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Draft LECTURE NOTES
AREN4525 STUCTURAL CONCEPTS AND SYSTEMS FOR ARCHITECTS VICTOR E. SAOUMA SPRING 1997
Dept. of Civil Environmental and Architectural Engineering University of Colorado, Boulder, CO 80309-0428 April 30, 1997
Draft 0{2
In order to invent a structure and to give it exact proportions, one must follow both the intuitive and the mathematical paths. -Pier Luigi Nervi
Victor Saouma
Structural Concepts and Systems for Architects
Draft
Contents 1 INTRODUCTION 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12
Science and Technology Structural Engineering Structures and their Surroundings Architecture & Engineering Architectural Design Process Architectural Design Structural Analysis Structural Design Load Transfer Mechanisms Structure Types Structural Engineering Courses References
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2 LOADS
2.1 Introduction 2.2 Vertical Loads 2.2.1 Dead Load 2.2.2 Live Loads E 2-1 Live Load Reduction 2.2.3 Snow 2.3 Lateral Loads 2.3.1 Wind E 2-2 Wind Load 2.3.2 Earthquakes E 2-3 Earthquake Load on a Frame E 2-4 Earthquake Load on a Tall Building, (Schueller 1996) 2.4 Other Loads 2.4.1 Hydrostatic and Earth E 2-5 Hydrostatic Load 2.4.2 Thermal E 2-6 Thermal Expansion/Stress (Schueller 1996) 2.5 Other Important Considerations 2.5.1 Load Combinations 2.5.2 Load Placement 2.5.3 Load Transfer 2.5.4 Structural Response 2.5.5 Tributary Areas
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Draft 0{2
CONTENTS
3 STRUCTURAL MATERIALS 3.1 Steel 3.1.1 Structural Steel 3.1.2 Reinforcing Steel 3.2 Aluminum 3.3 Concrete 3.4 Masonry 3.5 Timber 3.6 Steel Section Properties 3.7 Joists
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4 Case Study I: EIFFEL TOWER 4.1 4.2 4.3 4.4 4.5
Materials, & Geometry Loads Reactions Internal Forces Internal Stresses
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5 REVIEW of STATICS
5.1 Reactions 5.1.1 Equilibrium 5.1.2 Equations of Conditions 5.1.3 Static Determinacy 5.1.4 Geometric Instability 5.1.5 Examples E 5-7 Simply Supported Beam E 5-8 Three Span Beam E 5-9 Three Hinged Gable Frame 5.2 Trusses 5.2.1 Assumptions 5.2.2 Basic Relations 5.2.3 Determinacy and Stability 5.2.4 Method of Joints E 5-10 Truss, Method of Joints 5.3 Shear & Moment Diagrams 5.3.1 Theory 5.3.1.1 Design Sign Conventions 5.3.1.2 Load, Shear, Moment Relations 5.3.1.3 Moment Envelope 5.3.1.4 Examples E 5-11 Simple Shear and Moment Diagram E 5-12 Frame Shear and Moment Diagram E 5-13 Frame Shear and Moment Diagram; Hydrostatic Load E 5-14 Shear Moment Diagrams for Frame E 5-15 Shear Moment Diagrams for Inclined Frame 5.3.2 Formulaes 5.4 Flexure 5.4.1 Basic Kinematic Assumption; Curvature 5.4.2 Stress-Strain Relations 5.4.3 Internal Equilibrium; Section Properties 5.4.3.1 x = 0; Neutral Axis
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Victor Saouma
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Structural Concepts and Systems for Architects
Draft CONTENTS
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6 Case Study II: GEORGE WASHINGTON BRIDGE 6.1 Theory 6.2 The Case Study 6.2.1 Geometry 6.2.2 Loads 6.2.3 Cable Forces 6.2.4 Reactions
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7 A BRIEF HISTORY OF STRUCTURAL ARCHITECTURE 7.1 7.2 7.3 7.4 7.5
Before the Greeks Greeks Romans The Medieval Period (477-1492) The Renaissance 7.5.1 Leonardo da Vinci 1452-1519 7.5.2 Brunelleschi 1377-1446 7.5.3 Alberti 1404-1472 7.5.4 Palladio 1508-1580 7.5.5 Stevin 7.5.6 Galileo 1564-1642 7.6 Pre Modern Period, Seventeenth Century 7.6.1 Hooke, 1635-1703 7.6.2 Newton, 1642-1727 7.6.3 Bernoulli Family 1654-1782 7.6.4 Euler 1707-1783 7.7 The pre-Modern Period; Coulomb and Navier 7.8 The Modern Period (1857-Present) 7.8.1 Structures/Mechanics 7.8.2 Ei el Tower 7.8.3 Sullivan 1856-1924 7.8.4 Roebling, 1806-1869 7.8.5 Maillart 7.8.6 Nervi, 1891-1979 7.8.7 Khan 7.8.8 et al.
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8 Case Study III: MAGAZINI GENERALI 8.1 8.2 8.3 8.4 8.5
Geometry Loads Reactions Forces Internal Stresses
: : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
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Victor Saouma
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CONTENTS
9 DESIGN PHILOSOPHIES and GUIDELINES 9.1 Safety Provisions 9.2 Working Stress Method 9.3 Ultimate Strength Method 9.3.1 y Probabilistic Preliminaries 9.3.2 Discussion 9.4 Example E 9-18 LRFD vs ASD 9.5 Design Guidelines
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10 BRACED ROLLED STEEL BEAMS
10.1 Nominal Strength 10.2 Failure Modes and Classi cation of Steel Beams 10.3 Compact Sections 10.3.1 Bending Capacity of Beams 10.3.2 Design of Compact Sections 10.4 Partially Compact Section 10.5 Slender Section 10.6 Examples E 10-19 for Rectangular Section E 10-20Beam Design
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11 REINFORCED CONCRETE BEAMS
11.1 Introduction 11.1.1 Notation 11.1.2 Modes of Failure 11.1.3 Analysis vs Design 11.1.4 Basic Relations and Assumptions 11.1.5 ACI Code 11.2 Cracked Section, Ultimate Strength Design Method 11.2.1 Equivalent Stress Block 11.2.2 Balanced Steel Ratio 11.2.3 Analysis 11.2.4 Design E 11-21Ultimate Strength Capacity E 11-22Beam Design I E 11-23Beam Design II 11.3 Continuous Beams 11.4 ACI Code
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12.1 Introduction 12.1.1 Materials 12.1.2 Prestressing Forces 12.1.3 Assumptions 12.1.4 Tendon Con guration 12.1.5 Equivalent Load 12.1.6 Load Deformation 12.2 Flexural Stresses E 12-24Prestressed Concrete I Beam 12.3 Case Study: Walnut Lane Bridge
11{1 11{1 11{2 11{2 11{3 11{3 11{4 11{4 11{5 11{6 11{7 11{8 11{9 11{9 11{10 11{10
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Victor Saouma
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12 PRESTRESSED CONCRETE
9{1 9{2 9{3 9{3 9{5 9{7 9{7 9{7
12{1 12{1 12{4 12{4 12{4 12{4 12{4 12{5 12{8 12{10
Structural Concepts and Systems for Architects
Draft CONTENTS 12.3.1 12.3.2 12.3.3 12.3.4
0{5
Cross-Section Properties Prestressing Loads Flexural Stresses
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13 Three-Hinges ARCHES
13.1 Theory 13.1.1 Uniform Horizontal Load E 13-25Design of a Three Hinged Arch 13.2 Case Study: Salginatobel Bridge (Maillart) 13.2.1 Geometry 13.2.2 Loads 13.2.3 Reactions 13.2.4 Internal Forces 13.2.5 Internal Stresses 13.3 Structural Behavior of Deck-Sti ened Arches
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14 BUILDING STRUCTURES
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14.1 Introduction 14.1.1 Beam Column Connections 14.1.2 Behavior of Simple Frames 14.1.3 Eccentricity of Applied Loads 14.2 Buildings Structures 14.2.1 Wall Subsystems 14.2.1.1 Example: Concrete Shear Wall 14.2.1.2 Example: Trussed Shear Wall 14.2.2 Shaft Systems 14.2.2.1 Example: Tube Subsystem 14.2.3 Rigid Frames 14.3 Approximate Analysis of Buildings 14.3.1 Vertical Loads 14.3.2 Horizontal Loads 14.3.2.1 Portal Method E 14-26Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads 14.4 Lateral De ections 14.4.1 Short Wall 14.4.2 Tall Wall 14.4.3 Walls and Lintel 14.4.4 Frames 14.4.5 Trussed Frame 14.4.6 Example of Transverse De ection 14.4.7 E ect of Bracing Trusses
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Victor Saouma
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14{1 14{1 14{1 14{2 14{5 14{5 14{5 14{7 14{8 14{9 14{10 14{10 14{11 14{13 14{14 14{15 14{27 14{27 14{28 14{28 14{29 14{30 14{32 14{34
Structural Concepts and Systems for Architects
Draft List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9
Types of Forces in Structural Elements (1D) Basic Aspects of Cable Systems Basic Aspects of Arches Types of Trusses Variations in Post and Beams Con gurations Di erent Beam Types Basic Forms of Frames Examples of Air Supported Structures Basic Forms of Shells
2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14
Approximation of a Series of Closely Spaced Loads Snow Map of the United States, ubc Loads on Projected Dimensions Wind Map of the United States, (UBC 1995) E ect of Wind Load on Structures(Schueller 1996) Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures Vibrations of a Building Seismic Zones of the United States, (UBC 1995) Earth and Hydrostatic Loads on Structures Load Placement to Maximize Moments Load Transfer in R/C Buildings Two Way Actions Load Life of a Structure, (Lin and Stotesbury 1981) Concept of Tributary Areas for Structual Member Loading
2{10 2{12 2{13 2{18 2{21 2{22 2{23 2{24 2{25
3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8
Stress Strain Curves of Concrete and Steel Standard Rolled Sections Residual Stresses in Rolled Sections Residual Stresses in Welded Sections In uence of Residual Stress on Average Stress-Strain Curve of a Rolled Section Concrete microcracking W and C sections prefabricated Steel Joists
3{2 3{2 3{4 3{4 3{5 3{7 3{8 3{17
4.1 4.2 4.3 4.4 4.5
Ei el Tower (Billington and Mark 1983) Ei el Tower Idealization, (Billington and Mark 1983) Ei el Tower, Dead Load Idealization; (Billington and Mark 1983) Ei el Tower, Wind Load Idealization; (Billington and Mark 1983) Ei el Tower, Wind Loads, (Billington and Mark 1983)
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4{1 4{3 4{4 4{5 4{5
Draft 0{2
LIST OF FIGURES
4.6 4.7 4.8 4.9
Ei el Tower, Reactions; (Billington and Mark 1983) Ei el Tower, Internal Gravity Forces; (Billington and Mark 1983) Ei el Tower, Horizontal Reactions; (Billington and Mark 1983) Ei el Tower, Internal Wind Forces; (Billington and Mark 1983)
5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16
Types of Supports Inclined Roller Support Examples of Static Determinate and Indeterminate Structures Geometric Instability Caused by Concurrent Reactions Bridge Truss A Statically Indeterminate Truss X and Y Components of Truss Forces Sign Convention for Truss Element Forces Shear and Moment Sign Conventions for Design Sign Conventions for 3D Frame Elements Free Body Diagram of an In nitesimal Beam Segment Shear and Moment Forces at Di erent Sections of a Loaded Beam Slope Relations Between Load Intensity and Shear, or Between Shear and Moment Deformation of a Beam un Pure Bending Elastic Curve from the Moment Diagram Approximate Analysis of Beams
6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9
Cable Structure Subjected to ( ) Longitudinal and Plan Elevation of the George Washington Bridge Truck Load Dead and Live Loads Location of Cable Reactions Vertical Reactions in Columns Due to Central Span Load Cable Reactions in Side Span Cable Stresses Deck Idealization, Shear and Moment Diagrams
7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16 7.17
Hamurrabi's Code Archimed Pantheon From Vitruvius Ten Books on Architecture, (Vitruvius 1960) Hagia Sophia Florence's Cathedral Dome Palladio's Villa Rotunda Stevin Galileo Discourses Concerning Two New Sciences, Cover Page \Galileo's Beam" Experimental Set Up Used by Hooke Isaac Newton Philosophiae Naturalis Principia Mathematica, Cover Page Leonhard Euler Coulomb Nervi's Palazetto Dello Sport
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p x
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8.1 Magazzini Generali; Overall Dimensions, (Billington and Mark 1983) 8.2 Magazzini Generali; Support System, (Billington and Mark 1983)
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Victor Saouma
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Structural Concepts and Systems for Architects
Draft LIST OF FIGURES
0{3
8.3 8.4 8.5 8.6 8.7
Magazzini Generali; Loads (Billington and Mark 1983) Magazzini Generali; Beam Reactions, (Billington and Mark 1983) Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983) Magazzini Generali; Internal Moment, (Billington and Mark 1983) Magazzini Generali; Similarities Between The Frame Shape and its Moment Diagram, (Billington and Mark 1983) 8.8 Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark 1983) 8.9 Magazzini Generali; E ect of Lateral Supports, (Billington and Mark 1983)
8{3 8{3 8{4 8{4
9.1 Load Life of a Structure 9.2 Frequency Distributions of Load 9.3 De nition of Reliability Index
9{2 9{4 9{4
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and Resistance
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Lateral Bracing for Steel Beams Failure of Steel beam; Plastic Hinges Failure of Steel beam; Local Buckling Failure of Steel beam; Lateral Torsional Buckling Stress distribution at di erent stages of loading Stress-strain diagram for most structural steels Nominal Moments for Compact and Partially Compact Sections
11.1 11.2 11.3 11.4 11.5
Failure Modes for R/C Beams Internal Equilibrium in a R/C Beam Cracked Section, Limit State Whitney Stress Block Reinforcement in Continuous R/C Beams
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8{5 8{5 8{6
10{1 10{3 10{3 10{4 10{4 10{5 10{7 11{2 11{3 11{4 11{5 11{11
Pretensioned Prestressed Concrete Beam, (Nilson 1978) Posttensioned Prestressed Concrete Beam, (Nilson 1978) 7 Wire Prestressing Tendon Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978) Determination of Equivalent Loads Load-De ection Curve and Corresponding Internal Flexural Stresses for a Typical Prestressed Concrete Beam, (Nilson 1978) 12.7 Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum Moment Section and Support Section, (Nilson 1978) 12.8 Walnut Lane Bridge, Plan View 12.9 Walnut Lane Bridge, Cross Section
12{7 12{11 12{12
13.1 Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Lin and Stotesbury 1981) 13.2 Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981) 13.3 Two Hinged Arch, (Lin and Stotesbury 1981) 13.4 Arch Rib Sti ened with Girder or Truss, (Lin and Stotesbury 1981) 13.5 Salginatobel Bridge; Dimensions, (Billington and Mark 1983) 13.6 Salginatobel Bridge; Idealization, (Billington and Mark 1983) 13.7 Salginatobel Bridge; Hinges, (Billington and Mark 1983) 13.8 Salginatobel Bridge; Sections, (Billington and Mark 1983) 13.9 Salginatobel Bridge; Dead Load, (Billington and Mark 1983) 13.10Salginatobel Bridge; Truck Load, (Billington and Mark 1983) 13.11Salginatobel Bridge; Total Vertical Load, (Billington and Mark 1983)
13{2 13{2 13{3 13{3 13{5 13{6 13{6 13{7 13{8 13{9 13{10
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LIST OF FIGURES
13.12Salginatobel Bridge; Reactions, (Billington and Mark 1983) 13.13Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983) 13.14Salginatobel Bridge; Live Load Moment Diagram, (Billington and Mark 1983) 13.15Structural Behavior of Sti ened Arches, (Billington 1979)
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14.1 Flexible, Rigid, and Semi-Flexible Joints 14.2 Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads, (Lin and Stotesbury 1981) 14.3 Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal Frames Subjected to Vertical and Horizontal Loads 14.4 Axial and Flexural Stresses 14.5 Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981) 14.6 Trussed Shear Wall 14.7 Design Example of a Tubular Structure, (Lin and Stotesbury 1981) 14.8 A Basic Portal Frame, (Lin and Stotesbury 1981) 14.9 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments 14.10Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces 14.11Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments 14.12Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear 14.13***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment 14.14Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force 14.15Example; Approximate Analysis of a Building 14.16Approximate Analysis of a Building; Moments Due to Vertical Loads 14.17Approximate Analysis of a Building; Shears Due to Vertical Loads 14.18Approximate Analysis for Vertical Loads; Spread-Sheet Format 14.19Approximate Analysis for Vertical Loads; Equations in Spread-Sheet 14.20Approximate Analysis of a Building; Moments Due to Lateral Loads 14.21Portal Method; Spread-Sheet Format 14.22Portal Method; Equations in Spread-Sheet 14.23Shear Deformation in a Short Building, (Lin and Stotesbury 1981) 14.24Flexural Deformation in a Tall Building, (Lin and Stotesbury 1981) 14.25De ection in a Building Structure Composed of Two Slender Walls and Lintels, (Lin and Stotesbury 1981) 14.26Portal Method to Estimate Lateral Deformation in Frames, (Lin and Stotesbury 1981) 14.27Shear and Flexural De ection of a Rigid Frame Subsystem, (Lin and Stotesbury 1981) 14.28Side-Sway De ection from Unsymmetrical Vertical Load, (Lin and Stotesbury 1981) 14.29Axial Elongation and Shortening of a Truss Frame, (Lin and Stotesbury 1981) 14.30Transverse De ection, (Lin and Stotesbury 1981) 14.31Frame Rigidly Connected to Shaft, (Lin and Stotesbury 1981) 14.32E ect of Exterior Column Bracing in Buildings, (Lin and Stotesbury 1981)
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Structural Concepts and Systems for Architects
Draft List of Tables 1.1 Structural Engineering Coverage for Architects and Engineers : : : : : : : : : : : : : : : : 1{12 1.2 tab:secae : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1{12 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13
Unit Weight of Materials : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{2 Weights of Building Materials : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{3 Average Gross Dead Load in Buildings : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{3 Minimum Uniformly Distributed Live Loads, (UBC 1995) : : : : : : : : : : : : : : : : : : 2{4 Wind Velocity Variation above Ground : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{7 Ce Coecients for Wind Load, (UBC 1995) : : : : : : : : : : : : : : : : : : : : : : : : : : 2{8 Wind Pressure Coecients Cq , (UBC 1995) : : : : : : : : : : : : : : : : : : : : : : : : : : 2{8 Importance Factors for Wind and Earthquake Load, (UBC 1995) : : : : : : : : : : : : : : 2{9 Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{10 Z Factors for Di erent Seismic Zones, ubc : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{12 S Site Coecients for Earthquake Loading, (UBC 1995) : : : : : : : : : : : : : : : : : : : 2{13 Partial List of RW for Various Structure Systems, (UBC 1995) : : : : : : : : : : : : : : : 2{15 Coecients of Thermal Expansion : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{19
3.1 3.2 3.3 3.4
Properties of Major Structural Steels : Properties of Reinforcing Bars : : : : : Joist Series Characteristics : : : : : : Joist Properties : : : : : : : : : : : : :
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5.1 Equations of Equilibrium : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5{3 5.2 Static Determinacy and Stability of Trusses : : : : : : : : : : : : : : : : : : : : : : : : : : 5{10 5.3 Section Properties : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5{41 9.1 9.2 9.3 9.4
Allowable Stresses for Steel and Concrete : : : : : : : : : : : : : : : : : : : : : : : : : : : 9{3 Selected values for Steel and Concrete Structures : : : : : : : : : : : : : : : : : : : : : : 9{5 Strength Reduction Factors, : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9{6 Approximate Structural Span-Depth Ratios for Horizontal Subsystems and Components (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9{8
14.1 Columns Combined Approximate Vertical and Horizontal Loads : : : : : : : : : : : : : : 14{26 14.2 Girders Combined Approximate Vertical and Horizontal Loads : : : : : : : : : : : : : : : 14{27
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INTRODUCTION 1.1
Science and Technology
1 \There is a fundamental di erence between science and and technology. Engineering or technology is the making of things that did not previously exist, whereas science is the discovering of things that have long existed. Technological results are forms that exist only because people want to make them, whereas scienti c results are informations of what exists independently of human intentions. Technology deals with the arti cial, science with the natural." (Billington 1985)
1.2
Structural Engineering
Structural engineers are responsible for the detailed analysis and design of: Architectural structures: Buildings, houses, factories. They must work in close cooperation with an architect who will ultimately be responsible for the design. Civil Infrastructures: Bridges, dams, pipelines, o shore structures. They work with transportation, hydraulic, nuclear and other engineers. For those structures they play the leading role. Aerospace, Mechanical, Naval structures: aeroplanes, spacecrafts, cars, ships, submarines to ensure the structural safety of those important structures. 2
1.3 3
Structures and their Surroundings
Structural design is a ected by various environmental constraints: 1. Major movements: For example, elevator shafts are usually shear walls good at resisting lateral load (wind, earthquake). 2. Sound and structure interact: A dome roof will concentrate the sound A dish roof will di use the sound 3. Natural light: A at roof in a building may not provide adequate light.
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A Folded plate will provide adequate lighting (analysis more complex). A bearing and shear wall building may not have enough openings for daylight. A Frame design will allow more light in (analysis more complex). 4. Conduits for cables (electric, telephone, computer), HVAC ducts, may dictate type of oor system. 5. Net clearance between columns (unobstructed surface) will dictate type of framing.
1.4 Architecture & Engineering Architecture must be the product of a creative collaboration of architects and engineers. 5 Architect stress the overall, rather than elemental approach to design. In the design process, they conceptualize a space-form scheme as a total system. They are generalists. 6 The engineer, partly due to his/her education think in reverse, starting with details and without sucient regards for the overall picture. (S)he is a pragmatist who \knows everything about nothing". 7 Thus there is a conceptual gap between architects and engineers at all levels of design. 8 Engineer's education is more specialized and in depth than the architect's. However, engineer must be kept aware of overall architectural objective. 9 In the last resort, it is the architect who is the leader of the construction team, and the engineers are his/her servant. 10 A possible compromise might be an Architectural Engineer. 4
1.5 Architectural Design Process Architectural design is hierarchical: Schematic: conceptual overall space-form feasibility of basic schematic options. Collaboration is mostly between the owner and the architect. Preliminary: Establish basic physical properties of major subsystems and key components to prove design feasibility. Some collaboration with engineers is necessary. Final design: nal in-depth design re nements of all subsystems and components and preparation of working documents (\blue-prints"). Engineers play a leading role. 11
1.6 Architectural Design Architectural design must respect various constraints: Functionality: In uence of the adopted structure on the purposes for which the structure was erected. Aesthetics: The architect often imposes his aesthetic concerns on the engineer. This in turn can place severe limitations on the structural system. Economy: It should be kept in mind that the two largest components of a structure are labors and materials. Design cost is comparatively negligible.
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1{3
Buildings may have di erent functions: Residential: housing, which includes low-rise (up tp 2-3 oors), mid-rise (up to 6-8 oors) and high rise buildings. Commercial: Oces, retail stores, shopping centers, hotels, restaurants. Industrial: warehouses, manufacturing. Institutional: Schools, hospitals, prisons, chruch, government buildings. Special: Towers, stadium, parking, airport, etc. 13
1.7 Structural Analysis 14 Given an existing structure subjected to a certain load determine internal forces (axial, shear, exural, torsional; or stresses), de ections, and verify that no unstable failure can occur. 15 Thus the basic structural requirements are: Strength: stresses should not exceed critical values: < f Sti ness: de ections should be controlled: < max Stability: buckling or cracking should also be prevented
1.8 Structural Design Given a set of forces, dimension the structural element. Steel/wood Structures Select appropriate section. Reinforced Concrete: Determine dimensions of the element and internal reinforcement (number and sizes of reinforcing bars).
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17 For new structures, iterative process between analysis and design. A preliminary design is made using rules of thumbs (best known to Engineers with design experience) and analyzed. Following design, we check for Serviceability: de ections, crack widths under the applied load. Compare with acceptable values speci ed in the design code. Failure (limit state): and compare the failure load with the applied load times the appropriate factors of safety. If the design is found not to be acceptable, then it must be modi ed and reanalyzed. 18 For existing structures rehabilitation, or veri cation of an old infrastructure, analysis is the most important component. 19 In summary, analysis is always required.
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INTRODUCTION
Figure 1.1: Types of Forces in Structural Elements (1D)
1.9 Load Transfer Mechanisms From Strength of Materials, loads can be transferred through various mechanisms, Fig. 1.1 Axial: cables, truss elements, arches, membrane, shells Flexural: Beams, frames, grids, plates Torsional: Grids, 3D frames Shear: Frames, grids, shear walls.
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1.10 Structure Types 21
Structures can be classi ed as follows:
Tension & Compression Structures: only, no shear, exure, or torsion. Those are the most ecient types of structures. Cable (tension only): The high strength of steel cables, combined with the eciency of simple
tension, makes cables ideal structural elements to span large distances such as bridges, and dish roofs, Fig. 1.2. A cable structure develops its load carrying capacity by adjusting its shape so as to provide maximum resistance (form follows function). Care should be exercised in minimizing large de ections and vibrations. Arches (mostly compression) is a \reversed cable structure". In an arch, exure/shear is minimized and most of the load is transfered through axial forces only. Arches are used for large span roofs and bridges, Fig. 1.3
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Figure 1.2: Basic Aspects of Cable Systems
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INTRODUCTION
Figure 1.3: Basic Aspects of Arches
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Trusses have pin connected elements which can transmit axial forces only (tension and compression). Elements are connected by either slotted, screwed, or gusset plate connectors. However, due to construction details, there may be secondary stresses caused by relatively rigid connections. Trusses are used for joists, roofs, bridges, electric tower, Fig. 1.4
Figure 1.4: Types of Trusses
Post and Beams: Essentially a support column on which a \beam" rests, Fig. 1.5, and 1.6. Beams: Shear, exure and sometimes axial forces. Recall that = MI c is applicable only for shallow beams, i.e. span/depth at least equal to ve.
Whereas r/c beams are mostly rectangular or T shaped, steel beams are usually I shaped (if the top anges are not properly sti ened, they may buckle, thus we must have sti eners). Frames: Load is co-planar with the structure. Axial, shear, exure (with respect to one axis in 2D structures and with respect to two axis in 3D structures), torsion (only in 3D). The frame is composed of at least one horizontal member (beam) rigidly connected to vertical ones1. The vertical 1 The precursor of the frame structures were the Post and Lintel where the post is vertical member on which the lintel is simply posed.
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INTRODUCTION
Figure 1.5: Variations in Post and Beams Con gurations
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1{9
VIERENDEEL TRUSS
OVERLAPPING SINGLE-STRUT CABLE-SUPPORTED BEAM
TREE-SUPPORTED TRUSS
BRACED BEAM
CABLE-STAYED BEAM
SUSPENDED CABLE SUPPORTED BEAM
BOWSTRING TRUSS
CABLE-SUPPORTED STRUTED ARCH OR CABLE BEAM/TRUSS
CABLE-SUPPORTED MULTI-STRUT BEAM OR TRUSS
GABLED TRUSS
CABLE-SUPPORTED ARCHED FRAME
CABLE-SUPPORTED PORTAL FRAME
Figure 1.6: Di erent Beam Types
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members can have di erent boundary conditions (which are usually governed by soil conditions). Frames are extensively used for houses and buildings, Fig. 1.7.
Figure 1.7: Basic Forms of Frames
Grids and Plates: Load is orthogonal to the plane of the structure. Flexure, shear, torsion.
In a grid, beams are at right angles resulting in a two-way dispersal of loads. Because of the rigid connections between the beams, additional sti ness is introduced by the torsional resistance of members. Grids can also be skewed to achieve greater eciency if the aspect ratio is not close to one. Plates are at, rigid, two dimensional structures which transmit vertical load to their supports. Used mostly for oor slabs. Folded plates is a combination of transverse and longitudinal beam action. Used for long span roofs. Note that the plate may be folded circularly rather than longitudinally. Folded plates are used mostly as long span roofs. However, they can also be used as vertical walls to support both vertical and horizontal loads. Membranes: 3D structures composed of a exible 2D surface resisting tension only. They are usually cable-supported and are used for tents and long span roofs Fig. 1.8.
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Figure 1.8: Examples of Air Supported Structures
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INTRODUCTION
Shells: 3D structures composed of a curved 2D surface, they are usually shaped to transmit compressive axial stresses only, Fig. 1.9.
Figure 1.9: Basic Forms of Shells Shells are classi ed in terms of their curvature. 1.11
Structural Engineering Courses
Structural engineering education can be approached from either one of two points of views, depending on the audience, ??.
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Architects Global Structure Approximate, \rules of thumbs" preliminary Structures Most Design Approximate Approach Emphasis Analysis
Engineers Elemental Component Exact, detailled Final Trusses, Frames Per code
Table 1.1: Structural Engineering Coverage for Architects and Engineers Table 1.2: tab:secae
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Architects: Start from overall design, and move toward detailed analysis. Emphasis on good under-
standing of overall structural behavior. Develop a good understanding of load transfer mechanism for most types of structures, cables, arches, beams, frames, shells, plates. Approximate analysis for most of them. Engineers: Emphasis is on the individual structural elements and not always on the total system. Focus on beams, frames (mostly 2D) and trusses. Very seldom are arches covered. Plates and shells are not even mentioned. 1.12
References
Following are some useful references for structural engineering, those marked by y were consulted, and \borrowed from" in preparing the Lecture Notes or are particularly recommended.
23
Structures for Architect
1. Ambrose, J., Building Structures, second Ed. Wiley, 1993. 2. Billington, D.P. Rober Maillart's Bridges; The Art of Engineering, Princeton University Pres, 1979. 3. yBillington, D.P., The Tower and the Bridge; The new art of structural engineering, Princeton University Pres,, 1983. 4. yBillington, D.P., Structures and the Urban Environment, Lectures Notes CE 262, Department of Civil Engineering, Princeton University, 1978 5. French, S., Determinate Structures; Statics, Strength, Analysis, Design, Delmar, 1996. 6. Gordon, J.E., Structures, or Why Things Do'nt Fall Down, Da Capo paperback, New York, 1978. 7. Gordon, J.E., The Science of Structures and Materials, Scienti c American Library, 1988. 8. Hawkes, N., Structures, the way things are built, MacMillan, 1990. 9. Levy, M. and Salvadori, M., Why Buildings Fall Down, W.W.Norton, 1992. 10. yLin, T.Y. and Stotesbury, S.D., Structural Concepts and Systems for Architects and Engineers, John Wiley, 1981. 11. yMainstone, R., Developments in Structural Form, Allen Lane Publishers, 1975. 12. Petroski, H., To Enginer is Human, Vintage Books, 1992. 13. ySalvadori, M. and Heller, R., Structure in Architecture; The Building of Buildings, Prentice Hall, Third Edition, 1986. 14. Salvadori, M. and Levy, M., Structural Design in Architecture, Prentice hall, Second Edition, 1981. 15. Salvadori, M., Why Buildings Stand Up; The Strength of Architecture, Norton Paperack, 1990. 16. ySandaker, B.N. and Eggen, A.P., The Structural Basis of Architecture, Whitney Library of Design, 1992. 17. ySchueller, W., The design of Building Structures, Prentice Hall, 1996.
Structures for Engineers
1. y Arbadi, F. Structural Analysis and Behavior, McGraw-Hill, Inc., 1991. 2. Biggs, J.M., Introduction to Structural Engineering; Analysis and Design, Prentice Hall, 1986.
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INTRODUCTION
3. Hsieh, Y.Y., Elementary Theory of Structures, Third Edition, Prentice Hall, 1988. 4. Ghali, A., and Neville, A.M., Structural Analysis, Third Edition, Chapman and Hall, 1989 5. White, R. Gergely, P. and Sexmith, R., Structural Engineering; Combined Edition, John Wiley, 1976. 6. y Nilson, A., and Winter, G. Design of Concrete Structures, Eleventh Edition, McGraw Hill, 1991. 7. Galambos, T., Lin, F.J., and Johnston, B.G., Basic Steel Design with LRFD, Prentice Hall, 1996. 8. y Salmon C. and Johnson, J. Steel Structures, Third Edition, Harper Collins Publisher, 1990. 9. y Gaylord, E.H., Gaylord, C.N. and Stallmeyer, J.E., Design of Steel Structures, Third Edition, McGraw Hill, 1992. 10. Vitruvius, The Ten Books on Architecture, Dover Publications, 1960. 11. Palladio, A., The Four Books of Architecture, Dover Publication.
Codes
1. ACI-318-89, Building Code Requirements for Reinforced Concrete, American Concrete Institute 2. Load & Resistance Factor Design, Manual of Steel Construction, American Institute of Steel Construction. 3. Uniform Building Code, International Conference of Building Ocials, 5360 South Workman Road; Whittier, CA 90601 4. Minimum Design Loads in Buildings and Other Structures, ANSI A58.1, American National Standards Institute, Inc., New York, 1972.
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LOADS 2.1
Introduction
1 The main purpose of a structure is to transfer load from one point to another: bridge deck to pier; slab to beam; beam to girder; girder to column; column to foundation; foundation to soil. 2 There can also be secondary loads such as thermal (in restrained structures), di erential settlement of foundations, P-Delta e ects (additional moment caused by the product of the vertical force and the lateral displacement caused by lateral load in a high rise building). 3 Loads are generally subdivided into two categories Vertical Loads or gravity load 1. dead load (DL) 2. live load (LL) also included are snow loads. Lateral Loads which act horizontally on the structure 1. Wind load (WL) 2. Earthquake load (EL) this also includes hydrostatic and earth loads.
This distinction is helpful not only to compute a structure's load, but also to assign di erent factor of safety to each one. 5 For a detailed coverage of loads, refer to the Universal Building Code (UBC), (UBC 1995).
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2.2
Vertical Loads
6 For closely spaced identical loads (such as joist loads), it is customary to treat them as a uniformly distributed load rather than as discrete loads, Fig. 2.1
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LOADS P1
P2
P3
P4
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P6
P7
REPETITIVE JOIST LOADS ACTUAL DISCRETE LOADS ON SUPPORT BEAM
w LB/FT = TOTAL LOAD / SPAN
SUPPORT BEAM SPAN
ASSUMED EQUIVALENT UNIFORM LOAD TYPICAL SYSTEM OF JOISTS
Figure 2.1: Approximation of a Series of Closely Spaced Loads 2.2.1
Dead Load
Dead loads (DL) consist of the weight of the structure itself, and other permanent xtures (such as walls, slabs, machinery). 8 For analysis purposes, dead loads can easily be determined from the structure's dimensions and density, Table 2.1
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Material
lb=ft3 Aluminum 173 Brick 120 Concrete 145 Steel 490 Wood (pine) 40
kN=m3 27.2 18.9 33.8 77.0 6.3
Table 2.1: Unit Weight of Materials For steel structures, the weight per unit length of rolled sections is given in the AISC Manual of Steel Construction. 10 For design purposes, dead loads must be estimated and veri ed at the end of the design cycle. This makes the design process iterative. 11 Weights for building materials is given in Table 2.2 12 For preliminary design purposes the average dead loads of Table 2.3 can be used:
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2.2.2
Live Loads
Contrarily to dead loads which are xed and vertical, live loads (LL) are movable or moving and may be horizontal. 14 Occupancy load may be due to people, furniture, equipment. The loads are essentially variable point loads which can be placed anywhere.
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2{3
Material
Ceilings Channel suspended system Acoustical ber tile Floors Steel deck Concrete-plain 1 in. Linoleum 1/4 in. Hardwood Roofs Copper or tin 5 ply felt and gravel Shingles asphalt Clay tiles Sheathing wood Insulation 1 in. poured in place Partitions Clay tile 3 in. Clay tile 10 in. Gypsum Block 5 in. Wood studs 2x4 (12-16 in. o.c.) Plaster 1 in. cement Plaster 1 in. gypsum Walls Bricks 4 in. Bricks 12 in. Hollow concrete block (heavy aggregate) 4 in. 8 in. 12 in. Hollow concrete block (light aggregate) 4 in. 8 in. 12 in.
lb=ft2 1 1 2-10 12 1 4 1-5 6 3 9-14 3 2 17 40 14 2 10 5 40 120 30 55 80 21 38 55
Table 2.2: Weights of Building Materials
Material
lb=ft2 Timber 40-50 Steel 50-80 Reinforced concrete 100-150 Table 2.3: Average Gross Dead Load in Buildings
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LOADS
In analysis load placement should be such that their e ect (shear/moment) are maximized. 16 A statistical approach is used to determine a uniformly distributed static load which is equivalent to the weight of the maximum concentration of occupants. These loads are de ned in codes such as the Uniform Building Code or the ANSI Code, Table 2.4.
15
Use or Occupancy
lb=ft2 Assembly areas 50 Cornices, marquees, residential balconies 60 Corridors, stairs 100 Garage 50 Oce buildings 50 Residential 40 Storage 125-250 Table 2.4: Minimum Uniformly Distributed Live Loads, (UBC 1995) For small areas (30 to 50 sq ft) the e ect of concentrated load should be considered separately. 18 Since there is a small probability that the whole oor in a building be fully loaded, the UBC code speci es that the occupancy load for members supporting an area A larger than 150 ft2 (i.e. a column with a total tributary area, including oors above it, larger than 150 ft2 ) may be reduced by R where
17
R = r(A ; 150) 23:1 1 + DL LL
(2.1)
where r = :08 for oors, A is the supported area ( 2 ) DL and LL are the dead and live loads per unit area supported by the member. R can not exceed 40% for horizontal members and 60% for vertical ones. ft
Example 2-1: Live Load Reduction In a 10 story oce building with a column spacing of 16 ft in both directions, the total dead load is 60 psf, snow load 20 psf and live load 80 psf. what is the total live load and total load for which a column must be designed on the ground oor
Solution: 1. 2. 3. 4.
p
The tributary area is 16 16 = 256ft2 > 150 The reduction R for the roof is is R = :08(16 16 ; 150) = 8:48% ; p 60 Maximum allowable reduction Rmax = 23:1 1 + 80 = 40:4% which is less than 60% The reduced cumulative load for the column of each oor is
Floor A A ; 150 R R% LL (100 ; R) LL=100 0
Victor Saouma
Roof 10 9 8 7 6 5 4 3 2 256 512 768 1024 1280 1536 1792 2048 2304 2560 106 362 618 874 1130 1386 1642 1898 2154 2410 8.48 28.96 49.44 69.92 90.40 110.88 131.36 151.84 172.32 192.8 8.48 28.96 40.4 40.4 40.4 40.4 40.4 40.4 40.4 40.4 20 80 80 80 80 80 80 80 80 80 18.3 56.83 47.68 47.68 47.68 47.68 47.68 47.68 47.68 47.68
Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{5
The resulting design live load for the bottom column has been reduced from (2.2) LL = (20) psf(256) ft2 + (9)(80) psf(256) ft2 = 189,440 lbs {z } | {z } | Roof 9 oors to LL = (18 :3) psf(256) ft2 + (9)(47:68) psf(256) ft2 = 114,540 lbs (2.3) | {z } | {z } Roof 9 oors 5. The total dead load is DL = (10)(60) psf(256) ft2 (1 000)k lbs = 153:6 Kips, thus the total reduction ;268 100= 22% . in load is from 153:6+189:4 = 343 k to 153:6+114:5 = 268:1 k a reduction of 343343 Bef ore
Reduced
;
2.2.3
Snow
Roof snow load vary greatly depending on geographic location and elevation. They range from 20 to 45 psf, Fig. 2.2.
19
Figure 2.2: Snow Map of the United States, ubc Snow loads are always given on the projected length or area on a slope, Fig. 2.3. 21 The steeper the roof, the lower the snow retention. For snow loads greater than 20 psf and roof pitches more than 20 the snow load p may be reduced by
20
R = ( ; 20) 2.3 2.3.1
p
40 ; 0:5
(psf)
(2.4)
Lateral Loads Wind
Wind load depend on: velocity of the wind, shape of the building, height, geographical location, texture of the building surface and sti ness of the structure.
22
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2{6
LOADS LIVE LOAD DEAD LOAD
LE
N
G TH
RISE
WIND LOAD
RUN
Figure 2.3: Loads on Projected Dimensions Wind loads are particularly signi cant on tall buildings1. 24 When a steady streamline air ow of velocity V is completely stopped by a rigid body, the stagnation pressure (or velocity pressure) qs was derived by Bernouilli (1700-1782) 1 (2.5) qs = V 2 2 where the air mass density is the air weight divided by the accleration of gravity g = 32:2 ft/sec2 . At sea level and a temperature of 15oC (59oF), the ai weighs 0.0765 lb/ft3 this would yield a pressure of
23
qs
or
= 21 (0:0765)lb/ft2 (32:2)ft/sec
3
qs
(5280)ft/mile V (3600)sec/hr
= 0:00256V 2
2
(2.6) (2.7)
where V is the maximum wind velocity (in miles per hour) and qs is in psf. V can be obtained from wind maps (in the United States 70 V 110), Fig. 2.4. 25 During storms, wind velocities may reach values up to or greater than 150 miles per hour, which corresponds to a dynamic pressure qs of about 60 psf (as high as the average vertical occupancy load in buildings). 1 The primary design consideration for very high rise buildings is the excessive drift caused by lateral load (wind and possibly earthquakes).
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{7
Figure 2.4: Wind Map of the United States, (UBC 1995) 26
Wind pressure increases with height, Table 2.5. Height Zone (in feet) 1,200
20 15 20 25 30 35 40
Wind-Velocity Map Area 25 30 35 40 45 50 20 25 25 30 35 40 25 30 35 40 45 50 30 40 45 50 55 60 40 45 55 60 70 75 45 55 60 70 80 90 50 60 70 80 90 100
Table 2.5: Wind Velocity Variation above Ground Wind load will cause suction on the leeward sides, Fig. 2.6 28 This magnitude must be modi ed to account for the shape and surroundings of the building. Thus, the design base pressure (at 33.3 ft from the ground) p (psf) is given by
27
p = Ce Cq Iqs
(2.8)
The pressure is assumed to be normal to all walls and roofs and Ce Velocity Pressure Coecient accounts for height, exposure and gust factor. It accounts for the fact that wind velocity increases with height and that dynamic character of the air ow (i.e the wind pressure is not steady), Table 2.6. l Cq Pressure Coecient is a shape factor which is given in Table 2.7 for gabled frames. I Importance Factor as given by Table 2.8. where
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2{8
LOADS
Figure 2.5: E ect of Wind Load on Structures(Schueller 1996) Exposure D Open, at terrain facing large bodies of water C Flat open terrain, extending one-half mile or open from the site in any full quadrant 0.62-1.80 B Terrain with buildings, forest, or surface irregularities 20 ft or more in height Ce
1.39-2.34 1.06-2.19
Table 2.6:
Ce
Coecients for Wind Load, (UBC 1995)
Windward Side Leeward Side Gabled Frames (V:H) Roof Slope 9:12 ;0 7 ;0 7 9:12 to 12:12 04 ;0 7 12:12 07 ;0 7 Walls 08 ;0 5 Buildings (height 200 ft) Vertical Projections height 40 ft 13 ;1 3 height 40 ft 14 ;1 4 Horizontal Projections ;0 7 ;0 7
>
>
>
29
For the preliminary design of ordinary buildings p
Ce
= (1 3) 020256 :
:
V
= 1 0 and
2
:
= 00333 :
Cq
V
= 1 3 may be assumed, yielding :
(2.9)
2
which corresponds to a pressure of 21 psf for a wind speed of 80 mph, Fig. 2.6, Table 2.9.
Example 2-2: Wind Load Determine the wind forces on the building shown on below which is built in St Louis and is surrouded by trees.
Solution:
1. From Fig. 2.4 the maximum wind velocity is St. Louis is 70 mph, since the building is protected we can take e = 0 7, = 1 . The base wind pressure is s = 0 00256 (70)2 = 12 54 psf. C
Victor Saouma
:
I
:
q
:
:
Structural Concepts and Systems for Architects
Draft 2{10
LOADS
Height Above Grade (ft) 0-15 20 25 30 40 60 80 100 120 160 200 300 400
Exposure B C Basic Wind Speed (mph) 70 10 11 12 12 14 17 18 20 21 23 25 29 32
80 13 14 15 16 18 22 24 26 28 30 33 37 41
70 17 18 19 20 21 25 27 28 29 31 33 36 38
80 23 24 25 26 28 33 35 37 38 41 43 47 50
Table 2.9: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures
400 Exposure B, 70 mph Exposure B, 80 mph Exposure C, 70 mph Exposure C, 80 mph
350
Height Above Grade (ft)
300
250
200
150
100
50
0
0
5
10
15 20 25 30 35 40 Approximate Design Wind Pressure (psf)
45
50
Figure 2.6: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{11
2. The slope of the roof is 8:15=6.4:12 which gives q = ;0 7 for both the windward and leeward sides. The vertical walls have q = 0 8 for the winward side and q = ;0 5 for the leeward one. C
C
:
:
C
:
3. Thus the applied pressure on the roof is = 0 7 (;0 7) 12 54 = -6.14 psf that is the roof is subjected to uplift. p
:
:
:
4. The winward wall, the pressure is 0 7 0 8 12 54 = 7.02 psf , and for the leeward wall 0 7 (;0 5) 12 54 = -4.39 psf (suction) , :
:
:
:
:
:
5. The direction of the wind can change and hence each structural component must be designed to resist all possible load combinations. 6. For large structures which may be subjected to large wind loads, testing in a wind tunnel of the structure itself and its surroundings is often accomplished.
2.3.2
Earthquakes
Buildings should be able to resist Minor earthquakes without damage Moderate earthquakes without structural damage but possibly with some nonstructural damages Major earthquakes without collapse but possibly with some structural damage as well as nonstructural damage This is achieved through an appropriate dynamic analysis. 31 For preliminary designs or for small structures an equivalent horizontal static load can be determined. 32 Actual loads depend on the following 1. Intensity of the ground acceleration (including soil/rock properties). 2. Dynamic properties of the building, such as its mode shapes and periods of vibration and its damping characteristics. 3. Mass of the building.
30
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2{12
LOADS
A critical factor in the dynamic response of a structure is the fundamental period of the structure's vibration (or rst mode of vibration). This is the time required for one full cycle of motion, Fig. 2.7. If the earthquake excitation has a frequency close to the one of the building, then resonance may occur. This should be avoided.
33
Figure 2.7: Vibrations of a Building Earthquake load manifests itself as a horizontal force due to the (primarily) horizontal inertia force (F = ma). 35 The horizontal force at each level is calculated as a portion of the base shear force V
34
V = ZIC RW W
(2.10)
where:
Z : Zone Factor: to be determined from Fig. 2.8 and Table 2.10. Seismic Zone 0 1 2A 2B 3 4 Z 0 0.075 0.15 0.2 0.3 0.4 Table 2.10: Z Factors for Di erent Seismic Zones, ubc
I : Importance Factor: which was given by Table 2.8. C : Design Response Spectrum given by
Victor Saouma
S 2:75 C = 1T:25 2 =3
(2.11)
Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{13
Figure 2.8: Seismic Zones of the United States, (UBC 1995) is the fundamental period of vibration of the building in seconds. This can be determined from either the free vibration analysis of the building, or estimated from the following empirical formula T
T
(2.12)
= t ( n )3=4 C
h
where: n is the building height above base in ft. h
and
0.035 steel moment resisting frames 0.030 reinforced concrete moment resisting frames and eccentrically braced frames 0.020 all other buildings t : Site Coecient given by Table 2.11 Note that most of the damages in the 1990? earthquake
Ct
Ct C S
Type Description A soil pro le with either rock-like material or sti /dense soil less 1 than 200 ft. Dense or sti soil exceeding 200 ft 2 70 ft or more soil containing more than 20 ft of soft to medium sti 3 clay but not more than 40 ft. of soft clay. Soil containing more than 40 ft of soft clay 4
S
S
S S
S
Table 2.11:
S
Factor 1.0 1.2 1.5 2.0
Site Coecients for Earthquake Loading, (UBC 1995)
in San Francisco occurred in the marina where many houses were built on soft soil. and C RW
Victor Saouma
0 075 :
(2.13)
Structural Concepts and Systems for Architects
Draft 2{14
LOADS
is given by Table 2.12. Load total structure load.
RW W
The horizontal force is distributed over the height of the building in two parts. The rst (applied only if 0 7 sec.) is a concentrated force 1 equal to
36
V
T
:
F
Ft
= 0 07 :
TV
0 25 :
(2.14)
V
is applied at the top of the building due to whiplash. The balance of the force ; t is distributed as a triangular load diminishing to zero at the base. 37 Assuming a oor weight constant for every oor level, then the force acting on each one is given by ( ; t) x = ( ; t) x (2.15) x= ni=1 i 1 + 2 ++ n where i and x are the height in ft above the base to level , or respectively. Note that it is assumed that all oors have also same width. V
V
F
h
h
F
V
h
h
F
F
h
h
h
h
i
x
Example 2-3: Earthquake Load on a Frame Determine the approximate earthquake forces for the ductile hospital frame structure shown below. The DL for each oor is 200 lb/ft and the LL is 400 lb/ft. The structure is built on soft soil. Use DL plus 50%LL as the weight of each oor. The building is in zone 3.
Solution: 1. The fundamental period of vibration is T
2. The
C
= t ( n )3=4 = (0 030)(24)3=4 = 0 32 C
coecient is C
use = 2 75. 3. The other coecients are: C
h
:
:
sec.
= (1 25)(22=0) = 5 344 2 75 = 1 25 2=3 (0 32) 3 :
:
S
T
:
:
:
>
:
(2.16) (2.17)
:
Victor Saouma
Z
=0.3; =1.25; I
RW
=12
Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{15
Structural System
Bearing wall system
Light-framed walls with shear panels Plywood walls for structures three stories or less All other light-framed walls Shear walls Concrete Masonry
Building frame system
using trussing or shear walls)
Steel eccentrically braced ductile frame Light-framed walls with shear panels Plywood walls for structures three stories or less All other light-framed walls Shear walls Concrete Masonry Concentrically braced frames Steel Concrete (only for zones I and 2) Heavy timber
Moment-resisting frame system
RW H (ft) 8 6
65 65
8 8
240 160
10
240
9 7
65 65
8 8
240 160
8 8 8
160 65
Special moment-resisting frames (SMRF) Steel 12 Concrete 12 Concrete intermediate moment-resisting frames (IMRF)only for zones 1 and 2 8 Ordinary moment-resisting frames (OMRF) Steel 6 Concrete (only for zone 1) 5 Dual systems (selected cases are for ductile rigid frames only) Shear walls Concrete with SMRF 12 Masonry with SMRF 8 Steel eccentrically braced ductile frame 6-12 Concentrically braced frame 12 Steel with steel SMRF 10 Steel with steel OMRF 6 Concrete with concrete SMRF (only for zones 1 and 2) 9
N.L. N.L. 160 N.L. 160 160-N.L. N. L. N.L. 160 -
Table 2.12: Partial List of RW for Various Structure Systems, (UBC 1995)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2{16
LOADS
4. Check
(2.18)
= 2 ((200 + 0 5(400)) (20) = 16000 lbs
(2.19)
25)(2 75) = 0 086 = (0 3)(1 12
(2.20-a)
R
5. The total vertical load is W
p
= 21275 = 0 23 0 075
C
W
:
:
>
:
:
6. The total seismic base shear is V
=
ZIC R
W
:
:
:
:
W
= (0 086)(16000) = 1375 lbs :
(2.20-b)
7. Since 0 7 sec. there is no whiplash. 8. The load on each oor is thus given by T
07
= 1 25 = (1 25)(12=5) = 1 12 2 75 2 =3 (2 16) 3 4. The other coecients are =0.4; =1, W =12 5. We check p = 11212 = 0 093 0 075 C
Z
:
S
:
T
:
:
:
I
:
:
sec.
p
p
:
C
:
(2.25)
:
ZIC R
:
W
:
W
:
(2.26-a)
W
(2.26-b)
:
T >
07 :
sec.
, the whiplash e ect must be considered Ft
= 0 07 = (0 07)(2 16)(3108) = 470 0 25 = (0 25)(3108) = 777
le
Victor Saouma
(2.23) (2.24)
6. The total seismic base shear along the critical short direction is 12) = 0 037 = = (0 4)(1)(1 (12) W = (0 037)(84000) = 3108 kip 7. Since
(2.22)
R
RW
V
k
:
TV
:
V
:
:
:
k
k
(2.27-a) (2.27-b)
Structural Concepts and Systems for Architects
Draft 2{18
LOADS
Hence the total triangular load is V
; Ft = 3108 ; 470 = 2638
k
(2.28)
8. let us check if wind load governs. From Table xx we conservatively assume a uniform wind pressure of 29 psf resulting in a total lateral force of PW = (0:029) psf(175 300) ft2 = 1523 k < 3108 k (2.29) The magnitude of the total seismic load is clearly larger than the total wind force.
2.4 2.4.1 38
Other Loads Hydrostatic and Earth
Structures below ground must resist lateral earth pressure. q
= K h
(2.30)
1;sin is the pressure coecient, h is the height. where is the soil density, K = 1+sin 39 40
For sand and gravel = 120 lb= ft3 , and 30. If the structure is partially submerged, it must also resist hydrostatic pressure of water, Fig. 2.9.
Figure 2.9: Earth and Hydrostatic Loads on Structures q
= W h
(2.31)
where W = 62:4 lbs=ft3 .
Example 2-5: Hydrostatic Load Victor Saouma
Structural Concepts and Systems for Architects
Draft 2.4 Other Loads
2{19
The basement of a building is 12 ft below grade. Ground water is located 9 ft below grade, what thickness concrete slab is required to exactly balance the hydrostatic uplift?
Solution:
The hydrostatic pressure must be countered by the pressure caused by the weight of concrete. Since p =
h we equate the two pressures and solve for h the height of the concrete slab (62:4) lbs=ft3 (12 ; 9) ft = |
:4) lbs=ft (3) ft(12) in/ft = 14:976 in ' 15.0 inch (150) lbs{z=ft3 h} ) h = (62 3 (150) lbs=ft | concrete 3
{z
water
}
2.4.2 Thermal If a member is uniformly heated (or cooled) without restraint, then it will expand (or contract). This expansion is given by
41
l = lT
(2.32)
where is the coecient of thermal expansion, Table 2.13 (/F ) Steel 6:5 10;6 Concrete 5:5 10;6
Table 2.13: Coecients of Thermal Expansion 42 43
If the member is restrained against expansion, then a compressive stress = E T is developed. To avoid excessive stresses due to thermal loading expansion joints are used in bridges and buildings.
Example 2-6: Thermal Expansion/Stress (Schueller 1996) A low-rise building is enclosed along one side by a 100 ft-long clay masonary ( = 3:6 10;6 in./in./oF, E = 2; 400; 000 psi) bearing wall. The structure was built at a temperature of 60oF and is located in the northern part of the United States where the temperature range is between -20o and +120oF.
Solution:
1. Assuming that the wall can move freely with no restraint from cross-walls and foundation, the wall expansion and contraction (summer and winter) are given by LSummer = T L = (3:6 10;6) in= in=oF (120 ; 60)o F (100) ft(12) in/ft = 0.26 (2.33-a) in in LWinter = T L = (3:6 10;6) in= in=oF (;20 ; 60)oF (100) ft(12) in/ft = -0.35(2.33-b)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2{20
LOADS
2. We now assume (conservatively) that the free movement cannot occur (L = 0) hence the resulting stress would be equal to = E" = E LL = E LTL = E T Summer = E T = (2; 400; 000) 2 (3:6 10;6 ) in= in=oF (120 ; 60)oF = 518 2 Tension(2.34-a) in in lbs ; 6 o o W inter (2.34-b) = E T = (2; 400; 000) 2 (3:6 10 ) in= in= F (;20 ; 60) F = -691 lbs2 Compression in in lbs
lbs
(2.34-c)
Note that the tensile stresses being beyond the masonary capacity, cracking will occur.
2.5 Other Important Considerations 2.5.1 Load Combinations
Live loads speci ed by codes represent the maximum possible loads. 45 The likelihood of all these loads occurring simultaneously is remote. Hence, building codes allow certain reduction when certain loads are combined together. 46 Furthermore, structures should be designed to resist a combination of loads. 47 Denoting D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature; H= soil: 48 For the load and resistance factor design (LRFD) method of concrete structures, the American Concrete Institute (ACI) Building design code (318) (318 n.d.) requires that the following load combinations be considered: 1. 1.4D+1.7L 2. 0.75(1.4D+1.7L+1.7W) 3. 0.9D+1.3W 4. 1.4D +1.7L+1.7H 5. 0.75(1.4D+1.4T+1.7L) 6. 1.4(D+T) whereas for steel structures, the American Institute of Steel Construction (AISC) code, (of Steel COnstruction 1986) requires that the following combinations be veri ed 1. 1.4D 2. 1.2D+1.6L+0.5(Lr or S) 3. 1.2D+0.5L (or 0.8W)+1.6(Lr or S) 4. 1.2D+0.5L+0.5(Lr or S)+1.3W 5. 1.2D+0.5L(or 0.2 S)+1.5E 44
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2.5 Other Important Considerations
2{21
6. 0.9D+1.3W(or 1.5 E) Analysis can be separately performed for each of the basic loads (L, D, W, etc) and then using the principle of superposition the loads can be linearly combined (unless the elastic limit has been reached). 50 Loads are often characterized as Usual, Unusual and Extreme.
49
2.5.2 Load Placement Only the dead load is static. The live load on the other hand may or may not be applied on a given component of a structure. Hence, the load placement arrangement resulting in the highest internal forces (moment +ve or -ve, shear) at di erent locations must be considered, Fig. 2.10.
51
Figure 2.10: Load Placement to Maximize Moments
2.5.3 Load Transfer Whereas we will be focusing on the design of a reinforced concrete or steel section, we must keep in mind the following: 1. The section is part of a beam or girder. 2. The beam or girder is really part of a three dimensional structure in which load is transmitted from any point in the structure to the foundation through any one of various structural forms.
52
Load transfer in a structure is accomplished through a \hierarchy" of simple exural elements which are then connected to the columns, Fig. 2.11 or by two way slabs as illustrated in Fig. 2.12.
53
2.5.4 Structural Response 54 Under the action of the various forces and loadings described above, the structure must be able to respond with proper behavior, Fig. 9.1.
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Structural Concepts and Systems for Architects
Draft 2{22
LOADS
Figure 2.11: Load Transfer in R/C Buildings
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2.5 Other Important Considerations
2{23
Figure 2.12: Two Way Actions
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2{24
LOADS
Figure 2.13: Load Life of a Structure, (Lin and Stotesbury 1981)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2.5 Other Important Considerations
2{25
2.5.5 Tributary Areas For preliminary analyses, the tributary area of a structural component will determine the total applied load.
55
0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111
00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111
0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111
0000011111 00000 11111 0000011111 00000 11111 00000 00000 11111 0000011111 00000 11111 0000011111 00000 11111 11111 00000 00000 11111 0000011111 00000 11111 0000011111 00000 11111 11111
0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111
0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111
001111 11 0000 00 11 11 00 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 00 001111 11 000011 1111 00 11 00 11 0000 1111 00 11
Figure 2.14: Concept of Tributary Areas for Structual Member Loading
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Structural Concepts and Systems for Architects
Draft 2{26
LOADS
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 3
STRUCTURAL MATERIALS Proper understanding of structural materials is essential to both structural analysis and to structural design. 2 Characteristics of the most commonly used structural materials will be highlighted.
1
3.1 3.1.1
Steel Structural Steel
3 Steel is an alloy of iron and carbon. Its properties can be greatly varied by altering the carbon content (always less than 0.5%) or by adding other elements such as silicon, nickle, manganese and copper. 4 Practically all grades of steel have a Young Modulus equal to 29,000 ksi, density of 490 lb/cu ft, and a coecient of thermal expansion equal to 0:65 10;5 /deg F. 5 The yield stress of steel can vary from 40 ksi to 250 ksi. Most commonly used structural steel are A36 (yld = 36 ksi) and A572 (yld = 50 ksi), Fig. 3.1 6 Structural steel can be rolled into a wide variety of shapes and sizes. Usually the most desirable members are those which have a large section moduli (S ) in proportion to their area (A), Fig. 3.2. 7 Steel can be bolted, riveted or welded. 8 Sections are designated by the shape of their cross section, their depth and their weight. For example W 27 114 is a W section, 27 in. deep weighing 114 lb/ft. 9 Common sections are: S sections were the rst ones rolled in America and have a slope on their inside ange surfaces of 1 to 6. W or wide ange sections have a much smaller inner slope which facilitates connections and rivetting. W sections constitute about 50% of the tonnage of rolled structural steel. C are channel sections MC Miscellaneous channel which can not be classi ed as a C shape by dimensions. HP is a bearing pile section.
Draft 3{2
STRUCTURAL MATERIALS
Figure 3.1: Stress Strain Curves of Concrete and Steel
Figure 3.2: Standard Rolled Sections
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.1 Steel
3{3
M is a miscellaneous section. L are angle sections which may have equal or unequal sides. WT is a T section cut from a W section in two. 10
The section modulus Sx of a W section can be roughly approximated by the following formula Sx
wd=10 or
Ix
Sx d2 wd2 =20
(3.1)
and the plastic modulus can be approximated by Zx
11
wd=9
(3.2)
Properties of structural steel are tabulated in Table 3.1.
ASTM Desig. A36
A500 A501 A529 A606 A611 A 709
Shapes Available Shapes and bars
Use
Riveted, bolted, welded; Buildings and bridges Cold formed welded and General structural purseamless sections; pose Riveted, welded or bolted; Hot formed welded and seam- Bolted and welded less sections; Plates and bars 12 in and less Building frames and thick; trusses; Bolted and welded Hot and cold rolled sheets; Atmospheric corrosion resistant Cold rolled sheet in cut Cold formed sections lengths Structural shapes, plates and Bridges bars
y
(kksi)
u
(kksi)
36 up through 8 in. (32 above 8.) Grade A: 33; Grade B: 42; Grade C: 46 36 42 45-50 Grade C 33; Grade D 40; Grade E 80 Grade 36: 36 (to 4 in.); Grade 50: 50; Grade 100: 100 (to 2.5in.) and 90 (over 2.5 to 4 in.)
Table 3.1: Properties of Major Structural Steels Rolled sections, Fig. 3.3 and welded ones, Fig3.4 have residual stresses. Those originate during the rolling or fabrication of a member. The member is hot just after rolling or welding, it cools unevenly because of varying exposure. The area that cool rst become sti er, resist contraction, and develop compressive stresses. The remaining regions continue to cool and contract in the plastic condition and develop tensile stresses. 13 Due to those residual stresses, the stress-strain curve of a rolled section exhibits a non-linear segment prior to the theoretical yielding, Fig. 3.5. This would have important implications on the exural and axial strength of beams and columns.
12
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{4
STRUCTURAL MATERIALS
Maximum compressive stress, say 12 ksi average
Compression (-) (-)
Tension (+) (+)
Figure 3.3: Residual Stresses in Rolled Sections
say 20 ksi say 12 ksi
_
+
+ +
+
say 40 ksi
20 ksi
-
say 35 ksi tension
+
-
Welded H say 20 ksi compression
Welded box
Figure 3.4: Residual Stresses in Welded Sections
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.1 Steel
.3
Fy Average stress P/A
3{5
Ideal coupon containing no residual stress
.2 Fp
Maximum residual compressive stress
.1
1
Members with residual stress
2 Average copressive strain
3
Shaded portion indicates area which has achieved a stress Fy
Figure 3.5: In uence of Residual Stress on Average Stress-Strain Curve of a Rolled Section 3.1.2
Reinforcing Steel
14 Steel is also used as reinforcing bars in concrete, Table 3.2. Those bars have a deformation on their surface to increase the bond with concrete, and usually have a yield stress of 60 ksi1 .
Bar Designation No. 2 No. 3 No. 4 No. 5 No. 6 No. 7 No. 8 No. 9 No. 10 No. 11 No. 14 No. 18
Diameter Area Perimeter Weight (in.) ( 2) in lb/ft 2/8=0.250 0.05 0.79 0.167 3/8=0.375 0.11 1.18 0.376 4/8=0.500 0.20 1.57 0.668 5/8=0.625 0.31 1.96 1.043 6/8=0.750 0.44 2.36 1.5202 7/8=0.875 0.60 2.75 2.044 8/8=1.000 0.79 3.14 2.670 9/8=1.128 1.00 3.54 3.400 10/8=1.270 1.27 3.99 4.303 11/8=1.410 1.56 4.43 5.313 14/8=1.693 2.25 5.32 7.650 18/8=2.257 4.00 7.09 13.60 in
Table 3.2: Properties of Reinforcing Bars Steel loses its strength rapidly above 700 deg. F (and thus must be properly protected from re), and becomes brittle at ;30 deg. F 16 Steel is also used as wire strands and ropes for suspended roofs, cable-stayed bridges, fabric roofs and other structural applications. A strand is a helical arrangement of wires around a central wire. A rope consists of multiple strands helically wound around a central plastic core, and a modulus of elasticity of 20,000 ksi, and an ultimate strength of 220 ksi. 17 Prestressing Steel cables have an ultimate strength up to 270 ksi.
15
1
Stirrups which are used as vertical reinforcement to resist shear usually have a yield stress of only 40 ksi.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{6
STRUCTURAL MATERIALS
3.2 Aluminum Aluminum is used whenever light weight combined with strength is an important factor. Those properties, along with its resistance to corrosion have made it the material of choice for airplane structures, light roof framing. 19 Aluminum members can be connected by riveting, bolting and to a lesser extent by welding. 20 Aluminum has a modulus of elasticity equal to 10,000 ksi (about three times lower than steel), a coecient of thermal expansion of 2:4 10;5 and a density of 173 lbs=ft3 . 21 The ultimate strength of pure aluminum is low (13,000 psi) but with the addition of alloys it can go up. 22 When aluminum is in contact with other metals in the presence of an electrolyte, galvanic corrosion may cause damage. Thus, steel and aluminum in a structure must be carefully separated by means of painting or a nonconductive material.
18
3.3 Concrete Concrete is a mixture of Portland cement2 , water, and aggregates (usually sand and crushed stone). An ideal mixture is one in which: 1. A minimum amount of cement-water paste is used to ll the interstices between the particles of aggregates. 2. A minimum amount of water is provided to complete the chemical reaction with cement. In such a mixture, about 3/4 of the volume is constituted by the aggregates, and the remaining 1/4 being the cement paste. 24 Smaller particles up to 1/4 in. in size are called ne aggregates, and the larger ones being coarse aggregates. 25 Contrarily to steel to modulus of elasticity of concrete depends on the strength and is given by 23
E
or
pf 0
= 57; 000
p
c
(3.3)
= 33 1 5 f 0 (3.4) 3 0 where both f and E are in psi and is in lbs=ft . 26 Typical concrete (compressive) strengths range from 3,000 to 6,000 psi; However high strength concrete can go up to 14,000 psi. 27 All concrete fail at an ultimate strain of 0.003, Fig. 3.1. 28 Pre-peak nonlinearity is caused by micro-cracking Fig. 3.6. 0 29 The tensile strength of concrete f is about 10% of the compressive strength. 3 3 30 Density of normal weight concrete is 145 lbs=ft and 100 lbs=ft for lightweight concrete. E
:
c
c
t
2 Portland cement is a mixture of calcareous and argillaceous materials which are calcined in a kiln and then pulverized. When mixed with water, cement hardens through a process called hydration.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.4 Masonry
3{7
εu linear .5f’c non-linear f’c
Figure 3.6: Concrete microcracking Coecient of thermal expansion is 0 65 10;5 /deg F for normal weight concrete. 32 When concrete is poured (or rather placed), the free water not needed for the hydration process evaporates over a period of time and the concrete will shrink. This shrinkage is about 0.05% after one year (strain). Thus if the concrete is restrained, then cracking will occur3. 33 Concrete will also deform with time due to the applied load, this is called creep. This should be taken into consideration when computing the de ections (which can be up to three times the instantaneous elastic de ection). 31
:
3.4 Masonry Masonry consists of either natural materials, such as stones, or of manufactured products such as bricks and concrete blocks4, stacked and bonded together with mortar. 35 As for concrete, all modern structural masonry blocks are essentially compression members with low tensile resistance. 36 The mortar used is a mixture of sand, masonry cement, and either Portland cement or hydrated lime. 34
3.5 Timber Timber is one of the earliest construction materials, and one of the few natural materials with good tensile properties. 38 The properties of timber vary greatly, and the strength is time dependent. 39 Timber is a good shock absorber (many wood structures in Japan have resisted repeated earthquakes). 40 The most commonly used species of timber in construction are Douglas r, southern pine, hemlock and larch. 41 Members can be laminated together under good quality control, and exural strengths as high as 2,500 psi can be achieved. 37
3 For this reason a minimum amount of reinforcement is always necessary in concrete, and a 2% reinforcement, can reduce the shrinkage by 75%. 4 Mud bricks were used by the Babylonians, stones by the Egyptians, and ice blocks by the Eskimos...
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{8 3.6 42
STRUCTURAL MATERIALS
Steel Section Properties
Dimensions and properties of rolled sections are tabulated in the following pages, Fig. 3.7.
Figure 3.7: W and C sections ==============
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.6 Steel Section Properties Designation
A in2
d
3{9 bf
2tf
hc tw
in
Ix in4
Sx in3
Iy in4
Sy in3
Zx in3
Zy in3
W 36x848
249.0
42.45
2.0
12.5
67400
3170
4550
501
3830.0
799.0
W 36x798
234.0
41.97
2.1
13.2
62600
2980
4200
467
3570.0
743.0
W 36x720
211.0
41.19
2.3
14.5
55300
2690
3680
414
3190.0
656.0
W 36x650
190.0
40.47
2.5
16.0
48900
2420
3230
367
2840.0
580.0
W 36x588
172.0
39.84
2.7
17.6
43500
2180
2850
328
2550.0
517.0
W 36x527
154.0
39.21
3.0
19.6
38300
1950
2490
289
2270.0
454.0
W 36x485
142.0
38.74
3.2
21.0
34700
1790
2250
263
2070.0
412.0
W 36x439
128.0
38.26
3.5
23.1
31000
1620
1990
235
1860.0
367.0
W 36x393
115.0
37.80
3.8
25.8
27500
1450
1750
208
1660.0
325.0
W 36x359
105.0
37.40
4.2
28.1
24800
1320
1570
188
1510.0
292.0
W 36x328
96.4
37.09
4.5
30.9
22500
1210
1420
171
1380.0
265.0
W 36x300
88.3
36.74
5.0
33.3
20300
1110
1300
156
1260.0
241.0
W 36x280
82.4
36.52
5.3
35.6
18900
1030
1200
144
1170.0
223.0
W 36x260
76.5
36.26
5.7
37.5
17300
953
1090
132
1080.0
204.0
W 36x245
72.1
36.08
6.1
39.4
16100
895
1010
123
1010.0
190.0
W 36x230
67.6
35.90
6.5
41.4
15000
837
940
114
943.0
176.0
W 36x256
75.4
37.43
3.5
33.8
16800
895
528
86
1040.0
137.0
W 36x232
68.1
37.12
3.9
37.3
15000
809
468
77
936.0
122.0
W 36x210
61.8
36.69
4.5
39.1
13200
719
411
68
833.0
107.0
W 36x194
57.0
36.49
4.8
42.4
12100
664
375
62
767.0
97.7
W 36x182
53.6
36.33
5.1
44.8
11300
623
347
58
718.0
90.7
W 36x170
50.0
36.17
5.5
47.8
10500
580
320
53
668.0
83.8
W 36x160
47.0
36.01
5.9
50.0
9750
542
295
49
624.0
77.3
W 36x150
44.2
35.85
6.4
52.0
9040
504
270
45
581.0
70.9
W 36x135
39.7
35.55
7.6
54.1
7800
439
225
38
509.0
59.7
W 33x619
181.0
38.47
2.4
15.2
41800
2170
2870
340
2560.0
537.0
W 33x567
166.0
37.91
2.6
16.6
37700
1990
2580
308
2330.0
485.0
W 33x515
151.0
37.36
2.8
18.2
33700
1810
2290
276
2110.0
433.0
W 33x468
137.0
36.81
3.0
19.7
30100
1630
2030
247
1890.0
387.0
W 33x424
124.0
36.34
3.3
21.7
26900
1480
1800
221
1700.0
345.0
W 33x387
113.0
35.95
3.6
23.8
24300
1350
1620
200
1550.0
312.0
W 33x354
104.0
35.55
3.8
25.8
21900
1230
1460
181
1420.0
282.0
W 33x318
93.5
35.16
4.2
28.8
19500
1110
1290
161
1270.0
250.0
W 33x291
85.6
34.84
4.6
31.2
17700
1010
1160
146
1150.0
226.0
W 33x263
77.4
34.53
5.0
34.5
15800
917
1030
131
1040.0
202.0
W 33x241
70.9
34.18
5.7
36.1
14200
829
932
118
939.0
182.0
W 33x221
65.0
33.93
6.2
38.7
12800
757
840
106
855.0
164.0
W 33x201
59.1
33.68
6.8
41.9
11500
684
749
95
772.0
147.0
W 33x169
49.5
33.82
4.7
44.7
9290
549
310
54
629.0
84.4
W 33x152
44.7
33.49
5.5
47.2
8160
487
273
47
559.0
73.9
W 33x141
41.6
33.30
6.0
49.6
7450
448
246
43
514.0
66.9
W 33x130
38.3
33.09
6.7
51.7
6710
406
218
38
467.0
59.5
W 33x118
34.7
32.86
7.8
54.5
5900
359
187
33
415.0
51.3
W 30x581
170.0
35.39
2.3
13.7
33000
1870
2530
312
2210.0
492.0
W 30x526
154.0
34.76
2.5
15.1
29300
1680
2230
278
1990.0
438.0
W 30x477
140.0
34.21
2.7
16.6
26100
1530
1970
249
1790.0
390.0
W 30x433
127.0
33.66
2.9
18.0
23200
1380
1750
222
1610.0
348.0
W 30x391
114.0
33.19
3.2
19.9
20700
1250
1550
198
1430.0
310.0
W 30x357
104.0
32.80
3.5
21.8
18600
1140
1390
179
1300.0
279.0
W 30x326
95.7
32.40
3.7
23.7
16800
1030
1240
162
1190.0
252.0
W 30x292
85.7
32.01
4.1
26.5
14900
928
1100
144
1060.0
223.0
W 30x261
76.7
31.61
4.6
29.0
13100
827
959
127
941.0
196.0
W 30x235
69.0
31.30
5.0
32.5
11700
746
855
114
845.0
175.0
W 30x211
62.0
30.94
5.7
34.9
10300
663
757
100
749.0
154.0
W 30x191
56.1
30.68
6.3
38.0
9170
598
673
90
673.0
138.0
W 30x173
50.8
30.44
7.0
41.2
8200
539
598
80
605.0
123.0
W 30x148
43.5
30.67
4.4
41.5
6680
436
227
43
500.0
68.0
W 30x132
38.9
30.31
5.3
43.9
5770
380
196
37
437.0
58.4
W 30x124
36.5
30.17
5.7
46.2
5360
355
181
34
408.0
54.0
W 30x116
34.2
30.01
6.2
47.8
4930
329
164
31
378.0
49.2
W 30x108
31.7
29.83
6.9
49.6
4470
299
146
28
346.0
43.9
W 30x 99
29.1
29.65
7.8
51.9
3990
269
128
24
312.0
38.6
W 30x 90
26.4
29.53
8.5
57.5
3620
245
115
22
283.0
34.7
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{10 Designation
STRUCTURAL MATERIALS A in2
d
bf
2tf
hc tw
in
Ix in4
Sx in3
Iy in4
Sy in3
Zx in3
Zy in3
W 27x539
158.0
32.52
2.2
12.3
25500
1570
2110
277
1880.0
437.0
W 27x494
145.0
31.97
2.3
13.4
22900
1440
1890
250
1710.0
394.0
W 27x448
131.0
31.42
2.5
14.7
20400
1300
1670
224
1530.0
351.0
W 27x407
119.0
30.87
2.7
15.9
18100
1170
1480
200
1380.0
313.0
W 27x368
108.0
30.39
3.0
17.6
16100
1060
1310
179
1240.0
279.0
W 27x336
98.7
30.00
3.2
19.2
14500
970
1170
161
1130.0
252.0
W 27x307
90.2
29.61
3.5
20.9
13100
884
1050
146
1020.0
227.0
W 27x281
82.6
29.29
3.7
22.9
11900
811
953
133
933.0
206.0
W 27x258
75.7
28.98
4.0
24.7
10800
742
859
120
850.0
187.0
W 27x235
69.1
28.66
4.4
26.6
9660
674
768
108
769.0
168.0
W 27x217
63.8
28.43
4.7
29.2
8870
624
704
100
708.0
154.0
W 27x194
57.0
28.11
5.2
32.3
7820
556
618
88
628.0
136.0
W 27x178
52.3
27.81
5.9
33.4
6990
502
555
79
567.0
122.0
W 27x161
47.4
27.59
6.5
36.7
6280
455
497
71
512.0
109.0
W 27x146
42.9
27.38
7.2
40.0
5630
411
443
64
461.0
97.5
W 27x129
37.8
27.63
4.5
39.7
4760
345
184
37
395.0
57.6
W 27x114
33.5
27.29
5.4
42.5
4090
299
159
32
343.0
49.3
W 27x102
30.0
27.09
6.0
47.0
3620
267
139
28
305.0
43.4
W 27x 94
27.7
26.92
6.7
49.4
3270
243
124
25
278.0
38.8
W 27x 84
24.8
26.71
7.8
52.7
2850
213
106
21
244.0
33.2
W 24x492
144.0
29.65
2.0
10.9
19100
1290
1670
237
1550.0
375.0
W 24x450
132.0
29.09
2.1
11.9
17100
1170
1490
214
1410.0
337.0
W 24x408
119.0
28.54
2.3
13.1
15100
1060
1320
191
1250.0
300.0
W 24x370
108.0
27.99
2.5
14.2
13400
957
1160
170
1120.0
267.0
W 24x335
98.4
27.52
2.7
15.6
11900
864
1030
152
1020.0
238.0
W 24x306
89.8
27.13
2.9
17.1
10700
789
919
137
922.0
214.0
W 24x279
82.0
26.73
3.2
18.6
9600
718
823
124
835.0
193.0
W 24x250
73.5
26.34
3.5
20.7
8490
644
724
110
744.0
171.0
W 24x229
67.2
26.02
3.8
22.5
7650
588
651
99
676.0
154.0
W 24x207
60.7
25.71
4.1
24.8
6820
531
578
89
606.0
137.0
W 24x192
56.3
25.47
4.4
26.6
6260
491
530
82
559.0
126.0
W 24x176
51.7
25.24
4.8
28.7
5680
450
479
74
511.0
115.0
W 24x162
47.7
25.00
5.3
30.6
5170
414
443
68
468.0
105.0
W 24x146
43.0
24.74
5.9
33.2
4580
371
391
60
418.0
93.2
W 24x131
38.5
24.48
6.7
35.6
4020
329
340
53
370.0
81.5
W 24x117
34.4
24.26
7.5
39.2
3540
291
297
46
327.0
71.4
W 24x104
30.6
24.06
8.5
43.1
3100
258
259
41
289.0
62.4
W 24x103
30.3
24.53
4.6
39.2
3000
245
119
26
280.0
41.5
W 24x 94
27.7
24.31
5.2
41.9
2700
222
109
24
254.0
37.5
W 24x 84
24.7
24.10
5.9
45.9
2370
196
94
21
224.0
32.6
W 24x 76
22.4
23.92
6.6
49.0
2100
176
82
18
200.0
28.6
W 24x 68
20.1
23.73
7.7
52.0
1830
154
70
16
177.0
24.5
W 24x 62
18.2
23.74
6.0
50.1
1550
131
34
10
153.0
15.7
W 24x 55
16.2
23.57
6.9
54.6
1350
114
29
8
134.0
13.3
W 21x402
118.0
26.02
2.1
10.8
12200
937
1270
189
1130.0
296.0
W 21x364
107.0
25.47
2.3
11.8
10800
846
1120
168
1010.0
263.0
W 21x333
97.9
25.00
2.5
12.8
9610
769
994
151
915.0
237.0
W 21x300
88.2
24.53
2.7
14.2
8480
692
873
134
816.0
210.0
W 21x275
80.8
24.13
2.9
15.4
7620
632
785
122
741.0
189.0
W 21x248
72.8
23.74
3.2
17.1
6760
569
694
109
663.0
169.0
W 21x223
65.4
23.35
3.5
18.8
5950
510
609
96
589.0
149.0
W 21x201
59.2
23.03
3.9
20.6
5310
461
542
86
530.0
133.0
W 21x182
53.6
22.72
4.2
22.6
4730
417
483
77
476.0
119.0
W 21x166
48.8
22.48
4.6
24.9
4280
380
435
70
432.0
108.0
W 21x147
43.2
22.06
5.4
26.1
3630
329
376
60
373.0
92.6
W 21x132
38.8
21.83
6.0
28.9
3220
295
333
54
333.0
82.3
W 21x122
35.9
21.68
6.5
31.3
2960
273
305
49
307.0
75.6
W 21x111
32.7
21.51
7.1
34.1
2670
249
274
44
279.0
68.2
W 21x101
29.8
21.36
7.7
37.5
2420
227
248
40
253.0
61.7
W 21x 93
27.3
21.62
4.5
32.3
2070
192
93
22
221.0
34.7
W 21x 83
24.3
21.43
5.0
36.4
1830
171
81
20
196.0
30.5
W 21x 73
21.5
21.24
5.6
41.2
1600
151
71
17
172.0
26.6
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.6 Steel Section Properties Designation
A in2
d
3{11 bf
2tf
hc tw
in
Ix in4
Sx in3
Iy in4
Sy in3
Zx in3
Zy in3
W 21x 68
20.0
21.13
6.0
43.6
1480
140
65
16
160.0
24.4
W 21x 62
18.3
20.99
6.7
46.9
1330
127
58
14
144.0
21.7
W 21x 57
16.7
21.06
5.0
46.3
1170
111
31
9
129.0
14.8
W 21x 50
14.7
20.83
6.1
49.4
984
94
25
8
110.0
12.2
W 21x 44
13.0
20.66
7.2
53.6
843
82
21
6
95.4
10.2
W 18x311
91.5
22.32
2.2
10.6
6960
624
795
132
753.0
207.0
W 18x283
83.2
21.85
2.4
11.5
6160
564
704
118
676.0
185.0
W 18x258
75.9
21.46
2.6
12.5
5510
514
628
107
611.0
166.0
W 18x234
68.8
21.06
2.8
13.8
4900
466
558
96
549.0
149.0
W 18x211
62.1
20.67
3.0
15.1
4330
419
493
85
490.0
132.0
W 18x192
56.4
20.35
3.3
16.7
3870
380
440
77
442.0
119.0
W 18x175
51.3
20.04
3.6
18.0
3450
344
391
69
398.0
106.0
W 18x158
46.3
19.72
3.9
19.8
3060
310
347
61
356.0
94.8
W 18x143
42.1
19.49
4.2
21.9
2750
282
311
56
322.0
85.4
W 18x130
38.2
19.25
4.6
23.9
2460
256
278
50
291.0
76.7
W 18x119
35.1
18.97
5.3
24.5
2190
231
253
45
261.0
69.1
W 18x106
31.1
18.73
6.0
27.2
1910
204
220
39
230.0
60.5
W 18x 97
28.5
18.59
6.4
30.0
1750
188
201
36
211.0
55.3
W 18x 86
25.3
18.39
7.2
33.4
1530
166
175
32
186.0
48.4
W 18x 76
22.3
18.21
8.1
37.8
1330
146
152
28
163.0
42.2
W 18x 71
20.8
18.47
4.7
32.4
1170
127
60
16
145.0
24.7
W 18x 65
19.1
18.35
5.1
35.7
1070
117
55
14
133.0
22.5
W 18x 60
17.6
18.24
5.4
38.7
984
108
50
13
123.0
20.6
W 18x 55
16.2
18.11
6.0
41.2
890
98
45
12
112.0
18.5
W 18x 50
14.7
17.99
6.6
45.2
800
89
40
11
101.0
16.6
W 18x 46
13.5
18.06
5.0
44.6
712
79
22
7
90.7
11.7
W 18x 40
11.8
17.90
5.7
51.0
612
68
19
6
78.4
9.9
W 18x 35
10.3
17.70
7.1
53.5
510
58
15
5
66.5
8.1
W 16x100
29.4
16.97
5.3
24.3
1490
175
186
36
198.0
54.9
W 16x 89
26.2
16.75
5.9
27.0
1300
155
163
31
175.0
48.1
W 16x 77
22.6
16.52
6.8
31.2
1110
134
138
27
150.0
41.1
W 16x 67
19.7
16.33
7.7
35.9
954
117
119
23
130.0
35.5
W 16x 57
16.8
16.43
5.0
33.0
758
92
43
12
105.0
18.9
W 16x 50
14.7
16.26
5.6
37.4
659
81
37
10
92.0
16.3
W 16x 45
13.3
16.13
6.2
41.2
586
73
33
9
82.3
14.5
W 16x 40
11.8
16.01
6.9
46.6
518
65
29
8
72.9
12.7
W 16x 36
10.6
15.86
8.1
48.1
448
56
24
7
64.0
10.8
W 16x 31
9.1
15.88
6.3
51.6
375
47
12
4
54.0
7.0
W 16x 26
7.7
15.69
8.0
56.8
301
38
10
3
44.2
5.5
W 14x730
215.0
22.42
1.8
3.7
14300
1280
4720
527
1660.0
816.0
W 14x665
196.0
21.64
2.0
4.0
12400
1150
4170
472
1480.0
730.0
W 14x605
178.0
20.92
2.1
4.4
10800
1040
3680
423
1320.0
652.0
W 14x550
162.0
20.24
2.3
4.8
9430
931
3250
378
1180.0
583.0
W 14x500
147.0
19.60
2.4
5.2
8210
838
2880
339
1050.0
522.0
W 14x455
134.0
19.02
2.6
5.7
7190
756
2560
304
936.0
468.0
W 14x426
125.0
18.67
2.8
6.1
6600
707
2360
283
869.0
434.0
W 14x398
117.0
18.29
2.9
6.4
6000
656
2170
262
801.0
402.0
W 14x370
109.0
17.92
3.1
6.9
5440
607
1990
241
736.0
370.0
W 14x342
101.0
17.54
3.3
7.4
4900
559
1810
221
672.0
338.0
W 14x311
91.4
17.12
3.6
8.1
4330
506
1610
199
603.0
304.0
W 14x283
83.3
16.74
3.9
8.8
3840
459
1440
179
542.0
274.0
W 14x257
75.6
16.38
4.2
9.7
3400
415
1290
161
487.0
246.0
W 14x233
68.5
16.04
4.6
10.7
3010
375
1150
145
436.0
221.0
W 14x211
62.0
15.72
5.1
11.6
2660
338
1030
130
390.0
198.0
W 14x193
56.8
15.48
5.5
12.8
2400
310
931
119
355.0
180.0
W 14x176
51.8
15.22
6.0
13.7
2140
281
838
107
320.0
163.0
W 14x159
46.7
14.98
6.5
15.3
1900
254
748
96
287.0
146.0
W 14x145
42.7
14.78
7.1
16.8
1710
232
677
87
260.0
133.0
W 14x132
38.8
14.66
7.1
17.7
1530
209
548
74
234.0
113.0
W 14x120
35.3
14.48
7.8
19.3
1380
190
495
68
212.0
102.0
W 14x109
32.0
14.32
8.5
21.7
1240
173
447
61
192.0
92.7
W 14x 99
29.1
14.16
9.3
23.5
1110
157
402
55
173.0
83.6
W 14x 90
26.5
14.02
10.2
25.9
999
143
362
50
157.0
75.6
24.1
14.31
5.9
22.4
882
Victor Saouma W 14x 82
Structural Concepts and Systems for Architects 123
148
29
139.0
44.8
Draft 3{12 Designation
STRUCTURAL MATERIALS A in2
d
bf
2tf
hc tw
in
Ix in4
Sx in3
Iy in4
Sy in3
Zx in3
Zy in3
W 14x 74
21.8
14.17
6.4
25.3
796
112
134
27
126.0
40.6
W 14x 68
20.0
14.04
7.0
27.5
723
103
121
24
115.0
36.9
W 14x 61
17.9
13.89
7.7
30.4
640
92
107
22
102.0
32.8
W 14x 53
15.6
13.92
6.1
30.8
541
78
58
14
87.1
22.0
W 14x 48
14.1
13.79
6.7
33.5
485
70
51
13
78.4
19.6
W 14x 43
12.6
13.66
7.5
37.4
428
63
45
11
69.6
17.3
W 14x 38
11.2
14.10
6.6
39.6
385
55
27
8
61.5
12.1
W 14x 34
10.0
13.98
7.4
43.1
340
49
23
7
54.6
10.6
W 14x 30
8.9
13.84
8.7
45.4
291
42
20
6
47.3
9.0
W 14x 26
7.7
13.91
6.0
48.1
245
35
9
4
40.2
5.5
W 14x 22
6.5
13.74
7.5
53.3
199
29
7
3
33.2
4.4
W 12x336
98.8
16.82
2.3
5.5
4060
483
1190
177
603.0
274.0
W 12x305
89.6
16.32
2.4
6.0
3550
435
1050
159
537.0
244.0
W 12x279
81.9
15.85
2.7
6.3
3110
393
937
143
481.0
220.0
W 12x252
74.1
15.41
2.9
7.0
2720
353
828
127
428.0
196.0
W 12x230
67.7
15.05
3.1
7.6
2420
321
742
115
386.0
177.0
W 12x210
61.8
14.71
3.4
8.2
2140
292
664
104
348.0
159.0
W 12x190
55.8
14.38
3.7
9.2
1890
263
589
93
311.0
143.0
W 12x170
50.0
14.03
4.0
10.1
1650
235
517
82
275.0
126.0
W 12x152
44.7
13.71
4.5
11.2
1430
209
454
73
243.0
111.0
W 12x136
39.9
13.41
5.0
12.3
1240
186
398
64
214.0
98.0
W 12x120
35.3
13.12
5.6
13.7
1070
163
345
56
186.0
85.4
W 12x106
31.2
12.89
6.2
15.9
933
145
301
49
164.0
75.1
W 12x 96
28.2
12.71
6.8
17.7
833
131
270
44
147.0
67.5
W 12x 87
25.6
12.53
7.5
18.9
740
118
241
40
132.0
60.4
W 12x 79
23.2
12.38
8.2
20.7
662
107
216
36
119.0
54.3
W 12x 72
21.1
12.25
9.0
22.6
597
97
195
32
108.0
49.2
W 12x 65
19.1
12.12
9.9
24.9
533
88
174
29
96.8
44.1
W 12x 58
17.0
12.19
7.8
27.0
475
78
107
21
86.4
32.5
W 12x 53
15.6
12.06
8.7
28.1
425
71
96
19
77.9
29.1
W 12x 50
14.7
12.19
6.3
26.2
394
65
56
14
72.4
21.4
W 12x 45
13.2
12.06
7.0
29.0
350
58
50
12
64.7
19.0
W 12x 40
11.8
11.94
7.8
32.9
310
52
44
11
57.5
16.8
W 12x 35
10.3
12.50
6.3
36.2
285
46
24
7
51.2
11.5
W 12x 30
8.8
12.34
7.4
41.8
238
39
20
6
43.1
9.6
W 12x 26
7.7
12.22
8.5
47.2
204
33
17
5
37.2
8.2
W 12x 22
6.5
12.31
4.7
41.8
156
25
5
2
29.3
3.7
W 12x 19
5.6
12.16
5.7
46.2
130
21
4
2
24.7
3.0
W 12x 16
4.7
11.99
7.5
49.4
103
17
3
1
20.1
2.3
W 12x 14
4.2
11.91
8.8
54.3
89
15
2
1
17.4
1.9
W 10x112
32.9
11.36
4.2
10.4
716
126
236
45
147.0
69.2
W 10x100
29.4
11.10
4.6
11.6
623
112
207
40
130.0
61.0
W 10x 88
25.9
10.84
5.2
13.0
534
98
179
35
113.0
53.1
W 10x 77
22.6
10.60
5.9
14.8
455
86
154
30
97.6
45.9
W 10x 68
20.0
10.40
6.6
16.7
394
76
134
26
85.3
40.1
W 10x 60
17.6
10.22
7.4
18.7
341
67
116
23
74.6
35.0
W 10x 54
15.8
10.09
8.2
21.2
303
60
103
21
66.6
31.3
W 10x 49
14.4
9.98
8.9
23.1
272
55
93
19
60.4
28.3
W 10x 45
13.3
10.10
6.5
22.5
248
49
53
13
54.9
20.3
W 10x 39
11.5
9.92
7.5
25.0
209
42
45
11
46.8
17.2
W 10x 33
9.7
9.73
9.1
27.1
170
35
37
9
38.8
14.0
W 10x 30
8.8
10.47
5.7
29.5
170
32
17
6
36.6
8.8
W 10x 26
7.6
10.33
6.6
34.0
144
28
14
5
31.3
7.5
W 10x 22
6.5
10.17
8.0
36.9
118
23
11
4
26.0
6.1
W 10x 19
5.6
10.24
5.1
35.4
96
19
4
2
21.6
3.3
W 10x 17
5.0
10.11
6.1
36.9
82
16
4
2
18.7
2.8
W 10x 15
4.4
9.99
7.4
38.5
69
14
3
1
16.0
2.3
W 10x 12
3.5
9.87
9.4
46.6
54
11
2
1
12.6
1.7
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.6 Steel Section Properties Designation
A in2
d
3{13 bf
2tf
hc tw
in
Ix in4
Sx Iy Sy in3 in4 in3
Zx in3
Zy in3
W 8x 67
19.7
9.00
4.4
11.1
272
60
89
21
70.2
32.7
W 8x 58
17.1
8.75
5.1
12.4
228
52
75
18
59.8
27.9
W 8x 48
14.1
8.50
5.9
15.8
184
43
61
15
49.0
22.9
W 8x 40
11.7
8.25
7.2
17.6
146
36
49
12
39.8
18.5
W 8x 35
10.3
8.12
8.1
20.4
127
31
43
11
34.7
16.1
W 8x 31
9.1
8.00
9.2
22.2
110
28
37
9
30.4
14.1
W 8x 28
8.2
8.06
7.0
22.2
98
24
22
7
27.2
10.1
W 8x 24
7.1
7.93
8.1
25.8
83
21
18
6
23.2
8.6
W 8x 21
6.2
8.28
6.6
27.5
75
18
10
4
20.4
5.7
W 8x 18
5.3
8.14
8.0
29.9
62
15
8
3
17.0
4.7
W 8x 15
4.4
8.11
6.4
28.1
48
12
3
2
13.6
2.7
W 8x 13
3.8
7.99
7.8
29.9
40
10
3
1
11.4
2.2
W 8x 10
3.0
7.89
9.6
40.5
31
8
2
1
8.9
1.7
W 6x 25
7.3
6.38
6.7
15.5
53
17
17
6
18.9
8.6
W 6x 20
5.9
6.20
8.2
19.1
41
13
13
4
14.9
6.7
W 6x 15
4.4
5.99
11.5
21.6
29
10
9
3
10.8
4.8
W 6x 16
4.7
6.28
5.0
19.1
32
10
4
2
11.7
3.4
W 6x 12
3.5
6.03
7.1
21.6
22
7
3
2
8.3
2.3
W 6x 9
2.7
5.90
9.2
29.2
16
6
2
1
6.2
1.7
W 5x 19
5.5
5.15
5.8
14.0
26
10
9
4
11.6
5.5
W 5x 16
4.7
5.01
6.9
15.8
21
9
8
3
9.6
4.6
W 4x 13
3.8
4.16
5.9
10.6
11
5
4
2
6.3
2.9
M 14x 18
5.1
14.00
7.4
60.3
148
21
3
1
24.9
2.2
M 12x 12
3.5
12.00
6.8
62.5
72
12
1
1
14.3
1.1
M 12x 11
3.2
11.97
7.3
63.6
65
11
1
1
13.2
1.0
M 12x 10
2.9
11.97
9.1
68.0
62
10
1
1
12.2
1.0
M 10x 9
2.7
10.00
6.5
58.4
39
8
1
0
9.2
0.8
M 10x 8
2.3
9.95
7.4
59.3
34
7
1
0
8.2
0.7
M 10x 8
2.2
9.99
7.8
65.0
33
7
0
0
7.7
0.6
M 8x 6
1.9
8.00
6.0
53.8
18
5
0
0
5.4
0.5
M 6x 4
1.3
6.00
5.4
47.0
7
2
0
0
2.8
0.3
M 5x 19
5.6
5.00
6.0
11.2
24
10
8
3
11.0
5.0
S 24x121
35.6
24.50
3.7
26.4
3160
258
83
21
306.0
36.2
S 24x106
31.2
24.50
3.6
34.1
2940
240
77
20
279.0
33.2
S 24x100
29.3
24.00
4.2
28.3
2390
199
48
13
240.0
23.9
S 24x 90
26.5
24.00
4.1
33.7
2250
187
45
13
222.0
22.3
S 24x 80
23.5
24.00
4.0
42.1
2100
175
42
12
204.0
20.7
S 20x 96
28.2
20.30
3.9
21.6
1670
165
50
14
198.0
24.9
S 20x 86
25.3
20.30
3.8
26.2
1580
155
47
13
183.0
23.0
S 20x 75
22.0
20.00
4.0
27.1
1280
128
30
9
153.0
16.7
S 20x 66
19.4
20.00
3.9
34.1
1190
119
28
9
140.0
15.3
S 18x 70
20.6
18.00
4.5
21.8
926
103
24
8
125.0
14.4
S 18x 55
16.1
18.00
4.3
33.6
804
89
21
7
105.0
12.1
S 15x 50
14.7
15.00
4.5
23.2
486
65
16
6
77.1
10.0
S 15x 43
12.6
15.00
4.4
31.0
447
60
14
5
69.3
9.0
S 12x 50
14.7
12.00
4.2
13.9
305
51
16
6
61.2
10.3
S 12x 41
12.0
12.00
4.0
20.7
272
45
14
5
53.1
8.9
S 12x 35
10.3
12.00
4.7
23.4
229
38
10
4
44.8
6.8
S 12x 32
9.4
12.00
4.6
28.6
218
36
9
4
42.0
6.4
S 10x 35
10.3
10.00
5.0
13.8
147
29
8
3
35.4
6.2
S 10x 25
7.5
10.00
4.7
26.4
124
25
7
3
28.4
5.0
S 8x 23
6.8
8.00
4.9
14.5
65
16
4
2
19.3
3.7
S 8x 18
5.4
8.00
4.7
23.7
58
14
4
2
16.5
3.2
S 7x 20
5.9
7.00
4.9
12.3
42
12
3
2
14.5
3.0
S 7x 15
4.5
7.00
4.7
21.9
37
10
3
1
12.1
2.4
S 6x 17
5.1
6.00
5.0
9.9
26
9
2
1
10.6
2.4
S 6x 12
3.7
6.00
4.6
19.9
22
7
2
1
8.5
1.9
S 5x 15
4.3
5.00
5.0
7.5
15
6
2
1
7.4
1.9
S 5x 10
2.9
5.00
4.6
17.4
12
5
1
1
5.7
1.4
S 4x 10
2.8
4.00
4.8
8.7
7
3
1
1
4.0
1.1
S 4x 8
2.3
4.00
4.5
14.7
6
3
1
1
3.5
1.0
S 3x 8
2.2
3.00
4.8
5.6
3
2
1
0
2.4
0.8
S 3x 6
1.7
3.00
4.5
11.4
3
2
0
0
2.0
0.7
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{14
STRUCTURAL MATERIALS A in2
Designation
d
bf
2tf
hc tw
in
Ix in4
Sx in3
Iy in4
Sy in3
Zx in3
Zy in3
C 15.x 50
14.7
15.
0
0
404.0
53.8
11.
3.78
8.20
8.17
C 15.x 40
11.8
15.
0
0
349.0
46.5
9.23
3.37
57.20
6.87
C 15.x 34
10.0
15.
0
0
315.0
42.0
8.13
3.11
50.40
6.23
C 12.x 30
8.8
12.
0
0
162.0
27.0
5.14
2.06
33.60
4.33
C 12.x 25
7.3
12.
0
0
144.0
24.1
4.47
1.88
29.20
3.84
C 12.x 21
6.1
12.
0
0
129.0
21.5
3.88
1.73
25.40
3.49
C 10.x 30
8.8
10.
0
0
103.0
20.7
3.94
1.65
26.60
3.78
C 10.x 25
7.3
10.
0
0
91.2
18.2
3.36
1.48
23.
3.19
C 10.x 20
5.9
10.
0
0
78.9
15.8
2.81
1.32
19.30
2.71
C 10.x 15
4.5
10.
0
0
67.4
13.5
2.28
1.16
15.80
2.35
C 9.x 20
5.9
9.
0
0
60.9
13.5
2.42
1.17
16.80
2.47
C 9.x 15
4.4
9.
0
0
51.0
11.3
1.93
1.01
13.50
2.05
C 9.x 13
3.9
9.
0
0
47.9
10.6
1.76
0.96
12.50
1.95
C 8.x 19
5.5
8.
0
0
44.0
11.0
1.98
1.01
13.80
2.17
C 8.x 14
4.0
8.
0
0
36.1
9.0
1.53
0.85
10.90
1.73
C 8.x 12
3.4
8.
0
0
32.6
8.1
1.32
0.78
9.55
1.58
C 7.x 15
4.3
7.
0
0
27.2
7.8
1.38
0.78
9.68
1.64
C 7.x 12
3.6
7.
0
0
24.2
6.9
1.17
0.70
8.40
1.43
C 7.x 10
2.9
7.
0
0
21.3
6.1
0.97
0.63
7.12
1.26
C 6.x 13
3.8
6.
0
0
17.4
5.8
1.05
0.64
7.26
1.36
C 6.x 11
3.1
6.
0
0
15.2
5.1
0.87
0.56
6.15
1.15
C 6.x 8
2.4
6.
0
0
13.1
4.4
0.69
0.49
5.13
0.99
C 5.x 9
2.6
5.
0
0
8.9
3.6
0.63
0.45
4.36
0.92
C 5.x 7
2.0
5.
0
0
7.5
3.0
0.48
0.38
3.51
0.76
C 4.x 7
2.1
4.
0
0
4.6
2.3
0.43
0.34
2.81
0.70
C 4.x 5
1.6
4.
0
0
3.8
1.9
0.32
0.28
2.26
0.57
C 3.x 6
1.8
3.
0
0
2.1
1.4
0.31
0.27
1.72
0.54
C 3.x 5
1.5
3.
0
0
1.9
1.2
0.25
0.23
1.50
0.47
C 3.x 4
1.2
3.
0
0
1.7
1.1
0.20
0.20
1.30
0.40
Designation
A in2
wgt k=ft
Ix in4
L 8.0x4.0x1.000
11.00
37.40
69.6
14.1
11.60
3.94
24.30
7.72
L 8.0x4.0x0.750
8.44
28.70
54.9
10.9
9.36
3.07
18.90
5.81
L 8.0x4.0x0.563
6.43
21.90
42.8
8.4
7.43
2.38
14.50
4.38
L 8.0x4.0x0.500
5.75
19.60
38.5
7.5
6.74
2.15
13.00
3.90
L 7.0x4.0x0.750
7.69
26.20
37.8
8.4
9.05
3.03
14.80
5.65
L 7.0x4.0x0.625
6.48
22.10
32.4
7.1
7.84
2.58
12.60
4.74
L 7.0x4.0x0.500
5.25
17.90
26.7
5.8
6.53
2.12
10.30
3.83
L 7.0x4.0x0.375
3.98
13.60
20.6
4.4
5.10
1.63
7.87
2.90
L 6.0x6.0x1.000
11.00
37.40
35.5
8.6
35.50
8.57
15.50
15.50
L 6.0x6.0x0.875
9.73
33.10
31.9
7.6
31.90
7.63
13.80
13.80
L 6.0x6.0x0.750
8.44
28.70
28.2
6.7
28.20
6.66
12.00
12.00
L 6.0x6.0x0.625
7.11
24.20
24.2
5.7
24.20
5.66
10.20
10.20
L 6.0x6.0x0.563
6.43
21.90
22.1
5.1
22.10
5.14
9.26
9.26
L 6.0x6.0x0.500
5.75
19.60
19.9
4.6
19.90
4.61
8.31
8.31
L 6.0x6.0x0.438
5.06
17.20
17.7
4.1
17.70
4.08
7.34
7.34
L 6.0x6.0x0.375
4.36
14.90
15.4
3.5
15.40
3.53
6.35
6.35
L 6.0x6.0x0.313
3.65
12.40
13.0
3.0
13.00
2.97
5.35
5.35
L 6.0x4.0x0.875
7.98
27.20
27.7
7.2
9.75
3.39
12.70
6.31
L 6.0x4.0x0.750
6.94
23.60
24.5
6.3
8.68
2.97
11.20
5.47
L 6.0x4.0x0.625
5.86
20.00
21.1
5.3
7.52
2.54
9.51
4.62
L 6.0x4.0x0.563
5.31
18.10
19.3
4.8
6.91
2.31
8.66
4.19
L 6.0x4.0x0.500
4.75
16.20
17.4
4.3
6.27
2.08
7.78
3.75
L 6.0x4.0x0.438
4.18
14.30
15.5
3.8
5.60
1.85
6.88
3.30
L 6.0x4.0x0.375
3.61
12.30
13.5
3.3
4.90
1.60
5.97
2.85
L 6.0x4.0x0.313
3.03
10.30
11.4
2.8
4.18
1.35
5.03
2.40
L 6.0x3.5x0.500
4.50
15.30
16.6
4.2
4.25
1.59
7.50
2.91
L 6.0x3.5x0.375
3.42
11.70
12.9
3.2
3.34
1.23
5.76
2.20
L 6.0x3.5x0.313
2.87
9.80
10.9
2.7
2.85
1.04
4.85
1.85
Victor Saouma
Sx in3
Iy in4
Sy in3
Zx in3
Zy in3
Structural Concepts and Systems for Architects
Draft 3.6 Steel Section Properties
3{15
Designation
A in2
wgt k=ft
Iy in4
Sy in3
Zx in3
Zy in3
L 5.0x5.0x0.875
7.98
27.20
17.8
L 5.0x5.0x0.750
6.94
23.60
15.7
5.2
17.80
5.17
9.33
9.33
4.5
15.70
4.53
8.16
L 5.0x5.0x0.625
5.86
20.00
8.16
13.6
3.9
13.60
3.86
6.95
L 5.0x5.0x0.500
4.75
6.95
16.20
11.3
3.2
11.30
3.16
5.68
L 5.0x5.0x0.438
5.68
4.18
14.30
10.0
2.8
10.00
2.79
5.03
5.03
L 5.0x5.0x0.375
3.61
12.30
8.7
2.4
8.74
2.42
4.36
4.36
L 5.0x5.0x0.313
3.03
10.30
7.4
2.0
7.42
2.04
3.68
3.68
L 5.0x3.5x0.750
5.81
19.80
13.9
4.3
5.55
2.22
7.65
4.10
L 5.0x3.5x0.625
4.92
16.80
12.0
3.7
4.83
1.90
6.55
3.47
L 5.0x3.5x0.500
4.00
13.60
10.0
3.0
4.05
1.56
5.38
2.83
L 5.0x3.5x0.438
3.53
12.00
8.9
2.6
3.63
1.39
4.77
2.49
L 5.0x3.5x0.375
3.05
10.40
7.8
2.3
3.18
1.21
4.14
2.16
L 5.0x3.5x0.313
2.56
8.70
6.6
1.9
2.72
1.02
3.49
1.82
L 5.0x3.5x0.250
2.06
7.00
5.4
1.6
2.23
0.83
2.83
1.47
L 5.0x3.0x0.625
4.61
15.70
11.4
3.5
3.06
1.39
6.27
2.61
L 5.0x3.0x0.500
3.75
12.80
9.4
2.9
2.58
1.15
5.16
2.11
L 5.0x3.0x0.438
3.31
11.30
8.4
2.6
2.32
1.02
4.57
1.86
L 5.0x3.0x0.375
2.86
9.80
7.4
2.2
2.04
0.89
3.97
1.60
L 5.0x3.0x0.313
2.40
8.20
6.3
1.9
1.75
0.75
3.36
1.35
L 5.0x3.0x0.250
1.94
6.60
5.1
1.5
1.44
0.61
2.72
1.09
L 4.0x4.0x0.750
5.44
18.50
7.7
2.8
7.67
2.81
5.07
5.07
L 4.0x4.0x0.625
4.61
15.70
6.7
2.4
6.66
2.40
4.33
4.33
L 4.0x4.0x0.500
3.75
12.80
5.6
2.0
5.56
1.97
3.56
3.56
L 4.0x4.0x0.438
3.31
11.30
5.0
1.8
4.97
1.75
3.16
3.16
L 4.0x4.0x0.375
2.86
9.80
4.4
1.5
4.36
1.52
2.74
2.74
L 4.0x4.0x0.313
2.40
8.20
3.7
1.3
3.71
1.29
2.32
2.32
L 4.0x4.0x0.250
1.94
6.60
3.0
1.0
3.04
1.05
1.88
1.88
L 4.0x3.5x0.500
3.50
11.90
5.3
1.9
3.79
1.52
3.50
2.73
L 4.0x3.5x0.438
3.09
10.60
4.8
1.7
3.40
1.35
3.11
2.42
L 4.0x3.5x0.375
2.67
9.10
4.2
1.5
2.95
1.16
2.71
2.11
L 4.0x3.5x0.313
2.25
7.70
3.6
1.3
2.55
0.99
2.29
1.78
L 4.0x3.5x0.250
1.81
6.20
2.9
1.0
2.09
0.81
1.86
1.44
L 4.0x3.0x0.500
3.25
11.10
5.1
1.9
2.42
1.12
3.41
2.03
L 4.0x3.0x0.438
2.87
9.80
4.5
1.7
2.18
0.99
3.03
1.79
Victor Saouma
Ix Sx in4 in3
Structural Concepts and Systems for Architects
Draft 3{16
STRUCTURAL MATERIALS
Designation
A in2
Iy in4
Sy in3
Zx in3
Zy in3
L 4.0x3.0x0.375
2.48
8.50
4.0
L 4.0x3.0x0.313
2.09
7.20
3.4
1.5
1.92
0.87
2.64
1.56
1.2
1.65
0.73
2.23
L 4.0x3.0x0.250
1.69
5.80
1.31
2.8
1.0
1.36
0.60
1.82
L 3.5x3.5x0.500
3.25
1.06
11.10
3.6
1.5
3.64
1.49
2.68
L 3.5x3.5x0.438
2.68
2.87
9.80
3.3
1.3
3.26
1.32
2.38
2.38
L 3.5x3.5x0.375
2.48
8.50
2.9
1.1
2.87
1.15
2.08
2.08
L 3.5x3.5x0.313
2.09
7.20
2.5
1.0
2.45
0.98
1.76
1.76
L 3.5x3.5x0.250
1.69
5.80
2.0
0.8
2.01
0.79
1.43
1.43
L 3.5x3.0x0.500
3.00
10.20
3.5
1.5
2.33
1.10
2.63
1.98
L 3.5x3.0x0.438
2.65
9.10
3.1
1.3
2.09
0.98
2.34
1.76
L 3.5x3.0x0.375
2.30
7.90
2.7
1.1
1.85
0.85
2.04
1.53
L 3.5x3.0x0.313
1.93
6.60
2.3
1.0
1.58
0.72
1.73
1.30
L 3.5x3.0x0.250
1.56
5.40
1.9
0.8
1.30
0.59
1.41
1.05
L 3.5x2.5x0.500
2.75
9.40
3.2
1.4
1.36
0.76
2.53
1.40
L 3.5x2.5x0.438
2.43
8.30
2.9
1.3
1.23
0.68
2.26
1.24
L 3.5x2.5x0.375
2.11
7.20
2.6
1.1
1.09
0.59
1.97
1.07
L 3.5x2.5x0.313
1.78
6.10
2.2
0.9
0.94
0.50
1.67
0.91
L 3.5x2.5x0.250
1.44
4.90
1.8
0.8
0.78
0.41
1.36
0.74
L 3.0x3.0x0.500
2.75
9.40
2.2
1.1
2.22
1.07
1.93
1.93
L 3.0x3.0x0.438
2.43
8.30
2.0
1.0
1.99
0.95
1.72
1.72
L 3.0x3.0x0.375
2.11
7.20
1.8
0.8
1.76
0.83
1.50
1.50
L 3.0x3.0x0.313
1.78
6.10
1.5
0.7
1.51
0.71
1.27
1.27
L 3.0x3.0x0.250
1.44
4.90
1.2
0.6
1.24
0.58
1.04
1.04
L 3.0x3.0x0.188
1.09
3.71
1.0
0.4
0.96
0.44
0.79
0.79
L 3.0x2.5x0.500
2.50
8.50
2.1
1.0
1.30
0.74
1.88
1.35
L 3.0x2.5x0.438
2.21
7.60
1.9
0.9
1.18
0.66
1.68
1.20
L 3.0x2.5x0.375
1.92
6.60
1.7
0.8
1.04
0.58
1.47
1.05
L 3.0x2.5x0.313
1.62
5.60
1.4
0.7
0.90
0.49
1.25
0.89
L 3.0x2.5x0.250
1.31
4.50
1.2
0.6
0.74
0.40
1.02
0.72
L 3.0x2.5x0.188
1.00
3.39
0.9
0.4
0.58
0.31
0.78
0.55
L 3.0x2.0x0.500
2.25
7.70
1.9
1.0
0.67
0.47
1.78
0.89
L 3.0x2.0x0.438
2.00
6.80
1.7
0.9
0.61
0.42
1.59
0.79
L 3.0x2.0x0.375
1.73
5.90
1.5
0.8
0.54
0.37
1.40
0.68
L 3.0x2.0x0.313
1.46
5.00
1.3
0.7
0.47
0.32
1.19
0.58
L 3.0x2.0x0.250
1.19
4.10
1.1
0.5
0.39
0.26
0.97
0.47
L 3.0x2.0x0.188
0.90
3.07
0.8
0.4
0.31
0.20
0.75
0.36
L 2.5x2.5x0.500
2.25
7.70
1.2
0.7
1.23
0.72
1.31
1.31
L 2.5x2.5x0.375
1.73
5.90
1.0
0.6
0.98
0.57
1.02
1.02
L 2.5x2.5x0.313
1.46
5.00
0.8
0.5
0.85
0.48
0.87
0.87
L 2.5x2.5x0.250
1.19
4.10
0.7
0.4
0.70
0.39
0.71
0.71
L 2.5x2.5x0.188
0.90
3.07
0.5
0.3
0.55
0.30
0.55
0.55
L 2.5x2.0x0.375
1.55
5.30
0.9
0.5
0.51
0.36
0.99
0.66
L 2.5x2.0x0.313
1.31
4.50
0.8
0.5
0.45
0.31
0.84
0.56
L 2.5x2.0x0.250
1.06
3.62
0.7
0.4
0.37
0.25
0.69
0.46
L 2.5x2.0x0.188
0.81
2.75
0.5
0.3
0.29
0.20
0.53
0.35
L 2.0x2.0x0.375
1.36
4.70
0.5
0.4
0.48
0.35
0.63
0.63
L 2.0x2.0x0.313
1.15
3.92
0.4
0.3
0.42
0.30
0.54
0.54
L 2.0x2.0x0.250
0.94
3.19
0.3
0.2
0.35
0.25
0.44
0.44
L 2.0x2.0x0.188
0.71
2.44
0.3
0.2
0.27
0.19
0.34
0.34
L 2.0x2.0x0.125
0.48
1.65
0.2
0.1
0.19
0.13
0.23
0.23
L 1.8x1.8x0.250
0.81
2.77
0.2
0.2
0.23
0.23
0.34
0.34
L 1.8x1.8x0.188
0.62
2.12
0.2
0.1
0.18
0.14
0.26
0.26
L 1.5x1.5x0.250
0.69
2.34
0.1
0.1
0.14
0.13
0.24
0.24
L 1.5x1.5x0.188
0.53
1.80
0.1
0.1
0.11
0.10
0.19
0.19
L 1.3x1.3x0.250
0.56
1.92
0.1
0.1
0.08
0.09
0.16
0.16
L 1.3x1.3x0.188
0.43
1.48
0.1
0.1
0.06
0.07
0.13
0.13
L 1.1x1.1x0.125
0.27
0.90
0.0
0.0
0.03
0.04
0.07
0.07
L 1.0x1.0x0.125
0.23
0.80
0.0
0.0
0.02
0.03
0.06
0.06
Victor Saouma
wgt Ix Sx k=ft in4 in3
Structural Concepts and Systems for Architects
Draft 3.7 Joists 3.7
3{17
Joists
43 Steel joists, Fig. 3.8 look like shallow trusses (warren type) and are designed as simply supported uniformly loaded beams assuming that they are laterally supported on the top (to prevent lateral torsional buckling). The lateral support is often pro ded by the concrete slab it suppors. 44 The standard open-web joist designation consists of the depth, the series designation and the chord type. Three series are available for oor/roof construction, Table 3.3
Series Depth (in) K 8-30 LH 18-48 DLH 52-72
Span (ft) 8-60 25-96 89-120
Table 3.3: Joist Series Characteristics [Design Length = Span – 0.33 FT.] 4"
4"
4"
Span Figure 3.8: prefabricated Steel Joists Typical joist spacing ranges from 2 to 4 ft, and provides an ecient use of the corrugated steel deck which itself supports the concrete slab.
45
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{18 46
STRUCTURAL MATERIALS
For preliminary estimates of the joist depth, a depth to span ratio of 24 can be assumed, therefore d L=2
(3.5)
where d is in inches, and L in ft. 47 Table 3.4 list the load carrying capacity of open web, K-series steel joists based on a amximum allowable stress of 30 ksi. For each span, the rst line indicates the total safe uniformly distributed load-carrying capacity in pounds per linear foot. Note that the dead load (including the one of the joist) must be substracted in order to determine the safe live load. The second line indicates the live load (pounds/linear foot) which will produce an approximate delection of L=360.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.7 Joists Joint 8K1 Desig. Depth (in.)
W
3{19 10K1 12K1 12K3 12K5 14K1 14K3 14K4 14K6 16K2 16K3 16K4 16K5 16K6 16K7 16K9
8
10
12
12
12
14
14
14
14
16
16
16
16
16
16
16
5.1
5
5
5.7
7.1
5.2
6
6.7
7.7
5.5
6.3
7
7.5
8.1
8.6
10.0
(lbs/ft) Span (ft.) 8
550 550
9
550 550
10
550 480
550 550
11
532 377
550 542
12
444 288
550 455
550 550
550 550
550 550
13
377 225
479 363
550 510
550 510
550 510
14
324 179
412 289
500 425
550 463
550 463
550 550
550 550
550 550
550 550
15
281 145
358 234
434 344
543 428
550 434
511 475
550 507
550 507
550 507
16
246 119
313 192
380 282
476 351
550 396
448 390
550 467
550 467
550 467
550 550
550 550
550 550
550 550
550 550
550 550
550 550
17
277 159
336 234
420 291
550 366
395 324
495 404
550 443
550 443
512 488
550 526
550 526
550 526
550 526
550 526
550 526
18
246 134
299 197
374 245
507 317
352 272
441 339
530 397
550 408
456 409
508 456
550 490
550 490
550 490
550 490
550 490
19
221 113
268 167
335 207
454 269
315 230
395 287
475 336
550 383
408 347
455 386
547 452
550 455
550 455
550 455
550 455
20
199 97
241 142
302 177
409 230
284 197
356 246
428 287
525 347
368 297
410 330
493 386
550 426
550 426
550 426
550 426
21
218 123
273 153
370 198
257 170
322 212
388 248
475 299
333 255
371 285
447 333
503 373
548 405
550 406
550 406
22
199 106
249 132
337 172
234 147
293 184
353 215
432 259
303 222
337 247
406 289
458 323
498 351
550 385
550 385
23
181 93
227 116
308 150
214 128
268 160
322 188
395 226
277 194
308 216
371 252
418 282
455 307
507 339
550 363
24
166 81
208 101
282 132
196 113
245 141
295 165
362 199
254 170
283 189
340 221
384 248
418 269
465 298
550 346
25
180 100
226 124
272 145
334 175
234 150
260 167
313 195
353 219
384 238
428 263
514 311
26
166 88
209 110
251 129
308 56
216 133
240 148
289 173
326 194
355 211
395 233
474 276
27
154 79
193 98
233 115
285 139
200 119
223 132
268 155
302 173
329 188
366 208
439 246
28
143 70
180 88
216 103
265 124
186 106
207 118
249 138
281 155
306 168
340 186
408 220
29
173 95
193 106
232 124
261 139
285 151
317 167
380 198
30
161 86
180 96
216 112
244 126
266 137
296 151
355 178
31
151 78
168 87
203 101
228 114
249 124
277 137
332 161
32
142 71
158 79
190 92
214 103
233 112
259 124
311 147
Table 3.4: Joist Properties
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{20
STRUCTURAL MATERIALS
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 4 Case Study I: EIFFEL TOWER Adapted from (Billington and Mark 1983)
4.1 Materials, & Geometry The tower was built out of wrought iron, less expensive than steel,and Ei el had more expereince with this material, Fig. 4.1
1
Figure 4.1: Ei el Tower (Billington and Mark 1983)
Draft 4{2
Case Study I: EIFFEL TOWER The structure is essentially a tower, subjected to gravity and wind load. It is a relatively \light" structure, so dead load is small compared to the wind load. M Mgravity 3 To avoid overturning had to be much higher than 1. wind This can be achieved either by: 1. Increase self weight (as in Washington's monument) 2. Increase the width of the base 3. Design support to resist tension. 4. Post-tension the support.
2
The tower is 984 feet high, and 328 feet wide at the base. The large base was essential to provide adequate stability in the presence of wind load. 5 We can assume that the shape of the tower is parabolic. If we take the x axis to be along the vertical axis of symmetry and y the half width, then we know that at x = 984 the (half) width y = 0 and at x = 0 the half width is 328=2 = 164, thus the equation of the half width is given by 4
y = 164
984 ; x 2
|
2
We recall from calculus that for y = v(x) dy dx
d 2 ax dx
dv = dy dv dx = 2ax
Thus for our problem dy dx
(4.1)
(4.2-a) (4.2-b)
(4.3-a)
;x = 984 2; 952
(4.3-b)
dv dx
dy = tan ) = tan;1 dx where is the angle measured from the x axis to the tangent to the curve. 3 We now can tabulate the width and slope at various elevations
Victor Saouma
}
1 = 2(164) 984984; x ; 984 | {z } | {z } dy dv
Also
{z984 av(x)2
dy dx
(4.4)
Structural Concepts and Systems for Architects
Draft 4.2 Loads
4{3
Width Location Height Width/2 Estimated Actual Support 0 164 328 .333 18.4 First platform 186 108 216 240 .270 15.1 second platform 380 62 123 110 .205 11.6 Intermediate platform 644 20 40 .115 6.6 Top platform 906 1 2 .0264 1.5 Top 984 0 0 0.000 0 dy
dx
o o o
o o
o
The tower is supported by four inclined supports, each with a cross section of 800 in2 . An idealization of the tower is shown in Fig. 4.2.
4
ACTUAL CONTINUOUS CONNECTION
IDEALIZED CONTINUOUS CONNECTION
ACTUAL POINTS OF CONNECTION
Figure 4.2: Ei el Tower Idealization, (Billington and Mark 1983) 4.2 5 6
Loads
The total weight of the tower is 18; 800 k. The dead load is not uniformly distributed, and is approximated as follows, Fig. 4.3: Location Height Dead Weight Ground- second platform 380 ft 15; 500 k Second platform-intermediate platform 264 ft 2; 200 k intermediate platform - top 340 ft 1; 100 k Total 984 ft 18; 800 k
7 From the actual width of the lower two platforms we can estimate the live loads (the intermediate and top platforms would have negligible LL in comparison):
1st platform: (50) psf(240)2 ft2 (1 000) lbs 2; 880 k 2nd platform: (50) psf(110)2 ft2 (1 000)k lbs 600 k Total: 3; 480 k Hence the total vertical load is Pvert = DL + LL = 18; 800 + 3; 480 = 22; 280 k. kip
; ;
Victor Saouma
Structural Concepts and Systems for Architects
Draft 4{4
Case Study I: EIFFEL TOWER
Figure 4.3: Ei el Tower, Dead Load Idealization; (Billington and Mark 1983) The wind pressure is known to also have a parabolic distribution (maximum at the top), the cross sectional area over which the wind is acting is also parabolic (maximum at the base). Hence we will simplify our analysis by considering an equivalent wind force obtained from a constant wind pressure (force/length) and constant cross section Fig. 4.4: The pressure is assumed to be 2.6 k/ft, thus the lateral wind force is, Fig. 4.5 984 = 492 ft (4.5) lat = (2 6) k/ft(984) ft = 2,560 k acting at 2
8
P
4.3
:
Reactions
Simplifying the three dimensional structure with 4 supports into a two dimensional one with two supports, the reactions can be easily determined for this statically determinate structure, Fig.4.6. Gravity Load 9
Pvert
grav Rvert
= 22 280 ? = 22 2280 = 11,140
(4.6-a) (4.6-b)
;
;
k
6
Lateral Load Lateral Moment (we essentially have a cantilivered beam subjected to a uniform load). The
moment at a distance from the support along the cantilevered beam subjected to a uniform pressure is given by ; ( ; )2 (4.7) = lat = ( ; ) | {z } 2 } 2 | {z Force Moment arm x
p
M
Victor Saouma
p L
x
L
x
p
L
x
Structural Concepts and Systems for Architects
Draft 4.3 Reactions
4{5
Figure 4.4: Ei el Tower, Wind Load Idealization; (Billington and Mark 1983) TOTAL LOADS LOADS P
P=2560k Q
Q=22,280k
L/2
1111 0000 0000 H 1111 0000 1111 111 000 0 0000 1111 000 111 000 000 111 V111 0 000 111 111 000 000 111 000 111
REACTIONS M0
Figure 4.5: Ei el Tower, Wind Loads, (Billington and Mark 1983)
+
=
WINDWARD SIDE
VERTICAL FORCES
WIND FORCES
LEEWARD SIDE
TOTAL
Figure 4.6: Ei el Tower, Reactions; (Billington and Mark 1983)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 4{6
Case Study I: EIFFEL TOWER
Thus the lateral moment caused by the wind is parabolic. At the base (x = 0), the maximum moment is equal to (L ; x)2 = (2:6) (984 ; 0)2 2 = 1,260,000 ; (4.8) Mlat = p 2 2 We observe that the shape of the moment diagram is also parabolic, just like the tower itself. This is not accidental, as nearly optimum structures have a shape which closely approximate their moment diagram (such as the varying depth of continuous long span bridges). To determine the resulting internal forces caused by the lateral (wind) moment, and since we have two supports (one under tension and the other under compression) we use 1; 260; 000 = 3,850 wind Rvert = M (4.9) = 6? d 328 Lateral Forces to be resisted by each of the two pairs. By symmetry, the lateral force will be equally divided among the two pairs of supports and will be equal to wind (4.10) Rlat = (2; 560) = 1,280 2 k/ft
ft
k.ft
k.ft
k
ft
k
k
4.4 10
Internal Forces
First, a biref reminder cos = FFx (4.11-a) sin = FFy (4.11-b)
θ F
Fy
tan = FFy (4.11-c) x
θ Fx
Internal forces are rst determined at the base. 12 Gravity load are rst considered, remember those are caused by the dead load and the live load, Fig. 4.7:
11
β=18.40 β=18.4 INCLINED INTERNAL FORCE: N CONSEQUENT HORIZONTAL COMPONENT: H
KNOWN VERTICAL COMPONENT: V
N
V
H FORCE POLYGON
Figure 4.7: Ei el Tower, Internal Gravity Forces; (Billington and Mark 1983)
Victor Saouma
V cos = N ) N = cosV
(4.12-a)
Structural Concepts and Systems for Architects
Draft 4.4 Internal Forces
4{7 ; 140 k = 11 cos 18:4 = 11,730 kip ) H = V tan tan = H V N
o
= 11; 140 k(tan 18:4 ) = 3,700 kip
H
o
(4.12-b) (4.12-c) (4.12-d)
The horizontal forces which must be resisted by the foundations, Fig. 4.8.
H
H 3700 k
3700 k
Figure 4.8: Ei el Tower, Horizontal Reactions; (Billington and Mark 1983) Because the vertical load decreases with height, the axial force will also decrease with height. 14 At the second platform, the total vertical load is Q = 1; 100 + 2; 200 = 3; 300 k and at that height the angle is 11:6 thus the axial force (per pair of columns) will be 13
o
Nvert Hvert
3;300
k
2 = 1; 685 k = cos 11 :6 k = 3; 300 2 (tan 11:6 ) = 339 k o
o
(4.13-a) (4.13-b)
Note that this is about seven times smaller than the axial force at the base, which for a given axial strength, would lead the designer to reduce (or taper) the cross-section. The horizontal force will be resisted by the axial forces in the second platform itself. 15 Wind Load: We now have determined at each pair of support the vertical and the horizontal forces caused by the wind load, the next step is to determine their axial components along the inclined leg, Fig. 4.9: = = = N = = =
Nc
t
Victor Saouma
wind wind ;Rvert cos ; Rlat sin ;(3; 850) k(cos 18:4 ) ; (1; 280) k(sin 18:40) o
-4,050
Leeward wind ;R cos + Rlat sin (3; 850) k(cos 18:4 ) + (1; 280) k(sin 18:40) 4,050 k Winward k
wind vert
o
(4.14-a) (4.14-b) (4.14-c) (4.14-d) (4.14-e) (4.14-f)
Structural Concepts and Systems for Architects
Draft 4{8
Case Study I: EIFFEL TOWER
18.4
3,850 k
3,850 cos 18.4
1,280 sin 18.4
18.4
1,280 k
Figure 4.9: Ei el Tower, Internal Wind Forces; (Billington and Mark 1983) 4.5 16
Internal Stresses
The total forces caused by both lateral and gravity forces can now be determined: NLTotal = ;(11; 730) ;(4; 050) = -15,780 Leeward side |
{z
k
}|
{z
k
k
}
gravity lateral Total 050) } = -7,630 NW = |;(11;{z730) } |+(4;{z gravity lateral k
k
k
Winward side
(4.15-a) (4.15-b)
We observe that even under wind load, the windward side is still under compression. 2. 17 In the idealization of the tower's geometry, the area of each pair of the simpli ed columns is 1; 600 and thus the maximum stresses will be determined from T ; 780 = -9.9 comp = NAL = ;1;15600 (4.16) 2 in
k
ksi
in
18
The strength of wrought iron is 45 ksi, hence the safety factor is stress 45 Safety Factor = ultimate actual stress = 9:9 = 4.5 ksi
ksi
Victor Saouma
(4.17)
Structural Concepts and Systems for Architects
Draft Chapter 5
REVIEW of STATICS To every action there is an equal and opposite reaction. Newton's third law of motion
5.1
Reactions
1 In the analysis of structures (hand calculations), it is often easier (but not always necessary) to start by determining the reactions. 2 Once the reactions are determined, internal forces are determined next; nally, internal stresses and/or deformations (de ections and rotations) are determined last1 . 3 Reactions are necessary to determine foundation load. 4 Depending on the type of structures, there can be di erent types of support conditions, Fig. 5.1. Roller: provides a restraint in only one direction in a 2D structure, in 3D structures a roller may provide restraint in one or two directions. A roller will allow rotation. Hinge: allows rotation but no displacements. Fixed Support: will prevent rotation and displacements in all directions.
5.1.1 Equilibrium 5 6 7
Reactions are determined from the appropriate equations of static equilibrium. Summation of forces and moments, in a static system must be equal to zero2. In a 3D cartesian coordinate system there are a total of 6 independent equations of equilibrium:
1 This is the sequence of operations in the exibility method which lends itself to hand calculation. In the sti ness method, we determine displacements rsts, then internal forces and reactions. This method is most suitable to computer implementation. 2 In a dynamic system F = ma where m is the mass and a is the acceleration.
Draft 5{2
REVIEW of STATICS
Figure 5.1: Types of Supports x = y = z = 0 x = y = z = 0 8
F
F
F
M
M
M
(5.1)
In a 2D cartesian coordinate system there are a total of 3 independent equations of equilibrium: (5.2)
x = y = z = 0 F
F
M
For reaction calculations, the externally applied load may be reduced to an equivalent force3 . 10 Summation of the moments can be taken with respect to any arbitrary point. 11 Whereas forces are represented by a vector, moments are also vectorial quantities and are represented by a curved arrow or a double arrow vector. 12 Not all equations are applicable to all structures, Table 5.1 13 The three conventional equations of equilibrium in 2D: x y and z can be replaced by the independent moment equations zA, zB , zC provided that A, B, and C are not colinear. 14 It is always preferable to check calculations by another equation of equilibrium. 15 Before you write an equation of equilibrium, 1. Arbitrarily decide which is the +ve direction 2. Assume a direction for the unknown quantities 3. The right hand side of the equation should be zero
9
F ;
M
3
M
F
M
M
However for internal forces (shear and moment) we must use the actual load distribution.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.1 Reactions
5{3
Structure Type
Beam, no axial forces 2D Truss, Frame, Beam Grid 3D Truss, Frame
Fx
Fy Fy
Fx
Fy
Alternate Set
Equations Fz Fz
Mz Mz
Mx My Mx My Mz
Beams, no axial Force MzA MzB 2 D Truss, Frame, Beam Fx MzA MzB MzA MzB MzC Table 5.1: Equations of Equilibrium If your reaction is negative, then it will be in a direction opposite from the one assumed. 16 Summation of external forces is equal and opposite to the internal ones. Thus the net force/moment is equal to zero. 17 The external forces give rise to the (non-zero) shear and moment diagram.
5.1.2 Equations of Conditions 18 If a structure has an internal hinge (which may connect two or more substructures), then this will provide an additional equation (M = 0 at the hinge) which can be exploited to determine the reactions. 19 Those equations are often exploited in trusses (where each connection is a hinge) to determine reactions. 20 In an inclined roller support with Sx and Sy horizontal and vertical projection, then the reaction R would have, Fig. 5.2.
Rx S y = Ry Sx
(5.3)
Figure 5.2: Inclined Roller Support
5.1.3 Static Determinacy 21 In statically determinate structures, reactions depend only on the geometry, boundary conditions and loads.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{4
REVIEW of STATICS
If the reactions can not be determined simply from the equations of static equilibrium (and equations of conditions if present), then the reactions of the structure are said to be statically indeterminate. 23 The degree of static indeterminacy is equal to the di erence between the number of reactions and the number of equations of equilibrium (plus the number of equations of conditions if applicable), Fig. 5.3.
22
Figure 5.3: Examples of Static Determinate and Indeterminate Structures Failure of one support in a statically determinate system results in the collapse of the structures. Thus a statically indeterminate structure is safer than a statically determinate one. 25 For statically indeterminate structures, reactions depend also on the material properties (e.g. Young's and/or shear modulus) and element cross sections (e.g. length, area, moment of inertia).
24
5.1.4 Geometric Instability The stability of a structure is determined not only by the number of reactions but also by their arrangement. 27 Geometric instability will occur if: 1. All reactions are parallel and a non-parallel load is applied to the structure. 2. All reactions are concurrent, Fig. 5.4.
26
Figure 5.4: Geometric Instability Caused by Concurrent Reactions
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.1 Reactions
5{5
3. The number of reactions is smaller than the number of equations of equilibrium, that is a mechanism is present in the structure. Mathematically, this can be shown if the determinant of the equations of equilibrium is equal to zero (or the equations are inter-dependent).
28
5.1.5 Examples Examples of reaction calculation will be shown next. Each example has been carefully selected as it brings a di erent \twist" from the preceding one. Some of those same problems will be revisited later for the determination of the internal forces and/or de ections. Many of those problems are taken from Prof. Gerstle textbok Basic Structural Analysis.
29
Example 5-7: Simply Supported Beam Determine the reactions of the simply supported beam shown below.
Solution:
The beam has 3 reactions, we have 3 equations of static equilibrium, hence it is statically determinate. (+ - ) Fx = 0; ) Rax ; 36 k = 0 (+ 6) Fy = 0; ) Ray + Rdy ; 60 k ; (4) k/ft(12) ft = 0 (+ ;) Mzc = 0; ) 12Ray ; 6Rdy ; (60)(6) = 0
or through matrix inversion (on your calculator) 9 8 9 8 9 8 9 2 38 1 0 0 < Rax = < 36 = < Rax = < 36 k = 4 0 1 1 5 Ray = 108 ) Ray = 56 k 0 12 ;6 : Rdy ; : 360 ; : Rdy ; : 52 k ; Alternatively we could have used another set of equations:
Check:
(+ ;) Mza = 0; (60)(6) + (48)(12) ; (Rdy )(18) = 0 ) Rdy = 52 (+ ;) Mzd = 0; (Ray )(18) ; (60)(12) ; (48)(6) = 0 ) Ray = 56
Victor Saouma
k k
6 6
p
(+ 6) Fy = 0; ; 56 ; 52 ; 60 ; 48 = 0
Structural Concepts and Systems for Architects
Draft 5{6
REVIEW of STATICS
Example 5-8: Three Span Beam Determine the reactions of the following three spans beam
Solution:
We have 4 unknowns ( and ), three equations of equilibrium and one equation of condition ( = 0), thus the structure is statically determinate. 1. Isolating ab: = 22.2 k 6 ; = 0; (9)( ) ; (40)(5) = 0 ) ; (+ ) = 0; (40)(4) ; ( )(9) = 0 ) = 17.7 k 6 = 0; ) = 30 k Rax ; Ray ; Rcy
Rdy
Mb
M
Ray
b
Ma
S
Fx
2. Isolating bd: (+ ;)
Md
(+ ;)
= 0; ;(17 7)(18) ; (40)(15) ; (4)(8)(8) ; (30)(2) + (12) = 0 ) = 1 12236 = 103 k 6 = 0; ;(17 7)(6) ; (40)(3) + (4)(8)(4) + (30)(10) ; (12) = 0 ) = 20112 3 = 16.7 k 6 :
Rcy
;
:
Rdy
Victor Saouma
S
Rax
Rcy
Mc
Ray
Rdy
:
Structural Concepts and Systems for Architects
Draft 5.1 Reactions 3. Check
5{7
Fy
p
= 0; 6; 22 2 ; 40 ; 40 + 103 ; 32 ; 30 + 16 7 = 0 :
:
Example 5-9: Three Hinged Gable Frame The three-hinged gable frames spaced at 30 ft. on center. Determine the reactions components on the frame due to: 1) Roof dead load, of 20 psf of roof area; 2) Snow load, of 30 psf of horizontal projection; 3) Wind load of 15 psf of vertical projection. Determine the critical design values for the vertical and horizontal reactions.
Solution: 1. Due to symmetry, we will consider only the dead load on one side of the frame. 2. Due to symmetry, there is no vertical force transmitted by the hinge for snow and dead load. 3. Roof dead load per frame is p 1 lbs/k = 20 2 k ? = (20) psf(30) ft 302 + 152 ft 1 000 DL
Victor Saouma
;
:
Structural Concepts and Systems for Architects
Draft 5{8
REVIEW of STATICS
4. Snow load per frame is SL
1 = (30) psf(30) ft(30) ft 1 000 ;
lbs/k
= 27
:
k
?
5. Wind load per frame (ignoring the suction) is WL
1 = (15) psf(30) ft(35) ft 1 000 ;
lbs/k
= 15 75 k:
6. There are 4 reactions, 3 equations of equilibrium and one equation of condition ) statically determinate. 7. The horizontal reaction due to a vertical load at midspan of the roof, is obtained by taking moment with respect to the hinge H
V
(+ ;) C = 0; 15( ) ; 30( ) + 35( ) = 0 ) Substituting for roof dead and snow load we obtain M
V
A = DL A HDL = A VSL = A HSL = V
V
H
:
:
:
V
H
:
:
F
:
V
V
:
:
:
F
:
:
:
8. The reactions due to wind load are B (+ ;) A = 0; (15 75)( 20+15 2 ) ; WL (60) = 0 ; B (+ ) C = 0; WL (35) ; (4 6)(30) = 0 A =0 (+ - ) x = 0; 15 75 ; 3 95 ; WL B A (+ 6) y = 0; WL ; WL = 0 M
= 1535V = 429
= 20 2 k 6 = ( 429)(20 2) = 8 66 k= 27 k 6 = ( 429)(27 ) = 11 58 k-
B DL B HDL B VSL B HSL V
M
H
H
V
:
) ) ) )
6 ?
B = 4:60 k WL B = 3:95 k H WL A HWL = 11:80 k A VWL = ;4:60 k
V
9. Thus supports should be designed for H V
5.2
= 8 66 k + 11 58 k + 3 95 k = 24.19 k = 20 7 k + 27 0 k + 4 60 k = 52.3 k :
:
:
:
:
:
Trusses
5.2.1 Assumptions Cables and trusses are 2D or 3D structures composed of an assemblage of simple one dimensional components which transfer only axial forces along their axis. 31 Trusses are extensively used for bridges, long span roofs, electric tower, space structures. 32 For trusses, it is assumed that 1. Bars are pin-connected 30
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.2 Trusses
5{9
2. Joints are frictionless hinges4 . 3. Loads are applied at the joints only. 33 A truss would typically be composed of triangular elements with the bars on the upper chord under compression and those along the lower chord under tension. Depending on the orientation of the diagonals, they can be under either tension or compression. 34 In a truss analysis or design, we seek to determine the internal force along each member, Fig. 5.5
Figure 5.5: Bridge Truss
5.2.2 Basic Relations
Sign Convention: Tension positive, compression negative. On a truss the axial forces are indicated as forces acting on the joints. Stress-Force: = PA Stress-Strain: = E" Force-Displacement: " = LL Equilibrium: F = 0
5.2.3 Determinacy and Stability 35 Trusses are statically determinate when all the bar forces can be determined from the equations of statics alone. Otherwise the truss is statically indeterminate. 36 A truss may be statically indeterminate with respect to the reactions or externally indeterminate and/or statically indeterminate with respect to the internal forces that is internally indeterminate. 37 a 2D truss is externally indeterminate if there are more than 3 reactions.
4 In practice the bars are riveted, bolted, or welded directly to each other or to gusset plates, thus the bars are not free to rotate and so-called secondary bending moments are developed at the bars. Another source of secondary moments is the dead weight of the element.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{10
REVIEW of STATICS
38 Since each joint is pin-connected, we can apply M = 0 at each one of them. Furthermore, summation of forces applied on a joint must be equal to zero. 39 For a 2D truss we have 2 equations of equilibrium FX = 0 and FY = 0 which can be applied at each joint. For 3D trusses we would have three equations: FX = 0, FY = 0 and FZ = 0. 40 If we refer to j as the number of joints, R the number of reactions and m the number of members, then we would have a total of m + R unknowns and 2j (or 3j ) equations of statics (2D or 3D at each joint). If we do not have enough equations of statics then the problem is indeterminate, if we have too many equations then the truss is unstable, Table 5.2.
2D
3D
Static Indeterminacy
External R>3 R>6 Internal m + R > 2j m + R > 3j Unstable m + R < 2j m + R < 3j Table 5.2: Static Determinacy and Stability of Trusses Fig. 5.6 shows a truss with 4 reactions, thus it is externally indeterminate. This truss has 6 joints (j = 6), 4 reactions (R = 4) and 9 members (m = 9). Thus we have a total of m + R = 9 + 4 = 13 unknowns and 2 j = 2 6 = 12 equations of equilibrium, thus the truss is statically indeterminate.
41
Figure 5.6: A Statically Indeterminate Truss 42
There are two methods of analysis for statically determinate trusses 1. The Method of joints 2. The Method of sections
5.2.4 43
Method of Joints
The method of joints can be summarized as follows 1. Determine if the structure is statically determinate
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.2 Trusses
5{11
2. Compute all reactions 3. Sketch a free body diagram showing all joint loads (including reactions) 4. For each joint, and starting with the loaded ones, apply the appropriate equations of equilibrium (Fx and Fy in 2D; Fx , Fy and Fz in 3D). 5. Because truss elements can only carry axial forces, the resultant force (F~ = F~x + F~y ) must be along the member, Fig. 5.7. F Fx Fy = l = l l x y
(5.4)
Always keep track of the x and y components of a member force (Fx , Fy ), as those might be needed later on when considering the force equilibrium at another joint to which the member is connected.
44
Figure 5.7: X and Y Components of Truss Forces This method should be used when all member forces should be determined. 46 In truss analysis, there is no sign convention. A member is assumed to be under tension (or compression). If after analysis, the force is found to be negative, then this would imply that the wrong assumption was made, and that the member should have been under compression (or tension). 47 On a free body diagram, the internal forces are represented by arrow acting on the joints and not as end forces on the element itself. That is for tension, the arrow is pointing away from the joint, and for compression toward the joint, Fig. 5.8. 45
Figure 5.8: Sign Convention for Truss Element Forces
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{12
REVIEW of STATICS
Example 5-10: Truss, Method of Joints Using the method of joints, analyze the following truss
Solution: 1. R = 3, m = 13, 2j = 16, and m + R = 2j 2. We compute the reactions
p
(+ ;) ME = 0; ) (20 + 12)(3)(24) + (40 + 8)(2)(24) + (40)(24) ; RAy (4)(24) = 0 ) RAy = 58 k 6 (+ ?) Fy = 0; ) 20 + 12 + 40 + 8 + 40 ; 58 ; REy = 0 ) REy = 62 k 6
3. Consider each joint separately: Node A: Clearly AH is under compression, and AB under tension.
(+ 6) Fy = 0; ) FAHy ; 58 = 0 FAH = lly (FAHy ) p ly = 32 l = 322 + 242 = 40 40 (58) = 72:5 Compression ) FAH = 32 (+ - ) Fx = 0; ) ;FAHx + FAB = 0 24 (58) = 43:5 Tension FAB = llxy (FAHy ) = 32
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.2 Trusses
5{13
Node B:
(+ - ) x = 0; ) (+ 6) y = 0; )
Node H:
(+ - ) x = 0; )
F
F
F
F
BC = 43:5 Tension BH = 20 Tension
AHx ; HCx ; HGx = 0 p 24 43 5 ; p2424 (I) 2 +322 ( HC ) ; 242 +102 ( HG ) = 0 (+ 6) y = 0; ) AHy + HCy ; 12 ; HGy ; 20 = 0 p 10 58 + p2432 2 +322 ( HC ) ; 12 ; 242 +102 ( HG ) ; 20 = 0 (II) F
F
F
F
:
F
F
F
F
F
F
F
F
Solving for I and II we obtain HC = ;7:5 Tension Compression HG = 52
F
F
Node E:
y = 0; ) x = 0; ) F
F
EFy = 62 FED = FEFx
F
) )
= 77:5 k EF = p2432 2 2 (62) 24 (F+32 24 (62) = 46:5 k FED = ) = EF y 32 32
F
The results of this analysis are summarized below
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{14
REVIEW of STATICS
4. We could check our calculations by verifying equilibrium of forces at a node not previously used, such as D
5.3 Shear & Moment Diagrams 5.3.1
Theory
5.3.1.1 Design Sign Conventions Before we derive the Shear-Moment relations, let us arbitrarily de ne a sign convention. 5 49 The sign convention adopted here, is the one commonly used for design purposes . With reference to Fig. 5.9 48
Figure 5.9: Shear and Moment Sign Conventions for Design 5
Later on, in more advanced analysis courses we will use a di erent one.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{15
2D: Load Positive along the beam's local y axis (assuming a right hand side convention), that is
positive upward. Axial: tension positive. Flexure A positive moment is one which causes tension in the lower bers, and compression in the upper ones. For frame members, a positive moment is one which causes tension along the inner side. Shear A positive shear force is one which is \up" on a negative face, or \down" on a positive one. Alternatively, a pair of positive shear forces will cause clockwise rotation. Torsion Counterclockwise Draftpositive 3D: Use double arrow vectors (and NOT curved arrows). Forces and moments (including torsions) are x 6M de ned with respect to a right hand side coordinate system, Fig. 5.10. * 6 y
* * Tx -Mz
y
6M y 6 6 >
--
- z
> Figure 5.10: for 3D Frame Elements M Sign Conventions Tx
z
5.3.1.2 Load, Shear, Moment Relations Let us (re)derive the basic relations between load, shear and moment. Considering an in nitesimal length dx of a beam subjected to a positive load6 w(x), Fig. 5.11. The in nitesimal section must also
50
Figure 5.11: Free Body Diagram of an In nitesimal Beam Segment be in equilibrium. 51 There are no axial forces, thus we only have two equations of equilibrium to satisfy Fy = 0 and Mz = 0. 52 Since dx is in nitesimally small, the small variation in load along it can be neglected, therefore we assume w(x) to be constant along dx. 53 To denote that a small change in shear and moment occurs over the length dx of the element, we add the di erential quantities dVx and dMx to Vx and Mx on the right face. 6
In this derivation, as in all other ones we should assume all quantities to be positive.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{16 54
or
REVIEW of STATICS
Next considering the rst equation of equilibrium (+ 6) Fy = 0 ) Vx + wx dx ; (Vx + dVx ) = 0 dV = w(x) dx
(5.5)
The slope of the shear curve at any point along the axis of a member is given by the load curve at that point. 55
Similarly
(+ ;) Mo = 0 ) Mx + Vx dx ; wx dx dx 2 ; (Mx + dMx ) = 0 Neglecting the dx2 term, this simpli es to dM = V (x) dx
(5.6)
The slope of the moment curve at any point along the axis of a member is given by the shear at that point. 56
Alternative forms of the preceding equations can be obtained by integration V =
Z
w(x)dx
V21 = Vx2 ; Vx1 =
Z
(5.7) x2 x1
w(x)dx (5.8)
The change in shear between 1 and 2, V21 , is equal to the area under the load between x1 and x2 . and M =
Z
V (x)dx
M21 = M2 ; M1 =
Z
(5.9) x2 x1
V (x)dx (5.10)
The change in moment between 1 and 2, M21, is equal to the area under the shear curve between x1 and x2 . Note that we still need to have V1 and M1 in order to obtain V2 and M2 respectively. 58 Fig. 5.12 and 5.13 further illustrates the variation in internal shear and moment under uniform and concentrated forces/moment. 57
5.3.1.3 Moment Envelope For design, we often must consider di erent load combinations. 60 For each load combination, we should draw the shear, moment diagrams. and then we should use the Moment envelope for design purposes.
59
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{17
Figure 5.12: Shear and Moment Forces at Di erent Sections of a Loaded Beam
Positive Constant
Negative Constant
Positive Increasing Positive Decreasing Negative Increasing Negative Decreasing
Positive Constant
Negative Constant
Positive Increasing Positive Decreasing Negative Increasing Negative Decreasing
Load
Shear
Shear
Moment
Figure 5.13: Slope Relations Between Load Intensity and Shear, or Between Shear and Moment
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{18
REVIEW of STATICS
5.3.1.4 Examples Example 5-11: Simple Shear and Moment Diagram Draw the shear and moment diagram for the beam shown below
Solution:
The free body diagram is drawn below
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{19
Reactions are determined from the equilibrium equations (+ ) Fx = 0; ) ;RAx + 6 = 0 ) RAx = 6 k (+ ;) MA = 0; ) (11)(4) + (8)(10) + (4)(2)(14 + 2) ; RFy (18) = 0 ) RFy = 14 k (+ 6) Fy = 0; ) RAy ; 11 ; 8 ; (4)(2) + 14 = 0 ) RAy = 13 k Shear are determined next. 1. 2. 3. 4. 5.
Moment 1. 2. 3.
At A the shear is equal to the reaction and is positive. At B the shear drops (negative load) by 11 k to 2 k. At C it drops again by 8 k to ;6 k. It stays constant up to D and then it decreases (constant negative slope since the load is uniform and negative) by 2 k per linear foot up to ;14 k. As a check, ;14 k is also the reaction previously determined at F . is determined last: The moment at A is zero (hinge support). The change in moment between A and B is equal to the area under the corresponding shear diagram, or MB;A = (13)(4) = 52. etc...
Example 5-12: Frame Shear and Moment Diagram Draw the shear and moment diagram of the following frame
Solution: Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{20
REVIEW of STATICS
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{21
Reactions are determined rst
(+ ) Fx = 0; ) RAx ; 45 (3)(15) = 0 | {z }
) (+ ;) MA = 0; ) ) (+ 6) Fy = 0; ) )
load
RAx = 36
k
;
(3)(30)( 302 ) + 53 (3)(15) 30 + 92 ; 54 (3)(15) 122 ; 39RDy = 0 RFy = 52:96 RAy ; (3)(30) ; 35 (3)(15) + 52:96 = 0 RAy = 64:06 k
k
Shear: 1. For A ; B , the shear is constant, equal to the horizontal reaction at A and negative according to our previously de ned sign convention, VA = ;36 2. For member B ; C at B , the shear must be equal to the vertical force which was transmitted along A ; B , and which is equal to the vertical reaction at A, VB = 64:06. 3. Since B ; C is subjected to a uniform negative load, the shear along B ; C will have a slope equal to ;3 and in terms of x (measured from B to C ) is equal to VB;C (x) = 64:06 ; 3x k
4. The shear along C ; D is obtained by decomposing the vertical reaction at D into axial and shear components. Thus at D the shear is equal to 35 52:96 = 31:78 and is negative. Based on our sign convention for the load, the slope of the shear must be equal to ;3 along C ; D. Thus the shear at point C is such that Vc ; 35 9(3) = ;31:78 or Vc = 13:22. The equation for the shear is given by (for x going from C to D) V = 13:22 ; 3x k
5. We check our calculations by verifying equilibrium of node C p (+ ) Fx = 0 ) 53 (42:37) + 45 (13:22) = 25:42 + 10:58 = 36 p (+ 6) Fy = 0 ) 54 (42:37) ; 35 (13:22) = 33:90 ; 7:93 = 25:97
Moment:
1. Along A ; B , the moment is zero at A (since we have a hinge), and its slope is equal to the shear, thus at B the moment is equal to (;36)(12) = ;432 2. Along B ; C , the moment is equal to k.ft
Z
x
Z
x
MB;C = MB + VB;C (x)dx = ;432 + (64:06 ; 3x)dx 0 0 2 = ;432 + 64:06x ; 3 x2 which is a parabola. Substituting for x = 30, we obtain at node C : MC = ;432+64:06(30) ; 302 3 2 = 139:8 3. If we need to determine the maximum moment along B ; C , we know that dMdxB;C = 0 at the point where VB;C = 0, that is VB;C (x) = 64:06 ; 3x = 0 ) x = 643:06 = 25:0 . In other k.ft
words, maximum moment occurs where the shear is zero. (25:0)2 Thus MBmax ;C = ;432 + 64:06(25:0) ; 3 2 = ;432 + 1; 601:5 ; 937:5 = 232
Victor Saouma
ft
k.ft
Structural Concepts and Systems for Architects
Draft 5{22
REVIEW of STATICS
4. Finally along C ; D, the moment varies quadratically (since we had a linear shear), the moment rst increases (positive shear), and then decreases (negative shear). The moment along C ; D is given by Rx Rx MC ;D = MC + 0 VC ;D (x)dx = 139:8 + 0 (13:22 ; 3x)dx 2 = 139:8 + 13:22x ; 3 x2 which is a parabola. 2 Substituting for xp= 15, we obtain at node C MC = 139:8 + 13:22(15) ; 3 152 = 139:8 + 198:3 ; 337:5 = 0
Example 5-13: Frame Shear and Moment Diagram; Hydrostatic Load The frame shown below is the structural support of a ume. Assuming that the frames are spaced 2 ft apart along the length of the ume, 1. Determine all internal member end actions 2. Draw the shear and moment diagrams 3. Locate and compute maximum internal bending moments 4. If this is a reinforced concrete frame, show the location of the reinforcement.
Solution: The hydrostatic pressure causes lateral forces on the vertical members which can be treated as cantilevers xed at the lower end. The pressure is linear and is given by p = h. Since each frame supports a 2 ft wide slice of the
ume, the equation for w (pounds/foot) is w = (2)(62:4)(h) = 124:8h lbs/ft At the base w = (124:8)(6) = 749 lbs/ft = :749 k/ft Note that this is both the lateral pressure on the end walls as well as the uniform load on the horizontal members.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{23
End Actions 1. Base force at B is FBx = (:749) 26 = 2:246 2. Base moment at B is MB = (2:246) 36 = 4:493 3. End force at B for member B ; E are equal and opposite. 4. Reaction at C is RCy = (:749) 162 = 5:99 Shear forces 1. Base at B the shear force was determined earlier and was equal to 2:246 . Based on the orientation of the x ; y axis, this is a negative shear. 2. The vertical shear at B is zero (neglecting the weight of A ; B ) 3. The shear to the left of C is V = 0 + (;:749)(3) = ;2:246 . 4. The shear to the right of C is V = ;2:246 + 5:99 = 3:744 Moment diagrams k
k.ft
k
k
k
k
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{24
REVIEW of STATICS
1. At the base: = 4 493 as determined above. 2. At the support , c = ;4 493 + (; 749)(3)( 23 ) = ;7 864 3. The maximum moment is equal to max = ;7 864 + ( 749)(5)( 25 ) = 1 50 Design: Reinforcement should be placed along the bers which are under tension, that is on the side of the negative moment7 . The gure below schematically illustrates the location of the exural8 reinforcement. B M C
:
M
k.ft
:
:
M
:
:
:
k.ft
:
k.ft
Example 5-14: Shear Moment Diagrams for Frame 7 That is why in most European countries, the sign convention for design moments is the opposite of the one commonly used in the U.S.A.; Reinforcement should be placed where the moment is \postive". 8 Shear reinforcement is made of a series of vertical stirrups.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{25
8’
12’
10’
30k
5k/ft
2k/ft
B
A
Vba C
Hbd
M ba
E
Vbc
10k
5’ 20k
VA
Vbd
G
52.5k
M bc M bd 30k
15’ 0
0 0
D
650’k 450’k
HD
4k/ft
200’k 82.5k
VD
CHECK
30k
10k
5k/ft B
A
M bc
M ba
Vba
17.5k
2k/ft
B
Hbc
Hba
C Vbc
(10)+(2)(10) 30k
17.5k 17.5-5*x=0
-22.5k
3.5’
-22.5+(-30)
10k
Vbc M bc -200’k
17.5-(5)(8)
(10)(10)+(2)(10)(10)/2
Vba
-52.5k 30.6’k
(17.5)(3.5)/2
-20’k
(17.5)(3.5)/2+(-22.5)(8-3.5)/2
(-52.5)(12)+(-20)
-650’k
M ba
Vbd M bd
50k
Victor Saouma
20k
450’k (50)(15)-[(4)(5)/2][(2)(15)/3)]
(50)-(4)(15)/2
450’k
Hbd
20k
4k/ft
50k 82.5k
Structural Concepts and Systems for Architects
Draft 5{26
REVIEW of STATICS
Example 5-15: Shear Moment Diagrams for Inclined Frame
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{27
26k
26k
10’
13’
10’ 20k
13’ 13’
C
13 5
5
3
12
15’
4
D
B 2k/ft
20’ Ha
A
E
36’
20’
Ve
48.8k
2k/ft
60k
Fx F
800’k
z y
Fy
20k
60-(2)(20)
20k
(20)(20)+(60-20)(20)/2
19.2k 800k’
(60)(20)-(2)(20)(20)/2
Va
0k’
x F/Fy=z/x F/Fx=z/y Fx/Fy=y/x
60k 2
3
8k
1k
11.1k
28.8k
20k
8k 12k
778k’
39.1k
19.2k
7
29.3k
48.9k
’
777k
’k
1130
9 B-C
.1
-16 k
11 C-D
(39.1)(12.5)
488’k 12 C-D
14
13
k
1k
2k’
112
777k’ 48
8’
k
800’k
+60k
+20k
0’k 113
-23 .1 k k k 8 6 . 5 . 6 0 2 -39 .
(20)(12)/(13)=18.46 (19.2)(5)/(13)=7.38 (19.2)(12)/(13)=17.72 (26)(12)/(13)=24 (26.6)(13)/(12)=28.8 (26.6)(5)/(12)=11.1 (28.8)(4)/(5)=23.1 (28.8)(3)/(5)=17.28 (20)(4)/(5)=16 (20)(3)/(5)=12 (39.1)(5)/(4)=48.9 (39.1)(3)/(4)=29.3
k’
1,130-(.58)(13) k’ 800+(25.4)(13) 1122
3.1
8 B-C
777
-
25.42
.1k -2 39
1,122-(26.6)(13)
k k -26.6 -0.58 26 -0.6-26
488+(23.1)(12.5)
+25
+25.4
10
-23
8.7 17.7+ .4k
800’k
0k’
0k
18.46k 7.38k
17.72k
(20)(15)/13=7.7
17.2
20k
24k
24k
0k
26.6k
800k’
6
16k
10k
10k
5
28.8k
778k’
26k 26k
48.8k
4
23.
1
20-10-10
19.2k
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{28
REVIEW of STATICS
5.3.2 Formulaes Adapted from (of Steel COnstruction 1986)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{29
1) Simple Beam; uniform Load L
x
w L R
R L / 2
R
L / 2
x
V V
Shear
at center
V
max
M
x
M
max x
M max.
= = = = = =
V w
wL
2;
L
x
2
8 25 ( ;4 ) 384 3 24 ( ; 2
wx
L
x
wL
wx
EI
L
EI
2
Lx
+ 3) x
Moment
2) Simple Beam; Unsymmetric Triangular Load
Max :
V1
R
V2
x Mmax Mx V
at = 577 x
= 2 =
R1
L
at = 5193 max x x
:
L
= = = = = = =
W
23 3 2 ; 2 3 1283 2 2 3 2( ; 3 ) 01304 3 4 180 2 (3 ; 10 W
Wx
W
L
:
WL
Wx
L
L
x
WL
:
EI
Wx
EI L
x
2 2
L x
+ 7 4) L
3) Simple Beam; Symmetric Triangular Load
R
=
V
for L2 x at center max for L2 x for L2 x max x
b max 3
Pb L Pa L Pab L Pbx L Pa b = 3EIL b ;x ) = 6Pbx EIL (L ; p Pab(a + 2b) 3a(a + 2b) = 27EIL = = = =
2 2
2
2
2
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{31
7) Simple Beam; Two Equally Concentrated Symmetric Loads
=
= max = max = 24 (3 x = 6 (3 x = 6 (3
R
V
P
M
Pa
Pa
when when
x < a
a < x < L
;
L
EI Px
La
EI Pa
a
2 ; 4 a2 )
Lx
EI
; 3 2 ; 2) ; 3 2 ; 2) a
x
x
a
8) Simple Beam; Two Equally Concentrated Unsymmetric Loads
max when max when when ; max when max when when when ;
a < b
R
1 = V1 =
b < a
R
2 = V2 =
a < x < L
b
b < a a < b
x < a
a < x < L
b
= = = = =
x M1 M2 Mx Mx V
P L P L P L
( ; + ) ( ; + ) ( ; ) L
a
b
L
b
a
b
a
1 2 R1 x R1 x ; P (x ; a)
R a R b
9) Cantilevered Beam, Uniform Load 3
1 = V1 = 8 wL R2 = V2 = 85 wL Vx = R1 ; wx 2 Mmax = wL 8 9 2 wL M1 = R
at = 83 x
L
128
x
M
x at = 4215 x
Victor Saouma
:
L
max
2
= 1 ; 2 = 48 ( 3 ; 3 + 2 3 ) 4 = 185 R x
wx
EI wL
wx
L
Lx
x
EI
Structural Concepts and Systems for Architects
Draft 5{32
REVIEW of STATICS
10) Propped Cantilever, Concentrated Load at Center
1 = V1 2 = V2
R
at = max L when 2 x when L2 x at = 4472 max x
L
< x
x
:
M
M
x
b at x = a when x < a when a < x at L < x < L + a
2
3
1
R = V = Pb L Pa R =V = L Mx = Pbx L p Pab (a + 2b) 3a(a + 2b) max = 27EIL Pa b = 3EIL a (L ; b ; x ) x = 6Pbx EIL L ; x) (2Lx ; x ; a ) x = Pa6(EIL = Pabx x 1 6EIL (L + a)
( +2 ) 3
1
1
2
2
2 2
2
2
2
2
1
21) Continuous Beam, Two Equal Spans; Concentrated Force
R =V R =V +V R =V V Mmax at x = L M 1
1
2
2
3
3
2
1
Victor Saouma
3
= = = = = =
Pb 4L ; a(L + a) 4Pa L 2 L + b ( L + a ) 2LPab ; 4L (L + a) Pa 4L + b(L + a) 4Pab L 4 L ; a ( L + a ) 4L Pab 4L (L + a) 2
3
2
3
3
3
3
2
2
2
Structural Concepts and Systems for Architects
2
Draft 5{36
REVIEW of STATICS
22) Simple Beam, Uniform Load, End Moments
+ M1 ;L M2 R1 = V1 = wL 2 M1 ; M2 R2 = V2 = wL 2 L; L M ; M Vx = w 2 ;x + 1 L 2 ;M2 at x = L2 + M1wL M wx 1 ; M2 x ; M1 Mx = 2 (L ; x) + L s 2 ; M2 2 b = L4 ; M1 +w M2 + M1 wl M1 ; 4M2 x2 x = 24wxEI x3 ; 2L + 3wL wL 12 M 8 M 4 1L 1 3 + w x + L ; w ; Mw2L 23) Simple Beam; Concentrated Force, End Moments R1 = V1 R2 = V2 at x = L2 M3 when x < L2 Mx when L2 < x Mx when x < L2 x
Victor Saouma
= = = = = =
P + M1 ; M2 P2 ; M1 ;L M2 2 ML+ M PL ; 1 2 2 4 P + M1 ; M2 x ; M 1 2 L P (L ; x) + M1 ; M2 x ; M 1 2Px L 2 2 488(EIL ;3Lx) ; 4x ; PL (M1 (2L ; x) + M2(L + x))
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{37
24) Beam Overhanging one Support; Uniform Load = V1 = 2wL (L2 ; a2 ) w 2 R2 = V2 + V3 = 2L (L + a) V2 = wa V3 = 2wL (L2 + a2 ) at 0 < x < L Vx = R1 ; wx = w(a ; x1 ) at L < x< L + a Vx1 M1 = 8wL2 (L + a)2 (L ; a)2 at x = L2 1 ; La22 R1
at 0 < x < L x at L < x < L + a x1 R1 = V1 R2 = V2 + V3 V2
at 0 < x < L at L < x< L + a 2 at x = L2 1 ; La 2 at x = L
V3 Vx Vx1 M1 M2
at 0 < x < L Mx at L < x < L + a Mx1 at 0 < x < L x at L < x < L + a x1 5.4
2
= wa2 2 2 at 0 < x < L Mx = wx 2wL (L ; a2 ; xL) = 2 (a ; x1 ) at L < x < L + a Mx1 wx (L4 ; 2L2x2 + Lx3 ; 2a2 L2 + 2a2 x2 ) 24 EIL wx1 (4a2 L; L3 + 6a2x1 ; 4ax21 + x31 ) 24 w EI 2 2 2wL (L ; a ) 2 2L (L + a) at x = L
M2
= = = = = wa = 2wL (L2 + a2 ) = R1 ; wx = w(a ; x1 ) = 8wL2 (L + a)2 (L ; a)2 2 = wa2 2 2 = wx 2wL (L ; a ; xL) = 2 (a ; x1 )2 = 24wx (L4 ; 2L2x2 + Lx3 ; 2a2 L2 + 2a2 x2 ) EIL wx1 (4a2 L; L3 + 6a2x1 ; 4ax21 + x31 ) = 24 EI
Flexure
5.4.1 Basic Kinematic Assumption; Curvature Fig.5.14 shows portion of an originally straight beam which has been bent to the radius by end couples M . support conditions, Fig. 5.1. It is assumed that plane cross-sections normal to the length of the unbent beam remain plane after the beam is bent. 62 Except for the neutral surface all other longitudinal bers either lengthen or shorten, thereby creating a longitudinal strain "x. Considering a segment EF of length dx at a distance y from the neutral axis, its original length is EF = dx = d (5.11) 61
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{38
REVIEW of STATICS O +ve Curvature, +ve bending
dθ -ve Curvature, -ve Bending
ρ M
M
Neutral Axis F’
E’
Y dA
E
F
Z
X
dx
Figure 5.14: Deformation of a Beam un Pure Bending and d 63
= dx
(5.12)
To evaluate this strain, we consider the deformed length E F 0
E0F 0
0
= ( ; y)d = d ; yd = dx ; y dx
(5.13)
The strain is now determined from: "x
=
;
E 0 F 0 EF EF
=
dx
; y dx ; dx dx
(5.14)
or after simpli cation "x
= ; y
(5.15)
where y is measured from the axis of rotation (neutral axis). Thus strains are proportional to the distance from the neutral axis. 64 (Greek letter rho) is the radius of curvature. In some textbook, the curvature (Greek letter kappa) is also used where 1 = (5.16)
thus,
"x
Victor Saouma
= ;y
(5.17)
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{39
5.4.2 Stress-Strain Relations 65 So far we considered the kinematic of the beam, yet later on we will need to consider equilibrium in terms of the stresses. Hence we need to relate strain to stress. 66 For linear elastic material Hooke's law states
x = E"x
(5.18)
where E is Young's Modulus. 67 Combining Eq. with equation 5.17 we obtain
x = ;Ey
(5.19)
5.4.3 Internal Equilibrium; Section Properties Just as external forces acting on a structure must be in equilibrium, the internal forces must also satisfy the equilibrium equations. 69 The internal forces are determined by slicing the beam. The internal forces on the \cut" section must be in equilibrium with the external forces. 68
5.4.3.1 Fx = 0; Neutral Axis The rst equation we consider is the summation of axial forces. 71 Since there are no external axial forces (unlike a column or a beam-column), the internal axial forces must be in equilibrium. Z (5.20) Fx = 0 ) x dA = 0
70
A
where x was given by Eq. 5.19, substituting we obtain Z
Z
A
x dA = ;
A
EydA = 0
(5.21-a)
But since the curvature and the modulus of elasticity E are constants, we conclude that Z
A
ydA = 0
(5.22)
or the rst moment of the cross section with respect to the z axis is zero. Hence we conclude that the
neutral axis passes through the centroid of the cross section. 5.4.3.2 M = 0; Moment of Inertia
The second equation of internal equilibrium which must be satis ed is the summation of moments. However contrarily to the summation of axial forces, we now have an external moment to account for,
72
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{40
REVIEW of STATICS
the one from the moment diagram at that particular location where the beam was sliced, hence
Z
M = ; x ydA Mz = 0; ;+ve; |{z} Ext. | A{z } Int. where dA is an di erential area a distance y from the neutral axis. 73 Substituting Eq. 5.19 9 Z M = ; x ydA = M = E Z y2 dA A ; A x = ;Ey 74
(5.24)
We now pause and de ne the section moment of inertia with respect to the z axis as
I def = and section modulus as
75
(5.23)
Z
A
y2 dA
S def = Ic
(5.25) (5.26)
Section properties for selected sections are shown in Table 5.3.
5.4.4 Beam Formula We now have the ingredients in place to derive one of the most important equations in structures, the beam formula. This formula will be extensively used for design of structural components. 77 We merely substitute Eq. 5.25 into 5.24,
76
Z M = E y2 dA Z A y2 dA I = a
9 > = M == 1 > ; EI
(5.27)
which shows that the curvature of the longitudinal axis of a beam is proportional to the bending moment M and inversely proportional to EI which we call exural rigidity. 78 Finally, inserting Eq. 5.19 above, we obtain x = ;Ey = ; My (5.28) x M I = EI Hence, for a positive y (above neutral axis), and a positive moment, we will have compressive stresses above the neutral axis. 79 Alternatively, the maximum ber stresses can be obtained by combining the preceding equation with Equation 5.26
x = ; M S
(5.29)
Example 5-16: Design Example Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{41
Y
Y
A x y Ix Iy
x
h
X y
= = = = =
x
bh
b 2 h 2 bh3 12 hb3
h h’
A x y Ix Iy
X y
12
b’ b
b
bh ; b0 h0
=
b 2 h 2 bh3 ;b0 h03 12 hb3 ;h0 b03
= = = =
12
c
Y
Y
a
h
A y Ix
X y
= = =
h(a+b) 2 h(2a+b) 3(a+b) h3 (a2 +4ab+b2 36(a+b)
x
h
X y
A x y Ix Iy
= = = = =
bh 2 b+c 3 h 3 bh3 36 bh (b2 36
; bc + c ) 2
b
b
Y
Y
r X
A Ix = Iy
= =
r42 = d4 42
t
r = d 4 64
r
X
A Ix = Iy
= =
rt = dt3 r3 t = d8 t
2
Y
b X b a
A Ix Iy
= = =
ab3 ab 3 ba3 4
a
Table 5.3: Section Properties
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{42
REVIEW of STATICS
A 20 ft long, uniformly loaded, beam is simply supported at one end, and rigidly connected at the other. The beam is composed of a steel tube with thickness t = 0:25 . Select the radius such that max 18 , and max L=360. in
ksi
1 k/ft r
0.25’
20’
Solution: wL , and I = r3 t. 1. Steel has E = 29; 000 , and from above Mmax = wL8 , max = 185 EI 2. The maximum moment will be wL2 (1) (20)2 2 = 50 Mmax = = 8 8 2
ksi
k/ft
4
(5.30)
ft
k.ft
3. We next seek a relation between maximum de ection and radius wL4 max = 185 EI I = r3 t
wL4 Er t 3 = (185)(29;000) ksi(3:14)r3 (0:=25)ft in = 65r:365
(5.31)
= rM2 t k.ft(12) in/ft = (50) (3:14)r 2 (0:25) in = 764 r2
(5.32)
=
4. Similarly for the stress = M S S = Ir I = r3 t
9 = ;
3 185 4 3 (1) k/ft(20)4 ft (12)3 in
5. We now set those two values equal to their respective maximum (12) = 0:67 = 65r:365 ) r = max = 360 = (20) 360 r 764 = 6.51 764 max = (18) = 2 ) r = r 18 L
ft
in/ft
in
ksi
in
r 3
65:65 = 4:61 0:67
in
(5.33-a) (5.33-b)
5.4.5 Approximate Analysis M = = 1 ), we recall that that the moment is directly proportional From Fig. 5.14, and Eq. 5.27 ( EI to the curvature . 81 Thus,
80
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{43
1. A positive and negative moment would correspond to positive and negative curvature respectively (adopting the sign convention shown in Fig. 5.14). 2. A zero moment correspnds to an in ection point in the de ected shape. 82
Hence, for
Statically determinate structure, we can determine the de ected shape from the moment diagram, Fig. 5.15.
Figure 5.15: Elastic Curve from the Moment Diagram
Statically indeterminate structure, we can: 1. 2. 3. 4.
Plot the de ected shape. Identify in ection points, approximate their location. Locate those in ection points on the structure, which will then become statically determinate. Perform an approximate analysis.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{44
REVIEW of STATICS
Figure 5.16: Approximate Analysis of Beams
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{45
Example 5-17: Approximate Analysis of a Statically Indeterminate beam Perform an approximate analysis of the following beam, and compare your results with the exact solution.
20k 16’ 12’ 28’
28’
Solution: 20k 16’ 12’ 28’
22’
6’
28’
28’
20k Approximate Location of IP A
B 22’
C 6’
D 28’
1. We have 3 unknowns RA , RC , and RD , all in the vertical directions, and only two applicable equations of equilibrium (since we do not have any force in the x direction), thus the problem is statically indeterminare. 2. We sketch the anticipated de ected shape, and guess the location of the in ection point. 3. At that location, we place a hinge, and we now have an additional equation of condition at that location (M = 0).
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{46
REVIEW of STATICS
4. If we consider AB, and take the moments with respect to point B: (+ ;) B = 0; (22)( A ) ; (20)(22 ; 16) = 0 ) A = 5.45 M
R
R
k
6
(5.34-a)
5. If we now consider the entire beam:
(+ ;) D = 0; ( A )(28 + 28) ; (20)(28 + 12) + ( C )(28) = 0 (5 45)(56) ; (20)(40) + ( C )(28) = 0 ) C = 17.67 k 6 (+ 6) y = 0; A ; 20 + c + D = 0 5 45 ; 20 + 17 67 + D = 0 ) D = ;3 12 3.12 k ? M
R
R
:
R
R
F
R
R
:
:
R
6. Check
R
R
:
(+ ;) A = 0; (20)(16) ; ( C )(28) + ( D )(28 + 28) = 320 ; (17 67)(28) + (3 12)(56) = p 320 ; 494 76 + 174 72 = 0 M
R
R
:
:
:
(5.36-a)
:
7. The moments are determined next M
max = M1
A a = (5:45)(16) = 87.2 = RD L = (3:12)(28) = 87.36
(5.37-a) (5.37-b)
R
8. We now compare with the exact solution from Section 5.3.2, solution 21 where: = 28, = 16, = 12, and = 20 L
b
a
P
R1
R2
R3
= A = 4 3 4 2; ( + ) 2 4(28) ; (16)(28 + 16) = 6.64 = (20)(12) 3 4(28) Pb
R
L
L
a L
a
(5.38-a)
= B = 2 3 2 2+ ( + ) 2 2(28) + 12(28 + 16) = 15.28 = (20)(16) 3 2(28)
(5.38-b)
= D = ;4 3( + ) = ; (20)(16)(12) 4(28)3 (28 + 16) = ;1.92
(5.38-d)
Pa
R
L
P ab
R
L
max =
M
M1
=
R1 a R3 L
L
b L
L
a
a
= (6 64)(16) = 106.2 = (1 92)(28) = 53.8 :
:
(5.38-c) (5.38-e) (5.38-f) (5.38-g)
9. If we tabulate the results we have
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{47
Value Approximate Exact % Error RA 5.45 6.64 18 RC 17.67 15.28 -16 RD 3.12 1.92 63 M1 87.36 53.8 62 Mmax 87.2 106.2 18 10. Whereas the correlation between the approximate and exact results is quite poor, one should not underestimate the simplicity of this method keeping in mind (an exact analysis of this structure would have been computationally much more involved). Furthermore, often one only needs a rough order of magnitude of the moments.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{48
REVIEW of STATICS
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 6
Case Study II: GEORGE WASHINGTON BRIDGE 6.1
Theory
Whereas the forces in a cable can be determined from statics alone, its con guration must be derived from its deformation. Let us consider a cable with distributed load p(x) per unit horizontal projection of the cable length (thus neglecting the weight of the cable). An in nitesimal portion of that cable can be assumed to be a straight line, Fig. 6.1 and in the absence of any horizontal load we have 1
T V
H
V θ
w
x w(x)
H
v(x)
dv
ds
v
dx L
dx
θ V+dV
H T+dT
Figure 6.1: Cable Structure Subjected to p(x) H
=constant. Summation of the vertical forces yields
(+ ?) Fy = 0 ) ;V + wdx + (V + dV ) = 0 dV + wdx = 0
(6.1-a) (6.1-b)
where V is the vertical component of the cable tension at x (Note that if the cable was subjected to its own weight then we would have wds instead of wdx). Because the cable must be tangent to T , we have V tan = H
(6.2)
Substituting into Eq. 6.1-b yields d(H tan ) + wdx
d (H tan ) = w = 0 ) ; dx
(6.3)
Draft 6{2 2
Case Study II: GEORGE WASHINGTON BRIDGE
But H is constant (no horizontal load is applied), thus, this last equation can be rewritten as d ;H dx (tan ) = w
(6.4)
dv which when substituted in Eq. 6.4 yields Written in terms of the vertical displacement v, tan = dx the governing equation for cables ;Hv = w (6.5)
3
00
For a cable subjected to a uniform load w, we can determine its shape by double integration of Eq. 6.5
4
;Hv = ;Hv =
wx + C1
0
(6.6-a)
wx2
(6.6-b) 2 + C1 x + C2 and the constants of integrations C1 and C2 can be obtained from the boundary conditions: v = 0 at x = 0 and at x = L ) C2 = 0 and C1 = ; wL 2 . Thus v
= 2wH x(L ; x)
(6.7)
This equation gives the shape v(x) in terms of the horizontal force H , L ) we can solve for the horizontal force 5 Since the maximum sag h occurs at midspan (x = 2 H
2
= wL 8h
(6.8)
we note the analogy with the maximum moment in a simply supported uniformly loaded beam M = 2 . Furthermore, this relation clearly shows that the horizontal force is inversely proportional Hh = wL 8 to the sag h, as h & H %. Finally, we can rewrite this equation as r wL H
6
h L
(6.9-a)
= 8r
(6.9-b)
def
=
Eliminating H from Eq. 6.7 and 6.8 we obtain
v
= 4h ;
x2 L2
+
x L
(6.10)
Thus the cable assumes a parabolic shape (as the moment diagram of the applied load). 7 Whereas the horizontal force H is constant throughout the cable, the tension T is not. The maximum tension occurs at the support where the vertical component is equal to V = wL 2 and the horizontal one to H , thus s 2 s 2 p 2 2 wL wL=2 + H2 = H 1 + (6.11) Tmax = V + H = 2 H
Victor Saouma
Structural Concepts and Systems for Architects
Draft 6.2 The Case Study
6{3
Combining this with Eq. 6.8 we obtain1 .
p
Tmax = H 1 + 16r2 H (1 + 8r2 )
(6.12)
8 Had we assumed a uniform load w per length of cable (rather than horizontal projection), the equation would have been one of a catenary2.
v = Hw cosh Hw L2 ; x + h
(6.13)
The cable between transmission towers is a good example of a catenary. 6.2
The Case Study
Adapted from (Billington and Mark 1983) 9 The George Washington bridge, is a suspension bridge spanning the Hudson river from New York City to New Jersey. It was completed in 1931 with a central span of 3,500 ft (at the time the world's longest span). The bridge was designed by O.H. Amman, who had emigrated from Switzerland. In 1962 the deck was sti ened with the addition of a lower deck.
6.2.1 Geometry A longitudinal and plan elevation of the bridge is shown in For simplicity we will assume in our analysis that the two approaching spans are equal to 650 ft. 11 There are two cables of three feet diameter on each side of the bridge. The centers of each pair are 9 ft apart, and the pairs themselves are 106 ft apart. We will assume a span width of 100 ft. 12 The cables are idealized as supported by rollers at the top of the towers, hence the horizontal components of the forces in each side of the cable must be equal (their vertical components will add up). 13 The cables support the road deck which is hungby suspenders attached at the cables. The cables are made of 26,474 steel wires, each 0.196 inch in diameter. They are continuous over the tower supports and are rmly anchored in both banks by huge blocks of concrete, the anchors. I 14 Because the cables are much longer than they are thick (large L ), they can be idealized a perfectly
exible members with no shear/bending resistance but with high axial strength. 15 The towers are 578 ft tall and rest on concrete caissons in the river. Because of our assumption regarding the roller support for the cables, the towers will be subjected only to axial forces. 10
6.2.2 Loads 16 The dead load is composed of the weight of the deck and the cables and is estimated at 390 and 400 psf respectively for the central and side spans respectively. Assuming an average width of 100 ft, this n n; Recalling that (a + b)n = an + nan; b + an; b + or (1 + b)n = 1 + nb + n n; b2 + n n; n; b3 + ; 1 1 + b = (1 + b) 2 1+ b Thus for b p b
t
p
and for the web:
=
F
hc
tw
p
= 41:8 p = p640y = p64036 = 107 F
5. Check the Strength by correcting the factored moment M to include the self weight. Self weight of the beam W12X22 is 22 lb./ft. or 0.022 kip/ft u
w w M M M
D u u
n
b
n
= = = = =
0:2 + 0:022 = 0:222 1:2(0:222) + 1:6(0:8) = 1:55 (1:55)(20)2=8 = 77:3 3 M = Z F = (29 3)(12) (36) = 87:9 p 0:90(87:9) = 79:1 > M k/ft
k/ft
k.ft
p
x
:
y
in
ksi
in/ft
k.ft
k.ft
u
Therefore use W12X22 section. 6. We nally check for the maximum distance between supports.
r =
rI r 5
A = 6:5 = 0:88 L = p300 r F 300 = p 0:88 = 43 y
y
p
y
y
36
Victor Saouma
ft
in
(10.18-a) (10.18-b) (10.18-c)
Structural Concepts and Systems for Architects
Draft 10{10
BRACED ROLLED STEEL BEAMS
Case 2: A572 Grade 65 Steel: 1. same as in case 1 2. same as in case 1 3. Required Zx = Mb Fuy = 076(12) :90(65) = 15:6 (14)(12) approximated by wd = 9 = 18:7. 9 4. Check compact section limits p :
3
in
Select W12X14: Zx = 17:4
in
3
Note that Zx is
= thwc = 54:3 p = p64065 = 79:4 p = p640 Fy = 2btff = 8:82 p = p65F = p6565 = 8:1 < Not Good y
In this case the controlling limit state is local buckling of the ange. Since p < < r , as above, the section is classi ed as non-compact. 5. Check the strength: Since the section is non-compact, the strength is obtained by interpolation between Mp and Mr . For the anges:
pF ; = p ; = 19:0 = Mp ; (Mp ; Mr )( ;; ) Mp
r =
141 y
10
Mn Mp = Zx Fy = Mr
141 65 10
p r p 3 (17:4) in (65) ksi
= 94:2
(12) in/ft 3 (14:9) in (65
k.ft
;10) ksi = 68:3 k.ft: = Sx (Fy ; Fr ) = (12) in/ft = 94:2 ; (94:2 ; 68:3) 198::80;;88::11 = 92:5 k.ft p
Mn b Mn = 0:90(92:5) = 83:25 k.ft > Mu
Therefore provide W12X14 section.
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 11
REINFORCED CONCRETE BEAMS 11.1
Introduction
Recalling that concrete has a tensile strength (ft ) about one tenth its compressive strength (fc ), concrete by itself is a very poor material for exural members. 2 To provide tensile resistance to concrete beams, a reinforcement must be added. Steel is almost universally used as reinforcement (longitudinal or as bers), but in poorer countries other indigenous materials have been used (such as bamboos). 3 The following lectures will focus exclusively on the exural design and analysis of reinforced concrete rectangular sections. Other concerns, such as shear, torsion, cracking, and de ections are left for subsequent ones. 4 Design of reinforced concrete structures is governed in most cases by the Building Code Requirements for Reinforced Concrete, of the American Concrete Institute (ACI-318). Some of the most relevant provisions of this code are enclosed in this set of notes. 5 We will focus on determining the amount of exural (that is longitudinal) reinforcement required at a given section. For that section, the moment which should be considered for design is the one obtained from the moment envelope at that particular point. 0
1
11.1.1 6
Notation
In R/C design, it is customary to use the following notation
0
Draft 11{2 As b c d fc fr fs ft fy h 0
0
0
0
REINFORCED CONCRETE BEAMS
Area of steel Width Distance from top of compressive bers to neutral axis Distance from the top of the compressive bers to the centroid of the reinforcement Concrete compressive strength Concrete modulus of rupture Steel stress Concrete tensile strength Steel yield stress (equivalent to Fy in AISC) Height Steel ratio, Abds
11.1.2 Modes of Failure
7 A reinforced concrete beam is a composite structure where concrete provides the compression and steel the tension. 8 Failure is initiated by, Fig. 11.5:
Steel Yielding
Concrete Crushing
Figure 11.1: Failure Modes for R/C Beams 1. Yielding of the steel when the steel stress reaches the yield stress (fs = fy ). This occurs if we do not have enough reinforcement that is the section is under-reinforced. This will result in excessive rotation and deformation prior to failure. 2. Crushing of the concrete, when the concrete strain reaches its ultimate value ("c = "u = 0:003), ACI 318: 10.2.3. This occurs if there is too much reinforcement that is the section is overreinforced. This is a sudden mode of failure. Ideally in an optimal (i.e. most ecient use of materials) design, a section should be dimensioned such that crushing of concrete should occur simultaneously with steel yielding. This would then be a balanced design. 10 However since concrete crushing is a sudden mode of failure with no prior warning, whereas steel yielding is often accompanied by excessive deformation (thus providing ample warning of an imminent failure), design codes require the section to be moderately under-reinforced. 9
11.1.3 Analysis vs Design 11
In R/C we always consider one of the following problems:
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11.1 Introduction
11{3
Analysis: Given a certain design, determine what is the maximum moment which can be applied. Design: Given an external moment to be resisted, determine cross sectional dimensions (b and h) as well as reinforcement (As ). Note that in many cases the external dimensions of the beam (b and h) are xed by the architect.
12
We often consider the maximum moment along a member, and design accordingly.
11.1.4 Basic Relations and Assumptions In developing a design/analysis method for reinforced concrete, the following basic relations will be used, Fig. ??:
13
Compatibility
Equilibrium
C d
εy
T
T=C M_ext=Cd Figure 11.2: Internal Equilibrium in a R/C Beam 1. Equilibrium: of forces and moment at the cross section. 1) Fx = 0 or Tension in the reinforcement = Compression in concrete; and 2) M = 0 or external moment (that is the one obtained from the moment envelope) equal and opposite to the internal one (tension in steel and compression of the concrete). 2. Material Stress Strain: We recall that all normal strength concrete have a failure strain u = :003 in compression irrespective of fc . 0
14
Basic assumptions used:
Compatibility of Displacements: Perfect bond between steel and concrete (no slip). Note that those
two materials do also have very close coecients of thermal expansion under normal temperature. Plane section remain plane ) strain is proportional to distance from neutral axis.
11.1.5 ACI Code The ACI code is based on limit strength, or Mn Mu thus a similar design philosophy is used as the one adopted by the LRFD method of the AISC code, ACI-318: 8.1.1; 9.3.1; 9.3.2 16 The required strength is based on (ACI-318: 9.2)
15
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11{4
REINFORCED CONCRETE BEAMS U
= 14 +17 (11.1) = 0 75(1 4 + 1 7 + 1 7 ) (11.2) : D :
: L
: D
: L
: W
11.2 Cracked Section, Ultimate Strength Design Method 11.2.1
Equivalent Stress Block
17 In determining the limit state moment of a cross section, we consider Fig. 11.3. Whereas the strain distribution is linear (ACI-318 10.2.2), the stress distribution is non-linear because the stress-strain curve of concrete is itself non-linear beyond 0 5 c. 18 Thus we have two alternatives to investigate the moment carrying capacity of the section, ACI-318: : f
0
10.2.6
1. Use the non-linear stress distribution. 2. Use a simpler equivalent stress distribution. 19 The ACI code follows the second approach. Thus we seek an equivalent stress distribution such that: 1. The resultant force is equal 2. The location of the resultant is the same. We note that this is similar to the approach followed in determining reactions in a beam subjected to a distributed load when the load is replaced by a single force placed at the centroid.
Figure 11.3: Cracked Section, Limit State
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11.2 Cracked Section, Ultimate Strength Design Method 20
11{5
It was shown that the depth of the equivalent stress block is a function of fc : 0
1 = :85 if f 4; 000 1 if 4c; 000 < f < 8; 000 = :85 ; (:05)(fc ; 4; 000) 1;000 c 0
0
0
(11.3)
Figure 11.4: Whitney Stress Block 11.2.2
Balanced Steel Ratio
Next we seek to determine the balanced steel ratio b such that failure occurs by simultaneous yielding of the steel fs = fy and crushing of the concrete "c = 0:003, ACI-318: 10.3.2 We will separately consider the two failure possibilities: Tension Failure: we stipulate that the steel stress is equal to fy : 21
= Abds As fy = :85fcab = :85fcb 1 c 0
) c = 0:85ffy d c 1
0
0
(11.4)
Compression Failure: where the concrete strain is equal to the ultimate strain; From the strain diagram
"c = 0:003
:003 c d = :003+"s
22
)c=
:003
d fs Es + :003
(11.5)
Balanced Design is obtained by equating Eq. 11.4 to Eq. 11.5 and by replacing by b and fs by fy :
9 > = b fy d = fy :003 d > : 85 f ; c 1 Es + :003
fy :003 0:85fc 1 d = Efss +:003 d fs = fy = b 0
When we replace Es by 29; 000
ksi
0
we obtain
; 000 b = :85 1 ffc 87;87000 +f 0
y
y
(11.6)
This b corresponds to the only combination of b, d and As which will result in simultaneous yielding of the steel and crushing of the concrete, that is an optimal design.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11{6
REINFORCED CONCRETE BEAMS
23 Because we need to have ample warning against failure, hence we prefer to have an under-reinforced section. Thus, the ACI code stipulates:
(11.7)
< :75b
In practice, depending on the relative cost of steel/concrete and of labour it is common to select lower values of . If < 0:5b (thus we will have a deeper section) then we need not check for de ection. 25 A minimum amount of reinforcement must always be used to prevent temperature and shrinkage cracks: (11.8) 200 24
min
26
fy
The ACI code adopts the limit state design method (11.9)
MD = Mn > Mu = b = 0:9
11.2.3 Analysis
Given As , b, d, fc , and fy determine the design moment: 1. act = Abds 0
87 2. b = (:85) 1 ffyc 87+ fy 0
3. If act < b (that is failure is triggered by yielding of the steel, fs = fy ) )
a = :A85sffcyb ;From Equilibrium MD = As fy d ; 0:59 Afsfby a MD = As fy d ; 2 c | {z
0
0
Mn
}
Combining this last equation with = Abds yields
MD = fy bd2 1 ; :59 ffy c 0
(11.10)
4. y If act > b is not allowed by the code as this would be an over-reinforced section which would fail with no prior warning. However, if such a section exists, and we need to determine its moment carrying capacity, then we have two unknowns: (a) Steel strain "s (which was equal to "y in the previous case) (b) Location of the neutral axis c. We have two equations to solve this problem Equilibrium: of forces c = :85Afs fb s (11.11) c 1 0
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11.2 Cracked Section, Ultimate Strength Design Method
11{7
Strain compatibility: since we know that at failure the maximum compressive strain "c is equal to 0.003. Thus from similar triangles we have
c = :003 d :003 + "s
(11.12)
Those two equations can be solved by either one of two methods: (a) Substitute into one single equation (b) By iteration Once c and fs = E"s are determined then
1c MD = As fs d ; 2 11.2.4
(11.13)
Design
27 We distinguish between two cases. The rst one has dimensions as well as steel area unknown, the second has the dimensions known (usually speci ed by the architect or by other constraints), and we seek As .
b, d and As unknowns and MD known:
1. We start by assuming , at most = :75b, and if de ection is of a concern (or steel too expensive), then we can select = 0:5b with b determined from Eq. 11.6
; 000 = 0:75 :85 1 ffc 87;87000 +f 0
y
2. From Eq. 11.10
y
MD = fy 1 ; :59 ffy bd2 | {z c } 0
or
R
f y R = fy 1 ; :59 f c 0
(11.14)
which does not depend on unknown quantities1 . Then solve for bd2 : D bd2 = M R
(11.15)
3. solve for b and d (this will require either an assumption on one of the two, or on their ratio). 4. As = bd 1
Note analogy with Eq. 10.6 Mp = Fy Z for stell beams.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11{8
REINFORCED CONCRETE BEAMS
y b, d and Md known, As unknown: In this case there is no assurance that we can have a design with b . If the section is too small, then it will require too much steel resulting in an over-reinforced section. We will again have an iterative approach 1. Since we do not know if the steel will be yielding or not, use fs . 2. Assume an initial value for a (a good start is a = d5 ) 3. Assume initially that fs = fy 4. Check equilibrium of moments (M = 0)
D ; As = f M s d ; a2 5. Check equilibrium of forces in the x direction (Fx = 0) a = :A85sffsb c 6. Check assumption of fs from the strain diagram "s d;c :003 d ; c = c ) fs = Es c :003 < fy where c = a1 .
(11.16) (11.17)
0
(11.18)
7. Iterate until convergence is reached.
Example 11-21: Ultimate Strength Capacity Determine the ultimate Strength of a beam with the following properties: b = 10 , d = 23 , 2 , fc = 4; 000 and fy = 60 .
As = 2:35
Solution:
in
in
0
psi
2:35 = :0102 = Abds = (10)(23) p f 87 87 = :02885 > act = :85 1 fyc 87+fy = (:85)(:85) 604 87+60 :35)(60) = 4:147 = :A85sffcyb = (:(285)(4)(10) ) = 2; 950 = (2:35)(60)(23 ; 4:147 2 = Mn = (:9)(2; 950) = 2; 660 Note that from the strain diagram c = 0:a85 = 40::414 85 = 4:87 Alternative solution Mn = act fy bd 2 1 ; :59actffyc = As fy d 1 ; :59act ffyc = 245 = (2:35)(60)(23) 1 ; (:59) 604 (:01021) = 2; 950 MD = Mn = (:9)(2; 950) = 2; 660
act b a Mn MD
in
ksi
0
in
0
k.in
k.in
in
0
0
k.in
k.ft
k.in
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11.2 Cracked Section, Ultimate Strength Design Method
11{9
Example 11-22: Beam Design I Design a 15 ft beam to support a dead load of 1.27 k/ft, a live load of 2.44 k/ft using a 3,000 psi concrete and 40 ksi steel. Neglect beam weight
Solution:
w = 1:4(1:27) + 1:7(22 :44) = 5:92 2 M = (5 92) 8 (15) (12) = 2; 000 = :85 1 yc 87+87 y 87 = 0:040 = (:85)(:85) 403 87+40 = :75 = :75(0:04) = 0:030 yc R = f 1 ; :59 = (0:03)(40) 1: ; (0:59)(02:03) 403 = 0:917 2 000 = 2; 423 3 bd2 = d = (0 9)(0 917) k/ft
u
k/ft
:
D
f
b
ft
in/ft
k.in
0
f
f
b
f
y
f0
;
M
R
:
q
k.in in
:
ksi
in
k
Assume b = 10 , this will give d = 2 10423 = 15:57 . We thus adopt b = 10 Finally, A = bd = (0:030)(10)(16) = 4:80 2 we select 3 bars No. 11 ;
in
in
and d = 16
in
in
.
in
s
Example 11-23: Beam Design II Select the reinforcement for a cross section with b = 11:5 ; d = 20 using f = 3 ; and f = 40
M = 1; 600
Solution: d
in
0
k.in
ksi
c
in
to support a design moment
ksi
y
1. Assume a = 5 = 205 = 4 and f = f 2. Equilibrium of moments: d
in
s
y
1; 600 A = f M = 2:47 == (d ; 2 ) (:9)(40) (20 ; 42 ) k.in
D
s
a
y
in
2
in
ksi
3. Check equilibrium of forces:
:47) 2 (40) a = :A85ff b = (:(2 85)(3) (11:5) = 3:38 s
y
0
c
in
ksi
ksi
in
in
4. We originally assumed a = 4, at the end of this rst iteration a = 3:38, let us iterate again with a = 3:30
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11{10
REINFORCED CONCRETE BEAMS
5. Equilibrium of moments:
1; 600 D As = f M (d ; a ) == (:9)(40) (20 ; 3:3 ) = 2:42 k.in
y
ksi
2
2
in
2
in
6. Check equilibrium of forces:
p :42) 2 (40) = 3:3 a = :A85sffyb = (:(2 85)(3) (11 : 5) c in
ksi
in
0
ksi
in
7. we have converged on a. :42 = :011 8. Actual is act = (112:5)(20) 9. b is equal to 3 87 = :037 b = :85 1 ffc 87 87 = ( : 85)( : 85) +f 40 87 + 40 0
y
y
p
10. max = :75 = (0:75)(0:037) = :0278 > 0:011 thus fs = fy and we use As = 2:42
in
2
11.3 Continuous Beams Whereas coverage of continuous reinforced concrete beams is beyond the scope of this course, Fig. ?? illustrates a typical reinforcement in such a beam.
28
11.4 ACI Code Attached is an unauthorized copy of some of the most relevant ACI-318-89 design code provisions. 8.1.1 - In design of reinforced concrete structures, members shall be proportioned for adequate strength in accordance with provisions of this code, using load factors and strength reduction factors speci ed in Chapter 9. 8.3.1 - All members of frames or continuous construction shall be designed for the maximum e ects of factored loads as determined by the theory of elastic analysis, except as modi ed according to Section 8.4. Simplifying assumptions of Section 8.6 through 8.9 may be used. p 8.5.1 - Modulus of elasticity Ec for concrete may be taken as Wc1:533 fc ( psi) forpvalues of Wc between 90 and 155 lb per cu ft. For normal weight concrete, Ec may be taken as 57; 000 fc. 8.5.2 - Modulus of elasticity Es for non-prestressed reinforcement may be taken as 29,000 psi. 9.1.1 - Structures and structural members shall be designed to have design strengths at all sections at least equal to the required strengths calculated for the factored loads and forces in such combinations as are stipulated in this code. 9.2 - Required Strength 9.2.1 - Required strength U to resist dead load D and live load L shall be at least equal to U = 1:4D + 1:7L 9.2.2 - If resistance to structural e ects of a speci ed wind load W are included in design, the following combinations of D, L, and W shall be investigated to determine the greatest required strength U U = 0:75(1:4D + 1:7L + 1:7W ) 0
0
Victor Saouma
Structural Concepts and Systems for Architects
Victor Saouma Stirrups
Straight top bar
(d) Straight and bent bar reinforcement
Exterior span
Bent bar at noncontinuous end
Straight bottom bar
No.3 stirrup support if necessary
(c) Straight bar reinforcement
Exterior span
Straight bottom bar
Stirrups
(b) Moment diagram under typical loading
(a) Deflected shape
Interior span
Interior column
Interior span
Interior column
Points of deflection
Straight bar
Bent bars
Cracks
Cracks
Interior span
Interior span
Reinforcement
Bottom bars
Bottom bars
Bent bar
Section through beam
Stirrups
Top bars
Section through beam
Stirrups
Top bars
Draft 11.4 ACI Code 11{11
Figure 11.5: Reinforcement in Continuous R/C Beams
Structural Concepts and Systems for Architects
Draft 11{12
REINFORCED CONCRETE BEAMS
where load combinations shall include both full value and zero value of L to determine the more severe condition, and U = 0:9D + 1:3W but for any combination of D, L, and W, required strength U shall not be less than Eq. (9-1). 9.3.1 - Design strength provided by a member, its connections to other members, and its cross sections, in terms of exure, axial load, shear, and torsion, shall be taken as the nominal strength calculated in accordance with requirements and assumptions of this code, multiplied by a strength reduction factor . 9.3.2 - Strength reduction factor shall be as follows: 9.3.2.1 - Flexure, without axial load 0.90 9.4 - Design strength for reinforcement Designs shall not be based on a yield strength of reinforcement fy in excess of 80,000 psi, except for prestressing tendons. 10.2.2 - Strain in reinforcement and concrete shall be assumed directly proportional to the distance from the neutral axis, except, for deep exural members with overall depth to clear span ratios greater than 2/5 for continuous spans and 4/5 for simple spans, a non-linear distribution of strain shall be considered. See Section 10.7. 10.2.3 - Maximum usable strain at extreme concrete compression ber shall be assumed equal to 0.003. 10.2.4 - Stress in reinforcement below speci ed yield strength fy for grade of reinforcement used shall be taken as Es times steel strain. For strains greater than that corresponding to fy , stress in reinforcement shall be considered independent of strain and equal to fy . 10.2.5 - Tensile strength of concrete shall be neglected in exural calculations of reinforced concrete, except when meeting requirements of Section 18.4. 10.2.6 - Relationship between concrete compressive stress distribution and concrete strain may be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction of strength in substantial agreement with results of comprehensive tests. 10.2.7 - Requirements of Section 10.2.5 may be considered satis ed by an equivalent rectangular concrete stress distribution de ned by the following: 10.2.7.1 - Concrete stress of 0:85fc shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross section and a straight line located parallel to the neutral axis at a distance (a = 1 c) from the ber of maximum compressive strain. 10.2.7.2 - Distance c from ber of maximum strain to the neutral axis shall be measured in a direction perpendicular to that axis. 10.2.7.3 - Factor 1 shall be taken as 0.85 for concrete strengths fc up to and including 4,000 psi. For strengths above 4,000 psi, 1 shall be reduced continuously at a rate of 0.05 for each 1000 psi of strength in excess of 4,000 psi, but 1 shall not be taken less than 0.65. 10.3.2 - Balanced strain conditions exist at a cross section when tension reinforcement reaches the strain corresponding to its speci ed yield strength fy just as concrete in compression reaches its assumed ultimate strain of 0.003. 10.3.3 - For exural members, and for members subject to combined exure and compressive axial load when the design axial load strength (Pn ) is less than the smaller of (0:10fcAg ) or (Pb ), the ratio of reinforcement p provided shall not exceed 0.75 of the ratio b that would produce balanced strain conditions for the section under exure without axial load. For members with compression reinforcement, the portion of b equalized by compression reinforcement need not be reduced by the 0.75 factor. 10.3.4 - Compression reinforcement in conjunction with additional tension reinforcement may be used to increase the strength of exural members. 10.5.1 - At any section of a exural member, except as provided in Sections 10.5.2 and 10.5.3, where positive reinforcement is required by analysis, the ratio provided shall not be less than that given by = 200 0
0
0
min
Victor Saouma
fy
Structural Concepts and Systems for Architects
Draft Chapter 12
PRESTRESSED CONCRETE 12.1
Introduction
Beams with longer spans are architecturally more appealing than those with short ones. However, for a reinforced concrete beam to span long distances, it would have to have to be relatively deep (and at some point the self weight may become too large relative to the live load), or higher grade steel and concrete must be used. 2 However, if we were to use a steel with fy much higher than 60 ksi in reinforced concrete (R/C), then to take full advantage of this higher yield stress while maintaining full bond between concrete and steel, will result in unacceptably wide crack widths. Large crack widths will in turn result in corrosion of the rebars and poor protection against re. 3 One way to control the concrete cracking and reduce the tensile stresses in a beam is to prestress the beam by applying an initial state of stress which is opposite to the one which will be induced by the load. 4 For a simply supported beam, we would then seek to apply an initial tensile stress at the top and compressive stress at the bottom. In prestressed concrete (P/C) this can be achieved through prestressing of a tendon placed below the elastic neutral axis. 5 Main advantages of P/C: Economy, de ection & crack control, durability, fatigue strength, longer spans. 6 There two type of Prestressed Concrete beams: Pretensioning: Steel is rst stressed, concrete is then poured around the stressed bars. When enough concrete strength has been reached the steel restraints are released, Fig. 12.1. Postensioning: Concrete is rst poured, then when enough strength has been reached a steel cable is passed thru a hollow core inside and stressed, Fig. 12.2.
1
12.1.1
Materials
P/C beams usually have higher compressive strength than R/C. Prestressed beams can have fc as high as 8,000 psi. 8 The importance of high yield stress for the steel is illustrated by the following simple example. If we consider the following:
7
0
Draft 12{2
PRESTRESSED CONCRETE
Vertical bulkhead
Harping hold-up point
Harping hold-down point Jacks
Anchorage
Prestressing bed slab
Continuous tendon
Precast Concrete element Tendon anchorage
Jacks
Support force
Casting bed
Jacks
Tendon
Casting bed
Hold-down force
Figure 12.1: Pretensioned Prestressed Concrete Beam, (Nilson 1978)
Anchorage
Anchorage
Intermediate diaphragms
Jack
Beam
Jack
Tendon in conduct
Anchorage Jack
Slab
Wrapped tendon
Figure 12.2: Posttensioned Prestressed Concrete Beam, (Nilson 1978)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12.1 Introduction
12{3
1. An unstressed steel cable of length l 2. A concrete beam of length l 3. Prestress the beam with the cable, resulting in a stressed length of concrete and steel equal to l0 = l0 . 4. Due to shrinkage and creep, there will be a change in length s
c
s
c
l = (" + " )l c
sh
cr
(12.1)
c
we want to make sure that this amout of deformation is substantially smaller than the stretch of the steel (for prestressing to be e ective). 5. Assuming ordinary steel: f = 30 , E = 29; 000 , " = 2930000 = 1:03 10;3 = ksi
s
ksi
s
s
in
;
in
6. The total steel elongation is " l = 1:03 10;3l 7. The creep and shrinkage strains are about " + " ' :9 10;3 8. The residual stres which is left in the steel after creep and shrinkage took place is thus s s
s
cr
sh
(1:03 ; :90) 10;3(29 103) = 4
(12.2)
ksi
Thus the total loss is 3030;4 = 87% which is unacceptably too high. 9. Alternatively if initial stress was 150 after losses we would be left with 124 10. Note that the actual loss is (:90 10;3)(29 103 ) = 26 in each case ksi
ksi
or a 17% loss.
ksi
Having shown that losses would be too high for low strength steel, we will use Strands usually composed of 7 wires. Grade 250 or 270 ksi, Fig. 12.3.
9
0000000 1111111 0000000 1111111 0000000 0000000 1111111 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 0000000 1111111 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 000000 111111 0000000 0000000 1111111 1111111 000000 111111 000000 111111 000000 111111 0000000 1111111 0000000 1111111 000000 111111 000000 111111 000000 111111 0000000 1111111 0000000 1111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 111111 1111111 000000 0000000 000000 111111 000000 111111 000000 111111 111111 1111111 000000 0000000 000000 111111 000000 111111 000000 111111 0000000 1111111 111111 1111111 000000 0000000 111111 1111111 000000 0000000 000000 111111 0000000 1111111 111111 1111111 000000 0000000 111111 1111111 000000 0000000 111111 1111111 000000 0000000 111111 1111111 000000 0000000 111111 000000 0000000 1111111
Figure 12.3: 7 Wire Prestressing Tendon
Tendon have diameters ranging from 1/2 to 1 3/8 of an inch. Grade 145 or 160 ksi. Wires come in bundles of 8 to 52. Note that yield stress is not well de ned for steel used in prestressed concrete, usually we take 1% strain as e ective yield. 10 Steel relaxation is the reduction in stress at constant strain (as opposed to creep which is reduction of strain at constant stress) occrs. Relaxation occurs inde nitely and produces signi cant prestress loss. If we denote by f the nal stress after t hours, f the initial stress, and f the yield stress, then p
pi
fp fpi
Victor Saouma
log t f
= 1 ; 10
pi
fpy
; :55
py
(12.3)
Structural Concepts and Systems for Architects
Draft 12{4
PRESTRESSED CONCRETE
12.1.2 Prestressing Forces 11
Prestress force \varies" with time, so we must recognize 3 stages: 1. j Jacking force. But then due to (a) friction and anchorage slip in post-tension (b) elastic shortening in pretension is reduced to: 2. i Initial prestress force; But then due to time dependent losses caused by (a) relaxation of steel (b) shrinkage of concrete (c) creep of concrete is reduced to: 3. e E ective force P
P
P
12.1.3 Assumptions 12
The following assumptions are made; 1. Materials are both in the elastic range 2. section is uncracked 3. sign convention: + tension, ; compression 4. Subscript 1 refers to the top and 2 to the bottom I I 5. 2 = c2 1 = c1 6. + if downward from concrete neutral axis ve
I; S
e
ve
; S
ve
12.1.4 Tendon Con guration Through proper arrangement of the tendon (eccentricity at both support and midspan) various internal
exural stress distribution can be obtained, Fig. 12.4.
13
12.1.5 Equivalent Load An equivalent load for prestressing can be usually determined from the tendon con guration and the prestressing force, Fig. 12.5.
14
12.1.6 Load Deformation 15
The load-deformation curve for a prestressed concrete beam is illustrated in Fig. 12.6.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12.2 Flexural Stresses
12{5 W 000 111 111 000 000 111 000 111 000 111
h
f’y
Q P
P
h/2
fc 00 11 11 00 00 11 00 11 00 11 fc
+
fc 000 111 111 000 000 111 000 111 000 111 fc =f t
=
2f c 00 11 11 00 00 11 00 11 00 11
2Q
P
P
2h/3
2Q P
P
h/2 h/3
2f c 0 11 00 0000000 1111111 00 11 0000000 1111111 11 1111111 00 +1111111 0000000 = 00 11 0000000 00 1111111 11 0000000 2f c 2f =2f t c
2f c 11 00 00 11 11 00 00 11 00 11 0
0 2f c 11 1111111 00 0000000 00 11 0000000 00 +1111111 11 0000000 = 1111111 00 11 0000000 1111111 00 1111111 11 0000000 2f c 2f t =2f c fc Midspan 000 111 000 111 + = 0 000 111 000 111 000 111 Ends fc
2f c 11 00 00 11 00 11 00 11 00 11 0 fc 00 11 00 11 00 11 00 11 00 11 fc
0 00 11 00 11 00 11 00 11 00 11 2f c
Q P
P
h/2 h/3
fc 11 00 00 11 00 11 00 11 00 11 fc
+
f c 111 000 000 111 000 111 000 111 000 111 ft =f c
=
fc
Midspan +
0
fc 11 00 00 11 00 11 00 11 00 11
=
Ends
fc 11 00 00 11 00 11 00 11 00 11 fc
Figure 12.4: Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978) 12.2
Flexural Stresses
We now identify the following 4 stages: Initial Stage when the beam is being prestressed (recalling that r2 = AIc 1. the prestressing force, Pi only
16
f1 = ; APi + PiIec1 = ; APi 1 ; ecr21 (12.4) c c P P ec P i i 2 i f2 = ; A ; I = ; A 1 + ecr22 (12.5) c
c
2. Pi and the self weight of the beam M0 (which has to be acconted for the moment the beam cambers due to prestressing) 0 f1 = ; APi 1 ; ecr21 ; M S1 (12.6) c 0 f2 = ; APi 1 + ecr22 + M S (12.7) c
2
Service Load when the prestressing force was reduced from Pi to Pe beacause of the losses, and the actual service (not factored) load is apllied 3. Pe and M0
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12{6
PRESTRESSED CONCRETE Member
Equivalent load on concrete from tendon Moment from prestressing
(a) P
P sin θ
P
θ
P sin θ
P cos θ
P cos θ 2 P sin θ
(b) P
P sin θ
P
θ
P sin θ
P cos θ
P cos θ
(c) Pe P
Pe P
P
e
P
(d) P
P θ M
P sin θ
P sin θ
M
e P cos θ
P cos θ
(e) P P
P
P sin θ
P sin θ None
P cos θ
P cos θ 2 P sin θ
(f) P P
None
(g) P
P
Figure 12.5: Determination of Equivalent Loads Load
Ultimate Steel yielding Service load limit including tolerable overload
Ru
ptu
re
Overload
Tn
Service load range
First cracking load fcr
Decompression
or higher cgs (f=0)
Balanced Full dead load
∆o ∆D
∆ pe ∆ pi
∆L
Deformation ∆ (deflection of camber) ∆ pi= Initial prestress camber ∆ pe= Effective prestress camber ∆ O= Self-weight deflection ∆ D= Dead load deflection ∆ L= Live load deflection
Figure 12.6: Load-De ection Curve and Corresponding Internal Flexural Stresses for a Typical Prestressed Concrete Beam, (Nilson 1978)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12.2 Flexural Stresses
12{7 0 f1 = ; APe 1 ; ecr21 ; M S1 (12.8) c 0 f2 = ; APe 1 + ecr22 + M S (12.9)
c
2
4. Pe and M0 + MDL + MLL f1 = ; APe 1 ; ecr21 ; M0 + MSDL + MLL (12.10) c 1 M0 + MDL + MLL ec P 2 e f2 = ; A 1 + r2 + (12.11) S2 c
The internal stress distribution at each one of those four stages is illustrated by Fig. 12.7. Pi Ac
c1 e
Pi e c 1 Ic
11 00 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11
c2
11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111
Pi Ac
Stage 1
Pi (1Ac
e c1 ) r2
-
Pi (1+ Ac
Pe (1Ac
Stage 4
e c2 ) r2
e c1 Mo )r2 S1
000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111
Pe (1+ Ac
Mo S1
111 000 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111
+
+
Pi (1Ac
Md + Ml S1
Md + Ml S2
e c2 ) r2
e c1 Mo )r2 S1
000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111
Pi (1+ Ac
00000000000 11111111111 11111111111 00000000000 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111
e c2 Mo )+ r2 S2
Pi (1+ Ac
Mo S2
-
e c1 ) r2
000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111
Pi e c 2 Ic
111111111 000000000 000000000 111111111 000000000 111111111 111111111 000000000 000000000 111111111 111111111 000000000 000000000 111111111 111111111 000000000 000000000 111111111 111111111 000000000 000000000 111111111 000000000 111111111 111111111 000000000 000000000 111111111 111111111 000000000 000000000 111111111
Stage 2
Pi (1Ac
Pe (1Ac
e c2 Mo )+ r2 S2
e c1 Mt )r2 S1
111111 000000 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 111111 000000 000000 111111 000000 111111
Pe (1+ Ac
e c2 Mt )+ r2 S2
Figure 12.7: Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum Moment Section and Support Section, (Nilson 1978) 17
Those (service) exural stresses must be below those speci ed by the ACI code (where the subscripts
c, t, i and s refer to compression, tension, initial and service respectively):
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12{8
PRESTRESSED CONCRETE
fci fti fcs fts
permitted concrete compression stress at initial stage :60fp ci permitted concrete tensile stress at initial stage < 3 fci permitted concrete compressive stress at service stage :45 pfc p permitted concrete p tensile stress at initial stage 6 fc or 12 fc Note that fts can reach 12 fc only if appropriate de ection analysis is done, because section would be cracked. 18 Based on the above, we identify two types of prestressing: Full prestressing (pioneered by Freysinet), no tensile stresses, no crack, but there are some problems with excessive camber when unloaded. Partial prestressing (pioneered by Leonhardt, Abeles, Thurliman), cracks are allowed to occur (just as in R/C), and they are easier to control in P/C than in R/C. 0
0
0
0
0
0
19 The ACi code imposes the following limits on the steel stresses in terms of fpu which is the ultimate strength of the cable: Pj < :80fpuAs and Pi < :70fpuAs . No limits are speci ed for Pe .
Example 12-24: Prestressed Concrete I Beam Adapted from (Nilson 1978)
The following I Beam has fc = 4; 000 , L = 40 ft, DL+LL =0.55 k/ft, concrete density = 150 lb/ft3 and multiple 7 wire strands with constant eccentricity e = 5:19 . Pi = 169 , and the total losses due to creep, shinkage, relaxation are 15%. 0
psi
in
k
12" 4" 5"
2"
7"
6"
4"
24" 6"
7"
2" 5"
4"
The section properties for this beam are Ic = 12; 000 4 , Ac = 176 2 , S1 = S2 = 1; 000 3 , r = AI = 68:2 2 . Determine exural stresses at midspan and at support at initial and nal conditions. Solution: 2
in
in
in
in
1. Prestressing force, Pi only
f1 = ; APi 1 ; ecr21 c 169 ; 000 1 ; (5:19)(12) = ;83 = ; 176 68:2 ec P f2 = ; Ai 1 + r22 c (5 : 19)(12) 169 ; 000 = 1; 837 = ; 176 1 + 68:21
(12.12-a)
psi
Victor Saouma
(12.12-b) (12.12-c)
psi
(12.12-d)
Structural Concepts and Systems for Architects
Draft 12.2 Flexural Stresses
12{9
2. P and the self weight of the beam M0 (which has to be acconted for the moment the beam cambers due to prestressing) (176) 2 (:150) = 3 = :183 (12.13-a) w0 = (144) 2= 2 2 M0 = (:183)(40) = 36:6 (12.13-b) 8 The exural stresses will thus be equal to: ; 000) = 439 f1 20 = SM0 = (36:6)(12 (12.14) 1 ; 000 12 i
in
k
in
ft
k/ft
ft
k.ft
w ;
psi
;
0 f1 = ; AP 1 ; ecr21 ; M S1 = ;83 ; 439 = ;522 p f = 3 f = +190p 0 f2 = ; AP 1 + ecr22 + M S2 = ;1; 837 + 439 = ;1; 398 f = :6f = ;2; 400p
(12.15-a)
i
c
(12.15-b) (12.15-c) (12.15-d)
psi
0
ti
c
i
c
(12.15-e) (12.15-f)
psi
0
ci
c
3. P and M0. If we have 15% losses, then the e ective force P is equal to (1 ; 0:15)169 = 144 e
e
0 f1 = ; AP 1 ; ecr21 ; M S 1 ; 000 1 ; (5:19)(12) ; 439 = ; 144176 68:2 e
c
= ;71 ; 439 = ;510
ec2
psi
0 f2 = ; AP 1 + r2 + M S 2 (5 : 19)(12) 144 ; 000 + 439 = ; 176 1 + 68:2 e
c
= ;1; 561 + 439 = ;1; 122
psi
k
(12.16-a) (12.16-b) (12.16-c) (12.16-d) (12.16-e) (12.16-f)
note that ;71 and ;1; 561 are respectively equal to (0:85)(;83) and (0:85)(;1; 837) respectively. 4. P and M0 + M + M 2 = 110 (12.17) M + M = (0:55)(40) 8 and corresponding stresses ; 000) = 1; 320 f1 2 = (110)(12 (12.18) 1; 000 Thus, f = ; P 1 ; ec1 ; M0 + M + M (12.19-a) e
DL
LL
DL
k.ft
LL
psi
;
1
Victor Saouma
e
A
c
r2
DL
S1
LL
Structural Concepts and Systems for Architects
Draft 12{10
PRESTRESSED CONCRETE = ;510 ; 1; 320 = ;1; 830 fcs = :45fc = ;2; 700p f2 = ; APe 1 + ecr22 + M0 + MSDL + MLL
(12.19-b) (12.19-c) (12.19-d)
psi
0
c
= ;1; 122 + 1; 320 = +198 p fts = 6 fc = +380p
2
(12.19-e) (12.19-f)
psi
0
4
-1122
+198
12.3
2
1 -1398
3
-1837
-83 -510 -522
-1830
5. The stress distribution at each one of the four stages is shown below.
Case Study: Walnut Lane Bridge
Adapted from (Billington and Mark 1983)
The historical Walnut Lane Bridge ( rst major prestressed concrete bridge in the USA) is made of three spans, two side ones with lengths of 74 ft and a middle one of length 160 feet. Thirteen prestressed cocnrete beams are placed side by side to make up a total width of 44 fet of roadway and two 9.25 feet of sidewalk. In between the beams, and cast with them, are transverse sti eners which connect the beams laterally, Fig. 12.8
20
12.3.1 21
Cross-Section Properties
The beam cross section is shown in Fig. 12.9 and is simpli ed
Ac = 2(8" :9)(52) + (7)(61:2) = 1; 354 2 # 2 3 3 I = 2 (52)(8:9) + (52)(8:9) 79 ; 8:9 + (7)(61:2) in
12 = = 1; 277 103 = h2 = 79 2 = 39:5
2
in
c1 = c2
Victor Saouma
4
in
2
12
(12.20-a) (12.20-b) (12.20-c) (12.20-d)
Structural Concepts and Systems for Architects
Draft 12.3 Case Study: Walnut Lane Bridge
12{11
80 ft CENTER LINE
ELEVATION OF BEAM HALF
9.25’
44 ’
ROAD
9.25’
SIDEWALK
BEAM CROSS SECTIONS
TRANSVERSE DIAPHRAGMS
CROSS - SECTION OF BRIDGE
52" 10" 3" 7"
TRANSVERSE DIAPHRAGM 10"
7"
3’-3"
6’-7" SLOTS FOR CABLES
6 1/2" 3 1/2" 7" 30"
CROSS - SECTION OF BEAM
Figure 12.8: Walnut Lane Bridge, Plan View
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12{12
PRESTRESSED CONCRETE 52"
8.9"
22.5"
7"
22.5" 6’-7" = 79"
61.2"
8.9"
SIMPLIFIED CROSS - SECTION OF BEAM
Figure 12.9: Walnut Lane Bridge, Cross Section
103 = 32; 329 S1 = S2 = Ic = 1; 277 39:5 1 ; 277 103 = 943: 2 I r2 = A = 1; 354
in
(12.20-e)
3
(12.20-f)
in
12.3.2
Prestressing
Each beam is prestressed by two middle parabolic cables, and two outer horizontal ones along the
anges. All four have approximately the same eccentricity at midspan of 2.65 ft. or 31.8 inch. 23 Each prestressing cable is made up 64 wires each with a diameter of 0.27 inches. Thus the total area of prestressing steel is given by: 2 2 Awire = (d=2)2 = 3:14( 0:276 (12.21-a) 2 ) = 0:0598 Acable = 64(0:0598) 2 = 3:83 2 (12.21-b) 2 2 Atotal = 4(3:83) = 15:32 (12.21-c)
22
in
in
in
in
in
in
Whereas the ultimate tensile strength of the steel used is 247 ksi, the cables have been stressed only to 131 ksi, thus the initial prestressing force Pi is equal to Pi = (131) (15:32) 2 = 2; 000 (12.22)
24
ksi
25
in
k
The losses are reported ot be 13%, thus the e ective force is Pe = (1 ; 0:13)(2; 000) = 1; 740 k
Victor Saouma
k
(12.23)
Structural Concepts and Systems for Architects
Draft 12.3 Case Study: Walnut Lane Bridge 12.3.3
12{13
Loads
The self weight of the beam is q0 = 1:72 . 27 The concrete (density=.15 = 3 ) road has a thickness of 0.45 feet. Thus for a 44 foot width, the total load over one single beam is 1 (44) (0:45) (0:15) = 3 = 0:23 q = 13 (12.24) 26
k/ft
k
ft
ft
r;tot
ft
k
ft
k/ft
Similarly for the sidewalks which are 9.25 feet wide and 0.6 feet thick: 1 (2)(9:25) (0:60) (0:15) = 3 = 0:13 (12.25) q = 13 We note that the weight can be evenly spread over the 13 beams beacause of the lateral diaphragms. 29 The total dead load is q = 0:23 + 0:13 = 0:36 (12.26)
28
ft
s;tot
ft
k
ft
k/ft
k/ft
DL
The live load is created by the trac, and is estimated to be 94 psf, thus over a width of 62.5 feet this gives a uniform live load of 1 (0:094) =ft2 (62:5) = 0:45 (12.27) w = 13
30
k
LL
31
ft
k/ft
Finally, the combined dead and live load per beam is
w + = 0:36 + 0:45 = 0:81 DL
12.3.4
LL
(12.28)
k/ft
Flexural Stresses
1. Prestressing force, P only i
f1 = ; AP 1 ; ecr21 6) (31:8)(39:5) (2 10 = 490: = ; 1; 354 1 ; 943: f2 = ; AP 1 + ecr22 106) 1 + (31:8)(39:5) = ;3; 445: = ; (21; 354 943:
(12.29-a)
i
c
(12.29-b)
psi
(12.29-c)
i
c
psi
(12.29-d)
2. P and the self weight of the beam M0 (which has to be acconted for the moment the beam cambers due to prestressing) 2 = 5; 504 (12.30) M0 = (1:72)(160) 8 i
k.ft
The exural stresses will thus be equal to:
; 000) = 2; 043 f1 20 = SM0 = (5; 50:4)(12 943: w ;
Victor Saouma
12 ;
psi
(12.31)
Structural Concepts and Systems for Architects
Draft 12{14
PRESTRESSED CONCRETE f1
i
S1 490 ; 2; 043 = ;1; 553 p p 3 f = +190 P 1 + ec2 + M0 A r2 S2 ;3; 445 + 2; 043 = ;1; 402: :6f = ;2; 400p
3.
P and M0 . 1:74 106 e
(12.32-d)
i
c
= =
ci
(12.32-c)
0
c
=
f
(12.32-b)
psi
=
ti
(12.32-a)
r
c
=
f f2
ec M 1 ; 21 ; 0
; AP
=
(12.32-e)
psi
(12.32-f)
0
c
If we have 13% losses, then the e ective force
P
e
is equal to (1 ; 0:13)(2 106 ) =
lbs
f1
= =
f2
= =
4.
P
e
0 ; AP 1 ; ecr21 ; M S 1 1:74 106 (31:8)(39:5) ; 1; 354 1 ; 943: ; 2; 043: = ;1; 616 P 1 + ec2 + M0 A r2 S2 1:74 106 (31:8)(39:5) ; 1; 354 1 + 943: + 2; 043: = ;954:
(12.33-a)
e
c
psi
e
c
psi
(12.33-b) (12.33-c) (12.33-d)
and M0 + MDL + MLL
M
DL
+ MLL =
(0:81)(160)2 = 2; 592 k.ft 8
(12.34)
and corresponding stresses
; 000) = 962: f1 2 = (2; 592)(12 32; 329 ;
Thus,
f1 f f2 cs
= = = = =
f
ts
Victor Saouma
=
(12.35)
psi
; AP 1 ; ecr21 ; M0 + MS + M 1 ;1; 616 ; 962: = ;2; 578: :45f = ;2; 700p P 1 + ec2 + M0 + M + M A r2 S2 ;954 + 962: = +8: p p 6 f = +380 e
DL
LL
c
psi
0
c
DL
e
c
psi
0
c
LL
(12.36-a) (12.36-b) (12.36-c) (12.36-d) (12.36-e) (12.36-f)
Structural Concepts and Systems for Architects
Draft Chapter 13
Three-Hinges ARCHES 13.1
Theory
13.1.1 Uniform Horizontal Load 1 In order to optimize dead-load eciency, long span structures should have their shapes approximate the coresponding moment diagram, hence an arch, suspended cable, or tendon con guration in a prestressed concrete beam all are nearly parabolic, Fig. 13.1. 2 Long span structures can be built using at construction such as girders or trusses. However, for spans in excess of 100 ft, it is often more economical to build a curved structure such as an arch, suspended cable or thin shells. 3 Since the dawn of history, mankind has tried to span distances using arch construction. Essentially this was because an arch required materials to resist compression only (such as stone, masonary, bricks), and labour was not an issue. 4 The basic issues of static in arch design are illustrated in Fig. 13.2 where the vertical load is per unit horizontal projection (such as an external load but not a self-weight). Due to symmetry, the vertical reaction is simply V = wL 2 , and there is no shear across the midspan of the arch (nor a moment). Taking moment about the crown, L L (13.1) M = Hh ; wL 2 2 ; 4 =0
Solving for H
H = wL 8h
2
(13.2)
We recall that a similar equation was derived for arches., and H is analogous to the C ; T forces in a beam, and h is the overall height of the arch, Since h is much larger than d, H will be much smaller than C ; T in a beam. 5 Since equilibrium requires H to remain constant across thee arch, a parabolic curve would theoretically result in no moment on the arch section. 6 Three-hinged arches are statically determinate structures which shape can acomodate support settlements and thermal expansion without secondary internal stresses. They are also easy to analyse through statics.
Draft 13{2
Three-Hinges ARCHES
2
M = w L /8
L w=W/L
IDEALISTIC ARCH SHAPE GIVEN BY MOMENT DIAGRAM
C RISE = h -C BEAM +T W/2
M-ARM small C C-T large BEAM T
-C +T SAG = h
W/2
T
IDEALISTIC SUSPENSION SHAPE GIVEN BY MOMENT DIAGRAM
NOTE THAT THE "IDEAL" SHAPE FOR AN ARCH OR SUSPENSION SYSTEM IS EQUIVILENT TO THE DESIGN LOAD MOMENT DIAGRAM
Figure 13.1: Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Lin and Stotesbury 1981)
w
wL/2
w
H h
h
H
H
H = wL2 /8h
L
L/2 R
R V = wL/2
V
R = V 2+ H
2
V = wL/2 2
MCROWN = VL/2 - wL /8 - H h = 0 M BASE
2
= wL /8 - H h = 0
Figure 13.2: Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13.1 Theory
13{3
7 An arch carries the vertical load across the span through a combination of axial forces and exural ones. A well dimensioned arch will have a small to negligible moment, and relatively high normal compressive stresses. 8 An arch is far more ecient than a beam, and possibly more economical and aesthetic than a truss in carrying loads over long spans. 9 If the arch has only two hinges, Fig. 13.3, or if it has no hinges, then bending moments may exist either at the crown or at the supports or at both places.
w
h’
h
h’ M base
h
H’=wl 2/8h’< 2
w
M
H’ wl /8h
APPARENT LINE OF PRESSURE WITH ARCH BENDING INCLUDING BASE
APPARENT LINE OF PRESSURE WITH ARCH BENDING EXCEPT AT THE BASE
H’