsway moments; (c) first-​and second-​order sway moments; and (d) total first-​and second-​order moments. (Second-order effects due to axial load in beam are neglected).

Sway Moment Magnification Factor δs ACI-​6.6.4.6.2 gives three approaches to computing the sway moment magnification factor, δs, as follows: 1. Q Method: Use a magnification factor δs as follows (ACI Formula 6.6.4.6.2a):

δs =



1 ≥ 1.0 1−Q

(13.10.2)

where Q is the story stability index given in ACI-​6.6.4.4.1 as Q=



∑ Pu ∆ 0 Vu Lc

(13.10.3)

where ΣPu = total vertical factored load in a story to be sway-​resisted by the frame action ∆0 = relative lateral deflection between the top and bottom of the story in question due to Vu, computed by using a first-​order elastic frame analysis and stiffness parameter values of ACI-​6.6.3.1.1 (see Section 13.8) Vu = factored shear in the story in question Lc = length of the compression member in question, measured from center to center of the joints in the frame When δs computed according to Eq. (13.10.2) exceeds 1.5 [i.e., when Q from Eq. (13.10.3) exceeds 1/3], either the more accurately computed sway magnifier of Eq. (13.10.4) (item 2 below) or a second-​order elastic analysis (item 3 below) must be used.

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489

2. Sum of P Method: The sway magnifier is computed according to ACI Formula (6.6.4.6.1b), as follows:

δs =



1 ≥ 1.0 ∑ Pu 1− 0.75 ∑ Pc

(13.10.4)

where Pc =



π 2 ( EI )eff (ksway Lu )2

(13.10.5)

and ΣPu = total vertical factored load in a story to be sway-​resisted by the frame action ΣPc = summation of critical buckling loads for all sway-​resisting columns ksway = effective length factor k computed for unbraced (sway) frames (see Section 13.11) Lu = unsupported length of the column taken as the clear length of the column (EI)eff = Eqs. (13.9.8), (13.9.9), or (13.9.10) with βdns substituted for βds βds = ratio of maximum factored sustained shear within a story to the maximum factored shear in that story associated with the same load combination. This factor is zero when sway moments are caused by loads of short duration, such as those induced by wind or earthquake. In the unusual case of sway moments that are due to sustained loads, such as those caused by lateral earth pressure or by appreciable sway due to sustained gravity loads βds will not be zero. 3. Use a second-​order elastic analysis based on the member stiffness parameters given in ACI-​6.6.3.1.1 (see Section 13.8). No magnification factor δs is used. The analysis gives the magnified moments directly. The typical resulting second-​order sway moments are shown by the dashed lines in Fig.  13.10.2(c). As shown, the moments are amplified along the length and at the ends, with the maximum values occurring at the member ends in this case.

Combination of Nonsway and Sway Moments In general, the total second-​order moment in the column may be computed, albeit approximately, by adding the nonsway and sway magnified bending moment diagrams along the member length as follows:

2nd M tot = δ ns M ns + δ s M s

(13.10.6)

where 2nd M tot  = total factored moment including second-order effects Mns, Ms = first-​order factored moments computed from a nonsway and a sway frame analysis, respectively δns, δs = nonsway and sway moment magnifiers, respectively

Equation (13.10.6) assumes that the moment magnification factor is the same along the length of the member, which is not true. Nonetheless, this approach provides in most cases a reasonable estimate of the approximate second-​order moment diagram for design purposes. Because most columns in sway frames do not have transverse loading and because of the presence of large double-​curvature moments due to lateral loads, the nonsway moment component including second-order effects is almost always maximum and equal to the

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first-​order moment at the column ends (i.e., δns = 1). As a result, the ACI Code requires the total moment including second-order effects to be computed as the summation of the nonsway and magnified sway moments at the column ends as follows (ACI-​6.6.4.6.1):

M12 nd = M1ns + δ s M1s

(13.10.7)



M 22 nd = M 2 ns + δ s M 2 s

(13.10.8)

where M12 nd, M 22 nd  = smaller and larger factored end moment (including second-​order effects), respectively. [Note that the ACI Code refers to these moments as M1 and M2 in ACI Formulas (6.6.4.6.1a) and 6.6.4.6.1b), respectively. Here, the superscript “2nd” emphasizes that these moments include second-​order effects, and serves to avoid confusion with M1 and M2 in ACI Formula (6.6.4.5.3a), and M2 in ACI Formula (6.6.4.5.1). The moments M1 and M2 in these last two ACI Code equations refer to first-​order moments.] M1ns, M2ns = smaller and larger, first-​order factored end moments, respectively, due to loads that cause no appreciable sidesway (i.e., computed from a nonsway frame analysis) M1s, M2s = smaller and larger first-​order factored end moments, respectively, due to loads that cause appreciable sidesway (i.e., computed from a sway frame analysis) δs = sway moment magnifier Equations (13.10.7) and (13.10.8) assume that the controlling combined effects for columns in sway frames occur at the ends of the members, as for the column of the frame of Fig. 13.10.2. This approach is reasonable when the column has no transverse loading along its length and when it is bent in double curvature, as discussed earlier. When the column is subjected to transverse loading along its length or, when the nonsway moments cause single curvature, the maximum, nonsway moment, including second-​ order effects, may occur along the length of the member rather than at the ends. Sway moments, on the other hand, are always maximum at the ends. In other words, the locations of the maximum magnified nonsway and sway moments do not coincide. In this case, a reasonable estimate of the maximum second-​order moment may be obtained by adding the second-​order nonsway and sway moments along the member length using Eq. (13.10.6), as suggested earlier. This approach is not entirely correct because the true moment amplification factor is not constant along the member length. Rather, it leads to conservative estimates of the maximum second-order moments. Otherwise, a second-​order analysis (see Section 13.12) should be used.

13.11 ALIGNMENT CHARTS FOR EFFECTIVE LENGTH FACTOR k The most commonly used procedure for obtaining the equivalent pin-​end (effective) length is to use the alignment charts from the Structural Stability Research Council Guide [13.27], originally developed by O. J. Julian and L. S. Lawrence, and presented in detail by T. C. Kavanagh [13.52]. The charts are shown in Fig. 13.11.1. The effective length factor k is a function of the end restraint factors ψA and ψB, for the top and bottom joints at the ends of the column, respectively, defined as

ψ=

∑ EI /L for the columns in the plane of bending ∑ EI /L for thee beams in the plane of bending

(13.11.1)

491



13.11  ALIGNMENT CHARTS FOR EFFECTIVE LENGTH FACTOR k

491

which for a hinged end gives ψ = ∞ and for a fixed end, ψ = 0. Since a frictionless hinge cannot exist in practical construction, ψ is to be taken equal to 10 for an end assumed as hinged in the analysis. Similarly, perfect fixity rarely, if ever, can be achieved in reality. Thus, it is recommended that ψ be taken equal to 1.0 for an assumed fixed end. The value to be used for L in computing the end restraint factors ψA and ψB in Eq. (13.11.1) should be the column or beam length measured from center to center of joints. One nomogram (or alignment chart), Fig.  13.11.1(a), is for braced (nonsway) frames where sidesway (joint translation) is prevented. The other nomogram, shown in Fig. 13.11.1(b), is for unbraced (sway) frames where sidesway is possible, being restrained only by the stiffness of interacting beams and columns. The assumptions inherent in the development of the alignment chart for braced frames [Fig. 13.11.1(a)] are as follows [13.52]: 1. All columns reach their respective critical loads simultaneously. 2. The structure is assumed to consist of symmetrical rectangular frames. 3. At any joint, the restraining moment provided by the beams is distributed among the columns in proportion to their stiffnesses. 4. The beams are elastically restrained at their ends by the columns; at the onset of buckling, the rotations of the beam at its ends are equal and opposite (i.e., the beams are deflected in single curvature). See Fig. 13.3.3(a) and (c). 5. The beams carry no axial loads. For the unbraced (sway) frame alignment chart [Fig. 13.11.1(b)], assumptions 1 through 3 and 5 are unchanged; however, the beams are assumed to be deflected in double curvature, where the rotations of the ends are equal in magnitude and direction [see Fig. 13.3.3(b) and (d)]. By means of the alignment charts, one may then determine the k factor for a column of constant cross section in a multistory, multibay frame. ψA ∞ 50.0 10.0 5.0 4.0 3.0 2.0

knonsway

ψB

1.0

0.9

ψA ∞ 50.0 10.0 5.0 4.0 3.0 2.0

0.8 1.0 0.9 0.8 0.7 0.6 0.5

0.7

1.0 0.9 0.8 0.7 0.6 0.5

0.4

0.4

0.3

0.3

0.2

0.6

∞ 100.0 50.0 30.0 20.0

ksway ∞ 20.0 10.0 5.0 4.0

ψB ∞ 100.0 50.0 30.0 20.0

10.0 9.0 8.0 7.0 6.0 5.0

3.0

10.0 9.0 8.0 7.0 6.0 5.0

4.0

2.0

4.0 3.0

3.0

2.0

2.0 1.5

0.2 1.0

1.0 0.1

0

0.1

0.5 (a) Sidesway prevented (braced frame)

0

0

1.0

0

(b) Sidesway not prevented (unbraced frame)

Σ EI /L,columns

Figure 13.11.1  Alignment charts for effective length factor for columns in continuous frames: ψ = . (From Σ EI /L, beams Ref. 13.52.)

492

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The effective length procedure has been adopted by ACI-​6.6.4.4.3 in the use of the moment magnifier method to evaluate slenderness effects. ACI-​6.6.4.4.3 says that for “nonsway members k shall be permitted to be taken as 1.0, and for sway members, k shall be at least 1.0.” Nonetheless, the use of the alignment charts for determining the k factor is implicitly endorsed by their inclusion in the ACI Commentary (ACI-​R6.6.4.4.3). With steel frames where the material is considered homogeneous and isotropic, the modulus of elasticity E is constant for all members, and the moment of inertia I is computed for the gross cross section. In reinforced concrete, E varies with concrete strength and magnitude of loading, while I also varies depending on the degree of cracking and the reinforcement percentage. ACI-​6.6.4.4.3 requires that the effective length factor k be determined using values of Ec in accordance with ACI-​19.2.2 and I given in ACI-​6.6.3.1.1 (see Table 13.8.1). As an alternative to using the nomograms of Figure 13.11.1, some approximate formulas for the effective length factor k have been proposed [13.54–​13.57]. For members in nonsway frames, the 1972 British Code of Standard Practice [13.54] suggests that an upper bound for k is obtained by using the smaller of the following two equations:

k = 0.7 + 0.05(ψ A + ψ B ) ≤ 1.0

(13.11.2)

k = 0.85 + 0.05ψ min ≤ 1.0

(13.11.3)

where ψA and ψB are the ψ values at the two ends of the member, and ψmin is the smaller of the two values. For members in sway frames, Furlong [13.55] proposed for members restrained at both ends: when ψavg < 2 (i.e., high-end restraint), k=



20 − ψ avg 20

1 + ψ avg

(13.11.4)

when ψavg ≥ 2 (i.e., moderate to low end restraint), k = 0.9 1 + ψ avg



(13.11.5)

where ψavg is the average of the ψ values at the two ends of the member. Equations (13.11.4) and (13.11.5) give k values that are within 2% of those obtained by the nomograms. For members in sway frames, when hinged at one end, the British Code of Standard Practice [13.54] proposes k = 2.0 + 0.3ψ



(13.11.6)

where ψ is the value at the restrained end. Hu, Zhou, King, Duan, and Chen [13.56] provide additional discussion on the effective length factor k, and Duan, King, and Chen [13.57] have proposed a partial-​fraction equation for k, which, they assert, achieves “both accuracy and simplicity for design … .” Their equation is as follows:

k = 1−

1 1 1 − − 5 + 9ψ A 5 + 9ψ B 10 + ψ A ψ B

(13.11.7)

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13.13  MINIMUM ECCENTRICITY IN DESIGN

493

13.12 SECOND-​O RDER ANALYSIS—​A CI CODE In a second-​order analysis, equilibrium is established in the final displaced position of the structure, ∆2u, shown by the dashed line in Fig. 13.6.1(b). Exact solutions of the problem are often too cumbersome for practical applications in design, and thus approximate solutions are used instead. The moment magnifier method presented in Sections 13.8 through 13.10 is one such approach. Alternatively, an approximate solution may be obtained by conducting several iterations of a first-​order analysis wherein the moment Vu Lc is incremented by ΣPu ∆0 in each iteration, or by using a bona fide second-​order-​analysis computer program. Today, a number of commercial computer programs allow approximate second-​order analyses of multistory frames. The reader is cautioned, however, to study and fully understand the assumptions and solution procedures used in these programs, which often apply to common, but not all, situations encountered in practice. In particular, the designer is cautioned to ensure that the software includes second-​order effects from both joint lateral end translation and the deflection between member ends. The ACI Code allows both elastic (ACI-​6.7) and inelastic (ACI-​6.8) second-​order analysis. In accordance with ACI-​6.7.1.1, an elastic second-​order analysis must include “the influence of axial loads, presence of cracked regions along the length of the member, and effects of load duration,” which are considered to be satisfied when the section properties are calculated in accordance with ACI-​6.6.3.1 (see Section 13.8). For an inelastic second-​order analysis, the model must take into account material nonlinearity, as well as the effects of “member curvature and lateral drift, duration of loads, shrinkage and creep, and interaction with the supporting foundation” (ACI-​6.8.1.1). Furthermore, the member dimensions used in the analysis must be “… within 10% of the specified member dimensions in construction documents…”; otherwise, the analysis must be repeated with the updated dimensions (ACI-​6.8.1.4). Also, the analysis procedure must “… have been shown to result in prediction of strength in substantial agreement with the results of comprehensive tests …” (ACI-​6.8.1.2). MacGregor and Hage [13.29], Wood, Beaulieu, and Adams [13.36, 13.37], Scholz [13.38, 13.39], and Furlong [13.58] have suggested various approaches to conducting second-​order analysis of reinforced concrete frames.

13.13 MINIMUM ECCENTRICITY IN DESIGN Nonsway Frames When computations indicate only a small moment to be acting on a member such that the eccentricity Mu /​Pu is less than (0.6 + 0.03h) in., the primary moment Mm in Eq. (13.9.1)3 is to be calculated as (ACI-6.6.4.5.4)

M m ,min = Pu (0.6 + 0.03h)

(13.13.1)4

In such cases, ACI-​6.6.4.5.4 requires Cm to be taken as 1.0 or computed from Eq. (13.9.6) based on the ratio of the computed first-​order end moments. Note that the minimum

3  As noted earlier, the ACI Code uses M2 instead of Mm in Eq. (13.9.1) [ACI Formula Eq. (6.6.4.5.1)]. Mm in Eq. (13.9.1) is the maximum moment occurring in the member in general, it may occur at either end or, if there is transverse loading, within the span. Mm is used here to distinguish it from the larger end moment, M2, in members with end moments only (see Section 13.5). 4  For SI with 15 and h in mm, ACI 318M-​14 gives Eq. (13.13.1) as



M m,min = Pu (15 + 0.03h)

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moment, when it governs, is to be applied about each axis separately, not as a case of biaxial bending.

Sway Frames Prior to the 1995 ACI Code, the minimum eccentricity requirement applied to members in both nonsway (braced) and sway (unbraced) frames. However, ACI-​6.6.4.5.4 applies only to members in nonsway (braced) frames. The ACI Commentary makes no reference to this change. In the design of sway (unbraced) frames, members of such frames only rarely have end moments smaller than that given by Eq. (13.13.1). The writers believe, however, that should an analysis give end moments less than that given by Eq. (13.13.1), the moment obtained from that equation should be used as the largest primary moment. When the minimum eccentricity requirements control, the provisions are intended to be applied to bending about only one axis at a time, not as a case of biaxial bending.

13.14 BIAXIAL BENDING AND AXIAL COMPRESSION With the advent of faster computers and the structural analysis software available today, three-​dimensional, elastic second-​order analyses have become more common in practice. Such analyses, if properly done, can adequately account for second-​order effects in both directions of bending. Alternatively, an estimate of the magnified moments may be obtained by considering the moment about each axis to be magnified by the factor δ computed from the restraint conditions for that axis. Note that, in general, the effective length factor k, the stiffness factor EI, and Cm will differ for each bending axis. Also, note that the member may be considered as part of a braced system in one direction and unbraced in the other. Additional discussion and design recommendations on biaxial bending of slender members may be found in Refs. 13.49 and 13.59, as well as Refs. 13.55 through 13.58, and 13.62.

13.15 ACI CODE—​S LENDERNESS RATIO LIMITATIONS Although the slenderness ratio is never zero in actual structures, there are certain limits for kLu/​r below which slenderness effects may reasonably be neglected. The ACI Code provisions are based on the assumption that a moment increase due to slenderness effects up to 5% can be tolerated without the designer having to consider these effects; thus a significant number of ordinary columns can be designed considering only the provisions of Chapter 10. ACI-​ASCE Committee 441 surveyed typical reinforced concrete buildings to determine the normal range of variables found in columns of such buildings, as reported by Mac Gregor, Breen, and Pfrang [13.32]. A  great variety of buildings were studied, including towers (braced frames) as high as 33 stories and an unbraced frame 20 stories high. The total number of columns exceeded 20,000. The following results were reported [13.32]. For braced frames, 98% of the columns had L /​h less than 12.5 (L /​r ≈ 42) and e/​h less than 0.64. For unbraced frames, 98% of the columns had L /​h less than 18 (L /​r ≈ 60) and e/​h less than 0.84. Furthermore, it was found that the practical upper limit on the slenderness ratio kL /​r is about 70 in building columns. In general, these limits provide a guide to the range of variables that are used in any approximate method. Taking the idea that a 5% increase in moment due to slenderness effects is acceptable, the effects of slenderness are permitted to be neglected under the following conditions.

495



495

13.17 EXAMPLES

Braced (Nonsway) Frame Members ACI-​6.2.5 permits neglect of slenderness effects when

kLu M ≤ 34 + 12 1ns r M 2 ns

(13.15.1)

where M1ns is the smaller and M2ns the larger of end moments on the member, and [34 + 12M1ns /​M2ns] ≤ 40. The subscript ns refers to nonsway (i.e., braced). For single-​curvature deformation, the ratio M1ns /​M2ns is negative, while for double curvature the ratio is positive. When the member is subject to large transverse loading (other than end moments alone), the ratio M1ns /​M2ns should probably be taken as –​1.

Unbraced (Sway) Frame Members ACI-​6.2.5 permits neglect of slenderness when

kLu < 22 r

(13.15.2)

Upper Limit on Second-​Order Moments When the limits given by Eqs. (13.15.1) or (13.15.2) are not satisfied, second-​order effects must be considered in accordance with the moment magnifier method of ACI-​6.6.4 (Section 13.8) or by conducting second-​order analyses in accordance with ACI-6.7 or ACI-​ 6.8 (Section 13.12). In no case, however, may the magnified moments exceed 1.4 times the computed primary or first-​order moments (ACI-​6.2.6). In such a case, the probability of stability failure is considered to be too high, so such a frame is not permitted and the structure must be redesigned.

13.16 AMPLIFICATION OF MOMENTS IN BEAMS When a second-​order analysis is conducted, the magnified column moments are readily balanced by those in the beams to satisfy equilibrium. When the moment magnifier method is used, however, the designer must recognize that the computed magnified moments acting at the ends of the column must be carried by the beams in a manner ensuring that equilibrium of the joint is maintained, as illustrated in Fig. 13.10.2(d). For this reason, ACI-​6.6.4.6.3 requires that flexural members (beams) “be designed for the total magnified end moments of the columns at the joint.”

13.17 EXAMPLES The preceding sections of this chapter discussed basic concepts underlying the ACI Code provisions relating to slenderness effects on columns. To facilitate easy reference to the procedures and formulas, most of which appear in the ACI Code or Commentary, Table 13.17.1 provides a summary of the most common formulas needed for solving practical problems. Examples 13.17.1 through 13.17.5 illustrate the use of these formulas.

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TABLE 13.17.1  SLENDERNESS EFFECTS ON COLUMNS—​SUMMARY OF USEFUL FORMULAS 1. Section Properties to Be Used in First-​Order Analysis Ec = 57, 000 fc′ for normal-​weight concrete Ec = wc1.5 33 fc′ for wc between 90 and 155 lb/​cu ft Moment of inertia [to be divided by (1 + βds) for sustained lateral loads] = 0.35Ig for beams and cracked walls = 0.70Ig for columns and uncracked walls = 0.25Ig for flat plates and flat slabs Area = 1.0Ag 2. Radius of Gyration r = 0.30h for rectangular sections r = 0.25h for circular sections 3. Nonsway (Braced) Frames (a) Slenderness effects may be ignored if kLu M ≤ 34 + 12 1ns r M 2 ns M1ns /​M2ns is negative for single curvature, and the kLu  /​r limit is not to be greater than 40. (b) M  ay be assumed as braced if second-​order end moments ≤ [1.05 × (first-​order end moments)], or if [Q = ∑ Pu ∆ 0 / (Vu Lc )] ≤ 0.05 (c) Moment magnifier δns Mmax = δnsMm

δ ns =

Cm ≥ 1.0; Pu 1− 0.75Pc

Cm = 0.6 − 0.4

Pc =

π 2 ( EI )eff (knonsway Lu )2

M1ns M , for members with end moments only and 1ns is negative M 2 ns M 2 ns

for single curvature Cm = 1.0 for members with transverse loads (EI)eff computed from Eqs. (13.9.8), (13.9.9), or (13.9.10) Mm,min = Pu(0.6 + 0.03h) 4. Sway (Unbraced) Frames (a) Slenderness effects may be ignored if  (b) Moment magnifier δs

kLu ≤ 22 r

M12 nd = M1ns + δ s M1s M 22 nd = M1ns + δ s M1s

δs =

ΣP ∆ 1 1 and Q = u 0 for Q ≤ or δ s ≤ 1.5 only 1−Q Vu Lc 3

or

δs =

1 π 2 ( EI )eff and Pc = ΣPu (ksway Lu )2 1− 0.75ΣPc

( EI )eff computed from Eqs. (13.9.8), (13.9.9), or (13.9.10) with β ds substituted for β dns

497



497

13.17 EXAMPLES

EXAMPLE 13.17.1 Use the moment magnifier method to compute the required strength in flexure, Mu, and axial load, Pu, for the columns and the beam of the frame shown in Fig. 13.17.1. The frame is one of several identical frames in the structure and it supports uniformly distributed as well as concentrated service-​level dead and live loads, as shown. In addition, a strength-​ level wind load is assumed to be acting at the top of the frame. Assume that only the dead loads are sustained. Preliminary dimensions for the columns and the beam are as shown. The beam is restrained against out-​of-​plane movement by prefabricated elements that are simply supported on the top of the beam. There are no walls or any other bracing elements in the building in the direction of the frame. Use fc′ = 5, 000 psi and Grade 60 steel. PD = 130k PL = 50k

6.25’

PD = 160k PL = 70k

PD = 100k PL = 50k wD = 1 k/ft wL = 0.8 k/ft

W = 25k 16” × 26” beam 20” × 20” columns

12’

25’

Figure 13.17.1  Unbraced frame of Example 13.17.1.

SOLUTION (a) Slenderness ratio limits. The first step is to determine whether second-​order effects need to be considered. Because the frame is permitted to sway (i.e., there are no walls or other bracing elements), the columns are not braced against sidesway. In such a case, ACI-​6.2.5 permits neglect of slenderness effects when kLu ≤ 22 (13.15.2) r



The unsupported height, Lu, of both columns is  1 Lu = 12 − 26   = 9.83 ft.  12 



The effective length factor, k, may be computed using the alignment charts shown in Fig. 13.11.1. This requires the computation of the end restraint factors, ψ, at the top and bottom of the columns from Eq. (13.11.1). For the purpose of computing the end restraint factors, the moments of inertia given in ACI-​6.6.3.1.1 (see Table 13.8.1) may be used. Therefore,



16(26)3 = 8202 Ec kip-in.2 12 20 4 2 = Ec (0.70 I g ) = Ec (0.7) = 9333Ec kip-in. 12

( EI )beam = Ec (0.35I g ) = Ec (0.35) ( EI )col

(Continued)

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Example 13.17.1 (Continued) End restraint factors,

ψ A (top) =



(9333)Ec /12 ∑ EI /L for cols = = 2.4 ∑ EI /L for beams (8202)Ec / 255

Since the concrete strength is assumed to be the same for the columns and the beam in this example, the value of the modulus of elasticity is not needed for the computation of the end restraint factor. Also note that the length L for the columns and the beam are taken as the center-​to-​center distance for this calculation. At the bottom, the columns are pinned. Thus, ψB (bottom), will be taken equal to the upper practical value of 10. Entering the alignment chart for an unbraced frame (i.e., sidesway not prevented) of Fig. 13.11.1(b), the effective length factor is estimated as ksway ≈ 2.2



Taking the radius of gyration, r, as 0.3 times the depth of the column per ACI-​6.2.5.1(b), the slenderness ratio for the columns is ksway Lu



r

=

2.2(9.83)12 = 43.3 > 22 0.3(20)

Therefore, the columns are slender and second-​order effects must be considered. (b) The frame is subjected to dead, live, and wind loads. Seven load combinations must be considered in design (Section 2.7). In this example, however, only the following three load combinations will be considered 1. U = 1.4D [2.7.1] 2. U = 1.2D + 1.6L [2.7.2] 3. U = 1.2D + 1.0W + 0.5L [2.7.4] where the load factor on the live load L has been reduced to 0.5 in Eq. (2.7.4) as permitted for areas that are not used as garages, places of public assembly, or areas that do not sustain live loads greater than 100 lbs/​sq ft. 1. Load combination: U = 1.4D. (i) Nonsway or Sway Frame Analysis For a first-​order, factored analysis, the ACI Code requires that the section properties given in ACI-​6.6.3.1.1 and the modulus of elasticity, Ec, given by ACI-​19.2.2, be used (see Table 13.8.1). Thus, for  fc′ = 5000 psi,

Ec = 57, 000 5000

1 1000

= 4030 ksi

and from part (a)

( EI )beam = 8202 Ec = 8202(4030) = 33, 050, 000 kip-in.2 ( EI )col = 9333Ec = 9333(4030) = 37, 610, 000 kip-in.2

In ACI 318-​14, columns are permitted to be analyzed as nonsway when (a) the increase in column end moments due to second-​order effects does not exceed 5% of the first-​ order end moments or (b) the story stability index, Q, is less than or equal to 0.05 (ACI-​ 6.6.4.3). At this point, the second-​order moments have not been computed; thus the story stability index, Q, will be used to check this requirement. (Continued)

49



499

13.17 EXAMPLES

Example 13.17.1 (Continued) The stability index, Q, is computed using Eq. (13.8.1) Q=



∑ Pu ∆ 0 [13.8.1] Vu Lc

where ΣPu = total vertical factored load in a story to be sway-​resisted by the frame action Δ0 = relative lateral deflection between the top and bottom of the story in question due to Vu, computed using a first-​order elastic frame analysis and section properties of ACI-​6.6.3.1.1. Vu = factored shear in the story in question Lc = length of the compression member in question, measured from center to center of the joints in the frame It is noted that the stability index, Q, is a function of the ratio Δ0 /​Vu, in other words, the lateral flexibility of the frame. Therefore, Q can be computed using any lateral load that produces the shear, Vu, and the corresponding story deflection, Δ0. In general, to compute Q in a multistory frame, an arbitrary lateral load is applied at the top of the story being considered while the bottom of the story is restrained from lateral movement. No other loads are applied for the computation of Q. In this example, a lateral load of 10 kips will be applied at the top of the columns. The current ACI Code is unclear as to how to consider the effects of sustained loads for computing the stability index, Q. Because a lateral load analysis is required to compute the story deflection, Δ0, it may be assumed that ACI-​6.6.3.1.1 (or ACI-​6.6.3.1.2) would apply. ACI-​6.6.3.1.1 requires that the moment of inertia of compression members be divided by (1 + βds) if sustained lateral loads, such as those induced by earth pressure, are present. In this example, there are no sustained lateral loads according to this definition, although the sustained dead load does induce sway (asymmetric loading) and could potentially induce lateral instability of the frame. In past editions of the ACI Code (2005 and earlier editions), it was required that the moments of inertia of all members be reduced for stability checks involving the stability index, Q. In this example, the stiffness parameter EI of all members will be divided by (1 + βds) to allow consideration of the possible instability and moment magnification due to sway of the frame under the sustained dead loads. This approach is conservative, but reasonable. For this load combination, all the load inducing sway is sustained (i.e., βds = βdns =1), and thus, the EI values for the beam and the columns computed earlier will all be divided by (1 + βds) = 2. Thus, ( EI )beam =

33, 050, 000 = 16, 525, 000 kips-in.2 2

( EI )col =

37, 610, 000 = 18, 805, 000 kip-in.2 2





Analysis of the frame using these properties and a lateral load of 10 kips yielded a relative lateral deflection Δ0 of 0.58 in. The total vertical factored load under this load combination is

∑ Pu = 1.4[130 + 100 + 160 + 1(25)] = 581 kips



Thus,

Q=

∑ Pu ∆ 0 581(0.58) = = 0.23 > 0.05 Vu Lc 10(12)12

Since the stability index, Q, exceeds 0.05, the story cannot be treated as nonsway and the frame must be treated as a sway frame. (Continued)

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Example 13.17.1 (Continued) (ii) Computation of Nonsway Moments, Mns The nonsway moments are computed by conducting a first-​order analysis of the frame with lateral end translation prevented [Fig. 13.10.1(b)] and subject to the factored loads acting in this load combination (i.e., the factored dead load). The nonsway moment diagrams are shown in Fig. 13.17.2. Because of the asymmetry of the loading, a horizontal reaction of 14.7 kips is calculated at the top. 401ft-k

224ft-k

14.7 k

382ft-k

left

Pns = 312 k

right

Pns = 269 k

Figure 13.17.2  Nonsway moments, Mns, for load combination U = 1.4D of Example 13.17.1.

(iii) Computation of Sway Moments, Ms To compute the sway moments, a first-​order analysis of the unbraced (sway) frame is computed by applying a lateral load equal to the horizontal reaction [Fig. 13.10.1(c)] of 14.7 kips computed in (ii) above. No other loads are applied. The calculated sway bending moment diagrams are shown in Fig. 13.17.3. 88ft-k

left

Ps

= 7.1k

88ft-k

right

Ps

= 7.1k

Figure 13.17.3  Sway moments, Ms, for load combination U = 1.4D of Example 13.17.1.

(iv) Nonsway Magnifiers, δns Because the columns do not have transverse loading and because the moment at the base is zero, the maximum moment will be equal to the first-​order moment at the top of the columns. Thus, there is no need to compute the nonsway magnification factor. In this (Continued)

501



501

13.17 EXAMPLES

Example 13.17.1 (Continued) example, however, the nonsway magnifier will be computed anyway to illustrate the procedure. The nonsway magnification factor, δns, is given by ACI-​6.6.4.5.2 as

δ ns =



Cm ≥ 1.0 [13.9.4] Pu 1− 0.75Pc

This factor is a member factor and must be computed individually for each column. Left Column  • Factor Cm The column does not have transverse loading and is subjected to end moments only. Accordingly, the factor Cm is computed from Eq. (13.9.6) Cm = 0.6 − 0.4



M1ns [13.9.6] M 2 ns

where M1ns is the smaller of the nonsway, first-​order end moments. Because the columns are pinned at the base, M1ns is zero and Cm is computed as 0.6 in this case. • Factored axial load, Pu The factored axial load is the total factored axial load acting on the column for the load combination being considered. Accordingly,

Puleft = Pnsleft + Psleft = 312 − 7.1 = 305 kips

• Critical buckling load, Pc The critical buckling load is computed in accordance with ACI-​6.6.4.4.2 as Pc =



π 2 ( EI )eff [13.9.7] (knonsway Lu )2

where the stiffness parameter, (EI )eff, may be computed using any of Eqs. (13.9.8), (13.9.9), or (13.910). Since the amount of reinforcement has not been computed yet, Eq. (13.9.9) cannot be used at this stage. Here, (EI )eff will be computed using Eq. (13.9.8) to illustrate the procedure. Thus, ( EI )eff =



0.4 Ec I g (1 + β dns )

=

0.4(4030)13, 333 = 10, 750, 000 kip-in.2 (1 + 1)

where βdns was taken equal to 1 since all of the load is sustained for the load combina­ tion being considered. Therefore, the critical buckling load Pc for the left column is Pcleft =



π 2 ( EI )eff π 2 (10, 750, 000) = = 9000 kips 2 (knonsway Lu ) [0.92(9.83)12]2

where the effective length factor, knonsway, was computed using the end restraint factors ψA (top) and ψB (bottom) computed in part (a), and the alignment chart for braced (sidesway prevented) frames of Fig. 13.11.1(a). Thus,

δ left ns =

0.6 = 0.63 < 1 305 1− 0.75(9000)

(Continued)

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Example 13.17.1 (Continued) Therefore, δ left and the maximum nonsway moment for the left column occurs at the ns = 1 end and is equal to the first-​order moment at the top. Right Column  The procedure for the right column is the same as that used for the left column. Because both columns have the same dimensions, have no transverse loading, have the same shape of the bending moment diagram, and the same support conditions at the ends, the parameters involved in the computation of the nonsway magnifier are all the same except for the level of axial load level acting on the column. The following are the parameter values for the right column. • Factor Cm

Cm = 0.6 (column with end moments only and M1ns = 0)

• Factored axial load Pu:

Puright = Pnsright + Psright = 269 + 7.1 = 276 kips

• Critical buckling load Pc:



Pcright = Pcleft = 9000 kips

Thus,

δ right = ns



0.6 = 0.63 < 1 276 1− 0.75(9000)

Therefore, the maximum moment also occurs at the end (top) for the right column. (v) Sway Moment Magnifier δs The sway moment magnifier δs is computed in accordance to ACI-​6.6.4.6.2 using Eq. (13.10.2), (13.10.4), or a second-​order elastic analysis. It is noted that the sway moment magnifier is a story amplification factor and it is the same for all of the columns in a given story. • Q method [ACI-​6.6.4.6.2(a)] In this method, the sway moment magnifier is computed by using Eq. (13.10.2):

δs =



1 ≥ 1.0 [13.10.2] 1−Q

Using the stability index Q of 0.23 computed above [see part 1(i)],

δs =



1 = 1.30 1 − 0.23

This value is rather large, but less than 1.5 and thus, it is acceptable per ACI-​6.6.4.6.2. • Sum of P method [ACI-​6.6.4.6.2(b)] In this approach, the sway moment magnifier is computed using Eq. (13.10.4) as

δs =

1 ≥ 1.0 [13.10.4] ∑ Pu 1− 0.75 ∑ Pc

where

Pc =

π 2 ( EI )eff [13.10.5] (ksway Lu )2 (Continued)

503



503

13.17 EXAMPLES

Example 13.17.1 (Continued) where the stiffness parameter (EI )eff may be computed using Eq. (13.9.8), (13.9.9), or (13.9.10) with βdns substituted for βds. Here, (EI )eff will be computed using Eqs. (13.9.8). Since both columns have the same cross section, (EI )eff will be the same. Thus, using Eq. (13.9.8),

( EI )eff =

0.4 Ec I g (1 + β ds )

=

0.4(4030)13, 333 = 10, 750, 000 kip-in.2 (1 + 1)

where βds was taken equal to 1 because the sway moments are all produced by the sustained dead load. Therefore, the critical buckling load Pc is for both columns

Pc =

π 2 ( EI )eff π 2 (10, 750, 000) = = 1575 kiips (ksway Lu )2 [2.2(9.83)12]2

Therefore, ΣPc = 1575 + 1575 = 3150 kips. The total vertical factored axial load was computed earlier in part 1(i) as ΣPu = 581 kips. Thus, 1 δs = = 1.33 581 1− 0.75(3150) It is seen that in this case, the Q method and the sum of P method yield very similar values for the sway moment magnifier δs. (vi) Combination of Nonsway and Sway Moments The total second-​order moment is computed as the summation of the nonsway and sway moments at the column ends using Eqs. (13.10.7) and (13.10.8). The moment at the bottom of both columns is zero (pin). At the top, the maximum second-​order moment is computed as

M 22 nd = M 2 ns + δ s M 2 s [13.10.8]

The moment magnifier obtained with the Q method (i.e., δs = 1.30) will be used in the following calculations. A comparison of the moments obtained with the two values for the moment magnifier δs is presented later (see Table 13.17.2). Left Column 



M 22 nd = 401 + 1.3( −88) = 287 ft-kips

For this column, the nonsway and sway moments counteract each other. Therefore, the second-​order moment thus computed is smaller than the first-​order moment. It is nonconservative and inappropriate to use the value computed above. For design purposes, the larger, first-​order moment should be used, that is,

M 2 = 401 + ( −88) = 313 ft - kips



Thus, for this load combination, the column should be designed for

Pu = 305 kips and Mu = 313 ft-kips



Right Column  For the right column, the nonsway and sway moments are additive. Thus,

M 22 nd = 224 + 1.3 (88) = 338 ft-kips

(Continued)

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Example 13.17.1 (Continued) and the column should be designed for Pu = 276 kips



and Mu = 338 ft-kips



2. Load combination: U = 1.2D +1.6L. The approach to computing the second-​order moments is the same as that for the preceding load combination; thus only the differences in the calculations are described next. (i) Nonsway or Sway Frame Analysis For this load combination, the stability index Q is computed using Eq. (13.8.1) as above Q=



∑ Pu ∆ 0 [13.8.1] Vu Lc

where that lateral deflection of the frame will be computed with the stiffness parameter EI divided by (1 + β ds ) for all members to account for the effects of sustained loads. Since this load combination includes gravity loading only, the amount of sustained load inducing sway will be in direct proportion of the ratio between total sustained vertical load and the total vertical load (i.e., βdns = βds). The total vertical factored load under this load combination is ∑ Pu = 1.2[130 + 100 + 160 + 1(25)] +1.6[50 + 50 + 70 + 0.8(25)] = 498 + 304



∑ Pu = 802 kips The total vertical factored sustained load (i.e., the dead load) under this load combination is ∑ Psustained = 1.2[130 + 100 + 160 + 1(25)] = 498 kips Thus,

βdns =

498 = 0.62 [ = β ds in this case] 802

and (1 + β ds ) = 1.62. The revised EI values are 33, 050, 000 = 20, 400, 000 kip-in.2 1.62 37, 610, 000 = 23, 220, 000 kip-in.2 = 1.62

( EI )beam = ( EI )col

Analysis of the frame using these properties and a lateral load of 10 kips yields a relative lateral deflection, ∆0, of 0.47 in. [= 0.58(1.62)/2]. Thus,

Q=

∑ Pu ∆ 0 802(0.47) = = 0.26 > 0.05 Vu Lc 10(12)12

The stability index, Q, exceeds 0.05, and thus the story must be treated as a sway frame. (Continued)

50



505

13.17 EXAMPLES

Example 13.17.1 (Continued) (ii) Computation of Nonsway Moments, Mns The nonsway moments computed from first-​order analysis of the frame subject to all the loads acting in this load combination and restrained against lateral movement are shown in Fig. 13.17.4. Because of the asymmetry of the loading, a horizontal reaction of 20.9 kips is calculated at the top. The axial forces acting on each column are also shown. 591ft-k

340 ft-k

20.9 k

554ft-k

left

Pns = 427k

right

Pns

= 375k

Figure 13.17.4  Nonsway moments, Mns, for load combination U = (1.2D + 1.6L) of Example 13.17.1.

(iii) Computation of Sway Moments, Ms To compute the sway moments, the unbraced (sway) frame is subjected to a first-​order analysis by applying a lateral load equal to the horizontal reaction of 20.9 kips computed in (ii) above. No other loads are applied. The calculated sway bending moment diagrams are shown in Fig. 13.17.5. 126 ft-k

126 ft-k

left

Ps = 10.1 k

right

Ps

= 10.1 k

Figure 13.17.5  Sway moments, Ms, for load combination U = (1.2D + 1.6L) of Example 13.17.1.

(iv) Nonsway Magnifiers, δns As shown for the preceding load combination, the nonsway magnifier will be less than one because the columns do not have transverse loading and because the moment at the base is zero. Thus, the nonsway moments will occur at the top of the columns and will be equal to the first-​order moments shown in Fig. 13.17.4. (Continued)

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Example 13.17.1 (Continued) (v) Sway Moment Magnifier δs The sway moment magnifier δs is computed in the same manner as before. The main difference in the computed values is due to the different β dns to account for the effects of sustained loads. • Q method [ACI-​6.6.4.6.2(a)] Using the stability index Q of 0.26 computed in part 2(i)

δs =



1 = 1.35 1 − 0.26

This value is also rather large, but it is less than 1.5, thus it is acceptable per ACI-​6.6.4.6.2. • Sum of P Method [ACI-​6.6.4.6.2(b)] In this approach, the sway moment magnifier is computed using Eq. (13.10.4) as

δs =

1 ≥ 1.0 ∑ Pu 1− 0.75 ∑ Pc

[13.10.4]

where Pc =



π 2 ( EI )eff (ksway Lu )2

[13.10.5]

Using Eq. (13.9.8) for (EI )eff

( EI ) eff =

0.4 Ec I g (1 + β ds )

=

0.4(4030)13, 333 = 13, 270, 000 kip-in.2 (1 + 0.62)

Therefore, the critical buckling load Pc is for both columns

Pc =

π 2 ( EI )eff π 2 (13, 270, 000) = = 1945 kiips (ksway Lu )2 [2.2(9.83)12]2

and ΣPc = 2(1945) = 3890 kips. The total vertical factored axial load was computed earlier in part 2(i) as ΣPu = 802 kips. Thus,

δs =

1 = 1.38 802 1− 0.75(3890)

Again, both the Q method and the sum of P method yield similar values. (vi) Combination of Nonsway and Sway Moments The moment magnifier obtained with the Q method (i.e., δs = 1.35) is used in the following calculations. At the top, the maximum second-​order moment is computed as

M 22 nd = M 2 ns + δ s M 2 s [13.10.8]

Left Column  For this column, the nonsway and sway moments counteract each other. For design purposes, the larger, first-​order moment without the sway magnifier should be used, that is,

M 2 = 591 + ( −126) = 465 ft-kips

(Continued)

507



507

13.17 EXAMPLES

Example 13.17.1 (Continued) Thus, for this load combination, the column should be designed for

Pu = 427 − 10.1 = 417 kips and Mu = 465 ft-kips

Right Column  For the right column, the nonsway and sway moments are additive. Thus,

M 22 nd = 340 + 1.35(126) = 510 ft-kips



and the column should be designed for

Pu = 375 + 10.1 = 385 kips and Mu = 510 ft-kips



3. Load combination: U = 1.2D +1.0W + 0.5L. For this load combination, two cases need to be considered: wind load acting to the right and wind load acting to the left. The stability index, Q, and the moment magnifiers are a function of the stiffness parameter EI and the applied loads for the load combination being considered; both are the same and are independent of the wind direction. The procedure will be illustrated in detail for wind acting to the right; the approach for wind acting to the left is the same. (i) Nonsway or Sway Frame Analysis. Most of the load-inducing sway of the frame in this load combination is due to wind, which is not sustained. Whether the member stiffness should be reduced to include the sway component due the sustained dead load in this load combination is somewhat a matter of opinion. In this example, the stiffness parameter EI is divided by (1 + βds) to account for the sway component due to the sustained dead loads. This approach is conservative. The horizontal reaction at the top of the frame obtained from a nonsway analysis under 1.2D alone is computed as 12.6 kips. This will be considered as sustained. On the other hand, the horizontal reaction obtained for the load combination being considered (1.2D +1.0W + 0.5L) is computed as 40 kips (see Fig. 13.17.6). Thus,

β ds =



12.6 = 0.32 40

and (1 + βds) = 1.32. The revised EI values are

( EI )beam =

33, 050, 000 = 25, 040, 000 kip-in.2 1.32

( EI )col =

37, 610, 000 = 28, 500, 000 kip-in.2 1.32

Analysis of the frame using these properties and a lateral load of 40 kips yields a relative lateral deflection, Δ0, of 1.52 in. [or (0.58(1.32)/2) × 4] The total vertical factored load under this load combination is ∑ Pu = 1.2[130 + 100 + 160 + 1(25)]

+ 0.5[50 + 50 + 70 + 0.8(25)] = 498 + 95 ∑ Pu = 593 kips

(Continued)

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Example 13.17.1 (Continued) Thus, Q=



∑ Pu ∆ 0 593(1.52) = = 0.16 > 0.05 Vu Lc 40(12)12

The stability index, Q, exceeds 0.05, and thus the story must be treated as a sway frame, as was expected for this load combination including wind load. (ii) Computation of Nonsway Moments, Mns The nonsway moments computed from first-​order analysis of the frame subject to all the loads acting in this load combination and restrained against lateral translation are shown in Fig. 13.17.6. The axial forces acting on each column, as well as the lateral restraining force, are also shown. 418 ft-k

238 ft-k

40.0 k

401 ft-k

left

Pns = 317 k

right

Pns = 276 k

Figure 13.17.6  Nonsway moments, Mns, for load combination U = (1.2D + 1.0W + 0.5L) of Example 13.17.1.

(iii) Computation of Sway Moments Ms The sway moments obtained from a first-​order analysis of the unbraced (sway) frame under a lateral load equal to the horizontal reaction of 40.0 kips are shown in Fig. 13.17.7. 240 ft-k

left

Ps

= 19.2 k

240 ft-k

right

Ps

= 19.2 k

Figure 13.17.7  Sway moments, Ms, for load combination U = (1.2D + 1.0W +0.5L) of Example 13.17.1.

(Continued)

509



509

13.17 EXAMPLES

Example 13.17.1 (Continued) (iv) Nonsway Magnifiers, δns As before, it can be shown that the nonsway magnifier will be less than one for this load combination. Thus, the nonsway moments will occur at the top of the columns and will be equal to the first-​order moments shown in Fig. 13.17.6. (v) Sway Moment Magnifier, δs • Q method [ACI-​6.6.4.6.2(a)] Using the stability index Q of 0.16 computed in part 3(i)

δs =



1 = 1.19 1 − 0.16

which is less than 1.5 and thus, acceptable per ACI-​6.6.4.6.2. • Sum of P method [ACI-​6.6.4.6.2(b)] Using Eq. (13.9.8) for (EI )eff ( EI )eff =



0.4 Ec I g (1 + β ds )

=

0.4(4030)13, 333 = 16, 280, 000 kip-in.2 (1 + 0.32)

Therefore, the critical buckling load Pc is for both columns Pc =



π 2 ( EI )eff π 2 (16, 280, 000) = = 2390 kiips (ksway Lu )2 [2.2(9.83)12]2

Thus, ΣPc = 2(2390) = 4780 kips. From part 3(i), the total vertical factored axial load under this load combination is ΣPu = 593 kips. Thus,

δs =

1 = 1.20 593 1− 0.75(4780)

(vi) Combination of Nonsway and Sway Moments The moment magnifier obtained with the Q method (i.e., δs = 1.19) will be used in the following calculations. Left Column  The nonsway and sway moments counteract each other for this column. For design purposes, the larger, first-​order moment without the sway magnifier should be used, that is,

M 2 = 418 + ( −240) = 178 ft-kips

Thus, for this load combination, the column should be designed for

Pu = 317 − 19.2 = 298 kips and Mu = 178 ft-kips

Right Column  For the right column, the nonsway and sway moments are additive. Thus,

M 22 nd = 238 + 1.19(240) = 524 ft-kips

(Continued)

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Example 13.17.1 (Continued) and the column should be designed for

Pu = 276 + 19.2 = 295 kips Mu = 524 ft-kips

(c) Summary of Results. Table  13.17.2 summarizes the first-​order and second-​order moments computed at the top of the columns using the moment magnifier method for this example problem. The second-​ to-​ last column in the table shows the moments obtained from an elastic second-​order analysis of the frame that included moment magnification due to deflections between member ends and due to lateral end translation. The section properties used in these analyses were the same as those used to compute the stability index Q for each load combination and accounted for the effect of sustained loads using the same procedure to compute the moment magnifiers. It can be seen that the Q method and Sum of P method yield very similar moments, as was expected since the calculated magnifiers were very close. It can also be seen that the column moments computed by using the moment magnifier approach, as presented here, are in very good agreement with those computed from an elastic second-​order analysis. The last column in Table 13.17.2 shows the ratio between the second-​order and first-​ order moments computed for the columns in each load combination: the second-​order moments are at most 1.1 times the first-​order moments, and thus the upper limit of 1.4 per the requirements of ACI-​6.2.6 is satisfied (see Section 13.15). (d) Required Strengths Columns  For each factored load combination, the maximum moment, Mu, occurring concurrently with the axial force, Pu, must be investigated. In no case, however, should Mu be taken less than the first-​order moment computed for that load combination. The ACI Code does not require columns in sway frames to be designed for minimum eccentricity (see Section 13.13), but it is instructive to compare the calculated moments with the minimum moment of ACI-​6.6.4.5.4:

M m ,min = Pu (0.6 + 0.03h)

[13.13.1]

In this example, h = 20 in., and thus

M m ,min = Pu [0.6 + 0.03(20)] = 1.2 Pu [ kip-in.]

The maximum computed axial force is 417 kips on the left column under the load combination U = 1.2D + 1.6L. In this case,

M m,min = 1.2(417) = 500 kip-in = 42 ft-kips

which is much less than any of the first-order moments computed for either the left or the right column and thus does not govern. Though not required in this example, the design interaction diagram for a 20 by 20 in. column reinforced with 4–​#11 corner bars and 8–​#10 side bars (two in each face) is shown in Fig. 13.17.8. It can be seen that the selected reinforcement satisfies the required strengths (Pu, Mu) [boldface in Table  13.17.2] for the load combinations considered. (Continued)

a–​c

478 178

1.19

1.19

1.20

1.20

1.38

Columns (2), (4), magnifier and corresponding second-​order moment computed from Eq. (13.10.4) and (EI)eff from Eq. (13.9.8) [Sum of P method]

60 −60

178 478

1.35

Boldface type indicates required strengths (Pu and Mu) for the load combination considered based on second-order moments computed using the Q method

418 238

322 271 593

−240 240

465 466

1.33

(2)

c

Left column Right column ΣPu =

1.2D+0.5L − 1.0W

418 238

298 295 593

−126 126

1.30

(1)

Columns (1), (3), magnifier and corresponding second-​order moment computed from Eq. (13.10.2) [Q method]

Left column Right column ΣPu =

1.2D+0.5L +1.0W

591 340

417 385 802

313 312

Mns + Ms

Sway Magnifiers, δs

b

Left column Right column ΣPu =

1.2D+1.6L

401 224

305 276 581 −88 88

Ms

Mns

Pu(kips)

Mtot

a

Left column Right column ΣPu =

1.4D

Load Combination

Sway

Nonsway

489 167

132 524

421 510

287 338

(3)

490 166

130 526

417 514

284 341

(4)

Mns + δs Ms

Moment Magnifier Method, Mu2nd

488 167

133 523

421 509

285 339

2nd Order Elastic

(Continued)

1.02 0.94

0.75 1.09

0.91 1.09

0.91 1.09

Mu2 nd Mu1st

Second-​Order Moments (ft-​kips)



First-​Order Moments (ft-​kips)

TABLE 13.17.2  FIRST-​AND SECOND-​ORDER MOMENTS FOR THE COLUMNS OF EXAMPLE 13.17.1

51

13.17 EXAMPLES 511

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Example 13.17.1 (Continued) Beam  For the frame of this example, the moments at the beam ends must be in equilibrium with the first-​and second-​order moments at the top of the columns. For this reason, ACI-​6.6.4.6.3 requires that flexural members in sway frames be designed for the total magnified moment of the columns at the joint. Therefore, the beam would have to be designed for a negative moment of 489 ft-​kip at the left end and a negative moment of 524 ft-​kip at the right end. φPn, Pu (kips) 1800

Required strengths: left column right column

1600

4 – #11 (corners) 8 – #10 ρg = 4.1%

1400 1200 fc’ = 5000 psi fy = 60,000 ksi

1000 800 600 400 200

0

200

100

300

400

500

600 φMn, Mu (ft-kips)

Figure 13.17.8  Design interaction diagram for the columns of Example 13.17.1.

EXAMPLE 13.17.2 Determine the adequacy of the interior top floor column (column A) of the braced frame of Fig. 13.17.9. The column is 10 × 10 with 4–​#8 and 4–​#9 bars (  fy = 60 ksi and fc′ = 3 ksi) and is to carry a service axial compression of 106 kips live load and 39 kips dead load. The bending moments that may act in combination with the axial load have been computed and found to be negligible. If the member is not adequate, revise the design so that it satisfies the moment magnifier method of the ACI Code. Wall

12 × 20

12 × 12 10 × 10

12 × 20

A

12 × 20 12 × 12 10 × 10

3 @ 12’–0”

12 × 20

10”

#3 tie 4 @ 24’–0” (a)

Figure 13.17.9  Frame of Example 13.17.2.

1.5 in. clear cover to tie

10”

4 – #9 (corner bars) 4 – #8 (b)

(Continued)

513



513

13.17 EXAMPLES

Example 13.17.2 (Continued) SOLUTION (a) Determine the slenderness ratio. Unless an evaluation of end restraint is made, ACI-​ 6.6.4.4.3 permits taking the effective length factor k for a braced frame equal to 1.0. The radius of gyration may be taken as 0.3h according to ACI-​6.2.5.1(b). The unsupported height Lu is  1 Lu = 12 − 20   = 10.33 ft.  12 

Then

kLu 1.0(10.33)(12) = = 41.3 r 0.3(10)



(b) Slenderness ratio limits. Since the end moments are negligible, the minimum eccentricity provisions of ACI-​6.6.4.5.4 govern the design. Accordingly, the deformation should be considered as single curvature with M1ns /​M2ns = –​1.0. The slenderness limit for a braced frame is [Eq. (13.15.1)]

 M   kLu  = 34 + 12 1ns  = [34 − 12] = 22  r  M 2 ns   limit

which is less than 41.3. Thus slenderness effects must be considered. (c) Braced frame moment magnifier, δns. For this example, only the load combination of 1.2D + 1.6L is used.

δ ns =

Cm Pu 1− 0.75Pc

where Cm = 0.6 − 0.4

M1ns = 0.6 − 0.4( −1.0) = 1.0 M 2 ns

(for single-curvature meember in a braced frame )



Pu = 1.2(39) + 1.6(106) = 216 kips Pc =

π 2 ( EI )eff (knonsway Lu )2

For the stiffness parameter, (EI )eff, Eqs. (13.9.8) and (13.9.9) will be used to compare the values and illustrate the procedure.



Ec = 57, 000 3000 Es = 29, 000 ksi,

1 1000

= 3120 ksi,

Ig =

10(103 ) = 833 in.4 12

I se = 2(2.79)(2.58)2 = 37.1 in.4

The proportion of factored axial load that is sustained, βdns, is

β dns =

39(1.2) = 0.216 106(1.6) + 39(1.2) (Continued)

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Example 13.17.2 (Continued) Thus, using Eq. (13.9.8)

( EI )eff =

0.4 Ec I g (1 + βdns )

=

0.4(3120)833 = 855, 000 kip-in.2 (1 + 0.216)

and using Eq. (13.9.9)



( EI )eff =

0.2 Ec I g + Es I se (1 + βdns )

=

0.2(3120)833 + 29, 000(37.1) (1 + 0.216)

= 1, 310, 000 kip-in.2 Using the larger of the two values, the critical buckling load Pc for the column is Pc =



π 2 ( EI )eff π 2 (1, 310, 000) = = 844 kips (knonsway Lu )2 [1.0(10.33)12]2

Thus,

δ ns =



1 = 1.52 216 1− 0.75(844)

Thus, the column must be designed for the minimum moment Mm,min magnified by 1.52 (see Section 13.13). In other words, Mu = 1.52 M m ,min



However, because the magnified moment exceeds the 1.4 times the first-​order moment (ACI-​6.2.6), the column would be considered inadequate. (d) Alternative approach for effective length factor k. Using the moments of inertia given in ACI-​6.6.3.1.1, ( EI )beam = Ec (0.35I g ) = 3120(0.35)(8000) = 8, 740, 000 kip in.2

( EI )col = Ec (0.70 I g ) = 1, 820, 000 kip in.2

(10 in. × 10 in. upper column)

( EI )col = Ec (0.70 I g ) = 3, 770, 000 kip in.2

(12 in. ×12 in. lower column)

End restraint factors, ∑ EI /L for cols 1820 /12 = = 0.21 ∑ EI /L for beams 2(8740) / 24 (1820 + 3770) / 12 ψ B (bottom) = = 0.64 2(8740) / 24

ψ A (top) =



From Fig. 13.11.1(a), knonsway = k ≈ 0.65. In this case, the beams are very stiff compared to the columns, which yields a lower value than 1.0. A more realistic effective slenderness ratio is

kLu 0.65(10.33)12 = = 26.9 r 0.3(10)

This value is still greater than the limit of 22 computed in part (a), and thus slenderness effects must be considered. (Continued)

51



515

13.17 EXAMPLES

Example 13.17.2 (Continued) The revised critical buckling load and corresponding magnification factor are Pc =



π 2 ( EI )eff π 2 (1, 310, 000) = = 1995 kips (knonsway Lu )2 [0.65(10.33)12]2

Thus,

δ ns =



1 = 1.17 216 1− 0.75(1995)

The column should then be designed for Pu = 216 kips Mu = M max = 1.17 M m ,min





Mu = 1.17(216)[0.6 + 0.03(10)]

1 12

= 19 ft-kips

In this case, the upper limit of 1.4 per the requirements of ACI-​6.2.6 is satisfied (see Section 13.15). (e) Check strength. The strength of the section may be checked by the methods of Chapter 10. The design interaction diagram for the column is shown in Fig. 13.17.10. It is seen that the required strength is well below the design strength for the column. In fact, the column appears to be overdesigned for this load combination and the reinforcement could be reduced. φPn, Pu (kips) 450 400 350 300 250 200 150 100 50

0

10

20

30

40

50

60

φMn, Mu (ft-kips)

Figure 13.17.10  Design interaction diagram for the column of Example 13.17.2.

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EXAMPLE 13.17.3 Determine the adequacy of the square tied column (17 in. square, with 10–​#9 bars, fc′ = 3000 psi, fy = 60,000 psi) which is an exterior first-​floor column in the frame of Fig. 13.17.11. Assume that this frame is braced sufficiently to prevent relative translation of its joints. Also assume 40% of the factored axial load is sustained. 30’–0”

30’–0”

14 × 24

14 × 24

30’–0”

12’–0” B

12’–0”

17“ square columns

12’–0”

A

Assume hinged

(a) Frame dimensions Pu = 525k Mu = M2ns = 105’k

1

2 2”

10’–0”

1

2 2”

Clear height

10 – #9

e=

Mu = 2.4 in. Pu

Min e = 0.6 + 0.03h = 1.11 in.

17” square (b) Cross section Pu (c) Member A

Figure 13.17.11  Braced frame of Example 13.17.3.

SOLUTION (a) Effective length. In accordance with the more conservative procedure of ACI-​ 6.6.4.4.3, which allows the option of not doing an analysis to determine a k value less than 1.0 for a braced frame, use

kLu = Lu = 12 − 2 = 10 ft



(b) Slenderness ratio limits. The column slenderness ratio is

kLu 120 = = 23.5 r 0.3(17)

Slenderness effects may be neglected when

kLu  M  ≤ 34 + 12 1ns  ≤ 40 r M 2 ns   (Continued)

517



517

13.17 EXAMPLES

Example 13.17.3 (Continued) In this case, M1ns /​M2ns = 0,  kLu  = 34 > 23.6  r  limit Slenderness effects may be neglected. In the following, the moment magnifier method will be illustrated even though it would not be required by the ACI Code. (c) Braced frame moment magnifier δns. Ec = 57 fc′ = 3120 ksi Ig =



1 12

(17)(17)3 = 6960 in.4

I se = 2(5)(6) = 360 in. 2



4

Using Eq. (13.9.8), EI = 0.4 Ec I g = 8, 690, 000 kip in.2 or Eq. (13.9.9) EI = 0.2 Ec I g + Es I se

= 0.2(3120)(6960) + 29, 000(360) = 4, 340, 000 + 10, 440, 000 = 14, 800, 000 kip in.

2

Using the larger value of EI and applying the factor (1 + βdns) to account for sustained load, ( EI )eff = Pc =

14, 800, 000 = 10, 500, 000 kip-in.2 (1 + 0.4)

π 2 ( EI )eff π 2 (10, 500, 000) = = 7200 kips (knonsway Lu )2 [1.0(10)12]2

Pu = 525 kips

[from Fig.13.17.11(c)]

Pu 525 = = 0.097 0.75Pc 0.75(7200) Cm = 0.6 − 0.4

δ ns =



M1ns = 0.6 M 2 ns

Cm 0.6 = = 0.66 < 1.0 1 − Pu / 0.75Pc 1 − 0.097

In this case, no moment magnification is required. The factored loads to be carried are Pu = 525 kips and Mu = 105 ft-​kips at the top of the column. The design strength φPn is found to be 616 kips at e = 2.4 in. Thus, the column is adequate for the load combination considered.

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EXAMPLE 13.17.4 Determine the adequacy of the square tied column (20 in. square, with 12–​#10 bars, fc′ = 4000 psi, fy = 60,000 psi) which is an exterior first-​story column in the unbraced frame of Fig. 13.17.12. Assume that 20% of the factored axial load is sustained. The ΣPu on the four columns in the lowest story is 3150 kips. Use the Sum of P method to compute the moment magnifier. 12’–0” 12’–0” 12’–0”

14 × 24

14 × 24

14 × 24

30’–0”

30’–0”

30’–0”

20” square columns

A

Assume bases fixed

(a) Frame dimensions Pns = 488k; Ps = 0k #3 ties

Mu = M2ns + M2s = 70 + 265 = 335’k

10’–0”

12–#10

2.50”

5”

5”

5”

Clear height

20”

2.50” Mu = M1ns + M1s = 35 + 145 = 180’k

20”

(c) Member A

(b) Cross section

Figure 13.17.12  Unbraced frame of Example 13.17.4.

SOLUTION (a) Factored load combinations. Assume that for this example, the forces given in Fig. 13.17.12 are from the factored load combination including wind, and that wind causes insignificant axial compression (assumed zero here) on the member being studied. At the top, the first-​order nonsway and sway actions on the column are M2ns = 70 ft-​kips with Pns = 488 kips, and M2s = 265 ft-​kips with Ps = 0 kips (assumed), respectively. At the bottom, the nonsway and sway moments are M1ns = 35 ft-kips and M1s = 145 ft-kips. Only the check of strength required for U = 1.2D + 1.0W + 0.5L is illustrated in this example. (b) Effective length and slenderness ratio. The end restraint factors ψ must be determined. Using the moment of inertia for the beam of 0.35Ig per ACI-​6.6.3.1.1,

I beam = 0.35

14(24)3 = 5645 in.2 12

For the 20-​in. column, use 0.70Ig per ACI-​6.6.3.1.1,

I col = 0.70

(20)4 = 9333 in.4 12 (Continued)

519



519

13.17 EXAMPLES

Example 13.17.4 (Continued) For the unbraced frame, the magnifier δs involves the sum of the buckling loads Pc for all columns participating in the sidesway resistance in the first story. Thus, the effective length factor k is needed for each of these columns. In this example, ∑ EI /L for columns ∑ EI /L for beams

ψ A (top of exterior column) = =



2 Ec (9333) /12 = 8.26 Ec (5645) / 30

ψ A (top of interior column) = 4.13 For this calculation use of center-​to-​center span distances is recommended as being consistent with the nominal frame analysis using those distances. Theoretically, at the fixed end ψ equals zero; however, for practical purposes ψ should not be taken smaller than 1.0. (see section 13.11). Thus, ψ B (bottom) = 1.0 Using Fig. 13.11.1(b), find ksway of exterior column ≈ 1.85, ksway of interior column ≈ 1..65 The effective slenderness ratio for the exterior column being investigated is kLu 1.85(10.0)(12) = = 37 r 0.3(20)



which exceeds the limit of 22 given by ACI-​6.2.5 for unbraced frames. Slenderness effects must be considered. (c) Magnification factors. For the unbraced (sway) frames, M 22 nd = M 2 ns + δ s M 2 s [13.10.8]



For this example, the end moment M2ns under factored gravity load is given as 70 ft-​kips and the end moment M2s caused by factored lateral load is given as 265 ft-​kips. Using the sum of P method, the unbraced frame magnifier, δs, is computed as

δs =



1 ∑ Pu 1− 0.75 ∑ Pc

Compute (EI )eff from Eq. (13.9.9) as ( EI )eff =



0.2 Ec I g + Es I se (1 + β dns )

[13.9.9]

In this example, there is no sustained lateral load and it is assumed that the dead loads cause no sway; thus βdns = βds = 0. Accordingly, 1 (20)(20)3 = 13, 333 in.4 12 I se = 1.27(2)[ 4(7.5)2 + 2(2.5)2 ] = 603 in.4 Ig =

Ec = 57 fc′ = 57 4000 = 3605 ksi

( EI )eff = 0.2 Ec I g + Es I se = 0.2(3605)(13, 333) + 29, 000(603)



= 9, 610, 000 + 17, 500, 000 = 27,110, 000 kiip in.2

(larger than 0.4Ec I g )

(Continued)

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Example 13.17.4 (Continued) For the 20-​in. square exterior column,

Pc =

π 2 ( EI )eff π 2 (27,110, 000) = = 5429 kipps (ksway Lu )2 [1.85(10)12]2

For the 20-​in. square interior column,

Pc =

π 2 ( EI )eff π 2 (27,110, 000) = = 6825 kipps (ksway Lu )2 [1.65(10)12]2

Then, ΣPc is in this case,

∑ Pc = 2(5429 + 6825) = 24, 508 kips



The total factored load ΣPu acting on the four columns in the first story is given as 3150 kips. Thus, the magnifier δs for the P-​Δ effect is

δs =

1 = 1.21 3150 1− 0.75(24, 508)

Using Eq. (13.10.8),

M 22 nd = M 2 ns + δ s M 2 s

= 70 + 1.21(265) = 391 ft-kips

The ratio of second-​order to first-​order moment is 391/​335 = 1.17, which is less than the upper limit of 1.4 of ACI-​6.2.6. Thus, the column must be designed for Pu = 488 kips and Mu = 391 ft-​kips, or for Pu = 488 kips and an eccentricity e = 391(12)/​488 = 9.61 in., for this load combination. (d) Compute the design column strength φPn and φMn for the member. Using the principles of Chapter 10, at an eccentricity e = 9.61 in., the design strength is φPn = 506 kips, which is greater than Pu = 488 kips.             OK

EXAMPLE 13.17.5 Determine the adequacy of the 14 × 20 in. compression member reinforced with 6–​#11 bars, fc′ = 4500 psi, and fy = 60,000 psi, as shown in Fig. 13.17.13. The member serves as an exterior column in a braced frame; the loading and factored moment diagram are as shown in the figure, and the clear height is 22 ft 6 in. SOLUTION (a) For possible instability in the plane of the frame,

kLu 1.0(22.5)12 = = 45 rx 0.3(20)

where k for this braced column has been conservatively assumed as 1.0. (Continued)

521



521

13.17 EXAMPLES

Example 13.17.5 (Continued) Pu1 x

Bending axis

#4 ties

2.71” 6 – #11

14”

7’–6”

ePu2

2.71”

Mm = 279’k

Pu2 2.71”

2.71” x

Moment

20” 15’–0”

Pu1 + Pu2 = Pu = 115k

Figure 13.17.13  Column cross section, loading, and factored, first-​order bending moment diagram of Example 13.17.5.

For a braced column without transverse loading having this irregular moment diagram, when the bending moment diagram has the largest moment at a location other than at an end, M1ns /​M2ns should be taken conservatively as –​1.0. Thus, the slenderness limit is [Eq. (13.15.1)]

 M   kLu  = 34 + 12 1ns  = [34 − 12] = 22  r  M 2 ns   limit

which is less than 45. Thus slenderness effects must be considered. For this column, Ig =

1 (14)(20)3 = 9330 in.4 12

I se = 2(3)(1.56)(7.29)2 = 497 in.4



Ec = 57 4500 = 3820 ksi Assuming no sustained load,

β dns = 0 EI = 0.4 Ec I g = 0.4(3820)(9330) = 14, 300, 000 kip in.2



or

EI = 0.2 Ec I g + Es I se = 0.2(3820)(9330) + 29, 000(497) = 21, 500, 000 kipps in.2

(Continued)

52

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Example 13.17.5 (Continued) The column is not subjected to end moments only and it does not have transverse loading. Conservatively, assume Cm =1.0. Using the larger value of EI,

Pc =

π 2 ( EI )eff π 2 (21, 500, 000) = = 2910 kips (knonsway Lu )2 [1.0(22.5)12]2

δ ns =

Cm 1.0 = = 1.06 1 − Pu / 0.75Pc 1 − 115 / [0.75(2910)]

and

The required eccentricity for the column is e= e=

Mu δ ns M m = Pu Pu

1.06(279)12 = 30.9 in. 115

Assuming φ = 0.90 for this rather large eccentricity (e  /​h  =  1.55), the required nominal strength of the member in the plane of the frame (i.e., the strong orientation of the member) is required Pn = required M n =

Pu 115 = = 128 kips φ 0.90

δ ns M n 1.06(279) = 329 ft-kips = φ 0.90

A statics analysis of this section using an eccentricity of 30.9 in. gives

Pn = 194 kips > [ Pu /φ = 128 kips]

OK

(b) Instability transverse to the plane of the frame. The slenderness ratio is kLu 1.0(22.5)12 = = 64.3 ry 0.3(14)



which exceeds the limiting value of 22 for which the effect of slenderness may be neglected. Again, as in part (a), M1ns /​M2ns may be conservatively taken as –​1.0. There are no moments in this direction, thus the minimum moment Mm,min must be considered and magnified by the factor δns (see Section 13.13). Ig =

1 (20)(14)3 = 4570 in.4 12

I se = 2(2)(1.56)(4.29)2 = 115 in.4 EI = 0.4 Ec I g = 0.4(3820)(4570) = 6, 980, 000 kip in.

2

or

EI = 0.2 Ec I g + Es I se = 0.2(3820)(4570) + 29, 000(115) = 6, 830, 000 kip in.2

(Continued)

523



523

SELECTED REFERENCES

Example 13.17.5 (Continued) Using the larger of these two values and Cm = 1.0,

Pc =

π 2 ( EI )eff π 2 (6, 980, 000) = = 945 kips 2 (knonsway Lu ) [1.0(22.5)12]2

and

δ ns =

Cm 1.0 = = 1.19 1 − Pu / 0.75Pc 1 − 115 / [0.75(945)]

Thus, in the weaker direction e = 1.19(0.6 + 0.03h) = 0.71 + 0.036h = 0.71 + 0.036(14) = 1.21 in.





and M max = Pu e = 115(1.21)

1 12

= 12 ft-kips.



Assuming φ = 0.65 for this low eccentricity (e/​h = 0.09), the member must have a strength Pn = Pu  /​0.65 = 177 kips for an eccentricity of 1.21 in. A statics analysis indicates that the nominal strength Pn is 1390 kips at this eccentricity with respect to the weaker axis. However, the strength Pn may not be taken greater than 0.80Po,

Pn(max) = 0.80[0.85(4.5)(280 − 9.36) + 60(9.36)] = 1277 kips

which is less than the capacity at the minimum eccentricity of 1.21 in. Therefore,

Pn = 1277 kips > [ Pu /φ = 177 kips]

OK

SELECTED REFERENCES 13.1. L. Euler. De Curvis Elasticis, Additamentum I, Methodus In-​veniendi Lineas Curvas Maximi Minimive Proprietate Gaudentes. Lausanne and Geneva: 1744 (pp. 267–​268); and “Sur la Force des Colonnes,” Mémoires de l’Académie de Berlin, Vol. 13. Berlin: 1759 (pp. 252–​282). 13.2. F. Engesser. “Über die Knickfestigkeit gerader Stäbe,” Zeitschrift des Architektenund Ingenieur-​ Vereins zu Hannover, 35 (1889), 445–​462. Also “Die Knickfestigkeit gerader Stäbe,” Zentralblatt der Bauverwaltung, Berlin, Dec. 5, 1891, p. 483. 13.3. T.  von Kármán. “Die Knickfestigkeit gerader Stäbe,” Physikalische Zeitschrift, Vol. 9, 1908 (p. 136). Also, “Untersuchungen über knickfestigkeit,” Mitteilungen über Forschungsarbeiten auf dem Gebiete des Ingenieurwesens, No. 81. Berlin: 1910. 13.4. Luis P. Sáenz and Ignacio Martín. “Test of Reinforced Concrete Columns with High Slenderness Ratios,” ACI Journal, Proceedings, 60, May 1963, 589–​616. 13.5. John E. Breen and Phil M. Ferguson. “The Restrained Long Concrete Column As a Part of a Rectangular Frame,” ACI Journal, Proceedings, 61, May 1964, 563–​587. 13.6. Edward O. Pfrang and Chester P. Siess. “Behavior of Restrained Reinforced Concrete Columns,” Journal of the Structural Division, ASCE, 90, ST5 (October 1964), 113–​135; Disc., 91, ST3 (June 1965), 280–​287. 13.7. J.  G. MacGregor. Discussion of “Behavior of Restrained Reinforced Concrete Columns,” by E. O. Pfrang and C. P. Siess, Journal of the Structural Division, ASCE, 91, ST3 (June 1965), 280–​287.

524

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13.8. Richard W. Furlong and Phil M. Ferguson. “Tests of Frames with Columns in Single Curvature,” Symposium on Reinforced Concrete Columns (SP-​13). Detroit: American Concrete Institute, 1966 (pp. 55–​73). 13.9. Phil M. Ferguson and John E. Breen. “Investigation of the Long Column in a Frame Subject to Lateral Loads,” Symposium on Reinforced Concrete Columns (SP-​13). Detroit: American Concrete Institute, 1966 (pp. 75–​119). 13.10. Ignacio Martín and Elmer Olivieri. “Test of Slender Reinforced Concrete Columns Bent in Double Curvature,” Symposium on Reinforced Concrete Columns (SP-​13). Detroit: American Concrete Institute, 1966 (pp. 121–​138). 13.11. J.  G. MacGregor and S.  L. Barter. “Long Eccentrically Loaded Concrete Columns Bent in Double Curvature,” Symposium on Reinforced Concrete Columns (SP-​13). Detroit: American Concrete Institute, 1966 (pp. 139–​156.) 13.12. Edward O. Pfrang. “Behavior of Reinforced Concrete Columns with Sidesway,” Journal of the Structural Division, ASCE, 92, ST3 (June 1966), 225–​252. 13.13. R. Green and John E. Breen. “Eccentrically Loaded Concrete Columns under Sustained Load,” ACI Journal, Proceedings, 66, November 1969, 866–​874. 13.14. John E. Breen and Phil M. Ferguson. “Long Cantilever Columns Subject to Lateral Forces,” ACI Journal, Proceedings, 66, November 1969, 884–​893. 13.15. Shriniwas N. Pagay, Phil M. Ferguson, and John E. Breen. “Importance of Beam Properties on Concrete Column Behavior,” ACI Journal, Proceedings, 67, October 1970, 808–​815. 13.16. Hajime Okamura, Shriniwas N. Pagay, John E. Breen, and Phil M. Ferguson. “Elastic Frame Analysis—​Corrections Necessary for Design of Short Concrete Columns in Braced Frames,” ACI Journal, Proceedings, 67, November 1970, 894–​897. 13.17. Phil M.  Ferguson, Hajime Okamura, and Shriniwas N.  Pagay. “Computer Study of Long Columns in Frames,” ACI Journal, Proceedings, 67, December 1970, 955–​958. 13.18. G. A. Blomier and J. E. Breen. “Effect of Yielding of Restraints on Slender Concrete Columns with Sidesway Prevented,” Reinforced Concrete Columns (SP-​50), Detroit: American Concrete Institute, 1975 (pp. 41–​65). 13.19. F.  N. Rad and R.  W. Furlong. “Behavior of Unbraced Reinforced Concrete Frames,” ACI Journal, Proceedings, 77, July–​August 1980, 269–​278. 13.20. J. S. Ford, D. C. Chang, and J. E. Breen. “Behavior of Concrete Columns under Controlled Lateral Deformation,” ACI Journal, Proceedings, 78, January–​February 1981, 3–​20. 13.21. J. S. Ford, D. C. Chang, and J. E. Breen. “Experimental and Analytical Modeling of Unbraced Multipanel Concrete Frames,” ACI Journal, Proceedings, 78, January–​February 1981, 21–​35. 13.22. J. S. Ford, D. C. Chang, and J. E. Breen. “Behavior of Unbraced Multipanel Concrete Frames,” ACI Journal, Proceedings, 78, March–​April 1981, 99–​115. 13.23. Bengt Broms and I.  M. Viest. “Ultimate Strength Analysis of Long Hinged Reinforced Concrete Columns,” Journal of the Structural Division, ASCE, 84, ST1 (January 1958) (Paper No. 1510). 13.24. Nestor R.  Iwankiw. “Note on Beam–​Column Moment Amplification Factor,” Engineering Journal, AISC, 21, 1(1st quarter 1984), 21–​23. 13.25. Charles Massonnet. “Stability Considerations in the Design of Steel Columns,” Journal of the Structural Division, ASCE, 85, ST7 (September 1959), 75–​111. 13.26. Charles G.  Salmon, John E.  Johnson, and Faris A.  Malhas. Steel Structures—​Design and Behavior:  Emphasizing Load and Resistance Factor Design (5th ed.) New Jersey:  Pearson-​ Prentice Hall Publishers, 2009. 13.27. Ronald D.  Ziemian Ed. Guide to Stability Design Criteria for Metal Structures (6th ed.) New York: John Wiley & Sons, 2010. 13.28. J. G. MacGregor. “Out-​of-​Plumb Columns in Concrete Structures,” Concrete International, 1, June 1979, 26–​31. 13.29. James G. MacGregor and Sven E. Hage. “Stability Analysis and Design of Concrete Frames,” Journal of the Structural Division, ASCE, 103, ST10 (October 1977), 1953–​1970. 13.30. J. G. MacGregor. “Design of Slender Concrete Columns—​Revisited,” ACI Structural Journal, 90, May–​June 1993, 302–​309. 13.31. Chu-​Kia Wang and Carlito P. Talaboc. “Second-​Order Analysis of Multistory Building Frames by Marching Technique,” Microcomputers in Structural Engineering, 1, 3 (December 1986), 209–​232. 13.32. James G.  MacGregor, John E.  Breen, and Edward O.  Pfrang. “Design of Slender Concrete Columns,” ACI Journal, Proceedings, 67, January 1970, 6–​28. 13.33. Eldon F.  Mockry and David Darwin. “Slender Column Interaction Diagrams,” Concrete International, 4, June 1982, 44–​50. 13.34. Alfred L. Parme. “Capacity of Restrained Eccentrically Loaded Long Columns,” Symposium on Reinforced Concrete Columns (SP-​ 13). Detroit:  American Concrete Institute, 1966 (pp. 325–​367).

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525

13.35. PCA. Design Constants for Rectangular Long Columns. Advanced Engineering Bulletin No. 12. Skokie, IL: Portland Cement Association, 1964, 36 pp. 13.36. Brian R. Wood, Denis Beaulieu, and Peter F. Adams. “Column Design by P–​Delta Method,” Journal of the Structural Division, ASCE, 102, ST2 (February 1976), 411–​427. 13.37. Brian R. Wood, Denis Beaulieu, and Peter F. Adams, “Further Aspects of Design by P–​Delta Method,” Journal of the Structural Division, ASCE, 102, ST3 (March 1976), 487–​500. 13.38. H.  Scholz. “A Novel P–​Delta Method for Reinforced Concrete Frames Designed by LRFD Approaches,” ACI Journal, Proceedings, 83, July–​August 1986, 633–​641. 13.39. H. Scholz. “Alternative P–​Delta Method for Prestressed Members Subjected to Axial Load and Bending,” ACI Structural Journal, 87, July–​August 1990, 464–​472. 13.40. James Colville. “Slenderness Effects in Reinforced Concrete Square Columns,” Reinforced Concrete Columns (SP-​50). Detroit: American Concrete Institute, 1975 (pp. 165–​191). 13.41. J. G. MacGregor, U. H. Oelhafen, and S. E. Hage. “A Reexamination of the EI Value for Slender Columns,” Reinforced Concrete Columns (SP-​50). Detroit: American Concrete Institute, 1975 (pp. 1–​40). 13.42. Ian C.  Medland and Donald A.  Taylor. “Flexural Rigidity of Concrete Column Sections,” Journal of the Structural Division, ASCE, 97, ST2 (February 1971), 573–​586. 13.43. Jian-​ Min Zeng, Lian Duan, Fu-​ Ming Wang, and Wai-​ Fah Chen. “Flexural Rigidity of Reinforced Concrete Columns,” ACI Structural Journal, 89, March–​April 1992, 150–​158. 13.44. M.  A. Diaz and J.  M. Roesset. “Evaluation of Approximate Slenderness Procedures for Nonlinear Analysis of Concrete Frames,” ACI Structural Journal, 84, March–​April 1987, 139–​148. 13.45. S. A. Mirza. “Flexural Stiffness of Rectangular Reinforced Concrete Columns,” ACI Structural Journal, 87, July–​August 1990, 425–​435. 13.46. Gregory R. Sigmon and Shuaib H. Ahmad. “Flexural Rigidity of Circular Reinforced Concrete Sections,” ACI Structural Journal, 87, September–​October 1990, 548–​556. 13.47. M. R. Ehsani and F. Alameddine. “Refined Stiffness of Slender Circular Reinforced Concrete Columns,” ACI Structural Journal, 84, September–​October 1987, 419–​427. 13.48. Brij B. Goyal and Neil Jackson. “Slender Concrete Columns under Sustained Load,” Journal of the Structural Division, ASCE, 97, ST11 (November 1971), 2729–​2750. 13.49. Robert G.  Drysdale and Mark W.  Huggins. “Sustained Biaxial Load on Slender Concrete Columns,” Journal of the Structural Division, ASCE, 97, ST5 (May 1971), 1423–​1443. 13.50. B.  Vijaya Rangan. “Lateral Deflection of Slender Reinforced Concrete Columns under Sustained Load,” ACI Structural Journal, 86, November–​December 1989, 660–​663. 13.51. J. S. Ford, D. C. Chang, and J. E. Breen. “Design Indications from Tests of Unbraced Multipanel Concrete Frames,” Concrete International, 3, March 1981, 37–​47. 13.52. Thomas C. Kavanagh. “Effective Length of Framed Columns,” Transactions ASCE, 127, Part II, 1962, 81–​101. 13.53. John E.  Breen, James G.  MacGregor, and Edward O.  Pfrang. “Determination of Effective Length Factors for Slender Concrete Columns,” ACI Journal, Proceedings, 69, November 1972, 669–​672. 13.54. BSI. Code of Practice for the Structural Use of Concrete, Part I.  Design, Materials and Workmanship. London: British Standards Institution, 1972. 13.55. Richard W. Furlong. “Column Slenderness and Charts for Design,” ACI Journal, Proceedings, 68, January 1971, 9–​18. 13.56. Yu-​Xia Hu, Ren-​Gen Zhou, Won-​Sun King, Lian Duan, and Wai-​Fah Chen. “On Effective Length Factor of Framed Columns in the ACI Building Code,” ACI Structural Journal, 90, March–​April 1993, 135–​143. 13.57. Lian Duan, Won-​Sun King, and Wai-​Fah Chen. “K-​Factor Equation to Alignment Charts for Column Design,” ACI Structural Journal, 90, May–​June 1993, 242–​248. 13.58. Richard W.  Furlong, “Rational Analysis of Multistory Concrete Structures,” Concrete International, 3, June 1981, 29–​35. 13.59. S.  I. Abdel-​Sayed and N.  J. Gardner. “Design of Symmetric Square Slender Reinforced Concrete Columns under Biaxially Eccentric Loads,” Reinforced Concrete Columns (SP-​50) Detroit: American Concrete Institute, 1975 (pp. 149–​164). 13.60. S. A. Mirza, P. M. Lee, and D. L. Morgan. “ACI Stability Resistance Factor for RC Columns,” Journal of Structural Engineering, ASCE, 113, No. 9, September, 1987, 1963–​1976. 13.61. Madhu Khuntia and S.  K. Ghosh. ”Flexural Stiffness of Reinforced Concrete Columns and Beams: Analytical Approach,” ACI Structural Journal, 101, No. 3, May-​June 2004, 351–​363. 13.62. Richard W.  Furlong, Cheng-​Tzu Thomas Hsu, and S.  Ali Mirza. “Analysis and Design of Concrete Columns for Biaxial Bending—​Overview,” ACI Structural Journal, 101, No. 3, May–​ June 204, 413–​423.

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PROBLEMS All problems are to be done in accordance with the ACI Code unless otherwise indicated. All dead and live loads given are service-​level loads, while wind loads are strength-​level loads computed per ASCE 7 unless otherwise indicated. Even though the slenderness effects might be neglected according to the ACI Code in a given problem, consider slenderness effects anyway. For design problems, use whole inches for member size. 13.1 Determine the adequacy of the section, including slenderness effects, for a 16-​in. square tied column that has a clear height of 18 ft and serves as an interior member of a braced frame. The member is designed as axially loaded with the following service loads:  live load, 130 kips; dead load, 200 kips. Use fc′ = 4000 psi and fy = 60,000 psi. Without actually selecting bars, assume about 2 1 2 % total reinforcement equally divided in the opposite faces of the member. Use the ACI moment magnifier method. 13.2 Determine the adequacy, including slenderness effects, for a 14-​in.-​diameter spirally reinforced 1 column (assume about 2  2 % reinforcement) which is an interior second-​floor column (column A) in the braced frame of the figure for Problems 13.2 through 13.8. The member is to carry an axial service load of 87 kips dead load and 133 kips live load, with only the dead load considered sustained. Use fc′ = 5000 psi and fy = 60,000 psi. Use the ACI moment magnifier method. 14” square 14 × 22

15’ – 0”

14” square Column B 14 × 22

15’ – 0”

16” square Column C

13.3 Assume that the frame of Problem 13.2 is unbraced and consists of 4 bays symmetrical about the middle. The gravity axial loads are 44 kips dead load and 66 kips live load. The largest moments in column A are 15 ft-​kips dead load, 12 ft-​kips live load, and 58 ft-​kips wind load. An exterior column carries one-​half the factored axial load of an interior column. Investigate the adequacy of the interior column A. If not adequate, redesign it to make it satisfactory according to the moment magnifier method of the ACI Code. 13.4 Determine the adequacy of the exterior square column (column B) of the figure for Problem 13.2 if the member is carrying a gravity axial compression of 217 kips dead load and 145 kips live load and primary bending moments of 27 ft-​kips dead load and 18 ft-​kips live load. Assume that the primary bending moment varies from +M at the top of the member linearly to –​M/​2 at the bottom, with joint translation

14” diameter 14 × 22

14”

14” diameter Column A 4 – #9

14 × 22 16” diameter

1

22” Column B 40’ – 0”

40’ – 0” Symmetrical about this line

16”

6 – #9

1

22” Column C

Problems 13.2 to 13.8 

527



527

PROBLEMS

adequately prevented (i.e., a braced frame). Use fc′ = 5000 psi and fy = 60,000 psi. Use the ACI moment magnifier method. 13.5 Repeat Problem 13.4, except consider the primary bending moment constant over the height of the column. 13.6 Assume that the member (Column B) in Problem 13.4 is part of an unbraced frame and that maximum nonsway moments M2ns are 28 and 20 ft-​kips due to dead and live loads, respectively. Also, a maximum M2s sway moment of 85 ft-​ kips due to strength-​level wind load must be carried. The axial compressive loads are 30 kips service dead load and 40 kips service live load. The total factored load ΣPu to all five columns in the story is 590 kips for the load combination 1.2D + 1.0W + 0.5L. Investigate the adequacy of the 14-​in. square exterior column (column B). If found to be inadequate, redesign the member assuming that the loadings are not affected by a change in member stiffness. 13.7 Determine the adequacy of the 16-​in. square first-story column (column C) of the figure for Problem 13.2, which is carrying a service axial load of 370 kips and a primary bending moment of 121 ft-​kips (assume loads 70% dead load). The primary bending moment varies from a maximum at the top to zero at the bottom and joint translation is adequately prevented. Use fc′ = 5000 psi and fy = 60,000 psi. Use ACI moment magnifier method. 13.8 Assume that the member (column C) of Problem 13.7 is part of an unbraced frame. The axial loads are 45 kips dead load and 60 kips live load. The bending moments are 30 ft-​kips dead load, 22 ft-​ kips live load, and 110 ft-​kips wind load. Investigate

the adequacy according to the moment magnifier method of the ACI Code. If not adequate, redesign assuming the loadings are not affected by changes in member stiffnesses. The total factored load ΣPu to all columns in the lower story is 885 kips for load combination 1.2D + 1.0W + 0.5L. 13.9 Design column A for the unbraced frame of the figure for Problem 13.9 for the load combination of 1.2D + 1.0W + 1.0L (Assume that the columns will support areas having live loads greater than 100 psf.) The axial compression is 112 kips dead load and 112 kips live load; bending moment is 35 ft-​kips dead load, 35 ft-​kips live load, and 118 ft-​kips wind load. Consider only the dead load to be sustained. The factored axial load Pu on the interior columns is 1.8 times that on the exterior columns. Select a square member to contain approximately 2 1 2 % reinforcement. Use fc′ = 5000 psi, fy = 60,000 psi, and the moment magnifier method of the ACI Code. 13.10  (a) Redesign the column used in Problem 13.9 except for service loads, use an axial compression of 132 kips dead load and 132 kips live load, and bending moments M2 of 26 ft-​kips dead load, 26 ft-​kips live load, and 123 ft-​kips wind load. The length for column A is 20 ft instead of 21 ft.   (b) How much smaller could the member have been if the frame had been adequately braced to prevent joint translation? 13.11  (a)  Redesign the columns used in Problem 13.9 except use 18 ft instead of 21 ft for the length of column A.   (b) How much smaller could the member have been if the frame had been adequately braced to prevent joint translation?

Unbraced frame 14’–0” h 16 × 28 beam Assume same size for all columns

1

h Place reinforcement symmetrically in two faces

21’–0” A

There are three interior columns

ψ = 1.0 30’– 0”

Problems 13.9 to 13.11 

1

2 2”

22”

CHAPTER 14 STRUT-​A ND-​T IE MODELS—​ DEEP BEAMS, BRACKETS, AND CORBELS

14.1 INTRODUCTION As discussed in Section 5.7, the zone in the vicinity of a concentrated load, or at an abrupt discontinuity in the member (such as an abrupt change in member depth), is called a “D-​region” to refer to a discontinuity or a disturbance [5.9]. A  typical example of a D-​region is that of a deep beam where the shear span-​to-​depth ratio, a  /​d, is less than about 2. In such cases, plane sections cannot be assumed to remain plane after bending (Bernoulli hypothesis), and thus the distribution of stresses and strains cannot be obtained by using ordinary beam theory, as presented in Chapters 3, 4, and 5. For many years, the design of D-​regions relied on experimental evidence, past experience, and good practice. Today, however, truss models or strut-​and-​tie models are accepted tools for the design of D-​regions. Intended to represent the flow of stresses at failure, a strut-​and-​tie model is an idealized truss that is in equilibrium with the externally applied loads and that satisfies the requirements of the lower-​bound theorem of plasticity theory. This means that the failure load calculated by using a strut-​and-​tie model will be less than or equal to the actual failure load, and thus it offers a safe value of the load-​carrying capacity of the structure. In design, either two-​or three-​dimensional models, or both, are used to represent the D-​regions in various structures or elements. Two-​dimensional models are used for planar structures such as deep beams, brackets or corbels, dapped-​end beams, walls, and beam-​column joints. Elements for which three-​dimensional models may be required include bridge piers and pile caps. Consider the simply supported deep beam with concentrated loads and the corresponding truss or strut-​and-​tie model shown in Fig. 14.1.1. The flow of stresses may be idealized by two inclined members in compression, or struts in the concrete that carry the load to the supports and a tension member or tie at the bottom of the beam represented by the longitudinal reinforcement. Also, a horizontal strut is used to represent the stresses that develop in the compression zone of the beam at midspan. In the model, struts and ties have in-​and out-​of-​plane finite dimensions (width, length, and thickness) that will depend on the truss geometry and the forces developed in each truss element. The intersection of the axes of struts and ties defines the nodes (or joints) of the model, and the region surrounding the node defines what is referred to as the nodal zone. In general, failure of a truss will occur when one or more truss elements, including the nodes, reach the design strength. In the strut-​and-​tie model of Fig. 14.1.1, failure is assumed to occur when the struts reach their crushing strength or when the tie reaches yielding. Failure may also occur at the joints by failure of the nodal zone.

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14.1 INTRODUCTION

Nodal zones

Bottle-shaped strut

Idealized tapered strut

Idealized prismatic, parallel stress field

Nodal zone

Nodal zone

Tie

Figure 14.1.1  Idealized flow of stresses and corresponding strut-​and-​tie model for a simply supported deep beam.

Struts Strut elements represent the compression stress field regions that develop in the member under the applied loads. In practice, the shape of the compression field may vary depending on the geometry, the load, and the support conditions of the member. Jörg Schlaich, Schäfer, and Jennewein [14.1] have proposed three basic types or shapes of compression stress fields commonly encountered in practice, as shown in Fig. 14.1.2. The “basic” type is a prism of uniform cross-​sectional area, where the compression stress field is constant over the length [Fig. 14.1.2(a)]. The prismatic, parallel stress field is commonly used to represent the compression zone of a beam, such as that at midspan of the beam shown in Fig. 14.1.1. The second type of strut, referred to as a bottle-​shaped strut, is meant to represent regions where the concrete compressive stresses spread out at some distance away from the loading or reaction point [Fig. 14.1.2(b)]. These stress fields may be found, for example, between the applied load and the reaction of the deep beam of Fig. 14.1.1, which may be modeled as a bottle-​shaped strut. For simplicity in design, however, bottle-​shaped struts are often idealized as tapered struts instead, as shown in Fig. 14.1.1. In the third type of stress field, the stresses “fan out,” as shown in Fig. 14.1.2(c). Fan-​shaped action will develop, for example, in a simply supported deep beam with a uniformly distributed load as shown in Fig. 14.1.3(a). In practice, such regions will often be modeled as a series of prismatic struts as illustrated in Fig. 14.1.3(b). The strength (stress capacity) of the strut will depend on its shape and on the presence of reinforcement (if any) crossing the strut axis. In a bottle-​shaped strut, the spread of stresses from the end to the middle of the strut will tend to split the strut near the end, which will weaken the strut [see Fig.  14.1.4(a)]. In fact, the flow of forces within a bottle-​shaped strut may be idealized as a strut-​and-​tie model as shown in Fig. 14.1.4(b). A number of researchers have suggested values for the strength of struts of different shapes (J. Schlaich et al. [14.1], Marti [14.2], Rogowsky and MacGregor [14.3], Bergmeister, Breen, and Jirsa [14.4], and Ramirez and Breen [14.5]). While the range of suggested values is diverse, researchers agree that the strength of compressive struts is less than the compressive cylinder strength fc′ with values ranging from approximately 0.5 to 1 times  fc′.

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b

b

a

a

(a)

(b)

α

a (c)

Figure 14.1.2  Basic compression fields: (a) parallel field; (b) bottle-​shaped field; (c) fan-​shaped field. (Adapted from J. Schlaich et al. [14.1].) CL a

CL a/4

w

a/2 wa/2

a/4 wa/2

C

C

T

T

(a)

(b)

Figure 14.1.3  Fan action in a deep beam: (a) fan-​shaped stress field; (b) strut-​and-​tie model. (Adapted from Marti [14.2].)

Crack

Tie

2

Strut

1

2

1 (a)

(b)

Figure 14.1.4  Bottle-​shaped strut: (a) splitting cracks near the strut ends; (b) idealized strut-​and-​ tie model of the strut. (Adapted from MacGregor [14.6].)

531



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14.1 INTRODUCTION

Ties Ties represent the regions where tensile stresses develop in the member and are the tension members in the model. They consist of the reinforcement (nonprestressed or prestressed) plus a portion of the concrete around it. A tie may consist of one or several layers of reinforcement over the width of the tie, as shown in Fig. 14.1.5. For better crack distribution, it is recommended that the reinforcement be distributed uniformly within the cross section of the tie. It is noted that the centroid of the reinforcement must coincide with the axis of the tie in the model. In practice, it is assumed that the concrete surrounding the reinforcement does not carry tension; the strength of the tie is given by the strength of the reinforcement alone, which is often assumed to be at yield.

Nodal Zones The intersection between the axes of two or more struts and ties defines the nodes in the model. The region surrounding the nodes is called a nodal zone. To satisfy equilibrium in a planar structural model, at least three forces must intersect at a node. Depending on the type of forces acting at a node intersection (compression or tension), nodes are often classified as C-​C-​C nodes when they resist forces from three or more struts or as C-​C-​T nodes when two or more struts and a tie intersect at the node. A third type of node is a C-​T-​T node, where two ties intersect a strut. Preliminary dimensions of a nodal zone may be determined from the estimated size of the struts, ties, and the bearing area under the applied loads or reactions, such as bearing plates or column base dimensions. These dimensions, however, may need to be revised if the strength of the nodal zone is exceeded at one or more of its faces. If the faces of the nodal zone are perpendicular to the axes of the struts, the stresses will be the same on all faces and equal to the in-​plane principal stress. In such a case, the dimensions of the nodal zone are in the same proportion as the acting forces, as shown in Fig. 14.1.6(a). Such nodal zones are often called hydrostatic nodal zones, though in reality the state of stresses is not truly hydrostatic because the out-​of-​plane stress in a planar model is zero. A node anchoring one or more ties may also be considered as “hydrostatic” by treating the tie force as a compressive force acting on the far side of the nodal zone, as shown in Fig. 14.1.6(b). Laying out a model with hydrostatic nodal zones to meet the strength requirements in all faces can be cumbersome in practice. In a simplified approach to sizing the nodal region, Schlaich and Schäfer [14.7] have suggested that the faces of the nodal zone need not be perpendicular to the axis of the strut (or tie), as shown in Fig.14.1.7. In such a case, the node region is simply defined by the width of the struts and ties intersecting at the node. Unlike hydrostatic nodes, both normal and shear stresses will act on the faces of the nodal zone. Schlaich and Schäfer [14.7] have also suggested that the nodal zone be considered safe if the stresses acting on the cross-​sectional area taken perpendicular to the strut (or tie) axis are below the nodal zone stress limit. Note that the normal stresses acting on the faces of the

C

T

Tie width

Figure 14.1.5  Tie consisting of several layers of reinforcement plus a portion of surrounding concrete.

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C1

C1 d1

T

C1 C2 C3 = = d1 d2 d3

d3

Nodal zone

C2

C1

C3 Nodal zone

C3 d2 C1

C2

C2

C2

(a) C–C–C node

T (b) C–C–T node

Figure 14.1.6  Example of hydrostatic nodal zones: (a) C-​C-​C node; (b) C-​C-​T node.

d3

C3

fc3

fc2

d2

C2

fc1 C1 d1

Figure 14.1.7  Simplified arrangement of a C-​C-​C node. (After M. Schlaich and Schäfer [14.7].)

nodal zone need not be equal in this case. This approach can be particularly useful for nodal zones with four or more struts (or ties) intersecting at a node, as shown in Fig. 14.1.8(a). In another approach, though somewhat more laborious, M. Schlaich and Anagnostou [14.8] have shown that the nodal zone can be treated by using a combination of hydrostatic “subnodal” zones and “transition stress fields.” Figure  14.1.8(b), which exemplifies this approach, shows the same nodal zone of Fig. 14.1.8(a), where the entire nodal zone has been subdivided into two hydrostatic nodal zones and one short strut (or transition stress field). In this situation, the stresses on all faces of the two “subnodal” zones and within the short strut are the same. If a nodal zone anchoring a tie is too small to develop the tie reinforcement within the nodal zone, an anchor plate may be required [see Fig. 14.1.9(a)]. The tie reinforcement may be developed, however, within the region defined by the intersection of the strut and tie widths outside of the nodal zone, as shown by the dotted area in Fig. 14.1.9(b). This region, referred to as the extended nodal zone in the ACI Code, may be used to compute the available development length for the tie reinforcement (ACI-​R23.2.6). It is assumed that within the extended nodal zone the compression stresses due to the reactions and the struts help

53



14.1 INTRODUCTION

533

transfer the forces from the tie to the node. In some situations, however, the development length computed with the inclusion of the extended nodal zone may still be insufficient to anchor the tie. Hooks, mechanical anchorage, and/​or a reduction in bar diameter will be needed in such cases. The strength of a nodal zone will depend on its shape and the types of elements framing into the node. In a “hydrostatic” C-​C-​C node [see Fig. 14.1.8(b)], the nodal zone is subjected to a biaxial or triaxial state of compressive stresses, a rather favorable condition for the node. On the other hand, if a node anchors one or more ties, the tension stresses in the tie reinforcement will weaken the node, and so a lower strength of the nodal zone can be expected. Very limited experimental data on the strength of nodal zones exist (Jirsa et al. [14.9]). Suggested values range from about 0.7 to 0.85 fc′ for C-​C-​C nodes and from 0.52 to 0.6 fc′ for C-​T-​T nodes (MacGregor [14.6]).

Figure 14.1.8  Example of nodal zone with four struts: (a) simplified arrangement; (b) “hydrostatic” nodal zones and a “transition stress field.”

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C

T

C (a) C

Extended nodal zone Nodal zone hn 2

T

hn 2 La

C (b)

Figure 14.1.9  Anchorage of tie reinforcement: (a) anchored by a plate; (b) anchored by bond.

ACI Code Provisions Chapter 23 of the 2014 ACI Code contains the provisions for the design of D-​regions using strut-​and-​tie models. The basic design approach consists of ensuring that the selected strut-​ and-​tie model (i.e., the idealized truss) is capable of transferring the loads to the supports and to the adjacent B-​regions. Accordingly, the strut-​and-​tie model must satisfy the following main requirements. 1. The strut-​and-​tie model must be in equilibrium with the applied factored loads and the reactions (ACI-​23.2.4). 2. Struts shall not cross or overlap each other except at nodal zones (ACI-​23.2.6). If struts overlap each other, then a portion of the struts would be overstressed. The strength of the strut is calculated based on its geometry (e.g., prismatic or bottle-​ shaped), the concrete strength, and the reinforcement crossing the strut. 3. Ties shall be permitted to cross struts or other ties (ACI-​23.2.5). A  tie crossing a strut will induce tensile strains in the transverse direction, which will reduce the strut strength. Thus, the ACI Code reduces the strength of struts crossed by ties or whenever the compressive stresses are transferred across cracks in a tension zone. 4. The angle between the axis of a strut and that of a tie at a node shall not be taken less than 25° (ACI-​23.2.7). This requirement is meant to avoid incompatibility resulting from shortening of a strut and lengthening of a tie in nearly the same direction.

53



535

14.1 INTRODUCTION

5. The design strength of each strut, tie, and nodal zone in the strut-​and-​tie model must satisfy the following (ACI-​23.3.1):

φFns ≥ Fus

(Struts)

(14.1.1)



φFnt ≥ Fut

(Ties)

(14.1.2)



φFnn ≥ Fun

(Nodal zones)

(14.1.3)

where Fns = nominal compressive strength of a strut Fnt = nominal tenssile strength of a tie Fnn = nominal strength on the face of a nodal zone Fus = factored compressive force in a strut Fut = factored tensile force in a tie

Fun = factored force on the face of a node

φ = strength reduction factor, taken equal to 0.75 for struts, ties, nodal zones, and bearing areas designeed in accordance with the strut-and-tie method (ACI-21.2)

Strength of Struts, Fns The nominal strength Fns of a strut is taken as the smaller of the strengths computed at the two ends of the strut as follows [ACI-​23.4.1(a)]:

Fns = Acs fce

(14.1.4)

where Acs is the area of the strut at one end and fce is the effective compressive strength of the strut. As mentioned earlier, the compressive strength of a strut will depend on its shape and the presence of reinforcement (if any) perpendicular to the strut axis. In the ACI Code, the effective compressive strength of a strut is computed as (ACI-​23.4.3)

fce = 0.85β s fc′

(14.1.5)

where βs is a factor that accounts for the effect of cracking and crack-​control reinforcement, as shown in Table14.1.1. For example, βs is taken as 1.0 for a prism of uniform cross section, where the compression stress field can be considered parallel and constant over the length. This results in fce = 0.85 fc′, which is consistent with the strength of the rectangular stress block in the compression zone of a beam according to the ACI Code.

TABLE 14.1.1  VALUES OF βS FOR COMPUTING STRUT STRENGTH ACCORDING TO ACI-​23.4.3 Strut Type Prism of uniform cross section over its length (parallel stress field) Bottle-​shaped struts: with reinforcement satisfying ACI-​23.5 without reinforcement satisfying ACI-​23.5 Struts in tension members or in the tension flanges of members All other cases a

λ = modification factor to account for the unit weight of concrete (see Section 1.8).

βs 1.0 0.75 0.60λa 0.40 0.60λa

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For bottle-​shaped struts, βs is reduced to 0.6λ because of the tendency to split the strut as the stresses spread out from the strut end toward the midlength of the strut [see Fig. 14.1.4(a)]. The parameter λ is the modification factor that accounts for the unit weight of the concrete as shown in the table. If a bottle-​shaped strut is crossed by reinforcement that resists the transverse tensile stresses, a value of 0.75 may be used for βs (see Table 14.1.1) However, such an increase in strut strength is permitted only when the reinforcement is provided in accordance with ACI-​23.5. This section allows designers to use a strut-​and-​tie model for the bottle-​shaped strut to compute the required amount of reinforcement crossing the strut by assuming that the compressive forces spread out at a 2:1 slope, as shown earlier [see Fig.  14.1.4(b)]. Alternatively, for fc′ ≤ 6000 psi, the required amount of reinforcement crossing the strut may be computed to satisfy Eq. (14.1.6) [ACI Formula (25.5.3)] as follows: Asi

∑bs



sin α i ≥ 0.003

s i



(14.1.6)

where Asi is the total area of reinforcement at a spacing si of a layer of reinforcement at an angle αi with the axis of the strut, and bs is the strut thickness. ACI-​23.5.3.1 allows the reinforcement to be provided in two orthogonal directions, as shown in Fig. 14.1.10, or in one direction. In the latter case, the angle between the reinforcement layer and the strut axis, α1, shall be at least 40° For struts crossing the tension region of a member, such as the flanges of inverted T-​sections, βs is reduced to 0.4 to account for the reduced strenth of struts carrying compressive stresses across a cracked tension zone. For all other cases, the ACI Code specifies a βs value of 0.6λ. Such cases include, for example, fan-​shaped struts [Fig. 14.1.3(a)] and inclined struts either parallel to cracks [Fig. 14.1.11(a)] or crossed by cracks [Fig. 14.1.11(b)] in the compression field of a beam web. The strength of a strut may be increased by using compression reinforcement along the length of the strut (ACI-​23.4.1). Such a reinforcement must be parallel to the strut axis and located within the strut width. In addition, the reinforcement must be properly anchored and enclosed by ties, hoops, or spirals in accordance with the detailing requirements in ACI-​ 23.6.3. The strength of a strut with compression reinforcement is computed as [ACI-​23.4.1(b)] Fns = Acs fce + As′ fs′ (14.1.7)



where As′ is the area of compression reinforcement along the length of the strut and fs′ is the stress in the compression reinforcement at the nominal axial strength of the strut. For Grades 40 and 60 reinforcement, fs′ may be taken equal to fy.

Axis of strut Strut boundary α1

Strut boundary

As1 s2

α2 As2

s1

Figure 14.1.10  Details of reinforcement crossing a strut according to ACI-​23.5.3.1. (From ACI-​ 318 Code and Commentary, 2014.)

537



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14.1 INTRODUCTION

Strut

Cracks Tie

Strut (a) Struts in a beam web with inclined cracks parallel to struts Strut

Cracks Tie

(b) Struts crossed by skew cracks

Figure 14.1.11  Struts in the compression field of a beam web: (a) struts parallel to cracks; (b) struts crossed by cracks. (From ACI 318-​05 Code and Commentary, 2005.)

Strength of Nodal Zones, Fnn The nominal strength of a nodal zone is taken as (ACI-​23.9.1)

Fnn = Anz fce

(14.1.8)

where fce is the effective compressive strength of concrete at a face of a nodal zone and Anz is the smaller of a. the area of the face of the nodal zone taken perpendicular to the line of action of the strut or tie force, or b. the area of a section through the nodal zone taken perpendicular to the line of action of the resultant force. This case may be encountered when more than three forces meet at a node, as shown in Fig. 14.1.12. In the figure, the struts acting on faces AD and DC may be replaced by the resultant acting on face AC. The effective compressive strength fce in the nodal zone is computed as (ACI-​23.9.2)

fce = 0.85β n fc′

(14.1.9)

where βn is a factor that accounts for the type of node as shown in Table 14.1.2. A node bounded only by struts (a C-​C-​C node) represents the most favorable stress condition for the node and thus is assigned βn = 1.0 and fce = 0.85 fc′. On the other hand, nodes anchoring one or more ties are assigned a lower strength to reflect the reduction in the capacity of the nodal zone caused by the tensile stresses induced by the ties.

TABLE 14.1.2  VALUES OF βn FOR CALCULATING THE STRENGTH OF THE NODAL ZONE ACCORDING TO ACI-​23.9.2 Nodal Zone Type Bounded by struts, bearing areas, or both (C-​C-​C nodes) Anchoring one tie (C-​C-​T nodes) Anchoring two or more ties (C-​T-​T nodes or bounded only by ties)

βn 1.0 0.80 0.60

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A D

B

C

Nodal zone A

Node B

C

Figure 14.1.12  Example of nodal zone with four struts. Struts acting on faces AD and DC may be resolved into a single strut acting on face AC to check the nodal zone.

Strength of Ties, Fnt Ties may consist of nonprestressed or prestressed reinforcement, or both. Accordingly, the nominal strength of a tie, Fnt, is computed as follows (ACI-​23.7.2): Fnt = Ats f y + Atp ( fse + ∆f p )



(14.1.10a)

where Ats = area of nonprestressed reinforcement in a tie fy = yield strength of nonprestressed reinforcement Atp = area of prestressing reinforcement in a tie fse = effective stress in the prestressing reinforcement after losses1 Δfp = increase in stress in the prestressing steel (see Chapter 20) due to factored loads. Unless justified by analyses, the ACI Code permits Δfp to be taken as 60,000 psi for bonded tendons and 10,000 psi for unbonded tendons. In no case, however, can the sum (fse + Δfp) exceed the yield strength fpy of the prestressing reinforcement. When tie reinforcement consists of only nonprestressed bars, Eq. (14.1.10a) reduces to Fnt = Ats f y



(14.1.10b)

Tie reinforcement must be anchored by mechanical devices, post-​tensioning anchoring devices, standard hooks, or straight bar development (ACI-​23.8.2). When straight bars are used, the available development length La (i.e., the length available to anchor the bar) is measured from the point of intersection of the centroid of the bars in the tie and the extension of the outlines of either the strut or the bearing area [ACI-​23.8.3(b)], as shown in Fig. 14.1.9(b)]. Also, note that the bar may be developed by extending the reinforcement beyond the nodal zone if enough room is available. If La is insufficient to anchor the bar, the reinforcement may be anchored using 90° hooks or mechanical anchors, such as an anchor plate as shown in Fig. 14.1.9(a). If 90° hooks are provided, the hooks should be confined by reinforcement to avoid splitting of the concrete within the anchorage region.   Prestressed concrete requirements are treated in Chapter 20.

1

539



539

14.1 INTRODUCTION

For nodes with ties on opposite faces, tie reinforcement must be developed within the nodal zone for the difference between the tie force on one face and the tie force on the opposite face [ACI-​23.8.3(a)].

Selecting a Strut-​and-​Tie Model The first and most important task is the selection of a statically admissible truss. Since a strut-​and-​tie model represents a lower-​bound solution of the theory of plasticity, more than one admissible truss (solution) may exist, provided equilibrium and strength requirements are satisfied. Although all admissible solutions will be safe if the above conditions are met, some are preferable to others. Since ties are generally more flexible than struts, a model with the minimum number and shortest ties is preferred [14.7]. Figure 14.1.13 shows examples of preferred and undesirable models for a simply supported beam with a uniformily distributed load. The model in Fig. 14.1.13(b) is not only more complex, but also has longer ties than the simpler model shown in Fig.  14.1.13(a). Furthermore, the model of Fig. 14.1.13(b) has struts crossing each other, a condition that is not permitted by the ACI Code (ACI-​23.2.6). The truss geometry may be selected by visualizing the stress fields that develop within the structure under the load combination being considered. For isolated elements or simple structures, such as simple supported beams, the flow of stresses may be easily visualized, as illustrated earlier (see Fig. 14.1.1). For more complex structural systems, however, the selection of a suitable truss may be considerably more difficult, even for an experienced design engineer. In such cases, Schlaich et  al. [14.1] recommend that the model be based on the principal stress trajectories obtained from linear elastic finite element analyses. This approach can be particularly useful in design because the need to consider multiple load factors and load combinations can result in a number of different strut-​and-​tie models. Figures 14.1.14 through 14.1.17 show various strut-​and-​tie models for elements of different types. In addition to finite element analyses, several computer-​ based tools have been developed to assist the engineer in the selection of an appropriate and optimal layout for the strut-​and-​tie model (Ali and White [14.10], Alshegeir and Ramirez [14.11], Liang, Uy, and Steven [14.12], Tjhin and Kuchma [14.13], and Herranz et al. [14.14]).

h=L

h=L

L

L

(a) Preferred

(b) Undesirable (not permitted by the ACI Code)

Figure 14.1.13  Strut-​and-​tie models for a deep beam with uniformly distributed load. (After Schlaich and Schäfer [14.7].)

540

540

C H A P T E R 1 4     S trut- and - T ie M odels Uniform field

Fan

Strut

Tie

(a) Simply supported beam

(c) Deep beam

(b) Dapped end beam on corbel

(d) Wall with concentrated loads

Figure 14.1.14  Examples of strut-​and-​tie models. (Adapted from Cook and Mitchell [14.16].)

(b)

(a)

(c) (d)

Figure 14.1.15  Strut-​and-​tie action: (a) corbel; (b) knee joint under closing moment; (c) knee joint under opening moment; (d) interior beam-​column joint. (Adapted from Marti [14.2].)

541



541

14.1 INTRODUCTION

Figure 14.1.16  Strut-​and-​tie action in shear wall with openings. (Adapted from Marti [14.2].)

Strut

Tie

(a) Pile cap supported on four piles

(b) Simple three-dimensional truss model for a four-pile cap

Figure 14.1.17  Strut-​and-​tie model for pile cap. (From Adebar, Kuchma, and Collins [14.15].)

Dimensioning Struts, Ties, and Nodal Zones Once a suitable model (i.e., one that satisfies equilibrium) has been selected, the strength of its elements (struts, ties, and nodal zones) must be checked. The size (width and thickness) of the struts, ties, and nodal zones will depend on the forces acting on the member and the geometry of members meeting at a node. The designer should realize that dimensioning of the struts, ties, and nodal zones is an iterative procedure and, in many cases, a revision of the selected strut-​and-​tie model or the truss geometry will be required. In a planar model, the out-of-plane dimension of the truss elements (strut, tie, and nodal zone) is often taken as the thickness of the member, which reduces the task to simply determining the width of the element. The size of the struts can be initially defined in proportion to the magnitude of the force in the element using the strength for each strut type [i.e., struts of uniform cross sections (prisms) or bottle-​shaped struts]. In many instances, however, the size of the strut may have to be increased at one or both ends to reduce the stresses and

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prevent failure of the adjoining nodal zones. Under externally applied loads or the reactions at supports, bearing strength requirements are often used to determine the width of one or more faces of the nodal zone at those locations. Additionally, the force resultants from adjacent B-​regions, for example, the depth of the compression zone within the B-​region of a beam, can be used (see Fig. 14.1.5). Ties may also be initially sized in proportion to the force in the element. Calulations of the strength of the tie require only the total area of reinforcement and its yield strength [assuming that only nonprestressed reinforcement is used; see Eq. (14.1.10b)]. To size and check the strength of the nodal zones at the ends of a tie, the cross-​ sectional area of the tie, including a portion of the surrounding concrete, is needed (see Fig. 14.1.5). When the tie reinforcement is anchored with a bearing plate behind the joint [see Fig. 14.1.9(a)], the area of the tie is often taken as the size of the plate required to meet the bearing strength requirement for the nodal zone on that face. On the other hand, when the tie reinforcement is anchored by hooks or straight bar development, a hypothetical bearing area behind the joint must be defined [see Fig. 14.1.9(b)]. When a single layer of reinforcement is provided, the ACI Code Commentary recommends that the width of the tie be taken as the diameter of the bars plus twice the clear cover to the surface of the bars [ACI-​R23.8.1(a)]. If more than one layer of reinforcement is required, the ACI Code Commentary recommends that the tie reinforcement be uniformly distributed over the width and thickness of the tie. The writers recommend that bar sizes be selected so that the maximum spacing between bars within the tie does not exceed more than two bar diameters and that the tie area be taken as the perimeter defined by the outermost layer of reinforcement plus the cover to the surface of the bar.

14.2 DEEP BEAMS When the shear span-​to-​depth ratio (a/​d) is lower than about 2, a simply supported beam tends to behave like a tied arch as shown in Fig. 5.4.5(a). In such beams, often referred to as deep beams, the region between the support and the concentrated load must be treated as a D-​region (“disturbed region”) [14.1], since the assumption of ordinary beam theory that plane sections remain plane does not apply. The ACI Code defines a deep beam as a member with a ratio of clear span, Ln, to overall member depth, h, equal to or less than 4 [ACI-​9.9.1.1(a)] or as the region of a beam where a concentrated load exists within a distance equal to twice the member depth, h, from the face of the support [ACI-​9.9.1.1(b)]. Note that a concentrated load may be located near a support, creating a deep beam (D-​region) between the load and the nearer support, but an ordinary beam (B-​region) between the load and the more distant support. In this case, the provisions for deep beams would apply to the beam region between the nearer support and the concentrated load if the concentrated load is located within a distance of 2 times the member depth, h [ACI-​9.9.1.1(b)]. It is noted that the provisions for deep beams in ACI-​9.9 apply only to “members that are loaded on one face and supported on the opposite face such that strut-​like compression elements can develop between the loads and the supports.” If the loads, whether uniformly distributed or concentrated, are applied through the bottom or the sides of the member, the reinforcement should be designed by using strut-​and-​tie models to transfer the loads to the top of the member and distribute them to the supports. The design of deep beams may be done by (1) conducting nonlinear analyses that account for the nonlinear distribution of the longitudinal strains over the depth of the member (ACI-​ 9.9.1.2) or (2) using strut-​and-​tie models in accordance with Chapter 23 of the ACI Code (ACI-​9.9.1.3). The latter are based on the strut-​and-​tie approach, presented in general by J.  Schlaich et  al. [14.1] and in detail for deep beams by Rogowsky and MacGregor [14.3], Adebar, Kuchma, and Collins [14.15], Cook and Mitchell [14.16], M. Schlaich and Anagnostou [14.8], Siao [14.17, 14.18, 14.20], Walraven and Lehwalter [14.19], Foster and Gilbert [14.21], Wight and Parra-​Montesinos [14.22]. The ACI Code also requires minimum beam cross-​sectional dimensions to “control cracking under service loads and to guard against diagonal compression failures …”

543



543

14.2 DEEP BEAMS

(ACI-​R9.9.2.1) by limiting the maximum design shear stress to φ10 fc′(psi). Accordingly, minimum beam dimensions are computed so that (ACI-​9.9.2.1) Vu ≤ φ10 fc′ bw d



(14.2.1)

with φ = 0.75 for shear.

Minimum Reinforcement The minimum amount of flexural tension reinforcement for deep beams is the same as that required for ordinary beams (see Section 3.7). In addition, a minimum amount of both vertical and horizontal reinforcement distributed along the side faces of deep beams is required. This minimum reinforcement is intended to control the growth of inclined cracks and is to be provided irrespective of the method used for the design of the member. The minimum amount of vertical (i.e., perpendicular to the beam axis) reinforcement, Av, required by ACI-​9.9.3.1(a) is 0.0025bws, where bw is the beam width and s is the spacing of the vertical reinforcement. Similarly, the minimum required amount of horizontal reinforcement, Avh, is 0.0025bws2 where s2 is the spacing of the horizontal reinforcement [ACI-​ 9.9.3.1(b)]. Neither the vertical nor the horizontal spacing of the reinforcement should exceed the lesser of d/​5 and 12 in. (ACI-​9.9.4.3).

EXAMPLE 14.2.1 Design the flexural and shear reinforcement for a simply supported beam that carries two concentrated service live loads of 126 kips each on a clear span of 12 ft, as shown in Fig. 14.2.1. The beam has a width of 14 in. and an overall depth of 42 in.; it is supported on 14 by 16 in. columns. Use fc′ = 3500 psi and fy = 60,000 psi. Service loads k

4’–0”

126k

126

4’–0”

#3 horizontal bars

42” Tie width 3”

23 @ 6” = 11–6” #3 stirrups

3”

16”

12’–0”

16”

14”

Figure 14.2.1  Deep beam of Example 14.2.1.

SOLUTION (a) Determine whether the provisions for deep beams of ACI-​9.9.1 are applicable. The beam is loaded on one face and supported on the other with a concentrated load at 4(12)/​42  =  1.14 times the member depth. Therefore, the provisions of ACI-​9.9.1 apply. (b) Verify minimum beam cross-​sectional dimensions per ACI-​9.9.2.1. Neglecting the self-​weight of the beam, which is small in comparison to the applied concentrated loads, the maximum factored shear is

Vu = 1.6(126) = 201.6 kips

(Continued)

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Example 14.2.1 (Continued) Assuming an effective depth d = 0.8h, the maximum permitted factored shear force from Eq. (14.2.1) is   φ10 fc′ bw d = 0.75(10)

(

)

3500 (14)(0.8)42

1 1000

= 208.7 kips [> Vu = 201.6 kips] OK

(c) Select strut-​and-​tie model and geometry. Assume a strut-​and-​tie model consisting of a single truss as shown in Figure 14.2.2. The horizontal location of the truss nodes (joints) may be assumed to coincide with the line of action of the concentrated loads (nodes B and C) and the centerline at the supports (nodes A and D). The width and location of strut BC and the tie AD are unknown at this stage, but they can be initially estimated from flexural considerations at midspan. Neglecting the self-​weight, the factored maximum moment at midspan is  1 Mu = 1.6(126)  4 + 8  = 941 ft-kips 12  



18” d

18” e hBC s

B

C

f

c α

A

htAD

13.5” 2

a

Tie

α

D 16.8” 2

b 8”

8”

Figure 14.2.2  Strut-​and-​tie model for the beam of Example 14.2.1.

Strength of Strut BC  The strength of a strut will be given by the smaller of the strengths computed at the two ends of the strut. Because of symmetry, the strength of strut BC at ends B and C will be the same. The effective compressive stress of the concrete in a strut is given by Eq. (14.1.5)

fce = 0.85β s fc′

Since strut BC is located in the compression zone of the beam, it is considered as a prism of uniform cross section over its length (parallel stress field); thus βs  =  1 (see Table 14.1.1). Therefore, the effective compressive stress of strut BC is fce = 0.85(1) fc′ = 0.85 fc′ Assuming a strut with no compression reinforcement, the design strength of strut BC is φFnsBC = φ fce Acs = 0.75(0.85) fc′ bhsBC where b is the strut thickness (equal to the beam width), hsBC is the width of strut BC (see Fig. 14.2.2), and φ is taken as 0.75 per ACI-​21.2. (Continued)

54



545

14.2 DEEP BEAMS

Example 14.2.1 (Continued)

Strength of Tie AD  The strength of the tie will be governed by the strength of the nodes at its ends or by the yield strength of the tie reinforcement. Usually, however, the strength of the nodal zones anchoring the tie controls, and thus it will be used here to size the width of the tie. Nodes A and D anchor one tie, and thus, they are classified as C-​C-​T nodes. Their nominal strength is given by Eq. (14.1.8) as

FnnA = FnnD = fce Anz = 0.85β n fc′Anz

where βn = 0.80 for C-​C-​T nodes (see Table 14.1.2), and Anz is the area of the face of the nodal zone perpendicular to the line of action of the tie. Thus, the design strength of node A is φFnnA = 0.75(0.85)(0.80) fc′ bhtAD where htAD is the tie width (see Fig. 14.2.2). Using the expressions for the strengths of the strut and the tie obtained above, compute the minimum required widths of strut BC and tie AD based on flexural strength requirements. Since C = T, then

C = φFnsBC = 0.75(0.85) fc′ bhsBC = 0.75(0.85)(0.80) fc′ bhtAD = T



and

hsBC = 0.8htAD



Since

φ M n = (C or T )(arm )



then

 h BC h AD  φ M n = 0.75(0.85)(3.5)(14)hsBC  42 − s − t  2 2  

or

 h BC h BC  φ M n = 31.24hsBC  42 − s − s  2 2(0.8)  

Then, equating the design strength and the factored moment gives

φ M n = 31.2 4 hsBC (42 − 1.125hsBC ) = 941 (12) = Mu

therefore

hsBC = 13.45 in.

and

hsAD =

hsBC 13.45 = = 16.81 in. 0.8 0.8

Therefore, the required force in strut BC and tie AD is

FusBC = FutAD =

Mu 941(12) = 421 kips = 13.5 16.81 arm  −  42 −  2 2 

(Continued)

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Example 14.2.1 (Continued) (d) Check angle between diagonal struts AB and CD and tie axes, and compute forces in diagonal struts AB and CD. Knowing the widths of strut BC and tie AD, and the line of action of the applied loads and the reactions, the locations of the nodes can be established. The angle α between the diagonal struts and the tie is 13.45 16.81   − 42 −  2 2  = 25.6° > 25°(min, per ACI-23.2.7)    α = arctan   56    

OK

Therefore, the force in diagonal struts AB and CD is FusAB = FusCD =



421 = 467 kips cos(25.6)

(e) Check strength of nodal zones. Since the strut-​and-​tie model was not built with hydrostatic nodal zones (i.e., zones with equal stresses on each face), the strength of each face of the nodal zone must be checked separately.

Nodal Zone A (same as Nodal Zone D)  This node anchors one tie and thus is classified as a C-​C-​T node, as noted earlier. The nominal strength of the nodal zone is

FnnA = 0.85β n fc′Anz



with βn = 0.80 (see Table 14.1.2). Face a-​c. The strength of the nodal zone at the far end of the tie (face a-​b) is already satisfied, since its width htAD was sized to satisfy the strength of the nodal zone. Face a-​b. On this face,

Anz = b × hcol



where b is the node thickness (equal to the beam width) and hcol is the width of the support (i.e., equal to the column depth). Thus,

Anz = 14(16) = 224 sq in.



and

φ FnnA = 0.75(0.85)(0.80)(3.5)224 = 400 kips



The reaction at the supports is

RA = 1.6(126) = 201.6 kips < [φ FnnA = 400 kips]

Face b-​c. On this face,

Anz = b × ds

OK



where b is the node thickness (equal to the beam width) and ds is the width at end A of strut AB taken perpendicular to the line of action of the strut force (see Fig. 14.2.3). From the geometry of the truss and dimensions of the nodal zone, ds = hn cos α + hcol sin α (Continued)

547



547

14.2 DEEP BEAMS

Example 14.2.1 (Continued) where hn = htAD = 16.8 in. and hcol = 16 in.; thus ds = 16.8 cos (25.6) + 16 sin (25.6) = 22.1 in. and     A φFnn = 0.75(0.85)(0.80)(3.5)(14)22.1 = 552 kips > [ 467 kips = FusAB ]

OK

Therefore, nodal zones A and D are adequate.

ds c 16.8” 2

α

hn

Centroid of tie 16.8” 2 a

b hcol = 16”

1.5”

La

Figure 14.2.3  Nodal zone at end A of strut AB of beam of Example 14.2.1.

Nodal Zone B (same as Nodal Zone C)  This nodal zone is bounded by struts AD and BC, and the bearing area under the concentrated loads. Thus, it is classified as a C-​C-​C node. The nominal strength of the nodal zone is

FnnB = 0.85β n fc′Auz



with βn = 1.0 (see Table 14.1.2). Face e-​f. On this face, the strength should be adequate since the design strength of the nodal zone and strut BC are the same. Check      B φFnn = 0.75(0.85)(1.0)(3.5)(14)13.5 = 422 kips ≈ [ 421 kips = FusBC ]

OK

Face d-​e. Underneath the concentrated load, the nodal zone strength requirements will be satisfied by providing a bearing plate of length  B, such that or

φFnnB = 0.75(0.85)(1.0)(3.5)(14) B > 201.6 kips  B ≥ 6.5 in





Face d-​f. The width taken perpendicular to the line of action of the force in strut AB is

ds = 13.5 cos (25.6) + 6.5 sin (25.6) = 15 in. (Continued)

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Example 14.2.1 (Continued) and

φFnnB = 0.75(0.85)(1.0)(3.5)(14)15 = 469 kips > [ 467 kips = FusAB ]     

OK

Therefore, nodal zones B and C are adequate, provided a bearing plate 6.5 in. long or longer is used underneath the concentrated loads. (f) Check strength of diagonal strut AB (same as strut CD). The nominal compressive strength of the struts is

FnsAB = fce Acs



where Acs is the smaller of the areas at the ends of the strut and

fce = 0.85β s fc′



where the value βs depends on the strut type according to Table 14.1.1. Strut AB can be considered as a bottle-​shaped strut because there is room for the stresses to spread out from the nodal zones toward midlength of the strut. For simplicity, however, the strut is idealized as a tapered strut in Fig. 14.2.2. Assuming that the web will be provided with reinforcement as required by ACI-​23.5 (this will be checked later), βs may be taken as 0.75 (see Table 14.1.1). Therefore, the effective compressive strength of the strut is

fce = 0.85 (0.75) 3.5 = 2.23 ksi



At end A, the width ds taken perpendicular to the line of action of the strut was computed earlier in part (e) as 22.1 in. Thus, the design strength of the strut at end A is       AB φFns = 0.75(2.23)(14)22.1 = 517 kips > [ 467 kips = FusAB ]

OK

At end B, the strut width ds taken perpendicular to the line of action of the strut is 15 in. when a 6.5-in. long bearing plate is provided under the concentrated loads. Thus, the design strength of the strut at end B is      AB φFns = 0.75(2.23)(14)15 = 351 kips < [ 467 kips = FusAB ]

NOT OK

This means that the strut width is too small and that the stresses will be too high at end B. To reduce those stresses, the design strength of the strut must be increased by (a) adding compression reinforcement parallel to the strut axis (impractical in this case) or (b) increasing the bearing area under the concentrated load, if possible, to increase the strut area. The latter approach will be followed in this example. The required width of the strut at end B is ds =



467 (15) = 19.9 in. 351

which results in a required bearing length at nodes B and C of

B =  C =

19.9 − (13.5) cos(25.6) = 17.9 in. sin (25.6)

Earlier it was shown that the nodal zones at ends B and C satisfied the strength requirements with a smaller bearing plate. Clearly, a larger plate will be satisfactory. Therefore, use a bearing plate 14 in. wide by 18 in. long below each concentrated load. (Continued)

549



549

14.2 DEEP BEAMS

Example 14.2.1 (Continued) (g) Compute the required area of steel and select bars for the tie. The required area for tie AD is

Ats ≥

FutAD 421 = = 9.36 sq in φ f y 0.75(60)

Select 12–​#8 bars, Ats = 9.48 sq in. Since fc′ = 3500 psi < 4444 psi, then the minimum required flexural reinforcement ratio will be given by

min ρ =

200 200 = = 0.0033 fy 60, 000

or    min As = 0.0033 (14 ) 33.6 = 1.55 sq in. < [ 9.48 sq in. = Ats provided ]

OK

Use 12–​#8 bars. These bars are to be uniformly distributed over the tie width 16.8 in., as shown in Fig. 14.2.1. (h) Check anchorage of tie AD. According to ACI-​R23.8.2, the available development length is defined as the extension of the bearing area or the assumed outlines of the struts, whichever is larger. In this example, the larger of these two values is given by the outlines of the strut (see Fig. 14.2.3). Assuming a 1.5-​in. cover to the reinforcement at the far end of the tie, the available length La to anchor the tie reinforcement is

La =

16.8 / 2 + 16 − 1.5 = 32 in. tan(25.6)

It may be shown that the available length, La, is insufficient to develop a straight #8 bar, and thus hooked bars are required. Using bars terminating in a 90° hook, the required anchorage length is [see Eq. (6.10.1) in Chapter 6]

 fy ψ e ψ c ψ r  Ldh =   db [6.10.1]  50 λ fc′ 

but not less than 8db nor less than 6 in. For uncoated bars, ψe is equal to 1.0. Similarly, λ is equal to 1.0 for normal-​weight concrete. Since the clear cover is less than 2 in. and assuming that confinement reinforcement is not provided, the modification factors ψc and ψr are both equal to 1.0. Thus, for a #8 bar

 60, 000(1.0)(1.0)1.0  Ldh =  1.0 = 20.3 in. < [32 in. = La ]  50(1.0) 3500 

OK

Should the available length La be insufficient, the bars could be anchored by using additional layers of smaller bars. While an end plate at the exterior face of the beam could be provided, it may not be practical to do so for the12–​#8 bars in this case. (i) Minimum distributed reinforcement. In accordance with ACI-​9.9.3.1, a minimum amount of vertical and horizontal reinforcement must be provided along the side faces of deep beams. The minimum required amount of horizontal reinforcement, Avh, per ACI-​9.9.3.1(a) is

Avh ≥ 0.0025(14) = 0.035 in. s2

(Continued)

50

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Example 14.2.1 (Continued) Using 2–​#3 bars (one in each face), the required spacing is s2 ≤



2(0.11) = 6.3 in. 0.035

This reinforcement will be uniformly distributed above the tie reinforcement. Check maximum spacing in accordance to ACI-​9.9.4.3:



d 33.6 = = 6.72 in. 5 5 ≤ 12 in.

max s2 ≤

Controls

To meet the required spacing requirement of 6.3 in. over a depth of (42  –​16.8)  = 25.2 in., use four layers of #3 @ 6 in., as shown in Fig. 14.2.1. Similarly, the minimum required amount of vertical reinforcement,  Av, per ACI-​9.9.3.2(b) is

Av ≥ 0.0025(14) = 0.035 in. s

Using #3 U stirrups, the required spacing is s≤



2(0.11) = 6.3 in. 0.035

which is less than the maximum permitted spacing of 6.72 in. (maximum spacing requirements are the same for both the vertical and the horizontal reinforcement). Use #3 U stirrups @ 6 in. for both the horizontal and vertical reinforcement, as shown in Fig. 14.2.1. (j) Check requirements for reinforcement crossing bottle-​shaped struts. The diagonal struts AB and CD were sized as bottle-​shaped struts assuming that they would be provided with reinforcement crossing the struts in accordance with the requirements of ACI-​23.5 [see item (f)]. The required amount of this reinforcement may be computed by using the strut-​and-​tie model for bottle-​shaped struts shown in Fig. 14.1.4(b) [ACI-​23.5.1] or, when fc′ ≤ 6000 psi, by satisfying Eq. (14.1.6) [ACI-​ 23.5.3]. The latter approach is used in this example. In accordance with ACI-​23.5.3.1, the reinforcement crossing the strut may be provided in two orthogonal directions. Thus, use the grid of horizontal and vertical reinforcement computed in item (i) to verify whether such reinforcement is sufficient to satisfy the requirements of ACI-​23.5. If not, additional horizontal and vertical reinforcement would have to be provided. Using Eq. (14.1.6) with reference to Fig. 14.1.10 and noting that

As1 = Av and As 2 = Avh and

s1 = s (i.e., stirrup spacing ) bs = strut thickness (i.e., beam width)

Then    ∑

Asi 0.22 0.22 sin(90 − 25.6) + sin(25.6) = 0.00035 > 0.003 sin α i = bs si 14(6) 14(6)

OK

Therefore, the provided reinforcement along the side faces of the beam satisfies the requirements of ACI-​ 23.5 for bottle-​ shaped struts; no additional reinforcement is required. The proposed reinforcement layout for the beam is shown in Fig. 14.2.1.

51



551

14.2 DEEP BEAMS

EXAMPLE 14.2.2 The continuous beam of Fig. 14.2.4 is to support a uniformly distributed, factored load of 23 k/​ft from heavy machinery. Design the flexural and shear reinforcement for the beam. The factored moment and shear diagrams are given in Fig. 14.2.4. Use fc′ = 3500 psi and fy = 60,000 psi. 23 k/ft #4 stirrups #3 stirrups

#4 stirrups

3–#7

2.5” 26”

#3 horizontal bars

3”

3”

1

2–#7

4 2 in. spacing Ln = 8’–0”

15”

2.5”

14”

15”

(a) 92k + (b)

Critical section face of support

Vu

–

39 ft-kips

+ –

–

Mu

145 ft-kips (c)

Figure 14.2.4  Continuous deep beam of Example 14.2.2.

SOLUTION (a) Compute the ratio of beam span to overall depth. Ln 8(12) = = 3.69 < 4 h 26



Therefore, the provisions of ACI-​9.9 for deep beams apply. (b) Check minimum beam cross-​ sectional dimensions (ACI-​ 9.9.2.1). Assuming an effective depth d  =  0.8h, the maximum permitted factored shear force from Eq. (14.2.1) is    φ10 fc′ bw d = 0.75(10)( 3500 )(14)(0.8)26

1 1000

= 129.2 kips [ > Vu = 92 kips]

OK

(c) Select top and bottom reinforcement based on flexural considerations. Assuming a cover dimension to the center of the reinforcement of 2.5 in., the required Rn at the beam ends is

Rn =

145(12) 1000 = 300 psi 0.75(14)(23.5)2 (Continued)

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C H A P T E R 1 4     S trut- and - T ie M odels

Example 14.2.2 (Continued) where a φ factor of 0.75 has been used per ACI-​21.2.1 for strut-​and-​tie models. Then Equation (3.8.5) is used to estimate ρ as 0.0053. Since fc′ = 3500 psi < 4444 psi, then min ρ =



200 200 = = 0.0033 < 0.0053 fy 60, 000

OK

Thus,

required As = 0.0053(14)23.5 = 1.7 sq.in.



Choose 3–​#7 bars, As = 1.8 sq in. Check flexural strength a=

1.8(60) = 2.59 in. 0.85(3.5)14

2.59  1  φ M n = 0.75(1.8)(60)  23.5 −  = 150 ft-kips > 145 ft-kips  2  12

OK

At midspan, the required Rn is Rn =



39(12) 1000 = 81 psi 0.75(14)(23.5)2

From Fig. 3.8.1, minimum reinforcement controls. Thus,

As = 0.0033(14)23.5 = 1.09 sq in.



Choose 2–​#7 bars, As = 1.20 sq in. Check flexural strength, 1.2(60) = 1.73 in. 0.85(3.5)14 1.73  1  φ M n = 0.75(1.2)(60)  23.5 − = 102 ft-kips 39 ft-kips >   2  12 a=



OK

(d) Select strut-​and-​tie model and geometry. Figure  14.2.5 shows a truss model for the beam. Parallel chords through the centroid of the top and bottom longitudinal reinforcement at a distance of (26 –​2.5 –​2.5) = 21 in. are assumed.2 The uniformly distributed load is applied as a nodal force using the tributary width to each side of the node. The end moment (145 ft-​kips) is applied as an internal couple of 145(12)/​21 = 82.9 kips. (e) Check angle between the strut and tie axes. Table 14.2.1 shows the values for the angle between the different struts and ties of the selected model. All values are greater than 25°, thus satisfying ACI-​23.2.7. (f) Compute truss member forces and check strengths of struts, ties, and nodal zones. With the geometry of the truss defined, all truss member forces are determined. These forces are then used to size the truss elements and to check the design strengths of the struts, ties, and the nodal zones. Table 14.2.2 shows the forces in all truss members, as well as the effective compressive strength φ fce of the struts and nodes, and the tie strength φ f y . For the horizontal struts in the compression zone of the beam, a value of βs = 1.0 was used (see Table 14.1.1). Bottle-​shaped struts with reinforcement satisfying ACI-​23.5 (βs = 0.75) were used for all diagonal struts. (Continued) 2  This distance is less than the internal lever arms based on external strength calculations at the ends and at midspan. The implications of this assumption are discussed later in this example.

53



553

14.2 DEEP BEAMS

Example 14.2.2 (Continued) 23.0 k

25.9 k

1’–0”

0’–6”

28.8 k 1’–3”

28.8 k 1’–3”

28.8 k 1’–3”

25.9 k 1’–3”

23.0 k 0’–6”

1’–0”

82.9 k

82.9 k A

26”

B

C

D

E

F

G

21”

21”

82.9 k

H

I

J

K

L

M

92.0 k

82.9 k

92.0 k 1.5”

1.5”

Figure 14.2.5  Strut-​and-​tie model for the continuous beam of Example 14.2.2.



Except for nodal zones D, H, and M, which are bounded by struts or a bearing area (βn = 1.0), all other nodal zones anchor two or more ties and, thus, βn = 0.6. The last column of Table 14.2.2 shows the minimum required widths of the truss elements. The strut widths are based on the smaller of the effective strengths (shown in boldface) computed for the strut itself or the nodal zones at the ends of the strut. Note that all struts have at least one end connected to a node with two or more ties (Fig. 14.2.5). Such nodes have βn = 0.6, whereas for the struts βs is either 1.0 or 0.75. Thus, the width of all struts is controlled by the strength of the nodal zones. The minimum required width for the ties is thus based on the effective compressive strength of the nodal zone. The strength φ f y of the tie itself is given by the tie reinforcement and has no bearing in the calculation of the tie width. TABLE 14.2.1  ANGLES BETWEEN STRUTS AND TIES FOR MODEL OF FIG. 14.2.5 Strut

Tie

Angle (degrees)

BH and FM

AB and FG BI and FL BC and EF CJ and EK JK CJ and EK

47.1 42.9 54.5 35.5 54.5 35.5

CI and EL DJ and DK

To illustrate the steps involved in selecting the dimensions of the truss elements, detailed calculations are shown below for nodal zone I (same as nodal zone L).

Nodal Zone I (same as nodal zone L)  Figure 14.2.6 shows a detailed view of nodal zone I where the minimum required widths of the truss elements obtained from Table 14.2.2 have been increased to the nearest half-​inch. (Continued)

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C H A P T E R 1 4     S trut- and - T ie M odels

Example 14.2.2 (Continued)

TABLE 14.2.2  FORCES, DESIGN STRENGTHS, AND MINIMUM REQUIRED WIDTHS FOR THE STRUT-​AND-​TIE ELEMENTS OF EXAMPLE 14.2.2 Strut or Tie Strength Member Tie AB (FG) Tie BC (EF) Strut CD (DE) Strut HI (LM) Tie IJ (KL) Tie JK Strut AH (GM) Strut BH (FM) Strut CI (EL) Strut DJ (DK) Tie BI (FL) Tie CJ (EK) a

Strength of Nodal Zones

Force (kips)

βs

φ fcea or φfy (ksi)

Node

βn

φfcea (ksi)

Node

βn

φ fcea (ksi)

dmin (in.)

74.6 10.6 20.2 10.6 20.2 30.5 24.0 94.2 53.0 17.7 43.1 14.4

—​ —​ 1.00 1.00 —​ —​ 0.75 0.75 0.75 0.75 —​ —​

45.00 45.00 2.23 2.23 45.00 45.00 1.67 1.67 1.67 1.67 45.00 45.00

A (G) B (F) C (E) H (M) I (L) J A (G) B (F) C (E) D B (F) C (E)

0.6 0.6 0.6 1.0 0.6 0.6 0.6 0.6 0.6 1.0 0.6 0.6

1.34 1.34 1.34 2.23 1.34 1.34 1.34 1.34 1.34 2.23 1.34 1.34

B (F) C (E) D I (L) J (K) K H (M) H (M) I (L) J (K) I (L) J (K)

0.6 0.6 1.0 0.6 0.6 0.6 1.0 1.0 0.6 0.6 0.6 0.6

1.34 1.34 2.23 1.34 1.34 1.34 2.23 2.23 1.34 1.34 1.34 1.34

4.0 0.6 1.1 0.6 1.1 1.6 1.3 5.0 2.8 0.9 2.3 0.8

φ fce = φ (0.85)β s f ′c for struts or φ (0.85)β n fc′ for nodal zones. φ = 0.75.

This node is intersected by two ties, and thus it is classified as a C-​T-​T node (βn = 0.6). Its strength is given by

φFnnI = φ (0.85)β n fc′Anz

= 0.75(0.85)(0.6)3.5 Anz = 1.34 Anz

It is noted that more than three forces act on this nodal zone. In the following, the simplified approach proposed by Schlaich and Schäfer [14.7] for nodal zones (Section 14.1), where the stresses acting on the cross-​sectional area taken perpendicular to the strut or (tie) axis on each face, is used to verify the adequacy of the node. An alternative approach where the four forces are resolved into three intersecting forces is presented later. Face a-​b. On this face, the nodal zone area taken perpendicular to the line of action of the strut HI is

Anz = 1(14) = 14 sq in.

thus

φFnnI = 1.34 Anz = 1.34(14) = 18.8 kips > [10.6 kips = FusHI ]



OK (Continued)

5



555

14.2 DEEP BEAMS

Example 14.2.2 (Continued) 43.1k Tie BI

Strut CI 53.0k 1

22” 3”

e

10.6k

d

a

Strut Hl

Tie IJ 1

1”

20.2k

12 ” b

c

2.5”

Bottom of beam

Figure 14.2.6  Details of nodal zone I with minimum required widths from Table 14.2.2.

Face c-​d. The nodal zone area taken perpendicular to the line of action of the tie IJ is

Anz = 1.5(14) = 21 sq in.

and

φFnnI = 1.34(21) = 28.1 kips > [20.2 kips = FutIJ ] 

OK

Face e-​d. Nodal zone area taken perpendicular to the line of action of the strut CI is Anz = 3(14) = 42 sq in. and

φFnnI = 1.34(42) = 56.3 kips > [53.0 kips = FusCI ]

OK

Face a-​e. Nodal zone area taken perpendicular to the line of action of the tie strut BI is

Anz = 2.5(14) = 35 sq in.

and

φFnnI = 1.34(35) = 46.9 kips > [ 43.1 kips = FutBI ]

OK

Similar calculations can be done for the rest of the nodes. Note that the maximum horizontal force at nodes A, G, H, and M is due to the internal force of 82.9 kips (see Fig. 14.2.5). At A and G, the required depth of the nodal zone is 82.9/​[14(1.34)] = 4.2 in., which is less than the available width of 5 in. (i.e., twice the assumed cover dimension to the tie axis). Similarly, at H and M, the required width is 82.9/​ [14(2.23)] = 2.7 in., which can be easily accommodated within the available 5 in. The vertical reaction at nodes H and M is assumed to be 1.5 in. from the inside column face to satisfy the minimum width of 92/​[14(2.23)] = 2.9 in. on the side of the node facing the column. (Continued)

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Example 14.2.2 (Continued) The dimensions of the struts, ties, and nodal zones based on the values computed in Table  14.2.2 (increased to the nearest half-​inch.) are shown as light gray shading in Fig. 14.2.5. Note that all elements can be accommodated within the available dimensions and geometry of the beam. Thus, the selected truss and elements can be considered adequate. As mentioned earlier (Fig. 14.1.12), when four or more forces act on a node in a two-​ dimensional model, it is possible to resolve some of the forces into three intersecting forces. With reference to Fig. 14.2.7(a), the four forces acting on node I may be conveniently resolved into a vertical force from tie BI, a diagonal force from strut CI, and a net horizontal force equal to the summation of the forces from strut HI and from tie IJ, as shown in Fig. 14.2.7(b). In this case, the required node area perpendicular to the net horizontal force is computed from

[ FunI = 30.8 kips] ≤ [φFnnI = 1.34 Anz ]

thus Anz =



30.8 = 23.0 in. 1.34

which results in a required width of 23.0/​14 = 1.6 in. This value is slightly larger than that computed earlier (i.e., somewhat more conservative), but it can be easily accommodated within the available dimensions and assumed geometry, as before. Note that the required minimum widths for the node faces perpendicular to tie BI and strut CI remain the same. A similar analysis could be done for other nodes with more than three forces. In node B, for example, the horizontal force acting on the node would be taken as the difference between the forces in ties AB and BC or (74.6 –​10.6) = 64.0 kips. This would result in a minimum required width of 64.0/​[1.34(14)] = 3.4 in., which is smaller than that given in Table 14.2.2 for tie AB. 43.1k Tie BI Strut HI

43.1k Strut CI

Strut CI

Tie BI

53.0 k Tie IJ 20.2 k

10.6 k (a)

53.0 k 30.8 k

(Tie IJ + Strut HI)

(b)

Figure 14.2.7  Resolution of nodal forces into three intersecting forces for nodal zone I.

(g) Design the web reinforcement. In Table 14.2.2, the effective compressive strength of the diagonal struts was computed using βs = 0.75, which requires that reinforcement in accordance with ACI-​23.5 be provided. Note that since the depth of all struts was governed by the strength of the nodal zones with βn = 0.6, the strength and size of the diagonal struts would conform to the requirements for bottle-​shaped struts without web reinforcement according to ACI-​23.4. Thus, reinforcement according to ACI-​23.5 need not be strictly satisfied in this case. Vertical reinforcement must be provided, however, to carry the forces in the vertical ties BI, CJ, EK, and FL. In addition, ACI-​9.9.3.1 requires that minimum vertical and horizontal reinforcement be provided along the side faces of this deep beam.

Minimum Vertical Reinforcement.  According to ACI-​9.9.3.1(a), the minimum required area of vertical reinforcement is min Av ≥ 0.0025bw s (Continued)

57



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14.2 DEEP BEAMS

Example 14.2.2 (Continued) where



 d 23.5  max s ≤  = = 4.7 in. 5 5 

Controls

≤ 12 in. Choose s = 4.5 in., then

min Av ≥ 0.0025(14)4.5 = 0.16 sq in.



Thus, #3 U stirrups @ 4 1 2 in. Av = 2(0.11) = 0.22 sq in., satisfy the minimum amount required by ACI-​9.9.3.1(a). Forces in Vertical Ties  Ties CJ and EK. These ties must carry a force of 14.4 kips (see Table 14.2.2). Thus the total required area of steel is

F  14.4 Ats ≥  ut = = 0.32 sq in.  φ f y 0.75(60) 

Assuming that this reinforcement is distributed over the adjacent panels and spaced at 4 1 2 in., the required amount of vertical steel is

Av = 0.32

4.5 = 0.096 sq in.< min Av 15

Use #3 U stirrups @ 4 1 2 in. in the center portion of the beam, about 2 ft 2 in. from the face of the column. Ties BI and FL. The total required area of steel for these ties is

Ast ≥

43.1 = 0.96 sq in. 0.75(60)

Assuming the reinforcement is also distributed at 4 1 2 in., the required vertical steel is

Av = 0.96

4.5 = 0.32 sq in. > min Av 13.5

Use #4 U stirrups @ 4 1 2 in., Av = 2(0.20) = 0.40 sq in., at the end portions of the beam to about 2 ft 2 in. from the face of each column.

Minimum Horizontal Reinforcement  The minimum required area of horizontal reinforcement per ACI-​9.9.3.1(b) is

min Avh ≥ 0.0025bw s2



where

max s2 ≤ 4.7 in. but not greater than 12 in.

Controls

(Continued)

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Example 14.2.2 (Continued) Using #3 bars (in each face), the required spacing is s2 =



2(0.11) = 6.3 in. > max s2 0.0025(14)

Thus, use #3 bars horizontally in each face spaced at 4 1 2 in. to satisfy max s2.

Minimum Reinforcement for Bottle-​Shaped Struts According to ACI-​23.5.  Although not required in this example, it is instructive to compare the minimum reinforcement requirements of ACI-​9.9.3.1(a) and 9.9.3.1(b) with those of ACI-​23.5.3. For struts CI, EL, DJ, and DK, ACI-​23.5.3 would require Asi

∑bs



sin α i ≥ 0.003 [14.1.6]

s i

Using the minimum reinforcement provided in the center portion of the beam per ACI-​9.9.3.1, 2(0.11) (sin 35.5 + sin 54.5) = 0.0049 ≥ 0.003 14(4.5)



OK

Therefore, the provisions of ACI-​23.5.3 are already satisfied. It can be shown that the reinforcement provided in the end regions of the beam also satisfies ACI-​23.5.3. (h) Check strength of horizontal ties. The reinforcement for the horizontal ties was computed in part (c) based on flexural considerations and should be adequate to carry the tie forces. It is assumed that the top and bottom reinforcement will be as follows. Bottom Longitudinal Reinforcement. The required area of steel for tie JK (the tie with the largest force in the bottom of the beam) is

Ast ≥

30.5 = 0.68 sq in. < [1.20 sq in. = As for 2 − # 7] 0.75(60)

OK

Although not required near the supports, this reinforcement will be extended into the joints. Top Longitudinal Reinforcement. Ties AB and FG carry the largest force in the top of the beam. The required area of steel for these ties is

Ast ≥

74.6 = 1.66 sq in. < [1.80 sq in. = As for 3 − # 7] 0.75(60)

OK

Note that this reinforcement will be extended into the joint; according to the strut-​and-​ tie model of Figure 14.2.5, it should be able to carry a force of 82.9 kips computed in part (d). The required area of steel to carry this force is

Ast ≥

82.9 = 1.84 sq in. ≈ [1.80 sq in. = As for 3 − # 7] 0.75(60) (Continued)

59



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14.3  BRACKETS AND CORBELS

Example 14.2.2 (Continued) The required area of steel is somewhat larger than that provided because the internal lever arm assumed in the model is smaller than that based on flexural strength calculations. It is possible to construct a model where the location of the tie and the internal lever arm are consistent with those obtained from flexural considerations, as shown in Fig. 14.2.8. Such a model requires a variable depth and location for the top and bottom chords, a refinement that is hardly justified in practice. It can be shown that the forces in the truss elements in the more refined model of Fig. 14.2.8 are nearly identical to those of the parallel chord truss of Fig. 14.2.5, and thus the design based on the parallel chord truss is adequate. The final design is shown in Fig. 14.2.4. 23.0 k

25.9 k

1’–0”

0’–6”

28.8 k 1’–3”

28.8 k 1’–3”

28.8 k 1’–3”

1’–3”

23.0 k

1’–0”

0’–6”

2.5”

78.4 k B

A 26”

25.9 k

D

C

E

78.4 k

F

G

22.6”

22.2” 78.4 k

I

H

22.2” K

J

L

M

2.5”

92.0 k 1.5”

78.4 k

92.0 k 1.5”

Figure 14.2.8  Strut-​and-​tie model of variable depth for the continuous beam of Example 14.2.2.

14.3 BRACKETS AND CORBELS Design of brackets and corbels in accordance to the provisions of ACI-​16.5 was discussed in Section 5.16. As mentioned in Chapter 5, the provisions of ACI-​16.5 are primarily based on test data and thus are limited to brackets and corbels with shear span-​to-​depth ratios a/​d ≤ 1.0 and with a factored horizontal tensile force Nu ≤ Vu. Alternatively, the ACI Code allows brackets and corbels to be designed using the provisions for strut-​and-​tie models (ACI-​R16.5.1.1). Unlike the provisions of ACI-​16.5, the strut-​and-​tie provisions of Chapter  23 of the ACI Code can be used for brackets and corbels with any a/​d and Nuc /Vu ratios. This section presents the use of strut-​and-​tie models as a tool for the design of brackets and corbels. The flow of forces in a corbel subjected to a factored shear Vu and a horizontal force Nuc may be visualized with the aid of the strut-​and-​tie model shown in Fig. 14.3.1. The shear force Vu must be transferred into the column by a diagonal strut (AB). The vertical component of strut AB must be balanced by the shear force Vu at node A and by the compression resultant C in the column at node B. A horizontal tie (AC) is required to equilibrate the horizontal force Nuc and the horizontal component of the strut AB. The moment generated by the applied forces (Vu and Nuc) must be balanced by the internal couple (C-​T) in the column. Thus, a tie will be required on the opposite face of the column to carry the tension resultant T. The force imbalance generated by the ties meeting at node C must be balanced by strut CB. Finally, the horizontal shear caused by Nuc, resisted by shear reinforcement in the column, is represented by tie BD in the model.

560

560

C H A P T E R 1 4     S trut- and - T ie M odels Vu α

Nuc

A

C

D

B

C

T

Figure 14.3.1  Strut-​and-​tie model for a reinforced concrete corbel.

EXAMPLE 14.3.1 Redesign the bracket of Example 5.16.1 using the ACI Code provisions for strut-​and-​tie models. Compare the final design with that obtained previously using the provisions of ACI-​16.5. SOLUTION (a) Factored loads.

Vu = 113 kips (from Example 5.16.1)



(b) Select a strut-​and-​tie model. Assume a model as shown in Fig. 14.3.2. (c) Determine preliminary bracket size. The nodal zone beneath the bearing plate is a C-​C-​T node; thus, its effective compressive stress is

fce = 0.85β n fc′



where βn = 0.8 (see Table 14.1.2). Therefore,

bearing plate width =

113, 000 = 3.17 in. 0.75(0.85)(0.8)(5000)14

where φ = 0.75 for strut-​and-​tie models.3 Use 3 1 2 in. for bearing plate width. Allowing a tolerance gap of 1-​in. clear between the face of the column and the beam for possible overrun in beam length and also because the beam might be 1 in. too short, then

1 a = 2 + (bearing plate width ) = 2 + 1.75 = 3.75 in. 2

(Continued)

3  Note that although a larger φ factor is used here (compared to φ = 0.65 in Example 5.16.1), the strength of the nodal zone is only 80% of that required by the bearing provisions of ACI-​22.8.3.2. Thus, a slightly larger bearing plate width is required in this example.

561



561

14.3  BRACKETS AND CORBELS

Example 14.3.1 (Continued) a = 3.75” 14” square column

113k 3½”

C

48k

A

1.5”

8”

65k

k

81 k

123

15”

178k

B

D

E 2”

dsBD C

T

Figure 14.3.2  Strut-​and-​tie model for the corbel of Example 14.3.1.

(d) Determine the depth of the bracket. Choose, say h = 15 in., so that d ≈ 13.5 in. (as before in Example 5.16.1). Also, ACI-​16.5.2.2 requires that the depth at the outer edge of the bearing area should not be less than 0.5d. Select a depth of 8 in. for the outer face. (e) Define the geometry of the strut-​and-​tie model. Using a 1-​in. clear cover and one layer of bars for tie AC, assume the center of the tie to be at 1.5 in. from the top. Similarly, assuming a 1.5-​in. clear cover for tie CE and one layer of bars, assume the center of the tie is 2 in. from the face of the column. The width dsBD of strut BD will be governed by the strength of the strut itself or by the strength of the nodal zone B; therefore, the design strength is the smaller of

φFnsBD = φ (0.85)β s fc′Acs



or

φFnnB = φ (0.85)β n fc′Anz where



φ = 0.75 Acs = Anz = area of the strut at end B taken perpendicular to the line of action of the strut (Continued)

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Example 14.3.1 (Continued) Strut BD is located in the compression zone of the column and may be considered a prism of uniform cross s​ ection (parallel stress field). Thus, βs = 1 (Table 14.1.1). On the other hand, nodal zone B is a C-​C-​C node and βn = 1 (Table 14.1.2). Therefore, the strength at the end of the strut is the same as that on that face of the nodal zone. Thus,

φFnsBD = 0.75(0.85(1)5 Acs = 3.19 Acs = 3.19(14)dsBD = 44.6 dsBD



By taking moments about the axis through tie CE,



 d BD  Mu = 113(3.75 + 14 − 2) = C  14 − 2 − s  2    d BD  1780 = 44.6 dsBD  12 − s  2  

which leads to a quadratic equation for dsBD . Solving this equation gives

dsBD = 3.99 ≈ 4 in.



This defines the location of the resultant C and node B. The geometry of the truss is now completely defined, and it can be analyzed. (f) Check the angle between the strut and tie axes at nodes A and C. Angle between Strut AB and Tie AC 

 15 − 1.5  α = arctan  = 67° > 25° (ACI − 23.2.7 minimum )  3.75 + 2 

OK

Angle between Strut CE and Tie BC 

 14 − 2 − 2  α = arctan  = 36.5° > 25°  15 − 1.5 

OK

and Angle between Strut AC and Tie BC  α = 90 − 36.5 = 53.5° > 25°

OK

(g) Compute truss member forces and check strength of struts, ties, and nodal zones. With the geometry of the truss defined, the truss member forces are computed from equilibrium. The calculated forces in each member of the truss are shown in Fig. 14.3.2. These forces must then be checked against the design strength of the struts, ties, and the nodal zones, as follows.

Strut AB  Strut AB needs to carry a compression force FusAB of 123 kips. Its strength will be governed by the strength of the strut itself or by the strength of the nodes at its ends. (Continued)

563



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14.3  BRACKETS AND CORBELS

Example 14.3.1 (Continued) Therefore, the minimum required width of the strut will be given by the condition that FusAB does not exceed the smallest of the following:

φFnsAB = strength of the strut

φFnnA = strength of node A at thee end of the strut φFnnB = strength of node B at the end of the strut

This strut may be assumed to be a bottle-​shaped strut because there is room for the stresses to spread out from the nodal zones toward midlength of the strut. Assuming that the corbel will be provided with reinforcement according to ACI-​23.5, βs may be taken as 0.75 (Table 14.1.1). Thus, the design compressive strength of the strut is

φ fce = 0.75(0.85)(0.75)5 = 2.39 ksi

Nodal zone A is a C-​C-​T node and βn = 0.8, while nodal zone B is a C-​C-​C node and βn = 1.0 (Table 14.1.2). As a result, design compressive strengths in the nodal zones A and B are

φ fce = 0.75(0.85)(0.8)5 = 2.55 ksi at node A φ fce = 0.75(0.85)(1.0)5 = 3.19 ksi at node B

AB The minimum required width dmin for the strut is governed by the smallest of these values, which in this example is the strength of the strut itself, φfce = 2.39 ksi. The strut AB width dmin (taken perpendicular to the line of action of the strut) is then computed from



AB [ FusAB = 123 kips] ≤ [φ FnsAB = Acs φ fce = dmin (14)2.39] AB dmin ≥ 3.7 in.

From the geometry of the bracket and the chosen strut-​and-​tie model, one can verify that the available space to accommodate strut AB at node B is

AB davailable = 2[2 cos(90 − 67)] = 3.7 in. = dmin



OK

Similar calculations can be made for struts BC and BD, and for ties AC and CE. These results are summarized in Table  14.3.1, which lists the design compressive strength φ fce of the struts and nodes, and the design strength φ fy of the ties. The last column of the table shows the minimum required depths for the members based on the smaller of the design strengths (shown in boldface in the table) computed for the strut itself or the nodal zones.4 Figure 14.3.3, which shows the minimum dimensions from Table 14.3.1 for the struts, ties, and nodal zones for the model, illustrates that all can be accommodated within the dimensions and geometry selected for the corbel. (h) Compute the required area of steel and select bars for the ties. Tie AC  The required area for this tie is

Ats ≥

FutAC 48 = = 1.07 sq in. φ f y 0.75(60)

Use 4–​#5 (Ats = 1.24 sq in.) (Continued) 4  The minimum depth shown for the ties ensures that the tie force can be carried safely in the nodal zone, and it is based on the design compressive strength of the nodal zone. The design strength φ fy of the tie itself depends only on the tie reinforcement and is unrelated to this calculation. It is shown in Table 14.3.1 for completeness and will be used later to compute the required tie reinforcement.

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Example 14.3.1 (Continued) This horizontal reinforcement must be anchored and extended as far as possible into the column so that strut BC can be developed as assumed in the model. In practice, these bars can often be developed using 90° hooks. Also note that these bars will fit all in one layer at the assumed location of the tie axis in the model. If, for example, more than one layer or larger diameter bars were required, resulting in a change in the tie axis, the model would have to be revised. Tie CE  This tie force will be resisted by longitudinal reinforcement provided at the far end of the column. As a minimum, however, this reinforcement amount should be Ats ≥



FutCE 65 = = 1.44 sq in. φ f y 0.75(60)

TABLE 14.3.1  FORCES, DESIGN STRENGTHS, AND MINIMUM REQUIRED WIDTHS FOR THE STRUT-​AND-​TIE ELEMENTS IN EXAMPLE 14.3.1 Strut or Tie Strength

Node Strength

Member

Force (kips)

βs

φ fce or φ fy (ksi)

Node

βn

φ fce (ksi)

Node

βn

φ fce (ksi)

dmin (in.)

Strut AB Strut BC Strut BD Tie AC Tie CE

123 81 178 48 65

0.75 0.75 1.00 —​ —​

2.39 2.39 3.19 45.00 45.00

A B B A C

0.8 1.0 1.0 0.8 0.6

2.55 3.19 3.19 2.55 1.91

B C —​ C —​

1.0 0.6 —​ 0.6 —​

3.19 1.91 —​ 1.91 —​

3.7 3.0 4.0 1.8 2.4

C 1.8”

A

3.7”

3”

2.4” B 4” D

E

Figure 14.3.3  Minimum required widths for struts and ties of corbel of Example 14.3.1.

(Continued)

56



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14.4  ADDITIONAL REMARKS

Example 14.3.1 (Continued) (i) Design closed stirrups or ties. In accordance with ACI-​16.5.5.2, closed stirrups or ties must be provided parallel to the main reinforcement (tie AC). From Eq. (5.16.20),

required Ah ≥ 0.5( Asc − An ) = 0.5(1.24 − 0) = 0.62 sq in.



Try 3–​#3 closed hoops [Ah = 2(3)0.11 = 0.66 sq in.]. This reinforcement must be distributed within two-​thirds of the effective depth [i.e., within 2(15 –​1.5)/​3 = 9 in. spaced at 3 in.]. Since struts AB and BC were sized using a βs value of 0.75, the horizontal reinforcement must also satisfy ACI-​23.5. The ACI Code allows designers to compute this reinforcement amount using the strut-​and-​tie model for bottle-​shaped struts shown in Fig. 14.1.4. In this example, however, the required amount of reinforcement is computed using Eq. (14.1.6), Asi

∑bs



sin α i ≥ 0.003 [14.1.6]

s i

where Asi is the total area of reinforcement at a spacing si of a layer of reinforcement at an angle αi with the axis of the strut, and bs is the strut thickness. Since only horizontal reinforcement is provided, ACI-​23.5.3.1 requires that the angle αi be greater than 40°. The angles between the struts AB and BC and the horizontal were computed earlier as 67° and 53.5°, respectively. Thus ACI-​23.5.3.1 is satisfied. Using the horizontal reinforcement amount calculated above in accordance with ACI-​16.5.5.2 and the smallest angle of 53.5° for strut BC

Asi

∑bs

sin α i =

s i

2(0.11) sin 53.5 = 0.0042 > 0.003 14(3)

OK

Use 3–​#3 @ 3 in. closed hoops.  ( j) Final design. In this case, the dimensions and detailing of the bracket are essentially the same as those designed using the traditional provisions of ACI-​16.5 shown in Fig. 5.16.7. The only difference is in the bearing plate size (3.5 in.) and location (3.75 in. from the face of the column). This discrepancy is not important within the accuracy and details of the methods. However, the strut-​and-​tie provisions are much more versatile and can be applied to corbels of any size and with any Nuc  / ​Vu ratio.

14.4 ADDITIONAL REMARKS The strut-​and-​tie approach is a powerful tool that can be used for the design of reinforced concrete members of virtually any shape under a myriad of loading conditions. The approach is particularly useful, for example, for members with unusual shapes or with openings (see Fig. 14.1.16). Even when alternative methods are used in design, strut-​and-​tie models are very useful aids in visualizing the load paths within the member from the point of application of the loads to the supports. As noted earlier, determining an appropriate geometry for the model can be a time-​consuming task, even for an experienced engineer. While several computer-​based tools have been developed [14.10–​14.14], the reader is referred to Refs. 14.6 and 14.23, which contain several examples for a variety of members using strut-​ and-​tie models. Additional examples may be found in Refs. 14.2 and 14.22.

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SELECTED REFERENCES 14.1. Jörg Schlaich, Kurt Schäfer, and Mattias Jennewein. “Toward a Consistent Design of Structural Concrete,” PCI Journal, 32, May–​June 1987, 74–​150. 14.2. Peter Marti. “Basic Tools of Reinforced Concrete Beam Design,” ACI Journal, Proceedings, 82, January–​February 1985, 46–​56. 14.3. D.  M. Rogowsky and J.  G. MacGregor. “Design of Reinforced Concrete Deep Beams,” Concrete International, 8, August 1986, 49–​58. 14.4. K. Bergmeister, J. E. Breen, and J. O. Jirsa. “Dimensioning of Nodal Zones and Anchorage of Reinforcement,” Structural Concrete, International Association for Bridge and Structural Engineering Colloquium on Structural Concrete, Stuttgart, Germany, 1991, 551–​564. 14.5. Julio A.  Ramirez and John E.  Breen. “Evaluation of a Modified Truss-​Model Approach for Beams in Shear,” ACI Structural Journal, 88, September–​October 1991, 562–​571. 14.6. James G. MacGregor. “Derivation of Strut-​and-​Tie Models for the 2002 ACI Code,” Examples for the Design of Structural Concrete with Strut-​and-​Tie Models, Karl-​Heinz Reineck, Editor. (SP-​208). Farmington Hills, MI: American Concrete Institute, 2002 (pp. 7–​40). 14.7. J. Schlaich and K. Schäfer. “Design and Detailing of Structural Concrete Using Strut-​and-​Tie Models,” The Structural Engineer, 69, 6 (March 1991), 113–​125. 14.8. Michael Schlaich and Georg Anagnostou. “Stress Fields for Nodes of Strut-​and-​Tie Models,” Journal of Structural Engineer, ASCE, 116, 1 (January 1990), 13–​23. 14.9. J. O. Jirsa, K. Bergmeister, R. Anderson, J. E. Breen, D. Barton, and H. Bouadi. “Experimental Studies of Nodes in Strut-​and-​Tie Models,” Structural Concrete, International Association for Bridge and Structural Engineering Colloquium on Structural Concrete, Stuttgart, Germany, 1991, 525–​532. 14.10. Mohamed A. Ali and Richard N. White. “Automatic Generation of Truss Model for Optimal Design of Reinforced Concrete Structures,” ACI Structural Journal, 98, July–​August 2001, 431–​442. 14.11. A. Alshegeir and J. A. Ramirez. “Computer Graphics in Detailing Strut-​Tie Models,” Journal of Computing in Civil Engineering, 6, 2 (April 1992), 220–​232. 14.12. Qing Quan Liang, Brian Uy, and Grant P. Steven. “Performance-​Based Optimization for Strut-​ Tie Modeling of Structural Concrete,” Journal of Structural Engineering, ASCE, 128, 6 (June 2002), 815–​823. 14.13. Tjen N.  Tjhin and Daniel A.  Kuchma. “Computer-​Based Tools for Design by Strut-​and-​Tie Method:  Advances and Challenges,” ACI Structural Journal, 99, September–​October 2002, 586–​594. 14.14. Juan Pablo Herranz, Hernan Santa Maria, Sergio Gutierrez, and Rafael Riddell. “Optimal Strut-​ and-​Tie Models Using Full Homogenization of Optimization Method,” ACI Structural Journal, 109, September-​October, 2012, 605–​614. 14.15. Perry Adebar, Daniel Kuchma, and Michael P. Collins. “Strut-​and-​Tie Models for the Design of Pile Caps: An Experimental Study,” ACI Structural Journal, 87, January–​February 1990, 81–​92. 14.16. William D. Cook and Denis Mitchell. “Studies of Disturbed Regions near Discontinuities in Reinforced Concrete Members,” ACI Structural Journal, 85, March–​April 1988, 206–​216. 14.17. Wen Bin Siao. “Strut-​and-​Tie Model for Shear Behavior in Deep Beams and Pile Caps Failing in Diagonal Tension,” ACI Structural Journal, 90, July–​August 1993, 356–​363. 14.18. Wen Bin Siao. “Shear Strength of Short Reinforced Concrete Walls, Corbels, and Deep Beams,” ACI Structural Journal, 91, March–​April 1994, 123–​132. 14.19. Joost Walraven and Norbert Lehwalter, “Size Effects in Short Beams Loaded in Shear,” ACI Structural Journal, 91, September–​October 1994, 585–​593. 14.20. Wen Bin Siao, “Deep Beams Revisited,” ACI Structural Journal, 92, January–​February 1995, 95–​102. 14.21. Stephen J. Foster and R. Ian Gilbert. “The Design of Nonflexural Members with Normal and High-​Strength Concretes,” ACI Structural Journal, 93, January–​February 1996, 3–​10. 14.22. James K.  Wight and Gustavo J.  Parra-​Montesinos. “Strut-​and-​Tie Model for Deep Beam Design,” Concrete International, 25, May 2003, 63–​70. 14.23. Further Examples for the Design of Structural Concrete with Strut-​and-​Tie Models (SP-​ 273). Karl-​Heinz Reineck and Lawrence C. Novak, Editors. Farmington Hills, MI: American Concrete Institute, 2010, 288 pp.

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PROBLEMS

PROBLEMS All problems are to be worked in accordance with the provisions of the ACI Code unless otherwise indicated. 14.1. Redesign the beam of Example  14.2.1, but instead of using a single diagonal strut, use a parallel chord, with two panel trusses between the concentrated loads and the supports, as shown in the figure for Problem 14.1. Compare your solution with that presented in Example 14.2.1 and discuss. 14.2. The beam shown in the figure for Problem 14.2 is to carry a uniform live load of 0.6 kip/​ft in addition to the concentrated load shown and the beam weight. Use fc′ = 4000 psi and fy = 50,000 psi. (a) Construct an acceptable strut-​and-​tie model for the beam in accordance with Chapter 23 of the ACI Code. (b)  Using the model constructed in part (a), verify the adequacy of the longitudinal reinforcement provided in the bottom of the beam. If inadequate, prescribe the number and location of the required longitudinal bars. Also, check the bar anchorage.

Center of column

2’– 4”

2’– 4”

(c)  Compute the minimum required distributed vertical and horizontal reinforcement between the concentrated load and the support. 14.3. Design the reinforcement for the beam shown in the figure for Problem 14.3. The rectangular continuous beam is to support heavy machinery; it carries a uniformly distributed, factored load of 65 kips/​ft. The factored moment Mu diagram is given in the accompanying figure. Use fc′ = 4000 psi and fy = 60,000 psi. 14.4. Redesign the bracket of Problem 5.21 in Chapter  5 using the provisions of Chapter  23 in the ACI Code for strut-​and-​tie models. The bracket projects from one side of a 16 × 16 column to support a vertical load of 35 kips dead load and 65 kips live load. Assume that suitable bearings are provided so that horizontal restraint is eliminated. The reaction is located 5 in. from the column face. Use  =  5000 psi and fy = 60,000 psi.

4’

2’– 4”

2’– 4”

Center of column

Problem 14.1 

5’–0”

Column load: 20k (LL) 30k (DL) (12 in. sq column)

30” 12” support

Symmetrical about CL of span 10’–0

Problem 14.2 

9 –#9

16”

d = 26”

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50”

18”

18”

12’–0”

24”

symmetric

+ –

900 ft-kips

–

Mu

900 ft-kips

Problem 14.3 

14.5. Repeat Problem 14.4 if the reaction is from a restrained beam that induces a horizontal tension equal to 50% of the total gravity reaction. 14.6. Redesign the bracket (corbel) of Example 14.3.1 considering that the supported prestressed girder

is welded to the bracket. Creep, shrinkage, and temperature effects on the restrained girder induce a horizontal force of 50 kips (unfactored) on the bracket.

CHAPTER 15 STRUCTURAL WALLS

15.1 GENERAL Reinforced concrete walls are common structural elements used in buildings and non-​ building structures to carry both vertical (gravity) and lateral loads. When properly designed and judiciously situated within the structure, walls can be very effective to resist a large portion of the loads (gravity or lateral load) and to augment the lateral stiffness of the structure. In practice, walls may be classified according to their intended function within the structure as follows (see Fig. 15.1.1): 1. Nonbearing walls. These walls are designed to carry no loads other than their own weight and are usually supported by other elements within the structure. Nonbearing walls, also called curtain walls, are often used as partitions to divide and separate space in the interior of the structure, as well as for the building façade. 2. Bearing walls. A bearing wall is used primarily to bear, or support, gravity loads, such as the weight of floors and other superimposed dead loads, as well as live loads. 3. Shear walls. The primary function of shear walls in a structure is to resist a large portion of the total horizontal lateral load (e.g., due to wind or earthquake) by means of in-​plane shear and flexure. 4. Retaining walls. These walls are designed to serve as a barrier and to hold back soil or other loose material. As such, they are designed to resist the lateral pressure exerted on the wall through shear and flexure about their minor or weak bending axis.

15.2 MINIMUM WALL DIMENSIONS AND REINFORCEMENT REQUIREMENTS—​A CI CODE Wall Thickness Minimum design and detailing requirements for reinforced concrete walls are given in Chapter  11 of the ACI Code. The minimum required wall thickness, h, is the smaller between the unsupported length, ℓw, and the unsupported height ℓc, multiplied by either 1 30 for nonbearing walls or 1 25 for bearing walls, but not less than 4 in. (see Fig. 15.2.1). For exterior basement and foundation walls, however, the minimum thickness is 7.5 in. These minimum thickness requirements only apply to bearing walls and exterior basement and foundation walls designed in accordance with the simplified method presented later (see Section 15.4.) Thinner walls may be permitted if adequate strength and stability can be shown by structural analysis (ACI-​11.3.1.1).

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Erection of core wall in Lincoln Square Expansion Project, Bellevue, WA. (Photo courtesy of Cary Kopczynski & Co.)

Wall Reinforcement Walls are required to have a minimum amount of uniformly distributed reinforcement in both the longitudinal (vertical) and the transverse (horizontal) directions. Reinforcement may consist of deformed bars or welded wire mesh placed parallel to the wall faces. The minimum required amounts of reinforcement for cast-​in-​place walls when the in-​plane shear Vu ≤ 0.5φVc are shown for common grades of steel in Table 15.2.1. Minimum reinforcement requirements when the in-​plane shear Vu > 0.5φ Vc are presented later (see Section 15.5). It is noted that the reinforcement ratios ρℓ and ρt in Table 15.2.1 are defined as the area of steel reinforcement per gross area of concrete perpendicular to the direction of the reinforcement being considered. Wall reinforcement may be provided in one or two layers depending on wall thickness. A single layer of reinforcement is permitted for walls 10 in. thick or less. Walls with thickness greater than 10 in. are required to have two layers of reinforcement placed parallel with the wall faces, except basement walls and cantilever retaining walls.

Vy-y

Mx-x

(c) Shear wall

x

x

Wall self-weight

y

Wall self-weight

Axial loads (e.g., weight of above floors)

y

y

Footing/foundation

Wall self-weight

Footing/foundation

My-y

x

Wall self-weight

y

Lateral pressure (e.g., from soil)

y

Gravity loads (e.g., weight of above floors)

(d) Retaining wall

Vx-x

x

(b) Bearing wall

x

x

15.2  DIMENSION AND REINFORCEMENT REQUIREMENTS

Figure 15.1.1  Forces and main actions on different types of walls: (a) nonbearing wall, (b) bearing wall, (c) shear wall, and (d) retaining wall.

Footing/foundation

Lateral load (e.g., from wind or earthquake)

y

(a) Nonbearing wall

x

x

y



Footing/foundation

y

571

571

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Minimum Wall Thickness, h Nonbearing walls: ℓc

h

hw

Largest of: a) 4 in.

ℓ x smaller of ℓw c Bearing walls*: b)

ℓc ℓw

1 30

Largest of: a) 4 in. b)

1 25

ℓ x smaller of ℓw c

Exterior basement and foundation walls*: 7.5 in. * Applies only to walls designed according to the simplified method.

Figure 15.2.1  Minimum wall thickness requirements (ACI 11.3.1.1).

TABLE 15.2.1  MINIMUM REINFORCEMENT FOR WALLS WHEN IN-​PLANE Vu IS 0.5φVc OR LESS (ACI-11.6.1) Reinforcement Steel Reinforcement Type and Grade

Bar Size

Minimum Longitudinal, ρℓ

Minimum Transverse, ρt

Deformed bars of Grades 60, 75 or 80

#5 or smaller #6 or larger

0.0012 0.0015

0.0020 0.0025

Deformed bars of Grades 40 or 50 Welded-​wire reinforcement of any Grade

Any size

0.0015

0.0025

W31, D31 or smaller

0.0012

0.0020

When two layers of reinforcement are used, wall reinforcement is commonly distributed into equal amounts between the two layers. The ACI Code, however, permits up to two-​ thirds of the total required amount in one layer, with the balance of the required amount in that direction placed in the other layer (ACI-​11.7.2.3). The layer with the larger amount of reinforcement must be placed at least 2 in., but not more than h/​3, from the exterior surface. The other layer must be placed at least 3 4 in., but no more than h/​3, from the interior surface. Additionally, minimum requirements of concrete cover for the reinforcement must be satisfied. The minimum clear cover for cast-​in-​place walls not exposed to weather or in contact with ground is 3 4 in. when #11 or smaller bars are used (ACI-​20.6.1.3.1). This is the most common situation for walls built above ground. For walls cast against and permanently in contact with ground, such as retaining walls, the minimum concrete cover for any bar size is 3 in. Concrete cover requirements for other bar sizes or exposure conditions are given in ACI-​20.6.1.3.1. The spacing of the longitudinal (vertical) and transverse (horizontal) reinforcement in cast-​in-​place walls is limited to 3 times the wall thickness or 18 in., unless shear reinforcement is required for in-​plane shear strength (ACI-​11.7.2.1 and ACI-​11.7.3.1). Spacing that is required for shear strength is further limited to ℓw /​3 for the longitudinal reinforcement and to ℓw /​5 for the transverse reinforcement (see Fig. 15.2.2).

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573

1 5 . 4   D E S I G N O F B E A R I N G   WA L L S 0.0012 (#5 or smaller bars) 0.0015 (#6 or larger bars)

min ρℓ =

shoriz ≤ 3h or 18 in.

svert ≤ 3h or 18 in.

min ρt =

0.0020 (#5 or smaller bars) 0.0025 (#6 or larger bars)

ℓw

h

Notes: i. A single layer may be provided when h < 10 in. ii. Shear walls need also satisfy shoriz ≤ ℓw/3 and svert ≤ ℓw/5 when in-plane Vu > 0.5φVc.

Figure 15.2.2  Minimum requirements for Grade 60 reinforcement in walls.

15.3 DESIGN OF NONBEARING WALLS Nonbearing walls are designed to carry no gravity loads other than their own weight. They may be designed to carry some lateral loads (e.g., from soil pressure), in which case they are designed as a retaining wall (treated in detail later: see Section 15.7). When used to divide space or separate areas, as façades in buildings, nonbearing walls are supported laterally by other members in the structure. Design of these walls is based primarily on experience and good practice; usually it involves satisfying the minimum requirements for reinforcement and wall dimensions.

15.4 DESIGN OF BEARING WALLS Walls used to carry, in addition to their own weight, gravity loads such as the weight of floors and other superimposed dead loads, as well as live loads, are referred to as bearing walls. The strength of bearing walls may be calculated by using the same principles (i.e., strain compatibility and equilibrium) presented in Chapter  10 for members subjected to combined bending and axial forces. Alternatively, for solid rectangular walls with no or small out-​of-​plane bending, the ACI Code allows the use of a simplified design procedure (ACI-​11.5.3).

Simplified Design Method—​ACI Code The simplified method can be used only for solid walls of rectangular cross section and only if the resultant of all factored axial loads is located within the middle third of the wall thickness in the out-of-plane direction (ACI-​11.5.3.1); that is, the eccentricity may not be greater than h /​6. Walls with nonrectangular cross section, with larger out-​of-​plane eccentricity, or subjected to in-​plane bending must be analyzed by using the procedures for members subjected to combined bending and axial forces (Chapter 10). Furthermore, walls

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subjected to significant in-​plane shear forces must be designed as shear walls, as discussed next, in Section 15.5.

Wall Design Strength Since the eccentricity of the factored axial load acting on the wall must be within h /​6 (i.e., the kern point), no tension will be induced in the wall under the design actions. The design strength of the wall may then be simplified by computing a nominal axial compressive strength, Pn, so that

φPn ≥ Pu



[2.6.3]

where φ is the strength reduction factor for a compression-​controlled section and Pu is the factored axial force. The nominal axial compressive strength, Pn, of the wall is computed by ACI Formula (11.5.3.1) as   kℓ  Pn = 0.55 fc′Ag 1 −  c    32h 



2

  

[15.4.1]

where Ag is the gross concrete area of the wall, ℓc is the unsupported length of the wall (see Fig. 15.2.1), and k is the effective length factor for the wall given in Table 15.4.1. The first term in Eq. (15.4.1) is obtained by assuming a rectangular stress block over a depth equal to two-​thirds of the wall thickness, h, which results in a net axial compression resultant of 0.85 fc′(2 / 3h)ℓ w or 0.57fc′ hℓ w , where the coefficient is taken as 0.55. The term in brackets accounts for slenderness effects on the axial load capacity. It is noted that Pn is independent of the amount of steel reinforcement. This result is based on analyses of walls with no or low eccentricity, which showed that the minimum amount reinforcement as required by the ACI Code (Table 15.2.1) does not substantially contribute to wall strength [15.1]. Test results as well as a comparison of test data with the simplified method of the ACI Code may be found elsewhere [15.2]. TABLE 15.4.1  EFFECTIVE LENGTH FACTORS, k, FOR WALLS (ACI-11.5.3.2) End Restraint Conditions

k

Braced against lateral translation at top and bottom: Restrained against rotation at one or both ends Unrestrained against rotation at both ends Not braced against lateral translation

0.8 1.0 2.0

EXAMPLE 15.4.1 A solid, rectangular, bearing wall with a length of 30 ft. is to be designed to carry a factored axial load, Pu, of 1350 kips with an eccentricity of 1.1 in. The wall supports a floor system made of hollow-​core slabs that may be assumed to restrain lateral movement at the top of the wall. At the base, the wall is supported on a concrete footing that prevents lateral translation. Wall height is 15 ft. Find the required wall thickness and the amount of vertical and horizontal reinforcement. Use fc′ = 4000 psi and fy = 60,000 psi. (Continued)

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Example 15.4.1 (Continued) SOLUTION (a) Because this is a solid wall with a rectangular cross ​section subjected to an axial force with a small eccentricity, it may be designed in accordance with the simplified method of the ACI Code. In such a case, the minimum required thickness, h, of the bearing wall is to be computed as the larger of: a)  h = 4 in. 1 b)  h = × smaller of 25 the unsupported wall length, ℓ w = 30 ft = 360 in.  the unsupported wall height, ℓ = 15 ft = 180 in.  c  

Governs!

or h=



1 (180) = 7.2 in. 25

Therefore, a minimum thickness of 7.2 in. is required. Use h = 8 in. (b) Compute the design strength, φ Pn For the simplified method of the ACI Code to be applicable, the eccentricity of the factored axial load must be within h/​6 = 1.3 in. which is greater than the eccentricity of 1.1 in. of the applied axial load. Therefore, the simplified method may be used. From Eq. (15.4.1), Pn is computed as   kℓ c  2  Pn = 0.55 fc′Ag 1 −    [15.4.1]   32h  



where Ag = hℓ w = 8(30 × 12) = 2880 sq in. The unsupported wall height or the vertical distance between lateral supports, ℓc, is 15 ft. The wall is braced against lateral translation at the top and at the bottom, but it will be assumed to be unrestrained against rotation. From Table 15.4.1, the effective length factor, k, is taken as 1.0. Thus,

  1.0(15)12  2  Pn = 0.55(4)2880 1 −    = 3204 kips   32(8)  

Assume that the wall will be provided with rectilinear ties (no spirals), as is the case in common practice. Thus, φ = 0.65 and

φPn = 0.65(3204) = 2083 kips > Pu = 1350 kips



OK

(c) Select the vertical and horizontal reinforcement Since the wall thickness is 8 in. < 10 in., the wall may be reinforced with a single layer of reinforcement. In this example, a single layer of uniformly distributed reinforcement will be provided. For this bearing wall, the maximum spacing of the vertical and the horizontal reinforcement is the smaller of • 3h = 3(8) = 24 in. and • 18 in.

Governs!

Using Grade 60, #5 or smaller bars, a minimum ratio of vertical reinforcement, ρℓ , of 0.0012 is required (Table 15.2.1). Also, a minimum amount of 0.0020 is required for the horizontal reinforcement ratio ρt (Table 15.2.1). (Continued)

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Example 15.4.1 (Continued) Thus, the total required area of vertical reinforcement per foot of wall length is As vert = ρℓ h × (12 in.)

As vert = 0.0012(8)12 = 0.12 sq in./ft



From Table 8.3.1, choose #4 bars at the maximum permitted spacing of 18 in. to provide a total area of 0.13 sq in./​ft. Similarly, the total required area of horizontal reinforcement per foot of wall length is:

As horiz = ρt h × (12 in.) As horiz = 0.0020(8)12 = 0.19 sq in./ft



From Table 8.3.1, choose #4 bars spaced at 12 in., which provide an average area of 0.20 sq in./​ft. (d) Check concrete cover requirements The clear cover for a single layer of #4 bars for the horizontal and vertical reinforcement centered in the 8 in. thick wall is

cover =

8 − [0.5 + 0.5] = 3.5 in. >> 3/4 in. required 2

OK

Use an 8-​in.-​thick wall with one layer of reinforcement at the center of the wall made with #4 vertical bars @ 18 in. and #4 horizontal bars @ 12 in.

15.5 DESIGN OF SHEAR WALLS In addition to carrying gravity loads, walls may be used to carry a large portion of the lateral load by in-​plane shear and bending [Fig. 15.1.1(c)]; consequently, these are commonly referred to as shear walls. This terminology, however, has often led to the misconception that all shear walls exhibit a brittle failure in shear. This is not so. Shear walls can be designed to be an excellent source of lateral stiffness and strength, and they can be effectively used to control lateral drift and to resist lateral loads in multistory buildings. Common wall configurations used in commercial and residential construction are shown in Fig. 15.5.1. In many tall buildings, shear walls are designed to carry most if not all of the lateral loads from wind and earthquakes. Gravity loads are often carried by a slab-​column frame system. These shear walls are commonly concentrated around elevator shafts and stairwells, and thus they are often referred to as core walls [see Fig. 15.5.1(a)]. A combination of shear walls and moment-​resisting frames, often referred to as a dual system, is shown in Fig. 15.5.1(b). In these dual systems, which are commonly used in office buildings, lateral loads are shared by the shear walls and the frames. A common configuration used in apartment buildings is shown in Fig. 15.5.1(c). In such cases, the walls are designed to carry most if not all of the gravity loads in addition to resisting the lateral loads as shear walls. The amount of wall cross-​sectional area with respect to the floor area, or wall-​to-​floor area ratio, may vary significantly depending on the configuration of the floor plan [15.3, 15.4]. When shear walls are provided as part of a core wall system [Fig. 15.5.1(a)] or as part of a dual system [Fig. 15.5.1(b)], the ratio of wall to floor area can be much lower

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577

(a) Shear wall system (reinforced concrete core)

(b) Shear wall and moment-resisting frame (dual system)

(c) Bearing and shear wall system

Figure 15.5.1  Common wall configurations in residential and commercial buildings.

than in buildings where walls form the primary gravity and lateral load resisting system [Fig. 15.5.1(c)]. In general, walls in buildings with low ratios of wall to floor area (less than about 0.5% of wall area in each direction) can be expected to be more heavily reinforced than those in buildings with high ratios (greater than 1.5% of wall area in each direction). Shear wall cross sections commonly found in practice are shown in Fig. 15.5.2. Walls of rectangular cross section are the most common shape. Other shapes such as C-​, I-​, L-​, and T-​shaped cross sections are also commonly used in building construction. In walls with high flexural strength demands, such as in the lower stories of a high-​rise building, vertical reinforcement in addition to that required to be uniformly distributed is often concentrated near the ends of the walls. These end regions are commonly referred to as wall boundary elements (see Fig. 15.5.3). In some cases, the wall thickness may be increased over the length of the boundary element region to accommodate the required amount of vertical reinforcement at the ends. Another reason to provide an enlarged section at the wall ends is to stabilize and to prevent buckling of the boundary element in thin walls (Fig. 15.5.4). Walls with an enlarged boundary element region are referred to as “barbell” walls. In low-​to medium-​rise buildings, walls are often designed with a constant cross section over the height; that is, they are prismatic. In such cases, variations in the required strength with wall height are accommodated by varying the amount of vertical and horizontal reinforcement, and by varying the concrete strength. In taller structures, wall cross section may

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Figure 15.5.2  Typical shear wall cross sections.

Boundary element region

Boundary element

Boundary element region

Boundary element

Figure 15.5.3  Boundary element regions. (Note: Transverse reinforcement not shown for clarity.)

be reduced in the upper stories by varying the wall thickness or the wall length. Changes in cross section must be done with judgment and caution, without causing drastic changes in stiffness or strength over the wall height, which might otherwise result in an undesired failure at the transition level. In practice, walls may be found as solid walls or with openings, usually to allow for windows or doorways, or both. A detailed treatment of walls with openings is beyond the scope of this chapter, but some basic guidelines and terminology are provided. The analysis and design of walls with openings depend primarily on the size and arrangement of the openings. In general, the influence of the openings may be neglected when the total area of opening is relatively small in comparison to the wall surface area [see Fig. 15.5.5(a)]. In such cases, the wall may be treated as a solid wall with special provisions for reinforcement around the openings (ACI-​11.7.5.1). In contrast, when the openings are relatively large or are located near a critical region [see Fig.  15.5.5(b)], such as the boundary elements at the wall ends, their effect on the wall flexural and shear strength must be considered. The ACI Code offers little or no specific guidance for the design of walls with openings. In most cases, however, walls with openings may be designed by using the strut-​and-​tie method presented in Chapter  14. Tests on walls with openings are scarce [15.5–​15.8]. While the studies cited focused primarily on the response of walls subjected to reversed cyclic loads, they provide good insight into the general behavior and design of walls with openings.

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Figure 15.5.4  Wall buckling during the Maule, Chile, earthquake in 2010. Maule, Chile, 2010. (Photo courtesy of Fernando Yañez, University of Chile.)

A common and efficient structural system for resisting lateral loads may be obtained when openings are arranged in a regular pattern, by connecting two or more solid walls with beams in every floor level, as shown in Fig. 15.5.6. Under lateral loads, the beams will interact with and affect the behavior of the connected walls. Because the behavior of the connected walls and that of the beams is coupled, the walls are commonly referred to as coupled walls. Accordingly, the connecting beams are referred to as coupling beams. The level of interaction or level of coupling depends on the relative flexibility and strength of the beams. In a structure with stiff and strong coupling beams, the level of coupling will be high. In such cases, the shear forces in the coupling beams will induce high axial forces in the walls that must be considered in addition to those resulting from gravity loads. For the direction of the lateral forces shown in Figure 15.5.6, summation of the coupling beam shear forces will induce axial tension on the left wall and axial compression on the right wall. Clearly, the reverse is true for lateral loads acting in the opposite direction. If the walls carry little or no gravity loads, each wall will have to be designed under combined bending and axial tension and under combined bending and axial compression.

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(a)

(b)

Figure 15.5.5  Walls with (a) small and (b) large openings.

Figure 15.5.6  Coupled wall geometry and actions under lateral loads.

If the beams are relatively flexible and weak, the level of coupling will be low and may be ignored in design. In those cases, each wall may be designed as an isolated cantilever, solid wall.

Behavior and Design of Cantilever Walls The basic failure mechanisms of cantilever shear walls as described by Paulay [15.9] are shown in Figure 15.5.7. In general, the behavior and load transfer mechanism of cantilever shear walls may be divided into two categories in accordance to their height-​to-​length, or aspect, ratio (hw /​ℓw). Walls with an aspect ratio hw /​ℓw greater than 2 are often considered tall walls. The behavior of cantilever tall walls has been under investigation for decades [15.5, 15.10–​15.11]. These walls exhibit a flexural-​dominated behavior [Fig.  15.5.7(b)], and thus they can be analyzed and designed by means of the same principles described for members subjected to combined flexure and axial loads in Chapter 10. Shear failures of tall walls involving diagonal tension or diagonal compression [Fig. 15.5.7(c)] or sliding shear [Fig. 15.5.7(d)] should and can be avoided by means of adequate proportion and distribution of the transverse and longitudinal reinforcement.

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H

V

Nf hv

N

V M

(a) Wall actions

Vf

H

T

V

Vf C

(b) Flexure

Nf (c) Diagonal tension or compression

(d) Sliding shear

Figure 15.5.7  Basic failure mechanisms of cantilever walls. (Adapted from Paulay [15.9].)

Walls with aspect ratios (hw  /​ℓw) less than 2, often called squat walls, exhibit a different load transfer and failure mechanism. Such walls, commonly found in low-​rise construction, can develop large flexural strength even when minimum amounts of vertical reinforcement are provided. As a result, relatively large shear forces will be expected to act concurrently with the development of their flexural strength [15.12]. Because of the low aspect ratio, the behavior of squat walls is akin to that of deep beams, where the horizontal shear is transferred primarily by diagonal struts in equilibrium with both the vertical and horizontal reinforcement. Wood [15.13] has provided an excellent review of experimental results in light of ACI Code design provisions for low-​rise shear walls. Experimental results on the behavior of low-​rise squat walls are reported elsewhere [15.14–​15.19].

Flexural Strength—​Solid Rectangular Walls In general, cantilever shear walls are subjected to shear forces and bending moments from lateral forces and to axial forces from gravity loads. The flexural strength of shear walls with aspect ratios hw  /ℓw > 2 can be evaluated by using the same principles (i.e., strain compatibility and equilibrium) presented in Chapter  10 for members subjected to combined bending and axial forces. For walls of rectangular cross section with uniformly distributed vertical reinforcement, Cardenas and Magura [15.20] have developed an approximate procedure to estimate flexural strength. In their approach, it is assumed that the total area of vertical reinforcement is continuous over the entire wall length, and that the strain distribution is such that the axial load at ultimate is less than that at the balanced strain condition (see Fig. 15.5.8). By ignoring the contribution of the reinforcement that remains elastic near the neutral axis (i.e., within βc in Fig. 15.5.8), Cardenas and Magura [15.20] proposed the following approximate equation:



 Pn   c M n ≈ 0.5 As vert f y ℓ w  1 +  . 1 +  As vert f y   ℓ w  

(15.5.1)

where Mn = nominal flexural strength As vert = total area of distributed vertical reinforcement Pn = nominal axial force (taken as positive for compression) c q+α = ℓ w 2q + 0.85β1 (15.5.2)

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q=

As vert f y

=

hℓ w fc′

ρℓ f y

fc′ (15.5.3)

α = axial force ratio =



Pn hℓ w fc′ (15.5.4)

Dividing Eq. (15.5.1) by hℓw 2 yields the following strength coefficient:

Mn hℓw2

   α   = 0.5ρℓ f y  1 + fy    ρℓ f ′  c

 c . 1 −   ℓw 

(15.5.1a)

Mn has been plotted in Fig.  15.5.9 as a hℓw2 function of the reinforcement ratio, ρℓ, for Grade 60 steel and for fc′ equal to 4000 psi and 5000 psi, respectively. The curves are shown for axial load ratios, α, of 0, 0.05, 0.1, 0.15, and 0.2. These charts can be used to quickly estimate the wall flexural strength for a given reinforcing ratio, ρℓ, and axial load ratio (see Example 15.5.1), or to obtain a preliminary estimate of the required amount of reinforcement for a required moment and axial load (see Example 15.5.2). It must be emphasized that Eqs. (15.5.1) and (15.5.1a), as well as the charts shown in Fig. 15.5.9, are limited to walls with an axial load smaller than that at the balanced strain condition. For design purposes, the strength coefficient

0.85fc’

0.003 εy As

ϕu εy

βc

0.5β1c Cc βc

β1c

c

ƒy

βc

βc

ℓw

Pn

ℓw-c ℓw/2

h (a) Cross section

ƒy (b) Strain distribution; β =

εy 0.003

(c) Concrete stress distribution and axial load

(d) Steel stress distribution

Figure 15.5.8  Assumed strain and stress distributions at nominal strength in a rectangular wall with uniformly distributed vertical reinforcement. (After Cardenas and Magura [15.20]).

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α=

0.7

Pn hℓw f’c

0.65 0.6 0.55

= 0.20 0.15 0.10 0.05 0

0.5

Mn 0.45 hℓw2 0.4

(ksi)

0.35 0.3 0.25 0.2 0.15

f’c = 4000 psi fy = 60,000 psi

0.1 0.05 0 0

0.005

0.8 α=

0.75

Pn hℓw f’c

0.7 0.65 0.6

0.01

0.015

0.02

0.025

= 0.20 0.15 0.10 0.05 0

0.55 0.5

Mn hℓw2 (ksi)

0.45 0.4 0.35 0.3 0.25 0.2 f’c = 5000 psi fy = 60,000 psi

0.15 0.1 0.05 0 0

0.005

0.01

ρℓ

0.015

0.02

0.025

Mn , for rectangular walls with uniformly distributed hℓw 2 vertical reinforcement as a function of the longitudinal reinforcement ratio, ρℓ , for axial load ratios, α, of 0, 0.05, 0.1, 0.15 and 0.2. Charts are shown for fc′ = 4,000 psi (top chart) and fc′ = 5,000 psi (bottom chart) and Grade 60 reinforcing bars. Figure 15.5.9  Strength coefficient,

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EXAMPLE 15.5.1 A shear wall in a multistory building has cross-​sectional dimensions and reinforcement layout as shown in Fig.  15.5.10. The wall is 12 ft. long and 12 in. thick. The vertical reinforcement consists of uniformly distributed #5 bars spaced at 15 1 2 in., while the horizontal reinforcement consists of #5 bars spaced at 18 in. Assume that the required in-​plane shear strength Vu < 0.5φVc . Compute the maximum factored moment Mu that can be applied when the concurrent axial compressive force Pu is 500 kips. Use fc′ = 4, 000 psi and fy = 60,000 psi. #5 @ 18” (horiz.)

#5 @ 15½” (vert.)

15½”

¾”(min)

12” ¾”(min) 12’ – 0”

Figure 15.5.10  Dimensions and reinforcement layout for wall cross section of Example 15.5.1.

SOLUTION 1. First, verify that the requirements for maximum spacing and minimum amount of vertical and horizontal reinforcement are satisfied: a) Since wall thickness of 12 in. is greater than 10 in., two layers of reinforcement placed parallel with the wall faces are required as provided. OK b) Clear cover of reinforcement is 3 4 in. This satisfies the minimum cover required by ACI-​20.6.1.3.1 for walls not exposed to weather or in contact with ground when #11 bars or smaller are used (see Section 15.2). OK c)  Vertical (Longitudinal) Reinforcement: i.  Since #5, Grade 60 deformed bars are used, and Vu < 0.5φVc , the minimum required amount of vertical (longitudinal) reinforcement, ρℓ min, is 0.0012 or 0.12% (see Table 15.2.1). For two layers of #5 bars spaced at 15 1 2 in., the provided reinforcement ratio is provided ρℓ =



As vert 2(0.31) = = 0.0033 > 0.0012 = ρℓ min h shoriz 12(15.5)

OK

ii. Per ACI 11.7.2.1, maximum spacing of vertical reinforcement is the smaller of (see Fig. 15.2.2) Governs!

• 18 in. or • three times the wall thickness h, or 3(12) = 36 in. The provided spacing of the reinforcement is 15 1 2 in. < 18 in.

OK

Since Vu < 0.5φVc the limit based on ℓw / ​3 need not be satisfied. d)  Horizontal (Transverse) Reinforcement: i.  Since #5, Grade 60 deformed bars are used, and Vu < 0.5φVc , the minimum amount of horizontal (transverse) reinforcement, ρt min, is 0.0020 or 0.20% (see Table 15.2.1). For two layers of #5 bars spaced at 18 in., the provided reinforcement ratio is:

provided ρt =

As horiz 2(0.31) = = 0.0029 > 0.0020 = ρt min h svert 12(18)

OK (Continued)

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Example 15.5.1 (Continued) ii. Per ACI 11.7.3.1, maximum spacing of horizontal reinforcement is the smaller of: Governs!

• 18 in. or, • three times the wall thickness h, or 3(12) = 36 in.

The provided spacing of the reinforcement is 18 in., equal to the maximum permitted, 18 in. OK Since Vu < 0.5φ Vc , the limit based on ℓw /​5 need not be satisfied. 2. Compute the nominal strength, Mn Use the Cardenas and Magura approach [Eq. (15.5.1a)] to estimate the flexural strength of the wall. Assuming a tension-​controlled section (φ = 0.9), use Eq. (15.5.4) to compute the axial load ratio α:

α=

Pn P /φ 500 / 0.9 = u = = 0.080 hℓ w fc′ hℓ w fc′ 12(144)4

with ρℓ = 0.0033 and α = 0.080, enter the top chart ( fc′ = 4000 psi; fy = 60,000 psi) of Fig. 15.5.9 to obtain by interpolation:



Therefore,

Mn ≈ 0.22 ksi hℓw 2

 1 M n = 0.22(12)144 2   = 4562 ft-kips  12 

Mu ≤ φ M n = 0.9(4562) = 4100 ft-kips



3. The maximum factored moment that can be applied on the cross section can also be computed from the design axial load–​moment interaction diagram. Analysis of the wall cross section by using equilibrium and strain compatibility results in the interaction diagram shown in Fig. 15.5.11. As shown in the figure, for an axial load Pu of 500 kips, the design flexural strength, φ M n, is 4300 ft-​kips. This value is about 5% larger than the factored moment obtained with the approximate method. In other words, the approximate approach is somewhat conservative in this case, but it provides a very good estimate of the flexural strength of the wall. Note that the wall is tension controlled, as assumed above in item 2.

Figure 15.5.11  Nominal and design strength interaction diagram for wall of Example 15.5.1.

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EXAMPLE 15.5.2 A shear wall 10 in. thick and 10 ft long is to carry a factored moment Mu of 4500 ft-​kips with a concurrent axial force Pu of 900 kips. Find the required amount of uniformly distributed longitudinal reinforcement to carry the required moment and axial force. Use fc′ = 5000 psi and f y  = 60,000 psi. Assume that Vu < 0.5φ Vc (i.e., shear reinforcement is not required for in-​plane strength). SOLUTION 1. Use the charts of Fig. 15.5.9 to compute a first estimate of the required longitudinal reinforcing ratio, ρℓ. a.  Compute the axial load ratio α [Eq. (15.5.4)] assuming the wall is tension controlled:

α=

Pn P /φ 900 = u = = 0.17 hℓ w fc′ hℓ w fc′ 0.9(10)(120)5

b.  Compute the required strength coefficient 



required

Mn  : hℓw2

M n M u /φ 4500(12) = = = 0.42 hℓw2 hℓw 2 0.9(10)120 2

c.  From Fig. 15.5.9 (bottom chart), estimate ρℓ ≈ 0.0055 or 0.55%. 2. Select reinforcing bars and spacing. Since Vu < 0.5φ Vc, the minimum required amount of vertical reinforcement from Table 15.2.1 is ρℓ = 0.0012 (assuming that #5 bars are to be used), which is less than the required amount for strength of 0.0055. The total required area of reinforcement per foot of wall length is: As vert = ρℓ h × (12 in.)

As vert = 0.0055(10)12 = 0.66 sq in./ft



Since wall thickness is 10 in., two layers of reinforcement are required. Therefore, the required amount of reinforcement per layer is As vert = 0.66 / 2 = 0.33 sq in./ft From Table 8.3.1, choose #5 bars spaced at 12 in. This amount, which will provide an average area of 0.31 sq in./​ft, is somewhat less than that required above. Given the conservative nature of Eq. (15.5.1), however, it should be adequate. Maximum spacing requirements are • 18 in. • three times the wall thickness, h, or 3(10) = 30 in.

Governs!

Since Vu < 0.5 φVc , the limit based on ℓw / ​3 need not be satisfied. Maximum permitted spacing is 18 in., which is greater than 12 in. OK Try two layers of #5 bars spaced at 12 in.  (Continued)

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Example 15.5.2 (Continued) 3. Compute the design interaction diagram Assuming a minimum clear cover of 3 4 in., and #5 bars for the horizontal, transverse reinforcement, the total required number of bars spaced at 12 in. per layer is No. of bars ≈

ℓ w − 2 × (clear cover ) − 2 × dbhoriz − dbvert +1 spacing (in.)

120 − 2 × 3 / 4 − 2 × 0.625 − 0.625 +1 12 ≈ 10.7 ≈



Try 11 bars per layer for a total of 22 equally spaced #5 bars at 11 1 2 in. The design interaction diagram for the wall cross section with 22–​#5 bars at 11 1 2 in., equally distributed in two layers, is shown in Fig. 15.5.12: the design strength (φ Mn, φ Pn) is greater than the required strength (Mu, Pu). Note that the wall is tension controlled, as previously assumed. Therefore, the provided reinforcement is adequate, although somewhat in excess of that required. For this example, an attempt will be made to optimize the amount of reinforcement by increasing the bar spacing without exceeding the maximum permitted of 18 in. A revised analysis of the cross section with #5 bars spaced at 16 1 2 in. in two layers (ρℓ ≈ 0.0038 > ρℓ min ) shows that the strength requirement is satisfied but with a design strength that is much closer to the required strength. In practice, however, members must be designed for several load combinations. Therefore, trying to optimize wall reinforcement or size for a single load combination is not warranted. Use two layers of #5 spaced at 16 1 2 in. A design sketch of the final reinforcement layout is shown in Fig. 15.5.13.

Figure 15.5.12  Design strength interaction diagram for wall of Example 15.5.2. #5 @ 16½” (vertical)

¾”(min)

10” ¾”(min) 10’–0”

Figure 15.5.13  Design sketch for the shear wall of Example 15.5.2.

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Shear Strength—​Solid Rectangular Walls The ACI Code shear strength provisions for walls are largely based on the results of a series of tests conducted by the Portland Cement Association (PCA) on large rectangular shear wall specimens representative of both high-​and low-​ rise building construction [15.21]. Experimental results from Muto and Kokusho [15.22], Ogura et  al. [15.23], Williams and Benjamin [15.15], Benjamin and Williams [15.16], and Antebi et al. [15.17] were also considered in the development of the provisions. Based on these studies, the in-​plane shear strength of cantilever shear walls may be computed by using the same approach for members subjected to combined bending and axial load, where the nominal shear strength Vn is computed as Vn = Vc + Vs



[5.8.1]

where Vc and Vs are the contributions of the concrete and shear reinforcement to shear strength. This approach may be used to compute the shear strength of both squat (hw /ℓw ≤ 2) and tall (hw /ℓw > 2) walls. Alternatively, ACI-​11.5.4.1 permits squat walls (hw /ℓw ≤ 2) to be designed with the strut-​and-​tie method described in Chapter 14.

Concrete Contribution to Shear Strength, Vc The current ACI Code provisions provide both a simplified and a detailed method for computing Vc. Simplified Method (ACI-​11.5.4.5) In this approach, Vc is approximately taken as the shear corresponding to the onset of diagonal cracking for beams, and may be used when the wall is subjected to flexure with axial compression or with no axial load, as follows Vc = 2 λ fc′hd (15.5.5)1



where λ is the modification factor for lightweight concrete in accordance with ACI-​19.2.4.2 (see Section 1.8), h is the wall thickness, and d is taken as 0.8ℓw. A larger value of d is permitted (ACI-​11.5.4.2) if the centroid of the tension force resultant is computed on the basis of strain compatibility. If the wall is subjected to axial tension, the concrete contribution to shear can be either neglected (i.e., Vc = 0) or computed as





 N u  Vc = 2 1 +  λ fc′hd  500 Ag 

1  For SI, ACI 318-​14M, with fc′ in MPa, gives

(15.5.6)2

Vc = 0.17 λ fc′hd (15.5.5) 

2  For SI, ACI 318-​14M, with fc′ , N u /Ag , N u /(hℓ w ) in MPa gives



 N u  Vc = 0.17 1 +  λ fc′ hd  3.5 Ag 

(15.5.6) 

 Nu  Vc =  0.27λ fc′ + hd 4hℓ w  

(15.5.7) 

  0.2 N u     0.1λ fc′ + hℓ  ℓ w  w   hd Vc = 0.05λ fc′ +  Mu ℓ w −   Vu 2  

(15.5.8)

 

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where Nu is the factored axial tensile force in pounds acting concurrently with the factored shear force, Vu, and Ag is the gross area of concrete of the wall in sq in. Note that Nu in Eq. (15.5.6) must be entered as a negative quantity. Detailed Method In this method, Vc is computed as the lesser of the following (ACI-​11.5.4.6)  N  Vc =  3.3λ fc′ + u  hd 4hℓ w  

(15.5.7)2

  0.2 N u     1.25λ fc′ + hℓ  ℓ w  w  hd Vc = 0.6 λ fc′ +   Mu ℓ w −   Vu 2  

(15.5.8)2

or



where Nu is the factored axial force (positive for compression and negative for tension) acting concurrently with the factored shear force, Vu. In these equations, the term N u /(hℓ w ) must be in psi. Note that Eq. (15.5.8) does not apply when (Mu  / ​Vu –​ ℓw /​2) is negative. Equation (15.5.7) approximately corresponds to the development of web-​shear cracking when the principal tensile stress reaches 4 fc′ at the centroid of a rectangular cross section subjected to combined axial load and shear [15.21]. This equation may govern for thin-​ web flanged, or barbell walls, and will usually control for low-​rise squat rectangular walls. Alternatively, the ACI Code permits squat walls (hw /ℓw ≤ 2)  to be designed for in-​plane shear in accordance with the strut-​and-​tie method (ACI-​11.5.4.1), as noted earlier. Equation (15.5.8) corresponds to the onset of flexure-​shear cracking and will usually govern for walls with hw  /ℓw > 2.  In developing Eq. (15.5.8) for computing the concrete contribution Vc, it was assumed that flexural diagonal cracking would initiate at a section located at ℓw  /​2 from the base of the wall [15.21]. In general, ACI-​11.5.4.7 permits sections located between the wall base and a distance ℓw /​2 or hw /​2 from the base of the wall, whichever is less, to be designed for Vc computed by using the Mu  / ​Vu ratio at ℓw  /​2 or hw /​2, whichever is less from the wall base, as shown in Fig. 15.5.14. Note that the term (Mu  / ​Vu –​ ℓw /​2) in Eq. (15.5.8) may become negative for walls with low Mu / ​Vu ratios. In such a case, Eq. (15.5.8) does not apply, and Eq. (15.5.7) must be used to compute Vc. In other words, web-​shear cracking instead of flexure-​shear cracking will be expected to occur. Figure 15.5.15 shows the variation in the shear strength attributed to the concrete, Vc, for walls computed with the simplified [Eq. (15.5.5)] and the detailed [Eqs. (15.5.7) and

ℓw

h V hw

Smaller of ℓw or hw 2 2

Critical section for M Critical section for V

Figure 15.5.14  Definition of the critical section when the detailed method of ACI-​11.5.4.6 [Eqs. (15.5.7) and (15.5.8)] is used for computing the concrete contribution to shear strength, Vc.

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8

Nu = 1000 psi hℓw

7 6 Vc fc’ hd

fc’ = 5000 psi λ = 1.0

Web-shear cracking [Eq. (15.5.7)]

9

Nu = 500 psi hℓw

5

Flexure-shear cracking [Eq. (15.5.8)]

Nu = 0 psi hℓw

4 3 2 1 0

Simplified Eq. (15.5.5) 0

1

2

3 hw ℓw

4

5

6

Figure 15.5.15  Variation in the shear strength attributed to concrete, Vc, as a function of the wall aspect ratio, hw  /ℓw, in accordance with Eqs. (15.5.5), (15.5.7), and (15.5.8), for a cantilever wall subjected to an inverted-​triangle load distribution and axial compressive stresses of 0, 500, and 1000 psi.

(15.5.8)] methods (ACI-​11.5.4.5 and ACI-​11.5.4.6). In the figure, Vc is shown as an average shear stress Vc  /​(hd) divided by fc′ , as function of the wall aspect ratio, hw  /ℓw. The Mu /​  Vu ratio in Eq. (15.5.8) was computed based on an inverted triangular force distribution over the wall height. The curves are shown for three values of axial compressive stress in the wall: 0, 500, and 1000 psi, and for normal-​weight concrete with a compressive strength fc′ of 5000 psi. As shown in the figure, web-​shear cracking [i.e., Eq. (15.5.7)] provides an upper limit to the concrete contribution to shear strength and will govern for walls with low hw /ℓw ratios. At larger aspect ratios, flexure-​shear cracking, Eq. (15.5.8), will govern and will result in a reduced contribution from the concrete to shear strength. However, Vc need not be taken less than 2 fc′hd [15.21], and thus, this value should be taken as the lower limit for Vc in walls with compressive or zero axial force. Shear walls in bending combined with axial tensile forces may be found in tall walls coupled with stiff and strong coupling beams (see Fig. 15.5.6). In such cases, it is conservative to assume no contribution from the concrete to shear strength under axial tension (i.e., Vc = 0). Alternatively, Vc may be computed by using Eq. (15.5.6) of the simplified method or, Eqs. (15.5.7) and (15.5.8) of the detailed method. The variation in shear strength attributed to the concrete as a function of the wall aspect ratio, hw /ℓw, is shown in Fig. 15.5.16 for two levels of axial tension (–​100 and –​400 psi). As before, the curves were obtained by assuming an inverted triangular force distribution for computing the Mu / ​Vu ratio in Eq. (15.5.8). At low levels of axial tension (e.g., –​100 psi), the simplified method [Eq. (15.5.6)] provides a lower bound to Vc for walls with relatively low aspect ratios (about 3 or less). For walls with higher aspect ratios, however, the flexure-​shear cracking strength computed with Eq. (15.5.8) provides a lower estimate of Vc,—​perhaps as much as 60% lower for walls with aspect ratios of 6. At higher levels of axial tension (e.g.,  –​400 psi), the simplified method [Eq. (15.5.6)] provides a lower bound over the entire range of aspect ratios shown in Fig. 15.5.16. Test data on the shear strength of walls with axial tensile forces are scarce; thus, the ACI Code equations provide only estimates of wall shear strength under axial tension, and these should be used with caution.

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f’c = 5000 psi λ = 1.0

Web-shear cracking [Eq. (15.5.7)] 3

Vc fc’ hd

Flexure-shear cracking [Eq. (15.5.8)] Nu = –100 psi hℓw

2

1

0

Nu = –400 psi hℓw

Simplified Eq. (15.5.6)

0

1

2

3 hw ℓw

4

5

6

Figure 15.5.16  Variation in the shear strength attributed to concrete, Vc, as a function of the wall aspect ratio, hw  /ℓw, in accordance with Eqs. (15.5.6), (15.5.7), and (15.5.8) for a cantilever wall subjected to an inverted-​triangle load distribution and axial tensile stresses of –​100 and –​400 psi.

Shear Reinforcement Contribution to Shear Strength, VS As noted earlier, all walls are required to have at least a minimum amount of uniformly distributed vertical and horizontal reinforcement (see Section 15.2). In shear walls, this reinforcement will help restrain the growth (width and length) of diagonal, shear cracks; thus, both the horizontal and vertical bars contribute to shear strength. In design, however, only the horizontal reinforcement is used to compute the contribution of the steel reinforcement to shear strength, Vs. Accordingly, the contribution of the reinforcement to shear strength is computed, per ACI-​11.5.4.8, as

Vs =

Av f yt d s



(15.5.9)

where Av is the area of horizontal reinforcement, As horiz spaced at a distance s over the wall height (svert in Figure 15.2.2), and f yt is the yield strength of the horizontal reinforcement. In other words, the steel reinforcement contribution to shear strength may be computed as

Vs = ρt f yt hd

(15.5.9a)

where ρt is the transverse (horizontal) reinforcement ratio and h is the wall thickness. The effective depth, d, is taken as 0.8ℓw unless a larger value is used, per ACI-​11.5.4.2 as discussed earlier.

Minimum Shear Reinforcement Requirements Minimum requirements for the horizontal shear reinforcement in walls were presented in Section 15.2. These requirements apply to shear walls when the in-​plane required shear strength Vu ≤ 0.5φVc (ACI-​11.6.1). If Vu > 0.5φVc, however, both the horizontal shear reinforcement ratio, ρt , and the longitudinal reinforcement ratio,  ρℓ , must be at least 0.0025 (ACI-​11.6.2), irrespective of the size and Grade of reinforcement. In squat walls, however, tests have shown that the horizontal

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shear reinforcement becomes less effective than the longitudinal (vertical) reinforcement [15.14]. Accordingly, when Vu > 0.5φVc and hw /ℓw ≤ 2.5, the required longitudinal reinforcement ratio, ρℓ , is computed as follows:

 h  ρℓ ≥ 0.0025 + 0.5  2.5 − w  (ρt − 0.0025) ℓw  

(15.5.10)

However,  ρℓ need not exceed the horizontal reinforcement ratio, ρt , required for shear strength, as computed from Eq. (15.5.9a). Also, note that when hw /​ℓw = 0.5, the amount of longitudinal reinforcement,  ρℓ , is equal to the amount of horizontal shear reinforcement,  ρt . In addition to the maximum spacing requirements for the horizontal and vertical reinforcement for walls with Vu ≤ 0.5φ Vc (smaller of 3h or 18 in.), the maximum spacing is further limited to ℓw /​3 and to ℓw  /​5 for the longitudinal and transverse reinforcement, respectively, when Vu > 0.5φVc (see Fig. 15.2.2).

Maximum Amount of Shear Reinforcement Similar to the behavior of reinforced concrete beams, a heavily reinforced wall in shear may reach failure by crushing of the concrete diagonal struts prior to yielding of the transverse reinforcement (see Sections 5.7 and 5.10). To prevent such a failure mode in shear walls, ACI-​11.5.4.3 requires that at any horizontal section, the nominal strength, Vn, be limited to 10 fc′ hd .

EXAMPLE 15.5.3 The factored axial force, shear force, and bending moment diagrams in a wall of a 9-​story building are shown in Fig. 15.5.17. The wall is 20 ft long and 10 in. thick and is one of several walls resisting gravity and wind loads. Compute the amount of vertical and horizontal reinforcement required in the first story. Use normal-​weight concrete with fc′ = 5000 psi and Grade 60 steel.

Figure 15.5.17  Factored axial force, shear force and bending moment diagrams due to gravity and wind loads in the wall of Example 15.5.3.

(Continued)

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Example 15.5.3 (Continued) SOLUTION 1. Compute the required amount of vertical reinforcement, ρℓ , based on flexural strength requirements. From Fig. 15.5.17, the required moment strength at the base of the wall is Mu = 15,162 ft-​kips, and the factored axial force acting currently with this moment is Pu = 1036 kips. a. Assuming a tension-​controlled section (φ = 0.9), compute the axial load ratio, α, from Eq. (15.5.4)

α=

Pn P /φ 1036 = u = = 0.096 hℓ w fc′ hℓ w fc′ 0.9(10)(240)5

b.  Compute the required strength coefficient 

required

Mn  : hℓw2

M n M u /φ 15,162(12) = = = 0.35 ksi hℓw2 hℓw2 0.9(10)240 2

c.  From Fig. 15.5.9 (bottom chart), estimate  ρℓ ≈ 0.0065 or 0.65% . d. Assuming #5 bars, the minimum amount of longitudinal reinforcement when Vu ≤ 0.5φ Vc is 0.0012 (see Table 15.2.1), which is less than the required amount computed above based on flexural strength requirements. If, however, Vu > 0.5φVc, a larger minimum amount may be required. Thus, minimum requirements for the amount of vertical reinforcement will be checked later after shear strength requirements have been computed. 2. Compute the amount of horizontal reinforcement, ρt , based on shear strength requirements. a. Concrete contribution Vc—​According to ACI-​11.5.4.5 and 11.5.4.6, Vc may be calculated by using the simplified or the detailed method. Here, both methods are used and then compared. i. Simplified Method—​Since the wall is subjected to axial compression under the given factored loads, use Eq. (15.5.5). Thus,

Vc = 2 λ fc′hd = 2(1) 5000 (10)0.8(240)

1 1000

= 272 kips

where d was taken as 0.8ℓw and λ was taken as 1.0 for normal-​weight concrete. ii.  Detailed Method—​According to ACI-​11.5.4.6, Vc is to be computed as the lesser of the values resulting from Eqs. (15.5.7) and (15.5.8.) The axial compressive stress in the wall is

N u 1036(1000 ) = = 432 psi hℓ w 10(240) and the wall-​to-​height ratio, hw /ℓw, is 114/​20 = 5.7. • Vc based on web-​shear cracking [Eq. (15.5.7)]:



 N  Vc =  3.3λ fc′ + u  hd 4hℓ w   1036(1000)     =  3.3(1) 5000 + 10(0.8)240  1  = 655 kips  1000   4(10)240 

(Continued)

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Example 15.5.3 (Continued) •  Vc based on flexure-​shear cracking [Eq. (15.5.8)]: In accordance with ACI-​11.5.4.7, for sections located between the wall base and the smaller of ℓw  /​2 or hw /​2, Vc is permitted to be computed for the Mu  / ​Vu ratio at a distance equal to ℓw /​2 or hw /​2 from the base of the wall, whichever is less. For this wall, ℓw /​2 = 10 ft (< hw /​2 = 57 ft) governs. At this location, the factored moment Mu is 13,175 ft-​kips and the corresponding factored shear is 198 kips, resulting in a Mu  / ​Vu ratio of 66.5 ft. This value is greater than ℓw /​2 and Eq. (15.5.8) applies. Thus,   0.2 N u     1.25λ fc′ + hℓ  ℓ w  w  hd Vc = 0.6 λ fc′ +   Mu ℓ w −   Vu 2     0.2(1036)(1000)     1.25(1) 5000 +  240  10 240 ( )  10(0.8)240  1  = 0.6(1) 5000 +         1000   (66.5 − 10)(12)     = 200 kips Therefore, Vc = 200 kips based on flexure-​shear cracking governs. Since Vc computed with the detailed method (200 kips) is less than that computed with the simplified method (lower limit), Vc will be taken as the lower limit of 272 kips. In other words, for this wall with an aspect ratio hw /ℓw of 5.7, there would be no need to compute Vc with the detailed method. b. Shear reinforcement contribution, Vs—​Because the simplified method was used to compute Vc, the critical section for shear is taken at the base of the wall where the factored shear force is maximum in the first story. At this location, Vu = 200 kips. Since

0.5φVc = 102 kips < Vu = 200 kips < φVc = 204 kips only minimum transverse, horizontal reinforcement is required. Also, since Vu > 0.5φ Vc, a minimum of horizontal reinforcement ratio, ρt , of 0.0025 is required (ACI-​11.6.2). Because this wall has an aspect ratio hw /​ℓw= 5.7 > 2.5, Eq. (15.5.10) does not apply; thus the minimum required amount of longitudinal, vertical reinforcement, ρℓ , is equal to ρt , or 0.0025. This amount is less than that required from flexural strength requirements (0.0065) and thus, it is satisfied.

Since this wall requires only the minimum amount of horizontal reinforcement, a check on the maximum permitted amount to satisfy the requirement that Vn ≤ 10 fc′ hd is clearly not necessary. The procedure, however, is illustrated below: Vn = Vc + Vs = 272 + 0.0025(60)(10)(0.8)240

Vn = 560 kips 1

10 fc′ hd = 10 5000 (10)(0.8)240 1000 = 1358 kips

Thus, Vn < 10 fc′ hd , as was expected.



OK (Continued)

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Example 15.5.3 (Continued) 3. Select reinforcement and bar spacing. a.  Vertical reinforcement—​Since Vu > 0.5φVc, spacing of the vertical reinforcement is the smaller of • 3h = 3(10) = 30 in. • ℓw  /​3 = 240/​3 = 80 in., or • 18 in.

Governs!

The required amount of vertical reinforcement per flexural strength requirements is ρℓ ≈ 0.0065. Thus, the total required area of reinforcement per foot of wall length is required As vert = ρℓ h × (12 in.) = 0.0065(10)12





= 0.78 sq in./ft.

Since the wall thickness is 10 in., two layers of reinforcement are required. Therefore, the required amount of reinforcement per layer is required As vert = 0.78 / 2 = 0.39 sq in./ft / layer From Table  8.3.1, choose #5 bars spaced horizontally at 10 in. to provide a total amount of 0.37 sq in./​ft per layer. This amount is slightly less than that computed above. Given the approximations involved in the calculations, however, it can be considered adequate as a first trial. Try 24 bars per layer for a total of 48 #5 bars equally spaced at 10 in.



The design interaction diagram of the wall with this amount of reinforcement shows that the section is tension controlled and yields a design flexural strength, φ M n , of 15,170 ft-​kips at a factored axial force Pu = 1036 kips. Therefore, Mu = 15,162 ft-kips < φ M n = 15,170 ft-kips OK Use two layers of #5 bar spaced at 10 in. b.  Horizontal reinforcement—​Since Vu > 0.5φ Vc spacing of the horizontal reinforcement is the smaller of • 3h = 3(10) = 30 in. • ℓw  /​5 = 240/​5 = 48 in., or • 18 in.

Governs!

The required amount of horizontal reinforcement is ρt = 0.0025, and thus the total required area of reinforcement per foot of wall height is required As horiz = ρt h × (12 in.)



= 0.0025(10)12 = 0.30 sq in./ft

Since two layers of reinforcement are required, the required amount of reinforcement per layer is required As horiz = 0.30 / 2 = 0.15 sq in./ft/layer This requirement is satisfied with two layers of #4 bars spaced vertically at 16 in. (see Table 8.3.1). Use two layers of #4 bars spaced at 16 in. A  final design sketch is shown in Fig. 15.5.18. (Continued)

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Example 15.5.3 (Continued) #5 @ 10” (vertical)

#4 @ 16” (horizontal)

¾”(min) 10” ¾”(min) 20’ – 0”

Figure 15.5.18  Design sketch for the shear wall of Example 15.5.3.

15.6 LATERAL SUPPORT OF LONGITUDINAL REINFORCEMENT As described in Section 10.8 for columns, longitudinal reinforcement may tend to buckle if subjected to high compressive stresses at or shortly after spalling of the concrete cover. In walls not subjected to seismic actions, ACI-​11.7.4.1 requires that the longitudinal reinforcement be laterally supported by transverse ties whenever the total area of longitudinal steel exceeds 1% of the gross cross-​sectional area of the wall, or when longitudinal reinforcement is required for axial strength. The condition involving axial strength is taken to apply when vertical reinforcement is required as compression reinforcement to resist wall actions. In practice, vertical reinforcement in walls with low amounts of reinforcement (i.e., less than 1% of the gross area of the wall) will rarely be needed as compression reinforcement and thus would seldom require lateral support. On the other hand, walls with larger reinforcement amounts may have some of their longitudinal reinforcement concentrated at the wall ends (boundary elements), as shown in Fig. 15.5.3. In such cases, the longitudinal reinforcement in the boundary elements is designed to act in both tension and compression and thus must be laterally supported by transverse ties (see Fig.  15.6.1). The remaining vertical reinforcement in the web is commonly designed to satisfy shear strength requirements only and would not need to be laterally supported unless web reinforcement is used as compression reinforcement. Transverse ties must satisfy the requirements of ACI-​25.7.2 and may be arranged as shown in Fig. 10.8.1. It must be emphasized that these requirements apply to walls not subjected to seismic actions; additional requirements must be met for walls subjected to earthquake-​induced actions.

Figure 15.6.1  Common transverse reinforcement details to provide lateral support of longitudinal reinforcement in wall boundary elements.

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15.7 RETAINING STRUCTURES Retaining structures hold back soil or other loose material and prevent assumption of the natural angle of repose at locations where an abrupt change in ground elevation occurs. The retained material exerts a push on a structure and thus tends to overturn or slide it or both. Several types of these structures are described below (see Fig. 15.7.1). 1. Gravity wall. A gravity wall is usually of plain concrete and depends entirely on its weight for stability. It is used for walls up to about 10 ft high. 2. Cantilever retaining wall. Cantilever walls, the most common type of retaining structure, are used in the height range of 10 to 25 ft. The stem, heel, and toe of such a wall each acts as a cantilever beam. 3. Counterfort wall. In a counterfort wall the stem and slab are tied together by transverse walls (counterforts) spaced at intervals, which act as tension ties to support the stem wall. Counterfort walls are often economical for heights over about 25 ft. 4. Buttress wall. A buttress wall is similar to a counterfort wall except that the transverse support walls are located on the side of the stem opposite the retained material, where they act as struts. Buttresses, as compression elements, are more efficient than the tension counterforts and are economical in the same height range. The counterfort is more widely used than the buttress, however, because the counterfort is hidden beneath the retained material, whereas the buttress occupies what could otherwise be usable space in front of the wall. 5. Bridge abutment. A wall-​type bridge abutment acts like a cantilever retaining wall except that the bridge deck provides an additional horizontal restraint at the top of the stem. Thus this abutment is designed as a beam fixed at the bottom and simply supported or partially restrained at the top. 6. Box culvert. The box culvert, with either single or multiple cells, acts as a closed rigid frame that must resist not only lateral earth pressure but also vertical load, either from the soil that it supports or from both the soil and highway vehicle loads.

15.7.1 FORCES ON RETAINING WALLS The magnitude and direction of the earth pressure that tends to overturn and slide a retaining wall may be determined by applying the principles of soil mechanics. Excellent texts such as those reported in Terzaghi, Peck, and Mesri [15.24] and Huntington [15.25] will be useful for any extensive study of how to determine the soil pressure to be used in a given situation. The pressure exerted by the retained material is proportional to the distance below the earth surface and to its unit weight. Analogous to the action of a fluid, the unit pressure p at a distance h below the earth surface may be expressed as

p = Cwh

(15.7.1.1)

where w is the unit weight of the retained material and C is a coefficient that depends on the physical properties of the material. There are two categories of earth pressure: (1) the pressure exerted as the earth moves in the same direction as the retaining structure deflects, known as active pressure and (2) the resistance developed as a structure moves against the earth, known as passive pressure. Passive pressure is several times larger than active pressure. Both active and passive pressures may be expressed in the form of Eq. (15.7.1.1), but using Ca and Cp as the active and passive pressure coefficient C, respectively. The force Pa caused by active pressure on a wall of height h may be expressed as

Pa = Ca w

h2 2

(15.7.1.2)

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Retained material

Retained material

Stem Toe

(a) Gravity wall

Heel

(b) Cantilever retaining wall

Retained material

Retained material Buttress

Stem

Counterfort

Stem

Slab

Slab (c) Counterfort wall Bridge deck

(d) Buttress wall

Approach pavement

Top

Retained material

Wall Floor

Stem Toe

Heel

(e) Bridge abutment

(f) Box culvert

Figure 15.7.1  Types of retaining structures.

and the force Pp developed in passive pressure may be expressed as

Pp = C p w

h2 2

(15.7.1.3)

where Caw and Cpw in Eqs. (15.7.1.2) and (15.7.1.3) may be considered equivalent fluid pressures. Typical values for Ca and Cp are 0.3 and 3.3, respectively, for granular material such as sand. Roughly, Cp = 1/​Ca. The proper evaluation of earth pressures on retaining structures is outside the scope of this book; the preceding brief comments are offered to provide some logic for the earth pressures that are used in the design example of this chapter. The following factors may affect the pressure (active pressure) on a wall: 1. Type of backfill used. 2. Seasonal condition of the backfill material, such as wet, dry, or frozen. 3. Drainage of backfill material. 4. Possibility of backfill overload, such as trucks and equipment near the wall. 5. Degree of care exercised in backfilling. 6. Degree of rotational restraint between various components of the retaining structure. 7. Possibility of vibration in the vicinity of the wall (especially in the case of granular soil). 8. Type of material beneath the footing of the retaining structure. 9. Level of the water table.

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Probably the most important single factor is the need to prevent water from accumulating in the backfill material. Walls are rarely designed to retain saturated material, which means that proper drainage must be provided. When vehicles may travel near by, exerting their loads, or when buildings are constructed near the top of a retaining wall, the lateral pressure against that wall is increased. In the case of a fixed static load such as a building, the weight of the building can be converted into an additional height (surcharge) of backfill soil material. The effect of a highway or railroad passing over the retained material near the wall causes a dynamic reaction that cannot accurately be converted into a static effect. However, some specifications have traditionally prescribed an equivalent static surcharge corresponding to a number of additional feet of backfill material.

15.7.2 STABILITY REQUIREMENTS The first step in retaining wall design is to establish the proportions such that the stability of the structure under active pressure is assured (see Fig. 15.7.2.1). Three requirements must be satisfied: (a) the overturning moment Pah(h′/​3) must be more than balanced by the resisting moment Wx1 + Pav L, so that an adequate factor of safety against overturning is provided, usually about 2.0; (b)  sufficient frictional resistance F in combination with any reliable passive resistance Pp against the toe must provide an adequate factor of safety (usually 1.5) against sliding caused by Pah; and (c) the base width L must be adequate to distribute the load R to the foundation soil without causing excessive settlement or rotation. Typically, referring to the pressure distribution of Fig. 15.7.2.1, the overturning factor of safety (FS) would be computed as follows: FS =



Wx1 + Pav L resisting moment = overturing moment Pah (h ′ / 3)

(15.7.2.1)

or, perhaps more frequently, the vertical component of Pa would be neglected: FS =



Wx1 Pah (h ′ / 3)

(15.7.2.2)

where W represents the weight of the concrete wall and footing and of the soil resting on the footing.

δ

h

Commonly accepted pressure distribution h’

W

x1

Pa δ

Pav

Pah h’/3

Pp Toe x2

F = µR R L

Figure 15.7.2.1  Forces on retaining wall.

Heel

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Using the notation of Fig. 15.7.2.1, the factor of safety against sliding may be computed as, FS =



µ R + Pp

(15.7.2.3)



Pah

where µ is the coefficient of friction between the soil and footing. Table 15.7.2.1 gives a range of coefficients of friction that may be used as a guide for typical values in lieu of more accurate ones. The inclusion of some passive resistance Pp on the toe of the footing may or may not be justified. Certainly, to actually develop passive pressure in the soil in front of the wall, the design should specify that the concrete be placed without using forms for the toe and without disturbing the soil against which the concrete is placed. Referring to Fig. 15.7.2.2, the ordinary passive resistance against the toe is Pp1 =



1 C p wh12 2

(15.7.2.4)

TABLE 15.7.2.1  VALUES OF COEFFICIENT OF FRICTION μ BETWEEN SOIL AND CONCRETE Soil

µ

Coarse-​grained soils (without silt)

0.5–​0.6

Coarse-​grained soils (with silt)

0.4–​0.5

Silt

0.3–​0.4

Sound rock (with rough surface)

0.60

Frequently, however, Pp1 is neglected, and in nearly all cases at least the value of h1 is reduced under the assumption that during construction, or after, the earth surface cannot be expected to remain undisturbed at its final designated elevation. If, when all reliable resistances have been included, the factor of safety remains inadequate, a base key (Fig. 15.7.2.2) may be used. Essentially, the base key, when placed in an unformed excavation against undisturbed material, may be expected to develop an additional passive force Pp2, shifting the possible failure plane from line 1 to line 2. The base key develops the additional resistance Pp 2 =



1 C p w(h22 − h12 ) 2

(15.7.2.5)

and also, an inert region, bced of Fig. 15.7.2.2, is created, which moves the friction plane from bd to ce. Thus the frictional force developed along ce is based on the angle α, the angle of internal friction of the soil, rather than on the friction angle between soil and

Earth surface

a 1 2

h1 h2

Pp1

Toe b

Pp2 c

Stem

d

Base key e

Figure 15.7.2.2  Passive resistance and effect of base key.

f

601



1 5 . 7 . 3   P R E L I M I NA RY P R O P O RT I O N I N G O F CA N T I L E V E R   WA L L S

L L/3

L/3 R

601

L L/3

L/3

L/3

R

L/3

3x2

ph pt e

pt

L/2

CL of footing

CL of footing x2 (a) Resultant within middle third

e

(b) Resultant outside middle third

Figure 15.7.2.3  Soil pressure distribution.

concrete. Normally, tan α > µ for granular material, so that, by making the base key deep enough to create an inert block, additional frictional resistance is developed. Finally, the magnitude and distribution of the soil pressure requires investigation. Usual practice is to require the resultant vertical force R to be inside the middle third of the footing for sand and gravel subbases and within the middle half for rock subbase. In addition, the maximum pressure may not exceed the allowable value. Comments regarding allowable bearing capacity as well as some typical safe bearing values are to be found in Chapter 19. Referring to Fig. 15.7.2.3(a), when the entire footing is under compression, the basic equation for combined bending and axial compression acting on a 1-​ft strip along the wall is

p=

R Re( L / 2) R  6e  ± 3 = 1 ±  L L L L /12

(15.7.2.6)

For the limiting condition of zero stress at the heel, e = L / ​6; thus Eq. (15.7.2.6) is valid for all positions of R within the middle third. When the resultant R is outside the middle third [Fig. 15.7.2.3(b)], vertical force equilibrium requires



R=

1 pt (3 x2 ) 2

pt =

2R 3 x2



(15.7.2.7)

for 0 < 3x2  15.7 kips

OK

The ACI Code is unclear concerning the out-​of-​plane shear reinforcement requirements for cantilever retaining walls. However, cantilever walls have been traditionally treated as a one-​way slab in applying any ACI Code limitations. Accordingly, shear reinforcement is not required for one-​way slabs when Vu < φVc (ACI-​7.6.3.1). Where appearance on the front face is important, a batter of that face should be provided to counteract the effect of deflection. The usual batter is about 1 4 in./​ft of wall height. Thus the minimum batter here is

1 1 (18) = 4 in. 4 2

In this case, the thickness increases by 9 in. from the top to the bottom of the wall, so batter the front face 5 in. and the rear face 4 in. When the wall is in place, it will deflect (Continued)

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Example 15.7.4.1 (Continued)

Figure 15.7.4.2  Dimensions for computing resultant soil pressure in the design example.

forward and become nearly vertical; therefore, the analysis from this point will consider the front face to be vertical. Economics may dictate having only one sloping face, or in some cases no sloping face. (f) Factor of safety against overturning. Using the dimensions given in Fig. 15.7.4.2, locate the resultant of vertical forces (without overload factors) with respect to the heel, as in Table 15.7.4.1: 105 = 3.95 ft 26.6 resisting moment = 26.6(11.25 − 3.95) = 194 ft-kips

resultant, from heel =

overturning moment = P1 (10) + P2 (6.67) = 5.12(10) + 6.40(6.67) = 93.9 ft-kips





194 FS against overturning = = 2.07 > 2.0 93.9

OK

TABLE 15.7.4.1  Force

Arm

Moment

W1 = 0.120(18 +8)(6.5)  = 20.3

3.25

66

6.25

1

5.63 7.00

19 19

1 1 W2 = 0.030   (18)(0.75) =  0.2     2 W3 = 0.150(11.25)(2.0)   =  3.4 W4 = 0.150(1.0)(18)    = 2.7

Totals            26.6 1

105

This value corresponds to the difference between the concrete and retained material unit weights.

(Continued)

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Example 15.7.4.1 (Continued) Alternatively, if a stability check is to be made using factored loads, the load combination with 0.9D should be used with the horizontal earth pressure H (P1 and P2 in Fig. 15.7.4.2) multiplied by a load factor of 1.6 in accordance with ACI-​5.3.8(a), that is,

U = 0.9 D + 1.6 H

In such a case, it would be required that resisting moment ≥ overturning moment 0.9(194) ≥ 1.6(93.9)

175 ≥ 150



OK

In effect, the required factor of safety against overturning under service load conditions is 1.6 / ​0.9 = 1.8, which is not much less than the traditional value of 2.0. (g) Location of resultant and footing soil pressures. Referring to Fig. 15.7.4.3, and using service loads because the maximum soil pressure limitation is given for that condition, R = 26.6 kips 105 + 93.9 = 7.48 ft 26.6 11.25 e = 7.48 − = 1.85 ft 2 6e 6(1.85) = = 0.99 < 1.0 base length 11.25 x=



The resultant lies just inside of the middle third. The service load soil pressure diagram is essentially a triangle; thus 1 ( pmax )(base length ) 2 1 26.6 = ( pmax )(11.25) 2 pmax = 4.73 ksf < 5 ksf R=



OK

Figure 15.7.4.3  Soil pressure and location of resultant under service load in the design example.

(h) Factor of safety against sliding. Neglecting passive pressure against the toe of the footing and using service loads, force causing sliding = P1 + P2 = 5.12 + 6.40 = 11.52 kips frictional force = µ R = 0.40(26.6) = 10.64 kips

factor of safety =

10.64 = 0.92 < 1.5 11.52



NG (Continued)

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Example 15.7.4.1 (Continued) Sliding resistance may also be checked using factored loads, U = 0.9D + 1.6H, in which case it is required that resisting force ≥ sliding force 0.9(10.64) ≥ 1.6(11.52)

9.58 kips < 18.4 kips



NG

Thus, ACI-​5.3.8 would require the same factor of safety against sliding as against overturning, that is, 1.6 / ​0.9 = 1.78. This exceeds the traditional factor to resist sliding of 1.5. Since the resistance provided does not give an adequate safety factor, a key (Figs. 15.7.4.4 and 15.7.4.5) against sliding is required. Such a key is intended to develop passive pressure in the region in front of and below the bottom of the footing. The procedure for determining the size of a key is one that is debated by designers. Generally it seems desirable to place the front face of the key about 5 in. in front of the back face of the stem. This will permit anchoring the stem reinforcement in the key. Various procedures may reasonably be used for granular cohesionless materials to estimate the effect of the key. The maximum effect of a key would be to develop the passive resistance over the depth BC of Fig. 15.7.4.4, with any resulting failure occurring along a curved path such as C’C. Fisher and Mains [15.26] have advocated the inert-​ block concept of frictional resistance. In this method, any key used must extend deep enough below the footing to develop an inert block of soil, BCDE of Fig. 15.7.4.5; it will then have a failure plane approximately as C’CDGFH, so that the passive resistance is developed over only the depth BC. In determining the passive resistance, the top 1 ft of overburden is usually neglected in the height h1, of Fig. 15.7.4.4 or Fig. 15.7.4.5. For this design, a factor of safety of 1.5 against sliding under service loads has been considered proper. When passive resistance against the toe is also included (Fig. 15.7.4.4), a higher factor of safety, say 2.0, should be used.

Figure 15.7.4.4  The passive resistance concept of frictional resistance.

Figure 15.7.4.5  The inert-​block concept of frictional resistance.

(Continued)

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Example 15.7.4.1 (Continued) Using the inert-​block concept, the passive force Pp developed over the distance BC of Fig. 15.7.4.5 is equivalent fluid pressure = 400 pcf (given) Pp =



0.400(h1 + a )2 0.400(h1 )2 − 2 2

which for h1 = 3 ft gives Pp = 0.200(6 a + a 2 )



When an inert block is induced, the frictional coefficient over the region CD becomes tan α = tan 35° = 0.7, while the coefficient of 0.40 applies on DG and FH. Thus the frictional resistance F = µ1 R1 + µ 2 R2



 1  1 = 0.7   (4.73 + 2.59)(5.08) + 0.4   (2.59)((6.17)  2  2

= 13.0 + 3.2 = 16.2 kips

Force equilibrium, incorporating a 1.5 factor of safety, gives ( P1 + P2 )1.5 = Pp + F 11.52(1.5) = 0.200(6 a + a 2 ) + 16.2

a 2 + 6 a − 5.40 = 0



a = 0.8 ft

Neglecting the frictional force in front of the key and considering the passive resistance developed below the toe as shown in Fig. 15.7.4.4, the depth of key required may also be computed as b = 5.08 tan α = 5.08(0.7) = 3.55 ft PP = 0.400

(h1 + a + b)2 0.400(h1 )2 − 2 2

= 0.200(a 2 + 13.1a + 33.9)



Force equilibrium, using a 1.5 safety factor, gives ( P1 + P2 )1.5 = Pp + µ 2 R2  1 11.52(1.5) = 0.200(a 2 + 13 + 1a + 33.9) + 0.4   (2.59)(6.17)  2

a 2 + 13.1a − 36.5 = 0 a = 2.4 ft



This would indicate that the latter method is more conservative. With this discussion and these computations to serve as a guide, make the key depth 18 in. The key may be made square; use 1 ft 6 in. × 1 ft 6 in. Reinforcement will rarely be required, but it is common practice to extend some of the stem steel down into the key. (Continued)

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Example 15.7.4.1 (Continued) (i) Design of heel cantilever. The loads considered are included in Fig. 15.7.4.6, where the effect of the base key is neglected. For bending moment, the critical section is taken at the center of the stem steel (2.5 in. from face of wall). It seems that any possible plane of weakness will occur at the stem steel rather than at the face of support. Thus the downward uniform loading due to earth overburden, surcharge, and footing concrete using an overload factor of 1.2 for the earth and footing concrete and 1.6 for the surcharge is, from Fig. 15.7.4.6, wu = 1.2(2.16 + 0.30) + 1.6(0.96) = 4.49 kips/ft

Mu (downward) =

1 (4.49)(5.96)2 = 79.7 ft-kips 2

2” cover Stem reinforcement

w = 0.120 (18) = 2.16 kips/ft (earth overburden) = 0.120 (8) = 0.96 kips/ft (surcharge)

Critical section for inclined cracking due to shear

2’–0”



w = 0.15(2) = 0.30 kips/ft (footing) 2’–0” Heel (revised to 2’– 8”)

Base key Neglect upward pressure for heel design

5.75 = 2.42 ksf 4.73 11.25 (under service loads)

5’–9”

Figure 15.7.4.6  Design of heel cantilever in the design example.

The upward soil pressure would reduce this value. However, the upward soil pressure may not actually act in the linear fashion assumed; in fact, it might not be there at all. Applying overload conditions under ACI-​5.3.8, the most critical situation results under the condition of 0.9D (i.e., gravity dead load) and 1.6H (i.e., horizontal earth pressure). This would eliminate pressure under the heel entirely. The critical section for shear may be taken at the distance d from the face when support reaction, in direction of applied shear, introduces compression into the end regions of the member. In this case, tension is induced in the concrete where the heel joins the stem, and an inclined crack could extend into the region ahead of the back face of the stem. Thus, the critical section for shear must be taken at the face of the stem, as shown in Fig. 15.7.4.6. The factored shear at the face without including upward soil pressure is

Vu = 4.49(5.75) = 25.8 kips

The design shear strength, unless shear reinforcement is used, is

φ Vc = φ (2 λ fc′)bd

= 0.75[2(1) 3000 ](12)(21.5)

1 1000

= 21.2 kips < 25.8 kips

NG (Continued)

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Example 15.7.4.1 (Continued) Shear appears to control. The effective depth must be increased in the ratio 25.8 /​ 21.2 unless shear reinforcement is to be used. required d =



21.5(25.8) = 26.2 in. 21.2

Although slightly larger than required, use a heel thickness of 32 in. (d ≈ 29.5 in.). For flexural strength, using Mu = 82 ft-​kips (corrected for the weight of the 32-​in. footing),

required Rn =

Mu 82(12, 000 ) = = 105 psi φ bd 2 0.90 (12)(29.5)2

This is less than the Rn corresponding to the minimum percentage of reinforcement required for beams (see Fig. 3.8.1). The heel cantilever may be considered as a shallow foundation that must be designed according to the applicable ACI provisions for beams and slabs (ACI-​13.3.2.1). If treated as a one-​way slab, the minimum reinforcement per ACI-​7.6.1.1 should be provided. However, the retaining wall is a major beamlike structure and use of the minimum reinforcement required for beams is recommended. Increasing the heel thickness from the preliminary 24 in. to 32 in. reduces the stem height and would permit reducing its thickness. The extra heel thickness will also improve stability against overturning. From Eq. (3.8.5), the required reinforcement ratio ρ to satisfy the factored loading is about 0.0018 for Rn = 105 psi. The minimum reinforcement requirement is, according to ACI-​9.6.1.2, As ,min =



3 fc′ fy

bw d

[3.7.9]



but not less than 200bw d /​fy. Thus,

ρmin =

 3 fc′ 200   3 3000 200  = or or =  bw d f y   60, 000 60, 000   f y

As ,min

= [0.0027 or 0.0033] = 0.0033



Less than 0.0033 may be used, provided the amount is at least one-​third more than required for strength (ACI-​9.6.1.3); that is, 1.33(0.0018) = 0.0024. required As = 0.0024(12)29.5 = 0.85 sq in./ft Use #6@6 (As = 0.88 sq in./​ft). Regarding the development length for the #6 bars, the lateral spacing of 6 in. on centers means clear spacing exceeding 2db. Also, the 2-​in. clear cover exceeds db. Therefore, condition 2 of Category A of the simplified equations applies (see Section 6.6). From Table 6.6.1, Ld = 32.9 in. Because these bars are cast with more than 12 in. of concrete beneath them, a casting position factor ψt = 1.3 must be used. Thus, Ld (for #6) = 1.3(32.9) = 42.8 in. (simplified method) (Continued)

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Example 15.7.4.1 (Continued) This is a very long development length. Use the general equation, Eq. (6.6.1), to calculate Ld :    f ψ t ψ e ψ s  3 y Ld =  d  40 λ fc′  cb + K tr   b   d     b



[6.6.1]

where Ld = development length db = nominal diameter of bar or wire cb = cover or spacing dimension = the smaller of (1) distance from center of bar being developed to the nearest concrete surrface, and (2) one-half the center-to-center spacing of baars being developed In this situation, ψt = 1.3, ψe = 1.0, and λ = 1.0. The ACI Code permits a size factor ψs of 0.8 for #6 and smaller bars, though ACI Committee 408 does not support the use of this factor (see Section 6.7). Here, the authors will adopt the latter, more conservative approach, and use ψs = 1.0. Since there is no shear reinforcement, Ktr = 0. The dimension cb for cover or spacing is controlled by the cover,

cb = 2 (i.e.,clear cover ) + 0.375 ( i.e., bar radius) = 2.375 in.

thus, (cb + Ktr)/​db = 2.375/​0.75 = 3.2 > 2.5 max. Using Eq. (6.6.1) with λ = ψe = ψs = 1.0 and ψ t = 1.3,



   ψ t ψ e ψ s   3 60, 000 1.3  3 fy 0.75 = 32 in. (2.7 ft) Ld =  db =   40 λ fc′  cb + K tr    40 1.00 3000 2.5    d     b

Thus, the #6 bars in the heel must be embedded 32 in. to develop their full strength. This is measured as the distance beyond the main vertical reinforcement in the back face of the wall. Use an embedment of 2.7 ft from back face of wall. ( j) Design of toe cantilever. The toe of the footing is also treated as a cantilever beam, with the critical section for moment at the front face of the wall and the critical section for shear (inclined cracking) at a distance d (approximately 20.5 in.) from the front face of the wall. Shear will usually control the required toe thickness. The thickness need not be the same as the heel, though many engineers would make them the same. In this example the heel is unusually thick because of the heavy surcharge. Try a toe thickness somewhat less than the heel, say, 2 ft. Referring to Fig. 15.7.4.7 and neglecting the earth on the toe, the factored shear and bending moment are  4.73 + 3.87  Vu = 1.6  − 0.300 (2.04) = 13.1 kips   2  1  2 1  1 1 Mu = 1.6  (4.73)(33.75)2   + (3.15)(3.75)2   − (0.300)(3.75)2   3 2  3 2  2

  (3.75)2 = 1.6  (9.46 + 3.15 − 0.90) = 43.9 ft-kips 6  

(Continued)

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Example 15.7.4.1 (Continued) Critical section for inclined cracking due to shear 2.04’ 3’–9” w = 0.30 kips/ft for concrete footing only

2’–0”

3” cover

3.15 ksf 3.87 ksf 4.73 ksf

Service load stresses

Figure 15.7.4.7  Design of toe cantilever in the design example.

where the toe soil pressure is considered primarily the result of horizontal earth pressure; hence the use of the 1.6 factor [ACI-​5.3.8(a)]. required Rn =



43.9(12, 000) = 116 psi 0.90(12)(20.5)2

From Eq. (3.8.5), the required ρ ≈ 0.002 is less than the minimum reinforcement ratio [computed in part (i)] of 0.0033. Applying ACI-​9.6.1.3, minimum ρ = 1.33(0.002) = 0.0027

required As = 0.0027(12)(20.5) = 0.65 sq in./ft

Use #6 @ 8 (As = 0.66 sq in./​ft). The shear strength is

φ Vc = φ (2 λ fc′)bd

= 0.75[2(1) 3000 ](12)(20.5)

1 1000

= 20.2 kips > 13.1 kips

OK

The #6 bars in the heel have wide lateral spacing and 3-​in. cover; clearly the maximum (cb + Ktr)/​db = 2.5 applies. Using Eq. (6.6.1) with ψt = ψe = ψs = λ = 1.0,



    3 60, 000 1.0  ψ t ψ e ψ s  3 fy Ld =  db =   0.75 = 24.6 in. (2.1ft)  40 λ fc′  cb + K tr    40 1.00 3000 2.5    d     b

Use embedment of 2.1 ft from front face of wall. (Continued)

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Example 15.7.4.1 (Continued) (k) Reinforcement for wall. The wall height (17.33 ft) is now 8 in. less than that used in the preliminary design calculations. Retain the 21-​in. thickness at the base of the stem. The factored moment diagram, Mu, is shown in Fig. 15.7.4.8, calculated using Eq. (15.7.4.1) with maximum y = 17.33 ft along with an overload factor of 1.6. Also shown in the figure is the design strength diagram, φ Mn, of the selected wall steel. The steel area required at the base of stem is required Rn =

106(12, 000) = 344 psi 0.90(12)(18.5)2

required As = 0.006(12)(18.5) = 1.33 sq in./ft

Use #6 and #7 bars alternated @4 1 2 in. on centers (As = 1.39 sq in./​ft). The shear has been checked and found satisfactory. The embedment of the bars into the footing must equal or exceed the development length Ld. If all of the wall steel is run the entire height of the wall, vertical dowels extending above the footing must be used, since the footing is placed and cast first and the bars cannot extend 18 ft or so out of it. When bars extend out too far, they are often bent or broken off during construction. Also, in an endeavor to economize, the quantity of steel per foot of wall should be reduced in the upper parts of the wall so that the design strength approximately equals that required over the height of the wall. In this design, it is proposed that the #7 bars be extended up to 8 to 10 ft out of the footing (dowel bars). These bars may be terminated at a shorter distance where they are no longer required to satisfy flexural strength requirements, provided they can be developed. The #6 bars, on the other hand, will be lap spliced above the footing into the wall. (k.1) Embedment length into the footing, Ld (all bars). Ld for embedment into the footing can be based on the clear spacing because cover is not involved. For the #7 bars having the adjacent #6 bars centered 4 1 2 in. away,



0.875 0.75   − 4.5 −  2 2  d = 4.21d clear spacing =  b  b 0.875    

Since no transverse reinforcement is provided, Ktr = 0, and thus (cb + Ktr)/​db = 4.21 > 2.5 max. Using Eq. (6.6.1) with λ = ψ t = ψ e = ψ s = 1.0 ,



   ψ t ψ e ψ s   3 60, 000 1.0  3 fy 0.875 = 28.8 in. (2.4 ft) Ld =  d =  40 λ fc′  cb + K tr   b  40 1.00 3000 2.5    d     b

Use an embedment of 2 ft 6 in. into the footing for both the #6 and #7 bars. The base key is available for embedment of stem bars if it might be necessary. (k.2) Embedment length into the wall (#7 bars). When dowels are used, the clear cover of 2 in. over the portion of the dowels extending above the footing into the wall will control the development length Ld. Thus, for the general equation, Eq. (6.6.1),

cb = 2 (i.e., clear cover) + 0.438 (i.e., bar radius) = 2.44 in.

Thus (cb + Ktr)/​db = 2.79 > 2.5 max, and Ld = 2.4 ft (same length as into the footing). (Continued)

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1 5 . 7 . 4   D E S I G N of C A N T I L E V E R R E T A I N I N G   W A L L s

Example 15.7.4.1 (Continued) (k.3) Splice length above the footing (#6 bars). Since it is proposed to lap the #6 bars at the base of the wall, ACI-​25.5 must be applied to determine the required lap distance. When no more than half of the bar area is to be lap spliced within the lap length (less than half of the total bar area is to be spliced in this case), the tension splice must meet the requirements for a Class A splice. Referring to Table 6.16.1 (which summarizes ACI-​ 25.5.2.1), the lap required is equal to Ld (or 12 in. min), where Ld is the development length required for unspliced bars. The #6 bars to be spliced are located 9 in. on centers. The #7 bars 4.5 in. away are not being developed within the splice region; thus they do not influence the development length for the bars being spliced. The clear spacing is clear spacing =



9 − 2(0.75) db = 10 db 0.75

allowing an extra bar diameter for the lap splice. Clearly, (cb + Ktr)/​db is controlled by the 2-​in. clear cover to the tension steel at the back face of the wall. Thus, for the #6 bars cb = 2 (i.e., clear cover) + 0.375 (i.e., bar radius) = 2.38 in.



Thus, (cb + Ktr)/​db = 2.38 < 2.5 max., and Ld = 25.9 in. Use a splice length of 2 ft 3 in. for the #6 bars into the wall terminating at point A in Fig. 15.7.4.8 above the footing. (k.4) Bar cutoff locations. Proceeding from the stem base, the #7 bars embedded in the footing should not be extended more than about 8 to 10 ft out of the footing, as noted earlier. Bar spacing and cutoff are done in accordance with the ACI Code, by drawing the φ Mn diagram as described in Chapter 6. For #6 and #7 bars alternated at 4 1 2 in., the design strength is C = 0.85(3)(12)a = 30.6 a T = 1.39(60) = 83.4 kips



a = 2.72 in.



At top of wall,

φ M n = 0.90(83.4) [ 9.5 − 0.5(2.72)]



1 12

= 50.9 ft-kips



At bottom of wall,

φ M n = 0.90(83.4) [18.5 − 0.5(2.72)]

1 12

= 107 ft-kips

(Note: It may be readily verified that both sections are tension controlled, and thus φ = 0.9.) The design strengths computed above are used to locate the outer dashed line in Fig. 15.7.4.8. Next, use the moment diagram (Fig. 15.7.4.8) to locate the point where the #7 bars may be terminated, leaving the remaining design strength, φ Mn, based on #6 @ 9 in. This design strength is shown in Fig. 15.7.4.8 by an inclined dashed line that is plotted by using the following φ Mn values: C = 30.6 a,

T = 0.59(60) = 35.4 kips, a = 1.16 in.

φ M n (at top) = 0.9(35.4) [ 9.5 − 0.5(1.16)]

1 12

φ M n (at bottom ) = 0.9(35.4) [18.5 − 0.5(1.16)]

1 12

= 23.7 ft-kips = 47.6 ft-kips (Continued)

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Example 15.7.4.1 (Continued) 23.7 16

#6 @ 18”

#6 and #7 1 alternate @ 4 2 ” #6 @ 9”

12 C

d = 1.14’ > 12 diam

10

0

B

d = 1.28’ > 12 diam d = 15.4”

Terminate #7 bars

Ld = 28.8” for #7

7’–3”

6

Terminate every other #6 bar

8

2

d = 13.6”

Ld = 25.9” for #6

10’–6”

Distance from base of stem, ft

14

4

d = 9.5”

50.9

Ld = 25.9” for #6 bars

Factored moment diagram 2’–3” End of lap splice

47.6 Factored bending moment Mu (ft-kips)

φMn diagram

A

107 106

d = 18.5”

Figure 15.7.4.8  Determination of stem reinforcement for the design example.

The termination point B is found by extending beyond the intersection of the remaining φ Mn line (#6 @ 9 in.) with the factored moment diagram a distance of either the effective depth d or 12 bar diameters, whichever is greater. In this case, the calculated termination point and the practical location in 3-​in. increments approximately coincide (they differ by less than  1 2 in.) Whenever tension bars are terminated in a tension zone, reduction in shear strength and ductility may occur. For this reason, the ACI Code does not permit flexural reinforcement in beams or one-​way slabs to be terminated in a tension zone unless additional conditions are satisfied. Chapter 11 of the ACI Code is silent with respect to this requirement in cantilever retaining walls. Since the cantilever wall is assumed to behave as a beamlike structure, the provisions for cutting bars in the tension zone of beams will be used (see Section 6.11). Since stirrups are not used in retaining walls, the conditions, one of which must be satisfied in this case, are (1) that the factored shear Vu at the cutoff point not exceed two-​ thirds of the shear strength φVn and (2) that the continuing bars provide at least twice the area required for the bending moment at the cutoff point and that the factored shear Vu at the cutoff point not exceed three-​fourths of the shear strength φVn. From inspection of Fig.  15.7.4.8, it is clear that since the continuing bars provide less than twice the area required at the cutoff point, condition (2) above is not satisfied. (Continued)

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Example 15.7.4.1 (Continued) Check to see if item (1) above is satisfied. Using a 1.6 overload factor and Eq. (15.7.4.2), the shear at y = 10.12 ft (i.e., 17.33 ft –​7.21 ft) from the top is Vu = 1.6 0.256(10.12) + 0.016(10.12)2  = 6.8 kips

(

φVc = φ 2λ fc′bd

)

= 0.75[2(1) 3000 ](12)(14.8)

1 1000

= 14.6 kips

2 (14.6) > 6.8 3

OK

Thus, terminate #7 bars at 7 ft 3 in. from the top of heel. Additional economy may be achieved by cutting every other #6 bar, leaving #6 @ 18 in. to extend to the top of the wall (According to ACI-​11.7.2.1, the longitudinal reinforcement in cast-​in-​place walls shall be spaced not farther apart than 3 times the wall thickness nor more than 18 in.). Figure 15.7.4.8 shows the actual cutoff point C for every other #6. In this case, the extension d gives a point slightly below point C, since the actual cut point is located to produce bar lengths in the usual 3-​in. increments. Terminate every other #6 at 10 ft 6 in. from the top of heel. A tension zone cutoff check at point C (y = 6.83 ft from top; i.e., 17.33 –​10.5) shows Vu = 1.6 0.256(6.83) + 0.016(6.83)2  = 4.0 kips

(

φVc = φ 2 λ fc′bd

)

= 0.75[2(1) 3000 ](12)(13.0)

1 1000

= 12.8 kips

2 (12.8) > 4.0 3



OK

The complete bar arrangement for the stem is shown later (see Fig. 15.7.4.9). The reader may note that the moment decreases rapidly from the base of the stem toward the top of the wall. At about 5 ft from the base, the factored moment has decreased about 50%. Somewhere in this vicinity, the reinforcement ratio ρ that is required for strength equals the minimum of ACI-​9.6.1.2 [ρmin = 0.0033 from part (i)]. For the upper portion of the wall, ACI-​9.6.1 would require the minimum ρ to be either 0.0033 or four-​ thirds of the required ρ based on the factored moment. For example, the design strength φ Mn at point C is indeed approximately equal to four-​thirds of the factored moment at C. Thus, in general, wherever actual ρ is less than ρmin, the requirement of ACI-​9.6.1 may be checked by observing whether the design strength diagram is offset at one-​third or more from the factored moment curve. It is further noted that the stem has been designed for bending moment and shear only. However, the weight of the stem causes compression in itself, so that, strictly speaking, it might be treated as a compression member (under large moment) in accordance with the concepts of Chapter 10. In this design problem the compressive force Pu in the wall under factored loads would be

 12 + 21 Pu = 1.2  (0.5)(17.33)(0.15) = 4.3 kips  12  (Continued)

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Example 15.7.4.1 (Continued) The combination of this small Pu with Mu = 106 ft-​kips will give a design near the bottom of the tension-​controlled region (see Chapter  10). In general, treatment of such walls as compression members is rarely justified but, if done, it might permit some slight reduction in reinforcement or thickness of the section. (l) Minimum Horizontal Reinforcement. Horizontal bars along the length of the wall must be provided in accordance with ACI-​11.1.4 and 11.6. Accordingly, the minimum required amount if #5 or smaller bars are used is (see Table 15.2.1)

As = 0.0020bh = 0.0020(12)(16.5) = 0.40 sq in./ft

where the average wall thickness is used for h. Since it is primarily the front face that is exposed to temperature changes, more of the horizontal reinforcement should be placed there. Thus, it is suggested that about two-​thirds of this reinforcement be placed in the front face and one-​third in the rear face. Accordingly,

2 2 As = (0.40) = 0.27 sq in./ft 3 3



1 1 As = (0.40) = 0.13 sq in./ft 3 3

Use on the front face #5 @ 12 (As = 0.31 sq in./​ft). Use on the rear face #4 @ 14 (As = 0.17 sq in./​ft). For vertical reinforcement on the front face, use any nominal amount that is adequate for supporting the horizontal temperature and shrinkage steel in that face; the minimum amount specified in ACI-​11.6.1 is a good guide. Using an average wall thickness of 16.5 in. and assuming #6 bars (see Table 12.5.1),

min As = 0.0015(16.5)(12) = 0.30 sq in./ft

Use #6 @ 1 ft 6 in. spacing. (m) Drainage and other details. Adequate drainage of backfill must be provided because the pressures used are for drained material. A common minimum provision is for weep holes (say, 4-​in. diameter tile) every 10 to 15 ft along the wall. Construction of a retaining wall is accomplished in at least two stages; the footing is placed first and then the wall. A shear key is usually desirable for positive shear transfer between wall and footing (see Fig. 15.7.4.9). Such a key is made by embedding a beveled 2 × 4 or 2 × 6 timber in the top of the footing. This key may be designed using the shear-​friction provisions of ACI-​22.9 (discussed in Section 5.15). In this situation, it is not appropriate to consider shear as a measure of diagonal tension. The nominal shear strength of the shear key is based on a nominal stress of 0.2 fc′ but not to exceed 800 psi, as given by ACI-​22.9.4.4. This design uses no bent bars. Frequently, the bar arrangement in the toe can be conveniently bent up into the stem and meshed with additional stem reinforcement to give an economical system. (n) Design sketch. The final details of this design are presented in a design sketch (Fig. 15.7.4.9) which must accompany any set of design computations. (Continued)

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SELECTED REFERENCES

Example 15.7.4.1 (Continued)

5” 1 12 ” clear cover

1’–0” 4” 2”clear cover #6 @ 18” × 17’–3”

12” #5 (typical)

14” #4 (typical)

#6 @ 1’–6”

#6 @ 18” × 10’–6” For main reinforcement, see typical pattern in 10’–6” projection

1’–9” Ground surface

#7 @ 9 × 9’– 9”

#6 @ 6 × 8’– 9” 2” × 6” keyway

3’–9”

5’–9”

4’–0” 2’–0”

7’–3” 2’–3”

4” Ø weep holes @ 10’–0” spacing along wall

2’–8”

#6 @ 8 × 5’–9” 3” clear cover 1’–6” 5’–1”

#4 as shown

1’–6” 4’–8” 11’–3”

2” clear cover

18’–0”

#6 @ 9”× 4’–9”

Figure 15.7.4.9  Design sketch for the cantilever retaining wall.

SELECTED REFERENCES 15.1. K. M. Kriparanayan. “Interesting Aspects of the Empirical Wall Design Equation,” ACI Journal, 74, May 1977, 204–​207. 15.2. Garold D. Oberlender and Noel J. Everard. “Investigation of Reinforced Concrete Walls,” ACI Journal, 74, June 1977, 256–​263. 15.3. Sharon L.  Wood. “Performance of Reinforced Concrete Buildings During the 1985 Chile Earthquake: Implications for the Design of Structural Walls,” EERI Earthquake Spectra, 7, No. 4, 1991, 607–​638. 15.4. Burcu Burak and Hakki G. Comlekoglu. “Effect of Shear Wall Area to Floor Area Ratio on the Seismic Behavior of Reinforced Concrete Buildings,” ASCE Journal of Structural Engineering, 139, No. 11, November 2013, 1928–​1937. 15.5. K. N. Shiu, J. I. Daniel, J. D. Aristizabal-​Ochoa, A. E. Fiorato, and W. G. Corley. “Earthquake Resistant Structural Walls—​Tests of Walls with and without Openings.” Skokie, IL. Construction Technology Laboratories, Portland Cement Association (PCA), July 1981, 123 pp. 15.6. Aejaz Ali and James K. Wight, “RC Structural Walls with Staggered Door Openings,” ASCE Journal of Structural Engineering, 117, No. 5, May 1991, 1514–​1531. 15.7. Christopher P. Taylor, Paul A. Cote, and John Wallace. “Design of Slender Reinforced Concrete Walls with Openings,” ACI Structural Journal, 95, No. 4, July–​August 1998, 420–​433. 15.8. F. V. Yanez, R. Park, and T. Paulay. “Seismic Behavior of Reinforced Concrete Structural Walls With Regular and Irregular Openings,” Pacific Conference on Earthquake Engineering, New Zealand, November 1991. 15.9. Thomas Paulay. “The Design of Ductile Reinforced Concrete Structural Walls for Earthquake Resistance,” Earthquake Spectra, 2, No. 4, 1986, 783–​823.

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15.10. R.  G. Oesterle, A.  E. Fiorato, L.  S. Johal, J.  E. Carpenter, H.  G. Russel, and W.  G. Corley. “Earthquake Resistant Structural Walls—​Tests of Isolated Walls.” Skokie, IL: Construction Technology Laboratories, Portland Cement Association (PCA), November 1976, 328 pp. 15.11. John H. Thomsen and John W. Wallace, “Displacement-​Based Design of Slender Reinforced Concrete Structural Walls—​Experimental Verification,” ASCE Journal of Structural Engineering, 130, No. 4, April 2004, 618–​630. 15.12. Thomas Paulay. “The Shear Strength of Shear Walls,” Bulletin of the New Zealand Society of Earthquake Engineering, 3, No. 4, December 1970, 148–​162. 15.13. Sharon L.  Wood. “Shear Strength of Low-​Rise Reinforced Concrete Walls,” ACI Structural Journal, 87, January–​February, 1990, 99–​107. 15.14. Felix Barda, John M.  Hanson, and W.  Gene Corley, “Shear Strength of Low-​Rise Walls with Boundary Elements,” Reinforced Concrete Structures in Seismic Zones, ACI SP53-​8. Detroit: American Concrete Institute, 1977 (pp. 149–​202). 15.15. H. A. Williams and J. R. Benjamin. “Investigation of Shear Walls, Part 3, Experimental and Mathematical Studies of the Behavior of Plain and Reinforced Concrete Walled Bents Under Static Shear Loading.” Department of Civil Engineering, Stanford University, Stanford, CA, June 1953, 142 pp. 15.16. J. R. Benjamin and H. A. Williams. “Investigation of Shear Walls, Part 6, Continued Experimental and Mathematical Studies of Reinforced Concrete Walled Bents Under Static Shear Loading.” Department of Civil Engineering, Stanford University, Stanford, CA, August 1954, 59 pp. 15.17. J. Antebi, S. Utku, and R. J. Hansen, “The Response of Shear Walls to Dynamic Loads,” MIT Department of Civil and Sanitary Engineering, (DASA-​1160). Cambridge, MA, August 1960, 290 pp. 15.18. Thomas Paulay, M.  J. Nigel Priestley, and A.  J. Synge, “Ductility in Earthquake Resisting Squat Shear Walls,” ACI Journal, 79, July–​August 1982, 257–​269. 15.19. Pedro A. Hidalgo, Christian A. Ledezma, and Rodrigo M. Jordan, “Seismic Behavior of Squat Reinforced Concrete Shear Walls,” EERI Earthquake Spectra, 18, No. 2, May 2002, 287–​308. 15.20. A.  E.Cardenas, and D.  D. Magura. “Strength of High-​ Rise Shear Walls—​ Rectangular Cross Sections,” Response of Multistory Concrete Structures to Lateral Forces, ACI SP-​36. Detroit: American Concrete Institute, 1973, (pp. 119–​150). 15.21. A. E., Cardenas, J. M., Hanson, W. G., Corley, and E. Hognestad. “Design Provisions for Shear Walls,” ACI Journal, Proceedings, 70, March 1973, 221–​230. 15.22. K.  Muto, and K.  Kokusho. “Experimental Study on Two-​Story Reinforced Concrete Shear Walls,” Muto Laboratory, University of Tokyo, Japan. Translated by T. Akagi, University of Illinois, Urbana, August 1959. 15.23. K. Ogura, K. Kokusho, and N. Matsoura. “Tests to Failure of Two-​Story Rigid Frames with Walls, Part 24, Experimental Study No. 6,” Japan Society of Architects Report No. 18, February 1952. Translated by T. Akagi, University of Illinois, Urbana, August 1959. 15.24. Karl Terzaghi, Ralph B. Peck and Gholamreza Mesri. Soil Mechanics in Engineering Practice (3rd ed.). New York: John Wiley, & Sons, 1996. 15.25. Whitney Clark Huntington. Earth Pressures and Retaining Walls. New York: John Wiley & Sons, 1957. 15.26. G. P. Fisher and R. M. Mains. “Sliding Stability of Retaining Walls,” Civil Engineering, July 1952, 490.

PROBLEMS Shear Walls 15.1 For the wall of Example  15.5.3, use the ACI Code provisions to compute the amount of vertical and horizontal reinforcement required in the first story if normal-​weight concrete with fc′ = 8, 000 psi and Grade 60 steel is used. 15.2 Since the required strength of shear walls varies significantly with height, it is common practice to vary the concrete strength or the amount of reinforcement, or both, while maintaining wall thickness constant over the height. Assuming that the wall thickness and the amount of vertical reinforcement will remain the same over the

height, determine the level at which the concrete strength may be reduced from 5,000 psi to 4,000 psi for the wall of Example 15.5.3. 15.3 Assuming that the wall thickness and the amount of vertical reinforcement will remain the same over the height, determine the level at which the concrete strength may be reduced from 8,000 psi to 6,000 psi for the wall of Problem 15.1. If the concrete strength were to be further reduced in the upper levels of the wall, determine the level at which concrete strength could be reduced to 4,000 psi.

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15.4 Use the ACI Code provisions to determine the adequacy of the barbell wall of the figure for Problem 15.4. The wall is subjected to an axial compressive, factored load, Pu, of 1350 kips acting concurrently with a factored moment, Mu, of 14,900 ft-​kips and an in-​plane, factored shear force, Vu, of 120 kips. Vertical reinforcement in the boundary elements consist of 8–​#8 bars. In the web, both the vertical and horizontal reinforcement consist of #5 bars spaced at 18 in. on centers. Use fc′ = 6, 000 psi and Grade 60 steel.

Cantilever Retaining Walls Note: For all problems assume that frost pen­ etration depth is 4 ft. 15.5 Given the cantilever retaining wall of the figure for Problem 15.5 with the active pressure coefficient Ca  =  0.27, the weight of retained soil  = 100 pcf, and the coefficient of friction between concrete and earth = 0.40, use service load conditions to determine the following. (a) The factor of safety provided against overturning. Would you consider this to be adequate? (b) The factor of safety provided against sliding. Neglect any passive pressure resistance on the toe. Would you consider this factor of safety adequate? (c) The location of the resultant bearing pressure under footing. Is it within the middle third of the base? Would the position you 8 #8

determined probably be permissible if the foundation material is rock? 15.6 Determine the adequacy of the retaining wall of the figure for Problem 15.6 with regard to stability (overturning, sliding, and soil pressure magnitude and distribution). Assume that the wall is on good granular soil with a maximum safe soil pressure of 5000 psf under service load conditions. 15.7 Reconsider the retaining wall conditions of Section 15.7.4 if the surcharge is changed to 10 ft to approximate the effect of a railroad located parallel to and near the top of the wall. Sometimes under such conditions the lateral effect of a surcharge is included but the beneficial stabilizing effect of the vertical surcharge weight is omitted. Compare the wall proportions for both conditions, using a minimum length of base and keeping the soil pressure resultant under service load conditions within the middle third of the base. Verify the adequacy of the dimensions, but do not design the reinforcement. Use fc′ = 4000 psi, fy = 60,000 psi, and the ACI Code. 15.8 Assuming that the overall proportioning of the retaining wall of Problem 15.6 is adequate for earth stability, design the reinforcement for the wall cantilever. Besides dowels at the base, use three changes of reinforcement over the 20-​ft wall height. Draw the resulting design strength φMn diagram superimposed on the factored moment Mu diagram. Use fc′ = 4000 psi, fy = 60,000 psi. 8 #8

#5 @ 18

Problem 15.4  12”

Level

Equivalent fluid active pressure = 29 pcf

Retained material

Weight of retained material = 100 pcf

20’–0’’ 22’–0’’

Assumption for computing stability 2’–0”

1’–6”

15’–0”

2’–6” 2’–0’’

6’–0”

Problem 15.5 

Equivalent fluid passive pressure = 300 pcf

2’–0”

2’– 8’’

1’–4’’ 2’– 0’’

Problem 15.6 

5’–5’’

6’–4’’ 11’–0’’

1’–4’’

CHAPTER 16 DESIGN OF TWO-​WAY FLOOR SYSTEMS

16.1 GENERAL DESCRIPTION In reinforced concrete buildings, a basic and common type of floor is the slab-​beam-​girder construction, which was treated in Chapters 8 and 9. As shown in Fig. 16.1.1(a), the shaded slab area is bounded by the two adjacent beams on the sides and portions of the two girders at the ends. When the length of this area is two or more times its width, almost all of the floor load is transferred to the beams, and very little, except some near the edge of the girders, is transferred directly to the girders. Thus the slab may be designed as a one-​way slab as treated in Chapter 8, with the main reinforcement parallel to the girder and the shrinkage and temperature reinforcement parallel to the beams. The deflected surface of a one-​way slab is primarily one of curvature in its short direction. When the ratio of the long span L to the short span S as shown in Fig. 16.1.1(b) is less than about 2, the deflected surface of the shaded area becomes one of curvature in both directions. The floor load is carried in both directions to the four supporting beams around the panel; hence the panel is a two-​way slab. Obviously, when S is equal to L, the four beams around a typical interior panel should be identical; for other cases, the longer beams carry more load than the shorter beams.

Flat slab with drop panels and capitals. Parking garage of Helen C. White Hall, University of Wisconsin–Madison. (Photo by José A. Pincheira.)

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16.1  GENERAL DESCRIPTION

Two-​way floor systems may also take other forms in practice. Figure 16.1.2 shows flat slab and flat plate floor construction, which are characterized by the absence of beams along the interior column lines; edge beams may or may not be used at the exterior edges of the floor. Flat slab floors differ from flat plate floors in that slab floors provide adequate shear strength by having either or both of the following: (a) drop panels (i.e., increased thickness of slab) in the region of the columns or (b) column capitals (i.e., tapered enlargement of the upper ends of columns). In flat plate floors a uniform slab thickness is used and the shear strength is obtained by the embedment of multiple-​U stirrups [Fig. 16.16.1(a)], structural steel devices known as shearhead reinforcement [Fig. 16.16.1(b)] or, more commonly used today, headed shear stud reinforcement [Fig. 16.16.2] within the slab of uniform thickness. Relatively speaking, flat slabs are more suitable for larger panel size or heavier loading than flat plates. Historically [16.2], flat slabs predate both two-​way slabs on beams and flat plates. Flat slab floors were originally patented by O. W. Norcross in the United States on April 29, 1902. Several systems of placing reinforcement have been developed and patented since then—​the four-​way system, the two-​way system, the three-​way system, and the circumferential system. C. A. P. Turner was one of the early advocates of a flat slab system known as the “mushroom” system. About 1908, the flat slab began being recognized as an acceptable floor system, but for many years designers were confronted with difficulties of patent infringements. L

B

G B

B

B

G

(a)

B B

(b)

Figure 16.1.1  One-​way and two-​way slabs.

Drop panel

Column capital

(b) Flat plate floor Column capital

Drop panel

(a) Flat slab floor

Figure 16.1.2  Flat slab and flat plate floor construction.

S

624

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The terms two-​way slab [Fig.  16.1.1(b)], flat slab [Fig.  16.1.2(a)], and flat plate [Fig. 16.1.2(b)] are arbitrary. Following tradition, the implication is that in two-​way slabs there are beams between columns, but no such beams, except perhaps for edge beams along the exterior sides of the entire floor area in flat plates. In flat slabs, on the other hand, the slab depth near the columns is increased to provide additional shear and bending resistance. If methods of structural analysis and design are developed for two-​way slabs with beams, many of these general provisions should apply equally well to flat slabs or flat plates. Historically, until 1971, the design of two-​way slabs supported on beams was treated separately from that of flat slabs or flat plates without beams. Various empirical procedures have been proposed and used [16.6–​16.8]. Chapter 8 of the 2014 ACI Code takes an integrated view and refers to two-​way slab systems with or without beams. In addition to solid slabs, hollow slabs with interior voids to reduce dead weight, slabs (such as waffle slabs) with recesses made by permanent or removable fillers between joists in two directions, and slabs with paneled ceilings near the central portion of the panel are also included in this category (ACI-​8.1.3). Thus the term two-​way floor systems (rather than the term two-​way slab systems as in the ACI Code) is used in this book to include all three types of floor systems: the two-​way slab with beams, the flat slab, and the flat plate.

16.2 GENERAL DESIGN CONCEPT OF THE ACI CODE The basic approach to the design of two-​way floor systems involves imagining that verti­ cal cuts are made through the entire building along lines midway between the columns. The cutting creates a series of frames whose width lies between the centerlines of the two adjacent panels as shown in Fig. 16.2.1. The resulting series of rigid frames, taken separately in the longitudinal and transverse directions of the building, may be treated for gravity loading floor by floor, as would generally be acceptable for a rigid frame structure consisting of beams and columns, in accordance with ACI-​6.3.1.2. A typical rigid frame would consist of (1) the columns above and below the floor and (2) the floor system, with or without beams, bounded laterally between the centerlines of the two panels (one panel for an exterior line of columns) adjacent to the line of columns. Thus, the design of a two-​ way floor system (including two-​way slab, flat slab, and flat plate) is reduced to that of an equivalent frame. As in the case of design of actual rigid frames consisting of beams and columns, approximate methods of analysis may be suitable for many usual floor systems, spans, and story heights. As treated in Chapter 7, the analysis for actual frames could be (a) approximate, using the moment and shear coefficients of ACI-​6.5, or (b) more accurate, using structural analysis after assuming the relative stiffnesses of the members. For gravity load only and for floor systems within the specified limitations, the moments and shears on these equivalent frames may be determined (a)  by approximately using moment and shear coefficients prescribed by the Direct Design Method of ACI-​8.10 or (b) by structural analysis in a manner similar to that for actual frames, using the special provisions of the Equivalent Frame Method of ACI-​8.11. It is noted that an elastic analysis (such as by the equivalent frame method) must be used for lateral loads even if the floor system meets the limitations of the direct design method for gravity load. The equivalent rigid frame is the structure being dealt with whether the moments are determined by the Direct Design Method (DDM) or by the Equivalent Frame Method (EFM). These two ACI Code terms describe two ways of obtaining the longitudinal variation of bending moments and shears. When the EFM is used for obtaining the longitudinal variation of moments and shears, the relative stiffness of the columns, as well as that of the floor system, can be assumed in the preliminary analysis and then reviewed, as is the case for the design of any statically indeterminate structure. Design moment envelopes may be obtained for dead load in combination with various patterns of live load, as described in Chapter 7 (Section 7.3). In lateral

625

625

Interior equivalent frame floor area

equivalent rigid frame

1 6 . 3   TOTA L FAC TO R E D S TAT I C   M O M E N T

Width of



Figure 16.2.1  Tributary floor area for an interior equivalent rigid frame of a two-​way floor system.

load analyses, moment magnification in columns due to sidesway of vertical loads must be taken into account as prescribed in ACI-​6.6.4, 6.7, or 6.8 (see Chapter 13). Once the longitudinal variation in factored moments and shears has been obtained, whether by the DDM or EFM, the moment across the entire width of the floor system being considered is distributed laterally to the beam, if used, and to the slab. The lateral distribution procedure and the remainder of the design are essentially the same whether the DDM or EFM has been used. The accuracy of analysis methods utilizing the concept of dividing the structure into equivalent frames has been verified for gravity load analysis by tests [16.12–​16.25] and analytical studies [16.26–​16.35]. For lateral load analysis, where there is less agreement on procedure, various studies have been made, including those of Pecknold [16.38], Allen and Darvall [16.39,16.47], Vanderbilt [16.32, 16.40], Elias [16.41–​16.43], Fraser [16.44], Vanderbilt and Corley [16.45], Lew and Narov [16.46], Pavlovic and Poulton [16.48], Moehle and Diebold [16.49], Hsu [16.50], Cano and Klingner [16.51], Hwang and Moehle [16.145], and Dovich and Wight [16.146].

16.3 TOTAL FACTORED STATIC MOMENT Consider ABCD and CDEF, two typical interior panels in a two-​way floor system, as shown in Fig. 16.3.1(a). Let L1 and L2 be the panel size in the longitudinal and transverse directions, respectively. Let lines 1-​2 and 3-​4 be centerlines of panels ABCD and CDEF, both parallel to the longitudinal direction. Isolate as a free body [see Fig. 16.3.1(b)] the floor slab and the included beam bounded by the lines 1-​2 and 3-​4 in the longitudinal direction and the transverse lines 1′-​3′ and 2′-​4′ at the faces of the columns in the transverse direction. The load acting on this free body [see Fig. 16.3.1(c)] is wu L2 per unit distance in the longitudinal direction. The total upward force acting on lines 1′-​3′ or 2′-​4′ is wu L2 Ln /​2, where wu is the factored load per unit area and Ln is the clear span in the longitudinal direction between faces of supports (as defined by ACI-​8.10.3.2.1). If Mneg and Mpos are the numerical values of the total negative and positive bending moments along lines 1′-​3′ and 5-​6, respectively, then equilibrium of the free body of Fig. 16.3.1(d) requires

M neg + M pos =

wu L2 L2n 8

(16.3.1)

For a typical exterior panel, the negative moment at the interior support would be larger than that at the exterior support, as has been shown in Section 7.5. The maximum positive

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moment would occur at a section to the left of the midspan, as shown in Fig. 16.3.2(c). In design, it is customary to use Mpos at midspan for determining the required positive moment reinforcement. For this case, M neg (left) + M neg (right)



2

+ M pos =

wn L2 L2n 8

(16.3.2)

A proof for Eq. (16.3.2) can be obtained by writing the moment equilibrium equation about the left end of the free body shown in Fig. 16.3.2(a), M neg (left) + M pos =



wu L2 Ln 2

 Ln   Ln   4  − Vmidspan  2 

and by writing the moment equilibrium equation about the right end of the free body shown in Fig. 16.3.2(b), M neg (right) + M pos =



wu L2 Ln  Ln  L  + Vmidspan  n     2  2  4

Equation (16.3.2) is arrived at by adding the two preceding equations and dividing by 2 on each side. Note that Eq. (16.3.2) may also be obtained, as shown in Fig. 16.3.2(c), by the superposition of the simple span uniform loading parabolic positive moment diagram over the trapezoidal negative moment diagram due to end moments.

L1 A

B

L2

1

2

C

L2

wu L2 per unit distance Mneg

Mneg

D

3

Clear span = Ln

4

E

wu L2 Ln

wu L2 Ln

2

2 (c)

F (a) Mneg

1

1’

5

2’

2

0

Centerline

wu L2 Ln wu L2 Ln 2

Factored floor load = wu per unit area 3

3’

6

Mpos

Ln 4

2 (d) Mpos

4’

4

M0 Mneg

(b)

Figure 16.3.1  Statics of a typical interior panel in a two-​way floor system.

Mneg (e)

627



627

1 6 . 3   TOTA L FAC TO R E D S TAT I C   M O M E N T

wu L2 Ln 2

Mneg(left)

Mpos Ln /2 Ri

wu L2 Ln

VCL

2

(a)

Mpos

Mneg(right) Ln /2 VCL

Mmax

(b)

RJ

Mpos

+ –

–

Mneg(left) (c)

Mneg(right)

Figure 16.3.2  Statics of typical exterior panel in a two-​way floor system.

ACI-​8.10.3 uses the symbol M0 to mean wu L2 L2u /8 and calls M0 the total factored static moment. Further, it requires that the “absolute sum” of positive and average negative factored moments in each direction be not less than M0, or M neg (left) + M neg (right)

2

 w L L2  + M pos ≥  M 0 = u 2 n  8  

(16.3.3)

in which wu = factored load per unit area Ln = clear span in the direction moments are being determined, measured face-​to-​ face of supports (ACI-​8.10.3.2.1), but not less than 0.65L1 L1 = span length in the direction that moments are being determined, measured center-​to-​center of supports L2 = span length in direction perpendicular to L1, measured center-​to-​center of supports Equations (16.3.1) and (16.3.2) are theoretically derived on the basis that Mneg(left), Mpos, and Mneg(right) occur simultaneously for the same live load pattern on the adjacent panels of the equivalent rigid frame defined in Fig.  16.2.1. If the live load is relatively heavy compared with dead load, then different live load patterns should be used to obtain the critical positive moment at midspan and the critical negative moments at the supports. In such a case, the “equal” sign in Eqs. (16.3.1) and (16.3.2) becomes the “greater than” sign. This is why ACI-​8.10.3 states that “absolute sum …” shall be at least M0 as the design requirement. The designer should keep this in mind when steel reinforcement is selected for positive and negative bending moment in two-​way floors when the direct design method is used for gravity load. To avoid the use of excessively small values of M0 in the case of short spans and large columns or column capitals, the clear span Ln to be used in Eq. (16.3.3) is not to be less than 0.65L1 (ACI-​8.10.3.2.1). When the limitations for using the direct design method are met, it is customary to divide the value of M0 into Mneg and Mpos if the restraints at each end of the span are

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identical (Fig. 16.3.1), or into [Mneg (left) + Mneg(right)]  /​2 and Mpos if the span end restraints are different (Fig. 16.3.2). Then the moments Mneg(left), Mneg (right), and Mpos must be distributed transversely along the lines 1′-​3′, 2′-​4′, and 5-​6, respectively. This last distribution is a function of the relative flexural stiffness between the slab and the included beam.

Total Factored Static Moment in Flat Slabs The ability of flat slab floor systems to carry load has been substantiated by numerous tests of actual structures [16.2]. However, the amount of reinforcement used, say, in a typical interior panel, was less than what it should be to satisfy an analysis by statics, as is demonstrated in this section. This led to some controversy [16.1]. After studies by Westergaard and Slater [16.3], however, a provision was adopted (about 1921) into the Code that allowed a reduction of the moment coefficient from the statically required value of 0.125 to 0.09. This reduction was not regarded as a violation of statics but was used as a way of permitting an increase in the usable strength. The reduction, moreover, was applicable only to flat slabs that satisfied the limitations then specified in the code. Over the years these limitations had been liberalized, but at the same time the moment coefficient was raised to values closer to 0.125. The present ACI Code logically stipulates the use of the full statically required coefficient of 0.125. The statical analysis of a typical interior panel was first made in 1914 by Nichols [16.1] and further developed later by Westergaard and others [16.3–​16.5]. Consider the typical interior panel of a flat slab floor subjected to a factored load of wu per unit area, as shown in Fig. 16.3.3(a). The total load on the panel area (rectangle minus four quadrantal areas) is supported by the vertical shears at the four quadrantal arcs. Let Mneg and Mpos be the total negative and positive moments about a horizontal axis in the L2 direction along the edges of ABCD and EF, respectively. Then, load on area ABCDEF = sum of reactions at arcs AB and CD LL π c2  = wu  1 2 − 8   2 Considering the half-​panel ABCDEF as a free body, recognizing that there is no shear at the edges BC, DE, EF, and FA, and taking moments about axis 1-​1, LL π c 2   c  wu L1 L2  L1  wu π c 2  2c  + M neg + M pos + wu  1 2 −  −   = 0 8   π  2  4  8  3π   2



1

A

c /2

F

wu L2 per linear ft

B (L1 –

2 3

L2

c) Mpos

D

E L1/2

Mneg

1 2

L1 1 2

c π

Mneg

1

L1/2

panel load Mpos

exact, c3 approximate

panel load (a)

(b)

Figure 16.3.3  Statics of a typical interior panel in a flat slab floor system.

M0

C

629



629

1 6 . 3   TOTA L FAC TO R E D S TAT I C   M O M E N T

Letting M0 = Mneg + Mpos, 2

  1 4c 2c  c3  1 + M 0 = wu L2 L21  1 − ≈ wu L2 L21  1 − 2 8  3L1   π L1 3L2 L1  8



(16.3.4)

Actually, Eq. (16.3.4) may be more easily visualized by inspecting the equivalent interior span as shown in Fig. 16.3.3(b). ACI-​8.10.1.3 states that circular or regular polygon-shaped supports shall be treated as square supports having the same area. For flat slabs, particularly with column capitals, the clear span Ln, computed from using equivalent square supports, should be compared with that indicated by Eq. (16.3.4), which is L1 minus 2c/​3. In some cases the latter value is larger and should be used, consistent with the fact that ACI-​8.10.3.2 does express its intent in an inequality.

Design Examples In an effort to present, explain, and illustrate the design procedure for the three types of two-​way floor systems, identified in this chapter as two-​way slabs (with beams), flat slabs, and flat plates, it will be necessary to assume that preliminary dimensions and sizes of the slab (and drop, if any), beams, and columns (and column capitals, if any) are available. In the usual design processes, not only the preliminary sizes may need to be revised as they are found unsuitable, but also designs based on two or three different relative beam sizes (when used) to slab thickness should be made and compared. Preliminary data for the three types of two-​way floor systems to be illustrated are as follows. Data for Two-​Way Slab (with Beams) Design Example Figure 16.3.4 shows a two-​way slab floor with a total area of 12,500 sq ft. It is divided into 25 panels with a panel size of 25 ft × 20 ft. Concrete strength is fc′ = 3000 psi and steel yield strength is fy = 60,000 psi. Consider a superimposed service dead load of 24.5 psf and a service live load of 120 psf. Story height is 12 ft. The preliminary sizes are as follows: slab thickness is 6 1 2 in., long beams are 14 × 28 in. overall; short beams are 12 × 24 in. overall; upper and lower columns are 15 × 15 in. The four kinds of panels (corner, long-​sided edge, short-​sided edge, and interior) are numbered 1, 2, 3, and 4 in Fig. 16.3.4.

W

2

3

4

f’c = 3 ksi fy = 60 ksi 5 @ 20’ = 100’

N

1

E S 3

4

1

2

Service LL = 120 psf Story height = 12 ft Assume: 1 Slab thickness = 6 2 ” Long beams 14 × 28 overall Short beams 12 × 24 overall Columns 15” × 15”

5 @ 25’ = 125’

Figure 16.3.4  Plan view and data for two-​way slab (with beams) design examples.

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EXAMPLE 16.3.1 For the two-​way slab (with beams) (see Fig. 16.3.4), determine the total factored static moment in a loaded span in each of four equivalent frames whose widths are designated A, B, C, and D in Fig. 16.3.5. SOLUTION The total service dead load per unit floor area is Slab

(6.5)(150)

1 12

= 81.25 psf



Superimposed

= 24.50 psf



Total service dead load:

= 105.75 psf

The factored load wu per unit floor area is then wu = 1.2 wD + 1.6 wL = 1.2(105.75) + 1.6(120)

= 127 + 192 = 319 psf

5 @ 20’ = 100’

5 @ 20’ = 100’



A

D

B

5 @ 25’ = 125’

C

5 @ 25’ = 125’

Figure 16.3.5  Equivalent frame notations in the two-​way slab (with beams) of Example 16.3.5.

for frame B,

1 1 wu L2 L2n = (0.319)(20)(25 − 1.25)2 = 450 ft-kips 8 8 M 0 = 225 ft-kips

for frame C,

M0 =

for frame A,



for frame D,

M0 =

1 1 wu L2 L2n = (0.319)(25)(20 − 1.25)2 = 350 ft-kips 8 8 M 0 = 175 ft-kips

Data for Flat Slab Design Example Figure 16.3.6 shows a flat slab floor with a total area of 12,500 sq ft. It is divided into 25 panels with a panel size of 25 × 20 ft. Concrete strength is fc′ = 3000 psi and steel yield strength is fy = 60,000 psi. Service live load is 140 psf. Story height is 10 ft. Exterior columns are 16 in. square and interior columns are 18 in. round. Edge beams are 14 × 24 in. overall. Thickness of slab is 7  1 2 in. outside of drop panel and 10  1 2 in. through the drop panel. Sizes of column capitals and drop panels are as shown in Fig. 16.3.6.

631

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f’c = 3 ksi fy = 60 ksi

Slab thickness outside of drop = 7 2 ” 1

Slab thickness within drop = 10 2 ”

Service LL = 140 psf Story height = 10 ft

L = 25’

3

4

5’ – 0”

6’ – 8”

4’ – 6”

8’ – 4”

16” sq column

(Revised to 7’ – 0”)

5 @ 20’ = 100’

18” diam column

S = 20’



1

2

14” × 24” edge beams

5 @ 25’ = 125’

Figure 16.3.6  Plan view and data for flat slab design examples.

EXAMPLE 16.3.2 Compute the total factored static moment in the long and short directions of an interior panel in the flat slab design example as shown in Fig. 16.3.6. Compare the results obtained by using Eqs. (16.3.3) and (16.3.4). SOLUTION Neglecting the weight of the drop panel, the service dead load is 150 (7.5/12) = 94 psf; thus

wu = 1.2 wD + 1.6 wL = 1.2(94) + 1.6(140) = 113 + 224 = 337 psf



Using Eq. (16.3.4), 2

2

 1 2c  1 2(5)   = 396 ft-kips wu L2 L21  1 − = (0.337)(20)(25)2 1 − 8 8 3 (25)   3L1   (in long direction) M0 =



2 2 1 2c  1 2(5)  2  2  M 0 = wu L2 L1  1 − = (0.337)(25)(20) 1 −  = 293 ft-kips 8 8  3L1   3(20)  (in short direction) (Continued)

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Example 16.3.2 (Continued) The equivalent square area for the column capital (ACI-​8.10.1.3) has its side equal to 4.43 ft; then, using Eq. (16.3.3), with Ln measured to the face of capital (i.e., equivalent square), 1 1 wu L2 L2n = (0.337)(20)(25 − 4.43)2 = 356 ft-kips 8 8 (in long diirection ) M0 =

1 1 wu L2 L2n = (0.337)(25)(20 − 4.43)2 = 255 ft-kips 8 8 (in short direction)



M0 =

As to flat slabs with column capitals, it appears that the larger values of 396 and 293 ft-​kips should be used because Eq. (16.3.4) is especially suitable; in particular, ACI-​8.10.3.2 states that the total factored static moment shall be at least that given by Eq. (16.3.3).

Data for Flat Plate Design Example Figure  16.3.7 shows a flat plate floor with a total area of 4500 sq ft. It is divided into 25 panels with a panel size of 15 × 12 ft. Concrete strength is fc′ = 4000 psi and steel yield strength is fy = 60,000 psi. Service live load is 72 psf. Story height is 9 ft. All columns are rectangular, 12 in. in the long direction and 10 in. in the short direction. Preliminary slab thickness is set at 5 1 2 in. No edge beams are used along the exterior edges of the floor.

No edge beams

Live load = 72 psf

5 @ 12’ = 60’

3

4 f’c = 4 ksi fy = 60 ksi

10” 12” 1

2

Service LL = 72 psf Story height = 9 ft 1 Slab thickness = 5 2 ” (preliminary)

5 @ 15’ = 75’

Figure 16.3.7  Plan view and data for flat plate design examples.

EXAMPLE 16.3.3 Compute the total factored static moment in the long and short directions of a typical panel in the flat plate design example as shown in Fig. 16.3.7. SOLUTION The dead load for a 5 1 2 -​in. slab is wD = (5.5 /12)(150) = 69 psf

(Continued)

63



633

1 6 . 4   R AT I O O F F L E X U R A L S T I F F N E S S E S O F B E A M T O S L A B

Example 16.3.2 (Continued) The factored load per unit area is

wu = 1.2 wD + 1.6 wL = 1.2(69) + 1.6(72) = 83 + 115 = 198 psf



Using Eq. (16.3.3), with clear span Ln measured face-​to face of columns, 1 M 0 = (0.198)(12)(15 − 1)2 8

= 58.2 ft-kips (in long direction)

1 M 0 = (0.198)(15)(12 − 0.83)2 = 46.3 ft-kips (in short directtion) 8

16.4 RATIO OF FLEXURAL STIFFNESSES OF LONGITUDINAL BEAM TO SLAB When beams are used along the column lines in a two-​way floor system, an important parameter affecting the design is the relative size of the beam to the thickness of the slab. This parameter can best be measured by the ratio αf of the flexural rigidity (called flexural stiffness by the ACI Code) Ecb Ib of the beam to the flexural rigidity Ecs Is of the slab in the transverse cross section of the equivalent frame shown in Fig. 16.4.1. The separate moduli of elasticity Ecb and Ecs, referring to the beam and slab, provide for different concrete strength (and thus different Ec values) for the beam and slab. The moments of inertia Ib and Is refer to the gross sections of the beam and slab, respectively, within the cross section of Fig. 16.4.1(c). ACI-​ 8.4.1.8 permits the slab on each side of the beam web to act as a part of the beam. This slab portion is limited to a distance equal to the projection of the beam above or below the slab, whichever is greater, but not greater than 4 times the slab thickness, as shown in Fig. 16.4.2. More accurately, the small portion of the slab already counted in the beam should not be used in Is, but ACI permits the use of the total width of the equivalent frame in computing Is. Thus,

αf =



Ecb I b Ecs I s

(16.4.1)

The moment of inertia of a flanged beam section about its own centroidal axis (Fig. 16.4.2) may be shown to be Ib = k



bw h3 12

(16.4.2a)

in which

k=

2 3  t  b  t  t b  t 1 +  E − 1    4 − 6   + 4   +  E − 1     h   bw   h    h  bw   h  

b  t 1 +  E − 1    bw   h 

(16.4.2b)

where h = overall beam depth t = overall slab thickness bE = effective width flange computed as shown in Fig. 16.4.2. (Note that th his parameter is used here to compute the beam-to-slab stiffnness ratio α f and is different from the effective flange width used for computing the flexural strength of T-sections in Chapter 4.) bw = width of web

634

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Equivalent frame

L2

634

L1

L1 (a) Plan L2

t

A

h

A

L1

(c) Cross section A–A

L1 (b) Elevation

Figure 16.4.1  Plan, elevation, and beam and slab cross ​section of equivalent frame in a two-​way floor system. bE

bE t

(h–t) ≤ 4t

t (h–t) ≤ 4t

h

ACI–8.4.1.8

(h–t) ≤ 4t

ACI–8.4.1.8

bw

h

ACI–8.4.1.8

bw

Figure 16.4.2  Cross s​ ections for moment of inertia of a flanged beam section.

Equation (16.4.2b) expresses the nondimensional constant k in terms of (bE /​bw) and (t /​h). Typical values of k are tabulated in Table 16.4.1 and three curves are plotted in Fig. 16.4.3. The values of k are about 1.4, 1.6, and 1.8, respectively, for bE /​bw values of 2, 3, and 4, when t /​h values are between 0.2 and 0.5. Thus b  k ≈ 1.0 + 0.2  E   bw 



for 2
1.0. Using the prescribed values in Table 16.12.1, the proportion of moment going to the column strip is determined to be 81% by linear interpolation. L2/​L1

0.5

0.8

1.0

αf 1L2/​L1 = 6.61

90%

81%

75%

Exterior frame B, L2 /​L1 = 0.80 and αf1L2 /​L1 = 11.1. The proportion of moment is again 81% for the column strip, the same as for Frame A. Interior frame C, L2 /​L1 = 1.25 and αf1L2 /​L1 = 4.38. Using the prescribed values in Table 16.12.1, the proportion of moment going to the column strip is determined to be 67.5% by linear interpolation: L2/​L1

1.0

1.25

2.0

αf 1L2/​L1 = 4.38

75%

67.5%

45%

Exterior frame D, L2 /​L1  =  1.25 and αf1L2 /​L1 = 7.45. The proportion of moment is again 67.5% for the column strip, the same as for Frame C. (c) Positive moments in exterior and interior spans. Since the percentages for αf 1L2/​L1 ≥ 1.0 are the same for positive moment and for negative moment at interior support, the percentages of column strip moment for positive moments in exterior and interior spans are identical to those for interior negative moments (see Table 16.12.1) as determined in part (b) of this example. (d) The computed moments in the beams, the column strip slabs, and the middle strip slabs for frames A, B, C and D are summarized in Table 16.12.4.

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TABLE 16.12.4  TRANSVERSE DISTRIBUTION OF LONGITUDINAL MOMENTS (FT-​KIPS) IN TWO-​WAY SLAB (WITH BEAMS) DESIGN EXAMPLE (ALSO SEE FIG. 16.10.1) Interior frame A Total Width = 20 ft, Column Strip Width = 10 ft, Middle Strip Width = 10 ft Exterior Span

Total moment Moment in beam Moment in column strip slab Moment in middle strip slab

Interior Span

Exterior Negative

Positive

Interior Negative

Negative Positive

–​72 –​57 –​10 –​5

+257 +177 +31 +49

–​315 –​217 –​38 –​60

–​291 –​200 –​36 –​55

+158 +109 +19 +30

Exterior frame B Total Width = 10 ft, Column Strip Width = 5 ft, Half Middle Strip Width = 5 ft Exterior Span

Total moment Moment in beam Moment in column strip slab Moment in middle strip slab

Interior Span

Exterior Negative

Positive

Interior Negative

Negative Positive

–​36 –​28 –​5 –​3

+128 +88 +16 +24

–​158 –​109 –​19 –​30

–​146 –​101 –​17 –​28

+79 +54 +9 +15

Interior frame C Total Width = 25 ft, Column Strip Width = 10 ft, Middle Strip Width = 15 ft Exterior Span

Total moment Moment in beam Moment in column strip slab Moment in middle strip slab

Interior Span

Exterior Negative

Positive

Interior Negative

Negative Positive

–​56 –​39 –​7 –​10

+200 +115 +20 +65

–​245 –​140 –​25 –​80

–​228 –​131 –​23 –​74

+123 +71 +12 +40

Exterior frame D Total Width = 12.5 ft, Column Strip Width = 5 ft, Half Middle Strip Width = 7.5 ft Exterior Span

Total moment Moment in beam Moment in column strip slab Moment in middle strip slab

Interior Span

Exterior Negative

Positive

Interior Negative

Negative Positive

–​28 –​20 –​3 –​5

+100 +57 +10 +33

–​123 –​71 –​12 –​40

–​114 –​65 –​12 –​37

+61 +35 +6 +20

61



16.12  TRANSVERSE DISTRIBUTION OF LONGITUDINAL MOMENT

661

EXAMPLE 16.12.2 Divide the five critical moments in each of the equivalent frames A, B, C, and D in the flat slab design ­example (see Example 16.10.2), as shown in Fig. 16.10.2, into two parts: one for the half column strip (for frames B and D) or the full column strip (for frames A and C), and the other for the half middle strip (for frames B and D) or the two half middle strips on each side of the column line (for frames A and C). SOLUTION The percentages of the longitudinal moments carried by the column strip width are shown in lines 10 to 12 of Table 16.12.5. Note that the column strip width shown in line 2 is one-​half of the shorter panel dimension for frames A and C, and one-​fourth of this value for frames B and D. Note also that the sum of the values on lines 2 and 3 should be equal to that on line 1, for each respective frame. The moment of inertia of the slab equal in width to the transverse span of the edge beam is I s in βt for frames A and B =



240(7.5)3 = 8440 in.4 12

and I s in βt for frames C and D =



300(7.5)3 = 10, 600 in.4 12

These values are shown in line 5 of Table 16.12.5. The percentages shown in lines 10 to 12 are obtained from Table 16.12.1, by interpolation (as illustrated in Tables 16.12.2 and 16.12.3) if necessary. Having these percentages, the separation of each of the longitudinal moment values shown in Fig. 16.10.2 into two parts is a simple matter and thus is not shown further. TABLE 16.12.5  TRANSVERSE DISTRIBUTION OF LONGITUDINAL MOMENT FOR FLAT SLAB DESIGN EXAMPLE Line Equivalent Frame Number 1 2 3 4 5 6 7 8 9 10 11 12

Total transverse width (in.) Column strip width (in.) (see Fig. 16.12.1) Half middle strip width (in.) C (in.4) from Example 16.11.2 Is (in.4) in βt βt = EcbC/​(2Ecs Is) αf 1 from Example 16.6.2 L2/​L1 αf 1L2/​L1 Exterior negative moment, percent to column strip (Table 16.12.1) Positive moment, percent to column strip (Table 16.12.1) Interior negative moment, percent to column strip (Table 16.12.1)

A

B

C

D

240 120

120 60

300 120

150 60

2 @ 60 18,500 8440 1.10 0 0.80 0

60 18,500 8440 1.10 5.42 0.80 4.34

2 @ 90 18,500 10,600 0.87 0 1.25 0

90 18,500 10,600 0.87 4.34 1.25 5.43

89.0%

91.6%

91.3%

88.7%

60.0%

81.0%

60.0%

67.5%

75.0%

81.0%

75.0%

67.5%

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EXAMPLE 16.12.3 Divide the five critical moments in each of the equivalent frames A, B, C, and D in the flat plate of Example 16.10.3, as shown in Fig. 16.10.3, into two parts: one for the half column strip (for frames B and D) or the full column strip (for frames A and C), and the other for the half middle strip (for frames B and D) or the two half middle strips on each side of the column line (for frames A and C). SOLUTION The percentages of the longitudinal moments carried by the column strip width are shown in lines 10 to 12 of Table  16.12.6. Explanations are identical to those for the preceding example. TABLE 16.12.6  TRANSVERSE DISTRIBUTION OF LONGITUDINAL MOMENT FOR FLAT PLATE DESIGN EXAMPLE Line Equivalent Frame Number 1 2 3 4 5 6 7 8 9 10 11 12

Total transverse width (in.) Column strip width (in.) Half middle strip width (in.) C (in.4) from Example 16.11.3 Is (in.4) in βt βt = EcbC/​(2Ecs Is) αf1 L2 /​L1 αf1L2/​L1 Exterior negative moment, percent to column strip Positive moment, percent to column strip Interior negative moment, percent to column strip

A

B

C

D

144 72 2@36 474 2000 0.119 0 0.80 0

72 36 36 474 2000 0.119 0 0.80 0

180 72 2@54 362 2500 0.072 0 1.25 0

90 36 54 362 2500 0.072 0 1.25 0

98.8%

98.8%

99.3%

99.3%

60%

60%

60%

60%

75%

75%

75%

75%

16.13 DESIGN OF SLAB THICKNESS AND REINFORCEMENT Slab Thickness Ordinarily the minimum thickness specified in ACI-​8.3.1 to control deflections governs the thickness for design. Of course, reinforcement for bending moment must be provided, but the reinforcement ratio ρ required is usually well below 0.5ρtc; thus, it does not dictate slab thickness. Slab thickness based on shear strength requirements must also be investigated. The shear requirement for two-​way slabs (with beams) may be investigated by observing strips 1-​1 and 2-​2 in Fig. 16.13.1. Beams with αf1L2/​L1 values larger than 1.0 are assumed to carry the loads acting on the tributary floor areas bounded by 45° lines drawn from the

63



16.13  DESIGN OF SLAB THICKNESS AND REINFORCEMENT

663

1

45° E 2

2

S H

G F 1 L

Figure 16.13.1  Load transfer from floor area to beams.

corners of the panel and the centerline of the panel parallel to the long side (ACI-​8.10.8.1). If this is the case, the loads on the trapezoidal areas E and F of Fig. 16.13.1 are carried by the long beams, and those on the triangular areas G and H are carried by the short beams. The shear per unit width of slab along the beam is highest at the ends of slab strips 1-​1 and 2-​2, which, considering the increased shear at the exterior face of the first interior support, is approximately equal to  w S Vu = 1.15  u   2 



(16.13.1)

where S is the span length in the short direction. If αf1L2 /​L1 is equal to zero, there is, of course, no load on the beams (because there are no beams). When the value of αf1L2 /​L1 is between 0 and 1.0, the percentage of the floor load carried by the beams should be obtained by linear interpolation. In such a case, the shear expressed by Eq. (16.13.1) would be reduced, but the shear around the column due to the portion of the floor load transferred directly to the columns by two-​way action should be investigated as for flat plate floors. The shear strength requirement for flat slab and flat plate systems (without beams) is treated separately in Sections 16.15, 16.16, and 16.18.

Reinforcement When the nominal requirements for slab thickness as discussed in Section 16.6 are satisfied, no compression reinforcement will likely be required. The tension steel area required within the strip being considered can then be obtained by the following steps: 1. required M n = 2. m =

fy 0.85 fc′

factored moment Mu in the strip (φ = 0.90 — assumed, but reasonable for slabs)

, Rn =

Mn , bd 2

ρ=

2 mRn  1 1 − 1 − , m fy 

As = ρ bd



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Instead of using the equation for ρ in Step 2, the curves in Fig.  3.8.1 may be used. Note also that the values of b and d to be used in Step 2 for negative moment in a column strip with drop are the drop width for b and, for d, the smaller of the actual effective depth through the drop and that provided by a drop thickness below the slab at no more than one-​fourth the distance between the edges of the column capital and the drop. For positive moment computation, the full column strip width should be used for b, and the effective depth in the slab for d. After obtaining the steel area As required within the strip, a number of bars may be chosen so that they provide either the area required for strength, but not less than the minimum amount, As,min, equal to 0.0018bt for Grade 60 steel (see ACI-​8.6.1.1). The spacing of reinforcing bars must not exceed the smaller of 2 times the slab thickness or 18 in. at critical sections, and the smaller of 3 times the slab thickness or 18 in. at other sections (ACI-​8.7.2.2). In slabs of cellular or ribbed construction (i.e., two-​way joist systems), where the requirement for shrinkage and temperature reinforcement governs, maximum spacing must not exceed the smaller of 5 times the slab thickness or 18 in. (ACI-​24.4.3.3).

Reinforcement Details in Slabs without Beams ACI-​8.7, in particular ACI Fig.  8.7.4.1.3a, provides detailed dimensions for minimum extensions required for each portion of the total number of bars in the column and middle strips. Since the forces acting in the bars were empirically determined, there is no practical means to evaluate the distances required to develop the reinforcement. Thus, past practice and engineering judgment were used in preparing ACI Fig. 8.7.4.3.1a. In that figure, no details for bent bars are shown because they are seldom used, although their use is still permitted by ACI-​8.7.4.1.3c when the depth-​to-span ratio allows bends at 45° or less. For unbraced frames, reinforcement lengths must be determined by analysis but may not be less than those prescribed in ACI Fig. 8.7.4.1.3a. Also, ACI-​8.7.4.2 requires the use of “integrity steel,” which consists of a minimum of two of the column strip bottom bars passing continuously (or spliced with Class B splices or anchored within support) through the column core in each direction. The purpose of this integrity steel is to provide some residual strength following a single punching shear failure. The ACI Code also allows the use of full mechanical and full welded splices as alternative methods of splicing the reinforcement.

Corner Reinforcement for Two-​Way Slab (with Beams) It is well known from plate bending theory that a transversely loaded slab simply supported along four edges will tend to develop corner reactions as shown in Fig. 16.13.2, for which reinforcement must be provided. Thus, in slabs supported by edge walls or on one or more edge beams having a value of αf greater than 1.0, reinforcement shall be provided at exterior corners in both the bottom and top of the slab (see Fig. 16.13.3). This reinforcement in both the top and bottom of the slab must be sufficient to resist a moment equal to the maximum positive moment per foot of width in the slab (ACI-​8.7.3.1). The reinforcement is to be provided for a distance in each direction from the corner equal to one-​fifth the longer span (ACI-​8.7.3.1.2), and it may be placed in a single layer parallel to the diagonal from the corner in the top of the slab and perpendicular to that diagonal in the bottom of the slab, or in two layers parallel to the sides of the slab (ACI-​8.7.3.1.3), as shown in Fig. 16.13.3. The latter is the preferred option in practice, as providing reinforcement along the diagonal can be confusing in the field and would require bars of varying lengths. Reinforcement provided for flexure in the primary directions can be used to satisfy this requirement (ACI-R8.7.3.1). Since the maximum positive moment in the slab will exceed the negative moment at the exterior slab support, t​he corner reinforcement amount computed per ACI-​8.7.3.1 will govern whenever it is required. It is also noted that the moment per foot of width of slab, whether computed using the Direct Design Method or the Equivalent Frame Method (Section 16.20), must be distributed within the column strip width, which for a corner panel is one-​quarter of the shorter span. The corner reinforcement, however, is to be distributed over one-​fifth of the longer span. It is possible that the computed column strip width is larger than the distance where the corner reinforcement is to be distributed (e.g., for square panels). In such cases, it is recommended that the remaining portion of the column strip be

65



16.13  DESIGN OF SLAB THICKNESS AND REINFORCEMENT

665

L

S

R

Edge shear distribution

R

Edge shear distribution

R

Figure 16.13.2  Edge reactions for a simply supported two-​way slab. Llong

(Llong)/5

(Llong)/5

As top per 8.7.3

Lshort

As bottom per 8.7.3

OPTION 1 Llong

As in two layers per 8.7.3 top and bottom

Lshort

(Llong)/5

(Llong)/5

OPTION 2

Figure 16.13.3  Corner reinforcement in two-​way slabs supported by edge walls or where one or more edge beams have a value αf greater than 1.0. (Adapted from ACI 318-​14 Commentary.)

reinforced with the same amount computed for the corner reinforcement. It is also recommended that 50% of the top corner reinforcement be extended to one-​fifth of the longer span or 0.3Ln, whichever is larger, to be consistent with the required minimum bar extensions at the exterior supports (ACI Table 8.7.4.1.3a). The remainder must be extended at least to one-​fifth of the longer span in accordance with ACI-​8.7.3.1.2.

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Crack Control In addition to deflection control, crack control is the other major serviceability requirement usually considered in the design of flexural members. ACI-​24.3 gives criteria for distributing the flexural tension reinforcement in beams and one-​way slabs to minimize crack width under service loads. No ACI Code provisions are given for two-​way floor (or roof) systems; however, ACI Committee 224, Cracking, has suggested a formula to predict the possible crack width in two-​way acting slabs, flat slabs, and flat plates. The recommendations are based on the work of Nawy et al. [16.72–​16.75]. When the calculated crack width is considered excessive (there are no ACI Code limits for slabs), the distribution (size and spacing) of flexural reinforcement may be adjusted [16.75] to reduce the width of the expected cracks. Ordinarily, crack width is not a problem on two-​way acting slabs designed with fy less than or equal to 60,000 psi.

EXAMPLE 16.13.1 Investigate if the preliminary slab thickness of 6 1 2 in. in the two-​way slab (with beams) (see Fig. 16.3.4) is sufficient for resisting flexure and shear. SOLUTION For each of the equivalent frames A, B, C, and D, the largest bending moment in the slab occurs at the exterior face of the first interior support in the middle strip slab. From Table 16.12.4, this moment is observed to be 60/​10, 30/​5, 80/​15, or 40/​7.5 ft-​kips per ft of width in frames A, B, C, and D, respectively. Taking the effective depth to the contact level between the reinforcing bars in the two directions, and assuming #5 bars with 3 -​in. clear cover, 4 average d = 6.50 − 0.75 − 0.63 = 5.12 in. Assuming φ = 0.90, the largest Rn required is Rn =



Mu 6000(12) = = 254 psi φ bd 2 0.90(12)(5.12)2

From Eq. (3.8.5), the reinforcement ratio ρ for this value of Rn is about 0.0045, which is well below 0.35ρb = 0.0075, recommended for beams and one-​way slabs not supporting elements likely to be damaged by large deflections (Section 12.10). For two-​way slabs, deflections should be smaller. Hence excessive deflection should not be expected; this is further verification of the minimum thickness formulas given in ACI-​8.3.1. The factored floor load wu is

wn = 1.2 wD + 1.6 wL = 319 psf



Since all af 1L2 /​L1 values are well over 1.0, take V from Eq. (16.13.1) as Vu =



1.15wu S 1.15(0.319)(20) = = 3.67 kips 2 2

Vc = 2 λ fc′bw d = 2(1.0) 3000 (12)(5.12)

1 1000

= 6.73 kips

φVc = 0.75(6.73) = 5.05 kips > [Vu = 3.67 kips]

OK

Note that the factored shear 3.67 kips is maximum at strip 1-​1 of Fig. 16.13.1; actually, the average for all such strips will be lower.

67



667

16.13  DESIGN OF SLAB THICKNESS AND REINFORCEMENT

EXAMPLE 16.13.2 Design the reinforcement in the exterior and interior spans of a typical column strip and a typical middle strip in the short direction of the flat slab (see Fig. 16.3.6). As described earlier, in Section 16.3, fc′ = 3000 psi and fy = 60,000 psi. SOLUTION (a) Moments in column and middle strips. The typical column strip is the column strip of equivalent frame C of Fig. 16.10.2; but the typical middle strip is the sum of two half middle strips, taken from each of the two adjacent equivalent frames C. The factored moments in the typical column and middle strips are shown in Table 16.13.1. (b) Design of reinforcement. The design of reinforcement for the typical column strip is shown in Table  16.13.2; for the typical middle strip, it is shown in Table 16.13.3. Because the moments in the long direction are larger than those in the short direction, the larger effective depth is assigned to the long direction wherever two layers of steel (one in each direction) are required. This occurs in the top steel at the intersection of column strips and in the bottom steel at the intersection of middle strips. Assuming #5 bars and 3 4 -​in. clear cover, the effective depths provided at various critical sections of the long and short directions are shown in Fig. 16.13.4. (c) Slab thickness for flexure. For fc′ = 3000 psi and fy = 60,000 psi, the reinforcement ratio corresponding to ρ(εt = 0.005) = 0.0135 (Table 3.6.1). The actual percentages used (line 6 of Tables 16.13.2 and 16.13.3) are nowhere near this maximum. This result was expected since the slab thickness was governed by the minimum thickness required for deflection control.

TABLE 16.13.1  FACTORED MOMENTS IN A TYPICAL COLUMN STRIP AND MIDDLE STRIP, EXAMPLE 16.13.2 (FLAT SLAB) Exterior Span Line Number 1

2 3 4

Moments at Critical Section (ft-​kips) Total M in column and middle strips (Fig. 16.10.2, frame C) Percentage to column strip (Table 16.12.5) Moment in column strip Moment in middle strip

Negative Positive Moment Moment

–​88

+147

Interior Span

Negative Moment

Negative Positive Moment Moment

–​205

–​190

+103

Negative Moment

–​190

91.3% –​80

60% +88

75% –​154

75% –​143

60% +62

75% –​143

–​8

+59

–​51

–​47

+41

–​47

(Continued)

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Example 16.13.2 (Continued)

TABLE 16.13.2  DESIGN OF REINFORCEMENT IN COLUMN STRIP OF FRAME C, EXAMPLE 16.13.2 (FLAT SLAB) (fy = 60,000 psi,  f c′ = 3000 psi) Exterior Span Line No. 1 2 3 4 5 6 7 8 9 10 11 a

Item

Interior Span

Negative Positive Negative Moment Moment Moment

Moment, Table 16.13.1, line 3 (ft-​kips) Width b of drop or strip (in.) Effective depth d (in.) Mu /​φ (ft-​kips) Rn (psi) = Mu /​(φbd2) ρ, Eq. (3.8.5) As = ρbd a As,min = 0.0018bt  N = larger of (7) or (8)/​0.31 N = width of strip/​(smaller of 2t or 18 in.) N required, larger of (9) or (10)

Negative Positive Moment Moment

Negative Moment

–​80 100 8.81 –​89 138 0.24% 2.11 2.16 7 5

+88 120 6.44 +98 236 0.41% 3.17 1.62 11 8

–​154 100 8.81 –​171 264 0.47% 4.14 2.16 14 5

–​143 100 8.81 –​159 246 0.43% 3.78 2.16 13 5

+62 120 6.44 +69 166 0.29% 2.24 1.62 8 8

–​143 100 8.81 –​159 246 0.43% 3.78 2.16 13 5

7

11

14

13

8

13

bt = 100(10.5) + 20(7.5) = 1200 sq in. for negative moment region.

TABLE 16.13.3  DESIGN OF REINFORCEMENT IN MIDDLE STRIP OF FRAME C, EXAMPLE 16.13.2 (FLAT SLAB) (fy = 60,000 psi,  fc′ = 3000 psi) Exterior Span Line No. 1 2 3 4 5 6 7 8 9 10 11

Item Moment, Table 16.13.1, line 4 (ft-​kips) Width of strip, b (in.) Effective depth d (in.) Mu/​φ (ft-​kips) Rn (psi) = Mu/​(φbd2) ρ, Eq. (3.8.5) As = ρbd As,min = 0.0018bt N = larger of (7) or (8)/​0.31a N = width of strip/​(smaller of 2t or 18 in.) N required, larger of line 9 or 10

Interior Span

Negative Positive Negative Moment Moment Moment

Negative Positive Negative Moment Moment Moment

–​8 180 6.44 –​9 14 0.02% 0.23 2.43 8 12

+59 180 5.81 +66 130 0.22% 2.30 2.43 8 12

–​51 180 6.44 –​57 92 0.16% 1.85 2.43 8 12

–​47 180 6.44 –​52 84 0.14% 1.62 2.43 8 12

+41 180 5.81 +46 91 0.15% 1.57 2.43 8 12

–​47 180 6.44 –​52 84 0.14% 1.62 2.43 8 12

12

12

12

12

12

12

a

A mixture of #5 and #4 bars could have been selected.

(Continued)

69



669

16.14  SIZE REQUIREMENT FOR BEAM

Example 16.13.2 (Continued)

d = 6.44”

d = 9.44”

d = 6.44”

d = 6.44”

d = 6.44”

d = 9.44”

d = 6.44”

d = 9.44”

d = 6.44” d = 8.81” d = 8.81”

d = 9.44”

d = 6.44” d = 6.44” d = 5.81”

Short direction d = 8.81” d = 8.81” d = 6.44”

Long direction

Figure 16.13.4  Effective depths provided at critical sections in the flat slab of Example 16.13.2.

16.14 SIZE REQUIREMENT FOR BEAM (IF USED) IN FLEXURE AND SHEAR The size of the beams along the column centerlines in a two-​way slab (with beams) should be sufficient to provide the bending moment and shear strengths at the critical sections. For approximately equal spans, the largest bending moment should occur at the exterior face of the first interior column where the available section for strength computation is rectangular because the effective slab projection is on the tension side. Then, with the preliminary beam size, the required reinforcement ratio ρ may be determined. Although unlikely to be a problem with T-​sections, deflection must be investigated if excessive deflections are of concern. The maximum shear in the beam should also occur at the exterior face of the first interior column. The shear diagram for the exterior span may be obtained by placing the negative moments already computed for the beam at the face of the column at each end and loading the span with the percentage of floor load interpolated (ACI-​8.10.8.1) between αf 1L2/​L1 = 0 and αf 1L2 /​L1 ≥ 1.0. As discussed in Section 9.2, the stem (web) bw d should for practicality be sized such that nominal shear stress vn = Vu  /​(φ bwd) does not exceed about 6 fc′ at the critical section d from the face of support.

EXAMPLE 16.14.1 Investigate if the preliminary overall sizes of 14 × 28 in. for the long beam and 12 × 24 in. for the short beam are suitable for the two-​way slab (with beams) design example. SOLUTION Since the values of αf , or of αf 1L2 /​L1, are considerably larger than 1.0 for all beam spans, there is to be no reduction of the floor load going into the beams from the tributary areas (ACI-​8.10.8.1). As shown in Fig. 16.14.1, the most critical span is B1 for the long direction and B5 for the short direction. Actually, the load acting on the clear span of the beam should include the floor load (including the weight of the beam stem itself or any other load) directly over the beam stem width plus the floor load on the tributary areas bounded by the 45° lines from the corner of the panel. Also, for practical purposes, it is acceptable to consider the shear due to floor load at the face of column equal to one-​half of the floor load on the tributary areas between column centerlines, as shown in Fig. 16.14.1. (Continued)

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Example 16.14.1 (Continued) B1

B2

10’

5’

6.38k/ft 57 ft-fk

10’ 6.38k/ft 216 ft-fk

B6

B6

B8

Factored floor load = 319 psf

20’–0”

23’– 9” B1

25’– 0”

B2

Beam B1 6.38k/ft

B5

B5

B7

39 ft-k

140 ft-k 18’–9” 20’–0”

B3

B4

Beam B5

Figure 16.14.1  Tributary areas for the beams around the two-​way slab panel of Example 16.14.1.

(a) Size of long beam B1. The negative moments at the face of supports, 57 and 216 ft-​kips, are taken from Table 16.12.4, frame A. weight of beam stem = 14(21.5)

1 144

(150) = 314 lb/ft

1 (1.2)(0.314)(23.75)2 + 216 10 = 21 + 216 = 237 ft-kips

maximum negative moment =

b = 14 in,

d = 28 − 2.5 (assume one layer of steel) = 25.5 in. Rn =

Mu 237(12, 000 ) = = 347 psi 2 0.90(14)(25.5)2 φ bw d

From Eq. (3.8.5), ρ = 0.006, which is well below ρ (εt = 0.005) = 0.0135. Perhaps the beam size should be reduced. From Fig. 16.14.1, total factored floor load on B1 = 6.38(15) = 95.7 kips max Vu = 1.115(1.2)(0.314)

216 − 57 23.75 1 + (95.7) + 2 2 23.75

= 5.1 + 47.9 + 6.7 = 59.7 kips



vn =



Vu 59, 700 = = 223 psi = 4.1 fc′ < 6 fc′ φ bw d 0.75(14)(25.5)

OK

(b) Size of short beam B5. The negative moments at the face of supports, ​39 and ​140 ft-​kips, are taken from Table 16.12.4, frame C. weight of beam stem = 12(17.5)

1 144

(150) = 219 Ib/ft

1 (1.2)(0.219)(18.75)2 + 140 10 = 9 + 140 = 149 ft-kips

maximum negative moment =

b = 12 in. Rn =

d = 24 − 2.5(assume one layer of steel) = 211.5 in.



Mu 149(12, 000) = = 358 psi 2 0.90(12)(21.5)2 φ bw d (Continued)

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Example 16.14.1 (Continued) From Eq. (3.8.5), ρ =  0.007, which is well below ρ(εt = 0.005) = 0.0135. From Fig. 16.14.1,

total factored floor load on B5 = 6.38(10) = 63.8 kips max Vu = 1.15(1.2)(0.219)





18.75 1 140 − 39 + (63.8) + 2 2 18.75

= 2.8 + 31.9 + 5.4 = 40.1 kips V 40,100 vn = u = = 207 psi = 3.8 fc′ < 6 fc′ φ bw d 0.75(12)(21.5)

OK

As mentioned above, the size of beams in both the long and short directions should probably be reduced; the nominal stress vn is well below 6 fc′ at the face of support and is even lower at d therefrom.

16.15 SHEAR STRENGTH IN TWO-​W AY FLOOR SYSTEMS The shear strength of a flat slab or flat plate floor around a typical interior column under dead and full live loads is analogous to that of a square or rectangular spread footing subjected to a concentrated column load, except each is an inverted situation of the other. The area enclosed between the parallel pairs of centerlines of the adjacent panels of the floor is like the area of the footing, because there is no shear force along the panel centerline of a typical interior panel in a floor system. Consequently the discussion here is essentially identical to what is included in Chapter 19 on footings. The shear strength of two-​way slab systems without shear reinforcement has been studied by many investigators [16.76–​16.92, 16.142]. An excellent summary is provided by ASCE-​ACI Task Committee 426 [16.83].

One-​Way Action The shear strength of a flat slab or flat plate should be computed for one-​way action (ACI-​ 8.5.3.1.1) and for two-​way action (ACI-​8.5.3.1.2). To compute the one-​way shear strength (one-​way action), the slab is treated as a wide beam, where the critical section is parallel to the panel centerline in the transverse direction and extends across the full distance between two adjacent longitudinal panel centerlines. As in beams, this critical section of width bw times the effective depth d is located at a distance d from the face of the equivalent square column capital or from the face of the drop panel, if any. The nominal strength in usual cases where no shear reinforcement is used may be computed as

Vn = Vc = 2 λ fc′bw d (16.15.1)1

according to the simplified method of ACI-​22.5.5.1.

1  For SI, with fc′ in MPa and bw and d in mm, ACI-​318-​14M gives



Vc = 0.17λ fc′bw d

(16.15.1) 

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Figure 16.15.1  Punching shear failure along a truncated pyramid around the column. (Photo courtesy of James O. Jirsa.)

Two-​Way Action A second failure mode may occur by diagonal cracking along a truncated cone or pyramid around columns, concentrated loads, or reactions (see Fig. 16.15.1). This failure mode is commonly called “punching” shear. The critical section is assumed to be located so that its periphery b0 is at a distance d/​2 (i.e., half the effective depth) from the edges or corners of a column, concentrated load, or reaction areas and from changes in slab thickness, such as edges of capitals or drop panels. The ACI Code states that b0 is a “minimum but need not approach closer than d/​2  …” (ACI-​22.6.4.1). Some confusion may arise as to whether the “minimum” at d/​2 would require using a curved-​corner perimeter around a square or rectangular column. Since exactness is neither improved nor reduced by calculating the critical section b0 by such elaborate procedures, ACI-​22.6.4.1.1 permits the critical section for square or rectangular loaded areas to have four “straight sides.” For slabs with changes in thickness, such as slabs with capitals or drop panels, shear must be checked at several sections to determine the critical section. When shear reinforcement is not used, the nominal shear strength Vn = Vc, which is given by ACI-​22.6.5.2 as the smallest of



Vc = 4 λ fc′b0 d

(16.15.2a)2

 4 Vc =  2 +  λ fc′b0 d  β

(16.15.2b)

 α d Vc =  2 + s  λ fc′b0 d b0  

(16.15.2c)2

2

where b0 = perimeter of critical section β = ratio of long side to short side of the column αs = 40 for interior columns, 30 for edge columns, and 20 for corner columns d = average of effective depths in the two orthogonal directions

2  For SI, with fc′ in MPa and b0 and d in mm, ACI-​318-​14M gives



Vc = 0.33λ fc′b0 d

(16.15.2a) 

 2 Vc = 0.17  1 +  λ fc′b0 d  β

(16.15.2b) 

 α d Vc = 0.083  2 + s  λ fc′b0 d b0  

(16.15.2c) 

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Equation (16.15.2b) recognizes that there should be a transition between, say, a square column (β = 1) where Vc might be based on 4 λ fc′ for two-​way action, and a wall (β = ∞), where Vc should be based on the 2 λ fc′ used for one-​way action, as for beams. Note that unless β is larger than 2.0, Eq. (16.15.2b) does not control. For shapes other than rectangular (e.g., L-shaped), β is taken as the ratio between the longest and the largest perpendicular dimensions of the effective loaded area (ACI-​R22.6.5.2). The two-​way shear strength may be reduced below 4 λ fc′, even for a square concentrated load area, both (1) as the distance to the critical section from the concentrated load increases, such as for drop panels, and (2) as the perimeter becomes large in comparison to the slab thickness [e.g., a 6-​in. slab supported by a 30-​in square column (b0 /​d ≈ 28)]. Equation (16.15.2c) accounts for this reduced strength. In the application of Eqs. (16.15.2), b0 is the perimeter of the critical section at a distance d/​2 from the edge of column capital or drop panel. For Eq. (16.15.2c), αs for an “interior column” applies when the perimeter is four-​sided, for an “edge column” when the perimeter is three-​sided, and for a “corner column” when the perimeter is two-​sided. As shown by Fig.  16.15.2, Eq. (16.15.2c) will give a Vc smaller than 4 λ fc′b0d for large columns (or very thin slabs), such as a square interior column having side larger than 4d, a square edge column having side larger than 4.33d, and a square corner column having side larger than 4.5d. Thus, the nominal shear strength Vc in a two-​way system is generally set by Eq. (16.15.2a):  that is, Vc = 4 λ fc′b0 d , unless either of Eq. (16.15.2b) or Eq. (16.15.2c) gives a lesser value. Unlike the design of beams, a minimum amount of shear reinforcement is not required for two-​way slabs because there is the possibility of load sharing between the weak and strong areas. For deep, lightly reinforced one-​way slabs, however, shear failure may occur at loads less than Vc, especially if made of high-​strength concrete (ACI-​R7.6.3.1). Thus it would be prudent to provide a minimum amount of shear reinforcement even if not required by the code in these cases. The investigation for concentric shear (without moment) transfer from slab to column is shown in Examples 16.15.1 and 16.15.2, for the flat slab and flat plate design examples, respectively. When both shear and moment must be transferred from the slab to the column, ACI-​8.4.4.2 applies, as will be discussed in Section 16.18.

d 2

d 2

s

d 2

s b0 = 4(s + d) 40 If s = 4d, =2 b0 /d

b0 = 3s + 2d 30 =2 If s = 4.33d, b0 /d

(a) Interior column

(b) Edge column

b0 = 2s + d 20 =2 If s = 4.5d, b0 /d (c) Corner column

Figure 16.15.2  Minimum size of square columns for Vc = 4 λ fc′b0 d .

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EXAMPLE 16.15.1 Compute the shear strength for one-​way and two-​way actions in the flat slab design of Example 16.3.2 for an interior column with no bending moment to be transferred. Note that slab thickness is 7.5 in. and that normal-​weight concrete with fc′ = 3000 psi is used (i.e., λ = 1.0) SOLUTION (a) One-​way action. Investigation for one-​way action is made for sections 1-​1 and 2-​2 in the long direction, as shown in Fig. 16.15.3(a). The short direction has a wider critical section and a shorter span; thus it does not control. For section 1-​1, if the entire width of 20 ft is conservatively assumed to have an average effective depth of (6.44/2 + 5.81/2) = 6.12 in. (see Fig. 16.3.4), Vu = 0.337(20)(9.52) = 64 kips Vn = Vc = 2 λ fc′(240)(6.12)



1 1000

(section 1-1) = 161 kips



φVn = 0.75(161) = 121 kips > Vu

OK

If, however, bw is taken as 84 in. and d as 9.12 in. on the contention that the increased depth d is only over a width of 84 in.,

Vn = Vc = 2 λ fc′(84)(9.12)

1 1000

= 84 kips

1

2 avg d = 6.12”

7.82’

Drop 7’–0”

20’– 0”

Equivalent square column

9.52’

20’– 0”

avg d = 9.12”

4.43’ 8’– 4” 1

2

25’–0”

25’–0”

avg d = 4.56” 2

7’–0”

7.51’

avg d = 3.06” 2

25’– 0”

Section 2–2 Drop 20’–0”

avg d 5’– 0” = 4.56” 2 5.76’

Capital

20’– 0”

Section 1–1

avg d = 3.06” 2

(a) One-way action (long direction)

8’– 4” 8.84”

avg d = 3.06” 2

25’– 0” (b) Two-way action

Figure 16.15.3  Critical sections for shear in the flat slab of Example 16.15.1.

(Continued)

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Example 16.15.1 (Continued) This latter value is probably unrealistically low. For section 2-​2, the shear resisting section has a constant d of 6.12 in.; thus

Vu = 0.337(20)(7.82) = 53 kips

(section 2-2) φVn = 121 kips > Vu at sections 1-1 and 2-2

OK

In general, it will be rare that one-​way action will govern. (b) Two-​way action. The critical sections for two-​way action are the circular section 1-​1 at d/​2 = 4.56 in. from the edge of the column capital and the rectangular section 2-​2 at d/​2 = 3.06 in. from the edge of the drop, as shown in Fig. 16.15.3(b). Since there are no shearing forces at the centerlines of the four adjacent panels, the shear forces around the critical sections 1-​1 and 2-​2 in Fig. 16.15.3(b) are



  π (5.76)2  π (5.76)2  Vu = 0.337 500 −   + 1.2(0.038) 7(8.33) − 4 4     = 159.2 + 1.5 = 161 kips

(section 1-1)

In the second term, the 0.038 is the weight of the 3-​in. drop in ksf

Vu = 0.337[500 − 8.84(7.51)] = 146 kips

     (section 2-2)

Compute the shear strength at section 1-​ 1 around the perimeter of the capital [Fig. 16.15.3(b)],

b0 = π (5.76)12 = 217 in.,

b0 217 = = 23.8 d 9.12

Since b0 /​d > 20, and β = 1, Eq. (16.15.2c) controls. Thus, 40(9.12)   φVn = φVc = φ  2 +  λ fc′b0 d = φ (3.68λ fc′b0 d )  217 

= 0.75 3.68λ fc′(217)(9.12)

1 1000

= 299 kips (section 1-1)

At section 2-​2, Fig. 16.15.3(b),

b0 = [(2(8.84) + 2(7.51)]12 = 392 in.:

b0 392 = = 64.1 d 6.12

and since b0/​d > 20, Eq. (16.15.2c) controls. Thus, 40(6.12)   φVn = φVc = φ  2 +  λ fc′b0 d = φ (2.62 λ fc′b0 d )  392 

= 0.75  2.62 λ fc′(392)(6.12)

1 1000

= 258 kips

(section 2-2)

Though both sections 1-​1 and 2-​2 have φVn significantly greater than Vu, the section around the drop panel is loaded to a slightly higher percentage of its strength (57% for section 2-​2 vs. 54% for section 1-​1). In any case, shear reinforcement is not required at this interior location.

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EXAMPLE 16.15.2 Compute the one-​way and two-​way shear strengths in the flat plate of Example 16.3.3 for an interior column with no bending moment to be transferred. Note that slab thickness is 5.5 in. and that normal-​weight concrete of fc′ = 4000 psi is used (i.e., λ = 1.0). SOLUTION (a) One-​way action. Assuming 3 4 -​in. clear cover and #4 bars, the average effective depth when bars in two directions are in contact is avg d = 5.50 − 0.75 − 0.50 = 4.25 in. Referring to Fig. 16.15.4(a), Vu = 0.198(12)6.65 = 15.8 kips 12 φVn = φVc = 0.75 2(1.0) 4000  (12) 4.25 1000 = 58.1 kips > Vu



OK

1 Avg d = 4.25”

6.65’ 2.125”

10” 12’–0” 12”

Section 2–2

10”

14.25” 12” 2.125”

2.125” 2.125”

12’– 0”

16.25” 1 15’–0”

15’–0”

(a) One-way action (long direction)

(b) Two-way action

Figure 16.15.4  Critical sections for shear in the flat plate of Example 16.15.2.

(b) Two-​way action. Referring to Fig. 16.15.4(b),

Vu = 0.198 [15(12) − 1.35(1.19)] = 35.3 kips

The perimeter of the critical section at d/​2 around the column is

b0 = 2(16.25) + 2(14.25) = 61.0 in.,

b0 61.0 = = 14.4 < 20 d 4.25

Since b0 /​d < 20, and β = 1.2, Eq. (16.15.2a) controls. Thus,

1 φVn = φVc = 0.75  4(1.0) 4000  (61.0)4.25 1000 = 49.2 kips > V u

OK

Shear reinforcement is not required at this interior location.

16.16 SHEAR REINFORCEMENT IN FLAT PLATE FLOORS In flat plate floors, where neither column capitals nor drop panels are used, shear reinforcement is frequently necessary. In such cases, two-​way action usually controls. The shear reinforcement may take the form of single-​or multiple-​leg stirrups placed in vertical sections around the column [Fig. 16.16.1(a)]. Alternatively, they may consist of shearheads,

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which are steel I-​or channel-​shaped sections welded to form four (or three for an exterior column) identical arms at right angles and uninterrupted within the column section [Fig. 16.16.1(b)]. Stirrups can be difficult to install in the slab around the column because the region is commonly congested with the column and slab reinforcement. Shearheads can be used instead, but sometimes they are more expensive to fabricate and install. Alternatively, headed shear studs have been used in Canada and Europe as shear reinforcement in slabs for many years [16.101, 16.144], and today are also widely used in the United States. This type of reinforcement consists of headed steel studs welded to a steel strip as shown in Fig. 16.16.2. The strips may be arranged in orthogonal directions for rectangular and square columns or in radial directions for both rectangular (Fig. 16.16.2) and circular columns.

Critical section through slab shear reinforcement at d/2 from face of column

Critical section outside slab shear reinforcement d 2

Critical section

c1 d 2

Critical section

d 2

d 2

c2

d 2 Stirrups

d 2

Double U-stirrup

(a) Bar reinforcement

3 (L – c1 ) 4 v 2 Lv

(b) Shearhead reinforcement

Figure 16.16.1  Bar and shearhead reinforcement in flat plate floors.

Figure 16.16.2  Radial-type arrangement of headed shear stud reinforcement. (Photo courtesy of Cary Kopczynski & Co.)

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The strength of two-​way slab systems with shear reinforcement has been summarized by Hawkins [16.94]. Corley and Hawkins [16.93, 16.110] have studied shearhead reinforcement. Other studies of shear reinforcement in flat plates can be found in Refs. [16.95–​ 16.102, 16.115, and 16.148], Yamada, Nanni, and Endo [16.102], and Pillai, Kirk, and Scavuzzo [16.115].

Shear Strength Provided by Stirrups When stirrups are used in accordance with ACI-​8.7.6, their nominal shear strength is computed as

Av f yt d   Vn = Vc + Vs = 2 λ fc′b0 d +  ≤ 6 fc′b0 d s  

(16.16.1)

where b0 is the perimeter of the critical section, Av is the total stirrup bar area around b0, s is the stirrup spacing in the direction perpendicular to the column face, and fyt is the yield strength of the bars used for stirrups. Note that shear reinforcement is required wherever Vu exceeds φVc based on Vc of Eqs. (16.15.2), which can be greater than 2 λ fc′b0 d . However, in the design of stirrups, Vc for Eq. (16.16.1) may not be taken greater than 2 λ fc′b0 d (ACI-​22.6.6.1). Furthermore, according to ACI-​22.6.6.2, the nominal strength Vn (i.e., Vc + Vs) must not exceed 6 fc′b0 d . When stirrups are provided, shear stresses must also be checked at a critical section located at a distance d/​2 beyond the point at which shear reinforcement is to be terminated. The shape of this critical section is a polygon that minimizes b0, as shown by the outermost critical section in Fig. 16.16.1(a). At this critical section, the factored shear stress, vu, cannot exceed φ 2 λ fc′ . The requirements for the geometry and anchorage of stirrups in slabs (ACI-​8.7.6.2) may be difficult to meet in slabs thinner than 10 in. Furthermore, single-​or multiple-​leg stirrups are permitted as shear reinforcement in slabs (and footings) only when the effective depth d is greater than or equal to 6 in., but not less than 16 times the diameter of the stirrups (ACI-​22.6.7.1). In accordance with ACI-​8.7.6.3, the first stirrup must be located at a distance d/​2 from the face of the column, and the spacing between stirrups cannot exceed d/​2. In addition, the spacing between the vertical legs of stirrups (measured parallel to the face of the column) must not exceed 2d. Details of typical bar arrangements for stirrups in slabs are given in ACI-​R8.7.6.

Shear Strength Provided by Shearheads Shear strength may be provided by shearheads under ACI-​22.6.9 whenever Vu /​φ at the critical section is between that permitted by Eqs. (16.15.2) and 7 fc′b0 d . These provisions, based on the tests of Corley and Hawkins [16.93], apply only where shear alone (i.e., no bending moment) is transferred at an interior column. When there is moment transfer to columns, ACI-​22.6.9.12 applies, as is discussed in Section 16.18. With regard to the size of the shearhead, it must furnish a ratio αv of 0.15 or larger (ACI-​22.6.9.5) between the flexural stiffness for each shearhead arm (Es Ix) and that for the surrounding composite cracked slab section of width (c2 + d), or

min α v =

Es I x = 0.15 Ec ( composite I s )

(16.16.2)

where c2 is the dimension of the column measured perpendicular to the span for which the moments are being calculated. The steel shape used must not be deeper than 70 times its web thickness, and the compression flange must be located within 0.3d of the compression

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surface of the slab (ACI-​22.6.9.2 and 22.6.9.4). In addition, the plastic moment capacity Mp of the shearhead arm must be at least (ACI-​22.6.9.6).



min M p =

Vu  c1    hv + α v  Lv −   2 ηφ  2 

(16.16.3)

where η = number (usually 4) of identical shearhead arms Vu = factored shear around the periphery of column face hv = depth of shearhead Lv = length of shearhead measured from column centerline c1 = dimension of the column measured in the direction of the span for which the moments are being calculated φ = 0.90, strength reduction factor for tension-​controlled members The purpose of Eq. (16.16.3) is to ensure that the required shear strength of the slab is reached before the flexural strength of the shearhead is exceeded. The length of the shearhead, Lv, should be such that the nominal shear strength Vn will not exceed 4 fc′b0 d computed at a peripheral section located at 3 4  ( Lv − c1 / 2) along the shearhead, but no closer elsewhere than d/​2 from the column face (ACI-​22.6.9.8 and 22.6.9.10). This length requirement is shown in Fig. 16.16.1(b). When a shearhead is used, it may be considered to contribute a resisting moment (ACI-​22.6.9.7)

Mv =

φ α vVu 2η

c1    Lv − 2 

(16.16.4)

to each column strip, but not more than 30% of the total moment resistance required in the column strip, nor more than the change in column strip moment over the length Lv , and nor more than the required Mp given by Eq. (16.16.3).

Shear Strength Provided by Headed Shear Studs Shear reinforcement in the form of headed steel studs welded to a steel strip (Fig. 16.16.2), often called headed shear studs, are a common alternative to stirrups or shearheads. The steel studs provide the same function as the vertical legs of stirrups, while the head of the stud and the bottom strip provide anchorage to the stud. Design requirements for headed shear stud reinforcement are given in ACI-​8.7.7 and ACI-​22.6.8, The nominal shear strength is computed as



Av f yt d   Vn = Vc + Vs = 3λ fc′b0 d +  ≤ 8 fc′b0 d s  

(16.16.5)

where b0 is the periphery around the critical section, Av is the total area of shear studs around b0, s is the stud spacing in the direction perpendicular to the column face, and fyt is the yield strength of the studs. Shear reinforcement is required wherever Vu exceeds φVc based on Vc of Eqs. (16.15.2). However, in the design of headed shear studs, Vc for Eq. (16.16.5) may not be taken greater than 3λ fc′b0 d (ACI-​22.6.6.1), and the nominal strength Vn (i.e., Vc + Vs) must not exceed 8 fc′b0 d (ACI-​22.6.6.2). Furthermore, a minimum amount of shear stud reinforcement must be provided, such that

Av b ≥ 2 fc′ 0 s f yt

(16.16.6)

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Outermost peripheral line of studs

Column S d/2

≤ d/2 (typ.)

≤ 2d (typ.)

d/2 Shear critical sections

Figure 16.16.3  Layout requirements (plan view) for headed shear studs at an interior column. (Adapted from ACI 318-​14 Commentary.)

Although the ACI code allows higher shear stress limits for the concrete contribution (3 fc′ ) and for the maximum shear stress (8 fc′ ), the authors recommend that the same limits that apply to stirrups be used (i.e., 2 fc′ and 6 fc′ , respectively). As for slabs with stirrups, the shear studs must be extended over a distance such that the factored shear stresses, vu, at the outermost critical section do not exceed φ 2 λ fc′. The shape of this critical section is defined as a polygon located at a distance d/​2 beyond the outermost peripheral line of shear studs, as shown by in Fig. 16.16.3. Shear stud location and spacing limits are given in ACI Table 8.7.7.1.2. The first line of shear studs is not to be located farther than d/​2 from the face of the column. Shear stud spacing in the direction perpendicular to the column face should not exceed d/​2, as with stirrups. However, stud spacing may be increased to 0.75d for subsequent lines of shear studs if the factored shear stress, vu, at the critical section located at d/​2 from the face of the column does not exceed φ 6 fc′b0 d. Stud spacing in the direction parallel to the column face cannot exceed 2d. A typical radial-​type arrangement of headed shear studs at an interior column is shown in Fig. 16.16.2. Research has shown that a radial-​type arrangement of headed studs leads to better ductility in slabs compared to a cruciform-​type layout [16.148].

EXAMPLE 16.16.1 For the flat plate shown in Fig. 16.16.4, compute the shear strength for one-​way and two-​ way actions around an interior column. Assume a slab thickness of 8 in. and that the service live load is 200 psf. If the required nominal shear strength Vn for two-​way action is between that permitted by Eqs. (16.15.2) and 6 fc′b0 d , determine the Av /​s requirement for shear reinforcement at the peripheral critical section and show the variation in the nominal shear stress (i.e., factored shear Vu divided by φ b0 d) from the critical section to the panel centerline. Use normal-​weight concrete with fc′ = 4000 psi and fy = 60,000 psi; assume #5 bars for the slab longitudinal reinforcement and #3 bars for the stirrups, if needed. SOLUTION (a) One-​way action. wu = 1.2 wD + 1.6 wL = 1.2(150)(8)

= 120 + 320 = 440 psf

1 12

+ 1.6(200) (Continued)

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Example 16.16.1 (Continued) The average effective depth in a column strip is avg d in column strip = 8.00 –​0.75 –​0.375 –​0.63 = 6.25 in. OK

which is greater than 6 in. and 16db of stirrup (ACI-​22.6.71). For a 12-​in.-​wide strip along section 1-​1 of Fig. 16.16.4,

vn =

)

(

Vu 440(7.98) = = 50 psi < 2 λ fc′ = 126 psi φ bw d 0.75(15)(6.25)

OK

(b) Two-​way action. Referring to Section 2-​2 of Fig. 16.16.4, b0 = 2(18.25) + 2(16.25) = 69 in.

b0 69 = = 11.0 < 20 d 6.25



With a rectangular perimeter b0 having a ratio of long to short side of less than 2 (i.e., β less than 2), and b0 /​d less than 20 for an interior column, Eq. (16.15.2a) controls; thus, the strength without shear reinforcement is Vc = 4 λ fc′b0 d . Using nominal stress vn = Vu /​φ b0d, vn =



440 [ 270 − 1.52(1.35)] Vu = = 365 psi > (4 λ fc′ = 253 psi) φ b0 d 0.75(69)(6.25) 1 7.98’

avg d = 6.25” 10”×12” column

15’–0”

16.25” Section 2–2

8” slab

18.25”

1 18’– 0”

2 3

4 5 6

Excess (vn –2λ fc’) diagram

365 psi

2 3 45 6

15’–0”

81.88” 16.25” 81.88”

Figure 16.16.4  Critical sections for shear Example 16.16.1.

2λ fc’ = 126 psi 91 psi

98.88”

98.88” 18.25” 18’–0”

2

41 psi

17 psi

3 4 5 98.88” in long direction

6

81.88” in short direction

Figure 16.16.5  Variation of two-​way nominal shear stress (Vu /​φ b0d), Example 16.16.1.

(Continued)

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Example 16.16.1 (Continued) Since the maximum nominal shear stress of 365 psi exceeds 4 λ fc′ psi, but not the maximum 6 fc′ = 380 psi permitted when bar or wire shear reinforcement is used, shear reinforcement is required to take the excess stress vn that exceeds 2 λ fc′ = 126 psi. The shear reinforcement in this case may consist of properly anchored bars or wires and need not be a shearhead. The Av /​s requirement around the critical section of 69-​in. periphery, from applying Eq. (16.16.1) with required Vn = Vu /​φ, is

(



)

Vu − 2 λ fc′ b0 d vn − 2 λ fc′ b0 Av φ = = s f yt d f yt =

)

(



(365 − 126)(69) = 0.275 in. 60, 000

Assuming s = d/​2 ≈ 3-​in. spacing (ACI-​8.7.6.3),

required Av = 0.83 sq in.



If a double #3 U stirrup is used at each of the four sides,

provided Av = 4(2)(0.11) = 0.88 sq in.



which is greater than the required Av of 0.83 sq in.

OK

The variation of the nominal shear stress vn from the maximum value of 365 psi to zero at the panel centerline over the equally spaced points 2 to 6 is shown in Fig. 16.16.5. The nominal shear stress (Vu /​φb0d) drops to 126 psi in a rather steep manner. The number and spacing of the U stirrups may be laid out with the aid of the excess (vn − 2 λ fc′) diagram.

EXAMPLE 16.16.2 Redesign the connection using shearhead reinforcement for the two-​way shear action of Example 16.16.1. SOLUTION (a) Two-​way action. Since the maximum nominal shear stress vn = Vu /​φ b0 d of 365 psi is between 4 λ fc′ = 253 psi and the maximum of 7 fc′ = 443 psi when a shearhead is used, shearhead reinforcement for the interior column (having no moment transfer to column) is permitted to be designed according to ACI-​22.6.9. (b) Length of shearhead. The length of shearhead should be such that the nominal shear stress is less than 4 fc′, computed around a periphery passing through points at 3  ( L − c / 2) from, but no closer than d/​2 to the column faces. Assuming a square 4 v 1 as the critical periphery, since the shearhead is to have four identical arms (ACI-​ 22.6.9.1), the required b0 (ft) may be computed from Fig. 16.16.6 so that

4 4000 ≤

440 270 − (b0 / 4)2   1   12  0.75(b0 ) 6.25 (Continued)

683



1 6 . 1 6   S H E A R R E I N F O R C E M E N T I N F L AT P L AT E   F L O O R S

683

Example 16.16.2 (Continued)

3 L – c1 ( ) 4 v 2 Lv

3.13”

b

0

/4

b

3.13” c

a

o 45°

c2 = 10”

c1 = 12” d Lv

Figure 16.16.6  Required length of shearhead in Example 16.16.2.

Neglecting the (b0  /​4)2 in the numerator,

b0 =

440(270)  1    = 8.3 ft (100 in.) 253(0.75)6.25  12 

Note that because shearhead reinforcement is used instead of stirrups, slab reinforcement could have been placed closer to the top and bottom of the slab thus resulting in a slightly larger average effective depth d and a somewhat smaller b0. Nonetheless, the same d computed for the slab with stirrups in Example 16.16.1 will be used in this example. The required distance Lv may be computed from the following geometric considerations. From right triangle oab,

b0 3  c c  4  Lv − 2  + 2  2 = 4  

which gives, based on leg ob, c = c2 = 10 in.,

 100 4 Lv =  − 5 + 5 = 21.9 in. 4 2 3

and, based on leg oa, c = c1 = 12 in.,

 100 4 Lv =  − 6 + 6 = 21.5 in. 4 2 3

For the periphery abcd not to approach closer than d/​2 (3.13 in.) to the periphery of the column section, oa = ob = (6 + 3.13) + (5 + 3.13) = 17.26 in. But

3 oa = 6 + ( Lv − 6) 4

3 and ob = 5 + ( Lv − 5) 4

(Continued)

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Example 16.16.2 (Continued) which gives

4 Lv = (17.26 − 6) + 6 = 21.0 in. 3



4 Lv = (17.26 − 5) + 5 = 21.3 in. 3

A

nAs = 8(0.84) = 6.7 sq in. nAs = 8(2.79) = 22.3 sq in.

4” 0.75″

#5 bars avg d = 6.25”

10 + d

A

c = 2.23″

12 + d

Column strip width = 90”

Use Lv = 22 in.

c2 + d = 16.25” Section A–A

Column strip width = 90”

Figure 16.16.7  Cracked slab section of width (c2 + d) in Example 16.16.2.

(c) Size of shearhead. The shearhead stiffness must be at least 0.15 that of the composite cracked slab section of width (c2 + d). It can be shown that 15–​#5 bars and 13–​#5 bars are required for negative slab reinforcement in the 90-​in.-​wide column strips of the long and short directions, respectively. The composite cracked section across width A-​A in Fig. 16.16.7 should be used because there is more steel in the slab in the long direction. The steel area As at the top of the slab in section A-​A, of width c2 + d is

As =

10 + d 16.25 (15)(0.31) = (15)(0.31) = 0.84 sq in. 90 90

with 3 4 -​in. cover at the top and bottom face of slab, and assuming #5 bars for top and bottom reinforcement in the two orthogonal directions, the clear distance between the top and bottom layers of reinforcement is 4 in. Assume an S4 × 9.5 section for the shearhead placed as shown in Fig. 16.16.7. Note that at least two bottom bars are required to pass within the region bounded by the longitudinal reinforcement of the column to satisfy the structural integrity requirements of ACI-​8.7.4.2. In addition, ACI-​22.6.9.4 requires that the compression flange of the steel shape be within 0.3d of the compression surface (i.e., the bottom of the slab in this case). For this reason, the S4 × 9.5 section is placed closer to the bottom of the slab with the same cover of the reinforcing bars. Because clearance under the shearhead is insufficient for the bottom bars to pass in the transverse direction, they will be made to pass through holes in the shearhead arms, as permitted by ACI-​8.7.4.2.3. The ratio between the depth and the web thickness for this section is 4.00/​0.326 = 12.3, which is much less than 70 in accordance with ACI-​22.6.9.2. (Continued)

685



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685

Example 16.16.2 (Continued) The centroidal axis of the composite cracked section may be obtained by equating the static moments of the compression-​and tension-​transformed areas (ignoring the bottom steel bars), 16.25c 2 = 22.3 (2.75 − c) + 6.7 (6.56 − c) 2 c = 2.23 in. composite I s =

16.25(2.23)3 + n ( I x of steel section ) + 22.3(0.52)2 + 6.7(4.71)2 3 OK

= 60.1 + 8(6.76) + 6.03 + 148.6 = 269 in.4 provided α v =

Es (6.76) 8(6.76) = = 0.20 > 0.15 269 Ec (composite I s )

The plastic section modulus of the S4 × 9.5 is given by the AISC Manual [16.147] as 4.04 cu in. Using A992 Grade 50 steel, the provided Mp is

provided Mp = 50 ( 4.04 ) = 202 kip-in.

The required Mp is computed from Eq. (16.16.3) as required Mp = =



Vu  c1    hv + α v  required Lv −   8φ  2  0.440(270) [ 4 + 0.20(21.9 − 5)] 8(0.90)

= 122 kip-in. < 202 kip-in.

OK

The section has about 66% more strength than required. Perhaps a smaller section, such as an S4 × 7.7, could be used. (d) Shearhead contribution to resisting negative moment in slab. It can be shown that the negative moments at the face of the column in the 90-​in. column strip width in the long and short directions are column strip moment in the long direction = 125 ft-​kips column strip moment in the short direction = 102 ft-​kips The resisting moment of the shearhead may be computed from Eq. (16.16.4), Mc = =

φ α vVu 2η

c1    Lv − 2 

0.90(0.20)(0.440)(270) [(22 − 6) or (22 − 5) ] 121 8

= 3.6 or 3.8 ft-kips



Thus the contribution is rather small, and a revision of slab reinforcement is unnecessary.

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16.17 DIRECT DESIGN METHOD—​M OMENTS IN COLUMNS The moments in columns due to unequal loads on adjacent panels are readily available when an elastic analysis is performed on the equivalent frame for the various pattern loadings. In the Direct Design Method, wherein the limitations listed in Section 16.7 are satisfied, the longitudinal moments in the slab are prescribed by the provisions of ACI-​8.10.4. In a similar manner, the code prescribes the moment at an interior column as follows [ACI Formula (8.10.7.2)]:   1  M sc = 0.07  wDu + wLu  L2 L2n − wDu ′ L2′ ( Ln′ )2   2  



(16.17.1)

where wDu = factored dead load per unit area wLu = factored live load per unit area wDu ′ , L2′ , Ln′ = quantities referring to shorter span

(wDu + 1 wLu) L2 2

(wDu + 1 wLu) L2 2 Span 1

Col. 2

Span 1 Col. 2

Span 2

Col. 1

w’DuL’2

Col. 1

The moment is yet to be distributed between the two ends of the upper and lower columns meeting at the joint. The rationale for Eq. (16.17.1) may be observed from the stiffness ratios at a typical interior joint shown in Fig. 16.17.1(a), wherein the distribution factor for the sum of the column end moments is taken as 7 8 and the unbalanced moment in the column strip is taken to be 0.080/​0.125 times the difference in the total static moments due to the dead load plus half the live load on the longer span and the dead load only on the shorter span. For an edge column, ACI-​8.10.7.3 requires using 0.3M0 as the moment to be transferred between the slab and an edge column.

(a) Interior joint

(b) Exterior joint

Figure 16.17.1  Direct design method—​moments in columns.

EXAMPLE 16.17.1 Obtain the factored moments in the interior and exterior columns in each direction for the flat plate of Example 16.3.3. SOLUTION (a) Exterior column, long direction (Frame A). The factored moment Msc to be transferred to the exterior column is (ACI-​8.10.7.3) 0.3M0,

M sc = 0.3 M 0 = 0.3(58.2) = 17.5 ft-kips

(Continued)

687



687

16.18  TRANSFER OF MOMENT AND SHEAR

Example 16.17.1 (Continued) where the 58.2 ft-​kips was obtained from Table  16.10.2. The moment Msc is to be divided between upper and lower columns in proportion to their stiffnesses (in this case, equally). In this example, nearly all (98.8% for Frame A and 99.3% for Frame C–see Table 16.12.6) of the exterior frame moment is taken by the column strip; the arbitrary use of 0.3M0 to be taken by the column strip seems appropriate. (b) Interior column, long direction. The factored moment to be transferred to the column is empirically the amount obtained from ACI Formula (8.10.7.2) [Eq. 16.17.1],



M sc = 0.07(0.083 + 0.058)(12)(15 − 1)2 − 0.083(12)(15 − 1)2

= 0.07(0.058)(12)(14)2 = 9.5 ft-kips

The moment Msc is to be divided between upper and lower columns. (c) Exterior column, short direction (Frame C). The factored moment to be transferred is

M sc = 0.3 M 0 = 0.3(46.3) = 13.9 ft-kips



where the 46.3 ft-​kips was obtained from Table 16.10.2. The moment Msc is to be divided between upper and lower columns. (d) Interior column, short direction. The factored moment to be transferred is

M sc = 0.07(0.058)(15)(12 − 0.83)2 = 7.6 ft-kips

The moment Msc is to be divided between upper and lower columns.

16.18 TRANSFER OF MOMENT AND SHEAR AT JUNCTION OF SLAB AND COLUMN Inasmuch as the columns meet the slab at monolithic joints, there must be moment as well as shear transfer between the slab and the column ends. The moments may arise out of lateral loads due to wind or earthquake effects acting on the multistory frame, or they may be due to unequal gravity loads, as considered in Section 16.17. The transfer of moment and shear at the slab-​column interface is extremely important in the design of flat plates and has been the subject of numerous research studies [16.103–​ 16.138, 16.143]. The current provisions are largely based on the ACI-​ASCE Committee 352 report, “Recommendations for Design of Slab-​Column Connections in Monolithic Reinforced Concrete Structures” [16.128], and the background explanation by Moehle, Kreger, and Leon [16.127]. Let Msc be the total slab factored moment that is to be transferred to the ends of the upper and lower columns meeting at an exterior or an interior joint. Test results by Hanson and Hanson [16.104] have shown that about 60% of the moment is transferred by flexure and the remainder by eccentricity of the shear around the critical periphery located at d/​2 from the column faces. The ACI Code requires that the fraction of the total slab factored

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moment resisted by the column at a joint, Msc, be transferred by flexure and computed as (ACI-​8.4.2.3.2)     1 M Mu f = γ f M sc =  sc  2 b1  1+  3 b2  



(16.18.1)

where Mu f = fraction of total slab factored moment transferred by flexure b1 = critical section dimension in the direction of the span for which moments are determined = c1 + d/​2 for exterior columns [Fig. 16.18.1(a)] = c1 + d for interior columns [Fig. 16.18.1(b)] b2 = critical section dimension in the direction perpendicular to b1 = c2 + d (Fig. 16.18.1) The remaining portion of the total slab factored moment is assumed to be transferred by eccentricity of shear and is computed as (ACI-​8.4.4.2.2) Muv = γ v M sc = (1 − γ f ) M sc



(16.18.2)

The moment Mu f is considered to be resisted within an effective slab width equal to (c2 + 3t) at the column (ACI-​8.4.2.3.3), where t is the slab or drop panel thickness. The moment strength for Mu f is achieved by using additional reinforcement and closer spacing within the width (c2 + 3t) [ACI-​8.4.2.3.5]. b1 = c1 + d 2

b1 = c1 + d

z c1

z b2 = c2 + d

b2 = c2 + d

Column c2

688

z

z b1

b1

Vu Ac

Stresses due to Vu

b2

Muv x1 Jc

b2

x2 x1

x1

Muv x1 Jc

Muv x2 Jc

vu1

(a) Exterior column

vu2

x2

Muv x2 Jc vu1

(b) Interior column

Figure 16.18.1  Transfer of moments to columns through shear stresses.

vu2

Stresses due to Muv

Total stresses

689



16.18  TRANSFER OF MOMENT AND SHEAR

689

Note that if b2 = b1, Eq. (16.18.1) becomes Mu f = 0.60 M sc



and that if b2 = 1.5b1, Eq. (16.18.1) becomes Mu f = 0.648 M sc It appears reasonable that when b2 is larger than b1, the moment transferred by flexure is greater because the effective slab width (c2 + 3t) resisting the moment is larger. Analysis of the available test data [16.126, 16.127] has shown that the interaction between flexure and shear is reduced when the ratio between the shear caused by gravity loads and the slab shear strength (in the absence of moment transfer) is relatively low. In other words, a greater proportion of the total slab factored moment Msc may be assumed to be transferred by flexure; i.e., a larger value of γf may be used in Eq. (16.18.1). Accordingly, ACI-8.4.2.3.4 allows γf to be increased when the applied-to-strength shear ratio does not exceed a threshold value that depends on the connection type (corner, edge, or interior), and when the slab flexural reinforcement provided within (c2 + 3t) satisfies a minimum value for the net tensile strain. When the conditions set forth in ACI-8.4.2.3.4 are satisfied, flexure and shear interaction may be neglected in some cases, which can greatly simplify slab design. These conditions and the allowed increase in the values of γf are described next.

Modifications to γf (ACI-8.4.2.3.4) Corner Columns At a corner support, ACI-​8.4.2.3.4 permits neglect of the interaction between shear and moment when the factored shear due to gravity loads, Vug, acting on the slab critical section does not exceed 0.5φVc. In other words, for such situations, the full slab moment can be considered to be transferred by flexure (i.e., γf = 1.0), and the factored shear Vu can be considered independently. This provision applies to moments being transferred in either direction, and is permitted only when the slab flexural reinforcement required to resist Msc within the effective slab width (c2 + 3t) has a net tensile strain, εt, greater than or equal to 0.004. Edge Columns For moments about an axis perpendicular to the edge, the full slab moment can also be considered to be transferred by flexure alone (i.e., γf = 1.0) when the factored shear due to gravity loads, Vug, acting on the slab critical section does not exceed 0.75φVc. As for corner columns, this is permitted only when the net tensile strain, εt, in the slab flexural reinforcement provided within the effective slab width (c2 + 3t) is greater than or equal to 0.004. On the other hand, for moments about an axis parallel to the edge, ACI-​8.4.2.3.4 permits increasing by as much as 25% the proportion γf of the full exterior moment transferred by flexure when the factored shear due to gravity loads, Vug, acting on the slab critical section does not exceed 0.4φVc. The 25% increase in γf , however, is permitted only when the net tensile strain, εt, in the slab flexural reinforcement provided within the effective slab width (c2 + 3t) is greater or equal to 0.01. Interior Columns For interior columns, a 25% increase in γf is permitted for moments in either direction when factored shear due to gravity loads, Vug, acting on the slab critical section does not exceed 0.4 φVc, provided the net tensile strain, εt, in the slab flexural reinforcement within the effective slab width (c2 + 3t) is greater than or equal to 0.01. The modifications to γf described above are not permitted for prestressed concrete systems.

Computation of Shear Stresses due to Flexure and Shear Interaction (γv > 0)— Eccentric Shear Stress Model When part of the slab factored moment is assumed to be transferred by eccentricity of shear; i.e., when γv > 0, the shear stresses due to Muv must be combined with those induced

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by the factored shear force Vu. The procedure to compute shear stresses has been traditionally based on the eccentric shear stress model [16.103, 16.104], where shear stresses are assumed to vary linearly about the centroid of the critical section (located at d/2 from the column faces), as shown in Fig. 16.18.1. Based on these assumptions, the minimum and maximum values of the factored shear stresses acting along the faces of the critical section may be computed as (see Fig. 16.18.1)

vu1 =

Vu Muv x1 − Ac Jc

(16.18.3)



vu 2 =

Vu Muv x2 + Ac Jc

(16.18.4)

By using a section property Jc analogous to the polar moment of inertia of the shear area along the critical periphery taken about the z-​z axis, it is assumed that there are both horizontal and vertical shear stresses on the shear areas having dimensions b1 by d as shown in Fig. 16.18.2. The z-​z axis is perpendicular to the longitudinal axis of the equivalent frame: that is, in the transverse direction, and located at the centroid of the shear area. For an exterior column, x1 and x2 are obtained by locating the centroid of the channel-​ shaped vertical shear area represented by the dashed line (b1 + b2 + b1) shown in Fig. 16.18.1(a), and Ac = (2b1 + b2 )d x2 =

(16.18.5)

b12 d Ac

(16.18.6)

 b3  b d3 J c = d 2 1 − (2b1 + b2 ) x22  + 1 6  3 



(16.18.7)

For an interior column, referring to Fig. 16.18.1(b), Ac = (2b1 + b2 )d

(16.18.8)

 b3 b b 2  b d 3 Jc = d  1 + 2 1  + 1 2  6 6

(16.18.9)

Equations (16.18.5) to (16.18.9) may be derived by letting the shear stress at any location resulting from Muv alone be proportional to the distance from the centroidal axis z-​z to x1

z x2

z

Muv x1 Jc Muv

Muv x2 Jc

vuh

d

z

Muv ( 2 ) Jc

z (a) Shear areas to resist Muv

(b) Vertical resisting shear stresses

(c) Horizontal resisting shear stresses

Figure 16.18.2  Resisting shear stresses due to Muv acting on an exterior column.

691



691

16.18  TRANSFER OF MOMENT AND SHEAR

the shear areas b1 by d, and either (a) to the one shear area b2 by d for an exterior column as shown in Fig. 16.18.2, or (b) to the two shear areas b2 by d for the interior column. For slabs without shear reinforcement, the larger factored shear stress vu2 shown in Fig. 16.18.1 must not exceed the stress φ vn = φVc /​b0d obtained from Eqs. (16.15.2); otherwise shear reinforcement is required.

EXAMPLE 16.18.1 For the flat plate design of Example  16.3.3, investigate the transfer of gravity load moments in the long direction, as already computed in Example 16.17.1, to the exterior and interior columns, respectively. SOLUTION (a) Exterior column (long direction). Investigate whether the full slab moment may be considered to be transferred by flexure alone; i.e., whether γf is permitted to be taken as 1.0 in accordance with ACI-8.4.2.3.4. From Example 16.17.1, the moment to be transferred is Mu = M sc = 17.5 ft-kips The factored shear due to gravity loads, Vug, is taken as wu times the floor area, 12 ft × 7.5 ft, tributary to the exterior column. Vug = 0.198(12)7.5 = 17.8 kips Using the dimensions shown in Fig.  16.15.4 as a reference, b0 for an exterior (edge) column is b0 = 2(12 + 2.125) + 14.25 = 42.5 in.



Using the average effective depth d = 4.25 in. for #4 slab reinforcement, the nominal shear strength Vc in accordance with ACI-​22.6.5.2 is the smallest of Vc = 4 λ fc′b0 d

Controls

  4 4   λ fc′b0 d  = 5.3λ fc′b0 d V = 2 +  λ fc′b0 d =  2 +     c   β 12 /10  



  α d 30(4.25)   Vc =  2 + s  λ fc′b0 d =  2 + λ fc′b0 d  = 5.0 λ fc′b0 d   42.5  b0    Thus,

 1  Vc = 4 λ fc′b0 d = 4(1.0) 4000 (42.5)4.25  = 45.7 kips  1000 

According to ACI-​8.4.2.3.4, for an interior (or edge) column the full slab moment may be considered to be transferred by flexure alone when

[0.75φ Vc = 0.75 ( 0.75)( 45.7) = 0.75 (34.3) = 25.7 kips ] > [ Vug = 17.8 kips]

Thus, the interaction between shear and flexure may be neglected (i.e., γf = 1.0), insofar as the reinforcement required to resist γf Msc within the effective slab width has a net tensile strain, εt, greater than or equal to 0.004. The effective slab width is equal to the column width plus three times the slab thickness (ACI-​8.4.2.3.3); that is, 26.5 in. Using the average effective depth d = 4.25 in. for #4 slab reinforcement,

Rn =

M sc 17.5(12, 000) = = 487 psi φ bd 2 0.9(26.5)(4.25)2 (Continued)

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Example 16.18.1 (Continued) From Eq. (3.8.5), with fc′ = 4000 psi and fy = 60,000 psi, the required ρ ≈ 0.008, which is much less than the maximum reinforcement ratio corresponding to a net tensile strain εt of 0.004. Thus, the net tensile strains in the tension steel will be larger than 0.004. Therefore, the full slab factored moment can be considered to be transferred by flexure alone and Mu f = γ f M sc = 1.0(17.5) = 17.5 ft-kips The shear can be considered independently. (b) Exterior column (long direction). Check moment transfer using the eccentric shear stress model [Eqs. (16.18.3) and (16.18.4)]. This procedure involves more calculations and is more conservative. (i) Transfer by flexure. From Eq. (16.18.1), using the average effective depth d = 4.25 in. for #4 slab reinforcement,



  1 Mu f = γ f M sc =   2 1+ 3    = 2   1 + 3

  M sc b1   b2    1  17.5 = 0.601(17.5) = 10.5 ft-kips 12 + 2.125   10 + 4.25 



As shown by Fig. 16.18.3, this moment is to be carried over a slab width equal to the column width plus 3 times the slab thickness, that is, 26.5 in. From Table 16.12.6 and Fig. 16.10.3 (Frame A), the total moment in the 72-​in.-​wide column strip is M in column strip = 0.988 (15.1) = 15 ft-kips If the slab reinforcement is placed at equal spacing in the column strip, additional reinforcement is needed in the 26.5-​in. width for a bending moment of  26.5  Mu f − 15  = 10.5 − 5.5 = 5 ft-kips  72 



(ii) Transfer by eccentricity of shear. From part (a), the factored shear force is Vu = Vug = 0.198(12)7.5 = 17.8 kips and from Eq. (16.18.2)

Muv = M sc − Muf = 17.5 − 10.5 = 7 ft-kip ps



From Fig. 16.18.3, 2(14.12)7.06 = 4.7 in. 28.24 + 14.25 Ac = 4.25(28.24 + 14.25) = 181 sq in. x2 =



 14.12(4.25)3  2(14.12)3 J c = 4.25  − 42.49(4.7)2  + 3 6   = 3990 + 181 = 4171 in.4 17, 800 7000(12)9.42 − = 98 − 190 = −92 psi 181 4171 17, 800 7000(12)4.70 vu 2 = + = 98 + 95 = +193 psi 181 4171 vu1 =

(Continued)

693



693

16.18  TRANSFER OF MOMENT AND SHEAR

Example 16.18.1 (Continued) 12 +

d = 14.12” 2

10 + 3t = 26.5”

10 + d = 14.25”

x2

x1 t = slab or drop panel thickness Vu Muv Muf

Figure 16.18.3  Transfer of moments at exterior column in Example 16.18.1.

The nominal stress limit based on strength in shear was determined in part (a) to be that based on 4 λ fc′. Thus, the limit to the above stresses is

limit vu = φ vc = φ (4 λ fc′) = 0.75(253) = 190 psi ≈ [vu2 = 192 psi]

The required and design shear strengths are very close; the slab may be considered adequate based on the shear-​flexure interaction procedure. Otherwise, shear reinforcement could be provided. The horizontal shear stress vuh at the upper or lower edge of the two shear areas b1 by d is vuh =



7000(12)(4.25/ 2) = 43 psi 4171

The vuh of 43 psi, vu1 of –​91 psi, and vu2 of +192 psi may be drawn on a sketch like that of Fig. 16.18.2, and by basic statics computation of the resultant upward force should equal Vu of 17.8 kips and the resultant moment about the z-​z axis should equal Muv of 7 ft-​kips. (c) Interior column (long direction). Investigate whether an increase in γf would be permitted in accordance with ACI-8.4.2.3.4. The factored shear force due to gravity loads Vug is computed as wu times the tributary floor area of 12 × 15 ft,

Vug = 0.198(12)15 = 35.6 kips



Using Eqs. (16.15.2), as shown in part (a) for the exterior column, will indicate that the shear strength based on 4 λ fc′ controls. Since the interior and exterior columns have the same size, Eq. (16.15.2b), involving the aspect ratio β, gives the same value as in part (a), that is, 5.3λ fc′. Regarding Eq. (16.15.2c), αs is taken as 40 for interior columns, and

b0 = 2(16.25) + 2(14.25) = 61.0 in.



Thus, Eq. (16.15.2c) gives

  α d 40(4.25)   Vc =  2 + s  λ fc′b0 d =  2 + λ fc′b0 d  = 4.8λ fc′b0 d   b0  61.0    (Continued)

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Example 16.18.1 (Continued) The strength Vc cannot exceed that based on 4 λ fc′ from Eq. (16.15.2a). Thus,

Vc = 4 λ fc′b0 d = 4(1.0) 4000 (61.0)(4.25)

1 1000

= 65.6 kips

For interior columns, a 25% increase in the value of γf may be permitted when [0.4φVc = 0.4(0.75) (65.6) = 0.4(49.2) = 19.7 kips] < [Vug = 35.6 kipss] Since the factored shear Vug is not less than 0.4φVc, the increase in γf is not permitted! (Note that for interior columns, γf may be increased only if the net tensile strain in the slab reinforcement provided within (c2 + 3t) is greater than or equal to 0.01, even if 0.4 φVc were to be greater than Vug). (i) Transfer by flexure. From part (b) of Example 16.17.1, the factored moment to be resisted by the column is Mu = M sc = 9.5 ft-kips and the fraction to be transferred by fleexure is     1 M Mu f = γ f M sc =  sc  2 b1  1+  3 b2  



    1 =  9.5 = 0.584(9.5) = 5.5 ft-kips 2 12 + 4.25    1 +  3 10 + 4.25  From Table 16.12.6 and Fig. 16.10.3, the total moment in the 72-​in.-​wide column strip is M in column strip = 0.75(37.8) = 28.4 ft-kips Assuming that the slab reinforcement is placed at equal spacing within the column strip, the moment resisted within the 26.5-​in. width is 26.5(28.4)/ ​72 = 10.5 ft-​kips, which is larger than 5.5 ft-​kips; no additional reinforcement is needed.



(ii) Transfer by eccentricity of shear. The factored shear Vu is 35.6  kips. From Eq. (16.18.2) Muv = M sc − Muf = 9.5 − 5.5 = 4.0 ft-kips

From Fig. 16.18.1(b), Ac = 4.25(32.50 + 28.50) = 259 sq in.  (16.25)3 14.25(16.25)2  16.25(4.25)3 J c = 4.25  + + 2 6   6 = 11, 040 + 210 = 11, 250 in.4 vu1 =

35, 600 4000(12)8.12 − = 137 − 35 = −102 psi 259 11, 250

vu 2 =

35, 600 4000(12)8.12 = 137 + 35 = +172 psi + 259 11, 250

(



)

both of which are less than the design strength φ vn = φ vc = φ 4 λ fc′ = 0.75(253) = 190 psi when no shear reinforcement is provided. (Continued)

695



695

16.18  TRANSFER OF MOMENT AND SHEAR

Example 16.18.1 (Continued) Again, by basic statics the sum of the factored load vertical shear stresses on the two areas b1 × d plus that on the two areas b2 × d should total Vu of 35.6 kips. Likewise, the moment of these shear stress resultants along with that of the horizontal shear stresses on the two faces b1 × d should add up to 4.0 ft-​kips. The horizontal shear stress at the upper or lower edge of the faces b1 × d is vuh =



4000(12)(4.25 / 2) = 9 psi 11, 250

Moment Transfer from Flat Plate to Column When Shearheads Are Used Tests [16.93, 16.110] have indicated that shear stresses computed for factored loads at the critical section distance d/​2 from the column face are appropriate for transfer of Muv = Mu –​ Muf as described above, even when shearheads are used. However, the critical section for Vu is at a periphery passing through points at 3 4 ( Lv − c1 / 2) from, but no closer than, d/​2 to the column faces. When both Vu and Mu are to be transferred, ACI-​22.6.9.12 requires  that the sum of the shear stresses due to Muv and due to vertical load Vu computed at their respective critical sections not exceed φ (4 λ fc′). The reason for this apparent inconsistency (ACI Commentary R22.6.9.11) is that these two critical sections are in close proximity at the column corners, where the failures initiate.

EXAMPLE 16.18.2 Compute the moment at an interior column of the flat plate with shearheads of Example 16.16.2 using Eq. (16.17.1), ACI Formula (8.10.7.2). Also, compute the shear stresses due to moment transferred by eccentricity of shear. SOLUTION (a) Referring to Example 16.16.1, part (a) of solution,

wu = 440 psf



Referring to Example 16.16.2, part (d) of solution, column strip moment in the long direction = 125 ft-​kips Using Eq. (16.17.1), the moment to be transferred to the column is,

M sc = 0.07 (0.120 + 0.320)(15)(18 − 1)2 − 0.120(15)(18 − 1)2  = 0.07(0.320)(15)(17)2 = 97.1 ft-kips



The effective slab width is

c2 + 3t = 10 + 3(8) = 34 in.



Column strip moment in 34-​in. width of Fig. 16.18.4 is

 34  125   = 47.2 ft-kips  90  (Continued)

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Example 16.18.2 (Continued) Additional reinforcement is needed to carry (97.1 –​47.2) = 49.9 ft-​kips—​say, 50 ft-​kips within the 34-​in. width unless 97.1/​125 = 78% of the total column strip reinforcement is concentrated in the 34-​in. width. (b) Compute factored load shear stress at critical section of Fig. 16.18.4 due to Mu v only. The moment to be transferred by flexure is     1 M Mu f = γ f M sc =  sc  2 b1  1+  3 b2  



    1 =  97.1 2 12 + 6.25    1 +  3 10 + 6.25 



Mu f = 0.586 M sc = 0.586 ( 97.1) = 56.9 ft-kips and

Mu v = 97.1 − 56.9 = 40.2 ft-kips



The maximum shear stress due to Muv (see Fig. 16.8.1) computed about the centroid of the critical section at d/​2 (ACI-​22.6.4.1) is given by (see Fig. 16.8.1) vu =



Muv x1 Jc

The critical section properties for an interior column are Ac = 2(b1 + b2 )d [16.18.8]

Thus,

Ac = 2(18.25 + 16.25)6.25 = 431.3 sq in.

and  b3 b b 2  b d 3 Jc = d  1 + 2 1  + 1 2  6 6

[16.18.9]

 (18.25)3 16.25(18.25)2  18.25(6.25)3 J c = 6.25  + + 6 2   6 = 24, 000 in.4

Thus, the maximum shear stress due to Muv is

vu =

40.2(9.13)(12, 000) = 184 psi 24, 000 (Continued)

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1 6 . 1 9   O P E N I N G S A N D C O R N E R C O N N E C T I O N S I N F L AT   S L A B S

697

Example 16.18.2 (Continued) These stresses alone are very high and by adding the Vu /​Ac component they will exceed the maximum permitted of φ (4 λ fc′) = 190 psi. Therefore, the slab thickness should be increased around the column: that is, a drop panel would be required. 12 + d = 18.25”

10 + 3t = 34”

10 + d = 16.25”

Vu Muv Muf

Figure 16.18.4  Transfer of moments in the long direction at interior column in Example 16.18.2.

16.19 OPENINGS AND CORNER CONNECTIONS IN FLAT SLABS When openings and corner connections are present in floor slabs, designers must make sure that adequate provisions are made for them. ASCE-​ACI Joint Task Committee 426 [16.83] has summarized available information. Tests by Roll, Zaidi, Sabnis, and Chuang [16.79] have provided additional data for treating openings, while Zaghlool and de Paiva [16.107, 16.108] have provided data for corner connections. ACI-​8.5.4.1 first prescribes in general that openings of any size may be provided if it can be shown by analysis that all strength and serviceability conditions, including the limits on the deflections, are satisfied. In common situations (ACI-​8.5.4.2), however, no special analysis need be made for slab systems not having beams when (1) openings are within the middle half of the span in each direction, provided the total amount of reinforcement required for the panel without the opening is maintained; (2) openings in the area common to two column strips do not interrupt more than one-​eighth of the column strip width in either span, and the equivalent of reinforcement interrupted is added on all sides of the openings; and (3) openings in the area common to one column strip and one middle strip do not interrupt more than one-​fourth of the reinforcement in either strip, and the equivalent of reinforcement interrupted is added on all sides of the openings. In regard to shear strength in two-​way action, when the opening is within a column strip or closer than 10 times the slab thickness from a concentrated load or reaction area [ACI-​ 8.5.4.2(d)], the critical section for slabs without shearheads must not include the part of the periphery that is enclosed by radial projections of the openings to the center of the column (ACI-​ 22.6.4.3). For slabs with shearheads, the critical periphery is to be reduced only by one-​half of what is cut away by the radial lines from the center of the column to the edges of the opening (ACI-​22.6.9.9). Some critical sections with cutaways by openings are shown in Fig. 16.19.1.

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Opening Ineffective

d (typical) 2 Critical section (a)

(b) Free corner

Regard as free edge

(c)

(d)

Figure 16.19.1  Effect of openings and free edges on critical perimeter of two-​way shear action. (From ACI Commentary R ​ 22.6.4.3.)

16.20 EQUIVALENT FRAME METHOD FOR GRAVITY LOAD ANALYSIS The design of two-​way floor systems under gravity loads has been presented in earlier sections of this chapter, wherein the longitudinal distribution of moments in the equivalent frame3 follows the coefficients of the ACI Code, within the limitations set forth in the Direct Design Method (DDM). When the limitations of the DDM are not met, an elastic analysis of the equivalent frame must be made for various gravity load patterns to obtain the longitudinal moment and shear envelopes. For lateral loads, however, an elastic analysis of the equivalent frame must be made regardless of whether the system is within the limitations of the DDM for gravity loads. According to ACI-​8.2.1, “A slab system shall be designed by any procedure satisfying equilibrium and geometric compatibility, provided that the design strength at every section is at least equal to the required strength, and all serviceability requirements are met.” For gravity load analysis, ACI-​8.11.2.5 permits a separate analysis of each floor, with the far ends of columns considered fixed. The elevation and plan of a subassembly containing a floor with the attached upper and lower columns in a two-​way floor system are shown in Fig. 16.20.1(a) and 16.20.1(b). The subassembly enclosed between the two parallel centerlines of two adjacent panels in a multistory two-​way floor system is a three-​dimensional structure that includes the slab, beams (longitudinal and transverse), columns, drop panels, and column capitals, if present. A difficult problem in the analysis of the subassembly shown in Fig. 16.20.1 is modeling of the transfer of the slab moment to the columns. In the Equivalent Frame Method (EFM), the three-​dimensional subassembly is idealized by a series of two-​dimensional frames using equivalent columns and slab-​beams. For gravity load analysis the EFM prescribed by the ACI Code differs from the DDM only in the way by which the longitudinal moments along the spans of the equivalent frame (as defined in Section 16.2) are obtained. In either method of analysis, the transverse distribution of the longitudinal moments may be carried out as described in Section 16.12 as

3  Note that the concept of “equivalent frame” is applicable in both the direct design method and the equivalent frame method.

69



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long as, in two-​way slabs supported on beams, the beams are sufficiently stiff and satisfy limitation No. 6 of the direct design method discussed in Section 16.7 per ACI-​8.11.6.6. It is noted that when the EFM is used for gravity load analysis of two-​way floor systems that meet the limitations of the DDM, the computed moments may be reduced such that the absolute sum of the positive and average negative moments is at least equal to the total static moment wu L2 L2n /8 (ACI-​8.11.6.5). This provision exists because the “Code should not require a value greater than the least acceptable value” (ACI-​R8.11.6.5); in other words, if the values obtained from the DDM are acceptable, there is no reason to design the slab for larger moments irrespective of the method used in the analyses. At this point, axial deformation of the columns, reduction of column stiffness due to heavy gravity load, and increase of column moments and shears due to P-​Δ effects (see Chapter 13) are neglected to focus attention on the characteristics of the equivalent rigid frame in a two-​way floor system.

G

H

K

A

B

C

D

E

F

(a) Elevation

1 2

Middle strip

1 2

Middle strip

Width of equivalent frame

CL of panel

Column strip

CL of panel (b) Plan G

A

torsional elements

slab-beam

A’ D

H

columns

B

torsional element

slab-beam

B’ E

K

columns

C

C’ F

(c) Frame model

Figure 16.20.1  Equivalent frame of two-​way slab subassembly for gravity load analysis.

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Section Properties The subsections that follow describe the procedure for computing the section properties of the equivalent columns and beams. These properties are computed based on extensive analytical and experimental test data on two-​way slabs under gravity loads. The EFM and thus computation of section properties of the equivalent columns and beams, were originally developed to be used with the method of moment distribution. The equivalent frame method, however, may also be used with virtually any standard analysis software, as discussed in Section 16.21. Flexural Stiffness of Slab-​Beams, Ksb Consideration of the flexural stiffness of the slab-​beams, Ksb, must incorporate the stiffness of the slab, including any drop panels and capitals around columns, as well as the stiffness provided by any beams spanning in the direction of the frame. As a result, the equivalent beams will have a variable moment of inertia along the beam axis; that is, they will be nonprismatic. Figure  16.20.2 shows cross sections for computing the flexural stiffness of the slab-​ beams for various slab systems. As an example, consider the slab system with column capitals shown in Fig. 16.20.2(c). The moment of inertia of the slab-​beam between the edges of the drop panels, I1, is computed assuming a rectangular section of width, L2, and depth equal to the slab thickness, h1, as shown by Section A-​A in Fig. 16.20.2(c). Where a drop panel exists, the moment of inertia, I2, is computed for the T-​shaped section (Section B-​B) shown in Fig. 16.20.2(c). The moment of inertia from the center of the column to the face of the capital is assumed to be equal to that of the slab-​beam at the face of the capital, I2, divided by the quantity (1 –​c2 /​L2)2, where c2 is the width of the equivalent square column capital (ACI-​8.11.3.1). For other slab systems, c2 is the width of the support measured transverse to the direction of the span in which the moments are being computed. The coefficient 1/​(1 –​ c2 /​L2)2 accounts for the increased stiffness of the slab-​beam over the support in a manner that is consistent with the results of three-​dimensional slab analyses as well as with test results [16.27]. The variation of moment of inertia along the axis of the slab-​beams is taken into account (as required by ACI-​8.11.3.2) as shown by the equivalent slab-​beam stiffness diagram shown for each slab system in Fig. 16.20.2. Note that ACI-​8.11.3.3 permits the use of the gross concrete area for computation of the moment of inertia of slab-​beams in gravity load analysis. For lateral load analyses, the effects of cracking and reinforcement must be taken into account, as discussed later in Section 16.22. Stiffness of Equivalent Columns, Kec Consider the moment transfer mechanism from the slab to the edge column shown in Fig. 16.20.3. It may be easily visualized that some of the moment will be transferred by flexure from the slab directly to the column, while the remainder will be transferred first to the transverse beams and then to the column by twisting of the transverse beams framing into the column. Note that a similar transfer mechanism would occur for slab systems without beams: that is, some of the moment will be transferred to the column by twisting of the slab in the transverse direction. Stiffness of Torsional Members To account for the moment transferred by twisting of the transverse elements, Corley, Sozen, and Siess [16.26] developed the idea of using a torsional member attached to the column (a cutaway from the three-​dimensional structure) in the transverse direction at one end and connected to the slab-​beam elements at the other end. The conceptual model is illustrated in Fig. 16.20.1(c). In this model, beams A′-​B′ and B′-​C′ represent the slab-​beams while elements AA′, BB′, and CC′ represent the torsional elements.

701



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Figure 16.20.2  Sections for computing the flexural stiffness of equivalent beams (slab-​beams), Ksb, and variation of moment of inertia for two-​way slab systems. (Adapted from ACI 318-​77 Commentary.)

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Based on tests results, Corley and Jirsa [16.27] developed a formula for the torsional stiffness Kt of the torsional member as follows [ACI Commentary R8.11.5]: Kt = ∑



9 Ecs C  c  L2  1 − 2   L 

3

 I sb   I  s

(16.20.1)

2

in which C = torsional constant of the transverse beam (see Section 16.11) Ecs = modulus of elasticity of slab concrete Is = moment of inertia of slab over width of equivalent frame Isb = moment of inertia of entire T-​section (if so) within the width of the equivalent frame L2 = span of member subject to torsion c1 = width of rectangular or equivalent rectangular column, capital, or bracket measured in the direction for which moments are being computed c2 = width of rectangular or equivalent rectangular column, capital, or bracket measured in the direction perpendicular to c1 for which moments are being computed The summation sign is for the transverse spans (denoted by L2) on each side of the column. Flexural Stiffness of Column The flexural stiffness of the columns, Kc, must account for the variation of moment of inertia along the column axis (ACI-​8.11.4.2). The height of the column is to be measured from

c1

CL of panel

c2

of

Rotation more than column rotation

eq ui va l

en

tf

ra m

e

Rotation equals column rotation

A

W id th

702

A

x1 CL of panel

Slab thickness, t z

Axis of rotation

c1 t

Section A–A

Figure 16.20.3  Torsional members for equivalent columns.

Cross section of torsional member

703



703

1 6 . 2 0   E QU I VA L E N T F R A M E M E T H O D

I= ∞ I = linear variation H Constant Ic

I= ∞

Figure 16.20.4  Basis of calculation of column stiffness.

middepth of slab above to middepth of slab below (ACI-​R8.11.4), as shown in Fig. 16.20.4. From the top to the bottom of the slab-​beam at the joint (ACI-​8.11.4.1), the moment of inertia is to be taken as infinite. In flat slabs having column capitals, the moment of inertia may be assumed to vary linearly from infinite at the top to its value based on gross cross section of the column at the bottom of the capital, as shown in Fig. 16.20.4. Outside of joints or column capitals, the moment of inertia may be based on the column gross cross section (ACI-​8.11.4.3). The model shown in Fig.  16.20.1(c) consists of three element types:  the equivalent beams (slab-​beams), the torsional elements, and the columns. In the traditional equivalent frame method, the columns and the torsional elements are combined into and replaced by equivalent columns in order to allow analysis of the slab system as a plane frame. The flexibility (inverse of stiffness) of this equivalent column is taken as the sum of the flexibilities of the actual columns above and below the slab-​beam and the flexibility of the attached torsion member, such that 1 1 1 = + K ec ∑ K c K t



(16.20.2)

from which the stiffness of the equivalent column, Kec, is K ec =



∑ Kc ∑ Kc 1+ Kt

(16.20.3)

where the summation sign is for the columns above and below the slab. With stiffness properties computed for the slab-​beams and the equivalent columns, the three-​dimensional subassembly shown in Fig.  16.20.1 may then be analyzed as a plane frame as shown in Fig. 16.20.5.

Ksb

Kec

Ksb

Kec

Kec

Figure 16.20.5  Two-​dimensional equivalent frame of a two-​way slab system for gravity load analysis.

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Arrangement of Live Load When the load pattern is known, analysis should be made for it (ACI-​6.4.3.1). When service live load does not exceed 3 4 of the service dead load, analysis is permitted to be made only for full factored dead and live load on all spans (ACI-​6.4.3.2). When load patterns in accordance with influence line concepts are used, only 3 4 of the full factored live load may be used in accordance with ACI-​6.4.3.3; however, factored moments used in design should not be less than those due to full factored dead and live loads on all panels (ACI-​6.4.3).

Reduction of Negative Moments Obtained at Column Centerlines from Structural Analysis Negative moments obtained at interior column centerlines may be reduced to the face of rectilinear or equivalent square (for circular or polygonal) supports, but not greater than 0.175L1 from the column centerline (ACI-​8.11.6.1). For exterior columns having capitals or brackets, reduction of negative moments can be made only to a section no greater than halfway between the face of the column and the edge of the capital or bracket (ACI-​ 8.11.6.3). For exterior columns without brackets or capitals, the critical section for negative moments is taken at the face of the column (ACI-​8.11.6.2).

Deflections When the deflection must be calculated for a two-​way slab system, the ACI Code (ACI-​ 24.2.3.3) provides little guidance other than the requirement to take into account “size and shape of the panel, conditions of support, and nature of restraints at panel edges.” The effective moment of inertia Ie [Eq. (12.9.1)] is required to be used in such calculations. Although a number of techniques have been proposed [16.52–​16.71], adoption of the equivalent frame concept seems to have the most promise of being relatively simple to apply and giving reasonable results. This equivalent frame application has been developed by Nilson and Walters [16.53] for essentially uncracked systems and extended by Kripanarayanan and Branson [16.55] for partially cracked load ranges. Scanlon and others [16.58, 16.63, 16.65, 16.66, 16.68, and 16.70] have also treated the subject in detail.

EXAMPLE 16.20.1 Assuming that the equivalent frame method is to be applied to the two-​way slab (with beams) design of Example 16.3.1, obtain the properties necessary for the analysis of the equivalent rigid frames A, B, C, and D as shown by the notations in Fig. 16.3.5. SOLUTION (a) The cross sections and the moment of inertia values for the slab-​beams for frames A, B, C, and D are shown in Fig. 16.20.6. Note that moments of inertia are based on the gross sections as permitted by ACI-​8.11.3.3 for gravity load analysis. The variations in the moment of inertia of the slab-​beam in the long and short directions are shown in Fig. 16.20.7. For the long slab-​beam, the ratio of moment of inertia between the center and the face of the column to the moments of inertia of the rest of the span is 1.0/​(1 –​ 15/​240)2 = 1.14; it is 1.0/​(1 –​15/​300)2 = 1.11 for the short slab-​beam (ACI-​8.11.3.1). (b) Compute flexure properties of columns. The values and the variations in the moment of inertia of the column section in the long and short directions are shown in Fig. 16.20.8. (c) Compute torsional stiffness of transverse torsional members. The torsional constants C for the transverse members are taken from Example 16.11.1. These values, needed to compute the torsional stiffness Kt [Eq. (16.20.1)] in each direction, are shown in Fig. 16.20.9. (Continued)

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Example 16.20.1 (Continued) 7”

240”

120”

0.98”

21.5”

0.49”

21.5”

1

6 2 in. slab

Isb = 57,730 in.4

Isb = 66,540 in.4 Is = 5490 in.4

14”

Is = 2910 in.4

14”

Isb /Is = 19.84

Isb /Is = 12.12 (a) For frames A and B 300”

150”

6” 2.09”

17.5”

1.19”

17.5”

1

6 2 in. slab

Isb = 33,980 in.4

Isb = 39,530 in.4 Is = 6870 in.4

12”

Is = 3570 in.4

12”

Isb /Is = 9.52

Isb /Is = 5.75 (b) For frames C and D

Figure 16.20.6  Slab-​beam cross s​ ections in the two-​way slab (with beams) of Example 16.20.1.

15”

15”

6.5”

15”

28”

15”

6.5”

20’

25’

Isb (1 – c2/L2)2

Isb

= 1.14 Isb

Isb =

(1 – c2/L2)2 66,540 in 4 (frame A) 57,730 in 4 (frame B)

25.00’ (a) Long slab-beam

= 1.11 Isb

Isb = 0.625’

23.75’ 0.625’

24”

0.625’

39,530 in 4 (frame C) 33,980 in 4 (frame D) 18.75’

0.625’

20.00’ (b) Short slab-beam

Figure 16.20.7  Flexure properties of slab-​beam in the two-​way slab (with beams) of Example 16.20.1.

(Continued)

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Example 16.20.1 (Continued) I=∞ (ACI-8.11.4.1)

6.5”

24.75”

21.5”

2.06’

Bottom of slab-beam 12’ – 0”

12’ –0” 9.67’

Ic = 4220 in.4

I=∞

3.25” 0.27’ (a) Column section in long direction (for frames A and B) 6.5” 20.75”

17.5”

1.73’

I=∞

12’–0”

12’ – 0”

10.00’

3.25”

Ic = 4220 in.4

I=∞ 0.27’

(b) Column section in short direction (for frames C and D)

Figure 16.20.8  Flexure properties of columns in the two-​way slab (with beams) of Example 16.20.1.

C = 11,930 in.4

C

C = 10,700 in.4

20’

D

C = 20,700 in.4 C = 11,930 in.4

C = 10,700 in.4

20’

C = 20,700 in.4

A

B

C = 19,100 in.4

C = 19,100 in.4

25’

25’

Figure 16.20.9  Torsional constants in the two-​way slab (with beams) in Example 16.11.1.

(Continued)

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Example 16.20.1 (Continued) For Frame A, using Isb/​Is = 12.12 for 14 × 21.5 projection below 240 × 6.5 slab, and noting that Ecs = E,



exterior K t =

18 E (10, 700) (12.12) = 974 E (12.12) = 11, 800 E 240 (1 − 15 / 240)3

interior K t =

18 E (11, 930) (12.12) = 1086 E (12.12) = 13, 200 E 240 (1 − 15 / 240)3



For Frame B, using Isb/​Is = 19.84 for 14 × 21.5 projection below 127 × 6.5 slab,

exterior K t = 487 E (19.84) = 9660 E interior K t = 543E (19.84) = 10, 800 E



For Frame C, using Isb/​Is = 5.75 for 12 × 17.5 projection below 300 × 6.5 slab,



exterior K t =

18 E (19,100) (5.75) = 1340 E (5.75) = 7700 E 300 (1 − 15 / 300)3

interior K t =

18 E (20, 700) (5.75) = 1450 E (5.75) = 8340E 300 (1 − 15 / 300)3



For Frame D, using Isb/​Is = 9.52 for 12 × 17.5 projection below 156 × 6.5 slab,

exterior K t = 670 E (9.52) = 6380 E interior K t = 725E (9.52) = 6900 E



EXAMPLE 16.20.2 Assuming the equivalent frame method is to be applied to the flat slab of Example 16.3.2, obtain the properties necessary for the analysis of equivalent rigid Frame A in the long direction. SOLUTION (a) Compute flexure properties of slab strip. The moment of inertia through the 7 1 2 -​in. slab is computed as that of a rectangular section of width L2 = 240 in. Thus,

I sb =

240(7.5)3 = 8440 in.4 12

The moment of inertia through the drop, where there is a T-​section of a 240 × 7.5 in. flange and an 84 × 3 in. web, is 14,720 in.4 or 1.75Isb. The moment of inertia between the column centerline and the face of the equivalent square column capital is 1.75Isb  /​ (1 –​4.43/​20)2 = 2.88 Isb. The variation in the moment of inertia along the interior span of the slab strip is shown in Fig. 16.20.10. (b) Compute flexure properties of columns. The column length is measured between the centerlines of slab thickness, as shown in Fig. 16.20.11. The moment of inertia is assumed to be infinite from the top of the slab to the bottom of the drop panel, and then is taken to vary linearly to the base of the column capital. (Continued)

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Example 16.20.2 (Continued) (c) Compute torsional stiffness of transverse torsional members. From Example 16.11.2, C (edge beam ) = 18, 500 in.4



C (interior beam ) = 9800 in.4



For the two members, one framing in from each side, K t (edge) =

2(9 E )C  c  L2  1 − 2   L 

3

=

2

K t (interior ) =

2(9 E )(18, 500)  4.5  240  1 −   20 

= 2980 E



2(9 E )9800  4.43  240  1 −   20 

3

3

= 1560 E

18”

18”

7.5”

10.5”

1.5” 21”

1/2 width of equivalent square column capital

Idrop (1– c2/L2)2

Isb = 8440 in.4 = 2.88 Isb

Idrop = 1.75Isb

2.21’

Idrop = 1.75Isb

16.66” 1.96’

25.00”

2.88Isb

2.21’ 1.96’

Figure 16.20.10  Flexure properties of frame A in the flat slab in Example 16.20.2.

(Continued)

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Example 16.20.2 (Continued) 11 “ 2

0.188L

29.25”

18”

I= ∞ Linear variation

Ic = 5150 in.4

0.725 L

21”

10’–0”

10’–0”

21”

0.056 L

3.75”

I= ∞ 0.031 L

(a) Interior column

11 “ 2

3.75”

0.742 L

0.171 L

27.25”

19” 16”

10’– 0”

10’– 0”

19”

0.056 L

I= ∞ Linear variation

Ic = 5460 in.4

I= ∞ 0.031 L

(b) Exterior column

Figure 16.20.11  Flexure properties of columns in the flat slab of Example 16.20.2.

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16.21 EQUIVALENT FRAME MODELS As noted in the preceding section, the equivalent frame method was originally developed to be used with the moment distribution method—that is, the usual method of structural analysis for plane frames at the time. The method, however, can be used with any standard analysis software by specifying appropriate values of the stiffness properties for the slab-​beams, columns, and torsional elements. This section briefly discusses several alternative models for implementing the EFM into virtually any structural analysis software. Perhaps the most sophisticated model would be to divide the slab into a mesh of finite elements supported on columns as shown in Fig. 7.2.1. This technique has been used in some investigations [16.32] and is sometimes used in design practice as well. One of the main challenges in using this technique is arriving at a definition of the appropriate stiffness of the slab elements (and any beams) that will adequately capture the moment transfer between the slab and the column observed in experiments. When a finite element analysis may not be efficient or is too costly for a given project, a two-​dimensional model, where the slab-​to-​column moment transfer is modeled via torsional or rotational springs attached to the columns or the slab-​beams, could be used. These concepts are presented in the following sections.

Model with Torsional Springs This model is essentially that shown in Fig.  16.20.1(c) and corresponds to the original model developed for the EFM: the ends of the slab-​beam are connected to the columns with zero length elements that are rigid except in torsion. This approach may require placing a special joint at the same location where the slab-​beams are connected to the columns to insert the torsional springs. The properties of the slab-​beams, Ksb, columns, Kc, and torsional elements, Kt, are those presented earlier (see Section 16.20).

Model with Rotational Springs An alternative to using torsional springs is to use equivalent rotational springs at the ends of columns or the slab-​beams. In Fig. 16.21.1, the torsion element is replaced by rotational springs at the column ends, one joined to the lower end of the upper column and the other joined to the upper end of the lower column. In such a case, the stiffness of the original torsional element is divided proportionally according to the column stiffness at that end. Alternatively, the torsion element may be replaced by rotational springs at the slab-​ beam ends, as shown in Fig. 16.21.2. At each interior column, the torsional element may be divided into two such elements in parallel, one joined to the right end of the left beam and the other to the left end of the right beam, with the total torsional stiffness proportioned according to the beam stiffness at that end.

Figure 16.21.1  Torsion elements replaced by rotational springs at column ends, distributed to upper and lower columns in proportion to the column flexural stiffnesses (same as Kec model of Vanderbilt [16.32]).

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SELECTED REFERENCES

711

Figure 16.21.2  Torsion elements replaced by rotational springs at slab-​beam ends (same as Keb model of Vanderbilt [16.32]).

Other Models As yet another alternative approach to using torsional or rotational springs, some researchers have proposed the use of an effective slab width to define an equivalent beam. In this method, a slab width factor is computed based on the assumed stiffness of the columns and beam-​column connection regions and the expected crack regions in the slab. Studies concerning the use of an effective slab width include those of Pecknold [16.38], Allen and Darvall [16.39, 16.47], Elias [16.41–​16.43], Luo and Durrani [16.140, 16.141], Hwang and Moehle [16.145], and Dovich and Wight [16.146].

16.22 EQUIVALENT FRAME METHOD FOR LATERAL LOAD ANALYSIS As noted in Section 16.20, the equivalent frame method was developed and validated using data from tests and analyses of two-​way slab systems under gravity loads. The method, however, can also be used for lateral load analysis with few modifications. First, the analysis must be conducted for the entire frame rather than just the subassembly used for gravity loads. In addition, rather than gross section properties, the moments of inertia must be computed in accordance with ACI-​6.6.3.1 (see Section 13.8). Note that because the stiffness properties are different, two separate models are required: one for gravity load analysis and another for lateral load analysis. The results of these analyses are permitted to be combined in accordance with ACI-​8.4.1.9. Clearly, a more convenient approach would be to construct a single model for the analysis of both gravity and lateral loads. While this is possible with the available structural analysis software, the agreement between the results of the analyses and the test data can be quite sensitive to the model type and to the stiffness properties assumed in the model. Cano and Klingner [16.51] have provided a good review and comparison of various analysis procedures for two-​way slab systems under combined gravity and lateral loads. Additionally, Hwang and Moehle [16.145] have provided good insight and guidance for the analysis of slab-​column frames under lateral loads.

SELECTED REFERENCES General and Historical 16.1. J. R. Nichols. “Statical Limitations Upon the Steel Requirement in Reinforced Concrete Flat Slab Floors,” Transactions ASCE, 77, 1914, 1670–​1736. 16.2. George A. Hool and Nathan C. Johnson. Concrete Engineers Handbook. New York: McGraw-​ Hill, 1918 (pp. 457–​486). 16.3. H. M. Westergaard and W. A. Slater. “Moments and Stresses in Slabs,” ACI Proceedings, 17, 1921, 415.

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 16.4. H.  M. Westergaard. “Formulas for the Design of Rectangular Floor Slabs and Supporting Girders,” ACI Proceedings, 22, 1926, 26.   16.5. Joseph A. Wise, “Design of Reinforced Concrete Slabs,” ACI Proceedings, 25, 1929, 712.  16.6. Joseph DiStasio and M.  P. Van Buren. “Slabs Supported on Four Sides,” ACI Journal, Proceedings, 32, January–​February 1936, 350–​364.   16.7. R.  L. Bertin, Joseph DiStasio, and M.  P. Van Buren. “Slabs Supported on Four Sides,” ACI Journal, Proceedings, 41, June 1945, 537–​556.   16.8. C. P. Siess and N. M. Newmark. “Rational Analysis and Design of Two-​Way Concrete Slabs.” ACI Journal, Proceedings, 45, December 1948, 273–​316.   16.9. Sidney H. Simmonds and Janko Misic. “Design Factors for the Equivalent Frame Method,” ACI Journal, Proceedings, 68, November 1971, 825–​831. 16.10. Alex E. Cardenas, Rolf J. Lenschow, and Mete A. Sozen. “Stiffness of Reinforced Concrete Plates,” Journal of the Structural Division, ASCE, 98, ST11 (November 1972), 2587–​2603. 16.11. C. K. Wang. Intermediate Structural Analysis. New York: McGraw-​Hill, 1983.

Gravity Load—​Tests 16.12. W. L. Gamble, M. A. Sozen, and C. P. Siess. “Measured and Theoretical Bending Moments in Reinforced Concrete Floor Slabs,” Civil Engineering Structural Research Series No. 246. Urbana: University of Illinois, June 1962. 16.13. M. A. Sozen and C. P. Siess. “Investigation of Multi-​Panel Reinforced Concrete Floor Slabs,” ACI Journal, Proceedings, 60, August 1963, 999–​1028. 16.14. S.  A. Guralnick and R.  W. LaFraugh. “Laboratory Study of a 45-​Foot Square Flat Plate Structure,” ACI Journal, Proceedings, 60, September 1963, 1107–​1185. 16.15. David S. Hatcher, Mete A. Sozen, and Chester P. Siess. “Test of a Reinforced Concrete Flat Plate,” Journal of the Structural Division, ASCE, 91, ST5 (October 1965), 205–​231. 16.16. James O.  Jirsa, Mete A.  Sozen, and Chester P.  Siess. “Test of a Flat Slab Reinforced with Welded Wire Fabric,” Journal of the Structural Division, ASCE, 92, ST3 (June 1966), 199–​224. 16.17. W. L. Gamble, M. A. Sozen, and C. P. Siess. “Tests of a Two-​Way Reinforced Concrete Floor Slab,” Journal of the Structural Division, ASCE, 95, ST6 (June 1969), 1073–​1096. 16.18 M. Daniel Vanderbilt, Mete A. Sozen, and Chester P. Siess. “Tests of a Modified Reinforced Concrete Two-​Way Slab,” Journal of the Structural Division, ASCE, 95, ST6 (June 1969), 1097–​1116. 16.19. D. S. Hatcher, Mete A. Sozen, and Chester P. Siess. “Test of a Reinforced Concrete Flat Slab,” Journal of the Structural Division, ASCE, 95, ST6 (June 1969), 1051–​1072. 16.20. E. Ramzy, F. Zaghlool, H. A. Rawdon de Paiva, and Peter G. Glockner. “Tests of Reinforced Concrete Flat Plate Floors,” Journal of the Structural Division, ASCE, 96, ST3 (March 1970), 487–​507. 16.21. Alex E.  Cardenas and Paul H.  Kaar. “Field Test of a Flat Plate Structure,” ACI Journal, Proceedings, 68, January 1971, 50–​ 58. 16.22. Donald D.  Magura and W.  Gene Corley. “Tests to Destruction of a Multipanel Waffle Slab Structure—​1964–​1965 New  York World’s Fair,” ACI Journal, Proceedings, 68, September 1971, 699–​703. 16.23. A.  S. Hall and B.  V. Rangan. “Moments in Edge Panels of Flat Plate Floors,” Journal of Structural Engineering, ASCE, 109, 11 (November 1983), 2638–​2650. 16.24. Abdel Wahid Hago and Prabhakara Bhatt. “Tests on Reinforced Concrete Slabs Designed by Direct Design Procedure,” ACI Journal, Proceedings, 83, November–​December 1986, 916–​924. 16.25. B. Vijaya Rangan and A. S. Hall. “Moment Redistribution in Flat Plate Floors,” ACI Journal, Proceedings, 81, November–​December 1984, 601–​608.

Gravity Load—​Analytical Methods 16.26. W. G. Corley, M. A. Sozen, and C. P. Siess. “The Equivalent Frame Analysis for Reinforced Concrete Slabs,” Structural Research Series No. 218. Urbana:  University of Illinois, Civil Engineering Department, June 1961. 16.27. W.  G. Corley and J.  O. Jirsa. “Equivalent Frame Analysis for Slab Design,” ACI Journal, Proceedings, 67, November 1970, 875–​884.

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713

16.28. Maurice P. Van Buren. “Staggered Columns in Flat Plates,” Journal of the Structural Division, ASCE, 97, ST6 (June 1971), 1791–​1797. 16.29. William L.  Gamble. “Moments in Beam Supported Slabs,” ACI Journal, Proceedings, 69, March 1972, 149–​157. 16.30. Donald J.  Fraser. “Equivalent Frame Method of Beam-​ Slab Structures,” ACI Journal, Proceedings, 74, May 1977, 223–​ 228. 16.31. S. K. Sharan, D. Clyde, and D. Turcke. “Equivalent Frame Analysis Improvements for Slab Design,” ACI Journal, Proceedings, 75, February 1978, 55–​59. 16.32. M. Daniel Vanderbilt. Equivalent Frame Analysis of Unbraced Reinforced Concrete Buildings for Static Lateral Loads. Structural Research Report No. 36, Civil Engineering Department, Colorado State University, Fort Collins, June 1981. 16.33. D.  J. Fraser. “The Equivalent Frame Method Simplified for Beam and Slab Construction,” Concrete International, 4, April 1982, 66–​73. 16.34. J. F. Mulcahy and J. M. Rotter. “Moment Rotation Characteristics of Flat Plate and Column Systems,” ACI Journal, Proceedings, 80, March–​April 1983, 85–​92. 16.35. D. J. Fraser. “Simplified Frame Analysis for Flat Plate Construction,” Concrete International, 6, September 1984, 32–​41.

Pattern Gravity Loading 16.36. J.  O. Jirsa, M.  A. Sozen, and C.  P. Siess. “Pattern Loadings on Reinforced Concrete Floor Slabs,” Journal of the Structural Division, ASCE, 95, ST6 (June 1969), 1117–​1137. 16.37. Jan C. Jofriet and Gregory M. McNeice. “Pattern Loading on Reinforced Concrete Flat Plates,” ACI Journal, Proceedings, 68, December 1971, 968–​972.

Lateral Load Analysis on Slab Frames 16.38. David A.  Pecknold. “Slab Effective Width for Equivalent Frame Analysis,” ACI Journal, Proceedings, 72, April 1975, 135–​137. 16.39. Fred Allen and Peter Darvall. “Lateral Load Equivalent Frame,” ACI Journal, Proceedings, 74, July 1977, 294–​ 299. 16.40. M. Daniel Vanderbilt. “Equivalent Frame Analysis for Lateral Loads.” Journal of the Structural Division, ASCE, 105, ST10 (October 1979), 1981–​1998. Disc., 106, ST7 (July 1980), 1671–​ 1672; 107, ST1 (January 1981), 245. 16.41. Ziad M.  Elias. “Sidesway Analysis of Flat Plate Structures,” ACI Journal, Proceedings, 76, March 1979, 421–​442. 16.42. Ziad M. Elias and Constantinos Georgiadis. “Flat Slab Analysis Using Equivalent Beams,” ACI Journal, Proceedings, 76, October 1979, 1063–​1078. 16.43. Ziad M.  Elias. “Lateral Stiffness of Flat Plate Structures,” ACI Journal, Proceedings, 80, January–​February 1983, 50–​54. 16.44. Donald J.  Fraser. “Elastic Analysis of Laterally Loaded Frames,” Journal of Structural Engineering, ASCE, 109, 6 (June 1983), 1479–​1489. 16.45. M. Daniel Vanderbilt and W. Gene Corley. “Frame Analysis of Concrete Buildings,” Concrete International, 5, December 1983, 33–​43. 16.46. I. Paul Lew and Fruma Narov. “Three-​Dimensional Equivalent Frame Analysis of Shearwalls,” Concrete International, 5, October 1983, 25–​30. 16.47. Peter Darvall and Fred Allen. “Lateral Load Effective Width of Flat Plates with Drop Panels,” ACI Journal, Proceedings, 81, November–​December 1984, 613–​617. 16.48. Milija N. Pavlovic and Steven M. Poulton. “On the Computation of Slab Effective Widths,” Journal of Structural Engineering, ASCE, 111, 2 (February 1985), 363–​377. 16.49. Jack P. Moehle and John W. Diebold. “Lateral Load Response of Flat-​Plate Frame,” Journal of Structural Engineering, ASCE, 111, 10 (October 1985), 2149–​2164. 16.50. Cheng-​Tzu Thomas Hsu. “Lateral Displacement for Unbraced Concrete Frame Buildings,” ACI Journal, Proceedings, 82, November–​December 1985, 853–​862. 16.51. Mary Theresa Cano and Richard E. Klingner. “Comparison of Analysis Procedures for Two-​ Way Slabs,” ACI Structural Journal, 85, November–​December 1988, 597–​608. Disc. 86, September–​October 1989, 624–​626.

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Two-​Way Slab Deflections 16.52. Mortimer D. Vanderbilt, Mete A. Sozen, and Chester P. Siess. “Deflections of Multiple-​Panel Reinforced Concrete Floor Slabs,” Journal of the Structural Division, ASCE, 91, ST4 (August 1965), 77–​101. 16.53. Arthur H. Nilson and Donald B. Walters, Jr. “Deflection of Two-​Way Floor Systems by the Equivalent Frame Method,” ACI Journal, Proceedings, 72, May 1975, 210–​218. 16.54. B. Vijaya Rangan. “Prediction of Long-​Term Deflections of Flat Plates and Slabs,” ACI Journal, Proceedings, 73, April 1976, 223–​226. 16.55. K. M. Kripanarayanan and D. E. Branson. “Short Time Deflections of Flat Plates, Flat Slabs, and Two-​Way Slabs,” ACI Journal, Proceedings, 73, December 1976, 686–​690. 16.56. P. J. Taylor and J. L. Heiman. “Long-​Term Deflection of Reinforced Concrete Flat Slabs and Plates,” ACI Journal, Proceedings, 74, November 1977, 556–​561. 16.57. B.  Vijaya Rangan and Arthur E.  McMullen. “A Rational Approach to Control of Slab Deflections,” ACI Journal, Proceedings, 75, June 1978, 256–​262. 16.58. Andrew Scanlon and David W. Murray. “Practical Calculation of Two-​Way Slab Deflections,” Concrete International, 4, November 1982, 43–​50. 16.59. John A.  Sbarounis. “Multistory Flat Plate Buildings:  Construction Loads and Immediate Deflections,” Concrete International, 6, February 1984, 70–​77. 16.60. John A. Sbarounis. “Multistory Flat Plate Buildings: Effect of Construction Loads on Long-​ Term Deflections,” Concrete International, 6, April 1984, 62–​70. Disc., 7, February 1985, 62. 16.61. John A.  Sbarounis. “Multistory Flat Plate Buildings:  Measured and Computed One-​Year Deflections,” Concrete International, 6, August 1984, 31–​35. 16.62. R. I. Gilbert. “Deflection Control of Slabs Using Allowable Span to Depth Ratios,” ACI Journal, Proceedings, 82, January–​February 1985, 67–​72. 16.63. Cameron J. Graham and Andrew Scanlon. “Deflections of Concrete Slabs Under Construction Loading,” Deflections of Concrete Structures (SP-​ 86). Detroit:  American Concrete Institute, 1985. 16.64. B.Vijaya Rangan. “Estimation of Slab Deflections in Flat Plate Buildings,” ACI Journal, Proceedings, 83, March–​April 1986, 269–​273. 16.65. K. S. Stephen Tam and Andrew Scanlon. “Deflection of Two-​Way Slabs Subjected to Restrained Volume Change and Transverse Loads,” ACI Journal, Proceedings, 83, September–​October 1986, 737–​744. 16.66. Cameron J. Graham and Andrew Scanlon. “Long-​Time Multipliers for Estimating Two-​Way Slab Deflections,” ACI Journal, Proceedings, 83, November–​December 1986, 899–​ 908. 16.67. N. J. Gardner and H. C. Fu. “Effects of High Construction Loads on the Long-​Term Deflections of Flat Slabs,” ACI Structural Journal, 84, July–​August 1987, 349–​360. Disc., 85, May–​June 1988, 359. 16.68. David P. Thompson and Andrew Scanlon. “Minimum Thickness Requirements for Control of Two-​Way Slab Deflections,” ACI Structural Journal, 85, January–​February, 1988, 12–​22. 16.69. Amin Ghali. “Prediction of Deflections of Two-​Way Floor Systems,” ACI Structural Journal, 86, September–​October 1989, 551–​562. 16.70. Andrew Scanlon and David P. Thompson. “Evaluation of ACI 318 Requirements for Control of Two-​Way Slab Deflections,” ACI Structural Journal, 87, November–​December 1990, 657–​661. 16.71. ACI Committee 435. “State-​of-​the-​Art Report on Two-​Way Slab Deflections,” ACI Structural Journal, 88, July–​August 1991, 501–​514.

Crack Control in Two-​Way Slab Systems 16.72. Edward G. Nawy. “Crack Width Control in Welded Fabric Reinforced Centrally Loaded Two-​ Way Concrete Slabs,” Causes, Mechanism, and Control of Cracking in Concrete (SP-​20). Detroit: American Concrete Institute, 1968 (pp. 211–​ 235). 16.73. Edward G. Nawy and G. S. Orenstein. “Crack Width Control in Reinforced Concrete Two-​Way Slabs,” Journal of the Structural Division, ASCE, 96, ST3 (March 1970), 701–​ 721. 16.74. Edward G.  Nawy and Kenneth W.  Blair. “Further Studies on Flexural Crack Control in Structural Slab Systems,” Cracking, Deflection, and Ultimate Load of Concrete Slab Systems (SP-​30). Detroit: American Concrete Institute, 1971 (pp. 1–​41). 16.75. Edward G.  Nawy. “Crack Control Through Reinforcement Distribution in Two-​Way Acting Slabs and Plates,” ACI Journal, Proceedings, 69, April 1972, 217–​219.

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Shear Strength 16.76. Johannes Moe. Shearing Strength of Reinforced Concrete Slabs and Footings Under Concentrated Loads, Development Department Bulletin D47. Chicago:  Portland Cement Association, April 1961, 130 pp. 16.77. ACI-​ ASCE Committee 326. “Report on Shear and Diagonal Tension,” ACI Journal, Proceedings, 59, January, February, and March, 1962, 1–​30, 277–​344, and 352–​396. 16.78. Neil M. Hawkins, H. B. Fallsen, and R. C. Hinojosa. “Influence of Column Rectangularity on the Behavior of Flat Plate Structures,” Cracking, Deflection, and Ultimate Load of Concrete Slab Systems (SP-​30). Detroit: American Concrete Institute, 1971 (pp. 127–​146). 16.79. Frederic Roll, S.  T. H.  Zaidi, Gajanan Sabnis, and Kuang Chuang. “Shear Resistance of Perforated Reinforced Concrete Slabs,” Cracking, Deflection, and Ultimate Load of Concrete Slab Systems (SP-​30). Detroit: American Concrete Institute, 1971 (pp. 77–​100). 16.80. M. Daniel Vanderbilt. “Shear Strength of Continuous Plates,” Journal of the Structural Division, ASCE, 98, ST5 (May 1972), 961–​973. 16.81. M.  E. Criswell and N.  M. Hawkins. “Shear Strength of Slabs:  Basic Principles and Their Relation to Current Methods of Analysis,” Shear in Reinforced Concrete, Vol. 2 (SP-​42). Detroit: American Concrete Institute, 1974 (pp. 641–​676). 16.82. N.  M. Hawkins, M.  E. Criswell, and F.  Roll. “Shear Strength of Slabs Without Shear Reinforcement,” Shear in Reinforced Concrete, Vol. 2 (SP-​42). Detroit:  American Concrete Institute, 1974 (pp. 677–​720). 16.83. Neil M. Hawkins, (Chairman) “The Shear Strength of Reinforced Concrete Members—​Slabs, by the Joint ASCE-​ACI Task Committee 426 on Shear and Diagonal Tension of the Committee on Masonry and Reinforced Concrete of the Structural Division,” Journal of the Structural Division, ASCE, 100, ST8 (August 1974), 1543–​1591. 16.84. Brian E.  Hewitt and Barrington de V.  Batchelor. “Punching Shear Strength of Restrained Slabs,” Journal of the Structural Division, ASCE, 101, ST9 (September 1975), 1837–​1853. 16.85. Timothy J. Ross and Helmut Krawinkler. “Impulsive Direct Shear Failure in RC Slabs,” Journal of Structural Engineering, ASCE, 111, 8 (August 1985), 1661–​1677. 16.86. Fernando Gonzalez-​Vidosa, Michael D.  Kotsovos, and Milija N.  Pavlovic. “Symmetrical Punching of Reinforced Concrete Slabs: An Analytical Investigation Based on Nonlinear Finite Element Modeling,” ACI Structural Journal, 85, May–​June 1988, 241–​250. 16.87. Ibrahim A. E. M. Shehata and Paul E. Regan. “Punching in RC Slabs,” Journal of Structural Engineering, ASCE, 115, 7 (July 1989), 1726–​1740. 16.88. Lionello Bortolotti. “Punching Shear Strength in Concrete Slabs,” ACI Structural Journal, 87, March–​April 1990, 208–​ 219. 16.89. Carl Erik Broms. “Punching of Flat Plates—​A Question of Concrete Properties in Biaxial Compression and Size Effect,” ACI Structural Journal, 87, May–​June 1990, 292–​304. 16.90. David I. McLean, Long T. Phan, H. S. Lew, and Richard N. White. “Punching Shear Behavior of Lightweight Concrete Slabs and Shells,” ACI Structural Journal, 87, July–​August 1990, 386–​392. Disc., 88, May–​June 1991, 382–​383. 16.91. John S.  Lovrovich and David I.  McLean. “Punching Shear Behavior of Slabs with Varying Span–​Depth Ratios,” ACI Structural Journal, 87, September–​October 1990, 507–​511. Disc., 87, July–​August 1991, 515–​516. 16.92. J. S. Kuang and C. T. Morley. “Punching Shear Behavior of Restrained Reinforced Concrete Slabs,” ACI Structural Journal, 89, January–​February 1992, 13–​19.

Shear Reinforcement in Slabs 16.93. W.  G. Corley and N.  M. Hawkins. “Shearhead Reinforcement for Slabs,” ACI Journal, Proceedings, 65, October 1968, 811–​824. 16.94. N.  M. Hawkins. “Shear Strength of Slabs with Shear Reinforcement,” Shear in Reinforced Concrete, Vol. 2 (SP-​42). Detroit: American Concrete Institute, 1974 (pp. 785–​816). 16.95. Paul H. Langohr, Amin Ghali, and Walter H. Dilger. “Special Shear Reinforcement for Concrete Flat Plates,” ACI Journal, Proceedings, 73, March 1976, 141–​146. 16.96. Amin Ghali, Mahmoud Z. Elmasri, and Walter Dilger. “Punching of Flat Plates under Static and Dynamic Horizontal Forces,” ACI Journal, Proceedings, 73, October 1976, 566–​576. 16.97. Walter Dilger, Mahmoud Z.  Elmasri, and Amin Ghali. “Flat Plates with Special Shear Reinforcement Subjected to Static Dynamic Moment Transfer,” ACI Journal, Proceedings, 75, October 1978, 543–​549.

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  16.98. Frieder Seible, Amin Ghali, and Walter H. Dilger. “Preassembled Shear Reinforcing Units for Flat Plates,” ACI Journal, Proceedings, 77, January–​February 1980, 28–​35.   16.99. Walter H. Dilger and Amin Ghali. “Shear Reinforcement for Concrete Slabs.” Journal of the Structural Division, ASCE, 107, ST12 (December 1981), 2403–​2420. 16.100. Adel A.  Elgabry and Amin Ghali. “Design of Stud-​Shear Reinforcement for Slabs,” ACI Structural Journal, 87, May–​June 1990, 350–​361. 16.101. ACI-​ASCE Committee 421. “Guide to Shear Reinforcement for Slabs,” ACI Manual of Concrete Practice, ACI 421.1R-​08. Farmington Hills, MI: American Concrete Institute, 2008 (27 pp.). 16.102. Tetsuya, Yamada, Antonio Nanni, and Katsushiko Endo. “Punching Shear Resistance of Flat Slabs: Influence of Reinforcement Type and Ratio,” ACI Structural Journal, 89, September–​ October 1992, 555–​563.

Shear-​Moment Transfer 16.103. Joseph DiStasio and M. P. Van Buren. “Transfer of Bending Moment between Flat Plate Floor and Column,” ACI Journal, Proceedings, 57, September 1960, 299–​314. 16.104. N. W. Hanson and J. M. Hanson. “Shear and Moment Transfer between Concrete Slabs and Columns,” Journal of the PCA Research and Development Laboratories, 10, (1)  January 1968, 2–​16. 16.105. Neil M. Hawkins and W. Gene Corley. “Transfer of Unbalanced Moment and Shear from Flat Plates to Columns,” Cracking, Deflection, and Ultimate Load of Concrete Slab Systems (SP-​ 30). Detroit: American Institute, 1971 (pp. 147–​176). 16.106. Adrian E. Long. “Punching Failure of Slabs—​Transfer of Moment and Shear,” Journal of the Structural Division, ASCE, 99, ST4 (April 1973), 665–​685. 16.107. E. Ramzy F. Zaghlool and H. A. Rawdon de Paiva. “Strength Analysis of Corner Column-​ Slab Connections,” Journal of the Structural Division, ASCE, 99, ST1 (January 1973), 53–​70. 16.108. E. Ramzy F. Zaghlool and H. A. Rawdon De Paiva. “Tests of Flat-​Plate Corner Column-​Slab Connections,” Journal of the Structural Division, ASCE, 99, ST3 (March 1973), 551–​572. 16.109. N.  M. Hawkins. “Shear Strength of Slabs with Moments Transferred to Columns,” Shear in Reinforced Concrete, Vol. 2 (SP-​ 42). Detroit:  American Concrete Institute, 1974 (pp. 817–​846). 16.110. N. M. Hawkins and W. G. Corley. “Moment Transfer to Columns in Slabs with Shearhead Reinforcement,” Shear in Reinforced Concrete, Vol. 2 (SP-​42). Detroit: American Concrete Institute, 1974 (pp. 847–​880). 16.111. Shafiqul Islam and Robert Park. “Tests on Slab–​ Column Connections with Shear and Unbalanced Flexure,” Journal of the Structural Division, ASCE, 102, ST3 (March 1976), 549–​568. 16.112. Robert Park and Shafiqul Islam. “Strength of Slab–​Column Connections with Shear and Unbalanced Flexure,” Journal of the Structural Division, ASCE, 102, ST9 (September 1976), 1879–​1901. 16.113. Hans Gesund and Harinatha B. Goli. “Local Flexural Strength of Slabs at Interior Columns,” Journal of the Structural Division, ASCE, 106, ST5 (May 1980), 1063–​1078. 16.114. V. W. Neth, H. A. R. de Paiva, and A. E. Long. “Behavior of Models of a Reinforced Concrete Flat Plate Edge-​Column Connection,” ACI Journal, Proceedings, 78, July–​August 1981, 269–​275. 16.115. S.  Unnikrishna Pillai, Wayne Kirk, and Leonard Scavuzzo. “Shear Reinforcement at Slab–​Column Connections in a Reinforced Concrete Flat Plate Structure,” ACI Journal, Proceedings, 79, January–​February 1982, 36–​42. 16.116. Harinatha B. Goli and Hans Gesund. “Flexural Strength of Flat Slabs at Exterior Columns,” Journal of the Structural Division, ASCE, 108, ST11 (November 1982), 2479–​ 2495. 16.117. B.  Vijaya Rangan and A.  S. Hall, “Moment and Shear Transfer Between Slab and Edge Column,” ACI Journal, Proceedings, 80, May–​June 1983, 183–​191. 16.118. Denby G. Morrison, Ikuo Hirasawa, and Mete A. Sozen. “Lateral-​Load Tests of R/​C Slab–​ Column Connections,” Journal of Structural Engineering, ASCE, 109, 11 (November 1983), 2698–​2714. 16.119. Denby G.  Morrison. “Dynamic Lateral-​ Load Tests of R/​ C Column–​ Slabs,” Journal of Structural Engineering, ASCE, 111, 3 (March 1985), 685–​698. 16.120. Paul F. Rice and Edward S. Hoffman. “Shear-​Moment Transfer,” Concrete International, 9, January 1987, 31–​35.

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16.121. P. R. Walker and P. E. Regan. “Corner Column–​Slab Connections in Concrete Flat Plates,” Journal of Structural Engineering, ASCE, 113, 4 (April 1987), 704–​720. 16.122. Scott D.  B. Alexander and Sidney H.  Simmonds. “Ultimate Strength of Slab–​ Column Connections,” ACI Structural Journal, 84, May–​June 1987, 255–​261. Disc., 85, March–​April 1988, 226–​232. 16.123. Sidney H.  Simmonds and Scott D.  B. Alexander. “Truss Model for Edge Column–​Slab Connections,” ACI Structural Journal, 84, July–​August 1987, 296–​303. Disc., 85, May–​June 1988, 352–​353. 16.124. Ahmad J. Durrani and Hikmat E. Zerbe. “Seismic Resistance of R/​C Exterior Connections with Floor Slab,” Journal of Structural Engineering, ASCE, 113, 8 (August 1987), 1850–​1864. 16.125. Adel A. Elgabry and Amin Ghali. “Tests on Concrete Slab–​Column Connections with Stud-​ Shear Reinforcement Subjected to Shear-​Moment Transfer,” ACI Structural Journal, 84, September–​October 1987, 433–​442. 16.126. Jack P. Moehle. “Strength of Slab–​Column Edge Connections,” ACI Structural Journal, 85, January–​February 1988, 89–​98. Disc., 85, November–​December 1988, 703–​709. 16.127. Jack P. Moehle, Michael E. Kreger, and Roberto Leon. “Background to Recommendations for Design of Reinforced Concrete Slab–​Column Connections,” ACI Structural Journal, 85, November–​December 1988, 636–​644. 16.128. ACI-​ASCE Committee 352. “Recommendations for Design of Slab–​Column Connections in Monolithic Reinforced Concrete Structures,” ACI Structural Journal, 85, November–​ December 1988, 675–​696. Disc., 86, July–​August 1989, 496–​499. 16.129. Amin Ghali. Discussion of “Recommendations for Design of Slab–​Column Connections in Monolithic Reinforced Concrete Structures,” ACI-​ASCE Committee 352 Report (ACI Structural Journal, 85, November–​December 1988, 675–​696), ACI Structural Journal, 86, July–​August 1989, 496–​499. 16.130. Austin Pan and Jack P. Moehle. “Lateral Displacement Ductility of Reinforced Concrete Flat Plates,” ACI Structural Journal, 86, May–​June 1989, 250–​258. 16.131. Neil M. Hawkins, Aibin Bao, and Jun Yamazaki. “Moment Transfer from Concrete Slabs to Columns,” ACI Structural Journal, 86, November–​December 1989, 705–​716. 16.132. B.  Vijaya Rangan. “Punching Shear Design in the New Australian Standard for Concrete Structures,” ACI Structural Journal, 87, March–​April 1990, 140–​144. Disc., 88, January–​ February 1991, 115. 16.133. Douglas A.  Foutch, William L.  Gamble, and Harianto Sunidja. “Tests of Post-​Tensioned Concrete Slab–​Edge Column Connections,” ACI Structural Journal, 87, March–​April 1990, 167–​179. Disc., 88, January–​February 1991, 116–​117. 16.134. Peter Marti. “Design of Concrete Slabs for Transverse Shear,” ACI Structural Journal, 87, March–​April 1990, 180–​190. Disc., 88, January–​February 1991, 117–​118. 16.135. Ian N. Robertson and Ahmad J. Durrani. “Gravity Load Effect on Seismic Behavior of Interior Slab–​Column Connections,” ACI Structural Journal, 89, January–​February 1992, 37–​45. 16.136. Scott D. B. Alexander and Sidney H. Simmonds. “Tests of Column—​Flat Plate Connections,” ACI Structural Journal, 89, September–​October 1992, 499–​502. 16.137. Adel A. Elgabry and Amin Ghali. “Transfer of Moments between Columns and Slabs: Proposed Code Revisions,” ACI Structural Journal, 93, January–​February 1996, 56–​61. 16.138. Homayoun H. Abrishami, William D. Cook, and Denis Mitchell. “The Effect of Epoxy-​Coated Reinforcement and Concrete Quality on Cracking of Flat Plate Slab–​Column Connections,” ACI Materials Journal, 93, March–​April 1996, 121–​128. 16.139. ACI Committee 340. Supplement to Design Handbook Volume 1, Design of Two-​ Way Slabs, in Accordance with the Strength Design Method of ACI 318–​83. [SP-​17(84)(S)]. Detroit: American Concrete Institute, 1985.

Other References 16.140. Y. H. Luo and A. J. Durrani. “Equivalent Beam Model for Flat-​Slab Buildings—​Part I,” ACI Structural Journal, 92, January–​February 1995, 115–​124. 16.141. Y. H. Luo and A. J. Durrani. “Equivalent Beam Model for Flat-​Slab Buildings—​Part II,” ACI Structural Journal, 92, March–​April 1995, 250–​257. 16.142. N. J. Gardner and Xiao-​yun Shao. “Punching Shear of Continuous Flat Reinforced Concrete Slabs,” ACI Structural Journal, 93, March–​April 1996, 218–​228. 16.143. Adel A. Elgabry and Amin Ghali. “Moment Transfer by Shear in Slab–​Column Connections,” ACI Structural Journal, 93, March–​April 1996, 187–​196. 16.144. Amin Ghali. “An Efficient Solution to Punching of Slabs,” Concrete International, 11, June 1989, 50–​54.

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16.145. Shyh-​Jiann Hwang and Jack P. Moehle. “Models for Laterally Loaded Slab-​Column Frames,” ACI Structural Journal, 97, March–​April 2000, 345–​353. 16.146. Laurel M. Dovich and James K. Wight. “Effective Slab Width Model for Seismic Analysis of Flat Slab Frames,” ACI Structural Journal, 102, November–​December 2005, 868–​875. 16.147. AISC. Manual of Steel Construction (14th ed.). Chicago:  American Institute of Steel Construction, 2016. 16.148. T. X. Dam, J. K. Wight, and G. J. Parra-​Montesinos, “Behavior of Monotonically Loaded Slab-​Column Connections Reinforced with Shear Studs,” ACI Structural Journal, Vol. 114, No. 1, 2017, pp. 221–​232.

PROBLEMS Two-​Way Slabs (With Beams) 16.1 Design the typical interior frame along columns 2-​5-​7 for the two-​way slab system shown in the figure for Problem 16.1. The 13-​ft-​long columns are connected by beams, and no column capitals or drop panels are used. As an initial trial, assume that all beams (interior) are 12 × 24 in. overall. Revise beam size as necessary during the design. Determine slab thickness based on ACI-​8.3.1; then use the Direct Design Method for longitudinal distribution of moments. Show a design sketch giving all your decisions, including dimensions, bar sizes, and stirrups for the two spans from column 2 to column 7. The live load is 150 psf, fc′ = 4000 psi, and fy = 60,000 psi.

Parapet extends from columns 1 to 10 and from columns 1 to 3 10”

10 27’– 0”

All columns 24 × 24 6

27’– 0”

16.2 Compute the moments for the column and middle strips, as well as the moments carried by the beams for the interior frame of Problem 16.1, except that a 12-​in. wall exists at the lower-​story level and contains the 24-​in.-​square columns at locations 1, 2, 3, 4, 6, and 10. 16.3 Compute the moments for the column and middle strips, as well as the moments carried by the beams for the typical interior frame along column lines 4-​5-​8 for the two-​way slab system of Problem 16.1. 16.4 Compute the moments for the column and middle strips, as well as the moments carried by the beams for the exterior half-​ frame along column lines 1-​2-​3 for the two-​way slab system of Problem 16.1.

A

7 A

B

27’–0”

4

9

D 5

A 1

2’– 0

Columns Section A–A above and below 8 Crosshatch indicates column only below the slab

C 2

23’– 0”

Problems 16.1 to 16.4 

3 23’– 0”

23’– 0”

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719

PROBLEMS

Flat Slabs 16.5 In the flat slab shown in the figure for Problem 16.5, the columns are 24 in. square with columns 1 through 6 existing only below the floor slab, while columns 7 through 9 exist both above and below the floor slab. All columns are 13 ft long center-​to-​center of floor slabs. The live load is 150 psf, fc′ = 4000 psi , and fy = 60,000 psi. Use rectangular column capitals and drop panels to design the flat slab. (a)  Determine the slab thickness based on ACI-​8.3.1. (b) Use the Direct Design Method for longitudinal distribution of moments in interior equivalent frame defined by columns 2, 5, and 7 along its centerline. (c) Determine transverse distribution and select reinforcement for the column strip (defined by columns 2, 5, and 7)  and adjacent half middle strips.





16.6 Investigate the moment and shear transfer at the exterior support of a flat slab structure as detailed in the figure for Problem 16.6. The exterior support has a 5-​ft flat-​sided column capital

on a 24-​in. square column, along with a 7 ft 8 in. width of drop panel that is 11 in. thick. The slab is 8 1 2 in. thick. Assume that there is no edge beam or wall at the exterior support location. The factored moment Mu to be transferred is 340 ft-​kips and the factored shear Vu is 115 kips. The negative moment reinforcement provided in the column strip is #5 at 10-​in. spacing. Use fc′ = 4000 psi and fy = 60,000 psi. 16.7 Rework Example  16.10.2, assuming that the service live load is 200 psf instead of 140 psf. 16.8 Rework Example  16.13.2, assuming that the service live load is 200 psf instead of 140 psf.

Flat Plates 16.9 Design a typical interior rigid frame (long direction) of a flat plate floor using the data for the design example of Section 16.3, assuming that the service live load is 120 psf and the columns are 16 × 14 in. 16.10 Assuming that the slab thickness is 6 1 2 in., investigate the moment and shear transfer in the flat plate design example (see Examples 16.17.1 and 16.18.1) for a service live load of 150 psf instead of 72 psf.

Columns above and below 7

9

27’–0” A

A

5

3’–6” 2’–0” 11”

8 7’–8”

27’–0”

4

4’– 10”

Live load = 150 psf fc’ = 4000 psi fy = 60,000 psi All columns: 24” square 13’– 0” long

Crosshatch indicates column below slab only

5’–0”

6

Section A–A 1

3

2 23’– 0”

Problem 16.5 

Section A–A

23’–0”

A

Problem 16.6 

A

8 2” 1

CHAPTER 17 YIELD LINE THEORY OF SLABS

17.1 INTRODUCTION Reinforced concrete design methods under the present ACI Code are based on the results of an elastic analysis of the structure as a whole, when subjected to the action of factored loads, such as 1.2D + 1.6L, where D and L refer to service dead and live loads. Actually, the behavior of a statically indeterminate structure is such that inelastic behavior develops at one or more regions along the structural members and, consequently, the results of an elastic analysis are no longer valid. These regions of concentrated inelastic behavior are commonly called “plastic hinges.” If sufficient ductility exists, redistribution of bending moments will occur until a sufficient number of plastic hinges has formed to change the structure into a mechanism, at which time the structure collapses or fails. It is then said that the structure has reached a “collapse mechanism.” The term “ultimate load analysis,” as opposed to “elastic analysis,” relates to the use of the bending moment diagram at the verge of collapse as the basis for design. Other than the provisions for redistribution of moments at the supports of continuous flexural members (ACI-​6.6.5), the present ACI Code has no explicit allowance for ultimate load analysis. The redistribution as described in ACI-​6.6.5 has been presented and illustrated in Section 9.12. The design and analysis of two-​way slab systems was treated in Chapter 16. The factored moments are based on the elastic analysis of an equivalent frame, which has been devised as a simple substitute for the elastic analysis of a three-​dimensional system. The chief concern of this chapter is to develop the yield line theory for two-​way slabs. Although not included in the ACI Code, slab analysis by yield line theory may be useful in providing the needed information for understanding the behavior of irregular or single-​ panel slabs with various boundary conditions.

17.2 GENERAL CONCEPT Although the study of flexural behavior of plates up to the ultimate load may date back to the 1920s [17.1], the fundamental concept of the yield line theory for the ultimate load design of slabs was expanded considerably by K. W. Johansen [17.2, 17.3]. In this theory the strength of a slab is assumed to be governed by flexure alone; other effects such as shear and deflection are to be separately considered. At collapse, it is assumed that the steel reinforcement has fully yielded and that the bending and twisting moments are uniformly distributed along the yield lines.

721



721

17.2  GENERAL CONCEPT

Flat plate building, Chicago. (Photo by Gustavo J. Parra-Montesinos.)

Yield line theory for one-​way slabs is not much different from the limit analysis of continuous beams. On a continuous beam, achievement of flexural strength at one location—​ say, in the negative moment region over a support—​does not necessarily mean that the ultimate load on the beam has been reached. If the section, having reached its flexural strength, can continue to provide a constant resistance while undergoing further rotation, then the flexural strength may be reached at additional locations. Complete failure theoretically cannot occur until yielding has occurred at several locations (or along several parallel lines in the case of one-​way slabs), so that a mechanism forms, giving a condition of unstable equilibrium. Consider, for example, the one-​way slab of finite width shown in Fig. 17.2.1. A uniform loading on the slab will cause uniform maximum negative bending moment along AB and EF and uniform positive bending moment along CD, which is parallel to the supports. When the uniform load is increased until the moments along AB, CD, and EF reach their respective ultimate moment capacities, rotation of the slab segments will occur, with

A

C

Free edge

E

Yield lines

B

D

AB

Free edge

F EF

CD

Figure 17.2.1  Collapse mechanism of a one-​way slab.

72

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yield lines AB, CD, and EF acting as axes of rotation. Assuming an elastic–​plastic moment–​ curvature relationship (see Fig. 9.12.1), angle change can occur without an increase in the resisting moment. Thus, under the limiting condition with the slab segments able to rotate with no change in resisting moment, the slab system is geometrically unstable. This condition is known as a collapse mechanism. Yield line theory for two-​way slabs requires a different treatment from limit analysis of continuous beams, because in this case the yield lines will not in general be parallel to each other but instead will form a yield line pattern. The entire slab area will be divided into several segments, which can rotate along the yield lines as rigid bodies at the condition of collapse or unstable equilibrium. Some yield line patterns for typical situations are shown in Fig. 17.2.2. The slab of Fig. 17.2.2(a) has nonparallel supports. At the collapse condition, this slab will break into two segments: one will have an edge rotating about line I and the other will have an edge rotating about line II. The positive moment yield line must then intersect lines I and II at their intersection, point 0. The exact position of yield line III will depend on the reinforcement amount and direction, in both the positive and negative moment regions. For the case of Fig. 17.2.2(b) where a rectangular panel is either simply supported or continuous over four linear supports, the collapse mechanism consists of four slab segments. The exact locations of points a and b will depend on the moment strengths at the supports and the positive moment reinforcement in each direction. The slab in Fig. 17.2.2(c) is supported along two edges and, in addition, is supported by two isolated columns. The rotational axes for the slab segments at collapse must occur along the supports (lines I  and II), and additional rotational axes must pass through the isolated columns. The critical position of the positive moment yield lines a, b, c, d, and e is a function of the reinforcement amount and direction; in the meantime, compatibility of deflection along the yield lines must be maintained during the rigid body rotations of the slab segments. For a concentrated load at a significant distance from a supported edge, the yield line pattern will be circular, as shown in Fig. 17.2.2(d). The circle pattern will be a yield line of negative bending moment, while the radial yield lines are due to positive bending moment. For concentrated loads near a free edge, a fan or partial circular pattern is typical.

Supported edges

Free edge

a

Supported edges

III

II b

e d

IV c Columns

Free edge I

II I

0

(a)

III (c)

Four supported edges P a

b

Positive moment yield line

(b)

Figure 17.2.2  Typical yield line patterns.

(d)

Negative moment yield line

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17.3 FUNDAMENTAL ASSUMPTIONS In applying the yield line theory to the ultimate load analysis of reinforced concrete slabs, the following fundamental assumptions are made. 1. The steel reinforcement is fully yielded along the yield lines at failure. In the usual case, when the slab reinforcement is well below that in the balanced condition, the moment-​curvature relationship [17.4] is as shown in Fig. 17.3.1. 2. The slab deforms plastically at failure and is separated into segments by the yield lines. 3. The bending and twisting moments are uniformly distributed along the yield line and correspond to the maximum values provided by the flexural strengths in two orthogonal directions (for two-​way slabs). 4. The elastic deformations are negligible compared with the plastic deformations; thus the slab parts rotate as plane segments in the collapse condition. Assumption No. 3 may be considered to be the yield criterion of orthotropic reinforced concrete slabs. It means that along a yield line as shown in Fig. 17.3.2, the bending moment strength Mnb and twisting moment strength Mnt, each per unit distance along the yield line, are exactly equal to the sum of the projected components along the yield line of the moment strengths Mnx and Mny per unit distance in the y-​ and x-​directions, respectively. It may be noted that Mnx is the strength contributed by the reinforcement in the x-​direction, and Mny is the strength contributed by the reinforcement in the y-​direction. Also, the sign convention is that the bending moments Mnx, Mny, and Mnb are positive when they induce tension in the lower portion of the slab, and the twisting moment Mnt is positive if its vector is directed away from the free body on which it acts.

Typical Moment, M

Idealized

Elastic deformation Plastic deformation Curvature, ϕ

Figure 17.3.1  Typical and idealized M-​ϕ relationship for reinforced concrete slab.

L

Steel for Mnx per unit width

M nb M nt ld Yie

line

θ x

L cos θ

Figure 17.3.2  Bending and twisting moments on yield line.

Mny

L sin θ

Steel for Mny per unit width

Mnx

y

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The bending moment strength Mnb and twisting moment strength Mnt along the yield line in Fig. 17.3.2 may be expressed in terms of Mnx and Mny. Taking equilibrium of moment vectors parallel to the yield line, M nb ( L ) = M nx ( L sinθ)sin θ + M ny ( L cos θ) cos θ M nb = M nx sin 2 θ + M ny cos2 θ

M nb =

M nx + M ny 2

−

M nx − M ny 2



cos 2θ

(17.3.1)

and, taking equilibrium of moment vectors perpendicular to the yield line, M nt ( L ) = M nx ( L sin θ) cos θ − M ny ( L cos θ)sin θ M nt = ( M nx − M ny )sin θ cos θ

=

( M nx − M ny ) 2



sin 2θ

(17.3.2)

In using Eqs. (17.3.1) and (17.3.2), it is important to note that θ is the counterclockwise angle measured from the positive x-​axis to the yield line.

17.4 METHODS OF ANALYSIS There are two methods of yield line analysis of slabs: the virtual work method and the equilibrium method. Based on the same fundamental assumptions, the two methods should give exactly the same results. In either method, a yield line pattern must be first assumed so that a collapse mechanism is produced. For a collapse mechanism, rigid body movements of the slab segments are possible by rotation along the yield lines while deflection compatibility is maintained at the yield lines between slab segments. There may be more than one possible yield line pattern, in which case solutions to all possible yield line patterns must be sought; the one giving the smallest ultimate load would actually happen and thus should be used in design. For instance, the failure pattern of the simply supported rectangular slab subjected to uniform load may be that shown either in Fig. 17.4.1(a) or in 17.4.1(b), depending on the aspect ratio of the rectangular panel and the moment strengths Mnx and Mny. After a yield line pattern has been assumed, the next step is to determine the position of the yield lines, such as that defined by the unknown x in Fig. 17.4.1(a) or 17.4.1(b). It is at this point that one may choose to use the virtual work method or the equilibrium method. In the virtual work method, the total external work done by the applied loads during simultaneous rigid body rotations of the slab segments (while maintaining deflection compatibility) must balance the total internal work done by the bending and twisting moments on all the yield lines. The value of x that gives the smallest ultimate load is then found by means of differential calculus. In the equilibrium method, the value of x is obtained by applying

a

b

c d x

x (a)

(b)

Figure 17.4.1  Yield line patterns of a simply supported rectangular slab.

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725

the usual equations of statical equilibrium to the slab segments, but the optimal position x is defined by predetermined nodal forces placed at the intersection of yield lines. Expressions for the nodal forces in typical situations, once derived, can be conveniently used to avoid the necessity of mathematical differentiation, as required in the virtual work method. In the following sections, yield line analysis for one-​way slabs is dealt with first in a manner similar to limit analysis of continuous beams. Then both the virtual work method and the equilibrium method are illustrated for two-​way slabs.

17.5 YIELD LINE ANALYSIS OF ONE-​W AY SLABS The continuous slab span shown in Fig. 17.5.1(a) has nominal moment strengths of Mni and Mnj provided by top reinforcement at the supports and a nominal strength Mnp provided by bottom reinforcement within the span. The nominal strengths Mni, Mnj, and Mnp are absolute values of the moment strength per unit slab width. For uniform loading, the only possible yield pattern consists of three parallel yield lines as shown in Fig. 17.5.1(a), one each along the left and right supports and one at a distance x from the left support. The moment strength per unit slab width when collapse is imminent should be as shown in Fig. 17.5.1(b). The problem is to determine the collapse load wu /​φ per unit slab area in terms of Mni, Mnj, and Mnp, and the span length L. It will be shown that the virtual work and equilibrium methods will give exactly the same results. Referring to Fig.  17.5.1(c), the rigid body rotations of the slab segments at collapse are measured from the original horizontal positions to those indicated by the dashed lines, where the rotation of the left segment is θ1 in the clockwise direction and that of the right segment is θ2 in the counterclockwise direction, all the while maintaining the compatible deflection Δ at the positive yield line. The total external work done by the uniform load on the left and right segments of unit slab width is total external work = (wu /φ) x



∆ ∆ + (wu /φ)( L − x ) 2 2

The total internal work done by Mni and Mnp on the left segment is (Mni + Mnp) θ1 and that done by Mnj and Mnp on the right segment is (Mnj + Mnp) θ2, both in absolute quantities. Thus

total internal work = ( M ni + M np )θ1 + ( M nj + M np )θ2

in which, for small deformations

θ1 =



∆ , x

θ2 =

∆ L − x

The principle of virtual work states that the total work done by a force system in equilibrium in going through a virtual rigid body displacement is zero. By means of this principle, one can equate the total external work to the absolute value of the total internal work; or

(wu / φ) x

∆ ∆ ∆ ∆ + (wu / φ)( L − x ) = ( M ni + M np ) + ( M nj + M np ) 2 2 x ( L − x )

Dividing out the compatible deflection Δ from the above equation and solving for wu  /​φ,

2 M np 2 M nj wu 2 M ni = + + L x x( L − x ) L ( L − x ) φ

(17.5.1)

Differentiating Eq. (17.5.1) for wu with respect to x and setting the derivative to zero, one obtains the quadratic equation

( M nj − M ni ) x 2 + 2( M ni + M np ) L x − ( M ni + M np )L2 = 0

(17.5.2)

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Free edge Top steel, Mnj

Top steel, Mni

Bottom steel, Mnp

Negative moment yield line

Negative moment yield line

Free edge

L

(a) Plan of continuous one-way slab Mnp

Mni Mnj (b) Moment strength per unit slab width when collapse condition is imminent wu /φ per unit distance Mni

θ1 wu x φ

Mnj

θ2

θ1 θ2 Mnp

V V

wu (L – x) φ

Mnp

(c) Virtual work method wu/φ

Mni

V=0 Ri =

1 wu L– 2 φ

Mnp

Mnp

wu/φ

Mnj

V=0

Mnj – Mni

Nodal forces

L

Rj =

1 wu L+ 2 φ

Mnj – Mni L

(d) Equilibrium method

Figure 17.5.1  Yield line analysis of one-​way slabs.

In the virtual work method, then, first the value of x is found by solving the quadratic equation (17.5.2) and the collapse load wu /​φ is determined from Eq. (17.5.1). In the equilibrium method, the value of x is not to be obtained by differential calculus. Instead, it is defined by the magnitude of the nodal forces [V and –​V shown in Fig. 17.5.1(d)] acting on the slab segments on either side of the positive yield line. In this particular instance, it is known from elementary beam theory that the shear at a section of maximum positive bending moment should be zero. From this point on, the unknown values of Ri, Rj, wu, and x in Fig. 17.5.1(d) are obtained from the four independent equilibrium equations for the entire slab and either of the two slab segments. The left and right reactions on the free bodies of Fig. 17.5.1(d) are found by applying the equilibrium equations to the entire slab; thus



M nj − M ni  1  w  Ri =   u  L − , L 2  φ  

M nj − M ni  1  w  Rj =   u  L +  L 2  φ  

72



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Then, summing the vertical forces on the left slab segment, Ri =



M nj − M ni wu 1  wu  L− = x   2 φ  L φ

from which wu ( M nj − M ni ) = L ( L /2 − x ) φ



(17.5.3)

and, taking moments about the positive moment yield line on the left segment, M nj − M ni  1  w  1  wu  2 M np = − M ni +   u  L − x−   x L 2 φ  2  φ  



(17.5.4)

Substitution of Eq. (17.5.3) into Eq. (17.5.4) results in the same quadratic equation as Eq. (17.5.2). This shows that the two methods, virtual work and equilibrium, give exactly the same results. In the solution of a numerical problem, the quadratic equation (17.5.2) is first solved for x. Although wu can then be computed from either Eq. (17.5.1) or Eq. (17.5.3), using Eq. (17.5.1) is far superior to Eq. (17.5.3), because Eq. (17.5.1) contains the sum of three absolute quantities but Eq. (17.5.3) involves the division by a sensitive quantity (L/​2 –​ x).

EXAMPLE 17.5.1 Given the nominal moment strengths Mni = 14 ft-​kips, Mnp = 16 ft-​kips, and Mnj = 22 ft-​ kips per ft width of a 20-​ft continuous slab span as shown in Fig. 17.5.2, determine the location x of the positive moment yield line and the collapse load wu /​φ in kips per ft per foot width of slab. SOLUTION Using the quadratic equation (17.5.2), ( M nj − M ni ) x 2 + 2( M ni + M np )L x − ( M ni + M np )L2 = 0 (22 − 14) x 2 + 2(14 + 16)(200 x ) − (14 + 16)(20)2 = 0 8 x 2 + 1200 x − 12, 000 = 0



x 2 + 150 x + 752 = 1500 + 5625



x = 7125 − 75 = 84.41 − 75 = 9.41 ft Substituting the value of x = 9.41 ft in Eq. (17.5.1), 2 M np 2 M nj wu 2 M ni = + + φ L x x( L − x ) L ( L − x )

=

2(14) 2(16) 2(22) + + 20(9.41) 9.41(10.59) 20(10.59)

= 0.149 + 0.321 + 0.208 = 0.678 kip/ft (Continued)

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Example 17.5.1 (Continued) Using wu /​φ = 0.678 kip/​ft and the end moments of 14 ft-​kips and 22 ft-​kips, the end reactions, the shear diagram, and the moment diagram are computed and shown in Fig. 17.5.2. Mnp = 16 ft-k/ft Mnj = 22 ft-k/ft

Mni = 14 ft-k/ft

1.0’

x

20’ 14 ft-k

22 ft-k

wu/φ = 0.678 k/ft

+6.78 –0.40 +6.38

+6.78 +0.40 +7.18

+6.38k Shear diagram at collapse condition x = 9.41’ –7.18k 16 ft-k + –

–

Moment diagram at collapse condition

14 ft-k 22 ft-k

Figure 17.5.2  One-​way slab for Example 17.5.1.

17.6 WORK DONE BY YIELD LINE MOMENTS IN RIGID BODY ROTATION OF SLAB SEGMENT Before taking up the virtual work method of yield line analysis of two-​way slabs, it is desirable to derive a general procedure for obtaining the absolute value of the internal work done by the bending and twisting moments acting on the yield line through rigid body rotation of the slab segment. Figure 17.6.1(a) shows the moments acting on the edges of a slab segment having yield line of length L with horizontal and vertical projections of Lx and Ly, respectively. Let this slab undergo a rigid body rotation, whose components are θx and θy, shown in vector notations in Fig. 17.6.1(b). It can be shown algebraically that the absolute

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L

Mnb (L)

Yield line

Mnx (Ly )

Mnt (L)

θx Ly

θ Lx

θy

Mny (Lx)

(a) Total moments on edges

(b) Right body rotation

Figure 17.6.1  Work done by yield line moments in rigid body rotation of a slab segment.

value of the work done by the moments (Mnb L) and (Mnt L) acting on the yield line is equal to the work done by the moments (Mnx Ly) and (Mny Lx) acting on the horizontal and vertical projections of the yield line—​for the same rigid body rotation, of course. This is obviously correct because the moments (Mnx Ly) and (Mny Lx) on Fig. 17.6.1(a) are the equilibrants of the moments (Mnb L) and (Mnt  L) on the same figure. Certainly then, the work done by one set of generalized forces should be equal, numerically, to the work done by the alternative set of equilibrating generalized forces.

17.7 NODAL FORCES AT INTERSECTION OF YIELD LINE WITH FREE EDGE In the equilibrium method of yield line analysis, the position of a yield line is characterized by the insertion of nodal forces at the intersection of a yield line with another yield line, or of a yield line with a free edge. Section 17.5 demonstrated that for one-​way slabs, on the basis of elementary beam theory, the nodal force on either side of a positive moment yield line is zero. This section derives the expression for the pair of equal and opposite nodal forces V acting on each side of the intersection of a yield line with a free edge in a two-​way slab. Shown in Fig. 17.7.1(a) is a two-​way slab with nominal moment strengths of Mnx and Mny provided by the bottom reinforcement in the x-​ and y-​directions, respectively. A positive moment yield line is assumed to intersect the free edge at an angle α. The upward nodal force V acts on the left segment and the downward nodal force V acts on the right segment, shown by a dot and a cross for the upward and downward forces in Fig. 17.7.1(a) according to the convention used by Johansen [17.2, 17.3]. For equilibrium the upward and downward nodal forces must be equal in magnitude. Consider the equilibrium of an infinitesimal slab element shown in either Fig. 17.7.1(b) or 17.1.1(c). By its own definition, a yield line is always at the optimal position. Inasmuch as the edge ac in Fig.  17.7.1(b), or the edge ad in Fig.  17.7.1(c), is at an infinitesimal angle ∆α from the yield line, the same yield line moments as those on the yield line ab act on edges ac or ad. By applying the principle of equivalent force systems as indicated by Fig. 17.6.1(a), the free-​body diagram shown on the left side of the equal sign in Fig. 17.7.1(b) or 17.7.1(c) is transformed to the equivalent free-​body diagram on the right side of the equal sign. Using the equivalent free body and summing the moments about the line ac in Fig. 17.7.1(b),

M ny ( ∆L x ) cos(α − ∆α ) = V ( ∆L x ) sin(α − ∆α )

(17.7.1)

and summing the moments about the line ad in Fig. 17.7.1(c),

M ny ( ∆L x ) cos(α + ∆α ) = V ( ∆L x ) sin(α + ∆α )

(17.7.2)

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•Up +Down

α V

V = Mny cot α

+V

Free edge (a) Two-way slab with free edge a

a

Mnb V b

c

Mnb

Mnt

Mnx Ly

Mnx Ly

Mnt

α

V

c

b

Ly

Mny Lx Lx

(b) Free-body diagram of infinitesimal piece to left of yield line Mny Lx

b

+V

d

Mnb

Mnt

b

a

Mnx Ly

a

Mnb

Mnx Ly

Mnt

Ly

+V d Lx

(c) Free-body diagram of infinitesimal piece to right of yield line

Figure 17.7.1  Nodal force at intersection of yield line with free edge.

Solving for V by using either Eq. (17.7.1) or Eq. (17.7.2),

V = M ny cot α

(17.7.3)

Note that on the left side of Eq. (17.7.1) or Eq. (17.7.2), Mny(∆Lx) represents the net moment along the edges ca and ab, or the edges ba and ad, respectively; while on the right side, the moment of the nodal force V about an axis coincident with ac or ad is involved. Equation (17.7.3) is used when applying the equilibrium method to a situation where a positive moment yield line intersects a free edge of a slab at an angle other than 90°. This is illustrated by the following example.

EXAMPLE 17.7.1 The triangular slab ABC shown in Fig. 17.7.2(a) is uniformly loaded and is simply supported along edges AC and BC but has a free edge along AB. The reinforcement in the x-​ and y-​directions in the lower face of the slab provides nominal moment strengths Mnx = 8 ft-​kips and Mny = 10 ft-​kips, each per foot width of slab. Determine the yield line pattern and the collapse uniform load wu /​φ. (Continued)

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Example 17.7.1 (Continued) C Simply supported edges

Bottom reinforcement for Mnx = 8 ft-k/ft 12’

A

Bottom reinforcement for Mny = 10 ft-k/ft

B

Free edge 16’

(a) Uniformly loaded triangular slab

E

θ A

V • D

10x ft-k C

96 ft-k

96 ft-k

C

α 10x ft-k

D

+V x

B

(b) Free-body diagrams

Figure 17.7.2  Yield line analysis of the triangular slab of Example 17.7.1.

SOLUTION (a) Yield line pattern. It was shown in Fig. 17.2.2(a) that a yield line should pass through the point of intersection of two nonparallel supported edges. In this example, a posi­ tive moment yield line CD will divide the slab into two segments ACD and BCD, wherein a common compatible deflection Δ exists at point D due to the rigid body rotations of the slab segment ACD about the supported edge AC and of the slab segment BCD about the supported edge BC. (b) Equilibrium method of finding BD = x. The nodal forces V and –​V acting on the left and right slab segments of Fig. 17.7.2(b) are computed from Eq. (17.7.3); thus  x 5 V = M ny cot α = (+10)   = x kips  12  6



Note that the nodal force acts upward in the obtuse angle and it acts downward in the acute angle. Equilibrium of moments about the edge AC of the left segment requires 1 1 (wu /φ)(16 − x )(12) ( DE ) = V ( DE ) + 8(12)sin θ + 10 x cos θ 2 3

1 5  1 (wu /φ)(16 − x )(12)   (16 − x )(0.6) = x(16 − x )(0.6) + 96(0.6) + 10 x(0.8 8)  3 2 6

from which

wu 576 + 160 x − 5 x 2 = φ 12(16 − x )2 (Continued)

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Example 17.7.1 (Continued) Similarly, the equilibrium of moments about the edge BC of the right segment requires 1  BD  + V ( BD) = 96 (wu / φ)( x )(12)   3  2

2 1  wu   x  5x ( x )(12)   + = 96    3 2 φ  6



from which wu 576 − 5 x 2 = φ 12 x 2



Equating the two expressions for w u /​φ, 576 + 160 x − 5 x 2 576 − 5 x 2 = 12(16 − x )2 12 x 2 x 2 + 14.4 x − 115.2 = 0



x = 5.72 ft



With the known value of x, the value of wu  /​φ may be computed, wu = 1.05 kips/sq ft φ



It can be shown that the same quadratic equation in x may be obtained via the virtual work method by following the procedure illustrated in Section 17.5.

17.8 NODAL FORCES AT INTERSECTION OF THREE YIELD LINES Figure  17.8.1 shows three possible yield line patterns for an irregular quadrilateral slab with four simply supported edges. The optimum positions of the yield lines in each pattern should be such as to give the lowest collapse load. These positions are defined by the locations of points a and b in Fig.  17.8.1(a), of points c and d in Fig.  17.8.1(b), and of point e in Fig. 17.8.1(c). In the equilibrium method, the yield lines are characterized by the insertion of predetermined nodal forces. This section derives the expressions for the nodal forces at the intersection of three yield lines, such as at points a, b, c, or d in Fig. 17.8.1. B

A

A

b a

A

B

c

d e D

D

D C (a)

B

C

C (b)

Figure 17.8.1  Yield line patterns of a simply supported quadrilateral slab.

(c)

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3 2

φ3

V23 0 V31

φ2 V12

φ1 (a)

1

x

Bottom reinforcement for Mnx1 – Mnx2 – Mnx3 under yield lines 1-2-3

Bottom reinforcement for Mny1 – Mny2 – Mny3 under yield lines 1-2-3 (c)

(b)

Figure 17.8.2  Nodal forces at the intersection of three yield lines.

The derivation shown below follows the works of Johansen [17.2, 17.3], Jones and Wood [17.4], and Wood [17.5]. Assume that three yield lines 1–​2–​3, intersecting at a common point 0, are situated at angles φ1–​φ2–​φ3 measured counterclockwise from the positive x-​axis, as shown in Fig. 17.8.2(a). The nominal moment strengths under the yield lines 1–​2–​3 are, as shown in Fig. 17.8.2(b) and 17.8.2(c), Mnx1–​Mnx2–​Mnx3 provided by the reinforcement shown horizontally and Mny1–​Mny2–​Mny3 provided by the reinforcement shown vertically, all reinforcement being near the lower face of the slab. The nodal forces V12, V23, and V31 are shown by the dots (which mean upward nodal forces) in Fig. 17.8.2(a). Note that for vertical equilibrium at the point of intersection, the sum of V12, V23, and V31 must be zero. It is important to emphasize at the outset that, by definition, the yield lines 1–​2–​3 are all situated at their respective optimal positions. The bending and twisting moments on any line passing through the point of intersection and deviating by an infinitesimal angle from a particular yield line should be equal to the bending and twisting moments on that line as provided by the orthogonal moment strengths. Consider the equilibrium of an infinitesimal slab segment 0AB [Fig. 17.8.3(a)], bounded by a differential length 0A on yield line 3 and an arbitrary length 0B on yield line 1. Since BA is at a differential angle ∆α from B0, the bending and twisting moments on both B0 and BA are those provided by Mny1 and Mnx1 on their respective horizontal and vertical projections. Likewise the bending and twisting moments on the differential length 0A are those provided by Mny3 and Mnx3 on its horizontal and vertical projections. Call the upward nodal force at point 0 inside triangle B0A and the force bounded by yield line 1 and a different length on yield line 3 by the name V1–​∆3. Next, write the equation of equilibrium for moments about AB as the axis of rotation for the slab segment 0AB in Fig. 17.8.3(a), noting that the moment of the uniform load on 0AB about any axis is a differential of the second order and may be neglected. − V1− ∆ 3 0 A sin(φ3 − φ1 + ∆α ) + ( M ny 3 − M ny1 ) 0 A cos φ3 cos(φ1 − ∆α )



+( M nx 3 − M nx1 ) 0 A sin φ3 sin(φ1 − ∆α ) = 0



Solving the above equation for V1–​∆3 and letting ∆α approach zero at the limit,

V1− ∆ 3 =

( M nx 3 − M nx1 )sin φ3 sin φ1 + ( M ny 3 − M ny1 ) cos φ3 cos φ1 sin(φ3 − φ1 )



(17.8.1)

Making a similar analysis for the infinitesimal slab segment 0AC [Fig. 17.8.3(b)] bounded by a differential length 0A on yield line 3 and an arbitrary length 0C on yield line 2 gives the expression for the upward nodal force V2–​∆3 in Fig. 17.8.3(b) as

V2 − ∆ 3 =

( M nx 3 − M nx 2 )sin φ3 sin φ2 + ( M ny 3 − M ny 2 ) cos φ3 cos φ2 sin(φ3 − φ2 )



(17.8.2)

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Figure 17.8.3  Determination of nodal force V12 between yield lines 1 and 2.

For vertical equilibrium at the point of intersection, the upward nodal force V∆3–​1 in Fig. 17.8.3(a) in the zone going counterclockwise from the differential length on yield line 3 to yield line 1 is

V∆ 3 −1 = −V1− ∆ 3

(17.8.3)

and, for the same reason, the upward nodal force V12 in Fig. 17.8.3(c) is

V12 = −V2 − ∆ 3 − V∆ 3 −1

(17.8.4)

735



1 7 . 8   N O D A L F O R C E S AT I N T E R S E C T I O N O F T H R E E Y I E L D   L I N E S

735

Substitution of Eqs. (17.8.1), (17.8.2), and (17.8.3) into Eq. (17.8.4) gives V12 =

( M nx 3 − M nx1 )sin φ3 sin φ1 + ( M ny 3 − M ny1 ) cos φ3 cos φ1 sin(φ3 − φ1 ) −

( M nx 3 − M nx 2 )sin φ3 sin φ2 + ( M ny 3 − M ny 2 ) cos φ3 cos φ2 sin(φ3 − φ2 )

(17.8.5)

Replacing each numerical subscript in Eq. (17.8.5) by its successor in the cyclic order of 1–​2–​3–​1 (counterclockwise around the point of intersection) and then once more in the same manner, the following expressions for the upward nodal forces V23 and V31 as shown in Fig. 17.8.2(a) are obtained. V23 =

( M nx1 − M nx 2 )sin φ1 sin φ2 + ( M ny1 − M ny 2 ) cos φ1 cos φ2 sin(φ1 − φ2 ) −

V31 =

( M nx1 − M nx 3 )sin φ1 sin φ3 + ( M ny1 − M ny 3 ) cos φ1 cos φ3 sin(φ1 − φ3 )

(17.8.6)

( M nx 2 − M nx 3 )sin φ2 sin φ3 + ( M ny 2 − M ny 3 ) cos φ2 cos φ3 sin(φ2 − φ3 ) −

( M nx 2 − M nx1 )sin φ2 sin φ1 + ( M ny 2 − M ny1 ) cos φ2 cos φ1 sin(φ2 − φ1 )

(17.8.7)

Equations (17.8.5), (17.8.6), and (17.8.7) are expressions for the upward nodal forces at the intersection of three yield lines.

Nodal Force at Intersection of Yield Line with Free Edge The nodal forces at the intersection of a yield line with a free edge, as shown by Fig. 17.8.4, may be obtained by substituting φ1 = 0, φ2 = α, φ3 = π, Mnx1 = Mnx3 = Mny1 = Mny3 = 0, Mnx2 = Mnx, and Mny2 = Mny in Eqs. (17.8.5) to (17.8.7); thus V12 = − V23 =

V31 =

( − M ny 2 )( −1) cos α sin(π − α )

( − M ny 2 )(+1) cos α sin(0 − α ) ( M ny 2 ) cos α( −1) sin(α − π)

−

= − M ny cot α

= + M ny cot α ( M ny 2 ) cos α(+1) sin(α − 0)

=0



The above results check with the findings in Section 17.7. Bottom reinforcement for Mnx 2

V23 3 V31

e lin eld i Y α V12

Free edge

Bottom reinforcement for Mny 1

V12 = –Mny cot α V23 = +Mny cot α

Figure 17.8.4  Nodal forces at the intersection of a yield line with a free edge.

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Nodal Forces at Intersection of Three Yield Lines Having Identical Mnx and Mny Nominal Moment Strengths From Eqs. (17.8.5) to (17.8.7) it can be observed that wherever the nominal moment strengths under three intersecting yield lines are identical (i.e., Mnx1 = Mnx2 = Mnx3 = Mnx and Mny1 = Mny2 = Mny3 = Mny), the nodal forces at the intersection are zero. This fact is of great convenience when using the equilibrium method for the yield line analysis of two-​way rectangular slabs.

17.9 YIELD LINE ANALYSIS OF RECTANGULAR TWO-​W AY SLABS A typical rectangular two-​way slab panel, as shown in Fig. 17.9.1, has two-​way reinforcement within the panel near the bottom face providing positive moment nominal strengths Mnpx and Mnpy, as well as two-​way reinforcement along the edges near the top face providing negative moment nominal strengths Mnnx and Mnny; these strengths are per unit width of slab. The uniform load to give the collapse condition based on the yield line theory may be determined in terms of the sides a and b, and the absolute values of Mnpx, Mnpy, Mnnx, and Mnny.

Yield Line Pattern Three possible yield line patterns are shown in Fig. 17.9.2. There is no unknown position in yield line pattern No. 1 of Fig. 17.9.2(a); consequently the nodal forces V need not be predetermined, and their value is dictated by statics alone. The unknowns x and y in yield line patterns Nos. 2 and 3 of Fig. 17.9.2(b) and 17.9.2(c) must be determined by means of differential calculus in the virtual work method. For the equilibrium method, in this particular case, however, the nodal forces are all zero because the moment strengths under a set of three intersecting yield lines are identical.

Capacity = Mnnx Edges supported and restrained

Capacity = Mnpy

b

Capacity = Mnny

a (a) Dimensions

Capacity = Mnpx

(b) Top reinforcement

(c) Bottom reinforcement

Figure 17.9.1  A rectangular two-​way slab panel.

V

V + + V

V y x

(a) Yield pattern No. 1

(b) Yield pattern No. 2

(c) Yield pattern No. 3

Figure 17.9.2  Yield line patterns for a rectangular two-​way slab panel.

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V

+ A

V

V

+ A

Mnpx

Mnpy

Mnpy

B

D

V

Mnpx

D V

Mnpx

b

C +V

Mnpy

b Mnnx

θ=

Mnpx



Mnny

θ=

a

Mnpy

Figure 17.9.3  Analysis of yield line pattern No. 1.

Analysis for Yield Pattern No. 1 Assuming a vertical deflection Δ at the intersection of the diagonal yield lines in Fig. 17.9.3, the deflection at the centroids of the four triangles A–​B–​C–​D is ∆ /​3. The work done at the collapse condition by the uniform load is the product of the total load on the entire panel and Δ  /​3; thus W=



wu  ∆  ab    3 φ

(17.9.1)

The work done by the yield moments on the boundaries of all four slab segments, referring to Fig. 17.9.3, is  2∆   2∆  W = 2( M nny + M npy )(a )   + 2( M nnx + M npx )(b)    b   a 



(17.9.2)

Equating Eq. (17.9.1) to Eq. (17.9.2) and solving for wu,  M nnx + M npx M nny + M npy  wu = 12  +  φ a2 b2 



(17.9.3)

Alternatively, the same solution is obtained using the equilibrium method. Taking moments about the lower edge of slab segment A in Fig. 17.9.3, 1  wu   b   b   b a     + V   = ( M nny + M npy )(a )  2 2  φ   2   6 



(17.9.4)

Taking moments about the left edge of slab segment D in Fig. 17.9.3, 1  wu   a   a   a b     = ( M nnx + M npx )(b) + V    2 2  φ   2   6 



(17.9.5)

Eliminating V between Eqs. (17.9.4) and (17.9.5) and solving for wu  /​φ yields the same expression for wu  /​φ as Eq. (17.9.3).

Analysis for Yield Pattern No. 2 Assuming a vertical deflection Δ at the two points of intersection of the yield lines in Fig. 17.9.4, the work done at the collapse condition by the uniform load on the entire panel is W = 2WD + 2WA1 + 4WA 2 1  w  b  ∆ 1  w    ∆ w   b  ∆  = 2   u  bx    + 2  u  (a − 2 x )     + 4   u  x     2  2   φ 2  φ    3  2  φ  2  3    

=

wu φ

 ∆ (3ab − 2bx )  6 



(17.9.6)

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The work done by the yield moments on the boundaries of all four slab segments is, referring to Fig. 17.9.4,  2∆   ∆ W = 2( M nny + M npy )(a )   + 2( M nnx + M npx )(b)    b   x



(17.9.7)

Equating Eq. (17.9.6) to Eq. (17.9.7) and solving for wu /​φ, 2 wu 12[b ( M nnx + M npx ) + 2 ax ( M nny + M npy )] = φ b 2 (3ax − 2 x 2 )



(17.9.8)

Setting to zero the derivative of Eq. (17.9.8) with respect to x gives the quadratic equation in x, 4 a( M nny + M npy ) x 2 + 4b 2 ( M nnx + M npx ) x − 3ab 2 ( M nnx + M npx ) = 0



(17.9.9)

Using the equilibrium method with V = 0 because there are three intersecting yield lines, and taking moments about the lower edge of slab segment A in Fig. 17.9.4, 1  w  b   b w  b  b 2   u  x    + u (a − 2 x )     = ( M nny + M npy )(a )  2  4 2  φ  2  6 φ



24 a( M nny + M npy ) wu = 2 φ 2b x + 3b2 (a − 2 x )



(17.9.10)

Taking moments about the left edge of slab segment D in Fig. 17.9.4, 1  wu   x  bx   = ( M nnx + M npx )(b) 2  φ   3  wu 6( M nnx + M npx ) (17.9.11) = φ x2



Equating Eq. (17.9.10) to Eq. (17.9.11) gives the same quadratic equation in x as Eq. (17.9.9). The condition for x = a/​2 in Eq. (17.9.9) can be shown to be M nnx + M npx

M nny + M npy

=

a2 b2

for x =

a 2

(17.9.12)

which means that if the sum of the positive and negative moments contributed by the reinforcement in the a direction, each per unit width of slab, is equal to (a2/​b2) times the sum of the positive and negative moments contributed by the reinforcement in the b direction, each per unit width of slab, yield pattern No. 1 prevails.

b

D A2 x

A1

A2 x

A2

A1 Mnny

Figure 17.9.4  Analysis of yield line pattern No. 2.

Mnpy

A2

Mnpx

Mnpy

B

Mnpy

θ=

D Mnpx

C

Mnpy b Mnnx

θ=

Mnpx

a

Mnpx

738

Mnpy

x

739



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The condition for x < a/​2 in Eq. (17.9.9) can be shown to be M nnx + M npx

M nny + M npy


3a 2 nx

(17.11.2b)

For the equilibrium method, the nodal force V in Fig. 17.11.2(c) may be predetermined by use of Eq. (17.7.3); or

 x V = M ny cot α = M ny    b

745



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1 7 . 1 1   A P P L I CAT I O N TO S P E C I A L   CA S E S

For equilibrium of the triangular segment in Fig. 17.11.2(c), 1  wu   x   x bx   + M ny   x = M nx b      b 2 φ  3

from which

2 2 wu 6(b M nx − x M ny ) = φ b2 x 2



(17.11.3)

For equilibrium of the trapezoidal segment in Fig. 17.11.2(c), w 1  wu   b   b  x xb   2 + u (a − 2 x )b   = 2 M ny x + 2 M ny   b  2  b 2  φ   3  φ

from which

24 x M ny wu = 2 φ b (3a − 4 x )



(17.11.4)

The same quadratic equation in x as Eq. (17.11.2) is obtained from equating Eq. (17.11.3) to Eq. (17.11.4). For yield pattern No. 2, referring to Fig. 17.11.2(d) and letting Δ be the deflection at the yield line perpendicular to the free edge, the virtual work done by the uniform load is



W=

w 1  wu   a   ∆  1  w   ∆  a  ∆ y     2 + u (b − y)     2 +  u  ay             2 φ   3 2 φ  2 3 φ 2 2

W=

wu  a ( ∆ )   (3b − y)  6 φ



The virtual work done by the yield line moments is

 ∆  2∆  W = 2 M nx b   + M ny a    a   y

Equating the two expressions for W and solving for wu /​φ, 2 wu 6(4byM nx + a M ny ) = φ a 2 y(3b − y)



(17.11.5)

Setting to zero the derivative of Eq. (17.11.5) with respect to y,

 M ny   M ny  =0 y − 3a 2 b  4by 2 + 2 a 2    M nx   M nx 

(17.11.6a)

For the root of Eq. (17.11.6a) to be less than b,  M ny   4b 2   M  <  a 2  nx



(17.11.6b)

Since the moment strengths at the three intersecting yield lines in Fig. 17.11.2(d) are identical, the nodal forces are all zero, based on the treatment presented in Section 17.8. For equilibrium of the trapezoidal segment in Fig. 17.11.2(d),

1  wu  2  φ 

 a a w  a a y   + u (b − y)   = M nx b  2 6 φ  2 4

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from which 24bM nx wu = 2 φ a (3b − 2 y)



(17.11.7)

For equilibrium of the triangular segment in Fig. 17.11.2(d), 1  wu  y ay = M ny a 2  φ  3

from which

wu 6 M ny = 2 φ y



(17.11.8)

The same quadratic equation in y as Eq. (17.11.6a) is obtained from equating Eq. (17.11.7) to Eq. (17.11.8). Thus Eqs. (17.11.1) to (17.11.4) apply to yield pattern No. 1, which cannot happen if (Mny /​Mnx) is smaller than 4b2/​(3a2), and Eqs. (17.11.5) to (17.11.8) apply to yield pattern No. 2, which cannot happen if (Mny /​Mnx) is larger than 4b2/​a2. When (Mny /​Mnx) is between 4b2/​(3a2) and 4b2/​a2, both yield patterns should be analyzed; the one giving the smaller collapse load controls. Although an exact value of (Mny /​Mnx) in terms of b2/​a2 (between 1.33 and 4.00) at the transition point where yield pattern No. 1 begins to control over yield pattern No. 2 may be determined, the complication in the algebra does not seem to warrant the effort. Table 17.11.1 shows the results of analysis for six different cases in which the values of (Mny /​Mnx) become progressively larger while Mny remains constant.

TABLE 17.11.1  YIELD LINE ANALYSIS OF A RECTANGULAR SLAB WITH THREE SUPPORTED EDGES AND ONE FREE EDGE (FIG. 17.11.2), a = 25 ft, b = 20 ft Case Mny (ft-​kips/​ft) Mnx (ft-​kips/​ft) Yield pattern No. 1 x (ft) [Eq. (17.11.2)] w (ksf ) [Eq. (17.11.1), (17.11.3), or (17.11.4)] Yield pattern No. 2 y (ft) [Eq. (17.11.6)] w (ksf ) [Eq. (17.11.5), (17.11.7), or (17.11.8)] Controlling yield pattern Collapse load wu /​φ (ksf )

1

2

3

4

5

6

16 24

16 18.75

16 16

16 8

16 6.25

16 4

—​

12.5

12

9.78

9.01

7.68

—​

0.480

0.427

0.262

0.222

0.167

13.22

14.41

15.20

18.75

20

—

0.549

0.462

0.415

0.273

0.240

​—​

No. 2 0.549

No. 2 0.462

No. 2 0.415

No. 1 0.262

No. 1 0.222

No. 1 0.167

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PROBLEMS

747

SELECTED REFERENCES 17.1. A.  Ingerslev. “The Strength of Rectangular Slabs,” Journal of the Institute of Structural Engineers, London, 1 (1), January 1923, 3–​14. Disc., 14–​19. 17.2. K.  W. Johansen. “The Ultimate Strength of Reinforced Concrete Slabs,” Final Report. Third Congress, International Association for Bridge and Structural Engineering, Liège, Belgium. September 1948 (pp. 565–​570). 17.3. K. W. Johansen. Yield Line Theory. London: Cement and Concrete Association, 1962. 17.4. L. L. Jones and R. H. Wood. Yield Line Analysis of Slabs. New York: Elsevier, 1967. 17.5. R.  H. Wood. “Plastic Design of Slabs Using Equilibrium Methods,” Flexural Mechanics of Reinforced Concrete, Proceedings of the International Symposium, Miami, 1964 (pp. 319–​336). 17.6. E. Hognestad. “Yield Line Theory for the Ultimate Flexural Strength of Reinforced Concrete Slabs,” ACI Journal, Proceedings, 49, March 1953, 637–​656.

PROBLEMS 17.1 Assuming a 6-​in. slab thickness for the continuous slab span described in Example 17.5.1, the uniform load wu due to the weight of the slab itself may be considered to be 1.2 wD = 1.2(75) = 90 psf. If the slab supports a transverse wall at 7 ft from the left support line, determine the maximum wall load per transverse foot that will cause a collapse mechanism to occur. 17.2 Solve Example 17.7.1, but use Mnx = 10 ft-​kips and Mny = 8 ft-​kips, each per foot width of slab. 17.3 For the regular yield pattern solution to Case 1 of Example  17.9.1, investigate the effect of a corner yield pattern in which the negative

 moment yield line intersects the edges at 5 and 4 ft from the corner along the 25-​and 20-​ft edges, respectively. 17.4 Same as Problem 17.3, except for Case 2 of Example 17.9.1. 17.5 Same as Problem 17.3, except for Case 3 of Example 17.9.1. 17.6  Verify the solution for Case 1 in Table 17.11.1. 17.7  Verify the solution for Case 2 in Table 17.11.1. 17.8  Verify the solution for Case 3 in Table 17.11.1. 17.9  Verify the solution for Case 4 in Table 17.11.1. 17.10  Verify the solution for Case 5 in Table 17.11.1. 17.11  Verify the solution for Case 6 in Table 17.11.1.

CHAPTER 18

TORSION

18.1 GENERAL Reinforced concrete members may be subjected to torsion, frequently in combination with bending and shear. The cantilever member in Fig. 18.1.1(a) is largely subjected to torsion, although some bending and shear also exist due to its own weight. The fixed-​ended beam of Fig. 18.1.1(b) is subjected to substantial amounts of bending, shear, and torsion. Spandrel beams at the edge of a building built integrally with the floor slab are subjected not only to transverse loads but also to a torsional moment per unit length equal to the restraining moment at the exterior end of the slab. Similarly, spandrel girders receive torsional moments from the exterior ends of the floor beams that frame into them. Torsion on structural systems may be classified into two types: statically determinate torsion (sometimes called “equilibrium torsion”), for which the torsion can be determined from statics alone, and statically indeterminate torsion (sometimes called “compatibility torsion”), for which the torsion cannot be determined from statics alone and a rotation (twist) is required for deformational compatibility between interconnecting elements, such as a spandrel beam, slab, or column. Both examples in Fig. 18.1.1 are cases of statically determinate torsion. In cases of statically determinate torsion, as in Fig. 18.1.2(a) and 18.1.2(b), the amount of torsion the member is required to resist is based on the requirement of statics and is independent of the stiffness of the member. Statically indeterminate torsion, as shown in Fig. 18.1.2(c) and 18.1.2(d), sometimes exists where there would be no torsion if the statistical indeterminacy were eliminated. For instance, if the support at A is eliminated in Fig. 18.1.2(c), the torsion is eliminated. Similarly, in Fig. 18.1.2(d), if a flexural hinge is put at B, the torsion is eliminated. For such statically indeterminate torsion situations, the amount of torsion in a member depends on the magnitude of the torsional stiffness of the member itself in relation to the stiffnesses of the interconnecting members. For an excellent overview of the entire subject of torsion as it affects structures, with specific reference to reinforced concrete design, the reader is referred to Tamberg and Mikluchin [18.3]. An extensive bibliography is also included. An extensive treatment of torsion for reinforced and prestressed concrete can be found in the 1983 book by Hsu [18.12]. Collins and Mitchell [18.9] have provided a unified rational treatment of the function of reinforcement to resist shear and torsion, and have made design recommendations for reinforced and prestressed concrete members. A number of general references are available [18.1–​18.18]. Joint ACI-​ASCE Committee 445 on Shear and Torsion also recently published a Report on Structural Concrete [18.19]. In this chapter, brief treatment is given to the computation of torsional stress and torsional rigidity of homogeneous sections, the torsional strength of reinforced concrete sections, the

749



18.2  TORSIONAL STRESS IN HOMOGENEOUS SECTIONS

749

Inclined cracks due to torsion; test by J. P. Klus at the University of Wisconsin–​Madison.

(a)

(b)

Figure 18.1.1  Reinforced concrete members subjected to torsion.

P P (a)

P

P (b)

B

A (c)

(d)

Figure 18.1.2  Comparison of statically determinate torsion (Cases a and b) and statically indeterminate torsion (Cases c and d). (Structures shown in plan view with load P perpendicular to plane of frame.)

development and background for the ACI Code requirements, design examples, and the effect of torsional stiffness in a continuity analysis.

18.2 TORSIONAL STRESS IN HOMOGENEOUS SECTIONS A torsional moment T acting on a shaft of homogeneous material as shown in Fig. 18.2.1 causes shear stresses υ over the cross section.

Circular Sections For a circular section, a plane transverse section before twisting remains plane after twisting. Consequently, the resultant shear stress υ at any point is proportional to its distance from the center and is in a direction perpendicular to the radius. Calling h the diameter of the circle [Fig. 18.2.1(b)] and υt the maximum torsional shear stress at the perimeter,

2υ  2υ  T = ∫ r υ dA = ∫ r  t r  dA = t A A  h  h

∫

A

r 2 dA =

2C υt h

750

750

C hapter   1 8     T orsion

dV = v dA

r h/2

Shear stress due to torsion vt (a)

(b)

y

x

(c)

Figure 18.2.1  Torsional stress in homogeneous sections.

in which C, the polar moment of inertia, is C = ∫ r 2 dA = ∫



h/2

0

A

h/2

 2 πr 4  πh 4 r 2 2 πr dr =  = 32  4  0

The torsional shear stress υt becomes

υt =



16T πh 3

(18.2.1)

Rectangular Sections The torsional shear stress distribution over a rectangular section of dimension x by y cannot be as easily derived as that for a circular section. Unlike the circular section, where plane transverse sections remain plane after twisting, the noncircular cross section warps under torsion. If plane sections were maintained after twisting, the maximum shear stress would exist at a point farthest from the axis of twist. Such is not the case for rectangular sections. From the mathematical theory of elasticity [18.116], it has been found that the maximum torsional shear stress υt occurs at the midpoint of the long side and parallel to it. The magnitude of υt is a function of α, the ratio of y to x (long to short sides) [Fig. 18.2.1(c)]; and

υt =



T α x2 y

(18.2.2)

The values for α are given in Table 18.2.1.

TABLE 18.2.1 VALUES FOR α y/​x

1.0

1.2

1.5

2.0

2.5

3

5

∞

α

0.208

0.219

0.231

0.246

0.256

0.267

0.290

0.333

751



751

18.3  TORSIONAL STIFFNESS OF HOMOGENEOUS SECTIONS

T-​, L-​, and I-​Sections The torsional shear stress distribution in T-​, L-​, or I-​sections may be approximated by dividing the section into several component rectangles assuming that each component rectangle has a large ratio y/​x so that the value for α may be assumed to be 1 3 [18.116]. The maximum shear stress υt occurs at the midpoint of the long side y of the rectangle having the greatest thickness xm and

υt =

Txm 1 ∑ 3 x3 y

(18.2.3)

in which x and y are the thickness and side dimension, respectively, of each component rectangle. Since the web of the sections considered is usually thicker than the flange, xm will usually be the web thickness.

18.3 TORSIONAL STIFFNESS OF HOMOGENEOUS SECTIONS The torsional stiffness Kt of a member is defined as the ratio of torsional moment T to the angle of twist φ over a length L. The torsional rigidity is usually represented by the symbol GC in which G is the modulus of elasticity in shear and C is the torsion constant. Thus if θ is the total angle of twist over a length L, Kt =



T GC = θ L

(18.3.1)

Circular Sections It was just shown (see Section 18.2, subsection on Circular Sections) that the torsion constant C of a circular section of diameter h is the polar moment of inertia C=



πh 4 32

(18.3.2)

Rectangular Sections The torsion constant C of a rectangular section of height y and width x may be expressed, per Timoshenko and Goodier [18.116] as C = βx3 y



(18.3.3)

in which β is a function of y to x. The values for β are given in Table 18.3.1.

TABLE 18.3.1 VALUES OF β y/​x

β

1.0

1.2

1.5

2.0

2.5

3

4

5

0.141

0.166

0.196

0.229

0.249

0.263

0.281

0.291

752

752

C hapter   1 8     T orsion

T-​, L-​, and I-​Sections The torsion constant C of a T-​, L-​, or I-​section may be approximated [18.116] by the expression

1 C = ∑ x3 y 3

(18.3.4)

in which y and x are, respectively, the side length and thickness of each of the component rectangles into which the section may be divided. The following is a more exact expression [18.116, p. 313] for the torsion constant, giving values closer to those in Table 18.3.1 for sections composed of rectangular elements having y/​x less than about 10,

 1 x C = ∑ x 3 y  1 − 0.63  3  y

(18.3.5)

In the ACI Code, torsion provisions are based on the thin-​walled tube, space truss analogy (see Section 18.7); thus, no explicit use is made of the torsion constant C. However, in the provisions for two-​way slab systems as discussed in Chapter 16 (the definition for C in ACI-8.10.5.2), Eq. (18.3.5) is used (see Section 16.11). The lower estimate of stiffness using Eq. (18.3.5) is desirable in structural analysis when one is determining the restraining effect of spandrel members in two-​way floor systems.

18.4 EFFECTS OF TORSIONAL STIFFNESS ON COMPATIBILITY TORSION General Treatment of Torsion on Statically Indeterminate Systems To perform a statically indeterminate structural analysis, it is necessary first to be able to determine the relative stiffnesses of interacting members. The “compatibility torsion” discussed in Section 18.1 is involved. For example, if a spandrel member is uncracked and its torsional stiffness GC/​L is computed as shown in Section 18.3, the torsional moment that the member will attempt to carry may be very large. As the member cracks, its torsional stiffness reduces drastically, the member will twist, and the torsional moment carried is likewise reduced. Postcracking stiffness and torsional moment have been studied by Lampert [18.4] and Collins and Lampert [18.5], who proposed an expression for the torsional rigidity of a cracked section. Since the stiffness is needed before the torsional moment can be determined, the cracked section stiffness is not available because it requires knowledge of the steel reinforcement. Alternatively, Collins and Lampert [18.5] have indicated that in cases of compatibility torsion (where torsional moment depends on torsional stiffness), analysis on the basis of zero torsional stiffness resulted in a design as satisfactory as an analysis using uncracked stiffness. In fact, added steel may increase the torsional moment in the member but have little effect on the twist (rotation). Consequently, it may be more effective to design for a twist (rotation) than for a torque (torsional moment). The purpose of the torsional reinforcement then is to provide ductility and distribute cracks caused by the twist. Such a procedure would be within the spirit of the ACI Code, where ACI-​6.3.1.1 states, “Relative stiffnesses of members within structural systems shall be based on reasonable and consistent assumptions.” Further, ACI Commentary R6.3.1.1 indicates that “torsional stiffness may be neglected” when one is dealing with compatibility torsion. Using the zero torsional stiffness assumption, the member resisting torsion (say, a spandrel beam) would be designed for flexure and shear, neglecting torsion; then the torsional stiffness based on the cracked section could be computed. The structure may then be analyzed to determine the torsional moments, and the section may be checked according to the ACI Code rules for torsion design.

753



18.4  EFFECTS OF TORSIONAL STIFFNESS

753

ACI Code Procedure The ACI Code (ACI-​22.7.3.3) provides an optional simple procedure to reduce the design complexity for cases involving compatibility torsion. When a statically indeterminate situation involves torsion, and an internal redistribution of forces can occur as a result of cracking, the factored torsional moment to be used in design is reduced to a minimum value sufficient to provide the necessary rotation capacity (ductility). In other words, if the torsional restraint is omitted in determining the bending moments and shears on the structural elements, the design of those elements may be more conservative than otherwise, but the difficulty of determining the torsional moments has been eliminated. The torsional members must, however, have the ductility to twist the necessary amount. To summarize, the ACI Code provides two options for the design of torsional members when the torsional moment is dependent on the relative stiffness of the interacting members (compatibility torsion). 1. Estimate the torsional and flexural stiffness of all interacting members by making “reasonable and consistent assumptions.” (ACI-​6.3.1.1). Determine the moments, shears, and torsional moments by statically indeterminate structural analysis using factored loads. Then apply the ACI Code provisions for torsion design. 2. Neglect torsional stiffness in the statically indeterminate structural analysis. This assumes that the torsional member will crack and that “a large twist occurs under an essentially constant torsional moment, resulting in a large redistribution of forces in the structure” (ACI-​R22.7.3). Since no torsional moment will then be available from the computation, the torsional members must be designed for an ACI Code–​specified minimum torsional moment intended to control the width of torsional cracks. Under ACI-​22.7.3.2, the factored torsional moment Tu is permitted to be reduced to a value approximating the torsional cracking moment. For nonprestressed members, the design factored torsional moment φTn is permitted to be taken at the critical sections of ACI-​9.4.4.3 as [ACI-​Table 22.7.5.1(a)]:



 Acp 2  φ Tn = φ 4 λ fc′    pcp 

(18.4.1)

For prestressed members (see Chapter  20), the design factored torsional moment φTn is permitted to be taken at the critical sections of ACI-​9.4.4.3 as [ACI Table 22.7.5.1(b)]:



 Acp 2  f pc φ Tn = φ 4 λ fc′   1+ p 4 λ fc′  cp 

(18.4.2)

For nonprestressed members subjected to an axial tensile or compressive force, the design factored torsional moment is permitted to be taken as [ACI Table 22.7.5.1(c)]:



 Acp 2  Nu φ Tn = φ 4 λ fc′   1+ 4 Ag λ fc′  pcp 

(18.4.3)

where Acp = area enclosed by outside perimeter of concrete cross section Ag = gross area of concrete pcp = outside perimeter of the concrete cross section fpc = compressive stress in concrete at the centroid of the cross section resisting externally applied loads Nu = factored axial force

754

C hapter   1 8     T orsion

2G1

2G4

2G3

2G2

2G2

2G1

282

2B1

2S3

282

2S2

2B2

2S1

2B3

2B4

2S3

3 @26’ = 78’

2B1

2B1

2G4

2B3

2G3

2B1

754

4 @39’ = 156’

Figure 18.4.1  Floor plan of typical slab-​beam-​girder construction.

The compressive stress fpc used in prestressed concrete is discussed in Section 20.10 [see Eq. (20.10.9)]. Equations (18.4.1), (18.4.2), and (18.4.3) can be viewed as representing the cracking torsional moment at a principal tensile stress of 4 λ fc′ psi.

Spandrel Beams and Girders Consider the spandrel beam 2B4 (Fig.  18.4.1) in the typical slab-​beam-​girder floor of Example 8.3.1. This beam receives a vertical load and a torsional moment per unit length from the slab 2S1, which are equal, respectively, to the reaction and restraining moment at the exterior end of slab 2S1. In addition, the beam supports the weight of whatever walls or windows may rest directly on it. Thus the spandrel beam 2B4 is subjected to a torsional moment per unit length in addition to bending and shear. A similar condition exists in the spandrel girder 2G4; in that case, however, the torsional moments are applied only at the junction points with the beams. The torsional moments in the spandrel beams or girders cause torsional shear stresses, which are additive to the bending shear stresses at the inside face of the member. The usual approach in design is to provide for the sum of the torsional shear and flexural shear requirements. Since torsional shear stress goes around the member, closed stirrups or hoops are necessary. The magnitude of the torsional moment acting uniformly along a spandrel beam (ACI-​ 9.4.4.1) such as 2B4 of Fig. 18.4.1 might be roughly approximated as equal to the restraining moment along the exterior edge of the slab, using a value such as 1 24 wL2, as given by ACI Table  6.5.2. Alternatively, the torsional moment may be neglected if the slab is designed assuming the absence of a restraining moment along the spandrel beam. In such a case, the spandrel beam must be designed for a minimum torsional strength corresponding to that which will provide adequate ductility to twist (see Section 18.11). The design and behavior of spandrel beams has been the subject of many studies [18.87–​ 18.101], the most extensive of which are those of Raths [18.99] and Klein [18.100].

EXAMPLE 18.4.1 Estimate the maximum factored torsional moment Tu in the spandrel beam 2B4 of Fig. 18.4.1 if the restraining moment at the exterior end of slab panel (5.5-​in. slab and a clear span of 11.92 ft) 2S1 is M = wL2 /24. The service live and dead loads are 100 1 and 69 psf, respectively. Assume an 18 × 18 in. column and a 13 × 22 in. overall size 2 beam. Use  f ′ = 3000 psi (normal weight). c

(Continued)

75



1 8 . 5   T O R S I O N A L M O M E N T S T R E N G T H T cr AT C R A C K I N G

755

Example 18.4.1 (Continued) SOLUTION (a) Compute the factored torsional loading. The factored restraining moment Mu along the edge of the slab is approximately wu = 1.2(69) + 1.6(100) = 243 psf

Mu =

1 (0.243)(11.92)2 = 1.44 ft-kiips/ft of width 24

The torsional moment is largest at the face of the column and decreases nearly linearly to zero at midspan. The factored torsional moment Tu at the face of the column is approximately

1  Tu =  clear span of spandrel Mu = 12.25(1.44) = 17.6 ft-kips 2 

(b) Compute the ACI Code–​specified design torsional moment φTn permitted to be used in lieu of a structural analysis for statically indeterminate torsion. The spandrel member would then be designed arbitrarily to have a design strength

 Acp 2  φ Tn = φ 4 λ fc′   (18.4.1)  pcp 

From the thin-​walled space truss concept, the section resisting torsion is the primary rectangular portion bwh without any of the slab included. Thus

Acp = bw h = 13(22.5) = 292.5 sq in.

and the outside perimeter ppc of the resisting section is

p pc = 2(13 + 22.5) = 71 in.



Thus

 Acp 2   292.52  1 φ Tn = φ 4 λ fc′  = 16.4 ft-kips  = 0.75(4)(1) 3000   71  12,000  pcp 

In this case, the design for ductility using Eq. (18.4.1) would be slightly less conservative than using the Tu = 17.6 ft-​kips computed in part (a).

18.5 TORSIONAL MOMENT STRENGTH T cr AT CRACKING From Eq. (18.2.2) it follows that the nominal torsional moment strength Tn of a plain concrete rectangular section may be expressed

Tn = α x 2 y [ ft (max)]

(18.5.1)

because torsional stress υt equals maximum principal tensile stress ft (max) in the situation of pure torsion (same as pure shear). In Eq. (18.5.1), α is an elastic theory coefficient as given in Table 18.2.1, ranging from 0.208 to 1 3 as y/​x varies from 1.0 to ∞, and up to 0.50 for plastic theory.

756

756

C hapter   1 8     T orsion

x = dimension of shorter side 90° Compression zone Outline of actual failure surface from test

y = dimension of longer side

Failure plane in tension zone φ1 = 45° NA for skew bending

Figure 18.5.1  Skew bending of plain concrete rectangular section. (According to Hsu [18.21].)

Hsu [18.21] has shown that when ft  (max) is taken at about 5 to 6 fc′ (i.e., representing the stress at which normal-​weight concrete cracks in tension), and α at 1 3 , Eq. (18.5.1) gives the torsional cracking moment strength Tcr, Tcr =



)

(

1 2 x y 5 to 6 fc′ 3

(18.5.2)

Hsu [18.21] has also shown that a torsion failure of a rectangular section does not occur in a spiral form, as might be expected from a circular shaft. Instead, a rectangular section in torsion cracks by bending about an axis parallel to the wider face of the section and inclined at about 45° to the axis of the beam, as shown in Fig. 18.5.1. This is called the skew bending theory (see Section 18.6). An alternative line of thinking makes the analogy to torsion in thin-​walled sections. It is well known from mechanics of materials1 that for closed thin-​walled sections, the shear flow υt t resulting from torsion is

υt t =



T 2 Ao

(18.5.3)

where υt is the shear stress, t is the thickness of the tube wall, T is the torsional moment, and Ao is the area enclosed by the tube (measured at mid-​thickness of the tube wall). When the section is actually solid rather than a tube or box, the wall thickness must be defined. According to MacGregor and Ghoneim [18.86], the Euro-​International Committee (CEB) approximates t as Acp /​pcp, where pcp is the perimeter of the concrete section and Acp is the area enclosed by that perimeter. The Canadian Concrete Code (see Chapter  5, Ref. 5.57) assumes that, prior to cracking, the equivalent wall thickness is 0.75Acp /​pcp and the area Ao enclosed by the tube centerline is 2Acp /​3. Substituting the Canadian values into Eq. (18.5.3) gives

υt =

pcp T pcp T T 3 = = 2 2 Ao t 2 2 Acp 0.75 Acp Acp

(18.5.4)

The principal tensile stress ft (max) resulting from pure torsion equals the shear stress in a thin-​walled tube, that is, Eq. (18.5.4). Then, setting maximum υt equal to ft (max), the

1  See, for example, Charles G. Salmon and John E. Johnson, Steel Structures, Design and Behavior Emphasizing Load and Resistance Factor Design (4th ed). New York: Harper Collins College Publishers, 1996 (pp. 458–​462).

75



757

1 8 . 6   S K E W B E N D I N G T H E O R Y

cracking tensile stress lower bound for concrete in biaxial tension-​compression, assumed to be 4 λ fc′, gives the torsional moment Tcr to cause cracking as

υt = ft (max) =

Tcr pcp Acp 2

= 4 λ fc′



(18.5.5)

 Acp 2  Tcr = 4 λ fc′    pcp  (18.5.6) The cracking stress 5 to 6 fc′ used in the skew bending model for normal-​weight concrete is somewhat higher than the 4 fc′ used in the thin-​walled tube model because the modulus of rupture for tensile strength in bending (i.e., skew bending model) is larger than the principal tensile stress causing cracking for the biaxial tension-​compression state of stress (i.e., pure torsion on a thin-​walled tube).

18.6 STRENGTH OF RECTANGULAR SECTIONS IN TORSION—​S KEW BENDING THEORY Once steel reinforcement, both longitudinal and transverse, has been placed in a rectangular section, the behavioral mechanism changes from that of plain concrete. The resisting action of transverse reinforcement in the form of closed stirrups or hoops is similar to that of stirrups resisting flexural shear. Prior to cracking, the reinforcement participates little if at all; but after cracking, the reinforcement carries a large portion of the total torsional moment. The contribution of concrete is only about 40% of the torsional strength of an unreinforced section. The failure mode according to this theory, however, does continue to be one of skew bending. The skew bending concept was proposed by Lessig [18.20] and extended by Goode and Helmy [18.25], Collins, Walsh, Archer, and Hall [18.26], and Below, Rangan, and Hall [18.27], all of whom applied it to the case of combined bending and torsion. Hsu [18.23] has applied the concept to the case of torsion alone and has developed the expression that formed the basis for the 1971–​1989 ACI Code procedures. Hsu [18.22, 18.24], Zia [18.1, 18.2], and Warwaruk [18.10] have provided summaries of the theories relating to rectangular sections in torsion. McMullen and Rangan [18.7] have reviewed the research to clarify the contradictions between the skew bending and space truss theories (see Section 18.7 for the space truss theory). The following development presents some of the ideas relating to the strength expression developed by Hsu [18.23]. Referring to Fig. 18.6.1, the failure section is assumed to be a plane that is perpendicular to the wider face of the member and inclined at 45° to the axis of the member. The failure plane may be as shown in Fig. 18.6.1, due to twisting moment in the direction indicated. Since a bending mode of failure is assumed, the compression zone is treated as in any beam analysis; that is, it has a depth a over which the compressive stress may be assumed uniform. On the tension side, where the concrete cracks and no resistance to tension is assumed, the reinforcing hoops carry tensile forces Pυ and the longitudinal bars resist shear across the cracked concrete via dowel action (see Section 5.4). The horizontal and vertical components of this dowel action are designated Qx and Qy. As long as the concrete is uncracked and the concrete itself transmits shear, no dowel action exists. On the compression side [Fig. 18.6.2(a)], the longitudinal bars contribute tensile force Pℓ; the concrete contributes shear resistance Ps in the failure plane and also compressive resistance Pc normal to the failure plane. The components of the resultant force P are shown in Fig. 18.6.2(b). The hoop reinforcement in compression is neglected because, as has been shown in Section 3.10, the nominal moment strength Mn is not significantly affected by the inclusion of compression reinforcement.

758

758

C hapter   1 8     T orsion

a

y

x

s Twi

90° Qy

P

Compression zone under skew bending

ent

om

m ting

Qx Pv Pv y√ 2

y

ist

Pv

f tw

so Axi ting

s Twi

45°

Qy

nt me

x

mo

Qx z1

s

s

Figure 18.6.1  Forces acting on skew bending failure section.

On the tension side, it is noted that no longitudinal force can exist. If such a force were to act, it would have to be counterbalanced by a component of resistance acting oppositely. Since only Pυ, Qx, and Qy are assumed to be acting, and the resultant force must be directed upward (opposite P on the compression side), no resultant tension or compression can exist in the longitudinal direction of the tension zone under skew bending. The reader is reminded that unless the direction of the twisting moment is definitely established by analysis, a potential failure plane can exist opposite to that in Fig. 18.6.1, having the compression and tension sides interchanged. Thus, the longitudinal forces, stirrup forces, and dowel forces must be resisted on each side of the section.

Strength Attributable to Concrete The shear resistance Ps (Fig. 18.6.2) may be expressed as

( )

Ps = υavg y 2 a

(18.6.1)

where υavg is the average shear stress acting over the compression zone, y 2 is the width of the compression zone, and a is the depth of the compression zone. Alternatively, the shear strength Ps may be considered proportional to the effective area xy 2 and to fc′ (or λ fc′ in the case of lightweight concrete) [see Eq. (5.5.6) omitting effect of reinforcement]. Thus

Ps = k1 xyλ fc′

(18.6.2)

where k1 is a proportionality constant. From Fig. 18.6.2(b), one may note that

P = 2 Ps + P

(18.6.3)

The first term of Eq. (18.6.3) represents the portion attributable to concrete. Thus the torsional strength Tc attributable to concrete equals the force 2Ps times the moment arm (say, 0.80x), Tc = 2 Ps (arm) = 2 k1 xyλ fc′(0.80 x )

= k2 λ fc′x 2 y



(18.6.4)

759



759

1 8 . 6   S K E W B E N D I N G T H E O R Y

Pl

Pc

Longitudinal bar

P

Ps

Ps

Pl 45°

45° 45°

P

Pc

(a)

45°

(b)

Figure 18.6.2  Components of resultant force P acting on compression zone of failure plane.

Equation (18.6.4) is of the same form as Eq. (18.5.2) but includes the lightweight concrete factor λ. If 6 λ fc′ is used in Eq. (18.5.2), and 40% of that equation is used in recognition that the concrete strength contribution to the total torsional strength of a reinforced section represents only about 40% of the torsional cracking strength of a plain concrete section, then Tc of Eq. (18.6.4) becomes 1  Tc = 0.4  x 2 y(6 λ fc′) 3 

= 0.8λ fc′x 2 y



(18.6.5)

Strength Attributable to Hoops and Longitudinal Reinforcement Referring to Figs. 18.6.1 and 18.6.2, the forces Pυ, Qx, and Qy on the tension side, and Pℓ on the compression side, have yet to be considered. Assuming both the transverse and longitudinal steel have yield strength fy, the contribution of the closed vertical stirrups (or hoops) is y  Pυ = At f y  1   s 



(18.6.6)

where y1/​s (see Fig. 18.6.3) is the number of hoops intercepted by the 45° failure plane. The tensile force Pℓ in the longitudinal bars intercepting the compression concrete zone is A  P = ξ    f y  2  (18.6.7)



where ξ is an efficiency factor to account for location of the longitudinal bars at two or more points within the compression zone, and Aℓ is the total area of all longitudinal bars (assumed to be Aℓ  /​2 in the compression zone). Pℓ contributes to the torsional strength, since from Eq. (18.6.3) it is part of P.

y2 y1

x2 x1

Figure 18.6.3  Cross s​ ection dimensions.

760

760

C hapter   1 8     T orsion

The dowel forces Qx and Qy act after the concrete has cracked and these forces may be assumed proportional to the bar cross-​sectional area and to the bar lateral displacement, which is proportional to the distance (either 0.5x2 or 0.5y2) from the center of twist to the bar. Thus Qx = k3 A y2 , Qy = k3 A x2



(18.6.8)

where k3 is a proportionality constant. Dowel action in beams subject to torsion has been reviewed by Youssef and Bishara [18.84]. Next let m equal the ratio of the volume of longitudinal bars to the volume of closed stirrups or hoops such that m=



A s 2 At ( x1 + y1 )

(18.6.9)

or  2 m( x1 + y1 )  A = At   s  



(18.6.10)

Substitution of Eq. (18.6.10) into Eq. (18.6.7) gives  y   x1 At f y  P = ξm  1 + 1   x1   s  



(18.6.11)

Substituting Eq. (18.6.10) into Eq. (18.6.8) gives  2 mAt ( x1 + y1 )  Qx = k3 y2    s

=2

k3  y2   y   x1 y1 At f y  m 1 + 1    (18.6.12)   f y  y1   x1   s

Similarly, Qy = 2



k3  x2   y   x1 y1 At f y  m 1 + 1      f y  y1   x1   s

(18.6.13)

The torsional resistance from reinforcement then is x  x  y  x  Ts = Pυ  1  + Pt  2  + 2Qx  2  + 2Qy  2   2  2  2  2 



(18.6.14)

Substitution of Eqs. (18.6.6), (18.6.11), (18.6.12), and (18.6.13) into Eq. (18.6.14) gives  x1 y1 At f y  Ts = α t   s 



(18.6.15)

where



αt =

  1 k  y  x  y  1 + ξm  1 + 1   2  + 2 3 m  1 + 1  ( x22 + y22 )   2 x1   2 y1  fy  x1    y1 

(18.6.16)

761



1 8 . 7   S P A C E T R U S S A N A L O G Y

761

Assuming x2 ≈ x1 and y2 ≈ y1, the quantity αt is essentially a function of two parameters, m and y1/​x1 and might be written as

y  α t = C1 + C2 m + C3  1   x1 

(18.6.17)

where the constants C1, C2, and C3 may be experimentally determined. Hsu [18.23] has shown that for equal volume of longitudinal bars to closed hoops (i.e., m = 1), αt may be expressed as

y  α t = 0.66 + 0.33  1   x1 

(18.6.18)

which was used in the 1989 ACI Code. Thus the full nominal strength Tn of rectangular reinforced concrete sections may be written by combining Eqs. (18.6.5) and (18.6.15),



 x1 y1 At f y  Tn = Tc + Ts = 0.8λ fc′x 2 y + α t   s 

(18.6.19)

The first term represents the strength attributable to one rectangle. When the cross section consists of several rectangular segments, Eq. (18.6.19) becomes



 x1 y1 At f y  Tn = Tc + Ts = 0.8λ fc′(∑ x 2 y) + α t   s 

(18.6.20)

where x2y is computed for each segment having its short dimension x and long dimension y.

18.7 STRENGTH OF RECTANGULAR SECTIONS IN TORSION—​S PACE TRUSS ANALOGY Evaluation of torsional strength using the space truss analogy was first suggested by Rausch [18.28] and developed by Lampert and Thürlimann [18.4, 18.29]. The concept was further explained by Lampert and Collins [18.30], Müller [18.33], and Rabbat and Collins [18.34]. Advances in knowledge of reinforced concrete in torsion led to the conclusion that the space truss analogy is likely the simplest way to give unified treatment to flexure and torsion. Collins and Mitchell [18.9] suggested a rational design approach based on the space truss analogy, which led to changes in the ACI Code provisions in 1995 for design to include torsion. The European Concrete Committee (CEB) [18.31] has since about 1978 used the space truss analogy as its basis for design. The torsional strength of reinforced concrete rectangular sections is almost entirely provided by the reinforcement and the concrete that immediately surrounds the steel. Then one may consider a rectangular section as a thin-​walled tube [Fig. 18.7.1(a)] or as a space truss [Fig. 18.7.1(b)]. The longitudinal bars in the corners contribute tensile forces, while the concrete strips between cracks provide compressive strength. The inclined compressive forces act in a spiral fashion around the box section, giving compressive forces on both vertical and horizontal sides. The current provisions for torsion design are well explained by MacGregor and Ghoneim [18.86] and the following presentation is largely from that source. After a solid rectangular concrete member has cracked in torsion, the concrete interior contributes little strength. Thus, such sections can be treated as equivalent tubular members. Before cracking, the section acts as a tube as in Fig.  18.7.1(a), and after cracking

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T

Diagonal compressive stress at angle θ

T

Stirrups

x0

Cracks

y0

Shear flow path

θ

θ A0 Longitudinal tensile force

Longitudinal bar

Shear flow, q

(a) Thin-walled tube analogy

Concrete compression diagonals

V1 V2

V4

V3

(b) Space truss analogy

Figure 18.7.1  Thin-​walled tube and space truss analogies. (From MacGregor and Ghoneim [18.86].)

it acts as a space truss as in Fig. 18.7.1(b). Tests reported by Hsu [18.24] comparing the strengths of solid and hollow sections show that once cracking has occurred, the concrete in the center of the solid member has little effect on torsional strength.

Thin-​Walled Tube Shear Stress As discussed in Section 18.5, the shear flow υt t resulting from torsion in closed thin-​wall sections is

υt t =



T 2 Ao

[18.5.3]

where υt is the shear stress, t is the thickness of the tube wall, T is the torsional moment, and Ao is the area enclosed by the tube (measured at mid-​thickness of the wall). When the section is actually solid rather than a tube or box, the wall thickness must be defined. After cracking has occurred, the torsional strength is provided by the closed stirrups, longitudinal bars, and outer concrete skin. According to MacGregor and Ghoneim [18.86], Ao is empirically taken as 0.85Aoh, where Aoh is the area enclosed by the closed stirrups, measured to the centerline of the outermost hoops. The thickness t is taken as Aoh /​ph, where ph is the perimeter of the centerline of the closed stirrups.

Space Truss Idealization After cracking, the reinforced concrete section becomes essentially a space truss consisting of longitudinal bars in the corners, closed stirrups, and concrete compression diagonals that spiral around the member between torsional cracks [see Fig. 18.7.1(b)]. The width x0 and height y0 of the truss are measured center-​to-​center of the sides of the closed stirrups. The angle θ of the crack to the horizontal is initially about 45° on formation of the crack for nonprestressed beams, but may vary from approximately 30° to 60° as the cracking increases.

Area of Closed Stirrups Equation (18.5.3) gives the shear force per unit length of perimeter, known as shear flow υt t. The total shear force, say V2 in Fig. 18.7.1(b), for a given side of the tube is the shear flow υt t times the length of the side, say y0, to obtain V2. Thus, the shear force in one vertical side and in one horizontal side is

V2 =

T y0 and 2 Ao

V1 =

T x0 2 Ao

(18.7.1)

Similar forces act on all four sides, as shown in Fig. 18.7.1(b), oriented to cause a torsional moment about the axis of the member. Figure 18.7.2 shows a portion of one of the vertical

763



763

1 8 . 7   S P A C E T R U S S A N A L O G Y

At fyt

At fyt

Longitudinal bar

Vertical hoops V2

s

y0

θ

y0 cot θ Longitudinal bar

Figure 18.7.2  Vertical forces on one side of space truss of Fig. 18.7.1(b). (From MacGregor and Ghoneim [18.86].)

sides, where s is the spacing of closed stirrups, θ is the angle of cracks with the member longitudinal axis, y0 is the vertical height to the center of stirrups, and y0 cot θ is the horizontal projection of the crack intercepted by m stirrups. Thus,

ms = y0 cot θ

(18.7.2)

The force in the stirrups crossing the diagonal crack must equal V2. Assume the stirrups yield at nominal torsional strength, then

V2 = mAt f yt

(18.7.3)

where At is the cross-​sectional area of one leg of a closed stirrup, and fyt is the yield stress of the stirrup steel. Substitution of Eq. (18.7.2) into Eq. (18.7.3) gives

V2 =

At f yt y0 s

cot θ

(18.7.4)

Similarly, the shear force V1 will involve the x0 dimension, as follows:

V1 =

At f yt x0 s

cot θ

(18.7.5)

Substituting Eq. (18.7.1) into Eq. (18.7.4) for V2 [or Eq. (18.7.1) into Eq. (18.7.5) for V1], and letting T be the nominal torsional moment strength, Tn , gives

Tn =

2 Ao At f yt s

cot θ

(18.7.6)

Equation (18.7.6) is ACI Formula (22.7.6.1a). Hsu [18.35] has shown how to determine θ by analysis. The value can vary from about 30 to 60°; the angle will be smaller for prestressed concrete (see Chapter 20) than for nonprestressed concrete. From Eq. (18.7.6) the required stirrup cross-​sectional area At per unit spacing s at each side of the section becomes

At Tn T /φ = tan θ = u tan θ s 2 Ao f yt 2 Ao f yt

(18.7.7)

where Tu /​φ is the factored torsional moment Tu divided by the strength reduction factor φ; the quantity Tu /​φ represents the required nominal strength Tn.

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Longitudinal Reinforcement Figure 18.7.3 shows the forces on the side of the space truss of Fig. 18.7.1(b) designated by the shear force V2. The torsional cracks have created inclined concrete struts crossed by one leg of a closed stirrup (or hoop). The shear force V2 can be represented by a diagonal compressive force D2 parallel to the concrete compressive struts along with an axial tensile force N2 (which is divided equally between the top and bottom longitudinal steel). The forces D2 and T2 are D2 =

V2 sin θ

(18.7.8)

N 2 = V2 cot θ



(18.7.9)

Similarly, for the force N1 on the face of width dimension x0 where V1 acts, N1 = V1 cot θ



(18.7.10)

Assuming that the shear flow is constant along the side of the thin-​walled tube, it is reasonable to assume that the resultant forces D2 and N2 act at midheight of a side. Since longitudinal bars are expected to be located in all four corners, the total longitudinal force N for all four sides would be N = 2( N1 + N 2 )



(18.7.11)

Substituting Eqs. (18.7.1) into Eqs. (18.7.9) and (18.7.10), and then into Eq. (18.7.11), gives   T T y0 cot θ + x0 cot θ N = 2 2 Ao  2 Ao 



(18.7.12)

Then replacing T by the nominal strength Tn and N by the nominal strength Nn gives Nn =



Tn 2( x0 + y0 ) cot θ 2 Ao

(18.7.13)

Noting that 2(x0 + y0) is the perimeter ph of the closed stirrups, Eq. (18.7.13) may be written Nn =



Tn ph cot θ 2 Ao

(18.7.14)

The total longitudinal steel force Nn at nominal strength is N n = Al f y



(18.7.15)

N2/2 D2

fcd

y0 y0 cos θ

θ θ

V2

N2 N2/2

Figure 18.7.3  Resolution of shear force on the side of space truss shown in Fig. 18.7.1(b). (From MacGregor and Ghoneim [18.86].)

765



1 8 . 8   C O M B I N E D B E N D I N G A N D T O R S I O N

765

Thus, combining Eqs. (18.7.14) and (18.7.15) gives the total required longitudinal steel area Aℓ as

Al =

Tn ph (T /φ) ph cot θ = u cot θ 2 Ao f y 2 Ao f y

(18.7.16)

Solving for the nominal torsional moment strength Tn, Tn =



2 Al Ao f y ph

tan θ

(18.7.17)

which is ACI Formula (22.7.6.1b).

18.8 STRENGTH OF SECTIONS IN COMBINED BENDING AND TORSION 2 In most practical situations, torsion will occur simultaneously with flexure. There have been many studies of the interaction between bending and torsion [18.4, 18.9, 18.27, 18.29, 18.36–​18.46]. Both the skew bending theory [18.20, 18.26, 18.38] and the space truss analogy [18.4, 18.29] are in general agreement on the interaction behavior. Hsu [18.18] has provided a summary of the interaction relationship. Assuming that the concrete does not participate in carrying flexure, the bending moment M gives rise to a tensile force M/​y0 in the bottom steel and an equal compressive force in the top steel (see Fig. 18.8.1). In addition, the torsional moment T causes a total tensile force N in the longitudinal steel, given by Eq. (18.7.14): Nn =



Tn ph cot θ 2 Ao

(18.7.14)

Assuming that the force N from torsion alone is divided equally between the top and bottom longitudinal steel, 

N top = N bottom =

Tn ph cot θ 4 Ao

(18.8.1)

According to Collins and Lampert [18.30, 18.81] and Hsu [18.18], two failure modes are possible depending on relative amounts of longitudinal reinforcement in the top (flexural compression zone) and in the bottom (flexural tension zone). The first mode involves yielding of both the bottom longitudinal steel and the transverse steel (i.e., closed stirrups or hoops); the second involves yielding of both the top longitudinal steel and the transverse steel. In general, the strength interaction relationship is given by Collins and Lampert [18.81] and explained further by Hsu [18.18].

Symmetrically Reinforced Sections, As′ = As When equal amounts of top and bottom longitudinal reinforcement are used and a positive bending moment acts with torsion, the bottom steel will always yield before the top steel because the compressive force resulting from flexure counteracts the tensile forces arising from torsion. In this case, the addition of a bending moment will reduce the torsional

2  The discussion in this section is for combined action of torsional moment and positive bending moment, which if acting alone would cause tension in the bottom steel As and compression in the top steel As′ . If the bending moment is negative, the reader should interpret As′ as being the bottom steel and As as the top steel.

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+

=

y0 Tph 4Ao

M y0

cot θ

(a) Torsion

(b) Bending moment

–

Tph M cot θ = Ntop + y0 4Ao

+

Tph M cot θ = Nbottom + y0 4Ao

(c) Torsion + bending moment

Figure 18.8.1  Superposition of longitudinal forces resulting from torsion and bending. 1.2 Closed stirrup

1.0 Tn Tn0

0.8 Tn 0.6 Tn0

2

+

Mn =1 Mn0

As’ As

0.4 As = As’

0.2 0

0.2

0.4

0.6

0.8

1.0

Mn Mn0

Figure 18.8.2  Bending-​torsion interaction diagram for equal tension and compression longitudinal steel.

strength. The interaction relationship between bending and torsion for this case may be approximated as follows (see Fig. 18.8.2): 2

 Tn  Mn  T  + M = 1 n0 n0



(18.8.2)

where Tn = nominal torsional moment strength in presence of flexure Tn0 = nominal torsional strength when member is subject to torsion alone Mn = nominal flexural strength in the presence of torsion Mn0 = nominal flexural strength when the member is subject to flexure alone

Unsymmetrically Reinforced Sections, As′ < As When pure torsion is applied to an unsymmetrically reinforced section with the top compression steel area less than the bottom tensile reinforcement area, the top steel will yield first and will govern the failure mode. Addition of a positive bending moment will induce compression in the top steel and allow it to carry larger tensile forces arising from torsion. As a result, the torsional strength of the cross section will increase when a bending moment is applied. One of two failure modes is then possible: (a) if the bending moment is small (i.e., low compressive force in the top steel), then the top steel will yield before the bottom steel or (b) if the bending moment is large, then the bottom steel will yield first. Regardless of the failure mode, the torsional strength of an unsymmetrically reinforced section will increase in comparison to the section under pure torsion. The strength interaction for unsymmetrically reinforced sections can be approximated by the following expressions. For yielding of the bottom (flexural tensile zone) and transverse steel: 2

T  M r n  + n =1 T M  n0  n0



(

)(

)

(18.8.3)

where r = As′ f y′ / As f y and f y′ and fy are the yield stresses of the top and bottom steel, respectively.

76

767

1 8 . 9   C O M B I N E D S H E A R A N D T O R S I O N

Tn Tn0

1 r

Mn Mn0

r = 1 m ste el yie l

1/3 elds yi el

ste

Bo

Tn Tn0

= 1+

=

1.0

2

r

m tto Bo

r = To 1/ p 3 yie ste e ld l s



tto

Tn Tn0

ds

2

= r=

1 r

1–

Mn Mn0

A’s f’y Asfy

1.0

Mn Mn0

Figure 18.8.3  Bending-​torsion interaction relationships. (From Collins and Lampert [18.81].)

For the case of the top steel yielding first, the expression is 2



 Tn  Mn 1  T  − M r = 1 n0 n0

(18.8.4)

Equations (18.8.3) and (18.8.4) are shown graphically in Fig. 18.8.3. Note that the beneficial effect of a bending moment reaches its peak when top and bottom steel yield simultaneously. Since the ratio r is normally less than 1.0—​that is, the tension steel area As required for bending moment strength is significantly larger than the compression steel area As′ used for ductility—​the torsional strength Tn in the presence of bending moment Mn is not reduced by the interaction effect; in fact, for r in the range of 1 3 to 1 2 , the torsional strength increases. For practical purposes, it is reasonable to design for the sum of the requirements for torsion and bending moment. Accordingly, ACI-​9.5.4.5 permits reducing the area of longitudinal torsion reinforcement Aℓ in the flexural compression zone by an amount equal to Mu /​(0.9dfy), where fy is the yield stress of the longitudinal steel. This procedure would be in agreement with the requirement for the top steel in Fig. 18.8.1(c). That reduced amount, however, is not permitted to be less than that required (ACI-​9.6.4.3) as a minimum amount of torsion reinforcement, or the minimum torsion reinforcement required (ACI-​9.7.5.1) to be distributed at a maximum spacing of 12 in. around the perimeter of the closed stirrups.

18.9 STRENGTH OF SECTIONS IN COMBINED SHEAR AND TORSION Rectangular and L-​shaped sections have been studied under combined shear and torsion by a few investigators [18.47–​18.50]. However, since shear usually accompanies flexure, the combination of flexure, shear, and torsion has received more attention [18.51–​18.74]. Generally, though, flexural shear and torsional shear are of significance in regions where bending moment is low. Thus, for design purposes it is most necessary to know the strength interaction relationship between shear and torsion. Test data have provided a wide range of points on the interaction relationship using torsion and shear coordinates. Because of the unknowns involved, some investigators have proposed

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a linear interaction equation [18.1, 18.2, 18.56, 18.65] for design purposes. However, a number of studies at the University of Texas [18.47, 18.49, 18.51, 18.55, 18.65] on rectangular, L-​shaped, and T-​shaped beams have indicated that a quarter-​circle interaction relationship is acceptable for members without web reinforcement. For members with web reinforcement, the interaction is curved but flatter than the quarter-​circle. The quarter-​circle expression is 2

2

 Tn   Vn   T  +  V  = 1 n0 n0



(18.9.1)

where Tn and Vn are the nominal strengths in torsion and shear, respectively, acting simultaneously; Tn0 is the nominal strength under torsion alone; and Vn0 is the nominal strength under shear alone. Equation (18.9.1) was used for the 1989 ACI Code expression for the strength in combined shear and torsion on sections without web reinforcement. However, for web reinforcement the separate requirements for shear and torsion were added together, rather than arrival at by following Eq. (18.9.1).

18.10 STRENGTH INTERACTION SURFACE FOR COMBINED BENDING, SHEAR, AND TORSION The effect of the simultaneous application of bending, shear, and torsion may be most easily examined by means of an interaction surface. Such a concept was used in Section 10.20 for biaxial bending of columns. Various investigators have proposed interaction surfaces. Some of these surfaces are shown in Figs. 18.10.1 and 18.10.2. The work of Collins, Walsh, and Hall [18.56] seems to provide the most complete rational approach by correlating the regions of the failure surface with modes of failure, as shown in Fig. 18.10.2. Tn Tn0

Tn Tn0 1.0

1.0 Discontinuity

1.0

1.0 Vn Vn0

0.5 1.0

Vn Vn0

Mn 1.0 Mn0

Mn Mn0 (a) Hsu interaction surface (Ref. 18.53) Tn Tn0

(b) Mirza-McCutcheon interaction surface (Ref. 18.54)

1.0

1.0 Mn Mn0

Vn Vn0

1.0

(c) Victor-Ferguson interaction surface (Ref. 18.55)

Figure 18.10.1  Interaction surfaces for combined bending, shear, and torsion (for members without transverse reinforcement).

769



18.10  COMBINED BENDING, SHEAR, AND TORSION

769

Since most members that are subjected to bending, shear, and torsion will have less longitudinal steel in the compression zone due to flexure alone (top steel in the positive moment zone) than in the tension zone (bottom steel in the positive moment zone), only that case is given in Fig. 18.10.2(a). When equal amounts of top and bottom longitudinal steel are used, the Mode 3 (“negative” yield or yield in top steel when a positive moment is applied) cannot occur; in such a case the surfaces of Modes 1 and 2 should extend upward until they intersect the Tn /​Tn0 axis. The following conclusions may be drawn regarding the interaction of bending, shear, and torsion. 1. The interaction between torsion and shear may be represented for most situations (with As′ < As) by a quarter-​circle and is relatively unaffected by the simultaneous application of bending moment of a magnitude equal to one-​third to one-​half of the nominal bending strength when shear and torsion are absent. 2. When equal amounts of top and bottom longitudinal steel are used, the quarter-​circle shear–​torsion interaction still seems acceptable, but there is a reduction in strength when bending moment is also applied. 3. Based on most test results, the straight-​line shear-​torsion interaction, while simple to use, appears overly conservative.

Tn Tn0

1.0

Negative yield (Mode 3) Side yield (Mode 2) Shear failure (Mode 4)

Positive yield (Mode 1)

1.0 Vn Vn0

Mn Mn0

1.0 (a) Interaction surface for beam with weaker top steel than bottom steel (As’ < As) Compression zone

(b) Mode 1 (bending and torsion)

(c) Mode 2 (low shear–high torsion)

Top steel

(d) Mode 3 (low bending–high torsion; weaker top steel)

(e) Mode 4 (high shear–low torsion)

Figure 18.10.2  Interaction surface and failure modes (for members with transverse reinforcement) according to Collins, Walsh, and Hall. (From Ref. 18.56.)

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18.11 TORSIONAL STRENGTH OF CONCRETE AND CLOSED TRANSVERSE REINFORCEMENT—​A CI CODE The traditional ACI Code design procedure for torsion prior to the 1995 ACI Code was based largely on the work of Hsu along with recommendations of ACI Committee 438 [18.75–​18.78, 18.80]. Since 1995, however, the design for torsion has been based on the space truss analogy presented in Section 18.7. References are available for special topics such as design of inverted T-​girders [18.102, 18.103], ledger beams [18.104], channel-​shaped sections [18.105–​18.107], and beams having openings [18.108–​18.113].

Threshold Torsion The ACI Code permits (ACI-​9.5.4.1) neglect of torsion in solid sections when the factored torsional moment is less than the threshold torsion, Tth , defined in ACI-​22.7.4.1, multiplied by the strength reduction factor φ. For nonprestressed members,  Acp 2  Tu < φ Tth = φ λ fc′    pcp 

(18.11.1)

 Acp 2  f pc Tu < φ Tth = φ λ fc′   1+ 4 λ fc′  pcp 

(18.11.2)

For prestressed members,



For nonprestressed members subjected to an axial tensile or compressive force,



 Acp 2  Nu Tu < φ Tth = φ λ fc′   1+ 4 Ag λ fc′  pcp 

(18.11.3)

where Acp = area enclosed by outside perimeter of concrete cross section Ag = gross area of concrete pcp = outside perimeter of the concrete cross section fpc = compressive stress in concrete at the centroid of the cross section resisting externally applied loads Nu = factored axial force (taken positive for compression and negative for tension) Comparing Eqs. (18.11.1) through (18.11.3) with Eqs. (18.4.1) through (18.4.3), it is clear that the threshold torsion is equal to one-fourth of the torsional cracking moment strength, assuming a concrete tensile strength of 4 λ fc′. Equations (18.11.1), (18.11.2), and (18.11.3) apply to hollow sections as well, which are defined in ACI-​R22.7.4 as those “having one or more longitudinal voids, such as a single-​ cell or multiple-​cell box girder.” For hollow sections, Acp must be replaced by the gross area Ag of concrete in the above equations (ACI Table 22.7.4.1b). The outer boundaries of the cross section must meet ACI-​9.2.4.4 (see Section 16.4). For isolated members with flanges and for members cast monolithically with a slab, ACI-​9.2.4.4a indicates that the overhanging flange width on each side of the beam web used to compute Acp and pcp shall be taken equal to the greatest projection of the beam above or below the slab, but not greater than four times the flange thickness. ACI-​9.2.4.4b indicates, however, that the “overhanging flanges shall be neglected in cases where the Ag 2 parameter Acp2 / pcp for solid sections and p for hollow sections calculated for a beam with cp flanges is less than that calculated for the same beam ignoring the flanges.”

71



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18.11  TORSIONAL STRENGTH—ACI CODE

Nominal Torsional Moment Strength Tn The nominal torsional moment strength Tn is given by ACI-​22.7.6.1 as the lesser of Tn =



2 Ao At f yt s

cot θ

[18.7.6]

tan θ

(18.7.17)

which is ACI Formula (22.7.6.1a), and Tn =



2 Al Ao f y ph



which is ACI Formula (22.7.6.1b). In Eq. (18.7.6), Ao “shall be determined by analysis” (ACI-​22.7.6.1) except that it is permitted to be taken equal to 0.85Aoh (ACI-​22.7.6.1.1), where Aoh is the area enclosed by the closed stirrups, measured to the centerline of the outermost hoops. The angle θ of potential cracking “shall not be taken less than 30 degrees nor greater than 60 degrees” (ACI-​22.7.6.1). ACI-​22.7.6.1.2 permits θ to be taken as 45° for nonprestressed concrete and 37.5° for prestressed members having an effective prestress force not less than 40% of the tensile strength of the longitudinal reinforcement. The value of 37.5° has been chosen arbitrarily at halfway between the minimum 30° angle and the 45° angle traditionally used in the design of stirrups. According to ACI-​9.7.5.1, torsion reinforcement must consist of longitudinal bars distributed around the perimeter of transverse reinforcement. Transverse torsional reinforcement must consist of closed stirrups or hoops (ACI-​9.7.6.3.1). Details of torsional reinforcement are given in ACI-​9.7.5 and ACI-​9.7.6.3.

Area of Closed Stirrups (Hoops) Required for Torsion As developed in Section 18.7, Eq. (18.7.7) gives the area At of closed stirrup (hoop) reinforcement per unit spacing s as

At Tn T /φ = tan θ = u tan θ s 2 Ao f yt 2 Ao f yt

[18.7.7]

The quantity Tu /​φ represents the required nominal strength Tn. Equation (18.7.7) is essentially ACI Formula (22.7.6.1a). As used traditionally in design, At is the area of one leg of a closed stirrup resisting torsion within a distance s.

Area of Longitudinal Reinforcement Required for Torsion From Section 18.7, the longitudinal reinforcement Aℓ required in addition to that needed for flexure is obtained from Eq. (18.7.16) as

Al =

Tn ph (T /φ) ph cot θ = u cot θ 2 Ao f y 2 Ao f y

(18.7.16)

The perimeter distance ph of the closed stirrups (hoops) is 2(x0 + y0) for a rectangular section, where x0 and y0 are the horizontal and vertical dimensions measured to the center of the closed stirrups (hoops). In general, ph is the perimeter of the centerline of the outermost closed transverse torsional reinforcement. The angle θ is the same as that used to compute the area of closed stirrups in Eq. (18.7.7). Note that to provide control of the width of diagonal cracks, the ACI Code limits the values of fy and fyt to 60,000 psi (ACI-​20.2.2.4).

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Critical Section for Torsion The critical section for torsion is taken essentially the same as for shear. ACI-​9.4.4.3 provides the location of the first critical section for both nonprestressed and prestressed concrete beams. For nonprestressed members, the critical section is to be taken at the effective depth d from the face of support unless a concentrated torsional moment occurs between the face of support and the critical section, in which case the value at the face of support is to be used. The portion of a member between the critical section and the face of support is to be designed for the same (or greater if a concentrated torsion occurs in that region) torsional moment as that at the critical section. The explanation for using the distance d from the face of support is given in Section 5.11 on shear. For prestressed members, the critical section is to be taken at h/​2 from the face of support. The same conditions of the above paragraph apply as for nonprestressed members.

18.12 COMBINED TORSION WITH SHEAR OR BENDING—​A CI CODE Combined Shear and Torsion When torsion and shear act on a member, the nominal strength is reached either when the closed stirrups reach their yield stress or when the longitudinal steel reaches its yield stress. The member may not be considered “serviceable,” however, if inclined cracks are large at service load. The interaction relationship used in the ACI Code relating shear V and torsion T has been traditionally based on limiting crack width. The average shear stress υu resulting from the factored shear Vu is

υu =



Vu bw d

(18.12.1)

where bw is the web width and d is the effective depth. The shear stress υtu resulting from factored torsion Tu using Eq. (18.5.4) taking Ao = 0.85Aoh and t = Aoh /​ph is

υtu = τ =

υtu =

T  1  ph T = u 2 Ao t 2  0.85 Aoh  Aoh

Tu ph 2 1.7 Aoh



(18.12.2)

In the hollow section of Fig. 18.12.1(a), the shear stresses resulting from shear and torsion are additive on the left wall. Thus, algebraic summation of the stresses υu and υtu seems appropriate to compare with a limit stress; thus,

(

)

Vu T p + u h2 ≤ φ υc + 8 fc′ bw d 1.7 Aoh

(18.12.3)

The limit stress on the right-​hand side of Eq. (18.12.3) is the same as the upper limit of ACI-​ 22.5.1.2 of 8 fc′ for the shear stress resisted by the transverse reinforcement. The stress υc is the nominal strength Vc from ACI-​22.5.5 and ACI-​22.5.6 divided by bw d to get it into stress units. Thus, for hollow sections the limit of ACI-​22.7.7.1(b) is given as



 V  Vu T p + u h2 ≤ φ  c + 8 fc′ bw d 1.7 Aoh  bw d 

(18.12.4)

which is ACI Formula (22.7.7.1b). For cases in which the wall thickness is less than Aoh  /​ph, the second term on the left-​hand side of Eq. (18.12.4) shall be taken as Tu / (1.7 Aoh t ) (ACI-​22.7.7.2).

73



1 8 . 1 3   M I N I M U M TO R S I O NA L R E I N F O R C E M E N T — AC I C O D E

Shear stresses due to torsion

Shear stresses due to shear

Shear stresses due to torsion

(a) Hollow section

773

Shear stresses due to shear

(b) Solid section

Figure 18.12.1  Shear stresses due to shear and torsion.

In the solid section of Fig. 18.12.1(b), the shear stresses resulting from direct shear are assumed to be distributed uniformly across the section width, while the torsional shear stresses exist only in the walls of the assumed thin-​walled tube. Thus, since the shear stress can redistribute through the core region, the solid section would be more conservatively treated than the thin-​walled section by using Eq. (18.12.4). Thus, the following square-​root summation is used for the limiting stress on solid sections, 2



2

 Vu   Tu ph    Vc  b d  +  1.7 A2  ≤ φ  b d + 8 fc′ w w oh

(18.12.5)

which is ACI Formula (22.7.7.1a).

Combined Bending and Torsion The ACI Code does not explicitly consider this combination of loadings. Torsion induces an axial force in the longitudinal reinforcement. In regions where bending moment also exists, the flexural reinforcement requirement is added to the longitudinal torsion reinforcement requirement. This is a conservative procedure for the flexural compression zone. In that region the torsion improves the resistance, as shown in Fig. 18.8.1(c). For this reason, ACI-​9.5.4.5 permits reducing the area of longitudinal torsion reinforcement Aℓ in the flexural compression zone by an amount equal to Mu  /​(0.9dfy), where fy is the yield stress of the longitudinal steel. That reduced amount is not permitted to be less than the minimum (ACI-​9.6.4.3) nor less than the minimum (ACI-​9.7.5.1) based on a bar in each closed stirrup corner and a maximum spacing of 12 in. around the perimeter of the closed stirrup. In prestressed concrete beams (ACI-​ 9.5.4.4), the total longitudinal reinforcement (including tendons and nonprestressed steel) shall resist the factored bending moment Mu at that section plus an additional concentric longitudinal tensile force equal to Aℓ fy, based on the factored torsion Tu at that section; ACI-​9.5.4.5, however, permits reducing the area of longitudinal torsion reinforcement in the compression zone below that required by ACI-​ 9.5.4.4 by Mu /​(0.9dfy).

18.13 MINIMUM REQUIREMENTS FOR TORSIONAL REINFORCEMENT—​A CI CODE The minimum area requirements for the transverse reinforcement At and longitudinal reinforcement Aℓ are to ensure that there is ductile behavior prior to failure. Hsu [18.24] and Collins [18.78] have found that the strength attributable to concrete when closed stirrups are present is only about 40% of the torsional strength of a plain concrete member. When

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torsional reinforcement is required, the area At of transverse torsion reinforcement in combination with area Aυ for shear reinforcement must satisfy (ACI-​9.6.4.2),

( Av + 2 At ) ≥ 0.75 fc′

bw s f yt

(18.13.1)

but not less than 50bw s/​fyt . For torsion alone, Eq. (18.13.1) becomes At ≥ 0.375 fc′



bw s f yt

(18.13.2)

but not less than 25bw s/​fyt . The background for the equation for minimum area Aℓ of longitudinal reinforcement is provided by MacGregor and Ghoneim [18.86], whose work is the basis for much of what follows here. Two rectangular reinforced concrete beams in the tests by Hsu [18.24] failed at the torsional cracking load. In those beams, the total volumetric ratio of the sum of closed stirrups and longitudinal reinforcement to the volume of the concrete, respectively, was 0.80 and 0.88%. A beam having volumetric ratio of 1.07% failed at 1.08 times the cracking torsional moment Tcr. All other beams had volumetric ratios exceeding 1.07% and achieved strengths at least 1.2Tcr. Those tests suggest that beams loaded in pure torsion should have a minimum volumetric ratio of 0.9 to 1%, which can be expressed [18.86] A,min s



Acp s

+

At ph ≥ 0.01 Acp s

(18.13.3)

or, solving for Aℓ, min, and recognizing that the yield stress of the stirrups and longitudinal steel may be different, gives

f y A, min = 0.01kAcp −

f yt At ph s



(18.13.4)

where k is a constant assumed to be a function of the concrete strength. Dividing by fy gives



A, min =

0.01k  A   f yt  Acp −  t  ph    s   fy  fy

(18.13.5)

MacGregor and Ghoneim [18.86] indicate that the constant 0.01k, which now has stress (psi) units, can be taken as 7.5 fc′, which then makes Eq. (18.13.5) become



A, min =

7.5 fc′ fy

 A   f yt  Acp −  t  ph    s   fy 

(18.13.6)

During the Committee 318 (ACI Code Committee) balloting process for the 1995 ACI code, various forms of Aℓ,min were discussed. An objective was greater simplicity than the equation of the 1989 ACI Code, while still providing the performance objective. Professors Alan Mattock of the University of Washington and Thomas T. C. Hsu of the University of Houston developed [18.86] the final simplified version of ACI 9.6.4 3(a) as follows,



A, min =

5 fc′Acp fy

f yt A −  t  ph  s fy

(18.13.7)

75



775

18.14 EXAMPLES

In the above equation [ACI-​9.6.4.3(b)],



At 25bw ≥ s f yt

(18.13.8)

where fyt refers to closed transverse torsional reinforcement, and fy refers to longitudinal torsional reinforcement. Equation (18.13.7) appears to give results less than Eq. (18.13.6); however, Hsu in developing Eq. (18.13.6) applied the stress ratio υtu  /​(υtu + υu) to the first term, taking the ratio as 2 3 . That would make Eq. (18.13.6) become the same as the ACI Code equation, using the coefficient 5.

Spacing Limitations Transverse torsion reinforcement (closed stirrups) must be spaced not farther apart than ph /​8 or 12 in. (ACI-​9.7.6.3.3). This reinforcement must be provided over a distance at least bt + d beyond the section where it is no longer required by analysis (ACI-​9.7.6.3.2), where bt is the width of the section containing the torsional transverse reinforcement. Longitudinal bars must have a diameter of at least 0.042 times the stirrup spacing, but not less than #3 in size (ACI-​9.7.5.2) and must be distributed around the perimeter of the closed stirrups with a maximum spacing of 12 in. (ACI-​9.7.5.1). The longitudinal bars or tendons must be inside the closed stirrups, and there must be at least one longitudinal bar or tendon in each corner of the closed stirrups. As in the case of transverse reinforcement, longitudinal torsional reinforcement must be extended at least bt + d beyond the section where it is no longer required by analysis (ACI-​9.7.5.3).

Termination of Torsion Reinforcement Closed stirrups are permitted to be terminated (ACI-​9.7.6.3.2) at a location (bt + d) beyond the section where it is no longer required by analysis (i.e., Tu < φTth; ACI-​9.5.4.1). For solid, nonprestressed beams, this section is located where Eq. (18.11.1) is satisfied.



 Acp 2  Tu = φ Tth = φλ fc′    pcp 

[18.11.1]

where bt is the width of the part of the cross section containing the closed stirrups that are resisting torsion. Similar equations are used for prestressed members [Eq. (18.11.2)] and for nonprestressed members having axial load [Eq. (18.11.3)]. Where the torsional moment exceeds the threshold torsion, torsion reinforcement must be provided.

18.14 EXAMPLES Several examples are presented to illustrate use of the ACI Code procedures. Table 18.14.1, which summarizes most of the ACI provisions on torsion, can help the reader in reviewing the computations in the examples.

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TABLE 18.14.1  ACI CODE PROVISIONS FOR TORSION 1.

For nonprestressed concrete:  Acp 2  Tu = 0 if computed Tu < φλ fc′  .  pcp 

9.5.4.1; 22.7.4.1

 Acp 2  Tu = φ 4 λ fc′   , with redistribution  pcp 

9.4.4.4; 22.7.3.2; 22.7.5.1

For prestressed concrete:  Acp 2  f pc Tu = 0 if computed Tu < φλ fc′   1+ 4 λ fc′  pcp 

9.5.4.1; 22.7.4.1

 Acp 2  f pc Tu = φ 4 λ fc′  , with redistribution  1+ 4 λ fc′  pcp 

9.4.4.4; 22.7.3.2; 22.7.5.1

For nonprestressed members subjected to an axial tensile or compressive force  Acp 2  Nu Tu = 0 if computed Tu < φλ fc′   1+ 4 Ag λ fc′  pcp 

9.5.4.1; 22.7.4.1

 Acp 2  Nu Tu = φ 4 λ fc′  , with redistribution  1+ 4 Ag λ fc′  pcp 

9.4.4.4; 22.7.3.2; 22.7.5.1

where Acp = area enclosed by outside perimeter of concrete cross section pcp = outside perimeter of the concrete cross section Ag = gross area of concrete Nu = factored axial force 2.

2.2

Limitations on cross-​sectional dimensions: For solid sections: 2

2

  Vu   Tu ph   Vc  b d  +  1.7 A2  ≤ φ  b d + 8 fc′     Formula (22.7.7.1a) w w oh

22.7.7.1(a)

where ph = perimeter of centerline of outermost closed transverse torsional reinforcement For hollow sections:  V  Vu T p + u h2 ≤ φ  c + 8 fc′           Formula (22.7.7.1b) bw d 1.7 Aoh  bw d 

22.7.7.1(b)

If wall thickness < Aoh /​ph, second term on left side above becomes  Tu   1.7 A t  oh 3.

φTn ≥ Tu

22.7.7.2

9.5.1.1(c) (Continued)

7



777

18.14 EXAMPLES

TABLE 18.14.1  (Continued) 4.

Nominal torsional strength: Tn = Tn =

2 Ao At f yt s 2 Al Ao f y ph

Formula (22.7.6.1a) cot θ               

22.7.6.1

tan θ              Formula (22.7.6.1b)

Ao = 0.85Aoh unless determined by analysis Aoh = area enclosed by centerline of outermost closed transverse torsional reinforcement 30° ≤ θ ≤ 60° Permitted: θ = 45° for nonprestressed members Permitted: θ = 37.5° for prestressed members having an effective prestress force not less than 40% of the tensile strength of the longitudinal reinforcement 5.

Transverse and longitudinal reinforcement: At T /φ tan θ ≥ u s 2 Ao f yt Al ≥

6.

9.5.1.1(c); 22.7.6.1

(Tu /φ) ph cot θ 2 Ao f y

9.5.1.1(c); 22.7.6.1

Minimum reinforcement: ( Aν + 2 At ) ≥ 0.75 fc′

9.6.4.2

bw s f yt

but not less than 50 bw s  /​fyt A,min = where 7.

5 fc′Acp fy

f yt A −  t  ph  s fy

9.6.4.3(a)

9.6.4.3(b)

At 25bw ≥ s f yt

Spacing of reinforcement: max s (closed stirrups) ≤

ph ≤ 12 in. 8

torsion reinforcement to terminate at (bt + d) beyond theoretical point  longitudinal bars, diameter at least 0.042 times  max s  ≤ 12 in.  stirrup spacing, #3 or larger, one in each corner 

9.7.6.3.3 9.7.6.3.2; 9.7.5.3 9.7.5.2; 9.7.5.1

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EXAMPLE 18.14.1 A reinforced concrete spandrel beam has overall dimensions of 10 × 18 in. and is joined integrally with a 6-​in. slab (based on Example in Ref. 18.53), as shown in Fig. 18.14.1(a). The section shown is that at the critical location a distance d from the face of support. At this section, the factored loads are negative bending moment Mu = 75 ft-​kips, shear force Vu = 18 kips, and torsional moment Tu = 7 ft-​kips. Assume that the torsional stiffness was estimated and used in a structural analysis to obtain these design loadings. Check the adequacy of this section and select the transverse reinforcement steel required, if any, according to the ACI Code using fc′ = 4000 psi (normal weight) and fy = fyt = 40,000 psi. SOLUTION (a) Flexural strength. The required coefficient of resistance Rn is required Rn =



required M n Mu = bd 2 φ bd 2

Assuming φ = 0.90, required Rn =



75(12, 000) = 410 psi 0.90(10)(15.6)2

From Eq. (3.8.5), the required reinforcement ratio ρ is required ρ = 0.011 required As = 0.011(10)15.6 = 1.72 sq in.



provided As = 3(0.79) = 2.37 sq in. > [ required As = 1.72 sq in.]

OK

Available steel area for longitudinal torsion reinforcement, available As = 2.37 − 1.72 = 0.65 sq in.



available As′ = 2(0.60) = 1.220 sq in. hf = 6”

18” d = 15.6”

3 – #8 fc’ = 4000 psi 2 – #7 10”



fy = 40,000 psi

Beam defined by Hsu

3hf = 18”

(a) Given data (from Hsu [18.53]) 3 – #8

#3 hoop @ 5”

#3 hoop @ 5” #5 2 – #7

#3 minimum size, slab design will determine this reinforcement

(b) Details of torsion reinforcement

Figure 18.14.1  Spandrel beam of Example 18.14.1.

(Continued)

79



779

18.14 EXAMPLES

Example 18.14.1 (Continued) (b) Compute maximum Tu to neglect torsion. The maximum factored torsional moment Tu that may be neglected (ACI-​9.5.4.1) is  Acp 2  Tu = φλ fc′    pcp  Acp = 10(18) + 18(6) = 288 sq in. pcp = 2(10 + 18) + 18 + 18 = 92 in.



Note that Acp, the area enclosed by outside perimeter of the concrete section, and pcp, the outside perimeter of the concrete cross section, have included the slab portion (Fig. 18.14.1). Note also that the maximum effective width of slab used is 3 times its thickness, per Hsu [18.53] and within the limits of ACI-​9.2.4.4. Thus,

 2882  1 limit Tu < 0.75(1) 4000  = 3.56 ft-kips  92  12,000

Since Tu = 7 ft-​kips exceeds limit Tu, torsion must be included in the design. (c) Determine whether the section is large enough for combined shear and torsion. From ACI-​22.7.7.1(a), for solid sections, 2

  Vu   Tu ph   Vc  b d  +  1.7 A2  ≤ φ  b d + 8 fc′ w w oh



Assuming 1.5-​in. cover to outside of closed #3 stirrups,

Aoh = [18 − 2(1.5) − 0.375][10 − 2(1.5) − 0.375] = 96.9 sq in.

Conservatively, the effect of the integral slab is neglected. The perimeter ph of the centerline of outermost closed transverse torsional reinforcement, again neglecting the integral slab, is ph = 2(14.625 + 6.625) = 42.5 in. Vu 18.0(1000) = = 115 psii bw d 10(15.6) Tu ph 7.0(12, 000)42.5 = = 224 psi 2 1.7 Aoh 1.7(96.9)2





2

2

 Vu   Tu ph  2 2  b d  +  1.7 A2  = (115) + (224) = 252 psi w oh  V  φ  c + 8 fc′ = 0.75(2 fc′ + 8 fc′) = 0.75(10 fc′) = 474 psi  bw d  Since 474 psi exceeds 252 psi, the section has adequate size. (d) Compute stirrup reinforcement required for shear.

φVn ≥ Vu φ (Vc + Vs ) ≥ Vu 1



φVc = φ 2 λ fc′bw d = 0.75(2)(1) 4000 (10)15.6 1000 = 14.8 kips φVs = φ

Av f yt d s

= 0.75



Av (40, 000)15.6 468 Av ( kips) = s(1000) s (Continued)

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Example 18.14.1 (Continued) Since Vu > φVc, the requirement for transverse reinforcement based on strength is Av Vu − φ Vc 18 − 14.8 = = = 0.0068 s φ d f yt 468



This amount will be combined with the torsion requirement, and the total Av + 2At must be at least 0.75 fc′ bw s/​fyt but not less than 50bw s/​fyt (ACI-​9.6.4.2). (e) Compute transverse torsional reinforcement requirement. Using Eq. (18.7.7), At T /φ tan θ = u s 2 Ao f yt



Ao = 0.85 Aoh = 0.85(96.9) = 82.4 sq in.



Taking θ = 45° as permitted by ACI-​22.7.6.1.2(a) gives 7.0(12,000) At Tu /φ 0.75 (1) = 0.017 = tan θ = s 2(82.4)(40, 000) 2 Ao f yt



(f) Compute transverse reinforcement for combined shear and torsion. The total transverse reinforcement required for strength is

( Av + 2 At ) = 0.0068 + 2(0.017) = 0.041 s



b 10  A + 2 At  = 0.75 fc′ w = 0.75 4000 min  v = 0.012    s f yt 40, 000



but not less than 50

Controls!

bw (10) = 50 = 0.013 f yt 40, 000

For #3 closed hoops, max s =



2(0.11) = 5.4 in. 0.041

Since this exceeds the spacing limitation of ACI-​9.7.6.3.3, max s =



ph 42.5 = = 5.3 in. 8 8

or 12 in., the 5.3-​in. maximum spacing controls. Use #3 hoops at 5-​in. spacing. (g) Longitudinal torsional reinforcement. Using Eq. (18.7.16) gives



7.0(12,000) (42.5) (Tu /φ) ph 0.75 (1) = 0.72 sq in. A = cot θ = 2(82.4)40, 000 2 Ao f y

Check Aℓ,min

 25(10)  At   bw  s = 0.017 ≥ 25 f = 40, 000 = 0.0063    yt 

(Continued)

781



781

18.14 EXAMPLES

Example 18.14.1 (Continued) The strength-​related value (0.017) exceeds 25bw  /​fyt; thus A, min =

=

5 fc′Acp fy

f yt A −  t  ph  s fy

5 4000 (288)  40, 000  = 2.28 − 0.72 = 1.56 sq in. − 0.017(42.5)   40, 000  40, 000

The amount of 1.56 sq in. is to be distributed around the perimeter of the section at a spacing not to exceed 12 in. In this case, bars must be placed at middepth, A, min



3

=

1.56 = 0.52 sq in. 3

Use 2–​#5 bars at middepth. The longitudinal steel at the top and bottom in excess of that required for flexure is more than adequate for the longitudinal torsion requirement (0.52 sq in.). Also, #3 transverse reinforcement should be placed in the effective flange portion of the slab and should be anchored within the main rectangle resisting torsion. Though the ACI code requires only standard 90° hooks if those are restrained from opening by a slab [ACI-​25.7.1.6(b)], Mitchell and Collins [18.83] have recommended the use of 105° hooks (the 105° is the amount of bend from the initial straight bar). Furthermore, when longitudinal steel is to carry torsion at the face of support, the steel should be embedded into the support by an amount Ld [18.83]. The reinforced section is shown in Fig. 18.14.1(b).

EXAMPLE 18.14.2 For the continuous spandrel beam shown in Fig. 18.14.2, design the longitudinal and transverse reinforcement for the factored moment, factored flexural shear, and factored torsional moment given in Fig.  18.14.2. Assume that these design loads were obtained after estimating the torsional stiffness and performing a structural analysis. Use fc′ = 4000 psi (normal weight) and fy = fyt = 60,000 psi. SOLUTION (a) Flexural strength. The effective depth d is approximately 21.5 in. for one layer of reinforcement. At midspan, neglecting the T-​beam effect, the required coefficient of resistance Rn (assuming φ = 0.90) is required Rn =

required M n Mu = 2 bd φ bd 2

83.3(12,000) = = 200 psi 0.90(12)(21.5)2

From Fig. 3.8.1, the required reinforcement ratio ρ ≈ 0.0048, which gives required As = 0.0048(12)21.5 = 1.24 sq in. The minimum reinforcement required (ACI-​9.6.1.2) is

As , min =

3 fc′ fy

bw d =

3 4000 (12)21.5 = 0.82 sq in. 60, 000

(Continued)

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Example 18.14.2 (Continued) but not less than 200bw d 200(12)21.5 = = 0.86 sq in. fy 60, 000

As , min =



Controls!

At the support,

required As = 1.24(61.7 / 83.3) = 0.92 sq in.

6”

A 24” 30’–0” 2’–0”

12”

A

Section A–A

2’–0” (a) Beam 100

CL of support

Factored moment (ft-kips)

83.3’k

Face of support

80 60 40

CL of span

20 0 20 40 60

61.7’k

80

Factored shear (kips)

(b) Factored bending moment envelope 21.0k

20

19.4k

15 10

d 5.6

5

Factored torsion (ft-kips)

(c) Factored shear envelope 60

39.2’k

40

35.5’k

20

10.2’k 2

4

6

8

10

12

14

(d) Factored torsion envelope

Figure 18.14.2  Spandrel beam including design actions, for Example 18.14.2. (Adapted from Mattock [18.75].)

(Continued)

783



783

18.14 EXAMPLES

Example 18.14.2 (Continued) The requirements for the top and bottom reinforcement along the span due to flexure are shown in Fig. 18.4.3(a) and 18.4.3(b), respectively. The reader may note that less reinforcement than 0.86 sq in. may be used if the amount used is one-​third more than required for strength (ACI-​9.6.1.3). Thus, the sloping straight line of Fig. 18.14.3(a) and 18.14.3(b) represents approximately 4 3 of the required As in the region between the controlling minimum requirements of ACI-​9.6.1.3 and 9.7.3.8.2 (or ACI-​9.7.3.8.4). (b) Factored shear Vu and factored torsion Tu at the critical section. Unless a concentrated load (or girder framing in) exists between the face of support and the distance d therefrom, the maximum shear and torsion for which strength must be provided is at d from the face of support (ACI-​9.4.3.2 and ACI-​9.4.4.3). From Fig. 18.14.2, Vu = 19.4 kips



Tu = 35.5 ft-kips



(c) Compute maximum Tu to neglect torsion. The maximum factored torsional moment Tu that may be neglected (ACI-​9.5.4.1) is  Acp 2  Tu < φλ fc′    pcp 

[18.11.1]

Acp = 12(24) = 288 sq in.



pcp = 2(12 + 24) = 72 in. Note that in calculating Acp, the area enclosed by the outside perimeter of the concrete section, and pcp, the outside perimeter of the concrete cross section, the slab portion has been neglected (Fig. 18.14.2). Thus,

 2882  1 limit Tu = 0.75(1) 4000  = 4.6 ft-kips  72  12, 000

Since Tu of 35.5 ft-​kips far exceeds limit Tu, torsion must be included in design. (d) Determine whether the section is large enough for combined shear and torsion. From ACI-​22.7.7.1(a), for solid sections, 2

2



  Vu   Tu ph   Vc  b d   1.7 A2  ≤ φ  b d + 8 fc′ w w oh

Assuming 1.5-​in. cover to outside of closed #3 stirrups, Aoh = [24 − 2(1.5) − 0.375][12 − 2(1.5) − 0.375] = (20.625)8.625 = 178 sqq in. Conservatively, the effect of the integral slab is neglected. The perimeter ph of the centerline of outermost closed transverse torsional reinforcement, again neglecting the integral slab, ph = 2(20.652 + 8.625) = 58.5 in. Vu 19.4(1000) = = 75 psi bw d 12(21.5) Tu ph 35.5(12, 000)58.5 = = 463 psi 2 2 1.7 Aoh 1.7(178)

2

2

 Vu   Tu ph  2 2  b d  +  1.7 A2  = (75) + (463) = 469 psi w oh (Continued)

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Example 18.14.2 (Continued)

Area (sq in.)

(a)

1.24 sq in. 1.0 0.5

Area (sq in.)

1.0

ACI–9.6.1.3 0.31 2

6

4

6

10

12 14 ph /8 = 7.3” 5”

Spacing of #3 closed stirrups

4”

4 3” 12@312 ”

12@4” 4’–0

3’–6” 4

6

(d)

15@5” 6’–3” 10

8 Feet

12

14

2 – #7,1 – #6, As = 1.64 1.5

Area (sq in.)

8

1.8’

2

Area of top reinforcement required for flexure

ACI–9.7.3.8.4

Face of support d

2

1.0 0.5

0.92 d 1.8’ 2

1.5 1.0 0.5

1.29

1.14 0.88 bt + d 2.8’

4

1.57

1.34

2 – #7, As = 1.20

(e)

Area (sq in.)

14

0.86 (Min As for ACI–9.6.1.2)

(c) Stirrup spacing (in.)

12 6 8 10 Min As (ACI–9.7.3.8.2)

4

0.92 sq in.

0.5

8

Area of bottom reinforcement required for flexure

Min A s ACI–9.6.1.2

0.31 2

(b)

0.86

ACI–9.6.1.3

Total area of bottom reinforcement required for beam

ACI–9.7.5.3

6

8

10

12

14

4 – #6, As = 1.76

1.52 1.47 d 1.8’

2

bt + d 0.88

4

ACI–9.7.5.3 2 – #6, As = 0.88

Total area of top reinforcement required for beam

0.81

6

0.43

0.35

0.28

8

10

12

Feet

0.33 14

Figure 18.14.3  Reinforcement requirements for spandrel beam of Example 18.14.2.

Using the simplified expression of AC1-​22.5.5.1 for Vc with λ = 1 gives

 V  φ  c + 8 fc′ = 0.75 2 λ fc′ + 8 fc′ = 0.75 10 fc′ = 474 psi  bw d 

(

)

(

)

Since 474 psi exceeds 469 psi, the section has adequate size. (e) Compute stirrup reinforcement required for shear.

1 φVc = φ2 λ fc′bw d = 0.75(2)(1) 4000 (12)21.5 1000 = 24.5 kips

(Continued)

785



785

18.14 EXAMPLES

Example 18.14.2 (Continued) Since Vu > φVc /​2 (ACI-​9.6.3.1), transverse shear reinforcement is required, but there is no computed strength requirement because Vu < φVc. The combined shear and torsion transverse reinforcement must be equal to or greater than the minimum requirement of ACI-​9.6.4.2. (f) Compute transverse torsional reinforcement requirement. Using Eq. (18.7.7), At T /φ tan θ = u s 2 Ao f yt





Ao = 0.85 Aoh = 0.85(178) = 151 sq in. Taking θ = 45°, as permitted by ACI-​22.7.6.1.2(a), gives at d from the face of support, At T /φ 35.5(12,000) / 0.75 = u tan θ = = 0.031 s 2 Ao f yt 2(151)60, 000



(g) Compute transverse reinforcement for combined shear and torsion. The total transverse reinforcement required for strength is



Av 2 At + = 0 + 2(0.031) = 0.062 s s b 12 = 0.0095 min( Av + 2 At ) / s = 0.75 fc′ w = 0.75 4000 f yt 60, 000

Controls!

but not less than

50

bw 12 = 50 = 0.010 f yt 60, 000

For #3 closed hoops,

max s =

2(0.11) = 3.5 in. 0.062

The upper limit on spacing is given by ACI-​9.7.6.3.3,

max s =

ph 58.5 = = 7.3 in. 8 8

but not to exceed 12 in. In this case, the strength requirement controls. Use #3 hoops at 3.5-​in. spacing. (h) Longitudinal torsional reinforcement. Using Eq. (18.7.16) gives at d from face of support



 35.5    (12, 000)58.5 (Tu /φ) ph 0.75  (1) = 1.83 sq in. At = cot θ = 2 Ao f y 2(151)(60, 000)

Check Aℓ,min:

  At   25bw 25(12)  s = 0.031 ≥  f = 60, 000 = 0.005   yt   (Continued)

786

786

C hapter   1 8     T orsion

Example 18.14.2 (Continued) The strength-​related value (0.031) for At /​s exceeds 25 bw  /​fyt. A,min =

5 fc′Acp fy

f yt A −  t  ph  s fy

5 4000 (288)  60, 000  = 1.52 − 1.81 = negative = − 0.031(588.5)   60, 000  60, 000

The strength requirement (1.83 sq in.) controls. This amount is to be distributed around the perimeter of the section at a spacing not to exceed 12 in. (ACI-​9.7.5.1). In this case, bars must be placed at middepth, At 1.83 = = 0.61 sq in. 3 3



Use 2–​#5 bars at middepth (one at each side face). (i) Total longitudinal reinforcement for flexure and torsion. At the face of support, top steel = 0.92 (flexure) + 0.61(torsion) = 1.53 sq in. 1  bottom steel = 0.31  of 1.24 flexure + 0.61 (torsion) = 0.92 sq in. 4 



At d from face of support, top steel = 0.86 (flexure) + 0.61(torsion) = 1.47 sq in. 1  bottom steel = 0.31  of 1.24 flexure + 0.61 (torsion) = 0.92 sq in. 4 



(j) Transverse steel requirement along the span. Using the same procedure as at the first critical section, the requirements are computed as shown in Table 18.14.2. The maximum spacing curve and selected spacings are shown in Fig. 18.14.3(c). Since the  Acp 2  factored torsional moment Tu is never less than φλ fc′   = 4.6 ft-kips [see part  pcp  (c) above], closed stirrups are required along the entire span. TABLE 18.14.2  TRANSVERSE STEEL REQUIREMENT ALONG THE SPAN FOR EXAMPLE 18.14.2 Location from Center of Support (ft) d (1.8) 4 6 8 10 12 15 a b

Requiredb Vu (kips)

Tu (ft-​kips)

At s (in.)

2At Av + s s (in.)

max s for #3 hoops (in.)

19.4 17.7 15.5 13.3 11.1 8.9 5.6

35.5 33.0 28.9 24.7 20.6 16.4 10.2

0.031 0.0291 0.0255 0.0218 0.0182 0.0145 0.0090

0.062 0.0582 0.0510 0.0436 0.0364 0.0290 0.0180

3.5 3.8 4.3 5.0 6.0 7.3a 7.3a

max ph /​8 controls. (Aυ /​s) = 0 for strength requirement because Vu < φ Vc. All values exceed the minimum (0.010) of ACI-​9.6.4.2.

(Continued)

78



787

18.14 EXAMPLES

Example 18.14.2 (Continued) (k) Longitudinal steel requirement along the span. The flexural requirements have already been shown in Fig. 18.14.3(a) and 18.14.3(b). The torsional requirements are computed in Table 18.14.3. Because the member depth exceeds 12 in., the longitudinal torsional steel is placed one-​third at each of top, bottom, and midheight. The sums of the flexural requirement and Aℓ /​3 are shown as the total requirement for longitudinal steel in Fig. 18.14.3(d) and 18.14.3(e). In determining the length of longitudinal bars, a conservative interpretation has been made of ACI-​9.7.5.3, which requires an extension of (bt + d) beyond the theoretical termination point, as in Fig. 18.14.3(d) and 18.14.3(e). The theoretical termination point is taken as that for combined flexure and torsion. It is noted that in some cases ACI-​9.7.3.5 regarding cutting bars in a tension zone may control. The crack control provisions of ACI-​24.3.2 must be checked, as well as the deflection if excessive deflection may cause damage. The design details are shown in Fig. 18.14.4. TABLE 18.14.3  LONGITUDINAL STEEL REQUIREMENT ALONG THE SPAN FOR EXAMPLE 18.14.2 Location from Center of Support (ft) d (1.8) 4 6 8 10 12 15 a b

Aℓa (sq in.)

Aℓ,minb (sq in.)

Required Aℓ /​3 (sq in.)

Required As(top) (sq in.)

Required As(bottom) (sq in.)

1.83 1.70 1.49 1.28 1.06 0.85 0.53

negative negative 0.03 0.24 0.46 0.67 0.99

0.61 0.57 0.50 0.43 0.35 0.28 0.33

1.47 0.88 0.81 0.43 0.35 0.28 0.33

0.92 0.88 1.14 1.29 1.34 1.42 1.57

Eq. (18.7.16); from ACI Formula (22.7.6.1b). Eq. (18.13.8); ACI-​9.6.4.3(a).

A

B

6’– 0” 4’– 3” 4 – #6

2 – #7 12@3 12 = 3’– 6”

2’– 0”

2 – #6

12@4 = 4’– 0”

2 – #7 + 1– #6 15@5 = 6’– 3”

#3 Closed stirrups

A

B 6”

4 – #6

2 – #6

#5

#5 12”

CL of span

2 – #7

Section A–A

Figure 18.14.4  Design details for Example 18.14.2.

#7 #6

Slab reinforcement to be well anchored within closed hoops (design not shown) Section B–B

78

788

C hapter   1 8     T orsion

EXAMPLE 18.14.3 Redesign the spandrel beam shown in Fig. 18.14.2 taking the option permitted in ACI-​ 22.7.3.2 to allow for redistribution of forces. The torsional member is then designed for  Acp 2  a factored torsion Tu of φ 4 λ fc′   (ACI-​9.4.4.4; ACI-​22.7.3.2; ACI-​22.7.5.1). Use  pcp  fc′ = 4000 psi (normal weight) psi and fy = fyt = 60,000 psi. SOLUTION (a) Flexural strength. Assume that the factored bending moment Mu [Fig. 18.14.2(b)] and factored shear Vu [Fig. 18.14.2(b)] are not significantly affected by a change in the assumption for torsional stiffness. Note that most traditional structural analysis methods do not include the effect of torsional stiffness; thus, this approach is simpler. On completion of the structural analysis, disregarding torsional stiffness, any member that needs to undergo torsional deformation is then designed to include the ACI Code–​specified Tu. For this example, the requirements for flexure are as in Example 18.14.2 and are shown in Fig. 18.14.3(a) and 18.14.3(b). (b) Factored shear Vu and factored torsion Tu for which strength must be provided. At d from the face of support, Vu = 19.4 kips

 Acp 2   2882  1 = 18.2 ft-kips Tu = φ 4 λ fc′   = 0.75(4)(1) 4000   12 000 , p 72    cp 

Note that the 18.2 ft-​kips is significantly lower than the maximum factored torsional moment Tu of 35.5 ft-​kips, which was designed for at d from the face of support in Example 18.14.2. (c) Determine whether the section is large enough for combined shear and torsion. From ACI-​22.7.7.1(a), for solid sections, 2

2



  Vu   Tu ph   Vc  b d  +  1.7 A2  ≤ φ  b d + 8 fc′ w w oh

This check was made in Example 18.14.2, where the combined stress was close to the maximum given by the above equation. From Example 18.14.2, Aoh = 178 sq in.



ph = 58.5 in.



The effect of the integral slab is neglected. Vu 19.4(1000) = = 75 psi bw d 12(21.5) Tu ph 18.2(12, 000) 58.5 = = 237 psi 2 1.7(178)2 1.7 Aoh

2

2

 Vu   Tu ph  2 2  b d  +  1.7 A2  = (75) + (237) = 248 psi w oh

(Continued)

789



789

18.14 EXAMPLES

Example 18.14.3 (Continued) and the limit for λ = 1 is

 V  φ  c + 8 fc′ = 0.75 2 λ fc′ + 8 fc′ = 0.75 10 fc′ = 474 psi  bw d 

)

(

)

(

The 474-​psi limit far exceeds the combined stress value of 237 psi. This would permit a smaller cross section. Under the redistribution procedure of ACI-​22.7.3.2, the section could clearly be smaller than 12 × 24. However, use the larger section for the remainder of this example. (d) Compute stirrup reinforcement required for shear. In Example 18.14.2, it was shown that there is no strength requirement for Vu alone because Vu < φVc. The combined shear and torsion transverse reinforcement must be equal to or greater than the minimum requirement of ACI-​9.6.4.2. (e) Compute transverse torsional reinforcement requirement. Using Eq. (18.7.7), At T /φ tan θ = u s 2 Ao f yt





Ao = 0.85 Aoh = 0.85(178) = 151 sq in. Taking θ = 45°, as permitted by ACI-​22.7.6.1.2(a), gives at d from the face of support, At T /φ 18.2(12, 000) / 0.75 = u tan θ = = 0.0161 s 2 Ao f yt 2(151)60, 000



(f) Compute transverse reinforcement for combined shear and torsion. The total transverse reinforcement required for strength is



Av 2 At + = 0 + 2(0.0161) = 0.032 s s b 12 = 0.0095 min( Av + 2 At ) /s = 0.75 fc′ w = 0.75 4000 f yt 60, 000

Controls!

but not less than

50

bw 12 = 50 = 0.010 f yt 60, 000

For #3 closed hoops,

max s =

2(0.11) = 6.9 in. 0.032

The upper limit on spacing is given by ACI-​9.7.6.3.3,

max s =

ph 58.5 = = 7.3 in. 8 8

and always less than 12 in. In this case, the strength requirement controls. Thus for this beam a practical spacing of 6 in. may be used for the entire beam. Use #3 hoops at 6-​in. spacing along the entire span. (Continued)

790

790

C hapter   1 8     T orsion

Example 18.14.3 (Continued) (g) Longitudinal torsional reinforcement. Using Eq. (18.7.16) gives at d from face of support  18.2    (12, 000)58.5 (Tu /φ) ph 0.75  (1) = 0.94 sq in. A = cot θ = 2 Ao f y 2(151)(60, 000)

Check Aℓ, min

  At   25bw 25(12)  s = 0.0161 ≥  f = 60, 000 = 0.005   yt  



The strength-​related value (0.0161) for At /​s exceeds 25bw /​fyt . A,min =

5 fc′Acp fy

f yt A −  t  ph  s fy

5 4000 (288)  60, 000  = 1.52 − 0.94 = 0.58 sq in. = − 0.0161(58.5)   60, 000  60, 000

The strength requirement (0.94 sq in.) controls. This amount is to be distributed around the perimeter of the section at a spacing not to exceed 12 in. (ACI-​9.7.5.1). In this case, bars must be placed at middepth, A 0.94 = = 0.31 sq. in 3 3



Area (sq in.)

(a)

Area (sq in.)

(b)

Face of support 1.5 0.95

1.0

1.17

1.30

1.45

1.55 Total area of bottom reinforcement required for beam

0.62 0.5

1.5

d 1.23

Total area of top reinforcement required for beam

1.17

1.0

0.62 0.5

0.31

2

4

6

8

0.31 10

12

14

Feet

Figure 18.14.5  Longitudinal steel requirements for Example 18.14.3.

Note that the amount Aℓ will be divided into three equal parts, top of beam, bottom of beam, and midheight in accordance with ACI-​9.7.5.1. Use 2–​#4 bars at middepth (one at each side face). (Continued)

791



SELECTED REFERENCES

791

Example 18.14.3 (Continued) (h) Total longitudinal reinforcement for flexure and torsion. At the face of support,

top steel = 0.92 (flexure) + 0.31 (torsion) = 1.23 sq in. bottom steel = 0.31 (flexure) + 0.31 (torsion) = 0.62 sq in.

At d from face of support,

top steel = 0.86 (flexure) + 0.31 (torsion) = 1.17 sq in. bottom steel = 0.31 (flexure) + 0.31 (torsion) = 0.62 sq in.

For the top reinforcement, note that at d from the face, the ACI-​9.6.1.2 minimum was conservatively used instead of the 4 3 of a lesser value at location d permitted by ACI-​9.6.1.3. (i) Longitudinal steel requirement along the span. The flexural requirements were shown in Fig. 18.14.3(a) and 18.14.3(b). The sums of the flexural requirement and Aℓ /​3 are shown as the total requirement for longitudinal steel in Fig. 18.14.5. Note that this alternate method requires many fewer hoops (i.e., #3 @ 6 throughout); likewise, less longitudinal reinforcement is needed. When an actual redistribution is made, the shear and flexure requirements will increase somewhat; however, for this example, there will still be considerable difference (primarily in closed stirrups).

SELECTED REFERENCES General   18.1 Paul Zia. “Torsion Theories for Concrete Members,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 103–​132).   18.2 Paul Zia. “What Do We Know about Torsion in Concrete Members?” Journal of the Structural Division, ASCE, 96, ST6 (June 1970), 1185–​1199.   18.3 K. G. Tamberg and P. T. Mikluchin. “Torsional Phenomena Analysis and Concrete Structure Design,” Analysis of Structural Systems for Torsion (SP-​35). Detroit:  American Concrete Institute, 1973 (pp. 1–​102).   18.4 Paul Lampert. “Postcracking Stiffness of Reinforced Concrete Beams in Torsion and Bending,” Analysis of Structural Systems for Torsion (SP-​35). Detroit: American Concrete Institute, 1973 (pp.  385–​433). (Also presented at 1971 Annual Convention, American Concrete Institute, Denver, March 1971.)   18.5 Michael P. Collins and Paul Lampert. “Redistribution of Moments at Cracking—​The Key to Simpler Torsion Design,” Analysis of Structural Systems for Torsion (SP-​35). Detroit: American Concrete Institute, 1973 (pp. 343–​383).   18.6 David J.  Victor, Narayanan Lakshmanan, and Narayanan Rajagopalan. “Ultimate Torque of Reinforced Concrete Beams,” Journal of the Structural Division, ASCE, 102, ST7 (July 1976), 1337–​1352.  18.7 Arthur E.  McMullen and B.  Vijaya Rangan. “Pure Torsion in Rectangular Sections—​ A Re-​Examination,” ACI Journal, Proceedings, 75, October 1978, 512–​519.   18.8 Madhusudan Chakraborty. “Ultimate Torque of Reinforced Rectangular Beams,” Journal of the Structural Division, ASCE, 105, ST3 (March 1979), 653–​668.   18.9 Michael P. Collins and Denis Mitchell. “Shear and Torsion Design of Prestressed and Non-​ Prestressed Concrete Beams,” PCI Journal, 25, September/​October 1980, 32–​100. Disc., 26, November/​December 1981, 96–​118. 18.10 J.  Warwaruk. “Torsion in Reinforced Concrete,” Significant Developments in Engineering Practice and Research (SP-​72). Detroit: American Concrete Institute, 1981 (pp. 247–​277). 18.11 Himat T.  Solanki. “Reinforced Concrete Beams in Pure Torsion,” Journal of the Structural Division, ASCE, 108, ST12 (December 1981), 2243–​2250.

792

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18.12 Thomas T. C. Hsu. Torsion of Reinforced Concrete. New York: Van Nostrand Reinhold, 1983. 18.13 Thomas T. C. Hsu and Y. L. Mo. “Softening of Concrete in Torsional Members—​Theory and Tests,” ACI Journal, Proceedings, 82, May–​June 1985, 290–​303. Disc., 83, July–​August 1986, 690–​693. 18.14 Peter Marti and Keith Kong. “Response of Reinforced Concrete Slab Elements to Torsion,” Journal of Structural Engineering, ASCE, 113, 5 (May 1987), 976–​993. 18.15 Peter Marti, Peter Leesti, and Waseem U. Khalifa. “Torsion Tests on Reinforced Concrete Slab Elements,” Journal of Structural Engineering, ASCE, 113, 5 (May 1987), 994–​1010. 18.16 Thomas T.  C. Hsu. “Softened Truss Model Theory for Shear and Torsion,” ACI Structural Journal, 85, November–​December 1988, 624–​635. 18.17 Atef H. Bakhsh, Faisal F. Wafa, and Ali A. Akhtaruzzaman. “Torsional Behavior of Plain High-​ Strength Concrete Beams,” ACI Structural Journal, 87, September–​October 1990, 583–​588. 18.18 Thomas T. C. Hsu. Unified Theory of Reinforced Concrete. Boca Raton, FL: CRC Press, 1993. 18.19 Joint ACI-​ASCE Committee 445. “Report on Torsion in Structural Concrete,” Report 445.1R-​ 12. Farmington Hills, MI: American Concrete Institute 2013, 92 pp.

Skew Bending Concept 18.20 N. N. Lessig. “Determination of Load Carrying Capacity of Reinforced Concrete Element with Rectangular Cross Section Subjected to Flexure with Torsion,” Work No. 5. Moscow: Institut Betonai Zhelezobetona (Concrete and Reinforced Concrete Institute), 1959 (pp. 5–​28). (Also available as Foreign Literature Study No. 371, PCA Research and Development Laboratories, Skokie, IL.) 18.21 Thomas T.  C. Hsu. “Torsion of Structural Concrete—​Plain Concrete Rectangular Sections,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 203–​ 238). (Also Portland Cement Association Development Department Bulletin D134.) 18.22 Thomas T. C. Hsu. “Torsion of Structural Concrete—​A Summary of Pure Torsion,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 165–​178). (Also Portland Cement Association Development Department Bulletin D133.) 18.23 Thomas T.  C. Hsu. “Ultimate Torque of Reinforced Rectangular Beams,” Journal of the Structural Division, ASCE, 94, ST2 (February 1968), 485–​ 510. (Also Portland Cement Association Development Department Bulletin D127.) 18.24 Thomas T.  C. Hsu. “Torsion of Structural Concrete—​ Behavior of Reinforced Concrete Rectangular Members,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 261–​306). (Also Portland Cement Association Development Department Bulletin D135.) 18.25 C. D. Goode and M. A. Helmy. “Ultimate Strength of Reinforced Concrete Beams in Combined Bending and Torsion,” Torsion of Structural Concrete (SP-​18). Detroit:  American Concrete Institute, 1968 (pp. 357–​377). 18.26 M.  P. Collins, P.  F. Walsh, F.  E. Archer, and A.  S. Hall. “Ultimate Strength of Reinforced Concrete Beams Subjected to Combined Torsion and Bending,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 379–​402). 18.27 Kevin D.  Below, B.  Vijaya Rangan, and A.  Stanley Hall. “Theory for Concrete Beams in Torsion and Bending,” Journal of the Structural Division, ASCE, 101, ST8 (August 1975), 1645–​1660.

Space Truss Analogy 18.28 E. Rausch. Berechnung des Eisenbetons gegen Verdrehung und Abscheren (Design of Reinforced Concrete for Torsion and Shear). Berlin: Springer-​Verlag, 1929. 18.29 Paul Lampert and Bruno Thürlimann. “Ultimate Strength and Design of Reinforced Concrete Beams in Torsion and Bending,” Publications, International Association for Bridge and Structural Engineering, 31-​1, 1971, 107–​131. 18.30 Paul Lampert and Michael P.  Collins. “Torsion, Bending and Confusion—​An Attempt to Establish the Facts,” ACI Journal, Proceedings, 69, August 1972, 500–​504. 18.31 CEB-​FIP. Model Code for Concrete Structures, CEB-​FIP International Recommendations (3rd ed.). Paris: Comité Euro-​International du Beton, 1978, 348 pp. 18.32 Denis Mitchell and Michael P.  Collins. “Diagonal Compression Field Theory—​A Rational Model for Structural Concrete in Pure Torsion,” ACI Journal, Proceedings, 71, August 1974, 396–​408.

793



SELECTED REFERENCES

793

18.33 P.  Müller. “Failure Mechanisms for Reinforced Concrete Beams in Torsion and Bending,” Publications, International Association for Bridge and Structural Engineering, 36-​II, 1976, 146–​163. 18.34 B. G. Rabbat and M. P. Collins. “A Variable Angle Space Truss Model for Structural Concrete Members Subjected to Complex Loading,” Douglas McHenry International Symposium on Concrete and Concrete Structures (SP-​55). Detroit:  American Concrete Institute, 1978 (pp. 547–​587). 18.35 Thomas T. C. Hsu. “Shear Flow Zone in Torsion of Reinforced Concrete,” Journal of Structural Engineering, ASCE, 116, 11 (November 1990), 3206–​3226.

Combined Bending and Torsion 18.36 Hans Gesund, Frederick J.  Schuette, George R.  Buchanan, and George A.  Gray. “Ultimate Strength in Combined Bending and Torsion of Concrete Beams Containing Both Longitudinal and Transverse Reinforcement,” ACI Journal, Proceedings, 61, December 1964, 1509–​1522. 18.37 John P. Klus and C. K. Wang. “Torsion in Grid Frames,” Torsion of Structural Concrete (SP-​ 18). Detroit: American Concrete Institute, 1968 (pp. 89–​101). 18.38 G. S. Pandit and Joseph Warwaruk. “Reinforced Concrete Beams in Combined Bending and Torsion,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 133–​163). 18.39 A. A. Gvozdez, N. N. Lessig, and L. K. Rulle. “Research on Reinforced Concrete Beams under Combined Bending and Torsion in the Soviet Union,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 307–​336). 18.40 David J. Victor and Phil M. Ferguson. “Reinforced Concrete T-​Beams without Stirrups under Combined Moment and Torsion,” ACI Journal, Proceedings, 65, January 1968, 29–​36. Disc., 65, 560–​566. 18.41 V.  Ramakrishnan and Y.  Ananthanarayana. “Tests to Failure in Bending and Torsion of Reinforced Concrete,” ACI Journal, Proceedings, 66, May 1969, 428–​431. Disc., 943–​944. 18.42 D. W. Kirk and S. D. Lash. “T-​Beams Subject to Combined Bending and Torsion,” ACI Journal, Proceedings, 68, February 1971, 150–​159. 18.43 Hota V. S. GangaRao and Paul Zia. “Rectangular Prestressed Beams in Torsion and Bending,” Journal of the Structural Division, ASCE, 99, ST1 (January 1973), 183–​198. 18.44 David J.  Victor and P.  K. Aravindan. “Prestressed and Reinforced Concrete T-​Beams under Combined Bending and Torsion,” ACI Journal, Proceedings, 75, October 1978, 526–​532. 18.45 P. D. Zararis and G. Gr. Penelis. “Reinforced Concrete T-​Beams in Torsion and Bending,” ACI Journal, Proceedings, 83, January–​February 1986, 145–​155. 18.46 Ming B. Leung and William C. Schnobrich. “Reinforced Concrete Beams Subjected to Bending and Torsion,” Journal of Structural Engineering, ASCE, 113, 2 (February 1987), 307–​321.

Shear and Torsion 18.47 Ugur Ersoy and Phil M.  Ferguson. “Behavior and Strength of Concrete L-​Beams under Combined Torsion and Shear,” ACI Journal, Proceedings, 64, December 1967, 797–​801. Disc., 65, 477–​479. 18.48 John P.  Klus, “Ultimate Strength of Reinforced Concrete Beams in Combined Torsion and Shear,” ACI Journal, Proceedings, 65, March 1968, 210–​215. Disc., 786–​791. 18.49 Huey Ming Liao and Phil M. Ferguson. “Combined Torsion in Reinforced Concrete L-​Beams with Stirrups,” ACI Journal, Proceedings, 66, December 1969, 986–​993. Disc., 67, 475–​478. 18.50 Ahmed A. Ewida and Arthur E. McMullen. “Concrete Members under Combined Torsion and Shear,” Journal of the Structural Division, ASCE, 108, ST4 (April 1982), 911–​928.

Torsion, Shear, and Bending 18.51 Larry E.  Farmer and Phil M.  Ferguson. “T-​Beams under Combined Bending, Shear and Torsion,” ACI Journal, Proceedings, 64, November 1967, 757–​766. Disc., 65, 417–​421. 18.52 E. L. Kemp. “Behavior of Concrete Members Subject to Torsion and to Combined Torsion, Bending, and Shear,” Torsion of Structural Concrete (SP-​18). Detroit:  American Concrete Institute, 1968 (pp. 179–​201).

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18.53 Thomas T. C. Hsu. “Torsion of Structural Concrete—​Interaction Surface for Combined Torsion, Shear, and Bending in Beams Without Stirrups,” ACI Journal, Proceedings, 65, January 1968, 51–​60. Disc., 566–​572. 18.54 M. S. Mirza and J. O. McCutcheon. Discussion of “Torsion of Structural Concrete—​Interaction Surface for Combined Torsion, Shear, and Bending in Beams Without Stirrups,” ACI Journal, Proceedings, 65, July 1968, 567–​570. 18.55 David J.  Victor and Phil M.  Ferguson. “Beams under Distributed Load Creating Moment, Shear, and Torsion,” ACI Journal, Proceedings, 65, April 1968, 295–​308. Disc., 892–​894. 18.56 Michael P.  Collins, Paul F.  Walsh, and A.  S. Hall. Discussion of “Ultimate Strength of Reinforced Concrete Beams in Combined Torsion and Shear,” ACI Journal, Proceedings, 65, September 1968, 786–​788. 18.57 D. L. Osburn, B. Mayoglou, and Alan H. Mattock. “Strength of Reinforced Concrete Beams with Web Reinforcement in Combined Torsion, Shear, and Bending,” ACI Journal, Proceedings, 66, January 1969, 31–​41. Disc., 593–​595. 18.58 M. S. Mirza and J. O. McCutcheon. “Behavior of Reinforced Concrete Beams under Combined Bending, Shear, and Torsion,” ACI Journal, Proceedings, 66, May 1969, 421–​427. Disc., 940–​942. 18.59 Alfred Bishara. “Prestressed Concrete Beams under Combined Torsion, Bending, and Shear,” ACI Journal, Proceedings, 66, July 1969, 525–​538. Disc., 67, 61–​63. 18.60 Arthur E. McMullen and Joseph Warwaruk. “Concrete Beams in Bending, Torsion, and Shear,” Journal of the Structural Division, ASCE, 96, ST5 (May 1970), 885–​903. 18.61 Umakanta Behera and Phil M. Ferguson. “Torsion, Shear, and Bending on Stirruped L-​Beams,” Journal of the Structural Division, ASCE, 96, ST7 (July 1970), 1271–​1286. 18.62 Einar Gausel. “Ultimate Strength of Prestressed I-​Beams under Combined Torsion, Bending, and Shear,” ACI Journal, Proceedings, 67, September 1970, 675–​678. 18.63 Priya R.  Mukherjee and Joseph Warwaruk. “Torsion, Bending, and Shear in Prestressed Concrete,” Journal of the Structural Division, ASCE, 97, ST4 (April 1971), 1063–​1079. 18.64 P.  K. Syamal, M.  S. Mirza, and D.  P. Ray. “Plain and Reinforced Concrete L-​Beams under Combined Flexure, Shear, and Torsion,” ACI Journal, Proceedings, 68, November 1971, 848–​860. 18.65 K.  S. Rajagopalan and Phil M.  Ferguson. “Distributed Loads Creating Combined Torsion, Bending, and Shear on L-​ Beams with Stirrups,” ACI Journal, Proceedings, 69, January 1972, 46–​54. 18.66 K.  S. Rajagopalan, Umakanta Behera, and Phil M.  Ferguson. “Total Interaction Method for Torsion Design,” Journal of the Structural Division, ASCE, 98, ST9 (September 1972), 2097–​2117. 18.67 Thomas G. Barton and D. Wayne Kirk. “Concrete T-​Beams Subject to Combined Loading,” Journal of the Structural Division, ASCE, 99, ST4 (April 1973), 687–​700. 18.68 Robert L. Henry and Paul Zia. “Prestressed Beams in Torsion, Bending, and Shear,” Journal of the Structural Division, ASCE, 100, ST5 (May 1974), 933–​952. 18.69 Lennart Elfgren, Inge Karlsson, and Anders Losberg. “Torsion-​Bending-​Shear Interaction for Concrete Beams,” Journal of the Structural Division, ASCE, 100, ST8 (August 1974), 1657–​1676. 18.70 D. Wayne Kirk and David G. McIntosh. “Concrete L-​Beams Subject to Combined Torsional Loads,” Journal of the Structural Division, ASCE, 101, ST1 (January 1975), 269–​282. 18.71 B.  Vijaya Rangan and A.  S. Hall. “Strength of Rectangular Prestressed Concrete Beams in Combined Torsion, Bending and Shear,” ACI Journal, Proceedings, 70, April 1973, 270–​278. 18.72 B.  Vijaya Rangan and A.  S. Hall. “Strength of Prestressed Concrete I-​Beams in Combined Torsion and Bending,” ACI Journal, Proceedings, 75, November 1978, 612–​618. 18.73 K.  S. Rajagopalan. “Combined Torsion, Bending, and Shear on L-​Beams,” Journal of the Structural Division, ASCE, 106, ST12 (December 1980), 2475–​2492. 18.74 G. Taylor and J. Warwaruk. “Combined Bending, Torsion, and Shear of Prestressed Concrete Box Girders,” ACI Journal, Proceedings, 78, September–​October 1981, 335–​340.

Design Methods 18.75 Alan H.  Mattock. “How to Design for Torsion,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 469–​495). 18.76 ACI Committee 438. “Tentative Recommendations for the Design of Reinforced Concrete Members to Resist Torsion,” ACI Journal, Proceedings, 66, January 1969, 1–​8. Disc., 66, 576–​588.

795



SELECTED REFERENCES

795

18.77 Thomas T. C. Hsu and E. L. Kemp. “Background and Practical Application of Tentative Design Criteria for Torsion,” ACI Journal, Proceedings, 66, January 1969, 12–​23. Disc., 591–​593. 18.78 Michael P. Collins. Discussion of “Tentative Recommendations for the Design of Reinforced Concrete Members to Resist Torsion,” ACI Journal, Proceedings, 66, July 1969, 577–​579. 18.79 Umakanta Behera and K.  S. Rajagopalan. “Two-​Piece U-​Stirrups in Reinforced Concrete Beams,” ACI Journal, Proceedings, 66, July 1969, 522–​524. 18.80 ACI Committee 438. Discussion of “Proposed Revision of ACI 318–​ 63:  Building Code Requirements for Reinforced Concrete,” ACI Journal, Proceedings, 67, September 1970, 686–​689. 18.81 Michael P. Collins and Paul Lampert. “Designing for Torsion,” Structural Concrete Symposium, Toronto: University of Toronto Civil Engineering Department, May 1971 (pp. 38–​79). 18.82 David J. Victor. “Effective Flange Width in Torsion,” ACI Journal, Proceedings, 68, January 1971, 42–​46. 18.83 Denis Mitchell and Michael P. Collins. “Detailing for Torsion,” ACI Journal, Proceedings, 73, September 1976, 506–​511. 18.84 Mahmoud A. Reda Youssef and Alfred G. Bishara. “Dowel Action in Concrete Beams Subject to Torsion,” Journal of the Structural Division, ASCE, 106, ST6 (June 1980), 1263–​1277. 18.85 Thomas T.  C. Hsu and Y.  L. Mo. “Softening of Concrete in Torsional Members—​Design Recommendations,” ACI Journal, Proceedings, 82, July–​August 1985, 443–​452. Disc., 83, July–​August 1986, 690–​693. 18.86 J.  G. MacGregor and M.  G. Ghoneim. “Design for Torsion,” ACI Structural Journal, 92, March–​April 1995, 211–​218.

Spandrel Beams  18.87 M.  A. Gouda. “Distribution of Torsion and Bending Moments in Connected Beams and Slabs,” ACI Journal, Proceedings, 56, February 1960, 757–​774. Disc., 1425–​1446.   18.88 Robert A. Shoolbred and Eugene P. Holland. “Investigation of Slab Restraint on Torsional Moments in Fixed-​ Ended Spandrel Girders,” Torsion of Structural Concrete (SP-​18). Detroit: American Concrete Institute, 1968 (pp. 69–​88).  18.89 Kolbjorn Saether and N.  M. Prachand. “Torsion in Spandrel Beams,” ACI Journal, Proceedings, 66, January 1969, 24–​30.   18.90 James O. Jirsa, John L. Baumgartner, and Nathan C. Mogbo. “Torsional Strength and Behavior of Spandrel Beams,” ACI Journal, Proceedings, 66, November 1969, 926–​932. Disc., 67, 434–​435.   18.91 James O. Jirsa. “Torsion in Floor Slab Structures,” Analysis of Structural Systems for Torsion (SP-​35). Detroit: American Concrete Institute, 1973 (pp. 265–​292).   18.92 Mario G. Salvadori. “Spandrel-​Slab Interaction,” Journal of the Structural Division, ASCE, 96, ST1 (January 1970), 89–​106.   18.93 Umakanta Behera, K. S. Rajagopalan, and Phil M. Ferguson. “Reinforcement for Torque in Spandrel L-​Beams,” Journal of the Structural Division, ASCE, 96, ST2 (February 1970), 371–​380.  18.94 Ugur Ersoy. “Distribution of Torsional and Bending Moments in Beam–​Slab Systems,” Analysis of Structural Systems for Torsion (SP-​35). Detroit:  American Concrete Institute, 1973 (pp. 293–​324).   18.95 E. L. Kemp and W. J. Wilhelm. “Influence of Spandrel Beam Torsion on Slab Capacity Based on Yield Line Criteria,” Analysis of Structural Systems for Torsion (SP-​35). Detroit: American Concrete Institute, 1973 (pp. 325–​341).   18.96 Thomas T. C. Hsu and Kenneth T. Burton. “Design of Reinforced Concrete Spandrel Beams,” Journal of the Structural Division, ASCE, 100, ST1 (January 1974), 209–​229.   18.97 Thomas T. C. Hsu and Ching-​Sheng Hwang. “Torsional Limit Design of Spandrel Beams,” ACI Journal, Proceedings, 74, February 1977, 71–​79.  18.98 Alfred G.  Bishara, Larry Londot, Peter Au, and Majety V.  Sastry. “Flexural Rotational Capacity of Spandrel Beams,” Journal of the Structural Division, ASCE, 105, ST1 (January 1979), 147–​161.   18.99 Charles H.  Raths. “Spandrel Beam Behavior and Design,” PCI Journal, 29, March–​April 1984, 62–​131. 18.100 Gary J.  Klein. “Design of Spandrel Beams,” PCI Journal, 31, September–​October 1986, 76–​124. 18.101 Y.  L. Mo and Thomas T.  C. Hsu. “Redistribution of Moments in Spandrel Beams,” ACI Structural Journal, 88, January–​February 1991, 22–​30.

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Inverted T-​Girders and Ledger Beams 18.102 Sher Ali Mirza and Richard W. Furlong. “Serviceability Behavior and Failure Mechanisms of Concrete Inverted T-​Beam Bridge Bentcaps,” ACI Journal, Proceedings, 80, July–​August 1983, 294–​304. 18.103 Sher Ali Mirza and Richard W. Furlong. “Strength Criteria for Concrete Inverted T-​Girders,” Journal of Structural Engineering, ASCE, 109, 8 (August 1983), 1836–​1853. 18.104 Ned M.  Cleland and Thomas T.  Baber. “Behavior of Precast Reinforced Concrete Ledger Beams,” PCI Journal, 31, March–​April 1986, 96–​117.

Channel-​Shaped Sections 18.105 Petar Krpan and Michael P. Collins. “Predicting Torsional Response of Thin-​Walled Open RC Members,” Journal of the Structural Division, ASCE, 107, ST6(June 1981), 1107–​1127. 18.106 Petar Krpan and Michael P. Collins. “Testing Thin-​Walled Open RC Structure in Torsion,” Journal of the Structural Division, ASCE, 107, ST6 (June 1981), 1129–​1140. 18.107 Ching-​Sheng Hwang and Thomas T. C. Hsu. “Mixed Torsion Analysis of Reinforced Concrete Channel Beams—​A Fourier Series Approach,” ACI Journal, Proceedings, 80, September–​ October 1983, 377–​386.

Beams with Openings 18.108 M. A. Mansur and A. Hasnat. “Concrete Beams with Small Opening under Torsion,” Journal of the Structural Division, ASCE, 105, ST11 (November 1979), 2433–​2447. 18.109 Mohammad A. Mansur, Seng Kiong Ting, and Seng-​Lip Lee. “Ultimate Torque of R/​C Beams with Large Openings,” Journal of Structural Engineering, ASCE, 109, 8 (August 1983), 1887–​1901. 18.110 M.  A. Mansur. “Combined Bending and Torsion in Reinforced Concrete Beams with Rectangular Openings,” Concrete International, 5, November 1983, 51–​58. 18.111 M.  A. Mansur and P.  Paramasivam. “Reinforced Concrete Beams with Small Opening in Bending and Torsion,” ACI Journal, Proceedings, 81, March–​April 1984, 180–​185. 18.112 Abul Hasnat and Ali A.  Akhtaruzzaman. “Beams with Small Rectangular Opening under Torsion, Bending, and Shear,” Journal of Structural Engineering, ASCE, 113, 10 (October 1987), 2253–​2269. 18.113 W. A. M. Alwis and Mohammad A. Mansur. “Torsional Strength of R/​C Beams Containing Rectangular Openings,” Journal of Structural Engineering, ASCE, 113, 11 (November 1987), 2248–​2258.

Torsion of Prestressed Concrete Beams 18.114 Alfred Bishara. “Prestressed Concrete Beams under Combined Torsion, Bending, and Shear,” ACI Journal, Proceedings, 66, July 1969, 525–​538. 18.115 Paul Zia and W. Denis McGee. “Torsion Design of Prestressed Concrete,” PCI Journal, 19, March–​April 1974, 46–​65. 18.116 S. Timoshenko and J. N. Goodier. Theory of Elasticity (3rd ed.). New York: McGraw-​Hill, 1970.

PROBLEMS All problems are to be worked in accordance with the ACI Code, and all loads given are service loads, unless otherwise indicated. 18.1 Determine the reinforcement required on a 12 × 22 in. overall size member to carry a torsional moment of 3 ft-​kips dead load and 12 ft-​kips live load. Use fc′ = 4000 psi (normal weight) and fy = fyt = 60,000 psi. 18.2 Determine the reinforcement required for the member in the figure for Problem 18.2 to carry a torsional moment of 20 ft-​ kips dead

load. Use fc′ = 4000 psi (normal weight) and fy = fyt = 50,000 psi. 18.3 For the beam of the figure for Problem 18.3, assume that 4–​#9 bars are used in the bottom for the main flexural reinforcement at midspan, with 2–​#9 bars extended into the support and properly anchored. Further assume that 2–​#7 are used in the top at the supports. What is the nominal

79



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PROBLEMS

bE = 90”

2 – #7 # 3 @ 5” 5”

24”

36” 4 – #9 4 – #10

2 – #9

15”

14”

Problem 18.2

Problem 18.3

torsional strength Tn of the section according to load, and also that the girder has uniform dead the ACI Code, assuming no simultaneous transload (including its own weight) of 1 kip/​ft. Use verse shear? What is the factored negative bending fc′ = 3000 psi (normal weight) and fy = 40,000 moment Mu that might be permitted to act at the psi. Show design sketch. supports simultaneously when Tu = φTn, according 18.6 Redesign the transverse reinforcement for the to the logic of Section 18.8? Use fc′ = 4000 psi beam of Problem 5.1, except consider the total (normal weight) and fy = fyt = 60,000 psi. loading to be acting along a line at a distance 18.4 Determine the reinforcement required (including of 2 in. from the midwidth of the beam. Use the torsion) at midspan and at the supports for the beam rough approximation that equilibrium torsion of Example 18.4.1 (2B4 of Fig. 18.4.1). Assume is acting and that the torsional moment per unit the shear is the same as that of a simply supported length equals the uniform loading times 2 in. beam carrying a uniform dead load of 0.65 kip/​ 18.7 Redesign the transverse reinforcement for the ft and a uniform live load of 0.65 kip/​ft. Assume beam of Problem 5.8, except consider the line that flexure alone requires longitudinal steel areas uniform loading to be acting at 3 in. from the of 1.60 and 2.30 sq in., for positive and negative centerline of the beam cross section. Assume moment regions, respectively. Use fc′ = 3000 psi equilibrium torsion and use the same approxi(normal weight) and fy  =  fyt  =  40,000 psi. Show mation as in Problem 18.6. sketches of the cross sections. 18.8 Design the reinforcement for the edge beam that 18.5 Design the reinforcement to include torsion on is continuous at both ends shown in the figure the spandrel girder 2G4 of Fig. 18.4.1. Assume for Problem 18.8. Assume the torsional moment simple beam shears for this span, which is conis 76 ft-​kips (50% live load) at the face of coltinuous at both ends. Assume the reactions to umn and that the torsional moment varies progirder 2G4 from beams 2B1 are concentrated portionally with the flexural shear (this problem loads of 12.3 kips dead load and 15.9 kips live is similar to that of Hsu and Kemp [18.77]). Dead load 1.2 kips/ft (both cantilevers)

12’–0”

120 psf (live load) 18” 4” 16”

30’–0” 4”

4”

14”

Typical joist cross section

28’–0”

28’–8” Joists and slab A

A

18 × 18 columns

2’–0” Max. effective = 3hf hf = 4” 16”

Edge beam

20”

16”

16”

Beam cross section (Section A–A)

Problem 18.8

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Spandrel beam (25-ft span) 8”

60’–0” span of double tees

6’–3”

Symmetrical about CL R1 R1 R1 R1 4’–0” 4’– 0” 4’– 0”

12” Reactions R1 from tees

12’– 6” R2 (reaction to the corbel)

1’–0”

16”

(a) Spandrel beam span and loading Corbel projecting from 24” square column

8” R2

10” (b) Spandrel beam cross section (trial)

Problems 18.9 and 18.10

18.9 Design the parking garage spandrel beam in the figure for Problems 18.9 and 18.10 (from Ref. 18.9). The floor system, consisting of double tees spanning 60 ft, carries a live load of 50 psf. The double tees are supported on the ledge of the spandrel beam. The contribution to R1 from the dead load of the floor system is 10.7 kips, and from the 50 psf live load is 6.0 kips. Consider the given cross section as a preliminary trial section, with the 8-​in. eccentricity of R1 with respect to R2 held as constant. The spandrel member will have an attachment to the supporting column, providing lateral support at each vertical support of the spandrel girder. Use fc′ = 4000 psi (normal weight) and fy = fyt = 60,000 psi and the ACI Code. (Note:  The preliminary section is from Ref. 18.9 and is for a prestressed concrete

member. The final design for a nonprestressed member may require changes from the preliminary given dimensions; those given dimensions may serve as guides for arriving at reasonable proportions.) 18.10 Design a spandrel beam of similar L-​shape to that of Problem 18.9 for loads R1 of 5.5 kips dead load and 4.0 kips live load, plus the spandrel girder weight. The span of the spandrel and the locations of the R1 loads are the same as in Problem 18.9. The eccentricity e is 8 in. (same as for Problem 18.9). Consider the 8 in. used in Problem 18.9 as the minimum thickness; however, the other dimensions should be appropriate for the loads given in this problem. Use fc′ = 4000 psi (normal weight) and fy = fyt = 60,000 psi.

CHAPTER 19

FOOTINGS

19.1 PURPOSE OF FOOTINGS Footings are structural elements that transmit to the soil column loads, wall loads, or lateral loads such as from retained earth. If these loads are to be properly transmitted, footings must be designed to prevent excessive settlement or rotation, to minimize differential settlement, and to provide adequate safety against sliding and overturning.

19.2 BEARING CAPACITY OF SOIL It is not within the scope of this text to discuss the details of arriving at the bearing capacity of the soil, but there must be reliable information on the safe bearing capacity of the soil prior to the design of a footing. The allowable bearing capacity of soil is usually determined by the ruling building code, by comparison with existing footings and with related information in the area, by close examination of the soil and study of logs of test borings, by the application of the science of soil mechanics, by load test, or by combinations of the various sources and methods mentioned here. The following are some of the most common reasons for the many uncertainties concerning soil behavior under a footing. 1. There may be wide variations in soil types, which depend on their geological source, mode of transportation, and sedimentation mechanism. 2. The physical properties and probable behavior under load are unknown and may require extensive testing. 3. Frost action may cause heaving or subsidence. 4. Vibration may cause consolidation of granular material, which results in nonuniform settlement. 5. Man-​made hazards may exist below the earth surface, such as rock heaps, old sewers, and questionable fill. Soil pressure is commonly used in design under a working stress philosophy (Section 2.4). The soil mechanics/​foundations specialist (geotechnical engineer) will establish the ultimate bearing capacity, applying the appropriate margin for safety, and will specify a service load bearing capacity (allowable bearing capacity) to be used in design. In general, rock is considered the best foundation material; graded sand and gravel are good materials; fine particles of sand and silt are generally questionable; and clay should be studied carefully. The allowable bearing capacity used for design may range from 12,000 psf (575 kN/​m2) or higher for rock to 2000 psf (96 kN/​m2) for soft clay or silty clay. Soils unable to carry 2000 psf generally require piling.

80

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C hapter   1 9     F ootings

Wall and square spread footings visible during construction of the Kurt F. Wendt Engineering Library (now Wendt Commons Engineering Library) at the University of Wisconsin–Madison. (Photo by C. G. Salmon.)

19.3 TYPES OF FOOTINGS Most building footings may be classified as one of the following types (Fig. 19.3.1). 1. Isolated spread footings under individual columns. These may be square, rectangular, or occasionally circular in plan. 2. Wall footings, either flat or stepped, which support bearing walls. 3. Combined footings supporting two or more column loads. These may be continuous with a rectangular or trapezoidal plan, or they may be isolated footings joined by a beam. The latter case is referred to as a strap, or cantilever, footing. 4. A mat foundation, which is one large continuous footing supporting all the columns of the structure. This is used when soil conditions are poor but piles are not used. 5. Pile caps, structural elements that tie a group of piles together. These may support bearing walls, isolated columns, or groups of several columns.

19.4 TYPES OF FAILURE The procedures used for the design of footings in the United States are based primarily on the work of Talbot in 1907 [19.1], Richart in 1946 [19.2], and Moe in 1957–​1959 [19.3]. Moe defines the several types of failure that may occur in a slab acted on by concentrated loads. These failure modes are related to the shear span to depth (a/​d) ratio, that is,

801



801

19.4  TYPES OF FAILURE

(a) Isolated spread footing

(b) Wall footing Property line

A

B

(c) Combined footing: rectangular, PA = PB

B

(d) Combined footing: rectangular, PB > PA Property line

Property line A

A

B

(e) Combined footing: trapezoidal, PB > PA

A

B

(f) Combined footing: trapezoidal, PA > PB

Property line

(g) Combined footing: strap or cantilever

Figure 19.3.1  Types of footings.

Mu /​Vud, and are similar to those described for beams in Section 5.4 (Fig. 5.4.4). The failure mechanisms may be reviewed as follows. 1. Shear-​compression failure [Fig. 19.4.1(a)]. Typical with deep sections of short span (low a/​d ratios), inclined cracks form that do not cause failure but do extend into the compression zone, thus reducing its size until finally the compression zone fails under the combined compressive and shear stresses. 2. Flexure failure after inclined cracks form. Also typical with low a/​d ratios, inclined cracks that form first do not cause failure or prevent the development of the nominal moment strength. If embedment of tension steel is adequate, and no failure in the compression zone occurs, the tension steel may reach its yield strength. 3. Diagonal tension failure [Fig. 19.4.1(b)]. Commonly called punching shear (Section 16.15), this failure mode is typical with medium-​span average depth sections (intermediate values of a/​d); the footing fails on formation of the inclined cracks around the perimeter of the concentrated load. Test results indicate that the critical section can reasonably be considered at d/​2 from the periphery of a column or concentrated load. 4. Flexure failure before inclined cracks form. Typical with large values of a/​d, no inclined cracks form before flexural strength is reached. In the design of a footing as well as of a beam, a shear failure should not occur prior to reaching the member flexural strength.

802

802

C hapter   1 9     F ootings

d/2

(a) Shear-compression failure

d/2

(b) Diagonal tension failure

Figure 19.4.1  Shear-​related failure mechanisms in footings.

19.5 SHEAR STRENGTH Strength requirements for footings are given in Chapter 13 of the ACI Code. Most of these requirements, however, are given in reference to Chapter 7 (one-​way slabs) or Chapter 8 (two-​way slabs) of the ACI Code as appropriate. The shear strength of footings may be computed as that of two-​way slab systems, as discussed in Sections 16.15 and 16.18. The six principal variables involved in the strength of slabs without shear reinforcement are (a) the concrete strength fc′; (b) the ratio of the side length c of the loaded area to the effective depth d of the slab; (c)  the relationship (V/​M) between shear and moment near the critical section; (d) the column shape in terms of the ratio β of the long side to the short side of the rectangular column; (e) lateral restraints such as may be provided by stiff beams along the boundaries of a slab (no such restraints would normally be acting in the case of footings); and (f) the rate of loading. As discussed in Section 16.15, the nominal punching shear strength Vc when no shear reinforcement is used (i.e., when Vn = Vc) is the smallest of



Vc = 4 λ fc′b0 d

(19.5.1a)*

 4 Vc =  2 +  λ fc′b0 d  β

(19.5.1b)*

 α d Vc =  2 + s  λ fc′b0 d b0  

(19.5.1c)*

where b0 = perimeter of critical section taken at d/​2 from the loaded area (ACI-​13.2.7.2) d = effective depth of slab β = ratio of long side to short side of the loaded area αs = 40 for interior columns, 30 for edge columns, and 20 for corner columns In the application of Eq. (19.5.1c), αs for an “interior column” applies when the perime­ ter is four-​sided, for an “edge column” when the perimeter is three-​sided, and for a “corner column” when the perimeter is two-​sided. *  For SI, with fc′ in MPa and b0 and d in mm, ACI-​318-​14M gives



Vc = 0.33λ fc′ b0 d

(19.5.1a) 

 2 Vc = 0.17  1 −  λ fc′ b0 d  β

(19.5.1b) 

 α d Vc = 0.083  2 + s  λ fc′ b0 d b0  

(19.5.1c) 

803



803

19.6  FLEXURAL STRENGTH Critical section c W

c d

d

L (a) One-way action

Critical section W

c+d

c

d/2

d/2

L (b) Two-way action

Figure 19.5.1  Critical section for shear (inclined cracking) in footings.

For footings in which the bending action is primarily in one direction, the procedure used for one-​way slabs should be applied as described in Chapter 8. The critical section for such an investigation is to be taken at a distance d from the column face (ACI-​13.2.7.2). Figure 19.5.1 summarizes the critical sections to be investigated in regard to shear.

19.6 FLEXURAL STRENGTH AND DEVELOPMENT OF REINFORCEMENT Research has shown that critical sections for flexural strength and development of reinforcement occur at the face of a reinforced concrete column or wall. The bending in each direction should be considered separately. Richart’s tests [19.2] showed that, under service loads, the moment is greater on a strip under the column load than it is out near the corners. However, failure in flexure does not occur until all of the steel has reached yielding—​that is, after some redistribution of load has taken place. Gesund [19.5, 19.6], Jiang [19.7], and Subba Rao and Singh [19.8] have presented yield line analyses (see Chapter 17) of footings. The critical sections for moment and development of reinforcement, however, are to be taken at (ACI-​13.2.7.1) (a) the face of column, pedestal, or wall for footings supporting a column, pedestal or wall, respectively, (b) halfway between the middle and the edge of the wall for footings under masonry walls, and (c) halfway between the face of the column and the edge of the base plate for footings under steel bases.

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19.7 PROPORTIONING FOOTING AREAS FOR EQUAL SETTLEMENT Differential settlement between footings should be minimized because it can adversely affect the strength of the structure as well as interfere with the fitting of such items as partitions, doors, and ceilings. It is generally assumed that there will be equal settlement if the unit soil pressures due to service loads are equal under all footings. However, the service loads in the columns consist of dead and live loads. The dead load is always there, but in a tall building there is little chance for a column to receive maximum live load from all floors. Therefore, most building codes allow a reduction of live loads in columns. Typically, the total live load would have to be carried by columns supporting the roof and one floor. As the number of floors carried by a column increases, the percentage of the total live load that the column must be designed to carry may be reduced from 100% progressively. This percentage might be a minimum of 50% when at least 8 or 9 floors are carried by the column. The maximum soil pressure under a footing is, then, due to the sum of the dead load in the column, the maximum reduced live load in the column, and the weight of the footing itself. Soil is a substance that may be relatively elastic, such as granular material like sand, or it may be a relatively plastic substance, such as clay exhibiting time-​dependent deformation under sustained load. Often, for design purposes, equal settlement of footings is presumed when the soil pressure due to sustained loads is the same under all footings. The sustained load may be taken as the sum of the dead load in the column, the weight of the footing itself, and a certain percentage of the maximum reduced live load in the column. In such cases, the relative areas of the footings should be so proportioned that the unit soil pressures under sustained loads would be the same under all footings. This requirement is additional to the requirement that the soil pressure under maximum possible load not exceed the allowable bearing capacity at each footing.

19.8 INVESTIGATION OF SQUARE SPREAD FOOTINGS The investigation of square spread footings can best be treated by an illustrative example in which the items considered are soil pressure under the footing, shear (inclined cracking), bending moment, development of reinforcement, and load transfer from column to footing.

EXAMPLE 19.8.1 Check the adequacy of the square footing of Fig. 19.8.1, according to the ACI Code. The column service level (unfactored) axial load is 300 kips dead load and 160 kips live load. Both the column and the footing have fc′ = 3000 psi (normal-weight concrete) and fy = 60,000 psi. The allowable soil pressure is 5000 psf. There is a 2-​ft earth overburden having a unit weight of 100 pcf. SOLUTION (a) Soil pressure. The action of soil on the footing is taken to be uniform for isolated footings under concentric loads. The base of the footing must have an area large enough so that the allowable soil pressure will not be exceeded under the action of the column service load, footing weight, and weight of overburden.



Column (300 + 160) Footing weight 10(10)(2.08)0.150) Earth (100 –​4)2(0.100)

460 kips 31 19

  Total weight on soil

510 kips

soli pressure p =

510 = 5.1 ksf ≈ 5 ksf (10)(10)

OK (Continued)

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Example 19.8.1 (Continued)

2’– 0” 10’– 0” 2’– 0”

#9 column bars and #8 dowels 3” clear

3” clear

2’– 1”

3” clear 8 – #8 bars each way (spacing ~ 16”) 10’– 0”

Figure 19.8.1  Square spread footing for Example 19.8.1.

(b) Shear (inclined cracking) strength. Two possible critical sections must be investigated:  one-​way action as a beam and two-​way action as a slab, as shown in Fig.  19.4.1. The shear to be used is the upward soil pressure less the downward overburden and the footing weight acting outside the critical section. Since the footing weight and the overburden are usually uniform, the forces acting on the footing may be obtained by using the net upward pressure, which is caused by the column load only. Since the action of a square concentrically loaded footing is symmetrical about both axes, the reinforcement in each direction is presumed to do the same work. However, the effective depth cannot be the same for both directions since the bars must cross each other. The average depth d is commonly used except for very shallow footings (say, less than 15 in. deep) where the more conservative value should probably be used. Using a 3-in. clear cover for concrete cast and permanently in contact with ground (ACI20.6.1.3.2), the average d is d = 25 − 3 (i.e.,clear cover ) − 1 (i.e., bar diameter ) = 21 in. For one-​way action, the net earth pressure acting upward due to factored loads is

pnet =

300(1.2) + 160(1.6) = 6.16 ksf 100 (Continued)

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Example 19.8.1 (Continued) Using the loaded area shown in Fig. 19.8.2(a) at a distance d from the face of the column, the factored shear is Vu = ( pnet )(effective area ) = 6.16(2.25)10 = 139 kips When no shear reinforcement is used, vc = 2 λ fc′ unless the more detailed procedure is used; thus Vn = Vc = 2 λ fc′bw d = 2(1.0) 3000 (120)(21)

1 1000

= 276 kips

[φVn = 0.75(276) = 207 kips] > [Vu = 156 kips]

OK

For two-​way action, the factored shear [Fig. 19.8.2(b)] is

Vu = ( pnet )(effective area ) = 6.16 [100 − 3.75(3.75)] = 529 kips



When no shear reinforcement is used, the strength using Eqs. (19.5.1) is based on vc = 4 λ fc′ when β ≤ 2 and b0 /​d ≤ 20 for a four-​sided critical section; thus b0 /d = 4(3.75) /1.75 = 8.6 < 20 Vn = Vc = 4 λ fc′b0 d = 4(1.0) 3000 (4)(3.75)(21)

12 1000

[φVn = 0.75(828) = 621 kips] > [Vu = 529 kips]

= 828 kips

OK

Since isolated footings are rarely designed with shear reinforcement, Vc will usually control the thickness. (c) Flexural strength. The critical section and loaded area are as shown in Fig. 19.8.2(a). The factored bending moment is



pnet b 2 6.16(10)(4.0)2 = = 493 ft-kips 2 2 Mu 493(12, 000 ) required Rn = = 124 psi = φbd 2 0.90(120)(21)2 Mu =

With reference to Rn, note that footing thickness is rarely controlled by flexure; thus, the reinforcement ratio ρ will be low enough that φ will be 0.90, except in rare cases where ρ exceeds ρtc. Referring to Section 3.8, Eq. (3.8.5), required ρ =

2 mRn  1 1 − 1 −  m fy 

[3.8.5]

1  2(23.5)124  1− 1− = 0.0021 = 23.5  60, 000 

For spread footings, the minimum reinforcement (As min = 0.0018Ag for fy = 60 ksi) for structural slabs of uniform thickness applies (ACI-​8.6.1.1); thus required As = 0.0021(120)21 = 5.3 sq in.



min required As = 0.0018(120)25 = 5.4 sq in.

Controls!

provided As = 8(0.79) = 6.32 sq in.[> min required As = 5.4 sq in.]

OK

(Continued)

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Example 19.8.1 (Continued) Load area for one-way shear action

Load area for two-way shear action

4’– 0” 2.25’

3.75’

10’– 0”

2’– 0” + d

1’– 0”

Load area for moment

d = 21” 1.75’ (a)

(b)

Figure 19.8.2  Critical sections and loaded area for square spread footing.

Note that using 7–​#8 bars would satisfy the requirement of ≈ 5.4 sq in.; however, they would be spaced at (120 –​6 –​1)/​6 = 18.8 in. which exceeds the maximum spacing of 18 in., per ACI-​8.7.2.2. (d) Development of reinforcement—​ general equation. The general equation, [Eq. (6.6.1)], is



   ψ t ψ e ψ s  3 fy Ld =  d  40 λ fc′  cb + K tr   b   d     b

[6.6.1]

The cover or spacing dimension cb is the smaller of (1) distance from center of bar being developed to nearest concrete surface, and (2) one-​half center-​to-​center spacing of bars being developed. The distance cb is the smaller of the following two values: bottom and side cover = 3.0 (i.e., clear) + 0.5 (i.e., bar radius) = 3.5 in.

one-half center-to-center spacing ≈

(120 − 6 − 1) = 8.1 in. 2(7)



Thus, cb = 3.5 in. There are no stirrups; thus, Ktr = 0. Thus,

  cb + K tr 3.5 + 0 = = 3.5 > 2.5 max  1.00   db

Thus, (cb + Ktr)/​db = 2.5, and noting that ψt = ψe= ψs = 1.0

 3 60, 000 1.0(1.0)1.0  Ld (for #8) =   1.00 = 33 in. (2.75 ft) 2.5  40 (1.0) 3000

Allowing a 3-​in. cover on the end of the #8 bars, the embedment provided is

actual embedment = 48 − 6 = 42 in. > [ Ld = 33 in.]

OK (Continued)

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Example 19.8.1 (Continued) (e) Load transfer from column to footing. ACI-​16.3.1.1 requires factored forces and moments acting at the column base to be transferred into the footing. Tensile forces, if any, must be transferred by developed reinforcement, such as bar reinforcement, dowels, or mechanical connectors; however, compressive forces may be transmitted directly by bearing. The nominal bearing stress fb that the base of the column can withstand is 0.85 fc′ (ACI-​22.8.3.2). The nominal strength Pn in compression based on the column concrete strength  fc′ is

Pn = 0.85 fc′ Ag = 0.85(3)(576) = 1470 kips Pu = 1.2(300) + 1.6(160) = 616 kips



Using φ = 0.65 per ACI-​21.2.1, for bearing stresses [φPn = 0.65(1470) = 956 kips] > [ Pu = 616 kips]

OK

Regarding the bearing on the footing concrete, the capacity is increased because the footing area is much larger than the column area, thus confining the loaded area and permitting a distribution of the concentrated load. The bearing strength for the footing concrete is thus based on 0.85 fc′ increased by a multiplier αb that varies between 1 and 2, as follows:

αb =

A2 ≤ 2 A1

where A1 is the loaded area, which in this example is the 576 sq in. column area; A2 is the area of the lower base of the largest frustrum of a pyramid, cone, or tapered wedge contained wholly within the support and having for its upper base equal to the loaded area, and having side slopes of 1 vertical to 2 horizontal (ACI-​2.2). In this case, A2 is the entire footing area, 14,400 sq in. Since both column and footing contain the same strength concrete in this example, only the check on bearing in the column was necessary. In general, however, the bearing strength is to be computed as the lesser of the nominal concrete bearing strengths of the supported member or the foundation surface (ACI-​16.3.3.4). The bearing stress on the footing may control when columns of high-​strength concrete rest on footings of low-​strength concrete. Bearing in the bottom of the column is important unless the longitudinal reinforcement is developed by extension into the footing or by embedding dowels lapped to the column bars. In this case, the load can be carried without using developed reinforcement. When the transfer is made by bearing, as in this case, ACI-​16.3.4.1 still requires for cast-​in-​place columns and pedestals a minimum amount of reinforcement across the joint between the column and footing. Extended longitudinal reinforcement or dowels at least equal to 0.005 times the gross cross-​sectional area of the supported member must be provided. When dowels are used, they can be any size #11 and smaller. Though #14 and #18 column bars may be lap spliced at footings, dowels larger than #11 bars cannot be used (ACI-​16.3.5.4). Dowels comparable in size to the bars being developed should generally be used. The minimum area of developed reinforcement required to cross the interface of cast-​in-​place construction is thus (ACI-​16.3.4.1) required As = 0.005(576) = 2.88 sq in. Using a practical minimum of four bars, one in each corner,

required As per bar =

2.88 = 0.72 sq in. 4

(Continued)

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Example 19.8.1 (Continued) Thus the #8 dowels shown are adequate. These dowels must be embedded into the footing a distance equal to the development length Ld in compression for #8 bars. Using the basic development length Ldc for compression bars not enclosed by transverse reinforcement (ψr = 1.0),



 fy ψ r  Ldc =   db  50 λ fc′   60, 000(1.0)  Ldc =  1.0 = 22 in.  50(1.0) 3000 

[6.8.1]

but not less than

Ldc = 0.0003 f y ψ r db = 0.0003(60, 000)(1.0)(1.0) = 18.0 in.



nor less than 8 in. Thus, Ld (# 8) = Ldc = 22 in. [ = (25 − 3) = 22 in. available]

[6.8.2]

OK

If the available footing thickness is inadequate, a greater number of smaller diameter bars should be used. Thus the footing of Fig. 19.8.1 satisfies all ACI requirements.

19.9 DESIGN OF SQUARE SPREAD FOOTINGS The design of square spread footings involves the determination of the size and depth of the footing and the amount of main reinforcement and dowels so that all the requirements described in the preceding sections are fulfilled. Example  19.9.1 illustrates the design procedures.

EXAMPLE 19.9.1 Design a square spread footing to carry a column dead load of 197 kips and a live load of 160 kips from a 16-in. square tied column containing #11 bars as the principal column steel. The allowable soil pressure is 4.5 ksf. Consider that there is a 2-​ft overburden weighing 100 pcf. Use fc′ = 3000 psi (normal-weight concrete) and  fy = 60,000 psi. SOLUTION (a) Estimate the footing weight and determine the plan of the footing. The total weight of the footing, plus any overburden, may be estimated and added to the column load or, as an alternative, the effect of these items in terms of the unit soil pressure may be estimated. In this case, the footing is estimated to be about 2 ft thick—​that is, 300 psf, frequently the minimum used by designers. This leaves the net allowable soil pressure that must carry the column load as pnet = 4500 − 200 − 300 = 4000 psf

required A =

357 = 89.3 sq ft 4.0

(Continued)

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Example 19.9.1 (Continued) Try 9 ft 6 in. square, A = 90.3 sq ft. Note that ACI-​13.3.1.1 requires the base area of a footing to be determined by using service loads (unfactored loads) with the allowable soil pressure. This is reasonable, since the allowable soil pressure should be determined by using principles of soil mechanics and may incorporate varying factors of safety depending on type of soil and condition of loading. For the design of the reinforced concrete member, factored loads must be used. Applying overload factors to the column load, Pu = 1.2(197) + 1.6(160) = 492 kips

pnet =

492 = 5.45 ksf 90.3



(b) Determine depth based on shear (inclined cracking) strength. In most cases the depth necessary for shear without using stirrups controls the footing thickness. For two-​way action [Fig. 19.9.1(a)], assuming a thickness of 24 in., average d = 24 − 3 (cover) − 1 (bar diameter) ≈ 20 in. Vu = ( pnet )(effective area) = 5.45[9.5(9.5) − 3.0(3.0)] = 443 kips From Fig. 19.9.1(a),

b 0 /d = 4(16 + 20) / 20 = 7.2

For a four-​sided critical section with β ≤ 2 and b0 /​d ≤ 20, 1 Vn = Vc = 4 λ fc′b0 d = 4(1.0) 3000 [ 4(16 + 20)](20) 1000 = 631 kips



φ Vc = 0.75(631) = 473 kips > [Vu = 433 kips]



OK

No shear reinforcement is required. For one-​way action [Fig. 19.9.1(b)], Vu = 5.45(2.42)(9.5) = 125 kips and

1 Vn = Vc = 2 λ fc′ bw d = 2(1.0) 3000 (9.5)(12)20 1000 = 250 kips

φ Vc = 0.75(250) = 188 kips > [Vu = 125 kips]

OK

No shear reinforcement is required. Note that Vc is computed from the simplified procedure. The 24-​in. thickness is satisfactory for shear. (c) Check transfer of load at base of column (ACI-​16.3.1.1). The compressive design strength φ Pn based on the nominal ultimate bearing stress 0.85 fc′ in the column is

φ Pn = φ (0.85 fc′ ) Ag = 0.65(0.85)(3)(256) = 424 kips

where φ = 0.65 for bearing strength. φ Pn < [ Pu = 492 kips]

NG

Thus the column load cannot be transferred by bearing alone. It may well be that the minimum reinforcement across the interface required by ACI-​16.3.4.1 will be adequate to transfer the excess load. min required As = 0.005(256) = 1.28 sq in. (Continued)

81



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Example 19.9.1 (Continued)

9’– 6” 16” + d = 36”

16” + d = 36” 20”

9’– 6”

29” 2.42’

8” (a) Two-way action

(b) One-way action

Figure 19.9.1  Critical sections for shear in square footing design.

Neglecting the area of concrete displaced by the reinforcing bars, the excess load to be carried by the dowels is excess Pu = 492 − 424 = 68 kips

required As = =

excess Pu φ fy



68 = 1.74 sq in. 0.65(60)

Logically, to compensate for the displaced concrete, a stress of 0.85 fc′ should be subtracted from fy in the dowels. Thus, more correctly, required As =

excess Pu = 1.82 sq in. φ ( f y − 0.85 fc′ )

Use 4–​#7 bars as dowels, As = 2.40 sq in. The #7 dowels must be developed above and below the junction of column and footing. The development length Ldc required for bars in compression not enclosed by transverse reinforcement (ψr = 1.0) is  fy ψ r  Ldc =   db  50 λ fc′ 

 60, 000(1.0)  Ldc =  0.875 = 19.2 in.  50(1.0) 3000 

[6.8.1]

Controls!

but not less than Ldc = 0.0003 f y ψ r db = 0.0003(60, 000)(1.0)(0.875) = 15.8 in. and final Ld cannot be less than 8 in. Thus, Ld = Ldc = 19.2 in. The 24-​in.-​thick footing is thus adequate for straight dowels. Hooks or bending of bars are not considered effective in adding to the compressive resistance of bars (ACI-​25.4.1.2). Very often engineers will specify bending of the dowels as shown in Fig. 19.9.2 to prevent their being pushed through the footing during construction and thus reducing the effective embedment distance L2. In such cases of bending of the dowels, full development of the compressive force is required over the distance L1. (Continued)

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Example 19.9.1 (Continued)

L2

L1

Figure 19.9.2  Dowel anchorage.

For this design, if such a bend is used, the available length L1 is

L1 = 24 − 3 (cover) − 2(1)(footing bars) − 0.875 (dowels) ≈ 18 in.



This is less than the 19.2 in. required and is therefore unacceptable. Alternatives would include a thicker footing, a larger number of smaller-​sized dowels, or the use of a pedestal. Try 6–#5 bars as dowels instead, As = 1.86 sq in.  60, 000(1.0)  Ldc =  0.625 = 13.7 in.  50(1.0) 3000 



Controls!

but not less than

Ldc = 0.0003 f y ψ r db = 0.0003(60, 000)(1.0)(0.625) = 11.3 in.

and final Ld cannot be less than 8 in. Thus, Ld = Ldc = 13.7 in., which is less than the available length L1 = 18 in. Use 6–​#5 bars as dowels. (d) Design for flexural strength. The critical section for moment is at the face of the column (Fig. 19.9.3). 9’– 6” 4’– 1”

1’– 4”

1 24”

Net upward pressure = 5.45 ksf 1

Figure 19.9.3  Critical section for bending moment and development of reinforcement.

(Continued)

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Example 19.9.1 (Continued) 1 (5.45)(9.5)(4.08)2 = 431 ft-kips 2



Mu =



required Rn =

Mu 431(12,000) = = 126 psi 2 φ bd 0.90(114)(20)2

Referring to Section 3.8, Eq. (3.8.5),

ρ= =



2 mRn  1 1 − 1 −  m fy 

[3.8.5]

1  2(23.5)126  1− 1− = 0.0022  23.5  60, 000 

required As = ρbd = 0.0022(114)(20) = 5.02 sq in. min required As = 0.0018(114)(24) = 4.92 sq in. Try 12–​#6, As = 5.28 sq in.; average d = 24 –​ 3 –​ 0.75 = 20.3 in. C = 0.85 fc′ ba = 0.85(3)(9.5)(12)a = 291a T = As f y = 5.28(60) = 317 kips C = T;

a = 1.09 in.

φ M n = 0.90(317)[20.3 − 0.5(1.09)]

1 12

= 470 ft-kips



φ M n > [ Mu = 431 ft-kips]

OK

Use 12–​#6 bars (As = 5.28 sq in.) each way. (e) Development of reinforcement. The general equation, Eq. (6.6.1), often indicates a smaller Ld than given by the simplified equations. Using Eq. (6.6.1),    ψ t ψ e ψ s  3 fy Ld =  d [6.6.1]  40 λ fc′  cb + K tr   b   d     b



The center-​to-​center spacing of the 12–​#6 bars is approximately 9 3 4 -​in. For the general equation, the distance cb is the smaller of the following two values: bottom and side cover = 3.0 (i.e., clear) + 0.375 (i.e., bar radius) = 3.38 in. one-​half center-​to-​center spacing ≈ 9.75/​2 = 4.9 in. Thus, cb = 3.38 in. There are no stirrups; thus, Ktr = 0. Thus,

  cb + K tr 3.38 + 0 = = 4.5 > 2.5 max  d 0 . 75 b  

Thus, (cb + Ktr)/​db = 2.5. (Continued)

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Example 19.9.1 (Continued) Evaluate, using Eq. (6.6.1):

 3 60, 000 1.0(1.0)1.0  Ld (for #6) =   0.75 = 24.6 in (2.05 ft) 2.5  40 (1.0) 3000

Note that ψs was conservatively taken as 1.0, although ACI-​25.4.2.4 allows a value of 0.8; ψt and ψe are both equal to 1.0. Allowing a 3-​in. cover on the end of the #6 bars, the available embedment is

available embedment = 49 − 3 = 46 in. > [ Ld = 24.6 in.]

OK

(f) Design sketch. A design sketch as shown in Fig. 19.9.4 is necessary to convey the designer’s decision properly.

12 – #6 (spacing ≈ 9¾”) each way

1’– 4” 1’– 4”

3” clear cover 12 – #6 bars each way

9’– 6”

2’– 0”

9’– 6”

Figure 19.9.4  Design sketch for spread footing of Example 19.9.1.

For practical design of square footings, design aids are available [2.23, 19.9].

19.10 DESIGN OF RECTANGULAR FOOTINGS Rectangular footings may be used in locations where restricted space prevents the use of a square footing. The procedure for their design is essentially identical to that of square footings, except that one-​way shear action and bending moment must be considered in both principal directions.

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EXAMPLE 19.10.1 Design a rectangular spread footing to carry 235 kips service dead load and 115 kips service live load from an 18-​in. square tied column that contains #9 bars. One dimension of the footing is limited to a maximum of 7 ft. The allowable soil pressure is 5500 psf. Neglect the effect of overburden. Use fc′ = 3000 psi (normal-weight concrete) and fy = 60,000 psi. SOLUTION (a) Determine plan of footing. Assume a footing depth of 2 ft, or 300 psf. net soil pressure = 5500 − 300 = 5200 psf

required A =

350 = 67.3 sq ft 5.2



Space limitation prevents one dimension from exceeding 7 ft; thus

length =

67.3 = 9.6 ft 7.0

Try 7 ft × 9 ft 8 in. (area = 67.7 sq ft). (b) Determine depth required for shear strength. This footing may be long enough for one-​way beam action to govern. In such a case, a direct solution for the effective depth d is reasonably practical. The factored column load is

Pu = 1.2(235) + 1.6(115) = 466 kips



The net upward pressure under factored load condition is

pnet =

466, 000 = 6880 psf 67.7

Using section A-​A in Fig. 19.10.1, and making the nominal shear strength Vn = Vc, so that shear reinforcement is not required, means Vn = Vc = 2 λ fc′bw d Vu = φ Vc

6880(7.0)(4.08 − d ) = 0.75[2(1.0) 3000 ](7.0)(d )(144) d = 1.5 ft (18.0 in.)

Total depth = 18.0 + 3 (cover) + 1 (estimated bar diameter) = 22.0 in. Try 22 in. for total depth. Check this depth for two-​way shear action as a slab, using critical section B-​B-​B-​B shown in Fig. 19.10.1 with d = 18 in.

Vu = 6880 [7.0(9.67) − 3.0(3.0)]

1 1000

= 404 kips

For β = 9.67/​7.0 = 1.38 < 2.0, and b0 /​d = 4(18 + 18)/​18 = 8 < 20 for a four-​sided critical section, Vn = Vc = 4 λ fc′b0 d = 4(1.0) 3000 [ 4(18 + 18)](18)

φ Vc = 0.75(568) = 426 kips > [Vu = 404 kips]

1 1000

= 568 kips

OK (Continued)

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Example 19.10.1 (Continued) 3” clear

A Critical section for one-way action

B

18”+18” = 36”

9–#7 bars (spacing ≈ 9½”)

Critical section for two-way action B

3.0’ 7’– 0” d/2

B

B d

4.08’ – d

2’– 9”

A 1’–6”

L1

3” clear

3” clear

4’–1”

1’– 10”

9 – #7 bars (spacing ≈ 12”)

3” clear

9’– 8”

Figure 19.10.1  Rectangular footing for Example 19.10.1.

(c) Check transfer of load at base of column. Assuming transfer without aid of dowels, the design strength is

φ Pn = φ (0.85 fc′ )Ag = 0.65(0.85)(3)(18)(18) = 537 kips φ Pn > [ Pu = 466 kips]



OK

Only the minimum dowels required by ACI-​16.3.4.1 are needed.

min required As = 0.005(18)(18) = 1.62 sq in.

Use 4–​#6 bars as dowels (As = 1.76 sq in.). Minimum embedment length L1 (Fig. 19.10.1) required for bars in compression [Eq. (6.8.1)],

min L1 = Ldc (# 6) = 16.5 in. available L1 = 22 − 3 − 1.5 − 0.75 = 16.75 in. > 16.5 in.



OK

(d) Design for flexural strength. The reinforcement in the long direction is distributed uniformly across the 7-​ft width, while that in the short direction is concentrated more heavily under the column in a band equal to the footing width and less heavily near the ends. ACI-​13.3.3.3 prescribes that the portion 2/​(β + 1) of the total reinforcement in the short direction be placed in the central band (of a width equal to the short side of the footing) wherein β is the ratio of the long side to the short side of the footing. The ratio 2/​(β + 1) may be derived on the basis that the intensity of reinforcement in the central band is twice that of the outer portions. In the long direction, Mu = 6.88(7.0)

required Rn =

(4.08)2 = 401 ft-kips 2

Mu 401(12, 000) = = 186 psi 2 φ bd 0.90(84)(18.5)2

(Continued)

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1 9 . 1 0   D E S I G N O F R E C TA N G U L A R F O OT I N G S

Example 19.10.1 (Continued) Since the longitudinal bars are placed below the transverse bars, d was taken as 18.5 in. in the above calculation. Using the trial moment arm method, rather than the formula, Eq. (3.8.5), for ρ, assume the moment arm as 0.95d = 17.6 in. since the value of required Rn is very low. required As =



Mu 401(12) = = 5.06 sq in. φ f y (arm) 0.9(60)(17.6)

Check: C = 0.85 fc′ ba = 0.85(3)(84)a = 214 a T = As f y = 5.06(60) = 304 kips 304 = 1.42 in. 214 arm = 18.5 − 0.71 = 17.8 in. ≈ 17.6 in. assumed a=





 17.6  revisedd required As = 5.06  = 5.0 sq in.  17.8  min required As = 0.0018(84)(22) = 3.3 sq in.

(ACI-8.6.1.1)

Use 9–​#7, As = 5.4 sq in. In the short direction, Mu = 6.88(9.67)

(2.75)2 = 252 ft-kips 2

assumed arm = 0.95d = 0.95(17.5) = 16.6 in.



Mu 252(12) required As = = = 3.4 sq in. φ f y (arm) 0.9(60)(16.6)

Check: C = 0.85(3)(116)a = 296 a T = 3.4(60) = 204 a = 0.69 in. arm = 17.5 − 0.35 = 17.2 in.



 16.6  revised required As = 3.4  = 3.3 sq in.  17.2  For minimum reinforcement, ACI-​8.6.1.1 requires ρg = 0.0018. For the short direction, min required As = 0.0018(116)(22) = 4.6 sq in. In this case, the minimum requirement controls. Try 8–​#7, As = 4.8 sq in.

reinforcement in band width 2 2 = = = 0.84 total reinforcement β + 1 9.67 / 7.0 + 1

Number of bars in the 7-​ft band = 8(0.84) = 6.7, say 7. If one bar is placed on each side outside the 7-​ft band, a total of 9 bars would be required. Use 9–​#7 bars. (spaced at 12 in. for equal bar spacing within the 7-ft band) (Continued)

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Example 19.10.1 (Continued) (e) Development of reinforcement. The general equation, Eq. (6.6.1), often indicates a smaller Ld than given by the simplified equations. Use Eq. (6.6.1):    ψ t ψ e ψ s  3 fy Ld =  d  40 λ fc′  cb + K tr   b   d     b



[6.6.1]

The 9–​#7 give an approximate 9 1 2 -​in. center-​to-​center spacing across the 7-​ft width, and 12 in. center-​to-​center spacing across the 9 ft 8 in. width. For the general equation, the distance cb is the smaller of the following values: bottom and side cover = 3.0 (i.e., clear) + 0.438 (i.e., bar radius) = 3.44. 9.5 = 4.75 in. 2 12 one-half center-to-center spacing ≈ = 6.0 in. 2 one-half center-to-center spacing ≈



(7 ft width) (9 ft 8 in. width)



Thus, cb = 3.44 in. for both directions. There are no stirrups; thus, Ktr = 0. Thus,   cb + K tr 3.44 + 0 = = 3.93 > 2.5 max  0.875   db

Thus, (cb + Ktr) /​db = 2.5.

Evaluate Eq. (6.6.1), where ψt, ψc, and ψs are all equal to 1.0,

 3 60, 000 1.0(1.0)1.0  Ld (for #7) =   0.875 = 28.8 in. (2.4 ft) 2.5  40 (1.0) 3000

Allowing a 3-​in. cover on the end of the #7 bars, the available embedment is

available embedment = 49 − 3 = 46 in. > [ Ld = 28.8 in.]

(long direction)



available embedment = 33 − 3 = 30 in. > [ Ld = 28.8 in.]

(short direction)



OK OK

The complete design is shown in Fig. 19.10.1.

19.11 DESIGN OF PLAIN AND REINFORCED CONCRETE WALL FOOTINGS Wall footings carrying direct concentric loads may be of either plain or reinforced concrete. Those that are required to carry moment, such as for cantilever retaining walls, are treated in Chapter 15. When the wall footing has bending in only one direction, it may be designed or investigated by considering a typical 12-​in. strip along the wall. Many typical walls carry relatively light loads, and the supporting footings are proportioned by using arbitrary minimums. Footings carrying light loads on good soil are often made of plain concrete and may be designed in accordance with Chapter 14, Plain Concrete, of the ACI Code.

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EXAMPLE 19.11.1 Determine the adequacy of the plain concrete wall footing of Fig. 19.11.1 to carry service loads of 20 kips/​linear ft dead load including the wall weight and 8 kips/​linear ft live load. Use fc′ = 3000 psi (normal-weight concrete), an allowable soil pressure of 6 ksf, and the ACI Code. SOLUTION total service load = 28 + 6(2.5)(0.145) = 30.2 kips/ft maximum soil pressure =



30.2 = 5.0 ksf < 6 ksf 6.0



OK

For a concrete wall, the factored bending moment is computed on the critical section at the face of the wall (ACI-​13.2.7.1). wu = 20(1.2) + 8(1.6) = 36.8 kips/ft net soil pressure under factored load = 36.8/​6 = 6.13 ksf Mu = 6.13



(2.5)2 = 19.2 ft-kips/ft 2

For computing the moment of inertia of the section, the bottom 2 in. of concrete placed against the ground is assumed to be of poor quality and must be neglected (ACI-​ 14.5.1.7). Neglecting the bottom 2 in., Ig =



12(28)3 = 21, 950 in.4 12

For bending of plain concrete, the design strength is based on a linear stress-​strain relationship for both tension and compression (ACI-​ 14.5.1.4). In accordance with ACI-​14.5.2.1, the nominal flexural strength is the smaller of the value calculated at the tension face as

M n = 5λ fc′Sm



and that computed at the compression face as M n = 0.85 fc′Sm where Sm is the elastic section modulus. Therefore,

M n = 5λ fc′Sm = 5(1.0) 3000

21, 950 1 = 35.8 ft-kips (28 / 2) 12,000

Controls!

or

M n = 0.85 fc′ Sm = 0.85(3000)

21, 950 1 = 333 ft-kips (28 / 2) 12,000

For plain concrete, a strength reduction factor φ of 0.6 is prescribed (ACI-​21.2). Thus, φ M n = 0.60(35.8) = 21.5 ft-kips > [ Mu = 19.2 ft-kips/ft] OK It may be readily verified that shear strength is adequate since the critical section at a distance d from the face of a wall falls near the edge of the footing. (Continued)

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Example 19.11.1 (Continued)

w = 28 kips/ft

12”

2’– 6”

6’– 0”

Figure 19.11.1  Wall footing for Example 19.11.1.

EXAMPLE 19.11.2 Design a reinforced concrete footing for a 12-​in. masonry wall carrying 10 kips/​linear ft service dead load including the wall weight and 5 kips/​ft service live load. Use fc′ = 3000 psi (normal-weight concrete), fy = 60,000 psi, and an allowable soil pressure of 4000 psf. SOLUTION (a) Determine the footing thickness. Assume the footing depth to be 10 in. at 125 psf. Allowable net soil pressure = 4000 –​125 = 3875 psf. Footing width = 15/​3.875 = 3.87 ft. Use 4 ft. It is probable that the thickness will be governed by one-way shear action (inclined cracking), which is taken to be critical at a distance d from the face of the wall (Fig. 19.11.2). Applying the load factors,

wu = 1.2(10) + 1.6(5) = 20.0 kips/ft



Net soil pressure under factored load = 20.0/​4 = 5.0 ksf. When no shear reinforcement is used, the nominal strength for one-​way action, using the simplified procedure, is

Vn = Vc = 2 λ fc′bw d



which requires that Vu = φVc

(

)

5000 (1.5 − d ) = 0.75 2(1.0) 3000 (12)(d )(12)

1.5 − d = 2.37d d = 0.45 ft (5.4 in.)

(Continued)

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Example 19.11.2 (Continued) Critical section for moment and development of reinforcement

12”

1’– 9”

Masonry wall

3”

#4 @ 7”

#4

9”

4’– 0”

Figure 19.11.2  Wall footing for Example 19.11.2.

Assuming #4 bars, total thickness = 5.4 + 3 (cover) + 0.5(bar diameter) = 8.9 in. Use 9-​in. thickness.

check weight = 113 psf [ < 125 psf assumed]

OK

(b) Reinforcement for flexural strength. The critical section for bending moment strength on footings under masonry walls is to be taken halfway between the middle and the edge of the wall (ACI-​13.2.7.1). Mu =

required Rn =

1 (5.0)(1.75)2 = 7.66 ft-kips/ft 2 7.66(12,000) = 281 psi 0.90(12)(5.5)2

The steel area may be obtained by formula, Eq. (3.8.5),

ρ = 0.005(> ρmin )



required As = ρ bd = 0.005(12)(5.5) = 0.33 sq in.

Try #4 @ 7 in. (As = 0.34 sq in./​ft). Check strength. C = 0.85 fc′ ba = 0.85(3)(12)a = 30.6 T = As f y = 0.34(60) = 20.4 kips C = T;

a = 0.66 in.

φ M n = 0.90(20.4)[5.5 − 0.5(0.66)] 121 = 7.9 ft-kips φ M n > [ Mu = 7.66 ft-kips]

OK

(c) Development of reinforcement. For the wide bar spacing 7 in., the general equation gives

Ld (# 4) = 16.4 in. (1.4 ft)

(Continued)

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Example 19.11.2 (Continued) (Note:  ψs was conservatively taken as 1.0 in the computation of Ld, although ACI-​ 25.4.2.4 allows a value of 0.8). The available embedment distance is 21 in. less 3-​in. cover, i.e., 18 in., which is larger than the 16.4 in. required, and is acceptable. A minimum temperature and shrinkage reinforcement must be provided per ACI-​ 7.6.4.1 and ACI-​24.4, As, min = 0.0018(12)(9) = 0.19 sq in. / ft This temperature and shrinkage reinforcement must have a spacing not greater than 5 times the slab thickness (45 in.) and 18 in. Thus, provide 4–#4 bars at 12 in. spacing (As = 0.20 sq in./ft) perpendicular to the slab flexural reinforcement. This would leave a 5.5 in. cover on the outer bars, which is greater than the minimum required of 3 in. See Fig. 19.11.2 for the details of the complete design.

19.12 COMBINED FOOTINGS A combined footing is one that usually supports two columns. These may be two interior columns [Fig. 19.12.1(a)], which are so close to each other that isolated footing areas would overlap. If a property line exists at or near the edge of an exterior column, a rectangular [Fig. 19.12.1(b)] or trapezoidal [Fig. 19.12.1(c)] combined footing may be used to support the exterior column and its adjacent interior column. The area of the combined footing may be proportioned for uniform settlement by making its centroid coincide with the resultant of the respective portions of the two column loads that are sustained for long duration. It may be noted that for footings of constant thickness, the centroid of the bearing area always coincides with the resultant of the weight of the footing itself. Referring to the frequently occurring situation in Fig. 19.12.1(b), the load P1 is close to the property line; however, there is adequate space to the right of P2. Whenever P2 exceeds P1, a rectangular combined footing can be used, since it may be made long enough to make the load resultant and the footing centroid coincide. It may be shown that if 1 2 < P2 /P1 < 1 approximately, a trapezoidal footing could be used. If P2 /P1 < 1 2 approximately, then either a strap (Fig. 19.12.2) or a T-​shaped spread footing would have to be used. In the strength computation for a combined footing, maximum loads in columns (full dead load plus reduced live load as discussed in Section 19.7) should be used. Since the resultant of the maximum column loads does not necessarily coincide with that of the sustained column loads producing uniform settlement, under the former loading condition the distribution of the net soil pressure along the footing is not uniform. The deviation from

Property line

Property line

P2

P1

(a)

(b)

Figure 19.12.1  Combined footings.

P1

P2

(c)

823



19.13  DESIGN OF COMBINED FOOTINGS

823

Property line Pint

Pext

Rext

Rint

Figure 19.12.2  Cantilever, or strap, combined footing.

uniformity is usually so small that certain approximate shortcut procedures may be used in determining the shear and moment diagrams in the longitudinal direction. The ACI Code (see Chapter 13) does not provide full recommendations for combined footings. Kramrisch [19.10] has provided a detailed treatment of footing design, with particular emphasis on combined footings. However, ACI Committee 336 [19.11, 19.12, 19.20] has given design procedures for combined footings and mats, and additional suggestions have been given by Kramrisch and Rogers [19.13], Szava-​Kovats [19.14], and Davies and Mayfield [19.15]. When a series of columns is supported by a single footing extending over a large area, the footing is referred to as a mat. Mats are treated by Bowles [19.16] as well as by ACI Committee 336. Basically, transverse steel under each column tends to distribute the column load in the transverse direction. This being considered accomplished, the combined footing itself becomes a beam in the longitudinal direction. Thus it is suggested that the provisions for isolated footings be applied to the transverse direction, and those for beams to the longitudinal direction. The cantilever or strap footing shown in Fig. 19.12.2 is an alternative design to prevent overturning of an exterior footing placed eccentrically under an exterior column, the edge of which is at or close to a property line. Overturning of the exterior footing is prevented by connecting it with the adjacent interior footing by a strap beam. This strap beam is subjected to a constant shearing force and a linearly varying negative bending moment. Thus it behaves like a cantilever beam; hence the name “cantilever footing.” In the strength computation for a cantilever footing (Fig.  19.12.2), the weight of the strap, the exterior footing, and the interior footing is each assumed to be balanced by the soil pressure caused by it; thus such weight causes no shears and moments in any part of the structure. In the longitudinal direction, the column loads Pext and Pint are in equilibrium with the total “net” upward soil pressures Rext and Rint under the exterior and interior footings; the upward soil pressure is assumed to be uniform over the entire area of each footing. The exterior footing may be considered as under one-​way transverse bending, although some reinforcement in the longitudinal direction is desirable, and the interior footing is under two-​way bending as in isolated footings. Use of a cantilever footing may be justifiable under conditions where the distance between columns is large and a large area of excavation must be avoided. It is usual practice that the bottom surfaces of the exterior footing, the strap, and the interior footing be at the same elevation, but the total thickness of each element may be different, depending on the strength requirements. Certainly it is desirable, unless there are good reasons to the contrary, to make all three elements of constant thickness.

19.13 DESIGN OF COMBINED FOOTINGS The design of two common types of combined footings will be shown. The first is a rectangular footing, and the second is a strap or cantilever footing.

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EXAMPLE 19.13.1 Design a combined footing to support two columns as shown in Fig. 19.13.1(a): PA = 350 kips (40% live load); PB = 400 kips (40% live load); column A is centered 1 ft 3 in. from the property line, and column B is centered 19 ft 3 in. from the property line. Use fc′ = 3000 psi (normal-weight concrete), fy = fyt = 60,000 psi, and a maximum soil pressure = 5000 psf. Assume that the ratio of maximum column loads as given is equal to that of long duration loads in the exterior and interior columns. SOLUTION (a) Length and width of footing. ACI-​13.3.1.1 indicates base area of footings is to be determined using service loads and allowable soil pressure.



x from property line =

350(1.25) + 400(19.25) = 10.85 ft 750

length of footing, L = 10.85(2) = 21.70 ft Use a 21 ft ​9 in.-long footing Since the design for strength of the footing involves factored loads, there will be an eccentricity no matter how “exact” the length determination. In the design for shear and bending moment, the soil pressure under factored loads might be taken as linearly varying to account for the eccentric loading; in the case of a small eccentricity, however, it is probably sufficient to assume a uniform soil pressure as is done in this example. Assume the footing thickness to be about 2 ft 6 in., or 375 psf.

net soil pressure = 5500 − 375 = 4625 psf 750, 000 footing width = = 7.46 ft 4625(21.75)

Try a 7 ft 6 in.-wide footing (b) Longitudinal factored shears and factored moments. For gravity loading, column A, Pu = 1.2(210) + 1.6(140) = 476 kips

column B, Pu = 1.2(240) + 1.6(160) = 544 kips net soil pressure under factored load =



1, 020, 000 = 6250 psf 21.75(7.5)

The factored shear Vu diagram is computed as for a beam and given in Fig. 19.13.1(c). For simplicity, the column loads are taken to be acting along the column centerlines, thus producing the dashed portions within the column widths on the shear diagram. In the computations, the 20-​in.-​diameter column is treated as an equivalent square of side 17.7 in. in accordance with ACI-​13.2.7.3. net upward uniform pressure = 7.5(6.25) = 46.9 kips/ft Vu at centerline of column A = +46.9(1.25) − 476 = +58.6 − 476 = −417.4 kips

Vu at centerline of column B = −46.9(2.5) + 544 = −117.3 + 544 = +426.8 kips



417.4   point of zero shear = 18   417.4 + 426.8  = 8.90 ft from centerline of column A (Continued)

825



825

19.13  DESIGN OF COMBINED FOOTINGS

Example 19.13.1 (Continued) The factored moment Mu diagram as computed for a beam is given in Fig. 19.13.1(d). Note that the numerical values on the two small end portions of the factored moment diagram are based on assuming all the column loads to be concentrated at the column centerlines. 46.9(10.15)2 − 476(8.90) 2 = −1821 ft-kips

max Mu (computed from left side) =

46.9(11.60)2 − 544(9.10) 2 = −1795 ft-kips

max Mu (computed from right side) =

1’– 3”

18’– 0”

16”

Column A, PA = 350k

2’– 6”

Column B, PB = 400k

22”

7’– 6”



20”

Property line (a) Footing plan 3” clear cover

Equivalent square = 17.7” 2’– 5”

20 – #9 spaced 4 21 ” ±

1’– 2”6 @ 12” = 6’– 0” 6 – #6 Spaced @ 9” ±

#4 stirrups 6 @ 12” = 6’– 0” 1’– 3” 8 – #6 3” clear cover (side and bottom) Spaced @ 8 21 ”±

(b) Footing elevation

426.8k CL of column

295k

184k 58.6k

d= φVc φVc 24.9” 2

8.90’ CL of column d = 2.08’

φVc φVc 2

(c) Factored shear Vu diagram

1.76’

117.3k

Vc 1820 ft-kips 417.4k

Face of column

Face of column

Mu 870 ft-kips –

+

37 ft-kips (d) Factored moment Mu diagram

Figure 19.13.1  Rectangular combined footing for Example 19.13.1.

147 ft-kips

(Continued)

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Example 19.13.1 (Continued) The moments as computed from both sides are not exactly the same because a footing length of 21 ft 9 in. is used instead of the computed 21.70 ft and because the distance 8.90 ft to the point of zero shear contains only three significant figures. Use Mu = 1820 ft-​kips. (c) Thickness of footing. For moment, the thickness may be based on a desired reinforcement ratio ρ. Select ρ = 0.012, that is, approximately one-​half of ρb. For this value of ρ, using Eq. (3.8.4b), 1 Rn = ρ f y (1 − ρ m) 2 Rn = 0.012(60, 000)[1 − 0.5(0.012)23.5] = 618 psi   reqquired d =

Mu 1820 (1000) = = 20.9 in. φ Rn b 0.90(618)(7.5)

( based on flexural strength requirements)

The footing is considered a beam in shear computations. One-​way action is assumed to control at the distance d from the face of the columns. The shear at a distance d from the face of the 17.7-​in. equivalent square column (column B) is

Vn = 426.8 − (8.85 + d )

1 12

46.9 = 392.2 − 3.91d

The nominal shear strength when no shear reinforcement is to be used is

Vn = Vc = 2 λ fc′ bw d



unless the more detailed procedure is used. Then, using λ  =  1.0 for normal-​weight concrete, Vu = φVc

(

)

392.2 − 3.91d = 0.75 2(1.0) 3000 (7.5) d

12 1000

       392.2 − 3.91d = 7.39d d = 34.7 in.

(based on shear strength requirements)

It seems desirable to make the footing deep enough at the desirable reinforcement ratio for the bending moment, but not deep enough to give the extra 14 in. that would be required to eliminate stirrups. Try a 29-​in.-​thick footing. Check the weight, 29(150)/​12 = 363 psf.

max soil pressure =

750, 000 + 363 = 4961 psf < 5000 psf 21.75(7.5)

OK

(d) Main longitudinal reinforcement. At the middle of the span, assuming #9 bars d ≈ 29 − 3 (cover) − 0.5 (stirrup) − 0.6 (bar radius) = 24.9 in. Thus, required Rn =

Mu 1820(12, 000) = = 435 psi φ bd 2 0.90(7.5)(12)(24.9)2

 435  required As = 0.012  (7.5)(12)(24.9) = 18.9 sq in.  618 

(Continued)

827



827

19.13  DESIGN OF COMBINED FOOTINGS

Example 19.13.1 (Continued) Try 20–​#9 (approximate 4 1 2 -​in. spacing); As = 20.0 sq in. Check strength. C = 0.85 fc′ ba = 0.85(3)(90) a = 229.5a T = 20.0(60) = 1200 kips

a = 5.23 in.

φMn =



1 0.90(1200)[24.9 − 0.5(5.23)] 12

= 2006 ft-kips > 1820 ft-kips

OK

The anchorage required from the point of maximum moment to the end of the bars is the development length Ld for #9 bars. The distance from the bar center to the nearest concrete surface is [3 (clear cover) + 0.5 (stirrup) + 0.6 (bar radius)] = 4.1 in., which is greater than half the center-​to-​center bar spacing of 2.2 in. Therefore, use cb = 2.2 in. Since the bars have more than 12 in. of concrete cast beneath them, a casting position multiplier ψt of 1.3 must be applied. Thus, using Eq. (6.6.1) with cb = 2.2 in. and assuming no contribution from stirrups (Ktr = 0),

Ld (for #9) = 47.5(1.3) = 61.8 in. (5.1 ft)

If all 20 of the #9 bars are extended into the centerlines of the columns, the development length requirement is more than satisfied because an anchorage distance of 108 in. is provided on both sides of the point of maximum moment. Use 20–​#9 bars at 18 ft long (spaced approximately at 4 1 2  in.) Check development length requirement at the points of inflection according to ACI-​ 9.7.3.8.3. The ACI Code provision is checked here even though the reinforcement is negative moment reinforcement rather than the positive moment reinforcement as prescribed in ACI-​9.7.3.8.3. The footing may be visualized for this purpose as an inverted beam with the soil pressure as loading and the columns as supports. The situation is similar to the positive moment requirement in the sense that the required embedment is into the support (column) rather than out into the span as for the negative moment requirement in an ordinary continuous beam. Furthermore, since the inflection points are inside the faces of the columns, the 1.3 factor might be used (see Section 6.14); however, since the column width (22 in.) is only a small fraction of the footing width (7.5 ft), the authors do not recommend using the 1.3 factor. Since all 20–​#9 bars extend through the inflection points, M n = (2006 / 0.9) = 2229 ft-kips Vu = 417.4 kips(at column A)

Vu = 426.8 kips(at column B)



La = 15 in. approx (near both column A and B) In this case, the actual distance La from the inflection point to the end of the bars is larger than the maximum La limits of effective depth d (24.9 in.) or 12 bar diameters (13.5 in.); thus 13.5 in. controls. Since Mu and La are the same at both inflection points, the one near the column B having the larger shear Vu controls. At the column B inflection point,

Mn 2229(12) + La = + 13.5 = 76 in. > [ Ld = 61.8 in.] Vu 426.8

OK (Continued)

82

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C hapter   1 9     F ootings

Example 19.13.1 (Continued) (e) Alternative design of the main longitudinal reinforcement using cutoff. If it seems desirable to cut off some of the tension bars, say 10–​#9, and extend the remaining ones into the support, the general anchorage requirements of ACI-​9.7.3 must be applied. The theoretical point at which the 10–​#9 bars are no longer required, indicated in Fig.  19.13.2, corresponds to the flexural design strength, φ Mn, of 1060 ft-​kips. The provision in ACI-​9.7.3.3 requires an extension of at least the effective depth d or 12 bar diameters, whichever is greater, beyond the theoretical cutoff point. In this case, the effective depth of 24.9 in. (or 2.1 ft) controls over 12 bar diameters = 13.5 in. (or 1.1 ft). CL of column

CL of column

20 – #9 φMn = 2006 ft-kips Ld = 4.0’

Ldh = 24.7”

2.1’ d

1’

Ldh = 24.7”

2.1’ d

10–#9 φMn = 1060 ft-kips

X1

φMn diagram

Ld = 4.0’

X2

Theoretical cutoff points

Development over length Ldh of hooked bars 2’–3”

Mu

18’– 0”

Figure 19.13.2  Bar cutoff alternative for Example 19.13.1.

In addition, for such a cutoff in the tension bars to be permitted, ACI-​9.7.3.5 must be satisfied. It is assumed here that stirrups will be provided sufficient to satisfy ACI-​ 9.7.3.5(b). The continuing bars must be embedded Ld beyond the cutoff of the 10–​#9 bars according to ACI-​9.7.3.4. The development length Ld for the bars being cut as well as for the continuing bars, was computed in part (d) as 5.1 ft without stirrups. Assuming closely spaced stirrups such that [cb + Ktr]/​db is 2.5 max, the required development length of the #9 bars would be reduced to at most 4.0 ft. At neither end the available embedment distance (≈ 2.1 ft from end of bars at the column A end of the footing to point X1 and ≈ 3.5 ft from end of bars at the column B end of footing to point X2) is adequate to develop (Ld = 4.0 ft) the continuing bars through straight embedment. Thus, hooks would be needed at both ends. When hooks are provided as anchorage beyond an inflection point (as would be the case here at both ends of footing), ACI-​9.7.3.8.3 need not be satisfied. Referring to Table 6.10.2 and Fig. 6.10.2 for #9 hooked bars,

Ldh = 24.7 in.



In Fig. 19.13.2, the development of the #9 hooked bars is shown by the sloping dashed line at each end of the footing, starting from the end of the bar at 3.0 in. from the side face of footing. (Continued)

829



829

19.13  DESIGN OF COMBINED FOOTINGS

Example 19.13.1 (Continued) Since the cut bars at X1 and X2 are still in the tension zone, ACI-​9.7.3.5 must be satisfied. By inspection of Fig. 19.13.2, φ Mn exceeds twice Mu at the cutoff points, thus, check Vu /​ (φ Vn) ≤ 0.75 for the cuts to be acceptable: Av f yt d   φVn = φ  2 λ fc′bw d + s   Using #4 multiple-loop stirrups with N = 8 spaced at 12 in. [(Fig. 19.13.4) in part (h)],

8(0.2)(60)24.9   φVn = 0.75  2(1.0) 3000 (90)24.9 1 +  1000  12

φVn = 184 + 149 = 333 kips Since Vu ≈ 368 kips at the proposed cutoff locations (see Figs. 19.13.1 and 19.13.2), the cuts cannot be made at locations X1 and X2. The 10–​#9 bars must be lengthened to reach the inflection points, in which case they are no longer in the tension zone and ACI-​9.7.3.5 does not apply. [Note that the proposed cutoff locations are closer to the face of the supports than the first critical section for shear. Thus, for evaluation of shear strength, Vu = 295 kips < φ Vn as shown later in part (h)]. It is clear in this example that cutting bars is impractical when about 4 ft of straight length is saved on 10 bars but the remaining 10 must be hooked. (f) Longitudinal reinforcement at bottom of footing beyond column centers. The bending moment at the face of column B is

Mu =

1 (46.9)(1.76)2 = 72.6 ft-kips 2

Though certainly not always the case, the moment here appears small enough to require no reinforcement. Assuming a plain concrete element for this check, the design strength is based on a linear stress-​strain relationship for both tension and compression (ACI-​ 14.5.1.4). In accordance with ACI-​14.5.2.1, the nominal flexural strength is the smaller of the value calculated at the tension face as

M n = 5λ fc′ Sm



and that computed at the compression face as

M n = 0.85 fc′ Sm



where Sm is the elastic section modulus. Neglecting the bottom 2 in. of thickness,

Sm =

(7.5)(12)(27.0)2 = 10, 935 cu in. 6

Therefore,

M n = 5λ fc′Sm = 5(1.0) 3000 (10, 935)

1 12,000

= 250 ft-kips

Controls!

or

M n = 0.85 fc′ Sm = 0.85(3000)(10, 935)

1 12,000

= 2324 ft-kips (Continued)

830

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C hapter   1 9     F ootings

Example 19.13.1 (Continued) For plain concrete, a strength reduction factor φ of 0.6 is prescribed (ACI-​21.2). Thus, φ M n = 0.60(250) = 150 ft-kips > [Mu = 72.6 ft-kips] OK No flexural reinforcement in the longitudinal direction is required for strength at the bottom of either cantilever. (g) Transverse reinforcement. Bending in the transverse direction may be treated in a manner similar to isolated spread footings. The 1940 ACI, Joint Committee [19.17] recommended that the transverse reinforcement at each column be placed uniformly within a band having a width not greater than the width of the column plus twice the effective depth of the footing. The procedure seems to be reasonable. Certainly the behavior of the footing depends on the overall length-​to-​width ratio as well as the spacing of the columns. In this design example, the large spacing of the columns and the relatively narrow footing mean that most of the footing between the columns will be subjected to longitudinal curvature only, while locally, in the vicinity of the concentrated loads, curvature in both directions will result. Thus, the transverse reinforcement is placed into bands as shown in Fig. 19.13.3. d = 29 − 3 (cover ) − 0.5 (stirrup) − 0.375 (#6 bar radius assumed) = 25.1 in. column A band width, WA = 1.25 + (8 + 25.1)

 

1 12

= 4.0 ft

net factored load pressure in transverse direction =

476 = 63.5 kips/ft 7.5

1 Mu = (63.5)(2.83)2 = 254 ft-kips 2



required Rn =



Mu 254(12,000) = = 112 psi φ bd 2 0.90(4.0)(12)(25.1)2

From Eq. (3.8.5), ρ = 0.002, thus required As = 0.002(48.0)(25.1) = 2.4 sq in. Check minimum area of flexural reinforcement (ACI-8.6.1.1) As,min = 0.0018(48)(29) = 2.5 sq in.



Controls!

Try 6–​#6 bars (As = 2.64 sq in.). Check strength. C = 0.85 fc′ ba = 0.85(3)(48)a = 122.4 a T = 2.64(60) = 158.4 kips

a = 1.29 in.

φ Mn =

0.90(158.4)[25.1 − 0.5(1.29)] 1 12

= 291 ft-kips > 254 ft-kips

OK

Since 2 ft 10 in. is available from face of column, embedment can be provided that exceeds the required development length Ld for #6 bars (Ld  =  32.9 in., Category A, Table 6.6.1, by the conservative simplified equation). Use 6–​#6 bars (approximate 9-​in. spacing). column B band width, WB = 2.50 + (8.85 + 25.1)

1 12

net factored load pressure in transverse direction=

= 5.3 ft 544 = 72.5 kips/ft 7.5 (Continued)

831



831

19.13  DESIGN OF COMBINED FOOTINGS

Example 19.13.1 (Continued) 15” d = 12.45” 2

Column A

Critical sections for shear in one-way action

2.7’ WA = 4.0’

d = 12.45” 2

Column B

3.01’

2’–10”

d = 24.9” Property line

Critical sections for shear in two-way action

2.8’

2.5’

WB = 5.3’ d +17.7” 2

Figure 19.13.3  Band width for transverse reinforcement.

1 (72.5)(3.01)2 = 328 ft-kips 2 Mu 328(12,000) = 109 psi required Rn = = 2 2 φ bd 0.90(5.3)(12)(25.1) Mu =



From Eq. (3.8.5), ρ ≈ 0.002, the same as for column A. required As = 0.002(5.3)(12)(25.1) = 3.19 sq in.



Check minimum area of flexural reinforcement (ACI-8.6.1.1) As,min = 0.0018(63.6)(29) = 3.32 sq in.



Controls!

Try 8–​#6 (As = 3.52 sq in.). Check strength. C = 0.85 fc′ ba = 0.85(3)(63.6)a = 162.2 a T = 3.52(60) = 211.2 kips

a = 1.30 in.

φ Mn =



0.90(211.2)[25.1 − 0.5(1.30)] 1 12

= 387 ft-kips > 367 ft-kips

OK

Anchorage of #6 bars is adequate since available embedment of 3.01 ft exceeds Ld of 32.9 in. Use 8–​#6 bars (approximately 8 1 2 -​in. spacing). (h) Shear reinforcement. The usual approach is to consider the footing as a beam and to provide shear reinforcement on the assumption that the shear (inclined cracking) effect is uniform across the width. This approach seems appropriate in this case, given the large distance between columns and the relatively narrow footing width. The maximum shear to be provided for is at the critical section a distance d from the face of the column. From Fig. 19.13.1, Vu = 295 kips. The design shear strength φVc of a beam without shear reinforcement, using the simplified procedure, is

(

φVc = φ 2λ fc′bw d

(

)

φVc = 0.75 2(1.0) 3000 (90)24.9

)

1 1000

reequired φVs = Vu − φVc = 295 − 184 = 111 kips

= 184 kips

(Continued)

832

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C hapter   1 9     F ootings

Example 19.13.1 (Continued) When shear reinforcement is required, there must be at least the minimum specified by ACI-​9.6.3.3, as the larger of

min φVs = φ (0.75 fc′ )bw d = 0.75(0.75 3000 )(90)(24.9)

1 1000

= 69 kips

and

min φVs = φ (50)bw d = 0.75(50)(90)(24.9)

1 1000

= 84 kips

Controls!

Thus, the required φVs = 111 kips controls the closest spacing for shear reinforcement. Even though this is a footing, the rules for beams are believed appropriate here. Thus, the required amount of shear reinforcement is

required

Av φVs 111 = = = 0.1 sq in./in. s φ f yt d 0.75(60)24.9

For design of shear reinforcement, use a multiple-​ loop stirrup as shown in Fig. 19.13.4. Thus, Av As N = s s



where As is the cross section of the stirrup bar, and N is the number of times the multiple-​ loop stirrup crosses the neutral axis of the beam. Try a #4 stirrup, with N  =  8 (see Fig. 19.13.4), spaced at 12 in. This gives

provided

Av 0.20(8) = = 0.133 sq in./in. 12 s

which is greater than the required amount.

OK

Figure 19.13.4  Multiple-​loop stirrup (N = 8).

Use #4 stirrups @ 12 in. with N = 8, as in Fig. 19.13.4. The 12-​in. spacing is adequate since it is the maximum permitted (ACI-​9.7.6.2.2) when Vs does not exceed 4 fc′ bw d , which corresponds to φVs = 368 kips. The first stirrup is placed at s/​2 = 6 in. from the face of the column, and the last one should be within s/​2  =  6 in. of the location where Vu  =  φVc /​2, at which point shear reinforcement is no longer theoretically required for beams (ACI-​9.6.3.1). The stirrup arrangement will be made identical at each column. To space and hold in position the stirrups and the transverse reinforcement, a few longitudinal bars are placed in the bottom of the footing. (i) Check shear strength based on two-​way action. At the exterior column A, the calculation based on a perimeter at d/​2 (12.45 in.) from the column face on all four sides would be unrealistic on the side nearest the property line. Instead, a three-​sided perimeter is used at column A in Fig. 19.13.3,

b0 = 2(15 + 8 + 12.45) + [22 + 2(12.45)] = 117.8 in. (Continued)

83



833

19.13  DESIGN OF COMBINED FOOTINGS

Example 19.13.1 (Continued) According to Eq. (19.5.1c), αs = 30 for the three-​sided perimeter (“edge column”), and when b0 /​d does not exceed 15 that equation does not control. In this case,

b0 /d = 117.8 / 24.9 = 4.7 < 15

Thus, with b0 /​d less than 15 and a loaded area with β < 2, shear strength is based on a nominal stress of 4 λ fc′ according to Eq. (19.5.1a). Thus,

φVn = φVc = φ 4 λ fc′ b0 d = 0.75[ 4(1.0) 3000 (117.8)(24.9)]

1 1000

= 482 kiips

Actually, the resisting section for shear is not symmetrical about the column; therefore, the shear distribution is nonuniform and eccentricity of shear should be considered in accordance with ACI-​8.4.2.3. In this example, where no moment is assumed at the base of column and the moment in the slab at the face of column is small, it may be reason­ able to neglect the eccentricity of shear. Thus, neglecting any eccentricity of shear, Vu = 476 − 6.25[35.45(46.90)]



1 144

= 404 kips

Neglecting any possible contribution of the stirrups used for φVs in one-​way shear action,

φ Vn = φ Vc = 482 kips > [Vu = 404 kips]

OK

A similar calculation for two-​way shear action at column B will show the thickness to be adequate without shear reinforcement. Again, it is prudent to consider the failure section to extend to the end of the footing (see Fig. 19.13.3), rather than merely the circular path at d /​2 from the column. (j) Design sketch. The complete design details are given in Fig. 19.13.1.

EXAMPLE 19.13.2 Design a cantilever or strap footing for the situation shown in Fig. 19.13.5, where the property line is at the exterior edge of the exterior column. Column data are given in Table 19.13.1. The distance between column centers is 18 ft. Equal settlement is assumed for DL plus 1 2 LL condition at a uniform pressure of 3.35 ksf. Use fc′ = 3000 psf for footing fc′ = 3750 psi for columns, fy = 40,000 psi. TABLE 19.13.1  COLUMN DATA FOR EXAMPLE 19.13.2 Column

Size

Reinforcement

LL

DL

Exterior Interior

12 × 12 in. 14 × 14 in.

4–​#7 8–​#8

  70 kips 130 kips

55 kips 80 kips

SOLUTION (a) Size of exterior and interior footings. According to ACI-​13.3.1.1, footing size is always determined using service loads, rather than factored loads. The size of (Continued)

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Example 19.13.2 (Continued)

exterior and interior footings is a function of the assumed width of the exterior footing (which affects the cantilever action and therefore the reactions required of each footing) and the assumed total thickness of each footing (which affects the available net soil bearing capacity). Should these assumed values be revised in the subsequent computation, the sizes of the footings must be revised accordingly. Assume that the total thickness of the entire footing is 25 in. The net uniform soil pressure for both exterior and interior footings is 3.35  –​2.08(0.150)  =  3.04 ksf. Referring to Fig. 19.13.5 for the dimensions, taking moments about the interior column using the service loads in the equal settlement condition gives Rext =



(55 + 35)18 = 99.7 kips 16.25

area of exterior footing required =

99.7 = 32.8 sq ft 3.04

Using the assumed exterior footing width of 4 ft 6 in., 32.8 length of exterior footing = = 7.29 ft (try 7 ft 4 in.) 4.50 Taking moments about the line of action of Rext gives

Rint =

(80 + 65)16.25 − (55 + 35)(2.25 − 0.50) = 135.3 kips 16.25

area of interior footing required =

135.3 = 44.5 sq ft 3.04

side of square interior footing = 44.5 = 6.67 ft (try 6 ft 8 in.)

(b) Factored shear and factored moment diagram for strap. Referring to Fig. 19.13.5, applying load factors U to the maximum load condition gives

[1.2(55) + 1.6(70)]18 (178)18 = = 197 kips 16.25 16.25 178(1.75) Rint = [1.2(80) + 1.6(130)] − = 304 − 19 = 285 kips 16.25

Rext =

Vu in strap = − 178 + 197 = +19 kips Mu at right end of strap = 19((3.33) = 63 ft-kips

Mu at left end of strap = 19(14) = 266 ft-kips

(c) Design of strap. For the shear requirement, assuming no shear reinforcement is to be used and the simplified expression for strength is used,

(

)

φ Vc = φ 2 λ fc′bw d

Estimate d ≈ 25 –​3 (cover) –​0.5 (est. bar radius) = 21.5 in. Thus,

width bw of strap required =

19, 000 0.75 2(1.0) 3000  21.5

= 10.88 in.

Assume desirable reinforcement, ρ ≈ 0.012, for which Rn = 435 psi. (Continued)

835



835

19.13  DESIGN OF COMBINED FOOTINGS

Example 19.13.2 (Continued) 18’ – 6”

12 × 12 col.

7’ – 4”

2’ – 9” 14 × 14 col.

27”

Section for two-way shear action

14” 6’ – 8”

1.38’ 35.5” Section for one-way shear action

1 2”

2’ – 6 2.54’

4’ – 6”

6’ – 8”

k

Pint = 304k (max)

Pext = 178 (max) 7 – #6 each way

Loads given include load factors U

7 – #7 each way

7 – #8 3” cover

2’ – 1” 43.8 kips/ft (max) 16’ – 3”

42.8 kips/ft (max) Rint = 285k (max)

k

Rext = 197 (max) (a) 10’ – 8”

≈67k

137k

19k Vu

Shear diagram, Vu 134k

117k 179 ft-kips

67’k Development over length Ldh of hook

Factored moment diagram, Mu

Ldh = 14.6”

Mu

63

– 329

+

143

226’k

Ldh = 14.6”

φMn diagram for negative moment

(b)

Figure 19.13.5  Cantilever, or strap, footing of Example 19.13.2.

At the junction with the interior footing,

width of strap required =

Mu 63 (12, 000) = = 4.2 in. 2 φ Rn d 0.90(435)(21.5)2

At the junction with the exterior footing,

width of strap by moment requirement =

266 (12,000) = 17.6 in. 0.90(435)(21.5)2

Try a strap width varying from 14 to 20 in. Mu 266 (12, 000) = = 384 psi 2 φ bd 0.90(20)(21.5)2 63 (12, 000) = psi 130 required Rn at narrow end = 0.90(14)(21.5)2 required Rn at wide end =



(Continued)

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Example 19.13.2 (Continued) Approximately,



 384  (20)(21.5) = 4.6 sq in. required As at wide end ≈ 0.012   435   130  ( 14 )( 21 . 5 ) = 1 . 1 sq in. required As at narrow end ≈ 0.012   435 

The minimum requirement of ACI-​9.6.1.2 or 9.6.1.3 applies here because the strap is a beam. Thus,

As , min =

3 fc′ fy

bw d [3.7.10]

but not less than 200bw d/​fy. Thus, at the narrow end ACI-​9.6.1.2 requires

As , min

3 fc′

3 3000 (14)21.5 = 1.24 sq in. 40, 000 200 bw d 200(14)21.5 = = = 1.51 sq in. fy 40, 000

As , min =

fy

bw d =

Controls!

Alternatively, according to ACI-​9.6.1.3, 4 4 (required As ) = (1.1) = 1.50 sq in. 3 3 It may be shown that at the wide end As, min does not govern. Thus, use 7–​#8 (As = 5.53 sq in.) at the wide end. Extend 3–​#8 through the narrow end. Two bars would satisfy As,min = 1.50 sq in. but to maintain symmetry, it is necessary to extend three bars. The development length Ld for these #8 bars depends on their clear spacing and cover. For the general equation, the distance cb is the smaller of the following two values:

As ,min =

top and side cover = 3.0 ( i.e., clear ) + 0.50 ( i.e., bar radius) = 3.5 in.

one-half center-to-center spacing =

20 − 3(2) − 1 = 1.08 in. 2(6)



Thus, cb = 1.08 in. and Ktr = 0. Clearly, development will be a problem with such a small cb. Increase the strap width from 20 to 27 in. This may permit using smaller bars in the top of the strap. Recomputing Rn gives

required Rn at wide end =

Mu 266 (12, 000) = = 284 psi φ bd 2 0.90(27)(21.5)2

From Eq. (3.8.5), ρ = 0.008. Thus, required As = ρ bd = 0.008(27)21.5 = 4.64 sq in. The 7–​#8 will be retained, because the factored moment diagram increases more steeply than the assumed linear increase in development of reinforcement. Even with the increased value of cb for the general equation, the Ld value will still be a significant portion of the roughly 54 in. available. Using the general equation, with

cb = one-half center-to-center spacing =

27 − 3(2) − 1 = 1.67 in. 2(6)

Ld (#8) = 32.8(1.3) = 42.6 in. (3.6 ft) (Continued)

837



837

19.13  DESIGN OF COMBINED FOOTINGS

Example 19.13.2 (Continued) The ψt = 1.3 is the modification factor for the bar casting position when more than 12 in. of concrete lies beneath the bars. Even though the available development length of about 54 in. exceeds Ld, the flexural design strength, (φ Mn diagram) would encroach on the Mu diagram. The 7–​#8 bars should be hooked at the outside edge of the exterior column footing. When hooks are provided as anchorage, the development length Ldh for a #8 bar is, from Table 6.10.2,

 40  Ldh (# 8) = 21.9   = 14.6 in.  60 

where the 40/60 ratio accounts for the Grade 40 reinforcement. In Fig.  19.13.5, the development of the #8 hooked bars is shown by the sloping dashed line at the exterior end of the footing, starting from the end of the bar at 3.0 in. from the side face of the footing. (d) Investigate one-​way shear at the distance d (i.e., 21.5 in.) from the face of the exterior column.

Vu = 134 − (43.8)(21.5)

1 12

= 56 kips

For no shear reinforcement and the simplified expression for strength,

φ Vc = φ (2 λ fc′ bw d )

1 = 155 kiips > 56 kips φ Vc = 0.75 2(1.0) 3000  (88)(21.5) 1000

OK

(e) Investigate two-​way shear action at the critical section d/​2 from the face of the exterior column. At this location there is an unsymmetrical three-​sided perimeter, and there is a bending moment on the section across the 7 ft 4 in. width at the column face. ACI-​8.4.4.2 applies. This shear transfer check is an empirical check of stresses under factored load similar to that for a flat plate at an exterior column. The floor slab treatment was discussed in Section 16.18, where the details of the computation are shown. At the face of the exterior column, the factored moment* is

Mu = 67 ft-kips ( at face of column)

ACI-​8.4.4.2.2 and ACI-​8.4.2.3.2 require that the fraction of slab (footing) moment to be transferred to the column by eccentricity of shear be (see Section 16.18):

γv = 1− γ f = 1−

1 2 b1 1+ 3 b2

= 1−

1 2 12 + 21.5 / 2 1+ 3 12 + 21.5

= 0.355

Thus,

Muv = 0.355 Mu = 0.355(67) = 23.8 ft-kips (Continued)

*  This longitudinal moment is taken from the factored moment diagram of Fig. 19.13.5(b). Where the strap joins the exterior footing there is a sudden discontinuity, which may cause an indeterminate redistribution of longitudinal and transverse moments.

83

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Example 19.13.2 (Continued) Using the symbols of Fig. 16.18.1, for this example b1 = 22.75 in. and b2 = 33.5 in. The factored shear that must be carried is

Vu =

197  1  4.5(7.33) − 22.75(33.5)  = 165 kips 144  4.5(7.33) 

The section properties for the three-​sided critical section for shear transfer, as shown in Fig. 19.13.6(b), are A = 21.5 [2(22.75) + 33.5] = 1699 sq in. x2 =

2(22.75)(11.375) = 6.55 in. 2(22.75) + 33.5

 22.75(21.5)3  2(22.75)3 J c = 21.5  − (45.5 + 33.5)(6.55)2  + 6 3   = 133, 600 in.4

Referring to Fig. 19.13.6, the forces to be considered in analysis of the floor slab and the footing are compared. Note that the direction of the shear in the floor slab is the opposite of that in the strap footing; however, the moment direction is the same. Thus, in the floor slab the maximum factored shear stress occurs at the interior side of the critical section [as shown in Fig. 16.18.1(b)]. In the footing slab, however, the maximum factored shear stress will occur at the exterior edge. Using Eq. (16.18.3), the factored shear stress at the exterior side of column is (note that at this section the stresses due to Vu and Muv are additive and will be maximum in the footing slab)

vu1 =

165, 000 23.8(12,000) (22.75 − 6.55) + = 97 + 35 = 132 psi 1699 133, 600

Similarly, using Eq. (16.18.4), the factored shear stress at the interior side of column (here the shear stresses due to Vu and Muv subtract from each other),

vu2 =

165, 000 23.8 (12,000) (6.55) − = 97 − 14 = 83 psi 1699 133, 600

For β not exceeding 2.0, and b0 /​d = [2(22.75) + 33.5]/​21.5 = 79/​21.5 = 3.7, which is less than 15 for a three-​sided critical section, the stress due to factored loads is limited to φ (4 λ fc′) as a maximum; for this example, 164 psi (ACI-​22.6.5.2). The computed maximum stress of 132 psi is below 164 psi, and therefore acceptable. (f) Investigate development of reinforcement at the inflection point (ACI-​9.7.3.8.3). The Code provision is checked here, even though the reinforcement is negative moment reinforcement rather than the positive moment reinforcement as prescribed in ACI-​9.7.3.8.3 because the footing may be visualized as an inverted beam, as discussed in Example 19.13.1, part (d). Tension reinforcement is 3–​#8 at the narrow end of the strap near the inflection point. C = 0.85 fc′ ba = 0.85(3)(14)a = 35.7a T = 3(0.79)40 = 95 kips a=

95 = 2.66 35.7

M n = 95(21.5 − 1.33)

Vu = 67 kips

1 12

= 160 ft-kips

(Continued)

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839

19.13  DESIGN OF COMBINED FOOTINGS

Example 19.13.2 (Continued)

d/2 = 10.75” (Example 19.13.2)

12 × 12 Column Example 19.13.2

Critical section for two-way shear action

d/2

d/2 Footing V

Floor slab V

M

M Column

(a) Floor slab

(b) Cantilever (strap) footing

Figure 19.13.6  Comparison of forces to be transferred between floor slab and column to those between exterior footing and column in cantilever (strap) footing.

The #8 bars are proposed to be terminated at the interior side of the interior column, giving about 2.5 ft of embedment from the inflection point. Thus, La = 2.5 ft, but not to exceed the larger of d = 1.79 ft and 12db = 1.0 ft Thus,

Mn  160  + La =  + 1.79 12 = 50 in. > [Ld =38.9 in.]   Vu 67

OK

If this does not work, the bars can be extended beyond the interior face of the interior column, where they can be hooked down. (g) Investigate development of reinforcement at a simple support (ACI-​9.7.3.8.3): that is, at the top of the exterior column footing. Since it was determined in part (c) that the #8 bars must be hooked at face of the exterior footing, ACI-​9.7.3.8.3 need not be satisfied. Recall the preceding assumption of zero earth pressure under the strap. Hence the region below the strap should be disturbed during construction. The strap should be formed on the bottom. Furthermore, liberal anchorage lengths (perhaps even hooks) should be provided into the exterior footing [as shown in Fig. 19.13.5(a)] to accommodate fully the tensile force in the top of the footing across to the exterior column. Often some of the steel extending into the exterior footing is flared to distribute more effectively the load from the 27-​in. width to the 88-​in. width. (h) Design of exterior footing. In the transverse direction, d = 25 − 3 (i.e.,cover) − 0.5 (i.e., est. bar radius) = 21.5 in. 2  197  (2.54) = 87 ft-kips Mu att edge of strap =    7.33  2



required Rn =

Mu 87(12,000) = = 46 psi 2 0.90(54)(21.5)2 φ bd



(Continued)

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Example 19.13.2 (Continued) From Eq. (3.8.5), required ρ = 0.0012 < [min ρg = 0.002 (ACI-​8.6.1.1)],

required As = 0.002(54)(25) = 2.70 sq in.

Use 7–​#6 bars (As = 3.08 sq in.). Since bars are to extend the full 7 ft 4 in. length (less minimum cover) of the exterior footing, development of reinforcement for #6 bars is satisfied; that is, available embedment = 30.5 in. > [Ld (#6) = 21.9 in. ( 4 6 of value shown in Table 6.6.1)]  OK Also, provide minimum shrinkage and temperature reinforcement; use 7–​#6 bars in the perpendicular direction. The beam shear (one-​way action) at d from the face of the column in the 7.33-​ft direction of the footing is 197 Vu = (1.38) = 37 kips 7.33

φVc = φ (2 λ fc′ bw d )

1 φVc = 0.75 2(1.0) 3000  (54)21.5 = 95 kips > 37 kips 1000

OK

(i) Design of interior footing. net soil pressure under overload =

285 = 6.4 ksf (6.67)2

(2.75)2 = 161 ft-kips 2 Mu 161(12, 000) required Rn = = 58 psi = φ bd 2 0.90(80)(21.5)2 Mu = 6.4(6.67)





From Eq. (3.8.5), required ρ = 0.0015 < [min ρg = 0.002 (ACI-​8.6.1.1)] required As = 0.002(80)(25) = 4.0 sq in. Use 7–​#7 bars each way (As = 4.20 sq in.). The 2 ft 9 in. from face of column to edge of footing provides adequate embedment to develop the #7 bars (Ld = 31.9 in.). Check two-​way action for shear. When β < 2 and b0 /​d < 20 for a four-​sided critical section, the nominal strength Vc is based on 4 λ fc′. Thus, using λ = 1.0 for normal weight concrete,

φVc = φ 4 fc′ b0 d = 0.75(4 3000 )(4)(35.5)21.5

1 1000

Vu = 6.4 [(6.67)2 − (2.96)2 ] = 229 kips < 502 kips

= 502 kips

OK

(j) Design sketch. The final details of the design are shown in Fig. 19.13.5(a). Note that the flexural design strength, φMn diagram, in Fig. 19.13.5(b) for the strap portion is approximate; the #8 bars are extended as far as possible toward the narrow end as the width narrows from 27 to 14 in.

841



SELECTED REFERENCES

841

19.14 PILE FOOTINGS Main provisions for the design of pile footings are given in ACI-​13.4.2. The principles and methods to be used in the design of pile caps are little different from those of spread footings. The following, however, may be noted. 1. Computations of factored moments and shears may be based on the assumption that the reaction from any pile is concentrated at the center of the pile (ACI-​13.4.2.2). 2. In computing the external shear on any section through a footing supported on piles, the portion of the pile reaction to be assumed as producing shear on the section shall be based on straight-​line interpolation between full value when the pile center is at half the pile diameter (i.e., dp /​2) outside the section and zero value when the pile center is at half the pile diameter (i.e., dp /​2) inside the section (ACI-​13.4.2.5). 3. In reinforced concrete pile footings, the overall depth of the pile cap above the bottom reinforcement shall not be less than 12 in. (ACI-​13.4.2.1). Pile caps frequently must be designed for shear considering the member as a deep beam. In other words, when piles are located inside the critical sections d (for one-​way action) or, d/​2 (for two-​way action) from the face of column, the shear cannot be neglected (ACI-​ 13.4.2.3). The problem requires the most attention when large loads are carried by a few piles, such as a two-​pile cap having 100-​ton piles. For guidance, the reader is referred to the CRSI handbook [2.21], Rice and Hoffman [9.2, 19.18], and Gogate and Sabnis [19.19]. Alternatively, ACI-​13.2.6.3 allows the design of foundations using strut-​and-​tie models (see Chapter  14). This approach may be used, in particular, for the design of pile caps. Guidance for the design of pile caps using strut-​and-​tie models is provided by Adebar, Kuchma, and Collins [14.15].

SELECTED REFERENCES   19.1. A. N. Talbot. Reinforced Concrete Wall Footings and Column Footings. Urbana, IL: Engineering. Experiment Station Bulletin No. 67, University of Illinois, March 1913.   19.2. Frank E. Richart. “Reinforced Concrete Wall and Column Footings,” ACI Journal, Proceedings, 45, October 1948, 97–​127; November 1948, 237–​260.   19.3. Johannes Moe. Shearing Strength of Reinforced Concrete Slabs and Footings Under Concentrated Loads (Bulletin No. D47). Chicago: Portland Cement Association Research and Development Laboratories, April 1961.   19.4. ACI-​ASCE Committee 426. Suggested Revisions to Shear Provisions for Building Codes. Detroit: American Concrete Institute, 1979, 82 pp.  19.5. Hans Gesund. “Flexural Limit Analysis of Concentrically Loaded Column Footings,” ACI Journal, Proceedings, 80, May–​June 1983, 223–​228.   19.6. Hans Gesund. “Flexural Limit Design of Column Footings,” Journal of Structural Engineering, ASCE, 111, 11 (November 1985), 2273–​2287.   19.7. Da Hua Jiang. “Flexural Strength of Square Spread Footings,” Journal of Structural Engineering, ASCE, 109, 8 (August 1983), 1812–​1819. Disc., 112, 8 (August 1986), 1925–​1926.   19.8. Kanakapura S.  Subba Rao and Shashikant Singh. “Lower-​Bound Collapse Load of Square Footings,” Journal of Structural Engineering, ASCE, 113, 8 (August 1987), 1875–​1979.   19.9. Richard W. Furlong. “Design Aids for Square Footings,” ACI Journal, Proceedings, 62, March 1965, 363–​371. 19.10. Fritz Kramrisch. “Footings,” Handbook of Concrete Engineering (Mark Fintel, Editor). New York: Van Nostrand Reinhold Company, 1974 (Chapter 5, pp. 111–​140). 19.11. ACI Committee 336. “Suggested Design Procedures for Combined Footings and Mats,” ACI Structural Journal, 85, May–​June 1988, 304–​324. Disc., 86, January–​February 1989, 111–​116. 19.12. ACI Committee 336. Design and Performance of Mat Foundations—​State-​of-​the-​Art (SP-​ 152). Farmington Hills, MI: American Concrete Institute, 1995, 274 pp. 19.13. Fritz Kramrisch and Paul Rogers. “Simplified Design of Combined Footings,” Journal of Soil Mechanics and Foundations Division, ASCE, 87, SM5 (October 1961), 19–​44.

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19.14. Leslie J. Szava-​Kovats. “Design of Combined Footings Using Support Reaction and Moment Influence Lines of Continuous Beam on Elastic Supports,” ACI Journal, Proceedings, 64, June 1967, 312–​319. 19.15. Gwynne Davies and Brian Mayfield. “Choosing Plan Dimensions for an Eccentrically Loaded Footing Slab,” ACI Journal, Proceedings, 69, May 1972, 285–​290. 19.16. Joseph E. Bowles. “Mat Design,” ACI Journal, Proceedings, 83, November–​December 1986, 1010–​1017. Disc., 84, September–​October 1987, 449. 19.17. Report of the Joint Committee of Standard Specifications for Concrete and Reinforced Concrete. Detroit: American Concrete Institute, 1940. 19.18. Paul F. Rice and Edward S. Hoffman. “Pile Caps—​Theory, Code, and Practice Gaps,” CRSI Professional Members’ Structural Bulletin No. 2, February 1978. 19.19. Anand B.Gogate and Gajanan M.  Sabnis. “Design of Thick Pile Caps,” ACI Journal, Proceedings, 77, January–​February 1980, 18–​22. 19.20. ACI Committee 336. Suggested Analysis and Design Procedures for Combined Footings and Mats. ACI 336.2R-​88 (Reapproved 2002). Farmington Hills, MI: American Concrete Institute, 2002, 20 pp.

PROBLEMS All problems are to be done in accordance with the ACI Code, and all loads given are service loads unless otherwise indicated. 19.1 Design a square spread footing to support a 14-​in. square tied column carrying a dead load of 120 kips and a live load of 90 kips. The column reinforcement consists of #8 bars. Use fc′ = 4000 psi for the column, fc′ = 3000 psi for the footing, and fy = 60,000 psi. Use a 6 ft 9 in. square footing. 19.2 Design a square spread footing to support a 20-​in. square tied column carrying a dead load of 400 kips and a live load of 264 kips. The column reinforcement consists of #11 bars. Use fc′ = 3000 psi for both the column and footing, fy  =  60,000 psi, and allowable soil pressure  = 5 ksf. Include a design sketch.

19.3 Investigate the transfer of load from column to footing for the two conditions of the figure for Problem 19.3. The column is 20 in. square ( fc′ = 4000 psi) containing 12–​#10 bars ( fy = 60,000 psi) spirally reinforced. The footing is 10 ft square and 28 in. thick ( fc′ = 3000 psi) and is adequately reinforced. Use the provisions of ACI-​16.3 and ACI-​14.5. 19.4 Design a spread footing to carry a load from an 18 × 32 in. tied column. Dead load and live load are each 230 kips. Because of the closeness of the property line (see the figure for Problem 19.4), the footing cannot exceed 8 ft

20”

20” 580k (60% live load)

580k

2

1

1

1

10’– 0”

10’– 0”

(a) Without pedestal

(b) With pedestal

Problem 19.3 Property line 4’– 0” 8’– 0” Maximum footing width

Problem 19.4

2

30” × 30” × 12” pedestal (fc’ = 3,750 psi)

18” 32”

843



843

PROBLEMS

perpendicular to that line. Use fc′ = 3000 psi , fy = 60,000 psi, and allowable soil pressure = 5 ksf. 19.5 Design a plain concrete footing to carry a long 12-in. concrete block wall that must transmit 6 kips/​ft (60% live load) to the footing. Use fc′ = 3000 psi and allowable soil pressure = 4 ksf. 19.6 Design a reinforced concrete footing to carry a 12-​in. concrete wall to carry 20 kips/​ft (60% live load). Use fc′ = 3000 psi , allowable soil pressure = 3 ksf, and fy = 60,000 psi. 19.7 Design a rectangular combined footing for the situation shown in the figure for Problem 19.7. Column data are in the following table. Column

Size

Reinforcement

Exterior 12 in. square Interior 12 in. square

#9 bars #9 bars

LL



Equal settlement is taken for DL plus LL conditions at a uniform soil pressure of 3 ksf. Use fc′ = 4000 psi and fy = 60,000 psi. 19.8 Design a rectangular combined footing for the conditions of the figure for Problem 19.8. Assume that the system is within a building basement and is to have a 4-​in. concrete slab over the footing. Use fc′ = 3000 psi , fy = 60,000 psi, and allowable soil pressure for the given loads = 5 ksf. 19.9 Rework Example  19.13.2 using fc′ = 5000 psi and fy = 80,000 psi.

DL

45 kips 120 kips 90 kips 155 kips

Property line 18’– 0”

1’– 4”

12” × 12” column

12” × 12” column

Problem 19.7 Property line 18” × 18” column P = 200k (50% live load)

9”

Problem 19.8

16’– 0”

20” diameter column P = 450k (40% live load)

CHAPTER 20 INTRODUCTION TO PRESTRESSED CONCRETE

20.1 INTRODUCTION In very simple terms, “prestress” means a stress that acts even though no externally applied loads are acting. The principle of prestressing has been used for centuries. For example, wooden barrels may be made by tightening metal bands or ropes around barrel staves. The tensile stress in the bands causes a compression between the staves, thus making the barrel tight. In the making of early automobile wheels, the wooden spokes and rim were first held together by a hot metal tire. Upon cooling, the tensile stress due to shrinkage in the metal would then compress the wooden rim and spokes together. In bolted joints, the bolt is pretensioned by tightening, which in turn precompresses the elements being joined. The primary application of prestressing on a large scale today is in concrete construction. In general, prestress involves the imposition of stresses opposite in sign to those caused by the subsequent application of service loads. For example, prestressing wires placed eccentrically in a simple beam, as shown in Fig. 20.1.1, produce in the concrete an axial compression as well as a negative bending moment. Thus it is possible to keep the entire section in compression when service loads are added. This is a great advantage, since concrete is weak in tension but strong in compression. Of course, steel is used to impose the prestress, though less is required for prestressed concrete than in ordinary reinforced concrete. In general, prestress provides a means for the most efficient use of material—​that is, steel in tension and concrete in compression.

20.2 HISTORICAL BACKGROUND The general concepts of prestressed concrete were first formulated in the period 1885–​1890 by C. F. W. Doehring in Germany and P. H. Jackson in the United States. These early applications were handicapped by the low steel strengths obtainable at the time. Steel stressed to low tension levels will not precompress concrete adequately to maintain its compression after shrinkage and creep have occurred. The theory of prestressed concrete was first propounded by J.  Mandl of Germany in 1896. It was further advanced by M. Koenen of Germany in 1907 (the first recognition of losses in prestress force from elastic shortening of concrete) and by G. R. Steiner in the United States in 1908. Steiner recognized prestress losses due to shrinkage and suggested retensioning after shrinkage had occurred.

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845

Prestressed concrete T-​girders erected on precast columns for a parking garage. (Photo by C. G. Salmon.)

In practical uses, R. E. Dill of the United States in 1928 produced prestressed planks and fence posts. Circular prestressing of storage tanks began about 1935, but no significant linear prestressing (beams, slabs, planks, etc.) was done until about 1950. The Walnut Lane Bridge in Philadelphia, built in 1949–​1950, was the first major use of linear prestressing in the United States. In Europe, however, linear prestressing began about 1928 and advanced rapidly with the work of F. Dischinger, E. Freyssinet, E. Hoyer, G. Magnel, Y. Guyon, P. Abeles, and F. Leonhardt. With the publication of Magnel’s work on the loss of stress in work-​hardened steels in 1944, the basic theory of prestressed concrete was sufficiently complete for successful economical applications. In the United States, T. Y. Lin was a leading proponent and practitioner. The use of prestress is now widespread in nearly every type of simple structural element, as well as in many statically indeterminate structures. Zollman [20.1] has presented his interesting “Reflections” on the beginnings of prestressed concrete in America. Methods of inducing prestress are ingenious and unique. Many textbooks are available that describe the methods and applicable theories [20.2–​20.9].

20.3 ADVANTAGES AND DISADVANTAGES OF PRESTRESSED CONCRETE CONSTRUCTION The original concept of prestressed concrete was that the material was crack free under service loads. Especially when a structure is exposed to the weather, elimination of cracks prevents corrosion. Also a crack-​free prestressed member has greater stiffness under service loads because its entire section is effective. Prestressed concrete in several respects is more predictable than ordinary reinforced concrete. It permits accommodation of both shrinkage and creep reasonably well. Also, high-​ strength concrete may be more efficiently utilized by merely adjusting the prestress force. Precompression of the concrete reduces the tendency for inclined cracking, and the use of curved tendons provides a vertical component to aid in carrying shear. Shear strength exhibits less variability than in ordinary reinforced concrete.

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C hapter   2 0     I ntroduction to P restressed C oncrete

c Mb = bending moment from applied loads

Stress distribution t

(a) Effect of service loads on beam

P

e

c

P Prestressed wire

Prestressing force

(b) Effect of prestress with eccentricity

P

Mb P

e

c

(c) Combined effects of service load and prestress

Figure 20.1.1  Opposite effects of service load and prestress on a simply supported beam.

Other features of prestressed concrete are its high ability to absorb energy (impact resistance); its high fatigue resistance, due especially to the low steel stress variation resulting from the high initial pretension; and its high live load capacity, arising from the ability to design the prestressing tendon to support the dead load. As a result, higher span-​to-​depth ratios can be achieved with prestressed concrete elements. Some of the disadvantages of prestressed concrete construction are as follows: (1) the stronger materials used have a higher unit cost; (2) more complicated formwork may be necessary; (3)  end anchorages and bearing plates are often required; (4)  labor costs are greater; and (5) more conditions must be checked in design, and closer control of every phase of construction is required. Short-​span members and single-​unit applications of any kind are likely to be uneconomical in prestressed concrete. However, economy is usually achieved when units can be standardized and the same unit repeated many times. For many situations, the desirability of achieving a certain advantage is sufficient to justify a higher initial cost. Currently, prestressing need not create a crack-​free structure at service load. In fact, prestressing to various levels of stress can provide the whole range of results, from “fully” prestressed, when at service load no cracking occurs, to nonprestressed, as discussed in preceding chapters. The level of prestressing can be used to accomplish the desired crack control or stiffness objective. “Partial” prestressing has become common in construction. In general, this chapter focuses on prestressed concrete where the section is uncracked at service load. Mohan Rao and Dilger [20.17] have discussed flexural crack control in cracked prestressed members.

20.4 PRETENSIONED AND POST-​T ENSIONED BEAM BEHAVIOR Since discussion in this chapter is limited to beams, it is well to consider at this stage the behavior of such a member as it relates to the method of inducing the prestress. The most commonly used procedure is to put a specified tensile force into the wires by stretching them between two anchorages prior to placing the concrete into the forms. The concrete is then placed and allowed to cure. As the concrete hardens, the wires become bonded to

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847

the concrete throughout their length. After the concrete has reached a minimum prescribed strength, the wires are cut at the anchorages. The immediate shortening of the wires transfers through bond (i.e., interaction between the steel and the surrounding concrete) a compressive stress to the concrete. Such a process is called pretensioning. The behavior associated with pretensioning will be described step by step. In addition, the terminology and allowable values will be given in accordance with Chapters 20 and 24 of the ACI Code. In Step 1, as shown schematically in Fig. 20.4.1(a), the wires are stretched between two anchorages in the casting yard sufficiently to introduce a tensile stress fsi into the wires, which according to ACI-​20.3.2.5.1 may not exceed 94% of the specified yield strength fpy, but not greater than the lesser of (1) 80% of the specified tensile strength fpu, or (2) the maximum value recommended by the manufacturer of prestressing tendons or anchorages. The quality-​controlled concrete is then placed in the forms and frequently is steam cured. Concrete strength must be adequately developed by the time the compression is to be introduced; thus high early-strength cement is usually used. Generally, the concrete strength fci′ at transfer is specified by designers to be 4000 to 4500 psi. Step 2 is to cut the wires. Acting through bond, the force T0 in the wires acts as a compressive force on the entire effective (transformed) section. The stress in the concrete goes from zero [Fig. 20.4.1(b)] before the wires are cut to that shown in Fig. 20.4.1(c) after they have been cut. Once the prestress has been introduced, certain losses of prestress begin to occur. Loss of prestress may arise from slip at the anchorage, elastic shortening of the concrete member, creep and shrinkage of the concrete, relaxation of steel stress, and frictional losses due to intended or unintended curvature in the tendons. It is true that some small portion of such losses may occur prior to the transfer of stress to the concrete; however, it is practical and conservative to assume that the losses occur after the transfer. Dead load of the flexural member will, of course, be acting simultaneously with the prestressing force once the wires have been cut and transfer of load to the concrete has been accomplished. Dead load combined with prestress is shown in Fig. 20.4.1(d), where the most severe stress situation occurs immediately after transfer and before most losses have taken place. Limiting values for this temporary situation are (a)  a tensile stress of 3 fci′ (approximately 40% of the cracking strength), except at the ends of simply supported members, where the maximum allowable tensile stress is 6 fci′ (a conservative value for the modulus of rupture; see Section 1.8) (ACI-​24.5.3.2) and (b) a compressive stress equal to 60% of the concrete compressive strength fci′ that has been developed at the time of transfer, except for ends of simply supported beams, where a compressive stress equal to 0.70 fci′ is allowed (ACI-​24.5.3.1). One reason for limiting the temporary tensile stress to such a low value is to prevent any possibility that an upward buckling of the beam will result from sudden cracking at the top. Step 3 is the service load condition. Since 2002, the ACI Code has classified prestressed members based on the extreme fiber tensile stress ft at service loads as follows (ACI-​24.5.2.1): If ft ≤ 7.5 fc′

the section is assumed to behave as uncracked—​Class U

If ft > 12 fc′

the section is assumed to behave as cracked—​Class C

If 7.5 fc′ < ft ≤ 12 fc′ the section is assumed to be in transition between uncracked and cracked—​Class T except for prestressed two-​way slab systems, which are classified as Class U members with ft ≤ 6 fci′ . Because cracking of the cross section changes its behavior and properties under service loads, the code establishes different serviceability requirements depending on whether the section is uncracked or cracked. These requirements are summarized in ACI Table R24.5.2.1. For Class U and T members, the stresses at service loads are permitted to be calculated using uncracked section properties (ACI-​24.5.2.2). For Class C members, however, the stresses should be computed using a cracked transformed section (ACI-​24.5.2.3) (see Section 12.5).

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C hapter   2 0     I ntroduction to P restressed C oncrete

Anchorage

Stress fsi

(a) Wires stretched to a stress fsi; then concrete placed

To

To = Asfsi (b) Before cutting wires; no stress in concrete < ft

ft e

Before losses

To

After losses

To

< fc

fc (c) After cutting wires; concrete compressed (transformed area effective)

[

wD

ft ≤ fit = 3 fci‘

Before losses

To

To

]

[

fc ≤ fic = 0.60fci‘ (d) Dead load plus prestress acting

]

[

]

fc ≤ ffc = 0.45fci‘ (Class U and T)

wD + sust loads

T < To

After losses

T < To

[f = 7.5 ft

f ‘ ](Class U) [ f ‘ ]< f ≤[ f = 12 f ‘ ] (Class T) c

ft ≤ fft = 7.5

c

ft

c

t

(e) Dead load, prestress, and sustained live load after all prestress losses

[

]

fc ≤ ffc = 0.60fc‘ (Class U and T)

wD + L

T < To

After losses

T < To

[f = 7.5 ft

f ‘ ](Class U) [ f ‘ ]< f ≤[ f = 12 f ‘ ] (Class T) c

ft ≤ fft = 7.5

c

ft

c

t

(f) Dead load, prestress, and live load after all prestress losses

Figure 20.4.1  Stages of behavior up to service load—​pretensioned beam.

Two service load conditions are recognized (ACI-​24.5.4.1): (1) dead load + prestress (after losses) + sustained loads and (2)  dead load + prestress (after losses) + all loads. Although live loads are generally transient, it is possible for a portion of the live load to act long enough to cause significant long-​term deflections. In such a case, that portion of the live load should be considered to be sustained in design.

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849

Jacking device

Hollow ducts for tendons

Figure 20.4.2  Post-tensioned member.

Class U and T Members For the condition of prestress (after losses) + sustained loads, ACI-​24.5.4 permits a compressive stress (based on uncracked section properties) not to exceed 0.45 fc′ . For the condition of prestress (after losses) + all loads, the permitted compressive stress is 0.60 fc′ . According to the ACI Commentary R24.5.4.1, the compressive stress limit (i.e., 0.45 fc′ ) was “originally established to decrease probability of failure … due to repeated loads.” However, fatigue tests showed that crushing failure in concrete does not control the strength of prestressed concrete members; thus, the limit was raised to 0.60 fc′ for situations having large transient live loads.

Class C Members There are no stress limit requirements for this class of members. However, since Class C members are expected to be cracked, they must satisfy the crack control requirements of ACI-​24.3.2 (see Section 12.18). The alternative to pretensioning is post-​tensioning. In a post-​tensioned beam, the concrete is first cast either with a hollow tube enclosed or with unstressed tendons coated with grease or mastic to prevent bond with the concrete, as shown in Fig. 20.4.2. An end plate or anchorage is placed against each end of the member; then, once the concrete has hardened and reached the desired strength, the wires are pulled by jacking against the end plates. Tensile stress in the prestressing steel immediately after anchorage is not to exceed 70% of specified tensile strength fpu [ACI-​20.3.2.5.1]. During the tensioning process, elastic shortening occurs, frictional losses take place, and the dead load moment becomes partially active due to the induced curvature. When the prestressing wires or strands are tensioned in sequence, some loss of prestress will also occur in the reinforcement tensioned first as further shortening (due to axial shortening and curvature of the member) takes place during tensioning of additional tendons. These loses can often be accounted for when determining the jacking force. For post-​tensioning, the stresses induced and the allowable values at the different stages are essentially the same as those described in detail for pretensioning. However, since the tendons are not bonded to the concrete, bending of the member will affect the strains and stresses in the tendons along their entire length rather than locally, as is the case in pretensioned members. Prior to the imposition of live load, the tendons are usually grouted (i.e., the space in the ducts is filled). If such grouting is properly done, the strains and stresses in the tendons will vary locally with changes in curvature caused by the live load, which may be computed on the transformed section, the same as for pretensioning.

20.5 SERVICE LOAD STRESSES ON FLEXURAL MEMBERS—​T ENDONS HAVING VARYING AMOUNTS OF ECCENTRICITY To demonstrate some of the attributes of prestressed concrete, in Example 20.5.1 the prestressing elements are placed at the centroid of the section, giving a uniform precompression of the concrete.

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EXAMPLE 20.5.1 For the section shown in Fig.  20.5.1 assume that the member is pretensioned by 2.30 sq in. of steel wire having an initial tensile stress of 175,000 psi. The prestress wires are centered at the centroid of the section. The concrete has fc′ = 5000 psi (modular ratio n = 7), and it is to be assumed that the concrete has attained a strength of fci′ = 4000 psi at the time of transfer. Assume that the section will behave as an uncracked section. 500 + fLL

1171 psi

1604 psi

(–) (–) 30”

(–)

(–) Aps = 2.30 sq in. 15”

(+)

(+)

20”

171 psi 671 psi

(a)

(b) Initial prestress

530 psi

500 + fLL (c) DL + LL stresses

(d) Temporary stress after transfer

(e) Final stress at full service load

Figure 20.5.1  Section and stresses for Example 20.5.1.

Determine (a) the stresses due to prestress immediately after transfer, (b) the temporary stresses when the member is used on a 40-​ft simple span, and (c) the service live load moment capacity according to the ACI Code, allowing for a 20% loss of prestress due to creep, shrinkage, and other sources. SOLUTION (a) Stress due to prestress immediately after transfer.

T0 = fsi Aps = 175(2.30) = 402.5 kips

This force T0 acts as a compressive force on the transformed section immediately after the cutting of the wires. The stress in the concrete immediately after transfer can be determined as

fc =

T0 T0 402, 500 402, 500 = = = = 656 psi Ac + nAps Ag + (n − 1) Aps 600 + 6(2.3) 614

The decrease in steel stress due to elastic shortening of the member is therefore (see also Section 20.7)

∆fs = nfc = 7(656) = 4590 psi



Thus elastic shortening may be considered to have caused a loss of tensile stress, so that the remaining tensile stress in the wires is 175,000 –​4590 = 170,400 psi. The loss in this case is 2.6%, but it could be as high as 5%. (Continued)

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Example 20.5.1 (Continued) Although it may be theoretically correct to use the transformed section, in ordinary practice it is common and sufficiently accurate in most cases to use the gross section. Since the prestressing force is applied at the centroid of the gross section in the present problem, fc is uniform over the entire section, or fc =



402, 500 = 671 psi 600

which is little different from the value of 656 psi determined above by using the transformed section. (b) Temporary stress-​prestress plus dead load. wD =

20(30) 144

(150) = 625 plf

1 M D = (0.625)(40)2 = 125 ft-kips 8

Using the approximate method with gross moment of inertia Ig and neglecting the transformed area of reinforcement, 1 (20)(30)3 = 45, 000 in.4 12 402, 500 125(12,000) (15) f (initial prestress + DL) = −  600 45, 000 Ig =

= −671  500



= −1171 psi (compressiion, top)



= −171 psi (compression, bottom) These stresses are acceptable based on temporary stress restrictions immediately after transfer and before losses, fc (max) = 0.60 fci′ = 2400 psi

ft (max) = 3 fci′ = 190 psi



(c) Service live load moment capacity. f (prestress − losses + DL) = − 0.8(671)  500 = −537  500

= −1037 psi (compression, top) = −37 psi

(compression, bottom)

Based on stress at service load (prestress + DL + LL) [ACI-​24.5.4.1 and ACI-​24.5.2.1] for uncracked Class U members, after allowance for all prestress losses,

fc (max) = 0.60 fc′ = 3000 psi ft (max) = 7.5 fc′ = 530 psi (Continued)

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Example 20.5.1 (Continued) The stress available for live load may then be computed. f (prestress − losses + DL + LL) = −1037 + fLL = −3000 psi (top) = −37 + fLL = +530 psi

(bottom)

fLL (max,top) = −1963 psi  



fLL (max,bottom)) = +567 psi

Controls!

Thus M L (15) = 567 psi 45, 000



ML =

45, 000(567) 1 = 142 ft-kips 12,000 15



The wide divergence between the maximum acceptable live load stresses of 1963 psi compression and 567 psi tension indicates the need for an unsymmetrical section or some arrangement to equalize the stresses better. A study of Fig. 20.5.1 will show that the most economical arrangement would be for the initial prestress variation to offset the pattern of final stress under full dead load plus live load.

EXAMPLE 20.5.2 Repeat the solution of Example 20.5.1, except locate the tendons 5 in. from the bottom of the section (Fig. 20.5.2). 500 + fLL

671

(+) 30”

Aps = 2.30

(–)

(–) 2013

20” (a)

(–) (–)

(+)

5”

(b) Initial prestress

1603 psi

+171

(+)

(+) 530 psi

1513

500 + fLL (c) DL + LL stresses

(d) Temporary stress after transfer

(e) Final stress at full service load

Figure 20.5.2  Section and stresses for Example 20.5.2.

SOLUTION (a) Temporary stress (prestress + DL immediately after transfer). Using properties of the gross section as for Example 20.5.1, the prestressing force T0 = 402.5 kips applied with an eccentricity of 10 in. gives f (initial prestress + DL) = − 671 ±

402, 500(10)(15)  500 45, 000

= −671 ± 1342  500 = +171 psi = − 1513 psi

(tension, top)



(compression, bottom) (Continued)

853



20.6  THREE BASIC CONCEPTS OF PRESTRESSED CONCRETE

853

Example 20.5.2 (Continued) Note that the temporary tensile stress of 171 psi at the top is nearly equal to the allowable value of 3 fci′ = 190 psi for such stress. Any significantly greater eccentricity, therefore, would require a reduction in the prestressing force. (b) Final stress (prestress + DL), after allowance for 20% prestress loss, is f (prestress + DL − losses) = +171 − 0.2( −671 + 1342) = +37 psi (top)



= −15513 – 0.2(–671 – 1342) = −1110 psi



(bottom)

(c) Service live load moment capacity (see Example  20.5.1 for allowable stresses). Based on final dead load plus live load conditions,

+37 + fLL = −3000 psi (top), −1110 + fLL = +530 psi

fLL = −3037 psi

(bottom), fLL = +1640 psi



Since the neutral axis for live load resistance is assumed to be at middepth, fLL = +1640 psi controls.

ML =

45, 000(1640) 1 = 410 ft-kips 12,000 15

Thus, increasing the eccentricity of the prestressing force increases the live load capacity until the limit is reached—​that is, when the temporary stress at transfer reaches its maximum permissible value, either at the top or at the bottom of the section. It is to be noted that the magnitude of the prestress over the concrete section is constant for the entire span when the tendons are straight, whereas the magnitude of dead and live load stresses is a maximum at only one point. For straight tendons, the complete stress situation near the supports on simple spans approaches that of Fig. 20.5.2(b), less losses, because the superimposed dead and live load stresses vanish. Because of this, tendons are frequently placed to have an eccentricity that varies from zero at points of low external bending moment to a maximum in the region of high external bending moment. This variation in eccentricity may be accomplished in pretensioning by holding down the stressed tendons at midspan, or at other locations such as at the one-​third points. In post-​tensioning, the ducts or greased tendons are simply draped (held at the ends and permitted to take a natural deflected shape) such that desired eccentricities at the ends and at midspan are achieved; the points in between will lie on a curved path.

20.6 THREE BASIC CONCEPTS OF PRESTRESSED CONCRETE When considering the stresses in prestressed concrete under service load conditions, three general patterns of thought may be applied.

Homogeneous Beam Concept Section 20.5 used the homogeneous beam concept, wherein the prestressing effectively eliminates cracking and the combined stress formula, P/​A ± Mc/​I, may be used to investigate the section. Two examples appear in Section 20.5.

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Internal Force Concept The internal force approach uses the equilibrium of internal forces; steel takes the tension and concrete takes the compression, as shown in Fig.  20.6.1. This approach is analogous to the internal-​couple method used for nonprestressed reinforced concrete. At service load (assuming linear elastic behavior) in reinforced concrete, the points of action of the forces C and T (where C = T) are independent of the magnitude of applied bending moment, depending only on the cross-​sectional dimensions and the modulus of elasticity ratio n; thus the magnitude of the forces is directly proportional to the applied bending moment. In prestressed concrete under service load, the magnitude of internal forces is basically independent of applied bending moment, depending only on the prestress and the percentage of losses. In this case the location of the force C must vary with the applied loading. The approach may be summarized by the following steps: 1. A known prestress force put into the steel defines T. 2. A moment M is applied on the beam. 3. For equilibrium, the moment arm = M/​T and C = T. 4. Knowing the magnitude and point of action of the force C, the stress in the concrete may be computed as f =



C Cec y ± A I

(20.6.1)

C.G. of concrete section ec

C Arm T

Figure 20.6.1  Internal force concept of prestressing.

EXAMPLE 20.6.1 Apply the dead load moment of 125 ft-​kips and the live load moment of 410 ft-​kips (as computed in Example 20.5.2) to the rectangular beam of Fig. 20.5.2, using the internal force concept. Determine the service load stresses at transfer and under final conditions. SOLUTION (a) At transfer, the prestress force T0 is 402.5 kips.

C = T0 = 402.5 kips



When the applied moment is 125 ft-​kips, the moment arm of the internal forces must be

arm =

M D 125(12) = = 3.73 in. C 402.5 (Continued)

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20.6  THREE BASIC CONCEPTS OF PRESTRESSED CONCRETE

Example 20.6.1 (Continued) This means the compressive force C is eccentric to the middepth by an amount, ec = 15 − 5 (to steel) − 3.73 (arm) = 6.27 in. (below middepth) f =−



402.5 402.5(6.27)15 ± 600 45, 000



f (top) = − 671 + 842 = +171 psi (tension) f (bottom) = − 671 − 842 = −1513 psi (compression) exactly the same as in Example 20.5.2, Fig. 20.5.2(d). (b) At the final condition, the prestress force Te is 0.8(402.5) = 322 kips after losses. C = Te = 322 kips M D + M L (125 + 410)12 = = 19.93 in. C 322 ec = 15 − 5 − 19.93 = −9.93 in. (above middepth)

arm =



f =−

322 322(9.93)15  600 45, 000



f (top) = − 537 − 1065 = −1602 psi (compression) f (bottom ) = − 537 + 1065 = +528 psi (tension) exactly as shown in Fig. 20.5.2(e).

Load Balancing Concept The load balancing approach visualizes prestressing primarily as a process of balancing loads on the member. The prestressing tendons are placed so that the eccentricity of the prestressing force varies in the same manner as the moments from applied loads, which if exactly done would result in zero flexural stress. Only the axial stress P/​A (P is the horizontal component of force in tendon) would act. Refer to Fig. 20.6.2(a), showing a parabolically draped prestressing tendon. Figure 20.6.2(b) shows the free body of forces acting on the concrete due to prestress alone. The prestressing effect may be considered as an upward uniform load if the tendon is parabolically draped. The maximum prestress moment of T emax at midspan can be equated to an equivalent uniformly loaded beam moment, wp L2/​8; thus

wp =

8Temax = equivalent uniform load (acting upward) L2

(20.6.2)

Let

wnet = w(actual downward load ) − w p

then

wnet L2 8

(20.6.3)

C M net y  A I

(20.6.4)

M net =

and

f =−

If the tendons are not parabolically draped, the actual net moment (applied load moment minus prestress moment) may be used for Mnet in Eq. (20.6.4).

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Parabolic tendon

T

T

emax

L/2

(a) Member having parabolically draped tendon

C

C

wp (b) Forces acting on concrete from prestress alone

Figure 20.6.2  Load balancing concept of prestressing.

EXAMPLE 20.6.2 Compute the stresses on the beam of Fig. 20.5.2 using dead load and live load moments of 125 and 410 ft-​kips, respectively. Use the load balancing concept. SOLUTION The maximum moment due to prestress at transfer is  10  M prestress = 402.5   = 335 ft-kips  12 





M net = M D − M prestress = +125 − 335 = −210 ft-kips (negative bending) f =−

402.5 ( −210)(12)15  600 45, 000

f (top) = −671 + 842 = +171 psi

(ttension)

f (bottom) = −671 − 842 = −1513 psi (compression)



At the final condition, M net = 125 + 410 − 322(10) /12 = 267 ft-kips (positive bending) f =−

322 267(12)(15)  600 45, 000

f (top) = −537 − 1068 = −1605 psi (comppression) f (bottom) = −537 + 1068 = +531 psi



(tension)

The results agree with those previously obtained.

20.7 LOSS OF PRESTRESS The amount of prestress actually existing in a prestressed concrete member is not easily measured. The total force in the tendons at the time of prestressing is all that may conveniently be determined. Various losses reduce the prestress force during and after initial prestressing of the tendons. The difference between the final available prestress and the initial value is referred to as the loss of prestress. In practice the initial prestress is usually determined by a pressure gauge on the jack and may be verified by a direct measurement of the tendon elongation. In pretensioned

857



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20.7  LOSS OF PRESTRESS

members, the uniformity of initial prestress may be verified at several points. In certain post-​tensioning procedures, the prestressing force will diminish owing to friction at points remote from the jacking source, and even to friction in the anchorage device.

Elastic Shortening The loss of prestress due to elastic shortening can be easily determined. For example, let T0 be the prestressing force that is applied at the centroid of the concrete section in a pretensioned member. If Tf is the final tensile force in the tendons just after elastic shortening has occurred, the strain (unit shortening) in the concrete may be expressed as

εc =



Tf fc = Ec Ac Ec

(20.7.1)

where Ac = Ag –​ Aps. The change in strain in the tendons as a result of losses is ∆ε s =



T0 − T f

(20.7.2)

Aps Es

Equating the expressions for εc and Δεs gives T0 Ac + nAps AT = = Tf Ac Ac



(20.7.3)

The loss of prestress Δ fs is

∆fs =

T0 − T f Aps

=

nT f Ac

=

nT0 AT

(20.7.4)

or ∆fs ≈



nT0 Ag

(20.7.5)

For beams with eccentric tendons, the loss in prestress due to elastic shortening and bending of the section is obtained as n times the computed compressive stress in the concrete adjacent to the tendons. In the post-​tensioning case, the tendons are not stretched simultaneously. Thus, tendons tensioned first will undergone losses due to elastic shortening that takes place as other tendons are stressed. The various methods of accounting for these gradual losses are adequately described elsewhere [20.2–​20.8].

EXAMPLE 20.7.1 Determine the percent loss of prestress due to elastic shortening and bending in the pretensioned member of Fig. 20.5.2. Use fc′ = 5000 psi with n = 7 and fsi = 175,000 psi. SOLUTION (a) Prestress loss due to elastic shortening, neglecting bending. In the “exact” method [see also part (a) of Example 20.5.1], AT = 20(30) + (7 − 1)2.30 = 614 sq in.

∆fs = percent loss =

nT0 7(402.5) = = 4.59 kssi 614 AT 4.59 = 2.62% 175



(Continued)

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Example 20.7.1 (Continued) In the approximate method, ∆fs =

nT0 7(402.5) = = 4.70 ksi Ag 600

4.70 percent loss = = 2.69% 175



There is no significant difference in the two results. Three percent is a typical value for loss due to elastic shortening in a pretensioned beam, whereas such loss in a post-​ tensioned beam would be on the order of 1.5%, average. (b) Prestress loss, including bending due to eccentricity of prestressing force and dead load. By the approximate method (using gross section instead of transformed section), the stress in the concrete adjacent to the tendons is (using the internal force concept), arm =

M D 125(12) = = 3.73 in. T0 402.5

C = T0 = 402.5 kips



fc = − =−

C C (15 − 5 − 3.73)10 − Ag Ig



402.5 402.5(6.27)10 − 600 45, 000

= −671 − 561 = −1232 psi This is the stress in the concrete 5 in. from the bottom of the beam. The change in steel stress is nfc. ∆fs = nfc = 7(1.232) = 8.6 ksi

percent loss =

8.6 = 4.9% 175



Loss Due to Creep in Concrete Creep is the time-​dependent deformation that occurs in concrete under stress, and has already been discussed (see Sections 1.11 and 12.11). The strain due to creep will vary with the magnitude of stress and in general may be assumed to vary with the elastic strain from about 100% in humid atmosphere to about 300% in very dry atmosphere. In Chapter 12, Eq. (12.11.1), the creep coefficient Ct is defined as

Ct =

creep strain, ε cp initial elastic strain, ε i

(20.7.6)

The elastic strain in the concrete at the centroid of the section is (  fc = stress at centroid)

εi =

fc Ec

 f  ε cp = Ct ε i = Ct  c   Ec 

(20.7.7)

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20.7  LOSS OF PRESTRESS

The strain in the concrete due to creep equals the decrease in strain in the steel; thus



 f  ∆ε s = ε cp = Ct  c   Ec 

(20.7.8)

also

∆fs Es

(20.7.9)

∆fs = Ct nfc

(20.7.10)

∆ε s =

Then, equating Eqs. (20.7.8) and (20.7.9),

The coefficient Ct may be determined using the general expressions given earlier (see Section 12.11), or as suggested by Zia, Preston, Scott, and Workman [20.10], using Ct = 2.0 for pretensioned members and Ct = 1.6 for post-​tensioned members. The stress fc in Eq. (20.7.10) is recommended [20.10] to be taken as (fcir –​ fcds), where fcir (the subscript r means residual) is the net compressive stress in the concrete at the center of gravity of tendons immediately after the prestress has been applied to the concrete, and fcds (the subscript s means superimposed) is the stress in concrete at the center of gravity of tendons due to all superimposed permanent dead loads that are applied to the member after it has been prestressed. For example, fcir would correspond to fc = 1232 psi computed in part (b) of Example 20.7.1 because the prestress and the beam dead load act simultaneously when the prestress is applied. Typical values for percentage loss of prestress due to creep are from 5 to 6%. Pretensioned beams will exhibit more creep than post-​tensioned beams because the prestress is imposed when the concrete is at an earlier age; age at loading is a major factor in determining the magnitude of creep.

Loss Due to Shrinkage in Concrete Shrinkage is the volume change in concrete that occurs with time, as discussed earlier (see Sections 1.11 and 12.12). The loss of prestress due to shrinkage may be expressed as

∆fs = ε sh Es

(20.7.11)

where εsh is the shrinkage strain in concrete (see Section 12.12). Chapter 12 gives general expressions for evaluating shrinkage strain. Zia et  al. [20.10] recommend computing εsh by starting with a value of 550 × 10–​6 as the basic ultimate shrinkage strain, multiplying by (1 –​0.06V/​S) to correct for the ratio of volume V to surface S, and then multiplying by (1.5 –​ 0.015H) to correct for the relative humidity H. For post-​tensioned members, an additional reduction factor is used to account for the time between the end of moist curing and the application of prestress.

Loss Due to Relaxation of Steel Stress Relaxation is taken to mean the loss of stress in steel under nearly constant strain at constant temperature. Loss due to relaxation varies widely for different steels, and such loss should be provided for in accordance with test data furnished by the steel manufacturers. The percentage loss of prestress relating to relaxation varies with the type of tendon and the ratio of initial prestress to tensile strength of tendon. This loss is generally assumed to be in the range of 2 to 3% of the initial steel stress for stress-​relieved strands and about 1% for low-​relaxation strands (see Section 1.13). Zia et al. [20.10] have provided a formula for this computation.

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Friction Losses in Post-​tensioned Members There will be frictional losses, which are generally small, in the jacking equipment; also present will be friction between the tendons and the surrounding material (either duct or actual concrete member), due to intended or unintended curvature in the tendons. The friction between tendons and surrounding material is not small and may be considered as partly a length effect and partly a curvature effect. Referring to Fig. 20.7.1, let dx be a segment of a curved tendon. Assume that the tendon is being jacked from the left end by a force Ps, which results in a force Px at some distance to the right; these forces define the limits for the tension t. The full angle enclosed within the arc is α . For equilibrium of the entire segment dx, refer to the force polygon of Fig. 20.7.1(b); the normal force dN is  

dt  dα    dα dN = t   +  t + dα  2   dα  2



(20.7.12)

and neglecting infinitesimals of higher order,  dα  dN = 2t   = td α  2 



(20.7.13)

The friction force developed along the length dx is µ dN = µt d α



(20.7.14)

Summation of forces along the tendon gives t − µt d α – (t + dt ) = 0 dt = −µ d α (20.7.15) t



Lx

dx dα

t

µdN = µtdα

t+

dt dα dα

t+

t

dα

dt dα dα

dα dN 2 (b) Force polygon

α

R = radius of curvature (a) Element of curved tendon

Figure 20.7.1  Friction losses in post-​tensioned member.

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20.7  LOSS OF PRESTRESS

Integrating to obtain the total effect over the entire curved portion included within the angle α,

∫

Px Ps

α dt = ∫ −µ d α 0 t

log e Px – log e Ps = – µα



Px = e − µα Ps

(20.7.16) (20.7.17) (20.7.18)

Replacing the friction force term µα with the following expression, which contains a friction part due to curvature and a length effect (wobble effect), µα + KLx,

Px = e − ( µα + KLx ) Ps

(20.7.19)

Ps = Px e( µα + KLx )

(20.7.20)

or

where α = L /​R; the length L of the curve divided by R, the radius of curvature; and Lx is the length of prestressing tendon element from jacking end to any point x along the tendon. Equation (20.7.20) was included in the ACI Code up to its 2008 Edition. In the 2014 ACI Building Code, however, ACI-​20.3.2.6.2 specifies that “Calculated friction loss in post-​ tensioning tendons shall be based on experimentally determined wobble and curvature friction coefficients”; a reference is given in Commentary ACI-​R20.3.2.6.2 to the Post-​Tensioning Manual [20.21], where the same procedure used in the 2008 ACI Code is discussed. As an approximation, when Ps –​ Px is small (such as not more than 15 to 20% of the jacking force Ps), the friction force may be assumed to be constant. If the friction force is assumed to be proportional to the force Px, then (see Fig. 20.7.2) µN = µPx α



(20.7.21)

and assuming the wobble effect KLx is also proportional to Px, equilibrium requires Ps = Px + Px (µα + KL x )

or

Ps = Px (1 + µα + KL x )



(20.7.22)

which is permitted in the Post-​Tensioning Manual when ( µα + KLx ) does not exceed 0.15. Lx

µPxα Ps Jacking force

Px α

R

Figure 20.7.2  Approximate procedure for friction losses in post-tensioned members.

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Two-​way post-​tensioned slab. (Photo courtesy of Cary Kopczynski & Co.)

Losses Due to Anchorage Seating In most post-​tensioned members, the prestress force is transferred through anchorage fixtures at the ends of the member (see Fig. 20.4.2). These anchorage devices often utilize wedge action to anchor the tendons, although other methods, such as threaded bars with nuts or cold-​ formed rivet heads, are also used. Friction wedges, such as the one shown in Fig. 20.7.3, will slip a small amount before the tendon can be fully gripped (seated) at the end. As a result, the initial prestress force applied by the jack will be reduced by an amount depending on the type of anchorage device. Typical values of slip range between 0.03 and 0.1 in. [20.6]. In practice, this loss in prestress can be compensated for by applying a higher jacking force at the time of prestress. Note that since the slip due to anchorage seating is a fixed amount that depends on the anchorage device, the loss of prestress will be higher for shorter tendons.

Practical Design Consideration—​Total Losses The total loss in prestress may be expressed in unit strains, total strains, or unit stresses, or in percentage of initial prestress. Although it is difficult to generalize the amount of prestress loss, Lin and Burns [20.6] have suggested that for average steel and concrete properties and for average curing conditions the values in Table 20.7.1 may be taken as representative. TABLE 20.7.1  AVERAGE PERCENTAGES OF LOSS OF PRESTRESS [20.6] Pretensioning (Percent) Elastic shortening and bending of concrete member Creep of concrete Shrinkage of concrete Relaxation (creep) in steel Totals

Post-​tensioning (Percent)

4

1

6 7 8 25

5 6 8 20

Notes: Zia et al. [20.10] have indicated that the upper limit for total loss in steel stress (not including friction loss) for stress-​ relieved strand in normal-​weight concrete may be assumed as 50,000 psi. More detailed treatment of losses of prestress is to be found in the textbooks by Collins and Mitchell [20.2], Libby [20.3], Nawy [20.4], Naaman [20.5], Lin and Burns [20.6], and Nilson [20.7], as well as in the PCI Committee on Prestress Losses Report [20.11] and Zia et al. [20.10].

863



863

20.7  LOSS OF PRESTRESS

Figure 20.7.3  Typical wedge-​and-​chuck assembly for 7-​wire strands. (Photo by José A. Pincheira.)

EXAMPLE 20.7.2 The post-​tensioned beam of Fig. 20.7.4 containing a cable of 72 parallel wires, Aps = 3.60 sq in., is to be tensioned 2 wires at a time. The jacking stress is to be measured by a pressure gauge. The wires are to be stressed from one end of the member to a value f1 to overcome friction loss, then released to a value f2, so that an initial prestress of 144 ksi is obtained immediately after anchoring. Compute f1 and f2, as well as the final design stress after all losses, according to the ACI Code. Assumptions are as follows: (a) (b) (c) (d) (e)

Coefficient of friction μ = 0.50 Wobble coefficient K = 0.0008 (per ft) Deformation at anchorages and slip of wires = 0.06 in. Steel relaxation = 5% of initial prestress (144 ksi) Es = 29,000 ksi; Ec = 5000 ksi

SOLUTION (a) Loss due to friction. Using the “exact” method, Eq. (20.7.20), Ps = Px e( µα + KLx )



 45  µα ≈ 0.5  = 0.1108  203.125  KL x = 0.0008(45) = 0..0360



Ps = Px e( 0.0360 + 0.1108 ) = 1.158 Px The above expression can also be used in terms of unit stress f ; in this case if fx is desired to be 144 ksi, then the initial stress f1 to overcome frictional loss is

f1 = 1.158(144) = 167 ksi

(Continued)

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Example 20.7.2 (Continued) Using the approximate expression, Eq. (20.7.22),

Ps = Px (1 + 0.1468) = 1.1468 Px



or f1 = 1.1468(144) = 165 ksi



(b) Loss due to anchorage seating. For tensioning from one end,



εs =

0.06 = 0.00011 45(12)

∆fs = ε s Es = 0.00011(29, 000) = 3.2 ksi



To allow for anchorage seating, tension to f1 = 167 ksi to overcome friction, then release to  f2 = 144 + 3.2 = 147.2 ksi. The minimum stress fsi = 144 ksi will then exist at both ends. (c) Elastic shortening due to post-​tensioning two wires at a time. The wires tensioned first will have the greatest loss, having some additional loss as each succeeding pair of wires is tensioned. The pair tensioned last will have zero loss. Thus Δ fs in the first pair, given by Eq. (20.7.5), is ∆fs = n= T0 =

nT0 Ag 29, 000 = 5.8, say 6 to nearest whole number 5000 35 (3.60)(144) = 504 kips 36

6(504) ∆fs (first pair ) = = 4.66 ksi 18(36)



∆fs (last pair) = 0 ksi average ∆fs = 2.33 ksi (1.6% of 144 ksi) (d) Creep loss. Using the procedure recommended by Zia et  al. [20.10], fc in Eq. (20.7.10) is computed as (  fcir –​ fcds). The net concrete stress at midspan at the centroid of the steel tendons is fcir =



T0 T0 e2 M D e + − Ag Ig Ig

where T0 = 3.60(144) = 518.4 kips

MD =

1 18(36) (0.15)(45)2 = 170.9 ft-kips 144 8

Then,



fcir =

518.4 518.4(12)12 170.9(12)12 + − 18(36) 18(36)3 /12 18(36)3 /112

= 0.80 + 1.07 − 0.35 = 1.52 ksi (at midspan) (Continued)

865



865

20.7  LOSS OF PRESTRESS

Example 20.7.2 (Continued) In this case, if there is no additional superimposed permanent dead load, fcds is zero. At the support, where there is zero eccentricity of the tendons and zero dead load moment, fcir =



T0 = 0.80 ksi Ag

The average value of (  fcir –​ fcds) should be used as fc in Eq. (20.7.10) for post-​ tensioned beams: fc =



1.52 + 0.80 = 1.16 ksi 2

Note again that in this example fcds is taken as zero.

ε c = elastic strain =

fc 1.16 = = 0.000232 Ec 5000

and for a post-​tensioned beam using Ct = 1.6,

ε cp = Ct ε c = 1.6(0.000232) = 0.000371

∆fs = ε cp Es = 0.000371(29, 000) = 10.8 ksi (7.5%)



(e) Shrinkage loss. Using the procedure of Zia et  al. [20.10], the basic shrinkage strain is

basic ε sh = 550 × 10 –6



The correction factor (CF) for volume /​surface (V/​S) ratio is (CF)V / S = 1 − 0.06



V = 1 − 0.06(6) = 0.64 S

where V 18(36) = = 6 S 2(18) + 2(36)



The correction factor (CF) for humidity H is

(CF )h = 1.5 − 0.015 H = 1.5 − 0.015(70) = 0.45

for 70% relative humidity. The adjusted shrinkage strain then becomes

ε sh = (basic ε sh )(CF )V / S (CF )h = (550 × 10 −6 )(0.64)(0.45) = 0.000158





∆fs = ε sh Es = 0.000158(29, 000) = 4.59 ksi (3.2%) (f) Relaxation in steel.

∆fs = 0.05(144) = 7.2 ksi (5.0%)

(Continued)

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Example 20.7.2 (Continued) (g) Total losses. Loss of Stress

Elastic shortening Creep in concrete Shrinkage Relaxation in steel

ksi

Percent

2.3 10.8 4.6 7.2

1.6 7.5 3.2 5.0

24.9 ksi

17.3%

The final design prestress under dead load plus live load, after losses, is

fse = 144 – 24.9 = 119.1 ksi R = 203’– 112 ” Assume cable drape is circular curve 0.1108 radians



Symbolic about midspan

1’– 6”

1’– 6” CG

3’– 0”

1’– 6” 45’– 0” c–c of supports

e = 12” midspan section

Figure 20.7.4  Post-tensioned beam of Example 20.7.2.

20.8 NOMINAL STRENGTH M n OF FLEXURAL MEMBERS The nominal strength Mn of a prestressed concrete flexural member is computed in a manner similar to that discussed in Section 3.4, using the six assumptions stated there. Unlike nonprestressed reinforced concrete members, where nominal strength is the primary criterion determining whether a design is or is not satisfactory, a prestressed concrete member must have adequate strength; however, this is usually not the governing factor in a design. Design must include consideration of all significant load stages. Primarily, these stages are (1)  initial stage, including the period before and during prestressing, as well as the transfer of prestress to the concrete; (2) intermediate stage during transportation and erection; (3) final stage under service load, after losses; and (4) overload stage, where cracking and nominal strength (so-​called ultimate strength) are important. Though the actual number of conditions to be investigated varies with the situation, ordinarily the initial service condition involves beam dead load plus prestress before losses; the final service condition involves full dead load plus live load after losses; and the nominal strength must be adequate for possible overload. The initial and final service load stages were considered in Section 20.5. A beam may be properly prestressed to carry service loads with little deflection and generally without cracking but, for example, because of a small moment arm to the centroid of the steel, its nominal moment strength Mn may not give an adequate margin of safety. On the other hand, if only the strength Mn were considered, service loads might cause excessive cracking, camber (upward deflection), or deflection.

867



20.8  NOMINAL STRENGTH Mn OF FLEXURAL MEMBERS

867

Most of the following development applies for members containing only prestressed reinforcement (so-​called fully prestressed members); thus, any nonprestressed steel effect is omitted. Also, the discussion is limited to pretensioned and post-​tensioned construction with grouted tendons. The flexural strength of post-​tensioned members in which tendons are ungrouted is, in general, less than that of a beam with bonded tendons. The nominal moment strength may be determined for rectangular sections, in accordance with the principles of Chapter 3 (Section 3.4), as follows: a  Mn = T  d p −   2



(20.8.1)

where T = Aps fps, C = 0.85 fc′ ba , and from C = T, a=



Aps f ps 0.85 fc′ b

=

ρ p bdd f ps 0.85 fc′ b

=

ρ p f ps 0.85 fc′

dp



(20.8.2)

with Aps, ρp, and fps referring to the area, reinforcement ratio, and tensile stress for the prestressed reinforcement, respectively, and dp to the effective depth measured to the centroid of the prestressing steel. Note that fps is the average stress in the prestressing steel when nominal moment strength Mn is reached; it is used instead of fy because the steel usually exhibits no well-​defined yield point (see Fig. 20.8.1). Substitution of Eq. (20.8.2) into Eq. (20.8.1) gives

ρ p f ps   M n = Aps f ps  d p − dp 1.7 fc′  



(20.8.3)

Thus for any concrete cross section, the nominal moment strength depends on ρp and fps. The actual stress in the prestressed reinforcement when the nominal strength is reached may not be easily determined, particularly when the specific stress-​strain curve for the steel used is not available. Referring to Fig. 20.8.1, for low percentages of steel and therefore higher stress fps, the strain may be nearly 0.05, whereas for higher percentages of reinforcement, the strain may be closer to 0.01 (approximately corresponding to yield stress). Thus fps is not the same for all beams. In the ACI Building Code, determination of tensile stress in bonded prestressing steel is addressed in ACI-​20.3.2.3.1. For sections in which all prestressed reinforcement is in the member tension zone and the effective prestress fse (stress in the prestressing steel after

250

1724

fpu = 250 ksi

1000

100

Unit stress, MPa

Unit stress, ksi

1500 200

500

0

1

2

3

4

5

Unit strain, %

Figure 20.8.1  Typical stress–​strain curve for 250-​ksi steel (stress-​relieved) wire.

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losses) is at least 0.5fpu, the stress fps may be determined as follows, omitting the terms for nonprestressed reinforcement.  γ p ρ p f pu  f ps = f pu  1 − β1 fc′  



(20.8.4)

Note the values of γp /​β1 in Table 20.8.1. For members with unbonded tendons, fse ≥ 0.5fpu, and with a span-​to-​depth ratio of 35 or less, ACI-​20.3.2.4.1 gives f ps = fse + 10, 000 +



fc′ 100ρ p

(20.8.5)

but not greater than the lesser of fpy and (fse + 60,000). For members with unbonded tendons, fse ≥ 0.5fpu, and with a span-​to-​depth ratio greater than 35, such as one-​way slabs, flat plates, and flat slabs, f ps = fse + 10, 000 +



fc′ 300ρ p

(20.8.6)

but not greater than the lesser of fpy and (fse + 30,000). Harajli and Kanj [20.16] provide a study on the ultimate flexural strength of members having unbonded tendons.

Balanced Strain Condition, Tension and Compression Control Limits, Strength Reduction Factors As first described in Section 3.5, the balanced strain condition (or compression control limit for the case of one layer of tension steel) occurs when the concrete strain εc at the extreme compression fiber is 0.003 at the instant the steel reaches its yield strain εy = fy /​Es. In prestressed concrete members, however, the steel used does not exhibit the well-​defined yield point that occurs with ordinary deformed bars, and so the concept of balanced failure is nebulous. Further, the steel and concrete are strained prior to the application of external load (i.e., prestrained), and so the calculation of the total strain in the prestressing steel is not as straightforward as the determination of reinforcement strains in nonprestressed members. In the ACI Code, the same measure of strain in the extreme layer of tension reinforcement is used in nonprestressed and prestressed reinforcement for the purpose of determining whether a section is tension controlled, compression controlled, or in the transition region. ACI-​21.2.2.1 and 21.2.2.2 permit εy =  εt = 0.002 for Grade 60 reinforcement and

TABLE 20.8.1  VALUES FOR γp  /​β1 FOR EQ. (20.8.4) Concrete Strength, fc′ (psi)

γp

5000

6000

7000

8000

0.55 (for fpy  /​fpu ≥ 0.80) 0.40 (for fpy /​fpu ≥ 0.85) 0.28 (for fpy /​fpu ≥ 0.90)

0.69

0.73

0.79

0.85

0.50

0.53

0.57

0.62

0.35

0.37

0.40

0.43

869



20.8  NOMINAL STRENGTH Mn OF FLEXURAL MEMBERS

869

for all prestressed reinforcement, respectively. For prestressed reinforcement, the strain εt is not the actual strain in the steel; rather, it is the strain excluding those “due to prestress, creep, shrinkage, and temperature” (ACI-​R21.2.2). Thus, for all sections, nonprestressed reinforced, prestressed reinforced, or any combination thereof, and for all shapes of cross section, and no matter where the steel is located, including compression steel, when the extreme tensile strain εt is greater than 0.005, the section is “tension controlled” and φ = 0.90; and when the extreme tensile strain εt is less than 0.002 (for Grade 60 and prestressed reinforcement), the section is “compression controlled,” and φ will be either 0.75 or 0.65 as required for columns (see Chapter 10). The strain limits were shown earlier (see Fig. 3.6.2). When the extreme tensile strain εt is between 0.002 and 0.005, the φ factor is required to be obtained by linear interpolation, as shown in Fig. 3.6.2. Equations (3.6.3) and (3.6.4) give φ in terms of the extreme tensile strain εt, and Eqs. (3.6.5) and (3.6.6) give φ in terms of c /​dt, where c is the location of the neutral axis measured from the compression face of the section and dt is the distance from the extreme compression fiber to the extreme layer of tension steel.

EXAMPLE 20.8.1 Determine the nominal moment strength Mn of the pretensioned bonded section investi­ gated in Example 20.5.2, and evaluate whether the strength is adequate for the given ser­ vice loads. The concrete has fc′ = 5000 psi, and the stress-​relieved prestressing strands have fpu = 250,000 psi. Compare the stress in the prestressing steel at nominal condition calculated according to ACI-​20.3.2.3.1 with that obtained using an average stress-​strain relationship for the steel, as given in Fig. 20.8.1, and assuming 20% prestress losses. Use Es = 29,000 ksi and Ec = 4000 ksi. SOLUTION (a) The stress in the prestressing steel when nominal strength is reached may be taken, according to ACI-​20.3.2.3.1, or Eq. (20.8.4), as f pu   γp f ps = f pu  1 − ρ p β1 fc′  



2.30 = = 0.0046 ρp = bd p 20(25) Aps

For fc′ = 5000 psi, β1 = 0.80; and for stress-​relieved strands, fpy  /​fpu ≥ 0.85 and γp = 0.40. Thus,

250   0.40 = 250(1 − 0.115) = 221 ksi f ps = 250 1 − (0.0046) 5   0.80

From Fig. 20.8.2(c), Cu = 0.85 fc′ ba = 0.85(5)(20)a = 85a Tu = Aps f ps = 2.30(221) = 508 kips

Cu = Tu 508 a= = 5.98 in. 85 c=

a 5.98 = = 7.47 in. 0.8 β1



(Continued)

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Example 20.8.1 (Continued) The additional strain εs2 at the centroid of prestressing steel that is due to the superimposed nominal moment Mn is



εs2 =

fse Aps  1 e2  dp − c +  + ε cu  Ec  Ag I g  c

ε s2 =

0.8(175, 000)(2.3) 1  1 10 2  17.53 + = 0.00735 + 0.003   1000 4000 7.47  600 45, 000 

The first term represents the compressive strain in the concrete at the level of the prestressing steel under the effective prestress. The second term, on the other hand, is equal to the increase in tensile strain relative to the strain in the prestressing steel at the time the concrete at that level has zero strain. The total strain in the prestressing steel is then the strain due to prestress after losses, εs1, plus the additional strain due to superimposed nominal moment, Mn. fsi (initial) = 175, 000 psi ∆fs (losses) = 0.2(175, 000) = 35, 000 psii

fse = 175, 000 − 35, 000 = 140, 000 psi

140 + 0.00735 = 0.0122 ε s = ε s1 + ε s 2 = 29, 000

Using a typical stress-​strain relationship for a steel with fpu  =  250 ksi, such as in Fig. 20.8.1, the strain εs of 0.0122 corresponds approximately to that for a stress fps of 225 ksi. The value of fps agrees closely with the starting assumption of 221 ksi based on Eq. (20.8.4) in this case. Thus, the nominal moment strength is

a  5.98   1   = 932 ft-kips M n = Tu  d p −  = 508  25 −    2  2   12

(b) The factored moment Mu based on the service loads of Example 20.5.2 is

Mu = 1.2(125) + 1.6(410) = 806 ft-kips

Evaluate the strength reduction factor φ to be used. The neutral axis distance c for the section is c = 7.47 in. This section has h = 30 in. and dp = 25 in. for, say, two layers of strands. The distance dt from the extreme compression fiber to the extreme tension steel will be larger than dp, say dt = 27 in. The strain εt at the extreme tension steel is

ε i = 0.003

dt − c 27 − 7.47 = 0.003 = 0.0078 c 7.47

Since the strain εt exceeds 0.005, the section is “tension controlled” and φ = 0.90. Had the strain εt been less than 0.002, the section would be “compression controlled” and φ would have been either 0.65 or 0.75, depending on whether the section was tied or spirally reinforced (typically tied for the case of beams). If εt is between 0.005 and 0.002, Eq. (3.6.3) or (3.6.4) would be used for the linear interpolation between tension-​ controlled and compression-​controlled sections. (Continued)

871



871

20.9 CRACKING MOMENT

Example 20.8.1 (Continued) As mentioned earlier, in checking the strain εt in the extreme layer of steel in tension against the 0.005 limit, only the strain change relative to the steel strain at the time the concrete at the level of the prestressing steel has zero strain is considered. Since the strain in the concrete due to prestress at the level of the prestressing steel is very small compared to the increase in strain due to externally applied loads, the strain εt is an approximation of the strains induced after loading. The intent is thus to provide the beam with sufficient ductility once the external loads have been applied. (c) Compare φ Mn with Mu.

[φ M n = 0.90(932) = 839 ft-kips] > [Mu = 806 ft-kips] 20”

0.003 c

OK

0.85fc’ a = β1c

dp = 25” 30”

Cu = 0.85fc’ ba

dp – a 2

Aps = 2.30 sq in. Tu = Aps fps (a) Section

(b) Strain when Mn is reached

(c) Internal forces

Figure 20.8.2  Strain and stress distributions, and internal force resultants at nominal strength Mn of section for Example 20.8.1.

20.9 CRACKING MOMENT One of the features of prestressed concrete is that under service load it is usually crack free. To be sure that an adequate reserve exists against cracking, ACI-​9.6.2.1 requires that for beams, the total amount of prestressed and nonprestressed reinforcement be adequate to develop a factored load in flexure at least 1.2 times the cracking load calculated using a modulus of rupture equal to 7.5λ fc′. The same concept applies for one-​way slabs (ACI-​ 7.6.2.1) and two-​way slabs (ACI-​8.6.2.2). The basic cracking moment requirement is a means of ensuring that cracking will not lead to failure and that a margin large enough exists between flexural and cracking strength so that significant deflection will occur to warn that the strength Mn is being approached. The typical member will have a fairly large margin between cracking strength and flexural strength, but the designer must be certain by checking it. The ACI Code (ACI-​9.6.2.2 for beams, ACI-​7.6.2.2 for one-​way slabs, and ACI-​8.6.2.2.1 for two-​way slabs) waives the above requirement when φVn and φ Mn is at least twice Vu and Mu, respectively.

EXAMPLE 20.9.1 Compute the cracking moment Mcr and check its acceptability according to the  ACI Code, for the beam of Example 20.8.1 (Fig. 20.8.2). Use fc′ = 5000 psi (normal-​weight concrete) and assume that the effective prestress, after losses, is 140 ksi. (Continued)

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Example 20.9.1 (Continued) SOLUTION (a) Use the homogeneous beam concept (Section 20.6). The effective prestress force is

Te = Aps fse = 2.30(140) = 322 kips

Using the following equation to find the cracking moment:   moment   external  cracking  axial − − + =  prestress  prestress  loads   stress 

−

Te Te eyt M cr yt − + = fr = 7.5 5000 = 530 psi Ag Ig Ig −



322 322(10)15 M cr (15) − + = 0.530 600 45, 000 45, 000

322(45, 000)  1  0.530(45, 000) M cr =  + 322(10) + 15 600(15)  12 



= 133 + 268 + 134 = 535 ft-kips (b) Use the internal force concept (Section 20.6). Consider that the cracking moment Mcr has two parts,

M cr = M1 + M 2



where M1 is the superimposed moment necessary to give zero stress in the precompressed tension zone (see Fig. 20.9.1), and M2 is the additional moment to cause cracking, assuming that zero stress exists at the tension face when M2 is applied (see Fig. 20.9.1). C = Te = 322 kips  10 + 5  M1 = Te (e + ec ) = 322  = 402.5 ft-kips  12 

M2 =

fr I g yt

=

0.530(45, 000) = 132.5 ft-kips 15(12)



M cr = M1 + M 2 = 403 + 133 = 535 ft-kips (c) Check ACI-​9.6.2.1. From Example  20.8.1, the nominal moment strength Mn provided is 932 ft-​kips. Since ACI-​9.6.2.1 requires reinforcement “to develop a factored load at least” 1.2 Mcr,

φ M n ≥ 1.2 M cr

0.90(932) > 1.2(535) 839 ft-kips > 642 ft-kips

OK

Thus cracking will occur soon enough before reaching nominal moment strength so that large deflection will give a warning of impending failure. (Continued)

873



873

20.10  SHEAR STRENGTH

Example 20.9.1 (Continued) fr Centroid of beam cross section

10” C ec = 5”

30”

(–)

(–)

(–) =

+

e = 10” Te

(+) fr

0 Midspan

M1

+

fr

M2

=

Mcr

Figure 20.9.1  Computation of cracking moment.

20.10 SHEAR STRENGTH OF MEMBERS WITHOUT SHEAR REINFORCEMENT In general, the ideas presented in Chapter  5 regarding shear strength for nonprestressed beams are also applicable to prestressed beams. An excellent summary of background for the ACI Code expressions for shear strength of prestressed concrete beams is given by MacGregor and Hanson [5.63]. More information on shear strength of prestressed concrete members is available in the ASCE–​ACI Committee 426 Reports [5.4, 5.16]. Collins and Mitchell [18.9] have presented an excellent unified treatment of shear and torsion behavior for both prestressed and nonprestressed concrete beams, along with proposed design rules. Use of the strut-​and-​tie model for prestressed concrete members has been presented by Alshegeir and Ramirez [20.18] and by Ramirez [20.20]. It is known, however, that a prestressed concrete beam generally performs better under high shear conditions than does an ordinary reinforced concrete beam. Consider Eq. (5.3.1) as derived in Chapter 5 for maximum principal stress, 2

ft (max) =



ft f  +  t  + v2  2 2

[5.3.1]

+v (+ft’, +v)

v v

v

ft

α

α

2α

α

ft

+ft

ft (max)

v

(0, –v) (a) Principal stress–reinforced concrete beam +v (–ft’, +v)

v v

–ft v

α

–ft

α

2α

+ft

α v ft (max)

(0, –v)

(b) Principal stress–prestressed concrete beam

Figure 20.10.1  Comparison of directions of principal tensile stress: (a) reinforced and (b) prestressed concrete beams.

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In nonprestressed concrete, the normal stress ft is a tensile stress in the tension zone of the beam. When the concrete is prestressed, ft is a compressive stress throughout the member depth, or at least over most of it. On replacing ft with –​ ft, it is apparent that the magnitude of the principal tensile stress decreases, 2

−f f  ft (max) = t +  t  + v 2  2 2



(20.10.1)

The angle α that the principal tensile stress makes with the beam axis is greater for a prestressed concrete beam than for an ordinary reinforced concrete beam, as shown in Fig. 20.10.1. Inclined cracking will be less likely to occur and, if it occurs, will be at a lower angle with respect to the longitudinal axis than in nonprestressed members. Inclined cracks of two types are possible in prestressed concrete beams: (1) the flexure-​ shear crack that occurs in a beam previously cracked owing to flexure and (2) the web-​ shear crack that occurs in the thin web of a previously uncracked beam [see Fig. 5.4.1(a)]. Whereas only the flexure-​shear crack is common in nonprestressed beams, both types represent potential cracks in prestressed concrete beams.

Flexure-​Shear Cracking Strength The flexure-​shear crack arises from high principal tensile stress near the interior extremity of a flexural crack. Experimental studies have shown that the shear corresponding to the flexure-​shear crack is that which causes flexural cracking at approximately a distance dp /​2 from the load point, in the direction of decreasing bending moment [20.13] (see Fig.  20.10.2). The 1963 ACI Code formula for flexure-​shear cracking strength was the experimentally determined linear relationship of Fig. 20.10.3. Because the relationship between moment and shear (M/​V) was not the same for both dead and live loads in the tests, the effects of dead load were not included in the linear relationship. The ordinate of the plot (see Fig. 20.10.3) is thus in terms of Vci –​ Vd, where Vd is the ser­ vice dead load shear force and Vci is the total nominal shear strength. The abscissa involves the net cracking moment Mcr. The net cracking stress used for computing Mcr equals the modulus of rupture (assumed conservatively at 6 λ fc′ psi), plus the compressive stress fpe provided by the prestressing force (after losses) occurring at the extreme fiber at which tensile stresses are caused by the applied loads, less the tensile stress fd due to service dead load. Thus

M cr =

I (6 λ fc′ + f pe − fd ) yt

(20.10.2)*

which is ACI Formula (22.5.8.3.1c). The abscissa in Fig. 20.10.3 involves Mcr  /​(M/​V –​ dp /​2), which represents the shear corresponding to flexural cracking developing at dp /​2 from the cross section being investigated. Thus the linear relationship for the flexure-​shear cracking strength resulting from high principal tensile stress in the vicinity of a flexural crack is, according to Fig. 20.10.3, Vct − Vd

b w d p fc′

= 0.6 +

M cr ( M /V − d p / 2) bw d p fc′

(20.10.3)

Since 1971, this expression has been simplified by eliminating the subtracted dp /​2 term.

*  For SI, ACI 318-​14M, for fc′ , fpe, and fd in MPa, Mcr in N∙mm, I in mm4, and yt in mm, Eg. (20.10.2) may be rewritten as



M cr =

I yt

1   λ fc′ + f pe − fd  2

(20.10.2)

 

875



875

20.10  SHEAR STRENGTH

Thus the nominal flexure-​shear cracking strength Vci, (ACI-​22.5.8.3.1), is Vci = 0.6 λ fc′bw d p +



Vi M cr + Vd ≥ 1.7λ fc′bw d p M max

(20.10.4)*

In Eq. (20.10.4), Mmax replaces M of Eq. (20.10.3) and represents the maximum moment that can occur at the section under consideration, due to externally applied factored loads (i.e., applied loads other than beam weight and prestress, unless the prestress causes an external reaction). Vi replaces V of Eq. (20.10.3) and represents the shear force at the section due to the factored loading that caused maximum moment. In other words, one uses the moment envelope values for Mmax along with the corresponding shears, rather than the shear envelope values which would be larger. Of course, where partial span loadings are not considered, the full-​span loading gives Vi and Mmax for each point along the span. When full-​span uniform loading is used in a simply supported span, Vi = w(L –​ 2x)/​2 and Mmax = wx(L –​ x)/​2, and Vi L − 2x = M max x( L − x )



(20.10.5)

In effect, Eq. (20.10.4) gives the flexure-​shear cracking strength as the sum of (1) the shear due to actual dead load (beam weight) without overload factor, (2)  the shear due to superimposed factored load that is sufficient to cause flexural cracking, and (3)  the additional shear ( 0.6λ fc′ bw d p) that will cause the flexural crack to initiate the inclined flexure-​shear crack.

Web-​Shear Cracking Strength The web-​shear crack arises in a beam previously uncracked due to flexure. Such a crack is typical near the support of a thin-​webbed section (see Fig. 5.4.1) as a result of high principal tensile stress. For this development the beam may reasonably be considered as homogeneous. Thus Eq. (20.10.1) may be directly applied, 2

ft (max) =



− ft f  +  t  + v2  2 2

[20.10.1]

Decreasing moment A

Flexure-shear crack Initiating flexural crack dp

A

2

Figure 20.10.2  Flexure-​shear type of inclined crack.

*  For SI, ACI 318-​14M, for fc′ in MPa, bw and dp in mm, and V in N, gives



Vci =

VM 1 1 λ fc′ bw d p + i cr + Vd ≥ λ fc′ bw d p 20 M max 7

[20.10.4] 

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5

bwdp fc’

4 3 2 1 0

0

1

2

(

3 Mcr M V

–

)

dp 2

4

5

6

bwdp fc’

Figure 20.10.3  Comparison of the shear corresponding to flexure-​shear cracking in

prestressed concrete beams with the ratio of the flexural cracking moment to the shear span (both axes nondimensionalized by dividing by bw dp fc′ ). (From Sozen and Hawkins [20.13].)

Since tests have demonstrated that cracks of this type usually originate near the centroid of the section, the stresses refer to that location. Solving Eq. (20.10.1) for v, 2

2

2

2

ft    ft  2  ft (max) + 2  =  2  + v   f  f  ft (max) + ft (max) ft +  t  =  t  + v 2  2  2 2



v = ft (max) 1 +

ft ft (max)

(20.10.6)

where ft is the compressive stress at the level of the centroid, and ft(max) is the principal tensile stress, which should be less than the tensile strength of concrete if no cracks are to form. Since Eq. (20.10.6) should agree with the criteria for shear strength of nonprestressed beams, ft (max) is taken as 3.5λ fc′. The equation then becomes



v = 3.5λ fc′ 1 +

ft 3.5λ fc′

(20.10.7)

A plot of Eq. (20.10.7) for λ = 1.0 is given in Fig. 20.10.4, illustrating that this equation may be approximated by a straight line,

v = 3.5λ fc′ + 0.3 ft

(20.10.8)

87



877

20.10  SHEAR STRENGTH 10 v = 3.5√ fc’

8

v fc’

1+

ft [Eq. (20.10.7)] 3.5√fc’

6 v = 3.5√fc’ + 0.3ft [Eq. (20.10.8)]

4 2

0

2

4

6

8

10

12

14

16

18

20

ft fc’

Figure 20.10.4  Comparison of theoretical maximum shear stress with straight-line approximation.

or, in ACI Code terminology, multiplying by bw dp to give nominal shear strength,

Vcw = (3.5λ fc′ + 0.3 f pc )bw d p

(20.10.9)

where fpc is defined as the compressive stress (psi) in the concrete, after losses, at the centroid of the section resisting the applied loads, or if the centroid lies within the flange on a T-​section, fpc is the stress at the junction of the flange and web. If prestressing tendons are draped, a vertical component Vp arises which assists in carrying the shear. Therefore, ACI-​22.5.8.3.2 gives

Vcw = (3.5λ fc′ + 0.3 f pc )bw d p + Vp

(20.10.10)*

Alternatively, the web-​shear cracking strength may be determined (ACI-​22.5.8.3.3) via the principal stress equation, Eq. (20.10.6), where the principal stress ft (max) is limited to 4 λ fc′. This would give Eq. (20.10.7) with coefficients 4 instead of 3.5. Thus using fpc for ft, Eq. (20.10.7) multiplied by bwdp with Vp added becomes



Vcw = 4 λ fc′ 1 +

f pc 4 λ fc′

bw d p + Vp

(20.10.11)*

which is an acceptable alternate to using Eq. (20.10.10). The nominal shear strength Vn at which inclined cracking is imminent is given by the lesser of Vci and Vcw. When applying Eqs. (20.10.4) and (20.10.10) or (20.10.11) for Vci and Vcw, the effective depth dp is to be taken as the distance from the extreme compression fiber to the centroid of the prestressing tendons, or as 80% of the overall depth of the member, whichever is greater. In computing the prestress effect fpc, full prestress after losses may be used only when the prestressing tendons are embedded a distance that exceeds their required transfer length from the section being investigated. The critical section for maximum shear is generally at

*  For SI, ACI 318-​14M, for fc′ and fpc in MPa, bw and dp in mm, and V in N, gives



Vcw = (0.29λ fc′ + 0.3 f pc )bw d p + Vp

(20.10.10)



3 f pc 1 bw d p + Vp Vcw = λ fc′ 1 + 3 λ fc′

(20.10.11) 



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h/​2 from the face of support (ACI-​9.4.3.2), so that the region between the face of support and h/​2 therefrom must be designed for the shear at the critical section.

ACI Code Simplified Alternative for Shear Strength When the member has an effective prestress fse at least equal to 40% of the tensile strength fpu of the flexural reinforcement, the nominal shear strength may be taken as (ACI-​22.5.8.2) Vu d p   Vc = 0.6 λ fc′ + 700  bw d p ≤ 5λ fc′bw d p Mu  



(20.10.12)*

but need not be taken less than 2 λ fc′bw d p . Supporting data for this alternate relationship are shown in Fig. 20.10.5 from MacGregor and Hanson [5.63]. When applying Eq. (20.10.12) from ACI-​22.5.8.2, Vu is the maximum shear due to factored loading at the section and Mu is the simultaneously occurring moment; Vudp/​Mu is also limited to a maximum value of 1.0; and dp is the actual effective depth to the centroid of prestressed reinforcement. Note that d in Eq. (20.10.12) is the larger of dp or 0.8h. Equation (20.10.12) represents essentially the flexure-shear cracking strength, expressed in a manner similar to that for nonprestressed concrete. Since the percentage of reinforcement is low in prestressed concrete, a constant has been used instead of the variable ρ.

14 12 10

vc fc’

=

fse ≥ 0.4 fpu

=

fse < 0.4 fpu

8 6 4 2

vc = 0.6 fc’ + 700 0

2

4

6 1000 Vdp M f’

8

Vdp M 10

12

c

Figure 20.10.5  Alternate equation for computing vc for prestressed beams (l = 1.0). (From MacGregor and Hanson [5.63].)

*  For SI, ACI 318-​14M, for fc′ in MPa, bw and dp in mm, gives



Vu d p  1 Vc =  λ fc′ + 4.8  b d ≤ 0.42λ fc′ bw d p Mu  w p  20

(20.10.12) 

879



879

20.10  SHEAR STRENGTH

EXAMPLE 20.10.1 Determine the nominal shear strength Vn = Vc at a section 4 ft from the supports of the 40-​ft simple span shown in Fig. 20.10.6. The effective prestress force Te (after losses) is 322 kips. The beam is to support a service live load of 1.50 kips/​ft in addition to the beam weight of 0.675 kip/​ft. The concrete has normal weight with  fc′ = 5000 psi. SOLUTION (a) Simplified alternate procedure (from ACI-​22.5.8.2), Vc = [0.6 λ fc′ + 700



Vu d p Mu

]bw d ≤ 5λ fc′bw d (20.10.12)

Aps = 2.30 sq in. 12

15”

1

30” 5”

4’– 0” 10’– 0”

20’– 0”

10’– 0”

15” 20”

Figure 20.10.6  Beam, for Example 20.10.1.

In the application of the above formula, the symbols Vu and Mu indicate use of factored shear and moment. Assume partial span loading is not considered, wu = 0.675(1.2) + 1.50(1.6) = 0.81 + 2.40 = 3.21 kips/ft Vu = 3.21(20 − 4) = 51.4 kips

Mu = Vu d p Mu

1 (3.21)(4)(36) = 231 ft-kips 2



51.4(30 − 11) = = 0.35 < 1.0 231(12)

where dp in this term is the actual dp = 19 in. at 4 ft from the support. The nominal unit shear stress capacity is

vc = 0.6(1.0) 5000 + 700(0.35) = 42 + 245 = 287 psi vc = (upper limit ) = 5(1.0) fc′ = 354 psi > 287 psi



vc = (lower limit ) = 2(1.0) fc′ = 141 psi < 287 psi



OK OK

Thus at this section the prestressed member has a contribution to the shear strength attributable to the concrete, vc, of 287 psi. The nominal shear strength Vc at 4 ft from the support would be

Vc = vc bw d = 0.287(20)(24) = 138 kips



where d is taken as the larger of 0.8h [i.e., 0.8(30) = 24 in.] and the actual dp (i.e., 19 in.). (b) Flexure-​shear cracking strength using the more elaborate procedure (ACI-​22.5.8.3.1),

Vci = 0.6 λ fc′bw d p +

Vi M cr + Vd ≥ 1.7λ fc′bw d p [20.10.4] M max (Continued)

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Example 20.10.1 (Continued) Using Eq. (20.10.2), M cr =

I (6 λ fc′ + f pe − fd ) yt

1 (20)(30)3 = 45, 000 in.4 (neglectiing steel) 12 yt = 15 in. I=

f pe = compressive stress due to prestressingn force =

322, 000 322, 000(4)(15) + = 537 + 429 = 966 psi 20(30) 45, 000

e = 4 in. at 4 ft from center of support fd = dead load stress = M cr =

My 675(4)(36)(12)(15) = = 194 psi 2(45, 000) I

45, 000 [6(1) 5000 + 966 − 194] 15(12, 000)

= 0.25(424 + 966 − 194) = 0.25(1196) = 299 ft-kips M max = maximum moment due to externally applied factored loads, except beam weight at section being investiigated 1 (1.50)(4)(36)(1.6) = 173 ft-kips 2 Vi = corresponding shear due to factored load at section investigated =

= 1.50(20 − 4)(1.6) =38.4 kips



Vi M cr 38.4(299) = = 66.0 kips M max 173 Vd = dead load shear = 0.675(20 − 4) = 10.8 kips d p = 30 − 11 = 19 in.

0.80 h = 0.80(30) = 24 in.

or

Vci = 0..6(1.0) 5000 (20)(24)

1 1000

(use larger value)

+ 66.0 + 10.8 = 97.2 kips

The flexure-shhear cracking strength is not to be taken less than min Vcii = 1.7(1.0) fc′bw d p min Vci = 1.7(1.0) 5000 (20)(24)

1 1000

OK

= 57.7 kipps < 97.2 kip

(c) Web-​shear cracking strength using the more elaborate procedure (ACI-​22.5.8.3.2),

(

)

Vcw = 3.5λ fc′ + 0.3 f pc bw d p + Vp [20.10.10]



fpc = compressive stress in concrete at centroid of section resisting live load, due to all applied loads (zero due to all except Te /​Ac in this case) =

322, 000 = 537 psi 20(30)

d p = 19 in.

or

0.80(30) = 24 in..

(use larger value)



322 1 Vcw = 3.5(1.0) 5000 + 0.3(537) (20)(24) + = 223 kips 1000 12

(Continued)

81



881

20.11  SHEAR REINFORCEMENT

Example 20.10.1 (Continued) Since Vci < Vcw, Vc = Vci = 97.2 kips at 4 ft from the support for the member without shear reinforcement. The simplified alternate equation gives a nominal shear strength 42% higher than obtained from the more elaborate procedure:  138 kips compared to 97.2 kips.

20.11 SHEAR REINFORCEMENT FOR PRESTRESSED CONCRETE BEAMS The computations for shear reinforcement are essentially the same as for nonprestressed concrete as developed in Chapter 5. The total nominal shear strength Vn may be represented as the sum of the amount Vc attributable to the concrete and the amount Vs attributable to the shear reinforcement; thus Eq. (5.8.1) still applies,

Vn = Vc + Vs

[5.8.1]

The nominal strength Vc attributable to the concrete may be determined by (1) Eq. (20.10.12) when the effective prestress is at least 40% of the tensile strength of the steel, or (2) the smaller of Eqs. (20.10.4) and (20.10.10) for Vci and Vcw, respectively. The second and more accurate method using Vci and Vcw may be used whatever the magnitude of prestress. For the shear reinforcement contribution, Eq. (5.10.7) is used,

Vs =

Av f yt d s



[5.10.7]

where A v = effective area of shear reinforcement at any section s = spacing of the shear reinforcement Just as for nonprestressed reinforced concrete beams, prestressed concrete beams must have at least a minimum amount of shear reinforcement whenever Vu > φVc  /​2 (ACI-​9.6.3.1). However, this requirement may be waived if tests are made showing that the required flexural and shear strengths can be developed when shear reinforcement is omitted. Since prestressed concrete members are usually of standardized shapes with many identical or similar members manufactured, manufacturers frequently have made tests demonstrating that no shear reinforcement is required. When minimum shear reinforcement is required, ACI-​9.6.3.3 requires using the lesser of the following items: 1. ACI Table 9.6.3.3 rows (a) and (b):

min Av = 0.75 fc′

bw s 50bw s ≥ f yt f yt

which is also used for nonprestressd concrete, and

[5.10.8]

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2. where the effective prestress force equals at least 40% of the tensile strength of the steel based on fpu, ACI Table 9.6.3.3 row (e), as follows: min Av =



Aps  f pu   s  d    80  f yt   d  bw

(20.11.1)

where Aps = area of prestressed reinforcement fpu = tensile strength of prestressed reinforcement fyt = specified yield strength of shear reinforcement s = stirrup spacing, which may not exceed 0.75 of the depth of the member, or 24 in., whichever is smaller, if Vs ≤ 4 fc′ bw d (ACI- 9.7.6.2.2) d = distance from the extreme compressive fiber to the centroid of the prestressed reinforcement (but need not be taken less than 0.8h)

EXAMPLE 20.11.1 Determine the maximum spacing of #3 U stirrups at a point 4 ft from the support on the beam of Example 20.10.1 (Fig. 20.10.6). The prestressing steel has an area of 2.30 sq in. and a tensile strength fpu = 250,000 psi. The yield strength of the stirrup reinforcement fyt = 60,000 psi, and the concrete (normal weight) has fc′ = 5000 psi. The service live load is 1.50 kips/​ft, and the dead load is 0.675 kip/​ft. SOLUTION The inclined cracking strength Vc = Vci as determined in Example 20.10.1 is Vci (controls) = 97.2 kips (more elaborate precedure) At 4 ft from centerline of support, Vu = [1.2(0.675) + 1.6(1.50)](20 − 4) = 3.21(20 − 4) = 51.4 kips

0.75(97.2)   φV Vu = 51.4 kips >  c = = 36 kips 2 2  



unless tests are made to justify its omission, a minimum amount of shear reinforcement must be provided. Thus using Eq. (5.10.8) min

Av b 50bw s = 0.75 fc′ w ≥ s f yt f yt

min

Av    20  = 0.75 5000  = 0.018 in.   60, 000  s  



  50(20) ≥ = 0.017 in.   60, 000 min



Av = 0.018 in. s

or, alternatively, using the more elaborate formula, Eq. (20.11.1), min

Av Aps  f pu   1  d =    s 80  f yt   d  bw

A 2.30  250   1  24 min v = = 0.0055 in.    s 80  60   24  20

(Continued)

83



20.12  DEVELOPMENT OF REINFORCEMENT

883

Example 20.11.1 (Continued) Using the smaller of 0.018 in. and 0.0055 in., the maximum spacing s for #3 U stirrups is

s=

2(0.11) = 40.0 in. 0.0055

but, according to ACI-​9.7.6.2.2, the spacing may not exceed 0.75h = 0.75(30) = 22.5 in. or 24 in., whichever is smaller. Thus #3 stirrups could be placed no farther apart than 22.5 in.

20.12 DEVELOPMENT OF REINFORCEMENT Following the basic concepts established for nonprestressed reinforced concrete in Chapter 6, development of reinforcement must also be considered in prestressed members. The prestressing force must be transferred into the concrete by bond (interaction between concrete and steel strand) in the pretensioned beam, and the length required to accomplish this is called the “transfer length.” This, of course, occurs in end regions and, in effect, anchors the tendons. The mechanism of transfer is summarized by Zia and Mostafa [20.14]. The high stress in the pretensioned tendon must, on cutting of the tendon, be transferred to the concrete so that equilibrium is achieved. The situation in the transfer zone is quite different from that in the anchorage zones of nonprestressed concrete. In prestressed concrete, the wires or strands are pretensioned to a high level of stress, thus initially reducing the wire diameter prior to the placing of the concrete. Once the concrete has cured and the wires have been cut at the ends, the wires tend to shorten and correspondingly increase in diameter. Thus a compression between the wires and concrete is produced. The stress in the cut wire must increase from zero at the free end to the prestress value at a certain distance from the free end. This distance is known as the “transfer length.” During the accomplishment of the transfer, probably in the first few inches from the free end, it is solely the friction that is developed as the slipping wires compress against the concrete. Farther from the end, adhesive bond—​that is, slip resistance—​certainly contributes to the transfer. The transfer length Lt typically is about 50 diameters for a strand, and about 110 to 120 diameters for a single wire [20.14]. These values assume a clean strand (or wire) surface, a gradual release of the prestressing force to the concrete, and a steel stress of 140 to 150 ksi after transfer. For strands with a slightly rusted surface, the transfer distance is certainly less, and for a sudden release of stress such as can occur by burning the strand, the transfer length may easily be 20% greater. The transfer length does not seem to be affected by variation in concrete strength [20.15]. The importance of the transfer length depends on the member under consideration. It seems to be of little importance except on short members and in situations where flexural cracking may occur in the transfer zone.

Development Length for Prestressing Strand Exactly as for nonprestressed reinforcement, the tension in the prestressing steel necessary to achieve nominal flexural strength Mn must be developed, typically by embedment length. The purpose is to prevent general slip prior to achieving the necessary moment strength Mn; thus the steel stress must increase from the effective prestress value fse to the value fps used in computing the moment strength. The net anchorage length available to accommodate a change in the tensile force is the distance to the end of the strand from the section in question, less the transfer length Lt.

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The following empirical relationship was established based on data by Hanson and Kaar [20.15],

( f ps − fse ) in ksi =

Ld − L t db

(20.12.1)



where db is the nominal strand diameter. Thus,

f ps ( ksi) ≤ fse (ksi) +

L d − Lt db

(20.12.2)

Next, it is necessary to obtain an expression for Lt /​db. Applying the concept used in deriving Eq. (6.2.1), that is, the tensile force Aps fse in the strand must be transmitted to the surrounding concrete by a “stress” us acting around the perimeter πdb over the embedment length Lt, Aps fse = us πdb Lt

 π db 2   4  fse = us πdb Lt  

(20.12.3)

Since the actual strand (3-​or 7-​wire) properties differ from those based on the nominal diameter, a correction must be applied. The actual circumference is taken as 4 / 3 πdb ; and the actual area Aps as 0.725πdb2/​4.* Thus

 πd 2   4 0.725  b  fse = us   πdb Lt  3  4 

(20.12.4)

Lt 2.175 fse = db 16us

(20.12.5)

If the average bond stress us in the transfer zone is taken as 400 psi for clean strands [20.15],

Lt 2.175 1 fse = fse = db 16(0.4) 2.94

say,

1 fse 3

(20.12.6)

where fse is measured in ksi. Substituting Eq. (20.12.6) in Eq. (20.12.2) gives

f ps ≤ fse +

Ld 1 − fse db 3

and solving for Ld, the necessary development length to the end of the strand,

2   Ld =  f ps − fse  db  3 

(20.12.7)

which is equivalent to ACI Formula (25.4.8.1) for 7-​wire pretensioning strands. Note that the expression in parentheses uses stresses fps and fse in ksi units, but the resulting expression ( f ps − 2 / 3 fse ) is treated as a dimensionless constant. Investigation may be restricted to cross sections nearest each end of the member that are required to develop their nominal strengths Mn.

* The relationship between the nominal diameter and the actual area used herein corresponds to that at the time the ACI Code provisions for transfer length were developed.

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EXAMPLE 20.12.1 Investigate the development of reinforcement for the beam of Example  20.10.1 (see Fig. 20.10.6). Assume fps =221 ksi as computed in Example 20.8.1. The effective prestress, after losses, may be assumed to be 0.8(175) = 140 ksi. SOLUTION Since only at midspan is the maximum moment strength Mn required, the total embedment from midspan must exceed the development strength Ld,



2   Ld = 221 − (140) db 3  

Ld = [(221 − (93.3)](0.5) = 64 in. < 240 in n.

OK

20.13 PROPORTIONING OF CROSS ​S ECTIONS FOR FLEXURE WHEN NO TENSION IS PERMITTED This section is intended to give further insight into the variables involved by discussing the proportioning of the cross section. All cross sections used here consist of rectangular flanges and web, though in practical design 90° junctions between flanges and web are avoided because of forming difficulties. This brief treatment presents only one case: that of no tension permitted, either at the initial condition (at transfer of prestress before losses) or at the final condition (full dead and live load after losses). Thus the desired stress distributions are shown in Fig. 20.13.1. It can be shown that on any cross section having an axis of symmetry, there are two points on the axis of symmetry, known as the “kern points,” that are located at distances kb and kt from the centroid, as shown in Fig. 20.13.1. Each kern point represents the farthest distance from the centroid at which a resultant force can act without inducing a stress of opposite sign at the extreme fiber in the opposite direction from the centroid. This means that the stress distribution varies from zero at the top to maximum at the bottom, when the resultant force acts on the lower kern point; or it varies from zero at the bottom to maximum at the top when the resultant force acts on the upper kern point. Referring to Fig. 20.13.1, and using Eq. (20.6.1) with the internal force concept,

f =

C Ckb yt − = 0 A I

(20.13.1a)

f =

C Ckt yb − = 0 A I

(20.13.1b)

kb =

I r2 = Ayt yt

(20.13.2a)

kt =

I r2 = Ayb yb

(20.13.2b)

and

Solving Eqs. (20.13.1) for kb and kt, and

in which r is the radius of gyration of the cross section.

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Te h Ac yb

Upper kern point

yt

Te/Ac

T0/Ac

h Lower kern point

C = Te

kt

CG kb

(arm)D+L

C = T0

e

yb

(arm)D T0

Aps

Te = ηT0

T0 h Ac yt

Section

Initial condition (stress distribution at transfer)

Final condition (stress distribution under full load after losses)

Figure 20.13.1  Stress distributions when no tension is permitted—​small girder moment.

The reader may recall from Section 20.6 that as the load applied to a prestressed concrete beam changes, the position of the internal force C must also change; that is, the moment arm (see Fig. 20.13.1) measured from the position of the prestressing steel must increase as the load increases. Thus when the moment MD due to dead load of the girder is acting at the initial condition (immediately after transfer of prestress), the moment arm of the internal couple is (arm)D; and when moment due to dead plus live load is acting, the moment arm is (arm)D+L. To achieve the triangular distribution at the initial condition, the prestressing steel must have its centroid below the lower kern point by the amount (arm)D  =  MD /​T0 (see Fig. 20.13.1). If the girder moment MD is small, it may be possible and practical to position the steel at e = kb + (arm)D below the centroid of the section. If the moment MD is large, the distance [kb + (arm)D] may extend so far below the centroid of the section that it is too close to the bottom for proper cover, if not outside the section. Thus for larger girder weight (large MD), the optimum condition of triangular stress distribution at the initial condition may be impossible to achieve.

Preliminary Design for Small Girder Moment When the girder moment MD is small (say, MD representing 0.2 or less of the total, MD + ML ), the (arm)D will be small and C can probably be located at the lower kern point; that is, the steel will be located at e = kb + MD / ​T0 from the centroid of the section. Observing Fig. 20.13.1 again, (arm ) D + L − (arm ) D = kt + kb

MD + ML MD − = kt + kb Te T0

If MD is small, MD /​T0 in the above equation may be assumed to be MD /​Te, then

ML ≈ kt + kb Te

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or Te ≈



ML kt − kb

(20.13.3)

The maximum stress in Fig.  20.13.1 may be obtained from the stress at the centroid (which is T/​Ac) by the linear relationship. Assuming kt + kb ≈ 0.5h, Eq. (20.13.3) becomes Te ≈



ML 0.5h

(20.13.4)

If the section is symmetrical, the stress at the centroid will be half of the maximum; thus the required area Ac of the section would be, for the initial condition at transfer,

required Ac =

T0 0.50 fic

(20.13.5)

and for the final condition of dead plus live load after losses,

required Ac =

Te 0.50 f fc

(20.13.6)

For the preliminary selection of cross section in case of small girder moment MD, the following steps may be followed: 1. Select the overall depth h of the section. This is somewhat arbitrary but in the absence of other limitations, the guidelines of Lin and Burns [20.6] may be followed: (a) Use 70% of the depth that would be used for nonprestressed reinforced concrete construction. (b) For slabs having light loading, h ≥ 1/​55 of the span L. (c) For slabs having heavy loading, h ≥ L  /​35. (d) On bridges, h ranges between L /​15 and L  /​25. (e) As a rule of thumb, h (inches) of beams can be approximated by the empirical formula, 1.5 M (ft-kips)  to 2.0 M . 2. Compute the approximate Te from Eq. (20.13.4). 3. Determine the approximate Ac from Eq. (20.13.6). 4. Proportion a symmetrical I-​shaped section. 5. With the preliminary section, compute the section properties and locate the desired distance e of the steel centroid from the section centroid (CG),

e = kb + (arm ) D = kb +

MD T0

(20.13.7)

where T0 = Te /​η and η is the proportion of initial prestress remaining after losses. 6. If the steel can be located at the desired e, then Te is more correctly determined:

Te =

MD + ML MD + ML = (arm) D + L e + kt

(20.13.8)

Then T0 = Te /​η and a new value of e is established; the iterative process is repeated until the desired accuracy has been obtained.

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7. Equations (20.13.5) and (20.13.6) are then used to determine the required Ac. When the equations give significantly different requirements, the section may be changed to become somewhat unsymmetrical with respect to the centroidal axis. In such a case, the average stress is not half of the maximum. In general, from Fig. 20.13.1,

required Ac =

T0 h fic yt

(20.13.9)



required Ac =

Te h f fc yb

(20.13.10)

The minimum area Ac is obtained when Eqs. (20.13.9) and (20.13.10) give the same result.

Preliminary Design for Large Girder Moment When the girder moment MD exceeds about 0.2 to 0.3 of the total moment MD + ML, the (arm)D will be too large to permit the steel distance e to be at kb + MD /​T0 from the centroid of the section. Thus the initial stress distribution at transfer cannot be triangular but instead will be trapezoidal (see Fig. 20.13.2). The final condition, which can still give a triangular stress distribution, will probably govern. Thus Te =



MD + ML e + kt

(20.13.11)

When the final condition controls, more of the area Ac should be located at the top where the highest stress occurs. Thus an unsymmetrical section is indicated—​for instance, a T-​shaped section. As an approximation, Lin and Burns [20.6] suggest (e + kt) ≈ 0.65h for use in Eq. (20.13.11). Generally e + kt will vary from 0.3h to 0.8h, with the average about 0.65h. For preliminary design, Eq. (20.13.11) then becomes Te ≈



MD + ML 0.65h

(20.13.12)

The required area Ac would then be

required Ac =

Te MD + ML = avg fc 0.65h(avg fc )

(20.13.13)

where avg fc is the compressive stress at the centroid of the section. For the unsymmetrical section, yb /​h in Eq. (20.13.10) will be greater than 0.5, say, 0.6. Equation (20.13.13) for design would then be

required Ac =

MD + ML 2.6( M D + M L ) = 0.65h(0.6 ff c ) h ff c

(20.13.14)

For the preliminary selection of cross section in case of large girder moment, the following steps may be followed: 1. When MD /​(MD + ML) > 0.2 to 0.3, establish the overall depth according to Step 1 for small girder moment and use Eq. (20.13.14) to estimate Ac. 2. Proportion an unsymmetrical section; a T-​section may be a practical choice.

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20.13  PROPORTIONING OF CROSS SECTIONS FOR FLEXURE

3. With the preliminary section, compute the section properties and establish the distance e from the centroid of the section to the centroid of the prestressing steel. For large girder moment, one will find  M  e <  kb + D  T0  



If e can be equal to kb + MD /​T0, then the procedure for the small girder moment case is to be followed. 4. With the steel located, Te can be determined using Eq. (20.13.8). From that, T0 = Te /​η. 5. The required area Ac based on the final condition is then determined using Eq. (20.13.10). 6. Check the required area based on the initial condition. Referring to Fig. 20.13.2, a trapezoidal stress distribution should occur. The maximum compressive stress f is, using Eq. (20.6.1) with the internal force concept, f =



C Cec y + A I

(20.13.15)

Since C = T0, A = Ac, I = Ac r2, and y = yb, f =

=

T0 T0 ec yb + ≤ fic Ac Ac r 2 T0  e − M D /T0  1 +  ≤ fic Ac  kt 

(20.13.16)

The required area Ac based on the initial condition is required Ac =



T0  e − M D /T0   1 + fic  kt 

(20.13.17)

Again the minimum area Ac will be obtained when Eqs. (20.13.10) and (20.13.17) give the same result.

ec = e – MD/T0

C = Te

yt

kt

CG h yb

kb

Te Ac

C = T0

e

(arm)D+L

(arm)D

Aps

Te

T0 f Initial condition

Final condition

Figure 20.13.2  Stress distributions when no tension is permitted—​large girder moment.

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EXAMPLE 20.13.1 Design a cross section for a 30-​in.-​deep girder whose girder moment MD = 45 ft-​kips. The live load moment to be carried is 300 ft-​kips. The initial prestress fsi = 175,000 psi; assume 20% losses. The allowable service load stresses are fit = 0, fic = 2400 psi, fft = 0, and ffc = 2250 psi. Omit checks of moment strength, cracking moment, shear strength, and development of reinforcement. SOLUTION (a) Preliminary design. The girder moment as a percent of the total moment is MD 45 = = 0.13 < 0.2 M D + M L 45 + 300



Approach as a small girder moment design. Using Eqs. (20.13.4) and (20.13.6),



Te ≈

ML 300(12) = = 240 kips 0.5h 0.5(30)

T0 =

Te 240 = = 300 kips η 0.8

required Ac =



T0 300 = = 250 sq in. 0.5 fic 0.5(2.40)

Since ffc /​fic = 2.25/​2.4 = 0.94 is greater than Te /​T0 = 0.8, the equation based on the initial condition controls if the section is symmetrical. Though a slightly unsymmetrical section (with the centroid below the middepth in this case) would give the section of minimum area, the procedure is illustrated using a symmetrical one. Try a 30-​in.-​deep section with flanges 5 × 17 in. and a 4-​in.-​thick web, Ac = 250 sq in. (Fig. 20.13.3). (b) Determine the section properties. Area (sq in.) 17 × 30 rectangle 13 × 20 sides



r2 =

I = 118.3 sq in., Ac

I (in.4)

510 –​260

38,250 –​8677

250

29,583

kt = kb =

r 2 r 2 118.3 = = = 7.89 in.. yb yt 15

(c) Locate centroid of prestressed reinforcement. Using approximate Te, find approximate T0. T0 ≈

Te 240 = = 300 kips η 0.8

M 45(12) desired e = kb + D = 7.89 + = 9.69 in. T0 300



(Continued)

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Example 20.13.1 (Continued) The available distance is 15 in., less appropriate cover. Thus e  =  9.69 in. would be acceptable. Recalculate Te using Eq. (20.13.8), Te ≈

M D + M L (45 + 300)12 = = 235 kips e + kt 9.69 + 7.89

235 = 294 kips 0.8 45(12) revised e = 7.89 + = 9.73 in. 294

revised T0 =





This is in close agreement with the previous value of 9.69 in. If the first estimate had been farther off, additional iterations might have been needed. (d) Check whether the area Ac is adequate. Using Eqs. (20.13.9) and (20.13.10).

required Ac (initial condition) =

T0 h 294(30) = = 245 sq in. fic yt 2.4(15)



required Ac (final condition) =

Te h 235(30) = = 209 sq in. f f c yb 2.25(15)

The section is adequate. Use 5 × 17 in. flanges with a 4-​in. web (Ac = 250 sq in.). If made slightly unsymmetrical, the area could be reduced to somewhere between 209 and 245 sq in. A final check of initial and final stresses should be made, as was done in Section 20.6, but the illustration is omitted here. 17” 5”

30” 4” 5” 17”

Figure 20.13.3  Section for design example 20.13.1.

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EXAMPLE 20.13.2 Redesign the cross section of Example  20.13.1 for MD  =  200 ft-​ kips and ML  = 145 ft-​kips. Note that the total MD + ML = 345 ft-​kips is the same as in Example 12.13.1. SOLUTION (a) Preliminary design. The girder moment as a percentage of the total moment is MD 200 = = 0.58 > 0.2 to 0.3 M D + M L 345



Approach as a large-​girder moment design. Using Eq. (20.13.14), required Ac =



2.6( M D + M L ) 2.6(345)12 = = 159 sq in. h f fc 30(2.25)

An unsymmetrical shape is to be selected; try flanges 5 × 14 in. and 5 × 8 in. with a 4-​in. web having Ac = 190 sq in. (Fig. 20.13.4). The larger the ratio of MD to (MD + ML), the larger may be the numerator coefficient (2.6, used above); thus an area somewhat larger than indicated by the formula was used. 14” 5”

yt = 13.03” 30” 4”

yb = 16.97”

5” 8”

Figure 20.13.4  Section for design example 20.13.2.

(b) Determine section properties. Referring to Fig. 20.13.4, first locate the centroid of the area measured from the top.

5 × 10 top flange projection I0 5 × 4 bottom flange projection I0 4 × 30 web (full depth) I0

Area, Ac (sq in.)

Arm, y (in.)

50

2.5

125

20

27.5

550

120 190

15

Ac y (in.3)

1,800 2475

I (in.4) 313 104 15,125 42 27,000 9,000 51,584 (Continued)

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Example 20.13.2 (Continued) yt = y =

∑ A y = 2475 = 13.03 in. ∑ A 190

I 0 = I − Ac y 2 = 51, 584 − 190(13.03)2 = 19, 300 in.4

I0 = 101.7 sq in. Ac

r2 =



2

kt =

r 101.7 = = 5.99 in. yb 16.97

kb =

r 2 101.7 = = 7.81 in. yt 13.03

(c) Locate centroid of prestressed reinforcement. Assume that with adequate cover the steel may be centered 4 in. from the bottom of the section. Then

e = yb – 4 = 16.97 – 4 = 12.97 in.

and using Eq. (20.13.8), Te =



MD + ML 345(12) = = 218 kips e + kt 12.97 + 5.99

then T0 =



218 = 273 kips 0.8

Check:

kb +

MD 200(12) = 7.81 + = 7.81 + 8.79 = 16.60 in. > e T0 273

This shows that the tendons cannot be located far enough from the centroid of the section to give a triangular stress distribution at the initial condition. (d) Check whether the area Ac is adequate. Using Eqs. (20.13.10) and (20.13.17),

required Ac (final condition) =

required Ac (initial condition) = =

Te h 218(30) = = 171 sq in. f fc yb 2.25(16.97) T0 fic

 e − M D / T0  1 +  kt  

273  12.97 − 8.79  1 + 193 = sq in.  2.4  5.99 

This is close enough but shows that the initial condition governs in this case, the reason being that area has been shifted from the bottom of the section to the top, where it is needed for the final condition. The minimum area section for this case would be slightly more symmetrical than the chosen one. To verify the result, make a final check of stresses for the initial and final conditions (see Section 20.6). The general line of reasoning presented here may also be used when tension is permitted at initial or final conditions, or both. Lin and Burns [20.6] and Naaman [20.5] provide a detailed treatment of design of sections when tension is permitted.

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20.14 ADDITIONAL TOPICS Many other topics have been omitted from this introductory treatment of prestressed concrete. Topics such as the practical design approaches, use of I-​shaped and nonsymmetrical sections, prestressing of continuous members, stresses in end blocks, partial prestressing, deflections, composite construction, and other specific applications are adequately and extensively treated in textbooks devoted entirely to the subject [20.2–​20.7].

SELECTED REFERENCES   20.1 Charles C. Zollman. “Reflections on the Beginnings of Prestressed Concrete in America,” PCI Journal, Part 1, 23, May/​June 1978, 22–​48, Part 2, 23, July/​August 1978, 29–​63; Part 9, 25, January/​February 1980, 124–​145, March/​April 1980, 94–​117, May/​June 1980, 123–​152.   20.2 Michael P.  Collins and Denis Mitchell. Prestressed Concrete Structures. Canada:  Response Publications, 1997, 766 pp.   20.3 James R. Libby. Modern Prestressed Concrete, Design Principles and Construction Methods (4th ed.). Princeton, NJ: Van Nostrand Reinhold, 1990, 872 pp.   20.4 Edward G. Nawy. Prestressed Concrete, A Fundamental Approach (2nd ed.). Upper Saddle River, NJ: Prentice-​Hall, 1996, 789 pp.   20.5 Antoine E. Naaman. Prestressed Concrete Analysis and Design—​Fundamentals. Ann Arbor; MI: Techno Press 3000, 2012, 1173 pp.   20.6 T. Y. Lin and Ned H. Burns. Design of Prestressed Concrete Structures (3rd ed.). New York: John Wiley & Sons, 1981, 646 pp.   20.7 Arthur H.  Nilson. Design of Prestressed Concrete. New  York:  John Wiley & Sons, 1978, 526 pp.   20.8 Yves Guyon. Prestressed Concrete, Vols. 1 and 2. New York: JohnWiley & Sons, 1960.   20.9 Gustave Magnel. Prestressed Concrete (2nd ed.). London: Concrete Publications, 1950. 20.10 Paul Zia, H. Kent Preston, Norman L. Scott, and Edwin B. Workman. “Estimating Prestress Losses,” Concrete International, 1, June 1979, 32–​38. 20.11 PCI Committee on Prestress Losses. “Recommendations for Estimating Prestress Loss,” PCI Journal, 20, July–​August 1975, 43–​75. 20.12 J. R. Janney, E. Hognestad, and D. McHenry. “Ultimate Flexural Strength of Prestressed and Conventionally Reinforced Concrete Beams,” ACI Journal, Proceedings, 52, January 1956, 601–​620. 20.13 M.  A. Sozen and N.  M. Hawkins. Discussion of “Shear and Diagonal Tension Report,” Report of ACI-​ ASCE Committee 326, ACI Journal, Proceedings, 59, September 1962, 1341–​1347. 20.14 Paul Zia and Talat Mostafa. “Development Length of Prestressing Strands,” PCI Journal, 22, September/​October 1977, 54–​65. 20.15 Norman W.  Hanson and Paul H.  Kaar. “Flexural Bond Tests of Pretensioned Prestressed Beams,” ACI Journal, Proceedings, 55, January 1955, 783–​802. 20.16 M.  H. Harajli and M.  Y. Kanj. “Ultimate Flexural Strength of Concrete Members Prestressed with Unbonded Tendons,” ACI Structural Journal, 88, November–​December 1991, 663–​673. 20.17 S. V. Krishna Mohan Rao and Walter H. Dilger. “Control of Flexural Crack Width in Cracked Prestressed Concrete Members,” ACI Structural Journal, 89, March–​April 1992, 127–​138. 20.18 A.  Alshegeir and J.  A. Ramirez. “Strut-​Tie Approach in Pretensioned Deep Beams,” ACI Structural Journal, 89, May–​June 1992, 296–​304. 20.19 Ravi K.  Devalapura and Maher K.  Tadros. “Critical Assessment of ACI 318 Eq. (18-​3) for Prestressing Steel Stress at Ultimate Flexure,” ACI Structural Journal, 89, September–​October 1992, 538–​546. 20.20 Julio A.  Ramirez. “Strut-​Tie Design of Pretensioned Concrete Members,” ACI Structural Journal, 91, September–​October 1994, 572–​578. 20.21 Post-​Tensioning Institute. Post-​Tensioning Manual (6th ed.) PTI TAB.1-​06, Phoenix, AZ, 2006.

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PROBLEMS

PROBLEMS All problems are to be worked in accordance with the ACI Code, assuming Class U members and using the allowable concrete stresses in ACI-​24.5 unless otherwise indicated. Flexural strength-​related provisions are to be in accordance with ACI-​20.3.2 and ACI-​22.3, as indicated by the instructor. 20.1 The rectangular beam of the figure for Problem 20.1 contains pretensioned steel with an initial tensile stress of 160 ksi (  fpu  =  250 ksi; low-​ relaxation strand having fpy  =  0.90 fpu). The concrete has fci′ = 4000 psi and fc′ = 6000 psi (n = 6.5). The beam is on a simple span of 35 ft. (a) Determine the concrete stresses at top and bottom, and the steel stress at transfer immediately after the wires are cut at the ends. (b) Recompute the stresses in part (a) after a 20% loss in prestress. What is the maximum service live load that can be superimposed on the beam? Consider only the section of maximum bending moment; omit consideration of flexural strength Mn.

Aps = 0.8 sq in. 12”

18”

2”

Problems 20.1, 20.2 and 20.9

20.2 Based on the midspan cross section of the figure for Problem 20.1, investigate whether it is pos­ sible to increase the live load moment capacity by either or both of the following: (a)  Increasing the initial prestress above 160 ksi. (b)  Decreasing the eccentricity.

Assume a 20% loss of initial prestress. Determine the maximum service live load capacity possible by adjusting the prestress or the eccentricity or both, but still not violating the ACI Code limitations. Omit consideration of flexural strength Mn. 20.3 The rectangular section of the figure for Problem 20.3 has been pretensioned by a force of 300 kips after all losses. If the concrete has fci′ = 4000 psi and fc′ = 6000 psi (n  =  6.5), what uniformly

distributed live load may be safely carried on a 40-​ft simple span? Omit consideration of flexural strength Mn.

5”

24”

Aps = 2.5 sq in. 12”

Problem 20.3 

20.4 For the live load determined in Problem 20.1, determine the maximum permissible eccentricity of the prestressing tendons at the 1 4 point of the span to satisfy service load allowable stress limits. Omit consideration of flexural strength Mn. 20.5 For the live load determined in Problem 20.3, determine the maximum permissible eccentricity of the prestressing tendons at the 1 8 point of the span to satisfy service load allowable stress limits. Omit consideration of flexural strength Mn. 20.6 A  straight pretensioned member 35 ft long is 18 in. square in cross section. It is concentrically prestressed with 2.24 sq in. of high tensile strength steel wire. The wires are stressed originally to 170 ksi and are anchored to end bulkheads. Use both the approximate and the “exact” methods to calculate the loss and percentage of loss of prestress in the wires due to elastic shortening of the concrete at transfer. Use fci′ = 6000 psi (n = 6.5). 20.7  An 18-​ in. square concrete member is post-​ tensioned by four cables, each with an area of 0.56 sq in. These cables are stressed one after another, each to a stress of 170 ksi. Without taking any account of the eccentricity of the cables, compute the loss and percentage of loss of prestress in each cable due to the elastic shortening of the concrete. Compute the average loss of prestress. Assume n = 6.5.

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R = 200 ft

20.8  The symmetrical double cantilever beam shown U stirrups for the beam of Problem 20.3 if in the figure for Problem 20.8 is to be post-​ the maximum service load computed in that tensioned by a single cable ABCDE. The cable problem is acting. Use and compare both the consists of 12 wires, each of 0.20 in. diameter, alternate (ACI-​ 22.5.8.2) and more accurate and is to be prestressed simultaneously from (ACI-​22.5.8.3.1 and ACI-​22.5.8.3.2) proceboth ends of the member. It is desired that the dures in the ACI Code. minimum stress in the cable immediately after 20.12  Design of section for small dead load moment. stressing and before any creep or shrinkage (a)  Make a preliminary design (use rectanlosses take place be 145 ksi. The cable is such gular flanges and a web) for a section of that the friction constant µ = 0.50 and the woba prestressed beam to resist a total bendble effect K = 0.0010. Determine the steel stress ing moment of 960 ft-​kips, assuming that at the jack, the percentage of friction losses, and the moment due to the girder weight is the extension that will be required at each jack. 70 ft-​kips. The overall depth of the section Solve by each of the following methods. is to be 42 in. and the effective prestress fse (a) Neglecting variation in tension throughout in the steel is 136 ksi. In selecting a section the length, using Eq. (20.7.22), Ps = Px (1 + assume the minimum thickness of compoµα + KLx). nents (flanges or web) is 5 in. (b)  Neglecting variation in tension at every (b) Make the final design for the preliminary point along the length of the curve but consection you selected for part (a), revising sider variation from segment to segment, as you find necessary, with the objective of using Eq. (20.7.22). obtaining a minimum cross-​sectional area. (c)  Using the “exact” expression, Eq. (20.7.20). The selected cross-​ sectional area should 20.9   Determine the nominal moment strength Mn not exceed 460 sq in.; however, the “best” and the cracking moment Mcr for the pretendesign will be presumed to be the one havsioned bonded section of the figure for Problem ing the least cross-​sectional area. 20.1. The concrete has fc′ = 6000 psi and the (c) Make a check of stresses at initial (transsteel has fpu  =  250 ksi (stress-​relieved strand). fer) and final conditions. Use the following Assume the average stress–​ strain curve of control stresses: Fig. 20.8.1 is to be used for the steel. fic = 2400 psi ffc = 2250 psi fsi = 160 ksi 20.10 Assuming that no special tests are to be made, fit = 0 psi fft = 0 psi fse = 136 ksi determine the number and spacing for #3 U stir- 20.13 Design of section for large load moment. The rups for the beam of Problem 20.1 if a service load data are the same as Problem 20.12, except that of 0.48 kip/​ft is acting. Use the alternate procedure the moment due to the girder weight is 650 ft-​ for Vc of ACI-​22.5.8.2, as well as the more accurate kips, instead of 70 ft-​kips (total moment is the procedure of ACI-​22.5.8.3.1 and ACI-​22.5.8.3.2. same); the minimum thickness for components 20.11 Assuming that no special tests are to be made, (flanges or web) is 4 in.; and the cross-​sectional determine the number and spacing for #3 area should not exceed 340 sq in.

A

3’– 0”

B 1’– 6” 30’– 0”

R = 100 ft

1’– 6”

10’– 0”

Symbolic about midspan

C

D

3’– 0” 1’– 6”

20’– 0” 100’– 0”

Problem 20.8 

E

20’– 0”

CHAPTER 21 COMPOSITE MEMBERS AND CONNECTIONS

21.1 INTRODUCTION In the context of reinforced concrete construction, the term composite members refers to those constructed with two or more materials, at least one of them being concrete (with or without steel bars), interconnected in a way that allows transfer of stresses between the different materials. Based on the materials used, two broad categories of concrete composite members can be defined: concrete-​concrete composite members and concrete-​steel composite members. Concrete-​concrete composite members, sometimes simply referred to as concrete composite members, are those made of concrete cast at different times. This practice is common in precast girder construction, in which a cast-​in-​place concrete slab or topping is added on site once the precast girder is in place. Concrete-​steel composite members, on the other hand, combine concrete (with or without reinforcing bars) and structural steel shapes. These are commonly found in floor systems of steel structures [Fig. 22.1.1(a)], in which a concrete slab, typically supported on a metal deck, is constructed to act compositely with steel beams. Concrete-​steel composite members are also used as columns, where either a structural shape is embedded in reinforced concrete [Fig. 22.1.1(b)] or a steel tube is filled with concrete [Fig. 22.1.1(c)]. This chapter deals with the behavior and design of concrete composite flexural members, as well as composite columns that consist of either concrete-​encased steel shapes or concrete-​filled steel tubes [Fig. 22.1.1(b) and 22.1.1(c)]. Connections between composite columns and steel beams are also discussed, as the force transfer mechanisms differ from those in reinforced concrete or steel construction. Design of composite floor systems consisting of steel beams and a concrete deck is well covered in the AISC Steel Construction Manual [21.1] and in other textbooks [21.30], and is therefore not included here.

21.2 COMPOSITE ACTION Before discussing the behavior and design of composite members, it is necessary to understand the fundamentals of composite action and the forces developed at the interface between the various materials in the composite member. Consider a simply supported beam made of two different materials, as shown in Fig. 21.2.1(a). Three cases are considered with respect to the interface between the two materials: frictionless interface, “rough” interface, and “slip-​free” interface.

89

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C hapter   2 1     C omposite M embers and C onnections Shear studs

Shear studs

Concrete

Bars

Hollow steel section

Hollow steel section

Concrete

Concrete

Metal deck

Steel section

Steel section

Ties (a) Composite beam

(b) Concrete-encased steel composite column

(c) Concrete-filled tube

Figure 21.1.1  Common concrete-steel composite members. b2 Material 2 Material 1 b1 Cross section

(a) Beam made of two materials Material 2

M2 M1

Material 1

Strains

(b) Beam with frictionless interface (no composite action)

C2

Material 2

T1 Material 1

M2 M1

Strains

(c) Beam with “roughened” interface (partial composite action)

Material 2

Material 1

C2

T1

M2 M1

Strains

(d) Beam with slip-free interface (full composite action)

Figure 21.2.1  Composite action in flexural members.

Under the action of transverse loads, the member with frictionless interface will behave as two beams in parallel, with relative sliding free to occur along the interface between the two materials [Fig. 21.2.1(b)]. Thus, only normal stresses develop between the two material components. Even though two materials are used in this structural element, the member is considered to be noncomposite because the two material components work separately to

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resist the applied loading. In this case, the moment acting on the member is equal to the summation of the moments of the individual components [M1 and M2 in Fig. 22.2.1(b)]. When some roughness is introduced at the interface between the two materials, sliding may still occur, but not freely, leading to the transfer of force parallel to the interface (shear). In this case, the moment is resisted by a combination of bending in each material component [M1 and M2 in Fig. 22.2.1(c)], as well as by a couple created by the transfer of shear between the two materials [T1 and C2 in Fig. 21.2.1(c)]. This member is said to be partially composite. The case depicted in Fig. 21.2.1(d) corresponds to an interface at which relative slip is prevented: the section will behave as a single unit, with continuity of normal strains across the depth of the section. The moment carried by the composite beam can be calculated as for the case of partially composite action, in which each component carries a moment and an axial force. The amount of shear transferred along the interface is that required for equilibrium when the two material components act as a single unit. These members are referred to as fully composite. A slip-​free interface in concrete composite members is not feasible in practice unless a “rough” interface is provided and the interface shear stresses are relatively low, up to approximately 200 to 250 psi [21.2, 21.3]. The requirement for full composite action, therefore, is typically based on the shear transfer along the interface between the material components, not on achieving a slip-​free interface. As will be discussed later, interface slip is often ignored in the calculation of flexural strength of composite members. Its effect on member stiffness, however, may be important enough in some cases to require explicit consideration.

Interface Shear Calculation of shear developed at the interface between two material components behaving as a unit can be done using statics. Let us consider again the beam shown in Fig. 21.2.1(a). Transverse forces are applied on the composite beam only (i.e., any loads acting separately on each material component are not considered). For simplicity, let us transform the composite beam into a beam made of the material used in the bottom part of the beam (material 1). This can be done by changing the width b2 to b2 (E2 /​E1), where E1 and E2 are the moduli for materials 1 and 2, respectively. Taking a slice of thickness dx at a distance x from the left support [Fig. 21.2.2(a)] leads to the free-​body diagram shown in Fig. 21.2.2(b). Only normal stresses are shown acting on the cut faces, as these are the only stresses needed in the calculation of interface shear. For the case of linear elastic behavior, normal stresses are assumed to increase linearly with the distance from the centroid of the transformed section, y. In the presence of shear, the moments at the two sections, M(x) and M(x + dx), will differ, and so will the magnitudes of the normal stresses. By taking a horizontal cut along the interface between the two material components (Fig. 21.2.2(b), it can be seen that a force Vh(x) is required to balance the resultant normal forces on the two cut faces, F(x) and F(x + dx). These forces can be calculated as follows [Fig. 21.2.2(b)], F ( x) = ∫





F ( x + dx ) = ∫

y2 y1

y2 y1

M ( x) y b( y) dy I tr

y2 [ M ( x ) + dM ( x )] y M ( x + dx ) y b( y) dy = ∫ b( y) dy y1 I tr I tr

Vh ( x ) = F ( x + dx ) − F ( x ) =

dM ( x ) y2 yb( y) dy I tr ∫y1

(21.2.1) (21.2.2)

(21.2.3)

where y is the distance from the centroid of the transformed section to the location where normal stress is calculated, b( y) is the section width at a distance y from the centroid, and Itr is the moment of inertia of the transformed section. The shear force Vh(x) can be expressed in terms of a shear stress vh(x) acting on a surface of area bv dx, where bv is the width of the

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contact surface. Also, the integral in Eq. (21.2.3) represents the first moment of the area above the interface with respect to the centroid of the transformed section, Q1. Thus, vh ( x )bv dx =



dM ( x ) y2 dM ( x ) yb( y) dy = Q1 I tr ∫y1 I tr

(21.2.4)

Solving for vh(x), vh ( x ) =



V ( x )Q1 dM ( x ) Q1 = dx I tr bv I tr bv

(21.2.5)

Equation (21.2.5) is valid for linear elastic behavior and over regions without concentrated forces. For composite members with the same cross section along their length, Q1, Itr and bv are constant; thus the shear stress at the interface varies in the same proportion as the shear force V. It should be noted that Vh(x) may also be calculated from the normal resultant forces below the interface between the two materials. For members behaving beyond the linear elastic range, the expressions above are not valid. The horizontal force Vh, however, may be calculated between two sections separated by a distance, say L1–​2, based on equilibrium (Fig. 21.2.3). In practice, the horizontal shear force Vh is calculated between a section of maximum moment and the inflection point (i.e., M = 0), and also between concentrated load points. Considering the composite beam shown in Fig. 21.2.3, the interface shear force Vh and average shear stress vh between the sections of maximum moment and M = 0 are calculated as follows, Vh = F ( x = L /2) − F ( x = 0) = F2 − F1 = F2



vh =



(21.2.6)

Vh F2 = bv L1- 2 bv L / 2

(21.2.7)

where F2 is the resultant normal force above the interface at the midspan section, which can be calculated based on the nominal stress condition. For T-​sections exhibiting a rectangular

b2 E2/E1

b1 dx

x

Cross section

(a) Composite beam (transformed into material 1) bv cgtr

b(y)

dy y

y2

M(x)

M(x+dx)

y1 dx

F(x) y1

F(x+dx) Vh(x)

(b) Normal stresses in differential beam segment (above) and interface shear force (below)

Figure 21.2.2  Interface shear force at a section in a composite beam.

901



901

21.3  CONCRETE COMPOSITE FLEXURAL MEMBERS 1

2

Material 2 Material 1 L1–2

F1

F2 Vh

Figure 21.2.3  Calculation of interface shear force between two sections in a composite beam.

section behavior (i.e., a ≤ hf ; see Chapter 4), and ignoring the effect of compression steel in the flange (if any), F2 is calculated as

F2 = 0.85 fc′ bE a

(21.2.8)

F2 = 0.85 fc′ bE h f

(21.2.9)

For T-​section behavior,

where bE and hf are the flange effective width (see Section 4.3) and thickness, respectively.

21.3 CONCRETE COMPOSITE FLEXURAL MEMBERS The discussion on concrete composite members will focus on those consisting of a precast (either nonprestressed or prestressed) beam and a cast-​in-​place slab, which are commonly found in precast construction. However, the same concepts apply to composite cast-​ in-​place concrete members in which concrete is cast in stages. Assuming sufficient shear can be transferred along the interface between the precast beam and the cast-​in-​place slab to achieve full composite action, the design at ultimate for flexure and shear of concrete composite beams is basically the same as for monolithic members. Special attention should be paid to the loads expected to be resisted by the precast beam during construction, however, as these will depend on whether the member is shored or unshored during construction. For the case of shored construction, any load in addition to the self-​weight of the precast beam is resisted by the composite member. In unshored construction, on the other hand, the precast beam will need to be capable of supporting, in addition to its self weight, the weight of the cast-​in-​place slab and any other load applied before the slab behaves compositely with the precast beam. The design of the precast beam will thus need to be adequate to resist the loads expected prior to the achievement of composite action. Stresses and deflections under service loads will also depend on the construction process.

Interface Shear Strength Shear strength along the interface between concretes cast at different times has been the subject of extensive research [21.2–​21.8]. Shear along the interface between two concretes cast at different times (e.g., precast beam and cast-​in-​place slab concretes) is transferred primarily through friction and dowel action of reinforcement crossing the interface [21.6]. Thus, interface shear strength depends primarily on interface roughness, aggregate type, and amount and strength of reinforcement crossing the interface. For several decades, the use of a shear-​friction analogy (see Section 5.15) has been used for calculating interface shear strength [21.5], where a normal or clamping force is

902

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C hapter   2 1     C omposite M embers and C onnections

calculated based on the yield strength of the steel crossing the interface. This analogy is supported by the fact that as relative slip occurs along the interface between the two concretes, the roughness of the concrete surfaces causes a vertical separation, which in turn produces an elongation of the steel crossing the interface, provided it is well anchored on both sides [21.5]. The tensile force developed in the reinforcement translates then into a normal or clamping force. Experimental data [21.3, 21.6], however, have indicated that even in the absence of transverse reinforcement, some amount of shear can be transferred when a rough interface is provided. ACI 318-​14 provides two design approaches for interface shear in concrete composite members (ACI-​16.4). One of these is a sectional method, wherein the interface shear stress is calculated based on the shear acting on the section (ACI-​16.4.3 and ACI-​16.4.4). The alternative method, on the other hand, applies to a segment of the member (ACI-​16.4.5); it is based on the difference between resultant normal forces above (or below) the interface between two sections. Sectional Method In the sectional method, the factored shear stress acting on the interface, vuh, is calculated as follows,

vuh =

Vu bv d

(21.3.1)

where Vu is the factored shear acting on the section, bv is the width of the contact surface, and d is the effective depth of the composite member, but not to be taken less than 0.8h for prestressed concrete members (ACI-​16.4.4.3). At any section, Eq. (21.3.2) must be satisfied,

φvnh ≥ vuh

(21.3.2)

where vnh is the nominal shear stress capacity of the concrete interface and φ  =  0.75 (ACI-​21.2). The nominal interface shear strength depends primarily on the roughness of the interface and the amount and yield strength of the transverse reinforcement crossing the interface (if any). At a minimum, the contact surface of the hardened concrete must be clean and free of laitance, and either intentionally roughened to an amplitude of approximately 1 4 in. (referred to simply as intentionally roughened) or crossed by transverse steel with area Av ≥ Av,min. Three cases are listed in ACI 318-​14 for determination of interface shear stress capacity as follows (ACI-​16.4.4.1 and ACI-​16.4.4.2). 1) For concrete placed against an intentionally roughened hardened concrete surface and Av < Av,min, or for concrete placed against not intentionally roughened hardened concrete and Av ≥ Av,min [ACI Table 16.4.4.2(c) and (d)],

vnh = 80 psi

(21.3.3)

2) For vuh ≤ φ (500 psi), concrete placed against an intentionally roughened hardened concrete surface and Av ≥ Av,min [ACI Table 16.4.4.2(a) and (b)],

0.6 Av f yt   vnh = λ (260 psi) +  ≤ 500 psi bv s  

(21.3.4)

where λ is the lightweight concrete strength reduction factor (see Section 1.8) to reflect the lower interface shear strengths exhibited by lightweight concrete [21.8], Av is the area of transverse reinforcement perpendicular to the concrete surface at a spacing s, and fyt is the yield strength of the transverse reinforcement.

903



21.3  CONCRETE COMPOSITE FLEXURAL MEMBERS

903

3) For vuh > φ (500 psi) (ACI-​16.4.4.1 and ACI-​22.9), vnh =



µ Av f yt bv d

(21.3.5)

where µ = 1.0λ for concrete placed against an intentionally roughened concrete surface and 0.6λ for concrete placed against a not intentionally roughened concrete surface. For cases in which normal-​weight concrete is cast against intentionally roughened hardened concrete, the interface nominal shear stress capacity in Eq. (21.3.5) is limited to the least of [ACI-​Table 22.9.4.4(a)–​(c)]: a) 0.2 fc′ b) (480 psi + 0.08 fc′ ) c) (1600 psi) For all other cases, the interface nominal shear stress capacity is limited to the lesser of 0.2 fc′ and 800 psi [ACI Table 22.9.4.4(d) and (e)]. When concretes with different compressive strengths are used, choose the lesser  fc′ . For cases where vuh ≤ φ (500 psi), an intentionally roughed surface is present and at least minimum transverse reinforcement is provided, the calculation of nominal interface shear strength in ACI 318-​14 [Eq. (21.3.4)] is based on a modified shear friction analogy, where a minimum shear stress of 260λ psi, equivalent to a cohesion term, is used. The second term in Eq. (21.3.4) represents the interface shear strength based on a clamping force equal to the yield strength of the transverse steel and a coefficient of friction µ = 0.6λ. For vuh > φ (500 psi), on the other hand, the nominal shear strength is based entirely on a shear friction analogy. Alternative (Segment) Method Calculation of factored shear stress vuh in the alternative method of ACI 318-​14 (ACI-​ 16.4.5) is based on the difference in resultant normal force between two sections, above (or below) the interface (Fig. 21.2.3). In this case, the interface shear stress is calculated as

vuh =

F2 − F1 bv L1−2

(21.3.6)

where F1 and F2 are the resultant normal forces at sections 1 and 2, respectively, and L1-​2 is the distance between the two sections. In general, sections 1 and 2 correspond to the sections where M = 0 (inflection point) and M = Mmax. Thus, for simply supported beams subjected to uniform distributed loading, sections 1 and 2 correspond to the support section and the midspan section, respectively, while L1-​2 is equal to half the span length. When concentrated forces are present, interface shear stresses should also be checked between sections of maxi­ mum moment and adjacent concentrated force, as well as between concentrated forces. The nominal shear stress capacity vnh in the alternative method is the same as that used in the sectional method, except for vuh > φ (500 psi), where Eq. (21.3.7) should be used instead of Eq. (21.3.5). This is because in the segmental method, the stress capacity is equal to the shear friction strength divided by the interface area between the two sections considered (i.e., bv L1-​2).

vnh =

µ Av f yt bv L1-2

(21.3.7)

where Av in this case is the total area of transverse reinforcement crossing the interface between sections 1 and 2. Minimum Area of Interface Transverse Reinforcement Transverse reinforcement used in the precast member may be extended into the cast-​in-​ place slab and used also for interface shear transfer. To take advantage of its contribution

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to interface shear transfer, the area of transverse reinforcement must be at least Av,min for transverse shear (see Section 5.10) as follows (ACI-​16.4.6.1),

min Av = 0.75 fc′

bw s 50bw s ≥ f yt f yt

[5.10.8]

Transverse reinforcement must be developed in the two elements being connected (e.g., precast beam and cast-​in-​place slab) and spaced no farther than 4 times the minimum dimension of the supported element, typically the slab thickness, or 24 in., whichever is smaller (ACI-​16.4.7.2).

Flexural and Shear Strength The sequence of loading and the construction process (shored vs. unshored) does not affect the calculation of required and nominal flexural and shear strengths in the composite beam. However, the design of the precast beam must be adequate to resist the loads applied to it prior to the achievement of composite action. The flexural and shear strength of concrete composite beams is calculated using the same principles discussed in Chapters 3 and 5 for monolithic reinforced concrete sections and in Chapter 20 for prestressed concrete sections. Although a discontinuity in the strain distribution is expected at the interface between the two materials owing to strains in the precast beam prior to composite action and interface slip, satisfactory results are obtained assuming a monolithic concrete section behavior. As discussed earlier, negligible slips have been reported along intentionally roughened concrete interfaces subjected to interface shear stresses below approximately 250 psi [21.2, 21.3]. For larger interface shear stresses, even though some interface slip will occur, flexural and shear strength will not be appreciably affected as long as the interface shear strength is adequate. Effective depth d used in the calculation of flexural and shear strength of concrete composite beams is that corresponding to the same cross section as if it had been cast monolithically. Typically, the concrete strength in the precast section is greater than the slab concrete strength. In that case, the lower concrete strength (slab) should be used in the calculation of positive flexural strength (slab in compression), while for negative flexural

Precast concrete girder, prior to casting of topping slab. Note the girder transverse reinforcement extending beyond top surface for composite action with cast-​in-​place slab. (Photo courtesy of Luis Fargier-Gabaldón.)

905



21.3  CONCRETE COMPOSITE FLEXURAL MEMBERS

905

strength (slab in tension), the precast concrete strength should be used. In most cases, the depth of the precast beam represents 3 4 or more of the depth of the composite beam. Thus, the use of the precast concrete strength is often adequate for calculation of shear strength. As an alternative, a weighted concrete strength, based on the depth of each component, may be used.

Deflections Calculation of deflections in concrete composite beams follow the same principles outlined in Chapter 12. However, some loads are first resisted by the precast beam while others by the composite member, and this condition requires careful consideration of the loading sequence, which will depend on whether shored or unshored construction is used. Further, differential shrinkage between the slab and precast beam will lead to additional deflections, which are not considered in monolithic construction. The following description of basic steps for calculating deflections in composite beams consisting of a precast reinforced concrete section supporting a cast-​in-​place slab for both unshored and shored construction is based on research by Branson [21.9, 21.10]. The same principles can be extended to precast prestressed concrete beams. Detailed information about calculation of deflections in composite beams consisting of prestressed concrete sections and cast-​in-​place slabs can be found in References 21.9 through 21.11. Unshored Construction In unshored construction, the precast beam must be capable of supporting, in addition to its self-weight, the weight of the cast-​in-​place slab and other construction-​related loads. By the time the slab concrete has hardened, the deflection in the precast beam will consist of an instantaneous deflection due to the self-​weight of the beam and slab, plus time-​dependent deflections caused by creep and shrinkage. Once the slab concrete has gained sufficient strength and the beam has become composite, there will be additional instantaneous deflection due to superimposed dead load, if any, and time-​dependent deflections caused by sustained dead load (i.e., precast beam and slab self-weight, plus any additional superimposed dead load) and differential shrinkage between the precast beam and the slab. Deflections will also occur as the composite beam is subjected to live load. Since live loads are transient, however, these deflections are only instantaneous, unless part of the live load may be considered to be permanent for a significant period of time. Section 9.3.2.2 of ACI 318-​14 specifies that for cases in which both the depth of the composite beam and the precast section satisfy the minimum depth in ACI-​9.3.1 (see Table 12.15.2), there is no need to check deflections unless the beam supports or is attached to partitions or other elements likely to be damaged by large deflections. On the other hand, composite beam deflections must be investigated when the depth of the composite beam does not satisfy the minimum depth in ACI-​9.3.1 or elements likely to be damaged by large deflections are supported by or attached to the beam. Further, for cases in which only the composite beam depth satisfies the minimum depth, deflections occurring before the beam became composite should be investigated. Basic steps for calculating deflections in unshored construction are outlined below, with application to the particular case of a simply supported beam of span L consisting of a precast beam supporting a cast-​in-​place slab. Attention should be paid to the use of different moments of inertia depending on whether the beam is uncracked or cracked and, for cracked beams, on the maximum moment for the case considered. 1. Instantaneous deflection of precast beam due to self-​weight,



( ∆ i ) pb -precast =

5[(wD ) pb ]L4 384 Ec ( I g ) pb

(21.3.8)

where (wD)pb is the load corresponding to the self-​weight of the precast beam, Ec is the modulus of elasticity of the precast concrete, and (Ig)pb is the gross moment of

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inertia of the precast section. If the beam is expected to be cracked under its own weight, a cracked moment of inertia must be used. 2. Time-​dependent deflection of precast beam due to self-​weight (up to just before casting the slab concrete), ( ∆ td ) pb = (λ ∆ ) pb ( ∆ i ) pb -precast



(21.3.9)

where (λΔ)pb is a time-​dependent factor that accounts for the effect of creep and shrinkage on deflections (see Section 12.13 and ACI-​24.2.4.1.1). The sustained load duration used in the calculation of (Δtd)pb should be the period between the time the precast beam is first subjected to its own weight and the time the slab is cast. 3. Instantaneous deflection of precast beam due to slab self-​weight, ( ∆ i ) pb-slab . In this case, it is assumed that the precast beam will be cracked under the action of its own weight plus the slab self-​weight. The effective moment of inertia of the precast beam, (Ie)pb, can be calculated by using Eq. 21.3.10 below (see Section 12.9) with Ma equal to the maximum moment due to the weight of the precast beam and slab.



( I e ) pb

 M  M  =  cr  ( I g ) pb + 1 −  cr   Ma    M a  3

3

  ( I cr ) pb 

(21.3.10)

where Mcr and (Icr)pb are the cracking moment and the cracked moment of inertia of the precast section, respectively. The instantaneous deflection of the precast beam due to slab self-​weight, (wD)slab, is then

( ∆ i ) pb -slab =

5[(wD )slab ]L4 384 Ec ( I e ) pb

(21.3.11)

It should be noted that flexural cracking in the precast beam will cause an increase in the deflection due to its own weight, previously calculated assuming uncracked behavior [see Eq. 21.3.8]. Thus, the total instantaneous deflection in the cracked precast beam due to its own weight and the slab self-​weight is

( ∆ i ) pb =

5[(wD ) pb + (wD )slab ]L4 384 Ec ( I e ) pb



(21.3.12)

4. Instantaneous deflection in composite beam due to additional dead load (i.e., superimposed dead load, wSD),

5(wSD )L4 384 Ec ( I e )cb -D

( ∆ i )cb -SD =

(21.3.13)

where (Ie)cb-​D is the effective moment of inertia of the composite beam when subjected to the total dead load (self-​weight plus superimposed dead load). In general, the effective moment of inertia of the composite beam, (Ie)cb, can be calculated by assuming a monolithic section with the same dimensions as the transformed composite beam section. From Eq. (21.3.10), (Ie)cb can be calculated as

M  ( I e )cb =  cr   Ma 

3

( ) Ig

cb

  M 3 + 1 −  cr   ( I cr )cb (21.3.14)   M a  

where (Ig)cb and (Icr)cb are, respectively, the gross and cracked moment of inertia of the composite section. For the particular case of (Ie)cb-​D, Ma in Eq. (21.3.14) is equal to the moment due to the self-​weight of the composite beam plus superimposed dead load. 5. Time-​dependent deflection in composite beam due to sustained load, ( ∆ td )cb. This deflection represents the summation of time-​dependent deflections due to the precast

907



907

21.3  CONCRETE COMPOSITE FLEXURAL MEMBERS

beam self-​weight (in addition to those that took place before the slab was cast), the slab weight, and any additional permanent load: ( I g ) pb   ( ∆ td )cb = [(λ ∆ )cb − (λ ∆ ) pb ] ( ∆ i ) pb -precast  ( I e )cb -D   ( I e ) pb   + (λ ∆ )cb ( ∆ i ) pb -slab + ( ∆ i )cb -SD  ( I e )cb -D  



(21.3.15)

where (λ ∆ )cb is a time-​dependent factor that accounts for the effect of creep and shrinkage on deflections in the composite beam, which is determined based on the time period of interest, typically 5 years or longer. In that case, (λ ∆ )cb = 2.0. Note that in Eq. (21.3.15) a single factor (λ ∆ )cb is used for simplicity, which is acceptable when sustained loads are applied for several years. 6. Deflection due to differential shrinkage between the slab and the precast beam, ( ∆ td )sh. Since the concrete in the slab shrinks soon after casting and most of the shrinkage has already occurred in the precast beam, the precast beam will provide some restrain against shrinkage of the slab, leading to additional bending (and deflection) of the composite beam. Different methods have been proposed to account for deflections induced by differential shrinkage [21.9]. It should be kept in mind, however, that only rough estimates of shrinkage-​induced deflections are possible, given the uncertainties involved. A simple procedure to estimate deflections due to differential shrinkage consists of applying an equivalent shrinkage-​induced force, Q, at the centroid of the slab (Fig. 21.3.1). For a differential shrinkage strain εsh, the force Q is calculated as Q = ε sh Ecs Aslab



(21.3.16)

where Ecs is the modulus of elasticity of the slab concrete and Aslab is the cross-​ sectional area of the slab, calculated as the product of the effective slab width, bE , and the slab thickness. The deflection due to differential shrinkage is then calculated as ( ∆ td )sh =



Qys L2 8 Ec ( I e )cb -D

(21.3.17)

where ys is the distance between the slab centroid and the centroid of the composite (transformed) section. 7. Deflection due to live load. Given the transient nature of live load, the deflection associated with it is instantaneous unless a portion of the live load can be considered as permanent for a period of time. The instantaneous deflection due to live load, wL, is calculated as ( ∆ i )L =



5wL L4 384 Ec ( I e )cb -L

(21.3.18)

where ( I e )cb -L is calculated from Eq. (21.3.14) with Ma equal to the moment due to the total dead load plus live load. btr Q

cgslab

Q ys

cgtr

L

Figure 21.3.1  Equivalent shrinkage-​induced force in composite section.

ys

908

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C hapter   2 1     C omposite M embers and C onnections

As discussed in Section 12.6, two deflection checks need to be performed, one involving the instantaneous live load deflection, ( ∆ i )L, and the other all deflections that occur after attachment of nonstructural elements, ∆ t . Assuming (conservatively) that all shrinkage-​ related deformations occur after attachment of nonstructural elements, ∆ t = ( ∆ i )cb -SD + ( ∆ td )cb + ( ∆ td )sh + ( ∆ i )L



(21.3.19)

Limits associated with ( ∆ i )L and ∆ t are listed in ACI Table 24.2.2 and discussed in Section 12.6. Shored Construction In shored construction, all the loads other than the self-weight of the precast beam are resisted by the composite member. The procedure discussed above for unshored construction can also be used for shored construction with the exception that Eq. (21.3.11) must be replaced by Eq. (21.3.20). Also, since the self-​weight of the slab is resisted by the composite beam, Eq. (21.3.12) is not applicable for shored construction. ( ∆ i )cb -slab =



5[(wD )slab ]L4 384 Ec ( I e )cb -slab

(21.3.20)

where ( I e )cb-slab is calculated from Eq. (21.3.14) with Ma equal to the moment due to the self-​weight of the precast beam and slab. There is no need to check deflections in shored construction if the depth of the composite beam satisfies the minimum depth in ACI-​9.3.1 (see Table 12.15.2) and supports or is attached to elements not likely to be damaged by large deflections.

EXAMPLE 21.3.1 The composite beam shown in Fig. 21.3.2 is part of a floor system that consists of a series of precast reinforced concrete beams spaced 10 ft apart that support a 5-​in.-​thick cast-​in-​place slab. The beam has a center-​to-​center span length of 30 ft and is supported on 12-​in.-​long supports (i.e., clear span length = 29 ft). The floor system is subjected to a superimposed dead load of 20 psf and a live load of 60 psf. Consider a 20-​psf live load during construction. Check the adequacy of the design in terms of flexural and shear strengths, as well as deflections. Assume unshored construction and that the floor system supports nonstructural elements not likely to be damaged by large deflections. For the precast beam: fc′ = 6000 psi; fy = fyt = 60 ksi. For the cast-​in-​place slab:  fcs′ = 3500 psi; fy = 60 ksi. Concrete is normal weight. Cover to center of longitudinal bars = 2.5 in. Cast-in-place slab

#3 @ 6.5 in.

5”

Precast beam

2 #9 + 2 #8

16” (Slab bars not shown)

10”

Figure 21.3.2  Cross section for composite beam of Example 21.3.1.

(Continued)

90



21.3  CONCRETE COMPOSITE FLEXURAL MEMBERS

909

Example 21.3.1 (Continued) SOLUTION (a) Determine required flexural and shear strengths. Precast beam self-​weight: (wD ) pb =



10(16) 144

(0.15) = 0.17 kips/ft

Slab weight: (wD )slab =

Superimposed dead load: Construction live load: Service live load:

5(120) 144

(0.15) = 0.63 kips/ft

wSD = 0.020(10) = 0.20 kips/ft



(wL )const = 0.020(10) = 0.20 kips/ft wL = 0.060(10) = 0.60 kips/ft





Factored loads: -​ Precast beam: (wu ) pb = 1.2[(wD ) pb + (wD )slab ] + 1.6(wL )const = 1.2(0.17 + 0.63) + 1.6(0.20)



= 1.27 kips/ft



-​ Composite beam: (wu )cb = 1.2[(wD ) pb + (wD )slab + wSD ] + 1.6 wL = 1.2(0.17 + 0.63 + 0.20) + 1.6(0.60)



= 2.16 kips/ft



Maximum factored moment (at midspan section): -​ Precast beam: ( Mu ) pb =



(wu ) pb L2 8

=

1.27(30)2 = 143 ft-kips 8

-​ Composite beam:

( Mu )cb =

(wu )cb L2 2.16(30)2 = = 242 ft-kips 8 8

Maximum factored shear (at one effective depth from support face): Note that the effective depth increases once the slab becomes composite with the precast beam. -​ Precast beam:

13.5 L  (Vu ) pb = (wu ) pb  − d pb  = 1.27(15 − ) = 17.6 kips 12 2 

(Continued)

910

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C hapter   2 1     C omposite M embers and C onnections

Example 21.3.1 (Continued) -​ Composite beam: 18.5  L   (Vu )cb = (wu )cb  − dcb  = 2.16  15 − = 29.0 kips 12  2   



Required flexural and shear strengths. Assume both precast and composite beam sections are tension controlled (i.e., φ = 0.90). For shear, φ = 0.75. ( Mu ) pb

φ

=

143 = 159 ft-kips 0.90

( Mu )cb 242 = = 269 ft-kips 0.90 φ

(Vu ) pb

φ

=

17.6 = 23.5 kips 0.75

(Vu )cb 29.0 = = 38.7 kips φ 0.75 (b) Check flexural strength. -​ Precast beam: As = 2(1.0) + 2(0.79) = 3.58 sq in. a pb =



As f y 0.85 fc′ b

=

3.58(60) = 4.21 in. 0.85(6)(10)

a pb   4.21 1  = 3.58(60)  13.5 − (M n ) pb = As f y  d pb −  = 204 ft-kips  2  2  12 

a pb   4.21   13.5 −  d pb − β   0.75  = 0.0042 > 0.004 1  = 0.003  ε t = 0.003  4.21   a pb       0.75  β1 

Section is in the transition region. From ACI Table 21.2.2(d), φ = 0.83.

( M n ) pb = 204 ft-kips >

( Mu ) pb

φ

=

143 = 172 ft-kips 0.83

OK

-​Composite beam (neglecting any reinforcement in slab): Effective slab width. According to ACI-​6.3.2.1, the effective slab width, bE, for an interior beam is the least of bw + 16t = 10 + 16(5) = 90 in. bw +

Ln 29(12) = 10 + = 97 in. 4 4

center-to-center spacing of beams = 120 in.

where t is the slab thickness and Ln is the clear span length. Thus, bE = 90 in. (Continued)

91



911

21.3  CONCRETE COMPOSITE FLEXURAL MEMBERS

Example 21.3.1 (Continued) Assuming rectangular section behavior, acb =



As f y 0.85 fcs′ bE

=

3.58(60) = 0.80 in. < t 0.85(3.5)(90)

OK

a  0.80  1   ( M n )cb = As f y  dcb − cb  = 3.58(60)  18.5 −  = 324 ft-kipps   2  2  12 a   0.80   d − cb 18.5 −  cb β   0.85  = 0.049 1 ε t = 0.003   = 0.003  0.880   acb      β  0.85  1

Section is tension controlled. Thus, φ = 0.90. ( M n )cb = 324 ft-kips >



( Mu )cb 242 = = 269 ft-kips φ 0.90

OK

(c) Check shear strength. -​ Precast beam: (Vc ) pb = 2 λ fc′bw d pb = 2(1) 6000 (10)(13.5) = 20.9 kips

(Vn ) pb

d pb

0.22(60)(13.5) = 27.4 kips 6.5 (Vu ) pb = (Vc ) pb + (Vs ) pb = 48.3 kips > = 23.5 kips φ

(Vs ) pb = Av f yt

s

=

Because the design is the same throughout the span and satisfies shear requirements near the support, where shear force is maximum, there is no need to check shear design at other beam regions. Check maximum spacing and minimum amount of transverse steel. Since

(Vu ) pb < 4 fc′bw d pb , smax = d pb / 2 = 6.75 in. > s min Av = 0.75 fc′

OK

bw s 0.75 6000 (10)(6.5) = f yt 60, 000

= 0.063 sq in. ≥





50bw s 50(10)(6.5) = = 0.054 sq in. f yt 60, 000

min Av = 0.063 sq in. < Av = 0.22 sq in.

OK

-​ Composite beam: It should be noted that in this case (Vn)pb > (Vu)cb /φ and thus, no further check is necessary. For illustration purposes, however, the procedure will be shown here. As discussed earlier, calculation of shear strength in composite beams with concretes of different compressive strengths may be performed using a weighted strength based on (Continued)

912

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C hapter   2 1     C omposite M embers and C onnections

Example 21.3.1 (Continued) the precast and composite beam depths. The use of such an approach in this case would lead to a weighted compressive strength of approximately 5300 psi. Use 5000 psi. (Vc )cb = 2 λ fc′bw dcb = 2(1) 5000 (10)(18.5) = 26.2 kips

(Vs )cb = Av f yt

dcb 0.22(60)(18.5) = = 37.6 kips 6.5 s

(Vn )cb = (Vc )cb + (Vs )cb = 63.7 kips >



(Vu )cb = 38.7 kips φ

Because the design is the same throughout the span and satisfies shear requirements near the support, where shear force is maximum, there is no need to check shear design at other beam regions. Because, further, (Vu )cb < 4 fc′bw dcb , smax = dcb / 2 = 9.25 sq in. > s OK Amount of transverse steel satisfies minimum required, as checked above for precast section. d) Check interface shear. -​Sectional method: vuh =

Vu 29.0(1000) = = 156 psi < φ (500 psi) bv dcb 10(18.5)

0.6 Av f yt   0.6(0.22)(60, 000)   = 1 260 + vnh = λ  260 +  = 382 psi  b s 10(6.5)     v



φvnh = 286 psi > vuh



OK

-​Segment method: In this case, the interface shear force is calculated between the support (M  =  0) and midspan (M = Mmax). Because the entire concrete compressive stress block falls within the beam flange, the horizontal force at midspan, F2, in Eq. (21.3.6) is equal to the yield strength of the longitudinal steel (As  fy) while F1 = 0. Thus,

vuh =

F2 − F1 As f y 3.58(60, 000) = = = 119 psi < φvnh (30)(12) L bv L1- 2 bv 10 2 2

OK

e) Check deflections. -​Instantaneous deflection of precast beam due to self-​weight: Check first whether the precast beam is uncracked or cracked under its own weight. ( M cr ) pb =

fr bw (h)2pb 6

= 7.5(1) 6000

(10)(16)2 6

1 12,000

= 20.7 ft-kipss (Continued)

913



913

21.3  CONCRETE COMPOSITE FLEXURAL MEMBERS

Example 21.3.1 (Continued) ( M a ) pb = ( I g ) pb =



(wD ) pb L2 8

bw (h)3pb 12

=

Ec = 57, 000 fc′ =

(∆ i ) pb--precast =

= 18.8 ft-kips < ( M cr ) pb

Beam is uncracked.

10(16)3 = 3413 in.4 12 57, 000 6000 1000

5(wD ) pb L4 384 Ec I g

= 4415 ksi

  5(0.17)  1  (30 × 12)4  12  = 0.20 in. = 384(4415)(3413)

-​Time-​dependent deflection of precast beam due to self-​weight (up to time of beam becoming composite): Assuming a period of 3 months between the time the precast beam is subjected to its own weight and the time the beam becomes composite, (λ ∆ ) pb = 1.0 (ACI-​24.2.4.1). Thus,

( ∆ td ) pb = (λ ∆ ) pb ( ∆ i ) pb -precast = 1.0(0.20) = 0.20 in.

-​Instantaneous deflection of precast beam due to slab self-​weight: The precast beam will experience flexural cracking once it is subjected to the slab weight. From Eq. (21.3.10), the effective moment of inertia is calculated as  M  M  =  cr  ( I g ) pb + 1 −  cr   Ma    M a  3



( I e ) pb

3

  ( I cr ) pb 

where

(wD ) pb + (wD )slab  L2 (0.17 + 0.63)(30)2 Ma =  = = 89.1 ft-kips 8 8

The cracked moment of inertia of the precast beam section, (Icr)pb, can be obtained by following the procedure in Section 12.5. The neutral axis depth c and cracked moment of inertia are obtained as follows:

bw

c2 = ηs As (d pb − c) = 0 2

where ηs is the reinforcing steel-​precast concrete modular ratio (ηs = 6.57). Solving for c leads to c = 5.96 in. 2

( I cr ) pb =



bw c 3  c + bw c   + ηs As (d pb − c)2  2 12

The cracked and effective moments of inertia for the precast section are then

( I cr ) pb =

(10)(5.96)3 10(5.96)3 + + 6.57(3.58)(13.5 − 5.96)2 = 2043 in.4 12 4 (Continued)

914

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C hapter   2 1     C omposite M embers and C onnections

Example 21.3.1 (Continued) and   20.7   20.7  (3413) + 1 −  ( I e ) pb =     89.1    89.1  3



3

 4  (2043) = 2060 in. 

The instantaneous deflection due to the slab weight is then (∆ i ) pb -slab =

5[(wD )slab ]L4 384 Ec ( I e ) pb

  5(0.63)  1  (30 × 122)4  12  = 384(4415)(2060) = 1.25 in. Because the beam exhibits flexural cracking when subjected to the weight of the slab, the deflection of the precast beam due to its own weight will increase. From Eq. (21.3.12), the total instantaneous deflection of the precast beam due to its own weight and that of the slab is (∆ i ) pb =

5[(wD ) pb + (wD )slab ]L4 384 Ec ( I e ) pb

  5(0.17 + 0.63)  1  (30 × 12)4  12  = 384(4415)(2060)



= 1.59 in.



-​Instantaneous deflection in composite beam due to superimposed dead load. Because the precast beam is already cracked, the effective moment of inertia of the composite beam under the action of dead load, ( I e )cb -D , needs to be calculated. This requires first the calculation of the properties of the transformed section (both uncracked and cracked) as follows. Effective slab width of transformed section:

(bE )tr = bE ηc = 90

(57, 000 3500 ) = 68.7 in. 4415 × 1000

For uncracked transformed section, it can be shown that the centroid is located at ytr = 14.68 in., measured from the bottom of the section. The transformed moment of inertia and cracking moment are ( I tr )cb = 19, 240 in.4 ≈ ( I g )cb

( M cr )cb =

fr ( I tr )cb (19, 240) = 7.5(1) 6000 = 63.4 ftt-kips ytr 14.7(12,000)

The moment of inertia for the transformed cracked section for use in Eq. (21.3.14) can be determined following the same procedure as for the precast section. The neutral axis depth and crack moment of inertia for the cracked transformed composite section are

c = 3.23 in.

(ytr = 17.77 in.)

( I cr )cb = 6255 in.4 (Continued)

915



915

21.3  CONCRETE COMPOSITE FLEXURAL MEMBERS

Example 21.3.1 (Continued) The acting moment Ma to be used in Eq. (21.3.14) is Ma =

[(wD ) pb + (wD )slab + wSD ]L2

8 (0.17 + 0.63 + 0.20)(30)2 = 8 = 112 ft-kips

From Eq. (21.3.14),



( I e )cb -D

3   63.4  3   63.4  (19, 240) + 1 −  =   (6255)  112    112   = 8644 in.4

The instantaneous deflection of the composite beam due to superimposed dead load is then (∆ i )cb -SD =

5(wSD )L4 384 Ec ( I e )cb -D

  5(0.20)  1  (30 × 12)4  12  = 384(4415)(8644)



= 0.096 in.



-​ Time-​ dependent deflection of composite beam due to sustained load. From Eq. (21.3.15) and assuming a period of 5 years or more [i.e., (λ ∆ )cb = 2.0.], ( ∆ td )cb = 0.87 in. -​Deflection due to differential shrinkage between the slab and the precast beam. Assuming a shrinkage strain εsh = 400 × 10–​6,

Q = ε sh Ecs Aslab = 400(10)−6 (57, 000 3500 )(5)(90) ( ∆ td )sh =

1 1000

= 607 kips

Qys L2 8 Ec ( I e )cb -D

5   607  21 − − 17.8 (30 × 12)2   2 = 8(4415)(8644) = 0.19 in.





-​ Instantaneous deflection due to live load. From Eq. (21.3.14) with Ma corresponding to the moment caused by the total dead load plus live load, Ma =



[(wD ) pb + (wD )slab + wSD + wL ] L2 8

(0.17 + 0.63 + 0.20 + 0.60)(30)2 = 8 = 179 ft-kips 3   63.4  3   63.4  4 ( 19 , 240 ) ( I e )cb -D + L =  + 1 −    (6255) = 6833 in.   179  179  

(Continued)

916

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C hapter   2 1     C omposite M embers and C onnections

Example 21.3.1 (Continued) (∆ i )L =

5(wL )L4 384 Ec ( I e )cb -D + L

  5(0.6)  1  (30 × 12)4  12  = 384(4415)(6833) = 0.36 in.





Deflection checks: Live load deflection (∆ i )L = 0.36 in.
4 fc′bwE d ,

s ≤ d / 4 ≤ 12 in.

Shear Strength of Reinforced Concrete Portion The nominal shear strength of the reinforced concrete portion of the concrete-​encased composite column can be calculated following the procedure in Section 10.21 and Chapter 5 as the summation of the contribution of the concrete, Vc, and the lateral ties, Vs , as follows,

Vn = Vc + Vs [5.8.1]

Accounting for the influence of axial compression on shear strength, Vc can be calculated by using Eq. (5.13.1), given in ACI-​22.5.6.1.



 Nu  Vc = 2  1 +  λ fc′bw d  2000 Ag 

[5.13.1]

The contribution of the lateral ties is calculated using Eq. (21.5.10). As discussed in the preceding section, the area of tie reinforcement must satisfy the minimum requirement for reinforced concrete columns (Section 10.21), and the selected tie spacing shall not be greater than either d/​2 or d/​4, depending on the shear force required to be resisted by the transverse reinforcement through a truss mechanism. The applicable shear strength reduction factor when the shear strength of the composite column is taken as that of the reinforced concrete component is the same φ factor used for shear in ACI 318 (φv = 0.75).

Load Transfer In the discussion above on determination of axial and bending strength of concrete-​ encased steel composite columns, it was assumed that load was introduced to the reinforced concrete portion and encased steel shape based on their relative contributions to column strength. In reality, however, load is often introduced into the composite column through the steel section. This is the case, for example, of a gravity load frame consisting of composite columns and steel girder-​composite deck floor systems, where gravity load is introduced into the column through a shear connection between the steel floor girder and the encased steel shape (Fig.  21.5.3). Another example corresponds to the

924

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C hapter   2 1     C omposite M embers and C onnections

case of a composite column that is part of a composite braced frame system in which the braces are made of steel. Part of the vertical component of the force in the braces, transferred into the steel column as an axial force, must then be transferred to the surrounding reinforced concrete portion. Attention must therefore be paid to ensuring that the load introduced to either the embedded steel shape or the concrete is transferred to the other component according to its relative contribution to column strength. Otherwise, it may be unconservative to assume that both components of the composite column contribute to resist the applied loads as calculated using either strain compatibility or a plastic stress distribution. The discussion that follows will focus on cases in which the load is introduced through the encased steel shape, as this is most common in practice. Design of the connections (bolted, welded, or a combination of both) between the steel components is outside the scope of this book. The reader is referred to References 21.1 and 21.30 for information about design of connections between steel components. Sharing of Axial Load between Reinforced Concrete and Steel Shape Components The relative contribution of the encased steel section and the reinforced concrete encasing to nominal axial column strength varies depending on column axial load. A reasonable estimation, however, can be made by assuming that their relative strength contribution is independent of column axial strength and equal to that corresponding to the case of pure axial load. This is the principle followed in the AISC Specification for determining the magnitude of the axial force required to be transferred, through longitudinal shear, from the steel shape to the surrounding reinforced concrete (AISC-​I6.2a). Taking the axial load transferred to the steel column via a floor girder (Fig. 21.5.3) or diagonal braces as Pr , the required longitudinal shear to be transferred to the surrounding reinforced concrete, Vr′ , is [AISC Eq. (I6-​1)]

As f y   Vr′ = Pr  1 − Pnso  

(21.5.11)

where Pnso is obtained from Eq. (21.5.1). Longitudinal Shear Transfer Transfer of forces between the embedded steel shape and the surrounding concrete is achieved through shear connectors attached to the steel shape (e.g., steel headed stud anchors or channels) or steel plates bearing on concrete. The following discussion focuses on the use of headed studs for longitudinal shear transfer, as these are typically used for shear transfer. Provisions for steel channel anchors can be found in AISC-​I8.3d and AISC-​I8.3e. Provisions for determining the shear strength of headed stud anchors welded to the encased steel shape are given in AISC-​I8.3a. When concrete breakout failures do not control, as is typically the case in concrete-​encased steel composite columns, the shear strength of a single stud anchor, Qnv, is calculated as

Qnv = fua Asa

(21.5.12)

where fua and Asa are the specified minimum tensile strength and cross-​sectional area, respectively, of the steel headed stud anchor. The applicable strength reduction factor for headed stud anchors, φv, is equal to 0.65. The required number of headed stud anchors required to transfer the axial force Vr′ is calculated from Eq. (21.5.13).

na ≥

Vr′ φ v Qnv

(21.5.13)

where na is the number of headed stud anchors. The minimum required stud length Lsa, measured from the base of the stud to the head, is 5 times the stud shank diameter (AISC-​ I8.3). Longer studs are specified when these are required to resist tension and shear, as well as in members constructed with lightweight concrete, which would be a rare case for composite columns. The procedure above is illustrated later (see Example 21.5.2).

925



21.5  CONCRETE-ENCASED STEEL COMPOSITE COLUMNS

925

In general, moments due to eccentricity of the girder shear force transferred into the column do not affect the design of the headed studs, either because girders from opposite faces frame into the column and moments cancel out, as in Example 21.5.2, or because the magnitude of the moment is small enough to be neglected. At times, however, the moment transferred into the composite column due to the eccentricity of shear is not negligible. In this case, the number of headed studs must be calculated such that these are capable of transferring not only the force Vr′ , but the additional force caused by the moment to be transferred from the encased steel shape to the reinforced concrete encasing. This will be illustrated in Example 21.5.3. Placement of Headed Stud Anchors Headed stud anchors must be placed on at least two faces of the encased steel shape in a symmetrical arrangement about the axis of the shape (AISC-​I6.4a). The studs must be placed over a distance referred to as the load introduction length, which is to be taken no greater than twice the minimum section dimension of the composite column above and below the region where load is transferred (Fig. 21.5.3). In some cases, however, the magnitude of the force Vr′ is such that it cannot be transferred over the load introduction length. In this situation, the column should be treated as noncomposite over the length required to transfer the force Vr′ (AISC Commentary I6.4). In terms of headed stud anchor placement, AISC-​I8.3e requires a minimum spacing between stud anchors, in any direction, of 4 times the diameter of the shank, dsa, and a maximum spacing of 32dsa. AISC-​I8.3e also refers to ACI 318 for minimum clear cover for headed stud anchors which, according to ACI-​20.6.1.3.5, is the same as that for any other reinforcement in the member. For cast-​in-​place columns not exposed to weather or in contact with ground, the minimum cover is then 1.5 in. (ACI-​Table 20.6.1.3.1). The writers recommend, however, that in addition to satisfying this minimum cover, the head of the stud be located within the core of the column. Even though the flexural strength of concrete-​encased steel composite columns under combined axial force and bending has been shown not to be affected by the presence of headed studs [21.24], AISC-​I6.4a requires the placement of headed studs outside the load introduction length to help maintain composite action in the column. As discussed above, stud spacing must not exceed 32 times the dimeter of the stud shank, dsa.

≤ 32dsa Headed stud ≤ 2b

≥ 4dsa ≤ 32dsa V = Pr’

Load introduction length

Steel beam ≤ 2b

Shear tab Encased steel shape

b: minimum column section dimension; reinforcing bars in column and concrete slab not shown for clarity.

Figure 21.5.3  Load introduction length and stud spacing requirements in concrete-​encased steel composite columns.

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EXAMPLE 21.5.1 Verify that the design of the concrete-​ encased steel composite column shown in Fig. 21.5.4 is adequate according to the AISC Specification for Structural Steel Buildings to resist the following factored loads obtained from a second-​order analysis satisfying the requirements of the Direct Analysis Method of AISC (not including the effect of initial imperfections): Pu = 935 kips; Mu = 1100 ft-​kips (about strong axis of steel shape); Vu = 130 kips. Unsupported column length for bending about strong axis L = 12 ft. (column is laterally supported in perpendicular direction). For the W14 × 132 shape: ds = 14.7 in.; As = 38.8 sq in.; tw = 0.645 in.; bf = 14.7 in.; tf = 1.03 in.; Is= 1530 in.4. Material properties:  fc′  = 6000 psi (normal weight); fys = 50 ksi (A572 Grade 50 steel); fyr = fyt = 60 ksi. #4 @ 10”

12 #7 2.5” 4”

24”

11”

4” 2.5” W14 × 132

24”

Figure 21.5.4  Cross section of concrete-​encased steel composite column of Example 21.5.1.

SOLUTION (a) Check that reinforcement ratios satisfy minimum requirements. Longitudinal bars:

Asr = 12(0.6) = 7.2 sq in. > 0.004 Ag = 0.004(24)(24) = 2.3 sq in.



OK

Steel shape:

As = 38.8 sq in. > 0.01Ag = 0.01(24)(24) = 5.76 sq in.

OK

(b) Check axial force and moment strength. The determination of the nominal axial strength of the column requires first the calculation of the equivalent flexural rigidity EIeff, Euler’s buckling load Pe, and axial strength reduction factor κ. The effective flexural rigidity, EIeff, is calculated using Eq. (21.5.4), with the variables calculated as follows, Es = 29, 000 ksi I s = 1530 in.4 I sr = 2[ 4(0.60)(12 − 2.5)2 + 2(0.60)(12 − 6.5)2 ] = 506 in.4

Ic =

24(24)3 − 1530 − 506 = 25, 612 in.4 12



 38.8 + 12(0.6)  C1 = 0.25 + 3   = 0.49 < 0.70  (24)(24)  Ec =

57, 000 6000 1000

= 4415 ksi (Continued)

927



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21.5  CONCRETE-ENCASED STEEL COMPOSITE COLUMNS

Example 21.5.1 (Continued) Thus, EI eff = (29, 000)(1530 + 506) + 0.49(4415)(25, 612)



= 1.14 × 108 kip-sq in.



Since Pu, Mu have been obtained from the Direct Analysis Method, k may be taken as 1.0 and Pe =



π 2 EI eff π 2 (1.14 × 108 ) = = 54, 450 kips ( Lc ) 2 [1(12)(12)]2

The pure axial load capacity of the column neglecting slenderness effects, Pnso, is obtained using Eq. (21.5.1). Pnso = 0.85 fc′( Ag − As − Asr ) + As f ys + Asr fsr = 0.85(6)[(24)(24) − 38.8 − 12(0.60)] + 38.8(50) + 12(0.60)(60)



= 5075 kips



Pnso /​Pe = 0.093 < 2.25. The axial strength reduction factor κ is then Pnso



κ = 0.658

Pe

 5075   

= 0.658 54, 450  = 0.96



The adequacy of the column section design will be checked by calculating the design moment strength, φb Mn, corresponding to a design axial strength φc Pn = φc  κPns = Pu. In other words, the design will be satisfactory if Mn ≥ Mu  /​φb at an axial force Pns = Pu  /​(φc κ). For this particular case, the neutral axis depth, c, is equal to 12.9 in. The resultant forces in the concrete and steel reinforcement are as follows (Fig. 21.5.5):

Cc = 0.85 fc′ bβ1c = 0.85(6)(24)(0.75)(12.9) = 1184 kips



The resultant axial force is then be determined as, i=4

Pns = Cc + ∑ Fsri + Fstf + Fsbf + Fsw i =1

= 1184 + 1.0 + 680 + 36.3 − 606



= 1295 kips



TABLE 21.5.1  STRAINS, STRESSES AND FORCES IN REINFORCING BARS OF EXAMPLE 21.5.1 Layer i

Asr,i (sq in.)

(di) (in.)

(yi) (in.)

εsr,i

1 2 3 4

2.4 1.2 1.2 2.4

2.5 6.5 17.5 21.5

10.4 6.40 –​4.60 –​8.60

0.00242 0.00149 –​0.00107 –​0.00200

a

fsri (ksi) a

60–5.1 a 43.2–​5.1 –​31.0 –​58.0

Fsri (kips) a

132 a 45.7 –​37.2 –​139

(h/​2 –​di) (in.) 9.5 5.5 –​5.5 –​9.5

Stress/​force adjusted to account for displaced concrete. Tensile strains, stresses and forces are negative.

(Continued)

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Example 21.5.1 (Continued)

TABLE 21.5.2  STRAINS, STRESSES AND FORCES IN STEEL SHAPE OF EXAMPLE 21.5.1

Top flange Web (top) Web (bottom) Bottom flange a

As (sq in.)

(y) (in.)

εs

fs (ksi)

15.1

7.74 7.23 –​5.42 –​5.93

0.00180 0.00168 –​0.00126 –​0.00138

50–​5.1 a 48.7–​5.1 –​36.6 –​40.0

8.15 15.1

Fs (kips)

a

(h/​2 –​di ) (in.)

a

680

6.83

a

36.3

—​b

–​606

–​6.83

Stress/​force adjusted to account for displaced concrete. Tensile strains, stresses, and forces are negative.

In this example, the web has been modeled as a continuous steel element, not as discrete equivalent bars. Thus, a single lever arm cannot be used to calculate its contribution to the section moment capacity. This is discussed below. See also Fig. 21.5.6. b

εcu = 0.003 εsr1

di

εstf

c = 12.9”

εsr2

0.85 f c’ = 5.1 ksi β1c = 9.68”

Fsr1 = 132 k Cc = 1184 k Fsr2 = 45.7 k

fstf = 50 ksi Fstf = 680 k Msw = 56 ft-k

y

24”

Fsr3 = 37.2 k

εsr3 εsr4 24”

Fsbf = 606 k

Fsr4 = 139 k

εsbf

Strains

Fsw = 36.3 k

fsbf = 40 ksi

Stresses/forces (reinforced concrete)

Stresses/forces (steel section)

Figure 21.5.5  Strains, stresses, and forces for calculation of Pns–​Mn in Example 21.5.1.

Taking into account slenderness effects due to initial imperfections, the nominal axial strength is calculated as

Pn = κ Pns = 0.96(1295) = 1246 kips

The design axial strength is then

φc Pn = 0.75(1246) = 935 kips = Pu



The corresponding nominal flexural strength can be calculated by taking moments about the column plastic centroid (i.e., column center) of the various forces. The moment contribution of the concrete, reinforcing bars and steel shape, Mc, Msr, and Ms, respectively, are calculated as follows:  h β c  24 0.75(12.9)  1 M c = Cc  − 1  = 1184  −  12 = 707 ft-kipss 2 2  2 2  i=4 h  M sr = ∑ Fsri  − di  = 253 ft-kips   2 i =1

(Continued)

92



929

21.5  CONCRETE-ENCASED STEEL COMPOSITE COLUMNS

Example 21.5.1 (Continued) εcu = 0.003 εstf

c =12.9”

β1c = 9.68”

0.85 fc’

dtf = 5.17” Fstf = 680 k

dbf = 17.84”

24”

Strains

Fsw = 36.3 k

fs(c,z)

dz εsbf

Msw = 56 ft-k

Fsbf = 606 k

z Forces/moments

Stresses

24”

Figure 21.5.6  Stresses, resultant forces, and moments for steel section in Example 21.5.1.

The moment contribution from the encased steel section is divided into the contributions from the section flanges, Msf , and that from the web, Msw (Fig. 21.5.6) h  h  M sf = Ftf  − dtf  + Fbf  − dbf  = 732 ft-kips 2  2  h + ds



M sw = ∫ h −2ds 2

−t f +t f

h  fs (c, z )tt w  z −  dz = 56.3 ft-kips  2

M s = M sf + M sw = 732 + 56.3 = 788 ft-kips where z is the vertical distance from the base of the section to the point on the web being evaluated (Fig. 21.5.6), and fs(c, z) is the stress at that point, which depends on the curvature (and thus, neutral axis depth c) and the distance between the point on the web being considered and the neutral axis [z –​ (h – ​c)]. For web locations within the concrete compressive stress block, the stress in the web is reduced by 0.85 fc′ . The total flexural strength for the design axial strength φPn = 935 kips is then M c = M c + M sr + M s = 707 + 253 + 788

= 1748 ft-kips

φb M n = 1573 ft-kips > Mu = 1100 ft-kips



The design is thus adequate to resist the factored axial force and moment. (c) Check shear capacity. For illustration purposes, the nominal shear strength of the composite column will be calculated following the three options given in AISC. -​Steel section alone: Vn = 0.60 f ys Aw Cv = 0.60(50)(0.645)(14.7)(1.0)

= 284 kips

φ vVn = (1.0)(284) = 284 kips >Vu = 130 kips

(Continued)

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C hapter   2 1     C omposite M embers and C onnections

Example 21.5.1 (Continued) -​Steel section plus lateral ties: Vn = 0.60 f ys Aw Cv +

Av f yt d s

0.20(2 + 2 )(60)(0.8)(24) 10 = 284 + 78.7 kips

Vn = 284 +

= 363 kips Note that a factor equal to 2 / 2 is applied to the area of shear reinforcement provided by the octagonal shaped hoop to account for the portions near the column corners oriented at 45° from the direction of shear. Also, d was taken as 0.8h (see Section 10.21). Verify that Vs is less than upper limit. Vs < 8 fc′bwE d =



8 6000 (24 − 14.7)(19.2) 1000

= 111 kips

OK

Thus,

φ vVn = (0.75)(363) = 272 kips > Vu = 130 kips



Verify that the selected reinforcement satisfies minimum shear reinforcement requirement. From Eq. (5.10.8) and noting that 0.75 fc′ > 50 psi for fc′ = 6000 psi,    min Av =

0.75 fc′ f yt

=

0.75 6000 bw s = 0.23 sq in. < 0.20(2 + 2 ) = 0.68 sq in. OK 60, 000

Verify maximum tie spacing. Using an effective web width bwE = (24 –​ 14.7) = 9.3 in., the tie shear contribution Vs corresponds to 5.7 fc′bwE d. This leads to a maximum tie spacing of d/​4 = 4.8 in., which is less than the spacing provided (10 in.). According to this procedure, the shear design would thus be inadequate. -​Reinforced concrete section:  Nu  Vc = 2  1 +  λ fc′bw d  2000 Ag  935, 000   1 (1.0) 6000 (24)(19.2) = 2 1 + 1000  2000(24)(24)  = 129 kips

Vs =

Av f yt d s

=



0.20(2 + 2 )(60)(19.2) = 79.8 kips 10

OK

Vs < 8 fc′bwE d



φ vVn = φ v (Vc + Vs ) = 0.75(144 + 79.8) = 168 kips > Vu = 130 kips

As for the case of shear strength assumed to be provided by the steel section and tie reinforcement, the area of tie reinforcement satisfies the minimum required. However, tie spacing is limited to d/​4, which is smaller than the spacing provided. In this example, the shear design is adequate according to the first procedure. The other two procedures, on the other hand, would require a reduction in tie spacing. Given that the shear strength of the bare steel section is adequate, the shear design is considered acceptable. (Continued)

931



21.5  CONCRETE-ENCASED STEEL COMPOSITE COLUMNS

931

Example 21.5.1 (Continued) (d) Check minimum tie requirements for lateral support of longitudinal bars. ACI-​ 10.7.6.1.2 and ACI-​25.7.2 (see Section 10.8) require the following. –​ Minimum #3 size tie for #10 or smaller longitudinal bars. Because #4 ties are used with #7 longitudinal bars in this case, this requirement is satisfied. –​ Lateral tie spacing shall not exceed 16 longitudinal bar diameters, 48 tie bar diameters, or the least dimension of the column, which correspond to 14 in., 24 in., and 24 in., respectively. Thus, the tie spacing of 10 in. satisfies this requirement. –​ Every corner and alternate longitudinal bar shall have lateral support provided by the corner of a tie having an included angle of not more than 135° and no bar shall be farther than 6 in. clear on either side from such a laterally supported bar. All longitudinal bars are laterally supported by the corner of a tie having an included angle of not more than 135°. This requirement is therefore satisfied. In addition to the requirements in ACI-​10.7.6.1.2 and ACI-25.7.2, the following minimum requirements apply (AISC-​I2.1b): –​ #3 ties at 12-​in. spacing or #4 ties at 16-​in. spacing. As #4 ties at 10 in. spacing are provided, this requirement is satisfied. –​ Tie spacing not greater than half the least cross-​sectional dimension. The provided spacing of 10 in. is less than half the least cross-​sectional dimension (12 in.) It is also recommended, based on ACI-​10.7.6.1.4, that the tie diameter be not less than 0.02 times the maximum column side dimension, but not smaller than #3. This provision would require a minimum diameter of 0.48 in., which is slightly less than the tie diameter (0.5 in.).

EXAMPLE 21.5.2 Two W14 × 53 floor girders, framing into opposite faces of the composite column of Example  21.5.1, are shear-​connected to the flanges of the encased steel shape (see Fig. 21.5.7). If the factored shear at the end of the floor girders is 55 kips, determine the number and spacing of headed stud anchors required to transfer the corresponding force Vr′ to the reinforced concrete encasing. Determine also the required stud spacing outside of the load introduction length. Tensile strength of studs, fua = 65 ksi.

8.5” 5” Vu = 55 k

Vu = 55 k

W14 × 53

W14 × 132 24”

Shear tab

Reinforcing bars in column and concrete slab not shown for clarity.

Figure 21.5.7  Connection between steel beam and composite column of Example 21.5.2.

(Continued)

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Example 21.5.2 (Continued) SOLUTION (a) Determine force to be transferred from encased steel shape to reinforced concrete encasing, Vr′ . From Eq. (21.5.11) and Example 21.5.1, As f y    (38.8)(50)  = 55(2) 1 − Vr′ = Pr  1 − = 68.0 kipps Pnso  5075   



(b) Determine diameter and required number of headed stud anchors to transfer force Vr′ . As required by AISC-​I8.3e, which refers to ACI 318-​14, at least a 1.5-​in. clear cover to the outside edge of the stud must be provided (assuming cast-​in-​place column not in contact with ground; ACI-​20.6.1.3.5). Further, as discussed above, it is desirable to keep the head of the studs within the core of the column. Assuming a clear cover of 1.5 in. to the column ties, the maximum possible headed stud length is [(24–​14.7 in.) /​2–​1.5 in.] = 3.15 in. The maximum stud diameter for a minimum stud length of 5dsa (AISC-​I8.3) is then 3.15 in./​5 = 0.63 in. Use 5 8 -​in.-​diameter headed stud anchors with a length Lsa = 5dsa = 3 1 8 in. The shear strength of a single stud is calculated form Eq. (21.5.12), where Asa = 0.31 sq in. and fua = 65 ksi. Qnv = fua Asa = 65(0.31) = 20.2 kips



The required number of shear studs is calculated from Eq. (21.5.13).

ηa ≥



Vr′ 68.0 = = 5.2 (Use 6 studs minimum) φ v Qnv 0.65(20.2)

A minimum of three studs on each flange of the encased steel shape is thus required to transfer the force Vr′ . To maintain symmetry about the beam axis also, use four studs on each flange, two above and two below the shear tab. (c) Headed stud layout. As only eight studs are required over the load introduction length (not to exceed 96 in. or twice the minimum column cross-​sectional dimension above and below the shear tab), four above and four below the shear tab, maximum spacing considerations will govern in this case. According to AISC-​I8.3e, headed stud spacing should not exceed 32dsa = 24 in. This same requirement applies to the column regions outside the load introduction length. The final headed stud layout is shown in Fig. 21.5.8.

48” 20”

Load introduction length ≤ 96” 20” 48”

Headed stud anchors at 20 in. spacing (dsa = 5/8”, Lsa = 3 1/8”) 24”

Reinforcing bars in column and concrete slab not shown for clarity.

Figure 21.5.8  Headed stud layout in composite column of Example 21.5.2.

93



21.5  CONCRETE-ENCASED STEEL COMPOSITE COLUMNS

933

EXAMPLE 21.5.3 Determine the number and layout of headed studs in the connection of Example 21.5.2 assuming factored shear forces of 65 and 45 kips for the left and right girder, respectively (Fig. 21.5.9).

Vr’/2 + Mva/ds

Vr’/2 – Mva/ds

Vu = 65 k

Vu = 45 k

8.5”

ds

8.5”

Reinforcing bars in column and concrete slab not shown for clarity.

Figure 21.5.9  Beam-​column connection and transfer of forces between steel shape and concrete encasing for Example 21.5.3.

SOLUTION (a) Headed studs required to transfer force Vr′ . The total axial force introduced into the column, Pr , and Vr′ , are the same as in Example 21.5.2 (110 kips and 68 kips, respectively). Thus, six studs are required for transferring the force Vr′ . (b) Determine number of studs required to transfer moment from encased steel shape to reinforced concrete encasing. The total moment transferred from the steel girders to the encased steel section, Mv, is

 14.7  M v = (65 − 45)  + 8.5 = 317 in.-kips  2 

The fraction of Mv to be transferred to the reinforced concrete encasing, Mva, can be calculated assuming the same proportion as for axial load strength between the steel shape and reinforced concrete encasing. Thus,

As f y    (38.8)(50)  = 317 1 − M va = M v  1 − = 196 in.-kiips Pnso  5075   

The transfer of a moment Mva, counterclockwise, will increase the force in the studs on the left flange of the encased shape while decreasing the force in the studs on the right flange (Fig. 21.5.9). It is recommended, however, that the same number of studs be added to both flanges to maintain symmetry as well as to eliminate the possibility of placement errors. The required number of additional studs on the left flange of the encased steel shape, (na)Mva, are



(na ) Mva ≥

 M va   d  s

φ v Qnv

 196    14.7  = 1.0 = 0.65(20.2) (Continued)

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Example 21.5.3 (Continued) A minimum of four studs on each flange (two above and two below the shear tab) is required within the load introduction length. The total number of studs within the load introduction length is therefore 8, two more than the minimum required to transfer the force Vr′ , while the number of studs in each flange is 4, which satisfies the required number of studs on the left flange to transfer both Vr′ and the moment Mva (2.6 studs for Vr′ and 1 stud for Mva). The headed stud layout shown in Fig. 21.5.8 satisfies both strength and spacing requirements.

21.6 CONCRETE-​F ILLED TUBE COLUMNS The ease of construction of concrete-​filled steel tube (CFT) columns has made these composite members a very attractive column design alternative. CFT columns have been increasingly used in the past four decades, finding applications in both low-​rise and high-​ rise construction [21.31]. As interest by structural designers and contractors in field applications of CFT columns grew, so did the need for research to understand their behavior under various loading conditions. Research on CFT columns covers a wide variety of areas, including strength, stiffness, concrete-​steel tube bond, stability, and long-​term behavior (see, e.g., Refs. 21.32–​21.49). For an excellent summary on research work conducted on CFT columns in the past several decades, the reader is referred to Gourley et al. [21.50]. From tests of square and circular CFT columns, Furlong [21.32, 21.33] indicated that the infill concrete and the steel tube tend to behave independently under loading and that adhesion between the concrete and the steel does not prevent relative sliding between the two materials. As reported in Roeder et al. [21.45], however, mechanical bond caused by interior roughness of the steel tube will develop in typical applications, but at a decreasing rate with an increase in the ratio between the tube diameter D (for circular tubes) and thickness t. Further, as the compressive stress in the concrete approaches its peak strength, lateral expansion of the concrete will induce lateral pressure on the tube walls, creating additional friction at the concrete-​steel interface and confinement of the concrete by the steel tube. As noted by Knowles and Park [21.34], however, such an expansion can develop only if the steel tube has not buckled by the time the stress in the concrete approaches its compressive strength. The friction that develops between the infilled concrete and surrounding steel tube seems to be sufficient to allow the use of a strain compatibility approach to calculate the strength of CFT columns under combined axial force and bending, as allowed in AISC-​I1.2b. Even though this approach is based on the assumption of perfect bond between the concrete and steel tube, its suitability has been demonstrated through experimental research, where the strength of CFT short columns subjected to combined axial force and bending has been calculated with reasonable accuracy by means of strain compatibility [21.32, 21.35]. For consistency with the treatment of reinforced concrete columns in Chapter 10, as well as concrete-​encased steel composite columns earlier in this chapter, the strain compatibility approach will be used for calculation of strength of CFT columns under combined axial force and bending.

Reinforcement Limits In CFT columns, the minimum cross-​sectional area of the steel tube must be at least 1% of the gross area of the column (AISC-​I2.2a). In typical CFT columns, no longitudinal reinforcing bars are used, and no minimum is required in the AISC Specification. When longitudinal reinforcement is used, as in CFT columns of high-​rise structures, a minimum area equal to 0.4% of the gross area of the column is recommended, as for concrete-​encased steel composite sections.

935



935

21.6  CONCRETE-FILLED TUBE COLUMNS

Material Strength Limits In the AISC Specification for Structural Steel Buildings, the same concrete strength limitations for concrete-​encased steel composite columns apply to CFT columns (AISC-​I1.3). Specified concrete compressive strength fc′ must be at least 3000 psi. For strength calculation purposes, fc′ is limited to 10,000 psi and 6000 psi for normal-​weight and lightweight concrete, respectively. No upper limit in concrete compressive strength is specified for stiffness calculations. The specified yield strength of the steel section and reinforcing bars used for strength calculations cannot exceed 75 ksi and 80 ksi, respectively.

Design Wall Thickness and Limits on Tube Width-​to-​Thickness Ratio The design wall thickness t for CFT sections is generally smaller than the nominal tube thickness. Section AISC-​B4.2 requires that the design wall thickness be taken as 0.93 times the nominal wall thickness for tube sections, except those produced according to ASTM A1065 or A1085, for which the design thickness may be taken equal to the nominal thickness. The behavior of a tube section under compression and/​or bending is dependent on the stability of the tube walls. Tube sections with thick walls relative to their width are likely to be able to yield in compression without buckling, while thin-​walled tubes are likely to exhibit buckling prior to yielding. The AISC Specification for Structural Steel Buildings thus establishes width-​to-​thickness limits for purposes of classifying the walls of CFT members in terms of local buckling as compact, noncompact, or slender (AISC-​I1.4). These limits are given in Table 21.6.1. For rectangular sections, the limits apply to the wall width-​to-​thickness (b/​t) ratio, where the wall width is taken as the clear distance between the perpendicular tube walls (e.g. clear distance between flanges when calculating web width) minus the inside-corner radius on each side. For circular sections, these apply to the diameter-​to-​thickness (D/​t) ratio. A compact section is expected to be able to undergo significant yielding prior to local buckling. Noncompact sections, although able to undergo compression yielding prior to buckling, are not expected to provide adequate confinement to the concrete at large compressive stresses. Slender sections will undergo local buckling without any yielding. The limits in Table  21.6.1 are higher than those applicable to hollow steel sections (HSS). This is because in hollow steel members, the tube walls may buckle inward and outward. In CFT members, on the other hand, the buckled shape is different because no

TABLE 21.6.1  WIDTH-​TO-​THICKNESS RATIO LIMITS (b/​t OR D/​t) FOR CFT MEMBERS

Members subjected to axial compression

Members subjected to bending

Element

Compact

Walls of rectangular sections

≤ 2.26

Circular sections

≤ 0.15

Flanges of rectangular sections

≤ 2.26

Es f ys

2.26

Es E to 3.00 s f ys f ys

Webs of rectangular sections

≤ 3.00

Es f ys

3.00

Es E to 5.70 s f ys f ys

Circular sections

≤ 0.09

Es f ys Es f ys

Es f ys

Noncompact Es E to 3.00 s f ys f ys

2.26 0.15

0.09

Es E to 0.19 s f ys f ys

Es E to 0.31 s f ys f ys

Slender Es f ys

> 3.00 > 0.19

Maximum limit

Es f ys Es f ys

5.00 0.31

Es f ys Es f ys

5.00

Es f ys

—​

5.70

Es f ys

—​

0.31

> 3.00

Es f ys

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inward displacement is possible, which increases the stability of the tube walls. The vast majority of the commercially available steel hollow sections made out of ASTM A500 steel in the United States qualify as compact sections. It should also be mentioned that by satisfying the limits for compact sections in Table 21.6.1, the required minimum area of steel of 1% of the gross column area is also satisfied. The discussion that follows will thus focus on compact sections.

Axial Compressive Strength (Short Columns–​Compact Sections) As for concrete-​encased steel composite columns, the strength of short CFT columns is first discussed, followed by slenderness effects. The strength of a compact CFT short column subjected to pure axial load is calculated using Eq. (21.6.1), which is the same as Eq. (21.5.1), but using a generic concrete stress equal to C2   fc′ in order to account for an increase in concrete compressive strength due to lateral confinement in circular tubes.

Pnso = C2 fc′( Ag − As − Asr ) + As f ys + Asr f yr

(21.6.1)

where C2 = 0.85 for rectangular sections and 0.95 for circular sections. For CFT sections without steel reinforcing bars, Asr  =  0. It should be noted that Eq. (21.6.1) differs from Eq. (I2-​9b) in AISC-​I2.2b in the treatment of the contribution from longitudinal reinforcing bars, where the reinforcing bar stress is taken equal to (C2  fc′ )Es  /​Ec. This wrongly implies that the concrete exhibits linear elastic behavior up to its peak stress. Assuming such elastic behavior leads to a significant underestimation of the strain in both the concrete and longitudinal steel at peak concrete stress. The writers thus disagree with this approach and recommend the use of Eq. (21.6.1) instead.

Combined Axial Compression and Bending (Short Columns–​Compact Sections) The strength of CFT sections under combined axial force and bending can be calculated by using strain compatibility, a plastic stress distribution, an elastic stress distribution, or an effective stress-​strain method (AISC-​I1.2). As mentioned above, good agreement has been obtained between experimental and calculated strengths when strain compatibility was used [21.32, 21.35]. Based on this experience, and for consistency with the treatment of reinforced concrete columns, the strain compatibility approach is used here to illustrate the calculation of the strength of CFT sections under combined axial force and bending. The strain and stress distribution at nominal condition for a circular CFT column section subjected to combined axial force and bending is shown in Fig. 21.6.1. The steel tube can be modeled as a series of discrete equivalent reinforcing bars or as a continuous element. The latter has been chosen in the calculation of forces below. The stress intensity in the concrete compression block is generally taken as 0.85 fc′ . In circular sections analyzed by using the plastic stress distribution method of AISC-​I1.2a, a stress intensity of 0.95 fc′ is allowed because of the concrete confinement provided by the steel tube. This implies that such an increase in stress is also allowed when strain compatibility is used. For uniformity of treatment in the analysis of circular and rectangular sections, however, a single concrete stress intensity of 0.85 fc′ will be used. The procedure can be easily modified to account for the increased concrete stress intensity in circular sections. The calculation of the resultant concrete compressive force requires the determination of the concrete core area within a depth a from the extreme concrete compressive fiber, Acomp, as discussed in Section 10.18 (see Figures 10.18.1 and 10.18.2). Note that the diameter of the concrete core Dc = D - ​2t, rather than the outside tube diameter D, should be used in

937



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21.6  CONCRETE-FILLED TUBE COLUMNS

fys

εsy

Acomp

θs

0.85fc’ *

εcu = 0.003

Location of Cc

a = β1(c–t)

c yc

Cc

α D

t dθs

Dc/2

εsy

R εs,max

Cross section

f ys

Strains

Stresses (concrete)

Stresses (steel)

* A stress intensity of 0.95fc’ is allowed in circular sections.

Figure 21.6.1  Strain and stress distribution at nominal for circular CFT column section.

the determination of the concrete area in compression. The resultant concrete compressive force and its moment about the column centroid are

Cc = 0.85 fc′Acomp

(21.6.2)

M c = Cc yc

(21.6.3)

The strain εs and stress fs at any point on the steel tube defined by the angle θs in Figure 21.6.1 can be determined as follows:

ε s (θ s ) = [c + R(cos θ s − 1)]

ε cu c−t

(21.6.4)



− f ys ≤ fs (θ s ) = ε s (θ s ) Es ≤ f ys

(21.6.5)

The resultant axial force in the steel tube and the contribution of the steel tube to the section moment capacity are π

Fs = 2∫ fs (θ s )tR dθ s 0

π



M s = 2∫ fs (θ s )tR( R cos θ s )dθ s 0

(21.6.6) (21.6.7)

The nominal axial strength and moment in the composite section are then obtained as the summation of the contributions from the steel tube and the concrete core.



Pns = Cc + Fs

(21.6.8)

Mn = Mc + Ms

(21.6.9)

Calculation of the nominal axial strength and moment in a rectangular CFT section is easier. In that case, the same procedure used for rectangular concrete-​encased steel sections can be followed, with the tube walls parallel to the axis of bending representing the flanges and the other two tube walls being the web. When longitudinal bars are used in the core of the CFT column, the contribution of the reinforcing bars to axial force and moment strength can be determined as for a reinforced concrete section.

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Slenderness Effects The effect of initial imperfections in CFT columns on axial compressive strength is accounted for in the same manner as for concrete-​encased steel composite columns. However, a different expression must be used for the calculation of effective flexural rigidity (AISC-​I2.2b):

EI eff = Es I s + Es I sr + C3 Ec I c

(21.6.10)

 A + Asr  C3 = 0.45 + 3  s  ≤ 0.90  Ag 

(21.6.11)

where



On comparing Eq. (21.6.11) with Eq. (21.5.5), it can be seen that a higher contribution of the concrete to the section flexural rigidity is used in the case of CFT columns because of the confinement provided by the steel tube. Once the effective flexural rigidity has been determined, the elastic buckling load Pe is calculated by using Eq. (21.5.3), and the strength reduction factor due to initial imperfections, κ, is determined (for compact CFT sections with Pnso /​Pe ≤ 2.25) by using Eq. (21.5.2). As was the case for concrete-​encased steel composite sections, AISC does not account for the effect of sustained load on the stiffness of CFT columns. Creep strains in CFT columns substantially lower than those in reinforced concrete columns, as well as small shrinkage-​ induced strains, have been reported [21.48]. Further, a negligible increase in steel strain due to concrete creep was reported in circular CFT columns under sustained loading [21.49].

Axial Tensile Strength The strength of a CFT column under pure tension is calculated as the summation of the yield strength of the steel shape and longitudinal bars (see Eq. 21.5.7).

Shear Strength Shear rarely controls the design of a CFT section. Thus, the shear strength of CFT column sections is typically conservatively taken as that of the bare steel section, as specified in AISC-​I4.1(a). Although AISC-​I4.1 also allows the calculation of the shear strength of a CFT column as that of the reinforced concrete portion or as the summation of the shear strength of the steel section and the reinforcing steel, these two options are often not appropriate, since traditional bar-​type shear reinforcement is not typically used in CFT columns, even when longitudinal bars are present. The nominal shear strength of a rectangular CFT section is calculated based on Section AISC-​G4 as follows,

Vn = 0.60 f ys Aw Cv

(21.6.12)

where Cv = 1.0 for all compact sections. The effective shear area Aw = 2ht, where h is the length of the straight portion of the tube walls parallel to the direction of shear, and t is the design thickness. The distance h can be calculated as the clear distance between flanges minus 2 times the inside corner radius of the tube. If the corner radius is not known, h is taken as the overall depth of the section in the direction of shear minus 3 times the wall thickness. For compact circular sections, the nominal shear strength is calculated as (AISC-​G5)

Vn = 0.60 f ys Aw

where Aw is taken as As /​2. The applicable shear strength reduction factor φv for CFT sections is 0.90.

(21.6.13)

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21.6  CONCRETE-FILLED TUBE COLUMNS

EXAMPLE 21.6.1 A composite column consisting of an HSS20 × 0.50 steel section filled with concrete is subjected to the following factored loads obtained from a second-​order analysis satisfying the requirement of the Direct Design Method (not including the effect of initial imperfections): Pu = 735 kips; Mu = 680 ft-​kips; Vu = 85 kips. Unsupported column length L = 12 ft. For the HSS20 × 0.50 steel shape: D = 20.0 in.; As = 28.5 sq in.; t = 0.465 in.; Is= 1360 in.4. Material properties:  fc′  = 8000 psi (normal weight); fys = 42 ksi (A500 Grade B steel). Verify that the concrete-​filled tube column is adequate to resist the factored loads according to the AISC Specification for Structural Steel Buildings. SOLUTION (a) Check compactness criterion. From Table 21.6.1, E D 20.0 = = 43.0 < 0.09 s = 62.1 t 0.465 Fys



Section is compact for both compression and bending. (b) Check minimum area of steel. As discussed earlier, all compact sections satisfy the minimum steel area requirement of 1% of the column area. For this example, As 28.5 = = 0.09 > 0.01 Ag π(20)2 4



OK

(c) Check axial force and moment strength. The determination of the nominal axial strength of the column requires first the calculation of the equivalent flexural rigidity EIeff , Euler’s buckling load, Pe, and axial strength reduction factor, κ. The effective flexural rigidity, EIeff, is calculated using Eqs. (21.6.10) and (21.6.11), with the variables calculated as follows: Es = 29, 000 ksi I s = 1360 in.4



Ic =

π Dc4 π [20 − 2(0.465)]4 = = 6492 in.4 64 64

  28.5 C3 = 0.45 + 3  2  π (20)  4 Ec = 57, 000 8000

   = 0.72 < 0.90  

1 1000

= 5098 ksi

Thus,

EI eff = (29, 000)(1530) + 0.72(5098)(6492) = 6.34 × 10 7 kip-sq in.



and

Pe =

π 2 EI eff π 2 (6.34 × 10 7 ) = = 30,180 kips (kL )2 [1(12)(12)]2

where k has been taken as 1.0.

(Continued)

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Example 21.6.1 (Continued) The pure axial load capacity neglecting slenderness effects, Pnso, is obtained using Eq. (21.6.1) with Asr = 0. Pnso = C2 fc′( A g − As ) + As f ys  π [20 − 2(0.465)]2  = 0.95(8)   + 28.5(42) 4 





= 3368 kips

Pnso /​Pe = 0.11 < 2.25. The axial strength reduction factor κ is then,

κ = 0.658

Pnso Pe

 3368 

  = 0.658 30,180  = 0.95

As in Example 21.5.1, the adequacy of the column section design will be checked by calculating the design moment strength, φb Mn, corresponding to a design axial strength φc Pn = φc κPnso = Pu. The strain and stress distribution corresponding to φc Pn = 735 kips is shown in Fig. 21.6.2. As can be seen, the maximum compressive strain in the concrete is taken as 0.003, which means that the strain in the extreme compression fiber of the steel tube is slightly larger than 0.003. For this condition, the neutral axis depth, measured from the extreme tube compression fiber is c = 12.85 in. Dc = 19.07”

0.85 fc’ = 6.8 ksi

εcu = 0.003

Acomp =115 in.2 c = 12.85”

εsy

42 ksi

a = 8.05”

4.68” Cc = 779 k

yc = 4.86”

D = 20” α = 1.414 rad

εsy = 0.00145 t = 0.465” Cross section

Strains

42 ksi Stresses (concrete)

Stresses (steel)

Figure 21.6.2  Strains and stresses for calculation of Pns–​Mn in Example 21.6.1.

The resultant concrete compressive force Cc can be determined by using Figure 10.18.1 with h = Dc and a depth of compressed concrete ηh = β1(c – ​t) = 8.05 in., where β1 = 0.65.  α − sin α cos α  Acomp = Dc2   4  

where



 D / 2 − β1 (c − t )  α = cos −1  c  Dc / 2  

 9.54 − 0.65(12.85 − 0.465)  = cos   = 81.0 º = 1.414 rad 9.54   −1

Thus,



1.414 − sin(1.414) cos(1.414)  Acomp = (19.07)2   = 115 sq in. 4   Cc = 0.85 fc′Acomp = 0.85(8)(115) = 779 kips

(Continued)

941



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21.6  CONCRETE-FILLED TUBE COLUMNS

Example 21.6.1 (Continued) The contribution of the resultant concrete compressive force to the section moment capacity is [Eq. (21.6.3)],

M c = Cc yc



From Fig. 10.18.2, yc =



Dc3 (sin α )3 (19.07)3 (0.988)3 = = 4.86 in. 12 Acomp 12(115)

Thus,

 4.86  M c = Cc yc = 779  = 316 ft-kips  12 



The resultant axial force in the steel tube and its contribution to the moment capacity of the section are calculated using Eqs. (21.6.6) and (21.6.7). π

π

0

0

Fs = 2∫ fs (θ s ) t R dθ s = Dt ∫ fs (θ s ) dθ s = 246 kips

π

M s = 2∫ fs (θ s )t R( R cos θ s ) dθ s = 0

π D2 t fs (θ s ) cos θ s dθ s = 575 ft-kips ∫ 2 0

where fs (θ s ) is calculated from Eqs. (21.6.4) and (21.6.5). The resultant axial force and moment are then

Pns = Cc + Fs = 779 + 246 = 1025 kips M n = M c + M s = 316 + 575 = 891 ft-kips



Taking into account slenderness effects due to initial imperfections, the nominal and design axial strengths are

Pn = κ Pns = 0.95(1025) = 978 kips

φc Pn = 0.75(978) = 733 kips ≈ Pu



and the design moment capacity is

φb M n = 0.9(891) = 801 ft-kips > Mu = 680 ft-kips

The design is thus adequate to resist the factored axial force and moment. (d) Check shear capacity. From Eq. (21.6.13), Vn = 0.60 f ys Aw = 0.60 f ys



As 2

 28.5  = 0.60(42)  = 359 kips  2 

φ vVn = (0.9)(359) = 323 kips > Vu = 85 kips The design is also adequate to resist the factored shear force.



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Load Transfer Typical shear connections between steel girders and CFT columns consist of a shear tab welded to the tube section. Thus, transfer of load from the steel tube to the concrete core must occur if the column is to behave as a composite member. Bond between steel and concrete in CFT columns has been studied experimentally, typically through push-​out tests with or without a shear tab [21.43–​21.47]. While some friction develops between the steel and concrete depending on the roughness of the interior surface of the tube, rotation of the shear tabs leads to high concrete bearing stresses at the bottom of the shear tab, increasing shear transfer. Also, as the top of the shear tab pulls the tube in one direction, the tube pushes against the concrete in the lateral direction, further increasing bond. Current bond strength provisions for CFT columns in AISC are based on the work by Zhang et al. [21.46]. The shear transfer strength along the load introduction length is (AISC-​I6.3c) Rn = pb Lin Fin



(21.6.14)

where pb is the perimeter of the steel-​concrete interface, Lin is the load introduction length, and Fin is the nominal bond stress. The load introduction length Lin is taken as twice the minimum cross-​sectional dimension (rectangular sections) or twice the diameter (circular sections) above and below the shear transfer region (AISC-​I6.4b). The nominal bond stress is

Fin =

12 t ≤ 0.1 ksi H2

(rectangular sections)

(21.6.15a)

Fin =

30 t ≤ 0.2 ksi D2

(circular sections)

(21.6.15b)

where H is the maximum cross-​sectional dimension, D is the tube diameter, and t is the design tube thickness. As can be seen, the bond stress equations reflect the trend observed in tests, where an increase in cross-​sectional dimensions is accompanied by a decrease in bond [21.46]. Also, larger bond stresses are developed in circular tubes than in rectangular sections. The strength reduction factor applicable to bond is 0.50. In most cases, bond between the steel tube and concrete core is sufficient to transfer the required force from the steel tube to the concrete. However, when the force transferred through bond is not sufficient, shear connectors must be used, as discussed for concrete-​ encased steel composite columns. In that case, the shear strength along the tube-​concrete interface over the load introduction length is taken solely as that provided by the shear connectors. This is because slips associated with peak bond stresses in tests are much lower than slips required to develop the strength of shear connectors [21.46].

EXAMPLE 21.6.2 For the CFT column in Example 21.6.1, calculate the maximum shear force that can be transferred by steel girders without the need for shear connectors. Assume the column extends above the steel girders. SOLUTION (a) Determine nominal bond strength along load introduction length. 30t 30(0.465) = = 0.035 ksi < 0.2 ksi D2 (20)2 Lin = 2(2 D) = 2(40) = 80 in. Fin =



pb = π ( D − 2t ) = π [20 − 2(0.465)] = 59.9 in. Rn = pb Lin Fin = 59.9(80)(0.035) = 167 kips



(Continued)

943



21.7  MOMENT CONNECTIONS WITH COMPOSITE COLUMNS

943

Example 21.6.2 (Continued) The maximum shear force that can be transferred by the steel girders, Vu, is calculated from Eq. (21.5.11) by setting Vu = Pr and Vr′  = φRn.

Vu =

φ Rn 0.50(167) = 130 kiips = As f y   28.5(42)   − 1  1 − P   3368  nso

21.7 MOMENT CONNECTIONS WITH COMPOSITE COLUMNS Extensive research on connections between steel girders/​beams and composite columns, either concrete-​encased steel composite columns or concrete-​filled tubes, was conducted in the late 1980s and 1990s, primarily under earthquake-​type loading [21.51–​21.61]. Connections between concrete-​encased steel composite columns and steel beams can be lumped in two main groups, through-​column and through-​beam connections. In the former, the encased steel column is continuous through the joint, while in the latter the steel beam passes continuously through the joint. Through-​column type connections are similar to moment connections in steel frames [21.55, 21.62], involving welding of the steel beam flanges to the encased steel shape. Transverse reinforcement is necessary to provide confinement to the joint, although research [21.55] indicates that lower amounts than those required for reinforced concrete joints (see Chapter 11) have led to adequate seismic performance. In some cases, additional steel plates, welded to the steel beam flanges and paral­ lel to the beam web, are used for increased shear resistance [21.55]. Through-​beam connections are simpler than through-​column connections and do not rely on welding for force transfer between the steel beam flanges and the encased steel column. Typical details in these connections consist of either face bearing plates in combination with transverse reinforcement for confinement, or face bearing plates and steel band plates wrapping around the column just above and below the steel beam (Fig. 21.7.1). Because the steel beam is continuous through the column, moments are transferred in one direction only, although perpendicular beams may transfer shear into the column. Joint shear strength in through-​beam-​type connections is provided by the steel web panel, an inner concrete strut activated through concrete bearing of the face bearing plates and beam flanges, and an outer concrete panel, activated through concrete bearing of the encased steel section in combination with steel hoops above and below the steel beam, or through concrete bearing of the steel band plates. For connection with no transverse beams, joint confinement can be provided through overlapping U-​shaped ties passing through holes drilled on the web of the steel beam [Fig. 21.7.1(a)]. For connections with transverse beams, on the other hand, joint confinement may be provided by steel band plates welded to the steel beam flanges (Fig. 21.7.1(b)]. Additional ties, however, are needed to provide lateral support to the longitudinal column reinforcement over the beam depth. These ties may be anchored in the column core and do not need to pass through the steel beam web. Design guidelines for through-​beam type connections were published by an ASCE Task Group in 1994 [21.63]. Additional design recommendations are available in References 21.58 and 21.62. To ensure adequate behavior in moment connections between steel girders/​beams and CFT columns, both the steel tube and the concrete core must participate in the transfer of forces. Tensile and compressive forces in the beam flanges must be transferred to the concrete through bearing, typically using a bolted connection. An effective connection detail that has been shown to behave well under large shear reversals is shown in Fig. 21.7.2 [21.60]. In this case, the bolts passing through the column are unbonded and

94

944

C hapter   2 1     C omposite M embers and C onnections Steel column

Hoops

Transverse beam

Steel beam

Steel column

Steel beam Face bearing plates

U-shaped ties

Ties

Column bars (a) Connection with no transverse beams

Steel band plates

Face bearing plates

Column bars

(b) Connection with transverse beams

Figure 21.7.1  Typical details for through-​beam connections between steel beams and concrete-​ encased steel composite columns.

CFT column Through rods (unbonded)

Split-tee

Shear tab

Headed studs

Steel beam

Figure 21.7.2  Steel beam-​to-​CFT column moment connection.

must be designed to behave elastically, thus avoiding an unacceptable decrease in stiffness. Although the connection shown in Fig. 21.7.2 includes shear connectors for shear transfer, these may not be necessary in a connection transferring gravity loads, as discussed in Section 21.6. Further information on the design of this type of connection is available in References 21.62 and 21.64.

SELECTED REFERENCES 21.1. American Institute of Steel Construction. Steel Construction Manual (14th ed.). Chicago: AISC, 2011. 21.2. J. C. Saemann and George W. Washa. “Horizontal Shear Connections Between Precast Beams and Cast-​in-​Place Slabs,” Journal of the American Concrete Institute, 61, November 1964, 1383–​1408. 21.3. Robert E. Loov and Anil K. Patnaik. “Horizontal Shear Strength of Composite Concrete Beams With a Rough Interface,” PCI Journal, 39, January–​February 1994, 48–​69. 21.4. N. W. Hanson (1960). “Precast-​Prestressed Concrete Bridges. 2—​Horizontal Shear Connections,” Journal, PCA Research and Development Laboratories, 2, May 1960, 38–​58. PCA Development  Department Bulletin D35, 1960, 21 pp. 21.5. Philip W. Birkeland and Halvard W. Birkeland. “Connections in Precast Concrete Construction,” Journal of the American Concrete Institute, 63, March 1966, 345–​367. 21.6. Alan H.  Mattock and Neil M.  Hawkins. “Shear Transfer in Reinforced Concrete—​Recent Research,” PCI Journal, 17, March–​April 1972, 55–​75.

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SELECTED REFERENCES

945

  21.7. Joost Walraven, Jerome Frénay, and Arjan Pruijssers. “Influence of Concrete Strength and Load History on the Shear Friction Capacity of Concrete Members,” PCI Journal, 32, January–​ February 1987, 66–​84.  21.8. Alan H.  Mattock, W.  K. Li, and T.  C. Wang. “Shear Transfer in Lightweight Reinforced Concrete,” PCI Journal, 21, January–​February 1977, 20–​39.  21.9. Dan E.  Branson. “Time-​Dependent Effects in Composite Concrete Beams,” Journal of the American Concrete Institute, 61, February 1964, 213–​229. 21.10. Dan E. Branson. “Design Procedures for Computing Deflections,” ACI Journal, 65, September 1968, 730–​742. 21.11 Antoine E. Naaman. Prestressed Concrete Analysis and Design—​Fundamentals. Ann Arbor, MI: Techno Press 3000, 2012, 1173 pp. 21.12 Holger Eggemann. “Development of Composite Columns. Emperger’s Effort,” Proceedings of the First International Congress on Construction History, Madrid, January 2003, 787–​797. 21.13 National Association of Cement Users, Philadelphia. Standard No. 4—​Standard Building Regulations for the Use of Reinforced Concrete. Adopted by the American Concrete Institute in February, 1910, 14 pp. 21.14 American Concrete Institute. Standard Specifications No. 23—​Standard Building Regulations for the Use of Reinforced Concrete, 1920, 20 pp. 21.15 Richard W. Furlong. “Design Rules for Steel-​Concrete Composite Columns: 1910 to 1963,” Concrete International, 34, February 2012, 41–​47. 21.16 Richard W. Furlong. “Design Rules for Steel-​Concrete Composite Columns: 1971 to 2011,” Concrete International, 34, February 2012, 61–​66. 21.17 American Institute of Steel Construction. Specification for Structural Steel Buildings, ANSI/​ AISC 360-​16, Chicago: AISC, July 2016, 620 pp. 21.18 Lawrence G.  Griffis. “Some Design Considerations for Composite-​ Frame Structures,” Engineering Journal, Second Quarter, 1986, 59–​64. 21.19 Oscar Faber (1956). “Savings to Be Effected by the More Rational Design of Cased Stanchions as a Result of Recent Full Size Tests,” The Structural Engineer, 34, March, 88–​109. 21.20 Royston Jones and A. A. Rizk (1963). “An Investigation on the Behaviour of Encased Steel Columns Under Load,” The Structural Engineer, 41, January, 21–​33. 21.21 Shosuke Morino, Chiaki Matsui, and Hidehiko Watanabe. “Strength of Biaxially Loaded SRC Columns,” in Composite and Mixed Construction, Reston, VA:  American Society of Civil Engineers, 1985), Charles W. Roeder, Editor, pp. 185–​194. 21.22 S. Ali Mirza, Ville Hyttinen, and Esko Hyttinen. “Physical Tests and Analyses of Composite Steel-​Concrete Beam Columns,” Journal of Structural Engineering, 122, November 1996, 1317–​1326. 21.23 Cengiz Dundar, Serkan Tokgoz, A. Kamil Tanrikulu, and Tarik Baran. “Behavior of Reinforced and Concrete-​Encased Composite Columns Subjected to Biaxial Bending and Axial Load,” Building and Environment, 43, 2008, 1109–​1120. 21.24 James M. Ricles and Shannon D. Paboojian. “Seismic Performance of Steel-​Encased Composite Columns,” Journal of Structural Engineering, 120, August 1994, 2474–​2494. 21.25 Cheng-​Chih Chen, Jian-​Ming Li, and C.  C. Weng. “Experimental Behaviour and Strength of Concrete-​Encased Composite Beam-​Columns with T-​Shaped Steel Section under Cyclic Loading,” Journal of Constructional Steel Research, 61, 2005, 863–​881. 21.26 Roberto T. Leon, Dong Keon Kim, and Jerome F. Hajjar. “Limit State Response of Composite Columns and Beam-​Columns. Part  1:  Formulation of Design Provisions for the 2005 AISC Specification,” Engineering Journal, Fourth Quarter, 2007, 341–​358. 21.27 Timo K.  Tikka and S.  Ali Mirza. “Nonlinear EI Equation for Slender Composite Columns Bending about the Minor Axis,” Journal of Structural Engineering, 132, October 2006, 1590–​1602. 21.28 Timo K. Tikka and S. Ali Mirza. “Effective Flexural Stiffness of Slender Structural Concrete Columns,” Canadian Journal of Civil Engineering, 35, 2008, 384–​399. 21.29 Ivan M.  Viest, Joseph P.  Colaco, Richard W.  Furlong, Lawrence G.  Griffis, Roberto T.  Leon, and Loring A.  Wyllie Jr. Composite Construction. Design for Buildings. New York: McGraw-​Hill, 1996. 21.30 Charles G.  Salmon, John E.  Johnson, and Faris A.  Malhas. Steel Structures. Design and Behavior (5th ed.). Upper Saddle River, NJ: Prentice Hall, 2008. 21.31 Bungale S.  Taranath. Steel, Concrete, & Composite Design of Tall Buildings, (2nd ed.). New York: McGraw-​Hill, 1998, 998 pp. 21.32 Richard W.  Furlong. “Strength of Steel-​Encased Concrete Beam Columns,” Journal of the Structural Division, ASCE, 93, October 1967, 113–​124.

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21.33 Richard W.  Furlong. “Design of Steel-​Encased Concrete Beam Columns,” Journal of the Structural Division, ASCE, 94, January 1968, 267–​281. 21.34 Robert B.  Knowles and Robert Park. “Strength of Concrete Filled Steel Tubular Columns,” Journal of the Structural Division, ASCE, 95, December 1969, 2565–​2587. 21.35 Amit H.  Varma, James M.  Ricles, Richard Sause, and Le-​Wu Lu. “Experimental Behavior of High Strength Square Concrete-​Filled Steel Tube Beam-​Columns,” Journal of Structural Engineering, 128, March 2002, 309–​318. 21.36 Kenji Sakino, Hiroyuki Nakahara, Shosuke Morino, and Isao Nishiyama. “Behavior of Centrally Loaded Concrete-​Filled Steel-​Tube Short Columns,” Journal of Structural Engineering, 130, February 2004, 180–​188. 21.37 Eiichi Inai, Akiyoshi Mukai, Makoto Kai, Hiroyoshi Tokinoya, Toshiyuki Fukumoto, and Koji Mori. “Behavior of Concrete-​Filled Steel Tube Beam Columns,” Journal of Structural Engineering, 130, February 2004, 189–​202. 21.38 Toshiaki Fujimoto, Akiyoshi Mukai, Isao Nishiyama, and Kenji Sakino. “Behavior of Eccentrically Loaded Concrete-​Filled Steel Columns,” Journal of Structural Engineering, 130, February 2004, 203–​212. 21.39 Amit H. Varma, James M. Ricles, Richard Sause, and Le-​Wu Lu. “Seismic Behavior and Design of High-​Strength Square Concrete-​Filled Steel Tube Beam Columns,” Journal of Structural Engineering, 130, February 2004, 169–​179. 21.40 Tiziano Perea. Analytical and Experimental Study on Slender Concrete-​Filled Steel Tube Columns and Beam-​Columns. Ph.D. thesis, Georgia Institute of Technology Atlanta, December 2010, 667 pp. 21.41 Tiziano Perea, Roberto T. Leon, Jerome F. Hajjar, and Mark D. Denavit. “Full-​Scale Tests of Slender Concrete-​Filled Tubes: Axial Behavior,” Journal of Structural Engineering, 139, July 2013, 1249–​1262. 21.42 Tiziano Perea, Roberto T. Leon, Jerome F. Hajjar, and Mark D. Denavit. “Full-​Scale Tests of Slender Concrete-​Filled Tubes: Interaction Behavior,” Journal of Structural Engineering, 140, No. 9, 04014054, 2014. 21.43 K.  S. Virdi and P.  J. Dowling. “Bond Strength in Concrete Filled Steel Tubes,” IABSE Proceedings P-​33/​80, 1980, 125–​139. 21.44 Michelle Ann Parsley. Push-​Out Behavior of Concrete-​Filled Steel Tubes, Master of Science in Engineering thesis, University of Texas at Austin, August 1998, 145 pp. 21.45 Charles W. Roeder, Brad Cameron, and Colin B. Brown. “Composite Action in Concrete Filled Tubes,” Journal of Structural Engineering, 125, May 1999, 477–​484. 21.46 Jie Zhang, Mark D.  Denavit, Jerome F.  Hajjar, and Xilin Lu. “Bond Behavior of Concrete-​ Filled Steel Tube (CFT) Structures,” Engineering Journal, Quarter, 2012, 169–​185. Also, Errata, Third Quarter, 2013, 201–​213. 21.47 M. H. Mollazadeh and Y. C. Wang. “New Insights into the Mechanism of Load Introduction into Concrete-​Filled Steel Tubular Column Through Shear Connection,” Engineering Structures, Elsevier, 75, 2015, 139–​151. 21.48 Gianluca Ranzi, Graziano Leoni, and Riccardo Zandonini. “State of the Art on the Time-​ Dependent Behaviour of Composite Steel-​Concrete Structures,” Journal of Constructional Steel Research, 80, 2013, 252–​263. 21.49. Dawn E. Lehman, Katherine G. Kuder, Arni K. Gunnarrson, Charles W. Roeder, and Jeffrey W.  Berman. “Circular Concrete-​ Filled Tubes for Improved Sustainability and Seismic Resilience,” Journal of Structural Engineering, 141, March 2015, B4014008. 21.50 Brett C. Gourley, Ceng Tort, Mark D. Denavit, Paul H. Schiller, and Jerome F. Hajjar. A Synopsis of Studies of the Monotonic and Cyclic Behavior of Concrete-​Filled Steel Tube Members, Connections, and Frames, Department of Civil and Environmental Engineering, University of Illinois at Urbana-​Champaign, Report No. NSEL-​008, April 2008, 364 pp. 21.51 T. M. Sheikh, J. A. Yura, J. O. Jirsa. Moment Connections Between Steel Beams and Concrete Columns, PMFSEL Report No. 87-​4, University of Texas at Austin, 1987. 21.52 G.  G. Deierlein, J.  A. Yura, and J.  O. Jirsa. Design of Moment Connections for Composite Framed Structures, PMFSEL Report No. 88-​1, University of Texas at Austin, 1988. 21.53 R. Kanno. Strength, Deformation, and Seismic Resistance of Joints between Steel Beams and Reinforced Concrete Columns, Ph.D. thesis, Cornell University, Ithaca, NY, 1993. 21.54 S. S. F. Mehanny. Modeling and Assessment of Seismic Performance of Composite Frames with Reinforced Concrete Columns and Steel Beams, Ph.D. thesis, Stanford University, Stanford, CA, 2000. 21.55 Chung-​Che Chou and Chia-​Ming Uang. “Cyclic Performance of a Type of Steel Beam to Steel-​ Encased Reinforced Concrete Column Moment Connection,” Journal of Constructional Steel Research, 58, 2002, 637–​663.

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PROBLEMS

21.56 G.  Parra-​Montesinos and J.  K. Wight. “Seismic Response of Exterior RC Column-​To-​Steel Beam Connections,” Journal of Structural Engineering, 126, October 2000, 1113–​1121. 21.57 G.  Parra-​Montesinos and J.  K. Wight. “Modeling Shear Behavior of Hybrid RCS Beam-​ Column Connections,” Journal of Structural Engineering, 127, January 2001, 3–​11. 21.58 G.  J. Parra-​Montesinos, X.  Liang, and J.  K. Wight. “Towards Deformation-​Based Capacity Design of RCS Beam-​Column Connections,” Engineering Structures, 25, 681–​690. 21.59 Atorod Azizinamini and Stephen P.  Schneider. “Moment Connections to Circular Concrete-​ Filled Steel Tube Columns,” Journal of Structural Engineering, 130, February 2004, 213–​222. 21.60 J. M. Ricles, S. W. Peng, and L. W. Lu. “Seismic Behavior of Composite Concrete Filled Steel Tube Column-​Wide Flange Beam Moment Connections,” Journal of Structural Engineering, 130, February 2004, 223–​232. 21.61 Isao Nishiyama, Toshiaki Fujimoto, Toshiyuki Fukumoto, and Kenzo Yoshioka. “Inelastic Force-​Deformation Response of Joint Shear Panels in Beam-​Column Moment Connections to Concrete-​Filled Tubes,” Journal of Structural Engineering, 130, February 2004, 244–​252. 21.62 American Institute of Steel Construction. Seismic Provisions for Structural Steel Buildings, ANSI/AISC 341-16. Chicago: AISC, 2016, 430 pp. 21.63 ASCE Task Committee on Design Criteria for Composite Structures in Steel and Concrete. “Guidelines for Design of Joints between Steel Beams and Reinforced Concrete Columns,” Journal of Structural Engineering, 120, August 1994, 2330–​2357. 21.64 Erica C. Fischer and Amit H. Varma. “Design of Split-​Tee Connections for Special Composite Moment Frames,” Engineering Journal, Third Quarter, 2015, 185–​202.

PROBLEMS FOR PROBLEMS 21.1 and 21.2, use the provisions in ACI 318-​14. 21.1 Consider the concrete composite beam shown in the figure for Problem 21.1. The beam is part of a floor system with parallel, simply supported beams spaced at 10 ft. The center-​to-​center span length of the composite beam is 40 ft with a clear span length of 38 ft. During construction, the beam must support a live load of 25 psf. Once the beam becomes composite, it must sustain, in addition to its self-​weight, a superimposed dead load of 20 psf and a live load of 100 psf. The slab is reinforced with #4 bars at 14 in. spacing in the direction of the beam span.   Check the adequacy of the design in terms of flexural and shear strengths, as well as deflections. Assume unshored construction and that the floor system supports nonstructural elements not likely to be damaged by large deflections. For the precast beam:  fc′  = 8000 psi; fy = fyt = 60 ksi. 6” Cast-in-place slab

#4 @ 12 in. Precast beam

30” 5 #9 16”

Problems 21.1 and 21.2 

For the cast-​in-​place slab:  fcs′  =  4000 psi; fy  = 60 ksi. Concrete is normal weight. Cover to center of longitudinal bars = 2.5 in. 21.2 Repeat Problem 21.1 but assuming shored construction. FOR PROBLEMS 21.3 through 21.8, use the 2016 AISC Specification for Structural Steel Buildings. 21.3 Consider the concrete-​encased steel composite column shown in the figure for Problem 21.3. For bending about the strong axis of the encased steel section, determine: (a) Nominal axial force–​moment strength interaction diagram; use the strain compatibility approach and neglect slenderness effects. (b) Nominal axial force–​moment strength interaction diagram; use the strain compatibility approach, including slenderness effects for an unsupported column length of 12 ft. for bending about the strong axis. (c) Design shear strength for shear parallel to the web of the steel shape. Compare the results using the three approaches allowed in the AISC Specification (i.e., bare steel section, steel section plus ties, reinforced concrete section).

Material properties:  fc′   =  8000 psi (normal weight); fys  =  50 ksi (A572 Gr. 50 steel); fyr  = fyt = 60 ksi.

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21.6 For the column of Problem 21.3, determine the diameter, length, number, and spacing of headed 3” stud anchors required to transfer the shear force carried by two W16 girders framing into the 6” two flanges of the concrete-​encased steel sec5” tion. The factored shear force in each girder is 28” 5” 72 kips. Tensile strength of studs, fua  =  65 ksi. Draw a sketch of the connection, indicating the 6” layout of headed studs. 3” 21.7 For architectural reasons, the steel tube of a 28” rectangular CFT composite column is to be an HSS18 × 12 section. Determine the lightest W14 × 145 tube section capable of resisting the following factored loads obtained from a second-​order Problems 21.3 and 21.6  analysis (not including the effect of initial imperfections): Pu = 270 kips; Mu = 225 ft-​kips; Vu  =  28 kips (section is subjected to bending 21.4 Repeat Problem 21.3 but for bending about the about its strong axis). Calculate the maximum weak axis of the steel section. shear force that can be transferred by steel 21.5 For the concrete-​encased steel composite column girders without the need for shear connectors. section in the figure for Problem 21.3, determine Assume the column extends above the steel girdthe lightest W14 steel shape required to resist ers. Unsupported column length L = 12.5 ft. for the following actions obtained from a second-​ bending about strong axis (column is fully suporder analysis (not including the effect of initial ported in the perpendicular direction). Material imperfections): Pu = 1550 kips; Mu = 1310 ft-​kips properties:  fc′ =  6000 psi (normal weight); (about strong axis of steel shape); Vu = 160 kips. fys = 42 ksi (A500 Grade B steel). Unsupported column length L = 12 ft. for bending about strong axis (column is fully supported 21.8 Repeat Problem 21.7 but using a circular HSS16 tube section. in the perpendicular direction). #4 @ 10”

12 #10

94

INDEX Note: Page references followed by a “t” indicate table; “f  ” indicate figure. active pressure, on retaining walls, 597 admixtures, 6–9, 7f accelerating, 8 air-entraining, 8 chemical, 8–9 for flowing concrete, 9 for self-consolidating concrete, 9 mineral, 7–8 set-retarding, 8–9 water-reducing, 8–9 aggregates, 6 aggregate interlock, 115–16, 118, 124, 128 all-lightweight concrete, 6 shear friction and, 154 shear strength and, 121 tensile strength of, 13 arch action for shear strength, 115, 117–18 axial compression bending moment and, 326–28, 470–79 biaxial bending and, 360–67, 360f–65f, 367f, 494 in CFT columns, 936–37 in concrete-encased steel composite columns, 919–20 shear strength and, 147–48, 588–89, 923 slenderness effect and, 328, 464–66, 479–80, 921, 938 axial tension bending moment and, 357–59 in brackets and corbels, 158–59 in CFT columns, 938 in concrete-encased steel composite columns, 921 shear strength and, 149 torsional stiffness and, 753 in walls, 588–90 balanced strain condition (compression control limit) for beams, 51–52, 51f for columns, 327, 333–36, 333f–35f, 336, 357 for prestressed concrete, 868–69

bar details in floor beams, 287 in one-way slabs, 264 beams. See also deep beams balanced strain condition for, 51–52, 51f behavior of beams without shear reinforcement, 114–22 behavior of beams with shear reinforcement, 122–29 compression-controlled sections of, 52–55, 78–80 crack control for, 451–53 deflections of, 411, 446–51, 446f, 449f flexural strength of, 43–88, 177, 182–85 minimum tension reinforcement for, 58–59 minimum shear reinforcement for, 131–135 moment-curvature response, 86–88 nominal flexural strength for, 48–50 post-tensioning of, 849, 849f, 862 of prestressed concrete, shear reinforcement for, 881–83 pretensioning of, 847–49, 858f reinforcing bars for, 64–67, 65t–67t shear failures in, 111, 117–18, 117f, 119f shear strength of, 136–46, 137f–39f, 143f, 144t shear stress on, 113–14, 113f, 114f skin reinforcement in, 455 tension-controlled sections of, 52–57, 78–84, 80f truss model for, 125–27, 126f, 127f, 128f two-way floors with, 624, 629–31, 629f, 630f, 633–43, 634f–36f, 634t, 638f, 639f, 642f, 669–71, 670f bearing capacity of soil, footings and, 799 bearing strength, 163, 542 for square spread footings, 808

bearing stress, 179 on square spread footings, 808 bearing walls, 569, 571f design of, 573–76 strength reduction factors for, 574 biaxial bending axial compression and, 360–67, 360f–65f, 367f, 494 biaxial strength of concrete, 14, 14f bond, See development of reinforcement braced frames, 470f. See also nonsway frames. brackets and corbels, 157–68, 158f, 161f, 162f, 165f, 167f, 559–65 B-regions, 127 cantilever retaining walls, 580–81, 581f friction in, 608–9, 608f preliminary proportioning of, 603–18, 604f, 606f–8f, 606t, 610f, 613f, 616f, 619f safety factors for, 606–8 service loads on, 608 shear reinforcement for, 605 shear strength of, 605 soil pressure on, 604, 606f, 607f transverse reinforcement for, 618 cellular concretes, 6 cement, 5 cement slag, as mineral admixture, 7 CFT columns. See concrete-filled tube columns circular columns, 354–57, 355f, 356f, 356t, 937–38 circular sections. See also circular columns torsional stiffness in, 751 torsional stress in, 749–50, 750f coarse aggregate (gravel), 6 coefficient of friction, 153–54, 600, 903 collapse mechanism, 312–14, 721f, 720–27 column capital, 623, 627, 629, 639–43, 640f

950

950 INDEX

columns, 321f. See also circular columns; composite columns; concrete-encased steel composite columns; concretefilled tube columns; CFT columns; spirally reinforced columns; tied columns alignment charts, 490–92 axial compression in, 326, 360–67, 360f–65f, 367f, 368 axial load on, 34, 322–24, 323f–25f, 333 axial tension in, 357–59 balanced strain condition for, 327, 333–36, 333f–35f, 357 bending moment in, 321, 326, 345, 357–59, 463, 464f, 470–76, 471f–73f, 475f, 477f biaxial bending in, 360–67, 360f–65f, 367f compressive strength of, 323, 333–34 concrete-steel composite, 916–17, 917f concrete-encased composite, 918–34, 943–44 concrete-filled tube, CFT, 934–943 critical buckling load of, 483, 484, 506, 509, 514–15 design strength for, 515 ductility of, 323 effective length for, 468–69, 469f, 470f flexural stiffness of, 702–3 lateral ties in, 329–30, 330f longitudinal reinforcement for, 326, 332 nominal flexural strength of, 357 plastic centroid in, 327, 334, 336, 347 shear in, 368–70, 369f shear stress in, 332 slenderness effects for, 463–523 slenderness ratio for, 326, 328–29, 516 stiffness reduction factor for, 481, 483 strength interaction diagram for, 326–27, 327f, 328f, 479–80, 480f strength reduction factors for, 325–26, 352, 481 column strips, in two-way floors, 656–62, 656f, 658t–62t, 664 combined footing, 800, 801f, 822–41, 823f, 825f, 828f, 831f, 832f, 833t, 835f, 839f compatibility torsion (statically indeterminate torsion), 313, 748, 752–53 composite columns, 322, 323f concrete filled tube or CFT columns, 916, 934–43, 937f, 940f

concrete-encased steel composite columns, 917–34, 920f, 922f, 925f–27f, 927t, 929f, 931f–34f concrete-steel composite columns, 916–17, 917f design strength of, 917 moment connections with, 943–44, 944f composite members, 897–944, 898f composite action in, 897–901 CFT columns, 916, 934–43, 937f, 940f concrete composite flexural members, 901–16 concrete-encased steel composite columns, 917–34, 920f, 922f, 925f–27f, 927t, 929f, 931f–34f concrete-steel composite columns, 916–17, 917f interface shear in, 899–901, 900f, 901f compression-controlled sections of beams, 52–57, 53f, 78–84, 80f nominal flexural strength of, 72–78, 73f, 75f, 77f compression control limit. See balanced strain condition compression lap splices, 216, 216t compressive strength, 9–12 biaxial strength and, 14 of columns, 323, 333–34 of concrete-encased steel composite columns, 919–20 of concrete filled tubes, 936 effect of confinement on, 323 of nodal zones, in strut-and-tie models, 537, 537t, 563 of struts, 535–37, 536f, 537f, 544, 548 concrete-encased steel composite columns, 917, 918, 918–34, 920f, 922f, 925f, 926f, 927f, 927t, 929f, 931f–34f axial tensile strength of, 921 compressive strength of, 919–20 flexural strength of, 919–20 headed shear studs for, 925 shear strength of, 921–23 slenderness effects of, 920–21 concrete-filled tube (CFT) columns, 916, 934–43, 937f, 940f axial compressive strength of, 936–37 axial tensile strength of, 938 shear strength of, 938 slenderness effects for, 938 width-to-thickness ratio for, 935–36, 935t confinement, 155 effect on compressive strength, 323

for monolithic beam-column joints, 383–87 spirally reinforced columns and, 323–25 effect on bond, 181, 186 conjugate beam method, for deflections, 412–13, 412f corbels. See brackets and corbels corner connections, in flat slab floors, 697, 698f corner effects, yield line theory and, 742–43, 742f corner reinforcement, for two-way floors, 664–65, 665f corrosion reinforcement protection for, 24 crack control and, 451–52 coupled walls, 579 coupling beams, 579 crack control for beams, 451–54 for one-way slabs, 451–54 side face, for deep beams, 455–56, 455f for two-way floors, 666 cracking moment, 416 minimum tension reinforcement and, 58–59 for prestressed concrete beams, 871–72, 873f creep, 16–17 beam curvature and, 428–29, 429f compression steel and, 430–31 deflection and, 323, 428–31, 428f, 429f, 430t, 436–39, 442, 448 humidity and, 429–30, 430t prestressed concrete and, 858–59 critical section for brackets and corbels, 161 for footings, 803, 803f for nominal shear strength, 135, 136f for torsion, 772 cutoff points, 198–203 DDM. See Direct Design Method deep beams, 151, 542–59 deflections. See also immediate deflection of beams, 410–413, 446–51, 446f, 449f of concrete composite flexural members, 905–08 compression steel and, 430–31, 435 conjugate beam method for, 412–13, 412f creep and, 17, 323, 428–31, 428f, 429f, 430t, 436–39 effective moment of inertia for calculating, 415–18 immediate (instantaneous) deflection, 417–19

951

951

INDEX

maximum reinforcement ratio for control of, 427–28 minimum depth and, 439–41, 440t, 441t, 446 moment area theorems for, 433–35 of one-way slabs, 256 shrinkage and, 431–39 of two-way floors, 637–39, 638f, 639f, 704 deformation capacity, 87, 324 development of reinforcement, 176, 177–80 ACI Code on, 185–92 bond stresses, 176–79 bond failure mechanisms, 180–81 for bundled bars, 195 for compression reinforcement, 194 for hooked bars, 195–97, 195f–97f, 198t modification factors for, 189–92 of shear reinforcement, 211–13 for prestressed reinforcement, 883–84 splitting failure, 181 for tension reinforcement, 195–97, 195f–97f, 198t diagonal tension. See shear strength dimensions, 40 Direct Design Method (DDM) moments and, 644–51, 646f, 647, 649f, 650t, 651f, 656–62, 656f, 657t–62t moments in columns and, 686–87 for two-way floors, 624–25, 644–51, 686–87 discontinuity. See D-regions D-regions, 125–26 of deep beams, 528, 542 drop panels, 622–23, 639–40 dual system, for shear walls, 576–77 effective compressive strength of nodal zones, 537, 538f of struts, 535–36 effective flange width (or effective slab width) for monolithic beam-column joints, 382 for T-beams, 95–97, 96f, 97f for two-way floor systems, 688–89, 711 effective length alignment charts for, 490–92, 491f factor for columns, 468–69, 469f, 470f effective moment of inertia, 415–17 for computation of deflections, 411, 413–18, 441–42, 451 for evaluation of slenderness effects, 463, 465, 481–82, 484, 492, 496 EFM. See Equivalent Frame Method

epoxy-coated reinforcing bars, 24 development length factors for, 190–91, 196 for cantilever beams, 205, 207 equilibrium torsion (statically determinate torsion), 748–49 Equivalent Frame Method (EFM), 624–25 for gravity loads, 698–704 models for, 710–711 lateral loads, 711 equivalent frame models, 710 fiber-reinforced concrete, 25–26 as shear reinforcement, 129, 130f fiber reinforcement, 24 fine aggregate (sand), 6 flat-plate floors, 623, 623f shear reinforcement in, 676–85, 677f, 680f, 681f, 683f, 684f shearheads for, 695–96, 697f stirrups for, 678 flat slab floors, 623, 623f column capital for, 623 with drop panels, 622, 623, 623f openings in, 697, 698f fly ash, as mineral admixtures, 7 footings, 800 bearing capacity of soil and, 799 combined, 800, 801f, 822–41, 823f, 825f, 828f, 831f, 832f, 833t, 835f, 839f critical section for, 803, 803f pile, 800, 841 proportioning for equal settlement, 804 rectangular, 814–18, 816f shear strength of, 802–3 types of, 800 types of failures in, 800–801 for walls, 800, 818–22, 820f, 821f friction, loss of prestress, 860–61 galvanized reinforcing bars, 24 gravel (coarse aggregate), 6 headed shear studs for concrete-encased steel composite columns, 924–25 two-way slabs, shear strength contributed by, 679–80, 680f heavyweight, high-density concrete, 6 high-strength concrete, 12 shear strength of members with, 122 high-performance fiber-reinforced concrete (HPFRC), 25 hollow steel sections (HSS), 916. See also CFT columns hooks, standard, 195–98. See also Development length development length for, 195–97, 195f–97f, 198t

anchorage with hooks in onolithic beam-column joints, 390, 391f for ties and stirrups, 211–12 humidity creep and, 16–17, 428, 429–30, 430t shrinkage and, 17–18, 432, 859 hydraulic cements, 5 immediate deflection, 417–19 influence lines, for continuous beams for bending, 241–242 for shear, 250–251 integrity steel for beams, 287–88 for two-way slabs, 664 interaction diagrams. See strength interaction diagram interface shear, of composite members, 899–901, 900f, 901f interface shear strength, of concrete com­ posite flexural members, 901–4 interface shear transfer in beams (aggregate interlock), 115 Joints, 380 beam-to-column, 380 confinement of, 383–87 core of, 385–86 development of reinforcement in, 390–91 effective joint width, 388–39 forces on, 381–83 shear in, 381–83 shear strength of, 387–89 column-to-beam minimum strength ratio, 389 transverse reinforcement in, 383–87 joist floor systems, 306–07 kern points, 885 lateral ties, in columns, 329–30, 330f lightweight aggregate concrete, 6 lightweight concrete. See also alllightweight concrete; sandlightweight concrete interface shear strength in members with, 902 tensile strength of, 12–13 modulus of elasticity of, 15 shear strength of members with, 121, 130, 134, 588 shear friction in members with, 154 maximum shear strength in brackets and corbels, 161 minimum reinforcement ratio for deflection control in members with, 428 modification factor for bond in members with, 191 torsional cracking strength of members with, 753

952

952 INDEX

load balancing concept, for prestressed concrete, 855–56, 856f load combinations, 36–37 load contour method, 360, 364–65, 365f L-sections torsional stiffness in homogeneous, 752 torsional strength of homogeneous, 751 mat foundation, 800 maximum reinforcement ratio for deflections, 421, 427–28 for tension-controlled sections, 53 for any maximum tensile strain, 53 mineral admixtures, 7–8 modulus of elasticity, 14–15, 15f, 15t of concrete, 14–15 of prestressing reinforcement, 24 of nonprestressed steel reinforcement, 24 ratio, 404 in calculation of deflections, 414 shear, 751 modulus of rupture, 13, 13f, 58, 416, 757, 847, 871, 874 moment area theorems, for deflections, 433–35 moment coefficients for continuous beams, 246–47, 247t moment connections, with composite columns, 943–44, 944f moment-curvature analysis, 86–87, 87f moment magnifier for nonsway frames, 497–512, 497f, 500, 505–6, 505f, 508f, 511t, 512f, 513 for sway frames, 497–512, 497f, 500, 502–3, 505f, 508f, 511t, 512f moment of inertia, 58, 267, 270–71. See also effective moment of inertia for two-way floor systems, 633–34, 637, 647, 700–04 moments in columns DDM and, 686–87 two-way floors and, 686–97, 688f, 690f, 693f, 697f moment strength ratio for monolithic beam-column joints, 389, 393 monolithic beam-column joints, 380–99, 381f, 392f actions of, 381–82, 382f, 383f anchorage of reinforcement, 390–91, 391f, 392 axial load in, 381, 395, 398 confinement for, 383–85 ductility of, 399 eccentricity in, 396 effective width for, 395

floor systems and, 391 hooks for, 390, 391f moment strength ratio for, 389, 393 nominal strength for, 393–94, 398 shear for, 392, 393f shear strength for, 387–88, 388f, 389f, 389t, 398 transfer of axial forces, 391 transverse reinforcement for, 383–87, 384f, 385f, 387f, 394, 396–97, 396f natural stone aggregates, 6 negative moment region of continuous beams, cutoff points in, 198–200, 199f, 200f neutral axis transformed section method and, 409–10, 409f, 410f nodal forces, yield line theory and, 729–36, 730f–35f nodal zones, in strut-and-tie model, 528, 531–33, 529f, 532f, 533f, 534f, 538f, 546–57, 555f nominal strength of, 537–38, 537t dimensioning of, 541–42 effective compressive strength of, 553 hydrostatic nodal zones, 531–32, 532f nodes, in strut-and-tie models, 528, 537–38 nominal flexural strength for beams, 48–50 for columns (strength interaction diagram), 326 for rectangular sections, 72–78, 73f, 75f, 77f of T-beams, 97–105 of walls, 581 nominal shear strength critical section for, 135, 136f for two-way floors, 672 nominal shear stress, for two-way floors, 669 nominal strength, 35 nonbearing walls, 569, 571f design of, 573 lateral load of, 573 noncomposites, 898 nonrectangular sections, 84–86, 85f nonsway frames, 470 effective length factor, 490–92 minimum design eccentricity, 493–94 moment magnification, 470–77, 482 slenderness ratio limits, 494–495 one-way slabs, 254–65, 623f analysis methods for, 254–55 bar details for, 264, 264f collapse mechanism of, 721f, 722 crack control for, 451–54

deflection in, 256 design of, 255–58, 256t, 257f flexural reinforcement for, 258–59 shear in, 256 shrinkage and temperature reinforcement for, 259, 260f yield line theory for, 721, 725–28, 726f, 728f openings, in flat slab floors, 697, 698f overloads, safety provisions and, 34, 36 overreinforced, 46 partial composite, 899 passive pressure, on retaining walls, 597 passive resistance, of retaining walls, 600, 600f pile footings, 800, 841 plastic analysis, 312–17, 313f–16f plastic centroid in columns, 327, 334, 336, 347 plastic flow, creep and, 16 plastic hinges, 313–14 inelastic deformation and, 313 yield line theory and, 720 plasticizing admixtures, 9 plasticizing and retarding admixtures, 9 plastic rotation capacity, 313 portland blast-furnace slag cement, 5 portland cement, 5, 5t positive moment of continuous beams bar cutoff points for, 201, 201f, 202f postcracking stiffness, torsional moment and, 752 post-tensioning, of beams, 849, 849f, 862 friction on, 859–62, 860f, 862f pozzolan, 5, 7 prestressed concrete members, 844–93 advantages and disadvantages of, 845–46 balanced strain condition for, 868–71 cracking moment for, 871–72, 873f creep and, 844, 845, 858–59 deflection of, 866 development length for, 883–84 elastic shortening of, 857–58 flexural behavior of, 885–93, 886f, 889f, 891f flexure-shear cracks in, 874–75 homogenous beam concept for, 854 inclined cracks in, 875, 875f internal force concept for, 854–55, 854f load balancing concept for, 855–56, 856f loss of prestress in, 856–66, 862t nominal strength of, 866–71 post-tensioned beams, 849, 849f, 859–62, 860f, 862, 862f

953

953

INDEX

pretensioned beams, 847–49, 858f proportioning of cross sections of, 885–93, 886f, 889f, 891f reinforcement for, 883–85 relaxation of steel in, 859 service loads for, 846f, 849–53, 852f, 866 shear reinforcement for, 881–83 shear strength of, 873–81 web-shear cracks in, 875–78, 876f yield points for prestressing reinforcement, 866 prestressing reinforcement, 23–24, 23f pretensioning, of beams, 847–49, 858f prismatic struts, 529, 530f pullout failures bundled bars and, 195 development length and, 186 with lightweight aggregate concrete, 191 pullout resistance, development length and, 185 pullout tests, for development length, 181 punching shear, of footings, 801 quality control, for concrete, 18–19 radiation, heavyweight, high-density concrete for, 6 radius of gyration, 463, 496 slenderness ratio and, 497 reciprocal load method, 360, 366–67, 367f rectangular footings, 814–18, 816f rectangular sections balanced strain condition for, 51–52, 333, 334f, 336–40, 336f, 339f flexural strength of, 44–50, 45f minimum tension reinforcement for, 59 nominal flexural strength of, 72–78, 73f, 75f, 77f skew bending theory for, 756, 756f space truss analogy for, 761–65, 762f–64f tension reinforcement for, 60–64, 61f, 64f torsional stiffness of, 751 torsional stress in, 750, 750f, 750t torsion and shear on, 767–68 for walls, 581–83, 582f, 583f rectangular walls, shear strength of, 588 redistribution of moments, 312–17, 313f–16f reinforcement. See also specific topics areas, tables of, 21, 65 reinforcement ratio, 53–54, 54t

reinforcing bars. See also epoxycoated reinforcing bars for beams, 64–71, 65t–67t, 70f, 72f concrete cover and, 40, 452 deformed, 19–20, 20f, 21t, 22t, 180 galvanized, 24 relaxation, of prestressing steel, 859 required average compressive strength, 18–19 reserve capacity, safety provisions and, 34 retaining walls, 569, 571f, 597–601. See also cantilever walls coefficient of friction for, 600, 600t safety factors for, 599–601 soil pressure on, 601, 601f safety provisions, 34–38 sand (fine aggregate), 6 sand-lightweight concrete shear friction and, 154 tensile strength of, 13 sand replacement, 6 sanitary structures, crack control for, 452 secant modulus, modulus of elasticity and, 14–15, 414 second-order analysis, ACI Code on, 493 self-consolidating concrete, admixtures for, 9 serviceability, 403–56 crack control and, 451–54 deflection and, 38–39, 410–11 deflection and creep and, 428–31, 428f, 429f, 430t effective moment of inertia and, 414–17, 414f, 415f flexural behavior and, 403–4 floor system vibration control, 456 immediate deflections and, 417–28, 418f–20f, 423f, 424f linear elastic behavior and, 407 modulus of elasticity and, 404, 404t, 406, 414 side face crack control and, 454–55, 455f time-dependent deflections, 428 transformed section method and, 407–10, 407f, 409f, 410f for two-way floors, 644 service-level, unfactored loads, 37, 39, 403–4. See also service loads crack control and, 452 service loads, 33 set-retarding admixtures, 8–9 shear axial tension and, 13 bending moment and, 772–73, 773f in columns, 368–70, 369f compressive stress and, 110–11, 110f, 111f

continuous beams and, 250–52, 251f corbels and, 559 fiber-reinforced concrete and, 26 in flat slab floors, 624 live loads and, 250 in monolithic beam-column joints, 381–82, 392, 393f in one-way slabs, 256 stirrups and, 789 strength interaction diagram for torsion, bending, and, 768–69, 768f, 769f torsion and, 767–68 transverse reinforcement and, 1, 124, 124f, 368, 785, 789 two-way floors and, 669–71, 687–97, 688f, 690f, 693f, 697f shear failure in beams, 111, 117–18, 117f, 119f of cantilever walls, 580, 581f of deep beams, 117–18, 117f of fiber-reinforced concrete beams, 129, 130f of footings, 801, 802f shear friction, 152–57, 152f, 153f, 156f for brackets and corbels, 159, 166 shearhead reinforcement, 623 shear reinforcement beams without, 114–22, 115f, 117f, 119f for cantilever walls, 605 for columns, 369 for combined footings, 831–32 development of, 211–13, 212f, 213f fiber reinforcement as, 129, 130f in flat-plate floors, 676–85, 677f, 680f, 681f, 683f, 684f for prestressed concrete beams, 881–83 shear strength from, 123, 128–29, 128f shear strength without, 114–22, 115f, 117f, 119f, 149t–51t for slab-beam-girder floor systems, 285–87 for square spread footings, 806 stirrups as, 211–13, 212f transverse reinforcement as, 123–24, 123f, 140–46, 140t, 143f, 144t, 146f truss model and, 122 for wall footings, 820 for walls, 591–95, 592f, 596f welded wire reinforcement and, 212–13, 213f shear-related inclined cracks, 180 shear resistance, 115–17. See also shear strength of deep beams, 151 of monolithic beam-column joints, 387

954

954 INDEX

shear span, 116, 116f brackets and corbels and, 161 of continuous beams, 121 shear span to depth ratio, 116–119 shear strength, 110–68. See also nominal shear strength, shear resistance axial compression and, 588 axial load and, 146–51, 149t bending moment and, 146–51, 149t of beams, 136–46, 137f–39f, 143f, 144t of CFT columns, 938 of columns, 369 of concrete composite flexural members, 904–5 of concrete-encased steel composite columns, 921–23 of continuous beams, 121, 122f of deep beams, 119 of footings, 802–3 from headed shear studs, 679–80, 680f from shear reinforcement, 123, 128–29, 128f of high-strength concrete, 122 for joist floor systems, 307 of monolithic beam-column joints, 387–88, 388f, 389f, 389t, 398 of prestressed concrete members, 873–81 of rectangular spread footings, 815 without shear reinforcement, 114–22, 115f, 117f, 119f, 149t–51t size effect and, 122 skew bending theory and, 758 from shearheads, 678–79, 695–96, 697f of square spread footings, 805 of two-way floors, 640, 671–76, 673f, 674f, 676f, 697 of walls, 588–90, 589f–91f shear stress axial load and, 148f in beams, 113–14, 113f, 114f in columns, 332 linear elastic behavior and, 111–13, 112f skew bending theory and, 758 in thin-wall sections, 762 torsion and, 756, 772–73, 773f shear walls, 569, 571f, 584–87, 587f buckling of, 579f flexural strength of, 581–87 in-plane bending in, 576 interaction diagrams for, 585, 585f level of coupling for, 579–80 longitudinal reinforcement for, 584 shear strength of, 588 transverse reinforcement for, 584–85

T-sections of, 577–78 wall-to-floor area ratio for, 576 shored construction, of concrete composite flexural members, 908–16 shrinkage, 17–18, 18f correction factors for, 432, 432t crack control and, 451 creep and, 16 effect on deflections, 431–37 multiplier procedure for, 435–39, 437t prestress loss due to, 859 and temperature reinforcement, 191, 215, 259, 306–308, 432, 822 warping from, 433–35, 434f shrinkage curvature, 432 shrinkage strain, 431–32 beam curvature and, 433 side face crack control, for deep beams, 455–56, 455f silica fume, for mineral admixtures, 7–8 simple supports, development of positive moment reinforcement, 209–10 single curvature, 329, 471, 474, 476–77, 483, 487, 490–491, 495–496 singly reinforced sections, rectangular, 48 T-beams, 98–99 size effect, for shear strength, 122 skew bending theory, 757 skin reinforcement, for side face crack control, 455–56, 455f slab-beam-girder floor systems, 240f, 266–305 bar details for, 287–94, 290f beam web size for, 267–70 continuous frame analysis for, 270–72, 272f–74f, 272t shear reinforcement design for, 285–87 slabs. See also flat slab floors; two-way floor systems; oneway slabs effective width of, 711 yield line theory of, 720–46 slenderness effects in CFT columns, 938 in columns, 463–523 in concrete-encased steel composite columns, 920–21 effective length alignment charts, 490–92, 491f interaction diagrams including, 479–80, 480f, 515, 515f nonsway frames and, 470–76, 472f, 473f, 475f, 477f, 481–85, 485f, 495, 496 second-order analysis, 493

sway frames and, 477–79, 478f, 485–90, 486f, 495, 496 slenderness ratio, 326, 328, 342, 463, 465, 467, 480, 494 slippage failure, 178 slump, 6–10, 16–18 soil pressure on cantilever retaining walls, 601, 601f, 604, 606f, 607f on square spread footings, 804 on wall footings, 819, 820 space truss analogy (torsion), 761–65, 762f–64f spandrel beams torsional moment and, 754 torsional stiffness of, 778–91, 778f torsion in, 748 span-to-depth ratio for brackets and corbels, 559 for deep beams, 528, 542 for prestressed concrete members, 846 shearhead reinforcement for flat-plate floors, 695–96, 697f shear strength from, 678–79, 695–96, 697f special moment frames, monolithic beam-column joints in, 386 specified compressive strength, 18–19 spirally reinforced columns, 322, 323, 323f, 324f axial load in, 330 clear cover for, 332 confinement and, 323–24, 330–31 ductility of, 324 lap splices for, 332 longitudinal reinforcement in, 330–32, 331f, 332f mechanical splices for, 332 stiffness reduction factor for, 483 strength reduction factor for, 352 splice length, 213 splices under compression and bending, 217 compression lap, 216, 216t lap, for spirally reinforced columns, 332 length, 213 mechanical, 215–16, 217, 332 tension lap, 213–15, 214f, 215t welded tension, 215–16, 217, 332 split-cylinder test, 12–13 splitting failure, 181f bearing area and, 178 bundled bars and, 195 concrete cylinder hypothesis and, 182, 182f hooks and, 195 lugs and, 180–81 splitting resistance, development length and, 185

95

955

INDEX

square spread footings, 804–14, 805f, 807f design of, 809–14, 811f, 812f, 814f shear strength of, 805 soil pressure on, 804 squat walls, 581, 589 stability index, 482, 488, 498–99, 505 for sway frames, 509 statically determinate torsion (equilibrium torsion), 312, 748 statically indeterminate torsion (compatibility torsion), 313, 748, 752–53 torsional stiffness and, 752–55, 754f steel reinforcement, 19–24, 226 corrosion of, 24 ductility (deformability) of, 20, 21f modulus of elasticity of, 49, 404 stress-strain relationship, 21f, 23f stiffness reduction factor for columns, 481, 483 stirrups, 3, 54–55, 65–67,114, 122–135 See also shear reinforcement, hoop reinforcement; transverse reinforcement for brackets and corbels, 160 cutoff points and, 199–200 development of, 211–213 for flat-plate floors, 676, 678, 677f for shear reinforcement, 122–135 for shear strength, 122–135 hooks for, 212 for thin-walled sections (torsion), 762 strap footing. See combined footings. strength design method, 32–34 strength interaction diagram combined bending and axial force, 326–27, 327f, 328f combined bending and torsion, 766–67 combined bending, shear, and torsion, 768–69, 768f, 769f strength reduction factors, 37–38 for brackets and corbels, 161 for development length, 188 for lightweight concrete, 12–13 for tension- and compressioncontrolled sections, 54–55, 55f for sections in the transition zone, 327, 352 for struts, ties, and nodal zones in strut and tie models, 535 stress, principal tensile, 110–11, 113–14, 116, 119–20, 148, 873–874, 876–877 stress-strain relationship, concrete in compression, 11–12, 11f, 12f, 428f Hognestad’s, 465–66, 467f fiber-reinforced concrete, 25f for design, 48f

nonprestressed steel reinforcement, 21f prestressing steel reinforcement, 23f welded wire reinforcement, 23f structural design process, 31 strut-and-tie models, 528–65. See also struts; ties, nodal zones ACI Code provisions, 534–39 beam-column joints and, 388f bottle-shaped struts, 529 529f, 530f compression fields in, 529, 530f fan-shaped action, 529, 530f for brackets and corbels, 559–65, 560f for deep beams, 528, 529f, 542–59, 543f, 544f, 547f, 551f, 553f, 553t, 554t, 555f, 556f, 559f nodal zones, 531–534 pile cap, 541 selection of, 539, 539f–41f for shear walls, 541f struts, 529–31 ties, 531 transition stress fields, 532–33, 533f struts, in strut-and-tie models, bottle-shaped, 529, 530f compressive strength of, 535–37, 536f, 537f, 544, 548 dimensioning of, 541–42 effective compressive strength of, 535 in flanges of T-sections, 536 nominal strength of, 535–37 prismatic, 529f reinforcement for, 536, 536f tapered, 529f superplasticizers, 8 sustained loads, creep under, 16 effect on deflections, 411, 418, 428–39 effect on stiffness, 481, 484–85, 489 stability index and, 499 sway frames, 477–79, 496, 470f effective length factor, 490–92 moment magnification factor, 488 slenderness ratio limits, 494–95 stability index, 482, 488 tall walls, 580 T-beams, 59 in bending, 94–108, 95f comparison of rectangular sections and, 95 design of, 105 effective flange width for, 95–97, 96f, 97f minimum tension reinforcement for, 59 nominal flexural strength of, 97–105

positive moment region of continuous beams and, 427 singly reinforced, 98–99 with compression reinforcement, 102 temperature and shrinkage reinforcement for joist floor systems, 308, 308t for one-way slabs, 259, 260f for two-way floor systems, 622 tensile strength, concrete, biaxial strength and, 14 split-cylinder test, 12 lightweight, 121 modulus of rupture, 13 steel reinforcement, 19–24 tension-controlled sections, 52–57, 53f, 78–84, 80f, 227 tension lap splices, 213–15, 214f, 215t tension zone, cutting bars, 199–200 ternary blended cement, 5 thermal expansion coefficients, 2 thermal stress, crack control and, 451 thin-wall sections shear stress in, 762 space truss analogy for, 762 torsion in, 756, 762 threshold torsion, 770 tied columns, 322, 323, 323f, 324f stiffness reduction factor for, 483 strength reduction factor for, 352 ties, in strut-and-tie models, 531, 531f dimensioning of, 541–42 nominal strength of, 538–39 tolerances, 40 torsion, 748–91 combined bending and, 765–67, 766f, 767f combined shear and, 767–68 combined bending, shear, and, 768 in cantilever beams, 748 compatibility torsion, 748, 752–53 critical section for, 772, 783 inclined cracks from, 749f longitudinal reinforcement and, 772, 780, 786, 791 strength reduction factor for, 37, 38 shear stress and, 756, 772–73, 773f skew bending theory for, 758 space truss analogy for, 761–65, 762f–64f in spandrel beams, 748 statically determinate, 312, 748 statically indeterminate, 313, 748, 752–55, 754f strength interaction diagram for, 768–69, 768f, 769f tensile stress and, 756 in thin-wall sections, 756, 762 transverse reinforcement and, 780, 785, 789

956

956 INDEX

torsional moment compressive stress and, 753 girders and, 754 postcracking stiffness and, 752 spandrel beams and, 754 stirrups and, 762–63, 763f tensile stress and, 753 torsional moment strength at cracking, 755–57, 756f nominal, 771, 777 torsional rigidity. See torsional stiffness (rigidity), 240, 749, 751–53 statically indeterminate torsion and, 752–55, 754f of transverse beams in two-way floors, 651–57, 652f, 653f, 655f, 700–702 torsional strength skew bending theory, 758 space truss analogy, 761 torsional stress in circular sections, 749–50, 750f in homogenous sections, 749–51, 750f, 750t in I-sections, 751 in L-sections, 751 in rectangular sections, 750, 750f, 750t in T-sections, 751 torsion reinforcement longitudinal reinforcement and, 771 minimum requirements for, 774–75 total factored static moment for flat slab floors, 628–29, 628f for two-way floors, 625–33, 626f, 627f, 644, 647–48 transfer length, 887, 883 transformed section method effective moment of inertia and, 442 neutral axis and, 409–10, 409f, 410f serviceability and, 407–10, 407f, 409f, 410f transition zone, 338 balanced strain condition and, 333 bending moment and, 327f design strength for, 351–54, 354f strength reduction factors and, 327, 352 transverse distribution, of longitudinal moments, 656–62, 656f, 657t–62t transverse reinforcement (vertical stirrups), 3 for beams, 125, 128, 128f for brackets and corbels, 159 for cantilever walls, 580, 618 center-to-center spacing for, 186, 206, 386, 387 in columns, 368 for combined footings, 830, 831f for concrete composite flexural members, 903–4,

confinement from, 181, 186, 383 for continuous beams, 781 development length and, 186 for monolithic beam-column joints, 383–87, 384f, 385f, 387f, 394, 396–97, 396f shear and, 1, 124, 124f, 368, 785, 789 shear friction and, 154 as shear reinforcement, 123–24, 123f, 140–46, 140t, 143f, 144t, 146f for shear walls, 584–85 splitting planes and, 186–87 for struts, 549, 557–58, 565 torsional strength from, 770–72, 775 torsion and, 780, 785, 789 for walls, 572, 593 yield strength of, 223, 368 trial mixes, 18 triaxial strength of concrete, 14 stress-strain relationship and, 14f truss model for beams, 125–27, 126f, 127f, 128f shear reinforcement and, 122 T-sections. See also T-beams beam web in, 267–70 effective flange width for, 95–7 effective moment of inertia for, 414, 414f nominal moment strength of, 97 torsional stiffness of, 752 torsional stresses in, 751 in two-way floors, 669 two-way floors, 622, 622–711, 623f with beams, 624, 629–31, 629f, 630f, 633–43, 634f–36f, 634t, 638f, 639f, 642f, 669–71, 670f column capital for, 639–43, 640f column strips in, 656–62, 656f, 658t–62t, 664 concentric shear in, 673 corner reinforcement for, 664–65, 665f crack control for, 666 direct design method.See Direct Design Method (DDM) deflections in, 637–39, 638f, 639f, 704 drop panels for, 639–43, 640f edge beams in, 639–43 edge columns supporting, 689 effective depth for, 674 equivalent frame method. See Equivalent Frame Method (EFM). factored moment in, 625, 627, 663, 667, 667t–68t, 686–88, 704 flexural behavior of, 669–71, 689–91

flexural rigidity of, 657 flexural stiffness of, 633–37, 634f–36f, 634t, 644, 647, 657, 699, 702–3 flexural strength of, 640 gravity load in, 624, 625, 691, 698–708, 701f–3f, 705f, 706f, 708f, 709f longitudinal moments in, 645–51, 646f, 649f, 650t, 651f, 656–62, 656f, 657t–62t moment of inertia for, 633 moments in columns and, 686–97, 688f, 690f, 693f, 697f moment strength ratio for, 689 negative moment in, 627, 640, 664, 704 nominal shear strength of, 672 nominal shear stress in, 669 openings in, 697–8 positive moment for, 625–26, 647 reinforcement for, 662–68, 663f, 665f, 667t–68t, 669f shear reinforcement for, 676–85, 677f shear strength of, 640, 671–76, 673f, 674f, 676f, 697 slab thickness in, 637–43, 638f, 639f, 642f, 662–68, 663f, 665f, 667t–68t, 669f temperature reinforcement for, 622 torsional stiffness of, 651–55, 652f, 653f, 655f, 700–702 total factored static moment for, 625–33, 626f, 627f, 644, 647–48 transverse beams in, 652–55, 652f, 653f, 655f transfer of moment and shear in, 687–95 T-sections in, 669 yield line theory for, 722, 722f, 736–41, 736f–38f, 740f ultimate load analysis, yield line theory and, 720 ultimate strength method. See strength design method unbraced frames. See sway frames. underreinforced beams, 46 unintentional roughening, 903 unshored construction, of concrete composite flexural members, 905–8 vertical stirrups. See transverse reinforcement vibration control, for floor systems, 456 virtual work method, for yield line theory, 724–27

957

957

INDEX

wall boundary elements, 577, 578f walls, 569–618, 571f. See also bearing walls; cantilever walls; nonbearing walls; retaining walls; shear walls; thin-wall sections axial tension in, 588, 590 barbell, 578 with coupling beams, 579 effective length for, 574t flexure-shear cracks in, 590, 591f footings for, 800, 818–22, 820f, 821f longitudinal reinforcement for, 576, 593, 596, 596f rectangular sections for, 581–83, 582f, 583f reinforcement for, 570–72, 572t shear reinforcement for, 591–95, 592f, 596f shear strength of, 588–90, 589f–91f squat, 581, 589 tall, 580 thickness of, 569, 572f transverse reinforcement for, 593 web-shear cracks in, 590, 591f wall-to-floor area ratio, for shear walls, 576–77 warping, from shrinkage, 433–35, 434f water to cement (or to cementitious) ratio, 10, 16, 10f water-reducing admixtures, 8–9

web reinforcement, 122–24, 128, 123f, 124f for bottle-shaped struts, 536 torsion and shear and, 768 web-shear cracks, 114, 115f in prestressed concrete, 875–878 in walls, 589–590, 590f-591f welded splices in compression, 217 in tension, 215–216 welded wire reinforcement, 22–23 shear reinforcement and, 212–13, 213f stress-strain relationship for, 23f Whitney rectangular stress distribution, 47–48, 47f for T-beams, 97 wide-beam connections, 380 wind load, 37, 39 wire reinforcement, 22–23 workability, 5, 7, 8, 10, 25 working load. See service loads working stress method, 32–33, 404 yield line theory, 720 analysis methods and, 724–25 collapse mechanism and, 722, 724, 743, 747, 721f corner effects and, 742–43, 742f equilibrium method and, 724–726, 729–732, 736, 742, 744, 726f

free edge effect and, 729, 735, 730f, 731f fundamental assumptions on, 723–24 intersection of three yield lines, 729, 733 nodal forces and, 729–36, 730f–35f one-way slabs and, 721, 725–28, 726f, 728f patterns and, 722, 722f, 724f, 732f, 736f, 742f, 743f plastic hinges and, 720 rectangular two-way slabs and, 736, 742 special cases, 743–46, 744f, 746t triangular slab and, 730, 731f twisting moment and, 720, 723–24, 723f virtual work method and, 724–27 yield point in columns, 323 of deformed bars, 22t of prestressing reinforcement, 23 of welded wire reinforcement, 22–23 yield strain, 44, 46, 52, 55, 87 yield strength, 20, 22, 23 yield stress, 20–22 zinc-coated reinforcing bars, 24

958

95

960

961

962