The Reinforced Concrete Design Handbook, A companion to ACI-318M-14 (Metric system) 9781641950121

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An ACI Handbook

The Reinforced Concrete Design Handbook A Companion to ACI 318M-14

SP-17M(14)  

Eng.Mohammed Ashraf

 

Eng.Mohammed Ashraf



 







 

     

    

 

Eng.Mohammed Ashraf

 

Eng.Mohammed Ashraf



  

   

 

Eng.Mohammed Ashraf

May 2018 ISBN: 978-1-64195-012-1 THE REINFORCED CONCRETE DESIGN HANDBOOK Ninth Edition Copyright by the American Concrete Institute, Farmington Hills, MI. All rights reserved. This material may not be reproduced or copied, in whole or part, in any printed, mechanical, electronic, film, or other distribution and storage media, without the written consent of ACI. The technical committees responsible for ACI committee reports and standards strive to avoid ambiguities, omissions, and errors in these documents. In spite of these efforts, the users of ACI documents occasionally find information or requirements that may be subject to more than one interpretation or may be incomplete or incorrect. Users who have suggestions for the improvement of ACI documents are requested to contact ACI via the errata website at http://concrete.org/Publications/DocumentErrata.aspx. Proper use of this document includes periodically checking for errata for the most up-to-date revisions. ACI committee documents are intended for the use of individuals who are competent to evaluate the significance and limitations of its content and recommendations and who will accept responsibility for the application of the material it contains. Individuals who use this publication in any way assume all risk and accept total responsibility for the application and use of this information. All information in this publication is provided “as is” without warranty of any kind, either express or implied, including but not limited to, the implied warranties of merchantability, fitness for a particular purpose or noninfringement. ACI and its members disclaim liability for damages of any kind, including any special, indirect, incidental, or consequential damages, including without limitation, lost revenues or lost profits, which may result from the use of this publication. It is the responsibility of the user of this document to establish health and safety practices appropriate to the specific circumstances involved with its use. ACI does not make any representations with regard to health and safety issues and the use of this document. The user must determine the applicability of all regulatory limitations before applying the document and must comply with all applicable laws and regulations, including but not limited to, United States Occupational Safety and Health Administration (OSHA) health and safety standards. Participation by governmental representatives in the work of the American Concrete Institute and in the development of Institute standards does not constitute governmental endorsement of ACI or the standards that it develops. Order information: ACI documents are available in print, by download, on CD-ROM, through electronic subscription, or reprint and may be obtained by contacting ACI. Most ACI standards and committee reports are gathered together in the annually revised ACI Manual of Concrete Practice (MCP). American Concrete Institute 38800 Country Club Drive Farmington Hills, MI 48331 USA +1.248.848.3700 Managing Editor: Khaled Nahlawi Staff Engineers: Daniel W. Falconer, Matthew R. Senecal, Gregory M. Zeisler, and Jerzy Z. Zemajtis Technical Editors: Shannon B. Banchero, Emily H. Bush, and Cherrie L. Fergusson Manager, Publishing Services: Barry Bergin Lead Production Editor: Carl Bischof Production Editors: Kelli Slayden, Kaitlyn Hinman, Tiesha Elam Graphic Designers: Ryan Jay, Aimee Kahaian Manufacturing: Marie Fuller www.concrete.org

 

Eng.Mohammed Ashraf

 This edition of The Reinforced Concrete Design Handbook, SP-17M(14), is dedicated to the memory of Daniel W. Falconer and his many contributions to the concrete industry. He was Managing Director of Engineering for the American Concrete Institute from 1998 until his death in July 2015. Dan was instrumental in the reorganization of Building Code Requirements for Structural Concrete (ACI 318-14) and Commentary (ACI 318R-14) liaison to ACI Committee 318, Structural Concrete Building Code; and ACI Subcommittee 318-SC, Steering Committee. His vision was to simplify the use of the Code for pracSP-17M. His oversight and review comments were instrumental in the development of this Handbook. An ACI member since 1982, Dan served on ACI Committees 344, Circular Prestressed Concrete Structures, and 373, Circular Concrete Structures Prestressed with Circumferential Tendons. He was also a member of the American Society of Civil Engineers. Prior to

licensed professional engineer in several states. tans basketball and football programs. Above all, Dan was known as a devoted family man dedicated to his wife of 33 years, Barbara, his children Mark, Elizabeth, Kathryn, and Jonathan, and two grandsons Samuel and Jacob. In his memory, the ACI Foundation has established an educational memorial. For more information visit http://www.scholarshipcouncil.org/Student-Awards. Dan will be sorely missed for many years to come.

 

Eng.Mohammed Ashraf

 The Reinforced Concrete Design Handbook provides assistance to professionals engaged in the design of reinforced concrete Building Code Requirements for Structural Concrete (ACI 318M-14). The layout and look of the Handbook have also been updated. The Reinforced Concrete Design Handbook now provides dozens of design examples of various reinforced concrete members, such as one- and two-way slabs, beams, columns, walls, diaphragms, footings, and retaining walls. For consistency, many of the examples not related to the design of the members in the seven story building that illustrate various ACI 318M-14 requirements. Each example starts with a problem statement, then provides a design solution in a three column format—code provision elaborating on a certain condition or comparing results of similar problem solutions. In addition to examples, almost all chapters in the Reinforced Concrete Design Handbook contain a general discussion of the related ACI 318M-14 chapter.

appreciates the support of Dirk Bondy and Kenneth Bondy who provided free software to analyze and design the post-tensioned

Pincheira, David Rogowski, and Siamak Sattar, who reviewed one or more of the chapters. their software to perform analyses of structure and members. The Reinforced Concrete Design Handbook available for free download from the following ACI webpage links: https://www.concrete.org/store/productdetail.aspx?ItemID=SP1714DAE https://www.concrete.org/store/productdetail.aspx?ItemID=SP1714DA Keywords: tural analysis; structural systems; strut-and-tie; walls.

Managing Editor

 

Eng.Mohammed Ashraf

  1.1—Introduction, p. 9 1.2—Building plans and elevation, p. 9 1.4—Material properties, p. 12  2.1—Introduction, p. 13 2.2—Materials, p. 13 2.3—Design loads, p. 13 2.4—Structural systems, p. 14 2.5—Floor subassemblies, p. 20 2.6—Foundation design considerations for lateral forces, p. 22 2.7—Structural analysis, p. 23 2.8—Durability, p. 23 2.9—Sustainability, p. 23 2.10—Structural integrity, p. 23 2.11—Fire resistance, p. 23 2.12—Post-tensioned/prestressed construction, p. 23 2.13—Quality assurance, construction, and inspection, p. 23  3.1—Introduction, p. 25 3.3—Hand calculations, p. 26 3.4—Computer programs, p. 26 3.5—Structural analysis in ACI 318M, p. 27 3.6—Seismic analysis, p. 29  4.1—Introduction, p. 31 4.2—Background, p. 33 4.3—Requirements for concrete in various exposure categories, p. 33 4.4—Concrete evaluation, acceptance, and inspection, p. 35 4.5—Examples, p. 35 5.1—Introduction, p. 39 5.2—Analysis, p. 39 5.3—Service limits, p. 39  5.4—Required strength, p. 40 5.5—Design strength, p. 40 5.6—Flexure reinforcement detailing, p. 40 5.7—Examples, p. 42  6.1—Introduction, p. 81 6.2—Analysis, p. 81 6.3—Service limits, p. 81 6.4—Shear strength, p. 82 6.5—Calculation of required shear strength, p. 83 6.6—Calculation of shear reinforcement, p. 84 6.7—Flexural strength, p. 84 6.8—Shear reinforcement detailing, p. 84

6.9—Flexure reinforcement detailing, p. 85 6.10—Examples, p. 88  7.1—Introduction, p. 133 7.2—Service limits, p. 133 7.3—Analysis, p. 134 7.4—Design strength, p. 134 7.5—Temperature and shrinkage reinforcement, p. 140 7.6—Detailing, p. 140 7.7—Examples, p. 143  8.1—Introduction, p. 281 8.2—Material, p. 281 8.3—Service limits, p. 281 8.4—Analysis, p. 281 8.5—Design strength, p. 283 8.6—Reinforcement detailing, p. 284 8.7—Summary steps, p. 286 8.8—Examples, p. 289  9.1—Introduction, p. 353 9.2—General, p. 353 9.3—Design limits, p. 353 9.4—Required strength, p. 354 9.5—Design strength, p. 356 9.6—Reinforcement limits, p. 357 9.7—Reinforcement detailing, p. 357 9.8—Design steps, p. 359 9.9––Examples, p. 362   10.1—Introduction, p. 391 10.2—General, p. 391 10.3—Required strength, p. 393 10.4––Design strength, p. 394 10.5––Detailing, p. 398 10.6––Summary, p. 399 10.7—Examples, p. 400  11.1—Introduction, p. 419 11.2—Footing design, p. 419 11.3—Design steps, p. 420 11.5—Combined footing, p. 423 11.6—Examples, p. 425  12.1—General, p. 489 12.2—Design limits, p. 490 12.3—Applied forces, p. 491 12.4—Design strength, p. 492 12.5—Reinforcement limits, p. 492 12.6—Detailing, p. 493  

Eng.Mohammed Ashraf

12.7—Summary, p. 493 12.8—Examples, p. 495  13.1—Introduction, p. 589

one-way slabs, p. 589

slabs and beams, p. 592 13.6—Shrinkage and temperature reinforcement: nonprestressed, p. 593 13.7—Shrinkage and temperature reinforcement – posttensioned, p. 593 ural members, p. 594 13.9—Permissible stresses at transfer of prestress, p. 594 13.10—Permissible concrete compressive stresses at service loads, p. 594 13.11—Examples, p. 595

 14.1—Introduction, p. 633 14.2—Concept, p. 633 14.3—Design, p. 633 14.4—Struts, p. 634 14.5—Ties, p. 636 14.7—Usual calculation steps and modeling consideration to apply strut-and-tie model, p. 638 14.8—Examples, p. 639  15.1—Introduction, p. 687 15.2—Materials, p. 687 15.3—Design assumptions, p. 687 8 15.5—Discussion on anchors resisting tension, p. 690 15.6—Discussion on anchors resisting shear, p. 690 15.8—Examples, p. 692

 

Eng.Mohammed Ashraf



 The building depicted in this chapter was developed to show how, by various examples in this Handbook, to design and detail a common concrete building according to ACI 318M-14. This example building is seven stories above ground and has a one story basement. The building has been removed along Grid C on the second level so that there is open space for the lobby. The building dimensions are: • • • • Basement height = 3 m The basement is used for storage, building services and mechanical equipment. It is ten feet high and has an extra column added in every bay along Grids A through F to

support a two-way slab at the second level. There are basement walls at the perimeter. The structural system is an ordinary concrete shear wall in the north/south direction and an ordinary concrete moment frame in east/west direction. These basic systems were chosen as a starting point for the examples. Member examples may be expanded to show how they may be designed in intermediate or special systems but a new structural analysis is not done. The following analysis results provide the moments, shears, and axial loads given in the examples in other chapters in the manual. Those examples may modify this initial

 The following building plans and elevation provide the illustration of the example building.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





 The following loads for the example building are generated in accordance with ASCE 7-10. The Risk Category is II. Gravity Loads D: • Self weight 2 • 2 •

The fc

•

members must be “puddled” with higher strength concrete. Usually this situation only becomes an issue for taller buildings. For this example, the building height is moderate and the loads are typical. The locally available aggregate is a durable dolomitic limestone. Thus, the concrete can readily have a higher fc of the durability requirements of Table 19.3.2.1 in ACI 318M-14 shows that 35 MPa will satisfy the minimum fc for all exposure classes. For this concrete, a check of Table 19.2.1.1 in ACI 318M-14 shows that all the code minimum ties are chosen: • fc 3 • wc • Ec = 27,800 MPa = 0.2 • • eth =10 × 10–66/°C The use of lightweight concrete can reduce seismic forces and foundation loads. Based on local experience, however,

2

•

Typical Floor: Private rooms and corridors serving them 2

2

• • Ground load, Pg • Thermal, Ct = 1.0 • Exposure, Ce = 1.0 • Importance, Is = 1.0 • Flat roof load, Pf Lateral Loads • • • • • • • • • • • • • •

•

2

2

Exposure category = C Wind directionality factor, Kd = 0.85 Topographic factor, Kst = 1.0 Gf GCpi = ±0.18 Directional Procedure Importance, Ie = 1.0 Site class = D SS = 0.15, SDS = 0.16 S1 = 0.08, SD1 = 0.13 Seismic design category = B Equivalent lateral force procedure Building frame system; ordinary reinforced concrete shear walls in the north-south direction R=5 Cs = 0.046 Moment-resting frame system; ordinary reinforced concrete moment frame in the east-west direction R=3 Cs = 0.032

 The material properties for any building should have a reasonable knowledge of locally available concrete and steel materials. As a preliminary value for this example, a specifc -

fc placement usually proceeds in two stages for each story;

keep the concrete strengths of the vertical members within in ACI 318M-14 states that if this ratio is exceeded, the

of lightweight. The modulus of elastic for concrete, Ec, is calculated according to 19.2.2 in ACI 318M. For normal19.2.2.1.b in ACI 318M is applicable. weight concrete, Eq. 19.2.2.1

material properties, but an average value for concrete is 0.2. eth, of concrete can be found in ACI 209R. The most common and most available nonprestressed reinforcement is Grade 60. Higher grades are available but 20.2.2.4 in ACI 318M-14 limits many uses of reinforcing steel to 420 MPa. The modulus of elastic for reinforcement, Es, is given in 20.2.2.2 in ACI 318M. Reinforcement Material Properties • fy = 420 MPa • fyt = 420 MPa • Es = 200,000 MPa  American Concrete Institute ACI 209R-92—Prediction of Creep, Shrinkage, and

  

Eng.Mohammed Ashraf



 A chapter on structural systems of reinforced concrete

tural systems. A structural engineer’s primary concern is to design buildings that are structurally safe and serviceable under design vertical and lateral loads. Prior to the 1970s, reinforced

 Volume no. ACI SP-17M(14)

Chapter name ACI SP-17M(14) Building system Structural systems Structural analysis Durability

I

Two-way slab Beams Diaphragm Columns Walls Foundations Retaining walls Serviceability Strut-and tie Anchoring to concrete

nonstructural masonry walls and partitions, were seldom

science, and improvements in analysis tools have allowed structural engineers to develop economical building designs with more predictable structural performance. Structural systems and their component members must structural integrity is maintained, design loads are resisted, and serviceability limits are met. The individual members of a build building’s structural system are generally assumed to be oriented either vertically or horizontally, with the common exception of identi parking structure ramps. Chapter 4 of ACI 318M-14 identi-

II

Chapter no. ACI 318M-14 ACI SP-17M(14) — 1 4 and 5 2 6 3 19 4 7 5 8 6 9 7 12 8 10 9 11 10 13 11 7 and 11 12 24 13 23 14 17

15

act in orthogonal directions. Two types of lateral loads are discussed in this chapter:

Wind and earthquake loads are dynamic in nature; common to reinforced concrete building structural systems

-

induced in a structure. Wind loads are externally applied loads and, hence, are related to the structure’s exposed surface. Earthquake loads are inertial forces related to the magnitude and distribution of the mass in the structure. 2.3.1 Wind loading loading—Wind kinetic energy is transformed into potential energy when it is resisted by an obstruction. Wind pressure is related to the wind velocity, building height, building surface, the surrounding terrain, and the location and size of other local structures. The structural response to mode of vibration. The quasi-static approach to wind load design has gener-

In Table 2.1, code chapters are correlated with the chapters

 The concrete mixture proportion needs to satisfy the design properties and limits in ACI 318M-14, Chapter 19, and the reinforcing steel needs to satisfy the design proper-

 ACI 318M-14 assumes that ASCE 7-10 design loads are applied to the building’s structural system and to individual vertically and horizontally. Horizontal loads are assumed to

for very tall buildings, especially with respect to the comfort of the occupants and the permissible horizontal movement, “or drift,” which can cause the distress of partitions and glass. Therefore, to determine design wind loads for very tall buildings, wind tunnel testing is not unusual. 2.3.2 Earthquake loading tural design is life safety; that is, preserving the lives of occupants and passersby. Serviceability and minimizing studying the results of previous earthquakes on various structural systems, improvements to code provisions and design practices have been achieved. These improvements have led to a reduction in damage of reinforced concrete structures that experience an earthquake. Some code improvements for

  

Eng.Mohammed Ashraf





contains provisions meant to ensure inelastic deformation capacity in regions where yielding is likely, which then protects the overall integrity and stability of the building. structures, important structures, or for structures with an irregFor very important and potentially critical structures—for example, nuclear power plants—inelastic dynamic analysis

 All structures must have a continuous load path that can be traced from all load sources or load application to

into columns and walls, thus creating a continuous load path. Fig. 2.3.2—Typical distribution of equivalent static lateral forces representing seismic forces “ASCE 7-10.” .”

of course, adequately resist the factored forces applied to detailing information. Engineers commonly refer to a structure’s gravity-load-

stronger than beams—the so-called “strong column-weak beam” concept 2.. Improve detailing to increase ductility and large energy

yielding before reaching nominal shear strength 4.. Designing and detailing the connections to be stronger than the members framing into them

members. As the building increases in height, the importance

For most structures, the equivalent lateral force procedure given in ASCE 7-10 is used. Based on this procedure, the distribution of design forces along the height of a building roughly approximates the

erations and construction requirements. There are several types of structural systems or a combination thereof to resist gravity, lateral, and other loads, with deformation behavior as follows: 1. Frames

Applying recorded earthquake motions to a structure through elastic dynamic analyses usually result in greater force demands than from the earthquake design forces specifor force reductions due to inelastic response. For example, ASCE 7-10 applies an R over-strength, and energy dissipation through the soil-founprocess such that linear static elastic analysis can be used for building designs. The R factor reduces the calculated lateral loads and assumes a building may be damaged during an earthquake event, but will not collapse. The higher the R-value, the lower the lateral design load on a structure. having low deformation capacity, to 8 for ductile systems, earthquake, it is expected that some building members will yield. To promote appropriate inelastic behavior, ACI 318M

on the horizontal shear applied at each story level. 2. Walls bending. The behavior predominate mode depends on the wall’s height-to-width aspect ratio. 3. Dual systems—Dual systems are a combination of moment-resisting frames and structural walls. The momentresisting frames support gravity loads, and up to 25 percent the lateral loading. 4. Frames with closely spaced columns, known as cantilevered column system or a tube system deformations are due to both shear and bending, similar to a wall. Wider openings in a tube, however, can produce a behavior intermediate between that of a frame and a wall. Regardless of the system, a height is reached at which the resistance to lateral sway will govern the design of the

  

Eng.Mohammed Ashraf





Fig. 2.4—Structural systems and optimum height limitations (Ali and Moon 2007). Gravity loads are resisted by reinforced concrete members controls the building design. ing’s Seismic Design Category increases, from A through seismic design and a more ductile system to maintain an acceptable level of seismic performance. M provides three categories of earthquake ACI 318M detailing; ordinary, intermediate, and special. These catego categories provide an increasing level of system toughness.

strength. The related deformations are exaggerated and shown in Fig. 2.4.1 2.4.1. 2.4.2 Lateral-load-resisting system system—A lateral-forceto maintain integrity during high wind loading and design earthquake accelerations. Buildings are basically cantile cantile-

ACI 318M-14, 318M-14 Section 18.2.1, lists the relevant code For buildings in SDC A and B, wind load will usually

not limited in height for most systems for this SDC, but

force-resisting system. The following lateral-force-resisting systems are addressed as follows. 2.4.2.1 Moment-resisting frames—Cast-in-place momentresisting frames derive their load resistance from member strengths and connection rigidity. In a moment-resisting

force-resisting system design. For buildings in SDC D, E, and F, seismic loads almost always control design forces, and increased seismic detailing

force in columns.

For buildings in SDC C, seismic loads are likely to control

based on assumed structural performance level behavior. structural systems. Table 2.4 provides ASCE limits for choosing a structural system for a particular building. The ranges of applicability -

2.4.1 Gravity-resisting systems—A gravity-load-resisting and vertical members that support the horizontal members.

secondary moments caused by column axial forces multiP In buildings, moment-resisting frames are usually arranged parallel to the principal orthogonal axes of the structure

economical solution up to a certain height. 2.4.2.2 Shear walls—Reinforced concrete shear walls are often introduced into multistory buildings because of their

  

Eng.Mohammed Ashraf





 Practical limit of system (ASCE 7-10 limit according to SDC) SDC Type of LFRS A and B C D

SSW with IMF SSW with SMF

E

F

50 m

50 m

30 m

50 m

50 m

30 m

50 m

30 m

30 m

Fig. 2.4.2.2a—Coupled shear walls. A shear wall building usually consists of a series of parallel shear walls in orthogonal directions that resists lateral loads and supports vertical loads. Bearing walls should be located close to the plan perimeter if possible and should preferably be symmetric in plan to reduce or when the building program is conducive to layout of frame, shear walls behave as vertical cantilevers. Walls can rate walls can be coupled to act together by beams/slabs or deep beams, depending on design forces and architectural requirements. Coupling of shear walls introduces frame the system. Reinforced concrete walls are often used around

2.4.2.3 Staggered wall-beam system—This system uses story-high solid or pierced walls extending across the entire width of the building and supported on two lines of columns

The staggered wall-beam building is suitable for multistory construction having permanent interior partitions such as apartments, hotels, and student residences.

For shear wall types and functions, refer to Table 2.4.2.2.   

Eng.Mohammed Ashraf





 Structural walls Short—height-to-length ratio does not exceed 2

Height-to-length ratio is greater than 2

Behavior

Reinforcement

Remarks

only with shear strength.

Bars evenly distributed horizontally and vertically.

Wall foundation must be capable of resisting the actions generated in the wall. Consider sliding resistance provided by foundation.

both the wall’s shear and moment strength.

Evenly distributed vertical and horizontal reinforcement. Part of the vertical reinforcement may be concentrated at

Wall foundation must be capable of resisting the actions generated in the wall. Consider overturning resistance provided by foundation.

reinforcement in the web contributes to Ductile structural wall

Flexural bar spacing and size should be

Acceptable ductility can be obtained with proper attention to axial load level,

reinforcement, treatment of construction failure.

buckling.

Coupled walls with shallow coupling beams or

riorates quickly during inelastic reversed loading.

Place coupling slab bars to limit slab cracking at the stress concentrations at the wall ends.

Punching shear stress around the wall ends in the slab needs to be checked.

Coupled walls with

Depending on span-to-depth ratio, link beams may be designed as deep beams.

Main reinforcement is placed horizontally or diagonally. For coupling beams with main reinforcement placed diagonally from the deep beam’s corner to corner

Properly detailed coupling beams can achieve ductility. Coupling beams should maintain their load-carrying capacity under reverse inelastic deformation.

directly. -

Frames behave as braced frames, increasing the lateral strength and between diagonally opposite frame corners, and creates high shear forces in the columns.

separated from the moment frame to be connected structurally with the moment frame.

An advantage of the wall-beam building is the large open parking, commercial use, or even to allow a highway to pass under the building. This system should be considered in low 2.4.2.4 Tubes—A tube structure consists of closely spaced columns in a moment frame, generally located around the

ties of the moment frame. If there are story acts as a weak or soft story that is vulnerable to concentrated damage and instability.

Frames parallel to direction of force act like webs to carry the shear from lateral loads, while frames perpendicular to from lateral loads. Gravity loads are resisted by the exterior frames and interior columns. A reinforced concrete braced tube is a system in which a

Because tube structures generally consist of girders and

and should be included in analytical models. Tubes are often thought of as behaving like a perforated diaphragm.

of shear lag. 2.4.3 Dual systems—Dual systems consist of combining two of the structural systems discussed in the previous

  

Eng.Mohammed Ashraf





teristics, particularly with respect to seismic behavior. Some of the more common dual systems are discussed in 2.4.3.1 through 2.4.3.6. 2.4.3.1 Wall-frame systems isolated or coupled structural walls can be combined to acteristics of moment frames and structural walls, careful attention to the interaction between the two systems can improve the structure’s lateral response to loads by reducing The wall’s overturning moment is greatly reduced by interaction with the frame. Because drift compatibility is forced on both the frame and the wall, and the frame-alone and

Fig. 2.4.2.2c—Example of shear wall layouts.

columns are assumed to be braced against sidesway by the walls. The wall-frame dual system permits the structure to be designed for a desired yielding sequence under strong ground motion. Beams can be designed to experience signifsignif icant yielding before inelastic action occurs at the bases of the walls. By creating a hinge sequence, and considering the relative economy with which yielded beams can be repaired, wall-frame structures are appropriate for use in higher seismic zones. However, note that the variation of shears and overturning moments over the height of the wall and conditions. 2.4.3.2 Outrigger systems systems—An outrigger system uses orthogonal walls, girders, or trusses, one or two stories in height, to connect the perimeter columns to central core

In addition to the outrigger girders that extend out from the core, girders or trusses are placed around the perimeter of the structure at the outrigger levels to help distribute lateral forces between the perimeter columns and the core walls. These perimeter girders or trusses are called “hat” or “top-hat” bracing if located at the top, and “belt” bracing if located at intermediate levels. Some further reductions in total drift and core bending moments can be achieved by increasing the cross section of the columns and, therefore, ness and, thus, resist wind loads with less drift. Design of outrigger-type systems for SDC D through F must consider inelastic response of the entire system. Members framing into the outriggers should be detailed for ductile response. 2.4.3.3 Tube-in-tube—For tall buildings with a reasonably large service core, it is generally advantageous to use Fig. 2.4.2.3—Staggered wall-beam system.

part of the lateral-force-resisting system. The outer tube is formed by the closely-spaced column-spandrel beam frame. A bundled tube system consists of several framed tubes

  

Eng.Mohammed Ashraf





Fig. 2.4.2.4—Tube systems.

Fig. 2.4.3.2—Outrigger system.

Fig. 2.4.3.1—Shear wall and moment frame system. bundled into one larger structure that behaves as a multicell The tube-in-tube system combines the advantages of both the perimeter framed tube and the inner shear walls. The inner shear walls enhance the structural characteristics of the perimeter framed tube by reducing the shear deformation of the columns in the framed tube. The tube-in-tube system can action type structure.

Fig. 2.4.3.3—Tube-in-tube and bundled tube systems. 2.4.3.4 Bundled tubes—A bundled tube system consists of several framed tubes bundled into one larger structure that behaves as a multicell perforated box. Individual tubes can

  

Eng.Mohammed Ashraf





2.4.3.5 Mixed concrete-steel structures—Mixed concretesteel systems consist of interacting concrete and steel assemblies. The resulting composite structure displays most or all -

and concrete to prevent uneven displacement. Because the building trades and equipment, engineers who design mixed construction should consider scheduling issues. 2.4.3.6 Precast structures—Precast concrete members are widely used as components in frame, wall, and wall-frame systems. Mixed construction, consisting of precast concrete assemblies connected to a cast-in-place concrete core, is also of standardization, the ease of manufacture, the simplicity of assembly, and the speed of erection. -

H- or T-units, and shear walls and cores are assembled from prefabricated single-story panels. Planning and designing appropriate connection details for panels, frame members, related to prefabricated structures. Three main types of connections are described as follows:

Fig. 2.5.2—Flat slab with drop panels and capitals.

may be an integral and necessary part of the lateral-forceresisting system. Concrete structures are commonly analyzed for lateral -

slab’s plan dimensions relative to the location of the lateralload-resisting members, slab thickness, locations of slab

precast members are made continuous by mechanical of a frame to resist lateral moments, engineers usually limit is used, the engineer should specify appropriate welding procedures to avoid brittle connections. members are bolted or welded together and the gaps are grouted. 3. The individual precast units are post-tensioned together

loading depends to a considerable degree on the characteristics of the connections. Connection details can be developed that ensure satisfactory performance under seismic loadings, provided that the engineer pays particular attention to steel

 ture’s cost as well as the performance of its lateral-forceto resist gravity load. Additional important functions in most buildings are: tains the plan shape of the structure, and distributes horizontal forces to the lateral-force-resisting system.

to between 25 and 50 percent of the bay width. 2.5.1 Flat plates

commercial and residential buildings. Simple formwork and reinforcing patterns, as well as lower overall building height, are advantages of this system. In designing and detailing plate-column connections, particular attention must be paid to the transfer of shear and unbalanced moment between

ment over the column area. 2.5.2 Flat slabs with drop panels, column capitals, or both by thickening the slab around columns with drop panels,

systems normally act as diaphragms transmitting lateral forces to columns and walls.

  

Eng.Mohammed Ashraf





Fig. 2.5.6—One-way banded one-way slab. The beams and slabs can be placed in a composite fashion are placed at the beam-slab interface to ensure composite action. This system can provide good lateral force resistance, provided that the shear connectors are detailed with improve the slab shear strength by increasing the slab thickness around the column. To improve the slab shear strength without increasing the slab thickness, engineers can provide closely spaced stirrups or shear studs radiating out from the column.

of slab system include: between the top of the beam and a cast-in-place concrete support the formwork for the cast-in-place slab. In this concrete beams framing directly into columns.

or other bracing members, are unsuitable in high seismic 2.5.3 —For longer spans, a slab system consisting of a grid of ribs intersecting at a constant spacing can be used to achieve an appropriate slab depth for the longer span with much less dead load than a The ribs are formed by standardized dome or pan forms that are closely spaced. The slab thickness between the form a solid concrete drop panel, to satisfy requirements for transfer of shear and unbalanced moment between the slab and columns.

place concrete slab. The slab formwork is supported from

govern the structural design of concrete slabs for composite steel decks. 2.5.5 One-way ribbed slabs (joists) spanning between beams, which span between columns. The size of pan forms available usually determines rib depth and spacing. As with a two-way ribbed system, the thickness of the thin slab between ribs is often determined by the buildThis system provides an adequate shear diaphragm and is used in a structure whose lateral resistance comes from a

strength can be increased by the addition of closely spaced stirrups radiating out from the column face in two direcshould investigate the actions induced in the ribs by building drift. 2.5.6 One-way banded slabs—A one-way banded slab is 2.5.4 One-way slabs on beams and girders slabs on beams and girders consist of girders that span between columns and beams that span between the girders.

columns, usually in the long-span direction, and a one-way

provides a satisfactory diaphragm, and uses the girdercolumn frames and beam-column frames to resist lateral

shallow beam can be reinforced with closely spaced stirrups near the support to increase the slab’s shear strength. This system is also sometimes referred to as wide-shallow beams with one-way slabs.

detailing of the beam and girder reinforcement.

laterally than a structure using a ductile moment frame with

  

Eng.Mohammed Ashraf





The soil type and strata usually dictate whether the foundation is deep or shallow. A soils report from a licensed geotechnical engineer provides the detailed information and foundation recommendations that the licensed design profesfootings, the geotechnical engineer provides an allowable soil bearing pressure for the soil at the foundation elevation. tion, and includes consideration of the anticipated use of the 2 , the soil is very soft and deep foundation options are usually Fig. 2.5.7—Two-way slab with edge beams around perimeter.

ductile frames, structural walls, or other bracing members, are not suitable in SDC D through F. 2.5.7 Two-way slabs with edge beams—As shown in Fig. 2.5.7, the slab is supported by beams in two directions on the perimeter column lines. This system is useful where a

though frame action for use in SDC D through F.

2.5.8 Precast slabs—Precast, —Precast, one-way slabs are usually supported by bearing walls, precast beams, or cast-in-place beams. Precast slabs may be solid, hollow-core slabs, or single- or double-T-sections. They are sometimes topped by a thin cast-in-place concrete layer, referred to as a “topping slab.” Welded connections are normally used to transfer in-plane shear forces between precast slabs and their supports. Because precast slabs are individual units interconnected act as a shear diaphragm must be examined by the engineer. Boundary reinforcement may be required, particularly where the lateral-force-resisting members are far apart. In areas of high seismicity, the connections between the precast slab system and the lateral-load-resisting system must be carefully detailed. A concrete topping bonded to the precast slab improves the ability of the slab system to act as a shear diaphragm, and can be used in SDC D through F.   A foundation design must consider the weight of the building, live loads, and the transmission of lateral forces to the ground. A distinction should be drawn between external forces, such as wind, and inertia forces that result from the building’s response to ground motions during an earthquake. External lateral forces can include static pressures due to

tions. If the building is below grade, concrete walls can be part of the foundation system.

caissons can transfer a column load directly to the bedrock. 2 . Caissons are large in diameter, usually starting at approximately 750 mm. Piles are generally smaller in diameter, starting at around 300 mm, and can be cast-in-place in augered holes or precast piles that are driven into place. Piles are usually designed for lighter loads than caissons are. Groups of piles may be used where bedrock is too deep for a caisson. Tops of piles or caissons are bridged by pile caps and grade beams to distribute column loads as needed. Shallow foundations are referred to as footings. Types of footings are isolated, combined, and mat. Isolated rectrect angular or square footings are the most common types. Combined footings are often needed if columns are too close together for two isolated footings, if an exterior column is transmittoo close to the boundary line, or if columns are transmit ting moments to the footing, such as if the column is part of a lateral-force-resisting system. If the column loads are uniformly large, such as in multistory buildings, or if column spacing is small, mat foundations are considered. 2.6.1 Resistance to lateral loads—The vertical foundation pressures resulting from lateral loads are usually of short duration and constitute a small percentage of the total Allowing a temporary peak in vertical bearing pressures preferred to making the footing areas larger. The geotechnical engineer should report the likelihood of liquefaction of sands or granular soils in areas with a high groundwater table, or the possibility of sudden consolidafriction piles founded in soils susceptible to liquefaction or consolidation should be checked. 2.6.2 Resistance to overturning—The engineer should investigate the safety factor of the foundation against overturning and ensure it is within the limits of the local building code.

factor is included to account for their dynamic nature.   

Eng.Mohammed Ashraf



 The analysis of concrete structures “shall satisfy compatibility of deformations and equilibrium of forces,” as stated



 The introduction of post-tensioning/prestressing to permanent force within the structural system. Because cast-

method of analysis as long as these conditions are met. This discussion is intended to be a brief overview of the analysis process as it relates to structural concrete design. For more detailed information on structural analysis, refer to Chapter 3 of this Handbook.  Reinforced concrete structures are expected to be durable. The design of the concrete mixture proportions should consider exposure to temperature extremes, snow and ice, and ice-removing chemicals. Chapter 19 of ACI 318M-14 provides mixture requirements to protect concrete and reinforcement against various exposures and deterioration. Chapter 20 of ACI 318M-14 provides concrete cover requirements to protect reinforcement against steel corrosion. For more information, refer to Chapter 4 of this Handbook.  Reinforced concrete structures are expected to be as sustainable as practical. ACI 318M M allows sustainability requirements to be incorporated in the design, but they must not override strength and serviceability requirements.  The ACI Code concept of structural integrity is to “improve the redundancy and ductility in structures so that in the event

relatively small area and the structure will have a better

continuity through design and detailing rules within the beam and two-way slab chapters.

the behavior of the entire system. The engineer should changes in length, and rotations due to post-tensioning/ be given to the connection of post-tensioned/prestressed members to other members to ensure the proper transfer of forces between, and maintain a continuous load path. Because the post-tensioning/prestressing force is permanent,

 

Chapter 26 of ACI 318M-14 contains requirements to facilitate the implementation of competent construction docu documents, material, construction, and inspection.  American Concrete Institute —Code Requirements for Determining ACI 216.1 216.1M-14—Code Fire Resistance of Concrete and Masonry Construction Assemblies ACI 352R-02—Recommendations 352R-02—Recommendations —Recommendations for Design of BeamColumn Connections in Monolithic Reinforced Concrete Structures ACI 442R-88—Response of Concrete Buildings to American Society of Civil Engineers

 other Structures International Code Council IBC 2015 International Building Code

ratings to be performed in accordance with ACI 216.1M for concrete, concrete masonry, and clay masonry members.

 Ali, M. M., and Moon, K. S., 2007, “Structural Development in Tall Buildings: Current Trends and Future Prospects,” Architectural Science Review 205-223.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf



 Structural engineers mathematically model reinforced concrete structures, in part or in whole, to calculate member moments, forces, and displacements under the design loads



Analysis type

Applicable member or assembly

Analysis tool

First-order

conditions, equilibrium of forces and compatibility of of individual members for input into the model, under both service loads and factored loads, are discussed in detail in ACI 318M-14, Chapter 6. The factored moments and forces resulting from the analysis are used to determine the required strengths for individual members. The calculated displacements and drift are also checked against commonly accepted serviceability limits.

Static load Hand calculations

Analysis tables Continuous one-way slab

Two-way slab

 3.2.1 General—The analysis of concrete structures “shall satisfy compatibility of deformations and equilibrium of forces,” as stated in Section 4.5.1 of ACI 318M-14. The

Beam

Analysis tables

Continuous beam

6.5 of ACI 318M-14

Column

Interaction diagrams

Wall

Alternative method for out-ofplane slender wall analysis in Section 11.8 of ACI 318M-14

Two-dimensional frame

Portal method in Section 3.3.3 of this chapter

First-order

3.2.2 Elastic analysis—Most concrete structures are modeled using an elastic analysis. The stability of columns

Static load Computer programs

the loads applied on the laterally deformed structure. A series of analyses are made where the secondary moment from each analysis is added to the subsequent analysis until equilibrium is achieved. 3.2.3 Inelastic analysis—Inelastic second-order analysis determines the ultimate capacity of the deformed structure. The analysis may take into account material nonlinearity, tion with the supporting foundation. The resulting strength must be compatible with results of published tests. This design of materials and systems not covered by the code; and evaluation of building performance above code minimum

Gravity-only systems

Spreadsheet program based on the analysis tools for hand calculations above Program based on matrix assemblies for gravity loads

Second-order Static or dynamic load Computer programs Second-order Inelastic

for second-order, or P

Equivalent frame method in Section 8.11 of ACI 318M-14

Interaction diagrams

of analysis as long as these conditions are met. ACI 318M-14,

order. In addition, ACI 318M M permits the use of strut-and-tie modeling for the analysis of discontinuous regions. Except as noted in Chapter 18 18,, ACI 318M M provisions state that the designer may assume that reinforced concrete members behave elastically under design loads. It is also generally acceptable to model concrete members with constant sectional properties along the member length. These assumptions simplify analysis models but they may

6.5 of ACI 318M-14 Direct design method in Section 8.10 of ACI 318M-14

Two-dimen Two-dimensional frames and walls

Program based on matrix methods without iterative capability

Two-dimensional frames and walls

Programs based on matrix methods with iterative capability

Three-dimensional structure Threedimensional structure

element methods with iterative capability Beyond the scope of this Handbook

Information can downloaded from the ACI website; refer to Table of Contents.

include discussion of the inelastic analysis approach. 3.2.4 Strut-and-tie—The strut-and-tie method in Chapter 23, ACI 318M-14, is another analysis method that is permitted by ACI 318M. This method does not assume that plane sections of unloaded members remain plane under loading. Because this method also provides design provisions, it is considered both an analysis and design method. This method is applicable where the sectional strength assumptions in ACI 318M, Chapter 22, do not apply for a discontinuity region of a member or a local area. 3.2.5 Analysis types and tools—

  

Eng.Mohammed Ashraf





Fig. 3.3.3—Frame analyzed by portal method.

order inelastic. Table 3.2.5 shows some common analysis

 3.3.1 General—Before computers became widely avail-

such as the portal method, to calculate frame moments and shears due to lateral forces. In very limited applications, the design of an entire building using hand calculations is still possible with today’s set of building codes. For the large design approach is not practical due to the large number and complexity of design load combinations necessary to fully meet ASCE 7-10 requirements. shear— 3.3.2 Code design equations for moment and shear shear—The The nary estimating or member sizing, for designing isolated members or subassemblies, and to complete rough checks of computer program output. Because these equations and expressions are easy to incorporate into electronic spreadsheets and equation solvers, they continue to be helpful. In the member chapters of this handbook, examples of hand calculations are provided. 3.3.3 Portal method—The portal method was commonly used before computers were readily available to calculate a frame’s moments, shears, and axial forces due to lateral forces a design tool with the widespread use of commercial design software programs. The portal method has limitations as stated in the assumptions and considerations that follow, but is still a useful tool for the designer. With complex, three-dimensional modeling becoming commonplace, there is always a chance of modeling error. The portal method allows the designer to

of column and midspan of beams

These assumptions reduce a statically indeterminate problem to a statically determinate one. The following should be considered when using this method: changes in story height, and large changes in member from those calculated by a computer analysis. deformation calculated by a computer analysis.

 General—Computing power and structural software 3.4.1 General—Computing

programs and specialized analysis tools have been devel developed, taking advantage of increasing computer speeds. design structures. A multistory building only takes minutes of computing time on a personal computer compared to the past, when it would take several hours on a large mainframe computer. Computer software has also greatly improved: user interfaces have become more intuitive; members can be automatically meshed; and input and output data can be reviewed graphically and tabular in a variety of preproAlthough three-dimensional models are becoming commonplace, many engineers still analyze the building as a series of two-dimensional frames. Matrix methods mentioned in Table 3.2.5 are programs based on the direct

shears in a frame. This can be very useful for spot-checking The basics assumptions to the portal method are:

divided into multiple elements to account for changes in member properties along its length. The two-dimensional

loads elements connected at nodes. Each member of the structure consists of multiple rectangular elements, which more   

Eng.Mohammed Ashraf





accurately determine the behavior of the member. The act of modeling members as an assembly of these discrete elements is called “meshing” and can be a time-consuming task. A large amount of data is generated from this type of analysis, which may be tedious for the designer to review and process.

and the strut-and-tie method, however, are more commonly used to analyze walls with openings. For lateral load analysis, all of the parallel plane frames in a building are linked into one plane frame to enforce lateral deformation compatibility. Alternately, two identical frames

analyze the structure by dividing it into parts and using less complicated programs to analyze each part. 3.4.2 Two-dimensional frame modeling—A building can be divided into parts that are analyzed separately. For example, structures are often symmetrical with regularly spaced columns in both directions. There may be a few isolated areas of the structure where columns are irregularly spaced. These columns can be designed separately for gravity load and checked for deformation compatibility

by doubling the modulus of elasticity. Structural walls, if

the structure. Buildings designed as moment-resisting frames can often The complete structure is modeled using orthogonal sets at crossing points is not required. The geometry of beams moment frames, it may be possible to model according to the equivalent frame method in Section 8.11 in ACI 318M. For beam-column moment frames, it is permitted to model T-beams, with the limits on geometry given in Section 6.3.2 M; however, it is often simpler to ignore the slab of ACI 318M; and model the beams as rectangles. For beams in intermeinterme diate or special moment frames, the assumption of a rectrect angular section may not be conservative; refer to Sections 18.4.2.3, 18.6.5.1, and 18.7.3.2 in ACI 318M. M. if the beam spans are assumed to extend between column centerlines and the beam is modeled as prismatic along the entire span. Many computer programs thus allow for the beam to be modeled as spanning between faces of columns. To do this, a program may add a rigid zone that extends from the face of the column to the column centerline. If the program does not provide this option, the designer can

Walls with aspect ratios of total height to width greater than 2 can sometimes be modeled as column elements. A thin wall may be too slender for a conventional column analysis, and a more detailed evaluation of the boundary elements and panels may be necessary. Where a beam frames into a wall that is modeled as a column element, a rigid link should be provided between the edge of the wall and centerline of the element, rotational compatibility should be assured. wall with openings can be modeled as a frame, but the rigidity

ered after the lateral deformation compatibility analysis is run. For seismic loads, rigid diaphragms are required to account for accidental torsion according to Section 12.8.4.2 in ASCE 7-10. For wind loads, a torsional moment should be applied according to Fig. 27.4-8 in ASCE 7-10. 3.4.3 Three-dimensional modeling—A three-dimensional model allows the designer to observe structural behavior of structural irregularities and torsional response can be directly analyzed. Current computer software that provides ysis with automatic meshing. These high-end programs are capable of running a modal response spectrum analysis, seismic response history procedures, and can perform a host of other time-consuming mathematical tasks. times modeled as rigid diaphragms, reducing the number horizontal translations and one rotation about a vertical rigid if the following conditions are met:

If a rigid diaphragm is assumed, the stresses in the diaphragm are not calculated and need to be derived from the diaphragm requires more computation power but provides a distribution of lateral forces and calculates slab stresses. A semi-rigid diaphragm can also be helpful in analyzing torsion commentary in Section 12.8 of ASCE 7.  3.5.1 Arrangement of live loads—Section 4.3.3 in ASCE 7 states that the “full intensity of the appropriately reduced live load applied only to a portion of a structure or member shall be accounted for if it produces a more unfavorable load or member.” This is a general requirement that acknowledges greater moments and shears may occur with a pattern load than with a uniform load. There have been a variety of methods used to meet this requirement. Cast-in-place concrete is inherently continuous, and Section 6.4 in ACI 318M provides acceptable arrangements of pattern live load

link should be modeled from the centerline of the wall to the

  

Eng.Mohammed Ashraf





Fig. 3.4.2a—Element and frame analogies.

Fig. 3.4.2b—Idealization for plane frame analysis. 3.5.2 continuous beams and one-way slabs—Section 6.5 in ACI 318M provides approximate equations for conservative

k

already applied. Section 6.6.3.1.2 of ACI 318M also allows Ig for all members in a lateral load analysis. This is helpful for hand-calculation methods such as the portal method.

used more often to estimate initial member sizes for computer 3.5.3 First-order analysis order analysis are provided in Section 6.6 in ACI 318M. 3.5.3.1 Section properties—Section properties for elastic k of 0.875 already applied. These properties are acceptable for the analysis of the structure for strength design. For service level load analysis, the moment of inertia values in Table

and concrete compressive strength. These equations can also

used for moment of inertia discussed previously is for global building behavior. The moment of inertia for second-order k of 0.75, as discussed in R6.6.4.5.2 of ACI 318M. 3.5.3.2 318M assumes that only primary stresses are calculated.

by the design loads are not calculated. First-order analysis is typical when hand-calculation methods are used or basic matrix analysis computer programs are used that are not programmed for iterative analysis. This method ignores the P order moments caused by vertical loads acting on the build-

by using the factored axial load and moment, as presented, for any given axial load and moment. These moment-of  

Eng.Mohammed Ashraf





ties are used with this method, but the results of a secondorder analysis is a nonlinear solution. This is referred to as “geometric nonlinearity.” This means that the load cases cannot be computed separately and then combined for the calculation of the secondary moments. Software should be checked to determine how it accounts for P moment due to building lateral deformation but some software does not calculate the secondary moments along the member length. The designer may have to model the column a column member is modeled by two elements, the designer

that illustrates the options to account for slenderness. In summary, slenderness can be neglected if the column or wall meets the requirement of Section 6.2.5 in ACI 318M. If slenderness cannot be neglected, the next step is to determine if the building story being analyzed is sway or nonsway. If the story is nonsway, the column or wall end moments are only P P

P P 3.5.3.3 Superposition position to be used when combining loads. This is very helpful when performing hand calculations. The designer calculates the member moment, shear, and axial load for each load. The reactions are then superimposed according to the applicable load combination. Many hand-calculation that the designer is performing a linear analysis with super super3.5.3.4 Redistribution of moments—ACI 318M allows for shears by taking advantage of the ductility provided through the code detailing requirements. Ductile detailing is required redistribution can be very helpful in creating economical tion may permit the designer to specify uniform beam sizes over multiple beam spans. If column spacing is not uniform, some beam design moments may be slightly lower than the at factored loads, however, the moments will redistribute to regions that have not yet yielded, and the beam design moments will satisfy the beam required moments throughout the multiple beam spans. 3.5.4 Second-order analysis—In Section 6.7 of ACI loads on the laterally deformed structure is included in the computer analysis. The initial P due to story drift are computed. A computer algorithm then automatically carries out a series of iterative analyses using

culty of appropriately capturing the secondary moment along the column length, many programs calculate the secondary moments due to lateral deformation and use 6.6.4.5 of ACI 318M in a post-processing program to account for the secondary moment along the length of the column. 3.5.5 Inelastic second-order analysis—The consideration of the nonlinear behavior of structures arising from nonlinearity of the stress-strain curve for concrete and steel reinforcement, particularly under large deformations, ties in structural response, whether arising from material

numerical iterative or step-by-step procedures. For inelastic second-order analysis, the principle of superposition should Handbook. Several references that provide further informa information on nonlinear analysis are ASCE 41-13, FEMA report, 3.5.6 Finite element analysis of concrete structures is permitted by ACI 318M and can be inelastic second-order analyses as long as the element types are compatible with the response required. Section 6.9 in ACI 318M was added to acknowledge that

analysis and have sophisticated auto-mesh capabilities. Finite element analysis is a tool that may be used for either linear or nonlinear analyses, but care should be exercised in selecting element types, numerical solver methods, and nonlinear element properties.  For seismic loads, the structure may go through multiple

even though the structure will actually behave inelastically.

  

Eng.Mohammed Ashraf





displacements and the estimated behavior. Detailed explanations of these factors and their application can be found in ASCE 7 and FEMA P-750. For reinforced concrete structures, ACI 318M provides structures with the ability to deform inelastically by enforcing special seismic detailing requirements. The seismic detailing requirements in Chapter 18 of ACI 318M are additive to the detailing requirements in the member chapters, or the seismic detailing requirements supersede the member chapter requirements. The detailing requirements loads do not govern the required strength of the structure. Fig. 3.6—Inelastic force-deformation curve (SEAOC Seismology Committee 2008).

 American Concrete Institute ACI Committee 442-77—Response of Buildings to

American Society of Civil Engineers loads must comply with several limitations. Depending on and type, and the type of structural irregularities, a Modal

tation of Existing Buildings

or Seismic Response History procedure, either linear or buildings in SDC A and B up to 50 m in height. Section 12.3 of ASCE 7 provides limitations related to

Federal Emergency Management Agency Analysis Procedures -

 Deierlein, G. G.; Reinhorn, A. M.; and Willford, M. R.,

-

NEHRP Seismic Design Technical Brief No. 4 (NIST GCR 10-917-5) Gaithersburg, MD, 36 pp. Hibbeler, R., 2015, Structural Analysis, ninth edition,

7 and ACI 318M have additional requirements to account for

R factor Cd

to Seismic Design Factors,” Structure Magazine, Sept., pp. 30-32. http://www.structurearchives.org/article. aspx?articleID=756

  

Eng.Mohammed Ashraf

  Durability of structural concrete is its ability, while in service, to resist possible deterioration due to the surrounding environment, and to maintain its engineering properties. This can be accomplished by proper proportioning and tion, detailing, and construction practices. The ACI 318M Code provides minimum requirements to protect the structure against early serviceability deterioration. Depending on

exposure conditions, structural concrete may be required to resist chemical or physical attack, or both. The attack mechanisms the Code covers include exposure to freezing and thawing, soil and water sulfates, wetting and drying, and reinforcement corrosion due to chlorides. All these failure mechanisms depend on transport of water through concrete. For this reason, it is essential to understand the mechaconcrete’s resistance to these mechanisms. 4.1.1 Permeability ability of concrete to resist penetration by water or other

resistant to resaturation, freezing and thawing, sulfate and chloride ion penetration, and other forms of chemical attack

duracapillary pores in cement paste are relevant to concrete dura bility, as they are responsible for the transport properties of ence of capillary porosity in cement paste on permeability was w/c

Fig. 4.1.1a—Permeability versus capillary porosity for pastes (Powers 1958).

w/c and duration of the moist curing on permeability is presented in Fig. 4.1.1b. 4.1.2 Freezing and thawing—When thawing water freezes in concrete, it causes cement paste to dilate destructively by generating hydraulic and osmotic pressure. While hydraulic capillary cavities, osmotic pressure is produced by water

w/c w/c and curing duration on permeability (leakage) of mortar (Kosmatka   

Eng.Mohammed Ashraf





w/c ratio on sulfate resistance for rating indicates better resistance) (Stark 1989). prior to freezing-and-thawing exposure, which is demondemon strated by data presented in Fig. 4.1.2. 4.1.3 Sulfates Sulfates—Sulfates —Sulfates present in soil and water can react with hydrated compounds in the hardened cement paste However, formation of new crystalline substances due to those reactions is partly responsible for the expansion. If the volume of growing crystals cannot exceed the space available to them. At the same time, however, the swelling w/c, air-entrainment, and curing/drying on resistance to freezing and thawing of concrete (Kosmatka et al. 2008). Hydraulic pressures in cement paste are generated by the 9 percent expansion of water when it freezes and changes to ice. For the freezing to take place, a capillary has to reach its

concentrations of alkali solutions in the paste. When pressure in concrete due to freezing exceeds the tensile strength of concrete, some damage occurs, especially if concrete is saturated with water and exposed to repeated cycles of freezing and thawing. The resulting damage is cumulative as it increases with additional repetitions of freezing and thawing cycles. Deterioration due to freezing and thawing can appear in the form of cracking,

teristics needed for concrete to be frost resistant, while airentraining admixtures are used to control the pressure generated in concrete paste during freezing-and-thawing cycles. In other words, high resistance to freezing and thawing is associated with entrained air, low w/c, and a drying period

into the gel pores, which disturbs the equilibrium between the gel and its surrounding liquid phase, resulting in the movement of more water from the outside into the gel pores Although ordinary portland cements are most susceptible to sulfate attack, the use of sulfate-resistant cements will to sulfate attack can be greatly increased by decreasing the permeability of concrete through reduction of the waterw/cm 4.1.4 Corrosion layer on reinforcing steel. If, however, chloride ions are present in concrete, they can reach and disrupt that layer and lead to corrosion of steel in the presence of water form and may cause cracking, spalling, or delamination of concrete. This allows for easier access of aggressive agents to the steel surface and increases the rate of corrosion. Cross-sectional area of the corroding steel will decrease and the load-carrying capacity of the member will be reduced Chlorides can be introduced to concrete with materials

  

Eng.Mohammed Ashraf





• reduce likelihood of corrosion initiation, total chloride-ion content should not exceed a certain concentration value, referred to as the chloride threshold. A literature review of reported chloride threshold values revealed “that there is no single threshold value, but a range based on the conditions and materials in use” and were found to vary from ACI 318M-14 limits water-soluble chlorides to 0.15 percent by mass of cement for concrete exposed to external chlorides

Corrosion of the reinforcing steel in concrete can be reduced or prevented by minimizing the w/cm sion resistant steel.  To produce durable structural concrete, concrete materials and mixture proportions are selected based on design strength requirements, anticipated exposure conditions, and required service life of the structure. The selection of materials and mixture proportions has to be accompanied by appropriate

As stated in Section 1.3.1 of ACI 318M-14, “The The purpose of this Code is to provide for public health and safety by estabestab lishing minimum requirements for strength, stability, serviceservice structures. Section 4.8 ability, and integrity of concrete structures.” of ACI 318M-14 addresses global durability requirements related to material selection for concrete mixtures and corrosion protection of reinforcement. ACI 318M-14, Chapters 19 and 20,, provide detailed durability requirements for concrete and reinforcing steel, respectively. ACI 318M-14, Chapter

The Code’s durability focus is mainly on concrete resisw/cm and the composition of those materials. The use of

drying creep and shrinkage, and freezing and thawing May reduce resistance to deicer scaling The Code does not cover all topics related to concrete durability. It does not include recommendations for extreme •

preventive maintenance; however, the topic is not explicitly addressed in the Code. Additionally, the Code does not cover service life prediction. Information related to these topics are found in other ACI documents, including: • ACI 201.2R—Guide to Durable Concrete • ACI/TMS 216.1M—Code Requirements for Determining Fire Resistance of Concrete and Masonry Construction Assemblies • ACI 221.1R—Report on Alkali-Aggregate Reactivity • ACI 362.1R—Guide for the Design and Construction of Durable Concrete Parking Structures • ACI 222R—Protection of Metals in Concrete Against Corrosion • ACI 222.2R 222.2R—Report on Corrosion of Prestressing Steels • ACI 222.3R—Guide to Design and Construction Practices to Mitigate Corrosion of Reinforcement in Concrete Structures • ACI 224.1R—Causes, 224.1 —Causes, Evaluation, and Repair of Cracks in Concrete Structures Guide for Concrete Inspection • ACI 311.4 311.4R—Guide • • ACI 515.2R—Guide 515.2 —Guide to Selecting Protective Treatments for Concrete • ACI 562—Code Requirements for Evaluation, Repair, and Rehabilitation of Concrete Buildings and Commentary   The Code addresses durability by requiring that four exposure categories be assigned to each concrete member. The four exposure categories are: 1. F: concrete exposed to moisture and cycles of freezing

silica fume, calcined shale, calcined clay or metakaolin, or S: concrete in contact with soil or water containing deleterious amounts of water-soluble sulfate ions; 3. W: concrete in contact with water but not exposed to freezing and thawing, chlorides, or sulfates; 4. C: concrete exposed to conditions that require additional protection against corrosion of reinforcement. Each exposure category is divided into exposure classes 2.

in many ways, depending on their type, dosage, and other mixture proportions and composition. In general, SCMs have the following impacts on hardened concrete properties • •

Increase long-term strength

• • •

Reduce permeability and absorption Improve resistance to corrosion

assigned exposure classes and the concrete mixtures for these members satisfy those various requirements, the Code’s minimum durability requirements are met. 4.3.1 Freezing and thawing (F)—The volume of ice is 9 percent larger than water. As water freezes in satu-

  

Eng.Mohammed Ashraf





to internal pressure, which then causes concrete tensile stresses. If those stresses are greater than the tensile strength of concrete, cracking will occur. The cumulative expansion after many cycles of freezing and thawing may lead to signiffreezing-and-thawing damage is to reduce moisture penetration so it does not become critically saturated; however, this is not always possible. The other method is to generate small air bubbles in fresh concrete by addition of an air-entraining admixture, which creates voids for the freezing water to expand into without creating internal stress. The Code requires concrete in structural members exposed to cycles of freezing and thawing to be protected by using airresistance of saturated concrete to freezing and thawing. ACI 212.3R provides an in-depth discussion on these mate-

on nominal maximum aggregate size and concrete compressive strength. To achieve similar freezing-and-thawing protection, higher air content is generally required for concrete mixtures with smaller nominal maximum aggre aggregate size. For example, concrete with 10 mm aggregate requires 50 percent higher air content than concrete with 50

the nominal maximum aggregate size for each concrete aggregate size depends on locally available aggregates, as well as construction issues such as size of formwork, member depth, and clear bar spacing. The criteria for maximum size selection are given in Section 26.4.2.1 of ACI 318M-14. Table 19.3.3.1 lists target air content for Classes F1, F2 and F3, depending on the nominal maximum aggregate size. compressive strength. An air content reduction of 1 percent

particular member. Interior members, foundations below the frost line, or structures in climates where freezing temperatures are not anticipated are assigned Exposure Class F0. These conditions, therefore, do not require air entrainment and there is no limit on maximum w/cm or on the use of cementitious materials. The minimum compressive strength for concrete in Exposure Class F0 is the Code minimum: 17 MPa. that is not critically saturated. Structural members exposed to freezing and thawing cycles, but with low likelihood of being saturated, are assigned exposure class F1. Concrete for

In addition, the concrete should have maximum w/cm of 0.55 and at least 24 MPa compressive strength. Exposure Classes F2 and F3 are assigned to concrete in structural members with a high likelihood of water saturation during freezing. The distinction between the two classes is that Class F2 anticipates no exposure to deicing chemicals or seawater, while Class F3 does. Concrete in F2 and w/cm of 0.45 and 0.40, respectively. The minimum concrete compressive strengths for F2 and F3 classes are 31 and 35 MPa, respectively. F3, also has a limit on The most severe class of exposure, F cementitious materials in concrete mixtures, given in ACI The summary of requirements for concrete in Exposure Category F is listed in Table 19.3.2.1 of ACI 318M. 4.3.2 Sulfate (S)—All (S)—All soluble forms of sulfate, sodium, on concrete. Depending on the sulfate form, they react with hydrated cement phases and result in formation of ettringite or gypsum. Depending on the reaction product, the concrete

ingress, is to use cements with a low content of tricalcium 3

reason for air content reduction is that concretes with higher strengths are characterized by lower w/cm and reduced porosity, which improve resistance to freezing-and-thawing cycles. For example, a structural member in Exposure Class F2 with 13 mm nominal maximum aggregate size requires

Exposure Category S applies to structural members that predominantly come from exposure to soil, groundwater, 2– 4

should be determined in accordance with ASTM C1580 for soil samples and with ASTM D516 or ASTM D4130 for -

allows tolerance for air content in as-delivered concrete of ±1.5 percentage points. This is consistent with the tolerances

Additional requirements or limitations, such as minimum compressive strength, minimum w/cm, or limits on cementitious materials, depend on the exposure class assigned to a

The reason for lower class for seawater is the presence of chloride ions, which inhibit expansive reaction due to sulfate attack. Class S0 is assigned to concrete in members not exposed to sulfates and there is no restriction on w/cm, or type or limit on cementitious materials. The only requirement for concrete

  

Eng.Mohammed Ashraf



least 17 MPa. Greater minimum compressive strength and maximum w/cm limits are imposed on concrete in Exposure Classes S1 through S3. For these exposure classes, the type A summary of all requirements for concrete in Exposure Category S is listed in Table 19.3.2.1 of ACI 318M-14. 4.3.3 In contact with water (W)—The durability of structural members in direct contact with water, such as founby water penetration into or through concrete. Apart from

concrete permeability is to keep the w/cm low. Concrete for members assigned to Exposure Class W0 has no unique requirements except that it has a minimum compressive strength of 17 MPa. Concrete in structural members assigned to Exposure Class W1 requires low permeability. Table 19.3.2.1 of ACI 318M-14 requires w/cm not to exceed 0.50 and compressive strength to be forcement corrosion, sulfate exposure, or exposure to cycles of freezing and thawing. Recommendations for the design and construction of water tanks and reservoirs are provided R, and ACI 372R. in ACI 350.4R, ACI 334.1R, Requirements for concrete in Exposure Class W are listed M-14 in Table 19.3.2.1 of ACI 318M-14. —Corrosion of reinforcement may 4.3.4 Corrosion (C)—Corrosion member. Reinforcement corrosion usually occurs as a result of the presence of chlorides or steel depassivation due to



of cement weight, are 1 percent for Class C0, 0.30 percent for Class C1, and 0.15 percent for Class C2. Chloride limit for prestressed concrete is 0.06 percent by cement weight, regardless of exposure class. Apart from chloride limits, Exposure Classes C0 and C1 have no additional requirements, as there is no limit on w/cm and the minimum compressive strength is 17 MPa. Class C2 requires concrete strength of at least 35 MPa, a maximum w/cm cover to satisfy the Code’s minimum concrete cover provisions. The minimum concrete cover depends on exposure to weather, contact with ground, type of member, type of reinthe member is prestressed. Tables 20.6.1.3.1, 20.6.1.3.2, and 20.6.1.3.3 of ACI 318M-14 provide cover provisions for castin-place nonpressessed, cast-in-place pressessed, and precast nonpressessed or pressessed concrete members, respectively. If the design requires bundled bars, check Section 20.6.1.3.4 requirements in corrosive environments or other severe exposure conditions are more stringent and are provided in Section 20.6.1.4 of ACI 318M-14. Requirements for concrete in Exposure Class C are listed in Table 19.3.2.1 of ACI 318M-14. 318   Durability requirements are met once concrete propor proportions and properties satisfy the minimums set by the Code. To assure that the delivered concrete achieves the desired acceptance criteria consistent with ACI 318M-14, Section

than the original steel and therefore exert internal pressure on the surrounding concrete, causing it to crack or delamito increased steel stresses under service load and reduced member nominal strength. Because moisture and oxygen must be present at the steel surface for corrosion to occur, the quality of concrete and the reinforcing bar cover are of great importance. Corrosion can be mitigated by proper mixture design and construction practices; application of sealers, coatings, or membranes that protect concrete from moisture and chloride penetration; use of corrosion resistant reinforcement; or inclusion of corrosion inhibitors in the mixture to elevate the corrosion threshold concentration. Refer to ACI 222R, ACI 222.3R, and ACI 212.3R for additional information. Each exposure class within the corrosion exposure category has a limit on water-soluble chloride-ion content in concrete. The chloride-ion content is measured in accordance with ASTM C1218/C1218M, which requires the sample be representative of all concrete-making ingredients—that is, admixtures. Because chloride limits are imposed even on concrete in Exposure Class C0, all structural concrete must comply with the Code’s maximum chloride ion limits. Chloride limits for nonprestressed concrete, expressed as percent

26.13. Section 26.13  The following examples illustrate one approach of implementing minimum durability requirements of the Code. In some cases, durability requirements for material properties may exceed those of the structural design. This is more likely for severe exposure conditions, which require a minimum compressive strength of 35 MPa. In some cases, SCMs may be required, which may extend setting time and tion schedule. For these reasons, cooperation with engineers experienced with concrete materials and mixture proportioning, and with concrete suppliers, is recommended. 4.5.1 moisture or freezing and thawing—Consider the design of building. It is located in a climate zone with frequent freezing-and-thawing cycles; however, the slab will be constructed during summer and the temperatures at night during construction are expected to remain above 4 to 7°C. It is desirable for the slab to quickly gain strength to meet the construction schedule. For this reason, calcium chloride was proposed as an accelerating admixture. The required

  

Eng.Mohammed Ashraf





minimum compressive strength, from structural analysis, is 28 MPa. The slab is 175 mm thick with top and bottom mats bility requirements? Answer: every exposure category to each structural member or group

Assign exposure classes within each exposure category

is to assign exposure classes within every exposure category minimum durability requirements. The step-by-step instructions are as follows: Step description/ action item

step-by-step instructions are as follows: Step description/ action item

Answer: Durability requirements are met once the most

Code reference

Selection and discussion

Assign exposure classes within each exposure category

F2 freezing-and-thawing cycles

Code reference Table 19.3.1.1

S0

Table 19.3.1.1

F0

Selection and discussion

S0

W0 requirements for low

W0 requirements for low

C1 moisture but not to an external

C0 Assign required minimum compressive strength

17 MPa

Assign maximum w/cm

Not limited

Assign minimum concrete cover

20 mm

Assign nominal maximum size of aggregate

-

31 MPa 31 MPa is greater than design strength of 28 MPa, the 31 MPa governs

Table 19.3.2.1

Table 19.3.2.1

Assign required minimum compressive strength

Table 19.3.2.1

Assign maximum w/cm

0.40

Table 19.3.2.1

Assign minimum concrete cover

40 mm

Table 20.6.1.3.1

Assign nominal maximum size of aggregate

50 mm thickness, 3/4 x 150 mm clear use 25 mm as readily available

Assign required air content

6% ± 1.5%

Table 20.6.1.3.1

50 mm spacing – top and bottom mat, or use 25 mm as readily available

Assign required air content

Not air entrained

Table 19.3.2.1

Assign limits on cementitious materials

No limits

Table 19.3.2.1

Assign limits on calcium chloride admixture

No restriction

Table 19.3.2.1

content in concrete.] Assign maximum water-soluble –

content in concrete, percent by weight of cement

1.00 soluble chloride-ion content from all concrete ingredients determined by ASTM C1218/ C1218M at age between 28 and

be substituted with 20 mm aggregate with no air content change] Assign limits on cementitious materials

No limits

Table 19.3.2.1

Assign maximum water-soluble

0.30 ion content from all concrete ingredients determined by ASTM C1218/1218M at age

Table 19.3.2.1

Consult ASTM C94/C94M, ACI 306R, and ACI 301 for guidance on temperature limits for concrete delivered in cold weather.

Section 26.5.4.1

–

Table 19.3.2.1

4.5.2 freezing and thawing located in a climate zone with frequent freezing-and-thawing cycles. It is anticipated that the balconies will be exposed to moisture, but not deicing salts. The required minimum compressive strength, from structural analysis, is 28 MPa.

Table 19.3.3.1 and Section

content in concrete, percent by weight of cement Provide guidance on cold weather construction

4.5.3 soil and deicing salts while in service cast-in-place, nonprestressed foundation wall of a partially underground parking structure. The structure is located in a northern climate zone with frequent freezing-and-thawing 2– 4

13 bars spaced at 150 mm What additional information is needed for balcony concrete to meet durability requirements?

the nearby streets and a sidewalk. It is desirable for the foun-

  

Eng.Mohammed Ashraf



dation wall to quickly gain strength to reduce possible frost damage and to meet the construction schedule. The required minimum compressive strength, from structural analysis, is 28 MPa. The foundation wall is 200 mm thick with inside tion wall concrete to meet durability requirements? Answer: every exposure category to each structural member or group

The step-by-step instructions are as follows:



Assign limits on cementitious materials

Table 19.3.2.1 Table Table be tested in accordance with ASTM C1012/C1012M and meet the maximum expansion

Assign limits on calcium chloride admixture

Not permitted

Table 19.3.2.1

Assign maximum water-soluble

0.15 soluble chloride ion content from all concrete ingredients determined by ASTM C1218/ C1218M at age between 28 and

Table 19.3.2.1

–

Step description/ Action item Assign exposure classes within each exposure category

Selection and discussion F3 freezing-and-thawing cycles with frequent exposure to water and exposure to deicing

Code reference Table 19.3.1.1

 American Concrete Institute ACI 201.2R-08—Guide to Durable Concrete ACI 212.3R-10—Report on Chemical Admixtures for Concrete ACI 222R-01 222R-01—Protection of Metals in Concrete Against Corrosion 222.3 Guide to Design and Construction Prac ACI 222.3R-11—Guide Practices to Mitigate Corrosion of Reinforcement in Concrete Structures

S3 in direct contact with soluble W1 water and low permeability is C2 moisture and an external source of chlorides from deicing Assign minimum compressive strength

35 MPa because 35 MPa is greater than design strength of 28 MPa, 35 MPa governs

Table 19.3.2.1

Assign maximum w/cm

0.40

Table 19.3.2.1

Assign minimum concrete cover

50 mm – outside face of wall

Table 20.6.1.3.1 20.6.1.4.1

Table 20.6.1.3.1 for exposure to weather or in contact with

content in concrete, percent by weight of cement

ACI 306R306R-10—Guide Guide to Cold Weather Concreting ACI 334.1R-92—Concrete 334.1 —Concrete Shell Structures-Practice and Commentary ACI 350.4R-04—Design Environ350.4R-04—Design 04—Design Considerations for Environ mental Engineering Concrete Structures 372R-13— ACI 372R-13—Guide to Design and Construction of Circular Wire- and Strand-Wrapped Prestressed Concrete Structures ASTM International

smaller; cover increased to 50

Ready-Mixed Concrete ASTM C1012/C1012M-13—Standard Test Method for

20 mm – inside face of wall to weather or in contact with Assign nominal maximum size of aggregate

Assign required air content

40 mm thickness, 3/4 x 85 mm clear bar spacing – between interior and exterior mats of reinforcing use 40 mm

Section

5.5% ± 1.5%

Table 19.3.3.1

1. Changing to lower nominal maximum aggregate size will require higher air content; able if concrete compressive strength exceeds 35 MPa; refer to 19.3.3.3]

Sulfate Solution ASTM C1218/C1218M-99—Standard Test Method for Water-Soluble Chloride in Mortar and Concrete Portland Cement ASTM D516-11—Standard Test Method for Sulfate Ion in Water

Mixing ASTM C1580-09-Standard Test Method for WaterSoluble Sulfate in Soil ASTM D4130-15—Standard Test Method for Sulfate Ion in Brackish Water, Seawater, and Brines

  

Eng.Mohammed Ashraf





 Hewlett, P. C., ed., 1998, Lea’s Chemistry of Cement and Concrete Toronto, 1057 pp.

Powers, T. C., 1958, “Structure and Physical Properties of Hardened Portland Cement Paste,” Journal of the American Ceramic Society Stark, D., 1989, “Durability of Concrete in Sulfate-Rich

Design and Control of Concrete Mixtures (EB001), 14th

14 pp.

Design and Control of Concrete Mixtures (EB001), 15th edition, Port“Threshold Chloride Content for Corrosion of Steel in nation of Issues in Concrete Practice, American Concrete Institute, publisher, Farmington Hills, MI, 510 pp. -

  

Eng.Mohammed Ashraf



 A one-way slab is generally used in buildings with vertical supports (columns or walls) that are unevenly spaced, creating a long span in one direction and a short span in the perpendicular direction. One-way slabs typically span in the short direction and are supported by beams in the long direction. During preliminary design, the designer determines the loads and spans, reinforcement type (post-tensioned [PT] or nonprestressed), and slab thickness. The designer determines the concrete strength based on experience and the Code’s exposure and durability provisions. This chapter discusses cast-in-place, nonprestressed, and PT slabs. The Code allows for either bonded or unbonded tendons in a PT slab. Because bonded tendons are not usually used in slabs in the United States, this chapter will address PT slabs with unbonded tendons. At times, the design of a one-way slab will require point load considerations, such as wheel loads in parking garages. These result in local shear forces on the slab, requiring verishear is addressed in Chapter 6 for two-way slabs in this Handbook. For relatively small slab openings, trim bars can be used concentra to limit crack widths caused by geometric stress concentrations. For larger openings, a local increase in slab thickness, as well as additional reinforcement, may be necessary to provide adequate serviceability and strength.  ACI 318M allows for the designer to use any analysis ibility, as long as design strength and serviceability require-

calculate moments and shears.  5.3.1 Minimum thickness—For nonprestressed slabs, the Code allows the designer for slabs not supporting or attached to partitions or other construction likely to be or simply satisfy a minimum slab thickness (Section 7.3.1, ACI 318M-14). In the case where loads are heavy, nonuni-

Fig. 5.3.3—Load balancing concept. very heavy live load, superimposed dead load, or both, and

by any appropriate method, such as classical equations or software results. Note that the spacing of slab reinforcing bar to limit crack width, timing of form removal, concrete quality, timing of construction loads, and other construction variables all

calculations. In addition, creep over time will increase the

small. If the designer limits the maximum net concrete tensile stress to below cracking stress under service loads, stress—Nonprestressed slabs 5.3.3 Concrete service stress are designed for strength without reference to a concrete

stresses is a critical part of the design. In Section 8.3.4.1 of tive moment areas at columns in PT slab is limited to 0.50 f c . At positive moment sections, Section 8.6.2.3 of ACI 318M-14 requires slab reinforcing bar if the concrete tensile stress exceeds 0.17 f c limits are below the concrete cracking stress of 0.62 f c , thus 8.6.2.1 of ACI 318M-14 requires a PT slab’s axial compressive stress in both directions, due to PT, to be at least 0.9 MPa.

and tendon force are directly related to slab forces and moments created by PT. A common approach to calculate PT slab moments is the use of the “load balancing” concept.

limits (Section 24.2.2, ACI 318M-14). The Code does not provide a minimum thickness-tospan ratio for PT two-way slabs, but Table 9.3 of The PostTensioning Manual (Post-Tensioning Institute (PTI) 2006),

the tendon is at the minimum cover requirements at midspan and over supports; this maximizes the parabolic drape. Anchors are typically placed at mid-depth at the slab edge (Fig. 5.3.3). The tendon exerts a uniform upward force along its length that counteracts a portion of the gravity loads, usually 60 to 80 percent of the slab self-weight according to Libby (1990);

been found from experience to provide satisfactory structural performance. 5.3.2 —For nonprestressed slabs that are thinner than the ACI 318M minimum, or if the slab resists a

prestressing force in the tendon is then combined with the stresses.

  

Eng.Mohammed Ashraf





To achieve Code stress limits, the designer can use an iterative or direct approach. In the iterative approach, the

depending on results and design constraints. In the direct approach, the designer determines the highest tensile stress permitted, then rearranges equations so that the analysis calculates the tendon force required to achieve the stress limit. The Code does not impose a minimum concrete compressive stress due to PT, but majority of engineers use 0.9 MPa as a general minimum for cast-in-place slabs. For slabs exposed to aggressive environments, engineers usually design slabs with an increased minimum concrete compressive stress.

As,min  Reinforcement type

fy, MPa

As,min, mm2

Deformed bars

< 420

0.0020Ag

Deformed bars or welded wire reinforcement

Greater of:

0.0018 420 Ag fy 0.0014Ag

    Lesser of:

380

280 fs

2.5cc

300(280/fs)

— The design of one-way slabs are typically controlled by moment strength, not concrete stress or shear strength. Assuming a uniform load, the designer calculates the unit (usually a 1 m width) factored slab moments. The required lated with the same assumptions as a beam. 5.4.1 Calculation of required moment strength strength—For nonprestressed reinforced slabs, a quick way to calculate a slab’s gravity design moments (if the slab meets the speci speci318M-14 permits other, more exact analysis methods. (secondary moments) should be added to the factored gravity moments per Section 5.3.11 of ACI 318M-14 to calculate Mu. The slab’s secondary moments are a result of the beam’s vertical restraint of the slab against the PT “load” at each support. Because the PT force and drape are determined during the service stress checks, secondary moments can be quickly calculated by the “load-balancing” analysis concept. A simple method for calculating the secondary moment is to subtract the tendon force multiplied by the tendon eccentricity (distance from the neutral axis) from the total balance moment, expressed mathematically as M2 = Mbal – Pe. The critical locations to calculate Mu along the span are usually at the support and midspan. Section 7.4.2.1 of ACI 318M-14 allows Mu to be calculated at the face of support rather than the support centerline. 5.4.2 Calculation of required shear strength—Assuming a uniform load, the designer calculates the unit (usually a 1 method or more exact calculations. — One-way slabs must have adequate one-way shear strength and moment strength in all design strips. 5.5.1 Calculation of design moment strength—The and PT unit slab width are calculated with the same assumptions as for a beam, Chapter 7, of this Handbook.

5.5.2 Calculation of design shear strength—Discussion for nominal one-way shear strength is the same as for a beam, Chapter 7, of this Handbook.  ment be placed in tension regions to ensure that the slab defor deformation and crack widths are limited when the slab’s cracking strength is exceeded. If more than the minimum area is required by analysis, that reinforcement area must be provided. Reinforcement in one way slabs is usually uniformly spaced, 5.6.1 Nonprestressed reinforced slab – Flexural reinplacing—For nonprestressed slabs, the forcement area and placing As,min, is given in Table 7.6.1.1 of ACI 318M-14 (Table 5.6.1a of this Handbook). 24.3.2 of ACI 318M-14 (Table 5.6.1b of this Handbook). Because fs is usually taken as 280 MPa, the maximum spacing will not exceed 300 mm. The bar termination rules in Section 7.7.3 of ACI 318M-14 cover general conditions that apply to beams, but because one-way slab bars are usually spaced close to the maximum, bars generally cannot be terminated without violating the maximum spacing. This usually results in all bottom bars extending full length into the beams. 5.6.2 Post-tensioned slab – Flexural tendon area and placing—For PT one-way slabs, the Code does not have a limit for minimum tendon area or a minimum compressive on service stresses. The Code limits the maximum tendon spacing to 8h or 1.5 m. 5.6.3 Post-tensioned slab – Flexural reinforcing bar area and placing—The Code requires the reinforcing bar area, As,min, to be placed close to the slab face at the bottom at midspan and the top at the support. For one-way PT slabs As,min = 0.004Act. Because the one-way slab strip is rectangular, Act = 0.5Ag. This minimum is independent of service stress level. The maximum spacing of reinforcing bar in a PT one-way slab is the lesser of 3h and 450 mm.

  

Eng.Mohammed Ashraf



the tendons alone, the termination length of As,min bars for bottom bars is (a) and for top bars is (b): (a) At least n/3 in positive moment areas and be centered in those areas (b) At least n/6 on each side of the face of support The termination length for bars that are required for strength are the same as for nonprestressed slabs. 5.6.4 Temperature and shrinkage reinforcement and placing—Shrinkage and temperature (S&T) slab reinforcement is required and could be either reinforcing bar or If the designer uses reinforcing bar, the minimum area of temperature and shrinkage Grade 420 bar is 0.0018Ag. -



tive compression force due to temperature and shrinkage tendons is 0.7 MPa. The purpose of this reinforcement is to restrain the size and spacing of slab cracks, which can occur due to volume variations caused by temperature changes and slab shrinkage over time. In addition, if the slab is restrained against movement, the Code requires the designer to provide reinforcement that accounts for the resulting tension stress in the slab.  Libby, J., 1990, Modern Prestressed Concrete: Design Principles and Construction Methods, fourth edition, Springer, 871 pp. Post-Tensioning Institute (PTI), 2006, Post-Tensioning Manual, sixth edition, PTI TAB.1-06, 354 pp.

  

Eng.Mohammed Ashraf





 One-way Slab Example 1: Nonprestressed one-way slab—

Given: Load— Service live load L

2

Concrete— fc’ fy Geometry—

  

Eng.Mohammed Ashraf





ACI 318M-14 Discussion Step 1: Geometry

Calculation

7.3.1.1

Determine slab thickness using ratios from Table 7.3.1.1.

h

Determine cantilever thickness:

h

24

4400 mm 24

183 mm, say, 175 mm

10

1800 mm 10

180 mm, say, 175 mm

Because the slab and cantilever do not satisfy ACI 318M-14 span-to-depth ratios (Table 7.3.1.1), the Note: Architectural requirements specify a 20 mm step at the cantilever. Detail to maintain 175 mm slab thickness. Self-weight Slab: ws = (175 mm/1000 mm/m)(23.5 kN/m3) = 4.1 kN/m2 Step 2: Loads and load patterns The service live load is 4.8 kN/m2 in assembly 5.3.1 areas and corridors per Table 4-1 in ASCE 7-10. For cantilever use 4.8 kN/m2. To account for weights from ceilings, partitions, HVAC systems, etc., add 0.72 kN/m2 as miscellaneous dead load. The required strength equations to be considered are: U = 1.4D ((5.3.1a) U = 1.4(4.1 kN/m2 + 0.72 kN/m2) = 6.75 kN/m2 U = 1.2D + 1.6L ((5.3.1b) U = 1.2(4.82 1.2 4.82 kN/m2) + 1.6(4.8 kN/m2) = 5.8 kN/m2 + 7.7 kN/m2 = 13.5 kN/m2 Controls The slab resists gravity only and is not part of a lateral force-resisting system, except to act as a diaphragm. Both ASCE 7 and ACI provide guidance for addressing live load patterns. Either approach is acceptable. ACI 318M allows the use of the following two patterns, Fig. E1.2: 6.4.2 Factored dead load is applied on all spans and factored live load is applied as follows: (a) Maximum positive Mu near midspan occurs with factored live load on the span and on alternate spans. (b) Maximum negative Mu at a support occurs with factored live load on adjacent spans only.

Fig. E1.2—Live load loading pattern.

  

Eng.Mohammed Ashraf





Step 3: Concrete and steel material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements of ACI 318M-14. The designer determines the durability classes. Please refer to Chapter 4 of this design Handbook for an in-depth discussion of the categories and classes.

By specifying that the concrete mixture must be in accordance with ACI 301M-10 and providing the exposure classes, Chapter 19 (ACI 318M-14)

Based on durability and strength requirements, and experience with local mixtures, the compressive strength

coordinated with ACI 318M. ACI encourages

7.2.2.2

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. The reinforcement must satisfy Chapter 20 of ACI 318M-14. The designer determines the grade of bar and if the reinforcing bar should be coated by epoxy or galvanized, or both.

By specifying the reinforcing bar grade and any coatings, and that the reinforcing bar must be in accordance with ACI 301M-10, Chapter 20 420 bar and no coatings.

Step 4: Slab analysis 6.3 Because the building relies on the building’s other for braced frame assumptions, as discussed in the commentary. 6.6

The analysis should be consistent with the overall assumptions about the role of the slab within the building system. Because the lateral forceresisting-system only relies on the slab to transmit

Step 5: Required moment strength 7.4.2 The slab’s negative design moments are taken at the face of support as is permitted by the Code (Fig. E1.3).

Modeling assumptions: length of the slab. Only the slab at this level is considered. Analysis approach: The connection to the beams is monolithic; however, soften the joint.

The moments are calculated per meter width perpendicular to the direction of the analysis.

Fig. E1.3—Moment envelope.

  

Eng.Mohammed Ashraf





The negative moment at the centerline of the end right support is 0.0 kN·m.

centerline, column line (CL) E.

centerline. Because of pattern loading, a small negative moment can exist across all spans with the exception of the last span. The maximum negative moment at the exterior left support, CL A, is 26.7 kN·m because of the cantilevered slab.  Location from left to right along the span

Required strength

Exterior support

First midspan

Second support

Second midspan

Third support

Third midspan

Mu, kN·m

–26.7

+12.1

–20.9

+17.1

–25.9

+16.1

Continue: Location from left to right along the span

Required strength

Fourth support

Fourth midspan

Fifth support

Fifth midspan

End support

Mu, kN·m

–23.6

+14.1

–27.9

21.7 +21.7

0

Step 6:: Required shear strength The shear forces are calculated per meter width per7.4.3.1 The slab’s maximum shear is taken at the support per centerline for simplicity. The maximum shear under pendicular to the direction of analysis. all conditions is 35 kN (Fig E1.4).

Fig. E1.4—Shear envelope.

  

Eng.Mohammed Ashraf





Step 7: Design moment strength 7.5.1 The two common strength inequalities for oneway slabs, moment and shear, are noted in Section 7.5.1.1. 7.5.2

The one-way slab chapter refers to Section 22.3 for

7.3.3.1

To ensure a ductile failure usually the Code requires slabs to be designed such that the steel strain at ultimate strength exceeds 0.004 mm/mm In most reinforced slabs, such as this example, reinforcing bar strain is not a controlling issue.

21.2.1(a)

The design assumption is that slabs will be be checked later.

22.2.2.1 and 20 mm cover (Fig. E1.5):

  

Eng.Mohammed Ashraf



7.7.1.1 20.6.1.3.1

One row of reinforcement d = t – cover – db/2

22.2.2.1

The concrete compressive strain at nominal moment cu = 0.003

22.2.2.2

able property and is approximately 10 to 15 percent of the concrete compressive strength. ACI 318M neglects the concrete tensile strength to calculate nominal strength.

22.2.2.3

Determine the equivalent concrete compressive stress at nominal strength:

22.2.2.4.1



d = 175 mm – 20 mm – 16 mm/2 = 147 mm

The concrete compressive stress distribution is inelastic at high stress. The Code permits any stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular 0.85 c with a compressive stress distribution of 0.85f depth of: a compres1c 1 is a function of concrete compres sive strength and is obtained from Table 22.2.2.4.3.

22.2.2.4.3

For fc

22.2.1.1

Find the equivalent concrete compressive depth, a,, by equating the compression force to the tension force within the beam cross section:

1

0.85

0.05(35 MPa 28 MPa) 7 MPa

0.8

C=T 0.85fc’ba = Asfy

0.85(35 MPa)(b)(a) = As(420 MPa) a=

As (420 MPa) 0.85(35 MPa)(1000 mm)

0.014 As

  

Eng.Mohammed Ashraf





moments obtained from the approximate method above. 7.5.1.1 larger of the two moments. The beam’s design strength must be at least the required strength at each section along its length: Mn Vn

u u

Calculate the required reinforcement area: a M n M u = As f y d 2

Maximum positive moment: 21.7 kN·m

(0.9)(420 MPa) As 147 mm

0.014 AS 2

A No. 16 bar has a db = 16 mm and an As = 199 mm2 A+s’req’d = 398 mm2/m Use No. 13 at 300 mm on center or No. 16 at 450 mm No. at 450 mm on center center bottom. Try No.16 As,provd. = (199 mm2/m)(1000 mm/450 mm) = 442 mm2 As,provd. = 442 mm2 > A+s,req’d = 398 mm2

OK

Maximum negative moment: 27.9 kN m

A

s,req’d

((0.9)(420 0.9)(420 MPa) As 147 mm

0.014 As 2

= 515 mm2/m

Use No. 16 at 300 mm on center top As,provd. = 663 mm2/m > A

s,req’d

= 515 mm2/m

OK

Check if the calculated strain exceeds 0.005 mm/mm (tension controlled). Form similar triangles (Fig. E1.6). As f y

a

0.85 f c b 1

t

=

cu

c

and c = a

= 0.8 for fc ( d c)

1

Top reinforcement a = 0.014As = (0.014)(663 mm2) = 9.3 mm c = 9.3/0.8 = 11.6 mm t

=

0.003 (150 mm 11.6 mm) 11.6 mm

0.036

0.005

OK

Fig. E1.6—Strain distribution

  

Eng.Mohammed Ashraf





Step 8: Design shear strength 7.5.3 Assuming a one-way slab won’t contain shear reinforcement, Vn is equal to Vc. Assuming negligible axial force, the Code provides the following expression, 22.5.5.1 21.2.1

Vc 0.17 f c bd Shear strength reduction factor:

Vc = 0.17 35 MPa (1000 mm)(147 mm) = 147,843 kN/m = 0.75

Vc = (0.75)(147,843 kN) = 110,882 kN > 35,000 kN OK This exceeds the maximum Vu (35 kN); therefore, no shear reinforcement is required. 7.6.1

Check if design reinforcement exceeds the minimum required by the Code.

As,min = 0.0018 × 147 × 1000 = 265 mm2/m At all critical sections, the required As is greater than the minimum.

Step 10: Shrinkage and temperature reinforcement 7.6.4 For one-way with Grade 420 bars, the minimum 24.4.3.2 area of shrinkage and temperature (S+T) bars is 0.0018Ag. The maximum spacing of S+T reinforc reinforcing bar is the lesser of 3h and 450 mm

S+T steel area = 0.0018 × 1000 × 175 = 315 mm2 Based on S+T steel area, solutions are No. 13 at 400 mm or No. 16 at 450 mm; use No. 13 at 400 mm reinforcement.

7.7.2.1 25.2.1

7.7.2.2 24.3.2

The minimum spacing between bars must not be less than the greatest of: (a) 25 mm (b) db (c) 4/3dagg

(a) 25 mm (b) 16 mm (c) (4/3)(25 mm) = 33.3 mm

Assume 25 mm maximum aggregate size. For reinforcement closest to the tension face, the spacing between reinforcement is the lesser of (a) and (b): (a) 300(280/fs) (b) 380(280/fs) – 2.5cc

(a) 300(280/280) = 300 mm Controls (b) 380(280/280) – 2.5(20 mm) = 330 mm

24.3.2.1

fs = 2/3fy = 280 MPa

7.7.2.3

The maximum spacing of deformed reinforcement is the lesser of 3h and 450 mm

Controls

3(175 mm) = 525 mm > 450 mm Therefore, Section 24.3.2 controls; 300 mm

  

Eng.Mohammed Ashraf





Step 12: Select reinforcing bar size and spacing Based on the above requirement, use No. 16 bars. Spacing on top and bottom bars is 300 mm Note that there is no point of zero negative moment along all spans except the last bay, so continue the top bars across all spans. Also, No. 13 bars can be used instead of No. 16. While this solution is slightly conservative (No. 16 versus No. 13 bars), the engineer may desire consistent spacing and reinforcing bar use for easier installation and inspection. Step 13: Top reinforcing bar length at the exterior support The top bars have to satisfy the following provisions: is 1.5 m from support centerline. 7.7.3.3

Reinforcement shall extend beyond the point at distance equal to the greater of d and 12db, except at d = 147 mm or (12)(16 mm) = 192 mm supports of simply-supported spans and at free ends Therefore, use 192 mm of cantilevers.

7.7.3.8.4

reinforce At least one-third the negative moment reinforcement at a support shall have an embedment length d, 12db, and n/16. 16 point at least ((4.4 m – 0.45 m)(1000)/16 m)(1000)/ = 246.9 mm > 12db = 192 mm > d = 147 mm Because the reinforcing bar is already at maximum spacing, no percentage of bars (as permitted by zone.

Step 14: Development and splice lengths 7.7.1.2 ACI provides two equations for calculating 25.4.2.3 example, the detailed equation is used:

d

1 1.1

fy

t

fc

cb

e

s

db

Ktr

d

db

1 420 MPa (1.0)(1.0)(0.8) (16 mm) 1.1 (1.0) 35 MPa 1.7 486 mm

where t = bar location; not more than 300 mm of fresh concrete below horizontal reinforcement e = coating factor; uncoated s = bar size factor; No. 22 and larger However, the expression: taken greater than 2.5.

The development length of a No. 16 black bar in a 175 mm slab with 20 mm cover is:

cb

K tr db

must not be

t = 1.0, because not more than 300 mm of concrete is placed below bars. e = 1.0, because bars are uncoated s = 0.8, because bars are smaller than No. 22

27 + 0 = 1.7 16

  

Eng.Mohammed Ashraf



7.7.1.3 25.5 25.5.1.1

Splice The maximum bar size is No. 16, therefore, splicing is permitted.

25.5.2.1

Tension lap splice length, st, for deformed bars in tension must be the greater of: 1.3 d and 300 mm

7.7.3.3

st



= (1.3)(486 mm) = 631.8 mm; use 900 mm

The bottom bars have to satisfy the following provisions: Reinforcement must extend beyond the point at distance equal to the greater of d and 12db, except at from exterior support centerline at CL F and 675 mm supports of simply-supported spans and at free ends of cantilevers.

7.7.3.4 have an embedment length not less than d beyond the point where bent or terminated tensile

7.7.3.5

the beam span.

Flexural tensile reinforcement must not be terminatVu

Vn

35,000 N < (2/3)(110,882 110,882 kN) = 73,921 kN

OK

Note that (b) and (c) do not apply. 7.7.3.8.2

At least one-fourth the maximum positive moment reinforcement must extend along the slab bottom into the continuous support a minimum of 150 mm

7.7.3.8.3

db for positive moment tensile reinforcement shall be limited such that d

compressive reaction. Mn/Vu + a a is the greater of d and 12db = 192 mm a M n As f y d 2 d

The elastic analysis indicates that Vu point is 26,200 N.

Check if bar size is adequate Mn for an 175 mm slab with No. 16 at 300 mm, 20 mm cover is: Mn = (663 mm2)(420 MPa)(147 mm – 9.3 mm/2) = 39.64 × 106 N·mm 39.64 × 106 N·mm 486 mm 192 mm 1705 mm d 26, 200 N Therefore, No. 16 bar is

OK

Step 16: Bottom bar length the Code provisions because of reduced required strength along the span do not apply before the

The Code requires that at least 25 percent of bottom bars be full length, extending 150 mm into the support.

into both supports. would be violated. Because all bottom bars extend past the tensile zone, Section 7.7.3.5 does not apply. All bottom bars need to extend at least 175 mm (refer to Section 7.7.3.3) beyond the positive   

Eng.Mohammed Ashraf





Step 17: Slab detailing

Fig. E1.7—One-way slab reinforcement.

  

Eng.Mohammed Ashraf





One-way Slab Example 2: Assembly loading— Design and detail a one-way nonprestressed reinforced concrete slab both for service conditions and factored loads. The one-way slab spans 6000 mm and is supported by 300 mm thick walls on the exterior, and 300 mm wide beams on the interior. Given: Load— Live load L = 4.8 kN/m2 2 s = 23.5 kN/m Geometry— Span = 6000 mm Slab thickness t = 230 mm Material properties— fc fy = 420 MPa

Fig. E2.1—One-way slab framing plan.

  

Eng.Mohammed Ashraf





Calculation

Discussion

ACI 318M-14

Step 1: Geometry 7.3.1.1 (Table 7.3.1.1), the designer does not need to check tions or other construction likely to be damaged by

(6000 mm) 222 mm, say, 230 mm 27 27 This ratio is less than the table value for “both ends h

checked. Step 2: Loads and load patterns 5.3.1 For hotel lobbies, the live load is assembly occupancy; the design live load is 4.8 kN/m2 per Table 4-1 in ASCE 7-10. A 230 mm slab is a 5.4 kN/m2 dead load. To account for loads due to ceilings, partitions, HVAC systems, etc., add 0.5 kN/m2 as miscellaneous dead load. U = 1.4D (5.3.1a) U = 1.2D + 1.6L (5.3.1b) The slab resists gravity only and is not part of a lateral force-resisting system, except to act as a diaphragm.

The required strength equations to be considered are: U = 1.4(5.4 kN/m2 + 0.5 kN/m2) = 8.3 kN/m2 U = 1.2(5.9) + 1.6(4.8) = 7.1 + 7.7 = 14.8 kN/m2 Controls

Both ASCE 7 and ACI provide guidance for addressing live load patterns. Either approach is acceptable. M allows the use of the following two ACI 318M 2.2 patterns, Fig. E2.2: 6.4.2

Factored dead load is applied on all spans and factored live load is applied as follows: (a) Maximum positive Mu near midspan occurs with factored live load on the span and on alternate spans. (b) Maximum negative Mu at a support occurs with factored live load on adjacent spans only.

Fig. E2.2—Live load loading pattern.

  

Eng.Mohammed Ashraf



Step 3: Concrete and steel material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements of ACI 318M-14. The designer determines the durability classes. Please refer to Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.



By specifying that the concrete mixture must be in accordance with ACI 301M-10 and providing the

Based on durability and strength requirements, and experience with local mixtures, the compressive strength

coordinated with ACI 318M. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. 7.2.2.2

The reinforcement must satisfy Chapter 20 of ACI 318M-14. The designer determines the grade of bar and if the reinforcing bar should be coated by epoxy or galvanized, or both.

By specifying the reinforcing bar grade and any coatings, and that the reinforcing bar must be in accordance with ACI 301M-10, Chapter 20 420 bar and no coatings.

Step 4: Slab analysis 6.3 Because the building relies on the building’s other for braced frame assumptions, as discussed in the commentary.

6.6

Modeling assumptions: length of the slab.

Only the slab at this level is considered. The analysis should be consistent with the overall assumptions about the role of the slab within the building analysis is adequate. Analysis approach: soften the joint. Rather than attempting to estimate an appropriate level of softening, the slab is simply modeled twice:

support lengths are increased to 30 m long columns. Assume fully connected slab to exterior wall, with uncracked moment of inertia, modelled by a 3 m, 300 mm x 300 mm columns, and the middle three supports are slender, 30 m columns. Note: The moments resulting from analysis maximize the moments and may be used to design the exterior wall. Step 5: Required moment strength 7.4.2 The slab’s negative design moments are taken at the face of support, as is permitted by the Code.

  

Eng.Mohammed Ashraf





Analysis (a): Note: The moment at the exterior support is near zero (refer to Fig. E2.3):

Fig. E2.3—Moment envelope. The negative moment at the centerline of the exterior support is 0.0 kN·m.

across the span.

centerline.

 Location from left to right along the span Required strength Mu

Exterior support

First midspan

Second support

Second midspan

Middle support

0.0

+48.5

–55

+32.5

–42.3

  

Eng.Mohammed Ashraf





point is 1160 mm from the support centerline.

the support centerline.

line. The maximum negative moment at the face of the second interior support is 44 kN·m. The negative mm from the support centerline. Following are the maximum moments from a combination of Analysis (a) and (b) (conservative approach).  Location from left to right along the span

Required strength

Exterior support

First midspan

Second support

Second midspan

Middle support

Mu, kN·m

–38.7

+30.2

–44

+ +32 32

–44

Step 6: Required shear strength 7.4.3.1 The slab’s maximum shear is taken at the support centerline for simplicity. The maximum shear under any condition is 49.6 kN. Step 7: Design moment strength The two common strength inequalities for one7.5.1 way slabs, moment and shear, are noted in Sec21.2.1

reduction factor in Section 7.5.1.2 is assumed to be 0.9, which will be checked later.

7.5.2

The one-way slab chapter refers to Section 22.3

  

Eng.Mohammed Ashraf





22.2.1 22.2.2.2

Chapter 22 (ACI 318M-14) provides the design assumptions for reinforced concrete members.

To calculate As in terms of the depth of the compression block, a, set the section’s concrete compressive strength equal to steel tensile strength:

Generate the minimum area of steel for the required T=C moment.

Asfy = 0.85fc b)a

As

22.2.2.4.1

0.85(35 MPa)(1000 mm)a 420 MPa

70.8a

d, is the overall slab height minus the cover (20 mm) minus half the bar diameter To calculate the minimum required As, set the design (for a single layer of reinforcing bar). Assuming a strength moment equal to the required strength moment. No. 5 bar, therefore, Asfy(d – a/2) = Mu

d = 202 mm

0.9 As (420 MPa) 202 mm

As 2(70.8)

Mu

Table 2.3 following shows the required area of steel corresponding to the maximum moments from a conservative combination of Analysis (a) and (b).  Location from left to right along the span Mu, kN·m 2

Req’d As, mm /m

Exterior support

First midspan

Second support

Second midspan

Middle support

–38.7 38.7

+48.5

55 –55

+32.5

–44

514

647

737

430

586

To ensure a ductile failure mode, the steel strain at ultimate strength must be at least 0.004 mm/mm For usual reinforced slabs, such as this example, bar strain does not usually control the design.

To calculate reinforcing bar strain, begin with force equilibrium within the section: T=C 0.85fc ba Asfy = 0.85f

A strain diagram is drawn (Fig. E2.5).

where b = 1000 mm/m; fy = 420 MPa; and the slab’s maximum reinforcement is As = 737 mm2 From above calculations: As = 70.8a or a = As/70.8

22.2.2.1

Maximum strain at the extreme concrete cu

= 0.003 mm/mm

Therefore, a = 10.4 mm where a = 1c and 1 = 0.80 for fc so c = 13 mm From similar triangles (Fig. E2.5): 0.003(202 mm 13 mm) 0.044 t (13 mm) Therefore, the assumption of

0.004

= 0.9 is correct.

  

Eng.Mohammed Ashraf





Fig. E2.5—Strain distribution. Step 8: Design shear strength 7.5.3 Assuming a one way slab won’t contain shear reinforcement, Vn is equal to Vc. Assuming negligible axial force, the Code provides a simple expression: 22.5.5.1

Vc

f c bd

Vc = 0.17 35 MPa (1000 mm)(202 mm) = 203,158 N

21.2.1

Shear strength reduction factor:

= 0.75 Vc = (0.75)(203,158 N) = 152,369 N > 49,600 N OK This exceeds the maximum Vu (49.6 kN); therefore, no shear reinforcement is required.

0.17

Check if design reinforcement exceeds the minimum required reinforcement by Code. As, min = 0.0018 x 230 mm x 1000 mm = 414 mm2/m. At all critical sections, the required As is greater than the minimum. Step 10:: Shrinkage and temperature reinforcement For one-way slabs with Grade 420 bars, the mini 7.6.4 mini24.4.3.2 mum area of shrinkage and temperature (S+T) bars is 0.0018 Ag. The maximum spacing of S+T S+T steel area = 0.0018 x 1000 mm x 230 mm = 414 mm2 reinforcing bar is the lesser of 3h and 450 mm

7.6.1

Based on S+T steel area, solutions are No. 13 at 300 mm or No. 16 at 450 mm; use No. 16 at 450 mm placed ment. 7.7.2.1 25.2.1

The minimum spacing between bars must not be less than the greatest of: (a) 25 mm (b) db (c) 4/3dagg Assume 25 mm maximum aggregate size.

(a) 25 mm (b) 16 mm (c) (4/3)(25 mm) = 33.3 mm

Controls

  

Eng.Mohammed Ashraf





7.7.2.2 24.3.2

For reinforcement closest to the tension face, the spacing between reinforcement is the lesser of (a) and (b): (a) 300(280/fs) (b) 380(280/fs) – 2.5cc

(a) 300(280/280) = 300 mm Controls (b) 380(280/280) – 2.5(20 mm) = 330 mm

24.3.2.1

fs = 2/3fy = 280 MPa

7.7.2.3

The maximum spacing of deformed reinforcement is the lesser of 3h and 450 mm

3(175 mm) = 525 mm > 450 mm Therefore, Section 24.3.2 controls; 300 mm

Step 12: Select reinforcing bar size and spacing  Location from left to right along the span Bar size

Exterior support

First midspan

Second support

Second midspan

Middle support

No. 13 at spacing, mm

230

175

150

250

200

No. 16 at spacing, mm

300

275

250

300

300

No. 19 at spacing, mm

300

300

300

300

300

Refer to Fig. E2.6 for provided bar spacing.

Based on the above, use No. 16 bars. Spacing of top bars at exterior support is 300 mm, interior supports is 250 mm. Note that there is no point of zero negative moment along the second and third span, so continue the top bars across both spans. While this solution is slightly conservative (250 ( mm versus 300 mm spacing), the engineer may desire consistent spacing for easier installation and inspection.

  

Eng.Mohammed Ashraf





Step 13: Top reinforcing bar length at the exterior support The top bars have to satisfy the following provisions:

7.7.3.3

from support centerline.

Reinforcement shall extend beyond the point at d distance equal to the greater of d and 12db, except at supports of simply-supported spans and at free ends of cantilevers.

7.7.3.8.4

At least one-third the negative moment reinforcement at a support shall have an embedment length d, 12db, and n/16.

point at least –

d

db Because the reinforcing bar is already at maximum tension zone.

Bars at the wall connection It is assumed that the wall is placed several days

Solution

restrain the slab from shrinking as it cures. Many designers place extra reinforcement along the slab are close together, use a 375 mm extension for all bars. edge, parallel to the wall, to limit widths of possible A practical length for top bars is: cracks due to this restraint.

Step 14: Development and splice lengths 7.7.1.2 ACI provides two equations for calculating devel25.4.2.3 detailed equation is used:

d

1 1.1

fy

t

fc

e

cb

s

K tr

db

d

db

1.1

1.7

486 mm

where t

t

concrete below horizontal reinforcement

placed below bars.

e

e

s

s

But the expression: greater than 2.5.

cb

K tr db

must not be taken 16 mm

1.7

  

Eng.Mohammed Ashraf





7.7.1.3 25.5 25.5.1.1

Splice The maximum bar size is No. 16, therefore, splicing is permitted.

25.5.2.1

Tension lap splice length, st, for deformed bars in tension must be the greater of: 1.3 d and 300 mm

7.7.3.3

st

= (1.3)(486 mm) = 639 mm; use 900 mm

The bottom bars have to satisfy the following provisions: Reinforcement must extend beyond the point at distance equal to the greater of d and 12db, except at supports of simply-supported spans and at free ends of cantilevers.

from the exterior support centerline (analysis (a)) and

have an embedment length not less than d beyond the point where bent or terminated tensile rein-

the beam span.

(b)).

7.7.3.4

7.7.3.5

Flexural tensile reinforcement shall not be terminated in a tensile zone unless (a), (b), or (c) is (a) Vu

7.7.3.8.2

7.7.3.8.3

Vn

Note that (b) and (c) do not apply. At least one-fourth the maximum positive moment reinforcement must extend along the slab bottom into the continuous support a minimum of 150 mm. db for positive moment tensile reinforcement must be limited such that d

compressive reaction. d

Mn

n/Vu

As f y d

Check if bar size is adequate Mn for an 230 mm slab with No. 16 at 300 mm, 20 mm cover is:

a

a 2

The elastic analysis indicates that Vu point is 40.9 kN. The term a is 250 mm.

Mn = (663 mm2/m)(420 MPa)(202 mm – 10.4 mm/2) = 54.8 × 106 N·mm d

54.8 10 6 N·mm 409,000 N

250 mm 1590 mm > 486 mm

Therefore, No. 16 bar is OK

  

Eng.Mohammed Ashraf





Step 16: First span bottom bar length prior Code provisions because of reduced required strength along the span do not apply before the

The Code requires that at least 25 percent of bottom bars be full length, extending 150 mm into the support. 150 mm into both supports.

would be violated. Because all bottom bars extend past the tensile zone, Section 7.7.3.5 does not apply. All bottom bars need to extend at least 175 mm (refer to Section 7.7.3.3) beyond the positive Step 17: Interior span bottom reinforcing bar lengths Create a partial length bar that is symmetrical within mm from the left support centerline and 1280 mm from the right support centerline.

mm. The minimum length is 6 m – 1.1 m – 1.1 m + (0.6 m)(0.5) = 4.1 m, say, 4300 mm

In a repeating pattern, use 3 No. 16 at 4.3 m long and 1 No. 16 at 6.3 m long. Code requires at least 25 percent of bottom bars be full length, extending 150 mm into the support. Step 18:: Top reinforcing bar length at the middle support The required reinforcing bar is No. 16 at 300 mm in frame into the middle support. support is No. 16 at 250 mm, extend the No. 16 at 250 mm top over the middle support for simplicity.

Because the top bars will be continuous, no bars are Step 19: Detailing

Fig. E2.6—Slab reinforcement detailing.

  

Eng.Mohammed Ashraf





One-way Slab Example 3: One-way slab post-tensioned – Hotel loading There are four spans of 6000 mm each, with a 900 mm cantilever balcony at each end. The slab is supported by 300 mm walls on the exterior, and 300 mm wide beams on the interior (Fig. E3.1). This example will illustrate the design and detailing of a one-way post-tensioned (PT) slab, both for service conditions and factored loads.

Fig. E3.1—One-way slab. ACI 318M-14 Discussion Step 1: Geometry 7.3.2 The ACI 318M-14 span-to-depth ratios do not apply to PT slabs. The Post-Tensioning Manual, 2006, sixth edition Chapter 9, Table 9.3, suggests a ratio limit of /48.

Calculation

For this example, this ratio gives a slab thickness of (6000 mm)/48 = 125 mm This example uses a 150 mm thick slab.

  

Eng.Mohammed Ashraf





Step 2: Loads and load patterns 7.4.1.1 For hotel occupancy, the design live load is 1.9 kN/m2 per Table 4-1 in ASCE 7-10. A 150 mm slab is a 3.5 kN/m2 dead load. To account for weights from ceilings, partitions, HVAC systems, etc., add 0.5kN/m2 as miscellaneous dead load. D = 3.5 kN/m2 + 0.5 kN/m2 = 4 kN/m2 The slab resists gravity only and is not part of a lateral-force-resisting system, except to act as a diaphragm. 5.3.1

U = 1.4D U = 1.2D+1.6L

7.4.1.2

Both ASCE 7 and ACI provide guidance for addressing live load patterns. Either approach is acceptable.

The required strength equations to be considered are: U = 1.4(4) = 5.6 kN/m2 U = 1.2(4) + 1.6(1.9) = 4.8 + 3 = 7.8 kN/m2, say, 8 kN/m2 Controls

ACI 318M allows the design to use the following two patterns (Fig. E3.2): 6.4.2

Factored dead load is applied on all spans and factored live load is applied as follows: (a) Maximum positive Mu+ near midspan occurs with factored L on the span and on alternate spans. (b) Maximum negative Mu– at a support occurs with factored L on adjacent spans only.

Fig. E3.2—Live load pattern used in elastic analysis of slab for unbalanced loads.

  

Eng.Mohammed Ashraf





Step 3: Concrete and steel material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements. The designer determines the durability classes. Please refer to Chapter 4 of this design Handbook for an in-depth discussion of the Categories and Classes.

coordinated with ACI 318M. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. 7.2.2.2

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the ex-

Based on durability and strength requirements, and experience with local mixtures, the compressive strength of

The reinforcement must satisfy Chapter 20. In this example, unbonded, 13 mm single-strand tendons are assumed.

By specifying the reinforcement shall be in accordance with ACI 301M-10, the PT type and strength, and reinforcing bar grade (and any coatings), Chapter In this example, assume Grade 420 bar and no coatings.

The designer determines the grade of bar and if the reinforcement should be coated by epoxy or galva galvanized, or both. 20.3

The Code requires strand material to be 1860 MPa, low relaxation (ASTM A416a). A

20.3.2.5.1

The U.S. industry usually stresses, or jacks, monostrand to impart a force equal to the least of 0.80fpu and 0.9fy. immediate and long-term losses will reduce this force.

The jacking force per individual strand is: (1860 MPa)(0.8)(99 mm2) = 147.3 kN. This is immediately reduced by seating and friction losses, and elastic shortening of the slab. Long term losses will further reduce the force per strand. Refer to Commentary R20.3.2.6 of ACI 318M-14.

where fy = 0.94fu (Table 20.3.2.5.1) 0.80fpu controls, the maximum allowed by the Code.

A PT force design value of 117.8 kN per strand is common.

20.3.2.5.1

  

Eng.Mohammed Ashraf



Step 4: Slab analysis 6.3 Because the building relies on the building’s other for braced frame assumptions, as discussed in the commentary.



Modeling assumptions: Slab will be designed as Class U. Consequently, use the gross moment of inertia of the slab in the analysis. Assume the supporting beams have no torsional resistance and act as a knife edge support. Only the slab at this level is considered.

6.6

Analysis approach: The analysis performed should be consistent with the overall assumptions about the role of the slab within the building system. Because the lateral force- the concrete slab under service loads, the tendon drape resisting system only relies on the slab to transmit is assumed to be parabolic with a discontinuity at the support centerline shown as follows, which imparts a uniform uplift over each span when tensioned. The Although gravity moments are calculated indepen- magnitude of the uplift, wp, or “balanced load,” in dent of PT moments, the same model is used for each span of a prismatic member is calculated as: both. wp = 8Fa/ 2 where F a is the tendon drape (average of the two high points minus the low point). In this example the PT force is assumed constant for all spans, but the uplift force varies due to

20.6.1.3.2 resistance to dead load and live load. At the exterior support an eccentricity of 6 mm is chosen to balance the cantilever load. At the interior supports and midspans, the maximum possible eccentricity is chosen (25 mm cover)

  

Eng.Mohammed Ashraf





Step 5: Slab stress limits 7.3.4.1 This example assumes a Class U slab; that is, a slab under full service load with a concrete tension 19.2.3.1 stress not exceeding 0.62 f c . The slab analysis model for the service condition is the same as for the nominal condition. 19.2.3.1

0.62 35 MPa

3.7 MPa

To verify that the concrete tensile stresses are less than 0.62 35 MPa , the net service moments and tensile stresses at the face of supports are needed. This example assumes two parameters: (a) The PT force provides a F/A slab compressive stress of at least 0.9 MPa (131.3 kN/m), and provides an uplift force wp to balance at least 75 percent of the slab weight, or 2.7 kN/m2 (Fig. E3.4).

ws = (150 mm)(23.5 kN/m3)/1000 = 3.5 kN/m2 wp = (0.75)(3.5 kN/m2) = 2.63 kN/m2, say, 2.7 kN/m2

The basic equation for concrete tensile stress is: ft = M/S – F/A, where M is the net service moment. At the exterior support, the drape is 91 mm ). The required force is calculated (Fig. E3.3). from: wp = 8Fa/ 2 where wp = 2.7 kN/m2

S = (1000 mm)(150 mm)2/6 = 3.75 × 106 mm3 (section modulus), A = (1000 1000 mm)(150 mm) = 150 × 103 mm2 (gross slab area per foot). F

(2.7 kN/m 2 )(6 m) 2 8(91 mm)/(1000 mm/m)

134 kN/m

Fig. E3.4—Uplift force due to tendon layout. 7.3.4.2 The service gravity load is 6 kN/m3. The PT uplift is subtracted from the gravity load. Location from left to right along the span Service loads 2

Gravity uniform load, kN/m PT uniform uplift, kN/m

2

Net “unbalanced load,” kN/m2

7.3.4.2

Cantilever

First span

Second midspan

6

6

6

2.3

2.7

3.3

3.7

3.3

2.7

The slab stresses are determined from an elastic analysis using the “net” load, minus the slab’s axial compression. Slab axial compression force:

F/A = (134 N/mm)/(150 mm) = 0.9 MPa.

  

Eng.Mohammed Ashraf



7.3.4.2



Location from left to right along the span Net service unbalanced moments, kN·m/m

Exterior support

First midspan

Interior support

Second midspan

0.89

6.7

9.8

4.9

Net tensile stress, MPa

7.4.2.1 7.3.4.2

19.2.3.1

The cantilever moment at the support centerline is: wnet 2/2 = (3.7 kN/m)(0.9 m)2/2 = 1.50 kN·m/m. ACI 318M-14 permits to calculate the design slab moment at the face of the support: 0.89 kN·m/m. The moments for the interior supports are calculated at the faces of interior supports. The aforementioned results show the maximum slab tensile stress (1.6 MPa) calculated for an average PT force of 134 kN/m is less than 0.62 f c = 3.7 MPa. Therefore, the slab is uncracked and the aforementioned assumption is correct.

  

Eng.Mohammed Ashraf





7.3.2.1

3

Ig:

Ig

6

4

12

24.2.3.8

2

2

max

4

EI

4

max

6

4

Concrete International 24.2.2

2

2

2

2 2

. 833

7.4.2

Fig. E3.5—Moment envelope due to factored loads.   

Eng.Mohammed Ashraf



7.4.1.3

7.4.1.3

7.4.2.1



The Code requires that moments due to reactions induced by prestressing (secondary moments) be included with a load factor of 1.0. Secondary moments are calculated at each support as: M2 = Mpt – Fe. The secondary moment diagram is linear between supports (Fig. E3.6). The PT secondary moments are:

Fig. E3.6—Secondary moments due to post-tensioning. The combined factored and secondary moments, which are the required moment strengths, are shown in Fig. E3.7:

Location from left to right along the span Required strength (Gravity only)

Face of exterior support

Factored mom. at face, kN·m

First midspan

Face of second support

Second midspan

Face of middle support

24.4

–32

16

–25.3

M2, kN·m

0.0

1.8

3.6

2.9

2.2

Mu, kN·m

–3.5

26.2

–28.4

18.9

–23.1

Step 8: Calculate required As 7.7.4.2 If the PT tendons alone provide the design strength, Mn,, then the Code permits the reinforcing bar to be a reduced length. If the PT tendons alone do not provide the design strength, then the reinforcing bar The depth of the equivalent stress block, a, is calculength is required to conform to standard lengths. lated by a

Aps f ps 0.85 f c (1000 mm/m)

where Aps is the tendon area perfoot of slab. Mn. Section 22.3 refers to 22.2 for calculation of Mn. Section 22.2.4 refers to 20.3.2.4 to calculate fps. The span-to-depth ratio is 6000/150 = 40, so the following equation applies:

f ps

f se

70

fc 300

p

The reinforcing bar and tendons are usually at the same height at the support and at midspan.

  

Eng.Mohammed Ashraf





20.3.2.4

Each single unbonded tendon is stressed to the value prescribed by the supplier. Friction losses cause a variation of fse along the tension length, but for design purposes, fse is usually taken as the average value.

The tendon supplier usually calculates fse, and 1200 MPa is a common value. The force per strand is therefore: 1200 MPa × 99 mm2 = 118,800 N

so the spacing of tendons is:

118.8 kN/134 kN/m × 1000 mm/m = 887 mm, say, 850 mm Aps = (99 mm2)(1000 mm/m)/(850 mm) = 116.5 mm2/m. 2 2 p = (116.5 mm /m)/125,000 mm = 0.00093 35 MPa = 1395 MPa f ps = 1200 MPa + 70 MPa + 0.279

The value of Aps is therefore, The value of p in Aps/(b × dp) b = 1000 mm; dp = 125 mm 20.3.2.4.1

fps limit as follows: (a) fse + 210 and (b) fpy = 0.9fpu

(a) 1200 MPa + 210 MPa = 1410 MPa and (b) (0.9)(1860 MPa) = 1674 MPa So the design value of fps = 1395 MPa

The compression block depth is therefore: Aps f ps a 0.85 f c (1000 mm/m) 22.2.2.4.1

a

locations, except at the exterior joint. Therefore, the Code permits a minimum d of 0.8 0.8h, or 120 mm For 20 mm cover, Mn Apsfps(d – a/2)

(116.5 mm 2 )(1395 MPa) (0.85)(35 MPa)(1000 mm/m)

4.64 mm

Mn = (0.9 (0.9)(116.5 116.5 mm/m2)(1395 MPa)(125 mm – 4.64/2 mm) = 17.94 × 106 N·mm/m = 17.94 kN·m/m.

Location from left to right along the span

Mn, only tendons, kN·m Mu, kN·m

Face of exterior support

First midspan

Face of second support

Second midspan

Face of middle support

17.1

17.9

17.9

17.9

17.9

26.2

18.9

Mn when considering the tendons alone, standard reinforcement lengths must be used.

As, min = 0.004 × 1000 mm × 75 mm = 300 mm2/m.

7.6.2.3 meter is a function of the slab’s cross-sectional area. Step 10: Design moment strength 7.5.1 The two common strength inequalities for oneway slabs, moment and shear, are noted in Section 7.5.1.1 of ACI 318M-14. The strength reduction factor in Section 7.5.1.2 is assumed to be 0.9.

  

Eng.Mohammed Ashraf



7.5.2



Determine if supplying the minimum area of reinthat exceeds the required strength. Comparing this value with the required moment strength Mu indicates that the minimum reinforcement plus the tendons supply enough tensile reinforcement for slab to resist the factored loads at Set the section’s concrete compressive strength equal all locations. to steel tensile strength, and rearrange for compression block depth a: Aps f ps As f y (116.5 mm 2 )(1395 MPa) (300 mm 2 )(420 MPa) a a 0.85 f c (1000 mm/m) (0.85)(35 MPa)(1000 mm/m) a = 9.7 mm For 20 mm cover: Mn

Aps f ps

As f y

d

a 2

= 0.9[116.5 × 1395 + 300 × 420](125 – 9.70/2) = 34.7 × 106 N·mm = 34.7 kN·m Therefore, minimum reinforcement provides adequate strength to resist the applied moment. OK Step 11: Design shear strength 7.5.3.1 The slab’s maximum shear is taken at the support centerline for simplicity. 21.2.1

Vc

0.17( 35 MPa )(1000 mm)(125 mm) 125,717 N 0.17

Shear strength reduction factor: Vc = (0.75 (0.75)(125,717 125,717 N) = 94,287 N The maximum Vu = 29,200 kN Note: Shear does not typically control the thickness support. Therefore, OK. of one-way post-tensioning slab system.

  

Eng.Mohammed Ashraf





Step 12: Shrinkage and temperature reinforcement 7.6.4.2 To control shrinkage and temperature stresses in the direction to the span, it is typical to use tendons rather than mild reinforcement in one-way post tensioning slabs. To calculate the number and spacing of temperature tendons, the Code allows on the slab. Assuming the beam is 300 mm x 750 mm, the

A

= (150 mm)(6000 mm) + (300 mm)(600 mm) = 1.08 × 106 mm2

area is: 6

mm2) = 0.78 MPa.

force of 840 kN, which results in an average compression of: Code minimum of 0.7 MPa. The Code also has three spacing requirements which apply: Provide at least one tendon on each side of the beam. If temperature tendon spacing does not exceed 1.4 m, additional reinforcing bar is not needed; but if temperature tendon spacing exceeds 1.4 m, supplemental reinforcement is required along the edge of the slab adjacent to tendon anchors. Spacing above 1.8 m is prohibited. In this example, temperature tendons, starting at 1.2 No supplemental edge reinforcement is needed. 7.7.2.3 a PT slab is the lesser of 3h and 450 mm

mm2/m, use No. 13 bar at 400 mm on center, which also

  

Eng.Mohammed Ashraf





Step 14: Top reinforcing bar length at the exterior support 7.7.3 Reinforcing bar length and details at the exterior support from exterior support centerline. 7.7.3.3

The top bars have to satisfy: Reinforcement must extend beyond the point at distance equal to the greater of d and 12db, except at supports of simply-supported spans and at free ends of cantilevers.

7.7.3.8.4

At least one-third the negative moment reinforcement at a support shall have an embedment length d, 12db, and n/16.

Section 7.7.3.3 requires all bars to extend beyond the d (125 mm) or 12 x 13 mm; therefore 160 mm.

In addition, Section 7.7.3.8.4 requires 33 percent of (5700 mm)/16 = 356 mm. The top bars at the exterior joint will extend from the end of the cantilever, past the support and into the span. Because the reinforcing bar is at wide spacing, no percentage of bars (as tension zone.

Balcony considerations

Solution The top bar length is 900 mm (balcony) plus 450 m

about 20 mm at the exterior wall of residential units to guard against water intrusion. In addition, the arar length for top bars is 1800 m. mm/m. These two details result in a slab thickness at the edge of 115 mm. Balcony considerations are discussed in detail by Suprenant in, “Understanding Balcony Drainage,” Drainage, 2004, Concrete International International, Jan.

Trim bar at the outside edge The PT suppliers usually require two No. 13 continuous “back-up” bars behind the anchorages, about 50 to 75 mm from the edge. These bars can also limit widths of possible cracks due to unexpected restraint, drying shrinkage, or other local issues. At the edge of the balcony, it is recommended to hook the top

  

Eng.Mohammed Ashraf





7.7.3

The bottom bars have to satisfy the following provisions:

7.7.3.3

Reinforcement must extend beyond the point at distance equal to the greater of d and 12db, except at supports of simply-supported spans and at free ends of cantilevers.

7.7.3.4 have an embedment length not less than d beyond the point where bent or terminated tensile reinNote: the development length of a No. 13 black bar in an 150 mm slab with 20 mm cover is:

d

7.7.3.5

1 1.1

fy

t

fc

cb

e

s

K tr db

db

d

=

1 420 MPa 1.0 1.0 0.8 13 = 344 mm 1.1 1.0 35 MPa 25.4 + 0 13

Flexural tensile reinforcement must not be terminated in a tensile zone unless (a), (b), or (c) is Vu Vu

Vn Vn = 94,287 N (refer to Step 11)

Vn = 62,858 N > Vu = 29,200 N, therefore, OK Note that (b) and (c) do not apply. 7.7.3.8.2

7.7.8.3

At least one-fourth the maximum positive moment reinforcement must extend along the slab bottom into the continuous support a minimum of 150 mm. Extend bottom reinforcement minimum 150 mm into the supports. db for positive moment, tensile reinforcement must be limited such that d ACI 318M-14. d

Mn/Vu +

a

150 mm from the exterior support centerline and

  

Eng.Mohammed Ashraf





-

Mn Mn

[ As f y ] d

a 2

[ Aps f ps ] d p

a 2

2

Mn

2

Vu a

d

Mn/Vu +

a

d

Mn Vu

d

  

Eng.Mohammed Ashraf





Step 17: Second span bottom reinforcing bar lengths

mm from the left support centerline and 750 mm from the right interior support centerline.

reinforcing bar spacing would be violated. Because all bottom bars extend past the tensile zone, Section 7.7.3.5 doesn’t apply. All bottom bars need to extend at least 125 mm (Refer to Section 7.7.3.3) beyond the The Code requires that at least 25 percent of bottom bars be full length, extending 150 mm into the support. Solution The minimum bottom bar length is (6000 mm minus

bottom bars is one at 6000 mm and three at 5000 mm. Step 18:: Top reinforcing bar length at the middle support, CL C 7.7.3

The top bars have to satisfy the following provisions: mm from the support centerline on both sides.

7.7.3.3

Reinforcement must extend beyond the point at distance equal to the greater of d and 12db, except at supports of simply-supported spans and at free ends of cantilevers.

7.7.3.8.4

At least one-third the negative moment reinforcement at a support must have an embedment length

Section 7.7.3.3 requires all bars to extend beyond the d (125 mm) or 12 × 16 mm; therefore, 200 mm. In addition, Section 7.7.3.8.4

d, 12db, and n/16.

tion point at least 5700/16 = 356 mm, say 380 mm. Because the reinforcing bar is already at maximum spacing, no percentage of bars (as permitted by Section 11.7.3.5 or 7.7.3.8 of ACI 318M-14) can be cut

Solution plus an extension of 380 mm at each end) A practical length for top bars is 4500 mm.

  

Eng.Mohammed Ashraf





Step 19: Tendon termination There are requirements both for anchorage zones (the reinforced concrete around the anchorage) and for the anchorages themselves. 7.7.4.3.1

Post-tensioned anchorage zones must be designed and detailed in accordance with Section 25.9 of ACI 318M-14.

The concrete around the anchorage is divided into a local zone and a general zone. For monostrand anchorages, the local zone reinforcement, according to Code, “shall meet the bearing resistance requirements of ACI 423.7.” ACI 423.7 limits the bearing stresses an anchorage can impose on the concrete, unless the monostrand anchorage is tested to perform, as well as those meeting those stresses. All U.S. manufacturers’ supply tested anchorages. For the general zone, Section 25.9.3.1 (a) of ACI 318M-14 requires two “back up” bars for monostrand anchorages at the edge of the slab, and Section 25.9.3.2 (b) is not applicable to this example.

7.7.4.3.2

Post-tensioning anchorages and couplers must be designed and detailed in accordance with 25.7.

The information in Section 25.7 of ACI 318M-14 provides performance requirements for the design of PT anchorages. These only apply to the anchorage design, so the engineer rarely (if ever) is concerned about Section 25.7 25.7.

Step 20:: Shrinkage and temperature tendons 7.7.6.3 There are 4 temperature tendons evenly spaced per span (at 1200 mm O/C). Because the spacing is less than 1400 mm, no additional edge reinforcement recomis required by the Code. The commentary recom mends placing temperature tendons so that the resultant is within the kern of the slab (middle 50 mm). Anchors are usually attached to the outside forms at mid-height of the slab and longitudinally manner as to meet this recommendation. Step 21: Detailing

Fig. E3.8—Slab reinforcement.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf



 A two- way slab is usually used in buildings with columns that are approximately evenly spaced, creating a span length in one direction that is within a factor of 2 to the perpendicular direction. Structural concrete two-way slabs, which have been constructed for over 100 years, have taken many forms. The basic premise for these forms is that the slab system transmits the applied loads directly to the supporting

6.2.2 Equivalent Frame Method 14, Section 8.11) can be used for a broader range of slab geometries than are allowed for DDM use, as well as PT slabs.

This chapter discusses cast-in-place, nonprestressed, and post-tensioned (PT) slabs. The Code (ACI 318M-14) allows for either bonded or unbonded tendons in a PT slab. Because bonded tendons are not usually placed in two-way slabs in the U.S., this chapter only discusses PT slabs with unbonded tendons. At the preliminary design level, with spans given by the architect, the designer determines the loads, reinforcement type (prestressed or nonprestressed), and slab thickness. The preliminary concrete strength is based on experience and the Code’s exposure and durability provisions.

system by a series of two-dimensional frames that are then analyzed for loads acting in the plane of the frames. The

 proce ACI 318M-14 allows the designer to use any analysis proce-

met. The Code includes detailed provisions for the Direct slab system is important, the design results should not deviate reliability of the calculations used in the analysis. 6.2.1 Direct Design Method—The DDM (ACI 318M-14,

at each joint are based on papers by Corley, Jirsa, Sozen, and Siess (Corley et al. 1961; Jirsa et al. 1963, 1969; Corley

distribution method; however, any linear elastic analysis will work. The analysis calculates design moments and shears

slab strip is from the middle of one bay to the middle of the 6.2.3 Finite Element Method computer software programs are available, including those that perform static, dynamic, elastic, and inelastic analysis. element models could have beam-column elements that model structural framing members along with plane stress elements; plate elements; and shell elements, brick elements, tions, diaphragms, walls, and connections. The model mesh size selected should be capable of determining the structural -

 6.3.1 Minimum thickness -

several geometric and loading limitations. Nonprestressed designed by this method. The Code does not permit PT slabs to be designed by DDM. The results of the DDM are the approximate magnitude and distribution of slab moments, distribute the total static moment in the design panel to the column and middle strips are based on papers by Corley, Jirsa, Sozen, and Siess (Corley et al. 1961; Jirsa et al. 1963, 1969; Corley and Jirsa 1970). The total static moment is determined assuming that the reactions are along the faces of the support perpendicular to the span considered. Once the total static moment is determined, it is then distributed to

plate designs simply conform to the minimum thickness criteria and, therefore, designers do not usually calculate The Code does not provide a minimum thickness-to-span ratio for PT two-way slabs, but the ratio for usual conditions is in the range of 37 to 45. 6.3.2

it is further distributed to the column strip and middle strips. reinforcement area in the direction being designed. The designer needs to perform calculations in both directions to determine two-way slab reinforcement. The DDM also provides the design shear at each column.

(ACI 435R)).

  

Eng.Mohammed Ashraf





of Note that the spacing of slab reinforcing bar to limit crack Fig. 6.3.3—Load balancing concept. construction loads, and other construction variables all can

calculations. In addition, creep over time will increase the Typically, with a PT slab thickness-to-span ratios in the Code allowable limits. The Code limits the maximum service concrete tensile stress to below cracking stress, so

Fig. 6.4.1—Punching shear failure.

6.3.3 Concrete service stress—Nonprestressed slabs are designed for strength without reference to a pseudo-concrete

stresses is a critical part of the design. In ACI 318M-14, Section areas at columns PT slab is limited to 0.50

f c . At positive

bar if the concrete tensile stress exceeds 0.17

stress of 0.62

fc

,

f c . This

-

compressive stress in both directions due to post-tensioning to be at least 0.9 MPa.

Fig. 6.4.2—Critical 6.4.2—Critical section geometry. 

forces and moments created by the PT. A common approach to calculate PT slab moments is to use the “load balancing” bolic, the parabola has an angular “break” at the column centerlines, and that the tendon terminates at middepth at the The load balancing concept assumes the tendon exerts a uniform upward “load” in the parabolic length, and a point load down at the support. These loads are then combined with the gravity loads, and the analysis is performed with a

reverse parabolas over the interior supports rather than cusps. To conform to the Code stress limits, the designer can use an iterative approach or a direct approach. In the iterative

or both, depending on results and design constraints. In the direct approach, the designer determines the highest analysis calculates the tendon force needed to achieve the stress limit.

in each design strip (assuming the slab is a wide, shallow The discussion for the nominal one-way shear strength are the same as provided in Chapter 7 (Beams) of this Handbook and is not reproduced here. 6.4.1 Punching shear strength—Two-way shear strength, also called punching shear strength, is considered a critical strength for two-way slabs. ACI 318M calculates nominal punching shear strength based on the slab’s concrete strength

usually a truncated cone or pyramid-shape surface around the column. 6.4.2 Critical section 318M assumes a critical section. This is a vertical section extended from the column at a distance d/2, where d is the bo = 2[(c1 + d) + (c2 + d)]. The critical section to calculate concrete shear stress is bod. 6.4.3 Calculation of nominal shear strength—In ACI 318M, punching shear strength limits are given in terms of stress. As shown below, the shear stress limit for a nonprestressed reinforced slab is the least of three expressions: The Code punching shear strength limit for PT slabs are usually slightly higher than those for nonprestressed rein-

  

Eng.Mohammed Ashraf





   vc fc

0.33

0.17 1

Least of:

0.083 2

2

fc

sd bo

fc

Note: is the ratio of long side to short side of the column and columns, 30 for edge columns, and 20 for corner columns.

s

is 40 for interior

the column is closer to a discontinuous edge than four times the slab thickness h shear strength to be calculated by Table 6.4.3. vc is permitted to be the

(a) vc

(0.29

fc

(b) vc

0.083 1.5

0.3 f pc ) s

d

bo

Fig. 6.5—Assumed distribution of shear stress (ACI 318-14, R8.4.4.2.3). Commentary Section R

Vp bo d fc

0.3 f p c

Vp bo d

where s is the same as in Table 6.4.3, the value of fpc is the average of fpc in the two directions, limited to 3.5 MPa, Vp crossing the critical section, and the value of f c is limited to 5.8 MPa. Because of the shallow depth of most PT slabs, many engineers conservatively ignore the Vp/bod component when calculating vc. openings close to the column. Such openings, which are commonly used for heating, ventilating, and air conditioning (HVAC), and plumbing chases, will reduce the shear bo enclosed by straight lines projecting from the centroid of the column and tangent to the boundaries of the opening to be considered

 The factored punching shear stress at a column, vu, is the total of two components: 1) direct shear stress, vug; and 2) shear stresses due to moments transferred from the slab to the column. The two stress diagrams are added and the total Direct shear stress vug is calculated by vug = Vu/bod. To calculate the shear stresses due to slab bending, ACI 318M moment at the column, Msc a limited width over the column. The remaining percentage of Msc is transferred to the slab by eccentricity of shear. The follow two sections of Chapter 8 (ACI 318M-14) state that:

8.4.2.3.2 The fraction of factored slab moment resisted by the column, fMsc, shall be assumed to be transferred by f shall be calculated by:

1 f

1

b1 b2

2 3

8.4.4.2.2 The fraction of Msc transferred by eccentricity of shear, vMsc, shall be applied at the centroid of the critical section in accordance with Section 8.4.4.1 (ACI 318M-14), where v

f

Under certain circumstances given in Table 8.4.2.3.4 of f can be increased, which then decreases the fraction of Msc eccentricity of shear. The slab shear stresses due to the unbalanced moment transferred to the column by eccentricity of shear is calcuvMscc/Jc, where c is the distance from b0 to the critical section centroid, and Jc is the polar moment of inertia of the critical section about its centroidal axis. When vug is If the maximum total factored shear stress does not exceed the design shear stress, the slab’s concrete shear strength is the design shear stress, the slab thickness near the column can be increased by using, for example, shear capitals (ACI 318M-14, Section 8.2.5) or shear reinforcement can be added.

  

Eng.Mohammed Ashraf





   Type of shear reinforcement

Maximum vc at critical

Maximum vc at critical

318-14, Section 22.6.4.1

318-14, Section22.6.4.2

Stirrups

0.17

fc

0.17

fc

Headed shear stud reinforcement

0.25

fc

0.17

fc

 After the designer calculates the factored slab moments, is calculated with the same behavior assumptions as a beam. 6.7.1 Calculation of required moment strength—There are for two-way over the entire panel in the positive and negative moment areas.

The contribution of shear reinforcement is calculated by: vs = Av fyt/bos (ACI 318-14,

   Maximum vu at critical sections Type of shear reinforcement Stirrups

0.5 fc

Headed shear stud reinforcement

0.66 fc

to be considered, such as wheel loads in parking garages. These result in local shear slab stresses, and the slab’s

 Shear reinforcement can be provided to increase the slab’s nominal shear strength close to a column. Assuming the shear reinforcement is uniformly spaced, shear strength is dd/2 d/ 2 beyond the column face including the contribution of shear reinforcereinforce ment. The shear strength is then checked at d/2 beyond the outermost peripheral line of shear reinforcement, without the contribution of shear reinforcement. In slabs without shear reinforcement, vc is usually 0.33 fc ; however, for slab sections with shear reinforcement, the concrete contribution to shear strength is limited to the values in Table 6.6a. There is an upper limit to a slab’s nominal shear strength even with shear reinforcement, as shown in Table 6.6b. The Code states this limit in terms of the maximum factored two-way shear stress, vu, calculated at a critical section. Note that the use of stirrups as slab shear reinforcement d that satisfy (a) and (b): (a) d is at least 150 mm (b) d is at least 16db, where db is the diameter of the stirrups The use of shear studs is not limited by the slab thickness, overall height of the shear stud assembly needs to be at least the thickness of the slab minus the sum of (a) through (c): (b) Concrete cover on the base rail

should provide the distribution of panel factored moments to the column strip and middle strip. (secondary moments) need to be included. The slab’s secondary moments are a result of the column’s vertical restraint of the slab against the PT load at each support. Because the PT force and drape are determined during the calculated by the load-balancing analysis concept. A simple way to calculate the secondary moment is to subtract the tendon force times the tendon eccentricity (distance from the NA) from the total balance moment, expressed mathematically as M2 = Mbal – P × e. The second calculation is to determine fMsc at each slabcolumn joint. The value of Msc design moments on either side of the column. 6.7.2 Calculation of design moment strength strength—In a As resisting the column and middle strip’s negative and posi positive Mu is usually placed uniformly across each strip. The As fMsc must be placed within a width bslab. tendons are placed together in a line that follows the column lines in one direction and uniformly distributed in the other. fps determined from Section 20.3.2.4 of ACI 318M-14 substituted for fy in the Mn the uniform direction are the same, regardless of the tendon’s horizontal location within the slab. As resisting the panel’s negative Mu is usually placed only at the column region. The As plus Apt resisting fMsc must be placed within a width bslab per ACI 318M-14, Section 8.4.2.3.3. If the panel reinforcement already within bslab add only As Apt provided to limit concrete service panel’s positive Mu.  6.8.1 Stirrups—If stirrups are provided to increase shear strength, ACI 318M provides limits on their location and spacing in Table 6.8.1.

reinforcement includes the two critical section locations:

  

Eng.Mohammed Ashraf



6.8.2 Shear studs—If shear studs are provided to increase shear strength, ACI 318M-14 provides limits on shear stud locations and spacing in Table 6.8.2.



two critical section locations.  6.9.1 Nonprestressed reinforced slab reinforcement area and placing reinforcement in tension regions, with the area as shown in Table 6.9.1 (ACI 318M-14, Table 8.6.1.1). If more than the area must be provided.

-

8.7.4.1.3(a) provides the minimum reinforcing bar extensions, lap locations, and the minimum As at various sections. outer layer is usually placed parallel to the longer span. 6.9.2 Corners beams, induces slab moments in the diagonal direction and perpendicular to the diagonal. These moments are in addi addithis condition. 6.9.3 Post-tensioned slab – Reinforcing bar area and placing Acf in each direction, placed within 1.5h of the outside of the column.

areas (usually midspan at the bottom) exceeds 0.17 Fig. 6.8.1—Arrangement Arrangement of stirrup shear reinforcement, interior column (ACI 318-14, Commentary Fig. R8.7.6d).   Direction of measurement Perpendicular to column face Parallel to column face

Description of measurement

Maximum distance or spacing, mm

Distance from column

d/2

Spacing between stirrups

d/2

Spacing between vertical legs of stirrups

2d

fc

discretion of the designer. The Code allows for a reduced top and bottom minimum bar lengths if the design strength, calculated with only the bars must extend at least /6 on each side of the column. The bottom bars (if needed) must be at least /3 and be centered at the maximum moment. The shorter lengths often control under typical spans and loadings. If the sectional strength strength, then the minimum top and bottom bar lengths are the same as a nonprestressed reinforced slab.

 Direction of measurement

Maximum distance or spacing, mm

Description of measurement

Perpendicular to column face

Parallel to column face

Constant spacing between peripheral lines of shear studs

Spacing between adjacent shear studs on peripheral line nearest to column face

d/2

All

peripheral line of shear studs Nonprestressed slab with

vu

fc

3d/4

Nonprestressed slab with

vu

fc

d/2

Prestressed slabs conforming to Section 22.6.5.4 of ACI 318-14.

3d/4

All

2d

  

Eng.Mohammed Ashraf





Fig. 6.8.2—Typical arrangements of headed shear stud reinforcement and critical sections (ACI 318M-14, Commentary Fig. R8.7.7)    Reinforcement type Deformed bars Deformed bars or welded wire reinforcement

As,min, mm2 0.0020Ag

fy, MPa < 420 Greater of:

0.0018 420 Ag fy 0.0014Ag

6.9.4 Post-tensioned slab – Tendon area and placing—A minimum of 0.9 MPa axial compression in each direction is ration, the Code allows banding of tendons in one direction

and in the other direction the tendon spacing is uniform across the design panel, within the spacing limits of 8h and 1.5 m. at least two tendons to be placed within the column reinforcement cage in either direction for overall building integrity. 6.9.5 Slab openings trim reinforcing bar usually limits crack widths that can be caused by geometric stress concentrations and provides slab thickness as well as additional reinforcement may be

 American Concrete Institute (ACI) tures (Reapproved 2000)

  

Eng.Mohammed Ashraf





 Structural Research Series No. 218, Civil Engineering Studies, University of Illinois, June, 166 pp. Analysis for Slab Design,” ACI Journal Proceedings, V. 67, No. 11, Nov., pp 875-884.

Structural Research Series No. 269, Civil Engineering Studies, University of Illinois, Urbana, IL, July. Jirsa, J. O.; Sozen, M. A.; and Siess, C. P., 1969, “Pattern Proceedings, ASCE, V. 95, No. ST6 June, pp 1117-1137.

Fig. 6.9.1—Arrangement of minimum reinforcement near the top of a two-way slab (ACI 318-14, Commentary Fig. R8.6.1.1).

  

Eng.Mohammed Ashraf





 Two-Way Slab Example 1: Two-way slab design using direct design method (DDM) – Internal Frame This two-way slab is nonprestressed without interior beams between supports. This example designs the internal strip along locally available materials. Lateral loads are resisted by shear walls; therefore, the design is for gravity loads only. Diaphragm design is not considered in this example. Given: Uniform loads: Self-weight dead load is based on concrete density including reinforcement at 23.5 kN/m3 Superimposed dead load D = 0.75 kN/m2 Live load L = 4.8 kN/m2 Material properties: fc = 35 MPa fy = 420 MPa

  

Eng.Mohammed Ashraf



Discussion Step 1: Geometry This slab is designed using the Direct Design 8.2.1 Method (DDM) in Section 8.10. 8.10 8.10.1.1 8.10.2.1 8.10.2.2 8.10.2.3 8.10.2.1 through 8.10.2.4, which allows the use of 8.10.2.4 DDM. 8.10.2.5 8.10.2.6 The uniform design loads satisfy the limits of Sec8.10.2.7 tions 8.10.2.5 and 8.10.2.6 to allow use of DDM. 8.2.4 8.2.5



There are at least three continuous spans in each direc-

The successive spans are the same lengths so Section

The ratio of the longer to the shorter panel dimension

All design loads are distributed uniformly and due to

Ratio of unfactored live load to unfactored dead load is approximately 100/102.5 = 0.98. This ratio is less

There are no supporting beams so Section 8.10.2.7 is not applicable. This example does not include drop panels or shear caps so Sections 8.2.4 and 8.2.5 are not applicable. Using Table 8.3.1.1 with fy = 420 MPa, without drop

8.3.1.1

beam, the minimum thickness for the external panel is:

n

= 5500 – 2(300) = 4900 mm

n

33

4900 mm 33

148 mm

The minimum thickness for the internal panels is calculated using the same table and the result is the same as that for the external panel. The remainder of the building is using a slab thickness of 175 mm; therefore, use 175 mm. The slightly and shear strength. 8.3.1.3 8.3.2 of ACI 318M-14. Step 2: Load and load patterns 8.4.1.1 The load factors are provided in Table 5.3.1 of ACI 318M-14.

The load combination that controls is 1.2D + 1.6L. loading need not be checked.

  

Eng.Mohammed Ashraf





Step 3: Initial two-way shear check Before performing detailed calculations, it is often

vn, the calculations here are discussed in Step 10 more fully:

shear check. This check should reduce the probability of having to repeat the calculations shown in this example. This check uses the following limits on the ratio

vn

0.33 f c

vn

0.17 1

stress based on direct shear stress alone ( vn/vug):

vn

0.083 2

*

vn/vug

0.33 35 MPa 2

fc s

d

bo

controls vn 0.75 1.91 MPa

1.95 MPa

0.51 35 MPa fc

3 MPa

0.32 35 MPa

1.91 MPa *

1.43 MPa

vn/vug vn/vug If these ratios are not exceeded, it is possible that the slab will not satisfy two-way shear strength drop panels added, or other options for adding twoway shear strength may be considered. wu = 1.2(0.175 mm × 23.5 kN/m3 + 0.75 kN/m2) + 1.6(4.8 kN/m2) = 13.6 kN/m2

vug, the calculations here are discussed in Step 7 more fully: Vu vug bo d Vu

(4.4 m 5.5 m 0.742 m 0.742 m) 13.6 kN/m 2

Vu

321 kN

vug

321,000 N = 0.76 MPa 2968 mm 142 mm

vn/vvug

proceed.

Note that due to the basement wall supporting the exterior perimeter of the slab, the punching shear for the edge and corner columns will not need to be checked in this example. Step 4: Analysis – Direct design method moment determination 8.4.1.3 8.4.1.4 8.4.1.5 8.4.1.6 8.10.3.1

The design strip is bounded by the panel center line on each side of the column line and consists of a column strip and two half-middle strips.

  

Eng.Mohammed Ashraf



8.4.1.9

8.4.2, 8.4.2.1, 8.4.2.2 8.10.1, 8.10.2 8.10.3.1, 8.10.3.2.1, 8.10.3.2.2, 8.10.3.2.3 8.10.3.2

The lateral loads in this building are assumed to be resisted by shear walls with the slabs only acting as diaphragms between the shear walls. The slab lateral loads. The slab’s factored moments are calculated using the DDM of Section 8.10.

Because all lateral loads are assumed to be resisted combined with the lateral load analysis.

Section 8.10 provides the DDM to determine the slab’s factored moments.

The slab is eligible for design by DDM as shown in Step 1. directions and column dimensions in plan.

The DDM calculates a total panel Mo and then uses Mo negative design moments. In this example, all spans have n = 4.9 m. If spans vary, Mo must be calculated for each span length.

8.10.4



qu

2 2 n

8

Long span n = 4.9 m 2 = 4.4 m qu = 1.2 × (DLsuper + DLslab sw) + 1.6 × LL = 13.6 kN/m2 qu 2 2n Mo 180 kN·m 8

Distribute Mo in the end span, from A to B.

  

Eng.Mohammed Ashraf



8.10.4.1 8.10.4.2 8.10.4.3 8.10.4.4 8.10.4.5



Table 8.10.4.2 gives the Mo for the slab panel. In Table 8.10.4.2, this example uses the fully restrained column of the table. The reason for this is that the combined member of and little rotation is expected at the slab-to-wall connection.

In Table 8.10.4.2, for the exterior edge being fully restrained: Negative Mu at face of exterior column = 0.65Mo = 117 kN·m Maximum positive Mu = 0.35Mo = 63 kN·m Mo Negative Mu = 117 kN·m

Section 8.10.4.3 gives the option of modifying the factored moments by up to 10 percent, but that allowance is not used in this example. Section 8.10.4.4 indicates the negative moments are at the face of the supporting columns. Section 8.10.4.5

8.10.5

8.10.5.1

8.10.5.2

controls the design of the slab. Proportion the total panel factored moments from 8.10.4 to the column and middle strips for the end span, from A to B. After distributing the total panel negative and positive Mu as described earlier in Section 8.10.4, Table 8.10.5.1 then proportions the interior negative Mu assumed to be resisted by the column strip. After distributing the total panel Mu as described earlier in Section 8.10.4, 8.10.4, Table 8.10.5.2 then proportions the exterior negative Mu assumed to be resisted by the column strip.

In Table 8.10.5.1, 2/ 1 f1 = 0. Therefore, the top line of the table controls: Mu,int.neg, cs = 0.75 × 117 kN·m = 88 kN·m In Table 8.10.5.2, 2/ 1 f1 = 0. Assuming the wall behaves as a beam, C is calculated to t t

C

Ecb C 2 Ecs I s 1 0.63

x

250 mm

y

3000 mm

x x3 y y 3

C 1.48 1010 mm 4 Ecb Ecs Is t

8.10.5.5

8.10.6

After distributing the total panel Mu as described earlier in Section 8.10.4, Table 8.10.5.5 then proportions the positive Mu assumed to be resisted by the column strip. The total panel Mu from 8.10.4 is distributed into column strip moments and middle strip moments. The middle strip Mu is the portion of the total panel Mu not resisted by the column strip.

bh3 4400 (250 mm)3 12 12 10 1.48 10 3.8 2 1.97 109

1.97 109 mm 4

In Table 8.10.5.5, 2/ 1 f1 = 0. Therefore, the top line of the table controls. Mu, pos. cs = 0.60 × 63 kN·m = 38 kN·m Determine the amounts distributed to the middle strips. Subtract the amounts distributed to the column strips in Section 8.10.5 from the panel Mu calculated in Section 8.10.4. Mu, int. neg, ms = 117 kN·m – 88 kN·m = 29 kN·m Mu, ext. neg, ms = 117 kN·m – 88 kN·m = 29 kN·m Mu, pos., ms = 63 kN·m – 38 kN·m = 25 kN·m

  

Eng.Mohammed Ashraf



8.10.7.3

The gravity load moment transferred between slab and edge column by eccentricity of shear is 0.3Mo.

8.10

Repeat the Mu calculations for the interior span.



If there was no wall supporting the exterior edge of the slab, this moment would be used to calculate the twoway shear in the slab at the exterior column in 8.5. However, because of the wall, two-way shear does not apply to the design at the exterior column. The results are shown for interior panels, with the same negative Mu at either end of the panel. Mu, neg, cs = 88 kN·m Mu, neg, ms = 29 kN·m Mu, pos, cs = 38 kN·m Mu, pos, ms = 25 kN·m

column line. The middle strip moments are split into two half-middle strips, one on either side of the column strip.

Fig. E1.3—Final moment distribution in kN m Step 5: Required strength – Factored slab moment resisted by the column Slab negative moments at a column can be un 8.4.2.3 unThe columns are square, so b1/b2 = 1. 8.4.2.3.1 1 0.6 8.4.2.3.2 Msc, must f 2 b1 be transferred into the column, usually by a combi8.4.2.3.3 1 3 b2 8.4.2.3.4 8.4.2.3.5 a factor that determines the fraction of Msc trans8.4.2.3.6 8.4.2.3.3, 8.4.2.3.5

fMsc is the width of the column plus 1.5h of the slab on either side

resist fMsc.

tive slab width is considered during the detailing of the column slab joint in Section 8.5. Figure E1.4 shows undistributed total panel moments. The moment diagram is symmetric about the axis of that using this moment diagram will result in a net zero Msc condition in Section 8.10.7 to avoid an unconservative design for two-way shear.

  

Eng.Mohammed Ashraf





Fig. E1.4—Total panel moments 8.10.7 Msc to satisfy the DDM provisions at an interior 8.10.7.1 8.10.7.2 Msc = 0.07[(qDu + 0.5qLu) 2 n2 – qDu 2 n 2]

8.10.7.3

8.4.2.3.2, 8.4.4.2.2

indicates the next span. Msc to satisfy the DDM provisions at an exterior column is calculated by Section 8.10.7.3: Msc = 0.3Mo Msc two-way shear calculations are discussed in Step 7. fMsc into

s

in this -

ment from this step will be checked.

At an interior column: Msc = 0.07[(qDu + 0.5qLu) 2 n2 – qDu 2 n 2] Msc = 0.07[(5.9 kN/m2 + 0.5(7.7 kN/m2)) 4.4 m(4.9 m)2 – 5.9 kN/m2(4.4 m)(4.9 m)2] = 28.5 kN·m At an exterior column: Msc = 0.3Mo Msc = 0.3(180 kN·m) = 54 kN·m At interior columns: f = 0.6 Msc = 28.5 kN·m (0.6)(28.5 28.5 kN·m) fMsc = ( fMsc = 17.1 kN·m Using the method described in Step 8, the amount of h of the column is: As = 335 mm2/1140 mm or 294 mm2/m At exterior columns: f = 0.6 Msc = 54 kN·m fMsc = (0.6)(54 kN·m) fMsc = 32.4 kN·m Using the method described in Step 8, the amount of h of the column is: As = 643 mm2/1140 mm or 564 mm2/m

8.4.3 8.4.3.1 8.4.3.2

One-way shear slab rarely controls over two way shear in the design of a two-way slab, but it must be checked. In this section, Vu is determined. In Vn, is Vu.

is symmetric about the axis of the column at the center of the building (33 m). Vu = 147 kN

  

Eng.Mohammed Ashraf





Fig. E1.5—Shear diagram Step 7: Required strength — Factored two-way shear 8.3.1.4 Stirrups are not used as shear reinforcement in this example. 8.4.4, 8.4.4.1, 22.6.4, Determine the critical section for two-way shear 20.6.1.3.1 without shear reinforcement. Calculate bo at an interior column: bo = 2 × (c1 + d) + 2 × (c2 + d)

bo = 2 × (600 mm + 142 mm) + 2 × (600 mm + 142 mm) bo = 2968 mm

where d and this example assumes No. 16 bars when deterdeter mining d. Cover is assumed to be 20 mm per Table 20.6.1.3.1

bo, at an interior column.

  

Eng.Mohammed Ashraf



8.4.4.2 8.4.4.2.1



Determine vug due to direct slab shear stress.

Calculate the direct shear stress at the interior column with full factored load on all spans: Vu vug bo d Vu

(4.4 m 5.5 m 0.742 m 0.742 m) 13.6 kN/m 2

Vu

321 kN 321 kN = 0.76 MPa 2968 mm 142 mm

vug

8.4.4.2.1 8.4.4.2.2

Determine the slab shear stress due to moment.

Calculate the shear stress due to moments at an interior column: 0.4

v

M sc c AB Jc v

8.4.4.2.3

Calculate vu by combining the two-way direct shear stress and the stress due to moment transferred to the column via eccentricity of shear.

28.5 kN·m 376 mm 4.06 1010 mm 4

M sc c AB Jc

0.11 MPa

M sc cAB Jc Calculate the design shear stress at an interior column: vu = 0.76 MPa + 0.11 MPa = 0.87 MPa Note that these calculations are conservative. Msc assumes that some live load is not present to produce unbalanced moments, but vug assumes that full live load and dead load are present. vu

vug

v

  

Eng.Mohammed Ashraf



8.5.1 8.5.1.1 8.5.1.2 8.5.2 8.5.2.1 8.4.2.3.5



There are many methods available to determine within the span in each direction.

E1.9 for the column strip and middle strip, respeclocation. Reinforcement in an exterior panel

Mn

As f y d

Mn

As f y d

Column strip at the columns: Mu = 88 kN·m

a 2 As f y 2 0.85bf c

As f y

As

bdf c Mn

(bdf c (d 0.59 d ))

Mn

(bd 2 f c (1 0.59 ))

As As

Mn = Mu

0.0657 bdf c fy (0.0657)(2200 mm)(142 mm)(35 MPa) 420 MPa 1710 mm 2

Column strip at midspan: Mu = 38 kN·m

0.0275 As

bdf c fy

(0.0275 )(2200 mm)(142 mm)( mm)(35 MPa) (0.0275)(2200 420 MPa As 716 mm 2 Using the same method, the following can be found: As

Exterior Panels: Column strip at column line: As = 1710 mm2 Middle strip at column line: As = 549 mm2 Column strip at midspan: As = 716 mm2 Middle strip at midspan: As = 472 mm2

  

Eng.Mohammed Ashraf





Interior panels: Column strip at column lines: As = 1710 mm2 Middle strip at column lines: As = 549 mm2 Column strip at midspan: As = 716 mm2 Middle strip at midspan: As = 472 mm2

Fig. E1.8—Column Column strip moment diagram

Fig. E1.9—Middle strip moment diagram Step 9: Design strength – One-way shear 8.5.3.1.1 One-way design shear strength is calculated in 22.5 accordance with 22.5.

0.75 Vn

0.17 f c bd

Vn

(0.75)(0.17)( 35 MPa )(4400 mm)(142 mm)

Vn

471 kN

1 kN 1000 N

from Step 6; therefore, one-way shear is okay.

  

Eng.Mohammed Ashraf



Step 10: Design strength – Two-way shear 8.5.3.1.2 Two-way design shear strength is calculated in ac22.6 cordance with Section 22.6. Shear reinforcement is



Interior column: vn 22.6.5.2 of ACI 318M-14 vn

0.33 f c

vn

0.17 1

vn

0.083 2

*

0.33 35 MPa 2

fc sd bo

1.95 MPa

0.51 35 MPa fc

3 MPa

0.32 35 MPa

1.91 MPa *

controls vn

0.75 1.91 MPa

1.43 MPa

columns of 0.87 MPa from Step 7; therefore, two-way shear at interior columns is okay. The assumption that two-way shear reinforcement is

8.6 8.6.1 8.6.1.1

are provided at locations where tension is calculated in the slab.

fy = 420 MPa As,min = 0.0018 × Ag As,min = 0.0018 × 175 mm × 4400 mm As,min = 1386 mm2 This minimum area of reinforcement is split evenly be between the column and middle strips; therefore 693 mm2 per strip.

middle strip of all panels. 8.7.1 8.7.1.1 20.6.1

Concrete cover, development lengths, and splice lengths are determined in these sections.

20.6.1.3.1. The slab is not exposed to weather or in contact with the ground. Assuming No. 16 bars for

  

Eng.Mohammed Ashraf



8.7.1.2 25.4



Development length is needed to determine splice length.

Using normalweight concrete with No. 19 and smaller uncoated bars and the casting position with less than 300 mm of fresh concrete placed below the horizon25.4.2.4 are as follows: = 0.8 = 1.0 t = 1.0 = 1.0 s

e

The bar spacing is larger than the distance from the center of the bottom bar to the concrete surface. cb = 20 mm + (16 mm/2) = 28 mm db = 16 mm Ktr is assumed 0 as permitted by 25.4.2.3. 1 1.1

d

fy

t

fc

cb

e

s

K tr

db

db = 519 mm; use 560 mm - Lap splice lengths are determined in accordance with 8.7.1.3, 25.5 struction. Allowable locations for splices are shown Table 25.5.2.1 25.5.2.1.. The provided As does not exceed the As by a substantive amount. Therefore, class d

st

= 1.3 × 519 mm = 675 mm

use

st

= 700 mm

  

Eng.Mohammed Ashraf



Step 13: Reinforcement detailing – Spacing requirements Minimum and maximum spacing limits are deter8.7.2 mined. The bar spacing for design strength is also 8.7.2.1 reviewed. 25.2.1 8.7.2.2



Minimum spacing is determined in accordance with Section 25.2.1. Minimum spacing is 25 mm, db, and (4/3)dagg. Assuming that the maximum nominal aggregate size is 25 mm, than minimum clear spacing is 34 mm.With a No. 16 bar, this equates to a minimum spacing of approximately 50 mm Maximum spacing is limited by Section 8.7.2.2: At critical sections, the maximum spacing is the lesser of 2h (2 × 175 mm) and 450 mm, so 350 mm controls. All other sections, the critical spacing is the lesser of 3h (3 × 175 mm) and 450 mm, so 450 mm. However, because all of the bars cross a critical section, use a maximum spacing of 350 mm for all sections. Assuming No. 16 bars are used, the spacing for the

All spans: Column strip at column line: 1710 mm2/199 mm2 = nine No. 16 bars over 2.1 m – spacing is 230 mm Middle strip at column line: 549 mm2/199 mm2 = six No. 16 bars over 2.1 m – spacing is 350 mm (maximum spacing controls over minimum area in the middle strip) Column strip at midspan: 716 mm2/199 mm2 = six No. 16 bars over 2.1 m – spacing is 350 mm (maximum spacing controls over strength requirements at this location)

8.4.2.3.2

Middle strip at midspan: 472 mm2/199 mm2 = six No. 16 bars over 2.1 m – spacing is 350 mm (maximum spacing controls over minimum area in the middle strip) This is a check to verify that the reinforcement The minimum requirements for all column strips is a amounts required to transfer the fraction of factored 350 mm spacing of No. 16 bars. This equates to 569 mm2/m. This meets or exceeds the 277 mm2/m and design slab reinforcement. 532 mm2/m required from Step 5, therefore, Section met, additional steel would have been required to be Section 8.4.2.3.3.

Step 14: Reinforcement detailing – Reinforcement termination 8.7.4 Reinforcement lengths and extensions are at least 8.7.4.1 that required by Fig. 8.7.4.1.3 of ACI 318M-14. 8.7.4.1.1 8.7.4.1.2 8.7.4.1.3

Use ACI 318M-14, Fig. 8.7.4.1.3 to determine reforcement in these panels shows the design lengths.

  

Eng.Mohammed Ashraf





Step 15: Reinforcement detailing – Structural integrity 8.7.4.2 Structural integrity for a two-way slab is met by 8.7.4.2.1 satisfying ACI 318M-14 detailing provisions. 8.7.4.2.2

Section 8.7.4.2.1 is met when reinforcement is detailed

strip bottom bars pass through the column inside the column reinforcement cage. Step 16: Slab-column joints 8.2.7 Joints are designed to satisfy Chapter 15 of ACI 15.2.1 318M-14. 15.2.2 15.2.3 15.2.5 15.3.1 15.4.2

columns are identical and therefore, 15.2.1 and 15.3 are met. Section 15.2.2 is met in Steps 7 and 10 of this example. Section 15.2.3 is met by satisfying the detailing Sections in 15.4. Section 15.2.5 states that interior columns are restrained because they are laterally supported on four sides by the slab. Section 15.4.2 applies to columns along the exterior of the building.

(a) (15.4.2(a)) (a)) s

258 mm 2

420 MPa 0.062 35 MPa

600 mm

s = 492 mm

(b) (15.4.2(b)) s

258 mm 2

420 MPa 0.35 600 mm

s = 516 mm Because the spacing is larger than the joint depth column joint in each exterior and corner column.

  

Eng.Mohammed Ashraf





As As required, column strip, mm2 External bays

Internal bays

Column lines

Midspan

Column lines

Strength

1710

716

1710

716

Minimum

693

693

693

693

1194

1194

1194

1194

Midspan

Column lines

Maximum spacing, assuming No.16 bars

Midspan

As required, middle strip, mm2 Internal Bays

External bays Column lines

Midspan

Strength

549

472

549

472

Minimum

693

693

693

693

Maximum spacing, assuming No. 16 bars

1194

1194

1194

1194

Fig. E1.10—Summary sketch of required bars, column strip

  

Eng.Mohammed Ashraf





Fig. E1.11—Summary Summary sketch of required bars, middle strip

  

Eng.Mohammed Ashraf





Two-way Slab Example 2: Equivalent Frame Method (EFM) – Internal column line design This example is an interior column strip along grid line B in a nonprestressed two-way slab without beams between supports. This example uses the moment distribution method to determine design moments, but any method for analyzing a statically members analyzed. Given: Uniform loads: Self-weight dead load is based on concrete density including reinforcement at 23.5 kg/m3 Superimposed dead load D = 0.75 kN/m2 Live load L = 4.8 kN/m2 Material properties: fc fy = 420 MPa Thickness of slab t = 175 mm

Discussion Step 1: Geometry The moments in the slab are calculated using the 8.4.2 8.4.2.1 8.4.2.2 with Section 8.11 of ACI 318M-14.

  

Eng.Mohammed Ashraf



8.11.1 8.11.1.1 8.11.1.2 8.11.1.3 8.11.1.4 8.11.2 8.11.2.1 8.11.2.2 8.11.2.3 8.11.2.4 8.11.2.5



lent frame model.

and column frame, in a slab and column frame some of the unbalanced moments can redistribute around the column into the next span regardless of

Fig. E2.2 E2.2—Equivalent 2.2—Equivalent Equivalent frame strip 8.11.3 8.11.4 8.11.5

factor for use in the moment distribution method. sis for Slab Design,” ACI Journal Proceedings, V. 67, No. 11, Nov. pp. 875-884).

  

Eng.Mohammed Ashraf





Kec,

8.11.3, 8.11.5

9 Ecs C

Kt

with the column, Kt, is needed at each intersection. Kt on Kt are neglected in ACI 318M-14.

This example uses the same concrete strength throughout the structure, so the modulus of the calculations.

2

1

c2

3

2

2 = 4400 mm Ecs = Ecc c2 = 600 mm

Interior column torsional members the slab portion of the torsional member at the exterior columns, x = 175 mm y = 600 mm and C

1 0.63

x x3 y y 3

175 mm (175 mm)3 600 mm 600 mm 3 C = 875 × 106 mm4 Kt = 2.778 × 106 Ecc C

1 0.63

for each side of the column. Therefore, the total

The basement wall is monolithic with the column

greater than the column rotation so the torsional

Kt = 2 × 2.778 × 106 Ecc = 5.556 × 106 Ecc External column torsional members exterior columns, x = 300 mm y = 2875 mm And the total C for the exterior column torsional members is

sion, y = 2875 mm, in the calculations here is the distance from the bottom of the slab being designed C to the top of the mat foundation. C

1 0.63 1 0.63

x x3 y y 3 300 mm (300 mm)3 2875 mm 875 106 mm 4 2875 mm 3

C = 2.505 × 1010 mm4 Kt = 79.53 × 106 Ecc for each side of the column. Therefore, the total Kt = 2 × 79.53 × 106 Ecc = 159.06 × 106 Ecc

  

Eng.Mohammed Ashraf





Kec,

8.11.4

kc

Kc and below the slab are needed at each intersection. Because the slab thickness, column heights, and foundation thickness geometry is uniform, Kctop and Kcbot are consistent at each interior joint in this design strip. Kctop and Kcbot are determined using the Hardy column analogy. (K. Wang, Intermediate Structural Analysis, McGraw-Hill, New York, 1983).

Ecc

Ic

c

The following values are used in the calculations for Kctop ttop = 175 mm hbeam = 760 mm col = 4740 mm h = 5500 mm tbottom = 175 mm Kctop

it provides a section cut through the basement slab being designed in this example. The bottom-

kctop

c

Aa

Mc Ia

1 Aa

col

4740 mm

3

Ia

structure. Please refer to the short discussion at the end of this example regarding an alternate method for and slabs.

M bot c

Aa (4740 mm)3 8.874 109 mm 4 12 12 1.0cbottom 2282.5 mm

cbottom

kctop K ctop

2282.5 mm

5500 mm kctop

I col Ecc c

2282.52 8.874 109

1 4740

4.39 6004 Ecc 12 5500

8.62 106 Ecc

Fig. E2.3—Hardy column analogy for the columns above the slab being designed

  

Eng.Mohammed Ashraf



8.11.4



The following values are used in the calculations for Kcbot ttop = 175 mm col = 2825 mm h = 3000 mm tbottom = 1100 mm (assumed mat foundation thickness) Kcbot kcbot

c

Aa

1 Aa

Mc Ia

2825 mm

col 3

Ia M bot c

Aa (2825 mm)3 1.879 109 mm 4 12 12 1.0ctop 1524 mm

ctop

kcbot K cbot

1524 mm 1 15242 2825 1.879 109

3462.5 kcbot

I col c

5.51

5.51 (600) 4 Ecc 12 (3462.5)

17.186 106 Ecc

Fig. E2.4—Hardy column analogy for the columns below the slab being designed

  

Eng.Mohammed Ashraf





8.11.4 Kec

External column 1 K ec 1 1 K c Kt K ec 6

8.62 10 Ecc K ec

1 1 17.186 106 Ecc

159.06 106 Ecc

1 1 17.186 106 Ecc

1 5.556 106 Ecc

1

22.2 106 Ecc

Internal column 1 K ec 1 1 K c Kt K ec 8.62 106 Ecc K ec

4.571 106 Ecc

  

Eng.Mohammed Ashraf





8.11.2 column analogy.

c1 = c2 = 600 mm s = 5500 mm 2 = 4400 mm n = 4900 mm M = 5500 mm/2 = 2750 mm c = 5500 mm/2 = 2750 mm 1 1 I col 2 600 c 1 1 2 4400 2 Is

bh3 12 n

12

600 600 4900 2 1.34 2

Ia

49003 12

Ia

1.283 1010 mm 4

Aa

c1 1 2 I col

n

4900

Aa

5347.6

ks

s

5500

ks

4.27 ks

2

c2 1 2 I col

Mc Ia

1 Aa

ks

600/2 2

600 600 (0.746) ((0.746) 2 2

Aa

Ks

1.34

4400 1753 1.965 109 mm 4 12 c1 c2 2 c1 /2 n 2 I col 2 2

3

Ia

2

1 27502 5347.6 1.283 1010 Is

Ecs s

4.27 1.965 109 5500

Ecc

1.526 106 Ecc

  

Eng.Mohammed Ashraf





s

C.O.F .

Mc Ia

1 Aa ks

5500 C.O.F .

1 27502 5347.6 1.283 1010 ks

2.213 0.518 4.27 A1 A2 2 A3

C.O.F . Am

2

A1

2 3

A1

2 4900 3

A1

9.804 109

s n

8

c1 2 2

55002 8

s

300 2

c1 2 5500 300

Am

c1 /2 c1 s n 2 2 300 (5500 300) 4900 3.822 109 2 c1 2 c1 c1 1 s 2 2 2 2 300 1 (5500 117 106 5500 300) 300 300 2 2 9.804 109 3.822 109 2 117 106

Am

1.386 106

A2 A2 A3 A3

FEM

Am Aa s 2

1.386 1010 5347.6 55002

0.086

  

Eng.Mohammed Ashraf



Fig. E2.5—Section properties Step 4: Analysis – Moment distribution 8.11.1 This example uses moment distribution with pattern live load in accordance with Section 6.4.3 of ACI 318M-14. Loading all spans simultaneously



four column lines when full live load is applied to all spans. The structure is symmetrical and repeats from column line 2.5 through column line 5.5. The moment

stresses in the slab. Therefore, in Section 6.4.3, live not included here but have been incorporated into the example. live load patterns considered in the code. load patterns per Section 6.4.3.3 of ACI 318M-14. The

When reduced to face of support, these results are comparable to the DDM analysis in Example 1.

determined using known moments at the end of the slab along with known loads on the slab. The numerical values shown on these diagrams are the maximums determined at each location from the live load patterns discussed in Section 6.4.3.3 of ACI 318M-14. Note that the numerical values shown on these diagrams are the moments and shears reduced to the moments at the face of the columns, not at the midline of the columns as shown in the moment distribution

  

Eng.Mohammed Ashraf





Fig. E2.6—Code live load patterns, example uses 6.4.3.3 (a) and (b).

  

Eng.Mohammed Ashraf





Fig. E2.7—Moment distribution: example partial distribution with all spans with full live load.

Fig. E2.8—Moment diagram maximum values at face of support and midspan

  

Eng.Mohammed Ashraf





Fig. E2.9—Shear diagram maximum values at face of support Step 5: Design 8.11.6.6 The Code permits the moments determined from Refer to the DDM in the Two-way Slab Example 1 of the EFM to be distributed to the column and middle this Handbook for this procedure. strips in accordance with the Direct Design Method (DDM) in Section 8.10 of ACI 318M-14. Continuing on, the design solution follows a similar method as the direct design method.

  

Eng.Mohammed Ashraf





ACI 318M-14, Section 6.3.1.1 states that:

suggested in the following discussion.

in Table 6.6.3.1.1, detailed calculations for kc kc = 4.0 and : Columns: I c

0.7 I g

0.7

600(600) 12

3

7.56 109 mm 4

4400(300)3 12

3.465 109 mm 4

4400(175)3 12

0.491

Walls: I w

0.35I g

0.35

Slabs: I s

0.25 I g

0.25

109 mm 4

Kc to use for moment distribution calculations. Note that because all of the Kc

kc Ecc I c c

Upper column: Kc Kc

(4)(1)(7.56 7.56 109 ) 5500 kc Ecc I c

5.498 106

c

Lower column: Kc

(4)(1)(7.56 7.56 106 ) 3462.5

8.734 106

Kc

c

Walls (neglecting torsion action with the column): Kc

Ks

kc Ecc I w (4)(1)(3.465 109 ) 3462.5

4 106

ks Esc I s 1

Slabs: Ks

(4)(1)(0.491 109 ) 5500

0.357 106

Combining these values with Kt 2

  

Eng.Mohammed Ashraf





Two-way Slab Example 3: Post-tensioning This two-way slab is a prestressed solid slab roof without beams between supports. The strength of the slab is checked and ments of Chapters 5 and 6, engineering judgment, and known available materials. Given: Load: Superimposed dead load D = 0.75 kN/m2 Roof live load L = 1.92 kN/m2

Material properties— fc fy = 420 MPa

Fig. E3.1—Roof plan.

Discussion Step 1: Geometry 8.3.2

In the direction taken, there are six spans of 11 m. 45 The ACI 318M-14 span-to-depth ratios do not apply to post-tensioned (PT) slabs. Span/depth ratios between 40 and 50 are typically reasonable for two-way slab designs (Nawy G., 2006, Prestressed Concrete: A Fundamental Approach, Fifth edition, Pearson Prentice Hall, New Jersey, 945 pp).

t

t 11, 000 mm 244 mm

Use a thickness of 250 mm

Use a ratio of 45 to set the initial thickness of the slab.

  

Eng.Mohammed Ashraf





Step 2: Load and load patterns 8.4.1.2 Loading all spans simultaneously does not necesslab. Therefore, in Section 6.4.3 of ACI 318M-14,

using Section 6.4.3.2 of ACI 318M-14. Section 6.4.3.2 is applicable for this example because the roof live load is less than 75 percent of the combined dead loads.

Fig. E3.2—Code live load patterns, example uses 6.4.3.2

  

Eng.Mohammed Ashraf



8.2.6.1



The mixture proportion must satisfy the durabil-

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the -

determines the durability classes. Please refer to Chapter 4 of this Handbook for an in-depth discussion of the categories and classes. lates with ACI 318M. ACI encourages referencing

experience with local mixtures, the compressive least 35 MPa.

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which

8.2.6.2

suggested by the contractor. The reinforcement must satisfy Chapter 20 of ACI 318M-14. In this example, unbonded, 13 mm single strand tendons are assumed.

accordance with ACI 301M-10. This includes the PT type and strength, and reinforcing bar grade and any coatings for the reinforcing bar.

The designer determines the grade of bar and if the reinforcing bar should be coated by epoxy or galvanized, or both. In this case, assume Grade 420 bar and no coatings. 20.3 low relaxation (ASTM A416a). a). The U.S. industry A usually stresses, or jacks, monostrand to impart a al fpu, which is the maximum allowed by the Code.

The jacking force per individual strand is: 1860 MPa × 0.8 × 99 mm2 = 147.0 kN This is immediately reduced by seating and friction losses, and elastic shortening of the slab. Long term losses will further reduce the force per strand. Refer to 20.3.2.6 of the Code. R20.3.2.6

low relaxation strands. (64%fpu) per strand is common.

  

Eng.Mohammed Ashraf



Step 4: Analysis 6.6, 8.11 The analysis performed should be consistent with the overall assumptions about the role of the slab within the building system. Because the lateral force resisting system relies on the slab to transmit



Modeling assumptions: Assume a single moment of inertia for the entire length of the slab. Refer to Section 24.5.2.2 of the Code regarding cracked vs uncracked. Prestressed load limits of: ft

Although gravity moments are calculated independent of PT moments, the same model is used for both.

0.5 f c

The direct design method is not permitted for use with method is used. In practice, computer analysis software is typical. Analysis approach: the concrete slab under service loads, the tendon drape is assumed to be parabolic with a discontinuity at the support centerline as shown below, which imparts a uniform uplift over each span when tensioned. The magnitude of the uplift, wp, or “balanced load,” in each span of a prismatic member is calculated as: w =

8Fa Fa

where F a is tendon drape (average of the two high points minus the low point). In this example the PT force is assumed constant for

assumed in this example.

a1 a2

(125 mm 25 mm) (225 mm 25 mm) 150 mm 2 (225 mm 25 mm) (225 mm 25 mm) 200 mm 2

  

Eng.Mohammed Ashraf





Step 5: Analysis – Slab stress limits 8.3.4.1 is, a slab under full service load with a concrete this limit is 0.5 35 MPa 2.96 MPa tension stress not exceeding 0.5 . The slab analysis model for the service condition is the same as for the nominal condition. To verify that the concrete tensile stresses are less than 0.5 35 MPa , the net service moments and ft = M/S – F/A, where M is the net service moment, tensile stresses at the face of supports are needed. This example assumes two parameters: bt 2 (1000 mm)(250 mm) 2 S 1.04 107 mm3 6 6 (section modulus), 8.6.2.1

and A = bt = 1000 mm × 250 mm = 250,000 mm2 (gross slab area per meter).

(a) the PT force provides a F/A slab compressive stress of at least 0.9 MPa (225 kN/m) provides a uplift force wp of at least 75 percent of the slab weight, or 4.4 kN/m2 = 4.4 × 10–6 kN/mm2 Solve for F:

At the exterior support, the drape is a = 150 mm The

3

4.4 10

kN/mm/m =

8Fa l2

8F (150 mm) (11,000 mm) 2

F = 444 kN/m Location from left to right along the span Service loads

First midspan

Second midspan

Third midspan

Fourth midspan

Fifth midspan

Sixth midspan

Gravity uniform load, kN/m2

8.6

8.6

8.6

8.6

8.6

8.6

4.4

6

6

6

6

4.4

4.1

2.6

2.6

2.6

2.6

4.1

PT Uniform uplift, kN/m 2

Net load, kN/m

2

Using the above information and performing an Example 2), the following maximum service moments in the slab are determined: Negative moment is maximum at the face of the Positive moment is maximum at midspan of the

Use these moments to determine the stresses at service load:

ft ft ft

P A

M S 444 kN 250,000 mm 2

1.69 MPa

36 kN·m 1.04 × 107 mm3

2.96 MPa

OK

by Section 8.6.2.3. To avoid this, the tensile stresses in the slab should not exceed 0.17 35 MPa 1 MPa . P M ft A S 444 kN 23.6 kN·m ft 250,000 mm 2 1.04 × 107 mm3 ft

0.5 MPa 1 MPa

OK

  

Eng.Mohammed Ashraf



8.3.2



The two-way slab chapter refers the user to Section are identical and combines them to give the maximum due to service-level gravity loads…” for allowable due to the uniform live load are checked in Section Section 24.2.2 provides maximum allowed span-to- 24.2.2, therefore, the uniform live load only is applied Ig mate due to several simplifying assumptions, but it Commentary Section R24.2.3.3 of the Code alerts provides a reasonable result. of two-way slabs is challenging. This example

This is not an accurate assumption, but it should give conservative and reasonable results. Note that in PT slabs.

max

0.0065w EI

4

0.0065(1.9 N/mm 2 )(11,000 mm) 4 27,800 MPa(1.3 109 mm 4 )

5 mm

Assuming twice this to account for the two-way action of the slab, max = 2 × 5 mm = 10 mm /1100. This is Expressed as a ratio, much less than the limit of /180 >

Step 7: Analysis – Balanced, secondary, factored, and design moments Balanced moments are determined using the PT The moment curve below the table is the full design Uniform uplift load from the table of service loads moment curve with critical section moments shown on the curve. The design moments shown are at the analysis (refer to Two-way Slab Example 2). face of supports and at the point of maximum positive Secondary moments are determined using balanced moment in the span. moments and primary moments.

Two-way Slab Example 2). Design moments are determined by subtracting the secondary moments from the factored moments. The following table gives the balanced, secondary, factored, and design moments at the face of sup supports across the slab section.

The slab is symmetrical about the third column.

  

Eng.Mohammed Ashraf





Balanced moment, Mbal, kN·m/m

28

44.5

49.4

51.6

51.6

Eccentricity, e, mm

0

100

100

100

100

100

Primary moment, M1, kN·m/m

0

45.4

45.4

45.4

45.4

45.4

Secondary moment, Ms = Mbal M1, kN·m/m

28

–0.9

4

6.2

6.2

5.3

71.6

104.1

96.1

93.4

94.3

94.7

43.6

105

92.1

87.2

88.1

89.4

Mu , kN·m/m Design moment, Mu = Mu – Ms, kN·m/m

50.7

Fig. E3.4—Moment Moment diagram (negative moment at face of column). As 8.7.5.2 a, is If the PT tendons alone provide the necessary calculated by Mn,, then the code permits Aps f ps a 0.85 fc lengths. If the PT tendons alone do not provide the where Aps is the tendon area per foot of slab. to conform to standard lengths. Section 8.5.2.1 refers to Section 22.3 of the Code for The reinforcing bar and tendons are usually at the Mn. Section 22.3 refers to Section same position near the support and midspan. 22.2 for calculation of Mn. Section 22.2.4 refers to Section 20.3.2.4 to calculate fps. The span-to-depth applies: f ps

f se

70

fc 300

p

  

Eng.Mohammed Ashraf



20.3.2.4

Each single unbonded tendon is stressed to the value prescribed by the supplier. The value of fse



The tendon supplier usually calculates fse, and 1200 MPa is a common value. The force per strand is therefore 1200 MPa × 99 mm2 = 118,800 N. The

don length due to friction losses (ACI 423.10R-15), but for design purposes, fse is usually taken as the so the spacing of tendons is 118.8 kN/444 kN/m)(1000 mm/m) = 268 mm The value of Aps is therefore 99 average value. mm2 × 1000/268 × 369 mm2/m. The value of p is Aps/ (b × dp) = 369/225,000 mm2 = 0.00164. f ps = 1200 + 70 +

35 0.492

1341 MPa

This value has upper limits of fse + 210 (= 1410 MPa) and fpy (= 0.9fpu, or 1674 MPa from commentary), so the design value of fps is 1341 MPa. The compression block depth is therefore:

22.2.2.4.1 locations, except at the exterior joint. There, the Code permits a minimum d of 0.8h, or 200 mm

a

Aps f ps 0.85 f c

369 1341 0.85 35 1000

16.6 mm

Mn Apsfps(d – a/2) = 0.9 × 369 × 1341 × (225 – 8.3) = 96.5 × 106 N·mm/m = 96.5 kN·m/m. Location from left to right along the span Face of exterior support

First midspan

Face of second support

Second midspan

Face of third support

Third midspan

Face of fourth support

Mn, only tendons, kN·m/m

87.7

96.5

96.5

96.5

96.5

96.5

96.5

Mu, kN·m/m

44

74.6

105

57

92

62

87.0

Mn considering the tendons alone are greater than the design moments except for at the face of the second support. The ments of Section 7.7.3 of the Code while minimum reinforcing bar lengths can be used at all other locations. 8.6.2.3 foot is a function of the slab’s cross sectional area. Acf is based on the greater cross sectional area of lent frames intersecting at a column of a two-way slab.

As, min = 0.00075 × Acf = 0.00075 × 250 mm × 1000 mm = 187.5 mm2/m.

  

Eng.Mohammed Ashraf



8.5.2



Determine if supplying the minimum area of reinto steel tensile strength, and rearrange for compression block depth a: a

Aps f ps

As f y

0.85 f c

369 1341 187.5 420 0.85 35 1000 a 19.3 mm a

Mn

[ Aps f ps

As f y ] d

0.9 369 1341

a 2

187.5 420 (225 19.3 / 2)

6

111.2 × 10 N·mm = 111.2 kN·m strength Mu indicates that the minimum reinforcement plus the tendons supply enough tensile reinforcement for the slab to resist the factored loads at all locations. Step 11: Analysis – Distribute moments to column and middle strips Tests and research have shown that for uniformly 8.7.2.3 loaded structures variations in tendon distribution not dependent upon the distribution of tendons. It can be extrapolated that distribution of moments to the for the same total prestressing steel percentage. column and middle strips is unnecessary. tendon distribution that allows the use of banded tendon distribution in one direction. 8.4.3 8.4.3.1 8.4.3.2

One-way shear rarely controls thickness design of a two-way slab, but it must be checked. In this section, one-way shear load on the structure is determined.

one-way shear reduced to the face of support. Check maximum factored shear. Vu = 258 kN/4.4 m = 58.6 kN/m

Fig. E3.5—Shear diagram.   

Eng.Mohammed Ashraf



8.3.1.4 8.4.4 8.4.4.1 22.6.4



No stirrups are to be used as shear reinforcement. Exterior columns: d 250 mm 25 mm Determine the location and length of the critical section for two-way shear assuming that shear this examples critical sections. Note that only the exterior and interior columns are calculated in this example.

d 2

bo

2

c1

bo

2

600 mm

bo

2250 mm

(c2

225 mm d)

225 mm 2

(600 mm 225 mm)

Interior columns:

bo

d 2   2 (600 mm 225 mm) 2 (600 mm 225 mm)

bo

3300 mm

bo

2

c1

d 2

c2

Fig. E3.6—Two-way shear critical section locations.

  

Eng.Mohammed Ashraf



8.4.4.2, 8.4.4.2.1



Determine factored slab shear stress due to gravity loads vug. wu = 1.2(0.25 m(23.5 kN/m3) + 0.75 kN/m2) + 1.6(1.92 kN/m2) = 11 kN/m2

Direct slab shear stress on slab critical section at the exterior columns: Vu vug bo d Vu

(4.4 m 5.5 m 0.825 m 0.7125 mm) 11 kN/m 2

Vu

260 kN 260 kN 2500 mm 225 mm

vug

0.51 MPa

Direct slab shear stress on slab critical section at the interior columns: Vu vug bo d Vu

11 m 4.4 m (0.825 m)2

Vu

525 kN 525 kN 3300 mm 225 mm

vug 8.4.4.2.1, 8.4.4.2.2

Determine the slab shear stress due to factored slab moment resisted by column.

11 kN/m 2

0.7 MPa

Shear stress on slab due to moments at exterior columns: 0.4

v

M sc

((44 kN·m/m)(4.4 m)

cCD

225.6 mm 2.984 1010 mm 4

Jc v

M sc c AB Jc

0.59 MPa

Shear stress on slab due to moments at interior columns: 0.4 v M sc

(105 92) kN·m/m(4.4 m)

c AB

412.5 mm

Jc v

8.4.4.2.3

Determine vu by combining results from the twoway direct shear and the moment transferred to the column via eccentricity of shear.

vu

8.78 1010 mm 4 M sc c AB Jc vug

0.108 MPa

v

M sc cAB Jc

Exterior columns: vu = 0.51 MPa + 0.59 MPa = 1.1 MPa Interior columns: vu = 0.7 MPa + 0.108 MPa = 0.81 MPa

  

Eng.Mohammed Ashraf



Step 14: Design strength – One-way shear 8.5.3.1.1, Nominal one-way shear strength is calculated in 22.5 accordance with Section 22.5.



0.75 Vn

0.17 f c bd

Vn

0.75 0.17

Vn

170 kN/m

35 MPa 1000 mm 225 mm

from Step 12, therefore, one-way shear is okay. Step 15: Design strength – Two-way shear 8.5.3.1.2, Nominal two-way shear strength is calculated in 22.6 accordance with Section 22.6. Assume shear rein-

Exterior column: vn from Table 22.6.5.2 (ACI 318M-14) vn

0.33 fc

vn

0.17 1

vn

0.083 2

vn

0.33 35 MPa 2

fc d

s

bo

0.75 1.95 MPa

1.95 MPa

(0.17)3 35 MPa fc

3 MPa

(0.083)5 35 MPa

2.46 MPa

1.46 MPa

columns of 1.1 MPa from Step 13, therefore, two-way shear at exterior columns is okay. Interior column: vn from Table 22.6.5.2 (ACI 318M-14) vn

0.33 f c

vn

0.17 1

vn

0.083 2

vn

0.33 35 MPa 2

fc s

d

bo

0.75 1.95 MPa

1.95 MPa

(0.17)3 35 MPa fc

3 MPa

(0.083)4.7 35 MPa

2.31 MPa

1.46 MPa

columns of 0.81 MPa from Step 13, therefore, twoway shear at interior columns is okay. The assumption that two-way shear reinforcement is

8.7.1 8.7.1.1 20.6.1

Concrete cover, development lengths, and splice lengths are determined in these sections.

Concrete cover is determined using Table 20.6.1.3.2 (ACI 318M-14). The bottom of this slab is not exposed to weather or in contact with the ground. The

  

Eng.Mohammed Ashraf





8.7.1.2, 25.4 Development length is used for splice length determination assuming No. 16 bars.

Use Eq. (25.4.2.3(a)) of ACI 318M-14 to determine the development length. Using normal weight concrete with No. 19 and smaller uncoated bars and the casting position of less than 300 mm of fresh concrete placed below the horizontal re(ACI 318M-14) are: = 0.8 = 1.0 t = 1.0 s

e

Spacing of the bars is larger than the distance from the center of the bottom bar to the nearest concrete surface.

cb

16 mm 2

20 mm

db = 16 mm

28 mm

Ktr is assumed 0 as allowed by 25.4.2.3. 1 1.1

d

fy

t

fc

cb

e

s

K tr

db

db 8.7.1.3 25.5

d = 410.5 mm construc Lap splice lengths are determined in accordance with It is likely that splices are required during construction. Allowable locations for splices are shown in As is not more than two times larger than the required As. ACI 318M-14, Fig. 8.7.4.1.3.

st

= 1.3 × 410.5 mm = 534 mm

use Step 17: Reinforcement detailing – Spacing requirements 8.7.2 Minimum and maximum spacing requirements are 8.7.2.1 25.2.1 reinforcement are also reviewed. 8.7.2.3

st

= 600 mm

Minimum spacing is determined in accordance with Section 25.2.1. Minimum spacing is 25 mm, db, and (4/3)dagg. Assuming that the maximum nominal aggregate size is 25 mm, then the minimum clear spacing is 34 mm With a No. 16 bar, this equates to a minimum center-to-center spacing of approximately 50 mm Maximum spacing is controlled by Section 8.7.2.3. Assuming that this direction is banded, the maximum spacing requirements of Section 8.7.2.3 are not applidirection are limited to a maximum spacing of 1.5 m.

  

Eng.Mohammed Ashraf





Step 18: Reinforcement detailing – Reinforcement termination 8.7.5.2, Reinforcement termination is controlled by Section 8.7.5.5 8.7.5.2. termination of the minimum bonded reinforcement in that location. When the termination location is determined per Section 8.7.5.2, it is approximately 150 mm beyond the face of support. This termination location is within the minimum lengths of Section 8.7.5.5. Therefore, the termination locations indicated in Sec-

Step 19: Reinforcement detailing – Structural integrity 8.7.5.6 Structural integrity is met using detailing. two of the PT tendons pass through the column inside the column reinforcement cage. In this direction, banding of the post-tensioning tendons makes this a simple Step 20: Slab-column joints 8.2.7 Joints are designed to satisfy Chapter 15 of ACI 15.2.1 318M-14. 15.2.2 15.2.3 15.2.5 15.3.1 15.4.2

The concrete strength of the slab and columns are identical and therefore, Sections 15.2.1 and 15.3 are met. this example. provisions in Section 15.4 15.4. rior columns are restrained because they are laterally supported on four sides by the slab. columns along the exterior of the structure. Assuming that No. 13 bars are used as column ties, spacing of the column ties in the slab-column joint meets the following: (15.4.2(a)) s

(258 mm 2 )(420 MPa) (0.062) 35 MPa (600 mm)

492.4 mm

(15.4.2(b)) s

(258 mm 2 )(420 MPa) (0.35)(600 mm)

516 mm

the spacing is larger than the joint depth of 250 mm, exterior and corner column.

  

Eng.Mohammed Ashraf





Note: Design post-tensioning is 444 kN/m.

Assuming (118.8 kN/strand)(1 strand/x mm) = (444 kN/m)/(1000 mm)

mum bonded reinforcement required is 187.5 mm2/m No. 13 bars: 129 mm2/(x mm) = 187.5 mm2 x

Fig. E3.7—Reinforcement detailing.

  

Eng.Mohammed Ashraf



 Structural beams resist gravity and lateral loads, and any combination thereof, and transfer these loads to girders, columns, or walls. They can be nonprestressed or prestressed, cast-in-place, precast, or composite. This chapter discusses cast-in-place, nonprestressed, and post-tensioned (PT) beams. The Code allows for either bonded or unbonded tendons in a PT beam. This chapter does not cover precast, composite, or deep beams. Deep beams are also addressed in ACI 318M-14 Chapter 23, Strut-and-Tie. Beams are designed in accordance with Chapter 9 of ACI 318M-14 for strength and serviceability. Beams are assumed to be approximately horizontal, with rectangular or teeof tee-shaped beams are geometrically limited by Section 6.3.2 and 9.2.4.4, respectively, of ACI 318M-14 and the torsional strength. Beams, either nonprestressed or prestressed, that are ACI 318M-14 provides guidance on the spacing of lateral bracing.  7.2.1 Beam depth concrete strength, steel strength, and other material charchar acteristics to achieve the design performance criteria for strength and service life.

These are either provided by architectural constraints, attained from experience, or reached by assuming a depth and width and then adjusting as required through trial and error. Beam depth is addressed in Table 9.3.1.1 of ACI 318M14, which applies if a beam is nonprestressed, not supporting concentrated loads along its span, and not supporting or For prestressed or posttensioned beams, the code does not provide a minimum depth-to-span ratio, but for a superimposed live load in the range of 3 to 4 kN/m2 a usual spanto-depth ratio is in the range of 20 to 30. Table 9.3 of The Post-Tensioning Manual (Post-Tensioning Institute (PTI) have been found from experience to provide satisfactory structural performance. The slab thickness is considered as part of the overall beam depth if the beam and slab are monolithic or if the slab is composite with the beam in accordance with Chapter 16 of ACI 318M-14. —For nonprestressed beams that have 7.2.2 depths less than the ACI 318M-14, Table 9.3.1.1 minimum, or those that resist a heavy load—usually one- or two-way slabs subjected to above 4.8 kN/m2—and for PT beams, the

method, can be found in the supplement to this Handbook, ACI Reinforced Concrete Design Handbook Design Aid tions should not exceed the limits in Table 24.2.2 of ACI

for T- and L- cross section beams.

1) uncracked, U, 2) cracked, C, and transition between stress. Class U in Table 24.5.2.1, ACI 318M-14 limits the maximum beam tension stress to less than the cracking stress of concrete, 0.62 f beams, therefore, can use the gross moment of inertia. 7.2.3 stress 7.2.3.1 Strain limits limits—For —For nonprestressed beams with design axial force less than ten percent of gross sectional reinstrength (< 0.1f Ag), the minimum strain of the tension rein forcement is 0.004.. This limit is to ensure yielding behavior in case of overload. Nonprestressed beams have no concrete or steel service stress limits, only strength requirements. For pre- and- posttensioned beams, the permissible concrete service stresses are addressed in Section 24.5.3 of ACI 318M-14. For post-tensioned (PT) beams, the analysis of concrete

presented above, as U (uncracked), C (cracked), or T (transition). Beams with bonded post-tensioning, the combined area of post-tensioning and deformed bars, must be able to resist a factored load that results in moment of at least 1.2 times the section cracking moment. —Before the beam 7.2.3.2

related to the beam forces and moments created by the post-tensioning. A common approach to calculate PT beam moments is to use the “load balancing” concept, where the

has an angular “break” at the column centerlines, and that the tendon anchors are at middepth at the exterior (refer to Fig. 7.2.3.2). The load balancing concept assumes the tendon exerts a uniform upward “load” in the parabolic length, and a point load down at the support. These loads are then combined with the gravity loads, and the analysis is performed with a net service load.

  

Eng.Mohammed Ashraf





To conform to Code stress limits, the designer can use an iterative approach or a direct approach. In the iterative

both, depending on results and design constraints. In the direct approach, the designer determines the highest tensile stress, then rearranges equations so that the analysis calculates the tendon force needed to achieve the stress limit.  Loads and load combinations are obtained from Chapter 5 of ACI 318M-14. Beams can be analyzed by any method satisfying equilibrium and geometric compatibility, provided Chapter 6 of ACI 318M-14 allows for nonprestressed beams satisfying the conditions of Section 6.5.1 to use a simplishear forces in beams at the face of support and at midspan. Redistribution of design moments calculated by this method is not permitted. length are commonly calculated from classic elastic strucstruc tural analysis. The supplement to this Handbook, ACI ReinRein forced Concrete Design Handbook Design Aid – Analysis Tables (ACI SP-17DA) DA) has tables that provide moment and shear forces at beam supports and midspan for various boundary and loading conditions. The moment of inertia and modulus of elasticity values used in the classic elastic analysis are addressed in ACI 318M. Redistribution of elastic moments calculated by a classical method is permissible. -

Fig. 7.4.1—Assumed strain and stress at the nominal (a) Strains in reinforcement and concrete are directly proportional to the distance from neutral axis (plane sections remain plane after loading). (b) Maximum concrete compressive strain in the extreme (c) Stress in reinforcement varies linearly with strain up fy. The stress remains constant beyond this point as strains continue increasing. The strain hardening of steel is ignored. (d) Tensile strength of concrete is neglected. (e) Concrete compressive stress distribution is assumed to be rectangular (Fig. 7.4.1). 7.4.1.1 Nominal ((Mn ( Mn) Mn is calculated from internal forces

reinforcedepending on the strain level in the extreme tension reinforce s s

y

y

-

sition, 0.002 < s < 0.005. Reinforced concrete beams behave in a ductile manner by limiting the area of reinforcement such that the tension reinforcement yields before concrete crushes. Tension controlled beam sections produce ductile behavior at the nominal condition, which allows redistribution of stresses

The moment of inertia and modulus of elasticity values used

method is permissible.  Beams resist self-weight and applied loads, which can

forcement strain (if factored axial compression is less than 0.1f Ag) to be at least 0.004 (Section 9.3.3.1, ACI 318M14). This strain corresponds to a reinforcement ratio of about 0.75 . The Code lowers the strength reduction ( )-factor for transition-level strains to account for reduced ductility strain is shown in Fig. 7.4.1.1 and the corresponding strain 14). The basic design inequality is that the factored moment

Sn Su. 7.4.1 Flexure—Reinforced concrete beam design for the conditions of static equilibrium and strain compatibility across the depth of the section. Following are the assumptions for strength design method

expressed as Mu Mn. 7.4.1.2 ment—Nominal moment strength of a rectangular section with nonprestressed and prestressed tension reinforcement is calculated from the internal force couple shown in Fig. 7.4.1. The area of reinforcement is calculated from the equilibrium of forces. It is assumed that tension steel yields

  

Eng.Mohammed Ashraf





before concrete reaches the assumed compression strain limit of 0.003 mm/mm Accordingly, from equilibrium, set

T=C

(7.4.1.2a)

As f y 1

Ap f ps

(7.4.1.2c)

0.85

Take moments about the concrete resultant, and Mn is

Substituting the corresponding components for T and C Asfy + Apfps = 0.85f

1

(7.4.1.2b)

Mn

T di

1

2

As f y

d

a 2

where fps is calculated in Section 20.3 of ACI 318M-14. Assume that a 1

Ap f ps

a 2

dp

(7.4.1.2d) For reinforced concrete sections with single layer tension reinforcement, d = dt and s = t (7.4.1.3). The stress block 1 is between 0.85 and 0.65. For 1

Ibrahim and MacGregor 1997). For nonprestressed beams, the Apsfps term in Eq. (7.4.1.2c) and (7.4.1.2d) of this Handbook is deleted. 7.4.1.3 —Generally, beams are designed with tension reinforcement only. To add moment strength, designers can increase the tension reinforcement area or the beam depth. The cross-sectional dimensions of some applications, however, are limited by architectural or func functional considerations, and additional moment strength can be provided by adding an equal area of tension and comprescompres sion reinforcement. The internal force couple adds to the ductility. In such cases, the total moment strength consists

moment strength from the additional tension reinforcementcompression reinforcement couple, assuming both sets of reinforcement yield, as illustrated in Fig. 7.4.1.3. Mn = M1 + M2

(7.4.1.3a)

M1 is given by Eq. (7.4.1.3a) and M2 is obtained by taking the moment about the tension reinforcement. M2 = As fs d – d Assuming fs

fy,

  

Eng.Mohammed Ashraf





M2 = As fy(d – d rectangular beam forms the web. Precast double T-sections Because the steel couple does not require an additional concrete force, adding more tension steel does not create an over-reinforced section as long as an equal area is added in the compression zone. The underlining assumption in calculating the steel force couple is that the steel in compression yields at nominal strength, developing a force equal to the tensile yield strength. This assumption is true in most heavily reinforced sections because the compression Grade 420 steel (0.002 mm/ will strain to 0.003 mm/mm at nominal strength. Depending on the location of the compression reinforcement within the overall strain diagram, it is possible that the compression reinforcement has less strain than 0.002 at nominal strength and, therefore, does not yield. The designer in this case, increases the compression reinforcement area proportional to the ratio of yield strain to compression steel strain. The strain in compression steel, s from Fig. 7.4.1.3 as s d d ), once s is deters( mined for sections with tension reinforcement to assess if the compression steel yields at nominal strength. —Cast-in-place and many precast 7.4.1.4 concrete slabs and beams are monolithic, so the slab partici partici-

construction. than the web width (Fig. 7.4.1.4a). For a lightly reinforced section, this often places the neutral axis of the nominal strain same as rectangular sections, with section width equal to the In heavily reinforced T-sections, the area of tension reinforcement in the web (required by the applied moment) sion zone in the web. In such a case, the total moment strength

equal to the web concrete compression force. The condition for T-section behavior is expressed mathematically as Mu

Mn = (Mnf + M )

(7.4.1.4a)

where hf

0.85

M

A f

d

a 2

A f

2

dp

a 2

(7.4.1.4b)

(7.4.1.4c)

Many engineers calculate Mnf concrete. The M is then calculated assuming a rectangular cross section as shown in Fig. 7.4.1.4b. For continuous, statically indeterminate, PT beams, moments) need to be included per Section 5.3.11 of ACI

at each support. Because the post-tensioning force and drape are determined during the service stress checks, secondary moments can be quickly calculated by the “load-balancing” analysis concept. A simple way to calculate the secondary moment is to subtract the tendon force times the tendon eccentricity

  

Eng.Mohammed Ashraf





(distance from the neutral axis) from the total balance moment, expressed mathematically as M2 = M – P × e. 7.4.1.5 —Nonprestressed R22.5.8.3 of ACI 318M-14). provide a nominal strength larger than the cracking moment, the section cannot sustain its loads upon cracking. This level of reinforcement can be calculated under light loads or beams that are, for architectural and other functional reasons, much larger than required for strength. To protect against potentially brittle behavior immediately after cracking, ACI 318M-14 requires a minimum area of tension reinforcement (refer to Section 9.6.1.1 of ACI 318M-14). 0.25 f ,

fy

(7.4.1.5a)

but As,min needs to be at least 200 d/fy. For statically determinate beams, where the T-section a nominal strength above the cracking moment is approxi approximately twice that required for rectangular sections. Therefore, in Eq. (7.4.1.5a) is replaced by the smaller of 2 However, when the steel area provided in every section of one-third greater than required by analysis, the minimum steel area need not apply (Section 9.6.1.3 of ACI 318M-14). This exception prevents requiring excessive reinforcement in overlarge beams. For prestressed beams with bonded strands, the minimum reinforcement area is that required to develop a design moment at least equal to 1.2 times the cracking moment

Mn

M

(7.4.1.5b)

For prestressed beams with unbonded strands, abrupt because there is no strain compatibility between the unbonded strands and the surrounding concrete. Therefore, for unbonded tendons, the code only requires a minimum steel area of 0.004A . These bars should be uniformly distributed over the precompressed tensile zone, close to the 7.4.2 Shear—Unreinforced concrete shear failure is brittle. This behavior is prevented by providing adequate

shear reinforcement that intercepts the assumed inclined support, and decreases as it moves away from the support, usually at a slope equal to the magnitude of the unit uniform to the longitudinal tension reinforcement. Principal tension stresses are approximately horizontal at the longitudinal reinforcement, then change direction gradually to approxi approximately vertical at the location of maximum compression stress (the support). Consequently, cracks in concrete tend to “point” toward the region of maximum compression stress, as indicated by the cracks in Fig. 7.4.2. strength—Concrete beams are designed 7.4.2.1 Shear strength to resist shear and torsion and to ensure ductile behavior at the nominal condition. Shear strength at any location along a beam is calculated as the combination of concrete shear strength, V , and the steel shear reinforcement, Vs (Section 22.5.1.1 of ACI 318M-14). The nominal concrete shear strength V is often calculated using the simple ACI (22.5.5.1 V f d, instead of the 318M-14 Eq. (22.5.5.1), more detailed equations in Table 22.5.5.1 (ACI 318M-14). strength is calculated per ACI 318M-14, Table 22.5.6.1 and Eq. (22.5.7.1). Note that for a beam resisting tension, V cannot be smaller than zero. The nominal concrete shear strength, V , for prestressed Apsfps Apsfpu + Asfy) can be calcuTable 22.5.8.2, but need not be less than V calculated by Eq. (22.5.5.1) of ACI 318M-14. The more detailed approach to calculate V for prestressed beams is to use the lesser of shear diagonal cracking, V (Section 22.5.8.3.1 of ACI 318M-14) and shear web cracking, V (Section 22.5.8.3.2). The factored shear (or required shear strength), Vu, can be calculated at a distance d from a support face for the usual support condition (Section 9.4.3.2 of ACI 318M-14 and Fig. 7.4.2.1a of this Handbook). For other support conditions, or if a concentrated load is applied within the distance d from a support, the required shear strength is taken at the support face (refer to Fig. 7.4.2.1b and ACI 318M-14, Fig. R9.4.3.2 (e) and (f)). Beam shear reinforcement are usually U-shaped stirrups (Fig. 7.4.2.1c of this Handbook) or closed stirrups. Shear cracking is assumed to occur at 45 degrees from horizontal.

  

Eng.Mohammed Ashraf





There are limited exceptions to the above general rules given in Section 9.6.3.1 of ACI 318M-14. For example, a beam having width more than twice the thickness h does not require minimum shear reinforcement as long as the design concrete shear strength exceeds the required shear strength. Note that because the longitudinal beam bars requires support, it is impractical to eliminate beam stirrups, so it is recommended to provide stirrups in all cast-in-place beams, with spacing not exceeding d/2. often constructed without shear reinforcement in the joist thickness, rib width, and rib spacing, are provided in ACI conform to all the code limits (such as a skip joist system), the system needs to be designed as a beam and slab system. Section 9.6.3.3 (ACI 318M-14) sets lower limits on the A to ensure that stirrups do not yield upon shear crack formation. / The value of A must exceed the larger of 0.062 and 0.35 s/fyt f 318M-14 has an upper limit of f to 0.7 MPa, which corresponds to 70 MPa concrete strength. Section 22.5.3.2 of ACI 318M-14 allows the value of f to be greater than 0.7 MPa reinforceif the reinforced and prestressed beam has shear reinforce ment per Section 9.6.3.3 and 9.6.4.2 of ACI 318M-14. Refer to the Code commentary on this section for more information. 7.4.3 Torsion Torsion—Beam —Beam torsion (or twisting) creates sectional sectional center to the maximum at the section perimeter. stress occurs only around the section perimeter. Empirical expressions for torsional strength are provided in Section 9.5.4.1 of ACI 318M-14 318M-14. The torsion shear stress adds to the gravity shear stress on one vertical face, but subtracts from it on the opposite vertical face (Fig. 7.4.3a). Refer also to Fig.

rups are vertical tension ties in the truss with concrete acting as diagonal struts. The longitudinal reinforcement is the tension chord and concrete is the compression chord. For design, the tension force in each stirrup leg is assumed to be its yield strength times the leg area, and beam stirrups usually have two vertical legs. A U-stirrup has an area A = 2× calculated by Vs = A fytd/s. Designers usually calculate the required Vs and then determine the stirrup size and spacing, so the equation is rearranged as A /s = Vs/(fyd). 7.4.2.2 —In general, if V /2 is less than Vu, shear reinforcement, A , is

s

A fytd)/(Vu

V)

(7.4.2.2)

Vu/ – V represents the nominal shear strength provided by shear reinforcement Vs.

When designing for torsion, the engineer needs to distinguish between statically determinate (an uncommon condition) and statically indeterminate torsion (most common condition). Statically determinate (or equilibrium) torsion is the condition where the equilibrium of the structure requires cannot be reduced by internal force redistribution to other members. If inadequate torsional reinforcement is provided to resist this type of torsion, the beam cannot resist the applied factored torsion. Statically indeterminate (or compatibility) torsion exists is the condition where, if the beam loses its ability to resist torsion, the moment is able to be redistributed, equilibrium is maintained, and the torsion load is safely resisted by the rest of the structural system. Torsional moments can be redistributed after beam cracking if the member twisting is resisted by compatibility of deformations with the connected members. In Fig. 7.4.3b(a), the determinate beam must resist the eccentric load ( ue) on the ledge to columns through beam torsion. In Fig. 7.4.3b(b), the eccentric load can be resisted by

  

Eng.Mohammed Ashraf





 Condition Vu

V /2

V /2 < Vu

Spacing

Provision in ACI 318M-14

No shear reinforcement required

9.6.3.1

Minimum shear reinforcement s

V

d 2

600 mm

9.6.3.1

s

d 2

Prestressed

s

3h 4

600 mm

Nonprestressed

s

d 4

300 mm

s

3h 8

Nonprestressed

600 mm

+ 0.33

+ 0.33

9.7.6.2.2

0.66

9.7.6.2.2 Prestressed

0.66

300 mm

Increase cross section

22.5.1.2

1.4( A ) 2

T T , is calculated without consideration of torsion reinforcement. T

0.33

f ( A ) 2 /p

(7.4.3a)

f /p

(7.4.3b)

where Aoh is concrete area enclosed by centerline of the outermost closed transverse torsional reinforcement (Fig. 7.4.3c). —Concrete beams reinforced 7.4.3.1 for torsion per ACI 318M-14 are ductile and thus will continue

ACI 318M-14 assumes that torques less than 1/4 of T shear strength and thus is ignored. ACI 318M limits f to a maximum of 0.7 MPa, which corresponds to 70 MPa concrete strength. This limit is based on available research. ACI 318M-14, Eq. (22.7.7.1a), provides an upper limit to

beam reinforcement that resists torsion be closed stirrups and longitudinal bars located around the section periphery. so the torsion strength from closed stirrups is calculated as Tn

2 Ao At f yt s

cot

(7.4.3.1a)

  

Eng.Mohammed Ashraf





T

2 Ao A f y Ph

tan

(7.4.3.1c)



 Aoh

bar size(s), distribution around the perimeter, lengths, and

where Ao path, mm2, At is the area of one leg of a closed stirrup, mm2, and fyt is the yield strength of transverse reinforcement, MPa.

Vu and torsion shear Tu within the upper limits given by ACI 318M-14, 318M- , Eq. (22.7.7.1a): 2

Vu bw d

Tu ph 1.7 A2 oh

2

Vc bw d

fc

(7.4.3.1b)

7.6.1 is preferable to use a larger number of small bars, as opposed to fewer large bars. 7.6.1.1 t—Longitudinal reinforcement should be placed at

minimum spacing requirements for beam reinforcement. 7.6.1.2 reinforcement should be protected against corrosion and

Where stirrups are required for torsion in addition to shear, legs of a closed stirrup (Av + 2At bws/fyt) bws/fy. Longitudinal spacing of the closed stirrups must not ph -

ACI 318M, Eq. (22.7.7.1b) requires that the longitudinal bar area, A be the lesser of (a) and (b): tudinal reinforcement A (a)

(b)

f c Acp fy

7.6.1.3

—Where a beam’s -

7.6.1.4

—

f yt At ph s fy

f c Acp fy

ering cover, reinforcement should be placed as close to the

bw f yt

ph

f yt fy

  

Eng.Mohammed Ashraf



cracking between the bars, even when the required tension reinforcement area is provided and the sectional strength is adequate. To limit crack widths to acceptable limits for various exposure conditions, ACI 318M-14 Section 24.3.2 s, for reinforcement closest to the tension face. The spacing limit is the lesser of the two

280 380 fs s

280 300 fs

2.5 (7.6.1.4)

In the above equation, is the least distance from the reinforcement surface to the tension face of concrete, and fs is the service stress in reinforcement. The service stress, fs, can be calculated from strain compatibility analysis under unfac-

h



tored service loads or may be taken as 2/3fy. Note that Eq. subject to very aggressive exposure conditions or designed to be watertight. For such conditions, further investigation is warranted (Section 24.3.5 of ACI 318M-14). 7.6.1.5 —In deep beams, cracks may and the tension face. Therefore, the Code requires beams with a depth h a maximum spacing of s trated in ACI Reinforced Concrete Design Handbook Design Aid – Analysis Tables (ACI SP-17DA) (refer to Fig. 7.6.1.5 and Section 9.7.2.3 of ACI 318M-14). For this case, is the least distance from the skin reinforcement surface to the side face. ACI 318M does not specify a required steel area as skin reinforcement. Research indicates that No. 10 to No. 16 bar sizes or welded wire reinforcement with a minimum area of 0.1 mm2 7.6.2 —Stirrup bar size is usually a No.10, No. 13, or No. 16, because larger bar sizes can be

engineers increase the stirrup spacing by doubling the stir stirFor wider beams, stirrups should rups (refer to Fig. 7.6.2(d)). 7.6.2 be distributed across the cross section to engage the full beam width and thereby improve shear resistance (Fig. 7.6.2). 7.6.3 —The detailing requirements for beams resisting torsion are listed in ACI 318M14,, Sections 9.7.5 and 9.7.6, 9.7.6 for longitudinal and transverse reinforcement, respectively. The longitudinal bars are distributed around the stirrup perimeter, with at least one longitudinal bar is placed in each corner (Section 9.7.5.2 of ACI 318M-14). 14). To resist torsion, the stirrup ends are closed with 135-degree hooks (Fig. 7.6.2(c) and (e) and 7.6.3). A 135-degree hook may be replaced by a 90 degree hook where

stirrups.   

Eng.Mohammed Ashraf





aspx?ItemID=SP1714DA

(d)). Splicing stirrups is not acceptable for torsion reinforcement (Fig. 7.6.3).

 Frosch, R. J., 2002, “Modeling and Control of Side Face Beam Cracking,” , V. 99, No. 3, MayJune, pp. 376-385. tion of the ACI Rectangular Stress Block for High-Strength , V. 94, No. 1, Jan.-Feb., Concrete,” pp. 40-48. Ozbakkaloglu, T., and Saatcioglu, M., 2004, “Rectangular Stress Block for High-Strength Concrete,” , V. 101, No. 4, July-Aug., pp. 475-483. Post-Tensioning Institute (PTI), 2006, Post-Tensioning Manual, sixth edition, PTI TAB. 1-06, 354 pp.

 ACI

SP-17DA-14—Reinforced

Concrete

  

Eng.Mohammed Ashraf





 Beam Example 1: Continuous interior beam Design and detail an interior, continuous, six-bay beam, built integrally with a 175 mm slab. Given: Load–– Service additional dead load D = 0.75 kN/m2 Service live load L = 3.0 kN/m2 Beam and slab self-weights are given below. Material properties–– fc fy

Span length: 11 m Beam width: 450 mm Column dimensions: 600 mm x 600 mm Tributary width: 4.4 m

Fig. E1.1—Plan and partial elevation of 6-span interior beam. (Note: All dimensions in mm.)

  

Eng.Mohammed Ashraf





9.2.1.1

Calculation

Discussion

ACI 318M-14

The mixture proportion must satisfy the durability requirements of Chapter 19 (ACI 318M-14) and structural strength requirements. The designer determines the durability classes. Please refer to Chapter 4 of SP-17 for an in-depth discussion of the Categories and Classes.

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the exposure classes, Chapter 19 (ACI 318M-14)

Based on durability and strength requirements, and experience with local mixtures, the compressive

dinated with ACI 318M. ACI encourages refer-

least 35 MPa.

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. 9.3.1.1

Beam depth 318M-14 permits a beam to be designed without The beam has four continuous spans, so the control controlis not supporting or attached to partitions or other 9.3.2 of ACI 318M-14. 318MSelf-weight

h

18.5

(11,000 mm) 18.5

595 mm

Use 760 mm = [(450 mm)(760 mm)/(10 mm)/( 6)](23.5 kN/m3) = 8.1 kN/m 3 s = (4.4 m – 0.45 m)(0.175 m)(23.5 kN/m ) = 16.2 kN/m

9.2.4.2

Flange width The beam is placed monolithically with the slab on each side of the beam is obtained from Table 6.3.2.1.

6.3.2.1

Each side of web is the least of

f

=

n/8

+

8h s /2 n

/8

+ n/8

8(175 mm) = 1400 mm (4400 mm)/2 = 2200 mm (11,000 mm – 600 mm)/8 = 1300 mm

f

Controls

= 1300 mm + 450 mm + 1300 mm = 3050 mm

  

Eng.Mohammed Ashraf





The service live load is 2.4 kN/m2 3.8 kN/m2 in corridors per Table 4-1 in ASCE 7-10. This example will use 3.0 kN/m2 as an average as the actual layout is not provided. A 175 mm slab is a 4.1 kN/m2 service dead load. To account for the weight of ceilings, partitions, HVAC systems, etc., add 0.75 kN/m2 as miscellaneous dead load. 5.3.1

The beam resists gravity only and lateral forces are not considered in this problem. U = 1.4D U = 1.2D + 1.6L

U = 1.4(8.1 kN/m + 16.2 kN/m + (0.75 kN/m2)(4.4 m) = 39 kN/m U = 1.2(39 kN/m)/1.4 + 1.6(3.0 kN/m2)(4.4 m) = 55 kN/m Controls

9.4.3.1 therefore, the factored moments and shear forces (required strengths) are calculated at the face of the supports. 9.4.1.2

Chapter 6 of ACI 318M-14 M-14 permits several analysis procedures to calculate the required strengths.

6.5.1 using approximations per Table 6.5.2 of ACI 318M(a) Members are prismatic (b) Loads uniformly distributed (c) L D

Beams are prismatic 3.0 kN/m2 < 3(4.1 kN/m2 + 0.75 kN/m2 + Beam SW)

(d) Three spans minimum Beams have equal lengths percent. mate procedure is used.

  

Eng.Mohammed Ashraf





Using

n

= 10.4 m for all bays results in the following and moment and shear forces at face of columns.

6.5.2

6.5.4

Fig. E1.2—Shear Shear and moment diagrams. 6.5.3 The moments calculated using the approximate method cannot be redistributed in accordance with Section 6.6.5.1. Moment diagram drawn on the tension side of the beam.

  

Eng.Mohammed Ashraf





9.3.3.1 with steel strain less than 0.004 mm/mm at nominal strength. The intent is to ensure ductile behavior. In most reinforced concrete beams, such as this example, bar strain is not a controlling issue. 21.2.1(a)

20.6.1.3.1

The design assumption is that beams will be tensioned controlled with a moment reduction factor

One row of reinforcement d = h – cover – dtie – d /2

d = 760 mm – 40 mm – 10 mm – 22 mm/2 = 699 mm Use d = 700 mm

The concrete compressive strain at nominal 22.2.2.1 = 0.003

22.2.2.2

able property and is approximately 10 to 15 percent of the concrete compressive strength. ACI 318M-14 neglects the concrete tensile strength to calculate nominal strength. Determine the equivalent concrete compressive

22.2.2.3

The concrete compressive stress distribution is inelastic at high stress. The Code permits any stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonreason able agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85f

22.2.2.4.1 22.2.2.4.3

22.2.1.1

a 1 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3. For f

1

0.85

0.05(35 MPa 28 MPa) 7 MPa

0.8

Find the equivalent concrete compressive depth a by equating the compression force to the tension C=T 0.85f

= Asfy 0.85(35 MPa)( )(a) = As(420 MPa) =

=

f = 3050 mm

a

As (420 MPa) 0.85(35 MPa)(3050 mm)

a

As (420 MPa) 0.85(35 MPa)(450 mm)

= 450 mm

0.005 As

0.031As

  

Eng.Mohammed Ashraf





moments obtained from the approximate method above.

larger of the two moments. 9.5.1.1 to the required strength at each section along its Mn Mu Vn Vu 9.5.2.1

Beam is not subjected to axial force, therefore, assume Pu < 0.1ff Ag

 Mu, kN·m

Req’d

Prov.

M

1

372

1453

3.8

4

M

2

Location

22.3

Calculate the required reinforcement area (refer to Fig. E1.2 for design moment values and Fig. E1.4 for moment location). Mu

Mn

a 2

As f y d

Number of No. 22 bars

As,req’d, mm2

595

2373

6.2

7

M

541

2147

5.6

6

ME+

425

1616

4.2

5

MI+

372

1413

3.7

4

21.2.1a A No. 22 bar has a d = 22 mm and an As = 387 mm2 for the larger of M 2 and M refer to Fig. E1.4. a has been calculated above as a function of As  21.2.2 9.3.3.1

Check if the calculated strain exceeds 0.005 mm/ mm (tension controlled (Fig. E1.5), but not less than 0.004 mm/mm As f y 0.85

and

a 1

= 0.8 (calculated above) = 450 mm for negative moments and 3050 mm for positive moments. (

As,prov, mm2

a, mm

s, mm/mm

s> 0.005?

1

372

1548

48.6

0.0315

Y

2

595

2710

85

0.0168

Y

M

541

2322

77.3

0.0187

Y

ME+

425

1935

9

0.184

Y

MI+

372

1548

7

0.230

Y

M

1

t

Mu, kN·m

Location

M

)

  

Eng.Mohammed Ashraf



9.6.1.1 9.6.1.2

Minimum reinforcement The provided reinforcement must be at least the minimum required reinforcement at every section along the length of the beam. 0.25 f

((9.6.1.2a)

fy 1.4 fy

( (9.6.1.2b)

Because f

9.5.1.1 9.5.3.1 22.5.1.1 9.4.3.2



As

As

0.25 35 MPa mm)(700 mm) 1110 mm 2 (450 mm)( 420 MPa

Controls

1.4 (450 mm)( mm)(700 mm) 1050 mm 2 420 MPa

Required reinforcement areas exceed the minimum required reinforcement area at all locations. External spans

Shear strength Vn Vu Vn = V + Vs

Because conditions (a), (b), and (c) of Section at critical section at distance d from the face of the support (Fig. E1.6). Vu@d = (329 kN) – (55 kN/m)(0.7 m) = 291 kN

22.5.5.1

0.17

V

21.2.1(b)

9.5.1.1

0.17(1.0) 35 MPa (450 mm)(700 mm)/1000

317 kN

shear = 0.75 V = (0.75)(317 kN) = 238 kN

V

Vu

V = 238 kN < Vu@d = 291 kN NG Therefore, shear reinforcement is required.

  

Eng.Mohammed Ashraf



22.5.1.2



Prior to calculating shear reinforcement, check if the cross-sectional dimensions satisfy Eq. (22.5.1.2): Vu

0.66 f c bw d )

(Vc

Vu

21.2.1(b) Vu OK, therefore, section dimensions are satisfactory. 22.5.10.1

22.5.10.5.3 22.5.10.5.5

Shear reinforcement Transverse reinforcement satisfying Eq. (22.5.10.1) is required at each section where Vu Vc Vu

Vs

Vc

where Vs

Vs

Av f yt d

2

s

Calculate maximum allowable stirrup spacing:

s s checked against the maximum allowed.

First, does the beam transverse reinforcement value need to exceed the threshold value of Vs 0.33 f c bw d ?

f c bw d

Vs

f c bw d

Because the required shear strength is below the threshold value, the maximum stirrup spacing is the d lesser of d Use s

OK

sreq’d

OK

In the region where Vu Vc is not required. In this example, shear reinforcement, however, is provided over the full length. Av , min s

0.062 f c

bw f yt

Av , min

450 mm 420 MPa

s

2

and Av , min s

Controls 0.35

bw f yt

Av , min s

Av s

450 mm 420 MPa

2

2 2

300 mm

Av , min

2

s

OK

  

Eng.Mohammed Ashraf



9.7.2.1 25.2.1



Minimum bar spacing The clear spacing between the horizontal No. 22  25 mm

25 mm

Clear spacing the greater of  d  4/3(d agg )

22 mm

Assume maximum aggregate size is 19 mm

Therefore, clear spacing between horizontal bars must be at least 25 mm

9.7.2.2 24.3.4

f

4/3(19 mm) = 25 mm

= 3050 mm

n/10.

n/10

= (10,400 mm)/10 = 1040 mm < 3050 mm

n/10, additional bonded reinforcement is required in the outer

Use No. 16 for additional bonded reinforcement. Location

This requirement is to control cracking in the slab

*

No. 22 in web

No. 22 in * n/10

No. 16 in outer portion*

M

1

4

4

—

5

M

2

7

5

1

3

6

4

1

3

M

is concentrated within the web width.

Prov. No. 22

Bars on both sides of the web (Refer to Fig. E E1.12––Sections).

forcement per the higher design moment. For moment locations refer to Fig. E1.4. E Exterior span positive moment reinforcement.

Reinforced Concrete Design Handbook Design Aid – Analysis Tables, which can be downloaded aspx?ItemID=SP1714DA. The spacing is calculated below as a demonstration. ,req’d

= 2(cover + dstirrup + 20 mm) + 4d + 4(25 mm) (25.2.1)

,req’d = 2(40 mm + 10 mm + 20 mm) + 88 mm + 100 mm

= 328 mm < 450 mmOK in the 450 mm beam web (Fig. E1.7).

where dstirrup = 10 mm and d = 22 mm

OK

  

Eng.Mohammed Ashraf



9.7.2.2 24.3.1



Maximum bar spacing at the tension face must not

24.3.2

380

280 MPa fs

300

280 MPa fs

2.5

s

380

s

300

280 MPa 280 MPa

2.5(50 mm)

255 mm

Controls

and s

280 MPa 280 MPa

300 mm

where fs = 2/3fy = 280 MPa 24.3.2.1 width. Note that to the tie.

is the cover to the No.22 bar, not If bars are not bundled, 55 mm spacing is provided (Fig. E1.7), therefore OK

  

Eng.Mohammed Ashraf





Fig. E1.8––Moment diagram of exterior span.

Assume the maximum positive moment occurs at is obtained from the following free body diagram

Mmax

u(x)

2

/2 = 0

kN/m)(x)2/2 = 0 (425 425 kN·m) – (55 55 kN/m)( x = 3.93 m, say, 4 m

moment.

Mmax

u(x)

2

/2 + Vux = 0

x)2/2 + (286 kN)x = 0 x = 1.52 m, say, 1.5 m

moment.

  

Eng.Mohammed Ashraf





support

Mmax

u(x)

2

/2 + Vux = 0

x)2/2 + (329 kN)x = 0 x = 2.22 m, say, 2.3 m

moment. Development length of No.22 bar 9.7.1.2 fy t

25.4.2.2

1.7

e

f

(420 MPa)(1.3)(1.0)

d

d

where 25.4.2.4

25.4.10.1

(22 mm) 1194 mm

Say, 1200 mm

= 1.3 for top bars, because more than 300 mm of fresh concrete is placed below them and t = 1.0 for bottom bars, because not more than 300 mm of fresh concrete is placed below them. e e = 1.0,, because bars are uncoated t

(1.7 (1.7)(1.0) 35 MPa

t

((420 MPa)(1.0)(1.0 MPa)(1.0)(1.0) d

1.0)) 35 MPa (1.7)(1.0)

(22 mm)

920 mm

Say, 1000 mm

The calculated development lengths could be Areq d/A . except as required by Section 25.4.10.2. In this example, development reduction is not applied.

  

Eng.Mohammed Ashraf





First span top bars 9.7.3.2

Exterior support Bars must be developed at locations of maximum stress and locations along the span where bent or terminated tension bars are no longer required to

Four No. 22 bars are required to resist the factored negative moment at the exterior column interior face. Calculate a distance x from the column face where two No. 22 bars can resist the factored moment.

x2 (286 kN)( x) 2 2(387 mm 2 )(0.9)(420 MPa) (372 kN·m) (55 kN/m)

700 mm

2(387 mm 2 )(420 MPa) 2(0.85)(35 MPa)(450 mm)

1 10002

x = 636 mm, say, 660 mm At 660 mm from the column face, two No. 22 can be

9.7.3.3

Bars must extend beyond the location where distance equal to the greater of d or 12d .

1. d = 700 mm Controls 2. 12dd = 12(22 mm) = 264 mm Therefore, extend the middle two No. 22 bars the ( greater of the development length (1200 mm) and the d from column face

9.7.3.8.4

At least one-third of the bars resisting negative

660 mm + 700 mm = 1360 mm, say, 1400 mm Controls

22) must have an embedment length beyond the d, 12d , and n/16. d = 700 mm Controls 12d = 12(22 mm) = 264 mm n/16 = (11,000 mm – 660 mm)/16 = 646 mm Extend the remainder outside two No.22 bars, the greater of the development length (1200 mm) beyond 1200 mm + 660 mm = 1860 mm and d = 700 mm

Controls Therefore, extend bars minimum 2200 mm, increase to, say, 2300 mm from column face. Refer to Fig. E1.10. nate the two interior No. 22 bars at a distance 1400 mm from column face and extend the two exterior No. 22 bars the full span length of the beam to support the shear reinforcement stirrups as shown in Fig. E1.11.

  

Eng.Mohammed Ashraf





First span top bars

9.7.3.2 9.7.3.3

Interior support Following the same steps above, seven No. 22 bars are required to resist the factored moment at the

Calculate a distance x from the column face where four No. 22 bars can be terminated.

x2 (329 kN)( x) 2 2 3(387 mm )(0.9)(420 MPa) 3(387 mm 2 )(420 MPa) 700 mm 2 2(0.85)(35 MPa)(450 mm) 1000

(595 kN·m) (55 kN/m)

x = 0.980 m, say, 1000 mm Therefore, extend four No. 22 bars the greater of the development length (1200 mm) from column face and d 1000 mm + 700 mm = 1700 mm; increase to 1800 mm 1800 mm > 1200 mm, therefore, extend four No. 22 bars 1800 mm 9.7.3.8.4

d = 700 mm > ln/16 = 646 mm > 12ddb = 264 mm

Extend the remaining three No. 22 bars the larger ( of the development length (1200 mm) beyond the d = 700 mm controls (Fig. E E1.10).

2300 mm + 700 mm = 3000 mm > 1000 mm + 1200 mm = 2200 mm OK Note: These calculations are performed to present the code requirements. In practice, the engineer may terminate the four interior No. 22 bars at a distance d (700 mm) beyond the development length from column for a length of (1000 mm + 700 mm = 1700 mm, say, 1800 mm). Terminate one No. 22 at 3000 mm from the support and extend the remaining two exterior No. 22 bars the full span length of the beam to support the shear reinforcement stirrups as shown in Fig. E1.11. First span bottom bars 9.7.3.2 9.7.3.3

are required to resist the factored moment at the midspan of the exterior span. Calculate a distance x from the midspan where two No. 22 bars can resist the factored moment.

x2 2(387 mm 2 )(0.9)(420 MPa) 2 2(387 mm 2 )(420 MPa) 1 2(0.85)(35 MPa)(3050 mm) 10002

(425 kN·mm) (55 kN/m) 700 mm

x = 2.83 m Therefore, extend three No. 22 bars the larger of the development length (1000 mm) and a distance d maximum moment at midspan 2830 mm + 700 mm = 3530 mm, say, 3600 mm from maximum positive moment at midspan (Fig. E1.10). Extend the remaining two No. 22 bars at least the longer of 150 mm into the column or d = 1000 mm

  

Eng.Mohammed Ashraf



9.7.3.8.2



At least one-fourth of the positive tension bars must extend into the column at least 150 mm OK

9.7.3.8.3

d for positive moment tension bars must be limited such that d for that bar face (Fig. E1.9a).

d

Mn Vu

Vu = 286 kN – (55 kN/m)(1.2 m) = 220 kN a

where Mn is calculated assuming all bars at the section are stressed to fy. Vu is calculated at the section. At support, a is the embedment length beyond the center of the column. The term a is the

Mn

2(387 mm 2 )(420 MPa)  700 mm

2(387 mm 2 )(420 MPa) (2)(0.85)(35 MPa)(3050 mm)

limited to the greater of d and 12d . Mn = 227 × 106 N·mm = 227 kN·m

9.7.3.5 a bar stress discontinuity occurs. Therefore, the terminated in a tensile zone unless (a), (b), or (c) is

d

227,000 kN·mm 220 kN

This length exceeds (a) Vu

Vn

700 mm 1732 mm, say, 1800 mm d

= 1000 mm, therefore OK

(a) At 3 m and n/2 = 5.2 m m – 3 m) = 165 kN Vu = 286 kN – (55 kN/m)(5.2 kN/m)(

(b) Continuing bars provides double the area Vn

V + Vs

V is calculated in Step 6

Vu Vn . Vn (c) Stirrup or hoop area in excess of that required for shear and torsion is provided along each terminated bar or wire over a distance 3/4d from the termination point. Excess stirrup or hoop area shall be at least 0.41 s/fyt. Spacing s shall not exceed d/(8 ).

0.75 317 kN

2(71 mm 2 )(420 MPa)(700 mm) (300 mm)(1000)

Vn = 342 kN Vn = 2/3(342 kN) = 228 kN OK. Because only one of the three conditions needs to be

9.7.7.2 At least one-quarter the maximum positive moment No. 22 bars into the support. Also, two bars are more bars, but at least two bars, must be continuous. than 1/4 of the provided reinforcement.

9.7.7.3

Beam longitudinal bars must be enclosed by closed stirrups along the clear span.

Open stirrups are provided, therefore, the second

Beam structural integrity bars shall pass through the region bounded by the longitudinal column bars.

At least two No. 22 bars are extended through the column longitudinal reinforcement. Therefore, satisfying this condition.

  

Eng.Mohammed Ashraf



9.7.7.5



Splices are necessary for continuous bars. The bars splice length = (1.3)(development length) (a) Bottom bars (positive moment) shall be spliced at or near the support

st

= 1.3(1000 mm) = 1300 mm

(b) Top bars (negative moment) shall be spliced at or near midspan

st

= 1.3(1200 mm) = 1560 mm, 1600 mm

Flexural bars were calculated above in Step 5. Six No. 22 top bars are required at supports Four No. 22 bottom bars are required at midspan 9.7.6.2.2

Stirrup size and spacing were calculated following tice to maintain stirrups at 300 mm on center.

  

Eng.Mohammed Ashraf





2. The contractor may prefer to extend two No. 22 top reinforcement over the full beam length rather than adding two No. 16 hanger bars. Bars should be spliced at mid-length.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





Beam Example 2 Design and detail a one-span Beam B1 built integrally with a 175 mm slab of a seven-story building. The beam frames into girder Beam B2 at each end as shown in Fig. E2.1.

Load— Service dead load D = 0.75 kN/m2 Service live load L = 4.8 kN/m2 Material properties— f fy = 420 MPa

  

Eng.Mohammed Ashraf





Discussion

Calculation

The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements of ACI 318M-14. The designer determines the durability classes. Please refer to Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the exposure classes, Chapter 19 (ACI 318M-14) require-

ACI 318M-14

9.2.1.1

Based on durability and strength requirements, and experience with local mixtures, the compressive

dinated with ACI 318M. ACI encourages refer-

least 35 MPa.

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor.

9.3.1.1

Beam depth ACI 318M-14 permits a beam whose size satis-

supporting or attached to partitions or other -

For a simple supported beam the recommended depth The one-span beam is built integrally with the slab and the girders that it frames into.

h

16

(11, 11, 000 mm) 16

688 mm

Use 720 mm Self-weight = 450 mm t = 175 mm thick

9.2.4.2

= (0.45 m)(0.72 m)(23.5 kN/m3) = 7.6 kN/m 3 s = (0.763 m)(0.175 m)(23.5 kN/m ) = 3.1 kN/m

Tributary width = 1525 mm/2 = 763 mm (Fig. E2.1) Flange width The beam is poured monolithically with the slab on one side and will behave as an L-beam. The obtained from Table 6.3.2.1.

6.3.2.1

One side of web is the least of

(6)(175 mm) = 1050 mm (1525 mm)/2 = 763 mm Controls (11,000 mm)/12 = 917 mm

6h s /2 n

/12

f

=

n/12

+

f

= 763 mm + 450 mm = 1213 mm

  

Eng.Mohammed Ashraf





The service live load for public assembly is 4.8 kN/ The superimposed dead load is applied over a tributary m2 per Table 4-1 in ASCE 7-10. To account for the width of 1.525 m/2 + 0.45 m width of B1 = 1.21 m weight of ceilings, partitions, HVAC systems, etc., (refer to Fig. E2.1). add 0.75 kN/m2 as miscellaneous service dead load. The beam resists gravity load only and lateral forces are not considered in this example. 5.3.1

U = 1.4D

U = 1.4(7.6 kN/m + 3.1 kN/m + (0.75 kN/m2)(1.21 m) = 16.3 kN/m

U = 1.2D + 1.6L

U = 1.2(16.3 kN/m)/1.4 + 1.6(4.8 kN/m2)(1.21 m) = 23.3 kN/m Controls

9.4.3.1 therefore, the factored moments and shear forces (required strengths) are calculated at the face of the supports. 9.4.1.2

Chapter 6 permits several analysis procedures to calculate the required strengths. For this example, assume an elastic analysis results u

2

Mu =

/16.

The total moment is is u 2/16.

u

u

2

/16 = (23.3 kN/m)( kN/m)(11 m)2/16 = 176 kN·m

2

/8,, so the midspan moment

This distribution assumes the girder remains greatly reduced, which results in a higher moment at midspan. To be conservative, this example assumes the total beam moment, u 2/8, is resisted by the positive moment reinforcement and the supports resist u 2/16. Mu = u 2/8 = (23.3 kN/m)(11 m)2/8 = 352 kN·m Beam B1 frames into girders on both ends. Because girders are not as rigid as columns or walls and girders

(1/12) u 2 terms, the girder may tend to slightly rotate or may endure cracking which would reduce the rigidity and the 2 to account for a lower u positive moment at the mid-span approaches the moment of a simple support beam. So conservatively, use a positive midspan moment of (1/8) u 2.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf



9.3.3.1



The code requires a beam to be designed with steel strain at design strength of at least 0.004 mm/ mm The intent is to ensure ductile behavior at the nominal strength condition. For reinforced beams, such as this example, reinforcing bar strain is usually not a controlling issue.

21.2.1(a)

Assume the beam will be tensioned controlled with assumption will be checked later.

20.6.1.3.1

reinforcement is d = h – cover – d ,stirrup – d 22.2.2.1

d = 720 mm – 40 mm – 10 mm – 19 mm/2 = 660.5 mm say, d = 660 mm

,long/2

The concrete compressive strain at which nominal = 0.003 -

22.2.2.2 able property and its value is approximately 10 to 15 percent of the concrete compressive strength. For calculating nominal strength, ACI 318M neglects the concrete tensile strength. Determine the equivalent concrete compressive

22.2.2.3

22.2.2.4.1

22.2.2.4.3

The concrete compressive stress distribution is inelastic at high stress. The Code permits any stress distribution to be assumed in design if shown to result in predictions of nominal strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distria bution of 0.85f 1 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3. For f

1

0.85

0.05(35 MPa 28 MPa) 7 MPa

0.8

  

Eng.Mohammed Ashraf



22.2.1.1



Find the equivalent concrete compressive depth, a, by equating the compression force to the tension

C=T 0.85f

At midspan 0.85(35 MPa)( )(a) = As(420 MPa)

= Asfy =

= 1213 mm

f

a

As (420 MPa) 0.85(35 MPa)(1213 mm)

0.012 As

At support =

f

= 450 mm

a

As (420 MPa) 0.85(35 MPa)(450 mm)

0.031As

  

Eng.Mohammed Ashraf



at the midspan and the face of supports.

9.5.1.1



Mu,support = Mu,midspan =

u u

2

/16 = 176 kN·m /8 = 352 kN·m

2

The beam strength must satisfy the following equaMn Mu Vn Vu Calculate required reinforcement area based on the Midspan Mu

As f y d

Mn

a 2

No. 19 bars d = 19 mm and As = 284 mm2

352 kN·m

(0.9)(420 MPa) As 660 mm 10002

0.012 As 2

As,req’d = 1430 mm2 Supports 176 kN·m

(0.9)(420 MPa) As 660 mm 10002

0.031As 2

As = 716 mm2 21.2.2 9.3.3.1

Check if calculated strain is greater than 0.005 mm/mm (tension controlled), but not less than 0.004 mm/mm At midspan. As f y

and

0.85 1

a 1

mm2) = 20.4 mm a = 0.012A 0.012 s = (0.012)(6)(284 (0.012)(6)( = a/0.8 = 26 mm

= 0.8 < hf, therefore, beam section behaves as an L-shape.

t

=

(

)

t

0.003 (660 ( mm 26 mm) 26 mm

0.073

Note that = 450 mm for negative moments and 1213 mm for positive moments.

As f y

and

0.85 1

t

a 1

a = 0.031As = (0.031)(3)(284 mm2) = 26.4 mm = a/0.8 = 33 mm

= 0.8 (

) t

0.003 (660 mm 33 mm) 33 mm

0.057

  

Eng.Mohammed Ashraf



9.6.1.1 9.6.1.2



Minimum reinforcement ratio The provided reinforcement must be at least the minimum required reinforcement at every section along the length of the beams. (a)

(b)

0.25 f fy 1.4 fy

0.25 35 MPa (450 450 mm)(660 mm) 1046 mm 2 420 MPa Controls

As

1.4 (450 mm)( mm)(660 mm) 420 MPa

As

990 mm 2

Because f A

A

.

= (6)(284 6)(284 mm2) = 1704 mm2

As,min = 1046 mm2 OK As = 716 mm2 < As(min) = 1046 mm2 NG

= 1136 mm2

As(min) = 1046 mm2 OK

  

Eng.Mohammed Ashraf





Shear strength 21.2.1(b)

shear

9.5.1.1 9.5.3.1 22.5.1.1

Vn Vu Vn = V + Vs

9.4.3.2

Design shear force is taken at the face of the support because the vertical reaction causes vertical tension rather than compression (Fig. E2.5). Condi-

= 0.75

note at end of this step). Vu = 128 kN 22.5.5.1

0.17

V

V 9.6.3.1

0.17( 35 MPa )(450 mm)(660 mm)/1000

299 kN

V = (0.75)(299 kN) = 224 kN V Vu = 128 kN OK

Vu

Minimum area of shear reinforcement is required in V Vu all regions where Vu

V = 224 kN/2 = 112 kN

However, good engineering practice calls for providing minimum shear reinforcement over the full beam span.

300 mm < d/2 = 660 mm/ mm/2 = 330 mm OK 22.5.1.2

Cross-sectional dimensions are selected to satisfy

(

0.66

)

d from the face of the girder if hanger reinforcement is provided in the girder as outlined in a paper by A. H. Mattock and J. F. Shen, 1992, “Joints between Reinforced , V. 89, No. 3, May-June, pp. 290-295. Concrete Members of Similar Depth,”

  

Eng.Mohammed Ashraf





Step 7: Torsion

Fig. E2.6—Forces transferred from slab to edge beam. Calculate the design load at face of slab to beam connection:

22.7.4.1(a)

wu = 1.2(3.2 kN/m + (0.75 kN/m2)(1.525 m)/2) + 1.6(4.8 kN/m2)(1.525 m)/2 wu = 10.4 kN/m

Calculate the design unit torsion at beam center:

tu = (10.4 kN/m)(0.45 m/2) = 2.34 kN·m/m

Design torsional force: Therefore, check threshold torsion Tth:

Tu = (2.34 kN·m/m)(5.5 m) = 12.9 kN·m

Tth

fc

Acp2 pcp

hw = 720 mm – 175 mm = 545 mm

Fig. E E2.7––L-beam geometry to resist torsion. where Acp bihi is the area enclosed by outside perimeter Acp = (450 mm)(720 mm) + (545 mm)(175 mm) = 419,375 mm2 of concrete. pcp area.

bi + hi)pc is the perimeter of concrete gross

pcp = 2(450 mm + 545 mm + 175 mm + 545 mm) = 3430 mm

(419,375 mm 2 ) 2 smaller of the projection of the beam below the slab Tth (0.083)( 35 MPa ) 3430 mm (545 mm) and four times the slab thickness (720 6 10 T = 25.2 × N·mm = 25.2 kN·m mm). Therefore, use 545 mm (refer to Fig. E2.7). th 21.2.1c

Tth = (0.75)(25.2 kN·m) = 18.9 kN·m Tth = 18.9 kN·m OK Tu

  

Eng.Mohammed Ashraf



9.7.2.1 25.2.1

Minimum bar spacing Minimum clear spacing between the horizontal No.  25 mm Greatest of  d  4/3(d agg ) Assume maximum aggregate size 19 mm Check if six No. 19 bars can be placed in the

,req’d

25 mm 19 mm 4/3(19 mm) = 25 mm Therefore, clear spacing between horizontal bars must not be less than 25 mm

= 2(cover + dstirrup + 19 mm) + 5d + 5(25 mm)

where dstirrup = 10 mm and d = 19 mm

9.7.2.2 24.3.1 24.3.2



,req’d

= 2(40 mm + 10 mm + 19 mm) + 95 mm + 125 mm = 358 mm < 450 mm OK

Therefore , six No. 19 bars can be placed in one layer in the 450 mm beam web with 43 mm spacing between bars (Fig. E2.8).

Maximum bar spacing at the tension face must not exceed the lesser of

s

380

280 fs

300

280 fs

2.5

s

380

280 MPa 280 MPa

s

300

280 MPa 280 MPa

2.5(50 mm)

255 mm

Controls

300 mm

The maximum spacing concept is intended to is the 250 mm spacing is provided, therefore OK

  

Eng.Mohammed Ashraf





Development length of No. 19 bar 9.7.3 9.7.1.2 25.4.2.2

Top fy t 2.1

e

(420 MPa)(1.3)(1.0)

d

f

d

(2.1)(1.0) 35 MPa say, 900 mm

where 25.4.2.4

9.7.1.3 25.5.2.1

t t = 1.3 for top bars, because more than 300 mm of fresh concrete is placed t =1.0 for bottom bars, because not more than 300 mm of fresh concrete is placed below them. e e = 1.0, because bars are uncoated

Splice length of No. 19 reinforcing bar Per Table 25.5.2.1 splice length is ( st) = 1.3( d) Top tension reinforcement

Mmax

u(x)

/2 + Vux = 0

835 mm ,

(19 mm)

642 mm ,

Bottom (420 MPa)(1.0)(1.0) d

(2.1)(1.0) 35 MPa say, 750 mm

d

2

(19 mm)

= (1.3)(835 mm) = 1086 mm, say, 1.1 m d = (1.3)(642 mm) = 835 mm, say, 0.9 m

( 176 kN·m) 23.3 kN/m

x2 2

((128 kN)x

0

x = 1.61 m, say, 1.7 m

and the remainder two No. 19 bars will be extended over the full beam span to support stirrups. 9.7.3.8.4

At least one-third of the bars resisting negative moment at a support must have an embedment d, 12d , and n/16.

Two of three No. 19 bars are extended over full beam length and one No. 19 bars is terminated

d = 660 mm 12d = 12(19 mm) = 228 mm n/16 = 11,000 mm/16 = 688, say, 700 mm Controls 1700 mm + 688 mm = 2298 mm Therefore, the middle two No. 19 bar will be terminated at 2400 mm from face of support.

embedment length (700 mm).

  

Eng.Mohammed Ashraf



9.7.3.2 9.7.3.3



Bottom tension reinforcement Bars must be developed at points of maximum stress and points along the span where bent or terminated tension bars are no longer required to

Six No. 19 bars are required to resist the factored moment at the midspan.

x2 2(284 mm 2 )(0.9)(420 MPa) 2 2(284 mm )(420 MPa) 1 2(0.85)(35 MPa)(1213 mm) 106

(352 kN·m) 23.3 kN/m

2

Two No. 19 bars can resist a factored moment located at a section x from midspan.

9.7.3.3

660 mm

x = 4.25 m, say, 4.3 m

A bar must extend beyond the point where it is no to the greater of d or 12d .

1) d = 660 mm Controls 2) 12d = 12(19 mm) = 228 mm Therefore, extend four No. 19 bars the greater of the development length (750 mm) from the maximum moment at midspan and a distance d = 660 mm from

d= 660 mm Therefore, extend four No. 19 bars 5 m from midspan. 9.7.3.8.2

A minimum of one-fourth positive tension bars must extend into the support minimum 150 mm

19 a minimum of 150 mm into the support, but not less than d (Fig. E2.7 E2.7).

bars are extended into the support rather than terminating them 500 mm from the support as shown by calcu calculations. Integrity reinforcement 9.7.7.2 but not both. At least one-fourth the maximum positive moment No. 19 bars into the girders. bars, but not less than two bars must be continuous. Longitudinal bars must be enclosed by closed stirrups along the clear span of the beam.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





Beam Example 3 Determine the size of a one-span beam (B2) built integrally with a 175 mm slab of a seven-story building. The beam frames into two girder beams (B3) as shown in Fig. E3.1. Design and detail the beam.

Load–– Service dead load D = 0.75 kN/m2 Service live load L = 4.8 kN/m2 Concentrated loads Pu = 128 kN (PD = 64 kN and PL = 32 kN) located 1975 mm south and north of Column Lines B and D, respectively. (Refer to Example 2.) Material properties–– f fy = 420 MPa

Fig. E3.1—Plan of Beam B2.

  

Eng.Mohammed Ashraf





9.2.1.1

Calculation

Discussion

ACI 318M-14

The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements of ACI 318M-14. The designer determines the durability classes. Please refer to Chapter 4 of SP-17 for an in-depth discussion of the Categories and Classes. -

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the expo-

Based on durability and strength requirements, and experience with local mixtures, the compressive least 35 MPa.

dinated with ACI 318M. ACI encourages refer-

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor.

9.3.1.1

Beam depth Beam depth cannot be calculated using Table 9.3.1.1,, because two beams frame into it (concentrated loads). For framing simplicity, choose a to allow for ease of construction and placement of reinforcement.

h = 760 mm

24.2.2

9.2.4.2

to Table 24.2.2. Flange width The beam is monolithic with the slab on one side in the middle over a 4.4 m long section and slab on both sides for the remainder of the beam. At the maximum positive moment, the beam will behave on one side of the beam is the least from Table 6.3.2.1.

6.3.2.1

One side of web is the least of

6h

(6)(175 mm) = 1050 mm

s /2

(5050 mm)/2 = 2525 mm

n

/12

(8800 mm – 450 mm)/12 = 696 mm f

=

n/12

+

f

Controls

= 696 mm + 450 mm = 1146 mm

On both sides of the opening, the beam is placed monolithically with the slab and will behave as a is obtained from Table 6.3.2.1. 6.3.1

Each side of web is the least of

8h

8(175 mm) = 1400 mm

s /2

(5050 mm)/2 = 2525 mm

n

((8800 mm – 450 mm)/8 = 1044 mm

/8 f

= n/8 +

+

n/8

f

Controls

= 1044 mm + 450 mm + 1044 mm = 2538 mm

  

Eng.Mohammed Ashraf





Self-weights of B2 = [(0.45 m)(0.76 m)(23.5 kN/m3) = 8 kN/m

= 450 mm

Tributary load between Girder B2 and Column Line 3 (refer to Fig. E3.2). The load is transferred to B2 through Beam B4 along Column Line C spanning (4750 mm clear span).

For lobbies and assembly areas, the uniform design live load is 4.8 kN/m2 per Table 4-1 in ASCE 7-10. To account for weights from ceilings, partitions, and HVAC systems, add 0.75 kN/m2 as miscellaneous dead load.

Fig. E3.2 E3.2—Beams Beams B1 and B B4 framing into B2. Slab self-weight ((175 mm thick) supported by Beam

450 mm wide by 760 mm deep.

Ps = (0.175 m)(4.4 4.4 m)(5.05 m/2)(23.5 kN/m3) = 46 kN

PB

(0.45 0.45 m)(0.76 m – 0.175 m)

5.05 m 0.3 m (23.5 kN/m3 ) 2

14.7 kN

Superimposed dead load of 0.75 kN/m2

PSD = (0.75 kN/m2)(4.4 m)(5.05 m)/2 = 8.3 kN P = 46 kN + 14.7 kN + 8.3 kN = 69 kN

Concentrated load between Column Line 3 and girder transferred at midspan Beams B1 frame into Beam B2 at 1975 mm and 6825 mm from Column Line B (Fig. E3.2).

PL = (4.8 kN/m2)(4.4 m)(5.05 m)/2 = 53.3 kN

Pu = 128 kN The beam resists gravity load only. Lateral forces are not considered in this problem.

5.3.1a

u

= 1.4D

5.3.1b

u

= 1.2D + 1.6L

= 1.4(8 kN/m + (0.75 kN/m2)(0.45 m) = 11.7 kN/m 2 u = 1.2(11.7 kN/m)/1.4 + 1.6(4.8 kN/m )(0.45 m) = 13.5 kN/m Controls u

5.3.1a

Pu = 1.4PD

Pu = 1.4(69 kN) = 97 kN

5.3.1b

Pu = 1.2PD + 1.6PL

Pu = 1.2(69 kN) + 1.6(53.3 kN) = 168 kN

Controls

  

Eng.Mohammed Ashraf





Beam B2 is monolithic with supports. 9.4.1.2

Chapter 6 permits several analysis procedures to calculate the required strengths. For this example, calculate beam moment at Reinforced Concrete Design Handbook Design Aid – Analysis Tables, which can be downloaded aspx?ItemID=SP1714DA Mu =

u

2

/12+Pu /8 + Pua2 / 2+Pu

2

/

2

and the analysis shows the beam shear is Vu = u /2 +Pu/2

Mu = (13.5 kN/m)(8.8 m)2/12 + (168 kN)(8.8 m)/8 + (128 kN)(1.975 m)2(6.825 m)/(8.8 m)2 + (128 kN)(1.975 m)(6.825 m)2/(8.8 m)2 = 468 kN·m Vu = (13.5 kN/m)(8.8 m)/2 + (168 kN)/2 + 2(128 kN)/2 = 272 kN Note that the (2)(128 kN) shear force represents the two beams framing into the girder beam.

forced Concrete Design Handbook Design Aid –

aspx?ItemID=SP1714DA, aspx?ItemID=SP DA,, assuming maximum DA Mu = 363 kN·m This distribution assumes the girder remains uncracked. If the girder, does crack, however, its moments occurs. Assume that the moments at supports are reduced Mu = (0.85)(468 kN·m) = 398 kN·m Accordingly, the moment at midspan must be Mu = (363 kN·m) + (0.15)(468 kN·m) = 433 kN·m Alan Mattock states that, “…, and it is concluded that redistribution of design bending moments by up to 25% does not result in performance inferior to that of beams designed for the distribution of bending moments predicted by the elastic theory, either at working loads or at failure.” (Mattock, A. H., 1959, , “Redistribution of Design Bending Moments in Reinforced Concrete Continuous Beams,” Institution of Civil Engineers (London), V. 13, pp. 35-46.) imposed to avoid excessive cracking at elastic service moments.” (Scholz, H., 1993, “Contribution to Redistribution of Moments in Continuous Reinforced Concrete Beams,” , V. 90, No. 2, Mar.-Apr., pp. 150-155.)

  

Eng.Mohammed Ashraf





assume that (B1) reaction of 128 kN and 169 kN are applied at the face of Beam B2, but in opposite directions. Ignoring the distributed load

Refer to the torsion diagram in Fig. E3.3. Tu = (128 kN)(0.225 m) = 28.8 kN·m and from B4 Tu = (168 kN)(0.225 m) = 38 kN·m

Fig. E3.3—Shear, moment, and torsion diagrams.

  

Eng.Mohammed Ashraf



9.3.3.1



The Code does not allow a beam to be designed with steel strain at design strength less than 0.004 mm/mm The intent is to ensure ductile behavior at the nominal condition. For usual reinforced beams, such as this example, reinforcing bar strain is not a controlling issue.

21.2.1(a) 0.9. This assumption will be checked later. 9.7.1.1 20.6.1.3.1 d = h – cover – dtie – d /2

22.2.2.1

d = 760 mm – 40 mm – 10 mm – 19 mm/2 = 700 mm

The concrete compressive strain at which nominal = 0.003

22.2.2.2 variable property and its value is approximately 10 to 15 percent of the concrete compressive strength. For calculating nominal strength, ACI 318M neglects the concrete tensile strength. Determine the equivalent concrete compressive

22.2.2.3

The concrete compressive stress distribution is inelastic at high stress. The actual distribution of concrete compressive stress is complex and usually not known explicitly. The Code permits any stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85f a 1 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3.

22.2.2.4.1

For f

22.2.2.4.3

Find the equivalent concrete compressive depth, a, by equating the compression force in the section to C=T 0.85f

1

= Asfy

0.85

0.05(35 MPa 28 MPa) 7 MPa

0.8

0.85(35 MPa)( )(a) = As(420 MPa)

  

Eng.Mohammed Ashraf



22.2.1.1

f

=

= 1146 mm

= 450 mm

a



As (420 MPa) 0.85(35 MPa)(1146 mm)

0.0123 As

At support 0.85(35 MPa)(450 mm)(a) = As(420 MPa) a

As (420 MPa) 0.85(35 MPa)(450 mm)

0.0314 As

  

Eng.Mohammed Ashraf





at the midspan and the face of supports. 9.5.1.1 Mn Mu Vn Vu

Midspan Mu

Mn

a 2

As f y d

0.0123 As 2

As

No. 19 bars; db = 19 mm and As = 284 mm2

As = 1661 mm2; use six No. 19

Note that b = 450 mm for negative moments and b = 1146 mm for positive moments.

Supports 0.0314 As 2

As 21.2.2 9.3.3.1

As = 1559 mm2 use six No. 19

a

As f y 0.85 f c b 1

cu t

c

and c

a 1

Midspan a = 0.0123As c=a thickness

2

= 0.8

d

c

t

0.003 26.2 mm

Supports a = 0.0314A 0.0314 s c=a t

2

0.003 66.9 mm

Fig. E3.5—Strain distribution across beam section.

  

Eng.Mohammed Ashraf



9.6.1.1 9.6.1.2



Minimum reinforcement area The reinforcement area must exceed the minimum required at every section along the length of the beam. 0.25 f

(a)

As

fy

0.25 35 MPa (450 mm)(600 mm) 1117 mm 2 420 MPa

1.4 fy

(b)

Because f

A A

= 1704 mm2

As(min) = 1117 mm2 OK As(min) = 1117 mm2 OK

= 1704 mm2

Shear strength 21.2.1(b) 9.5.1.1 9.5.3.1 22.5.1.1

shear

Vn

= 0.75

Vu

Vn = V + Vs

9.5.1.2 21.2.1b 9.4.3.1 9.4.3.2

Design shear force is taken at the face of the support because the vertical reaction causes vertical E3.5). tension rather than compression (Fig. E 3.5). Condi.

Vu = 268 kN 22.5.5.1

V

0.17

Vn

0.17 35 MPa (450 mm)(700 mm)/1000

317 kN

V = (0.75)(317 kN) = 238 kN Vn = 238 kN < Vu = 262 kN NG

Vu

Therefore, shear reinforcement is required. Cross-sectional dimensions are selected to satisfy

22.5.1.2

(

0.66

)

Vu

(317 kN 0.66 35 MPa (450 mm)(700 mm)/1000) 1160 kN

Section dimensions are satisfactory. d from the face of the girder if hanger reinforcement is provided in the girder, as outlined in a paper by A. H. Mattock and J. F. Shen, “Joints between Reinforced Concrete Members of Similar Depth,” , V. 89, No.3, May-June 1992, pp. 290-295.

  

Eng.Mohammed Ashraf



22.5.10.1

22.5.10.5.3 22.5.10.5.6



Shear reinforcement Transverse reinforcement is required at each section where Vu V Section AB Vs Vu V where Vs A s

Vs

A f d s

V f yt d

9.7.6.2.2

Vs

24.5 kN 0.75

A s

(32 kN) (420 MPa)(700 mm)

0.33

?

0.109

0.33( 35 MPa )(450 mm)(700 mm)

0.33

Is

32 kN

32 kN

0.33

615 kN

615 kN

OK

d/2 = 700 mm/2 = 350 mm or 600 mm Try No. 10 stirrups at 300 mm on center. A s

2(71 mm 2 ) 300 mm

0.473 mm 2 /mm

0.109 mm 2 /mm

OK 9.6.3.3

A,

/s

0.062 f

A, f yt

s

0.062 35 MPa

450 mm 420 MPa

0.393 mm 2 /mm Controls

and

As , min /s

0.35

A, f yt

s

A s

0.35

450 mm 420 MPa

2(71 mm 2 ) 300 mm

0.375 mm 2 /mm

0.473 mm 2 /mm

A, s

0.393 mm 2 /mm

OK

  

Eng.Mohammed Ashraf





Determine if girder torsion can be neglected. 22.7.4.1(a)

Check threshold torsion Tth T

0.083

f

A2 p

Determine portion of slab to be included with the

T-section between a and and between d and e and

where A ihi is the area enclosed by outside perimeter of concrete p area

i

A = (450 mm)(760 mm) + 2(585 mm)(175 mm) = 546,750 mm2

+ hi)is the perimeter of concrete gross p = 2(450 mm + 585 mm + 585 mm + 585 mm + 175 mm) = 4760 mm

Tth 9.5.1.1 9.5.1.2

Refer to Fig. E3.3 for torsional value Tu near supports.

(0.083)(1.0) (0.083)(1.0 35 MPa

(546, 750 mm 2 ) 2 4760 mm

Tth = 30.84 × 106 N·mm = 30.8 kN·m

21.2.1c Tth = (0.75 (0.75)(30.8 30.8 kN·m) = 23.1 kN·m Tth Tu = 9.8 kN·m OK Torsion reinforcement is not required between a and and between d and e. Determine portion of the slab to be included with the beam for torsional design. L-section between and d

smaller of the projection of the beam below the slab (585 mm) and four times the slab thickness (700 mm). Therefore, use 585 mm (refer to Fig. E3.8).

A = (450 mm)(760 mm) + (585 mm)(175 mm) = 444,375 mm2 p = 2(450 mm + 585 mm + 175 mm + 585 mm) = 3590 mm (444,375 mm 2 ) 2 3590 mm Tth = 27 × 106 N·mm = 27 kN·m Tth

Refer to Fig. E3.3 for torsional value Tu at midspan.

(0.083)(1.0)( 35 MPa )

Tth = (0.75)(27 kN·m) = 20.2 kN·m Tth = 20.2 kN·m Tu = 19 kN·m OK Torsion reinforcement is not required between

  

Eng.Mohammed Ashraf

and d.



9.7.2.1



Minimum top bar spacing From Appendix A of SP-17(14)) Reinforced Concrete Design Handbook Design Aid – Analysis Tables, , six No. 19 bars can be placed in one layer within an 450 mm wide beam.

9.7.2.1 25.2.1

Bar spacing can also be calculated as shown as follows for bottom bars. Minimum bottom bar spacing Minimum clear spacing between the longitudinal

25 mm

 25 mm

Controls

Clear spacing  d  4/3(d agg )

0.75 mm

Assume 19 mm maximum aggregate size.

Therefore, clear spacing between horizontal bars must not be less than 25 mm

4/3(19 mm) = 25 mm

Controls

Check if six No. 19 bars can be placed in the

= 2(cover + dstirrup + 19 mm) + 5d + 5(25 mm)

(25.2.1)

,req’d = 2(40 mm + 10 mm + 20 mm) + 5(19 mm) + 5(25 mm) ,req’d = 360 mm < 450 mm OK

Refer to Fig. E3.10 for reinforcement placement in Beam B2.

  

Eng.Mohammed Ashraf



9.7.2.2 24.3.1 24.3.2



Maximum bar spacing at the tension face must not exceed the lesser of

s

380

280 fs

300

280 fs

2.5

s

380

s

300

280 MPa 280 MPa

280 MPa 280 MPa

2.5(50 mm)

255 mm

Controls

300 mm

where = 50 mm is the least distance from surface 250 mm spacing is provided, therefore OK of deformed reinforcement to the tension face. Development length of No. 19 reinforcing bar 9.7.1.2 Top bars 25.4.2.2

25.4.2.4

9.7.1.3 25.5.2.1 9.7.3

fy t 2.1

e

f

d

(420 MPa)(1.3)(1.0) d

(2.1)(1.0) 35 MPa say, 900 mm

(19 mm)

t t = 1.3 for top horizontal bars, because more than 300 mm of fresh concrete Bottom bars t = 1.0 for bottom hori(420 MPa)( MPa)(1.0)(1.0) zontal bars, because not more than 300 mm of fresh (19 mm) d (2.1)(1.0 35 MPa (2.1)(1.0) e is say, 700 mm e = 1.0 because bars are uncoated

Splice length of No. 19 reinforcing bar 25.5.2.1, splice length is 1.3( d). Per Table 25.5.2.1,

d

835 mm

642 mm

= (1.3)(835 1.3)(835 mm) = 1086 mm, say, 1.2 m d = (1.3)(642 mm) = 835 mm, say, 0.9 mm

Bottom tension reinforcement Four No. 16 bottom bars are terminated beyond Beam B1 a distance equal to the development length of 700 mm = 4.4 m + 2(0.45 m) + 2(0.7 m) = 6.7 m Extend two No. 19 bottom bars the full length of the beam and develop into the girder beams along Column Lines B and D at each end with a hook.

  

Eng.Mohammed Ashraf



9.7.3.2



Top tension reinforcement Reinforcement must be developed at points of maximum stress and points along the span where terminated tension reinforcement is no longer Six No. 19 bars are required to resist the factored moment at the support.

(398 kN·m) 135 kN/m

point. 9.7.3.8.2

x2 2

271 kN( x )

At least one-third of the negative moment reinforcement at a support must have an embedment

x = 1.53 m, say, 1.6 m

d, 12db, and n/16.

For No. 19 bars: 1) d = 700 mm Controls 2) 12db = 12(25 mm) = 300 mm 3) n/16 = 522 mm

0

Therefore, extend four No. 16 bars a distance d

1600 mm + 700 mm = 2300 mm Place four No. 19 bars within the Beam B2 B web and two No. 19 bars on either side of the beam web Extend the remaining two No. 19 bars over the full over 900 mm (refer to Fig. E E3.10, Section B) length of the beam as hanger bars for the stirrups. Step 9:: Integrity reinforcement Integrity reinforcement 9.7.7.2 but not both.

At least one-fourth the maximum positive moment reinforcement, but not less than two bars must be continuous.

No. 19 bottom reinforcement bars into the support. 2 No. 19 > (1/4)6 No. 19

Longitudinal reinforcement must be enclosed by closed stirrups along the clear span of the beam.

the full length of the beam. Refer to Fig. E3.9.

  

Eng.Mohammed Ashraf





9.3.2 24.2.3.1

19.2.2.1

E

4700 f MPa

24.2.3.4

The beam is subjected to a factored distributed force of 13.2 kN/m or service dead load of 8.04 kN/m and 0.75 kN/m service live load.

(19.2.2.1b)

E

4700 35 MPa /1000 = 27.8 GPa

The beam is also subjected to concentrated loads from Beam B1 at 1975 mm and 6825 mm from Column Line B framing into it having the following 64 kN and 32 kN service live load. Also, Beam B2 is subjected to a concentrated load at midspan from Beam B4 of 169 kN factored or 69.9 kN dead service load and 53.3 kN service live load.

4

/384EI

3

/192EI and P are service loads.

Ie

24.2.3.5

I

For simplicity, assume that the beam is rectangular for

M Ma

3

I

1

M Ma

3

3

I

Ig

12

(450 mm)(760 mm)3 12

1.646 1010 mm 4

where

M

fr I g yt

(24.2.3.5b)

M

0.62( 35 MPa )(1.646  1010 mm 4 ) (380 mm)(1000)2

159 kN·m

Ma is the moment due to service load.

Reinforced Concrete Design Handbook Design Aid – Analysis Tables, which can be downloaded aspx?ItemID=SP1714DA are used to calculate the moments.

the moment of inertia of the cracked section, I .

  

Eng.Mohammed Ashraf





2 s(

)

( 1) s ( ) 2 where is the uncracked remaining concrete depth n = Es/E

For values, refer to the table below

n = 200 GPa/27.8 GPa = 7.2

Cracking moment of inertia, I 3

3

(

1)

)2

(

1704 mm2

As

568 mm2

2

For I values, refer to the table below

)2

(

2

1704 mm2 568 mm2

As

568 mm 96 mm

168 mm

96 mm

I

1.64 × 109 mm4

4.22 × 109 mm4

1.64 × 109 mm4

52 kN·m

26 kN·m

52 kN·m

14 kN·m

7 kN·m

14 kN·m

175 kN·m

137 kN·m

175 kN·m

108 kN·m

87 kN·m

108 kN·m

1704 mm

Distributed load Ma,dist

Concentrated load M

Total Ma

349 kN·m

257 kN·m

349 kN·m

Ie

3.04 × 109 mm4

7.12 × 109 mm4

3.04 × 109 mm4

Use

3 × 109 mm4

7 × 109 mm4

3 × 109 mm4

2 where Ma,dist i kN/m)(28 m) for concentrated load at midspan and Ma = Pa2 / the left support

M PL , where a is the distance of the concentrated load to

2

24.2.3.6 (positive and negative moments), the Code permits to take Ie as the average of values obtained from Eq.(24.2.3.5a) for the critical positive and negative moments. I

I e,left @supp ,

I e, right @supp

I e, midspan

3

I

3 109 mm 4 ,

7 109 mm 4 3

3 109 mm 4

I

= 4.33 × 109 mm4

Ie,

= 0.7(7 × 109 mm4) + 0.15(3 × 109 mm4 + 3 × 109 mm4) = 5.8 × 109 mm4

ACI Committee 435 recommends alternate equations to calculate the average equivalent moment of (Eq. (2.15a) of ACI 435R-95). I

= 0.7Ie,midspan + 0.15(Ie,left@supp + Ie,right@supp)

Ie,

  

Eng.Mohammed Ashraf





distr

(8.0 kN/m + 0.75 kN/m)(8.8 m) 4 384(27.8 GPa)(4.33  109 mm 4 ) (122.6 kN)(8.8 m)3 (1000) 2 192(27.8 GPa)(4.33  109 mm 4 )

PD+L = 69.3 kN + 93.3 kN = 122.6 kN

6825 mm from Column Line B. PD+L = 64 kN + 32 kN = 96 kN

C .L.

1.14 mm

3.6 mm

2(96 kN)(1.975 m)2 (4.4 m) 2 (1000)3 6(27.8 GPa)(4.33  109 mm 4 )(8.8 m)3 (3(6.825 m)(8.8 m) 3(6.825 m)(4.4 m) (1.975 m)(4.4 m)) 2.4 mm

Equation is obtained from Reinforced Concrete Design Handbook Design Aid – Analysis www.concrete.org/store/productdetail. aspx?ItemID=SP1714DA 24.2.2

D+L

= 1.14 mm + 3.6 mm + 2.4 mm = 7.14 mm

all.

= (8800 mm)/480 mm)/480 = 18.3 mm

all.

elements and not likely to be damaged by large /480 480

OK

A shallower beam could have been selected. But because of Beam B B1 (720 mm deep) that is framing into it, a 760 mm deep beam was chosen.

  

Eng.Mohammed Ashraf





4

3

distr

4

3

3

conc

2

2

3

C .L.

4.4 m)

D

L

24.2.4.1.1 2

24.2.4.1.3

T

i

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





Beam Example 4 Determine the size of a continuous six-bay edge beam built integrally with a 175 mm slab on the exterior of the building. Design and detail the beam. Ignore openings at Column Lines 3 and 5. Given: f fy = 420 MPa

ACI 318M-14

9.2.1.1

Calculation

Discussion The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements of ACI 318M-14. The designer determines the durability classes. Please refer to Chapter 4 of SP-17 for an in-depth discussion of the Categories and Classes.

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the expo-

Based on durability and strength requirements, and experience with local mixtures, the compressive least 35 MPa.

dinated with ACI 318M. ACI encourages refer-

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor.

  

Eng.Mohammed Ashraf



9.3.1.1



Beam depth 318M permits a beam design without having to

The beam has six continuous spans. Taking the

attached to partitions or other construction likely to

(11,000 mm) 18.5

h

595 mm

Use 760 mm Self-weight = [(450 mm)(760 mm)/(1000)2](23.5 kN/m3) = 8 kN/m 3 s = [((4.4 m – 0.45m)/2)(10.175m)](23.5 kN/m ) = 8.1 kN/m 2

9.2.4.2

spanning 3600 mm vertically Flange width The beam is placed monolithically with the slab

= (1.7 kN/m2)(3.6 m) = 6.1 kN/m

one side of the beam is obtained from Table 6.3.2.1. 6.3.2.1

One side

6h

of web is the least of

s /2 n

f

(6)(175 mm) = 1050 mm [(4400 – 450 mm)/ mm)/2 = 1975 mm 10,400 mm)/12 = 867 mm, say 880 mm (10,400

/12 12 n/12

f

Controls

= 880 mm + 450 mm = 1330 mm

The service live load is 2.4 kN/m2 kN/m2 in corridors per Table 4-1 in ASCE 7-10. This example will use 3.0 kN/m2 as an average as the actual layout is not provided. To account for the weight of ceilings, partitions, and mechanical (HVAC) systems, add 0.75 kN/m2 as miscellaneous dead load. The beam resists gravity load only and lateral forces are not considered in this problem. 5.3.1

U = 1.4D

U = 1.2D + 1.6L

u

= 1.4(8 kN/m + 8.1 kN/m + 6.1 kN/m + (0.75 kN/m2) ((4.4 m)/2)) = 33.4 kN/m

u

= 1.2(33.4 kN/m)/1.4 + 1.6(3.0 kN/m2)(4.4 m)/2 = 39.2 kN/m, say 40 kN/m Controls

  

Eng.Mohammed Ashraf





9.4.3.1 therefore, the factored moments and shear forces (required strengths) are calculated at the face of the supports. 9.4.1.2

n

= 11 m – 0.6 m = 10.4 m

Chapter 6 permits several analysis procedures to calculate the required strengths. The beam required strengths can be calculated using approximations per Table 6.5.2, if the condi-

6.5.1 Members are prismatic Loads uniformly distributed L D

Beams are prismatic

3.0 kN/m2 < 3(4.2 kN/m2 + 0.75 kN/m2 + beam selfweight OK

Three spans minimum Beams have equal clear span lengths 10,400 mm percent. mate procedure is used. 6.5.4

6.5.2

Fig. E4.2—Shear and moment diagrams.

6.5.3

The moments calculated using the approximate method cannot be redistributed in accordance with Section 6.6.5.1.

  

Eng.Mohammed Ashraf





The slab load is eccentric with respect to the edge beam center. Therefore, the beam needs to resist a torsional moment (Fig. E4.3).

= [((1.2)((0.175 mm)(23.5 kN/m3) + (0.75 kN/m2)) + (1.6 (1.6)(3.0 kN/m2))][(4.4 m – 0.45 m)/2 + 0.075 m] u = 21.8 kN/m u

tu =

u(

/2) = (21.8 21.8 kN/m)(0.45 mm/2) = 4.9 kN·m/m Tu = (4.9 kN·m/m)(10.4 kN·m/m)( m)/ = 25.5 kN·m, say, 26 kN·m m)/2

Fig. E4.4—Torsion diagram.

  

Eng.Mohammed Ashraf





Step 5: Moment design The Code does not permit a beam to be designed 9.3.3.1 with steel strain less than 0.004 mm/mm at design strength. The intent is to ensure ductile behavior at the factored condition. In most reinforced beams, such as this example, reinforcing bar strain is not a controlling issue. 21.2.1(a)

The design assumption is that beams will be checked later.

9.7.1.1 20.6.1.3.1 One row of reinforcement d

h – cover – dtie – db/2

d

The concrete compressive strain at nominal moment strength is: 22.2.2.1

cu

22.2.2.2

able property and is approximately 10 to 15 percent of the concrete compressive strength. ACI 318M neglects the concrete tensile strength to calculating nominal strength.

22.2.2.3

Determine the equivalent concrete compressive stress at nominal strength:

22.2.2.4.1

The concrete compressive stress distribution is inelastic at high stress. The Code permits any stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85fc

22.2.2.4.3 22.2.1.1

a 1c 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3. For fc

1

0.85

0.8

Find the equivalent concrete compressive depth, a, by equating the compression force to the tension force within the beam cross section (Fig. E4.5): C T 0.85fc ba Asfy For positive moment: b

bf

For negative moment: b = bw

b)(a a

a

As

As

As

0.0106 As

0.0314 As

  

Eng.Mohammed Ashraf





moments obtained from the approximate method above. Mmax = 423.6 kN·m

9.5.1.1

The beam design strength must be at least the required strength at each section along its length Mn Mu Vn Vu

  Mn

Mu

Number of No. 19 bars

a 2

As f y d

Each No. 19 bar has a d = 19 mm and an As = 284 mm2

M M

Mu, kN·m

As,req’d, mm2

Req’d

Prov.

1

270

1050

3.7

4

2

433

1709

6

6

393

1545

5.4

6

M +

309

1183

4.2

5

MI+

270

1033

3.6

4

ME

21.2.2 9.3.3.1

Check if the calculated strain exceeds 0.005 mm/mm  Mu, As,prov, s, (tension-controlled—Fig. E4.7) but not less than kN·m mm2 a, mm mm/mm 0.004 mm/mm As f y 0.85 1

and

a

M

1

270

1136

35.7

0.055

Y

M

2

433

1704

53.5

0.036

Y

1

= 0.8 (calculated above)

393

1704

53.5

0.056

Y

+

309

1420

15

0.136

Y

MI+

270

1136

12

0.171

Y

M ME

Note that = 450 mm for negative moments and 1330 mm for positive moments (refer to Fig. E4.5). t

(

s> 0.005?

)

  

Eng.Mohammed Ashraf



9.6.1.1 9.6.1.2



Minimum reinforcement The reinforcement area must be at least the minimum required reinforcement area at every section along the length of the beams. 0.25 f As

fy

Equation (9.6.1.2a) 9.6.1.2a) a) controls because f

0.25 35 MPa (450 450 mm)(697 mm) 1105 mm 2 420 MPa Controls

All calculated reinforcement areas exceed the minimum required reinforcement area. Therefore, OK

  

Eng.Mohammed Ashraf





External spans Shear strength The shear forces in the external and internal spans fore, the continuous beam will be designed for 239 kN of shear force 9.4.3.2

Because conditions (a), (b), and (c) of 9.4.3.2 are taken at a distance d from the face of the support (Fig. E4.8) Vu@d = (239 kN) – (40 kN/m)(0.697 m) = 211 kN The controlling factored load combination must

9.5.1.1 9.5.3.1 22.5.1.1 22.5.5.1

Vn u Vn = V + Vs V

0.17

21.2.1(b)

shear = 0.75 V = (0.75 (0.75)(315.4 kN) = 236.6 kN

0.17 9.5.1.1b

9.6.3.1

Vn

(0.17) 35 MPa(450 mm)(697 mm) = 315.4 kN

V Vu = 211 kN OK Therefore, shear reinforcement is not required.

Vu

Code requires that minimum shear reinforcement must be provided over sections where Vu V

Vu

V = 1/2(236.6 kN) = 118.3 kN

Check if the cross-sectional dimensions satisfy Eq.

22.5.1.2

(

0.66

)

Therefore, provide minimum shear reinforcement over full beam length. (Refer to torsion calculation Step 7).

dimensions are satisfactory.

  

Eng.Mohammed Ashraf



9.4.4.3



Torsion design Calculate the torsional moment at d from the face tu = 5 kN·m/m and Tu = 26 kN·m (Step 4, Fig. E4.3)

9.2.4.4

Tu@d = (26 kN·m) – (5 kN·m/m)(0.697 m) = 22.5 kN·m

Determine the concrete section resisting torsion.

smaller of the projection of the beam below the slab (585 mm) and four times the slab thickness (700 mm). Therefore, 585 mm controls (Fig. E4.9). 22.7.4.1

T

0.083

f

A2 P

whereA is the area enclosed by outside perimeter A = (450 mm)(760 mm) + (585 mm)(175 mm) = 444,375 mm2 p is the outside perimeter of concrete cross section p = 2(450 450 mm + 2( 2(585 585 mm) + 175 mm) = 3590 mm Tth

(444,375 mm 2 )2 3590 mm

27 kN-m

Tth = (0.75 (0.75)(27 27 kN·m) = 20.3 kN·m

21.2.1(c) 9.5.4.1

(0.083)(1.0) (0.083 1.0)) 35 MPa

Tu@d

Tth?

Tu@d

Tth = 20.3 kN·m

NG

ment and detailing requirements for torsion must be considered.

  

Eng.Mohammed Ashraf



9.5.1.1 9.5.4.2 22.7.5.1

22.7.3.2



Torsion reinforcement Calculate cracking torsion: Tcr

0.33

fc

Acp2

Tcr

Pcp

(44,375 mm 2 ) 2 3590 mm

0.33(1.0) 35 MPa

Check if cross section will crack under the torsional moment. Tu = 22.5 kN·m < Tcr = 107.4 kN·m Reducing Tu to Tcr is not required.

107.4 kN·m

OK

Fig. E4.10— E4.10—Aoh area. 450 mm – 2(40 2(40 mm) – 13 mm) ph = 2[(450 + (760 mm – 2(40 mm) mm)– 13 mm)] = 2048 mm ( Aoh = (357 mm)(667 mm) = 238,119 mm2 Check if cross section is adequate to resist the torsional moment.

22.7.7.1

Vu bw d

2

Tu ph 1.7 Aoh2

2

Vc 8 fc bw d

211,000 N mm)( (450 mm)(697 mm)

where ph is the perimeter of centerline of outermost closed transverse torsional reinforcement; and Aoh is the area enclosed by centerline of the outermost closed transverse torsional reinforcement 9.7.5 9.7.6

Calculate required transverse and longitudinal torsion reinforcement:

22.7.6.1

Transverse: Tn

22.7.6.1.1

s

cot

(22.7.6.1a)

where Ao = 0.85Aoh is the gross area enclosed by

Longitudinal: Tn 22.7.6.1.2

2 Ao At f yt

2 Ao A f y ph

tan

(22.7.6.1b)

(0.75)

2

((22.5 106 N·mm)(2048 mm) 1.7(238,1192 ) 2

2

315,400 N 0.66 35 MPa (450 mm)(697 mm)

0.83 MPa < 3.7 MPa

OK

Therefore section is adequate to resist torsion.

Tu @ d

22.5 106 N·mm 0.75

Tn

2(202,401 mm 2 ) At (420 MPa) cot 45 s

At/s = 0.176 mm2/mm Ao = 0.85(238,119 mm2) = 202,401 mm2 Tu @ d

A

22.5 106 N·mm 0.75

Tn

2(202,401 mm 2 )A (420 MPa) tan 45 2048 mm

2

  

Eng.Mohammed Ashraf



9.5.4.3



The required area for shear and torsional transverse

A s

A s

2

A s

A s

A = 0 mm2 At multiplied by 2.

0 mm2 /mm 2(0.176 mm 2 /mm)

0.352 mm 2 /m

At is

Calculate the maximum spacing of stirrups at d from the column face. 9.7.6.3.3

Maximum spacing of transverse torsional reinforceAssume No. 13 stirrup ment must not exceed the lesser of ph/8 and 300 mm ph = 2048 mm calculated above ph

9.6.4.2 (A + 2At)min/s 0.062 f

f yt

( A ) min s

0.062 35 MPa

( A ) min s

0.35

450 mm 420 MPa

0.393 mm 2 /mm

and 0.35

f yt

450 mm 420 MPa

0.375 mm 2 /mm

A = 0 mm2 because calculations showed that shear reinforcement is not required. Minimum shear reinforcement is, however, provided. At = (2)(129 mm2) = 258 mm2 A s A s

9.6.4.3

The torsional longitudinal reinforcement A

A , min

0.42 f A fy

f At ph s fy

(2)(129 mm 2 /mm) = 1.03 mm 2 /mm 250

1.03 mm 2 /mm

( A ) min s

0.0393 mm 2 /mm OK

must 0.42 35 MPa (446,400 mm 2 ) 420 MPa

A , min

(0.51 mm 2 /mm)(2048 mm)

= 1596 mm2

A , min

0.42 f A fy

0.175 f yt

ph

f fy

420 MPa 420 MPa

Controls

0.42 35 MPa (446,400 mm 2 ) 420 MPa 0.175(450 mm) 420 MPa (2048 mm) 420 MPa 420 MPa

A ,min

= 2257 mm2 Ap = 446,400 mm2 calculated above

A

.

= 361 mm2 < Al,req’d = 1596 mm2

OK

The longitudinal reinforcement must be added to the

  

Eng.Mohammed Ashraf



9.5.4.3

Torsion longitudinal reinforcement, A , must be distributed around the cross section and the portion of A that needs to be placed where As is needed is added to As found in Step 5, Table 4.1. Assume that two No. 19 bars will be added at each side face and the remainder will be divided equally between top and bottom of beam with one in each corner. A =A ANo.19 Add 460 mm2/2 = 230 mm2 to M and M+ from Step 5, Table 1.



A = (1596 mm2

9.7.5.2

Number of No. 19 bars

As,req’d, mm2

A /2, mm2

As A, mm2

Req’d

Prov.

1280

4.5

5

M

1

1050

230

M

2

1709

230

1939

6.8

7

1545

230

1775

6.2

7

+

1183

230

1413

4.98

5

MI+

1033

230

1263

4.4

5

ME

The spacing of longitudinal torsional reinforcement should not exceed 300 mm on center and the minimum diameter must be greater than 0.042 times the transverse reinforcement spacing, but not less than 10 mm

)) = 460 mm2

 

M

9.7.5.1

2

300 mm spacing between longitudinal reinforcement

d d

= ((0.042)(250 250 mm) = 10.5 mm d = 10.5 mm No.19

OK

Typical span reinforcement due to torsion moment The torsional moment varies from maximum at the face of the support to zero at span mid-length. Theoretically, torsional reinforcement is required

x

9.7.5.3

9.7.5.4

Tu

Tth Tu

n

/2

Longitudinal torsional reinforcement must be developed beyond this length a minimum of xo +d (Fig. E.11). Develop longitudinal reinforcement at face of support.

((26 kN·m) (20.5 kN·m))(10.4 m/2) 26 kN·m from the face of the support x

1.1 m

xo +d = 357 mm + 697 mm = 1054 mm Therefore, bars due to torsion moment must extend a 1100 mm + 1054 mm = 2154 mm, say, 2300 mm from the face of the support on both sides of the span.

full length of the beam. Extend stirrups No. 13 at 250 mm on center over 2300 mm from supports. Remainder of span provide No. 13 stirrups at 300 mm on center.

  

Eng.Mohammed Ashraf





Minimum top bar spacing 9.7.2.1

At maximum and interior negative moments

25.2.1

The clear spacing between the horizontal bars must 25 mm

 25 mm Clear spacing greater of  d  4/3((d agg )

19 mm

Assume 19 mm maximum aggregate size

Therefore, clear spacing between horizontal bars must be at least 25 mm

4/3(19 mm) = 25 mm

Check if seven No. 19 bars can be placed in one = 2(cover (cover + dstirrup + 20 mm) + 6d 6 + 6(25 mm)

(25.2.1)

= 2(40 mm + 13 mm + 20 mm) + 114 mm + 150 mm = 410 mm < 450 mm OK Therefore seven No. 19 bars can be placed in one layer in the 450 mm beam web.

Note improved concrete placement (Fig. E4.12b).

  

Eng.Mohammed Ashraf





Minimum bottom bar spacing 25.2.1

The clear spacing between the horizontal bars must

Clear spacing greater of

25 mm d

25 mm

4/3( d agg )

4/3(19 mm) = 25 mm

19 mm

Therefore, clear spacing between horizontal bars must be at least 25 mm = 2(cover + dstirrup + 20 mm) + 4d + 4(25 mm)

(25.2.1)

,req’d = 2(40 mm + 13 mm + 20 mm) + 76 mm + 100 mm = 322 mm < 450 mm OK

Refer to Fig. E4.13 for steel placement in beam web. in the 450 mm beam web.

9.7.2.2 24.3.1 24.3.2

Maximum bar spacing at tension face must not exceed  s

380

280 fs

300

280 fs

the lesser of  

2.5

where fs = 2/3fy = 280 MPa

s

380

280 MPa 280 MPa

2.5(50 mm)

255 mm Controls

s

300

280 MPa 280 MPa

300 mm

24.3.2.1 width, where = 50 mm is the least distance from the No. 19 bar surface to the tension face.

OK

  

Eng.Mohammed Ashraf





9.7.3 Assume the maximum moment occurs at midspan -

Mmax

Mmax

u(x)

2

/2 = 0

u(x)

2

/2 + Vux = 0

(309 kN·m) – (40 kN/m)( kN/m)(x)2/2 = 0 x = 3.93 m, say, 4 m

Mmax

u(x)

2

/2 + Vux = 0

x)2/2 + (208 kN)x = 0 x = 1.52 m, say, 1600 mm

M= x = 2.2 m, say, 2300 mm

x2/2 + (239 kN)x = 0

  

Eng.Mohammed Ashraf





Development length of No. 6 bar

fy t

25.4.2.2

25.4.2.4

2.1

e

f

(420 MPa)(1.3)(1.0)

d

d

(2.1)(1.0) 35 MPa say, 900 mm

t t = 1.3, because more than 300 mm of fresh concrete is placed below top t = 1.0, because not more than 300 mm of fresh concrete is placed below bottom horizontal bars

(420 MPa)(1.0)(1.0) d

(19 mm)

(19 mm)

835 mm

642 mm

(2.1)(1.0) 35 MPa

say, 700 mm e=

e

1.0, because bars are

uncoated First span top reinforcement 9.7.3.2

Lengths at the exterior support Reinforcement must be developed at sections of maximum stress and at sections along the span where bent or terminated tension reinforcement is Four No. 19 bars are required to resist the beam factored negative moment at the exterior column face. Calculate a distance x from the face of the column

( 270 kN·m) 40 kN/m

x2 2

0.9)(420 MPa) 697 mm (0.9)(420

208 kN(x)

2(284 mm 2 )

2(284 mm 2 )(420 MPa) 2(0.85)(35 MPa)(450 mm)

x = 0.625 m, say 650 mm factored moment. 9.7.3.3

Reinforcement must extend beyond the section at

1) d = 697 mm Controls 2) 12d = 12(19 mm) = 228 mm

distance equal to the greater of d or 12d . Therefore, extend two No. 19 bars the longer of the development length (900 mm) and the sum of 625 mm + 697 mm = 1322 mm, say, 1400 mm from the face of the column. d controls—extend two No. 19 bars 1400 mm from the interior face of the exterior support shown bold in Fig. E4.16. 9.7.3.8.4

At least one-third of the negative moment reinforcement at a support must have an embedment d, 12d , and n/16.

1) d = 697 mm Controls 2) 12d = 12(19 mm) = 228 mm 3) n/16 = (11 m – 0.6 m)/16 = 0.65 m = 650 mm

from the face of the column. The remaining two No. 19 bars are extended over the full length of the beam.

must be extended a minimum of 1600 mm + 697 mm = 2297 mm They are, however, spliced at midspan with the bars from the opposite support to act as hanger bars for stirrups.

  

Eng.Mohammed Ashraf





First span top reinforcement

9.7.3.2 9.7.3.3 9.7.3.8.4

Lengths at the interior support Following the same steps above, seven No. 19 bars are required to resist the factored moment at the

(433 kN·m) 40 kN/m

x2 2

(0.9)(420 MPa) 697 mm

Calculate a distance x from the face of the column the factored moment. (Four No. 19 bars will be discontinued).

239 kN( x)

3(284 mm 2 )

3(284 mm 2 )(420 MPa) 2(0.85)(35 MPa)(450 mm)

x = 0.97 m, say, 1000 mm Therefore, extend four No. 19 bars the greater of the development length (900 mm) and the sum of theod. 1000 mm + 697 mm = 1697 mm The distance of 1697 mm shown bold in Fig. E4.16 from the exterior face of the exterior support controls. Say, 1700 mm as shown in Fig. E4.16. Extend the remaining three No. 19 bars the longer of the development length (900 mm) from where the four d = 697 mm beyond the face of the exterior support. The longer length is the distance d E4.16. One of the three tion point shown bold in Fig. E No. 19 bars will be terminated at 2300 mm + 697 mm extended and spliced at midspan.

  

Eng.Mohammed Ashraf





First span bottom reinforcement Lengths from midspan toward the column 9.7.3.2 9.7.3.3

are required to resist the factored moment at the midspan. Calculate a distance x from the face of the column where two No. 19 bars can resist the factored

(309 kN·m) 40 kN/m

x2 2

697 mm

2(284 mm 2 )(0.9)(420 MPa) 2(284 mm 2 )(420 mm) 2(0.85)(35 MPa)(1330 mm)

x = 2.83 m, say, 2850 mm Therefore, extend the three No. 19 bars the longer of the development length (700 mm) and 2850 mm + 697 mm = 3547 mm 3600 mm from maximum positive moment at midspan. 3600 mm is longer—shown bold in Fig. E4.16 from midspan. 9.7.3.8.2

A minimum of one-fourth of the positive tension reinforcement must extend into the support minimum 150 mm The 150 mm requirement is superseded by the integrity reinforcement requirement to develop the bar at the column face.

No. 19) the greater of the development length (700 d = 697 mm into the support. The controlling bottom bar length is the distance 150 mm into the support shown in bold in Fig. E4.16. E

9.7.3.8.3

d for positive moment tension reinforcement must be limited such that

d

Mn Vu

d

column. kN/m)( Vu = 208 kN – (40 kN/m)(1.6 m) = 144 kN

a

Mn

where Mn is calculated assuming all reinforcement at the section is stressed to fy. Vu is calculated at the section. At the support, a is the embedment length beyond the center of the support. At the point of a is the embedment length beyond the d and 12d .

284 mm 2 )(420 MPa) 697 mm 2(284

2(284 mm 2 )(420 MPa) (2)(0.85)(35 MPa)(1330 mm)

Mn = 165.6 × 10–6 N·mm d

165.6  106 N·mm 144,000 N

This length exceeds

d

697 mm 1847 mm

= 700 mm, therefore OK

9.7.3.5 then stress discontinuity in the continuing bars will tensile reinforcement must not be terminated in a (a) Vu Vn (a) At 2850 mm (b) Continuing reinforcement provides double the Vu = 208 kN – (40 kN/m)(5.2 m – 2.85 m) = 114 kN Vu Vn. (c) Stirrup or hoop area in excess of that required for shear and torsion is provided along each terminated bar or wire over a distance 3/4d from the termination point. Excess stirrup or hoop area shall be at least 0.41 s/fyt. Spacing s shall not exceed d ).

Vn V + Vs) where V is calculated in Step 6 Vn = 0.75(315.4 kN + 0) = 236.6 kN Vn = 2/3(236.6 kN) = 157.7 kN Vu OK Vn = 157.7 kN Because only one of the three conditions needs to be

  

Eng.Mohammed Ashraf



9.7.7.1 9.7.7.1a

9.7.7.1b

9.7.7.1c



Integrity reinforcement for the perimeter beam At least one-fourth the maximum positive moment reinforcement, but at least two bars must be continuous.

25.4.3.1

E4.16).

At least one-sixth the maximum negative moment reinforcement at the support, but at least two bars must be continuous.

Four No. 19 bars are extended into the column region

Longitudinal reinforcement must be enclosed by closed stirrups along the clear span of the beam.

Longitudinal reinforcement is enclosed by No. 13 stirrups at 300 mm on center along the full beam length—

Longitudinal structural reinforcement must pass through the region bounded by the longitudinal reinforcement of the column. 9.7.7.4

Two No. 19 bars are extended into the column region

Integrity reinforcement must be anchored to develop fy at the face of the support. Therefore, development length for deformed bars in tension terminating in a

0.24 f    (a)

f

d

Fig. E4.16).

19 top and bottom bars full length and through the column cores. At the exterior support, the No. 19 bars must be developed at the face. Calculate if a standard hook will allow a No. 19 bar to develop within the column.

(a)

0.24(420 MPa)(1.0)(0.7)(1.0) 0.24 dh

(1.0 (1.0) 35 MPa

(19 mm)

227 mm

Controls (b) 8d (c) 150 mm 25.4.3.2

(b) 8(19 mm) = 152 mm (c) 150 mm

e

e

= 1.0 because bars

OK

are uncoated t = 0.7 because bars are smaller than No. 36 and terminate with a 90-degree hook with cover on bar extension

r

9.7.7.5

r

spaced at s d. Splices are necessary in continuous structural reinforcement shall be spliced in accordance with (a) Positive moment reinforcement shall be spliced at or near the support (b) Negative moment reinforcement shall be spliced at or near midspan

9.7.7.6

Splice length = (1.3)(development length) = 1.3(700 mm) = 910 mm, say, 1000 mm Refer to Fig. E 4.17

Refer to Fig. E 4.17

Use Class B tension lap splice

  

Eng.Mohammed Ashraf





Flexure reinforcement was calculated above in Step 5. Six No. 19 top bars are required at supports Five No. 19 bottom bars are required at midspan 9.7.6.2.2

Shear and torsion reinforcement following the same calculation in Steps 6 and 7, No. 13 at 250 mm are required for minimum 7 m 0 mm from face of each column. Space No. 13 stirrups at maximum 300 mm on center (d/2) for the remainder of the span.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





Beam Example 5 Continuous transfer girder Design and detail an interior, continuous, four-bay beam, built integrally with a 175 mm slab. The span between Column

Load— Service additional dead load D = 0.75 kN/m2 Service live load L = 3.0 kN/m2 Girder, beam and slab self-weights are given below. Material properties— f fy = 420 MPa

Span length—

  

Eng.Mohammed Ashraf





Discussion

ACI 318M-14 9.2.1.1

Calculation

The mixture proportion must satisfy the durability requirements of Chapter 19 (ACI 318M) and structural strength requirements. The designer determines the durability classes. Please refer to Chapter 4 of SP-17 for an in-depth discussion of the Categories and Classes. dinated with ACI 318M. ACI encourages refer-

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the

Based on durability and strength requirements, and experience with local mixtures, the compressive least 35 MPa.

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. 9.3.1.1

Girder depth The transfer girder supports a column at midspan with tributary loads from the third level, four stories, and a roof. Therefore, the depth limits in Assume 1250 mm deep transfer girder. Beams on either side of the girder beam are subjected to uniform load. Therefore, the depth limits in Table 9.3.1.1 are used and the controlling

h

18.5

4400 mm 18.5

238 mm

Because the beams are intended to provide continuity this purpose. Use a beam depth of 760 mm 9.2.4.2

Flange width The transfer girder and beams are poured monolithically with the slab and will behave as a T-beam. transfer girder is obtained from Table 6.3.2.1.

6.3.2.1

Each side 8h of web is S /2 the least of n /8

f

=

n/8

+

Each side of web is the least of f

=

n/8

+

+ n/8

8h S /2 n

/8

+ n/8

(8)(175 mm) = 1400 mm (does not control) (8800 mm – 600 mm)/8 = 1025 mm Controls

f

= 1025 mm + 600 mm + 1025 mm = 2650 mm

(8)(175 mm) = 1400 mm (does not control) ((4400 mm – 600 mm)/8 = 475 mm

f

Controls

= 475 mm + 600 mm + 475 mm = 1550 mm

  

Eng.Mohammed Ashraf





Applied load on transfer girder The service live load is 2.4 kN/m2 3.8 kN/m2 in corridors per Table 4-1 in ASCE 7-10. This example will use 3.0 kN/m2 as an average live load as the actual layout is not provided. To account for the weight of ceilings, partitions, and HVAC systems, add 0.25 kN/m2 as miscellaneous dead load.

Concentrated load on girder from column above Wgir = [(0.6 m)(1.25 m)](23.5 kN/m3) = 17.6 kN/m P = (0.175 m)(4.4 m)(11 m)(23.5 kN/m3) = 200 kN P

Typical beam weight framing into girder at

PBM

(0.6 m)(0.6 m) 3.6 m 0.175 m (23.5 kN/m3 )

29 kN

0.45 m 0.76 m – 0.175 m (10.4 m)(23.5 kN/m 3 ) 65 kN

To account for the weight of ceilings, partitions, and mechanical (HVAC) systems, add 0.75 kN/m2 as miscellaneous dead load. PSDL = ((4.4 m)(11 11 m)(0.75 kN/m2) = 37 kN PD = (200 kN + 65 kN + 37 kN)(6) + (29 kN)(5) = 1957 kN The service live load is 2.4 kN/m2 3.8 kN/m2 in corridors per Table 4-1 in ASCE 7-10. This example will use 3.0 kN/m2 as an average, as the actual layout is not provided. A 175 mm slab weighs 4.2 kN/m2 service dead load.

s = [(600 mm)(760 mm) + (1500 mm – 600 mm)(175 mm)] 3 6

× [(23.5 kN/m )/10 ] = 14.5 kN/m

Roof live load 1.7 kN/m2 Total live load—per ASCE 7 live load with excepPRoof = (4.4 m)(11 m)(1.7 kN/m2) = 82.3 kN L

Lo 0.25

4.57 K LL AT

where L Lo is unreduced live KLL is live load element factor = 4 for internal columns and 2 for interior beam (ASCE 7 Table AT—tributary are 2 and KLLAT 2 2 KLL

L

(3 kN/m 2 ) 0.25

L = 1.74 kN/m2

4.57 193.6

1.74 kN/m 2

Lo = 1.2 kN/m2

OK

  

Eng.Mohammed Ashraf





KLL and L

2

Lo = 1.2 kN/m

2

2

L

(3 kN/m 2 ) 0.25

2

96.8 m Lo = 1.2 kN/m2

L = 2.1 kN/m2

P

4.57

2.1 kN/m 2 OK

= (4.4 m)(11 m)(1.74 kN/m2) = 84.2 kN

PL.3rd = (4.4 m)(11 m)(2.1 kN/m2) = 101.6 kN

PL = 4P 5.3.1

+ PL,3rd + PRoof

PL = (101.6 kN) + (84.2 kN)(4 levels) + (82.2 kN) = 520.6 kN

The transfer girder resists gravity only and lateral forces are not considered in this problem. Transfer girder U = 1.4D

= 1.4(17.6 kN/m) = 24.6 kN/m Controls Pu = 1.4(1957 kN) = 2740 kN

u

The superimposed dead load is calculated above and is included in the concentrated load. Live load is applied over the width of the girder ((600 mm) U = 1.2D + 1.6L

= 1.2(17.6 kN/m) + 1.6(3 kN/m2)(0.6 m) = 24 kN/m Pu = 1.2 1.2(1957 kN) + 1.6( 1.6(520.6 kN) = 3182 kN Controls u

Beams 14.5 kN/m + ((0.75 kN/m2)(1500 mm)/1000) u = 1.4(14.5 = 21.9 kN/m 2 u = 1.2(21.9 kN/m)/1.4 + 1.6(3 kN/m )(1500 mm)/(1000) = 26 kN/m Controls

  

Eng.Mohammed Ashraf





9.4.3.1 therefore, the factored moments and shear forces (required strengths) are calculated at the face of the supports. The beams were analyzed as part of a frame. The moment and shear diagram obtained from a software are presented below (Fig. E5.3).

1. Factored moments and shear forces are shown at faces of columns. 2, Span BD is subjected to large concentrated load at midspan and relatively small distributed beam selfweight. Therefore, the appearance of a straight line moment diagram. Fig. E5.3—Shear and moment diagrams.

  

Eng.Mohammed Ashraf



9.3.3.1



The code does not permit a beam to be designed with steel strain less than 0.004 mm/mm at nominal strength. The intent is to ensure ductile behavior. In most reinforced beams, such as this example, reinforcing bar strain is not a controlling issue.

21.2.1(a)

The design assumption is that beams will be

later. 20.6.1.3.1 stirrups and No. 36 bars for the transfer girder positive moment and No. 29 bars for the transfer girder negative moment. Assume No. 13 stirrups and No. 19 and No. 29 bars for beams positive and negative moments, respectively. Girder beam and

Assume that the transfer girder requires two rows of reinforcement with one bar spacing between the two rows within the span and one layer of bars at the supports. Beams require one row only d = h – cover – dtie – 3d /2

positive moment

d = 1250 mm – 40 mm – 16 mm – 3 (36 mm)/2 = 1140 mm

negative moment

d = 1250 mm – 40 mm – 16 mm – 29 mm/2 = 1179.5 mm use d = 1179 mm

positive moment

d = 760 mm – 40 mm – 13 mm – (19 mm)/2 = 697.5 mm use d =697 mm

negative moment

d = 760 mm – 40 mm – 13 mm – 29 mm/2 = 692 mm

  

Eng.Mohammed Ashraf



22.2.2.1



The concrete compressive strain at nominal = 0.003 mm/mm

22.2.2.2

able property and is approximately 10 to 15 percent of the concrete compressive strength. ACI 318M neglects the concrete tensile strength to calculate nominal strength.

22.2.2.3

Determine the equivalent concrete compressive

22.2.2.4.1

22.2.2.4.3 22.2.1.1

The concrete compressive stress distribution is inelastic at high stress. The Code permits any stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85f a 1 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3. For f

0.85

0.05(35 MPa 28 MPa) 7 MPa

0.8

Find the equivalent concrete compressive depth a by equating the compression force to the tension C=T 0.85 f

= Asfy

0.85 0.85(35 MPa)( )(a) = As(420 MPa)

=

f

= 2650 mm

a

As (420 MPa) 0.85(35 MPa)(2650 mm)

0.005 As

=

f

= 1550 mm

a

As (420 MPa) 0.85(35 MPa)(1550 mm)

0.009 As

= 600 mm

a

As (420 MPs) 0.85(35 MPa)(600 mm)

=

0.024 As

  

Eng.Mohammed Ashraf





The transfer girder and beams are designed for the moment diagram (Step 4).

required strength at each section along their 9.5.1.1

Mn Mu Vn Vu

Mn

Mu

As f y d

a 2 Mu, kN·m

d = 29 mm, As = 645 mm2, and d = 1179 mm

d = 36 mm, As = 1006 mm2, and d =

9.6.1.1 9.6.1.2

Number of bars

As,req’d, mm2

Req’d

Select

Max. M

2588

6198

9.6

10

Max. M+

4648

11,054

11

12

Max. M

2594

6214

9.6

10

1140 mm Minimum reinforcement The provided reinforcement must be at least the minimum required reinforcement at every section along the length of the beam. d@midspan = 1140 mm Use d@support will yield higher required minimum reinforcement 0.25 f As

fy Because f 9.7.2.3 24.3.2

0.25 35 MPa (600 mm)(1179 mm) 420 MPa

2491 mm 2

Provided reinforcement area exceeds the minimum required. Therefore, OK

The girder is 1250 mm deep. To control cracks within the web, ACI 318M requires skin reinforcement to be placed near the vertical faces of the 1250 mm/2 = 625 mm Spacing of skin reinforcement in girder must not

s

24.3.2.1

380

280 fs

300

280 fs

2.5

and

where is the clear cover from the skin reinforcement to the side face. = 40 mm + 16 mm = 56 mm and fs = 2/3fy = 280 MPa

s

380

280 280

2.5(56 mm)

s

300

280 280

300 mm

240 mm

Controls

Place two No. 25 skin reinforcement at girder at 850 mm from the top of the girder.

  

Eng.Mohammed Ashraf



9.7.2.3



Skin reinforcement can be used in the strength calculation of the girder. 870 mm Mn

2

(0.9)(420 MPa)(2)(510 mm ) 620 mm

Using strain compatibility, try two No. 25 bars in two layers on both sides of the girder. Assume reinforcement is yielding. It will be checked later. Re-evaluating the positive tension reinforcement at

66 mm 2 66 mm 2

Mn = 549 × 106 N·mm = 549 kN·m

Calculated design moment 4648 kN·m (Step 4). Mn = (4648 kN·m) – (549 kN·m) = 4099 kN·m Required reinforcement area by solving the

Mn

Mu

As f y d

a 2 As,No.11 = 9720 mm2 Required No. 36 bars = 9720 mm2/1006 mm2 = 9.7 Choose 10 No. 36 bars.

Negative reinforcement at the supports Provide three layers of skin reinforcement on both sides in the top half of the girder. Extend the two No. 25 middepth skin bars over the full length of the girder and use for the top half at the support two layers of No. 19 bars on both sides of the girder. Assume that skin reinforcement reach yielding (will be checked later).

Mn

MPa)(2)(510 mm 2 ) 640 mm ((0.9)(420 420 MPa)(2)(

134 mm 2

+ (0.9)(420 MPa)(2)(284 mm 2 ) 825 mm )(420 MPa)(

134 mm 2

+ (0.9)( (0.9 420 MPa)(2)(284 mm 2 ) 1010 mm

134 mm 2

Calculate the provided moment from the skin reinMn = 586 × 106 N·mm = 586 kN·m Re-evaluating the negative tension reinforcement at

Calculated design moment 2594 kN·m (Step 4).

Mn = (2594 kN·m) – (586 kN·m) = 2008 kN·m As,No.11 = 4764 mm2

Required reinforcement area by solving the

Mn

Mu

As f y d

Required No. 29 bars = 4764 mm2/645 mm2 = 7.4 Choose 8 No. 29 bars.

a 2

  

Eng.Mohammed Ashraf



21.2.2 9.3.3.1



Check if the calculated strain exceeds 0.005 mm/mm (tension controlled), but not less than 0.004 mm/mm As f y

1

Mu, kN·m

As,prov, mm2

a, mm

c, mm

s, mm/mm

s> 0.005?

2588

5160

124

155

0.020

Y

M

4648

10,600

53

66

0.0488

Y

M

2594

5160

124

155

0.020

Y

1

= 0.8

0.003 s

a

and

0.85 f

M +

(

)

Note that = 600 mm for negative moments and 2650 mm for girder positive moments. Place eight No. 29 in one layer with d = 1179 mm

Positive half, strain at middepth

Check strain in the upper skin layer at 610 mm

s

0.003 (610 mm 66 mm) 66 mm

0.025

0.005 OK

By inspection the other skin reinforcement layer has reinforcement layers is yielding and the assumption of using = 0.9 is correct.

  

Eng.Mohammed Ashraf





Negative half, strain at supports B and D (Fig. E5.6)

Check strain in the lower skin layer at 640 mm

0.003 (640 mm 155 mm) 155 mm

s

0.009

0.005

By inspection the other skin reinforcement layers have forcement layers is yielding and the assumption made of using = 0.9 is correct. Beams Design beams for the maximum load condition. Extend No. 29 top bars from Span BD to resist the 898 kN·m and 909 kN·m moments at Column Lines B and D in Spans AB and DE, respectively, and No. 19 bars to resist the rest of the moments.

Span Column Line AB

M (No. 29)) 2

As,req’d mm2

Req’d

Select

898

3513

5.5

6

1000

3.5

4

d = 19 mm, As = 284 mm , and d = 697 mm M (No.19) 259 d = 29 mm, As = 645 mm2, and d = 692 mm for one layer. Span Column Line DE +

M (No. 29) +

M (No. 19)

Number of bars

Mu, kN·m

Number of bars

Mu, kN·m

As,req’d mm2

Req’d

Select

909

3557

5.5

6

283

1095

3.9

4

Span Column Line EF Number of bars

Mu, kN·m

As,req’d mm2

Req’d

Select

M (No. 19)

70

275

0.97

2

M+ (No. 19)

64

244

0.9

2

  

Eng.Mohammed Ashraf



9.6.1.1 9.6.1.2



Minimum reinforcement The provided reinforcement must be at least the minimum required reinforcement at every section along the length of the beam. 0.25 f s , min

d

fy

Check to see if the required reinforcement areas exceeds the code minimum reinforcement area at all locations.

As , min

Because f

0.25 35 MPa (600 mm)(697 mm) 1473 mm 2 420 MPa

Provide a minimum six No. 19 bars at all beam tension locations. Except at Column Lines B and D, where transfer girder top reinforcement is extended over adjacent spans to resist the negative moment and Span DE positive moment, where six No. 19 longitudinal bars is required.

eight No. 29 by extending the two top No. 19 skin bars from the girder into Beam DE to resist part of the negative moment. Calculate strain in No. 19 bars (Fig. E5.7). is obtained from table below.

Fig. E5.7 E5.7—Strain E Strain diagram. 0.003 (520 mm 116.1 mm) 116.1 mm

s

0.010

0.005

Therefore, reinforcement in the two No. 19 skin bars is yielding and the assumption made of using = 0.9 is correct. 21.2.2

Check if the calculated steel strain exceeds 0.005 mm/mm (tension controlled), but not less than As f y

1

M 1

(

+

M

= 0.8

0.003 s

a

and

0.85 f

Beams (only maximum moments are checked) Mu, kN·m

As,prov, mm2

a, mm

c, mm

s, mm/mm

s> 0.005?

909

3870

92.9

116.1

0.011

Y

283

1136

10.2

12.8

0.119

Y

)

Note that = 600 mm for negative moments and 2650 mm and 1500 mm for girder and beam positive moments, respectively.

< hf the T-section members assumption for positive moments is correct.

  

Eng.Mohammed Ashraf





Transfer girder Shear strength 9.4.3.2

Because conditions a), b), and c) of 9.4.3.2 are section at distance d from the face of the support (Fig. E5.9). use d = 1179 mm at support The controlling factored load combination must

9.5.1.1 9.5.3.1 22.5.1.1 22.5.5.1

Vn Vu Vn = V + Vs

Vu@d = (1700 kN) – (24.6 kN/m)(1.179 m) = 1671 kN

f

0.17

V

21.2.1(b)

shear

0.17 V

9.5.1.1b

0.17( 35 MPa )(600 mm)(1179 mm) 1000 N/kN

711 kN

= 0.75

V = (0.75)(711 kN) = 533 kN Vu

V = 533 kN < Vuu@d @d d = 1671 kN

NG

Therefore, shear reinforcement is required. 22.5.1.2

Prior to calculating shear reinforcement, check if

(

0.66

)

Vu

711 kN

0.66 35 MPa (600 mm)(1179 mm) 1000 N/kN

2605 kN OK, therefore, section dimensions are satisfactory.

  

Eng.Mohammed Ashraf



22.5.10.1



Shear reinforcement Transverse reinforcement satisfying equation 22.5.10.1 is required at each section where Vu V

Vu

V

V

1671 kN 0.75

Vs

711 kN 1517 kN

Try a No. 16 bar, two legged stirrup 22.5.10.5.3 22.5.10.5.6 where Vs

(2)(199 mm 2 )(420 MPa)(1179 mm) s (1000 N/kN) s 130 mm

A f d

1517 kN =

s

This is a relatively tight spacing. Use two No. 16 double stirrups side by side. This will

9.7.6.2.2

0.33 f

0.33( 35 MPa )(600 mm)(1179 mm) 1000 N/kN

1381 kN

First, does the beam transverse reinforcement value need to exceed the threshold value? 0.33 9.7.6.2.2

1517 kN

?

0.33 f

Because the required shear strength is higher than the threshold value, the maximum stirrup spacing is 4 and 300 mm d = 1179 mm/4 d/ the lesser of d/4 d/4

over the length of the transfer girder (Fig. E5.3a), E use two No. 16 stirrups equally spaced over the

1381 kN

OK

mm < 300 mm

s s = 202.5 mm < 295 mm < 300 mm OK

50 mm from face of column and distribute the remaining 39 stirrups equally over the span. 9.6.3.3 0.062 f

f yt

and 0.35

A, f yt

s

0.062 35 MPa

450 mm 420 MPa

0.393 mm 2 /mm

Because f A, s

4(199 mm 2 ) 200 mm

3.98 mm 2 /mm

OK

  

Eng.Mohammed Ashraf





Beams Shear strength 21.2.1(b) 9.5.1.1

shear

Vn

= 0.75

Vu

9.5.3.1 22.5.1.1

Vn = V + Vs

9.4.3.2

Because conditions a), b), and c) of 9.4.3.2 are section at distance d from the face of the support (Fig. E5.10). Vu@d = Vu – (26 kN/m)(0.692 mm) = Vu – 18 kN f

0.17

22.5.5.1

V

(0.17)( 35 MPa )(600 mm)(692 mm)

417.6 kN

V = (0.75)(417.6 kN) = 313 kN V

9.6.3.1

Vu

Vu@d, kN Beam 1

However, ACI 318M M requires that minimum shear

Beam 3

Vu

V = 1/2(313 kN) = 156.5 kN Beam 4

Vc

Left

206

Y

Right

320

N

Left

328

N

Right

214

Y

Left

27

Y

Right

88

Y

Vu ?

however, provide minimum shear reinforcement for Therefore, shear reinforcement is required where integrity. concrete strength is less than the factored shear force. Beam AB DE EF

Is Vu

Vu@d, kN

Vc ?

Left

206

Y

Right

320

Y

Left

328

Y

Right

214

Y

Left

27

N

Right

88

N

Provide minimum shear reinforcement over all beams spans. Shear reinforcement V

Vu

where Vs

Vs

V A f d s

and V = 417.6 kN

s

328 kN 0.75

417.6 kN

20 kN

(2)(71 mm 2 )(420 MPa)(692 mm) 20,000 N

2064 mm

Using No. 10 stirrups The spacing exceeds the maximum allowed d therefore, use 300 mm spacing over the full length of beam. By inspection provisions, 9.7.6.2.2 and 9.6.3.3 are

  

Eng.Mohammed Ashraf





Minimum bar spacing 25.2.1

The clear spacing between the horizontal No. 36  25 mm d  4/3(d agg ) Assume 19 mm maximum aggregate size.

25 mm 36 mm Controls 4/3(19 mm) = 26 mm Therefore, clear spacing between horizontal bars must be at least 36 mm, say, 40 mm

Fig. E5.11. = 2(cover + dstirrup + 25 mm) (25.2.1) + 4d + 4(40 mm)

,req’d

,req’d = 2(40 mm + 16 mm + 25 mm) + 144 mm + 160 mm = 466 mm < 600 mm OK in the 600 mm transfer girder web.

sp

600 mm [2(40 mm + 16 mm + 25 mm) + 4(36 mm)] 4 = 73.5 mm, say, 73 mm

Beams AB, DE, and EF are reinforced with six No. 19 bottom bars uniformly spaced. The calculated spacing is 73 mm Therefore, OK.

  

Eng.Mohammed Ashraf





Top reinforcement 24.3.4

f

= 2650

n/10. n/10, additional bonded reinforcement is required in the outer

Use No. 19 placed in slab over

f

for additional

n/10

= (8200 mm)/10 = 820 mm < 2650 mm

n/10

= (3800 mm)/10 = 380 mm < 1550 mm

This requirement is to control cracking in the slab

AB BD

is concentrated within the web width. 600 mm [2(81 mm) 5(25 mm)] 5

Bar spacing

Prov. No. 29

Span

DE EF

= 58.6 mm, say, 58 mm

No. 29 in web

No.29 in * n/10

No. 19 in outer portion*

L

4

4

—

4

R

8

6

2

4

L

8

6

2

8

R

8

6

2

8

L

8

6

2

4

R

4

4

—

4

L

†

5

5

†

—

4

R

5†

5†

—

4

*

Bars to be divided equally on both sides of the web (refer to Fig. E5.12-sections)

†

Section reinforced with No. 19 bars

Development length of No. 19, No. 29, and No. 36 bar

fy t

25.4.2.2

25.4.2.4

1.7

e

f

d and

fy t  2.1

e

f

d

(420 MPa)(1.0)(1.0) ,19

t t = 1.3, if more than 300 mm of fresh concrete is placed below top horit = 1.0, if not more than 300 mm of fresh concrete is placed below bottom horizontal bars.

e

are uncoated

e

= 1.0, because bars

(1.7)(1.0) 35 MPa

(d )

41.76d

d = 1.3(41.76d ) = 54.3d

Top Bottom

d,

mm

Use d, mm d, mm

Use d, mm

No. 19

No. 29

No. 25

No. 36 —

835

1575

1358

900

1600

1400

—

642

—

—

1503

700

—

—

1600

  

Eng.Mohammed Ashraf





lated at both supports and at midspan (Fig. E5.13(a)).

(a) Girder moment diagram Bottom bar length along girder

4648 kN·m – (24.6 kN/m)(x)2/2 – 1592x = 0 x = 2.86 m, say, 3 m

4648 kN·m Assume the maximum positive moment occurs at

Mmax

u(x)

2

/22 – P/2x = 0

Top bar length along transfer span

Mmax

u(x)

2

/2 + Vux = 0

x)2/2 + (1699 kN)x = 0 x = 1.54 m, say, 1.6 m

Mmax

u(x)

2

/2 + Vux = 0

x)2/2 + (1700 kN)x = 0 x = 1.545 m, say, 1.6 m

  

Eng.Mohammed Ashraf





The moment diagram envelop shows that all three maximum moment at midspan. The moment varies from maximum negative moment at one support and increases to maximum positive moment at the other support.

Mmax

u(x)

2

/2 + Vux = 0

x)2/2 + (320 kN)x = 0 x = 3.22 m, say, 3.3 m

Mmax

u(x)

2

/22 + Vux = 0

Following the same concept for spans three and

(259 259 kN·m) – 26 kN/m(x) kN/m( 2/2 – (206 kN)x = 0 x = 1.2 m, say, 1.3 m

Span

x-left

x-right

DE

3.2 m, say, 3.3 m

1.25 m, say, 1.3 m

EF

3.43 m, say, 3.5 m

0.93 m, say, 1 m

  

Eng.Mohammed Ashraf





Transfer girder

9.7.3.2 9.7.3.3

Support Bars must be developed at locations of maximum stress and locations along the span where bent or terminated tension bars are no longer required to

Eight No. 29 bars and four No. 19 and two No. 25 skin bars are required to resist the factored moment at Column Lines B and D. The two end moments at the supports are close, so the calculation will be applied at one end only. Calculate a distance x from the column face where four No. 29 bars can be discontinued and the contribution from the skin reinforcement is ignored.

2594 kN·m 24.6 kN/m

x2 2

(0.9)(420 MPa) 1179 mm

1700 kN( x)

4(645 mm 2 )

4(645 mm 2 )(420 MPa) 2(0.85)(35 MPa)(600 mm)

x = 0.87 m, say, 1000 mm from column face full girder length and are properly developed or extended into the adjacent spans at the supports.

1) d = 1179 mm Controls d = 12(29 mm) = 348 mm 2) 12d Therefore, extend the four No. 29 bars the greater of the development length ((1600 mm – Step 8) from the column face and d d = 1600 mm Therefore, 2279 mm Controls Four No. 29 bars can be terminated 2300 mm from the E face of the column, shown bold in Fig. E5.15.

9.7.3.8.4

At least one-third of the bars resisting negative moment at a support must have an embedment d, 12d , and n/16.

1) d = 1179 mm Controls 2) 12d = 12(29 mm) = 348 mm 3) n/16 = (8.8 m – 0.6 m)/16 = 0.51 m = 510 mm Extend the remaining four No. 29 bars the larger of the development length (1600 mm) beyond the theoretical d = 1179 mm beyond the E5.15). 1600 mm + 1000 mm = 2600 mm 1600 mm + 1179 mm = 2779 mm Controls Extend the remaining four No. 29 bars 2800 mm from the face of the column. The four No. 29 bars will be, however, extended over the full length of the girder.

  

Eng.Mohammed Ashraf



9.7.3.2 9.7.3.3



Transfer girder bottom bars Following the same steps above, ten No. 36 bars and four No. 25 skin bars are required to resist the factored moment at midspan. Calculate a distance x from the midspan where four No. 36 bars can resist the factored moment.

(4648 kN·m) (24.6 kN/m) (0.9)(420 MPa) 1140 mm

length.

x2 2

(1592 kN)x = 4(1006 mm 2 ) 4(1006 mm 2 )(420 MPa) 2(0.85)(35 MPa)(2650 mm)

x = 1.83 m, say, 1900 mm from midspan Therefore, extend six No. 36 bars the larger of the development length (1600 mm – Step 8) and a distance d Controls) 1900 mm + 1140 mm = 3040 mm, say, 3100 mm from maximum positive moment at midspan (Fig. E5.15). Extend the remaining four No. 36 bars at least the longer of 150 mm into the column or d = 1600 mm

1600 mm + 1900 mm = 3500 mm < (3800 mm) + 150 mm = 3950 mm The 150 mm into the column controls, however, it is recommended to extend bars to the far face of the column and develop them. 9.7.3.8.2

At least one-fourth of the positive tension bars must extend into the column at least 150 mm

OK

  

Eng.Mohammed Ashraf



9.7.3.8.3



d for positive moment tension bars must be limited such that d for that bar from the column face. Vu = 1700 kN – (24.6 kN/m)(1.1 m) = 1673 kN

Mn

4(1006 mm2 )(420 MPa) 1140 mm

4(1006 mm 2 )(420 MPa) (2)(0.85)(35 MPa)(600 mm)

Mn = 1847 × 106 N·mm = 1847 kN·m d

Mn Vu

a

1,847,000 N·mm 1140 mm 2244 mm 1673 kN This length exceeds d = 1600 mm, therefore, OK d

where Mn is calculated assuming all bars at the section are stressed to fy. Vu is calculated at the section. The term a is the embedment length of d and 12d . 9.7.3.5 then a bar stress discontinuity occurs. Therefore, be terminated in a tensile zone unless (a), (b), or (c)

(a) Vu Vn Continuing bars provides double the area required

(b) Vu

Vn = V + Vs = 688 + 1554.6 = 2243 kN Vn = 1495 kN

Vn .

(c) Stirrup or hoop area in excess of that required for shear and torsion is provided along each terminated bar or wire over a distance 3/4d from the termination point. Excess stirrup or hoop area shall be at least 0.41 s/fyt. Spacing s shall not exceed d/(8 ), where

(c) Therefore, over a distance of 3/4(1179 mm) = 884 mm from the end of the terminated bars space stirrups at 1179 mm/(8(4/12)) = 492 mm on center and excess mm/( A,

= 0.41(600 mm)(200 mm)/420 MPa = 117 mm2

area of tension reinforcement at section.

Vu@7ft

1648 kN 516 kN

(0.75) A ,

'

(420 MPa)(1179 mm) 200 mm

2

A

=A

–A

d

A = 610 mm A = 4(199 mm2) – 610 mm2 = 186 mm2 2 2 186 mm , therefore, OK Because only one of the three conditions needs to be

  

Eng.Mohammed Ashraf



9.7.7.2

9.7.7.3



Integrity reinforcement At least one-fourth of the maximum positive moment bars, but at least two bars, must be continuous and developed at the face of the column.

No. 36 bars into the support. OK

Beam longitudinal bars must be enclosed by closed stirrups along the clear span.

stirrups over the full length of the beam.

Beam structural integrity bars shall pass through the region bounded by the longitudinal column bars.

Four No. 36 bars are extended through the column longitudinal reinforcement, therefore satisfying this condition.

dimensions (600 mm). Therefore, beam longitudinal forcement.

  

Eng.Mohammed Ashraf



9.7.3.2



Spans AB and DE Beams spanning between column A and B and between D and E are subjected to comparable two beams will be designed for the same loads (larger of the two beams). To simplify detailing, simply extend eight No. 29 bars and two No. 19 skin bars (span DE only) from the transfer girders to resist the negative moment in the adjacent beams (Step 5). Calculate a distance x from the column face where four No. 29 bars can resist the factored moment.

(909 kN·m) (26 kN/m)

x2 2

(0.9)(420 MPa) 692 mm

328 kN( x )

4(645 mm2 )

4(645 mm 2 )(420 MPa) 2(0.85)(35 MPa)(600 mm)

x = 0.83 m, say, 1 m At 1 m from the column face, four No. 29 can be cut factored moment. 9.7.3.3

9.7.3.8.4

location where they are no longer required to resist d or 12d .

1) d = 692 mm Controls 1 d = 12(29 mm) = 348 mm 2) 12d

At least one-third of the bars resisting negative moment at a support must have an embedment

Therefore, extend four No. 29 bars the greater of the development length ((1600 mm – Step 8) from the d,

12d , and n/16.

and d Controls Say, 1700 mm (a) d = 692 mm Controls (b) 12d = 12(29 mm) = 348 mm (c) n/16 = (8.8 m – 0.6 m)/16 = 0.51 m = 510 mm Extend the remainder four No. 29 bars, the greater of the development length (1600 mm) beyond the theod = 692 mm beyond the = 1.6 m + 1 m = 2.6 m 3.3 m + 0.692 m = 3.992 m Controls Therefore, extend the remainder bars over the full length

At the Support E, where there is no negative moment, provide minimum reinforcement of six No. 19 bars extending minimum the development length (1100 mm) on both sides of the column. Of the six No. 19 bars, extend two bars over the full Fig. E5.16.

  

Eng.Mohammed Ashraf





bars are extended over the full span length for Beams 1, 3, and 4. 9.7.7.5

9.7.2.2 24.3.1 24.3.2

Splices are necessary for continuous bars. The bars (a) Positive moment bars shall be spliced at or near the support (b) Negative moment bars shall be spliced at or near midspan Maximum bar spacing at the tension face must not exceed the lesser of 380

280 fs

2.5

Splice length = (1.3)(development length) = 1.3(1600 mm) = 2080 mm, say 2100 mm = 1.3(700 mm) = 910 mm, say, 1000 mm = 1.3(1600 mm) = 2080 mm, say, 2100 mm = 1.3(900 mm) = 1170 mm, say, 1200 mm

s

380

280 MPa 280 MPa

255 mm

2.5(50 mm)

Controls and 24.3.2.1

s = 300(280/fs) where fs = 2/3fy = 280 MPa

s

300

280 MPa 280 MPa

300 mm

width. Note that is the cover to the longitudinal bars, not to the tie. Spacing, mm

No. 19

No. 29

No. 36

73

58

73

All longitudinal bar spacing satisfy the maximum bar OK

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





Fig. E5.18—Sections. (Note: All dimensions in mm.)   

Eng.Mohammed Ashraf





Beam Example 6 f = 28 MPa. Assume the will be composed of 13 mm diameter individually coated and sheathed seven-wire prestressing strands.

Load–– Service additional dead load D = 0.75 kN/m2 Service live load L = 3 kN/m2 Girder, beam and slab self-weights are given below. Material properties–– f f = 28 MPa fy = 420 MPa fpu = 1860 MPa

Span length––

600 mm Area of 13 mm diameter strand = 99 mm2

  

Eng.Mohammed Ashraf



9.2.1.1

Calculation

Discussion

ACI 318M-14



The mixture proportion must satisfy the durability requirements of Chapter 19 (ACI 318M-14) and structural strength requirements. The designer determines the durability classes. Please refer to Chapter 2 of SP-17 for an in-depth discussion of the categories and classes.

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the exposure classes, Chapter 19 requirements are satis-

Based on durability and strength requirements, and experience with local mixtures, the compressive

dinated with ACI 318M. ACI encourages refer-

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor.

9.3.1.1

least 35 MPa. The engineer must specify the transfer stress—in this case, 28 MPa.

Girder depth Assume 1250 mm deep transfer girder. The transfer girder supports a column at midspan. The column load includes self-weight and its tributary loads from the third level, the four stories concenabove it, and the roof. Because of this large concen trated load, the depth limits in Table 9.3.1.1 cannot

9.2.4.2 6.2.3.1

Flange width The transfer girder is poured monolithically with the slab and will behave as a T-beam.

R6.3.2.3

It is allowed per ACI 318M-14 comment to ignore and past performances.

prestressed T-beams is therefore left to the experience and judgment of the licensed design professional. 6.3.2.1

Use

f

= 8h +

+ 8h

f

= (8)(175 mm) + 600 mm + (8)(175 mm) = 3400 mm

  

Eng.Mohammed Ashraf





Fig. E6.2—Transfer girder geometry.

Ay, × 107 mm3

A(y – yt)2, × 1010 mm4

Ix, × 1010 mm4

87.5

4.29

5.32

0.13

5.45

625

46.875

3.2

9.77

12.97

51.165

8.52

9.91

18.42

A, mm2

y, mm

1

(2)(1400 mm)(175 mm) = 490,000

2

(1250 mm)(600 mm) = 750,000 1,240,000

yt

51.165  107 mm3 1,240,000 mm 2

Ii, × 1010 mm4

417 mm from top of girder

y = 1250 mm – 417 mm = 833 mm from bottom of girder

STop

18.42  1010 mm 4 417 mm

4.4 108 mm3

STop

18.42  1010 mm 4 833 mm

2.2 108 mm 3

  

Eng.Mohammed Ashraf





Applied load on transfer girder The service live load is 2.4 kN/m2 3.8 kN/m2 in corridors per Table 4-1 in ASCE 7-10. This example will use 3.0 kN/m2 as an average live load, as the actual layout is not provided. To account for the weight of ceilings, partitions, and HVAC systems, add 0.75 kN/m2 as miscellaneous dead load.

= [(600 mm)(1250 mm) [(23.5 kN/m3)/106] = 17.6 kN/m P = (0.175 m)(4.4 m)(11 m)(23.5 kN/m3) = 199 kN

= 450 mm × (760 mm – 175 mm) × (11 m – 0.6 m)

P

(0.6 m)(0.6 m)(3.6 m 0.175 m)(23.5 kN/m3 )

P

(0.45 m)(0.585 m)(10.4 m)(23.5 kN/m3 )

29 kN

84.3 kN

PSDL = (4.4 m)(11 m)( m)(0.75 kN/m2) = 36.3 kN PD = (199 kN + 64.3 kN + 36.3 kN)(6) + (29 kN)(5) = 1943 kN

2

PL,Roof = ((4.4 m)(11 m)(1.7 kN/m2) = 82.3 kN

Total live load—per per ASCE 7 live load with excepexcep

L

Lo 0.25

4.57 K LL AT

L

(3 kN/m 2 ) 0.25

4.57 193.6 m 2 Lo = 1.2 kN/m2

1.74 kN/m 2

L = 1.74 kN/m2 OK where L Lo is unreduced live KLL is live load element factor = 4 for internal columns and 2 for interior beam (ASCE 7 Table 4-2) 4.57 AT is tributary area L (3 kN/m 2 ) 0.25 2.1 kN/m 2 2 2 96.8 m and KLLAT 2 4(11 m)(4.4 m) = 193.6 m2 L = 2.1 kN/m2 Lo = 1.2 kN/m2 OK 2 2(11 m)(4.4 m) = 96.8 m2 2 and L Lo = 1.2 kN/m PL = (4.4 m)(11 m)(1.74 kN/m2) = 84 kN

P

= (4.4 m)(11 m)(2.1 kN/m2) = 102 kN L = (84 kN)(4) + (102 kN) + (82.3 kN) = 520 kN

  

Eng.Mohammed Ashraf





20.3.1.1 20.3.2.2

fpu = 1860 MPa

20.3.2.5.1

0.7fpu = 1302 MPa

20.3.2.6.1 0.65fpu = 1209 MPa R20.3.2.1

Modulus of elasticity is assumed for design and checked against test results after a supplier is contracted. Ep = 196,500 MPa f f = 28 MPa

Concrete compressive strength Concrete strength at initial stressing f ,

24.5.2.1 Class T, at service.

0.62 f

3.7 MPa

ft

f

5.9 MPa

Concrete compressive and tensile stresses immedi-

24.5.3.1 24.5.3.2

The transfer girder will be reinforced with nonpre nonprestressed bonded reinforcement. Therefore, the 0.25 f can be exceeded. Use 0.62 f

Location All

Type

Stress limit, MPa

Compression

0.60 f = 16.8

Tension

24.5.4.1 Load condition Prestress plus sustained load Prestress plus total load

0.62

f

= 3.28 MPa

Concrete compressive stress limits 0.45f 0.60f

The girder is designed using unbonded single-strand tendons, as it is the preferred construction method in the United States. Bonded tendons may be preferable in other parts of the world. The tendons center of 9.4.3 factored moments and shear forces (required strengths) are calculated at the face of the supports. The

reaches a concrete compressive strength of minimum 28 MPa and subjected only to its self-weight. Before stressing the remaining tendons, the girder will be supporting three levels—dead load only. At the second

  

Eng.Mohammed Ashraf





Find the total number of strands required to resist the total load The required prestress force to support the total dead and live load is calculated at service load condition to satisfy the limit in the selected Class in Table 24.5.2.1. The service load is 1943 kN concentrated dead load and 520 kN concentrated live load from all levels above the girder—refer to Step 3. Assume that approximately 73 percent of the dead load is balanced by the harped post-tensioned tendons.

0.73(1943 kN) = 1418 kN 1418 kN = 2Fsin tan

F

Calculate number of strands in the tendon. (Step 4) fpe = 0.65fpu = 1209 MPa Asp,strand = 99 mm2

1

740 mm 4100 mm

1418 kN 2sin10.2

10.2 degrees

4004 kN

Required tendon area = (4004 ( kN)/(1209 MPa) = 3312 mm2 Number of strands = 3312 mm2/(99 mm2) = 33.5

F = (34 ((34)(99.1 mm2 OK

R20.3.2.6.2 of losses. Estimation of friction losses in posttensioned tendons is addressed in PTI TAB.1-06. The lump-sum losses were used herein to determine

Moments at support and midspan for dead load, live load, and post-tensioning are obtained from PTData, Fig. E6.4, 6.5, and 6.6, respectively.

  

Eng.Mohammed Ashraf





Fig. E6.6—Post-tensioning moment diagram.

Tension (+) lute values are used.

Assume zero dead load present and all 34 diameter

M

stress losses in the strands incurred over time. ftop

P A

f top

M Stop

P A

6

(2510 kN·m)

2928 kN·m

(7/6)(4069 kN  103 ) 1, 240, 000 mm 2

(2928 kN·m)(106 ) 4.4  108 mm 3

ftop = 2.8 MPa < fall = 3.88 MPa

Check if actual stress is less than allowable stress

f

7

M S

f top

f

(7/6)(4069 kN  103 ) 1, 240, 000 mm 2

OK

(2928 kN·m)(106 ) 2.2  108 mm3

fall = 16.8 MPa

2.8 MPa

–17.1 MPa

OK

Check if actual stress is less than allowable stress

  

Eng.Mohammed Ashraf



P A

f top



M

7 ( 480 kN·m) 6

f top

(7/6)(4069 kN  103 ) 1,240,000 mm 2

M Stop

560 kN·m (560 kN·m  106 ) 4.4  108 mm 3

5.1 MPa

Check if actual stress is less than allowable stress ftop = 5.1 MPa < fall = 16.8 MPa P A

f

M S

(7/6)(4069 kN  103 ) 1,240,000 mm 2

f

OK

(560 kN·m  106 ) 2.2  108 mm 3

1.28 MPa

Check if actual stress is less than allowable stress f

= 1.28 MPa < fall = 3.28 MPa

OK

No stage stressing is required. Stress all tendons after girder beam concrete attained 28 MPa compressive strength. At service

MTL = MD + ML

M = MTL

MTL = (3711 kN·m) + (845 kN·m) = 4556 kN·m MBal

MBal P A

f top

M M Stop

f top

4069 kN  103 1,240,000 mm 2

2046 kN·m  106 4.4  108 mm3

7.9 MPa

Check if actual stress is less than allowable stress ftop = 7.9 MPa < fall = 15.75 MPa P A

f

M S

4069 kN  103 1,240,000 mm 2

f

OK

2046 kN·m  106 2.2  108 mm3

6 MPa

Check if actual stress is less than allowable stress – f

fall = 5.9 MPa

OK

MTL MBal = +480 kN·m M = MTL ftop

MBal

M

P

M

A

Stop

f top

4069 kN  103 1,240,000 mm 2

176 kN·m  106 4.4  108 mm3

–2.9 MPa

Check if actual stress is less than allowable stress ftop = 2.9 MPa < fall = 15.75 MPa

f

P

M

A

S

f

4069 kN  103 1,240,000 mm 2

OK

176 kN·m  106 2.2  108 mm3

–4.1 MPa

Check if actual stress is less than allowable stress – f

= 4.1 MPa < fall = 15.75 MPa

OK

  

Eng.Mohammed Ashraf





All service load stresses are acceptable. Initial stressing 24.5.2.1 24.5.3.1 24.5.3.2 24.5.3.2.1 25.5.4.1

Support Midspan

Support Midspan

Location

Stress, MPa

Allowable stress, MPa

Status

Top

–5.1

–16.8

OK

Bottom

–1.28

3.28

OK

Top

2.8

3.28

OK

Bottom

17.1

–16.8

~OK

Location

Stress, MPa

Allowable stress, MPa

Status

–15.75

OK

Top Bottom

–4.1

–15.75

OK

Top

–7.9

–15.75

OK

Bottom

6

5.9

OK

(a) Flexure Factored loads Shear and moment diagrams are obtained from

5.3.1

The beam resists gravity only and lateral forces are not considered in this problem. U = 1.4D U = 1.2D + 1.6L

E6.4 From Moment diagrams, Fig. E 1.4(3711 kN·m) = 5195 kN·m Mu = 1.4 Mu = 1.2(3711 1.2 3711 kN·m) + 1.6(845 kN·m) = 5804 kN·m Controls M2 = 459 kN·m

Mu = 5804 kN·m + 459 kN·m = 6263 kN·m Pu = 1.2(1943 kN) + 1.6(520 kN) = 3164 kN

Fig. E6.7—Shear and moment diagrams.

  

Eng.Mohammed Ashraf



21.2.1(a)

It is assumed that the girder is tension controlled, = 0.9. This assumption will be checked later.

20.6.1.3.1

d = h – cover – dtie – d /2

22.2.2.1

The concrete compressive strain at nominal



d = 1250 mm – 40 mm – 13 mm – (29 mm)/2 = 1182 mm

= 0.003 22.2.2.2

able property and is approximately 10 to 15 percent of the concrete compressive strength. ACI 318M neglects the concrete tensile strength to calculate nominal strength.

22.2.2.3

Determine the equivalent concrete compressive

22.2.2.4.1

The concrete compressive stress distribution is inelastic at high stress. The code permits any stress distribution to be assumed in design if shown to reason result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distri distribution of 0.85f a = 1 , where 1 is a function of concrete comprescompres

22.2.2.4.3

For f

1

0.85

0.05(35 MPa 28 MPa) 7 MPa

0.8

  

Eng.Mohammed Ashraf



20.3.2.4.1



For unbonded tendons and as an alternative to a more accurate calculation, the stress in postthe least of: (a) fse + 70 + fc (b) fse + 420 (c) fpy

p)

Controls (c) fsy = 0.9fpu

if fse = 1209 MPa > 0.5fpu = 930 MPa 3 n h = (8.2 m)(10 and where Aps (34)(99 mm 2 ) p

bd p

Therefore, use fps

(3400 mm)(1150 mm)

The strands are unbonded. A minimum area of -

At midspan: Act the minimum area: As,min = ((0.004)(501,000 mm2) = 2004 mm2 As,min = 0.004Act where Act is the area of that part of the cross section Try four No. 25: )(510 510 mm2) = 2040 mm2 OK As,prov = ((4)(510 midspan with only PT tendons: C=T 0.85 fc ba = Apsfps + Asfy

2

a

2

2

(0.85)( (0.85)(35 MPa)(3400 mm)

a

hf = 175 mm

c Check that section is tension controlled: Is the strain in bars closest to the tension face are

Deformed bars: Alternatively: cd

s

d 1 c

cu

s

1184 mm 78.4 mm

Therefore, use

= 0.9.

dp = 1150 mm d No. 13 stirrup db

Fig. E6.9—Strain distribution over girder depth.

  

Eng.Mohammed Ashraf

)



Mn

Aps f ps d p

a 2

As f y d

a 2

Mn

Aps = (34)(99 mm2) = 3366 mm2



(3366 mm 2 )(1629 MPa) 1150 mm

62.7 mm 2

(4)(510 mm 2 )(420 MPa) 1184 mm

62.7 mm 2

Mn = (0.9)(6123 kN·m + 988 kN·m) = 6400 kN·m Mn Mu = 6263 kN·m OK

= 0.9 calculated above

The beam resists gravity only and lateral forces are not considered in this problem.

5.3.1

From moment diagrams, Fig. E6.4 Mu = 1.4(420 kN·m) = 588 kN·m Mn = 1.2(420 kN·m)+ 1.6(236 kN·m) = 882 kN·m Controls

U = 1.4D U = 1.2D + 1.6L

M2 = 459 kN·m

Mu 20.3.2.4.1 (a) fse + 70 + f

p)

(a) 1209 MPa + 70 MPa + 35 MPa/(100(0.0067)) = 1331 MPa Controls (b) 1209 MPa + 420 MPa = 1629 MPa (c) fpy = 0.9fsu = (0.9)(1860 MPa) = 1674 MPa

(b) fse + 420 (c) fpy if fpu = 930 MPa fse /h = 8200 m/1250 mm = 6.6 < 35 and where Aps p p

9.6.2.3

(34)(99 mm 2 ) (600 mm)(835 mm)

Therefore, use fps = 1331 MPa.

0.0067

Minimum area of deformed reinforcement at

As,min = 0.004A

A = (3400 mm)(175 mm) + (600 mm)(240 mm) = 739,000 mm2 As,min = (0.004)(739,000 mm2) = 2956 mm2

where A is the area of that part of the cross section of the gross section A

= (4)(819 mm2) = 3276 mm2

OK

Calculate design moment strength of section at C=T 0.85f

= Apsfps + Asfy

a

(34)(99 mm 2 )(1331 MPa) (4)(819 mm 2 )(420 MPa) (0.85)(35 MPa)(600 mm)

a = 328 mm

C = 328/0.8 = 410 mm

  

Eng.Mohammed Ashraf





Is the strain in bars closest to the tension face are greater than 0.005? d

1

s

d = 1250 mm – 40 mm – 13 mm – 32 mm/2 = 1181 mm

Mn

Aps f ps d p

a 2

As f y d

a 2

1181 mm 1 (0.003) 410 mm

Therefore, use

Mn

0.0056 mm/mm

= 0.9.

)(1331 MPa) 835 mm (3366 mm 2 )( ((4 (4)( 819 mm 2 )(420 MPa) 1181 mm (4)(819

= 0.9 calculated above

328 mm 2 328 mm 2

Mn = ((0.9)(3006 3006 kN·m + 1399 kN·m) = 3964 kN·m Mn Mu = 423 kN·m OK

  

Eng.Mohammed Ashraf



(b) Shear design Shear strength Vu = wu /2 + Pu/2



Vu

2 2

2 Vu

9.4.3.2

Because conditions a), b), and c) of 9.4.3.2 are section at distance h/2 from the face of the support (Fig. E6.11).

Fig. E6.11—Factored shear at the critical section.

21.2.1(b) 9.5.1.1

Shear strength reduction factor: Vn Vu

9.5.3.1 22.5.1.1

Vn = Vc + Vs

22.5.5.1 22.5.2.1

Vc 0.17 f c bw d d = dp = 0.8h = 0.8(1250 mm) = 1000 mm Vc

9.5.1.1

Vu

Vu@h/2 shear = 0.75

Vc

Vc

0.75(0.17)( 0.75 0.17)( )( 35 MPa )( )(600 mm)(1000 mm) = 45

Vu@h/2

NG

Therefore, shear reinforcement is required. 22.5.1.2

Vu

Vu

(Vc

0.66 f c bw d )

Vu

Vc

f c bw d

OK

  

Eng.Mohammed Ashraf





22.5.8.2 Apsfse

22.5.8.2

Therefore, V is the least of the following three Vu d p

(a) 0.05

f

4.8

(b) (0.05

f

4.8) d

(c) 0.42 22.5.5.1

2 (99 mm2 ) 2 (34)(1860 MPa) + (3276 mm )(420 MPa)) = 3055 kN

Apsf pu + Asfy)

and

Mu

f V

0.17

0.17(1.0) 35 MPa (600 mm)(1000 mm)

603 kN

For the calculation of the three equations, refer to the

x, mm

Vu, kN

Mu, kN·m

dp, mm

Vudp/M /Mu

Vc (a), kN

Vc (b), kN

Vc (c), kN

300

1680

–882

1000

1.905 –1.905

–5308

3057

1491

600

1673

–379

1000

–4.413

–12,531

3057

1491

925

1665

163

1000

10.195

29.540

3057

1491

1200

1658

620

1000

2.674

7878

3057

1491

1500

1651

1117

1000

1.479

4436

3057

1491

1800

1644

1611

1000

1.020

3116

3057

1491

2100

1637

2103

1000

0.778

2419

3057

1491

2400

1630

2593

1000

0.628

1987

3057

1491

2700

1622

3081

1000

0.527

1694

3057

1491

3000

1615

3567

1000

0.453

1482

3057

1491

3300

1608

4050

1000

0.397

1321

3057

1491

3600

1601

4531

1039

0.367

1283

3177

1549

3900

1594

5010

1096

0.349

1294

3350

1634

4200

1586

5487

1152

0.333

1310

3523

1718

4400

1582

5804

1190

0.324

1323

3638

1774

(22.5.8.2c), shown shaded in table above. Both equations are greater than Eq. (22.5.5.1).

  

Eng.Mohammed Ashraf



Vu

9.5.3 22.5.10.1

V at all sections along the girder length. Therefore, shear reinforcement is required. Try No. 13 As = 2(129 mm2) = 258 mm2 and dp,min = 1000 mm

x, mm

Vu, kN

Vc , kN Eq. (22.5.8.2a) and (22.5.8.2c)

300

1680

1118

562

144.6

140

600

1673

1118

555

146.5

140

925

1665

1118

547

148.6

140

1200

1658

1118

540

150.4

140

1500

1651

1118

533

152.5

140

1800

1644

1118

526

154.5

140

2100

1637

1118

519

156.7

140

2400

1630

1118

511

158.9

140

2700

1622

1118

504

161.2

140

3000

1615

1111

504

161.3

140

3300

1608

991

617

131.7

120

3600

1601

962

638

127.3

120

3900

1594

971

623

130.5

120

4200

1586

983

604

134.. 134..6

120

4400

1582

992

590

137.8

120

s

Vs, kN

A f d V

V

, mm s, mm

Shear reinforcement Transverse reinforcement satisfying Eq. (22.5.10.1) ( V is required at each section where Vu V

22.5.10.5.3 22.5.10.5.5



Vu

V

where Vs = A fytd/s; Vs = 590 kN @ midspan

Vs = 590 kN/0.75 = 787 kN

9.7.6.2.2 First, does the beam transverse reinforcement value need to exceed the threshold value? 0.33 f

?

0.33 f

The required shear strength is less than the

V

stirrup spacing as the lesser of 3h/8 and 300 mm

3h

787 kip

0.33( 35 MPa )(600 mm)(1190 mm) = 1394 kN

0.33 f

= 1394 kN

OK

d/4 = 300 mm Controls

support. Place No. 13 stirrups at 120 mm on center in the middle 2.2 m of the girder. Place No. 13 stirrups at 140 mm on center on both sides of the midsection.

  

Eng.Mohammed Ashraf





9.6.3.3 (c) and (d) and (e) (c) 0.062 f

(d) 0.35

(e)

f yt

, and

A ,min s A ,min

f yt

Aps f pu

s d

80

0.062 35 MPa

0.35

600 mm 420 MPa

600 mm 420 MPa

0.524 mm 2 /mm

0.5 mm 2 /mm

(99 mm 2 )(34)(1860 mm) 1000 mm 80(420 MPa)(1000 mm) 600 mm

0.254 mm 2 /mm

Controls to develop ductile behavior. A ,min s

2(129 mm 2 ) 140 mm

1.84 mm 2 /mm

OK

  

Eng.Mohammed Ashraf



25.2.1



Minimum longitudinal bar spacing The clear spacing between longitudinal No. 32 25 mm 32 mm Controls 4/3(19 mm) = 25 mm assuming a 19 mm maximum aggregate size

 25 mm Clearing spacing greater of  d  4/3(d agg ) Check if four No.32 bars (resisting positive

Therefore, clear spacing between horizontal bars must be at least 32 mm, say, 40 mm = 2(cover + dstirrup + 25 mm) + 3d +3(40 mm)

(25.2.1)

,req’d = 2(40 mm + 13 mm + 25 mm) + 96 mm + 120 mm = 372 mm < 600 mm OK Therefore, four No. 32 bars can be placed in one layer in the 600 mm transfer girder web.

24.3.4 f

= 136 mm

n/10..

addi n/10,, additional bonded reinforcement is required in the outer Use No. 16 placed at slab middepth for additional bonded reinforcement. This requirement is to control cracking in the slab

is concentrated within the web width.

= 2(cover + dstir + 25 mm) + 3d + 3(40 mm)

,req’d = 2(40 mm + 13 mm + 25 mm) + 75 mm + 120 mm = 351 mm < 600 mm OK

  

Eng.Mohammed Ashraf



9.7.2.2 24.3.1 24.3.2



Maximum bar spacing at the tension face must not exceed the lesser of 380

280 fs

2.5

s

380

280 MPa 280 MPa

2.5(50 mm)

s

300

280 MPa 280 MPa

300 mm

255 mm Controls

and s = 300(280/fs)

width. Note that is the cover to the longitudinal bars, not to the tie. Skin reinforcement

Longitudinal bar spacing satisfy the maximum bar OK

9.7.2.3

24.3.2

mm Although the Code does not require skin reinforcement for Class T prestressed beams, many engineers provide it. Skin reinforcement is placed a Use two No. 16 bars each face side as shown in Fig. distance h/2 from the tension face. E6.15

Development length Extend top and bottom deformed bars over the full length of the beam. This will ensure better resistance to creep stresses and control of cracks in the girder. Therefore, development length calculation is not required into the girder. Bars, however, must be developed within the support at each end.

fy t

25.4.2.2

25.4.2.4

1.7

e

f

(420 MPa)(1.0)(1.0)

d

(1.7)(1.0) 35 MPa

t = 1.0 top bar location with not more than 300 mm of fresh concrete below horizontal reinforcement. Otherwise, use 1.3.

t

d

d

(d )

41.8d

= (1.3)(41.8d )

= 41.8(25 mm) = 1044 mm, say, 1100 mm

t

1.3 d = (1.3)(41.8 mm)(32 mm) = 1739 mm, say, 1800 mm 9.7.7.5 reinforcement splicing is not required.

  

Eng.Mohammed Ashraf





Integrity reinforcement 9.7.7.2 four No. 32 bars into the support.

OK

one-quarter of the positive moment bars, but not less than two bars continuous. 9.7.7.3

25.8.1

25.9

Beam structural integrity bars must pass through the region bounded by the longitudinal column bars.

Two No. 25 bottom bars are extended and placed between column reinforcement. Because the columns and girder are of the same dimension (600 mm), the two inner girder bars are selected to be the integrity bars.

Post-tensioning detailing Anchorages for tendons must develop 95 percent of fpu when tested in an unbonded condition. Post-tensioning anchorage design and detailing is usually provided by the post-tensioning supplier, as well as the detailed tendon layout.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





Beam Example 7: Design and detail an interior, simply supported precast beam supporting factored concentrated forces of 70 kN located at 1400 mm from each end and a continuously distributed factored force of 66 kN/m. The beam is supported on a 150 mm ledge. Given Material properties— f fy = 420 MPa Load— Pu1 = 70 kN at 1400 mm from each support u = 66 kN/m

ACI 318M-14

9.2.1.1

Discussion

Calculation

The mixture proportion must satisfy the durability requirements of Chapter 19 (ACI 318 318M-14) and structural strength requirements. The designer determines the durability classes. Please refer to Chapter 3 of this Handbook for an in-depth discussion of the categories and classes.

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the exposure classes, Chapter 19 (ACI 318M-14) require-

coordinated with ACI 318M. M. ACI encourages

least 28 MPa.

Based on durability and strength requirements, and experience with local mixtures, the compressive

There are several mixture options within ACI M, such as admixtures and pozzolans, which 301M, the designer can require, permit, or review if suggested by the contractor.

9.3.1.1

Beam depth Since beam resists concentrated loads, the beam depth limits in Table 9.3.1.1 of ACI 318M-14 cannot be used.

Assume 560 mm deep beam with

  

Eng.Mohammed Ashraf





The beam is simply supported and the loads are symmetrical. Therefore, the maximum shear and moment are located at supports and midspan, respectively.

Fig. E7.2—Shear and moment diagrams. ( ) 2

u

Vu , max

u

M u , max

16.2.6.2

( )2 8

( Pu )( x1 )

Vu , max

(66 kN·mm)(5.5 m) 2

M u , max

(66 kN/m)(5.5 m) 2 8

(70 kN) = 251.5 kN

(70 kN)(1.35 m) = 344 kN·m

The minimum seating length of the precast beam n/180

22.8.3.2

( Pu )

and 75 mm

180 = 5500 mm/180 = 31 mm < 75 mm n/180 Provided 150 mm, therefore OK.

Bearing strength Check bearing strength at seat and concentrated

The supporting surface (ledge) is wider on three of

0.85f A1

0.85(28 MPa)(350 mm)(150 mm)/1000 = 1249.5 kN OK 0.85(28 MPa)(350 mm)(250 mm)/1000 OK

  

Eng.Mohammed Ashraf



9.3.3.1



The Code does not permit a beam to be designed with steel strain less than 0.004 mm/mm at nominal strength. The intent is to ensure ductile behavior. In most reinforced beams, such as this example, reinforcing bar strain is not a controlling issue.

21.2.1(a)

The design assumption is the beams will be tensioned controlled, = 0.9. This assumption will be checked later.

20.6.1.3.3 and 25 mm cover. For precast concrete beam, the minimum cover is the greater of 16 mm and d and need not exceed 40 mm One row of reinforcement d = h – cover – dtie – d /2 22.2.2.1

Use d = 25 mm cover

d = 560 mm – 25 mm – 10 mm – 25 mm /2 = 512.5 mm, say, 510 mm

The concrete compressive strain at nominal = 0.003

22.2.2.2

able property and is about 10 to 15 percent of the concrete compressive strength. ACI 318 M neglects 318M the concrete tensile strength to calculate nominal strength. Determine the equivalent concrete compressive

22.2.2.3

22.2.2.4.1 22.2.2.4.3

The concrete compressive stress distribution is inelastic at high stress. The Code permits any stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85f a = 1 , where 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3.

22.2.1.1 For f

1

= 0.85

Find the equivalent concrete compressive depth a by equating the compression force to the tension C= T 0.85f

= Asfy

0.85(28 MPa)( )(a)=As (420 MPa)

= 350 mm

a

As (420 MPa) 0.85(28 MPa)(350 mm)

0.05 As

  

Eng.Mohammed Ashraf





9.5.1.1 Mn Vn

Mu u

Mn

Mu

As f y d

a 2

344 kN·m 106

0.05 As 2

(0.9)(420 MPa)As 510 mm

A No. 25 bar has a d = 25 mm and an As = 510 mm2 As, req’d = 1976 mm2 Use four No. 25 2 A = 2040 mm2 s, req’d = 1976 mm

OK

Per Reinforced Concrete Design Handbook Design Aid – Analysis Tables, which can be downloaded aspx?ItemID=SP1714DA, four No. 25 bars require a minimum of 295 mm wide beam. Therefore, 350 mm width is adequate. Check if the calculated strain exceeds 0.004 mm/ mm (tension controlled). 9.3.3.1

As f y

and

0.85 1

a 1

a = 0.05As = ((0.05)(4)(510 mm2) = 102 mm = 120 mm = a/0.85 /0.85 = 102 mm/0.85 mm/

= 0.85 (

)

t

=

0.003 ((510 510 mm 120 mm) 120 mm

Therefore, assumption of using

0.00975

= 0.9 is correct.

  

Eng.Mohammed Ashraf



9.6.1.1 9.6.1.2



Minimum reinforcement The provided reinforcement must be at least the minimum required reinforcement at every section along the length of the beam. 0.25 f ,

As , min

fy 1.4 fy

,

As , min

A

0.25 28 MPa (350 mm)(510 mm) 420 MPa 1.4 (350 mm)(510 mm) 420 MPa

= 2040 mm2

562 mm 2

595 mm2 Controls

As,min = 595 mm2

OK

Required positive moment reinforcement areas exceed the minimum required reinforcement area at all positive moment locations. Top reinforcement While not required by Code, top bars are needed continuous bars. Shear strength 21.2.1(b) 9.5.1.1

shear

Vn

= 0.75

Vu

9.5.3.1 22.5.1.1

Vn = V + Vs

9.4.3.2

Because conditions (a), (b), and (c) of 9.4.3.2 are d from the face of the support (Fig. E E7.4).

Vu = (251.5 kN) – (66 kN/m)(0.510 mm) = 217.8 kN, say, 218 kN 22.5.5.1

0.17 Check if V

22.5.1.2

f

and

1.0

Vu

0.66 f

(0.17)( 28 MPa )(350 mm)(510 mm) = 160 kN

V = (0.75)(160 kN) = 120 kN < Vu = 218 kN NG Therefore, shear reinforcement is required.

Determine required Vu on each side of Pu Left of Pu V = Vu – ux1 Right of Pu Vu,r = Vu – ux1 – Pu Prior to calculating shear reinforcement, check if the cross-sectional dimensions satisfy Eq.

(

V

)

V = 251.5 kN – (66 kN/m)(1.4 m) = 159.1 kN Vu,r = 251.5 kN – (66 kN/m)(1.4 m) – 70 kN = 89.` kN

Vu

(160 kN 0.66( 28 MPa )(350 mm)(510 mm)) = 587 kN

OK Section dimensions are satisfactory.

  

Eng.Mohammed Ashraf



9.5.3 22.5.10.1 22.5.10.5.3



Shear reinforcement Transverse reinforcement satisfying equation 22.5.10.1 is required at each section where Vu Vs

Vu

V Vs

where

V Vs

A f d

kN) – (120 kN) = 98 kN

Assume a No. 10 bar, two legged stirrup

s

98 kN

(0.75)(2)(71 mm 2 )(420 MPa)(510 mm) s  103

s = 232 mm 22.5.10.5.6 First, does the required transverse reinforcement value exceed the threshold value? 0.33

?

98 kN 0.75

Vs

131 kN

0.33( 28 MPa )(350 mm)(510 mm) = 312 kN 131 kN

0.33

312 kN

OK

Because the required shear strength is below the threshold value, the maximum stirrup spacing is the d/2 = 510 mm/2 d mm/ = 255 mm lesser of d/2 and 600 mm Use s = 180 mm < d/2 2 = 255 mm, therefore, OK 9.7.6.2.2

It is unnecessary to use No. 10 stirrups at 180 mm on center over the full length of the beam. Since the maximum spacing is 255 mm, determine

Vn V Vs with s = 255 mm

Vn

120 kN

(0.75)(2)(71 mm 2 )(420 MPa)(510 mm) (0.75)(2 255 mm

Vn = 209 kN Determine distance x from face of support to point at which Vu = 209 kN Determine distance are not required. Find

1

1,

1

0.64 m

beyond x1 at which stirrups

= (Vu@x1 – V /2)/

Compute x1 +

251.5 kN 209 kN 66 kN/m

x

u

1

x1 +

89.1 kN 120 kN/2 66 kN/m 1

0.44 m, say, 0.5 m

= 1.4 m + 0.5 m = 1.9 m

  

Eng.Mohammed Ashraf



use s = 180 mm until Vu < 44.5 kN and use s = 250 mm until Vu < 0.5 V



From face of support use 50 mm space then Six spaces at 180 mm on center (1080 mm) and Five spaces at 250 mm on center (50 mm+ 1080 mm + OK

5500 mm – 2(2380 mm) = 740 mm does not require shear reinforcement. However, extend No. 10 stirrups over the remaining length of 740 mm at 250 mm on center as good practice.

concentrated load. Place six No. 10 stirrups at 100 mm centered on each concentrated load.

  

Eng.Mohammed Ashraf





9.3.2.1 24.2.3.1

24.2.3.4 19.2.2.1

E

4700 f MPa

(19.2.2.1b)

E

4700 28 MPa = 24,870 MPa

The beam resists a factored distributed force of 66 kN/m or service dead load of 32 kN/m and 17 kN/m service live load. The beam also resists a concentrated factored load of 70 kN or service dead load of 34 kN and 18 kN service live load.

5 4 384EI P 3 28.23EI · and P are service loads. · I is the equivalent moment of inertia given by

24.2.3.5

M Ma

I

3

I

3

M Ma

1

3

I

Ig

fr I g

M

(24.2.3.5b)

yt

M

12

mm)3 (350 mm)(560 mm)( 12

5.12 109 mm4

0.62( 28 MPa )(5.12  109 mm 4 ) 280 mm

60 106 N·mm

M = 60 kN·m Ma is the moment due to service load. The beam is assumed cracked, therefore, calculate the moment of inertia of the cracked section, I . Calculate the concrete uncracked depth,

n

200,000 MPa 24,870 MPa

8

2 s(

)

where n

(

2

1)

s(

Es and As E

(8)(2040 mm 2 )(510 mm

)

)

(350 mm) 2

2

2

Solving for

= 176.4 mm

Cracking moment of inertia, I 3

3

(

1)

(

)2

(

)2

I = 2.46 × 109 mm4

  

Eng.Mohammed Ashraf





4 distr

9

3 conc

24.2.2

12 4

12 9

4

-

all.

/240

OK

all.

24.2.4.1.1 2.0 2

1 50

1.8

1 50

24.2.4.1.3

T

i

T

  

Eng.Mohammed Ashraf





Beam Example 8: Determination of closed ties required for the beam shown to resist shear and torque Design and detail a simply supported precast edge beam spanning 9 m. The beam is subjected to a factored load of 69 kN/m. Structural analysis provided a factored shear and torsion of 271 kN and 72 kN·m, respectively. The torsional moment is determinate and cannot be redistributed back into the structure. Given: fc fy d Vu Tu wu

Fig. 8.1—Beam subjected to determinate torque. ACI 318M-14 Discussion Step 1: Section properties Determine section properties for torsion. Acp bwh 22.7.6.1 h Aoh bw 22.7.6.1.1 Ao Aoh pcp bw + h 22.7.6.1 ph bw h

Step 2: Cracking torsion Calculate cracking torsion Tcr. Acp2 Tcr fc pcp

21.2.1c

2

Acp Aoh 2 2

Ao pcp ph

2

2 2

Tcr

torsion strength reduction factor Calculate threshold torsion Tcr Tth

22.7.7.1

Calculation

Because Tu are therefore required.

Is section large enough? Calculate fv Vu bwd Calculate fvt Tuph Aoh2

Calculate limit Is

f v2

f vt2 < limit?

fc

fv fvt

6

2 2

]

fc 2

2

Therefore, section is large enough.

  

Eng.Mohammed Ashraf



9.5.1.1 22.5.1.1 22.5.5.1 22.5.10.5.3 21.2.1b

22.7.6.1a 22.7.6.1.2 9.6.4.2

A s

At s

(

0.17

(

)

fyd

and

1.0

Tu 2 Ao f y cot

Calculate total tie area/spacing (A /s + 2At/s)



A s

(271 kN 0.17(0.75)( 35 MPa )(400 mm)(510 mm) (0.75)(420 MPa)(510 mm)

A s

0.72 mm 2 /mm

At s

72 kN·m  106 2(0.75)(134,385 mm 2 )(420 MPa)cot45°

A s

2

A s

0.85 mm 2 /mm

0.72 mm 2 /mm 2(0.85 mm 2 /mm) 2.42 mm 2 /mm

s = 258 mm2/(2.42 mm2/mm) = 106 mm Use No. 13 ties for which (A + 2At)= 258 mm Calculate s = 0.40/(A /s + 2At/s).

Use 100 mm

Check minimum transverse reinforcement.

Is

0.062 f yt

A s

2A s

?

0.062(( 35 MPa )(400) 258 mm 2 2(129 mm 2 ) < + 420 MPa 100 mm 100 mm 0.35 mm < 5.16 mm

OK

  

Eng.Mohammed Ashraf





Step 4: Longitudinal reinforcement 22.7.6.1 Calculate torsional longitudinal reinforcement from 22.7.6.1.2 Eq. (22.7.6.1(b)):

2 Ao A f y

Tn

ph

Set Tn 9.6.4.3

72 kN·m 106 2(134,385 mm 2 )A (420 MPa) = tan 45 0.75 1640 mm

tan

Tu

A = 1395 mm2

The torsional longitudinal reinforcement A be the lesser of: (a)

0.42 f c Acp fy

must 0.42( 35 MPa )(240,000 mm 2 ) 420 MPa 420 MPa (0.85 mm 2 /mm)(1640 mm) 420 MPa

f yt At ph s fy

25.9 mm 2 Controls

(b)

0.42 f c Acp fy

f yt 0.175bw ph f yt fy

0.42( 35 MPa )(240,000 mm 2 ) 420 MPa 0.175(400 mm) 420 MPa (1640 mm) 420 MPa 420 MPa A

9.7.5.1

Distribute torsional longitudinal reinforcement around the perimeter of closed stirrups that satisfy Section 25.7.1.6 (ACI 318M-14)) (ends of stirrups are terminated with 135-degree -degree standard hooks around a longitudinal bar).

9.7.6.3.3

Tie spacing must not exceed the lesser of ph/8 and 300 mm

.

= 1395 mm2 > A

1147 mm 2

= 25.9 mm2

OK

ph/8 = 1640 mm/ mm/8 = 205 mm < 300 mm Use ten No. 16 longitudinal bars are required, three top and bottom, and two in each vertical face. Use ten No. 16 bars. 2 2 A . = (10)(199 mm ) = 1990 mm At a distance d from support; Mu decreases by the amount of: Mu = Vud – wud2/2

OK

Mu = (271 kN)(0.510 m) – (69 kN/m)(0.51 m)2/2 Mu = 129 kN·m

resist 9.7.5.2

Mu

Mu

fyAs(d – a

fyAs(0.9d)

No. 16 bars must be at least equal to 0.042 times the transverse reinforcement spacing, but not less than 10 mm

Mu = (0.9)(420 MPa)As(510 mm)(0.9) = (129 kN·m × 106) As = 744 mm2

(0.042)(100 mm) = 4.2 mm < 16 mm = No. 16 No. 16 > 10 mm OK

OK

bottom, two No. 16 in each side face, and three No. 16 in top.

  

Eng.Mohammed Ashraf





Step 5: Beam detailing

Notes: 2. Side reinforcing bar due to torsional moment

  

Eng.Mohammed Ashraf





Beam Example 9: Use the same data as that for Example 8, except that the factored torsion of 72 kN·m is not an equilibrium requirement, but because the structure is indeterminate, can be redistributed if the beam cracks.

f = 1.0 fy = fyt = 420 MPa = 400 mm h = 600 mm Vu = 271 kN Tu = 72 kN·m

Discussion

ACI 318M-14

9.2.4.4 22.7.6.1 22.7.6.1.1 9.2.4.4 22.7.6.1

Calculation

Determine section properties for torsion. A = h Aoh = ( – 90 mm)( mm)(h h – 90 mm) Ao = 0.85Aoh p = 2( + h) ph = 2( – 90 mm + h – 90 mm)

9.5.4.1 22.7.4.1a T

f

0.083

A2

A = (400 mm)(600 mm) = 240,000 mm2 Aoh = (400 mm – 90 mm)(600 mm)( mm – 90 mm) = 158,100 mm2 Ao = 0.85 0.85(158,100 158,100 mm2) = 134,385 mm2 p = 2( 2(400 mm + 600 mm) = 2000 mm 400 mm – 90 mm + 600 mm – 90 mm) = 1640 mm ph = 2(400

Tth

p

(0.75)(0.083)( 35 MPa )

(240,000 mm 2 ) 2 2000 mm

10.6 106 N·mm 9.5.1.2 21.2.1c

Torsion strength reduction factor

9.5.1.1

Check if Tu

22.7.5.1

Calculate cracking torsion T

9.5.1.1

0.33

Check if Tu

= 0.75

Tth

f

Tth = 10.6 kN·m Tth = 10.6 kN·m OK Tu Design section to resist torsional moment. A2 p

T ?

In statically indeterminate structures where Tu T , a reduction of Tu in the beam can occur due to redistribution of internal forces after torsion cracking. Therefore, reduce Tu to T .

T = 42.2 kN·m

Tu

T = 42.2 kN·m

Use Tu = 42.2 kN·m and design for torsional reinforcement.

  

Eng.Mohammed Ashraf



22.7.7.1



Check if cross-sectional dimensions are large enough. V

2

T p 1.7 Aoh2

2

V

271,000 N (400 mm)(510 mm)

0.66 f

2

+

(42.2  106 N·mm)(1640 mm) 1.7(158,000 mm 2 ) 2

(0.75)(0.17 35 MPa

(1.33 MPa) 2

(1.63 MPa) 2

2.1 MPa

2

0.66 35 MPa )

3.7 MPa OK

Therefore, section is large enough. Find area/spacing of ties due to shear and torsional

9.5.4.1 22.7.6.1a 22.7.6.1.2 9.5.1.1 2.5.1.1 22.5.5.1 22.5.10.5.3 21.2.1b

Tn

2 Ao At f y s

cot

Tu

42.2 kN·m

At s

42.2  106 N·mm 2(0.75)(134,385 mm 2 )(420 MPa)(cot45°) 0.498 mm 2 /mm

Vn = V + Vs and Vs

Vu = 271 kN

A f d s

Calculate total tie area/spacing (A /s + 2At/s)

A s A s A s

271,000 N 0.75

0.17 35 MPa (400 mm)(510 mm) (420 MPa)(510 mm)

0.729 mm 2 /mm

2

A s

0.729 mm 2 /mm 2(0.498 mm2 /mm) 1.725 mm 2 /mm

9.6.4.2

Try No. 13 ties and calculate s

s

258 mm 2 1.725 mm 2 /mm

149.6 mm 150 mm

Use s = 150 mm Torsional longitudinal reinforcement 22.7.6.1b

Tn

2 Ao A f y ph

tan

42.2  106 N·mm 2(134,385 mm 2 ) A (420 MPa) = tan 45 0.75 1640 mm

A = 817 mm2

  

Eng.Mohammed Ashraf



9.6.4.3



The minimum torsional longitudinal reinforcement min

(a)

0.42 f A fy

0.42( 35 MPa )(240,000 mm 2 ) 420 MPa 420 MPa (0.498 mm 2 /mm)(1640 mm) 420 MPa

f At ph s fy

603 mm 2 Controls

(b)

0.42 f A fy

0.175 f yt

ph

f fy

0.42( 35 MPa )(240,000 mm 2 ) 420 MPa 0.175(400 mm) 420 MPa (1640 mm) 420 MPa 420 MPa A

.

= 817 mm2

A

= 603 mm2

9.7.5.1

Distribute torsional longitudinal reinforcement around the perimeter of closed stirrups that satisfy Section 25.7.1.6 (ends of stirrups are terminated -degree standard hooks around a longituwith 135-degree dinal bar).

9.7.6.3.3

Transverse torsional reinforcement spacing must not exceed the lesser of ph/8 and 300 mm

ph/88 = 1640 mm/8 = 205 mm < 300 mm

bottom and two No. 13 in each side face. Excess d from support can serve in place of three No. 13 in top.

Refer to Beam Example 8.

1146 mm 2

OK

1. Bottom bars summation of moment and torsion reinforcement requirements. 2. Side bar due to torsional moments

  

Eng.Mohammed Ashraf



 Building diaphragms are usually horizontal, reinforced concrete one-way or two-way slabs spanning between columns or walls, or both columns and walls. They can be

 Concrete compressive strength for diaphragms and collec-

with CIP topping, interconnected precast elements without CIP topping, or precast elements with end strips formed of Building slabs are designed to resist gravity loads and also

 The minimum diaphragm slab thickness must satisfy the

forces to the lateral-force-resisting system, such as moment For dual system structures such as shear walls and special moment frames, special moment frames deform in a shear

Diaphragms maintain compatible deformations between the

 Diaphragm slabs must resist gravity loads and lateral in-plane force combinations simultaneously. For concrete

shear walls and columns. As a rule of thumb, approximately 2 to 5 percent of a column axial force must be resisted by the diaphragm to provide adequate lateral support to the walls and columns. This force is easily achieved in lowrise building diaphragms, but must be checked for columns with high axial force, such as those in high-rise buildings

of a rigid diaphragm if the diaphragm aspect ratio, which is the span-to-depth ratio, is 3 or less for seismic design and 2

into columns at edge connections

supports for the deep beam. The diaphragm reinforcement resisting tension due to

structures are expected to behave inelastically during an earthquake, it is expected that rigid diaphragms will perform elastically under all load conditions. Diaphragm slabs are commonly designed as a deep beam that resists lateral forces

bracing forces

Fig. 8.1—Shear wall and moment frame dual system deformation.   

Eng.Mohammed Ashraf





as chords. Because earthquake and wind forces are reversible,

the diaphragm in-plane bending moment resulting from the lateral load. If edge beams are not provided, the slab acts as a deep rectangular beam resisting bending in the plane of the slab, with the chord tension reinforcement placed within h/4 of the tension face, where h is the diaphragm width in The diaphragm shear forces are resisted by the system columns and walls. The beams or slab sections that transfer connection to the columns and walls must be appropriately designed and detailed to achieve shear transfer.

is distributed to the columns and walls according to their Fig. 8.4a—Diaphragm tension-compression and shear forces due to lateral forces.

  

Eng.Mohammed Ashraf





Fig. 8.4.1a—Collector having same width as shear wall— forces are reversible.

Fig. 8.4c—Rigid Rigid diaphragm lateral force distribution. lateral resistance, the lateral force can be distributed to the columns and walls according to their tributary areas.

8.4.1 Collectors—Shear —Shear walls do not usually extend the full length of a building. Collectors, also called drag members or distributers, are designed to collect lateral forces from the diaphragm and transfer them to the seismicforce-resisting system, or to transfer lateral loads from a shear wall into the diaphragm. Collectors can be the full

within the depth of the diaphragm or as a beam as part of the diaphragm. Collectors, as part of a rigid diaphragm, are expected to perform elastically during an earthquake event. Collectors parallel to a shear wall can have the same width

Fig. 8.4.1b—Collector wider than the shear wall—forces are reversible. eccentricity creates secondary stresses in the slab transfer

Collectors, like rigid diaphragms, are expected to behave elastically under axial and compression forces. Reinforcement is usually placed at mid-depth in collectors. Shear reinforcement perpendicular to the walls is needed as shear friction reinforcement for eccentric collectors, and is placed

b than the thickness of the wall plus one-half the length of the 

will simply transfer the slab lateral forces by axial compression or tension to the shear wall. Collectors having a width wider than the shear wall will transfer part of the diaphragm lateral force by axial compression or tension and the balance will be transferred along the wall length through shear friction. An eccentricity results between the resultant force in

through F are designed in accordance with Chapter 18 of ACI 318M-14. ness under factored load combinations; its thickness must

  

Eng.Mohammed Ashraf





simultaneously. Diaphragms are designed to resist the design seismic force calculated from the structural analysis, Fpx, which must be at n

Fpx

i x n i x

Fi W px

Wi

where Fpx is the diaphragm design force at level x. The design force applied to level xi is Fi; Wi is the weight tributary to level xi; and Wpx is the weight tributary to the diaphragm at level x. The force calculated from this equation need not exceed 0.45SDSIWpx, but needs to be at least 0.25SDSIWpx Collectors in SDCs C through F are designed for the Fx obtained from structural analysis using load combio of ASCE/SEI 7-10, Section 12.4.3.2 Fpx using load combinations with overstrength factor o of ASCE/SEI 7-10, Section 12.4.3.2 Fpx,min = 0.25SDSIWpx using load combinations of Fx are applied to

Fig. 8.6a—Reinforcement detail around opening within diaphragm.    Reinforcement

Forces Fpx and Fpx,min “are applied one level at a time

Vn

Acv

fc

t

Splice

Requirement

Spacing

3db

Cover

2.5d 2.5 b 0.062 fc

fy Transverse

Vn 0.66 f c Acv In ACI 318M-14, Sections 12.5.3.3 and 12.5.3.4, Acv is the gross area of concrete section bounded by web thickness and section length in the direction of shear force considered, t is the ratio of area of distributed transverse reinforcement to gross concrete area positioned perpendicular to the

seismic-force-resisting system it is part of if the nominal shear strength of the member is less than the shear corresponding to the development of the nominal strength of the member. The

Greater of

ACI 318M-14

bw s f yt

0.35bw s f yt

-

transfer forces between the diaphragm and columns or walls, chord bar splices should be Type 2 and chord bar spacing should satisfy the requirements of Table 8.6a. Chord bars in

forces in the diaphragm, the local design strengths should be within the shear and torsion strength of the diaphragm.

spirals per Table 8.6b. Chords at openings need to be proportioned to resist the sum of the factored axial forces acting in the plane of the diaphragm and the force obtained by dividing the factored moment at the section by the distance between the chords at the section. A collector parallel to a shear wall has its critical connection at the face of the shear wall. Collector longitudinal bars must extend deep enough into the shear wall to develop and

 Generally, chord and collector reinforcement is placed

The collector reinforcement is in addition to the horizontal diaphragm reinforcement required to resist the shear force

At diaphragm discontinuities, such as openings and reentrant corners, the design needs to consider the dissipation or

with Section 20.2.1 of ACI 318M-14 with two exceptions: approximately twice the slab thickness with only the displaced reinforcement, but at least one bar on any side.

must satisfy ASTM A706/706M, Grade 420. Reinforcement complying with ASTM A615/A615M is permitted if:

  

Eng.Mohammed Ashraf







Compressive stress

Transverse reinforcement requirements

ACI 318M-14

Details Single or overlapping spirals per Sections 25.7.3.5 and 25.7.3.6 of ACI 318M-14 Circular hoops or rectangular hoops with or without crossties spaced not more than 350 mm

18.12.7.5

> 0.2fc fc if forces are

Transverse reinforcement spacing along length of the diaphragm is the smallest of: db of smallest longitudinal reinforcement so

350 hx 100 3

Rectilinear hoop

Spiral or circular hoop

Ash

18.12.7.5 and 18.7.5.3

so

0.09

sbc f c f yt

s

0.45

s

0.12

Ag Ach

1

fc f yt

18.12.7.5

Greater of: fc f yt

> Fu >> Fu

OK OK

t

Vn Vn

Acv

fc

Vn

Acv OK

  

Eng.Mohammed Ashraf





refer to Fig. E1.4: Diaphragm is idealized as rigid. Design moments, shear, and axial forces are calculated assuming the diaphragm is a beam supported by idealized rigid supports with a depth equal to full diaphragm depth. The wall and frame forces and the assumed direction of torque due to the eccentricity are shown in Fig. E1.4. 12.4.2.4

The diaphragm force distribution is calculated by using qL and qR as the left and right diaphragm Fig. E1.4— Seismic forces in the lateral-forceresisting systems in the N-S direction.

Force equilibrium qL

L 2

qR

L 2

Fpx des

qL

NS

66.6 m 2

qR

66.6 m 2

Moment equilibrium qL

L 2

L 3

qR

L 2

2L 3

Fpx des

NS

L 0.05 L 2

qL and qR:

qL

66.6 m 6

qL + qR qL + 2qR qL

2

qR

2 66.6 m

2

6

66.6 m 2

qL

  

Eng.Mohammed Ashraf



derivative of the moment equation expressed as a dM/dx = 0 function of x



x = 34.9 m Mmax

917-4, “Seismic Design of Cast-in-Place Concrete Diaphragms, Chords, and Collectors,” by Moehle et al. states that, “This approach leaves any moment due to the frame forces along column lines A and F unresolved. Sometimes this is ignored or, alternatively, it too can be incorporated in the trapezoidal loading.” In this example the small moment due to the frame

Fig. E1.5 E1.5—Shear —Shear Shear and moment diagrams.

the calculations by distributing the load uniformly:

Resulting in a maximum moment of:

q

66.6 m max

-

Shear diagram for the second approach is a straight line with equal shear force at both ends. In this example, the detailed approach is presented.

12.5.2.3

Maximum moment is calculated above:

Mu

Chord reinforcement resisting tension must be located within h/4 of the tension edge of the diaphragm.

h/4 = 22.6 m/4 = 5.6 m

Placing chord reinforcement along the exterior frame is a simpler and cleaner load path for the forces in the diaphragm. Crack control reinforcement should be added in the balcony slab for crack control.

  

Eng.Mohammed Ashraf





Assume tension reinforcement is placed in a 1 m strip at both north and south sides of the slab edges 1 m < h/4 = 5.6 m

OK

Chord force The overhang is placed monolithic with the rest of the slab. Chord forces are usually highest furthest from the geometric centroid, in this case, edge of the overhang. To prevent cracking, place chord reinforcement at the outside edge of the overhang. The maximum chord tension force is calculated at

Tu 12.5.2.2

Mu B 1m

Tu

22.6 m 1 m

Tension due to moment is resisted by deformed bars conforming to Section 20.2.1 of ACI 318M-14.

12.5.1.5 strength and 420 MPa.

12.5.1.1

fy = 420 MPa

Required reinforcement Tn = fyAs Tu 433 mm 2

As , req ' d 22.4.3.1

The building is assigned to SDC B. Therefore, Chapter 18 requirements for chord spacing and transverse reinforcement of Section 18.12.7.6 of ACI 318M-14 do not apply.

18.12.7.5

re o for chord design is not required. Therefore, use the compression stress limit in provision 18.12.7.5 of 0.2fc Required chord width: wchord

CChord 0.2 f c hdiaph

wchord

134 mm OK

Choose reinforcement: As,prov. Check if provided reinforcement area is greater than required reinforcement area:

2

2

As,prov. = 568 mm2 > As,req’d = 433 mm2

  

Eng.Mohammed Ashraf





The engineer has two options for providing chord reinforcement along the exterior frames: 1. Excess amount of reinforcement used in the beams to resist gravity loads could be used to resist part of 2. The chord force is resisted with additional reinforcement.

engineer has several options: 1. Place chord reinforcement outside the web width 2. Place chord reinforcement within the web width

Fig. E1.6—Chord Chord reinforcement along CLs A and F.

  

Eng.Mohammed Ashraf



12.5.4.1



Collector elements transfer shear forces from the diaphragm to the vertical walls at both east and Collector elements extend over the full width of the diaphragm. Unit shear force: vu

Fu F

F

From Step 6: Fu

B

In diaphragm: vu

In wall: vu

Fpx F

Ldiaph

vu

F

22.6 m

vu

F

8.5 m

Fpx F

Lwall

Force at diaphragm to wall connection East wall south end: F7/D.5 East wall north end: F7/B.5 At diaphragm end: F7/A

Per collector force diagram, the maximum axial force on the collector is Tu = Cu force must be transferred from the diaphragm to the

The building is assigned to SDC B. Therefore, the collector force and its connections to the shear wall will not be multiplied by the system overstrength factor, o 12.5.4.2

Collectors are designed as tension members, compression members, or both.

Fig. E1.7—Collector force diagram.

  

Eng.Mohammed Ashraf



12.5.1.1 22.4.3.1

Tension is resisted by reinforcement as calculated above. Required reinforcement: Tn = fyAs Tu

As , req ' d



Tu 0.9 f y

72,700 kN (0.9)(420 MPa)

192 mm 2

provided to maintain symmetry of load being transferred from the slab into the wall. 18.12.7.5

Check if collector compressive force exceeds 0.2fc . Calculate minimum required collector width using 0.2fc wcoll .

CColl . 0.2 f c tdiaph

This results in compressive stress on the concrete diaphragm collector being relatively low. The section is adequate to transfer shear stress without additional reinforcement.

wcoll

72,700 N (0.2)(35 MPa)(175 mm)

59 mm

Use 300 mm wide collector (same width as shear wall). Provide two No. 16 bars at mid-depth of slab to prevent additional out-of-plane bending stresses in the slab. Space the two No. 16 bars at 200 mm on center starting at 50 mm from the edge of the diaphragm within the 300 mm wide collector/shear wall (Fig. E E1.8).

The collector compression and tension forces are transferred to the lateral force-resisting system within its width (shear wall). Therefore, no eccen eccentricity is present and no-in-plane bending occurs. 12.5.4.1

ACI 318M-14 permits to discontinue the collector along the length of the shear wall where transfer of design collector is not required.

Fig. E1.8—Collector reinforcement. 12.4.2.4

12.5.3.3 21.2.4.2

Check slab shear strength along shear walls Slab shear strength along walls: L = 8.5 m and slab thickness t = 175 mm From Vc Acv 0.17 fc

Vc

(0.75)(0.17)(1.0)( 35 MPa )(8500 mm)(175 mm)

Vc

= 0.75

12.5.1.1

Is the provided shear strength adequate?

12.5.3.4

By inspection, the diaphragm shear design force

Vc = 1122 kN > Vu = 233 kN (from Step 7)

OK

318M-14. Vn

Acv 0.66 f c   

Eng.Mohammed Ashraf





Design moments, shear, and axial forces are calculated assuming a simply supported beam with depth

Fig. E1.9—Seismic forces in the lateral force resisting systems in the E-W direction. qL torsion, the inertial force is uniformly distributed across the diaphragm width.

22.6 m

Shear force: V x = 11.3 m Maximum moment is located at midlength. M max

w 2 8

2

8

Fig. E1.10—Shear and moment diagrams.

  

Eng.Mohammed Ashraf



Step 12: Chord reinforcement E-W Calculate chord reinforcement Maximum moment is calculated above (Fig. E1.8). 12.5.2.3



Mu = 1724 kN·m

Chord reinforcement must be located within h/4 of the tension edge of the diaphragm.

h/4 = 66.6 m/4 = 16.65 m

Assume tension reinforcement is placed within a 0.3 m strip of the slab edge at both east and west sides of the slab.

0.3 m < h/4 = 16.65 m

OK

Chord force The maximum chord tension force is at midspan: Tu

12.5.2.2

Mu L 1 ft

Tu

1724 kN·m 66.6 m 0.3 m

26 kN

Chord reinforcement Tension due to moment is resisted by deformed

12.5.1.5

12.5.1.1 22.4.3.1

strength and 420 MPa.

fy = 420 MPa

Required reinforcement Tn = fyAs Tu

As , req ' d

Tu 0.9 f y

26,000 N (0.9)(420 MPa)

68.8 mm 2

Along column lines 1 and 7, two No. 16 bars collector reinforcement are provided to resist inertia force in the N-S direction. These bars can be used for chord reinforcement in the E-W direction (refer to Fig. E1.8). As,prov. = (2)(199 2)(199 mm2) = 398 mm2 > 68.8 mm2 Maximum shear in the E-W direction occurs at CLs 1 and 7: Unit shear force in frame: vu @1,7

Fu @1,7

vu @1,7

L

306 kN 66.6 m

4.6 kN/m

Step 13: Collector reinforcement Collector along CLs A and F: The continuous reinforced concrete frame over the full length of the building acts as a collector. Note: Provide continuous reinforcement with tension splices (Step 15). 12.5.3.7 transferred from the diaphragm to a collector, or from the diaphragm or collector to a shear wall, temperature and shrinkage reinforcement is usually adequate to transfer that force.

  

Eng.Mohammed Ashraf

OK





Step 14: Shrinkage and temperature reinforcement 12.6.1 The minimum area of shrinkage and temperature 24.4.3.2 reinforcement, AS+T: AS+T 24.4.3.3

Ag

Spacing of S+T reinforcement is the lesser of 5h and 450 mm:

2

AS+T

/m

be part of the reinforcing bars resisting diaphragm in-plane forces and gravity loads. If provided

h Controls

reinforcing bars to resist positive moments at midspans and top reinforcing bars to resist negative bottom reinforcing bars can sometimes be achieved by providing adequate splice lengths between them.

Step 15: Reinforcement detailing Reinforcement spacing Minimum and maximum spacing of chord and 12.7.2.1 collector reinforcement must satisfy 12.7.2.1 and 12.7.2.2. 25.2.1

Section 25.2 limits minimum spacing of Minimum spacing 25 mm

Controls

dagg. dagg db 18.12.7.6a

The minimum collector reinforcement spacing at a splice must be at least the largest of: db cc

12.7.2.2

16 mm

db, 50 mm]

Maximum spacing is the smaller of 5h or 450 mm

40 mm 50 mm 450 mm

Controls Controls

  

Eng.Mohammed Ashraf





Step 16: Details

Fig. E1.11—Diaphragm chord and collector reinforcement.

  

Eng.Mohammed Ashraf





Rigid Diaphragm Example 2: Reinforced concrete diaphragm with opening—Refer to Diaphragm Example 1 for structure E2.1. For diaphragm building elevation, material properties, and design criteria, refer to Diaphragm Example 1.

Fig. E2.1—Eight Eight story building. ACI 318M-14

Discussion

Calculation

Step 1: Material requirements Refer to Diaphragm Example 1. Step 2: Slab geometry

For lateral forces and design forces calculations, refer to Diaphragm Example 1, Step 3.

this example.

  

Eng.Mohammed Ashraf





Design second-level diaphragm:

Fig. E2.2—Mass Mass center and rigidity center location excluding accidental torsion.

yCOM

11.85 m

xCOR yCOR = y y = 11.9 m – 11.85 m = 0.05 m, say, 0.1 m y COR COM Accidental torsion forces in addition to the actual eccentricity is considered, referred to as accidental eccentricity. ex ey1 ey2

  

Eng.Mohammed Ashraf





Force in walls and moment-resisting frames are given by the following equations: Fuxi

Fuyi

kix kix kiy kiy

Fpx

ki d i Fpx e y ki di2

Fpy

ki d i Fpy ex ki di2

where di

xi or yi

Fpx,y Fpx,x directions, respectively. Mass accidental eccentricities are ex = 3.33 m; ey1 = 1.22 m + 0.1 m = 1.32 m; and ey2 = –1.22 m + 0.1 m = –1.12 m are calculated in Step 4 of this example. The torsional force is calculated by multiplying the lateral inertia force by the corresponding eccentricity: Ty = Fpyex EW: Tx1 = Fpxey1 Tx2 = Fpxey2

Fpy Fu , wall , max

2

Fu , wall , min

Fu , MF

Fpx

2

2

ey1 = 1.32 m 2

Fu , MF ,min

Fu , wall

2

2

2

2

2

2

2

ey2 = 9.44 m

Fu , MF ,max

Fu , MF ,min

Fu , wall

2

2

Fu , MF ,max

Fpx

2

2

2

2

2

  

Eng.Mohammed Ashraf





Step 7: Check shear force in diaphragm In-plane shear in diaphragm Nominal shear strength in E-W direction The diaphragm slab is cast-in-place concrete, therefore, shear force is calculated from Eq. (12.5.3.3) Vn , N (66,600 mm)(175 mm)(0.17(1.0) 35 MPa 12.5.3.3

Vn

Acv (0.17

fc

t

= 11,721,825 N 11,722 kN

fy )

Ignoring the strength contribution of reinforcement; Nominal shear strength in N-S direction t=0 Acv is the diaphragm gross area less the 1.8 m over- V (22,600 mm)(175 mm)(0.17(1.0) 35 MPa n, E = 3,977,676 N 12.5.3.2 21.2.4.2

12.5.3.2 21.4.2.1

0.75 at the north and south ends along column lines A and F.

0)

3978 kN

Design shear strength in E-W direction Vn,N = (0.75)(11,722 kN) = 8791 kN

At the east and west ends along Column Lines 1 not exceed the least value for shear used for the vertical components of the primary seismic-force-

12.5.1.1 22.5.1.2

0)

Check if factored shear force is less than design shear strength calculated in Step 7.

Design shear strength in N-S direction Vn,E = (0.75)(3978 kN) = 2984 kN Vn = 2984 kN >> Fu = 167.5 kN OK Vn = 8791 kN >> Fu = 219 kN OK Therefore, diaphragm has adequate strength to resist the lateral inertia force and shear reinforcement is not t = 0..

12.5.3.4

The nominal shear strength, Vn, must not exceed: Vn

0.66 Acv

fc

Vn

0.66(22,600 0.66 22,600 mm)( mm)(175 mm) 35 MPa 1000 N/kN 15, 443 kN

Acv is the diaphragm gross area less the 1.8 m overhang.

OK

  

Eng.Mohammed Ashraf





Diaphragm is idealized as rigid. Design moments and shear axial forces are calculated based on a beam with depth equal to full diaphragm depth satisfying equilibrium requirements. The wall forces and the assumed direction of torque due to the eccentricity are shown in Fig. E2.3. 12.4.2.4

The distribution of the applied force on the diaphragm Fig. E2.3—Forces in the structural resisting systems is calculated by using qL and qR as the left and right due to a seismic force in the N-S direction. Force equilibrium qL

L

qR

2

L 2

Fpx des

qL

NS

66.6 m 2

qR

66.6 m 2

Moment equilibrium qL

L 2

L 3

qR

L 2

2L 3

Fpx des

NS

L 2

2

0.05 L

qL and qR:

derivative of the moment equation expressed as a function of x dM/dx = 0

qL

6

2

qR

qL + qR qL + 2qR qL

6

2

qL

x = 34.9 m; Mmax

Draw the shear and moment diagrams and determine

917-4, “Seismic Design of Cast-in-Place Concrete Diaphragms, Chords, and Collectors,” by Moehle et al. states that, “This approach leaves any moment due to the frame forces along column lines A and F unresolved. Sometimes this is ignored or, alternatively, it too can be incorporated in the trapezoidal loading.” In this example the small moment due to the frame

Fig. E2.4—Shear and moment diagrams.   

Eng.Mohammed Ashraf





the calculations by using uniformly distributed load:

q

Resulting in a maximum moment of:

M max

Maximum moment obtained from moment diagram:

Mu

Chord reinforcement resisting tension must be located within h/4 of the tension edge of the diaphragm.

h/4 = 22.6 m/4 = 5.7 m

66.6 m 2

12.5.2.3

8

Placing chord reinforcement along the exterior frame is a simpler and cleaner load path for the forces in the diaphragm. Crack control reinforcement should be added in the balcony slab for crack control. Assume tension reinforcement is placed in a 600 mm strip at both north and south sides of the slab edges 600 mm < h/4 /4 = 5650 mm OK Chord force Maximum chord tension force that must be resisted by the chord at midspan is: Tu

Mu B 0.6 m

Tu

22.6 m 0.6 m

  

Eng.Mohammed Ashraf





Chord forces at opening The opening in the diaphragm results in local bending of the diaphragm segments on either side of the opening (refer to Fig. E2.5). 1. The diaphragm sections above and below the 2. The applied loading on the sections above and below the opening are based on the relative mass of each section (1.64:1). 3. The secondary chord forces are calculated based on the internal moment in the diaphragm sections adjacent to the opening. 4. The calculated tension and compression secondary chord forces are added to the primary Fig. E2.5—Idealization of sections above and below tension and secondary chord forces. opening. The opening is located at midlength of the building

Force at east and west ends of opening

forces at left and right edges of the opening are:

qu @ beg

(5.9 kN/m 3.3 kN/m)(27.8 m) 66.6 m

The load on the north and south section of the diaphragm bound by the opening is distributed accord(5.9 kN/m 3.3 kN/m)(38.9 m) ing to the ratio of the masses north and south of the q u @ end 66.6 m opening. Therefore, 38 percent and 62 percent of the overall applied trapezoidal load will be distrib distributed to the north and south section over this portion Force north of opening of the diaphragm, respectively. q u@bN @bN bN q u@eN @eN eN The unit forces magnitude at the east and west ends of the opening are close (1.67 kN/m versus 1.83 ( Force south of opening kN/m north of opening) and (2.72 ( kN/m versus q u@bS @bS bS 2.99 kN/m south of opening). Therefore, the averq u@eS age unit force of 1.75 kN/m and 2.85 kN/m will be used for calculating the diaphragm moment segments north and south of the opening (Fig. E2.6). er-aided design software programs or from the Reinforced Concrete Design Handbook Design Aid – Analysis Tables, which can be downloaded from: https://www.concrete.org/store/productdetail.

Fig. E2.6—Moment diagram of sections at opening.   

Eng.Mohammed Ashraf





The secondary chord forces are obtained from the moment B end of the span between opening and diaphragm edge: Secondary chord force north of opening: Tu ,opening

Mu D

Tu ,op , N

6.6 m Secondary chord force south of opening: Tu ,op , S

12.5.1.1 22.4.3.1

Required reinforcement Tn = fyAs Tu

11 m 7.1 mm 2

As , req ' d

Refer to the following discussion. The engineer has two options for providing chord reinforcement at the opening: 1. The beams are designed for 1.2D + 1.6L SDS D + 0.5L + E. The demand from 1.2D + 1.6L is usually higher than the gravity portion of the moments under SDS D + 0.5L. The two loads are proportioned and then the balance reinforcement is used to carry seismic chord/collector forces.

continuous. Diaphragm edge Total moment to be resisted is the sum of the main chord force and the secondary chord force: Tu,Total = Tu1 + Tu2

Tu,total,N = Cu,total,N Tu,total,S = Cu,total,S

12.5.2.2

Tu,Total = Tu1 + Tu3 Chord reinforcement: Tension due to moment is resisted by deformed bars conforming to Section 20.2.1 of ACI 318M-14.

12.5.1.5

12.5.1.1 22.4.3.1

strength and 420 MPa.

fy = 420 MPa

Required reinforcement Tn = fyAs Tu

As , req ' d

315 mm 2

The building is assigned to SDC B. therefore, ACI 318M-14 Chapter 18 requirements for chord spacing and transverse reinforcement of 18.12.7.6 do not apply. 18.12.7.5 o, for chord design is not required. Therefore, use the compression stress limit in Provision 18.12.7.5 of 0.2fc Required chord width:

wchord

CChord 0.2 f c hdiaph

wchord

97 mm OK 2

2

Choose reinforcement:

As,prov.

Check if provided reinforcement area is greater than required reinforcement area:

As,prov. = 398 mm2 > As,req’d = 315 mm2

  

Eng.Mohammed Ashraf



12.5.4.1



Collectors transfer shear forces from the diaphragm to the vertical walls at both east and west ends extend over the entire diaphragm width. Unit shear force: vu

Fu F

F

From Step 6: Fu

B Fpx

In diaphragm: vu

In wall: vu

F

Ldiaph

vu

F

22.6 m

vu

F

8.5 m

Fpx F

Lwall

Force at diaphragm to wall connection Wall west end: F7/D.5 East wall north end: F7/B.5 At diaphragm end: F7/A Per collector force diagram, the maximum axial force on the collector is Tu = Cu force must be transferred from the diaphragm to the

The collector force and its connections to the shear over wall will not be multiplied by the system overo

12.5.4.2

12.5.1.1 22.4.3.1

Collectors are designed as tension members, com compression members, or both. Tension is resisted by reinforcement as calculated above.

Fig. E2.7—Collector force diagram.

Required reinforcement: Tn = fyAs Tu

As , req ' d

Tu fy

139 mm 2

provided to maintain symmetry.

  

Eng.Mohammed Ashraf



18.12.7.5



Check if collector compressive force exceeds 0.2fc Calculate minimum required collector width using 0.2fc wcoll .

CColl . 0.2 f c't diaph

This results in compressive stress on the concrete diaphragm collector being relatively low. The section is adequate to transfer shear stress without additional reinforcement. The collector compression and tension forces are transferred to the lateral force-resisting system within its width (shear wall). Therefore, no eccentricity is present and no in-plane bending occurs.

wcoll

52,400 N (0.2)(35 MPa)(175 mm)

43 mm

Use 300 mm wide collector (same width as shear wall). Provide two No. 16 bars at mid-depth of slab to prevent additional out-of-plane bending stresses in the slab. Space the two No. 16 bars at 200 mm on center starting at 50 mm from the edge of the diaphragm within the 300 mm wide collector/shear wall (Fig. E2.8).

ACI 318M permits to discontinue the collector along the length of the shear wall where transfer of design collector is not required.

Fig. E2.8—Collector reinforcement. 12.4.2.4 12.5.3.3 21.2.4.2

Check slab shear strength along shear walls Slab shear strength along walls: L = 8.5 m and slab thickness t = 175 mm From Vc Acv 0.17 fc

12.5.1.1

Is the provided shear strength adequate?

12.5.3.4

By inspection, the diaphragm shear design force Vn

Vc = (0.75)(0.17)(1.0)( )(8500 mm)(12(175 mm)) Vc = 1,122,022 N ~ 1122 kN Vc = 1122 kN > Vu = 167.5 kN (from Step 7)

Acv 0.66 f c

  

Eng.Mohammed Ashraf

OK





Design moments, shear, and axial forces are calculated based on a beam with depth equal to full diaphragm length satisfying equilibrium requirements. The wall forces and the assumed direction of torque due to the eccentricity are shown in Fig. E2.9. The distribution of the applied force on the dia-

Fig. E2.9 Forces in the structural resisting systems E2.9—Forces E due to a seismic force in the E-W direction. qL

22.6 m

Shear force: V x = 22.6 m/2 = 11.3 m; Mmax

  

Eng.Mohammed Ashraf





Maximum moment is taken at midspan.

Fig. E2.10—Shear Shear and moment diagrams. Step 12:: Chord reinforcement E-W Maximum moment is calculated above: 12.5.2.3 Chord reinforcement resisting tension must be located within h/4 of the tension edge of the diaphragm.

Mu

h/4 = 66.6 m/4 h 4 = 16.6 m

Assume tension reinforcement is placed in a 300 mm strip at both north and south sides of the slab edges 0.3 m < h/4 = 16.6 m

OK

Chord force Maximum chord tension force that must be resisted by the chord at midspan: Tu

Mu B 1 ft

Tu ,1

  

Eng.Mohammed Ashraf





Chord forces at opening The opening in the diaphragm results in local bending of the diaphragm segments on either side of the

1. The diaphragm sections to the east and west of 2. The applied loading on the sections east and west the opening are based on the relative mass of Fig. E2.11—Idealization of sections above and below 3. The secondary chord forces are calculated opening. based on the internal moment in the diaphragm sec4. The calculated tension and compression secondary chord forces are added to the primary tension and secondary chord forces.

Force east and west of opening qu

bE

qu

bW

2

The opening is located at mid-length of the buildnorth and south sections of the diaphragm bound by the opening are equal to one half of the overall applied trapezoidal load over this portion of the

Because forces at both ends of openings are close, a uniform load is assumed. Fixed end moment can be obtained from computeraided aided design software programs or from Reinforced Concrete Design Handbook Design Aid – Analysis Tables, which can be downloaded at: https://www.concrete.org/store/productdetail. aspx?ItemID=SP1714DA Fig. E2.12—Moment diagram of sections at opening. The secondary chord forces are obtained from the each end of the span between opening and diaphragm edge: Tu,opening = Mu/D

Tu ,op ,S

27.8 m 0.6 m

Total moment to be resisted is the sum of the main chord force and the secondary chord force: Tu,total = Tu1 + Tu,O2

Tu,total,N = Cu,total,N

  

Eng.Mohammed Ashraf



12.5.2.2



Chord reinforcement: Tension due to moment will be resisted by deformed bars conforming to Section 20.2.1 of ACI 318M-14.

12.5.1.5 strength and 420 MPa.

12.5.1.1 22.4.3.1

Required reinforcement Tn = fyAs Tu

As

56 mm 2

The chord forces north of and south of opening are equal. forcement is adequate and no additional reinforcement is required. Step 13: Collector reinforcement E-W Continuous reinforced concrete frame over the full length of the building will act as a collector.

In cast-in-place diaphragms, where shear is transferred from the diaphragm to a collector, or from the diaphragm or collector to a shear wall, temperature and shrinkage reinforcement is usually adequate to transfer that force. Step 14:: Shrinkage and temperature reinforcement 12.6.1 The minimum shrinkage and temperature 24.4.3.2 Reinforcement, AS+T:

12.5.3.7

AS+T 24.4.3.3

Ag

Spacing of S+T reinforcement is the lesser of 5h and 450 mm 5h 450 mm Controls

2

AS+T

be part of the main reinforcing bars resisting diaphragm in-plane forces and gravity loads. If provided inforcing bars to resist positive moments at midspans and top reinforcing bars to resist negative moments at ing bars may be achieved by providing adequate splice lengths between them.

  

Eng.Mohammed Ashraf





Step 15: Reinforcement detailing Reinforcement spacing 12.7.2.1 Chord and collector reinforcement minimum and maximum spacing must satisfy 12.7.2.1 and 12.7.2.2. 25.2.1

Section 25.2 requires minimum spacing of Minimum spacing 25 mm Controls dagg. db

18.12.7.6a

Collector reinforcement spacing at a splice must be at least the larger of: db cc

12.7.2.2

16 mm

db, 50 mm]

Maximum spacing is the smaller of 5h or 450 mm Edge reinforcement The opening has four beams around its perimeter. Therefore, the beams reinforcement is adequate to resist the tension forces due to inertial forces and additional reinforcement is not required.

40 mm 50 mm 450 mm

Controls Controls

recommended around the opening as shown in the Fig. E2.13 E2.13—Two Two No. 16 reinforcement around opening. develop Fig. E2.13 and extended a minimum of its development length.

  

Eng.Mohammed Ashraf





Step 16: Details

Fig. E2.14—Typical diaphragm to wall section.

Fig. 2.15A—Collector reinforcement in shear walls along CL 1 and 7.

  

Eng.Mohammed Ashraf





Fig. 2.15B—Chord reinforcement at midspan.

Fig. 2.15D—Chord Chord reinforcement at opening.

Fig. 2.15C—Chord reinforcement at supports.

Crack control reinforcement at balcony edge. Fig. 2.15E—Crack

  

Eng.Mohammed Ashraf





Rigid Diaphragm Example 3: Lateral force distribution of a rigid diaphragm to shear walls—A three-story wood apartment building is built on a normalweight reinforced concrete one-story slab. The slab is 60 m x 27 m with fc fy supporting the wood structure is 250 mm thick and the wall lengths, height, and thicknesses are shown as follows. Assume the 2 2

m miscellaneous dead load to the slab. Refer to Fig. E3.1 for geometric information. This example will determine the seismic forces that are resisted by the shear walls, design the diaphragm, chords, and collec-

Given: Project data— Diaphragm size 60 m x 27 m Wall 1: 27 m x 200 mm Wall 2: 9 m x 250 mm Wall 3: 9 m x 250 mm Wall 4: 8.4 mm x 250 mm Wall 5: 12 m x 250 mm Slab thickness: t = 250 mm 3.6 m above the foundation Concrete— fc fy = 420 MPa

Fig. E3.1—Slab —Slab Slab that supports a four-story wood building.

Seismic criteria— SDC D CS = 0.316

ACI 318MDiscussion 14 Step 1: Material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength

The designer determines the durability classes. Please refer to Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

Calculation

By specifying that the concrete mixture shall be in accordance with ACI 301M and providing the exposure

Based on durability and strength requirements, and experience with local mixtures, the compressive 28 MPa.

coordinated with ACI 318M-14. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Step 2: Slab geometry 12.3.1.1 under factored load combinations.

Given: h = 250 mm

  

Eng.Mohammed Ashraf





The lateral force is obtained by multiplying the self-weight of the reinforced concrete slab, wood frame building dead load, miscellaneous dead load, and the contribution

Gravity loads The reinforced concrete slab self-weight: Wslab L B h c

3

Wslab

Weight of wood frame building dead load and miscellaneous dead load:

WD

Total gravity dead load:

W

2

2

Shear wall self-weight contribution to diaphragm lateral force calculation is half the wall height.

Wi

L H

tw

c

W1 W1

3

3

W4 W4

3

W5 W5 W

E-W direction Wi L H tw

c

W2 W2 W3 W3

Total gravity dead load in the E-W direction:

Base shear is obtained from ASCE7-10 Section 12.8.1: V = CsW Cs is calculated using ASCE 7-10 Section 12.8.1.1; not shown here for brevity:

3

3

W

Cs = 0.316 given

The equivalent lateral force distribution over the

The diaphragm design forces Fpx are calculated per

Fpx and Fpy must be in accordance with ASCE 7-10 Calculations not shown here as it is outside the scope of this Handbook. Equivalent lateral force at the concrete level is:

Fx

  

Eng.Mohammed Ashraf





Diaphragm design forces: Fpy Fpx

E-W:

example, the contribution of wall weights parallel to the applied seismic force is considered in the calculation of diaphragm shears. Walls perpendicular to the applied seismic force are included in determining the lateral force of concrete diaphragms. Determine center of mass Assume that the diaphragm is rigid.

 Weight, Length, m

Area, m2 L x 1/2H

Weight, kN Col. 2 × Col. 4

Direction

x, m

xW, kN·m Col. 5 × Col. 7

y, m

yW, kN·m Col. 5 × Col. 9

4.70

27

48.6

228.4

x

30

6853

26.90

6144

5.88

9

16.20

95.2

y

0.125

12

13.5

1285

5.88

9

16.20

95.2

y

59.875

5699

13.5

1285

5.88

8.4

15.12

88.8

x

16.2

1439

3

266

5.88

12

21.6

126.9

x

42.0

5330

3

2

Wall No.

635

19,332

381 9361

The values of xcg and ycg are the center of mass of each wall. For example: Wall 1 has the following coordinates: xcg = 16.5 m + 27 m/22 = 30 m and y Wall 2 has the following coordinates: xcg y = 9 m + 9 m/2 = 13.5 m Center of mass of all walls: x1

y1

Wi xcg ,i Wi Wi ycg ,i Wi

30.5 m

14.8 m

Center of mass of the slab is: x2 = 60 m/2 = 30 m and y2 = 27 m/2 = 13.5 m

xm

Wi xi Wi

30 m

and ym

Wi yi Wi

13.54 m

  

Eng.Mohammed Ashraf





Determine center of rigidity

Flexure

Ph3 3EI

Shear

1.2 Ph where G AG Ph3 3EI

Flexure

Ph3 L3t 3E 12

Ph AG

Shear

Rigidity ki

4P

Ph E Lt

0.4E and E = 24,870 MPa h L Et

3

h L Et

3P

i

 Height h, m

L, m

h/L

t, mm

i

× 10–3, mm

ki

i

× 103, mm–1

1

3.6

27

0.1333

200

0.082

12.147

2

3.6

9

0.4000

250

0.234

4.270

3

3.6

9

0.4000

250

0.234

4.270

4

3.6

8.4

0.4286

250

0.257

3.885

5

3.6

12

0.3000

250

0.162

6.168

 Direction

x, m

y, m

kix

kiy

kiy x

kix y 327

1

x

—

26.90

12.15

—

—

2

y

0.125

—

—

4.27

0.5

—

3

y

59.88

—

—

4.27

255.7

—

4

x

—

3.0

3.88

—

—

12

5

x

—

3.0

6.17

—

—

19

22.20

8.54

256.2

357

  

Eng.Mohammed Ashraf





Calculate the system’s center of rigidity: kiy xi 256.2 m/m xr 30 m kiy 8.54 1/m yr

kix yi kix

358 m/m 22.2 1/m

16.1 m

Torsional eccentricity ex = xr – xm = 30 m – 30 m = 0 m, which is negligible ey = yr – ym = 16.1 m – 13.5 m = 2.6 m

Fig. E3.3—Locations Locations of the system’ss center of rigidity and center of mass. ASCE 7-10 requires shifting the center of mass by a minimum of 5 percent of the building dimension, referred to as accidental eccentricity, in addition to the calculated eccentricity. ex ey ey1 = 2.6 m – 1.35 m = 1.25 m ey2 = 2.6 m + 1.35 m = 3.95 m

  

Eng.Mohammed Ashraf





In-plane wall forces due to direct lateral shear force are calculated by: Fvx

Fpx

Fvy

Fpy

kix kix kiy kiy

In-plane wall forces due to torsional moment are calculated by: Ftx

ki xi Tx ki xi 2

Fty

k i yi Ty k i yi 2

The torsional moment is the lateral shear force multiplied by the corresponding eccentricity: Ty = Fpyex EW: Tx = Fpxey1 Tx = Fpxey2 The in-plane diaphragm force is the sum of the direct lateral shear force, Fvi, and the torsional moment, Fti Fu = Fvi + Fti  kix

kiy

xi, m

yi , m

kid

kid2

1

12.15

0

—

10.82

131.46

1423

0

121

–121

121

–121

2

0

4.27

–29.9 29.9

—

–127.574 –127.574

3811

1655.0

–118

118

1537

1773

3

0

4.27

29.9

—

127.574

3811

1655.0

118

–118

1773

1537

4

3.88

0

—

–13.08

––50.7994 50.7994

664

0

–47

47

–47

47

5

6.17

0

—

–13.08

–80.6634

1055

0

–74

74

–74

74

22.20

8.54

0.00

10,765

Fvi

Ftiti-1 -1

Fti-2

Ftotal-1

Ftotal-2

Example on calculating dyi: Wall 1: dyi Walls 4 and 5: dyi = 16.1 m – 3 m = 13.1 m  Fti-2 ey2 = 3.95 m

kix

kiy

xi , m

yi, m

kid

kid2

1

12.15

0

—

10.82

131.4629

1423

1767

–47

1720

2

0.00

4.27

–29.9

—

–127.574

3811

0

46

46

3

0.00

4.27

29.9

—

127.574

3811

0

–46

–46

4

3.88

0

—

–13.08

–50.7994

664

565

18

583

5

6.17

0

—

–13.08

–80.6634

1055

897

29

926

22.20

8.54

0.00

10,765

Fvi

Ftotal-2

 kix

kiy

xi, m

yi, m

ki d

kid2

1

12.15

0

—

10.82

131.4629

1423

1767

–153

1613

2

0.00

4.27

–29.9

—

–127.574

3811

0

149

149

3

0.00

4.27

29.9

—

127.574

3811

0

–149

–149

4

3.88

0

—

–13.08

–50.7994

664

565

59

624

5

6.17

0

—

–13.08

–80.6634

1055

897

94

991

22.20

8.54

0.00

10,765

Fv

Fti-1

Ftotal-1

  

Eng.Mohammed Ashraf





where d dxi or dyi Fx1 is the additional shear force due to eccentricity of 3.95 m. Fx2 is the additional shear force due to eccentricity of 1.25 m. • dxi and dyi are the distances of a wall from the center of rigidity in the x- and y-direction. • If torsional moment reduces the magnitude of the direct lateral shear on a wall, then it is ignored. The wall design shear forces are summarized in Table E.7.  Wall length, m 1

27

1720

121

1720

2

9

149

1773

1773

3

9

149

1773

1773

4

8.4

624

47

624

5

12

991

74

991

Step 7: Diaphragm shear strength In-plane shear in diaphragm The diaphragm nominal shear strength is calculated 12.5.3.3

Vn

Acv

fc

t

Vn

fy

without reinforcement; therefore, ignore the strength contribution of reinforcement: t = 0.

East and west Vn

Assume collector length is the full length of diaphragm in both directions. 12.5.3.2 21.2.4.2

Applying the reduction factor

18.2.4.1b

Per ACI 318M-14, must not exceed the least value for shear used for the vertical components of the primary seismic-force resisting system:

12.5.1.1 22.5.1.2

Check if design shear strength exceeds the factored shear force.

18.12.9.2

The nominal shear strength, Vn, must not exceed: 0.66 Acv f c

= 0.6.

Vn EW: Vn

Vn Vn

Vu Vu

Vn, NS Vn, EW

OK OK

OK OK

  

Eng.Mohammed Ashraf





Step 8: Diaphragm lateral force distribution N-S 12.4.2.4 12.5.1.3 moments, shears, and axial forces are calculated assuming a simply supported beam with depth equal

The wall forces and the assumed direction of the torsional moment are shown in Fig. E3.4. Refer to previous Step 5 for calculation of seismic force location. The seismic force on the diaphragm is distributed Fig. E3.4—Shear wall forces due to seismic force in the N-S direction. Force equilibrium qL

L 2

qL

L 2

qR

L 3

L 2

qR

Fpx des

L 2

2L 3

qL

NS

Fpx des

NS

L 2

60 m 2

qR

60 m 2

2

qL

0.05 L

qL and qR:

6

2

qR

6

2

qL qR -

“Seismic Design of Cast-in-Place Concrete Diaphragms, Chords, and Collectors,” by Moehle et al. states that, “This approach leaves any moment due F unresolved. Sometimes this is ignored or, alternatively, it too can be incorporated in the trapezoidal loading.” Vmax Mmax

Fig. E3.5—Shear and bending moment diagrams due to a lateral seismic force in the N-S direction. tions by using uniformly distributed load:

q

Calculate a maximum moment:

  

Eng.Mohammed Ashraf





• Shear diagram for the second approach is a straight line with maximum shear force: V R12.1.1

Assume the slab behaves like a beam with compression and tension forces at the near and far edges, respectively: Cchord = Tchord = M/d

12.5.2.3

ACI 318M suggests placement of chord reinforcement within an arbitrary width of h/4 of h/4 = 27 m/4 = 6.75 m d The maximum chord tension force is:

Tu

23.625 m

Tension due to moment is resisted by deformed bars conforming to Section 20.2.1 of ACI 318M-14. strength and 420 MPa.

fy = 420 MPa

Required chord reinforcement area: Tn = fyAs Tu

As

3150 mm 2

It is, however, recommended to place tension rein reinforcement close to the tension face. Assume tension reinforcement moment arm is approximately 0.95B at both north and south sides of the slab edges: The calculated tension force is: Tu

Mu 0.95B

Tu

Required tension reinforcement is: Tn = fyAs Tu 18.12.7.5

As

25.65 m 2902 mm 2

Per provision 18.12.7.5, the required chord width for the concrete compressive strength limit of 0.2fc wchord

Cchord 0.2 f c hdiaph

by the overstrength factor.

wchord

784 mm

and wchord = 784 mm < h/4 = 27 m/4 = 6.75 m Say, 800 mm h

OK

-

  

Eng.Mohammed Ashraf





Fig. E3.6—Suggested chord reinforcement at the north and south edges of the diaphragm. Collectors transfer shear forces from the diaphragm

of the diaphragm. Partial depth collectors can be considered, but a cacomplete force path should be designed that is ca pable of transmitting all forces from the diaphragm to the collector and into the vertical elements. 12.5.4.1

Unit shear force is the maximum diaphragm shear divided by the diaphragm depth, B = 90 m:

vu

Fu F

F

Fu

B

In slab:

vu

F

27 m

In wall:

vu

F

9m

Check if the concrete shear strength excluding reinforcement exceeds the factored shear: vc

0.17 f c btdiaph

vc

vc

Shear reinforcement is, therefore, required. Use

vc 199 mm 2

t

NG

0.00199

shrinkage reinforcement in each direction to invn vu OK

  

Eng.Mohammed Ashraf





Force at diaphragm to wall connection The proportional diaphragm force that the collector

Wall south end:

Wall north end:

Slab end:

Fig. E3.7—Collector Collector forces due to inertial forces in N-S direction. The collector factored force that is transferred to the walls is shown in Fig. E3.7: E 18.12.2.1

This collector force is then multiplied by the system overstrength factor o = 2.5 for building systems

12.5.4.2

Collectors are designed as both tension and compression members.

Tu =

R12.5.4

oTColl

=

oCColl

the slab is used as a collector. The collector width for tension reinforcement is provides in the commentary that a collector width cannot exceed approximately one-half the contact length between the collector and the vertical element measured from the face of the vertical element. b = 9000 mm/2 + 250 mm = 4750 mm

18.12.7.5

22.4.3

The collector width, however, is chosen such that 2.5Ccoll the limiting stresses are not exceeded. When the 422 mm wcollector fct tension and compression collector forces are increased by the overstrength factor, then the limiting wcollector = 422 mm < 4750 mm concrete compression stress is 0.5fc but wcollector = 422 mm > tw = 250 mm required compressive collector width: Therefore, part of the slab, b , is needed to resist the collector force. Required reinforcement area to resist collector force: oTcoll As 3910 mm 2 Tn = fyAs Tu fy

where not required. In this example, the reinforcement is extended over the full length of the diaphragm.   

Eng.Mohammed Ashraf





A number of bars are placed in line with the wall. The balance is distributed across the width of the collector element. In this case, for the 250 mm thick in a reasonable bar spacing of 150 mm Therefore,

As,line = 1020 mm2 The balance of the required reinforcement is:

2 As,bal = 3910 mm2 = 2890 mm2 distributed over the 4.5 m wide collector; 2890 mm2/4.5 m = 642 mm2/m

2 2 As,prov. As,prov. = 3184 mm2 > As,req’d = 2890 mm2

OK

Design collector region: The collector geometry results in a moment in the the eccentricity between the collector and the wall. This moment is solved through shear forces in the diaphragm perpendicular to the collector and bend bending in the plane of the diaphragm due to eccentric

Mu = Tdisteten + Cdistecomp –

wall

where Tdist is the portion of the tension collector force resisted by As,dist; Cdist is the portion of the comcom pression collector force resisted by slab outside the wall; and V is the shear strength of the diaphragm.

Where the collector element is in tension, the concrete contribution to V is neglected, Vc = 0. Vs fy ttw wcomp – twall

For more in-depth understanding refer to: Structural Engineer Association of California

Fig. E3.8—Diaphragm segment plan. Vs

199 mm 2 t

0.00199

0.0018

crete Slab Collectors,” from the Aug. 2008 SEAOC Compilation, Structural Engineers Association of California, Sacramento.   

Eng.Mohammed Ashraf



Tdist

The moment arm:

etens Cdist



2890 mm 2 3910 mm 2

4500 mm 2

250 mm 2

2375 mm

422 mm 250 mm 422 mm

Moment arm:

Mu = Tdisteten + Cdistecomp –

Mu Mu

Assume 6

Required reinforcement:

As , req ' d As,req’d = 840 mm2

Refer to Fig. E E3.9. 3.9.

  

Eng.Mohammed Ashraf





Shear transfer design: A number of bars are placed in line with the wall, which results in transferring portion of the diaphragm force in tension and direct bearing of slab against the wall in compression. The diaphragm and shear transfer interface is designed for the balance of the collector element. Assuming tension forces are distributed in proportion to collector area, Vu for diaphragm and shear transfer design is then calculated as follows:

Vu = 1092 kN + 602 kN + (65.7 kN/m)(9 m) = 2285 kN

3910 mm2 is the required reinforcement area to resist the collector force in prior calculation. 1020 mm2 is the area of two No. 25 bars placed inline with the shear wall. 12.5.3.7

Shear from the diaphragm is transferred by shear friction to the wall with dowels placed perpendicular to the wall-slab interface: Vn

22.9.4.2

vffy

Assume that diaphragm slab is placed against hard hardened wall concrete that is clean, free of laitance, and intentionally roughened to a full amplitude of

Vn = (0.75)( (0.75)(1.0)Avf(420 MPa)

21.2.1(b)

The wall is

wall

= 9 m long.

The required diaphragm strength is:

Avf = 7254 mm2 or Avf/ wall = 7254 mm2/(9 m) = 806 mm2/m Try No. 19 at 300 mm on center. As,prov. = 947 mm2/m OK vu = 2285 kN/9 m = 254 kN/m

The diaphragm shear strength is the contribution of concrete and reinforcement in prior calculation of this step:

22.9.4.4

vn vc + vs) The value of Vn across the assumed shear plane must not exceed the lesser of the following limits: (a) 0.2fc Ac (b) (3.3 + 0.08fc Ac (c) 11Ac

vn = 260 kN/m > vu = 254 kN/m

(0.2)(28 MPa) = 5.6 MPa (3.3 + 0.08(28 MPa)) = 5.5 MPa 11 MPa

OK

Therefore, Vn must not exceed:

  

Eng.Mohammed Ashraf





Collector reinforcement at Shear Walls 2 and 3

Enlarged collector reinforcement at Wall 2 Fig. E3.9—Collector reinforcement for lateral force in the N-S direction.

  

Eng.Mohammed Ashraf





Step 11: Diaphragm lateral force distribution E-W The wall forces and the assumed direction of torque due to accidental eccentricity are shown in Fig. E3.10. The distribution of the diaphragm force is calculated by using qL and qR as the left and right

Case I: Eccentricity at ey = 1.25 m Refer to Step 5 of this example for calculation of eccentricity. Force equilibrium

Fig. E3.10 E3.10—Shear E Shear wall forces due to a seismic force in the E-W direction at ey = 2590 mm.

F2, F4, and F5 are per Table E.5. qL and qR:

qR qL

  

Eng.Mohammed Ashraf





Draw the shear and moment diagrams for the diaphragm assuming simply supported beam

The maximum moment is located at 12.15 m from the south end of the diaphragm. Vmax Mmax

number GCR 10-917-4 that, “For a rectangular diaphragm of uniform mass, a trapezoidal distributed force having the same total force and centroid is then applied to the diaphragm. The resulting shears and moments are acceptable for diaphragm design. This approach leaves any moment due to

or, alternatively, it too can be incorporated in the trapezoidal loading.”

Fig. E3.11 E3.11—Shear 3.11—Shear Shear and moment diagrams for Case I. Case II: Eccentricity at ey = 3.95 m Refer to previous Step 5 for calculation of eccentricity. Force equilibrium

F2, F4, and F5 are per Table E.6. qL and qR:

qR qL

  

Eng.Mohammed Ashraf





Draw the shear and moment diagrams for the diaphragm assuming simply supported beam

The maximum moment is located at 36 m from the south end of the diaphragm. Vmax Mmax

Fig. E3.12—Shear and moment diagrams for Case II. Case I

Case II

1720

1613

8729

9128

Taking the approach of equivalent uniformly distributed inertial force by ignoring the accidental torsion, the corresponding shear and moment forces are:

Experienced engineers will usually continue the design using the uniformly distributed inertia force. This example, howForce summary in the E-W direction Case I

Case II

Controlling case

8729

9128

II

W1

1720

1613

I

W4

584

625

II

W5

926

992

II

  

Eng.Mohammed Ashraf





Step 12: Chord reinforcement E-W R12.1.1 Assume the slab behaves like a beam with compression and tension forces at the near and far edges, respectively: Cchord = Tchord = M/d Chord reinforcement resisting tension must be located within h/4 of the tension edge of diaphragm.

h/4 = 60 m/4 = 15 m d = 52.5 m

Assume that tension reinforcement will be placed within wall thickness. Therefore, moment arm is both east and west sides of the slab edges:

d = 59.875 m

Chord force The maximum chord tension force is at midspan: T 18.12.7.5

Calculate the required chord width for the calculated concrete compressive strength limit of 0.2fc

the overstrength factor. 12.5.2.3

Chord reinforcement resisting tension must be lo located within h/4 of the tension edge of diaphragm. Check if required calculated width is less than wall thickness:

wchord = 109 mm < tw = 250 mm

OK

Tension due to moment is resisted by deformed bars conforming to Section 20.2.1 of ACI 318M-14. strength and 420 MPa.

fy = 420 MPa

Required reinforcement Tn = fyAs Tu

forcement. Therefore, OK. Step 13: Diaphragm shear strength Refer to Step 7 for diaphragm shear strength.

  

Eng.Mohammed Ashraf





Step 14: Collector design E-W Wall 1 12.5.4.1 Collectors transfer shear forces from the diaphragm In this example, assume collectors extend over the full length of the diaphragm. Fu

Unit shear force:

In slab:

In Wall W1:

12.5.4.1

Walls 4 and 5 Collectors transfer shear forces from the diaphragm

diaphragm. Unit shear force:

Slab: Fu

Wall 4: Fu In slab:

Wall 5: Fu = In wall:

  

Eng.Mohammed Ashraf





Force at diaphragm to wall connection The proportional diaphragm force that the collector

Wall 1 west end: Wall 1 east end:

Diaphragm end:

Wall 4 west end:

Wall 4 east end:

Wall 5 west end:

Wall 5 east end:

Diaphragm east end:

Fig. E3.13—Collector forces in the E-W direction. 18.12.2.1

The collector factored force that is transferred to the walls is shown in Fig. E3.13: This collector force is then multiplied by the system o = 2.5 for building systems

Wall 1

Wall 4

Wall 5

2.5Tu

1180

1180

807

188

860

810

2.5Cu

–1180

–1180

–807

–188

–860

–810

  

Eng.Mohammed Ashraf



12.5.4.2

18.12.7.5



Collectors are designed as tension and compression members. slab is used as a collector. The collector width is determined by engineerstresses are not exceeded. When the tension and compression collector forces are increased by the overstrength factor, then the limiting concrete compressive stress is 0.5fc collector width: wchord = 2.5Ccoll/0.5fc t Wall 1 wcoll, mm

421

Wall 4 421

231

Wall 5 54

246

231

is wcoll > tw?

mic force is resisted by reinforcement placed in-line with the shear wall to transfer the force directly to the end wall. The balance of seismic force is resisted by reinforcing bars placed along the sides of the wall and uses the slab shear-friction capacity at the wall-to-slab interface to transfer seismic forces to the wall.

Wall 1: Required area of collector reinforcement: Tn = fyAs

R12.5.4

Tu

As

oTcoll fy

3122 mm 2

and terminated when not required. In this example, the reinforcement is extended over the full length of the diaphragm. The collector width, as suggested by ACI 318M commentary, is an arbitrary width equal to half the wall width taken from the face of the wall plus the wall width. b = wall/2 + twall There are several options to detail the collector reinforcement:

b = 27 m/2 + 0.2 m = 13.7 m

1. Place bars within the arbitrary width of 13.7 m.

Spreading bars over more than half the diaphragm width is not practical.

2. Use the calculated chord reinforcement in the

This option is acceptable as the required chord and collector calculated reinforcement is approximately equal, 3122 mm2. The collector is wider than the wall, therefore, longitudinal and transverse reinforcement must be provided to transfer forces from the collector into the wall.

in the E-W direction over h/4.

3. Place bars in a 600 mm deep by 300 mm wide edge beam

Collector and chord reinforcement are placed in a beam. In this example, this option is used.

  

Eng.Mohammed Ashraf





The required reinforcement to resist the gravity dead and live load is calculated to be equal to 2375 mm2. Required collector reinforcement is 3122 mm2 calculated above. Therefore, a total of 5497 mm2 must be placed within the beam to resist gravity loads combined with either the calculated chord force or collector

Gravity load calculation is not provided in this example. D + 1.6L 0.2SDS D + 0.5L, which is usually less than the previous case. Therefore, it may be possible to count on a portion of provided gravity reinforcement for seismic collectors. Mu Therefore, the force transferred from the beam to the

Tension and compression forces at both ends of the wall are:

T

C

27,600 mm 6 mm 2

As Required reinforcement:

that the result of the eccentric force between beam and wall centerlines is resisted by the diaphragm. and placed and properly developed at both ends of the wall and extending into the diaphragm a minimum length equal to the development length of the bars. Wall 4: Required collector width is less than the wall thickthick ness. Reinforcement may be placed within the wall width:

As

oTcoll fy

2135 mm 2

As,prov. = 2580 mm2 > As, req’d = 2135 mm2

22.9.4.4

Shear friction reinforcement is not required as the collector force is already developed into the wall. The value of Vn across the assumed shear plane must not exceed the lesser of the following limits: fc Ac fc Ac Ac

OK 11 MPa Therefore, Vn must not exceed: Vn

Wall 5: Required collector width is less than the wall thickness. Reinforcement may be placed within the wall width:

As

oTcoll fy

2275 mm 2

As,prov. = 2580 mm2 > As, req’d = 2275 mm2 Shear friction reinforcement is not required as the collector force is already developed into the wall.

  

Eng.Mohammed Ashraf





Step 16: Shrinkage and temperature reinforcement 12.6.1 Shrinkage and temperature 24.4.3.2 reinforcement: AS+T 24.4.3.3

Ag

Spacing of S+T reinforcement is the lesser of 5h and 450 mm h Controls

2

AS+T

be part of the main reinforcing bars resisting diaphragm in-plane forces and gravity loads. If provided inforcing bars to resist positive moments at midspans and top reinforcing bars to resist negative moments between top and bottom reinforcing bars by providing adequate splice lengths between them.

Step 17: Reinforcement detailing Development Chord and collector reinforcement are extended 25.4.2.2 over full length and width of the edges of the 25.4.10.2 diaphragm. Therefore, development length will be 12.7.3.2 calculated only to determine splice lengths. Development length of shear transfer reinforcement fy d

25.5.2.1

t

2.1

d,

d

e

fc

Splices Because the building lengths are longer than a

mm

Use d, mm

604

650

718

750

420 MPa d

reinforcement, splices will be needed. Use Class B

Say, 1500 mm

d

18.12.7.6

12.7.2.1

25.2.1

The center-to-center spacing of the longitudinal bars for collector and chords at splices and anchorage zones is but not less than 40 mm and concrete db, but not less than 50 mm

3db

Therefore, transverse reinforcement is not required. Reinforcement spacing Chord and collector reinforcement minimum and maximum spacing must satisfy 12.7.3.2 and 12.7.3.3. Section 25.2 requires minimum spacing of dagg. db

18.12.7.6

12.7.2.2

Minimum spacing 29 mm Controls

Collector reinforcement spacing at splice must be at least the larger of: a. At least three longitudinal db b. 40 mm c. cc db, 50 mm] Maximum spacing is the smaller of 5h or 450 mm

Controls

450 mm

Controls

  

Eng.Mohammed Ashraf





Step 18: Detailing

Fig. E3.14—Diaphragm Diaphragm reinforcement detailing.

Section A—Edge beam gravity and collector or chord reinforcement.

  

Eng.Mohammed Ashraf





Section B—Chord/collector reinforcement at south end of diaphragm.

Section C—Collector reinforcement along Walls 4 and 5.

Section D—Collector reinforcement along Walls 2 and 3.

either side of the wall.

  

Eng.Mohammed Ashraf





Step 19: Discussion There is no consensus among engineers on how to distribute the diaphragm inertia force. Based on discussions with several respected engineers, the main approaches are as follows:

building dimension in either direction and perpendicular to the seismic loading, referred to as accidental eccentricity. This recommendation is located in the commentary, therefore, it is not mandatory. This example follows the ASCE 7-10 recommendation. The 5 percent eccentricity is excluded in the analysis. The diaphragm inertia force is uniformly distributed to the diaphragm. The shear forces in the lateral-force-resisting system due to the equivalent lateral force analysis are compared to the forces due to diaphragm inertia forces. The diaphragm is designed for the larger force.

  

Eng.Mohammed Ashraf





direction. Drawing the moment diagram shows a discontinuity at the center of rigidity. Moehle et al. also state in NIST Report No. GCR 10-917-4 that: “For a rectangular diaphragm of uniform mass, a trapezoidal distributed force having the same total force and centroid

sometimes this is ignored or, alternatively, it too can be incorporated in the trapezoidal loading.”

Fig. E3.15—shear and bending moment diagrams.

  

Eng.Mohammed Ashraf





In this example, the moment diagram is constructed by incorporating the shear force in the trapezoidal loading for the construction of the moment diagram. Since Case II resulted in a slightly higher moment, the calculations for that case follow:

Fig. E3.16—Free body diagram. Mx = (625 kN + 992 kN)(x

x2

x](x2

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf



 The column chapter, Chapter 10 in ACI 318M-14, follows the organization of the other member chapters: applicability; initial data; analysis to determine the required strength; design area of reinforcement needed to exceed the required strength; check against minimum reinforcement required; and detailing. The analysis must be consistent with ACI 318M-14, Chapters 4 through 6. A quick review of Sections 10.1 through 10.4 in ACI 318M-14, will remind the designer of limits and rules for columns that should be accounted for in their analysis model. The requirements for column design start in Section 10.5 in ACI 318M-14. A column is always part of the gravity-force-resisting system, and in cast-in-place construction is often part of the lateral design forces are seismic and wind. For buildings designed to resist seismic forces, the requirements of Chapter 10 in ACI 318M-14 apply, along with the additional seismic requirements in Chapter 18 in ACI 318M-14 for columns that are part of an ordinary, intermediate, or special moment frame system. There are also seismic requirements ments are intended to increase column ductility to accom accommodate the large displacements that are expected during a maximum design earthquake. For buildings that resist only wind forces, columns are designed by Chapter 10 in ACI not. There are no additional requirements for wind forces. This Handbook provides some explanation of ACI 318M-14 requirements and how they impact a column’s design and detailing. A review of basic engineering prin princolumns by hand or computer using only this Handbook.  The provisions of Chapter 10 in ACI 318M-14 apply for the design of nonprestressed, prestressed, and composite columns. Headings of a section are considered part of the code and care should be taken to notice when a heading limits the following requirements to a particular type of column. The word “nonprestressed” is typically in reference to castin-place columns and “prestressed” with precast columns. While ACI 318M-14 covers post-tensioned columns, they are not commonly used. Chapter 14 in ACI 318M-14 covers the design of plain concrete pedestals.  9.3.1 General opportunity to create various cross-sectional shapes for the aesthetic purposes. Unusual cross-sections, however, are

for various situations that allow for a simpler analysis. For example, Section 10.3.1.1 in ACI 318M-14 permits an ideal-

Fig. 9.3.1—Permitted cross-section for analysis. The key point is that a using a portion of an oversized column or wall that is easier to analyze is permitted. Section 10.3.1.2 in ACI 318M-14 allows an oversized lower limit of one-half the total area. The analysis and design

Ag based Ag cannot be less than half determined by the building architecture but it must always meet the requirements of ACI 318M-14. sizing—It is not economical to have a unique 9.3.2 Initial sizing—It design for each column in the building. The following guide guidelines help in economical column construction: 1 1.. Reuse formwork as much as possible. It is common to use only three or four columns sizes for the entire building. 2 2.. Use the same strength of concrete for all columns at a level. Structures under six stories or less commonly use one concrete strength for the full height of the

the concrete strength in the columns exceeds 1.4 3. Proportion the column cross sections and concrete strengths so that reinforcement ratios are in the range of 1 to 2 percent. 4. If higher axial strength is needed, increasing the increasing reinforcement area. The analysis and design of columns is an iterative process. To begin, the designer assumes sectional properties in order to run an analysis. An initial column area, Ag, can be estimated by dividing the maximum factored axial load by 0.4fc for ordinary columns or 0.3fc areas. Columns are usually rectangular, square, or round. tive moment of inertia, I , of concrete members is used in the analysis to account for cracking at the nominal condition. The simple I 318M-14 are generally used and the cross-section properties are assumed to be uniform for the length of the member. With these assumptions, an initial analysis is run and, subse-

  

Eng.Mohammed Ashraf





as necessary.  9.4.1 General—The required strength is calculated using the factored load combinations in Chapter 5 and analysis

Fig. 9.4.1b—

permitted as discussed in Chapter 3 of this Handbook. Regardless of the method chosen, all columns must be checked for slenderness. ness ratios, u/r.. Higher slenderness ratios occur for long columns, columns with a small cross-sectional dimension, or columns with limited end restraint. As a column becomes more slender, larger lateral deformations and deformation along the length occur due to the applied load. The column must support additional moment created by the column axial load acting on the deformed column, also known as second-order moments. There are two types of second-order

PPually increase due to this geometric nonlinearity until the column stabilizes. If the slenderness ratio is very high, the column becomes unstable and cannot resist the axial load, 9.4.2 Slenderness concepts—A column’s degree of slenderness is expressed in terms of its slenderness ratio, u/r, where u is unsupported column length; k

u.

cates the plane of the frame in which the stability of column is investigated. k restraint and lateral bracing conditions as shown in Fig. 9.4.2b. The factor k varies between 0.5 and 1.0 for latercolumns. Most columns have end restraints that are neither

and r of a column cross section. The column’s unsupported length u is the clear distance between the underside of the beam, slab, or column capital

ness. Jackson and Moreland alignment charts, given in Fig. R6.2.5 in ACI 318M-14, can be used to determine the factor k A

B

framing conditions and corresponding unsupported lengths x and y u

-

  

Eng.Mohammed Ashraf





and (d), (e), and (f) are for sway frames.

Fig. 9.4.2c—Radius of gyration for circular, square, and rectangular sections. sion dominant members. The reduced moment of inertia values given in Section 6.6.3.1.1 of ACI 318M-14 should be used to determine k. Tables D1.1 through D1.5 in the supplement to this Handbook, ACI Reinforced Concrete Design

It is permissible to use r = 0.3h for square and rectangular sections, and r = 0.25h for circular sections, where “h” is the dimension in the direction stability is being considered. This is shown in Fig. 9.4.2c. 9.4.3 First-order analysis ysis, Section 6.6.4 in ACI 318M-14 provides a moment slenderness. This method is shown in Example 9.1 of this

sectional size and shape to slenderness. A section with a higher moment of inertia per unit area produces a lower slenderness ratio and thus a more stable column. The radius of gyration, r

r

I A

frame is analysed once and results are used for input to the 9.4.3.1 Sway or nonsway frames—The designer needs to determine if the column is in a sway or nonsway frame. bracing, such as structural walls. Structural walls used for

  

Eng.Mohammed Ashraf





elevator shafts, stairwells, partial building enclosures, or and lateral bracing. In many cases, even a few structural walls can brace a multi-story, multi-bay building. Sway

1. Section 6.2.6, ACI 318M-14: Columns are nonsway if

nonlinear, so the principle of superposition does not apply for calculating the second-order moments. The forces for a load combination must be combined before running the analysis. Section 6.7.1.1 in ACI 318M-14 states, “An elastic loads, presence of cracked regions along the length of the k.

columns in that story. This is a simple, conservative hand calculation. nonsway if the increase in column end moments due to order end moments.

P- and P- , refer to R6.6.3.1.1, ness reduction factor is required for P- compared to P- . tion in its second-order analysis. These programs should be

nonsway if the stability index “Q” does not exceed 0.05 as 

Q

Pu Vus

o

0.05

c

combined moment and axial compression. As can be seen,

Pu is total factored axial load acting on all the columns in a story, Vus o is Vus o should be computed using section properties taking into account the presence of cracked regions along the member, refer to Section 6.6.3.1 in ACI 318M-14. 9.4.3.2 Column slenderness method states that for columns in sway or nonsway frames,

or compression controlled behavior. The combination of moment and axial force that result in a column’s nominal strength is traditionally presented by “column interac interaction diagrams.” Interaction diagrams are constructed by computing moment and axial force nominal strengths, as

Pn = Cc + Cs1 + Cs2 – Ts are below the limits given in Section 6.2.5 in ACI 318MM14. If these limits are exceeded in a nonsway frame, then PP6.6.4.5 in ACI 318M-14 need be considered. If the slenderPPPare then used in Section 6.6.4.5 in ACI 318M-14 to calculate P9.4.4 Second-order analysis—Many software analysis Such a program must be capable of the iterative calculations to determine: PPthe length of the column The column needs to be divided into at least two segments

Q&A, Using an Elastic Frame Model for Column Slender-

Mn = Ccx2 + Cs1x1 + Cs2 + Tsx3

compression to pure bending, a nominal strength curve can be created as shown by the outer curve in Fig. R10.4.2.1

for all tension-controlled sections is 0.9. This factor varies linearly from 0.65 or 0.75 to 0.9 through the transition zone shown in Fig. 9.5. The design strength curve is shown by the inner curve in Fig. R10.4.2.1 in ACI 318M-14. The interaction diagram is the plot of both the nominal and design strength curves shown on a graph. An electronic spreadsheet is provided as a supplement to this Handbook to demonstrate how to make an interaction

stress at 0.0fy, reinforcement stress at 0.5fy, and the balanced point where reinforcement stress is at 1.0fy and the concrete reaches it maximum usable strain.

  

Eng.Mohammed Ashraf





Fig. 9.5—Column section analysis. The previous version of this Handbook was a design manual that contained interaction diagrams. These diagrams have been retained in a supplement to this Handbook titled, Reinforced Concrete Design Handbook Design of how the diagrams are made is given in the supplement along with information on biaxial moments. Because shear design in columns is similar to beams, review Beam Chapter 8 of this Handbook for this informa-

development length, and splice lengths, are covered in that many detailing provisions of this chapter are related to how columns are constructed. 9.7.2 Longitudinal bars 9.7.2.1 Spacing—The minimum number of bars in a column is given in Section 10.7.3.1 in ACI 318M-14. Square and rectangular columns must have a minimum of four bars. Circular columns must have three or six depending if the tie is triangular or circular. Eight bars are suggested, however, to ensure that the design moment is achieved

For column torsion, the beam chapter should be reviewed.  The minimum column vertical reinforcement ratio is 0.01Ag. This amount is enough to keep the reinforcement from yielding due to concrete creep under sustained service reinforcement ratio is 0.08Ag. This amount is about the maximum that can be realistically provided at the perimeter of a concrete section that would also meet the minimum cover and spacing requirements. This percentage was set in ACI 318-63 when butt splices were common. For present day construction, lap splices are more common and for many ment is 4 percent, since the maximum percentage of reinforcement at a section is 8 percent. If the lap splices are staggered, the maximum percent of reinforcement can increase up to 6 percent. As stated earlier, usual reinforcing ratios are in the range of 1 to 2 percent. There are cases where more reinforcement is necessary, such as to meet a required strength. If the column design routinely requires reinforcement in the

The minimum bar spacing is given in Section 25.2 in ACI 318M-14. 14.. A maximum spacing requirement in Chapter 10 in ACI 318M318 318M-14,, however, states that for columns in a special moment frame, the spacing of longitudinal bars laterally supported by the corner of a crosstie or hoop leg, hx, shall not that this spacing is further reduced to 200 mm for conditions Typically, bars are evenly spaced around the perimeter, as this helps to create a more stable column cage during construction. Designers will often place bars only at the work necessary to place ties. Bundled bars are often needed that Section 18.7.5.2 in ACI 318M-14 makes this practice impractical for columns in special moment frames due to is necessary or desired, evenly spaced longitudinal bars are ular location, some bars can be added to the evenly spaced column bars. 9.7.2.2 Splicing—Splice locations, lengths, and types

section or higher-strength concrete.  9.7.1 General—Section 10.5 in ACI 318M-14 focuses on the calculation of reinforcement area needed for to resist design forces and moments. Section 10.6 in ACI 318M-14 provides minimum column reinforcement area. Section 10.7 in ACI 318M-14 provides limitations on the location, spacing, and splicing of longitudinal reinforcement and the location, spacing, geometry, and type of transverse reinforcement. General requirements such as concrete cover,

are the most common splice type due to their ease of fabrication and construction. Mechanical connectors and buttwelded splices are helpful where bar arrangement becomes congested but they can require additional erection time. End-bearing splices are not common today but are sometimes used in bundled bar arrangements that are only in compression under all load combinations. They are not permitted in high seismic applications. Columns splices are

  

Eng.Mohammed Ashraf





Fig. 9.7.2.2a—Column Column splice locations (Fanella 2007). 2007 in buildings assigned to SDC D, E, or F are required to be located in the center half of the column length according to Sections 18.7.4.3 and 18.14.3.3 in ACI 318M-14. Bottom and mid-height column splice locations are illustrated in Fig. 9.7.2.2a. For lap splices than are about one-third to one-half the story height, it may be more economical to lap-splice the

eter, concrete strength, bar spacing, concrete cover, position of the bar, distance from other bars, and if the bar is tension

To maintain bars in the corners of rectangular column, into the column above, whether there is change in column bent where columns size does not change. The slope of the

resist the lateral force created by the bend and one of the ties may be part of the regularly spaced ties. Separate splice bars and more ties may be necessary where the column section

Column bar area in the column above must be extended from the column below to lap bars above. Bundled bars are typically groups of larger bars that bars require staggering the individual bars. For this reason, bundled bars preassembly is more complicated and can create additional erection time to place longitudinal and transverse bars on the free-standing cage. 9.7.3 Transverse bars 9.7.3.1 Column ties ments of ties for various numbers of vertical bars are shown in Fig. 9.7.3.1. The one and two-piece tie arrangements shown provide maximum rigidity for column cages preassembled on the site before erection. The spacing of ties depends on the sizes of longitudinal bars, columns, and of ties. The maximum spacing of ties required for shear is shown in Table 10.7.6.5.2 in ACI 318M-14. 9.7.3.2 Spirals—Spirals are used primarily for circular columns, piers, and caissons. Spiral reinforcement can be plain or deformed bars or wire. The term “spiral” used in the code is more than a geometric description of a circular splicing, and termination, which are listed in Section 25.7.3 in ACI 318M-14. A continuously-wound bar or wire not meeting all of the requirements of 25.7.3 is simply a continuous circular tie. The spiral pitch is between 25 and 75 mm, inclusive, and is typically given in 6 mm increments. The spiral size and pitch should meet the volumetric reinforcement ratio, rs

are illustrated in Fig. 9.7.2.2b. Where there is a reduction of reinforcement, longitudinal bars from the column below are diameters to which standard spirals can be formed is given

  

Eng.Mohammed Ashraf





Fig. 9.7.2.2b—Column splice details (ACI 315-99).   Spiral bar diameter, mm

Minimum outside diameter that can be formed, mm

10

225

13

300

16

380

19

750

 1. Determine an initial size of the column and amount of reinforcement. Ag = Pu,max/0.4fc Ag = Pu,max/0.3fc

otherwise, square or round columns are common

2. Run initial analysis to determine column loads. program, go to Step 3

method in Section 6.6.4 in ACI 318M-14 to calcu3. Create a moment-interaction diagram using an electronic spreadsheet or commercial software. Check the required moment and axial load strengths against the design strength curve.

  

Eng.Mohammed Ashraf





Fig. 9.7.3.1—Standard column ties (ACI 315-99).

size and reinforcement. Rerun the analysis, if necessary, until design strength is greater than required strength for all load cases. 5. Check shear strength and minimum shear reinforcement requirements. 6. Detail column longitudinal and transverse requirements showing all bar locations, spacing, splices, and bar terminations. It is common to use typical column details and sections along with a column schedule table.

 American Concrete Institute (ACI) ACI 315-99—Details and Detailing of Concrete Reinforcement ACI 318M-63—Building Code Requirements for Structural Concrete and Commentary ACI 318M-14—Building Code Requirements for Structural Concrete and Commentary ACI SP-17DA-14—Reinforced Concrete Design Handbook Design Aid – Analysis Tables; h t t p s : / / w w w. c o n c r e t e . o rg / s t o r e / p r o d u c t d e t a i l . aspx?ItemID=SP1714DA ACI SP-17DAE-14— Interaction Diagram Excel spreadsheet;

  

Eng.Mohammed Ashraf





h t t p s : / / w w w. c o n c r e t e . o rg / s t o r e / p r o d u c t d e t a i l . aspx?ItemID=SP1714DAE  Fanella, D., 2007, Seismic Detailing of Concrete Buildings Frosch, R., 2011, “Using an Elastic Frame Model for Column Slenderness Calculations,” Concrete International,

“Detailing Concrete Columns,” Concrete International,

  

Eng.Mohammed Ashraf







Columns Example 1: Column analysis

elastic methods using commercial software for comparison purposes. The factored moments are from an analysis of the moment frame along Grid E. Three common controlling load combinations are considered. Given: Materials–– fc = 35 MPa fy = 420 MPa Modulus of elasticity of concrete, Ec = 27,800 MPa

Dead load D L Snow load S

Loading–– Dead + live + snow U = 1.2D 1.2 + 1.6L + 0.5SS 1.2D + 1.6L + 0.5S 0.5 0

Vu Pu Mu Mu

top bot

3860 2712 –4067

Dead + wind + live + snow D + 1.0W + 1.0L 1.0 + 0.5S S 1.2D +1.0L + 0.5S 1.0W 1.2D + 0 6

Dead + EQ + live + snow U* = 1.2D + 1.0E + 1.0L + 0.2S 1.2D +1.0L + 0.2S 1.0E 0 22

0 27,228 73,440

3460 1356 SDS

3447 1356

0 196,594 263,029

= 1.0 for buildings in SDC B.

Reference: SP-17 Supplement, Reinforced Concrete Design Handbook Design Aid – Analysis Tables, found at https://www. concrete.org/store/productdetail.aspx?ItemID=SP1714DA, in empirical units and results converted to metric in this example. ACI 318M-14

Discussion Step 1—Determine initial column size Estimate the maximum load at the interior column, 5.3.1

U = 1.2D + 1.6L + 0.5S Estimate the size of a square column by dividing force by 0.4fc

Calculation

Pu area h

0.4

35 MPa

275,714 mm 2

275,714 mm 2 525 mm

  

Eng.Mohammed Ashraf



6.2.5 6.2.5.1



Complete a rough check on slenderness: r Concrete columns become very slender when u/r exceeds 45. Since this column is likely to be in a sway frame, assume a k = 1.5 and determine a size u/r < 45.

0.3h and

k

u

r

45

u = 5500 mm – 750 mm = 4750 mm Rearranging terms and solve for h

1.5 4750 0.3 45

h

528 mm

It is common for an engineer to choose a large enough column size so the column design is permit- Try 600 mm x 600 mm ted to ignore slenderness. For a frame not braced against sidesway, a u/r limit of 22 would allow the engineer to ignore slenderness. This example, however, will consider slenderness and show the full set of calculations needed for a slender column. Step 2—Sway or nonsway moment frame Fig. R6.2.6 termine column analysis options. This example shows the full extent of the provisions when done by hand. mine if the structure is a sway or nonsway frame. A moment frame that is nonsway greatly reduces the required calculations. The Code provides three options to permit the frame to be considered as nonsway. 6.2.5 There are no bracing elements in the direction of the moment frame. columns in the direction of evaluation. Bracing elements generally means walls but braces are sometimes used. Second order moments are calculated in Step 4. The results of the calculation show that the second order

6.6.4.4.1

Q, in accordance with Section 6.6.4.4.1,, does not exceed 0.05. Q is determined for a single story and

Q

o c

where

Q values.

c

c

Pu Vus

5500 mm

750 mm 2

175 mm 2

5212.5 mm

is the height of the column from center-to-center

To calculate Q, the load case must consider lateral load and also impose the maximum gravity load. Pu is the sum all factored column and wall gravity

Chapter 1 of this Handbook. Vus is the total factored horizontal story shear for

Pu

Vus

from the lateral forces calculated in Chapter 1 of this Handbook.

  

Eng.Mohammed Ashraf





o

an elastic analysis. For this hand calculation ex-

o

to have a hinge at 0.67 c. The following equation to the hinge. Vus 3 EI 6.6.3.1.2

3

cracking. A value of 0.5 is commonly used for all members in an analysis calculated by hand.

Vus 3 EI

3

2

3

c

c

3

3

Ec = 27,800 MPa I = 600 × 6003/12 = 1.08 × 1010 mm4 EI = 37 columns × 0.5 × 27,800 × 1.08 × 1010 2 = 5.55 × 1015 o

= 10 mm

114,319 10 3447 5212.5

Q

0.064

0.05

Therefore, the frame is sway. analysis from software that accounts for the rela-

114,319 16.3 3447 5212.5

Q advantage of a more accurate calculation of lateral Step 3—Check Check to see if slenderness can be neglected Compute the slenderness ratio, u/r. The notation

u

0.104

= 5500 m – 750 mm = 4750 mm

u

is 750 mm deep. k for the bf bw = 450 mm hf = 175 mm h = 750 mm For rectangular sections, I = bh3/12.

y

moment of area area

then use transformation of sections to calculate I.

ybeam

I beam

211.3 mm 3000 1753 12 450 5753 12 3.28 1010 mm 4

2

2

See the plan in Chapter 1 for plan dimensions. h = 175 mm b = 4400 mm Islab = 4400 × 1753/12 = 1.965 × 109 mm4

h = 600 mm b = 600 mm Icol = 600 × 6003/12 = 1.08 × 1010 mm4   

Eng.Mohammed Ashraf



Table

EI values. For this part, a more detailed values for EI are used.



Ec = 27,800 MPa = 0.35 × 27,800 MPa × 3.28 × 1010 mm4 2 = 3.19 × 1014 EcI slab = 0.25 × 27,800 MPa × 1.965 × 109 mm4 2 = 1.37 × 1013 EcI col = 0.70 × 27,800 MPa × 1.08 × 1010 mm4 2 = 2.1 × 1014 EcI

R6.2.5

Factor k -

Joint at top of column: 2.1 1014 2.1 1014 5212.5 4400 A 3.19 1014 3.19 1014 11, 000 11, 000 A

EI/ EI/

Fig. R6.2.5

c col, above

beam, left

EI/

EI/

c col, below]/

beam

= 1.5

Joint at bottom of column:

beam, right]

2.1 1014 3000 B 1.37 1013 5212.5 B = 21.0

Read k from the nomograph.

2.1 1014 5212.5 1.37 1013 5212.5

For a sway frame, k k

6.2.5.1

Determine the radius of gyration, r

4

Ig = 1.08 × 10 mm Ag = 360,000 mm2

Ig

r

6.2.5

10

1.08 1010 360, 000 000

r

Ag

mation of 0.3 times the width of the column in the direction of the frame, which would be 0.3 × 600 = 180 in this case. Check to see if slenderness can be neglected for sway frame. Slenderness for a sway frame can be neglected if u/r

173

For a sway frame, 2.2 4750 173

60

22

Slenderness cannot be neglected. that must be met: k

u

r

34 12

k u M1 and M2 r

k

40

u

r M1 and M2.

24.7

0.9 4750 173

24.7

34 12

m

42.9 and 24.7

therefore, slenderness could be neglected if this condition occurred.

  

Eng.Mohammed Ashraf

40





P , sway For a sway frame, the secondary moments at the end ends of column must be calculated before calculating moments are often called “P “P 6.6.4.6.1

Mns

Ms

1.2D +1.0L + 0.2S

1.0E

M1

1356

196,594

M2

–2712

263,027

s

The Code provides a conservative method to esti-

M1 = M1ns M2 = M2ns

sM1s sM2s

M1ns M1s tor s. 6.6.4.6.2 determined one of two ways: Eq. s

1 1 Q

1

s

1 1 0.064

1.07

This expression is commonly used if software

Eq.

1 s

1

1

Pu 0.75

Pc

This expression is commonly used in hand calculations.

1 s

1

Pu 0.75

Pc

Pu

software that uses a second-order analysis.

  

Eng.Mohammed Ashraf



6.6.4.4.2 6.6.4.4.4

Calculate the critical load Pcr EI may be calculated from one of three equations

Pcr

reinforcement is not known at this point of the design. Key points about EI are

R6.6.4.6.2

EI

ds dns and is 0.0 for short term lateral loads Ise Table D.4.5 in the supplement to this Handbook, see reference at the start of this example.





2

EI

eff 2

k

u

0.4 Ec I g eff

1

ds

0.4 27,800 MPa 1.08 1010 mm 4 1 0 2 = 1.2 × 1014

2

k u

= 4750 mm 2

14 6

Pcr

2

1 s

1.61

1

bounds for stable results, and the engineer should consider increasing the column size. Discussion: o

Q becomes 0.104. Thus, the revised s, for Eq.

s

1 1 0.104

1.12

Q method. A few helpful suggestions from the reference are Q values between 0.05 and 0.2, the error in second-order moments will be less than 5 percent /H o factored loads Pu Pcr Pu Pcr Gregor and Hage; thus, the second order-moments calculated by the Q method should be within 5 percent of o

software-based Q Calculate the design moments M1 and M2

Pu Pcr < 0.2. sM1s = 1.12 × sM2s = 1.12 × M1 = M1ns M2 = M2ns

sM1s sM2s

  

Eng.Mohammed Ashraf



6.2.6



Check to see if the second-order moments exceed

First order M1 is OK. First order M2 OK.

P nonsway 6.6.4.6.4

6.6.4.5.1

column must be calculated for a sway or nonsway frame where slenderness cannot be neglected. The M1 and M2, from Section 6.6.4.6 are used in Section 6.6.4.5. The required moment for design is calculated by multiplying the larger end moment M2 by the Mc

M2 is calculated by Pu

6.6.4.5.2 Cm Pu 1 0.75 Pc

6.6.4.5.3

Cm

6.6.4.5.4

Check M2,min: M2,min = Pu

R6.6.4.5.3

6.6.4.4.2 R6.6.4.4.4

1.0

0.6 0.4

M1 M2

M1 M2

M2,min ,min

×

h

M1/M2, has been updated to follow the right hand rule; thus, the sign convention has been changed in ACI 318M-14.

Cm

+ M1/M2 is double curvature –M1/M2 is single curvature The critical buckling load was calculated in Step 4, but ds dns. It shall now be calculated using dns. The commentary states that dns may be assumed to be 0.6. Another common way of calculating dns is to divide the dead load by transient gravity loads for a given load combina-

Pcr

dns

Controls

1.2 D 1.2 D 1.0 L 0.2 S 1.2 2310 1.2 2310 1.0 672 0.2 48 0.8

EI

0.6 0.4



2

k

EI

221,541 297,304

0.30

eff 2

u

0.4 Ec I g eff

1

dns

0.4 27,800 MPa 1.08 1010 mm 4 1 0.8 2 × 1013

2

k u

= 4750 mm 2

13

2 6

Pcr

2

k

  

Eng.Mohammed Ashraf





0.3 3447 1 0.75 6030

the length of the column.

Mc

1.26

M2

Step 6—Summary and discussion This example calculates the second-order moment for one column, in one-direction, for one load case. It is easy to see that this is a time-consuming process. The following table shows a comparison between the moment and axial loads calculated

Pu Mu

Hand

Hand

Computer

Computer

3447

3447

3640

3640

265,741

315,793

245,688

271,277

The second-order moment increase for the computer is 10 percent which is lower than the 18 percent calculated by the Q o computed by a percent would have been required which is more than the 40 percent allowed. Thus, a larger column would have been selected as suggested in the discussion in Step 4.

  

Eng.Mohammed Ashraf





Column Example 2: Column for an ordinary moment frame from the example building given in Chapter 1 of this Handbook. The column is part of an ordinary moment frame. Example commercial software capable of second-order elastic analysis. Given: Materials–– fy = 420 MPa Modulus of elasticity of steel, Es = 200,000 MPa fc Modulus of elasticity of concrete, Ec = 27,800 MPa

Loading–– Pu 3960

U = 1.2D + 1.6L + 0.5S U = 1.2D + 1.0W + 1.0L + 0.5S U* = 1.2D + 1.0E + 1.0L + 0.2S U* = 0.9D + 1.0E + 1.0L + 0.2S

3560 3640 2160

Mu

Vu 0

0

73,550 271,300 271,300

22 80 80

Reference: SP-17 Supplement, Interaction Diagram Excel spreadsheet found at https://www.concrete.org/store/productdetail. aspx?ItemID=SP1714DAE, in empirical units and results converted to metric in this example. Calculation Discussion Find the required area of longitudinal reinforcement Step 1—Find The Code references Section 22.4 for the calculation of Pn and Mn. Section 22.4 provides an equation to calculate P0 and references Section 22.2 for strain limits. ACI 318M-14

The interaction of Pn and Mn is evaluated by making an interaction diagram. A tutorial spreadsheet is pro provided with this manual that demonstrates how to make a diagram for a given section and reinforcement ratio This example used the spreadsheet to create an interactive diagram. The generated interaction diagram shows if the assumed longitudinal reinforcement is satisfactory. The section properties and geometry were determined in the analysis from Column Example 1. The next step is to assume a quantity, size, and location of the longitudinal reinforcement. The design of columns is in which case, the analysis will need to be redone. The spreadsheet analyzes rectangular or square The designer inputs the number of steel layers in the column cross section. The spreadsheet places

10.6.1.1

The remaining layers are evenly spaced between the outer layers. The spacing of the bars within each layer is similar to the placement of the layers. The permitted and the remaining bars are evenly spaced between the outer bars.

  

Eng.Mohammed Ashraf



10.7.3.1

The design moments are not very large, so try the minimum area of reinforcement and assume a uniform distribution of bars around the perimeter.



As,min = 0.01Ag = 0.01 × 6002 = 3600 mm2

and greater, to make stable column cages for erection Area bar needed = 3600 mm2/8 = 450 mm2 and to reduce the number of ties at a section.

22.2.2.4.3

2

At least four bars are needed in a rectangular column. The spreadsheet does a sectional strength analysis using an equivalent rectangular stress distribution

Section properties and geometry 3

fc’ which is automatically calculated and displayed for the user’s information.

fc’ =

35

MPa

0.80

25.7.2.2

provide good initial shear strength and are rigid enough to provide column cage stability during erection. Concrete cover protects the reinforcement from

20.6.1.3.1

-

b=

600

mm

h=

600

mm

fy =

420

MPa

Es =

200,000

MPa

Tie bar size:

13

Clear cover to tie =

40

mum cover is provided in Section 20.6. The spreadsheet calculates the distances to the layers of reinforcement, which is needed for later calculations, by using the tie bar size, cover to tie, and longitudinal bar size.

mm

25

Asi, mm2

di, mm 3

65

3

1530

2

300

2

1020

1

535

3

1530 4080

25.2.3

The minimum bar spacing is calculated and displayed for the user’s information.

Bar spacing checks d1 =

535

mm

240

mm

240

mm

40

mm

Key variables needed for design are displayed for 21.2 change as the column interaction diagram transitions from compression-controlled sections to tensioncontrolled sections.

fy/E= cu

0.00207

=

0.003

Ductile strain =

–0.005

Brittle strain =

–0.002

tension-controlled

=

compression-controlled

0.9 =

0.65

The interaction diagram is made by calculating the Pn and Mn for incremental changes in the net tensile strain in the extreme layer of longitudinal tensile reinforcement at nominal strength, t. A random strain of 0.0007 was chosen to illustrate the calculation of Pn and Mn in the following steps. Find Pn and Mn for t = 0.0007 t from pure compression, t equal to fy/Es at all layers, to pure tension, –fy/Es at all layers.

  

Eng.Mohammed Ashraf





2. Calculate c, the distance from the extreme comt by using similar triangles. d1

cu

c t

c

0.003 535 0.0007 0.003

a

min 

cu

698 mm

3. Calculate a, the depth of the equivalent stress block. a

min 

1

c

h

h

0.80 698 mm 600 mm

558 mm

Use a = 558 mm 4. Calculate Cc, the force at concrete compression block. Cc = 0.85 × a × b × fc 5. Calculate si, the strain at each layer of bars.

if c

if c

d i then

d i then

si

si

 min   

 max   

Cc s1

c di cu

c

s1

0.003

s2

0. 0 003 0.003

s3

0.003

fy Es c di cu

c fy

fy

Es

Es

420 200, 000

If

si

> 0, then Fsi =

si

× Asi × Es – 0.85 × Asi × fc

c

di

698 698

698

0.0007 0.00171

0.00272 and

0.0021 s2

s3

= 0.0021

Fss11 = 0.0007 × 1530 × 200,000 – 0.85 × 1530 × 35 Fs2 = 0.00171× 1530 × 200,000 – 0.85 × 1530 × 35 = 478 Fs3 = 0.0021 × 1530 × 200,000 – 0.85 × 1530 × 35 = 597

displaced concrete. Fsi = If si 7. Calculate Po. Fsi P o = Cc

s3,

= 0.0017 0.0017, and For Fs1, Fs22, and Fss33, s s1

6. Calculate Fsi, the force at each layer of bars.

s2

si

× Asi × Es Po

Pn,max is not applied in this spreadsheet until the design strength curve is calculated. 8. Calculate di, the moment arm to forces Cc and each Fsi. d Cc

di

h 2

a 2

h di 2

600 558 21 mm 2 2 d1 = 300 – 535 = –235 mm d2 = 300 – 300 = 0 mm d3 = 300 – 65 = 235 mm dCc

  

Eng.Mohammed Ashraf





9. Calculate Mi, the moment for each force about the center of the section. MCc = dCc × Cc

M Cc

21

Mi = di × Fsi

M1

235

9960 1000 168.7 1000

478 1000

M2

0

M3

235

209 kN·m

597 1000

39.6 kN·m

0 kN·m

140.3 kN·m

10. Calculate Mn.

21.2

22.4.2.1

Mn = MCc Mi The spreadsheet changes the strain in small increments to create a smooth plot of a nominal strength interaction diagram. The values are then multiplied by and the limits on axial strength are applied to create the design strength interaction diagram. The for compression-controlled sections is

Mn = 209 – 39.6 + 0 + 140.3 = 310 kN·m Design Strength Interaction Diagram:

sections is 0.9.. This factor varies linearly from 0.65 to 0.9 through the transition zone shown which crecre ates the sharp change in the lower part of the curve. The diagram on the right is for this column example and it has the loads from the four given load combinations plotted. Note that points for load combinations (i), (ii), and (iii) are to the left of the “Stress = 0 fy” Line, meaning all the bars remain in compression. Point iv) is for a load combination with lighter gravity loads. It is just to the right of the line, meaning a few bars will be in tension. Thus, a tension splice is required.

Use eight No. 25 bars evenly spaced around the perimeter.

Another line could be drawn on the diagram showing where the tensile bar stress is 0.5fy. That line delineates whether a Class A or B splice is 10.7.5.2.2 required. For this example, all the bars are going to be spliced at one location so a Class B splice is always required. Step 2—Find the required area and geometry of transverse reinforcement The Code references Section 22.5 for the calculation Load Combination (iii): of Vn. The Pu given in Load Combination (iii) has Vu = 80 kN 22.5 the expression 0.2SDS is applied in the downward Pu = 3280 kN positon as required by ASCE 7-10. The Pu value SDS in the upward position for the lowest axial load associated with this load combination.

18.3.3

Note that u (4750 mm) is greater than 5c1 (3000 mm); therefore, the additional shear requirement for ordinary moment frame columns does not apply.   

Eng.Mohammed Ashraf



Eq.



The shear force resisted by the column is minimal the compression load on shear resistance. Use Eq. Vc. Vc

10.6.2

0.17

f c bw d

535 1000

Vc Vc

Minimum shear reinforcement is required when Vu Vc

Check for minimum shear reinforcement

21.2 Column ties are not necessary for shear resistance. 25.7.2 10.7.6.2

Ties are not needed for shear but they are necessary for lateral support of longitudinal bars. Thus, tie requirements must meet the geometry requirements of 25.7.2 and location requirements of 10.7.6.2 -

25.7.2.2 was chosen as discussed in Step 1. 25.7.2.1

dagg; however, the maximum tie requirements controls for this example. The maximum spacing shall not exceed the least of: db of the longitudinal bar db of the transverse bar h or b

× 25.4 mm = 406.4 mm Controls. × 12.7 mm = 609.6 mm

Use s 25.7.2.3

Fig. R25.7.2.3a

Section 25.7.2.3 requires that lateral support from ties is provided for bars at every column corner and also bars with greater than 150 mm clear on each side. Thus, every vertical bar needs lateral support in this example Ties with 90-degree standard hooks are acceptable for this case. A diamond-shaped tie is used to support bars along the sides. This is desirable because it provides a strong column cage for erection; however, it becomes a fabrication problem when the column is rectangular and this tie becomes oblong. It is common to use alternative tie geometry, such as cross ties, for columns that are not square.

15.3 building uses the same fc’ are necessary. 15.2.2, 15.2.3, 15.2.4, 15.2.5, 15.4.1

moment is being transferred to the column. Section 15.4 does not require minimum shear reinforcement

This column is part of an ordinary moment frame; therefore, it is part of the seismic-force-resisting system.

  

Eng.Mohammed Ashraf



15.4.2

Section 15.4.2 applies which requires that miniregion even if minimum shear reinforcement is not required in the column. Av shall be at least the greater of: 0.062 f c

0.35

15.4.2.1 15.4.2.2

ACI 352R



bs fy

At each tie, Av = 129 mm2 × 4 legs = 516 mm2 Check s = 400 mm 35 600 400 420

0.062

bs fy

0.35 600 400 420

The maximum spacing is one-half the depth of the shallowest beam or slab framing into the element and shall be distributed along the column for the full depth of the deepest member.

750 mm 2

s

210 mm 2

200 mm 2

516 mm 2

516 mm 2

375 mm Controls

at a spacing not greater than 375 mm Section 15.2.2 of the Code states that, “the shear resulting from moment transfer shall be considered in the

Section R18.18.4.. The shear is so low for this example that a check on shear is not necessary. See Example 4 for Design of Beam-Column Connections in Monolithic Reinforced Concrete Structures.”

Since all the splices are at one location, a Class B tension lap length is provided and checked against the compression lap length as a minimum. st

is the greater of: d

25.5.2 10.7.5.2.2 where,

d

25.4.2

1 1.1

fy

t

fc

cb

e

db and K tr

s

K tr

db

Determine st t= e= s db = 25.4 mm cb = 40 + 12.7 + 25.4/2 = 65 mm n = 3, number of longitudinal bars along the splitting plane s = 400 mm, spacing of ties Atr = 4 tie legs × 129 mm2 = 516 mm2 K tr cb

40 516 400 3 K tr db

40 Atr sn d

st

17.2 65 17.2 25.4

1 420 1.0 25.4 1.1 1.0 35 2.5

3.24

2.5

656 mm

= 1.3 × 656 mm = 853 mm

  

Eng.Mohammed Ashraf





For fy equal to 420 MPa, fy db = 30db

sc

is the greater of:

25.5.5.1 10.7.5.2.1

Check compression lap splice length. sc

= 30 × 25.4 = 762 mm

For ease of construction, use st = 800 mm for all A common way of expressing this splice on the structural drawings is to make a lap splice table and splices and splice at every level. splice at every other story to save time on labor in

10.7.4.1 10.7.6.4

column ties at bend. Ties need to resist 1.5 times the horizontal tension component of the computed horizontal component was determined for the bar

fyAst Horizontal tension at the bend: Pu

fyAst

The current tie detail meets this requirement. 10.7.6.2 to be within, s/2,, of the top of the slab. It is good construction practice to start the tie at 50 or 75 mm

15.4.1 crete on all four sides. In this case, only two beams

requires the last tie to be within s/2 of the lowest Again, it is good construction practice to place a tie

spacing. The tie requirements are not additive in this case. Step 4—Discussion and summary

in formwork may not overcome the cost of the small amount of concrete that is saved. For this building, functionality and expense.

  

Eng.Mohammed Ashraf





Column Example 3: Column for an intermediate moment frame (IMF) The column is part of an IMF. results of an analysis from commercial software capable of second-order elastic analysis and for Seismic Design Category C. Given: Materials–– fy = 420 MPa Modulus of elasticity of steel, Es = 200,000 MPa fc Modulus of elasticity of concrete, Ec = 27,800 MPa

Loading–– U = 1.2D + 1.6L + 0.5S U = 1.2D + 1.0W + 1.0L + 0.5S U* = 1.2D + 1.0E + 1.0L + 0.2S U D + 1.0E + 1.0L + 0.2S

Pu 3960 3560 3770 2030

Mu 0 73,553 364,715 364,715

Vu 0 22 111 111

Reference: SP-17 Supplement, Interaction Diagram Excel spreadsheet found at https://www.concrete.org/store/productdetail. aspx?ItemID=SP1714DAE, in empirical units and results converted to metric in this example. ACI 318M-14

Discussion

Calculation

This example demonstrates the column design requirements for an IMF. The design requirements are o,

-

R base shear. An IMF is permitted to be used for structures assigned to SDC B and required for SDC C. For the column designed in Examples 1 and 2, the shear is so low that it is not economical to require

Step 2—Find the required area of longitudinal reinforcement The same column from Example 2 is checked using the interaction diagram spreadsheet referenced at the start of this example. See Example 2 for a detailed discussion on the calculation of nominal and design interaction diagrams.

Design Capacity Interaction Diagram:

The diagram on the right is for this column example and it has the loads from the four given load combinations plotted.

-

  

Eng.Mohammed Ashraf





Step 3—Find the required area of transverse reinforcement 22.5

but it is still very low. The concrete design strength

10.6.2 18.4.3.1

load. For columns in IMFs, there are additional shear requirements. Vn shall be at least the lesser of: Shear calculated for reverse curvature from the maximum Mn over the range of factored axial

Vc 2

Vu

2

The range of nominal moments is shown in the fol-

Mn Design of Reinforced Concrete Special Moment

In this example, the range of Mn does not contain The interaction diagram spreadsheet allows the user the balance point of the curve. Thus, the maximum to input an axial load value and it will calculate the value for Mn is at the load combination with the greatest Mn. For Pu Mn For Pu

Mn

Use, Mn Vu

M nt

M nb u

Shear with the overstrength factor

o

applied.

2 1, 059, 000 4750

o

Vu

Controls

  

Eng.Mohammed Ashraf



22.5.10.5.3

Shear reinforcement is required. The equation for shear reinforcement is

10.7.6.5.2

Vu

NG

2

s

When shear reinforcement is required, there is a limit on spacing. The maximum spacing is the lesser of d/2 or 600 mm for Vs

Vn From Example 2

Av f yt d

Vs



f c bw d

Av = 4 legs × 129 = 516 mm2 d = 600 – 65 = 535 mm fyt = 420 MPa Assume maximum spacing of d 2

smax

535 2

267.5; use s = 250 mm

Calculate Vs Vs

516 420 535 smax assumption is OK.

Vn

22.5.1.1 10.6.2.2

Check the Av,min. Av,min shall be at least the greater of: 0.062 f c

0.35

bs fy

bs fy

Vs + Vc

OK Av = 4 legs × 129 = 516 mm2 0.062

35 600 250 420

0.35 600 250 420

131 mm 2

125 mm 2

516 mm 2 OK

516 mm 2 OK

Find the required geometry and spacing of transverse reinforcement Step 4—Find The tie geometry from Example 2 is acceptable for Typical section along o. 18.4.3.3 all locations except in the plastic hinge region, o. Section 18.4.3.3 requires that hoops are used in the plastic hinge region, o.

25.7.4

Hoop geometry is similar to ties but the tie must be closed with seismic hooks at each end. Hoop is that the bend must not be less than a standard 135-degree hook.

18.4.3.3

Another common tie arrangement is one exterior hoop with one cross tie for the middle bars in each direction. The plastic hinge o is the greatest of u

Find

o

× 4750 = 792 mm

Controls.

h or b

  

Eng.Mohammed Ashraf



18.4.3.3

18.4.3.5

15.2.4 18.4.4



The maximum spacing, so, along the plastic hinge region is the smallest of db, smallest longitudinal bar db, hoop h or 0.5b

Find the maximum spacing for so × 25 = 200 mm Controls. × 12.7 = 304.8 mm × 600 = 300 mm

Thus, the transverse reinforcement is o the column transverse reinforcement is provided as shown in Example 2 but with a 250 mm o and spacing for shear. o

Section 18.4.4 refers the engineer to the Chapter 15 so there are no additional requirements according to the Code for an IMF. The shear required for design is larger for the IMF. As discussed in Step 3 of Example 2, the provisions

18.4.2

a Type 1 connection in ACI 352R. Calculate column shear force Vu associated with the nominal moments and related shears calculated in accordance with Section 18.4.2. The column shear can be approximated with the following the T-beam stem. column midheights:

M nl

M nr

Vu,col

Vul Vur

h 2

Mnl Mnr Vul Vur

Where is the distance between the mid-height of

Also,

is unconservative to ignore the slab for this check; bf, of the T-Beam in the calculations.

5500 mm 4500 mm 2 h = 600 mm

600 2 

 Vu , col

5000 mm

5000

24

T1 T2 Vj = T2 + C1 – Vu C1 = T1 C2 = T2 Vj

  

Eng.Mohammed Ashraf



18.8.4.3

Aj, is calculated by the multiplying the column depth, h tive width which is the lesser of:



bw,beam = 450 mm hcol = bcol = 600 mm 600 mm 450 mm 2

x

bw,beam + hcolumn bw,beam + 2x bcolumn

Fig. R18.8.4

600 mm

75 mm

Controls

where x is the smaller distance between the edge of the beam and edge of the column.

Aj = 600 × 600 = 360,000 mm2

18.8.4.1 provides Vn for several conditions. This

Vn

18.8.4.1 21.2.4.3

1.2 1.0

35 360, 000 1000

Vn Vn

1.2

f c Aj

higher shear strength is permitted. Vn

1.7

f c Aj

18.4.3.4 within so

so next level. The splice is permitted to start at the top of the slab for an IMF, which is desired for ease of construction.

Step 7—Discussion and summary This example is an extension of Example 2 but the seismic loads where increased for a Seismic Design requirement to use hoops only resulted in a 135-degree hooks instead of 90-degree hooks. The column size and longitudinal reinforcement remained the same.

  

Eng.Mohammed Ashraf





Column Example 4: Column for a special moment frame (SMF) The column is part of an SMF. results of an analysis from commercial software capable of second-order elastic analysis and for Seismic Design Category D. Given: Materials–– fy = 420 MPa Modulus of elasticity of steel, Es = 200,000 MPa fc Modulus of elasticity of concrete, Ec = 27,800 MPa

Loading–– Pu 3960 3560 3880 1920

U = 1.2D + 1.6L + 0.5S U = 1.2D + 1.0W + 1.0L + 0.5S U = 1.2D + 1.0E + 1.0L + 0.2S U= 0.9D + 1.0E + 1.0L + 0.2S

Mu 0 73,553 507,415 507,415

Vu 0 22 151 151

Reference: SP-17 Supplement, Interaction Diagram Excel spreadsheet found at https://www.concrete.org/store/productdetail. aspx?ItemID=SP1714DAE, in empirical units and results converted to metric in this example. Calculation

Discussion

ACI 318M-14

This example demonstrates the column design requirements for an SMF. The additional design and detailing requirements for a SMF increases the

The following properties are from the intermediate information.

an IMF to 8 for an SMF. A SMF is permitted to be Column: h = b = 600 mm used for structures assigned to SDC B and C and required for SDC D, E, and F. For this example, the example building is analyzed for a region assigned are revised. This example starts with the column designed in Example 3.

Beam: h = 750 mm bf = 3000 mm bw = 450 mm

Step 2—Check dimensional limits and axial load 18.7.2.1 The column cross section shall satisfy the following:

b/h

h

b

h

b/h

OK

OK

  

Eng.Mohammed Ashraf



18.8.2.3



The largest longitudinal reinforcement in the beam prevent slip of the longitudinal bar from the beam. to the beam must be 20 times the largest longitudinal reinforcement in the beam.

20db = 20 × 22.2 = 444 mm h

18.8.2.4

hbeam 2 The arrangement of the longitudinal reinforcement

Pu > 0.3Agfc fc The value hx shall not exceed 200 mm if this occurs; otherwise, it shall not exceed 350 mm hx for the cross section in Examples 2 and 3 is

OK 750 2

375 mm

hcolumn Pu Ag = 360,000 mm2

OK

0.3 360, 000 35 1000 35 MPa is not greater than 70 MPa Therefore, hx shall not exceed 200 mm The area of steel in the current column is Ast = 8 × 510 mm2 = 4080 mm2

235 mm > 200 mm

Ast = 12 × 387 mm2 = 4644 mm2 Thus, either the cross-section can be increased or the hx reinforcement should be rearranged to satisfy this requirement.

This column is at a lower level and the axial load is slightly greater than 0.3Agfc columns will not exceed this limit. It is therefore recommended to rearrange the reinforcement. The reinforcement for the upper levels can switch back to the previous cross-section if it is found to be more economical. Step 3—Find the required area of longitudinal reinforcement Using the Interaction Diagram spreadsheet referenced at the start of this Example, the revised column section is analyzed for the four load combinations given for this example.

around the perimeter.

Design Capacity Interaction Diagram:

18.7.4.1

For SMFs, the maximum amount of longitudinal reinforcement, Ast, is reduced to 0.06Ag. The minimum amount of Ast stays the same at 0.01Ag.

0.01Ag = 0.01 × 600 × 600 = 3600 mm2 0.06Ag = 0.06 × 600 × 600 = 21,600 mm2 Ast = 12 × 387 mm2 = 4644 mm2

OK

  

Eng.Mohammed Ashraf





18.7.3.2 satisfy:

M nc 18.6.3.2

6 5

M nb

The beam design is not part of this example. The

From Example 3, the nominal moments in the beam are: Mnl Mnr

so that the positive moment at least one-half the nega-

and IMF does not meet all the requirements for a

Mn is the minimum nominal moment in the range of the interaction curve related to the minimum and maximum axial loads for the seismic load combinations. For more information on how to calculate Mn, refer to Step 2 in Example 3.

The Interaction Diagram Spreadsheet allows the cal user to input an axial load value and it will calculate the nominal moment, see the “Select Axial

Mnl Mnr Column:

For Pu For Pu

Mn Mn

Use the lowest value, Mn 6 5 OK Determine the geometry of the transverse reinforcement Step 4—Determine 18.7.4.3 Hoops for rectangular columns or spirals for 18.7.5.1 circular columns are required for the entire column 18.7.5.5 height. The hoops at the plastic hinge, o, and the splice must meet geometry requirements of Section 18.7.5.2 18.7.5.2. There are six conditions that geometry of the section must satisfy. The cross-section shown on the right meets these conditions.

  

Eng.Mohammed Ashraf





Note that hoops are closed ties with standard hooks at the end that are at least 135 degrees. Crossties are permitted to support the longitudinal bars between the corners. Where crossties are used they shall be alternated end for end along the longitudinal bar. For this example, every bar needed support. Since there are an even number of bars, overlapping hoops provide the least number of pieces that still provide the

Notice that Section 18.7.5.2(f) was checked early in this example. The maximum longitudinal spacing requirements of Section 18.7.5.2(e) and (f) can impact the design of the column as it did in this example. Step 5—Determine the maximum spacing of the transverse reinforcement 18.7.4.3 The maximum spacing, so, along the plastic hinge 18.7.5.1 length, o, and splice regions is the smallest of: Maximum spacing for so (a) 0.25 × 600 = 150 mm 18.7.5.3 (a) 0.25h or 0.25b (b) 6db, smallest longitudinal bar (b) 6 × 22.2 = 133.2 mm Controls 350 hx 3

(c) 100

(c) 100

350 157 3

164.3 mm

Also, so shall not exceed 150 mm and need not be taken less than 100 mm 18.7.5.5

The maximum spacing, s, between o and the column splice regions is the smallest of: (a) 6db, smallest longitudinal bar (b) 150 mm

Maximum spacing for s (a) 150 mm (b) 6 × 22.2 = 133.2 mm Controls. Use s = 125 mm along the column height unless noted otherwise from the following checks. Check the minimum amount of transverse reinforcement required along the plastic hinge length, o Step 6—Check 18.7.5.4 Since Pu is greater than 0.3Agfc Ash = 129 mm2 × 4 = 516 mm2 Ash/sobc shall be greatest of: bc = 600 – (2 × 40) = 520 mm Ag = 600 × 600 = 360,000 mm2 Ach = 5202 = 270,400 mm2 kf kn

Ag

(a) 0.3

1

Ach

(b) 0.09

fc f yt

fc f yt

(c) 0.2k f kn

35 0.6 0.8 1.0; use 1.0 175 12 1.2 12 2

(a) 0.3

360,000 35 1 270, 400 420

(b) 0.09 35 420 Pu f yt Ach

0.0083

0.0075

(c) 0.2 1.0 1.2

3880 420 270, 400

0.0082 Controls

where kf

kn

fc 25, 000

0.6 1.0 and

nl nl

2

nl is the number of longitudinal bars around the perimeter of the column core that are laterally supported by the corner of hoops or seismic hooks.   

Eng.Mohammed Ashraf





Find the required maximum spacing for this amount of reinforcement Ash so bc so

0.0082 or Ash 0.0082bc

516 0.0082 520

121 mm

Use s = 120 mm along o Step 7—Check the minimum amount of transverse reinforcement required for shear 18.7.6.1.1 Section 18.7.6.1.1 states: “The design shear force Ve shall be calculated from considering the maximum forces that can be generated at Mpr, at each end of the column associated with the range of factored axial forces, Pu on Mpr analysis of the structure.”

Ve be less than the factored shear calculated by

Using the Interaction Diagram Spreadsheet, generate to how Mn is calculated in Step 3 above, except that the curve for fy = 520 MPa. 1.25fy is used to generate the interaction diagram

Special Moment Frames: A Guide for Practicing moment is called the probable moment, Mpr.

Mpr. For Pu For Pu

Mpr Mpr

Use Mpr Ve

M prt

M prb u

2 1,161,900 4750

  

Eng.Mohammed Ashraf



The second part of Section 18.7.6.1.1 is similar to



The probable Mpr1 Mpr2 Ve1 Ve2

in Step 3 of Example 3. Calculate column shear force Ve associated with the probable moments and related shears of the beam. The column shear can be approximated with the following equation

- Also, heights:

5500 mm + 4500 mm 2 h = 600 mm

 M pr1 M pr 2 Ve,col

Ve1 Ve 2

h 2

5000 mm

600 2 

 Ve,col

5000

where is the distance between the mid-height of

22.5.10.5.3

is unconservative to ignore the slab for this check; bf, of the T-beam in the calculations. The last part of Section 18.7.6.1.1 states that Ve cannot be less than Vu from the analysis. calThe requirement permits the use of the shear cal culated for the second part of Section 18.7.6.1.1, 18.7.6.1.1 but try to develop the shear related to Mpr of the column. Determine the maximum spacing of hoops required for shear

The maximum Vu from the load combinations Ve Use Ve

From Example 2: Vc

2 2

Av = 4 legs × 129 = 516 mm d = 600 – 62 = 538 mm 22.5.1.1

The equation for shear reinforcement is

Vs Vn

Av f yt d

s = 125 mm from Step 5

Vs

s Vs + Vc

Ve

516 420 538 125

Vn

Ve

Step 8—Summarize the amount and spacing of transverse reinforcements along the height of the column 18.7.5.1 The plastic hinge length, o, is the greatest of: Find o h Controls u

From Step 6, s = 120 mm o is starts at the top of the slab and bottom of the beam and extends toward the middle of the column. Along o 18.7.4.3 of the column. The hoop spacing along the splice length must meet Sections18.7.5.2 and 18.7.5.3 but not 18.7.5.4. In the calculation of hoop spacing for o

so small in this case that the same spacing used.   

Eng.Mohammed Ashraf





st

is the greater of:

Determine st t= e= s db = 22.2 mm cb = 40 + 12.9 + 22.2/2 = 64 mm n = 4, number of bars along the splitting plane s = 120 mm, spacing of ties Atr = 4 tie legs × 129 mm2 = 516 mm2

d

25.5.2 10.7.5.2.2 where

d

1 1.1

fy

t

fc

cb

e

s

K tr

db

K tr

db cb

40 516 120 4 Ktr db

43 64 43 22.2

4.82

2.5

and K tr

40 Atr sn

1 420 1.0 22.2 1.1 1.0 35 2.5

d

573 mm

= 1.3 × 573 = 745 mm Check compression lap splice length. st

For fy equal to 420 MPa, fy db = 30db

sc

is the greater of:

25.5.5.1

sc

10.7.5.2.1

= 30 × 22.2 = 666 mm

For ease of construction, use st = 720 mm for all splices and splice at mid-height at every level. The end result is hoop spacing of 120 mm at o regions and along the splice length. The remainder of the column has a spacing of 130 mm

18.7.4.3 18.2.7 18.2.8

18.8 18.8.2.1

permitted. Type 1 mechanical splices and welded splices are not permitted within a distance equal to twice the column depth from the bottom of beam or top of slab. Type 2 mechanical splices do not have a location restriction for cast-in-place SMFs.

Ve,col in Step 7. The column shear is calculated from the tensile reinforcement is 1.25fy.

  

Eng.Mohammed Ashraf



Vj = T2 + C1 – Vu



T1 T2 C1 = T1 C2 = T2 Vj

18.8.4.3

Fig. R18.8.4

Aj, is calculated by the multiplying the column depth h width which is the lesser of: bw,beam + hcolumn bw,beam + 2x bcolumn

bw,beam = 450 mm hcol = bcol = 600 mm x

600 mm – 450 mm 2

75 mm

where x is the smaller distance between the edge of the beam and edge of the column. Controls Aj = 600 × 600 = 360,000 mm2 18.8.4.1 18.8.4.1 provides Vn for several conditions. This

Vn

21.2.4.3

1.2 1.0

35 360, 000 1000

Vn Joint is OK. Discussion and summary Step 10—Discussion This example is an extension of Examples 2 and 3 but the seismic loads where increased for a Seismic

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf

  The scope of ACI 318M-14, Chapter 11 addresses nonprestressed and prestressed cast-in-place and precast reinforced concrete walls. In addition to Chapter 11, Section 18.10 of ACI 318M-14 provides design and detailing requirements for special cast-in-place walls forming part of the seismicforce-resisting system. Reinforced concrete structural walls are common in buildings, and are almost always part of a building’s lateral-force-

behavior for in-plane loads. For further discussion and information regarding unique shear behavior for in-plane loads in

gravity and lateral forces.

10.2.1.1 Longitudinal reinforcement—In general, longitudinal bars require lateral support to prevent buckling of the bars due to axial compression. In a wall, if longitudinal reinforcement is required for axial strength, or if Ast exceeds 0.01Ag, then Section 11.7.4.1 of ACI 318M-14 states the longitudinal reinforcement must be supported by transverse ties. This requirement could be used as a practical limit to determine whether the member should be designed as a column or a wall. If the wall requires heavy reinforcement Ag intersection of longitudinal and transverse reinforcement,

of the total lateral forces acting on the structure through in-plane shear. In ACI 318M, all walls are referred to as structural walls. ACI 318M and ASCE 7 have coordinated requirements

construction labor. Designing this same member as a column could be more practical. 10.2.1.2 Shear aspect ratios—The next limit to consider is shear. Most walls have a length-to-thickness ratio of at least 6. For these aspect ratios, it is easy to see how the shear

Although structural walls are also part of the gravity-forceresisting system, they are often lightly axially loaded. They

cast-in-place walls: 1.. An ordinary cast-in-place structural wall, permitted designed and detailed in compliance with Chapter 11 of ACI 318M-14. 2.. A special structural wall, which is designed and detailed in compliance with Chapters 11 and 18 of ACI 318M318M-14. 14. Special walls are required in SDCs D, E, and F, but can be constructed in all seismic categories. Seismic requirements are intended to increase wall strength and ductility to accommodate the large displacements demands expected during a maximum design earthquake. Walls are laterally connected to diaphragms and vertically to foundations or support elements. In seismic and nonseismic design, the connections to diaphragms are designed to remain elastic, and the energy from the lateral forces is absorbed by the structural wall. In seismic design, connection to the foundation can be the point of maximum wall moment and yielding of vertical reinforcement is expected.  10.2.1 Distinguishing a column from a wall—The design of a wall can be so similar to a column that the question of when a rectangular column becomes a wall is often deliberated.

14. Although this limit is necessary to achieve the expected behavior, it might not be the best limit for the consideration of column or wall design. Expected behavior and construction constraints of the member are often the keys to answering the question of wall or column. Columns usually have high axial loads and their shear behavior is similar to beams. Walls usually have low axial loads and their shear behavior is similar to one-way slabs for out-of-plane loads, with unique shear

of-plane loads. For smaller aspect ratios of approximately 2.5 to 6,, the member is designed either as a wall or column, depending on the shear force applied and the direction of aspect ratios under 2.5, the member is likely to be designed as a column. Further discussion on this topic is given in 10.2.2 Wall layout layout—Shear walls should be located within shear walls in the center half of each building is generally a arrangement, however, can restrict architectural use of space. Although shear walls are commonly located at the ends of a building, such wall locations will increase slab restraint and shrinkage stresses, especially in long buildings and buildings such as parking structures that are exposed to large

tively. Walls away from the perimeter, however, could have a higher tributary area and, consequently, larger gravity axial force to resist uplift or overturning. They are, however, less An unsymmetrical arrangement, however, does not

be designed explicitly for torsion. A symmetrical arrangement is preferable to avoid designing walls for torsion. 10.2.3 —Shear walls could have or channel sections. A plane wall section is rectangular with

concrete walls around building elevator core shafts and stair-

  

Eng.Mohammed Ashraf





Fig. 10.2.2—Shear wall layouts. 10.2.4 Wall type type—The selection of a shear wall type is based on several factors, including functionality, constructability,

hw/

w

hw/

w

Fig. 10.2.3a—Shear Shear wall cross sections. opening percentage, the wall strength could be reduced. A row of vertically aligned openings in a slender wall results in dividing the wall in two sections, termed “coupled walls” because they behave as two individual continuous wall 10.2.5 Design limits—Minimum wall thicknesses are as shown in Table 10.2.5. For walls designed by the alternative method for out-ofplane loads in Section 11.8 of ACI 318M-14. ACI 551.2R provides the following slenderness limits: c/h

wall, c/h ACI 318M does not provide separate thickness limits for

following minimum wall thickness:

i. Boundary element: 300 mm ii. Wall: 250 mm

in Fig. 10.2.3b; where h is the wall thickness, hw is the wall height, and w in the wall length.

i. Coupling beam designed as a special moment beam: 350 mm ii. Coupling beam designed with diagonal reinforcement: 400 mm

  

Eng.Mohammed Ashraf





Fig. 10.2.4—Shear wall types.  Wall type

Minimum thickness h 100 mm Greater of:

1/25 the lesser of unsupported length and unsupported height

Greater of:

1/30 the lesser of unsupported length and unsupported height

100 mm

Exterior basement and foundation

190 mm

There is no code limit on the overall building drift. should not exceed 1/500 of the story height.  Chapter 5 in ACI 318M-14 provides the load combinations necessary to design a shear wall for moment, shear, and axial force. Section 12.4 in ASCE 7 has additional seismic o. 10.3.1 Methods of analysis—ASCE 7 allows for three

ASCE 7-10 Table 12.2-1 provides the required response R factors Cd are listed for shear walls in Tables 10.3.1a through 10.3.1c. Sections 6.6, 6.7, and 6.8 in ACI 318M-14 address and second-order inelastic analysis, respectively. Section 18.2.2.1 in ACI 318M-14 requires that the interaction of linear and nonlinear response of the structure to earthquake motions be considered in the analysis. varies depending on the anticipated deformation level and

forces: lent lateral force analysis is the simplest method of analysis an approximated fundamental period Ta, which can be conservative. For long periods greater than 3.5 seconds, this method is not acceptable. response spectrum analysis accounts for the elastic dynamic behavior of the structure and determines the building period. The calculated base shear can be less than the base shear

base shear. response history method of a three-dimensional model is used to analyze the building.

  

Eng.Mohammed Ashraf





  Seismic-forceresisting system

R

factor Cd

Special reinforced concrete shear walls

5

5

forced concrete shear walls

4

4

Intermediate precast shear walls

4

4

3

3

walls

  Seismic-forceresisting system

R

factor Cd

Special reinforced concrete shear walls

6

5

forced concrete shear walls

5

4.5

Intermediate precast shear walls

5

4.5

4

4

walls

    Seismic-forceresisting system

R

10.4.1 Design for axial load—Wall design for axial load is similar to column design. The wall slenderness is consid-

factor Cd

Special reinforced concrete shear walls

7

5.5

forced concrete shear walls

6

5

a computer program that accounts for P second-order analysis. The design is completed according to Section 22.4 of ACI 318M-14, which can be quickly evaluated using an interaction diagram generated by software or an electronic spreadsheet. If the resultant of all axial loads is located in the middle third of the wall thickness h, then a where moment can be ignored and Pn is directly calculated. trated gravity load is the width of the bearing plus four

10.4.2 Axial and out-of-plane loads—Walls should be analyzed for combined axial and out-of-plane loads. For walls that are part of a multistory building lateral load system, combined axial and out-of-plane loads rarely control the design of the wall. The design of these walls can be completed according to Section 22.4 of ACI 318M-14 and slenderness checked using Section 6.6.4 of ACI 318M-14 that accounts for P For a tall one-story building with long shear walls, such as a warehouse, combined axial and out-of-plane loads typically control the wall design. There are many one-story commer commercial buildings that use exterior concrete walls to support roof plane and in-plane loads. These buildings are typically 12 to 18 m in height to allow for rack storage or second-story mezzanines. The wall thickness is usually made as slender typi as possible for economy. The design of these walls is typically completed in accordance with Section 11.8 of ACI 318M-14. Alternative methods for out-of-plane slender wall analysis include ACI 318M-14, or by using a Finite Element P The alternative method has several limitations as stated control this method use are: Pu at midheight cannot exceed 0.06fc Ag

 Walls are a versatile building element used in a variety of ways that determine how the wall is approached for design. A wall is typically very long in one plan dimension, compared to the orthogonal dimension making the wall slender. This slenderness can control the design if there are large loads applied laterally along the smaller wall dimension h applied in this direction are commonly called out-of-plane sion w are commonly called in-plane loads. Rarely do out-of-plane and in-plane loading have to be considered at the same time, though axial loads are always present. The designer typically designs a wall for the two conditions discussed for axial load with out-of-plane and those with

c/150

A comprehensive discussion of this method, including its derivation, limitations, use, and worked examples, is given in ACI 551.2R. 10.4.3 Axial and in-plane loads, squat walls—Walls are hw/ w mode is diagonal shear. Shear applied to the top of the wall is delivered to the base through compressive struts. Diagonal cracks form along the struts at inclined angles of approxi-

tion. Separation at the crack engages the vertical reinforce-

  

Eng.Mohammed Ashraf





ment in tension, creating a clamping force and increased hw/ w reinforcement begins to provide a portion of the resistance. At a height-to-length ratios exceeds 2.5, the horizontal reinforcement provides most of the shear strength and the shear behavior is more like a beam. This change in shear behavior is accounted for in the minimum reinforcement requirement according to Eq.

hw t w

t t is calculated according to Section 11.5.4.8 of ACI 318M-14. If this value is less than a reinforcement ratio of 0.0025, then the minimum reinforcement ratio for both horizontal and vertical reinforcement is 0.0025. If the required shear reinforcement exceeds 0.0025, -

resistance in squatter walls. At hw/ w computed from 11.5.4.8. However, t required by 11.5.4.8 t calculated by 11.5.4.8. At hw/ w minimum amount of longitudinal reinforcement that changes 2.5. At t required at an hw/ w of 0.5 to hw/ w of 2.5. hw/ w of 0.0025 is provided. strength is calculated according to Section 22.4 of ACI 318M-14, as discussed in 10.4.2 of this Handbook. Concrete shear strength Vc is calculated according to Table 11.5.4.6 Vs is The maximum nominal shear strength Vn is, according to Section 11.5.4.3 Vn

0.83 f c hd

Fig. 10.4.3—Shear 10.4.3—Shear in squat walls. typically the length of the wall, the wall web, h

w,

Acv instead of hd. Acv is multiplied by the width of

i. For ordinary structural walls, the minimum reinforce reinforceVu

Vc

hd



ii. For special structural walls, the minimum reinforcement is according to Table 11.6.1 for Vu less than 0.083Acv  t

may also be calculated according to Section 22.4 of ACI 318M-14, which is discussed in 10.4.2 of this Handbook. Section 18.10 in ACI 318M-14 implies that a wall is squat when the height-to-length ratio is less than 2.0. For walls with hw/ w < 2.0, the displacement method of Section

hw/

w

t

Vu greater than 0.17Acv

 or hw/

w

w

beyond

determine if boundary elements are necessary. If boundary elements are not necessary, the design of a special structural wall is similar to an ordinary structural wall. A more detailed discussion about boundary elements is given in 10.4.4 of this

ment is expected, development lengths shall be 1.25fy in

special structural walls versus ordinary structural walls are

10.4.4 Axial and in-plane loads, slender walls—The term “slender walls” is used if the predominant failure mode is

  

Eng.Mohammed Ashraf





axial load for the given lateral seismic force. The limit of u/hw limit provides increased deformation capacity for a range elements. This method also has an additional requirement for the termination of the transverse reinforcement at special boundary elements, if required. The transverse reinforcement must extend vertically above and below the critical section at least the greater of w and Mu/4Vu, except at the foundation, the boundary element ties or hoops must extend into the foundation 300 mm, or if the edge of the boundary element is within one-half the foundation depth from an edge of the footing, ties or hoops must extend into the foundation or support a distance equal to the development length of the

element must be adequately developed in the foundation. The stress-based method is used for irregular walls or walls with disturbed regions, for example, around openings, Fig. 10.4.4—Calculation of neutral axis depth c (Moehle et al. 2011).

ings exceeds 0.2f 0.2fc

of ACI 318M-14,, as discussed in 10.4.2 of this Handbook. 318MM-14 For special structural walls, Chapter 18 of ACI 318M-14 requires special boundary elements at prescribed limits. There are two methods for determining the need for special boundary elements: the displacement method of Section 18.10.6.2 and the stress-based method of Section 18.10.6.3 The displacement method is the preferred method of

single critical section height hw/ w u/hw A special boundary element is required if the neutral axis depth c

c

w

600 1.5 u /hw

This equation was derived from the relationship shown in u

was added in ACI 318M-14 to more accu-

u is the design displacement found through a linear elastic analysis and

fc model is based on a linear elastic analysis. 10.4.5 Boundary elements—Special boundary elements elements— are required if the limits in Sections 18.10.6.2 or in 18.10.6.3 of ACI 318M-14 318M- are not met. Special boundary elements are in 18.10.6.2 or 18.10.6.3 are met, boundary elements are still

boundary reinforcement is required if the longitudinal fy. Boundary reinforcement is also required for the stress-based 0.15fc fc for size and detailing of these requirements are described in Fig. 10.4.5 of this Handbook. Horizontal reinforcement in structural walls with boundary elements must be anchored into the core of the boundary element with hooks, headed bars, or straight embedment and also has to extend to within 6 mm of the end of wall Section 18.10.6.4. 10.4.6 Vertical wall segments and wall piers—A vertical wall segment is any portion of a wall that is bound by the outer edge of a wall and an edge of an opening, or the portion of a wall bound by the vertical edges of two openings. According to Chapter 11 of ACI 318M-14, the design of nonseismic vertical wall segments is the same as that of walls. For special structural walls designed according to Chapter 18 of ACI 318M-14, there are additional requirements. The nominal shear strength is reduced for the total cross section of a wall at the vertical wall segments. The calculated Vn may not exceed 0.66Acv  for the total Acv, as shown in Fig. 10.4.6 of this Handbook. For an individual vertical wall segment, Vn may not exceed 0.83Acv  .

  

Eng.Mohammed Ashraf





Fig. 10.4.5—Summary of boundary element requirements for special walls (Fig. R18.10.6.4.2 of ACI 318M-14). or wall piers according to the segment geometry as summarized in Table 10.4.6 of this Handbook. In many cases, the special structural wall requirements apply. If the wall segment is designed as a column, Section 18.10.8.1 requires that the special detailing of 18.7.4, 18.7.5, and 18.7.6 for

be designed as special columns or by alternative requirements given in Section 18.10.8 in ACI 318M-14. Wall piers designed to these alternative requirements require: Ve that can develop Mpr at the ends of the column or o

boundary element

10.4.7 Horizontal wall segments and coupling beams—A horizontal wall segment is any portion of a wall that is bound by the outer edge of a wall and an edge of an opening, or the portion of a wall bound by the horizontal edges of two openings. According to Chapter 11 of ACI 318M-14, the design of nonseismic horizontal wall segments is the same as that of walls. Horizontal wall segments in special structural walls are designed as special structural walls according to Chapter 18 of ACI 318M-14. If horizontal wall segments are part of a coupled special structural wall system, the segment is called a coupling beam. The coupling beam is separated into three categories in ACI 318M-14. n/h a special moment frame Acv  , design the beam n/h < 2 and Vu shear through the member special beam or with diagonal placed bars

segments   

Eng.Mohammed Ashraf





The design of a coupling beam is beyond the scope of this Handbook. For more information, reference Moehle et al.

 Structural walls are, in general, thin, tall, wide members

detailed shear walls in buildings have resisted seismic forces

transfer moments, shear, and axial forces to the foundation.

shear wall locations. In these cases, the shear wall will are much less due to frame action. Reinforcement placed in the horizontal and vertical directions resists in-plane shear forces and limits cracking. For tional reinforcement is placed at the ends of a wall or within Fig. 10.4.6—Shear strength for vertical wall segments.    Clear height of vertical wall segment/length of vertical wall segment (hw/ w) hw/

w

Length of vertical wall segment/wall thickness ( w/bw)

( w/bw

< 2.0

w

hw is the clear height, the wall segment.

the concrete. The casting position of transverse reinforcement assumes more than 300 mm of concrete below the bar for the t

Wall Wall pier required to

hw/

Although one curtain of reinforcement is permitted for ordinary shear walls, two curtains of reinforcement are recommended. It is advantageous to place the transverse rein reinreinforcement as the exterior layer to prevent longitudinal rein

column design requirements; refer to Section

w

2.5 < ( w/bw Wall

( w/bw) > 6.0 Wall

Wall pier required column design requirements or alternative requirements; refer to Section

immediately above the foundation, where wall longitudinal reinforcement laps with foundation dowel bars. These dowels, lapped with the wall bars, provide the critical mechanism of foun transferring tension and shear forces from the wall to the foundation. The dowels should extend into the foundation with

Wall

is the horizontal length, and bw is the width of the web of

For constructability purposes, dowels with 90-degree hooks should extend to the bottom of the foundation where they can Structural walls in SDCs D, E, and F have an additional requirement. Section 18.10.2.3 of ACI 318M-14 requires that longitudinal reinforcement be developed or spliced

  

Eng.Mohammed Ashraf





Fig. 10.5b––Boundary element requirements for special walls. for 1.25fy if the foundation connection is where yielding of wall reinforcement is likely to occur as a result of lateral displacements. The amount of required transverse reinforcement is found transverse reinforcement is tighter at the base of the wall for a distance w or Mu/4Vu at the lesser of 150 mm or 6db. Above this region, the spacing widens to 200 mm or 8db until the reinforcement ratio drops below 2.8/fy, where transverse

 Structural walls have two main advantages: 1.. They are relatively easy to construct because reinforce reinforcement detailing of walls is straightforward. mize sway and damage in structural and nonstructural elements, such as glass windows and building contents, for buildings exposed to high lateral loads. Structural walls have disadvantages, including these two: 1. Shear walls can create interior barriers that interfere with architectural and mechanical requirements. 2. Shear walls carry large lateral forces resulting in the possibility of large overturning moments. Attention is required at the wall-foundation interface and foundation design.

 American Concrete Institute (ACI) ACI 318M-14—Building Code Requirements for Structural Concrete and Commentary ACI 551.2R-10 551.2R-10—Design Guide for Tilt-Up Concrete Panels American Society of Civil Engineers (ASCE)

 Barda, F.; Hanson, J. M.; and Corley, W. G., 1977, “Shear Reinforced Concrete Structures in Seismic Zones Zones, SP-53, American Concrete Institute, Farmington Hills, MI, pp. 149-202. tions,” Concrete International Moehle, J. P., 2015, Seismic Design of Reinforced Concrete Buildings Moehle, J. P.; Ghodsi, T.; Hooper, J. D.; Fields, D. C.; and Gedhada, R., 2011, “Seismic Design of Cast-in-Place Concrete Special Structural Walls and Coupling Beams: A Guide for Practicing Engineers,” NEHRP Seismic Design Technical Brief No.6 Technology, Gaithersburg, MD, 37 pp. Wallace, J. W., 1996, “Evaluation of UBC-94 Provisions for Seismic Design of RC Structural Walls,” Earthquake Spectra 10.1193/1.1585883

  

Eng.Mohammed Ashraf





 Shear Wall Example 1: Seismic Design Category B/wind—The reinforced concrete shear wall in this example is nonpre-

The resultant maximum factored moments and shears over the height of the wall are given for the load combination selected. This example provides the shear wall design only at the base. Given: Pu In-plane–– Vu Mu Out-of-plane–– Vu Mu Material properties–– fc fy = 420 MPa

This example uses the Interaction Diagram spreadsheet aid found at https://www.concrete.org/store/productdetail. aspx?ItemID=SP1714DAE in U.S. customary units and results converted to metric in this example. ACI 318M-14 Discussion Calculation Step 1: Geometry 11.3.1 This wall design example follows the requirements of Section 11.5.2, and, therefore, the wall thickness does not need to meet the requirements of Table

an indication that the thickness chosen is appropriate. From Table 11.3.1.1, the wall thickness must be at least the greater of 100 mm and the lesser of 1/25 the

20.6.1.3.1

The unsupported height controls; 5.5 m < 8.5 m hreq’d = 5500 mm/25 = 220 mm Example shear wall h = 300 mm > hreq’d = 220 mm OK

A 300 mm wall is used in this design and the wall is assumed to be exposed to weather on the exterior of the structure. Concrete cover is 40 mm, which is in

  

Eng.Mohammed Ashraf



11.4 6.9



The structure is analyzed using the assumptions and requirements of 11.4. The structure was analyzed using 3D elastic Finite and shear force at the base of the wall are listed in the analysis requirements of Section 11.4 and Chapters given section at the start of this example. 5 and 6 of ACI 318M-14 for loading and analysis, respectively refer to Fig. E1.2 and E1.3 for in-plane wall, respectively.

Fig. E1.3—In-plane shear along the height of the wall. Step 3: Concrete and steel material requirements 11.2.1.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength

The designer determines the durability classes. Please refer to Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

By specifying that the concrete mixture shall be in accordance with ACI 301M and providing the exposure

Based on durability and strength requirements, and experience with local mixtures, the compressive least 35 MPa.

coordinated with ACI 318M-14. ACI encourages

11.2.1.2

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. The reinforcement must satisfy Chapter 20 of ACI 318M-14. The designer determines the grade of bar and if the bar should be coated by epoxy, galvanized, or both.

By specifying the rebar grade and any coatings, and that the rebar shall be in accordance with ACI 301M, In this case, assume Grade 420 bar and no coatings.

  

Eng.Mohammed Ashraf





11.5, 11.5.2 a shear wall can be determined using an interaction diagram similar to a column interaction diagram.

The wall interaction diagram is generated using the from the end of the wall, the second pair is placed at 300 mm from the end of the wall, and the remaining Refer to Column Example 9.2 in this Handbook for additional information about the Interaction Diagram spreadsheet.

end of the wall is to allow for cover on the end pairs of bars in the wall and to force the reinforcement to be symmetrical. The reinforcement is symmetrical about the center of the wall and this bar layout is applied to both ends of the wall.

To estimate an initial reinforcement area, the wall is Fig. E1.4 shows the resulting design strength interaction diagram. The design strength interaction diagram reinforcement necessary to resist the moment is calculated. The Interaction Diagram spreadsheet contains a sheet Pn, the sheet calculates the associated maximum Mn on the interaction diagram curve and plots a point on the interaction diagram to show that point. This point is named the “Input Point” on the interac interaction diagram. The input point of Pn corresponding to a Pu interaction diagram Mu point is plotted as a solid triangle. The open triangle indicates where the example load resultants are and shows that this iteration does satisfy required strength. Further iterations are unnecessary.

Fig. E1.4––Design strength interaction diagram.

  

Eng.Mohammed Ashraf



Step 5: Shear design strength (in-plane) Shear walls resist both in-plane and out-of-plane 11.5.4 shear. The out-of-plane shear is very small, 11.5.4.4 therefore, by inspection it will not control the wall 11.5.4.5 design. 21.2.1 In-plane shear, Vu is given as 1045 kN. 11.5.1.1 Equation gives: Vn = Vc + Vs Vc 0.17 f c ( h)(d ) where h = 300 mm, wall thickness and d = 0.8 w = (0.8) (8500 mm) = 6800 mm From Table 21.2.1(b) use shear strength reduction

Vc Vu Step 6: Flexure design strength (out-of-plane) 11.5.1 As shown in Step 4, the layers of No. 16 vertical 11.5.3.1 for in-plane bending. 21.2.2 The resultant of the out-of-plane moment, Mu = 81 kN·m is within the middle third of the wall. This allows Section 11.5.3.1 to be used to check the outof-plane strength of the wall.

Pn

0.55(( f c )( Ag ) 1

k c 32h 32h

Vc

0.17



35 MPa (300 mm)(6800 mm)

2052 kN

Vc = (0.75)(2052 kN) = 1539 kN

Vc = 1539 kN > Vu = 1045 kN

OK

Eccentricity of the resultant load: 4515 kN)(e) = 81 kN·m (4515 e = 18 mm e < 50 mm

2

= 5325 mm From Table 21.2.2(b), use axial strength reduction

Pn

(0.8)(5325 mm) 32(300 mm)

0.55(35 MPa)(300 mm)(8500 mm) 1 0.55 mm)(

Pn = 39,421 kN Pn = (0.65)(39,421 0.65)(39,421 kN) = 25,624 kN OK

  

Eng.Mohammed Ashraf

2





Step 7: Reinforcement limits 11.6 applies. 11.6.2 Vc Vu

Vc, 11.6.2

must be the greater of the three values below: 0.0025 hw/ w t t

Assuming 300 mm for spacing in both directions, the minimum reinforcement results in: = 0.0025

in accordance t

t

Minimum reinforcement in both the longitudinal and horizontal directions is 0.0025sh. 2 . As,min gives 398 mm2 in both directions. As,prov. = 398 mm2/m > As,req’d = 225 mm2/m

OK

11.5.4.8 strength because the concrete strength is adequate in shear. Step 8: Reinforcement detailing 11.7 rection at 300 mm meets the detailing requirements 11.7.2.1 11.7.3.1 in the horizontal direction for ease of construction.

11.7.2.3

h > 250 mm, therefore two layers are required.

11.7.4.1

If the area of vertical reinforcement exceeds 0.01Ag, or if the reinforcement is needed to resist

Reinforcement is not required for shear strength; therefore, the maximum spacing for vertical bars h cannot exceed 3h spacing of vertical bars are less than these limits. Similarly, the maximum spacing for horizontal bars cannot exceed 3h spacing of horizontal bars are less than these limits. Two layers are provided having equal reinforcement area.

ural vertical reinforcement at the wall ends needs to mm on center spacing. The Ag within this length is be calculated to determine if ties are required. 9 ,000 mm2, Ast is 398 mm2 The ratio of Ast to Ag is 0.0044. This is less than the 0.01 and therefore ties are not required by 11.7.4.1.

The factored axial stress on the concrete due to the combined loads is:

106

13

mm4 = 16.2 MPa.

This is below the design strength of concrete and thus steel is not needed to resist the axial load. Therefore, ties are not required by Section 11.7.4.1. Refer to Fig. E1.5 and E1.6 for wall elevation and section cut at the ends of the wall.

  

Eng.Mohammed Ashraf





Step 9: Detailing

Fig. E 1.5—Vertical bar distribution.

Fig. E1.6—Plan reinforcement layout.

  

Eng.Mohammed Ashraf





Shear Wall Example 2: Seismic Design Category D The reinforced concrete shearwall in this example is nonprestressed. This shearwall is part of the lateral force-resisting-

software model that includes shearwall-frame interaction. The resultant maximum factored moments and shears over the height of the wall are given for the load combination selected. This example provides the shearwall design and detailing at the base of the wall. Given: Forces and moments at the wall base–– Pu In-plane–– Vu Mu Material properties–– fc fy = 420 MPa Out-of-plane–– Vu Mu

This example shows the design and detailing of a special structural shear wall due to in-plane forces, including a seismic one loading condition is checked. In a typical design, several load combinations require checking. This example uses the Interaction Diagram spreadsheet aid found at https://www.concrete.org/store/productdetail. aspx?ItemID=SP1714DAE in U.S. customary units and results converted to metric in this example.

  

Eng.Mohammed Ashraf



Calculation

ACI 318M-14 Discussion Step 1: Geometry 11.3.1 This wall design example follows the requirements of Chapter 18 and, therefore, does not need to meet

Table 11.3.1.1 can provide an indication that the thickness chosen is an appropriate design starting are required, the special boundary element will be thicker. From Table 11.3.1.1, the wall thickness must be at least the greater of 100 mm and the lesser of 1/25 of



hreq’d Example shear wall h = 300 mm > hreq’d = 220 mm OK

20.6.1.3.1

A 300 mm wall is used in this design and the wall is assumed to be exposed to weather on the exterior of the structure. Concrete cover is 50 mm, which is in

11.4

The structure is analyzed using the assumptions and requirements of Section 11.4. and shear force at the base of the wall are listed in the The structure was analyzed using 3D elastic Finite given section at the start of this example. the analysis requirements of Section 11.4 of ACI 318M-14 and Chapter 5 and 6 for loading and

  

Eng.Mohammed Ashraf





Fig. E2.3—In-plane In-plane shear along the height of the wall. Step 3: Concrete and steel material requirements 11.2.1.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength

The designer determines the durability classes. Please refer to Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

By specifying that the concrete mixture shall be in acac cordance with ACI 301M and providing the exposure

Based on durability and strength requirements, and experience with local mixtures, the compressive least 35 MPa.

coordinated with ACI 318M-14. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. 11.2.1.2

The reinforcement must satisfy Chapter 20 of ACI 318M-14. The designer determines the grade of bar and if the rebar should be coated by epoxy or galvanized, or both.

By specifying the rebar grade and any coatings, and that the rebar shall be in accordance with ACI 301M-

coatings.

  

Eng.Mohammed Ashraf





of a shearwall is determined using an interaction diagram similar to a column interaction diagram. The wall interaction diagram is generated using the

Refer to Column Example 9.2 in this Handbook for additional information about the Interaction Diagram spreadsheet. 11.1.2, 18.10.5

Section 11.1.2 requires that special structural walls be designed in accordance with Chapter 18 of ACI 318M-14. Chapter 18 covers all requirements necessary to design the wall.

25 bars at 300 mm spacing throughout the wall. It is assumed that all of the longitudinal reinforcement is

To estimate an initial reinforcement area, the wall is reinforcement necessary to resist the moment is calculated.

mm from the end of the wall, the second pair is placed at 300 mm from the end of the wall, and the remaining pairs at 300 mm spacing. The wall is symmetrical about the center of the wall and this bar layout is ap applied at both ends of the wall. Fig. E2.4 shows the resulting design strength interac interaction diagram. The design strength interaction diagram Pu, Mn on the design strength interaction diagram curve and plots a point on the design strength interaction diagram. It also generates a corresponding maximum Mn on the This point is called the “Input Point” on the interaction diagram. The input point of Pu Mn point on the interaction diagram of triangle along the interaction curve. The example has a Pu Mu open triangle indicates where the example Pu and Mu are and shows that this iteration does satisfy required strength; therefore, further iterations are unnecessary.

  

Eng.Mohammed Ashraf





Fig. E2.4—Design strength interaction diagram.

11.5.1 for in-plane bending. 11.5.3.1

The resultant of the out-of-plane moment, Mu = 163 Eccentricity of the resultant load: e allows Section 11.5.3.1 to be used to check the out- e = 36 mm of-plane strength of the wall. e < 50 mm = 300 mm/6

Pn

fc Ag

k c 32h

Assume slab 175 mm thick mm = 5325 mm

2

2

Pn

c

Pn Pn OK Vu The axial and moment design strength over a one foot section of the wall is adequate.

21.2.2 factor: 22.5.5.1, 21.2.1

d say, d = 225 mm

For out-of-plane shear, the design strength calculation is: Vc f c bw d factor:

= 224.5mm,

Vc Vc

OK

  

Eng.Mohammed Ashraf



Step 5: Seismic reinforcement detailing 18.10.1 This structure is using a special structural wall to resist the lateral loads applied.

18.10.2

t, for structural walls must be at least 0.0025, except that if Vu does not exceed 0.083Acv fc and t are permitted to be reduced to the values in Section 11.6 of ACI 318M-14.



Two curtains of steel are used, the distributed reinforcement ratios are met, and the forces are determined within code allowed analysis methods.

direction in each face at 300 mm This provides 258 mm2 per 300 mm in the horizontal direction. The transverse reinforcement ratio is: 2 2 × = 0.0029 > 0.0025 OK t = 258 mm direction in each face at 300 mm. This provides 1020 mm2 per 300 mm in the vertical direction. 2 = 1020 mm2 = 0.011 > 0.0025 OK

18.10.2.2

The code requires two curtains of distributed reinforcement if; hw/ w Vu > 0.17Acv fc

hw/ w Two curtains are required and are provided.

18.10.3

The code allows the Vu calculated from the factored Vu loads to be the design shear. Mu

18.10.4

The shear strength of special structural walls is af affected by the height to length ratio of the wall. The code limits Vn to: Vn

Acv

c

fc

t

fy

c is 0.17 for hw/lw 0.25 and 0.17 for hw/ w between 1.5 and 2.0.

Vn

OK

2

where Acv

2

  

Eng.Mohammed Ashraf



21.2.4, 21.2.4.1

18.10.4.4



mined by Sections 21.2.4 and 21.2.4.1. From the analysis of the structure, the maximum axial load under seismic loading combinations for this wall is

Pu From the interaction diagram, Mu corresponding to Pu

In this example, the code limits on shear strength based on concrete strength of Vn Acw fc

Maximum shear: Vn

Mu Vu Vu 2

The Vu calculated from the nominal moment strength of the shear wall is greater than the maximum code al-

Vn calculated on previous page.

18.10.5

Vn

OK

Requirements of Section 18.10.5 are met through the for a non-seismic structural wall. in Step 4b. This wall is rectangular in plan and does not have end

18.10.6, 18.10.6.1

Special boundary elements are often required in special structural walls to resist the large com compression forces at the ends of the walls during an earthquake event. Boundary elements are required in accordance with

  

Eng.Mohammed Ashraf



18.10.6.2



Section 18.10.6.2 applies to walls satisfying the following: hw/ w of structure to top of wall, and are designed to have

OK hw/ w Wall is continuous from bottom of structure to top of wall. OK Wall is designed having a single critical section for OK Therefore, Section 18.10.6.2 will be used.

Section 18.10.6.2 requires a special boundary element if the neutral axis depth calculated for the factored axial force and factored moment is greater

The interaction diagram spreadsheet calculates the neutral axis depth c, which is 1723 mm The software that analyzes the structure presents the u from the software is 61 mm and does not include the Cd factor of 5 for a special structural concrete wall from ASCE 7-10.

Check if.

c

w

600 1.5 u /hw

If so, a special boundary element is needed.

u/hw

= 0.0109

is: c Because c of 1723 mm is greater than 866 mm, Section 18.10.6.2 does require a boundary element. reinforcement must extend vertically above and below the critical section at least the greater of w and Mu/4Vu except as permitted by Section w

= 8.5 m

Controls

not delve into the calculations regarding how high the special boundary element would need to be. It was conservatively assumed that the w requirement w

selected as the top of the special boundary element. Further calculations would require this height.

  

Eng.Mohammed Ashraf





metric requirements upon the special boundary elements. Section 18.10.6.4 states the following: Where special boundary elements are required by

ment extend a minimum from the extreme compres-

c – 0.1 w or least the greater of c – 0.1 w and c/2, where c is the c/2 = 1723 mm/2 = 861.5 mm largest neutral axis depth calculated for the factored The example rounds the length of the boundary eleaxial force and nominal moment strength consistent ment to 890 mm u. b, over hu/16.

hw

over the boundary element to meet the requirement of hu/16 = 5500 mm/16 = 344 mm Therefore, the wall thickness will be 350 mm at the boundary elements.

w

assumes that the wall and boundary element will be top of wall, designed to have a single critical sec secw

of steel required; therefore, the wall is considered

b over the axial loads. greater than or equal to 300 mm

and shall extend at least 300 mm into the web. on the transverse reinforcement. The boundary element transverse reinforcement

spacing of the vertical and crosstie reinforcement shall meet the requirements of a special structural column.

Section 18.7.5.3, except the value hx in Section 18.7.5.2 must not exceed the lesser of 350 mm and two-thirds of the boundary element thickness, and the transverse reinforcement spacing limit of dimension of the boundary element.

  

Eng.Mohammed Ashraf



Transverse reinforcement shall be in accordance



Section 18.7.5.2 requires that the transverse boundary element reinforcement satisfy essentially the same requirements as those of a special concrete column.

18.7.5.2 single or overlapping spirals, circular hoops, or rectilinear hoops with or without crossties.

limit of hx The permitted hx is less than 2/3 of the 350 mm width of the boundary element. 2/3 × 350 mm = 233 mm This permitted hx is for both across the width of the element and along the length of the element.

engage peripheral longitudinal reinforcing bars.

of Section 25.7.2.2 of ACI 318M-14. Consecutive crossties shall be alternated end for end along the longitudinal reinforcement and around the perimeter of the cross section.

they shall provide lateral support to longitudinal reinforcement in accordance with Sections 25.7.2.2 and 25.7.2.3 of ACI 318M-14. spacing hx of longitudinal bars laterally supported by the corner of a crosstie or hoop leg shall not ex exceed 350 mm around the perimeter of the column.

in a 350 mm thick wall is approximately: Therefore, for the thickness of the boundary element, there is no need to add vertical reinforcement. To satisfy this requirement for hx along the length of the element, reinforcement is spaced at 185 mm for mm

necesHowever, this added precision may not be neces

18.7.5.3 Spacing of transverse reinforcement shall 18.7.5.3

determines a maximum spacing of the transverse rein reinforcement in the special boundary element. dinal bar so, as calculated by:

so

100

350 hx 3

The value of so 150 mm and need not be taken less than 100 mm

350 mm/3 = 117 mm

so

100

350 219 3

143.7 mm

The example uses a spacing of 100 mm for the transverse reinforcement in the special boundary element and all longitudinal bars in the special boundary element are engaged by a crosstie.

  

Eng.Mohammed Ashraf





forcement for the ties in the boundary element.

0.3

Ag Ach

1

fc f yt 2

Ag Ach

Since this is greater than the minimum required by

Ach = 197,500 mm2 Ash = 433 mm2

reinforcement extend into the wall support base when the critical section occurs at the wall base.

reinforcement a minimum of 300 mm into the foundation element below the base of the wall.

reinforcement in the wall be developed within the core of the boundary element.

for lateral reinforcement is much less than the 890 mm lateral bars through the boundary element to within 150 mm of the end of the wall will satisfy this require requirement.

18.10.6.5

Does not apply.

18.10.7, 8, and 10

Do not apply.

18.10.9 and contact surfaces shall be roughened consistent Final sketch of structural wall using the special boundary elements

proximately a 6 mm amplitude in the construction documents. wall if special boundary elements were required.

  

Eng.Mohammed Ashraf





Fig. E2.5––Final layout of special boundary element reinforcement.

Fig. E2.6––Elevation of wall with special boundary elements.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf



 The foundation is an essential building system that transfers column and wall forces to the supporting soil. Depending on the soil properties and building loads, the engineer may choose to support the structure on a shallow or deep foundation system. Because ACI 318M-14 provides design and detailing provisions for shallow foundations and pile caps only, deep foundations are not covered in this Handbook. Shallow foundation systems include isolated footings that

1. Determine the necessary soils parameters. This step is often completed by consulting with a geotechnical engineer who furnishes information in a geotechnical report. Important information that a geotechnical report should include are the: acteristics of soil, groundwater, rock, and other soil elements of sloped soil below frost penetration level per unit volume, used to determine the additional load

In this chapter, isolated, combined, and continuous footing examples are presented. A pile cap design is presented in the Strut-and-Tie Chapter of this Handbook.  Footing design typically consists of four steps:

able pressure that a footing is permitted to exert on the supporting soil; the size of the footing is based on allowable loads movement of a footing over time

Fig. 11.1—Shallow foundation types.   

Eng.Mohammed Ashraf





Fig. 11.3.1a—One-way shear critical section in footings. (g) Liquefaction, which is an important soil character characteristic if the building is located in an active seismic area. 2. Analyze the building’ss structure under service loads 13.2.6.1)) and factored loads (ACI (ACI 318M-14, Section R13.2.6.1) 318M-14, Section 5.3.1)) to calculate moments and forces on the columns and walls at the footing level; the service load analysis is used to calculate footing bearing areas and the factored load analysis to design the footing. 3.. Select the footing geometry so that the soil parameters are not exceeded. The following are typical parameters: (a) Calculated bearing pressures are assumed to be uniform or to vary linearly; bearing pressure is measured in units of force per unit area, such as pounds per square foot ment between adjacent footings on the superstructure are considered (c) Footings need to be able to resist sliding caused by any horizontal loads (d) Shallow footings, assumed not to be able to resist tension, should be able to resist overturning moments from compression reactions only; overturning moments are commonly caused by horizontal loads (e) Local conditions or site constraints, such as proximity to property lines or utilities, are adequate. 4. Design and detail the footing in accordance with ACI 318M-14, Chapter 13. During this step, the previously selected geometry is checked against strength requirements of the reinforced concrete sections. The step-by-step structural design process for concentrically loaded isolated footings follows:  1. Find service dead and live column loads: ACI 318M-14, Section R13.2.6

The footing geometry is selected using service loads. D = service dead load from column L = service live load from column P=D+L Areq = P/ P P/q qall For square footings,  For rectangular footings, choose one of the sides from site constraints and calculate the other such that: b × Areq 2.. Calculate the design (factored) column load U: ACI 2 318M-14,, Section 5.3.1 3.. Obtain the allowable soil pressure qnet. Because soil and concrete unit weights are close ((18.9 and 23.5 kN/m3, respectively), the footing self-weight may initially be ignored. 4. Calculate the soil pressure based on initial footing base dimensions: Square footing: qu = U/ 2 Rectangular footing: qu = U/ 5. Check one-way (beam) shear: The critical section for one-way shear extends across the width of the footing and is located at a distance d from the face of a column or wall (Fig. 11.3.1a and Fig. 11.3.1a(b) left side), ACI 318M-14, Section 8.4.3.2. The shear is calculated assuming the footing is cantilevered away from the column or wall ACI 318M-14, Section 8.5.3.1.1. For masonry walls the area is halfway between the wall center and the face of the masonry wall (Fig. 11.3.1a(b) right side). NOTE: If the calculated one-way factored shear exceeds the one-way shear design strength, then increase the footing thickness. Footings are typically not designed with shear reinforcement. 6. Check two-way (slab) shear: (a) Determine the dimensions of loaded area for: i) Rectangular concrete columns, the loaded area coincides with the column area (Fig. 11.3.1b(a))

  

Eng.Mohammed Ashraf





Fig. 11.3.1b—Two-way Two-way shear critical section in footings. to be halfway between the faces of the steel column and

beff

bf

bbp

bf 2

where bf plate side.

bbp is base

d eff

df

d bp

Fig. 11.3.1c—Column load distribution in footing.

df 2

where df plate side.

dbp is base

of d

stress, vu vu

vn

shear stress, then increase footing thickness and repeat steps

Square footings are designed and detailed for moment in one direction and the same reinforcing is placed in the other direction. For rectangular footings the reinforcing must be designed and detailed in each direction. The critical section for moment extends across the width of the footing at the

face of the column. ACI 318M-14, Sections 13.2.6.4 and 13.2.7.1. x, x = /2 – c/2, where c is the smaller dimension of the column for a square footing. For a rectangular footing, c is the dimension perpendicular to the critical section in each direction Mu, at the critical section As. ACI 318M-14, Sections 13.3.2.1 and 7.6.1.1, specify a As must be met, and 7.7.2.3 8. Check the load transfer from the column to the footing

318M-14, Section 22.8.3.2

Bdowels = Pu

Bn

  

Eng.Mohammed Ashraf





Fig. 11.3.1f—Bar distribution in short direction.

For rectangular footings, As must be furnished in each direction. Bars in long direction should be uniformly spaced. Bars in the short direction should be distributed rein Fig. 11.3.1d—Moment Moment critical section in footing at reinforced concrete column face.

Bs centered on column: #bars in Bs

2 /b 1

minimum reinforcement requirements of ACI 318M-

Calculate the bar’s development length, d, in tension per ACI 318M-14, Section 25.4. The development Lm length, d, Lm d, use bars of a smaller diameter.

Fig. 11.3.1e–Column/wall dowels into footing. Bn Pu only a minimum area of reinforcement is required ACI 318M-14, Section 16.3.4.1.

 In addition to vertical loads, footings often resist lateral loads or overturning moments. These loads are typically from seismic or wind forces.

size and number. pressure under the footing, where soil-bearing pressure is ACI 318M-14, Section 25.4.9 dowels in compression, dc tive in compression and are used to stabilize the dowels during construction. in compression,

sc: ACI

318M-14, Section 25.5.5

For square footings, As must be furnished in each direction. The same size and number of bars should

soil bearing can also be caused by a column located away from the footing’s center of gravity. If overturning moments are small in proportion to vertical loads, that is, the total applied load is located within the kern e and a P/A ± M/S analysis is appropriate to calculate the soil P

=

the total vertical service load, including any applied loads along with the weight of all of the foundation

  

Eng.Mohammed Ashraf





components, and also including the weight of the soil located directly above the footing. A = the area of the footing bottom. M = the total overturning service moment at the footing bottom. S = the section modulus of the footing bottom. If overturning moments are larger, that is, the total applied load falls outside the kern, e > /6, then P/A – M/S analysis

2. Calculate the total service overturning moment M, measured at the footing bottom 3. Determine whether P/A exceeds M/S 4. If P/A exceeds M/S, then the maximum bearing pressure equals P/A + M/S and the minimum bearing pressure equals P/A – M/S 5. If P/A is less than M/S, then the soil bearing pressure is as shown in Fig. 11.4. Such a soil-bearing pressure distri-

This soil is only able to transmit compression. The following are typical steps to calculate footing bearing pressures if nonuniform bearing pressures are present. These steps are based on a footing that is rectangular in plan and assumes that overturning moments are parallel to one of the footing’s principal axes. These steps should be completed for as many load combinations as required by the applicable design criteria. For instance, the load combination with the maximum P usually causes the maximum bearing pressure while the load combination with the minimum P usually is critical for overturning. 1. Determine the total service vertical load P

maximum bearing pressure = 2 P B X e = M/P. –e

X

/2

 If a column is near a property line or near a pit or a mechanical equipment in an industrial building, a footing may not be able to support a column concentrically and the eccentricity is very large. In such a case the column footing

The combined footing is sized to have the resultant force of the two columns within the kern, or preferably to coincide with the center of the footing area. The combined footing can be rectangular, trapezoidal, or having a strap, connecting Design steps: 1. 1. Calculate the total service column loads, P1 and P2 2. Calculate service column load resultant location Center for rectangular footing:

xc

P1 x1

P2 x2 Pi

If P1 is much larger than P2, then trapezoidal combined footing may be used. Fig. 11.4—Footing under eccentric loading.

Fig. 11.5—Common types of combined footing geometries.   

Eng.Mohammed Ashraf





Determine combined footing length from construction constraints. Calculate the widths B1 and B2 such that the center of the footing coincides with the force resultant or is at least within /6 of the force resultant.

x

B1 2 B2 3 B1 B2

c 2

3. Determine combined footing dimensions assuming uniform bearing Combined rectangular footing length: = 2xc Combined rectangular footing width:

B

P1

P2 qa

4. Steps to design a combined footing to resist one-way and two-way shear and moment is similar to the isolated footing design steps presented above.

  

Eng.Mohammed Ashraf





 Foundation Example 1: Design of a square spread footing of a seven-story building

Given: Column load— Service dead load D Service live load L Seismic load E Material properties— Concrete compressive strength fc Steel yield strength fy = 420 MPa 3

Allowable soil-bearing pressures— 2 D only: qall.D 2 D + L: qall,D+L 2 D + L + E: qall.Lat

ACI 318M-14

Fig. E1.1—Rectangular foundation plan.

Discussion

Calculation

Step 1: Foundation type 13.1.1 The bottom of the footing is 900 mm below the basement slab. Therefore, it is considered a shallow foundation. 13.3.3.1

The footing will be designed and detailed with the and Chapter 8, Two-way slabs of ACI 318M-14.

  

Eng.Mohammed Ashraf





Step 2: Material requirements Concrete compressive strength 19.2.1.1 The value of concrete compressive strength at a 19.2.1.3 ments. Table 19.2.1.1 provides a lower concrete compressive strength limit of 17 MPa.

19.3.1

Exposure categories and classes The engineer must either assign exposure classes to the footing with respect to Table 19.3.1.1

19.3.2

proportion the concrete mixture, or use the classes to directly specify mixture proportions in the contract documents. Based on the exposure classes, the concrete mixtures must satisfy to the most restrictive requirements of Table 19.3.2.1.

f c,min = 17 MPa

Provided: fc

OK

Concrete exposure categories There are four categories: F, S, W, and C.

19.3.1.1

Category F The foundation is placed below the frost line, therefore, it is not exposed to external elements—freezing and thawing cycles. Therefore, class F F0 applies.

19.3.2.1 are listed in Table 19.3.2.1. 19.3.3.1

F Class F0 w/cm w/ Minimum fc

max

Air content is not required and there are no limits on cementitious materials

Requirements of Table 19.3.3.1 do not apply. Category S w/cm cm

19.3.1.1 19.3.2.1

requirements for S0 are listed in Table 19.3.2.1.

19.3.1.1 19.3.2.1

Category W The footing may be in contact with water and low permeability is not required.

19.3.1.1 19.3.2.1

Category C The concrete is exposed to moisture and there is no external source of chlorides; therefore the class is C1. Mixture requirements for C1 are listed in Table 19.3.2.1.

w/cm

w/cm

fc

max

max

max

= none and fc

= none and fc

Therefore, there is no restriction on w/cm and f c 28 MPa

Conclusion: w/cm. Therefore, in the

with other exposure limits. fc

  

Eng.Mohammed Ashraf



Step 3: Determine footing dimensions 13.3.1.1 To calculate the footing base area, divide the service load by the allowable soil pressure.

area of footing =

P allowable soil pressure qa



3 3

; close. Therefore, footing self-weight will be ignored for initial sizing of footing: D qall ., D

2400 19

2

D L qall ., D L

2

D L E qall ., Lat

Assuming a square footing. The footing thickness is calculated in Step 5, footing design.

12.6 m 2

12.6 m 2

Controls

12.1 m 2

11.4 m 2

2

3.55 m

Therefore, try a 3.6 x 3.6 m square footing.

  

Eng.Mohammed Ashraf





Step 4: Soil pressure Footing stability Because the column doesn’t impart a moment to the footing, the soil pressure under the footing is assumed to be uniform and overall footing stability is assumed. Calculate factored soil pressure. This value is needed to calculate the footing’s required strength. qu

Pu area

Calculate the soil pressures resulting from the applied factored loads. U = 1.4D

U = 1.4D qu

U = 1.2D + 1.6L

U = 1.2D + 1.6L 1.6

qu U = 1.2D D+E+L

13.3.2.1

Controls

2

3.6 m 2

U = 0.9D + 1.0E qu

The load combinations include the seismic uplift force. In this example, uplift does not occur.

2

3.6 m 2

U = 1.2 1.2D + 1.0E E + 1.0L 1.0L

qu U = 0.9D + E

2

3.6 m 2

2

3.6 m 2

E includes not only earthquake loads dues to overturning, but also earthquake loads due to vertical acceleration of ground as per ASCE 7-10, Section 12.4.2.

Because the footing is square, it will only be designed in one direction.

  

Eng.Mohammed Ashraf





Step 5: One-way shear design

Fig. E1.2—One-way shear in longitudinal direction. 21.2.1(b) 7.5.1.1

Shear strength reduction factor: Vn Vu

7.5.3.1 22.5.1.1

Vn = Vc + Vs

shear

= 0.75

Assume Vs = 0 (no shear reinforcement) Vn = Vc

Therefore: 22.5.5.1 7.4.3.2

Vc

0.17 f c bw d Vc

20.6.1.3.1

Vu

The Code allows the critical section for one-way shear at a distance d from the face of the column 1.2 (refer to Fig. E1.2). The engineer could either assume a value for d that d.

Assume that the footing is 750 mm thick. The cover is 75 mm to bottom of reinforcement. Assume that No. 25 bars are used in the both directions and design for the more critical case (upper d: d = 750 mm – 75 mm – 25 mm – 25 mm/2 = 637.5 mm say, d = 637 mm Vn

Vu

2

c d bqu 2

3600 mm 2

Vu

600 mm (3.6 m)(328 kN/m 2 ) 637 mm 2 1000 mm/m

= 1019 kN

Vc

0.75(0.17) 28 MPa (3600 mm)(637 mm)

= 1.547 × 106 N 1574 kN Vc = 1547 kN > Vu = 1019 kN OK

h = 750 mm

  

Eng.Mohammed Ashraf





Step 6: Two-way shear design The footing will not have shear reinforcement. Therefore, the nominal shear strength for this twoway footing is simply the concrete shear strength: v n = vc 22.6.1.2 22.6.1.4 22.6.4.1

Under punching shear theory, inclined cracks are assumed to originate and propagate at 45 degrees away and down from the column corners. The area of concrete that resists shear is calculated at an average distance of d/2 from column face on all

Fig. E1.3—Two-way shear. bo

c+d

bo

where bo is the perimeter of the area of shear resistance. 22.6.2.1

ACI 318M-14 permits the engineer to take the directions when calculating the shear strength of the footing, but in this example the smaller

8.4.2.3.4 22.6.5.1 22.6.5.2

The two-way shear strength equations for

vc

fc

0.33

vc

Controls 2 1

vc

vc 22.6.5.3

s

0.083

sd bo

2

fc

vc

4948 mm

= 40, considered interior column

Vc

f c bo d

0.33

vc = 1.75 MPa Vc

Use a shear strength reduction factor of 0.75: Vc Vu = qu a 8.5.1.1

f c bo d 2

c + d 2]

Check if design strength exceeds required strength: Vc Vu?

Vc

Vu

OK

2

600 mm + 637 mm 1000 mm/m

2

Vc Vu OK Two-way shear strength is adequate.

  

Eng.Mohammed Ashraf





Step 7: Flexure design The code permits the critical section to be at the 13.2.7.1 face of the column (refer to Fig. E1.4).

Fig. E1.4—Flexure in the longitudinal direction.

Mu

c

qu

2

2

(b)/2

22.2.1.1

Set concrete compression force equal to the steel tension force at the column face: C = T

22.2.2.4.1

C = 0.85fc ba and T = Asfy a

7.5.2.1

22.3.1.1 22.2.2.2 21.2.1(a) 8.5.1.1(a)

As f y 0.85 f c' b Mn

As (420 420 MPa) 0.85(28 0.85 28 MPa)(3600 mm)

a

and

As f y d

8.6.1.1 21.2.1(a)

As

a 2

Substitute for a in the equation above. Use moment strength reduction factor from Table 21.2.1. Mn

Mu = 1328 kN·m and solving for As:

Mn As

13.3.3.3(a)

2

((328 kN/m2 ) 3.6 m 0.6 m (3.6 m)/2 1328 kN·m

Mu

Distribute bars uniformly across the entire 3.6 m width of footing: l

As

As 2

2

Use 13 No. 25 bars (13 × 570 = 6630 mm2) distributed uniformly across the entire 3.6 m width of footing.

mm)(750 mm) = 4860 mm2 = 0.0018 As,min = 0.0018(3600 2 As,prov = 6630 mm > As,min = 4860 mm2

Check if the assumption of tension controlled

  

Eng.Mohammed Ashraf



21.2.2



To answer the question, the calculated tensile strain in reinforcement is compared to the values in Table 21.2.2. The strain in reinforcement is calculated

c t

22.2.2.4.1 22.2.2.4.3

c

2

a c=

32.5 mm 0.85

38.2 mm

0.003 38.2 mm = 0.047 > 0.005 t

d c

t

where: c = a 1 and a = 0.28As = 0.9 is correct

Fig. E1.5 E1.5—Strain E Strain distribution across footing cross section.

when including footing self-weight and slab selfweight and live load above footing:

Footing self-weight less soil self-weight:

2

live load:

3

WF

3

Ws

Total weight on supporting soil:

WT

Calculate actual soil pressure:

qa

2

3

2

2

qall

OK

  

Eng.Mohammed Ashraf





Step 8: Transfer of column forces to the base 13.2.2.1 Factored column forces are transferred to the footing 16.3.1.1 by bearing on concrete and through reinforcement. 22.8.3.2

The foundation is wider on all sides than the loaded area. Therefore, the nominal bearing strength, Bn, is the smaller of the two equations. A1 A2

Bn and Bn

f c A1

fc A1

A2 2.0 A1 where A1 is the bearing area of the column and A2 is the area of the part of the supporting footing that is geometrically similar to and concentric with the loaded area. Check if

The bearing strength reduction factor is 0.65:

2

A2 A1

bearing

2

6

2

= 0.65 2

Bn Bn 16.3.4.1

16.3.5.4

Column factored forces are transferred to the founfoun dation by bearing and through reinforcement, usu usually dowels. Provide dowel area of at least 0.005A 0.005 g and at least four bars.

OK

2

As,dowel

= 1800 mm2

Bars are in compression for all load combinations. Therefore, the dowels must extend into the footing a compression development length, dc, the larger of the

0.24 f y r db  fc   f y r db 

dc

25.4.9.2

dc

Controls 28 MPa

dc

where r r

The footing depth must satisfy the following inoped within the provided depth: h 25.3.1

dc

+r+d

where r

hreq’d

+ 2db,bars + 75 mm db

+ 75 mm = 611 mm hreq’d = 611 mm < hprov. = 750 mm

OK

  

Eng.Mohammed Ashraf





Fig. E1.6—Reinforcement development length. Step 9: Footing details Development length 13.2.8.3 Flexural reinforcement bar development is required cb K tr 90 mm + 0 3.5 13.2.7.1 at the critical section. This is the point of maximum db 25.4 mm factored moment, which occurs at the column face. Bars must extend at least a tension development cb K tr 2.5 Use maximum length beyond the critical section. db

25.4.2.2 25.4.2.4

d

1 fy t e s db c 1.1 f c b K tr db t

d

= casting position;

1.1

2.5

db

28.9d b

d

Therefore, OK. OK t = 1.0 because not more than 300 mm of fresh concrete below horizontal reinforcement

d

in the longitudinal direction:

d,prov. e

e

= 1.0, because bars are

d,prov.

= 1425 mm >

d,req’d

= 734 mm

OK

uncoated s

s

cb = spacing or cover dimension to center of bar, whichever is smaller Ktr = transverse reinforcement index It is permitted to use Ktr = 0. However, the expression:

cb

K tr must not exceed 2.5. db

  

Eng.Mohammed Ashraf





Step 10: Detailing

Fig. E1.7—Footing reinforcement detailing.

  

Eng.Mohammed Ashraf





Foundation Example 2: Design of a continuous footing Seismic Category D and is 900 mm Given: Wall load— Service dead load D Service live load L W E

Material properties— Concrete compressive strength fc Steel yield strength fy = 420 MPa 3

Allowable soil-bearing pressures— 2 D only: qall.,D 2 D + L: qall,D+L. all,W all,E

2 2

Fig. E2.1— E2.1—Plan Plan and elevation of continuous footing. ACI 318M-14

Discussion

Calculation

Step 1: Foundation type 13.1.1 Therefore, it is considered a shallow foundation. 13.3.2.1

The footing will be designed and detailed with the and Chapter 9, Beams, of ACI 318M-14.

13.2.3.1

Foundation resisting earthquake forces must comply with Section 18.2.2.3 of ACI 318M-14.

  

Eng.Mohammed Ashraf



Step 2: Material requirements 13.2.1.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength

The designer determines the durability classes. Please see Chapter 4 of this Handbook for an indepth discussion of the categories and classes.



By specifying that the concrete mixture shall be in accordance with ACI 301M and providing the exposure

Based on durability and strength requirements, and experience with local mixtures, the compressive strength

coordinated with ACI 318M-14. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 1 of this chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes. Step 3: Determine footing dimensions 13.3.1.1 To calculate the footing width, divide the service Ignoring the footing self-weight: combinations are obtained from ASCE ASCE7-10, Section 2.4.

area of footing

P allowable soil pressure qa

The footing thickness is calculated in Step 4, footing design.

D qall , D

2.5 m

2

D L

2.85 m

Controls

all D L

D

D

L qall , Lat

L qall , Lat

W

2

2.27 m

2

2.3 m

E

Use B = 3 m

  

Eng.Mohammed Ashraf





Step 4: Footing design Wall stability Because there is no out-of-plane moment, the soil pressure under the footing is assumed to be uniform and overall wall stability is assumed. 7.4.2.1

5.3.1

The footing cantilevers on both sides of the wall are designed as one-way slabs. Calculate soil pressure Factored loads— Calculate the soil pressures resulting from the applied factored loads. U = 1.4D U = 1.2D + 1.6L

2

U = 1.4D U = 1.2D + 1.6L 2

U = 1.2D + W + L

Controls

U = 1.2D + 1.0W + 1.0L 2

U = 0.9D + W

U = 0.9D + 1.0W W 2

U = 1.2D + E + L

U = 1.2 1.2D + 1.0E E + 1.0 1.0L L 2

U = 0.9D 0.9 + E

U = 0.9 0.9D + 1.0E E 2

  

Eng.Mohammed Ashraf



7.5.1.1

One-way shear design— Shear strength reduction factor: Vn Vu

shear



= 0.75

7.5.3.1 22.5.1.1 22.5.5.1

Vn = Vc + Vs

Assume Vs

Therefore: Vn 0.17 f c bw d

Vn = Vc

22.5.1.2

And satisfying: Vu

7.4.3.2

Vu is calculated at d from the face of the wall.

Vn

The engineer can either assume a value for d that d.

13.2.7.1 13.2.7.2

is located halfway between center and face of masonry wall. Fig. E2.2—Shear —Shear critical section.

7.4.3.2

In this example the second approach is followed:

Vu

B 2

t wall 2

d qall .

Vc

Vu

3m 2

0.3 m 2

d

Vc

d in m

d , where d is in mm

of more than 150 mm

Solving these two equations for d = 360 mm, use 400 mm > 150 mm

Concrete cover must satisfy Table 20.6.1.3.1. Use c = 75 mm Total footing depth: h = d + db + c

h = 400 mm + 13 mm + 75 mm = 468 mm, say, 500 mm

13.3.1.2

20.6.1.3.1

2

  

Eng.Mohammed Ashraf





Flexure design E2.3 to indicate that for masonry walls, the critical section for shear or moment are not the face of the masonry wall, but is tw/4 from the face. 13.2.7.1

For concrete walls, the factored moment and moment strength are calculated at the face of the wall. From maximum factored load Mu is:

22.2.2.4.1

Set the concrete compression strength equal to the steel tension strength: C = T; 0.85fc

7.5.1.1

Mn

= Asfy

Fig. E2.3—Moment critical sections. 2

Mu

Mu

C

2

a = As

a = 0.0176As mm As

0.0176 As 2

2

As

use of straight bars. 7.6.1.1

Check if As exceeds the minimum: As,min 0.0018Ag

As,min = 900 mm2/m < 1515 mm2/m

OK

Check if the section is tension controlled and the

21.2.2

22.2.1.2

To answer the question, the tensile strain in reinforcement must be calculated and compared to the values in Table 21.2.2. The strain in reinforcement is assumed to be proportional to the distance from

c t

c

d

2

a = 0.0176As c=

26.7 mm 0.85

31.4 mm

0.003 31.4 mm = 0.035 > 0.005 t

c

t

where: c = a and a = 0.0176As

  

Eng.Mohammed Ashraf





Fig. E2.4—Strain distribution across footing. exceeded when including footing self-weight and soil self-weight above footing: Footing self-weight:

WF

Soil weight above footing:

Ws

Total weight on supporting soil:

WT

Calculate actual soil pressure:

qa

3

3

3

2

3m

OK

qall

  

Eng.Mohammed Ashraf





Step 5: Footing details Shrinkage and temperature reinforcement along length of footing 7.6.4.1 7.6.1.1

The area of shrinkage and temperature reinforcement: AS+T

Ag

2

AS+T

requirement for shrinkage and temperature reinforcement in the long direction. Development length Check if the width of the footing provides adequate length for the bottom tension reinforcement beyond the critical tension section.

25.4.2.3

d

1 1.1

fy

t

e

f c cb

s

Ktr

db

db = 25.4 mm

d

1.1

db

25.4.2.4

db

2.5 db

where t = casting position; t

concrete is placed below horizontal reinforcement e = coating factor; e s

= bar size factor;

s

cb whichever is smaller 25.4.2.3

Ktr = transverse reinforcement index It is permitted to use Ktr

cb

K tr db

must not exceed 2.5.

Provided length: B/2 – twall/2 – 75 mm

cb

K tr db

25.4 mm

3.5

Use maximum value of 2.5 avail. avail.

d

= 734 mm

OK

bars for development and does not require hooks at both ends.

  

Eng.Mohammed Ashraf





Step 6: Earthquake requirements 13.2.3.2

The foundation is in SDC D therefore, ACI 318M-

18.13.2

The requirements listed in 18.13.2 for structural uplift occurs:

18.13.2.1 extend into the footing and must be fully developed for tension at the interface. 18.13.2.3 that have an edge within one-half the footing depth from an edge of the footing shall have transverse reinforcement in accordance with 18.7.5.2 through 18.7.5.4 provided below the top of the footing. This reinforcement must extend into the footing and be developed a length equal to the development length, calculated for fy in tension, of the boundary element longitudinal reinforcement. This condition does not apply for this problem. 18.13.2.4 in boundary elements of special structural walls, of the footing to resist actions resulting from the design factored load combinations, and must be less than required by 7.6.1 or 9.6.1.. This condition does not apply for this problem. Step 7: Detailing

Fig. E2.5—Continuous footing reinforcement.

  

Eng.Mohammed Ashraf





Foundation Example 3: Design of a continuous footing with an out-of-plane moment

Given: Wall load— Horizontal wind shear V

Material properties— Concrete compressive strength fc Steel yield strength fy = 420 MPa 3

Soil data— qall qu,permitted

2 2 3

Fig. E3.1— E3.1—Plan Plan and elevation of continuous footing.

ACI 318M-14 Discussion

Calculation

Step 1: Foundation type 13.1.1 Therefore, it is considered a shallow foundation. 13.3.2.1

The footing will be designed and detailed with the and Chapter 9, Beams, of ACI 318M-14.

  

Eng.Mohammed Ashraf



Step 2: Material requirements 13.2.1.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength

The designer determines the durability classes. Please see Chapter 4 of this Handbook for an indepth discussion of the categories and classes.



By specifying that the concrete mixture shall be in accordance with ACI 301M and providing the exposure

Based on durability and strength requirements, and experience with local mixtures, the compressive strength

coordinated with ACI 318M-14. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 1 of this chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes. Step 3: Determine footing dimensions 13.3.1.1 calculations. Iterations may be needed. Try B = 2.1 m footing width. 2 Area: A /m Section modulus: S

3

/m

The footing thickness is also assumed and then design. 13.3.1.2

The footing thickness must be such that the bottom reTry 380 mm footing thickness.

  

Eng.Mohammed Ashraf





Step 4: Footing design Wall stability lateral force on the stem wall, the overall wall stability must be checked. To calculate the stability of a footing, the total vertical load is calculated and the resisting moment MR MOTM Commonly, engineers require MR consider a footing stable.

MOTM to

Weights on bearing soil below footing: Weight of footing:

Wftg

Weight of soil above footing:

Wsoil

Weight of concrete wall pier:

Wconc pier

3

2

3

3

2

Total vertical dead load:

2

2

2 2

of applied lateral wind shear.

H = 900 mm + 300 mm = 1200 mm

The overturning moment, MOTM, is measured at base of footing. The lateral wind force must be convert to service load level.

MOTM

W H

The resisting moment, MR, is calculated as the product of vertical load by distance from the centerline to edge of footing: MR = P B

MR

To ensure footing stability, the following inequality

MR > 1.5MOTM OK

  

Eng.Mohammed Ashraf





Step 5: Calculate soil pressure 13.3.1.1 Service loads The maximum soil pressure is calculated from service forces and moments transmitted by foundation to the soil. To calculate soil pressure, the location of the vertical service resultant force is determined. The distance to the resultant from the front face of stem:

e

M otm P

e

0.12 m

Check if resultant falls within the middle third B/6 = 2.1 m/6 = 0.35 m > e = 0.12 m

OK

Because e B/6, the footing imposes compression to the soil across the entire width. The resulting soil pressure must be less than the allowable bearing pressure provided by the geotechnical report. Maximum and minimum soil pressures are calculated by:

q1,2

P A

M OTM S

q1,2

2

qmax

2

2

2

< qall =

2

2

2

>

2

qmin 2

OK

  

Eng.Mohammed Ashraf





Step 6: Factored loads 13.3.2.1 The footing is designed as one-way slab. Calculate the soil pressures resulting from the applied factored loads. 5.3.1a U = 1.4D Use D = P MOTM = W

U qu

U = 1.2D/A + 1.0W/S + 0.5L/A where S

1.2D/A 1.0W/S 0.5L = 0

e

e

2 2

< qu

OK

5.3.1d

W

P

2 3

2

= 0.35 m

Because e < B/6, the footing bearing pressure varies

qu

D/A

2

qu,max qu,min qu,max

W/S

2

2 2

2 2

2 2

< qu,permitted

OK

5.3.1f U = 0.9D/A+ 1.0W/S 1.0 e

W

P

Because e < B/6, bearing pressure qu

2

0.9D/A 0.9D/A / 1.0W/S WS W/ e

D/A

W/S

q1,2 qu,max qu,min

3

2

2

2 2

< qu,permitted

2

2

2

OK

Fig. E3.2—Soil pressure distribution under factored loads, critical shear section, and critical moment section.

  

Eng.Mohammed Ashraf



Step 7: Shear strength One-way shear design 21.2.1(b) Shear strength reduction factor: 7.5.1.1

Vn

Vn = Vc + Vs

22.5.1.1 22.5.5.1

Vc

Assume Vs = 0 (no shear reinforcement) Vn = Vc f c bw d

0.17

d db/2 Assume db = 25.4 mm and c = 75 mm Therefore: Vc 0.17 f c bw d Calculate factored soil pressure at distance d from face of wall: qu , d

= 0.75

Vu

7.5.3.1

20.6.1.3.1a

shear



qu,min

qu,max

qu,min B

B 2

t wall 2

d

d = 380 mm – 75 mm – 13 mm = 292 mm Vc

qu , d

0.75(0.17)( 28 MPa )(1 m)(292 mm) 197 kN/m

74 kN/m 2 2.1 m 2

218 kN/m 2 74 kN/m 2 2.1 m 0.3 m 0.292 m = 176 kN/m 2 2

Note that qu,max and qu,min are from Step 6, Load Case II. Calculate factored shear force at d from face of wall: Vu

Vu

218 kN/m 2 176 kN/m 2 2 ((1.05 m 0.15 m 0.292 m) = 120 kN/m

created by qu. 7.4.3.2 22.5.1.2

Vc

Vu

Vc = 197 kN/m > Vu = 120 kN/m OK

  

Eng.Mohammed Ashraf





Step 8: Flexural strength Flexure design 13.2.7.1 The footing factored moment is calculated at the

Fig. E3.3—Moment critical section. 5.3.1a

5.3.1d

Calculate Mu at face of wall from U = 1.4D/A. 2 qu

2 2

Mu

2

Calculate Mu at face of wall: U = 1.2D/A + 1.0W/S + 0.5L/A

qu , wall

qu,max

qu ,min

qu , min B B 2

t wall 2

2

2

2

qu

2.1 m 2

Factored moment: Mu

1 B qu , wall 2 2 1 qu , max 3

5.3.1f

twall 2 qu , wall

2

2

Mu

B 2

t wall 2

2

2 2

2

2 2

3

Controls

By inspection load condition U = 0.9D/A + 1.0W/S does not control.

and soil weight are conservatively neglected. 22.2 Set concrete compression strength equal to steel tension strength

22.2.2.4.1 22.2.2.3 22.2.2.4.3 7.5.1.1 22.3.1.1

C=T 0.85fc

= Asfy

a = As a = 0.0176As

Mu

Mn = 0.9Asfy d – a

Mn = 0.9As

As

As

2

/m

  

Eng.Mohammed Ashraf



9.6.1.2



Check As against the minimum: 1.4 bd fy

As , min

1.4 420 MPa As,min = 973 mm2/m > As,req’d = 743 mm2/m As , min

As,prov. = 1018 mm2/m > As, req’d = 973 mm2/m

2

OK

Check if the section is tension controlled and the

21.2.2

22.2.1.2

To answer the question, the tensile strain in reinforcement is calculated and compared to the values in Table 21.2.2. The strain in reinforcement is cal-

c t

c

2

d c

where: c = a

c=

1

and a = 0.0176As

0.85

21 mm

0.003 21 mm t = 0.039 > 0.005 Section is tension-controlled t

Fig. E3.4—Strain distribution through depth of footing.

  

Eng.Mohammed Ashraf





Step 9: Footing details 7.6.4.1 Shrinkage and temperature reinforcement: 24.4.3.2 13.2.8.3 The area of shrinkage and temperature reinforcement: AS+T Ag

2

AS+T

mm2 temperature reinforcement placed perpendicular to the

13.2.7.1

25.4.2.3

25.4.2.4

Development length Reinforcement development is calculated at the maximum factored moment and the code permits the critical section to be located at the wall face. Bars must extend at least a tension development length beyond the critical section.

d

1 1.1

fy

t

e

s

db

f c c K tr db

d

1.1

2.5

db

db

where t = casting location; t = 1.0,, because not more than 300 mm of fresh concrete is placed below horizontal reinforcement e = coating factor; e = 1.0,, because bars are uncoated s = bar size factor; s

cb = spacing or cover dimension to center of bar, whichever is smaller Ktr = transverse reinforcement index It is permitted to use Ktr = 0. cb

K tr

25.4.2.3

But the expression: greater than 2.5.

25.4.2.1

The development length is the greater of the

db

must not be taken

cb

K tr db

87.4 mm 22.2 mm

3.94

Use maximum value of 2.5 OK d = 642 mm > 300 mm

bars, without hooks. d

provided perpendicular to the wall:

d,prov. d,prov.

= 825 mm >

d,req’d

= 642 mm

OK

  

Eng.Mohammed Ashraf





Step 10: Final design

Fig. E3.5—Footing reinforcement detailing.

  

Eng.Mohammed Ashraf





Foundation Example 4—Design of a rectangular spread footing Given: Column load— Service dead load D Service live load L Factored wind W

Material properties— Concrete compressive strength fc Steel yield strength fy = 420 MPa 3 3

Allowable service level soil bearing pressures— 2 D only: qall,D 2 D + L: qall,D+L 2 D + L + W: qall.Lat

Fig. E4.1—Rectangular footing plan.

Computation

ACI 318M-14 Procedure

Step 1: Foundation type 13.1.1 Therefore, it is considered a shallow footing. The footing will be designed and detailed with the 13.3.3.1 and Chapter 8, Two-way slabs, of ACI 318M-14. Step 2: Material requirements 13.2.1.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength

The designer determines the durability classes. Please see Chapter 4 of this Handbook for an indepth discussion of the categories and classes.

By specifying that the concrete mixture shall be in accordance with ACI 301M-10 and providing the exposure classes, Chapter 19 requirements are

Based on durability and strength requirements, and experience with local mixtures, the compressive least 28 MPa.

coordinated with ACI 318M-14. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 1 of this chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes.

  

Eng.Mohammed Ashraf



Step 3: Determine footing dimensions 13.3.1.1 To calculate the footing area, divide the service load by the allowable soil pressure.

area of footing

5.3.4

Wservice

P allowable soil pressure qa

W



3 3

; close. Therefore, footing self-weight will be ignored for initial sizing. Actual soil pressure is checked end of Step 6: D qall , D

3

4.64 m 2

D L qall , D L

2

4.8 m 2 Controls

D L W qall , Lat

2

4.7 m 2

The lateral wind force must be multiplied by 0.6 load level. Assume that there is a constraint on the width of B The footing thickness is calculated in Step 4, footing design.

2

xB Aprov. = 5.1 m2 > Areq’d = 4.8 m2

  

Eng.Mohammed Ashraf





Step 4: Factored soil pressure Footing stability 13.3.3.2 Because there is no out-of-plane moment, the soil pressure under the footing is assumed to be uniform and overall footing stability is assumed.

Calculate soil pressure qu

Pu area

Factored loads Calculate the soil pressures resulting from the column factored loads. U = 1.4D

U qu

U = 1.2D + 1.6L

U qu

U = 1.2D D+W+L

2

Controls

5.1 m 2

U qu

The load combinations include the possibility of wind uplift force. In this example, uplift does not occur.

2

5.1 m 2

U qu

U = 0.9D + W

2

5.1 m 2

2

5.1 m 2 2

Assume that the calculated qu is acceptable per the geotechnical report.

have larger moments and thus is the more critical condition.

  

Eng.Mohammed Ashraf





Fig. E4.2—One-way shear in longer direction. Shear strength reduction factor:

shear

= 0.75

Assume Vs Vn = Vc

7.5.1.1 7.5.3.1 22.5.1.1

Vn Vu Vn = Vc + Vs

13.2.7.2 7.4.3.2

The critical section for one-way shear is permitted at a distance d

Vc

Vc

The engineer assumes a value of d then checks d is selected. 13.3.1.2

Assume that the footing is 700 mm thick.

20.6.1.3.1

The cover is 75 mm to bottom of reinforcement. d is: d = h – cover – db/2

Vu

2

c d bqu 2

d say, d = 615 mm Vu

3m

615 mm

450 mm 2

22.5.5.1

7.5.1.1

and Vc

0.17 f c bw d

Vn > Vu?

Vc

Vc

Vu

OK

  

Eng.Mohammed Ashraf





Short direction Check one-way shear.

20.6.1.3.1

13.2.7.2 7.5.1.1

d = h – cover – db – db/2 d = 700 mm – 75 mm – 19.1 mm – 19.1/2 mm d = 596.35 mm, say, d = 596 mm Vn

Vu

b 2

c d 2

qu

Fig. E4.3—One-way shear in short direction. c

b 2 7.4.3.2

d

450 mm

596 mm

0.821 m

1.7 m = 0.85 m 2

Distance of critical shear plane from center of by inspection because the critical shear plane is at the edge of the footing. Half of footing width:

  

Eng.Mohammed Ashraf





Step 6: Two-way shear design 13.3.3.1 The footing will be without shear reinforcement. 13.2.7.2 Therefore, the nominal shear strength for two-way punching shear is equal to the concrete strength: 8.5.1.1

Vn = Vc + Vs Footings are usually designed with Vs = 0

22.6.1.2 22.5.1.1 22.6.2.1

13.2.7.1 22.6.1.4 22.6.4.1

v n = vc ACI 318M permits the engineer to take the average tions when designing the footing.

615 mm + 596 mm 2

605.5 mm

bo

c+d

Two-way nominal shear strength is the least value

vc

fc

vc

fc

sd bo

vc

s

vc

2

vc

22.6.5.3

d

Under punching shear theory, inclined cracks are assumed to originate and propagate at 45 degrees away and down from the column corners. The critical section location, bo, is calculated at an average distance of d/2 = 12 mm from face of the column

bo 8.4.2.3.4 22.6.5.1 22.6.5.2

Fig. E4.4—Two-way shear.

fc

vc

Controls 2 1

4222 mm

vc = 1.75 MPa

= 40, considered interior column

Vc

f c bo d

0.33

Vc

Use a shear strength reduction factor of 0.75: Vc

f c bo d

Vu = qu

8.5.1.1

B

c+d

2

Vc Vu

2

 

450 mm 605.5 mm 1000 mm/m

Check if design strength exceeds required strength: Vc

Vu?

Vc Vu OK Two-way shear strength is adequate.   

Eng.Mohammed Ashraf

2

 





Calculate the service-level soil pressure: B = 1.7 m, L = 3 m, h = 700 mm Weight of displaced soil by the footing. Weight of soil above footing:

Wftg

The footing weight is added to the dead load:

Wsoil

3

3

3

D D

L W Atotal

2

5.1 m 2

Allowable soil pressure: qall

2

OK

  

Eng.Mohammed Ashraf





Step 7: Flexure design 13.2.7.1

The long direction in the rectangular footing will generate larger moments because of the longer moment arm. The critical section is permitted to be at

Fig. E4.5—Flexure in the long direction.

Mu

22.2.2.4.1

22.2.2.4 22.2.2.4.3

7.5.2.1 22.3.1.1

c

qu

2

2

B

Set the concrete compression strength equal to the steel tension strength: C = T C = 0.85fc ba and T = Asfy and fc

Mn

As f y d

2

Mu

As f y

a

1

3 m 0.45 m 2

2

0.01As ; b = 1700 mm

0.85 f c b = 0.85

a 2

Substitute 0.28As for a in the equation above.

Mn

Mu and solving for As, where Mu =

As

As

8.6.1.1

Check the minimum area: As,min = 0.0018Ag

As As,min

2

2

= 1882 mm2 < As,req’d = 2781 mm2

OK

width of footing. Reinforcement in the longitudinal direction is uniformly distributed. 21.2.2 22.2.2.4.1 22.2.2.4.3

Check if the section is tension controlled and the

To answer the question, the tensile strain in reinforcement is calculated and compared to the values in Table 21.2.2. The strain in reinforcement is cal-

c t

c

2

c=

d c

t

where c = a 1 and a = 0.01As

t

0.85

33.4 mm

0.003 33.4 mm = 0.052 > 0.005

  

Eng.Mohammed Ashraf





Fig. E4.6—Strain distribution across footing section.

  

Eng.Mohammed Ashraf





Short direction Calculate moment in the short direction, at the column face.

for the long direction: d = 596 mm

Fig. E4.7—Flexure in the short direction.

b c 2

Mu

22.2.2.4

22.2.2.4.1

Set compression force equal to tension force at the column face: C=T C = 0.85fc ba and T = Asfy

22.2.2.4.3

fc

7.5.2.1

22.3.1.1

qu

2

13.2.7.1

As f y

a

As f y d

Mn

0.006 As ; b

0.85 f c b

1

1.7 m 0.45 m 2

2

Mu

2

3000 mm

= 0.85

a 2

Substitute 0.006A 0.006 s for a

0.006As 2

As

Mn = Mu and solving for As: As = 1160 mm2 8.6.1.1

Check minimum reinforcement area: As,min = 0.0018Ag

As,min = 3780 mm2 > As,req’d = 1160 mm2 Use minimum required reinforcement: As = 3780 mm2

In the short direction, a portion of the reinforcesAs centered on the column. 2 s

s

1

The band width is equal to the length of the short

2 3m 1 1.7 m

0.72

Reinforcement area in 1.7 m band width = 2

2

1.7 m band width s As is distributed equally on both sides outside the band width. The remaining area of reinforcement must be at least the minimum reinforcement with the bars spacing not exceeding the smaller of 3h or 450 mm

Reinforcement area outside the central band 2

2

2

  

Eng.Mohammed Ashraf



7.6.1.1



The area of reinforcement outside the band width ural reinforcement: As,min = 0.0018Ag

As,min = 819 mm2 > As,req’d = 940 mm2/2 = 470 mm2 outside the band width. As,prov = 852 mm2 As,min = 819 mm2

OK

Step 8: Column-to-footing connection 16.3.1.1 the footing by bearing on concrete and the reinforcement, usually dowels. 22.8.3.2

The footing is wider on all sides than the loaded area. Therefore, the nominal bearing strength, Bn, is the lesser of the two equations.

Bn

f c A1

and Bn

fc

Check if

A2 A1

A2 A1

1

2

A2 A1

2.0 where

2

3.8

2

A1 is the area of the column and A2 is the area of the footing that is geometrically similar to and concenconcen tric with the column. The reduction factor for bearing is 0.65:

bearing

= 0.65 2

Bn Bn 16.3.4.1

16.3.5.4

OK

Provide minimum dowel area of 0.005Ag and at least four bars. This requirement is to ensure ductile As,dowel behavior between the column and footing.

2

= 1012.5 mm2

Bars are in compression for all load combinations. Therefore, the bars must extend into the footing a compression development length, dc, the larger of the two and at least 200 mm:

25.4.9.2

0.24 f y r db  fc   f y r db 

dc

dc

dc

Controls 28 MPa 2

OK

dc

25.3.1

The footing depth h must satisfy the following inequality so that the vertical reinforcement can be developed: h

dc

where r

+ r + db,dwl + 2db,bars + 75 mm

hreq’d + 75 mm = 611 mm hreq’d = 611 mm < hprov. = 700 mm

OK

db   

Eng.Mohammed Ashraf





Assume that the column is reinforced with four

25.5.5.4

25.4.9.1 25.4.9.2

As mentioned above, bars are in compression for all load cases. Therefore, the compression lap splices is the larger of the two conditions: 1. The development length, dc, of the larger bar and

dc

2

dc

25.5.5.1

2. The compression lap splice of the smaller bar

Controls

28 MPa

sc

Use

sc

= 600 mm > 300 mm

OK

Step 9: Footing details Development length 13.2.8.3 Reinforcement development is calculated at the 13.2.7.1 maximum factored moment and the code permits 13.2.8.1 the critical section to be located at the column face. Bars must extend a tension development length beyond the critical section.

25.4.2.3

d

1 1.1

fy

t

e

s

db

f c c K tr db

d

1.1

d b = 28.9d b

2.5

where 25.4.2.4

t t = 1.0,, because not more than 300 mm of fresh concrete below horizontal reinforcement e e = 1.0, because bars are uncoated s

25.4.2.1

s

cb = spacing or cover dimension to center of bar, whichever is smaller Ktr = transverse reinforcement index It is permitted to use Ktr = 0. But the expression:

cb

K tr db

cb

must not exceed 2.5.

K tr db

87.4 mm + 0 19.1 mm

4.58

use maximum value of 2.5 The development length must be the greater of the Therefore, OK d

in the long direction:

d,prov. d,prov.

d

= 1200 mm >

d,req’d

= 552 mm

OK

in the short direction:

d,prov. d,prov.

d,req’d

= 552 mm

OK

  

Eng.Mohammed Ashraf





Step 10: Detailing

Fig. E4.8—Footing Footing reinforcement detailing. Square footing If the problem was solved as square footing, then in Step 3,, the following footing dimensions would have been selected: 2100 x 2100 mm

design. Distribution of reinforcement within a central band does not apply to square footings.

  

Eng.Mohammed Ashraf





Foundation Example 5: Design of a combined footing The columns only transmit axial force and neither shear nor moment is transmitted from the frame above into the footing. The soil reaction to column loads is assumed to be uniform across the footing bearing area. Given: Exterior column load— Service dead load D1 Service live load L1 c1 x c1 = 450 mm x 450 mm Interior column load— Service dead load D2 Service live load L2 c2 x c2 = 500 mm x 500 mm Material properties— Concrete compressive strength fc = 28 MPa Steel yield strength fy = 420 MPa 3

Allowable soil bearing pressure Fig. E5.1—Foundation 5.1—Foundation Foundation plan. 2 under all loads qa ACI 318M-14 Procedure Calculation Step 1: Foundation type 13.1.1 fore, it is considered a shallow footing. Step 2: Material requirements 13.2.1.1 The mixture proportion must satisfy the durability By specifying that the concrete mixture shall be requirements of Chapter 19 and structural strength in accordance with ACI 301M and providing the exposure classes, Chapter 19 requirements are determines the durability classes. Please refer Based on durability and strength requirements, and to Chapter 4 of this Handbook for an in-depth discussion of the categories and classes. experience with local mixtures, the compressive dinated with ACI 318M. ACI encourages referencing

least 28 MPa.

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 1 of this chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes.

  

Eng.Mohammed Ashraf





Step 3: Determine footing dimensions Service loads 13.1.1 To calculate the footing area, assume the columns 13.3.1.1 are supported on isolated square footings. Divide the column service loads by the allowable soil pressure. Exterior column: Areq’d D1 + L1 qa Interior column: Areq’d D2 + L2 qa

3 3

will be checked later:

2

Areq’d Use 2.15 m x 2.15 mm

= 463 m2

2

Areq’d Use 2.8 m x 2.8 m

= 7.8 m2

Because the column is in close proximity to the property line, the exterior column footing cannot be concentric with the column, and the footing needs external bracing to remain stable. This can be supplied by a moment connection between the exterior and the interior footing, but in this case, the two footings are simply combined.

13.2.6.2

The footing thickness is calculated in Step 5, Twoway shear design. Determine the location of the resultant of the two service column loads by taking the moments about the center of the exterior column.

xc

P1 x1

P2 x2 Pi

The distance of the resultant from the property line is: 13.3.1.1

The footing length, L, is taken equal to 2x so the soil pressure can be assumed as uniform under the two column loads:

13.3.4.3

Distribution of bearing pressure under combined footing must be consistent with the soils properties

B

P qa L

xc

1.9 m

x = 1.9 m + 0.3 m = 2.2 m

B

2

2.82 m

Use 2.8 m x 4.4 m

  

Eng.Mohammed Ashraf



Fig. E5.2—Combined footing dimensions. Step 4: Design forces Calculate soil pressure Factored loads Calculate the soil pressures resulting from the 13.3.4.3 applied factored loads including footing self13.2.6.1 weight. Assume 750 mm thick footing. Footing self-weight:

Soil self-weight above combined footing: 13.2.6.1 13.3.4.3



W

3

W

3

Total dead load:

Total live load:

5.3.1a

U = 1.4D

qu

5.3.1b

U = 1.2D + 1.6L

qu Controls

Distributed soil pressure per square area below qu

2

This is a conservative approach. The footing concrete displaces soil. Therefore, the actual load on soil

  

Eng.Mohammed Ashraf





Fig. E5.3—Shear Shear and moment diagrams.

  

Eng.Mohammed Ashraf





Step 5: Two-way shear design 13.3.4.1 Design of combined footing must satisfy the requirements of Chapter 8 for two-way slab of ACI 318M-14. Check footing two-way shear strength at both columns. 21.2.1b

22.6.1.4

Shear strength reduction factor:

shear

Exterior column The footing critical shear perimeter, bo, at the exterior column is three sided. From the free body diagram, the direct shear force, Vug, is the result of the factored column force less the factored soil pressure force within the critical shear perimeter

= 0.75

Fig. E5.4—Determining centroidal axis of shear perimeter.

Vug = Pu1 – quA1 22.6.4.1

where A1 c2 + d d say, 660 mm

c1 + d

e

450 mm + 660 mm 2

A1 949, 050 mm 2

e = 300 mm edge distance 5.3.1 Solving for Vug, where Pu1 = 1.2D1 + 1.6L1

Pu1

Fu = quA1

Fu

Substituting into: Vug = Pu1 – quA1

Vug

2 2 2

  

Eng.Mohammed Ashraf





The Code requires the footing moment at the critical shear centroid is transferred into the column by

b1 2 where 2b1 b2

x

c 2 b2 = c + d b1

2

x

d and 2

e

259 mm

450 mm 660 mm 855 mm 2 2 b2 = 450 mm + 660 mm = 1110 mm b1

300 mm

and soil pressure force about the critical section centroid: 8.4.4.2.1

8.4.4.2.3

Mu

Pu1 b1 e x

Fu

b1 2

Mu

x

The maximum shear stress due to direct shear and shear due to moment transfer is:

vu

Vug

v

Ac

Mu

Muc Jc

Ac = bod;; is the area of concrete within the critical section bo

8.4.4.2.3

855 mm 2

2

Ac

The shear perimeter moment of inertia Jc is: Jc

 b1

d3 12

d

b13 12

b1d

b1 2

2

x  b2 d x 

2

3

Jc



3

12

12 2

855 mm 2

  2

= 1.909 × 1011 mm 4 8.4.2.3.2 fMu

f

is:

1 f

1 f

2 b1 1 3 b2 vMu

8.4.4.2.2

2 855 mm 1 3 1110 mm

0.63

,

v is: v

f

v

= 1 – 0.63 = 0.37

  

Eng.Mohammed Ashraf



8.4.4.2.1

Solving for vu vu

where c b1 x c = 855 mm – 259 mm = 596 mm 22.6.5.1 22.6.5.2

1,861,200 mm 2 = 0.63 MPa 0.45 MPa

1.909 1011 mm 4 1.08 MPa

and the least value controls:

vu

vc

22.6.5.3

s

22.6.5.2

vc

21.2.1



 sd  0.083 b o   2  0.17 1   0.33  

 2         

0.083

fc

sd bo

0.17 1

2

2

2

0.17 1

Therefore, 0.33

2 1

2.7

0.51 0.33

0.33

NG

NG

Controls

= 30, edge column 0.33

fc

Shear strength reduction factor: 0.75

vc vc = 1.3 MPa > vu = 1.08 MPa

OK

The factored stress exceeds the footing design shear stress.

Fig. E5.5—Two-way shear at the exterior column.

  

Eng.Mohammed Ashraf



8.4.4.1.1



Interior column: The maximum factored shear force at the critical section is equal to the factored column load less the factored soil pressure within the critical section

22.6.1.4

Vu = Pu2 – qu c2 + d c2 + d

5.3.1

where Pu2 = 1.2D2 + 1.6L2

Vu 2 2

Vu 22.6.4.1 22.6.5.1 22.6.5.2

bo

Critical section, bo: Two-way shear is the least of:

Check if the

0.33 is used if the other factors are larger than 0.33.

0.33 vu

vc

0.17 1 0.083

22.6.5.3

f c factors are less than 0.33.

2 sd bo

fc sd/bo

2

s

Therefore, use the factor 0.33. vc Vc

8.5.1.1

Vc

Vu

OK

Vc > Vu

Fig. E5.6—Two-way shear at interior column.

  

Eng.Mohammed Ashraf



13.3.2.1 13.2.7.2 7.4.3.2



d where the maximum shear force is permitted to be calculated. Vc with d = 660 mm exceeds Vu

7.4.1.1 7.4.3.2

Calculate required strength from column factored loads less soil pressure Vu

Vu@face – qu c2/2 + d

660 mm 1000 mm/m

Vu

Calculate shear strength and verify that it exceeds the calculated required strength: 7.5.1.1 7.5.3.1 22.5.1.1

Vn Vu Vn = Vc + Vs

7.5.3.1 22.5.5.1 21.2.1

Shear strength is calculated from:

7.5.1.1

Check if Vu

Vs Vn = Vc

Vc

0.17 f c bw d Vc

Vc Vu

Vc

Therefore, shear reinforcement is not required.

Fig. E5.7—One-way shear.

  

Eng.Mohammed Ashraf





Step 7: Flexure design combined footing.

footing between the two columns and at the bottom of the footing at both interior and exterior columns

Top reinforcement between columns At the section of maximum moment, set the internal compression force equal to internal tension force to calculate the reinforcement area: 22.2.2.4

C=T

22.2.2.4.1 22.2.3.1

0.85fc ba = Asfy

22.3.1.1

Mu

a = As a = 0.006As

Mn

Asfy d – a

Mu

21.2.1a 21.2.1 9.5.1.1a

Mn tion above.

Mu and substitute for a in the equaequa 6

9.6.1.2

2

As

Solve for As

0.006 As

As

Check if the minimum reinforcement area controls: As , min

As , min

0.25 f c fy

bd

As , min

1.4 bd fy

0.25 28 MPa 420 MPa

2

1.4 420 MPa As,min = 6160 mm2 > As,req’d = 3176 mm2

2

As , min

sive strength fc

Therefore, minimum reinforcement controls.

over the width of the footing. 2

As,prov. 22.2.2.4.1 22.2.2.4.3

Check if section is tension controlled and the use of

21.2.2

c

a

a c 1

1 mm 38.7 mm 0.85

2

2

45.5 mm

  

Eng.Mohammed Ashraf





Calculate the strain in the tension reinforcement and compare to the minimum strain required for tension-controlled section:

7.7.2.3

t

0.003

d c c

t

0.003

660 mm 45.5 mm 45.5 mm

0.0405

0.005

Place bars such that the spacing between them does Therefore, section is tension controlled. not exceed 3h or 450 mm h = 2300 mm and

  

Eng.Mohammed Ashraf



13.2.7.1



Reinforcement at interior column The moment is taken at the interior face of the interior column:

22.2.2.4

At the section of maximum moment, set the internal compression force equal to internal tension force to calculate the reinforcement area:

22.2.2.4.1

C=T

22.2.3.1

0.85fc ba = Asfy

Mu

a = As a = 0.006As

22.3.1.1 22.2.2.1 22.2.2.2 21.2.1a

Mn fyAs d – a Substitute for a in the equation above.

21.2.1: Mn

9.5.1.1a

9.6.1.2

Mu and solving

6

0.006 As

As

2

Solve for As

As

This is less than the minimum reinforcement area calculated above.

Therefore, use As,min = 6160 mm2 > As,req,d = 1165 mm2

distributed over the width of the footing. 2

As 1 mm

a

2

2

2

Check if section is tension controlled and the use of

22.2.2.4.1 22.2.2.4.3

a

c

38.7 mm 0.85

c 1

45.5 mm

Calculate the strain in the tension reinforcement and compare to the minimum strain required for tension-controlled section:

21.2.2

t

0.003

d c c

t

0.003

660 mm 45.5 mm 45.5 mm

0.0405

0.005

Therefore, section is tension controlled. full length of combined footing and spaced at 300 mm on center < 3h = 2300 mm and 450 mm

  

Eng.Mohammed Ashraf





Transverse reinforcement In a combined footing, transverse moment distribution may be addressed similar to an isolated spread footing. A strip over the width of the footing is considered to resist the column load. This strip is, however, not independent of the footing itself. Calculate d to the center of the second layer. d = 750 mm – 75 mm – 28.7 mm – 25 mm = 621.3 mm, say, d = 620 mm

d face of columns.

w Interior column The factored column load distributed over the width of the footing is used to determine the transverse bending moment. 5.3.1

Factored distributed soil reaction is: Pu ,int

qu ,TI 13.2.7.1

qu ,TI

B

2.8 m

The factored moment at the column face is: 1 b qu 2 2

Mu

c 2

2

1 2

Mu

;

2

where b

c

2.8 m

500 mm

1.15 m

Refer to Fig. E5.8. 8.5.1.1

Calculate required reinforcement: Mn fyAsjd Mu

As As = 2830 mm2

d: j = 0.9 21.2.1

Flexural strength reduction factor:

9.6.1.2

Check if the minimum reinforcement area controls: 1.4 bd fy

As , min

As , min -

1.4 420 MPa

2

Required reinforcement is greater than the minimum

sive strength fc is less than 31 MPa. mm on center and placed within the calculated width Check if section is tension controlled and the use of 46.4 mm 2 2 As,prov > As,req’d = 2830 mm2 2

a 22.2.2.4.1 22.2.2.4.3

c

44 mm

a 1

c

44 mm 0.85

51.8 mm

  

Eng.Mohammed Ashraf





Calculate the strain in the tension reinforcement and compare to the minimum strain required for tension-controlled section: 21.2.2

t

0.003

d c c

t

0.003

660 mm 51.8 mm 51.8 mm

0.035

0.005

Therefore, section is tension controlled. Exterior column The factored column load distributed over the width of the footing is used to determine the transverse bending moment. 5.3.1

Factored distributed soil reaction is: Pu ,int

qu ,TI 13.2.7.1

qu ,TI

B

2.8 m

The factored moment at the column face is: 1 b qu 2 2

Mu

c 2

2

b c 2.8 m 0.45 m 2 2 2 2 Refer to Fig. E5.8..

8.5.1.1

d = 620 mm w = 300 mm + 450 mm/2 + 620 mm/2 = 835 mm

1 2

Mu

2

1.175 m

Calculate required reinforcement: Mn Mu fyAsjd d: j = 0.9

As As = 1768 mm2

21.2.1

Flexural strength reduction factor:

9.6.1.2

Check if the minimum reinforcement area controls: 1.4 bd fy

As , min

As , min

1.4 420 MPa

2

Minimum reinforcement is equal to the calculated spaced at 250 mm on center. 2 As,prov Check if section is tension controlled and the use of

2 2

a

22.2.2.4.1

a

c

21.2.2

36.8 mm 36.8 mm 0.85

c 1

22.2.2.4.3

> As,req’d = 1725.7 mm2

43.3 mm

Calculate the strain in the tension reinforcement and compare to the minimum strain required for tension-controlled section:

t

0.003

d c c

t

0.003

620 mm 43.3 mm 43.3 mm

0.04

0.005

Therefore, section is tension controlled.   

Eng.Mohammed Ashraf



exterior and interior columns, provide minimum reinforcement area.

As , min



1.4 420 MPa

2

h = 2300 mm or 450 mm

OK

Fig. E5.8—Footing width at columns for transverse reinforcement calculations.

  

Eng.Mohammed Ashraf





Step 8: Footing details Development length of No. 29 top bars From the moment diagram in Fig. E5.3(b), the

develop the top bars at midspan in tension.

(25.4.2.3a):

Fig. E5.9—Longitudinal reinforcement of combined footing. 25.4.2.2

25.4.2.4

fy d

t

e

fc

db

db

d

db

t

e

fc is less than 8.3 MPa

25.4.1.4 25.4.2.1

28 MPa

5.3 MPa

8.3 MPa

OK

OK

d

OK

provided development length,

d

25.4.3.1 fy dh

e

c

fc

r

db

dh

28 MPa

dh

Controls

b. 8db

e c

r

  

Eng.Mohammed Ashraf





Development of bottom bars Longitudinal bars No. 29 The factored moment at the exterior column is negligible, 3 kN·m at face of column, 32 kN·m at column centerline (refer to Fig. E5.3(b)). Calculate the development length at the interior column Mu = 506 kN·m at exterior face (Fig. E5.3(b)): fy 25.4.2.2

d

1.7

t

e

fc

(420 MPa)(1.0)(1.0)

db

d

where 25.4.2.4

= 1.0, because not more than 300 mm of fresh concrete is placed below horizontal reinforcement e e = 1.0, because bars are uncoated

25.4.2.1

Check if development is less than 300 mm

t

t

(1.7)(1.0) 28 MPa Bar size No. 29 No. 22

db

d, mm 1354 1027

47.6d b Use, mm 1400 1050

Both required development length exceeds 300 mm Therefore, OK

Interior column: Column is located 1100 mm from footing edge, which is less than the required calculated develop development length of 1475 mm. Therefore, provide a hook 1100 < 1400 mm + 75 mm = 1475 mm for the bottom No. 29 bars at the interior column. Refer to prior calculations for top bars. dh = 380 mm < 4400 mm – 3550 mm = 850 mm OK Note: That if the more detailed development length Eq. ((25.4.2.3a) 25.4.2.3a) is used, then adequate distance is availavail able to place the No. 29 bars without having to bend them. Transverse reinforcement: From Fig. E5.8,, the overhang at the interior column is 1150 mm, which is greater than the required calculated development length = 1050 mm. Therefore, No. 22 bars are placed straight.

  

Eng.Mohammed Ashraf





Step 9: Column-to-footing connection Interior column 16.3.1.1 Factored column forces are transferred to the footing by bearing and through reinforcement, usually dowels. 22.8.3.2

The footing is wider on all sides than the loaded area. Therefore, the nominal bearing strength, Bn, is the smaller of the two equations.

Bn

f c A1

Bn

fc A1

A2 A1

and

A1 is the bearing area of the column and A2 is the area of the part of the supporting footing that is geometrically similar to and concentric with the loaded area.

Center of column is located 1100 mm from the end of footing and 1150 mm of the combined footing long sides.

1100 mm/ mm/2 = 550 mm < 750 mm footing thickness 2 The sides of the pyramid tapered wedges are sloped A2 = 7.29 × 106 mm2 1 vertical to 2 horizontal. Check if

A2 A1

A2 A1

2.0 where

Bn 21.2.1

The bearing strength reduction factor is 0.65:

bearing

7.29 106 mm 2 2

Column factored forces are transferred to the footing by bearing and through dowels. The minimum dowel area is 0.005Ag and at least four bars across the interface between interior column and combined footing.

As,dowel

footing depth.

As

2

fc A1 = 0.65 2

Bn Bn 16.3.4.1

4.9

D + 1.6L 2

= 1250 mm2

16.3.5.4 2

2

> As,dowel = 1250 mm2

Bars are in compression for all load combinations. Therefore, the bars must extend into the footing at least a compression development length dc, which is the larger of the following two expressions:

25.4.9.2 dc

fy r  db  0.24 fc   f y r db 

dc

28

Controls

dc

  

Eng.Mohammed Ashraf



25.4.9.3



r r

The footing depth must satisfy the following inequality: h dc + r + db,dwl + db,#7 + db,#9 + 75 mm

hreq’d + 28.7 mm + 75 mm = 704.3 mm, say, 710 mm hreq’d = 710 mm < hprov. = 750 mm OK

20.6.1.3.2 Check development length of dowel reinforcement into the column. The length of dowels in the column is the greater of the development length and lap splice length. 25 bars. db,dowel < db,column

25.4.9.2

Therefore, the lap splice length must be the greater

Fig. E5.10 E5.10—Reinforcement E Reinforcement development length. fy r  db  0.24 fc sc = larger of   0.043 f d y b  where r r

= 1.0, because stirrup spacing is greater than

25.5.5.1 the larger of:

sc

  0.071 f y d b = larger of  300 mm

300 mm Use 670 mm long lap splice.

  

Eng.Mohammed Ashraf



16.3.1.1

22.8.3.2



Exterior column The column factored forces are transferred to the footing by bearing and through dowels. The footing is wider on all sides than the loaded area. Therefore, the nominal bearing strength, Bn, is the smaller of the following two equations.

(a) Bn

(0.85 f c A1 )

A2 A1

and (b) Bn = 2(0.85fc A1) A1 is the bearing area of the column and A2 is the area of the part of the supporting footing that is geometrically similar to and concentric with the loaded area.

Center of column is located 300 mm from the end of footing and 1175 mm of the combined footing long sides.

300 mm/2 = 150 mm < 750 mm footing thickness The sides of the pyramid tapered wedges are sloped A2 = [2(225 mm + 75 mm)]2 = 360,000 mm2 1 vertical to 2 horizontal. Check if

A2 A1

2.0 , where

360,000 mm 2 450 mm)2 (450

A2 A1

1.33

2

Therefore, Eq. ((22.8.3.2(a)) controls. 21.2.1

16.3.4.1

The bearing strength reduction factor for bearing is 0.65: Column factored forces are transferred to the footing by bearing and dowels. The minimum dowel area is 0.005Ag and at least four bars are needed across the interface between column and footing. The four No. 19 dowels must be developed in the footing.

16.3.5.4

25.4.9.2

The bars are in compression for all load combinations. Therefore, the bars must extend into the footing a compression development length dc, which is the larger of the two following expressions: 0.24

fy

r

fc

dc

(0.043 f y

bearing

= 0.65

Bn

(0.65)(0.85 (0.65)(0.85)(28 MPa)(450 mm) 2

Bn = 4177 kN > 1512 kN

OK

As,dowel = 0.005(450 mm)2 = 1012.5 mm2 Use one No. 19 bars in each corner of the column. As = (4)(284 mm2) = 1136 mm2 > As,dowel = 1012.5 mm2 OK

db

(0.24)(420)(1.0) dc

r

360,000 mm 2 (450 mm)2

db ) dc

28

(19.1)

364 mm

Controls = 0.043(420 MPa)(1.0)(19.1 mm) = 345 mm

where r r

The footing depth h must satisfy the following inequality: h dc + r + db,dwl + db,#7 + db,#9 + 75 mm 20.6.1.3

hreq’d = 364 mm + 6(19.1 mm) + 19.1 mm + 22.2 mm + 28.7 mm + 75 mm = 623.6 mm, say, 630 mm hreq’d = 630 mm < hftg,prov. = 750 mm OK

75 mm cover (refer to Fig. E5.10)   

Eng.Mohammed Ashraf





Check development length of dowel reinforcement into the column. The length of dowels in the column is the greater of the development length and lap splice length. 25 bars. db,dowel < db,column 25.4.9.2

Therefore, the lap splice length must be at least

dc

fy r  db  0.24 fc   f y r db 

Controls

where 25.5.5.1

r r

must be at least the larger of:

sc

  0.071 f y d b = larger of   300 mm 

Controls 300 mm Use 600 mm long lap splice.

Step 10: Details

Fig. E5.11—Combined footing dimensions and reinforcement.

 Design of Concrete Structures, McGraw-Hill Professional Publishing, Fanella, D., ed., 2011,

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf



 A retaining or cantilevered wall is a structural system that provides horizontal resistance to a soil mass and prevents it from assuming its natural slope. A retaining wall consists

geometry and the local soil properties. Traditional FS against these soil-related failures are shown as follows:

The stem provides horizontal resistance to the soil behind the wall, which is at higher elevation than the soil in front of the wall. The footing toe and heel transfers the lateral soil

These factors are usually provided by the geotechnical engineer and are not in the scope of this guide.

-

Reinforced concrete stems are designed to withstand horizontal soil pressures and surcharge loads. Earthquake loads for retaining walls are not addressed in this Handbook. For the purposes of this chapter, it is assumed that the geotechnical report states that the retaining wall geometry

shear design strength must be at least equal to the factored moments and shears determined from the analysis. Figure

In situations where the factor of safety against sliding failure is low, and there are site constraints against lengthening the heel, a “key” can be constructed below the footing to increase sliding resistance, as shown in Fig. 12.1d. Typical retaining walls vary in height between 1.5 and 6.0 m. For retaining walls beyond 6.0 m, buttresses or counterforts are usually provided. Counterforts are normally preferred as they create clean and an unobstructed view from

associated with the interaction between the retaining wall

Fig. 12.1a—Retaining or cantilever wall types.

Fig. 12.1b—Cantilever wall soil failure modes.   

Eng.Mohammed Ashraf





A small axial load tends to increase the moment strength of the wall according to the interaction equation, so ignoring the axial load is conservative. 12.2.2 Wall footings—for preliminary calculations, the footing length can be estimated as 40 to 70 percent of the total height h of the stem. Usually the footing extends on both sides of the stem unless there is a physical constraint, such as a property line or an existing utility. The footing thickthickness is usually taken as at least equal to the stem thick

known as the heel. It is designed to support the entire weight nent of the soil pressure. known as the toe. The toe length can be estimated as one

If the horizontal force applied to the stem exceeds the maximum allowable frictional force of the footing, the retaining wall resistance to sliding should be increased. In this case, either the overall footing length is increased or a key can be constructed at the underside of the footing. The shear key should be proportioned such that its width is at least twice its depth to develop the strength of the key in Fig. 12.1d—Retaining wall with key.  12.2.1 Wall stem—Retaining wall stems are designed as ness t is usually uniform. A tapered front face complicates the formwork construction and reinforcement layout, but requires less concrete material. The minimum recommended stem thickness is 200 mm, but many engineers prefer a 250 mm minimum. For preliminary calculations, choose a stem thickness at base from between 7 to 12 percent of the retained earth height h, as shown in Fig. 12.2.1a. Stem moment design strength requirements. Axial loads, including the wall stem weight and frictional considered in the stem design in addition to bending due to eccentric vertical loads, surcharge loads, and lateral earth

resisting the soil passive force Hp shown in Fig. 12.1d. for the toe and heel are the front and back face of the wall stem. The critical sections for calculating shear strength for the toe and heel are taken at a distance d from the front and back face of the wall stem. 12.2.3 Buttresses and counterforts—Buttresses or counterforts are analyzed and designed as rectangular beams. Horizontal reinforcement connects the stem to buttresses or counterforts, and vertical reinforcement connects the footing to the buttresses or counterforts. If stirrups are used to provide shear strength, they should be detailed in accordance with ACI 318M. If straight bars are used, their yield strength should be fully developed through straight embedment or hooks at the interface of the stem or footing and the buttress or counterfort.

  

Eng.Mohammed Ashraf





Fig. 12.1e—Cantilever wall with counterforts or buttresses.

acting on the wall stem. Fig. 12.2.1b—Forces 12.2.1 height and unit weight of the retained soil. The lateral pressure is calculated by the Rankine or Coulomb theory as

P = Ca sh P = CP sh

Fig. 12.2.1a—Retaining wall preliminary dimensions. The stem and buttress or counterfort are designed as area equal to the distance between individual buttresses or counterforts.  Retaining wall stems are designed to resist axial loads, including the weight of the stem and frictional forces due to to eccentric vertical loads, surcharge loads, and lateral earth pressure. The horizontal soil pressures acting on the rear and front faces of the stem wall are referred to as the active and passive soil pressures, respectively, and they are proportional to the

where Ca and CP properties and angle of inclination of retained earth. CoefCa and CP can vary from approximately 0.3 for loose granular soil to 1 for cohesive soil. The engineer should refer to the soils engineering report for values of Ca and CP to use in design. This chapter considers two locations for a uniform surcharge load: 1. The surcharge load is located immediately behind the wall. To calculate the force on the wall, an equivalent height of soil is determined from h2 = wa s, where wa is the 2

s

3

m

  

Eng.Mohammed Ashraf





Fig. 12.3—Surcharge Surcharge load behind a cantilever wall. adding the equivalent height of soil to the stem height as H2

H

H1

H2

Ca

s

h2 2

hhs

2. The surcharge load is located at a distance from the wall. The surcharge load will increase the pressure on the stem if a line at 45 degrees from the edge of the surcharge intersects the stem above the heel. The American Association of State

pressure due to point load and line load applied at a distance from the stem. For a uniform load strip at a distance away from and parallel to the wall, the soil pressure on the stem is

2w

h

For in depth discussion on calculating the lateral pressure against a retaining wall from surface surcharge, point, line,

 The stem, heel, and toe must satisfy design strength Sn U. The two commonly

Mn

Mu

Vn

Vu

Pn

Pu

H = H1 + H2 where H1 and

Ca

s

h2 2

 For wall stems 250 mm or greater in thickness, two layers of horizontal and vertical reinforcement are required. The maximum spacing of horizontal and vertical reinforcement is the smaller of 3t or 450 mm where t is the wall thickness.

  

Eng.Mohammed Ashraf





Retaining wall reinforcement limits for cast-in-place stems are determined per Table 12.5 (Table 11.6.1 of ACI 318M-14).      Reinforcement type

Bar/wire size

Deformed bars Welded wire reinforcement

or MMD200

Minimum

Minimum

420

0.0012

0.002

420

0.0015

0.0025

Any

0.0012

0.002

fy, MPa

t

tion are determined per 9.6.1.2 of ACI 318M-14.  Many engineers specify weakened plane to form contrac-

tion and allows the concrete to move. A contraction joint is usually a vertical 20 mm chamfer on both sides of the stem with horizontal reinforcement continuous through the joint. accom of appropriate thickness and waterstops capable of accommodating the anticipated range of joint movement (refer to Fig. 12.6). —Saturation increases the soil density and 12.6.1 therefore the pressure on the stem. If the retained earth is sure must be added to that of earth pressure for design of the retaining wall.

 1. Determine wall height. 2. Estimate stem thickness. 3. Determine base thickness. 4. Estimate length of base. 5. Estimate length of toe. 6. Calculate vertical and horizontal loads. 8. Check wall against overturning, sliding, and bearing pressure.

12.. Design a plain key for shear, if present. 12 reinforce14.. Detail reinforcement per ACI requirements; reinforce ment layout, development locations, and splices lengths and locations. 15.. Provide drainage system.  American Association of State Highway and Transporta Transporta-

American Association of State Highway and Transpor-

pipe drains, gravel drains, perforated drains, or geosynthetic drains, placed at suitable intervals and elevations. Counterfort walls are usually designed with at least one drain between counterforts.

cations Customary U.S. Units,” AASHTO-LRFD 2007, AASHTO, Washington, DC. Bowles, J., edition, The McGraw-Hill Co, Inc., 1241 pp.

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





 Retaining Walls Example 1: Reinforced concrete basement wall The eight-story building has a 3 m high, 250 mm thick normalweight reinforced concrete basement wall. The walls support 2 2

compressive strength is fc

fy = 420 MPa. The retained earth behind the basement wall is 3 , angle of soil

2

. Assume an earthquake lateral force at the diaphragm level of

Given: Soil data— s

µ = 0.5 qall

3

2

Concrete— fc a = 1.0 normalweight concrete fy = 420 MPa Loads— Superimposed dead load SDL 2 L

Slab— Span = 7.3 m Basement wall— h=3m w = 22 m

2

Fig. E1.1—Basement wall.

  

Eng.Mohammed Ashraf





ACI 318M-14 Discussion Step 1: Material requirements Concrete compressive strength 19.2.1.1 The value of concrete compressive strength at a 19.2.1.3 ments. Table 19.2.1.1 provides a lower concrete compressive strength limit not less than 17 MPa.

19.3.1 19.3.2

Calculation

Provided: fc

f

c,min=

17 MPa

OK

Exposure categories and classes The engineer must either assign exposure classes to the retaining wall with respect to Table 19.3.1.1 so the ready-mix supplier can proportion the concrete mixture, or use the classes to directly specify mixture proportions in the contract documents. Based on the exposure classes, the concrete mixtures must satisfy to the most restrictive requirements of Table 19.3.2.1. Concrete exposure categories There are four categories: F, S, W, and C.

19.3.1.1 19.3.2.1

Category F eleThe retaining wall is exposed to external ele ments—freezing and thawing cycles—in continu continuF2 ous contact with moisture, therefore, Class F applies.

Class F2 Maximum w/cm = 0.45 Minimum fc

19.3.3.1 Class F2 are listed in Table 19.3.2.1.

19.3.2.1 19.3.1.1 19.3.2.1

19.3.1.1

19.3.2.1

Category S The geotechnical report usually provides the soil sulfate content. Assuming the sulfate content is in the range of moderate severity, the class is S1. Mixture requirements for S1 are listed in Table 19.3.2.1.

Class S1 Maximum w/cm w/ is 0.50 Minimum fc Allowable cementitious materials are ASTM C150

Category W The retaining wall is in contact with water, but asClass W0 suming that a proper drainage system is provided to Minimum fc prevent water from being retained behind the wall, low permeability is not required. The maximum water soluble chloride ion content in nonprestressed concrete as percent by weight of cement is 0.3. Category C The concrete is exposed to moisture, but not to an external source of chlorides; therefore, the class is C1. Mixture requirements for Class C1 are listed in Table 19.3.2.1.

Class C1 w/cm max = none and fc

Conclusion: w/cm with other exposure limits. fc

max

is 0.45.

fc   

Eng.Mohammed Ashraf



Step 2: Equivalent lateral pressure The geotechnical report provides the equivalent

p



3

Step 3: Applied forces The cantilevered retaining wall is a continuous 6.2.2 member. It will be analyzed for the maximum load culated forces, moments, and reinforcement areas are based on 1 m length of the retaining wall. For in-plane shear, however, the full basement wall length is used.

The vertical weight is the self-weight of the retainweight of the elevated slab, and the tributary SDL and L

Fig. E1.2—Applied Applied forces on retaining wall.

Stem wall: P1 conc tstem h Base: P2 conc tbase b Curtain wall: P3

P1 P2 P3

P4 is the elevated slab self-weight and live load portions transferred to the basement wall:

3 3

2

P4

2

Total vertical load:

The retained soil behind the wall exerts lateral H H = Ca sh2/2

H

3

2

where Ca

1 sin 1 sin

and q = Ca sh

3

  

Eng.Mohammed Ashraf



6.2.3



The basement wall can be modeled as a simply supported member at the bottom and laterally determinate.

M

M Top

HB x

Ca

s

x

x 2

x 3

Fig. E1.3 E1.3—Applied E Applied forces on basement wall.

Calculate reactions: Summation of vertical forces less base weight of P2 Summation of moments about Point A:

P

PA PA

MA = 0 MA

3m 2 HB

MA

3m 3

HB

HB

The top lateral force is: Summation of horizontal forces: Moment, Shear, and Axial Diagrams shown in Fig. E1.4:

H=0 HA + HB HA

Moment equation: Mx

H B x M Top

q x x h 2

x 3

Shear equation: Vx

HB

q x x h 2

Axial equation: Px = PTop + wall height section

Fig. E1.4—Moment and shear diagrams.

  

Eng.Mohammed Ashraf





5.38 U = 1.6H controls; when lateral pressure acts alone. Step 4: Wall design

Mu

max

Basement concrete wall supported laterally at top and bottom is modeled as a hinged one-way slab as shown in Fig. E1.5. Therefore, the provisions in

11.7.2.3

Flexure: The basement wall is assumed 250 mm thick. Therefore, two layers of reinforcement will be provided although not required by code because it is a basement wall.

Fig. E1.5—Soil lateral force on stem.

The main reinforcement in the stem for resisting the lateral earth pressure is vertical and placed at the inside face of the stem wall, away from the retained earth. 20.6.1.3.1

Adequate concrete cover protects reinforcement against moisture in the soil. Concrete cover is measured from the concrete surface to the outermost surface of the steel to which the cover requirement applies. c = 50 mm cover

21.2.2

Assume that the member is tension controlled; steel t c = 0 and

22.2.2.1 22.2.2.2

The concrete compressive strain at which ultimate c = 0.003 ACI 318M, Section 22.2.2.2 requires that the tensile strength of concrete shall be neglected in

22.2.2.3

22.2.2.4.1

22.2.2.4.3

Choose maximum concrete compressive stress: The concrete compressive stress distribution is inelastic at high stress. The actual distribution of concrete compressive stress is complex and usually not known explicitly. The Code permits any particular stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. The Code allows the use of an equivalent rectangular compressive stress distribution of: 0.85fc a 1c 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3. For fc 1

0.85

fc 7 MPa

1

0.85

7 MPa

0.829

  

Eng.Mohammed Ashraf



22.2.1.1



Find the equivalent concrete compressive depth a by equating the compression force to the tension force within a unit length of the wall cross section: C=T C = 0.85fc ba and T = Asfy Calculate required reinforcement area:

a = As As

a

0.0159 As

The wall is designed for the maximum moment Mn Mu

7.5.2.1 22.3 21.2.1

Mn

f y As d

a 2

Mn

As d

0.0159 As 2

20.6.1.3.1 d = tstem – cover – db/2 use 50 mm cover Mn

7.5.1.1

d = 250 mm – 50 mm – 19 mm/2 = 190.5 mm

Mu

Mu Step 3.

0.0159 As 2

As

6

6

As = 217.3 mm2/m Solving for As To prevent a sudden failure, the Code requires that greater of: As,min

9.6.1.2

0.25 f c fy

bw d

As , min As , min

compressive strength fc = 31 MPa. 21.2.2

0.25 31 MPa 420 MPa 631 mm 2

As,provided = 645 mm2 > As,req’d = 631 mm2 and provide

Check if the tension controlled assumption and the

To answer the question, the tensile strain in reinforce2

values in Table 21.2.2. The strain in reinforcement is

22.2.1.2

c t

c

where c

0.825

6.2 mm

0.003 6.2 mm = 0.089 > 0.005 Section is tension controlled and, t

d c a

c=

t

and a = 1.31As derived previously.

1

  

Eng.Mohammed Ashraf





Fig. E1.6—Strain distribution across stem. 22.4

Axial In addition to the lateral soil pressure, the basement

3

Pu 2

2

Pu Calculate the factored axial force: 22.4.2.2

The wall’s maximum axial strength is calculated from 22.4.2.2 2

Po Po = 0.85fc Ag – Ast

Astfy

2

Po 22.4.2.1

Apply an axial strength limit factor of 0.8. 0.8.

0.8P Po 0.8P 0.8 Po

Pu

OK

Shear Diaphragm in-plane force: VEQ Factored diaphragm reaction at each end wall for load

Vu

VEQ Fig. E1.7—In-plane shear forces in basement walls.

22.5.1.1

Vn = Vc + Vs

7.5.3.1 22.5.5.1

Vn

Vc Vs

f c bw d

11.5.4.2

where: d

21.2.1

Shear strength reduction factor:

d

w

Vn 7.5.1.1

Vn > Vu?

Vs

Vc

Vn

6

Vn

Vu

6

OK

Therefore, shear reinforcement is theoretically not required.

  

Eng.Mohammed Ashraf



7.4.3.2



Out-of-plane shear The closest inclined crack to the support of the basement wall will extend upward from the face of the foundation reaching the compression zone about d from the face of the base. Accordingly, the Code permits design for a maximum factored shear force Vu at a distance d from the support for nonprestressed members when the conditions listed

Therefore, the critical section for shear strength is taken at a distance d from the bottom of the stem: 7.5.3.1 22.5.5.1 21.2.1

Vc

0.17

f c bd

Vn

(0.75)(0.17) 31 MPa (1000 mm)(190.5 mm) 135.2 kN

Shear strength reduction factor: Vc

7.5.1.1 11.6.1

Vn

Assume Vs = 0

Vc = 132.5 kN > Vu = (1.6)(18.7 kN) = 29.9 kN OK shear reinforcement is not required

Vu

Provided minimum reinforcement as required by Table 11.6.1 because Vc Vu = 0.0012

t

= 0.002 placing No. 16 or smaller

2 /m As,t Use No. 13 at 400 mm on center at each face 2 2 As,l(provided) provided = 645 mm /m > As,l(required) = 300 mm /m

As,t

2

/m

Use No. 13 at 450 mm on center at the moment face (interior face) and No. 13 at 450 mm on center at the exterior face As,t(prov’d) = 287 mm2 + 287 mm2 = 574 mm2/m OK As,t(prov’d) = 574 mm2 > As,t(required) = 500 mm2/m Step 5: Reinforcement detailing 11.7.2.1 Maximum spacing of vertical bars is the lesser of: a. 3h = 3(250 mm) = 750 mm or b. 450 mm (controls) 11.7.3.2 Maximum spacing of horizontal bars is the lesser of: c. 5h = 5(250 mm) = 1250 mm or d. 450 mm (controls) Step 6: Foundation 13.3.1.1 Even though reinforcement is provided between the wall and footing, the narrow basement wall footing consequently the wall is assumed to hinge at the foundation. Therefore, no moment is being transferred from the wall to the foundation and the foundation can be designed for transmitting unfactored forces to the soil.

Areq'd

P qall

64 kN/m 0.53 m 2 120 kN/m 2 say, 600 mm width of footing. Reinforce footing with four No. 13 continuous. Areq ' d

  

Eng.Mohammed Ashraf





Step 7: Details

Fig. E1.8—Retaining wall reinforcement.

  

Eng.Mohammed Ashraf





Retaining Walls Example 2—Reinforced concrete cantilever wall

3 2

.

Given: Soil data–– s

3

qall

2

Concrete–– a

fy = 420 MPa Cantilever wall–– h1 = 4600 mm h2 = 900 mm

Fig. E2.1—Cantilever retaining wall. ACI 318M-14

Discussion Step 1: Material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements. The designer determines the durability classes. Please refer to Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

Calculation By specifying that the concrete mixture shall be in accordance with ACI 301M and providing the exposure classes, Chapter 19 requirements are

Based on durability and strength requirements, and experience with local mixtures, the compressive strength

coordinated with ACI 318M. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 1 of this Chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes.

  

Eng.Mohammed Ashraf



Step 2: Equivalent lateral pressure The geotechnical report provides the equivalent lateral pressure the wall is required to resist. Step 3: Preliminary cantilever wall data General criteria The preliminary retaining wall dimensions are determined from guidelines presented by Bowles Foundation Analysis and Design, The McGraw-Hill Companies, Inc., 1996. Figure E2.2 shows the variables used in the example.

p



3

Fig. E2.2 E2.2—Retaining E Retaining wall required dimensions. Estimating stem thickness, tstem h h, but Estimating base thickness, tbase at least 300 mm Engineers commonly specify the base be at least as thick as the stem.

h = h1 + h2 = 4.6 m + 0.9 m = 5.5 m tstem ~ tbase ~

13.3.1.2 20.6.1.3.1

OK btoe +tstem + bheel btoe: ~ b/4 to b/3 Heel length: bheel = bbase– btoe – tstem

h

bbase btoe bheel = 3000 m – 800 mm – 375 mm = 1825 mm

  

Eng.Mohammed Ashraf





Step 4: Applied forces For out-of-plane moment and shear, the cantilevered retaining wall is assumed to be continuous, and a representative 1 m strip is analyzed for the

The vertical weight is the self-weight of the retain retainabove the heel. The soil weight over the toe is nene glected as it may erode away or be removed. Figure E2.3 shows the forces applied to the retaining wall: Fig. E2.3 E2.3—Applied E Applied forces on retaining wall. Stem wall: P1 = conc tstem h – tbase Base: P2 = gconc tbase b Soil: P3 s h – tbase bheel Total vertical load:

P1 P2 P3 P

3 3 3

Moments The self-weight of the retaining wall and the soil above the heel tend to counteract the overturning moment. Moments taken about the front edge of

Stem wall: M1 = P1 btoe +tstem Base: M2 = P2 bbase Soil: M3 = P3 b – bheel Restoring moment:

M1 M2 M3 MR

The retained soil behind the wall exerts lateral H H = Ca sh2/2

H

where Ca

3

2

1 sin 1 sin

Therefore, this lateral force tends to overturn the retaining wall about the front edge of the toe: MOTM = H h Summation of moments: M MR – MOTM

M OTM

5.5 m 3

M

  

Eng.Mohammed Ashraf





Step 5: Soil pressure 13.3.1.1 The cantilever wall base is checked using unfactored forces and allowable soil bearing pressure. To calculate soil pressure, the location of the vertical resultant force must be determined. The distance of the resultant to the front face of stem: a

M P

a

288.2 kN·m 230.6 kN

1.25 m

location and the base mid-length: e = bbase/2 – a

e = 3 m/2 – 1.25 m = 0.25 m

Check if resultant falls within the middle third (kern) of the base.

bbase 6

Maximum and minimum soil pressure:

Therefore, there is no uplift.

q1,2

P A

Pe S

q1,2

3m 6

e

0.5 m

230.6 kN (3 m)( m)(1 m)

0.25 m

(230.6 kN)(0.25 m) (3 m) 2 /6

q1 = qmax = 115.3 kN/m2 < qall. = 145 kN/m2 q2 = qmax = 38.4 kN/m2 > 0 kN/m2 Soil bearing pressure is acceptable. Step 6: Stability requirements Calculate the factor of safety against overturning: FS

MR M OTM

2.0

416 kN·m 3.26 127.8 kN·m FS = 3.26 > 2.0 OK FS

Calculate the factor of safety against sliding: FS

P H

1.5

FS

(0.5)(230.6 kN) (4.6 kN/m3 )(5.5 m)(5.5 m)/2

FS = 1.66 > 1.5

1.66

OK

This calculation neglects the passive pressure against the toe (conservative). Conclusion The retaining wall preliminary dimensions are adequate to resist overturning, sliding, preventing uplift, and limiting pressure on the soil to less than the allowable provided soil pressure in the geotechnical report. In the following steps, the retaining wall is designed for strength. If any of the aforementioned determined dimensions are not satisfactory, then all the previous steps must be revised. Note: Unfactored loads were used to determine the stability of the retaining wall and to calculate the soil pressure.

  

Eng.Mohammed Ashraf





Step 7: Stem design The cantilevered concrete stem is a determinate member and is modeled as a 1 m wide cantilever beam.

13.2.7.1

Flexure The maximum design moment in the stem is calcu-

earth pressure and is placed near the face of the stem wall that is against the retained soil. Adequate concrete cover protects bars against moisture changes in soil. Cover is measured from the concrete surface to the outermost surface of the reinforcing bar.

Fig. E2.4—Soil lateral force on stem.

Stem height:

hstem

20.6.1.3.1

From Table 20.6.1.3.1, use 50 mm cover

21.2.2

Assume that the stem wall is tension controlled; minimum steel strain t

7.4.1.1 Hu Mu

U = 1.6H;; when lateral pressure acts alone. The moment is taken at the bottom of the stem at the base. 7.5.1.1 22.2.2.1

Mn

3

Mu

The concrete compressive strain at which nominal moments are calculated is: c = 0.003

22.2.2.2

able property and its value is approximately 10 to 15 percent of the concrete compressive strength. For calculating nominal strength, ACI 318M neglects the concrete tensile strength.

22.2.2.3

Determine the equivalent concrete compressive stress for design: The concrete compressive stress distribution is inelastic at high stress. The actual distribution of concrete compressive stress is complex and usually not known explicitly. The Code permits any stress distribution to be assumed in design if shown to result in predictions of nominal strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85fc depth of:

  

Eng.Mohammed Ashraf



22.2.2.4.1

a 1c 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3.

22.2.2.4.3

For fc

22.2.1.1

1



0.85

0.829

7 MPa

Find the equivalent concrete compressive depth a by equating the compression force and the tension force within a unit length of the wall cross section: C = T, where C = 0.85fc ba and T = Asfy

a a

As

As 0.0159 As

Calculate required reinforcement area: 7.5.2.1 22.3

Mn

a 2

f y As d

Mn

0.0159 As 2

As d

21.2.1

20.6.1.3.1

d = tstem – cover – db/2 use 50 mm cover Mn

7.5.1.1

d = 375 mm – 50 mm – 25 mm/2 = 312.5 mm

Mu

Mu = 165 × 106

6

As,req’d = 1456 mm2

Solving for As 9.6.1.2

0.0159 As 2

As

The cantilevered retaining wall calculated required tensile reinforcement is usually very small compared to the member concrete section. The stem reinforcement is checked against the area rather than the one-way slab minimum reinforcement area because of the lack of redundancy. The Code requires that the beam reinforcement area be at least the greater of: As,min

0.25 f c fy

bw d

As , min

0.25 31 MPa 420 MPa

2

As,prov. = 1700 mm2/m > As,min = 1036 mm2/m As,prov. = 1700 mm2/m > As,req’d = 1456 mm2/m compressive strength fc 21.2.2

Check if the section is tension controlled and the

To answer the question, the tensile strain in reinto the values in Table 21.2.2. Assume concrete and nonprestressed reinforcement strain varying pro-

2

c=

0.829

32.6 mm

  

Eng.Mohammed Ashraf

OK



22.2.1.2



c t

c

0.003 32.6 mm t = 0.026 > 0.005

d c

where: c ously.

t

a

and a = 0.0159As derived previ-

1

Fig. E2.5—Strain distribution across stem. 7.4.3.2

Shear The closest inclined crack to the support of the cantilevered wall will extend upward from the face of the base reaching the compression zone approxiapproxi mately d from the face of the base. The lateral load applied to the cantilever between the face of the base and point d away from the face is transferred cantiledirectly to the base by compression in the cantile ver above the crack. Accordingly, the Code permits design for a maximum factored shear force Vu at a distance d from the support for nonprestressed members.

Vu = Hu

For simplicity, the critical section for design shear strength in this example is calculated at the bottom of the stem: Vn

f c bw d

Vc Vs

7.5.3.1 22.5.5.1

with Vs = 0

21.2.1

Shear strength reduction factor: Vn

7.5.1.1

Vn Vu

Vc is:

Vc Vn Vc Vu Shear reinforcement is not required.

OK

  

Eng.Mohammed Ashraf





Step 8: Heel design Shear The base heel is designed for shear caused by the superimposed weight of soil, including self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may

Fig. E2.6—Force on heel. 21.2.1

Shear strength reduction factor:

7.4.3.2

The critical section for shear strength is taken at a distance d from the bottom of the stem: Vn Vc Vs f c bw d

7.5.3.1

Vc 3

22.5.5.1

20.6.1.3.1

with Vs = 0 d = tbase – cover – db/2

d = 375 mm – 75 mm – 25/2 mm = 287.5 mm

20.6.1.3.1, use 75 mm cover to the From Table 20.6.1.3.1,

Vn

Vc is:

Vn

3

Vc

3

A load factor of 1.2 is used for the concrete self3

Vu 3

Vu 7.5.1.1 22.5.5.1

Vn Vn

Vu

Vn

Vc Vs

Vc

Vu

NG

f c bw d

To increase Vn, two options are possible: 1. Reduce heel length 2. Increase base thickness Reducing the heel length will reduce the weight of the retained earth on the heel and consequently reduce the shear demand. However, this will also reduce the cantilevered retaining wall resistance against sliding and will reduce the length of the base in contact with soil, thus increasing the soil pressure as the change in soil pressure is a function of d2. In this example, the base thickness is increased by the factor Vu Vn

Vu Vn

1.13

  

Eng.Mohammed Ashraf





The revised base thickness is: Minimum required base thickness: dreq’d = hreq’d – cover – db/2

20.6.1.3.1

Use 75 mm cover to the soil.

22.5.1.1

Revised shear strength of heel: The heel thickness increase will result in change to the calculations performed in Steps 4 through 8 above.

hreq’d

dreq’d = 500 mm – 75 mm – 12.5 mm = 412.5 mm, say, 412 mm

Vc

3

Vu 3

Below is a summary of the recalculated values. Shear, revised Balancing moment: Safety factor against overturning: Force causing sliding: Resisting force: Safety factor against sliding: Soil pressure:

Stem: Factored moment: Required tension reinforcement: Design shear strength: Factored shear: Heel: Design shear strength: Factored shear:

Vu P MBAL MOTM FS = 3.29 H Hfr FS = 1.68 qmax = 117 MPa qmin = 39 MPa

Mu As = 1375 mm2 Vn Vu

Vn Vu Vn > Vu

OK

  

Eng.Mohammed Ashraf





Flexure imposed weight of soil and self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may not be linear as assumed. Therefore, it is not included in

6.6.1.2

The cantilever heel maximum moment occurs at moments cannot occur. Fig. E2.7 – Heel critical moment section.

5.3.1 5.3.8

A load factor of 1.2 is used for the concrete self3

Mu

2

/2 2 /2

3

Mu 22.2.1.1

Setting C = T

a a

7.5.2.1

Mn

f y As d

7.5.1.1

Mn Mu

Mu

a 2

Mn

0.0159 As 0.0159 As 2

As

0.0159 As 2

As As,req’d = 1658 mm2

Solving for As: 9.6.1.2

As

As

The cantilevered retaining wall calculated tensile reinforcement is very small compared to the member concrete section. Because the wall is structurreinforcement area is at least the greater of:

As,min

0.25 f c fy

bw d

compressive strength fc

21.2.2

Check if the section is tension controlled and the

22.2.1.2

To answer the question, the tensile strain in reinto the values in Table 21.2.2. Concrete and nonprestressed reinforcement strain is assumed to vary proportionally from the neutral axis. From similar

As , min

0.25 31 MPa 420 MPa

2

As,prov. = 1700 mm2 > As,min = 1200 mm2 As,prov. = 1700 mm2 > As,req’d = 1658 mm2

OK

2

c=

0.829

32.6 mm

  

Eng.Mohammed Ashraf





c t

c

d c

0.003 32.6 mm t = 0.035 > 0.005 t

where: c

a

and a = 0.0159As

1

Fig. E2.8—Heel reinforcement. Step 9: Toe design the bearing pressure. The weight of the soil above the toe is usually

21.2.1 22.5.5.1

Shear Shear strength reduction factor:

Vc Vs

f c bw d

Vn:

5.3.1

Assume Vs = 0

The critical section for shear strength is taken at a distance d from the stem. Vn

7.4.3.2

Fig. E2.9 E2.9—Shear E Shear at toe side.

The applied force on the toe is from the soil-bearing pressure acting upwards and the applied shear is calculated at distance d from stem:

Vn Vn

p

2

2

For soil reaction on the toe portion, a load factor of 1.6 is taken. For the self-weight of concrete portion, Vu a load factor of 0.9 is used.

2.2 m 0.412 m 3m

2

2

2

2

2 3

Vu 7.5.1.1

Vn

Vu

Vn

Vu

OK

Steel reinforcement is not required.

  

Eng.Mohammed Ashraf





Flexure Soil bearing pressure at face of stem:

p

2

3

stem L

2.2 m 3m

2

2

7.4.1.1

2

Mu2

Moment due to soil:

2

2

5.3.1

20.6.1.3.1

22.2.1.1

Moment due to toe self-weight; a load factor of 0.9 is used: Resultant moment:

Mu1 Mu,req’d = Mu2

Concrete is placed against earth. Therefore, use 75 mm cover.

d = 412 mm

7.5.2.1

Mn

f y As d

7.5.1.1

Mn Mu

Mu

a 2

2

3

2

Mu1

Setting C = T to calculate required reinforcement: C = 0.85fc ba and T = Asfy

a a=

/2 +

2

As

As 0.0159 As

Mn

0.0159 As 2

As

0.0159 As 2

As

As,req’d = 618 mm2 As,provid = 1700 mm2/m > As,req’d OK

Solving for As:

Minimum required reinforcement is the greater of

9.6.1.2

As,min

0.25 f c fy

bw d

As , min

0.25 31 MPa 420 MPa

2

Provide same reinforcement area as in the stem: compressive strength fc = 31 MPa. 21.2.2

Check if the section is tension controlled assump-

21.2.2

To answer the question, the tensile strain in reinto the values in Table 21.2.2. The strain in concrete and reinforcement is assumed to vary proportionally from the neutral axis. From similar triangles

2

c=

0.829

32.6 mm

  

Eng.Mohammed Ashraf



22.2.1.2



c t

c

d c

where: c

a

0.003 32.6 mm

t

and a = 0.0159As

1

t

= 0.035 > 0.005. Section is tension controlled.

Fig. E2.10—Toe reinforcement. Step 10: Minimum transverse reinforcement Stem 11.6.1 transverse wall reinforcement. Per Table 11.6.1,

min = 0.002 Distribute transverse reinforcement equally between the front and back face of the stem wall.

As,min As,min

As , prov.

2 front

As,min

199 mm 2 As , req ' d

back

2

2

= 750 mm /2 = 375 mm

1000 mm 450 mm

442 mm 2

375 mm 2 /face

Provide vertical reinforcement at front face to support the transverse wall reinforcement. Base 24.4.3

ture and shrinkage reinforcement. The reinforcement can be located at the top, bottom, or allocated between the two faces.

As,min

2

  

Eng.Mohammed Ashraf





Step 11: Dowels 7.7.1.2 The development length concept is based on the 25.4.2 attainable average bond stress over the embedment length of reinforcement. Development lengths are required because of the tendency of highly stressed bars to split relatively thin sections of restraining concrete. In application, the development length concept requires minimum lengths or extensions of reinforcement beyond all points of peak stress in the reinforcement. Such peak stresses generally occur at the points of critical sections for development nates, or is bent. For the cantilevered wall, toe and heel reinforcement must be developed. Heel reinforcement is developed beyond the stem critical section. Toe reinforcement must be checked for: opment length in the toe section. 2.. Extension of toe reinforcement in the stem to satisfy the required development length.

checked against the stem reinforcement required splice length and the larger length controls. 25.4.2.1

Development length

25.4.2.2

1.

d

25.4.2.3

2.

d

fy 1.7 1 1.1

t

e

fc fy

d

is the greater of Eq.

db

t

f c cb

e

s

K tr

db

db

25.4.2.4

t = bar location; not more than 300 mm of fresh concrete below horizontal reinforcement e = coating factor; uncoated

25.4.2.3

cb = spacing or cover dimension to center of bar, whichever is smaller

s

t = 1.0, because not more than 300 mm of concrete is placed below bars e = 1.0, because bars are uncoated s

Ktr = transverse reinforcement index It is permitted to use Ktr = 0. But the expression: cb

Ktr db

63 mm 0 25 mm

2.5

2.52

2.5

must not be taken greater than 2.5.   

Eng.Mohammed Ashraf



25.4.2.3



Substitute in Eq. (25.4.2.3a) The available length for toe bars (800 mm) is long enough to accommodate the required development length of (686 mm) for No. 25.

Development length of heel and toe reinforcement:

d

Note: the development length of No. 25 bar extended d from the base into the stem must be checked against the splice length of stem reinforcement. The Less than the length available (refer to Fig. E2.10) OK larger length controls

25.5 25.5.1.1

25.5.2.1

25.2.3

Splice The maximum bar size is No. 25, therefore splicing is permitted. The required development length for No. 25 bar extended from the toe is calculated previously as 686 mm The deformed bars are in tension and the ratio of provided reinforcement area to required reinforcement area is less than 2. Therefore, per Table 25.5.2.1,, splice is Type B and the splice length is the greater of (1.3 d) or 300 mm Clear spacing between bars is the greater of: (a) 40 mm (b) 1.5db (c) 4/3dagg Assume 25 mm maximum aggregate size is used.

Use 1000 mm for splice length Therefore, extend No. 25 toe bar 1000 mm into the E stem. Refer to Fig. E2.11.

(a) 40 mm

Therefore, 40 mm controls

  

Eng.Mohammed Ashraf





Step 12: Details Weight of steel per linear meter in stem only:

Fig. E2.11—Retaining wall reinforcement.

  

Eng.Mohammed Ashraf





Step 13: Alternate detailing coming from the base/toe higher into the stem to a level where smaller bar diameters, larger bar spacing, or both can be used for the stem reinforcement. 1500 mm into the stem. The factored applied moment at that level can be calculated from: Mu = Ca ho3/6

Mu

The required reinforcement area:

As,req’d = 460 mm2/m

3

3

Use a bar size such that the required calculated As,prov = 663 mm2/m > As,req’d

OK

from the base. As,min As,prov = 663 mm2/m > As,min

11.6.1 minimum required in Table 11.6.1. 7.7.1.3

2

OK

Required splice length:

25.5.1.1 is permitted. 25.5.2.1

The deformed bars are in tension and the ratio rein of provided reinforcement area to required reinforcement area is less than 2.. Therefore, per Table 25.5.2.1,, splice is Type B and the splice length is ld d is the development length calculated previously.

25.4.2.3 d

1.1

2.5

= 436 mm, say, 450 mm Splice 7.7.1.3

st

Use 25.5 25.5.2.2

d

= 750 mm >

st

= 567 mm

Refer to Fig. E2.12. greater of d of the larger bar and bar is used for the splice length.

st

of the smaller

300 mm on center. Use overlap length of 750 mm

  

Eng.Mohammed Ashraf





Step 14: Details Weight of steel per linear foot in stem only:

Fig. E2.12—Retaining Retaining wall reinforcement (alternate solution). Conclusion: Alternate solution provides a saving of: 23.8 kg/m – 12.6 kg/m = 11.2 kg/m or 1

12.6 kg/n 23.8 kg/m

47 percent.

  

Eng.Mohammed Ashraf





Retaining Walls Example 3 Design a normalweight reinforced concrete cantilever wall that retains a 10-degree sloped earth bank 3.7 m high above

m3 2

.

Given: Soil data–– s

3

qall

2

Concrete–– a = 1.0 normalweight concrete fy = 420 MPa Cantilever wall–– h1 = 3700 mm h2 = 1100 mm

Fig. E —Cantilever retaining wall. E3.1—Cantilever ACI 318M-14

Discussion

Step 1: Material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements. The designer determines the durability classes. Please see Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

Calculation By specifying that the concrete mixture shall be in acac cordance with ACI 301M and providing the exposure

Based on durability and strength requirements, and experience with local mixtures, the compressive least 31 MPa.

coordinated with ACI 318M. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 1 of this Chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes. Step 2: Equivalent lateral pressure p The geotechnical report provides the equivalent lateral pressure the wall is required to resist.

3

  

Eng.Mohammed Ashraf





Step 3: Preliminary cantilever wall data General criteria The preliminary retaining wall dimensions are determined from guidelines presented by Bowles Foundation Analysis and Design, The McGraw-Hill Companies, Inc., 1996. Figure E3.2 shows the variables used in the example.

Fig. E3.2 E3.2—Determine Determine retaining wall dimensions. h = h1 + h2 = 3.7 m + 1.1 m = 4.8 m h h,, but at least 300 mm

tstem tbase

Engineers commonly specify the base to be slightly thicker than the stem. 13.3.1.2 20.6.1.3.1

inspection larger than 150 mm h b/4 to b/3 Heel length: bheel = bbase – btoe – tstem

bbase btoe bheel = 2.6 m – 0.7 m – 0.38 m = 1.52 m

heel end:

h

  

Eng.Mohammed Ashraf





Step 4: Applied forces For out-of-plane moment and shear, the cantilevered retaining wall is assumed to be continuous, and a representative 1 m strip is analyzed for the

Fig. E3.3 E3.3—Acting Acting forces on retaining wall. hss1 hs2

Soil height at stem: Soil height at base end:

The vertical weight is the self-weight of the soil above the heel. The soil weight over the toe is neglected as it may erode away or be removed: Stem wall: P1 tbase conc tstem h Base: P2 t b conc base Soil: P3 s h – tbase bheel Weight of sloped soil above stem: H: Total vertical load:

P1 P2 P3 P4 HV = P5 P

3 3 3 3

2 3

2

Moments The self-weight of the retaining wall and the soil above the heel tend to counteract the overturning moment. Moments taken about the front edge of Stem wall: M1 = P1 btoe + tstem Base: M2 = P2 bbase Soil: M3 = P3 bbase heel Sloped soil: M4 = P4 bbase – bheel

M1 M2 M3 M4

Moment of lateral soil pressure vertical component: M5 M5 = HV bbase

  

Eng.Mohammed Ashraf



Restoring moment:

The retained soil behind the wall exerts lateral pressure (H) on the wall: H = Ca sh2/2 where Ca is the Rankine’s equation obtained from any soils textbook.



MR = (34.5 kN·m) + (35.8 kN·m) + (231.1 kN·m) + (82 kN·m) + (31.7 kN·m) = 341.3 kN·m, say, 342 kN·m

H = (5.4 kN/m3)(5.1 m)2/2 = 70.2 kN

Horizontal component of lateral force: (H): Hh = H

Hh = (70.2 kN)(cos10°) = 69.2 kN

Therefore, this lateral force tends to overturn the retaining wall about the front edge of the toe: Summation of moments about the front of toe:

MOTM = (69.2 kN)(5.1 m)/3 = 118 kN·m

M

MR – MOTM

M = (342 kN·m) – (118 kN·m) = 224 kN·m

overturning. Step 5: Soil pressure 13.3.1.1 The aforementioned determined cantilever wall base is checked using unfactored forces and allowallow able soil bearing pressure. To calculate soil pressure, the location of the verti vertical resultant force must be determined. The distance of the resultant to the front face of stem: a

M P

a

224 kN·m 208 kN

1.08 m

location and the base mid-length: e = bbase/2 – a

e = 2.6 m/2 – 1.08 m = 0.22 m

Check if resultant falls within the middle third of the base.

bbase 6

2.6 m 6

0.43 m

e

0.22 m

OK

Therefore, there is no uplift. Maximum and minimum soil pressure: q1,2

P A

Pe S

q1,2

208 kN (208 kN)(0.22 m) ± (2.6 m)(1 m) (2.6 m) 2 /6

q1,2 = 80 kN/m2 ± 40.6 kN/m2 q1 = qmax = 120.6 kN/m2 < qall = 145 kN/m2 q2 = qmax = 39.4 kN/m2 > 0 kN/m2 no uplift Soil bearing pressure is acceptable. Use qmax = 121 kN/m2 qmin = 39 kN/m2

  

Eng.Mohammed Ashraf





Step 6: Stability requirements Calculate the factor of safety against overturning: FS

M Res M OTM

2.0

FS

2.9

FS = 2.9 > 2.0

OK

Calculate the factor of safety against sliding:

FS

P H

1.5

FS FS = 1.50

3



OK

This calculation neglects the passive pressure Conclusion The retaining wall preliminary dimensions are adequate to resist overturning, sliding, preventing uplift, and limiting pressure on the soil to less than the allowable provided soil pressure in the geotechnical report. In the following steps, the retaining wall is designed for strength. If any of the aforementioned determined dimensions are not satisfactory, then all the previous steps must be revised.

  

Eng.Mohammed Ashraf





Step 7: Stem design The cantilevered concrete stem is a determinate member and is modeled as a 1 m wide cantilever

13.2.7.1

Flexure The maximum design moment in the stem is calculated at the face of the base foundation. earth pressure and is placed near the face of the stem wall that is against the retained soil. Adequate concrete cover protects reinforcement against moisture changes in soil. Cover is measured from the concrete surface to the outermost surface Fig. E3.4—Soil lateral force on stem. of the reinforcing bar. Stem height: Soil height at heel end:

20.6.1.3.1 21.2.2

hstem hsoil

From Table 20.6.1.3.1, use 50 mm cover. Assume that the stem wall is tension controlled; minimum steel strain t

7.4.1.1 U = Hu = 1.6H;; when lateral pressure acts alone.

Hu

The moment is taken at the bottom of the stem at the base:

Mu

3

2

o

say, 140 × 106 7.5.1.1 22.2.2.1

Mn

Mu

The concrete compressive strain at which nominal c = 0.003

22.2.2.2 variable property and its value is approximately 10 to 15 percent of the concrete compressive strength. For calculating nominal strength, ACI 318M neglects the concrete tensile strength.

  

Eng.Mohammed Ashraf



22.2.2.3

22.2.2.4.1

22.2.2.4.3 22.2.1.1



Determine the equivalent concrete compressive stress for design: The concrete compressive stress distribution is inelastic at high stress. The actual distribution of concrete compressive stress is complex and usually not known explicitly. The Code permits any stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85fc a 1c 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3.

For fc

1

0.85

0.829

7 MPa

Find the equivalent concrete compressive depth a by equating the compression force to the tension force within a unit length of the wall cross section: C=T

As a

As

0.0159 As

Calculate required reinforcement area: 7.5.2.1 22.3

Mn

a 2

f y As d

Mn

0.0159As 2

As d

21.2.1

20.6.1.3.1

d = tstem – cover – db/2 use 50 mm cover Mn

7.5.1.1

d = 380 mm – 50 mm – 22 mm/2 = 319 mm

Mu

Mu

Mu = 140 × 106

As

0.0159As 2

6

As = 1197 mm2

Solving for As

As,prov. = 1290 mm2/m > As,req’d = 1197 mm2/m 9.6.1.2

OK

The cantilevered retaining wall calculated required tensile reinforcement is usually very small compared to the member concrete section. The stem reinforcement is checked against the beam minimum one-way slab minimum reinforcement area because of the lack of redundancy. The Code requires that the beam reinforcement area at least: As , min

fc

0.25 f c fy

bw d

As , min

0.25 31 MPa 420 MPa

2

As,prov. = 1290 mm2/m > As,req’d = 1197 mm2/m > As,min = 1005 mm2/m OK   

Eng.Mohammed Ashraf



21.2.2



Check if the tension controlled assumption and the

To answer the question, the tensile strain in reinto the values in Table 21.2.2. Assume concrete and nonprestressed reinforcement strain varying proportional to the distance from the neutral axis

22.2.1.2

c t

c

d c

where: c =

a 1

and a = 0.0159As

2

c=

t

t

0.828

24.8 mm

0.003 24.8 mm = 0.036 > 0.005

Fig. E3.5—Strain distribution across stem.

  

Eng.Mohammed Ashraf



7.4.3.2



Shear The closest inclined crack to the support of the cantilevered wall will extend upward from the face of the base reaching the compression zone approximately d from the face of the base. The lateral load applied to the cantilever between the face of the base and point d away from the face is transferred directly to the base by compression in the cantilever above the crack. Accordingly, the Code permits design for a maximum factored shear force Vu at a distance d from the support for nonprestressed members.

2

Vu

3



2

For simplicity, the critical section for design shear strength in this example is calculated at the bottom of the stem: f c bw d

7.5.3.1 22.5.5.1

Vn Vc Vs with Vs = 0

21.2.1

Shear strength reduction factor:

0.17

Vn Vn

7.5.1.1

Vc is:

Vn

Vn

Vu

Vn

Vu

OK

Shear reinforcement is not required. Step 8: Heel design Shear The base heel is designed for shear caused by the superimposed weight of soil, including self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may

Fig. E3.6—Force on heel. 21.2.1

Shear strength reduction factor:

7.4.3.2

The critical section for shear strength is taken at a distance d from the bottom of the stem:

7.5.3.1 22.5.5.1

Vn

20.6.1.3.1

d = tbase – cover – db/2

Vc Vs

with Vs

f c bw d

Vn

0 where d = 450 mm – 75 mm – 22 mm/2 = 364 mm

  

Eng.Mohammed Ashraf





From Table 20.6.1.3.1, use 75 mm cover to tension reinforcement and assume No. 22 bars. Vn

Vc is:

Vn

Vu = (1.2)(23.5 kN/m3)(1.52 m – 0.364 m)(0.45 m) + (1.6)(19 kN/m3)(4.35 m + (5.1 m – 0.45 m))/2 × (1.52 m – 0.364 m)(1 m) = 172.8 kN Vn = 258 kN > Vu = 172.8 kN OK

A load factor of 1.2 is used for the concrete self-

7.5.1.1

Vc = (0.75)(344.5 kN) = 258 kN

Vn Vu Flexure -

linear as assumed. Therefore, it is not included in

6.6.1.2

The heel maximum moment occurs at the stem face. Redistribution of moments cannot occur.

Fig. E E3.7–Heel critical moment section. 5.3.1 5.3.8

A load factor of 1.2 is used for the concrete self-

)(0.45 m) + 1.6(19 kN/m3) Mu = (1.2(23.5 kN/m3)( × ((4.35 4.35 m + (5.1 5.1 m – 0.45 m))/2(1 m))(1.52 m)2/2 = 172.7 kN·m

22.2.1.1

Setting C = T:

MPa)(1000 0.85 0.85(31 1000 mm)(a) mm)( = As(420 MPa) a

7.5.2.1

Mn

a 2

f y As d

Mn

As (420 420 MPa) 0.85 0.85(31 MPa)(1000 mm)

0.0159 As

(420 420 MPa) As 364 mm

0.0159 As 2

21.2.1(a) 7.5.1.1

Mn

(0.9)(420 MPa) As 364 mm

Mu

Mu = 172.7 × 106 N·mm calculated above.

Solving for As:

172.7

0.0159 As 2

106 N·mm

As = 1290 mm2

  

Eng.Mohammed Ashraf



9.6.1.2



The cantilevered retaining wall calculated tensile reinforcement is very small compared to the member concrete section. Because the wall is structur-

0.25 f c

As ,min

fy

bw d

As ,min

0.25 31 MPa 420 MPa

2

As,prov = 1290 mm2 = As,req’d = 1290 mm2 > As,min = 1206 mm2

OK

is only used. 21.2.2

Check if the tension controlled assumption and the

To answer the question, the tensile strain in reinto the values in Table 21.2.2. Concrete and nonprestressed reinforcement strain is assumed to vary proportionally from the neutral axis. From similar triangles:

22.2.1.2

c t

c

2

c=

d c

where: c =

a

t

t

0.828

24.8 mm

0.003 24.8 mm = 0.041 > 0.005

and a = 0.0159A 0.0159 s

1

Fig. E3.8––Heel reinforcement.

  

Eng.Mohammed Ashraf





Step 9: Toe design the bearing pressure. The weight of the soil above the toe is usually

Fig. E3.9—Shear at toe side. 21.2.1

Shear Strength reduction factor:

9.4.3.2

The critical section for shear strength is taken at a distance d from the stem:

7.5.3.1 22.5.5.1

Vn

f c bw d

Vc Vs

with Vs

Vn

0

Vn

Vn: The applied force on the toe is from the soil bearing pressure acting upwards. 7.4.3.2

Applied shear at distance d from stem: From Step 5: qmax qmin

5.3.1

7.5.1.1

p

d

2

=

2

1.9 m + 0.364 mm 2.6 m

2

For soil reaction on the toe portion a load factor of 1.6 is taken. For the self-weight of concrete portion, Vu a load factor of 0.9 is used.

Vn

Vu

2

Vn

2

Vu

2

=

2

OK

Shear reinforcement is not required

  

Eng.Mohammed Ashraf

2





Flexure Soil bearing pressure at stem:

p

2

2

stem

1.9 m 2.6 m

2

2

7.4.1.1

2

Mu2

Moment due to soil:

2

/2

2

5.3.1

20.6.1.3.1

22.2.1.1

3

2

2

Moment due to toe self-weight:

Mu1

Resultant moment:

Mu,req’d = Mu2 Mu1

Concrete is placed against earth. Therefore, use 75 mm cover.

d = 450 mm – 75 mm – 22 mm/2 = 364 mm a

Setting C = T to calculate required reinforcement: As

a

7.5.2.1

Mn

7.5.1.1

Mn

f y As d

a 2

Mn

As 0.0159 As 0.0159 As 2

As

0.0159

Mu

Mu Solving for As

9.6.1.2

2

As,req’d = 352 mm2 Provide same reinforcement area as in the stem: As,prov. = 1290 mm2 As,provided > As,req’d OK

Minimum required reinforcement: 0.25 f c

As,min

fy

bw d

As ,min

0.25 31 MPa 420 MPa

2

As,prov’d = 1290 mm2 > As,mm = 1206 mm2

OK

compressive strength, fc 21.2.1

Check if the tension controlled assumption of the

21.2.2

To answer the question, the tensile strain in reinto the values in Table 21.2.2. The strain in concrete and reinforcement is assumed to vary proportionally from the neutral axis. From similar triangles

22.2.1.2

c t

c

d c

where: c

a

2

c

0.828

t

and a = 1.31 As

24.8

0.003 24.8 mm

t = 0.041 > 0.005 Section is tension-controlled

1

  

Eng.Mohammed Ashraf





Fig. E3.10—Toe reinforcement. Step 10: Minimum transverse reinforcement Stem 11.6.1 for transverse reinforcement. Per Table 11.6.1, min = 0.002 Distribute shrinkage and temperature reinforcement equally between the front and back face of the stem wall.

2

As,min

As,min

As,min

front

back

2

As , prov. As , req ' d

= 760 mm2/2 = 380 mm2

1000 mm 450 mm

2

380 mm 2 /face

horizontally. Provide vertical reinforcement at front face to support the transverse wall reinforcement. Base

exterior face.

24.4.3 for temperature and shrinkage reinforcement. The reinforcement can be located at the top, bottom, or allocated between the two faces.

2

As,min

  

Eng.Mohammed Ashraf





Step 11: Dowels 7.7.1.2 The development length concept is based on the 25.4.2 attainable average bond stress over the embedment length of reinforcement. Development lengths are required because of the tendency of highly stressed bars to split relatively thin sections of restraining concrete. In application, the development length concept requires minimum lengths or extensions of reinforcement beyond all points of peak stress in the reinforcement. Such peak stresses generally occur at the points of critical sections for

reinforcement terminates, or is bent. For the cantilevered wall, toe and heel reinforcement must be developed. Heel reinforcement is developed beyond the stem critical section. Toe reinforcement must be checked for: opment length in the toe section. 2.. Extension of toe reinforcement in the stem to satisfy the required development length.

checked against the splice length and the larger length controls. 25.4.2.1

Development length

25.4.2.2

1.

d

25.4.2.3

2.

d

fy 1.7

1 1.1

t

e

fc

fy

d

db

t

f c cb

e

s

K tr

db

db

25.4.2.4

t = bar location; not more than 300 mm of fresh concrete below horizontal reinforcement e = coating factor; uncoated

25.4.2.3

cb = spacing or cover dimension to center of bar, whichever is smaller

s

t = 1.0 because not more of 300 mm of concrete is placed below it. e = 1.0, because bars are uncoated s

cb = 50 mm + 22 mm/2 = 61 mm < 63

  

Eng.Mohammed Ashraf





Ktr = transverse reinforcement index It is permitted to use Ktr = 0. However, the expression: taken greater than 2.5.

cb

Ktr must not be db

61 mm 0 22 mm

2.77

2.5

Development length of toe reinforcement: d

1.1

2.5

= 603 mm, say, 625 mm

enough to accommodate the required development

d

< 625 mm length available

OK

checked against the splice length and the larger length controls. Splice 7.7.1.3 25.5 25.5.1.1

is permitted. The required development length for the bars extended from the toe is calculated previously as 600 mm

25.5.2.1

The deformed bars are in tension and the ratio of provided reinforcement area to required reinrein forcement area is less than 2. Therefore, per Table 25.5.2.1, splice is Type B and the splice length is

Splice length: st

Controls

d

7.7.2 25.2.3

Clear spacing between bars is the greater of: into the stem. db dagg Assume 25 mm maximum aggregate size is used.

Therefore, 40 mm controls

  

Eng.Mohammed Ashraf





Step 12: Details Weight of steel per lineal foot in stem only:

= 15.82 kg/m

Fig. E3.11—Retaining Retaining wall reinforcement.

  

Eng.Mohammed Ashraf





Step 13: Alternate solution coming from the base/toe higher into the stem to a level where smaller bar diameters, larger bar spacing, or both can be used for the stem reinforcement. 1.5 m into the stem. The factored applied moment at that level can be calculated from: Mu = Ca ho3/6

Mu

The required reinforcement area:

As,req’d = 278 mm2/m

Use bar size such that the required calculated spacing matches the spacing of the reinforcement extended from the base.

As,prov = 663 mm2/m > As,req’d = 278 mm2/m

3

As,min As,prov = 663 mm2/m > As,min = 456 mm2

11.6.1 minimum required in Table 11.6.1. 25.5.1.1

3

OK

2

OK

Required splice length:

25.5.2.2 is permitted.

greater of d of the larger bar and bar is used for the splice length. 25.5.2.1

st

of the smaller

The deformed bars are in tension and the ratio of provided reinforcement area to required reinrein forcement area is less than 2. Therefore, per Table 21.5.2.1, splice is Type B and the splice length is d

25.4.2.3

d

1.1

2.5

t

Splice 25.5 25.5.2.2

d

= 625 mm >

st

= 600 mm; use 650 mm

mm on center. The overlap length is 650 mm.

  

Eng.Mohammed Ashraf





Step 14: Details Weight of steel per linear foot in stem only:

Fig. E3.12—Retaining Retaining wall reinforcement (alternate solution). Conclusion: Alternate solution provides a saving of: 15.8 kg/m – 10 kg/m = 5.8 kg/m or 1

10 kg/m 15.8 kg/m

36.7 percent.

  

Eng.Mohammed Ashraf





Retaining Walls Example 4—Reinforced concrete cantilever wall retaining earth with surcharge load 2 , as shown in Fig. E4.1. The concrete mixture must satisfy durability and strength requirements. The grade of reinforcing bar is assumed at 420 MPa. The soil data were obtained from the geotechnical 3 2

.

Given: Soil data–– 3

s

= 35 degrees = 0.55 2

qall Concrete–– a

fy = 420 MPa Cantilever wall–– h1 = 2700 mm h2 = 760 mm Load–– w

2

– surcharge

Fig. E4.1— E4.1—Cantilever Cantilever retaining wall.

ACI 318M-14

Calculation

Discussion

Step 1: Material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements. The designer determines the durability classes. Please see Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

By specifying that the concrete mixture shall be in acac cordance with ACI 301M and providing the exposure

Based on durability and strength requirements, and experience with local mixtures, the compressive strength

dinated with ACI 318M. ACI encourages referenc-

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 2 of this Chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes. Step 2: Equivalent lateral pressure The geotechnical report provides the equivalent p lateral pressure the wall is required to resist.

3

  

Eng.Mohammed Ashraf





Step 3: Preliminary cantilever wall data General criteria The preliminary retaining wall dimensions are determined from guidelines presented by Bowles Foundation Analysis and Design, The McGraw-Hill Companies, Inc., 1996. Figure E4.2 shows the variables used in the example.

Fig. E4.2—Retaining wall required dimensions. The uniform surcharge from storage area is con converted to an equivalent earth height above the top of the wall: hs

2

w

hs

3

0.9 m

s

Retaining wall height: Equivalent wall height due to surcharge: heq heq, but at

h = h1 + h2 = 2.7 m + 0.76 m = 3.46 m heq = h + hs = 3.46 m + 0.9 m = 4.36 m tstem tbase

least 300 mm Engineers commonly specify the base be at least as thick as the stem. 13.3.1.2

OK heq

20.6.1.3.1 b/4 to b/3 Heel length: bheel = bbase – btoe – tstem

bbase btoe ~ 2200 mm/3 = 733 mm, say, 700 mm bheel

  

Eng.Mohammed Ashraf





Step 4: Applied forces For out-of-plane moment and shear, the cantilevered retaining wall is assumed to be continuous, and a representative 1 m strip is analyzed for the

The vertical load is the self-weight of the retainabove the heel, including the storage area uniform load. The soil weight over the toe is neglected as it may erode away or be removed: tbase Stem wall: P1 conc tstem h Base: P2 t b conc base Soil: P3 tbase bheel s h + hs Total vertical load:

Fig. E4.3 E4.3—Applied forces on retaining wall. P1 P2 P3

3 3 3

P

The self-weight of the retaining wall and the soil and distributed load above the heel tend to counteract the overturning moment. Moments taken about the Stem wall: M1 = P1 btoe + tstem/ Base: M2 = P2 b Soil: M3 = P3 b – bheel

M1 M2 M3

Restoring moment:

·m ·m

MR

The retained soil behind the wall and the superimH Due to retained earth: H1 Ca s h2 Due to surcharge: H2 Ca s hs h Ca is calculated from Rankine’s equation. Ca

H1 H2

2

2

2

1 sin 1 sin

Therefore, this lateral force tends to overturn the retaining wall about the front edge of the toe: MOTM = H1 h

H2 h

MOTM ·m M

MR – MOTM

M turning.

  

Eng.Mohammed Ashraf





Step 5: Soil pressure 13.3.1.1 The aforementioned determined cantilever wall base is checked using unfactored forces and allowable soil bearing pressure. To calculate soil pressure, the location of the vertical resultant force must be determined. The distance of the resultant to the front face of stem: a

M P

a

0.93 m

location and the base mid-length: e = bbase/2 – a

e = 2.2 m/2 – 0.93 m = 0.17 m

Check if resultant falls within the middle third of the base. Maximum and minimum soil pressure: q1,2

P A

M S

bbase 6

2.2 m 6

0.36 m

e

0.17 m

OK

Therefore, there is no uplift. q1,2

q1 ,2

±

2

2

q1 = qmax q2 = qmin

2

2

2

< qall

2

2

no uplift

Soil-bearing pressure is acceptable. Step 6: Stability requirements Calculate the factor of safety against overturning: FS

MR M OTM

2.0

FS

2.94

FS = 2.94 > 2.0

OK

Calculate the factor of safety against sliding: FS

P H

1.5

This calculation neglects the passive pressure

FS FS = 1.56 > 1.5

1.56 OK

Conclusion The retaining wall preliminary dimensions are adequate to resist overturning, sliding, preventing uplift, and limiting pressure on the soil to less than the allowable provided soil pressure in the geotechnical report. In the following steps, the retaining wall is designed for strength. If any of the aforementioned determined dimensions are not satisfactory, then all the previous steps must be revised.

  

Eng.Mohammed Ashraf





Step 7: Stem design The cantilevered concrete stem is a determinate member and is modeled as a 1 m wide cantilever beam.

13.2.7.1

Flexure The maximum design moment in the stem is calculated at the top face of the base foundation.

earth pressure and is placed near the face of the stem wall that is against the retained soil. Adequate concrete cover protects reinforcement against moisture changes in soil. Cover is measured from the concrete surface to the outermost surface of the reinforcing bar. hstem

Stem height: 20.6.1.3.1

From Table 20.6.1.3.1, use 50 mm cover.

21.2.2

Assume that the member is tension controlled; minimum steel strain t

7.4.1.1

Equivalent lateral pressure due to surcharge load:

Psurcharge

U = 1.6H;; when lateral pressure acts alone.

U

3

2

2 3

The moment is taken at the bottom of the stem and

Mu ·m = 77 × 106

7.5.1.1 22.2.2.1

Mn

Mu

The concrete compressive strain at which nominal moments are calculated is: c = 0.003 -

22.2.2.2 able property and its value is approximately 10 to 15 percent of the concrete compressive strength. For calculating nominal strength, ACI 318M neglects the concrete tensile strength.

  

Eng.Mohammed Ashraf



22.2.2.3

22.2.2.4.1

22.2.2.4.3 22.2.1.1



Determine the equivalent concrete compressive stress for design: The concrete compressive stress distribution is inelastic at high stress. The actual distribution of concrete compressive stress is complex and usually not known explicitly. The Code permits any stress distribution to be assumed in design if shown to result in predictions of nominal strength in reasonable agreement with the results of comprehensive tests. Rather that tests, the Code allows the use of an equivalent rectangular compressive stress distribution of: 0.85fc a 1c, where 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3.

For fc

1

0.828

7 MPa

Find the equivalent concrete compressive depth, a, by equating the compression force to the tension force within a unit length of the wall cross section: C=T

a a=

7.5.2.1 22.3

0.85

As

As 0.0159 As

Calculate required reinforcement area: Mn

f y As d

a 2

Mn

As d

0.0159As 2

21.2.1 20.6.1.3.1

7.5.1.1

d = tstem – cover – db/2 use 50 mm cover Mn Mu = 77 × 106

d = 300 mm – 50 mm – 19 mm/2 = 240.5 mm, say, 240 mm Mu As

0.0159 As 2

6

Solving for As As,req’d = 874 mm2 As,prov’d = 947 mm2/m > As,req’d = 874 mm2/m

  

Eng.Mohammed Ashraf



9.6.1.2



The cantilevered retaining wall calculated required tensile reinforcement is usually very small compared to the member concrete section. The stem reinforcement is checked against the area rather than the one-way slab minimum reinforcement area because of the lack of redundancy. The Code requires that the beam reinforcement area is:

As , min

0.25 f c fy

bw d

0.25 31 MPa 420 MPa As,prov’d = 947 mm2/m > As,req’d = 795 mm2/m

2

As ,min

OK

compressive strength fc 21.2.2

Check if the tension controlled assumption and the

To answer the question, the tensile strain in

2

c= compared to the values in Table 21.2.2. Assume concrete and nonprestressed reinforcement strain varying proportional to the distance from the

22.2.1.2

c t

c

where: c

t

d c t

a

0.828

18.2 mm

0.003 18.2 mm = 0.0366 > 0.005

and a = 0.0159A 0.0159 s derived previously.

= 0.9

1

Fig. E4.5—Strain distribution across stem.

  

Eng.Mohammed Ashraf



7.4.3.2



Shear The closest inclined crack to the support of the cantilevered wall will extend upward from the face of the base reaching the compression zone approximately d from the face of the base. The lateral load applied to the cantilever between the face of the base and the point d away from the face are transferred directly to the base by compression in the cantilever above the crack. Accordingly, the Code permits design for a maximum factored shear force Vu at a distance d from the support for nonprestressed members. For simplicity, the critical section for design shear strength in this example is calculated at the bottom of the stem:

Vu

7.5.3.1 22.5.5.1

Vn

Vn

21.2.1

Shear strength reduction factor:

= 0.75

Therefore, Vn = Vc is:

Vn

7.5.1.1

Vc Vs

with Vs

Is

Vn

f c bw d

3 3

2

0

Vu

Vn

Vu

OK

Therefore, shear reinforcement is not required

  

Eng.Mohammed Ashraf





Step 8: Heel design Shear The base heel is designed for shear caused by the superimposed weight of soil, including self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may

Fig. E4.6—Force on heel. 21.2.1

Strength reduction factor:

7.4.3.2

The critical section for shear strength is taken at a distance d from the face of the stem: Vn

= 0.75

f c bw d

Vc Vs

7.5.3.1 22.5.5.1

with Vs

20.6.1.3.1

d = tbase – cover – db/22

0 d = 350 mm – 75 mm – 19 mm/2 = 265.5 mm, say, 265 mm

From Table 20.6.1.3.1, 20.6.1.3.1 use 75 mm cover to the ten tenVn

Then, Vn is: A load factor of 1.2 is used for the concrete self2

.

3

Vu

3

7.5.1.1

Is

Vn

Vu

Vn

Vu

2

OK

  

Eng.Mohammed Ashraf





Flexure imposed weight of soil and self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may not be linear as assumed. Therefore, it is not included in

6.6.1.2

The cantilever wall maximum moment and shear in the heel and toe of the base occur at the stem face. Redistribution of moments cannot occur. Fig. E4.7––Critical moment section in heel.

5.3.1 5.3.8

A load factor of 1.2 is used for the concrete and 1.6

22.2.1.1

Setting C = T

3

Mu1

2

3

2 2

/2

say, 95 × 106 a As

a=

7.5.2.1

Mn

a 2

f y As d

As 0.0159 As

Mn

0.0159As 2

As

= 0.9 7.5.1.1

Mn Mu Mu = 95 × 106

The cantilevered retaining wall calculated tensile reinforcement is very small compared to the member concrete section. Because the wall is structurally determinate, the Code requires that the

As , min

0.25 f c fy

bw d

As,min

Check if the tension controlled assumption and the use of = 0.9 is correct.

0.25 31 MPa 420 MPa

As,min

compressive strength fc 21.2.2

6

As = 977 mm2

Solving for As 9.6.1.2

0.0159 As 2

As

878 mm 2

As,prov’d = 1290 mm2 > As,min = 878 mm2 > As,req’d = 977 mm2 OK Refer to Fig. E4.8.

To answer the question, the tensile strain in 2

compared to the values in Table 21.2.2. Concrete and nonprestressed reinforcement strain is assumed to vary proportionally from the neutral axis. From

22.2.1.2

c t

c

where: c

1

0.828

24.8 mm

0.003 24.8 mm = 0.029 > 0.005 t

d c a

c=

t

and a = 0.0159As derived previously. Section is tension controlled and

= 0.9.

  

Eng.Mohammed Ashraf





Fig. E4.8—Heel reinforcement. Step 9: Toe design the bearing pressure. The weight of the soil above the toe is usually

21.2.1

Shear Shear strength reduction factor:

Fig. E4.9—Shear at toe side. = 0.75

The critical section for shear strength is taken at a distance d from the stem: 7.5.3.1 22.5.5.1

Vn

Vc Vs

with Vs

f c bw d

Vn

0

Therefore Vn:

Vn

The applied force on the toe is from the soil bearing pressure acting upward. 7.4.3.2

Applied shear at distance d

p

2

2

d

1.5 m 0.265 m 2.2 m 5.3.1

7.5.1.1

For soil reaction on the toe portion a load factor of 1.6 is taken. For the self-weight of concrete portion, Vu a load factor of 0.9 is used.

Is Vn

Vu

Vn

2

2

2 3

Vu

OK

Shear reinforcement is not required.

  

Eng.Mohammed Ashraf

2





Flexure Soil bearing pressure at face of stem (refer to Fig. E4.9):

p@ stem , L = (88 kN/m 2

32 kN/m 2 )

1.5 m 2.2 m

+ 32 kN/m 2

= 70 kN/m 2

Mu2 = 1.6[(70 kN/m2(0.7 m)2/2 2 + (88 kN/m2 )(0.7 m)2(2/3)] = 1.6[(17.2 kN·m) + (5.9 kN·m)] = 37 kN·m

7.4.1.1 5.3.8(a)

Moment due to soil:

5.3.1

For the self-weight of concrete portion, a load factor of 0.9 is used.

Mu,req’d = Mu2 Mu1 = (37 kN·m) – (1.8 kN·m) = 35.2 kN·m, say, 35 × 106 N·mm

Resultant moment:

20.6.1.3.1

22.2.1.1

Mu1 = 0.9(23.5 kN/m3)(0.35 m)(0.7 m)2/2 = 1.8 kN·m

Concrete is placed against earth. Therefore, use 75 mm cover.

d = 265 mm calculated above

Setting: C = T to calculate required reinforcement

a) = As As

a=

7.5.2.1 7.5.1.1 21.2.1(a)

Mn

a 2

f y As d

Mn

0.0159 As

Mn

0.0159As 2

As

Mu

0.0159As 2

As

6

Mu = 35 × 10 N·mm calculated above. E Solving for As (refer to Fig. E4.10): 9.6.1.2

6

As = 355 mm2

Minimum required reinforcement is calculated from Eq. (9.6.1.2(a)): (a) As , min

0.25 f c fy

bw d

As ,min

(1000 mm)(265 mm) = 878 mm 2

Use No. 19 at 300 mm on center Equation (9.6.1.2(a)) controls, because concrete compressive strength fc 21.2.2

Check if the section is tension controlled assump-

21.2.2

To answer the question, the tensile strain in rein-

As,prov’d = 947 mm2 > As,min = 878 mm2 (Refer to Fig. E4.10.)

c= to the values in Table 21.2.2. The strain in concrete and reinforcement is assumed to vary proportionally from the neutral axis. From similar triangles (refer to Fig. E4.5): 22.2.1.2

c t

c

(d c)

where: c

a 1

(0.0159)(947 mm 2 ) 0.825

OK

18.2 mm

0.003 (265 mm 18.2 mm) 0.041 18.2 mm t = 0.041 > 0.005 Section is tension controlled. Therefore, assumption of = 0.9 is correct. t

and a = 0.0159As

  

Eng.Mohammed Ashraf





Fig. E4.10—Toe reinforcement. Step 10: Minimum transverse reinforcement Stem 11.6.1 temperature and shrinkage reinforcement. Per Table As,min 11.6.1, min = 0.002 Distribute transverse reinforcement equally between the front and back face of the stem wall.

As,min

As , prov.

2

front

As,min

129 mm 2 As , req ' d

Provide vertical reinforcement at front face to support the transverse wall reinforcement. 24.4.3

back

= 600 mm2/2 = 300 mm2

1000 mm 400 mm

323 mm 2

300 mm 2 /face

Therefore, provided reinforcement is adequate.

Base for temperature and shrinkage reinforcement. The reinforcement can be located at the top, bottom, or allocated between the two faces.

2

As,min

  

Eng.Mohammed Ashraf





Step 11: Dowels 7.7.1.2 The development length concept is based on the 25.4.2 attainable average bond stress over the embedment length of reinforcement. Development lengths are required because of the tendency of highly stressed bars to split relatively thin sections of restraining concrete. In application, the development length concept requires minimum lengths or extensions of reinforcement beyond all points of peak stress in the reinforcement. Such peak stresses generally occur at the points of critical sections for

reinforcement terminates, or is bent. For the cantilevered wall, toe and heel reinforcement must be developed. Heel reinforcement is developed beyond the stem critical section. Toe reinforcement must be checked for:

development length in the toe section. 2.. Extension of toe reinforcement in the stem to satisfy the required development length.

checked against the splice length and the larger length controls. 25.4.2.1

Development length

fy

25.4.2.2

1.

d

25.4.2.3

2.

d

t

2.1

1 1.1

e

fc

fy

d

db

t

f c cb

e

s

K tr

db

db

25.4.2.4

t = bar location; not more than 300 mm of fresh concrete below horizontal reinforcement e = coating factor; uncoated s

t = 1.0, because not more than 300 mm of concrete is placed below bars e = 1.0, because bars are uncoated s

  

Eng.Mohammed Ashraf



25.4.2.3



cb = spacing or cover dimension to center of bar, whichever is smaller Ktr = transverse reinforcement index It is permitted to use Ktr = 0. But the expression: greater than 2.5.

cb

K tr must not be taken db

25.4.2.3

63 mm 0 19 mm

3.3

2.5

Development length of heel and toe reinforcement: d

1.1

2.5

= 520 mm, say, 600 mm space to accommodate the required development Available length: 700 mm + 300 mm – 75 mm = 925 mm > ment extended from the base into the stem must be checked against the splice length of stem reinforcement. The larger length controls.

d

Fig. E4.11, therefore, OK

Splice 7.7.1.3 25.5

is permitted.

25.5.1.1 reinforcement extended from the toe is calculated above as 5600 mm 25.5.2.1

The deformed bars are in tension and the ratio of provided reinforcement area to required reinforcement area is less than 2.. Therefore, per Table 25.5.2.1 splice is Type B and the splice

Use 900 mm for splice length

E4.11. into the stem. Refer to Fig. E

d

7.7.2 25.2.3

Clear spacing between bars is the greater of: db dagg From Step 1, 25 mm maximum aggregate size is used.

Therefore, 40 mm controls

  

Eng.Mohammed Ashraf





Step 12: Details

Fig. E4.11––Retaining Retaining wall reinforcement.

  

Eng.Mohammed Ashraf





Retaining Walls Example 5—Reinforced concrete cantilever wall retaining earth with surcharge load 2 as shown in Fig. E5.1. The concrete mixture must satisfy durability and strength requirements. The grade of reinforcing bar is assumed at 420 MPa. The soil data were obtained from the geotechnical 3

and soil and between soil and soil is 0.4 and 0.7, respectively. Allowable soil bearing pressure obtained from the geotechnical 2 .

Given: Soil data–– 3

s

= 0.4 = 0.7

concrete-soil soil-soil

2

qall

Concrete–– a = 1.0 normalweight concrete fy = 420 MPa Cantilever wall–– h1 = 3700 mm h2 = 760 mm Load–– w

2

Fig. E5.1— E5.1—Cantilever Cantilever retaining wall. ACI 318M-14

Discussion

Step 1: Material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements. The designer determines the durability classes. Please see Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

Calculation By specifying that the concrete mixture shall be in accordance with ACI 301M and providing the exposure classes, Chapter 19 requirements are

Based on durability and strength requirements, and experience with local mixtures, the compressive strength

coordinated with ACI 318M. ACI encourages

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 1 of this Chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes.

  

Eng.Mohammed Ashraf





Step 2: Equivalent lateral pressure The geotechnical report provides the equivalent lateral pressure the wall is required to resist. Step 3: Preliminary cantilever wall data General criteria The preliminary retaining wall dimensions are determined from guidelines presented by Bowles Foundation Analysis and Design, The McGraw-Hill Companies, Inc., 1996. Figure E4.2 shows the variables used in the example.

p

3

The storage area applies uniform loading behind the retaining wall that is converted to an equivalent, imaginary earth height above the top of the wall:

Fig. E5 E5.2—Retaining E —Retaining wall required dimensions. hs

2

w

hs

2

0.94 m

s

Retaining wall height is:

h = h1 + h2 = 3.7 m + 0.76 m = 4.46 m

Equivalent retaining wall height:

heq = h + hs = 4.46 m + 0.94 m = 5.38 m heq

heq, but at least 300 mm Engineers commonly specify the base be slightly thicker than the stem. 13.3.1.2 20.6.1.3.1

greater than 150 mm

tstem

tbase

OK h

b/4 to b/3 Heel length: bheel = bbase – btoe - tstem

bbase btoe ~ 2.5 m/4 = 0.625 m, say, 0.7 m bheel = 2.5 m – 0.7 m – 0.38 m = 1.42 m

  

Eng.Mohammed Ashraf





Step 4: Applied forces For out-of-plane moment and shear, the cantilevered retaining wall is assumed to be continuous, and a representative one meter strip is analyzed for

The vertical weight is the self-weight of the retainabove the heel, including the storage area uniform load. The soil weight over the toe is neglected as it may erode away or be removed: tbase Stem wall: P1 conc tstem h Base: P2 = gconc tbase b Soil: P3 s h + hs tbase bheel

Total vertical load:

Fig. E5.3 E5.3—Applied Applied forces on retaining wall. P1 P2 P3

3 3 3

P

The self-weight of the retaining wall and the soil and distributed load above the heel tend to counteract the overturning moment. Moments taken about

Stem wall: M1 = P1 btoe + tstem Base: M2 = P2 b Soil: M3 = P3 b – bheel Restoring moment:

M1 M2 M3 MR

The retained soil behind the wall and the superimH wall: Due to retained earth: H1 Ca s h2

H1

2

Due to surcharge: H2 Ca s hs h)

H2

2

2

  

Eng.Mohammed Ashraf





Ca is obtained from Rankine’s formula found in soils engineering handbooks. Therefore, this lateral force tends to overturn the retaining wall about the front edge of the toe: Hihi

MOTM

MOTM

Summation of moments about the front edge of stem: M

MR – MOTM

M

overturning. Step 5: Soil pressure 13.3.1.1 The aforementioned determined cantilever wall base is checked using unfactored forces and allowable soil bearing pressure. To calculate soil pressure, the location of the vertical resultant force must be determined. The distance of the resultant to the front face of stem: a

M P

a

location and the base mid-length: e = bbase/2 – a Check if resultant falls within the middle third of the base.

Maximum and minimum soil pressure: q1,2

P A

M S

0.91 m

e = 2.5 m/ m/2 – 0.91 m = 0.34 m bbase 6

2.5 m 6

0.42 m

e

0.34 m

Therefore, there is no uplift. q1,2

q1,2 q1 = qmax q2 = qmin

±

2

2

2 2

2

< qall

2

2

no uplift

Soil bearing pressure is acceptable.

  

Eng.Mohammed Ashraf





Step 6: Stability requirements Calculate the factor of safety against overturning: FS

MR M OTM

2.0

FS FS = 2.46 > 2.0

2.46 OK

Calculate the factor of safety against sliding: FS

P H

1.5

FS FS = 1.09 < 1.5

1.09 NG

This calculation neglects the passive pressure

There is not enough frictional resistance to

Two solutions:

loads and thus increase resistance to the sliding force. 2.. Provide shear key to engage the soil and to develop a passive pressure at the shear key large enough to resist the lateral force.

should be increased to 3150 mm and 500 mm to satisfy the1.5 factor against sliding and to satisfy the shear demand, respectively. In this example, the shear key solution approach is chosen.

  

Eng.Mohammed Ashraf





Shear key geometry A shear key is constructed at 900 mm from the toe and has the following dimensions: Depth: 380 mm Width: 500 mm The shear key is positioned such that reinforcement is extended from the stem into the shear key

Assume earth above toe side is neglected in the calculation of the passive pressure as it may erode or may be removed. Cp

1 sin 1- sin

1 Ca

1 0.271

3.69

Fig. E5.4 E5.4—Forces 5.4—Forces Forces acting on R-Wall with shear key.

  

Eng.Mohammed Ashraf



1 C p s tbase hkey 2 2 The sliding occurs at the bottom surface of the foundation along the path Hp



1 2

Hp

3

2

a-c + c-d + e-f a-c c-d

ac = 900 mm e-f cf = 1600 mm

Fig. E5.5—Sliding load path. The sliding along Path ac occurs within the soil soil-soil

= tan35° = 0.7

The sliding along Path c-d + e-f is between

2

2 2

qc

2.5 m

Calculate soil pressure at Point c.

2

qc The friction resistance force P: qa

Pac

Pac

ac

2

Pcd+ef

Pcd

2

qc

ef

2

qf

Pcd

cd ef

2

2

Pcd+ef

P – Pac

qc

2

ef

2

2

Along Path ac cf, sliding occurs between

P

1Pac

rounding of results.

P

2Pcd+ef

Due to the shear key, the imposed lateral pressure increases. Due to retained earth: H1 = Ca s h + hkey 2/2

H1

3

Due to surcharge: H2 = Ca shs h + hkey Safety factor against sliding:

H2

3

FS

P

1 AC

P

2 CF

H

Hp

FS =

2

1.60 1.5

OK

Hp   

Eng.Mohammed Ashraf





Conclusion The retaining wall preliminary dimensions are adequate to resist overturning, sliding, preventing uplift, and limiting pressure on the soil to less than the allowable provided soil pressure in the geotechnical report. But the length of the retaining wall is not adequate to prevent sliding. Therefore, a shear key is provided. In the following steps, the retaining wall is designed for strength. If any of the aforementioned determined dimensions are not satisfactory, then all the previous steps must be revised.

  

Eng.Mohammed Ashraf





Step 7: Stem design The cantilevered concrete stem is a determinate member and is modeled as a 1 m wide cantilever beam.

13.2.7.1

Flexure The maximum design moment in the stem is

earth pressure and is placed near the face of the stem wall that is against the retained soil. Adequate concrete cover protects reinforcement against moisture changes in soil. Cover is measured from the concrete surface to the outermost surface of the reinforcing bar.

Fig. E5.6—Soil lateral force on stem. hstem

Stem height: 20.6.1.3.1

20.6.1.3.1, use 50 mm cover. From Table 20.6.1.3.1,

21.2.1

Assume that the member is tension controlled; steel t

7.4.1.1 U = Hu alone.

3

Psurcharge

Equivalent lateral pressure due to surcharge load:

H1 + H2 when lateral pressure acts

Hu

2

2

3

Mu 6

7.5.1.1 22.2.2.1

Mn

Mu

The concrete compressive strain at which ultimate c = 0.003.

22.2.2.2

able property and its value is approximately 10 to 15 percent of the concrete compressive strength. For calculating nominal strength, ACI 318M neglects the concrete tensile strength.

22.2.2.3

Determine the equivalent concrete compressive stress for design.

  

Eng.Mohammed Ashraf



22.2.2.4.1

22.2.2.4.3

22.2.1.1



The concrete compressive stress distribution is inelastic at high stress. The actual distribution of concrete compressive stress is complex and usually not known explicitly. The Code permits any stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85fc a 1c 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3. For fc

0.85

1

0.828

7 MPa

Find the equivalent concrete compressive rectangle depth, a, by equating the compression force to the tension force within a unit length of the wall cross section: C = T where C = 0.85fc ba and T = AsTy

a = As As

a=

0.0159 As

Calculate required reinforcement area: a 2

f y As d

7.5.2.1 22.3

Mn

Mn

20.6.1.3.1

d = tstem – cover – db/2 use 50 mm cover.

As d

0.0159As 2

d = 380 mm – 50 mm – 22 mm/2 = 319 mm

21.2.1 Mn

7.5.1.1

Mu

0.0159As 2

As

6

Mu = 142 × 106 As,req’d = 1214 mm2

Solving for As:

As,provided = 1290 mm2/m > As,req’d = 1214 mm2 9.6.1.2

OK

The cantilevered retaining wall calculated required tensile reinforcement is usually very small compared to the member concrete section. The stem reinforcement is checked against the beam minimum one-way slab minimum reinforcement area because of the lack of redundancy. The Code requires that the beam reinforcement area is at least: As , min

0.25 f c fy

bw d

As ,min

0.25 31 MPa 420 MPa

2

  

Eng.Mohammed Ashraf



compressive strength fc 21.2.2

Check if the tension controlled assumption and the

22.2.1.2

To answer the question, the tensile strain in rein-



As,prov. = 1290 mm2 > As,min = 1057 mm2

to the values in Table 21.2.2. Assume concrete and nonprestressed reinforcement strain varying pro-

2

c=

c t

c

a

24.8 mm

0.003 24.8 mm t = 0.036 > 0.005

d c

where: c

0.828

t

and a = 0.0159As

1

Fig. E5.7—Strain distribution in stem.

  

Eng.Mohammed Ashraf



7.4.3.2



Shear The closest inclined crack to the support of the cantilevered wall will extend upward from the face of the base reaching the compression zone approximately d from the face of the base. The lateral load applied to the cantilever between the face of the base and the point d away from the face are transferred directly to the base by compression in the cantilever above the crack. Accordingly, the Code permits design for a maximum factored shear force Vu at a distance d from the support for nonprestressed members. For simplicity, the critical section for shear strength in this example is calculated at the bottom of the stem:

7.5.3.1 22.5.5.1

Vn

21.2.1

Shear strength reduction factor: Vn is:

7.5.1.1

(Vc Vs )

with Vs

Vn

0.17

f c bw d

Vu = 1.6[(4.7 kN/m3)(0.94 m)(4.06 m – 0.319 m) + (4.7 kN/m3)(4.06 m – 0.319 m)2/2] = 79 kN Vn

0.17 31 MPa (1000 mm)(319 mm) = 302 kN

0

Vu

Vn = (0.75)(302 kN) = 226 kN Vn = 226 kN > Vu = 79 kN

OK

Shear reinforcement is not required.

  

Eng.Mohammed Ashraf





Step 8: Heel design Shear The base heel is designed for shear caused by the superimposed weight of soil, including self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may not be linear as assumed or may not be present. Therefore, conservatively, it is not included in the calculation of shear strength.

Fig. E5.8—Force on heel. 7.4.3.2

The critical section for shear strength is taken at a distance d

7.5.3.1 22.5.5.1

Vn

f c bw d

Vc Vs

with Vs

Vn

Vc

0

d = tbase – cover – db/2 20.6.1.3.1

20.6.1.3.1, use 75 mm cover to tension From Table 20.6.1.3.1,

21.2.1

Shear strength reduction factor:

7.4.3.2

Vn is: A load factor of 1.2 is used for the concrete and 1.6

d = 400 mm – 75 mm – 22 mm/2 = 314 mm

Vn 3

Vu 3 2

7.5.1.1

Vn

Vu

Vn Vu OK Shear reinforcement is not required.

  

Eng.Mohammed Ashraf





Flexure imposed weight of soil and self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may not be linear as assumed or may not be present. Therefore, conservatively, it is not included in the calculation

6.6.1.2

The cantilever wall is a determined system. Therefore, maximum moment and shear in the heel and toe of the base occur at the stem face and at distance, d, from stem face, respectively. Redistribution of moments does not occur.

5.3.1 5.3.8

A load factor of 1.2 is used for the concrete self3

Mu1

2

/2

load.

22.2.1.1

6

in Step 4. Setting C = T C = 0.85fc ba and T = fyAs

a a=

a 2

Mn

21.21

Flexure strength reduction factor:

Mn

Mu

As

As

Mu = 149 × 106 Solving for As: 9.6.1.2

0.0159 As

Mn

7.5.2.1

7.5.1.1

f y As d

As

As

0.0159As 2

0.0159As 2

6

As,req’d = 1295 mm2/m

ural member. The tensile reinforcement amount is very small compared to the member concrete section. Also, the cantilevered retaining wall is a statically determinate member with no possibility for moment redistribution. Because the wall is structurreinforcement area is not less than the greater of: As,min

0.25 f c fy

bw d

As , min

0.25 31 MPa 420 MPa

2

  

Eng.Mohammed Ashraf



Moments were rounded up and the calculated reinforcement area is approximately equal to No. 22 at 300 mm. Therefore, use No. 22 at 300 mm o.c.



Use No. 22 at 300 mm on center (refer to Fig. E5.7). 2

As,prov’d

2

~ As,req’d

OK

2

21.2.2

Check if the tension controlled assumption and the

c

To answer the question, the tensile strain in reinto the values in Table 21.2.2. Concrete and nonprestressed reinforcement strain is assumed to vary proportionally from the neutral axis. From similar triangles (refer to Fig. E5.7): c

22.2.1.2

t

c

0.003 t

( d c) t

where: c

a

and a

As

1

the bearing pressure. The weight of the soil above the toe is usually neglected as it may erode or be removed.

21.2.1

7.5.3.1 22.5.5.1

Shear Shear strength reduction factor: The critical section for shear strength is taken at a distance d from the stem: Vn (Vc Vs ) 0.17 f c bw d with Vs

Vn

0

The applied force on the toe is from the soil bearing Shear calculation is similar to the shear calculation for pressure acting upward. Applied shear at distance d from stem:

p@ d

2

2

2

2.5 m 2

5.3.1

2

For soil reaction on the toe portion, a load factor of Vu 1.6 is taken. For the self-weight of concrete portion,

2

OK 7.5.1.1

Vn

Vu

Vn

Vu

OK

Shear reinforcement is not required.

  

Eng.Mohammed Ashraf



7.4.1.1



Flexure Soil bearing pressure at the face of stem: Moment due to soil load:

p

2

1.8 m 2.5 m

2

stem L

2

2

2

Mu2

2

/2

2

5.3.1

2

3

Moment due of toe self-weight:

Mu1

Resultant moment:

Mu,req’d = Mu2

2

2

Mu1 6

20.6.1.3.1

22.2.1.1

Concrete is placed against earth. Therefore, use 75 mm cover.

d = 400 mm – 75 mm – 22/2 mm = 314 mm

Setting C = T to calculate required reinforcement

a As

a=

Mn

21.2.1

Flexural strength reduction factor Mn

f y As d

a 2

7.5.2.1

Mu

0.0159 As

As

0.0159As 2

0.0159As 2

As

Mu = 54 × 106 Solve for As: 9.6.1.2

Mn

As

6

calculated above. As = 460 mm2/m

The cantilevered retaining wall behaves as a amount is very small compared to the member concrete section. Also, the cantilevered retaining wall is a statically determinate member with no possibility for moment redistribution. To prevent a reinforcement area is not less than the greater of:

As,min

0.25 f c fy

bw d

As , min

0.25 31 MPa 420 MPa

2

As,prov’d = 1290 mm2 > As,req’d = 1041 mm2

OK

  

Eng.Mohammed Ashraf



21.2.1

Check if the tension controlled assumption and the

21.2.2

To answer the question, the tensile strain in reinto the values in Table 21.2.2. The strain in concrete and reinforcement is assumed to vary proportionally from the neutral axis. From similar triangles

21.2.1.2

c t

2

c=

d c

t

24.8 mm

0.828 0.003 24.8 mm

t

c where: and a = 0.0159As



= 0.035 > 0.005

Shear key design strength, the soil above the toe is not considered, as this will apply higher force on the shear key and it may not be present. Cp

1 Ca

Hp

1 2

Hp

1 2

1 0.271

3.69

Passive pressure is calculated at top and bottom of shear key: Top of shear key: Hp

1 Cp 2

s

tbase

2

3

2

Bottom of shear key: Hp

1 Cp 2

s

hbase

hkey

2

3

2

7.4.3 5.3.1

Required shear design:

Vu

16.5.2.4

Check corbel dimensions if they satisfy the following requirements: Vu least of: fc bd fc bd bd

dcorbel = 500 mm – 75 mm – 22 mm/2 = 414 mm, say, 400 mm

2

Vu/

K

Shear strength: The shear key may be considered similar to a the shear key. The shear resistance is calculated

16.3.3.5 22.9.4.2 21.2.1 22.9.3.1

Vs

Avf fy

Vn

2

Vs

Vu

shear key. Vn

Vu

OK

  

Eng.Mohammed Ashraf





Step 10: Minimum transverse reinforcement Stem 11.6.1 temperature and shrinkage reinforcement. Per Table As,min min = 0.002. Distribute shrinkage and temperature reinforcement equally between the front and back face of the stem wall.

As,min As , prov.

2

front

As,min

199 mm 2 As , req ' d

back

= 760 mm2/2 = 380 mm2

1000 mm 450 mm

442 mm 2

380 mm 2 /face -

tally. Provide vertical reinforcement at front face to support the horizontal shrinkage and temperature reinforcement. Base As,min

24.4.3 for temperature and shrinkage reinforcement. The reinforcement can be located at the top, bottom, or allocated between the two faces.

2

= 1800 mm2

or anywhere within the base. Therefore, place three heel end.

Step 11: Dowels The development length concept is based on the 7.7.1.2 25.4.2 attainable average bond stress over the embedment length of reinforcement. Development lengths are required because of the tendency of highly stressed bars to split relatively thin sections of restraining concrete. In application, the development length concept requires minimum lengths or extensions of reinforcement beyond all points of peak stress in the reinforcement. Such peak stresses generally occur at the points of critical sections for

reinforcement terminates, or is bent. For the cantilevered wall, toe and heel reinforcement must be properly developed. Heel reinforcement is developed beyond the stem critical section. Toe reinforcement must be checked for: opment length in the toe section. 2. Extension of toe reinforcement in the stem to satisfy the required development length.

checked against the splice length and the larger length controls.

  

Eng.Mohammed Ashraf



25.4.2.1

Development length, d, is the greater of Eq 25.4.2.2 or 25.4.2.3 of ACI 318M-14 and 300 mm:

25.4.2.2

1.

d

25.4.2.3

2.

d

fy 1.7

1 1.1

t

e

fc

fy



db

t

f c cb

e

s

K tr

db

db

25.4.2.4

t

= bar location

t = 1.0, because not more than 300 mm of fresh concrete is placed below horizontal reinforcement e = 1.0; because bars are uncoated

= coating factor = bar size factor s e

s

cb = spacing or cover dimension to center of bar, whichever is smaller Ktr = transverse reinforcement index It is permitted to use Ktr = 0.. 25.4.2.3

However, the expression: taken greater than 2.5. 2.5

cb

K tr db

must not be

25.4.2.3

61 mm 0 22 mm

2.77

2.5

Development length of heel and toe reinforcement: d

1.1

2.5

= 603 mm, say, 625 mm

checked against the splice length of stem reinforcement and the larger length controls.

  

Eng.Mohammed Ashraf





Splice 7.7.1.3 25.5 25.5.5.1

25.5.2.1

is permitted. ed from the toe is calculated previously as 625 mm The deformed bars are in tension and the ratio of provided reinforcement area to required reinforcement area is less than 2. Therefore, per Table 25.5.2.1, splice is Type B and the splice length is d

25.2.3

Use 850 mm for splice length.

into the stem. Refer to Fig. E5.9.

Clear spacing between bars is the greater of: db dagg

Assume, 25 mm maximum aggregate size is used. Step 12: Details

Therefore, 40 mm controls.

Fig. E5.9––Retaining wall reinforcement.

  

Eng.Mohammed Ashraf





Retaining Walls Example 6—Reinforced concrete cantilever wall retaining earth built at property line level as shown in Fig. 6.1. The concrete mixture must satisfy durability and strength requirements. The grade of reinforcing bar 3 , angle of soil 2

.

Given: Soil data–– s

3

qall

2

Concrete–– a

fy = 420 MPa Cantilever wall–– h1 = 2400 mm h2 = 760 mm

Fig. E6.1—Cantilever 6.1—Cantilever retaining wall. ACI 318M-14 Discussion Step 1: Material requirements 7.2.2.1 The mixture proportion must satisfy the durability requirements of Chapter 19 and structural strength requirements. The designer determines the durability classes. Please see Chapter 4 of this Handbook for an in-depth discussion of the categories and classes.

Calculation By specifying that the concrete mixture shall be in accordance with ACI 301M and providing the exposure

Based on durability and strength requirements, and experience with local mixtures, the compressive strength

dinated with ACI 318M. ACI encourages referenc-

There are several mixture options within ACI 301M, such as admixtures and pozzolans, which the designer can require, permit, or review if suggested by the contractor. Example 1 of this Chapter provides a more detailed breakdown on determining the concrete compressive strength and exposure categories and classes.

  

Eng.Mohammed Ashraf





Step 2: Equivalent lateral pressure The geotechnical report provides the equivalent lateral pressure the wall is required to resist. Step 3: Preliminary cantilever wall data General criteria

p

3

3

experience and are presented in the text.

Fig. E6.2 E6.2—Retaining E Retaining L-Wall required dimensions. h = h1 + h2 = 2.4 m + 0.76 m = 3.16 m h h, but at least 300 mm Engineers commonly specify the base be slightly thicker than the stem.

tstem tbase

OK

20.6.1.3.1 13.3.1.2 h Heel length: bheel = bbase – btoe – tstem

bbase bheel

  

Eng.Mohammed Ashraf





Step 4: Applied forces For out-of-plane moment and shear, the cantilevered retaining wall is assumed to be continuous, and a representative 1 m strip is analyzed for the

The vertical weight is the self-weight of the retainabove the heel.

tbase Stem wall: P1 conc tstem h Base: P2 t b conc base Soil: P3 tbase b tstem s h

Fig. E6.3—Applied forces on retaining wall. P1 P2 P3

3 3 3

Total vertical load: P The self-weight of the retaining wall and the soil above the heel tend to counteract the overturning moment. Moments taken about the front edge of

Stem wall: M1 = P1 tstem Base: M2 = P2 bbase Soil: M3 = P3 b – bheel Restoring moment: The retained soil behind the wall exerts lateral pressure H on the wall: H1 Ca s h2

M1 M2 M3 MR

H1

3

2

Therefore, this lateral force tends to overturn the retaining wall about the front edge of the stem: MOTM

MOTM = H h Summation of moments: M MR – MOTM

M

  

Eng.Mohammed Ashraf





Step 5: Soil pressure 13.3.1.1 The aforementioned determined cantilever wall base is checked using unfactored forces and allowable soil bearing pressure. To calculate soil pressure, the location of the vertical resultant force must be determined. The distance of the resultant to the front face of stem: a

M P

a

0.61 m

location and the base mid-length: e = bbase/2 – a

e = 1.75 m/2 – 0.61 m = 0.265 m

Check if resultant falls within the middle third of the base.

bbase 6

Maximum and minimum soil pressure:

Therefore, there is no uplift.

q1,2

P A

Pe S

1.75 m 6

0.29 m

q1,2 q1,2 q1 = qmax q2 = qmin

e

0.265 m

±

2

2

2 2

2

< qall

2

2

, no tension

Soil bearing pressure is acceptable. Step 6: Stability requirements Calculate factor of safety against overturning: FS

MR M OTM

2.0

FS

3.5

FS = 3.5 > 2.0

OK

Calculate the factor of safety against sliding: FS

P H

1.5

FS FS = 1.74 > 1.5

1.74

3

OK

This calculation neglects the passive pressure Conclusion: The retaining wall preliminary dimensions are adequate to resist overturning, sliding, preventing uplift, and limiting pressure on the soil to less than the allowable provided soil pressure in the geotechnical report. In the following steps, the retaining wall is designed for strength. If any of the aforementioned determined dimensions are not satisfactory, then all the previous steps must be revised.

  

Eng.Mohammed Ashraf





Step 7: Stem design The cantilevered concrete stem is a determinate member and is modeled as a 1 m wide cantilever beam (refer to Fig. E6.4).

13.2.7.1

Flexure The maximum design moment in the stem is calculated at the face of the base foundation. Vertical reinforcement in the stem resists the lateral earth pressure and is placed near the face of the stem wall that is against the retained soil. Adequate concrete cover protects reinforcement against moisture changes in soil. Cover is measured from the concrete surface to the outermost surface of the reinforcing bar.

Fig. E6.4—Soil lateral force on stem. Stem height: 20.6.1.3.1

20.6.1.3.1, use 50 mm cover From Table 20.6.1.3.1,

21.2.2

Assume that the member is tension controlled;

hstem = ((2400 mm) + (760 mm) – (380 mm) = 2780 mm

t

7.4.1.1 5.3.8(a)

Load factor: U = Hu = 1.6H;; when lateral pressure acts alone. The moment is taken at the bottom of the stem and above the base; h1 = 2.78 m

7.5.1.1 22.2.2.1

Mn

1.6(5.1 kN/m3)(2.78 m)(2.78 m)/2 = 31.5 kN Hu = 1.6

Mu

6

Mu

The concrete compressive strain at which ultimate c = 0.003.

22.2.2.2 variable property and its value is approximately 10 to 15 percent of the concrete compressive strength. For calculating nominal strength ACI 318M neglects the concrete tensile strength.

  

Eng.Mohammed Ashraf



22.2.2.3

22.2.2.4.1

22.2.2.4.3 22.2.1.1



Determine the equivalent concrete compressive stress for design. The concrete compressive stress distribution is inelastic at high stress. The actual distribution of concrete compressive stress is complex and usually not known explicitly. The Code permits any stress distribution to be assumed in design if shown to result in predictions of nominal strength in reasonable agreement with the results of comprehensive tests. Rather than tests, the Code allows the use of an equivalent rectangular compressive stress distribution of 0.85fc a 1c 1 is a function of concrete compressive strength and is obtained from Table 22.2.2.4.3. For fc

1

0.85

Find the equivalent concrete compressive depth, a, by equating the compression force and the tension force within a unit length of the wall cross section: C=T C = 0.85fc ba and T = Asfy

a a

Calculate required reinforcement area:

7.5.2.1 22.3

0.828

7 MPa

Mn

a 2

As f y d

As

As 0.0159 As

Equating design moment strength and required mo moment strength, As is: As d

Mn

0.0159 As 2

21.2.1 20.6.1.3.1 d = tstem – cover – db/2 use 50 mm cover Mn

7.5.1.1

d = 250 mm – 50 mm – 16 mm/2 = 192 mm

Mu

Mu = 29.2 × 106 Solving for As 9.6.1.2

As

0.0159As 2

6

As = 409 mm2

The cantilevered retaining wall calculated required As,prov’d = 663 mm2/m > As,req’d = 409 mm2/m tensile reinforcement is usually very small compared to the member concrete section. The stem reinforcement is checked against the beam minimum one-way slab minimum reinforcement area because of the lack of redundancy. The Code requires that the beam reinforcement area at least the greater of: As , min

0.25 f c fy

bw d

As ,min

0.25 31 MPa 420 MPa

2

  

Eng.Mohammed Ashraf



compressive strength, fc = 31 MPa. 21.2.2

Check if the tension controlled assumption and the



As,prov’d = 663 mm2/m As,prov’d = 663 mm2/m > As,min = 636 mm2/m > As,req’d = 409 mm2

To answer the question, the tensile strain in reinto the values in Table 21.2.2. Assume concrete and nonprestressed reinforcement strain varying pro-

2

c= 22.2.1.2

c t

c

a

12.7 mm

0.003 12.7 mm t = 0.042 > 0.005

d c

where: c

0.828

t

and a = 0.0159As derived previously.

1

Fig. E E6.5—Strain distribution across stem. 7.4.3.2

7.5.3.1 22.5.5.1 21.2.1 7.5.1.1

Shear The closest inclined crack to the support of the cantilevered wall will extend upward from the face of the base reaching the compression zone approximately d from the face of the base. The lateral load applied to the cantilever between the face of the base and point d away from the face are transferred directly to the base by compression in the cantilever above the crack. Accordingly, the Code permits design for a maximum factored shear force Vu at a distance d from the support for nonprestressed members. For simplicity, the critical section for design shear strength in this example is calculated at the bottom of the stem: Vn

Vc Vs

f c bw d

with Vs 0 Shear strength reduction factor: Vn Vc is: Vn Vu

hstem = h – tbase = 3.16 m – 0.38 m = 2.78 m 3 Vu Vn

Vn Vn Vu OK Shear reinforcement is not required

  

Eng.Mohammed Ashraf





Step 8: Heel design Shear The base heel is designed for shear caused by the superimposed weight of soil, including self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may

Fig. E6.6—Force on heel. 21.2.1

Shear strength reduction factor:

7.4.3.2

The critical section for shear strength is taken at a distance d from the bottom of the stem:

7.5.3.1 22.5.5.1

Vn

20.6.1.3.1

d = tbase – cover – db/2

f c bw d

Vc Vs

with Vs

0 d = 380 mm – 75 mm – 19 mm/2 = 295.5 mm, say, 295 mm

From Table 20.6.1.3.1 20.6.1.3.1,, use 75 mm cover to tension reinforcement.

Vn

Vc is:

Vn

A load factor of 1.2 is used for the concrete selfVu

3 3

Vu 7.5.1.1

Vn

Vu

Vn Vu OK Shear reinforcement is not required

  

Eng.Mohammed Ashraf





Flexure superimposed weight of soil and self-weight of heel. The soil pressure counteracting the applied gravity loads is neglected as the soil pressure may not be linear as assumed. Therefore, it is not

6.6.1.2

The cantilever wall maximum moment and shear in the heel and toe of the base occur at the stem face. Redistribution of moments cannot occur.

5.3.1 5.3.8

A load factor of 1.2 is used for the concrete self-

Mu1

3

2

3

2

/2 /2

6

22.2.1.1

Setting C = T

a a=

7.5.2.1

a 2

f y As d

Mn

As

As 0.0159 As

Mn

As

0.0159As 2

21.2.1 7.5.1.1

Mn

As

Mu

0.0159As 2

6

Mu = 107 × 106 As = 986 mm2

Solving for As: 9.6.1.2

The cantilevered retaining wall calculated tensile reinforcement is very small compared to the member concrete section. To prevent a reinforcement area is at least the greater of: As , min

0.25 f c fy

bw d

compressive strength fc 21.2.2

As,min

0.25 31 MPa 420 MPa

2

As,prov’d = 1290 mm2/m > As,req’d = 986 mm2/m

OK

Check if the tension controlled assumption and the

  

Eng.Mohammed Ashraf





To answer the question, the tensile strain in

2

c=

c t

c

0.003 24.8 mm t = 0.0327 > 0.005

d c

where: c

a

24.8 mm

0.828

compared to the values in Table 21.2.2. Concrete and nonprestressed reinforcement strain is assumed to vary proportionally from the neutral axis. From

t

and a = 0.0159As derived previously.

1

Fig. E6.7—Heel reinforcement. Step 9: Toe design Because the cantilever wall is built at the property line, there is no toe at the base. Step 10:: Minimum transverse reinforcement Stem 11.6.1 temperature and shrinkage reinforcement. Per Table 11.6.1: min

As,min As,min

= 0.002

Distribute shrinkage and temperature reinforcement equally between the front and back face of the stem wall. Provide vertical reinforcement at front face to support the transverse wall reinforcement.

As , prov.

2 front

As,min

2

back

2

2

= 500 mm /2 = 250 mm

1000 mm 450 mm

2

> As,req’d = 250 mm2/face Base 2

24.4.3 for temperature and shrinkage reinforcement. The reinforcement can be located at the top, bottom, or allocated between the two faces.

bottom as continuous nose bars for the dowel rein-

  

Eng.Mohammed Ashraf





Step 11: Dowels 7.7.1.2 The development length concept is based on the 25.4.2 attainable average bond stress over the embedment length of reinforcement. Development lengths are required because of the tendency of highly stressed bars to split relatively thin sections of restraining concrete. In application, the development length concept requires minimum lengths or extensions of reinforcement beyond all points of peak stress in the reinforcement. Such peak stresses generally occur at the points of critical sections for development nates, or is bent. For the cantilevered wall, heel reinforcement must be properly developed. Heel reinforcement is developed beyond the stem critical section. 25.4.2.1

Development length,

25.4.2.2

1.

d

25.4.2.3

2.

d

fy 1.7

1 1.1

t

e

fc

fy

d

db

t

f c cb

e

s

K tr

db

db 25.4.2.4 = bar location; not more than 300 mm of fresh concrete below horizontal reinforcement e = coating factor; uncoated t

s

25.4.2.3

t = 1.0, because not more than 300 mm of concrete is placed below bars. e = 1.0, because bars are uncoated s

cb = spacing or cover dimension to center of bar, whichever is smaller

59 mm 0 16 mm

3.7

Ktr = transverse reinforcement index It is permitted to use Ktr = 0.

61 mm 0 22 mm

2.77

57 mm 0 13 mm

4.4

However, the expression: taken greater than 2.5.

cb

K tr db

must not be

2.5

2.5

2.5

checked against the splice length of stem reinforcement and the larger length controls. d

Because this wall does not have a toe, head bars are used to satisfy development.

1.1

2.5 d,req’d,

22 16 13

mm 603 439 357

db

d,prov.,

mm 625 450 380

  

Eng.Mohammed Ashraf





Splice 7.7.1.3 25.5 25.5.1.1 25.5.2.1

st,req’d,

16 13

is permitted.

mm 571 464

st,prov.,

mm 625 500

in the stem. The deformed bars are in tension and the ratio of provided reinforcement area to required reinforcement area is less than 2. Therefore, per Table 21.5.2.1, splice Type B is used and the splice length d

Clear spacing between bars is the greater of: 7.7.2 25.2.3

db dagg

Assume 25 mm maximum aggregate size is used. Step 12: Details

Therefore, 40 mm controls.

Fig. E6.8—Retaining wall reinforcement.

  

Eng.Mohammed Ashraf





Ma Ec I e

-

Ma



Ec Ie fc

fc fc

wc -

fc

Ec

wc

fc

fy Ec   Ec

h

wc

Ec

fc

fc Es

n

Es Ec -

wc wc fy

fy

E  M

Iy

  

Eng.Mohammed Ashraf





Fig. 13.3a––Stress-strain relationship for concrete compressive strength.

Fig. 13.3b––Stress-strain relationship for reinforcement Grade 60.

AE

Ac equiv Ec

As Es

Ac equiv

Es As Ec

Ac equiv

nAs

Es Ec

Fig. 13.3c––Moment at a distance from neutral axis.

n

M cr

Mcr

fr I g yt

yt fr



-

  

Eng.Mohammed Ashraf





a beam. kd

c -

c k k

n

M cr Ma

Ig

n

n

Fig. 13.3e––Rectangular Rectangular beam transformed cross section. Ie

M cr Ma

I cr

Ma Mcr Ig Icr Ma Mcr EcIg Ma

Ma

Mcr EcIcr c bc

nAs d c

n

As c a l

c I e avg I cr

bc

n

As c a

I e left

sup

I e right

sup

I e midspan

nAs d c

  

Eng.Mohammed Ashraf

l





I e sup

I e avg

I e midspan

-

Ie,avg

Ie,midspan Ie,avg

Ie,left-sup

Ie,midspan

Ie,right-sup

Ie,sup Fig. 13.4a––Change in curvature of a cracked reinforced section under sustained loading.

i

-

T

i

i

  Time



  

Eng.Mohammed Ashraf





Type of member *

L L

‡

§

† *

†

‡ §

n

n

ACI Structural Journal  

Ast Ag -

-

-

  -

cc fs fs

fs fs

cc

  

Eng.Mohammed Ashraf





    

-



ft     ft U*



ft  < ft

C *

ft fc

 

  

fc fc

-

 -



 ACI Structural Journal  

neering Practice and Research: A Tribute to Chester P. Siess

fci fci

ACI Structural Journal

  

Eng.Mohammed Ashraf





 Ie

b d h As fc n Ma

Fig. E1.1—Rectangular beam section.

Mcr h

fc

Kcr

Kcr Mcr

Mcr

Kcrb Icr

As As bd Ki

n Icr

Ki bd Ig

Icr

Ig

bh

Ig

Ki

Ie Ie Mcr Ma Ki Ie *

Ki Ig

Icr Ig

Icr Ig Mcr Ma Icr Ig Ki Ie Ie

Icr

Mcr Ma

Mcr Ma

Icr Ig

  

Eng.Mohammed Ashraf



fc n As M M b d (M



M

Fig. E2.1—Lintel and its cross section.

Ie Mcr M

Mcr

Mcr M

Ki

Mcr M

Icr Ig

Ki

Icr Ig Ie Ki Ig

Ig

Ie

Ka

fc Ka

fc

n

Es Ec

Ka

= (Ka Ie Ka M

c

Ka

c

Ie c

Ka Ie Ka M

c L

ac

ac

L

c

c L

  

Eng.Mohammed Ashraf





Icr

n b bw hf d As

Fig. E3.1—T-section.

w

As bwd

w

wn

wn

hf d

hf d

hf

b bw

d

c w

c

n

Ki c

wn

hf d c

Ki wn

hf d

Ki c

Icr

Ki bwd

Ki

Icr

  

Eng.Mohammed Ashraf





Moment of inertia of a cracked section with tension and compression reinforcement Icr

n b h d d As As

Fig. E4.1—Section of double reinforced beam.

As bd As bd n

n

n n

c

c

d

d

Ki c

n

d

Ki d d

c

Ki c

Icr

Ki bwd

Ki

Icr

  

Eng.Mohammed Ashraf





fc fv n wc

Fig. E E5.1—Moment —Moment Moment and cross sections of a continuous beam. Ie Ie

Ig

bh

Ig

  

Eng.Mohammed Ashraf





Icr As bd As bd n

n n

c

c

n

d d

d d

c

n

d d Icr

Ki

Ki

Ki bd Ki

Icr

As bd n Ki

Ki Icr

Ki bd

Icr fc Kcr

Kcr Mcr

Kcrb

h

Mcr

Ie A Mcr Ma Icr Ig Icr

Ma

Ki

B

C

Mcr Ma Icr Ig Ki

Ie

Ki Ig

Ie

Ie

*

Ie

Ie at C

Ie

Ie

  

Eng.Mohammed Ashraf





Ka

fc wc

Ka

Ka Ka

Ka Ied Mc

c

MA MB Ka

c

*

c

Ie A Mcr Ma

B

C

Mcr Ma

M

Use Icr Ig

Ki Ki Ie

Ie

Ki Ig Ie

*

c

L

Ie

Ie

Ie

Ie at C

Ka Ie Mc

MA

MB Ka

c

L

c

L

Ka

c L

=

c

c

c L

c L *

  

Eng.Mohammed Ashraf





Ie

As h d b fc fy Ma

Fig. E6.1—Beam cross section.

n As bd wc n

Es Ec

fc n

n

n dh Ig bh

n dh Ig t

t

Mayt Ig

yt

h

t

Ie Ki

Ki fc

-

Ie Ig n dh Ie Ig

Ie

Ki Ig

Ie Ig Ie

Ki

  

Eng.Mohammed Ashraf





Cracking moment for T-section Mcr

fc = wc = bw = b h hf Fig. E7.1—T-section.

Kcr h

fc

Kcr Kcrt b h

= b bw hf h

b h

Kcrt

Mcr

Kcrt Kcrt

Mcr bwKcrKcrt

Mcr

Mcr

  

Eng.Mohammed Ashraf





h

fc

c

m

max

c,x

m,y

max

c,y

m,x

  

Eng.Mohammed Ashraf





-

q EI e q Ie

L

L

qconst

Concrete International

Mo

q

n

n

M M M

  

Eng.Mohammed Ashraf





a

b a

b fr

fc

a

b

Ec t

f ci

at b a

b

t f cr



fcr M cr

Ig

f ci

fc

fr

Ig

M cr

yt

bt yt

I cr

d

bc

h cc

nAs d c

db

d

As bd

n c

bc

Es Ec -

n

c

nAs d c

c

c

  

Eng.Mohammed Ashraf



c

c

k

n

n



kd

n

k

n c

kd

c I cr

Icr

Ie

Ie

M cr Ma

I cr

M cr Ma

Ig

Ig

I cr

I cr

M cr Ma

Ie

Ie Ie

Ie

I e avg

I e left@supp

I e midspan

I e right@supp

I e avg

-

Ie,avg

Ie,midspan

Ie@left supp

Ie@right supp Ie,avg

  

Eng.Mohammed Ashraf





q n EI

q EI

-

C x L

C,x,L

-

  

Eng.Mohammed Ashraf



C,x,T

t



C,x,T

Ig

Ig Mcr

M M

mid-strip mid-strip

Mcr

M

mid-strip

>M

mid-strip

Mcr Ig Ie

Ig

q EI

M x L

'

M,x,T

i

M,x,T

  

Eng.Mohammed Ashraf





C,x MS,x

n

L

Mo

q

n

M

M M

Icr Ig Ig

Mcr Ma

d

h cc

db

< Mcr

Ma

d

As bd

n c

bc

Es Ec

n -

c

nAs d c

c

c   

Eng.Mohammed Ashraf



c

c

k

n

n



kd

n

k

n c

I cr

bc

kd

c

I cr

nAs d c

Icr

Ie

Ie

M cr Ma

I cr

M cr Ma

Ig

I cr

Ig

cr

I cr

Ie

Ie I e avg

Ie

q EI

  

Eng.Mohammed Ashraf





C y L

C,y,T

i

f cr



fcr M cr

fr

C,y,T

Ig yt

Ig

bt

Ig

M cr

yt Mcr > Ma

Mcr

Ma Ie

Ma

Ig

q n EI Ie

Ig

M y L

M,y,T

i

M,y,T

  

Eng.Mohammed Ashraf





LL C,y M,y

L,T

C,x,L

M,y,L

L,midspan

-

x

y

OK

L,midspan

all

all

  

Eng.Mohammed Ashraf





-

max

L LT

L

C,x,L C,x,T

M,y,L M,y,T

LT

L LT max

all

max

OK

  

Eng.Mohammed Ashraf





max

M K ec M Kec

rot

total

rot-left

max

rot-left

rot-right

rot-right

n

n

n

n

n

-

-

T midspan

  

Eng.Mohammed Ashraf





Mcr

Kcr

fr h 

h

h  Mcr Mcr Mcr

bKcr bKcr bKcr Mcr

bwKcrKcrt

Kcrt

  

Eng.Mohammed Ashraf





Mcr

M cr

f r bw h

K crt

bw K cr K crt

Kcr

  

Eng.Mohammed Ashraf





Mcr

M cr

f r bw h

b

K crt

K crt

h

bw K cr K crt h

h

h h

b

h

b

Kcr

  

Eng.Mohammed Ashraf





Mcr

M cr

f r bw h

K crt

h

bw K cr K crt

Kcr

  

Eng.Mohammed Ashraf





Icr

Icr K

Ki bd cd i

n

cd

cd

As bd

  

Eng.Mohammed Ashraf





Ig Ig

Ki

bwh h

Ki

b

h

b

h h

b

Ig b h

b bw hf h h

Ig

Ki

Ki (bwh

  

Eng.Mohammed Ashraf





Icr Ki

Icr

c

Ki bwd cd

Ki

n

cd

cd

c

c

n

n

cd

c

cd

d d

n

c

d d hf

d

d d b bw

hf d

w

n

Ki n

wn

  

Eng.Mohammed Ashraf



c

d d hf

d



Ki n

wn

  

Eng.Mohammed Ashraf





Icr Ki

Icr

c

Ki bwd cd

Ki

n

cd

cd

c

b bw

c

c

hf d

w

n

n

cd

c

cd

d d

n

wn

n

n

d d hf

d

d d

Ki

  

Eng.Mohammed Ashraf



c

d d hf

d



Ki n

wn

  

Eng.Mohammed Ashraf





Ie Ie

Ki

Ki Ig M cr Ma

Ki

M cr Ma

I cr Ig

Mcr Ig

Icr bh

Ki (bwh

  

Eng.Mohammed Ashraf





Ie Ie

Ki

Ki

K i Ig M cr Ma

M cr Ma

I cr Ig

  

Eng.Mohammed Ashraf





Ka

c

Mc

 Ka M c Ka Ie K

Ie

  

Eng.Mohammed Ashraf





Ka

Ka

Ec Ka M c Ka Ie

c

Ka

Mc

Ie

Ec

  

Eng.Mohammed Ashraf





c

  

Eng.Mohammed Ashraf





Ec

Ec

n

wc

fc

Es Ec

Es

wc

fc

Ec

n wc

fc Ec

n

n

  

Eng.Mohammed Ashraf





  

Eng.Mohammed Ashraf





concept and their work provided the basis for the current strut-and-tie model. The strut-and-tie model is usually applied to the design of individual members such as corbels, pile caps, beams, and girders, but it can also be applied to entire structures and is particularly advantageous in discontinuous or disturbed regions due to loading, geometry, or both. In beam, or B-, regions, the strains in reinforcement and concrete are proporciple does not apply. It is assumed that a D-region extends a distance equal to the overall height of the member on each side of a support, concentrated load, change in cross section, Fig. 14.1a—Geometric 14.1a discontinuous regions. 

truss are “ties,” the compression elements are “struts,” and the connection points are “nodes.” An engineer can review

Therefore, the engineer chooses the most appropriate truss model, usually the model that most closely follows the force

used to show how the assumed internal truss transfers load to the supports by a combination of concrete struts and steel ties. Beams designed by strut-and-tie will satisfy strength criteria,

 Design concrete strength for struts and the area of reinforcement in ties are calculated in accordance with ACI 318M-14, Chapter 23. The following discussion and design examples are based on the requirements of ACI 318M-14. However, the use of strut-and-tie is an evolving area of structural concrete design. Revisions and additions to the ACI 318M-14 are under consideration, and practice guidelines are emerging. Many of these developments are incorporated in the examples provided with this chapter as well

Fig. 14.1b—Loading and geometric discontinuous regions.

when using the strut-and-tie method is ACI 318M, Section 23.3.1. Fig. 14.1c—Simple supported beam divided into B- and D-regions. Fn

Fu   

Eng.Mohammed Ashraf





Fig. 14.2a—Truss Truss model for top- and bottom-loaded deep beams tested by Leonhardt and Walther (1966). members, the concrete strut forces can spread such that the strut is wider in the middle than at the nodes. Such a strut is are the predominant type of struts in strut-and-tie models. The nominal strength in a strut is given as ACI 318M, Eq.

Fns = fceAcs where Acs is the area of a strut taken at the face of the node, and fce strut. Fig. 14.2b—Simple strut-and-tie model of a deep beam. fce

where Fu is the factored force, and Fn is the nominal strength is 0.75 for struts, ties, or nodal zones.  Struts are compression concrete elements, generally symbolized as prismatic or uniform tapered elements that

s fc

s

compressive strength in a strut, as shown in Table 14.4. A prismatic strut is modeled as an element of a uniform cross section and its reduction factor is equal to 1.0. A bottleshaped strut assumes that the compressive force spreads

  

Eng.Mohammed Ashraf





Fig. 14.4a—Strut-and-tie Strut-and-tie model of a deep beam with prismatic and bottle-shaped strut elements.  Strut geometry and location Struts with uniform crosssectional area along length Struts located in a region of a member where the width of the compressed concrete at midlength of the strut can



Reinforcement crossing a strut

s

1.0 Satisfying 23.5

Struts located in tension members or the tension zones of members

0.75

0.40

All other cases

within the member, causing longitudinal tension splitting

stress spreads laterally. †

shaped strut is assumed to spread at a 1:2 slope away from the node, as shown in Fig. 14.4b. The resulting tension has s

value of 0.6, as shown

of the bottle-shaped strut. For beams with concrete strength not exceeding 40 MPa, the requirement to resist the splitting

in Table 14.4. s, reinforcement must resist the splitting tensile force resulting from the spreading compres-sion force. The reinforcement must cross the axis

  

Eng.Mohammed Ashraf





Fig. 14.4c—Reinforcement crossing a strut. where Asi is the total area of surface reinforcement at spacing -th layer of reinforcement crossing a strut at an si in the i-th i to the axis of the strut, and bs The reinforcement calculated from ACI 318M, M, Eq. 1

2

to the axis of the strut, or in one direction -

degrees.

Fnt = fceAcs + As fs where As fs strength.  Ties are tension elements extending between two nodal zones. The reinforcement axis must coincide with the axis of the tie in the model. The nominal strength of a tie is calculated by ACI 318M,

Fnt = Ats fy + Atp fse

fp

where Ats and Atp are the reinforcement areas in nonprestressed and prestressed ties, respectively; fy and fse are fp is the increase in stress in prestressing steel due to factored loads. For nonprestressed ties, fse fp are set equal to zero. For fse +

Fig. 14.5—Node element types. fp fp can be set equal to 420 MPa and 70 MPa for bonded and unbonded bars or strand, respectively. It is imperative to anchor nonprestressed reinforcement by mechanical devices, standard hooks, or straight bar devel-

  

Eng.Mohammed Ashraf





extended nodal zone.   Nodal zone Forces on condition the node n Bounded by struts or bearing 1.0 CCC areas anchored in nodal zone Two or more ties anchored in nodal zone

0.8

CCT

0.6

CTT

ACI 318M

Figure

Table

Table

Table

The strength of a nodal zone is given by ACI 318M, Eq.

Fnn = fce Anz Fig. 14.6a—Extended nodal zone. opment, and by anchorage devices for post-tensioning reinforcement. They must be anchored at the node as illustrated in Fig. 14.5.

where fce the nodal zone; and Anz is the area of the nodal zone face on which Fu acts and perpendicular to the line of action of Fu.

fce

in practice, it is preferable to apply Atpfse as an equivalent external load with a load factor of 1.0 to model the correct

nfc

n

 forces from struts, ties, or both intersect. A nodal zone is an assumed volume of concrete around a node that transfers forces from struts and ties through the node. At least three forces must intersect at a node to be in equilibrium.

compressive strength in a node as shown in Table 14.6. There are two nodal zones types—hydrostatic and nonhydrostatic. A hydrostatic nodal zone has faces that are perpendicular to the struts and ties acting on the node and has wn1:wn2:wn3 are proportional to the three incoming forces C1:C2:C3. A nonhy-

  

Eng.Mohammed Ashraf





  Usual steps to apply the strut-and-tie model:

For each controlling condition: 1. Calculate the member reactions from statics. 2. Check bearing mode stress at external loads and reactions. 3. Use appropriate analysis methods or test results to esti4. Create a trial internal truss model made up of struts and 5. Check that the strut-and-tie truss model is in equilibrium with the factored applied loads, and the reactions and cient to assume the axis of the struts and ties only. 6. Calculate or estimate the required size of each node based on the limiting compressive strength of the node or strut, whichever controls. Fig. 14.6c—Resolution of forces at nodal zone.

the struts, ties, and nodal zones exceed the factored forces.

ular to the struts and ties acting on the node, which results in  Before exiting the extended nodal zone in a CCT or CTT, ties must be anchored by a plate, hooks, or straight bar



are determined by outlines of the compressed zone at the intersection of struts or reactions with a prism of concrete concentric with the ties.

Bulletin für Stahlbeton, Wilhelm Ernst & Sohn, Berlin, 1966, 159 pp. Concrete-Steel Construction (Der Eisenbetonbau),, Translation of the third German Edition by Eisenbetonbau)

nodal zones and the light shaded areas are the extended nodal zones. The extended nodal zone area is stressed in compression due to the strut reactions. Forces are transferred from strut to face of the node is then calculated as illustrated in Fig. 14.6a ws is a function of the tie width wt, the bearing length b, and the angle between the

layers of reinforcement plus twice the distance from the face of the concrete section to the centerline of the closest bar. If more than three struts frame into a nodal zone, it is generally better to resolve some of the strut forces such that only three remain, as shown in Fig. 14.6c. If the support width in the direction perpendicular to the member is less than the member width, transverse forces may require reinforcement to restrain concrete splitting. The design of the transverse reinforcement can be designed using a transverse strut-and-tie model.

Ritter, W., 1899, 1899 “The Hennebique Design Method Schweizerische Bauzeitung (Züich) Reineck, K.-H., ed., 2002, Examples for the Design of Structural Concrete with Strut-and-Tie Models, SP-208, American Concrete Institute, Farmington Hills, MI, 2002, 242 pp. Further Examples for the Design of Structural Concrete with Strutand-Tie Models, SP-273, American Concrete Institute, Farmington Hills, MI, 250 pp. “Towards a Consistent Design of Structural Concrete,” PCI Journal Betonkalender 90

  

Eng.Mohammed Ashraf





 Strut-and-tie Example 1: Strut-and tie model of a deep beam without shear reinforcement Two transverse beams frame into a simply supported deep beam as shown in Fig. E1.1. Determine the required reinforcement to support the applied loading. Given: D L b = 600 mm h = 2300 mm fc = 28 MPa fy = 420 MPa = 6900 mm Support length:450 mm at reaction a bearing length of 450 mm

Fig. E1.1—Deep beam geometry.

ACI 318M-14 Discussion Step 1: Required strength 5.3.1 Calculate self-weight of beam:

Calculation 3

wc 3

U = 1.2D + 1.6L

U Pu = U

Hu = 0 Ru = Vu Ru Step 2: Strut-and-tie model 23.2.1 represent the compression elements, the solid line the tension element, and the dimensionless full round 23.2.2 circles represent the nodes—intersection of struts and ties. Because of the short distance between the ap ap23.2.3 plied loads and supports, only a single compression strut is used between the applied load and the closest 23.2.4 support. A tie is required at the bottom to maintain equilibrium.

Fig. E1.2—Deep beam strut-and-tie model without shear reinforcement.

  

Eng.Mohammed Ashraf





strength in struts: 23.4.3

fce

s fc

S1 s

fce S2

s

fce

Bottle-shaped strut without reinforcement: fce s strength in nodes: 23.9.2

fce

n fc

N2; N3 n

fce

n

fce

N1 and N4

Anchoring two or more ties—CTT: fce n Step 4:: Stress at support and applied load Calculate bearing plate area: Compressive stress at support and loading:

AA = 450 mm × 600 mm = 270,000 mm2

N1—CCT: N1, and the third face is bound by the bearing plate.

fce Check if applied stress is smaller than the fce = 14.3 MPa > fu = 7.4 MPa

OK

N2—CCC: N2, and the third side is bound by the applied load bearing plate.

fce Check if applied stress is smaller than the fce = 17.85 MPa > fu = 7.4 MPa

OK

  

Eng.Mohammed Ashraf





Step 5: Required strength Because of symmetry, only half the beam will be

N1 and N2 are located at plate centerlines. The nodes’ vertical position must be estimated or determined. They should be as close as possible to the top and bottom of the beam. The height of the top strut and bottom tie are estimated to be 280 mm and 380 mm, respectively. Usually, each estimate is based on “backof-the-envelope” calculation. The adequacy of these estimates is checked below.

Fig. E1.3—Nodes, struts, and ties geometry

T1

wn/2 = 190 mm S1 tioned at ws/2 = 140 mm from top of the beam. Calculate the forces in Strut S1 and Tie T1 from the free body diagram.

N1 : Mu = 0

Pu a – S1 q

Moment arm: q = 2300 mm – 190 mm – 140 mm = 1970 mm S1 S1 Calculate the angle and force of the diagonal strut, S2:

S2

Step 6: Strut-and-tie model geometry Strut S1 centerline is located 140 mm from top of beam. Tie T1 centerline is located 190 mm from bottom of beam.

Fig. E1.4—Final determined strut-and-tie geometry.   

Eng.Mohammed Ashraf





Step 7: Tie reinforcement 23.7.2 Calculate minimum tie reinforcement: 23.3.1 1

Choose reinforcement arrangement:

Ast = 7740 mm2 at 75 mm, 190 mm, and 300 mm from bottom Ast = 7740 mm2 at 75 mm and 250 mm from bottom

•

• •

Ast = 7824 mm2 at 65 mm, 160 mm, and 260 mm from bottom For better steel distribution and for ease of anchorage bars; Ast = 7740 mm2 N2 N2 S1 and S2 bearing plate of the applied load; CCC: From Step 3: fce = 17.85 MPa fce = 17.85 MPa

Fig. E1.5—Node —Node

Fns

fceAcs

Check if design strength is greater than the required strength.

2

geometry.

Fns

Fns

S1

OK

Struts S2 is expected to be a bottle-shaped strut. Calculate the width at top of strut from: wst = b wscos

Strut width top: wst = 505 mm, say, 500 mm

and resist the bursting force in the strut. The N1 fce = 13.4 MPa fce = 17.85 MPa

23.4.1

Fns

fceAcs

Check if design strength is greater than the required strength.

Fns

Fns

S2

OK

  

Eng.Mohammed Ashraf





N1 N1 S2 T1 framing into it. The third side is bound by the N1 is CCT. From Step 3: fce = 13.4 MPa fce = 14.3 MPa

Fig. E1.6—Node Calculate the width at top of strut from: wscos wst = b Fns

fceAcs

Check if design strength is greater than the required strength.

1

geometry.

Strut width bottom: wst Fns Fns

The tension reinforcement is anchored beyond the back face of the node, thus imposing a compression force on that face. Although not required by the Code, the height of the node can be checked based on the limiting compressive stress of the node.

Fnn

Check if design strength is greater than the required strength.

Fnn

S2

OK

T1

OK

  

Eng.Mohammed Ashraf





Step 10: Reinforcement anchorage Reinforcement should be developed in the node. The factored tie force must be developed at the point where the centroid of the reinforcement in a tie leaves the extended nodal zone and enters the

N1 and two longitudinal bars from the second layer will be headed deformed bars for anchorage. Conservatively, check anchorage length of

dh,anchorage

= 648 mm

25.4.4.1 Abg

Ab db db

Check if assumed beam width is adequate: bw db db db,stirrup db,cover

Provide Abg creq s

2

cprov = 65 mm

OK

bw bw

25.4.4.2

2

OK

Development length of headed bars:

dh,anchorage

= 648 mm >

dt

= 433 mm

OK

  

Eng.Mohammed Ashraf





Step 11: Minimum reinforcement for crack control Area of vertical web reinforcement provided shall center on each face over entire length. not be less than:

OK

Area of horizontal web reinforcement parallel to bars at 230 mm on center on each face over entire height.

Ash

bws2

OK

s and s2 cannot exceed d/5 or 300 mm Reinforcement must also satisfy the requirements for strut-and-tie minimum reinforcement. 23.5.3

9.9.2.1 Control cracking under service loads and guard against diagonal compression failure.

Vu Vu OK Step 12:: Reinforcement detailing Deep beam reinforcement based on strut-and-tie model without shear reinforcement

Fig. E1.7—Deep beam reinforcement.

  

Eng.Mohammed Ashraf





Strut-and-tie Example 2: Strut-and-tie model of a deep beam with shear reinforcement Two transverse beams frame into a simply supported deep beam, as shown in Fig. E2.1. Determine the required reinforcement to support the applied loading. Given: D L b = 500 mm h = 2300 mm fc = 28 MPa fy = 420 MPa = 6900 mm Support length: 450 mm at reaction Fig. E2.1—Deep beam geometry. bearing length of 450 mm ACI 318M-14 Discussion Step 1: Required strength 5.3.1 Calculate self-weight of beam

Calculation wc

3 3

U = 1.2D + 1.6L

U = Pu = U =

Hu = 0 Ru = Vu Ru Step 2:: Strut-and-tie truss model 23.2.1 23.2.2 lines represent the compression elements, solid lines the tension elements, and the dimensionless full 23.2.3 round circles represent the nodes—intersection of struts and ties. The distance between the applied loads 23.2.4 and supports has a tension element representing the shear reinforcement in the beam. A tie is required at the bottom to maintain equilibrium.

Fig. E2.2—Deep beam strut-and-tie model with shear reinforcement.

  

Eng.Mohammed Ashraf





Step 3: Sectional analysis Flexure: This step is not required for strut-and-tie models. It is shown for comparison purposes only. Calculate the moment arm to estimate the preliminary reinforcement area and depth of top compression and bottom tension chords.

reinforcement: Moment arm: d = h – cover – stirrup diameter – bar diameter – 1/2 distance between two layers.

d = 2300 mm – 165 mm = 2135 mm

Mn

Mu

Mu

4.6 × 106 the reduction factor in Chapter 23 is related to material, rather than member, behavior as stated in Chapter 21..

23.4.3

As

2

As

strength in struts: fce s fc

S1 s = 1.0

fce S2

s = 0.75

fce

Bottle-shaped strut without reinforcement: fce s = 0.6

23.9.2

strength in nodes: fce n fc

N4 n

= 1.0

fce

n

= 0.8

fce

n

= 0.6

fce

N1 and N2

N3

  

Eng.Mohammed Ashraf





Step 5: Stress at support and applied load Calculate bearing plate area: Compressive stress at support and loading:

AA = 450 mm × 500 mm = 225,000 mm2

N1—CCT: N1. The third face

fce Check if applied stress is smaller than the fce = 14.3 MPa > fu = 8.9 MPa

OK

N4—CCC: N4. The forth side is bound by the applied load bearing plate

fce Check if applied stress is smaller than the fce

fu = 8.9 MPa

OK

  

Eng.Mohammed Ashraf





Step 6: Required strength Because of symmetry, only half the beam will be considered. N1 and N4 are located at plate centerlines. The nodes’ vertical position must be estimated or determined. They should be as close as possible to the top and bottom of the beam. The height of the top strut and bottom tie are estimated to be 280 mm and 330 mm, respectively. Usually, each estimate is based on “back-of-theenvelope” calculation. The adequacy of these estimates is checked below. T1 S2 wn/2 = 165 mm and ws/2 = 140 mm from bottom and top of beam, respectively. z

S T2 T3 S1 S S2 T1 Step 7: Member forces

  

Eng.Mohammed Ashraf





Step 8: Tie reinforcement Calculate tie reinforcement, T1: 23.7.2 1 23.3.1 Choose reinforcement arrangement:

•

Ast 7740 mm2 at 75 mm and 250 mm from bottom

• Ast = 7824 mm2 at 65 mm, 160 mm, and 260 mm from bottom For better steel distribution and for ease of anchorage

Calculate tie reinforcement, T2: 23.7.2 23.3.1

bars; Ast = 7740 mm2

2

2

Ast Extend bottom layer reinforcement of T1.

.

Tie T3 represents the shear required reinforcement in the region between the support and the applied load. 23.7.2

Fnt = Ats fy ,

Wight and Parra-Montesinos’ “Strut-and-Tie Model for Deep Beam Design,” in Concrete International,, May 2003, pp. 63-70,, provides a discussion on stirrup distribution.

2

Ats,prov

> Ats,prq’d = 6349 mm

2 2

“Some engineers might prefer to concentrate the stirrup reinforcement at approximately the location of tie” T3 N2 is determined as half the disN1 and N4 s However, “it is reasonable to assume that struts” S3 and S4 “will fan out and engage several stirrups.” In this example, the second approach will be followed and the reinforcement will be distributed between the support and the applied force. Assume 100 mm stirrup spacing:

For stirrup distribution, refer to Fig. E2.8.

  

Eng.Mohammed Ashraf





N1 This node has one strut, a bearing force, and a tie. N1 The node has a horizontal dimension equal to the

From Step 4: fce = 14.3 MPa fce = 13.4 MPa

Bottle-shaped strut:

Fig E2.3—Node N1 geometry.

Determine width of strut S3 at N1: ws: ws =

b

wtcos

ws

Fns

fceAcs

Fns

Check that the design strength is larger than the required strength:

Fns

S3

OK

N2 This node has two struts and one tie framing into

From Step 4: fce = 14.3 MPa fce fce

Stress at vertical face: Strut S1 2550 MPa; however, the controlling stress is N2 fce = 14.3 MPa. Therefore, the N2 is calculated from:

N2

Fig. E2.4—Node N2 geometry.

N2. The hori-

mined in Step 8.

  

Eng.Mohammed Ashraf





N4 This node is an all-compression node—three struts and one applied compression force framing into it

fce = 17.85 MPa fce = 17.85 MPa Bottle-shaped strut: fce = 13.4 MPa

The node has a horizontal length equal to the bearing plate and height equal to 280 mm assumed above. The stress at bearing plate was checked in Step 5. S1 is divided into two equal forces as shown in S2 and the second balances the horizontal component of S4.

Fig. E2.5 E2.5—Node Node N4 geometry.

Fns

fce Acs

Fns

Check that the design strength is larger than the required strength:

Fns

Required vertical face height for Strut S1 at N4 is based on the same compressive stress exerted by Strut S2. Both S1 and S2 are prismatic struts. Assume node depth facing S1 is 140 mm S2

Fns

Check that the design strength is larger than the required strength:

Fns

Strut S4—bottle-shaped strut: Determine width of Strut S4

S2

OK

S2

OK

S3

OK

N1: ws:

ws = Fns

b

wtcos fce Acs

Check that the design strength is larger than the required strength:

ws Fns

Fns

  

Eng.Mohammed Ashraf





N4: R23.2.2

There are four forces acting on a node and, per ACI 318M-14, Section R23.2.2, it is generally necessary to resolve some of the forces to end up

S1 and S4

N4 Fig. E2.6 E2.6—Revised Revised geometry of Node N4.

Check adequacy of nodal zone: ws =

23.3.1

b

wtcos

Fns fce Acs Fn Fu Check that the design strength is larger than the required strength:

Available length at node: ws Fns

Fns

S5

OK

  

Eng.Mohammed Ashraf





N3 This node has three ties and one strut framing

From Step 4: fce fce

There is load transfer between Ties T1 and T2. N3 are determined from the vertical and horizontal distribution of reinforcement. Because Tie T3 N3, assume that the Strut S4 fans out similar to Strut S3. Strut S4 needs only to be N4

Fig. E2.7—Node N3 geometry.

N3 due to the transfer of forces T1 and T2 is: 23.7.2

Strut S3 is fan-out shaped and engages several stirrups.

ws,prov = 330 mm > ws,req = 215 mm

OK

Step 13:: Reinforcement detailing 23.8.2 Anchor longitudinal reinforcement by hooks, headed bars, mechanical anchorages, or straight bar development. In this example, hooked bars are used, although

N1: Use 90-degree hook bar:

25.4.3.1

25.4.3.2

For hooks with side cover not less than 65 mm and for 90-degree hooks with cover on bar extension beyond hooks not less than 50 mm, the development length can be reduced by 0.7.

dh,anchorage

cover

N1: = length of extended nodal zone –

dh

= 547 mm × 0.7 = 383 mm

dh,anchorage = 175 mm + 450 mm + 75 mm/tan60.0° – 75 mm = 593 mm OK dh,anchorage > dh

  

Eng.Mohammed Ashraf





N3: T2 T1

As,prov = 3870 mm2 As,prov = 7740 mm2

For longitudinal reinforcement distribution, refer to Step 7.

height of 330 mm

Assume straight deformed bar development:

d

dh,anchorage

1150 mm/2 = 1875 mm dh,anchorage

d,required

>

dh

= 828 mm

OK

= 828 mm

  

Eng.Mohammed Ashraf





Check that reinforcement ratio is larger than the minimum required of 0.0025:

full length of beam: Avh

9.9.4.3

23.5.3

2

2

s

Maximum allowable spacing of horizontal reinforcement:

d/5 = 2100 mm/5 = 420 mm controls)

OK

Check horizontal reinforcement ratio based on

S3 and

i

uted vertically.

23.5.3

9.9.4.3

S3 and

2

Maximum allowable spacing of vertical reinforcement: Check if reinforcement ratio is larger than the minimum requirement of 0.0025:

23.5.3

d/5 = 2100 mm/5 = 420 mm d d/ controls)

OK

v

Check if sum of horizontal and vertical reinforcement exceeds the minimum required 0.003. h

v

= 0.0022 + 0.0068 = 0.009 > 0.003

OK

Provide minimum shear reinforcement between

with bw = 500 mm and s2 = 200 mm 9.9.2.1

To control cracking under service loads and guard center. against diagonal compression failure, the upper limit for shear force in deep beams should not exceed: Check that the design shear is greater than the required shear force. Vn

Vu

OK

  

Eng.Mohammed Ashraf





Stirrup distribution: Reinforcement required for crack control is Stirrups spacing is the smaller of: 9.9.4.3

d/5 = 420 mm > 300 mm > 100 mm

d/5, and 300 mm

OK

In the area between the two applied loads use beam. These stirrups shall have 135-degree bends alternatively around one or the other of longitudinal bars. Step 14: Arrangement of reinforcement

Fig. E2.8—Deep beam reinforcement based on strut-and tie design. Step 15: Conclusions 1. Comparing Examples 1 and 2, either of the two models presented is acceptable, provided that equilibrium and yield conditions 2. The predominant mechanism of shear transfer in a deep beam region is attributed to the arching action of a direct strut. a/d experimental observations and past research.

6. The capacity of a two-panel model is controlled by the capacity of the vertical tie or stirrups, thereby ignoring the contribution to shear strength provided by the concrete.

  

Eng.Mohammed Ashraf





Strut-and-tie Example 3: Design of one-sided corbel using strut-and-tie method Design a one-sided corbel supporting a crane girder beam reaction force acting at 125 mm from the face of the column. A in this example. Determine the required cross section of corbel and reinforcement. Given: D L HD b1 b2 fc fy

= = = =

400 mm 400 mm 35 MPa 420 MPa

Bearing plate size: 150 x 300 mm Fig. E3.1—Geometry and loading of one-sided corbel. ACI 318M-14 Step 1: Required strength 5.3.1 U = 1.2D + 1.6L

Discussion

Calculation U= Pu = U = Hu

23.4.3

strength in struts: fce s fc s

fce S1, S2, S3

s

fce

Bottle-shaped strut without reinforcement: fce s

  

Eng.Mohammed Ashraf



23.9.2

strength in nodes: fce n fc



Bound by struts or bearing area or both—CCC: fce n N1 and N3 n

fce

n

fce

N2 Step 3: Bearing Calculate bearing plate area: Compressive stress at loading:

Ac = 300 mm × 400 mm = 120,000 mm2

23.9.2 fce = 17.85 MPa > fu = 2.56 MPa Bearing plate area is adequate. Step 4: Corbel dimensions 23.2.9 Choose a span-to-depth ratio of maximum 2.

OK

Use an overall corbel depth of 500 mm

Depth at outside edge of bearing area is at least half the depth of the section at column face. Use a depth of 250 mm Step 5: Strut-and-tie model Consider erection tolerances and load eccentricities by shifting Pu one inch toward the outer edge of the corbel from center of bearing plate. 16.5.2.2

50 mm + 150 mm/2 + 25 mm = 150 mm Assume center of Tie T1 is at 45 mm from top of corbel—considering one layer of reinforcement and 25 mm of cover. 23.2.9

Moment arm: d = 500 – 45 = 455 mm N1 from plate centerline is obtained from geometry. It is the intersection of the force resultant line and T1. x Fig E3.2—Strut-and-tie model of one-sided corbel loading.

  

Eng.Mohammed Ashraf





The horizontal Tie T3 is assumed to be located on the horizontal line passing through the sloping end of the corbel. The position of Strut S3 is found by calculating the strut width ws N4 N1.

= S3

ws

The width of Strut S3 is controlled by the smaller S3 N3—Step 2: fce = 17.85 MPa Bottle-shaped strut: fce

S3 = fcebws

S3

S3 =

ws

ws = 6680ws

Solving for ws ws = 91 mm, say, 90 mm Step 6: Truss forces 23.7.2 From statics, the forces in the truss members can 23.3.1 be calculated. N1: 1:

Forces in Strut S1 and Tie T1.

S1 T1 N2:

Assume 2 mm cover. 2:

Forces in Strut S2 and Tie T2.

S2 = T1 T2 = S2

Forces in Strut S3 and Tie T3.

N3: T3 = S2cos56.2° – S1cos65.8° T3

S3 = S1sin65.8° + S2sin56.2° S3 S3

  

Eng.Mohammed Ashraf





Step 7: Force summary

Step 8: Tie reinforcement The area of reinforcement required for Tie T1 is

23.3.1 23.7.2

Fnt

Fu = T1

Tie strength: Fnt = Ats fy The provided steel area must be at least per ACI 318M prescriptive requirements:

T2 has larger tension force than T1. However, tension force T2-T1 should be resisted by column longitudinal reinforcement. Therefore, continue lap length. For Class B splice, 1.3

2

Ast,prov Anchor Tie T1

dh:

t

e

s

= 1.0

25.4.2.3

1.3

dh

The area of reinforcement required for Tie T3 is N2. 23.3.1 23.7.2

Fnt

Fu = T3

Tie strength: Fnt = Ats fy

slope; 2 Ast,prov spaced at 50 mm on center.

2

. These ties are

  

Eng.Mohammed Ashraf

2

.





N3 N3 at Strut S3 were determined N3 at Struts S1 and S2 is N3 and Struts S1 and S2; refer to Step 2: fce = 17.85 MPa fce = 16.7 MPa

Fns fceAcs and Acs = bwws Available width ws of nodal zone where S1 frames is:

ws = 90 mm, calculated in Step 5

1/2ws S1 = 1/2ws S3 sin 65.8° = 45sin 65.8° = 44 mm ws S2 , provided = 88 mm ws,provided = 88 mm > ws,required = 50 mm

OK

Use 88 mm strut width ws N3 for S2:

Required strut width ws

Use 54 mm for node zone face width at S2. Assume 100 mm node depth to accommodate the Although not required by code, the height of the node can be checked based on the limiting comcom pressive stress for the node: Fnn

23.9.1 Fnn ws =

Fu = T3 b

wscos

Fu = S1

Fnn Fnn

Fns Fns

fce Anz

T3

OK

ws,req fce Anz

Fns Fns

S1

OK

Fig. E3.3—Dimensions of strut-and-tie members.   

Eng.Mohammed Ashraf





N3 is to combine Struts S1 and S2 as shown in Fig.

The resultant force S1 + S2 N3 and has an angle of 6 degrees with the vertical. The resultant force can be simply calculated from statics: From geometry, ws can be calculated and is equal to 150 mm.

N3 forces Fig. E3.4 E3.4—Alternate E Alternate resolution of node forces.

Accordingly, the strut strength is:

Fns

fceAcs

Check that the design strength is larger than the required strength: Step 10:

S Fns

Fns

S1+2

OK

N1 N1 has one strut, one bearing, and a tension

N1 are the same N3. Therefore, the width of Strut S1 is the same. However, due to geometry, bearing plate, and tension reinforcement, a larger width is provided. N1 is adequate.

  

Eng.Mohammed Ashraf





N2 N2 has one strut and two ties framing into it—CTT; Step 2 fce = 13.4 MPa fce = 16.7 MPa

The available tie width is:

23.9.1

Width of node at T2: Assume vertical reinforcement is located 50 mm from the edge: ws Width of node at Strut S2: Strut S2, a bottle-shaped strut with reinforcement N3. At N2:

Fnn

wn

T1 and T2 bend may result in a potential failure of the concrete in the diagonal compression zone if the radius of the bent reinforcement is too small, which is discussed in Klein, “Curved-Bar Concrete International Sept. 2008, pp. 42-47. 42-47

25.3.1

db.

rmin

rb = 48 mm

OK

  

Eng.Mohammed Ashraf





Step 12: Minimum reinforcement for crack control 16.5.6.6 Closed stirrups are required parallel to the reinforcement T1 to be uniformly distributed T1 as recommended by the descriptive requirements of ACI 318M. The area of these ties or stirrups must exceed: Ah

16.5.5.1

Asc – An

where An is the area of reinforcement resisting the tensile force T3, and Asc is calculated from 16.5.51. Asc = 819 mm2 Area of vertical web reinforcement provided shall 2 2 Av not be less than: with average spacing of 300 mm/3 = 100 mm s was used as 0.75 when calculating the force in the diagonal Struts S1 and S2. Therefore, minimum reinforcement must be provided to satisfy:

23.5.3

mini i is the angle between the axis of minimum reinforcement and the axis of strut. 23.5.3.1

i must be greater than 40° because only horizontal reinforcement is provided.

  

Eng.Mohammed Ashraf





Step 13: Reinforcement detailing

Fig. E3.5—One-sided One-sided corbel reinforcement based on strut-and-tie design.

  

Eng.Mohammed Ashraf





Strut-and-tie Example 4: Design of double corbel

located 150 mm from the face of the column. Horizontal tensile forces are acting away from the column, accounting for creep

Determine the required reinforcement. Given: PD PL ND NL HD b1 = 400 mm b2 = 400 mm fc = 28 MPa fy = 420 MPa Bearing plate size: 150 x 300 mm Fig. E E4.1—Double-sided corbel geomtetry. ACI 318M-14 Discussion Step 1—Required strength Column factored load: 5.3.1 U = 1.2D + 1.6L Corbel factored load: U = 1.2D + 1.6L Horizontal factored load: Hu = 1.2Hd Step 2—Stress limits

23.4.3

Calculation

U= Pu = U = U= Nu = U Hu

strength in struts: fce s fc

S3, S6, S7 s

fce S1 and S2

s

fce S4 and S5

23.9.2

s

fce

n

fce

n

fce

strength in nodes: fce n fc

N3 and N4

N1 and N2

Anchoring two or more ties—CTT: fce n

  

Eng.Mohammed Ashraf





Step 3: Bearing at loading Calculate bearing area. Use full corbel width. Compressive stress at loading:

Ac = 400 mm × 150 mm = 60,000 mm2

23.9.2

The nodal zone at applied load location is a

fce = 14.3 MPa. N1:

fce = 14.3 MPa > fu = 5.5 MPa

OK

Step 4: Corbel dimensions 23.2.9 The corbel must satisfy the requirement a/d < 2. 16.5.2.2

Depth outside of the bearing area is at least 0.5d. For this problem, select corbel depth at column face = 500 mm and at free end of the corbel = 300 mm

Fig. E4.2 E4.2—Double-corbel Double-corbel geometry. Step 5: Strut-and-tie model Consider erection tolerances and load eccentricities by shifting Nu one inch toward the outer edge of the corbel from center of bearing plate.

75 mm + 150 mm/2 + 25 mm = 175 mm Assume center of Tie T1 is at 50 mm from top of corbel—considering two layers of reinforcement and 25 mm of cover. 23.2.9

Moment arm: d = 500 mm – 50 mm = 450 mm N1 from plate centerline is obtained from geometry. It is the intersection of the force resultant line and T1. x Fig E4.3—Double-corbel strut-and-tie model. The horizontal Strut S3 is assumed to be located on the horizontal line passing through the sloping end of the corbel.

  

Eng.Mohammed Ashraf





The column axial load Pu is resolved into two equal loads acting in line with Struts S6 and S7. The location of Strut S6 centerline can be found by calculating its strut width ws based on the

From Step 2: fce = 17.85 MPa fce = 17.85 MPa Strut and node both control. Due to symmetry, the force in S6 is equal to: S6 = Pu/2 + Vu Fns

fcebws

S6 = S7

S6

23.3.1

Step 6: Truss forces 23.7.2 From statics, the forces in the truss members can 23.3.1 be calculated. N1:

Forces in Strut S1 and Tie T1.

Forces in Struts S4 and S5. From geometry force in S3. Step 7: Force summary

S1 T1 Because of symmetry, S2 = S1 S 4 = S 5 = Pu S3 = S1

Step 8: Tie reinforcement The area of reinforcement required for Tie T1 is

Fnt 23.7.2

Fu = T1

Fnt = Ats fy The provided steel area must be at least per ACI 318M prescriptive requirements:

23.2.9



Ast,prov

2

Arranged in two layers as shown in Fig. E4.4.

  

Eng.Mohammed Ashraf

2

.





Step 9: Node N3 (N4) Nodes N3 and N4—CCC: Because of symmetry, only Node N3 will be addressed (Step 2). fce = 17.85 MPa fce = 17.85 MPa fce = 13.4 MPa (with reinforcement) fce = 10.7 MPa (without reinforcement) The width ws = 139 mm of Strut S6 (S7) was calof node face at prismatic Strut S3 is calculated from: 23.4.1 23.3.1(a)

Fns

fceAcs with Acs = bwws Fns S3

(23.4.1(a)) Say, 50 mm

Node width, N3 (N4), facing Strut S4 (S5): The column is reinforced with four No.25 longitudinal bars with 40 mm cover. The column a) run parallel to the axis of the strut, b) are located within the strut, and M c) are enclosed in ties satisfying ACI 318M (25.7).

23.4.1(b) R23.4.1

Therefore, Strut S4 (SS5) strength can be increased

Two bars are located within Strut S4 (S5).

Fns = fceAcs + As fs with fs fy = 420 MPa

As fs

2

)(420 MPa) = 428,400 N )(

Assume reinforcement is not provided to satisfy 23.5.1 in the column above the corbel section. The nominal concrete strength of Strut S4 (S5) is limited to: fce = 10.7 MPa (Step 2) 23.4.1(a)

Node N3 (N4) width, 139 mm, was calculated based on limit strength—Step 4.

ws,prov = 139 mm > ws,req = 80.3 mm

OK

Node width N3 (N4) facing Strut S1 (S2): Strut S1 has reinforcement satisfying 23.5.1 (Step 10), Step 2: fce = 17.85 MPa fce = 13.4 MPa (with reinforcement) controls Determine strut width: Fns fceAcs S1 Acs = bwws

(23.4.1(a))

  

Eng.Mohammed Ashraf





Step 10: Node N1 Node N1 (N2)—CCT: Because of symmetry, only Node N1 will be addressed. From Step 2: fce = 14.3 MPa fce = 13.4 MPa (with reinforcement) T1 is located at 50 mm from the top. Therefore, height of Node N1 facing T1 is 100 mm 23.4.1(a) 23.3.1(c) 21.2.1(g)

23.9.1 23.9.2

R23.2.6b(ii)

The tension reinforcement is anchored/welded to an edge steel angle, thus imposing a compression force on that face. Although not required by the Code, the height of the node can be checked based on the limiting compressive stress for the node. Fnn

fceAcs with Acs = bwws

(23.4.1(a))

Fnn = (14.3 MPa)(400 mm)(100 mm) = 572,000 N Fnn = 572,000 N > T1 = 269,000 N OK

Check that design strength is greater than the required strength: Determine strut width of S1 (S2): wtcos ws = b ws,req Fns fceAcs 23.4.1(a)) Fns = ((13.4 MPa)(193 mm)( ((23.4.1(a)) mm)(400 mm) = 1,034,480 N Fns = 1,034,480 N > S1 = 389,000 N OK Acs = bwws Check that design strength is greater than the required strength:

Anchor Tie T1 by welding eight No. 13 reinforcement to a steel angle of L100 x 100 x 13 mm Refer to Fig. E4.4 for details.

Fig. E4.4—Strut-and-tie member geometry.

  

Eng.Mohammed Ashraf





Step 11: Minimum reinforcement for crack control Closed stirrups are required parallel to the 16.5.6.6 reinforcement T1 to be uniformly distributed T1 as recommended by the descriptive requirements of ACI 318M. 16.5.5.2

The area of these ties or stirrups must exceed: Ah

16.5.5.1

Asc – An

where An is the area of reinforcement resisting the tensile force T1. Asc is the greatest of 16.5.5.1. Asc = 1381 mm2 Area of vertical web reinforcement provided shall 2 not be less than: Av with average spacing of 300/3 = 100 mm Minimum reinforcement must satisfy:

23.5.3 i is the angle between the axis of minimum reinforcement and the axis of strut. i must be greater than 40° because only horizontal reinforcement is provided. Step 12:: Reinforcement detailing

23.5.3.1

Fig. E4.5—Double-corbel reinforcement based on strut-and-tie design.   

Eng.Mohammed Ashraf





This problem is an idealized example where the forces on the left side are equal to the forces on the right side. While it may be reasonable to predict that the dead loads on both sides are equal, the same cannot be tie model for two-sided corbel with the load on the left side is not equal to the load on the right side. The in the column longitudinal reinforcement between the right side and left side.

Fig. E4.6—Proposed Proposed strut-and-tie model if forces are not equal on corbels. If the load on the right side of the corbel reduces further to reach the “no load” condition, the model will

  

Eng.Mohammed Ashraf





Strut-and-tie Example 5: Design a pile cap subjected to the dead and live load axial forces and to axial forces and overturning moment Determine the required pile cap depth and reinforcement. Given:

D L MD = ML

D L MD ML fc fy = 420 MPa Assume pile has 150 mm embedment into pile cap.

E5.1—Pile cap subjected to axial and Fig. E5.1— moment forces.

ACI 318M-14 Discussion Step 1: Calculate ultimate load and reaction Case I: Pile factored load 5.3.1 U = 1.2D + 1.6L Case II: U = 1.2D + 1.6L

Mu = 1.2MD + 1.6ML

Calculation

U= Pu = U = U= Pu = U = Mu

  

Eng.Mohammed Ashraf





Pile reaction: Case I: M = 0

2

R

R

Case II: M

Step 2: Pile reaction summary

  

Eng.Mohammed Ashraf





Step 3: Design pile cap per ACI 318M-14 sectional analysis Critical section at face of column calculated for

22.5.5.1

Shear: Vu = 2Ru Vu

bd

reduction factors are related to member behavior rather than material behavior. Solving for d: Flexure:

Vu Vu

d = 2034d

d Try: d

to: d = 985 mm Mu

Rue

Mu

e = c.l. of pile to c.l. of col. – col./2 = 750 mm = 0.75 m 24.4.3.2

Check reinforcement: Mu

Asfy

d As = 5100 mm2

Check minimum reinforcement ratio:

21.2.2

Check actual required reinforcement: Cc = 0.85fc ba T = A s fy Cc = T

As= 6450 mm2 a

2

a = 38 mm

2 As,req = 4417 mm2 < As,prov = 6450 mm2

OK

Refer to Fig. E5.8 for pile cap reinforcement.

  

Eng.Mohammed Ashraf





Reinforcement development: For straight bars:

25.4.2.3

d

d t=

e=

s

= 829 mm

cb + Ktr db = 2.5.

For bars with hooks:

25.4.3.1

There is enough distance to develop straight bars at the face of the column. In this example, hooked bars will be used as common in practice. Cracking:

Ma Pile cap section modulus:

Stress:

Cracking stress is acceptable < 0.62 Step 4: Establish strut-and-tie model and determine truss forces for Case I

.

column: 600mm2/5 = 72,000 mm2 or 268 x 268 mm Assume the forces are acting at the center of each of assume each of the four corner forces is applied at 125 mm from face of column. Thus, four corner forces are applied at 350 mm apart and located symmetrical to column axis of symmetry, and the

Fig. E5.2—Proposed truss model.

  

Eng.Mohammed Ashraf





Forces in struts and ties: Because the pile cap is symmetrical with respect to two axes, only one bottom node will be analyzed. The rest are similar. S1, S2, S3, and T1. N1 and N2 equal to 750 mm Assume node N2 is located below pile top surface at 125 mm From statics: S1 = R/sin T1 = S1 S2 = S1 S3 = P/5 Step 5: Struts and ties forces summary

23.4.3

S T1 S2 S3

strength in struts: fce s fc

S1 s

fce S2

s

fce

Bottle-shaped strut without reinforcement: fce s

23.9.2

strength in nodes: fce n fc

N4 n

fce

n

fce

n

fce

N1 and N2

N3

  

Eng.Mohammed Ashraf





N1 N1—CCT: N1 has one strut, S1, and one tie in one direction, T1, framing into it. The third side is N1 is CCT. From Step 6: fce = 14.3 MPa fce = 10.7 MPa

Fig. E5.3—Bottom node geometry.

23.9.1

Check bearing strength: Fnn fceAnz Check if design strength is greater than the required strength.

Anz = 350 mm × 350 mm = 122,500 mm2 2 Fnn

Fnn

Ru

OK

Assume tie resultant is located at 190 mm above top of pile. Therefore, the node depth is taken as 2 × 190 mm = 380 mm From geometry, calculate the width of strut from: wscos ws = b

Fns = fceAcs Check if design strength is greater than the required strength.

23.7.3

ws ws = 505 mm, say, 510 mm Fns

Fns

S1

OK

Calculate the required reinforcement area: Fnt Ats fy + Atp fse fp Atp = 0 mm2 and Fnt Fu = T1

2

Ast,prov = arranged in two layers at 125 mm

and 250 mm

  

Eng.Mohammed Ashraf





N2—CCC: fce = 17.85 MPa fce = 17.85 MPa fce = 10.7 MPa

Fig. E5.4—Top node geometry. Strut S3—Bottle-shaped strut without reinforcement: Calculate Strut S3 area: Fns fceAcs Remaining available area for the four identical S1 A2 Acolumn – A1

2

A2

2

Assume diagonal struts are square: Fns fce Acs fcewsws

ws = ws2

ws1cos

ws1cos 30.7° ws1 = 398 mm

Strut S2—Prismatic: Calculate strut strength: Fns fcews1ws

Fns

Check that strut design strength is greater than required strength:

Fns

S2

OK

Total pile cap depth: H = 750 mm + pile embedment depth + wt/2,bottom + ws1/2,top

H = 750 mm + 150 mm + 380 mm/2 + 398 mm/2 = 1289 mm, say, 1350 mm

a

Therefore, use a pile cap depth of 1350 mm, reinforced as shown in Fig. E5.9.   

Eng.Mohammed Ashraf





Step 8: Development Check development length: 

25.4.3.1

There is adequate space to develop the reinforcement. temperature reinforcement placed between piles each Step 9: Establish strut-and-tie model and determine truss forces for Case II From Case I: Use h = 1350 mm Truss depth = h – pile embedment depth – ws/2 – wn/2 Truss depth = 750 mm Angle calculated above = 30.7°

Fig. E5.5 E5.5—Strut-and-tie E Strut-and-tie forces.

  

Eng.Mohammed Ashraf





Step 10: Calculate forces in strut-and-tie model Forces on pile cap from column are calculated from simple static: P

P1,2 P3,4 P5

Pile forces are obtained from Table 5.1.

S T1

S T2 23.2.7

Angle between strut and tie shall not be less than 25°.

S

S

T3 T3 S3 Step 11: Force summary

  

Eng.Mohammed Ashraf





Step 12: Design nodal zones N1 has one strut, S1, and one tie in one direction, T1, framing into it. The third side is bound N1 is CCT. From Step 6: fce = 14.3 MPa fce = 10.7 MPa

Fig. E5.6—Bottom node geometry.

23.9.2

Check bearing strength: Fnn fceAnz Check if design strength is greater than the required strength.

Anz = 350 mm × 350 mm = 122,500 mm2 2 Fnn

Fnn

Ru

OK

Assume tie resultant is located at 210 mm above top of pile. Therefore, the node depth is taken as 2 x 210 mm = 420 mm From geometry, calculate the width of strut from: ws = bsin + wscos

23.4.1

Fns = fceAcs Check if design strength is greater than the required strength.

23.7.2

ws ws = 540 mm

OK

Fns

Fnn

S1

OK

Required reinforcement to resist Tie T1: Fnt Fu = T1 Fnt = Ats fy

2

Ast 2

2 2

Ast Arranged in two layers as shown in Fig. E5.8.

  

Eng.Mohammed Ashraf





N2—CCT: From Step 6: fce = 17.85 MPa fce = 14.3 MPa fce = 10.7 MPa

Assume the tension reinforcement T2 has 65 mm cover. The node width could then be calculated as twice the reinforcement depth = 130 mm, resulting in available node area to resist the compression forces of: Fig. E5.7—Top node geometry. 600 mm × 600 mm – 130 mm × 600 mm = 282,000 mm2 Assume that all struts are square:

23.4.3

Si fceArea From Step 6: fcu = 10.7 MPa w s1 =

23.9.1

23.9.2

A re a =

Si --------f ce

= 0.75 Fns fcews1ws2 > Sisin S i sin w s2 = ------------------f ce w s1

fce = 17.85 MPa

ws1 = ws2sin + ws3cos ws1 – ws2 sin ws3 = --------------------------------cos Step 13: Top nodal zone dimensions Code Strut S6 S1 S2 S3

B C C C

23.4.1 Si 836 1980 1496 274

Total column area available:

Total pile cap depth:

23.4.3 fce, MPa 10.7 10.7 10.7 10.7

ws1, mm 280 430 374 160

90 30.7 26.5 65

23.9.2 fce, MPa 17.85 17.85 17.85 17.85

ws2, mm 280 132 100 87

ws3, mm 0 422 368 192

ws1

ws2 78,400 mm2 56,760 mm2 37,400 mm2 13,920 mm2

Acolumn = 600 mm x 600 mm = 360,000 mm2 Anode = 360,000 mm2 Acolumn = Anode H = 750 mm + 150 mm + 400 mm/2 + 422 mm/2 = 1311 mm Therefore, use 1350 mm deep pile cap.

  

Eng.Mohammed Ashraf





Step 14: Reinforcement detailing

Fig. E5.8—Pile cap depth and reinforcement as obtained from sectional design.

Fig. E5.9—Pile cap depth and reinforcement as obtained from strut-and-tie design.

  

Eng.Mohammed Ashraf





While traditional sectional design places reinforcement between piles, strut-and-tie model design places reinforcement above piles to resist the tension forces resulting from the 3-D struts due to column loading. 2. It is recommended to add additional reinforcement around the perimeter of a pile cap at pile level to prevent spalling of concrete cover in case piles are resisting lateral forces. 3. It is important to simplify the assumptions: a. Use square struts rather than bottle-shaped or those varying in cross section along length of strut to simplify complex geometry where struts intersect in three dimensions.

  

Eng.Mohammed Ashraf







                                                                                                                                                                             

  

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

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

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                                                                                                                                         

  

Eng.Mohammed Ashraf





                           

                            

  

Eng.Mohammed Ashraf







  

      

        

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

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











                                                                                                 

                      





            

  

Eng.Mohammed Ashraf





                                                                                                                                                

                                                                                                                                

  

Eng.Mohammed Ashraf





       

   

   

     















                 



     







            

    

                                                 

  

Eng.Mohammed Ashraf





                    

     

       







  









   

       



   

    



  

Eng.Mohammed Ashraf





              

     



  

Eng.Mohammed Ashraf





                    

    

      





   

     

  

Eng.Mohammed Ashraf



     



        

 

    

 





      

 



     



          

   



            



             

  

Eng.Mohammed Ashraf





     

       



         





            

  

  



 

     

    

  

  

Eng.Mohammed Ashraf







   

       

 

 

  



        







  



  



   



            

Eng.Mohammed Ashraf











   



  

      

 

             



 



     



   

    

     

            

  

Eng.Mohammed Ashraf





     

      

      

             

      

 

  

Eng.Mohammed Ashraf





                  

       

        



   

      

     

   





       

Eng.Mohammed Ashraf





   

 

 

    

        

 



    

 

     

  





   

       

       

 

  

Eng.Mohammed Ashraf





 

     

 

 

  

 

       



   





     

  



  



   





 



               

  



                

  

  

  



  



   

                

  



  

Eng.Mohammed Ashraf





      

   

          

    



                    

     

       

             

 

   

 

 

 

   

 

           



 



  

Eng.Mohammed Ashraf





                    

    

    



 

       

    



  

Eng.Mohammed Ashraf





                  

 

                     

     



   





     

Eng.Mohammed Ashraf





    

     

            

    

    



                





                               

  

 

  

  

  







            

 

Eng.Mohammed Ashraf





      

   

  

        

         





            

 

 

   

     

 

   

   



 



  

       

                  

   

 

                  







 

Eng.Mohammed Ashraf





 

  

       

  

  

Eng.Mohammed Ashraf





    

     

 

 

  

  







     

  



  



    



         

 

   

  

     

     

    

          

     

    

  

Eng.Mohammed Ashraf





     





 

                



 

 

   

     

 



                

   

 

        





  



   



 



   

  

   

  

       

  

Eng.Mohammed Ashraf





         



 

 

  

  





     

  



  



  



         



 

  

        

     

    

          

     

         

  

Eng.Mohammed Ashraf





                        

        





  

  

        





      





   

  

  

Eng.Mohammed Ashraf





    

 

   

       

        



 



 

 

  

  

Eng.Mohammed Ashraf





         

       

 

 

   

 

           



    

 



    





 

  

Eng.Mohammed Ashraf





        







    





      



    



     





    



 

       



 

      





 

 

  

 

   

  

  

Eng.Mohammed Ashraf







  

       

    

     

 

 

  

                     



    





          

 

         

  



   



  

Eng.Mohammed Ashraf







    



  



          



 

 







 





  

         

     

  

          

     



  

  

Eng.Mohammed Ashraf





       

             

    





  

  

      







    



 

 





  

  

Eng.Mohammed Ashraf





    

 

   

       

  

 

       

 



   

  

  

Eng.Mohammed Ashraf





     



       

 

 

   

 

         





    





  



 



 

  

Eng.Mohammed Ashraf







        







    





     





    



 

      





       



 

      



    

 

 

  

 

   

  

  

Eng.Mohammed Ashraf







  

       



    

     

 

 

  

                   



    





          

 

         

  



   



  

Eng.Mohammed Ashraf







    



  



          



 









 





  

       

    

   

       

    

  

  

Eng.Mohammed Ashraf





                           

     





  

  

       







       



     

        

Eng.Mohammed Ashraf





    

 

   

       

        



 



   

  

  

Eng.Mohammed Ashraf





     



       

 

 

   

             

       



  



  



 

  

Eng.Mohammed Ashraf





        







    





      



    



     





    



 

       



 

      





 

 

  

 

 

 

 



  

Eng.Mohammed Ashraf







  

        



 

     

 



     

 



 

   

   

                     



     



          

 

         

  

   

Eng.Mohammed Ashraf







  





         





     

 

     

 

       

  

    

                     

 

                



   





         

   

 

       



              

  

Eng.Mohammed Ashraf







   



  



  

       

    

   

    

  

    

  

    

   

     

   

 

  

Eng.Mohammed Ashraf





                               

      

    

  

Eng.Mohammed Ashraf











           

         

    

 

     

 



    





  

        

 

   

   

    

  

Eng.Mohammed Ashraf



    





 



        





 

 

 

 



  

   

  

Eng.Mohammed Ashraf





         



 

 

  

         



 

    

 



  



             

    

 

  

         

 

        

  

Eng.Mohammed Ashraf









  



  





             

    



     

     

     





  



       

 

      

    

 

  

Eng.Mohammed Ashraf





      

     

         

     

     

     

  



     

 



   



  

  

  

Eng.Mohammed Ashraf





     



       

 

 

  





 





  

     



         





 



 

  

Eng.Mohammed Ashraf







        







    





     



    



     





    



 

      



 

 

     

 

 

  

 

         

 





     





 

   

               

     

         

     

     

  

Eng.Mohammed Ashraf





 

    

      



  

   

      



 

       



        

 

  

  

Eng.Mohammed Ashraf





                               

      

    

  

Eng.Mohammed Ashraf









            



         

    

 

     





    





 

        

 

    

   

    

  

Eng.Mohammed Ashraf





    



 



      





 

 

 

 

 



  

   

  

Eng.Mohammed Ashraf





         



 

 

  

         



 

    

 



  



             

    

   

Eng.Mohammed Ashraf









 



          

 

        

 











                





   

     

  

Eng.Mohammed Ashraf







 

   

  

   

    

       

   

            

    

    

 

       

    

    

  

Eng.Mohammed Ashraf





     

  



        

    



  

  

  

Eng.Mohammed Ashraf





     

  

     

 

 

   



 





  

     



    





  



 



 

  

Eng.Mohammed Ashraf







        







    





     





    



 

     



    



      



 

       

 

 

  

 

 

       

 





     





 

   

              

     

         

     

     

  

Eng.Mohammed Ashraf





 

    



        



   



       



      



        

 

  

  

Eng.Mohammed Ashraf





                               

      

    

  

Eng.Mohammed Ashraf









            



         

        

 

 





    





   

      

 

   

 

 

 

  

  

Eng.Mohammed Ashraf





  



                  





 

 

 

 

 



  

  

  

Eng.Mohammed Ashraf





         



 

 

 



         



 

    

 



  



             



  

       

          

   

         

  

Eng.Mohammed Ashraf











  



  



                

 

     

 

 

  

 

 

     

     

 



       

  

       

    

    

       

    

    

     

Eng.Mohammed Ashraf





     

  

 

       

   



  

  

  

Eng.Mohammed Ashraf





     



       

 

 

   



 



        



    



  







 



 

  

Eng.Mohammed Ashraf







        







    



        





    



 

         



    



      



 

 



     

 

 

  

 

 

        



      



 

 

  

        

  

Eng.Mohammed Ashraf





      

     

         

     

     

 

    



       



   



       



      



        

 

  

  

Eng.Mohammed Ashraf





                               

  



     

  

Eng.Mohammed Ashraf











           

         

        

 

 





    





    

      

 

    

      

  

  

Eng.Mohammed Ashraf



    

  





  

 

       





 

 

 

 

 



  

    

  

Eng.Mohammed Ashraf





         

 

 

 



   



            



     



    



 

           

      



  

 

        

 

         



 

Eng.Mohammed Ashraf





  



  



 



 

          

   



    

 



    

  

Eng.Mohammed Ashraf





   

     

 

   





      

                   

          

    

          

 

  

            

  

Eng.Mohammed Ashraf





   

 



          

 

        

   

      





      

 

   

 



 

             

    

       

    

    

    

  

Eng.Mohammed Ashraf





     

  

 

          

   



  

  

  

Eng.Mohammed Ashraf





      

       



 

 

   

  

 



        



     



  



  



 

  

Eng.Mohammed Ashraf

 

Eng.Mohammed Ashraf





     

    

       

    

    

 

    

      







   

      



 

      

        



 

      

   

     

       

       

     

       

       



   



  

Eng.Mohammed Ashraf





                          

 

    

   









  

 

  

Eng.Mohammed Ashraf





          





  

 

    

    

 

      

 



      



   

 

       



    

  

  

Eng.Mohammed Ashraf





   

 

 

       

  

    

  

Eng.Mohammed Ashraf





   

     

 

 

 

 





  

   



    

               



  



     



 





         

  

 

         



  

   

Eng.Mohammed Ashraf







               





 

  

  

      

 



    

      



 

       

     

  

  

Eng.Mohammed Ashraf





              

     

         

     

     

   

  

Eng.Mohammed Ashraf





                            



    

     

 



  

  

Eng.Mohammed Ashraf





           

    



                     

 

       

  

    



                

 

       



      

  

  

Eng.Mohammed Ashraf





   

  

 

      

  

    

  

Eng.Mohammed Ashraf





         

 

 

 

 







 

       





                           



  

          

  

 

         



   

   

Eng.Mohammed Ashraf







           







  

 

 

  

  

Eng.Mohammed Ashraf





        



 

 

  

 

      



 

         





  



        

         

 

        

  

Eng.Mohammed Ashraf







   

      



             



      





 

 

  

               

      

       

    



       

    

    

   

  

Eng.Mohammed Ashraf





                            

 

    

  

Eng.Mohammed Ashraf















        



  

  

 

        

 

  

Eng.Mohammed Ashraf





  





       

        

    

        

                                         

            

       

  

            





  

Eng.Mohammed Ashraf





       





     





 

             

   

    

   

   

  

 

       

 

 

    

  

Eng.Mohammed Ashraf





         

 

 

 

 



   

      

     



     



     

        

 

           

                  

           

  

Eng.Mohammed Ashraf





   

    







          

 

        

 



  



         



      





                             

 

 

 

      

 



         

 

 

      

Eng.Mohammed Ashraf





                        

     

         

     

     

           

  



      



  

  



 



  

Eng.Mohammed Ashraf





         

       



 

 

 

    



      



 

              





       







       

  

 

       



  

 



  

 

 



    

  

Eng.Mohammed Ashraf





   

  



    





                         

 

     







    



      



    



      

     

 

 

 

   

  

 

   

  

Eng.Mohammed Ashraf





      

                



     

 

 

 

 



           













 

















 

 



    

                      

  

Eng.Mohammed Ashraf





      

     

         

     

     

 

     



      



      

 

    



      



          

  

Eng.Mohammed Ashraf





                       

 

     





                          

 

              



 

     

Eng.Mohammed Ashraf





                                      

        

                    



            

     



             

         

  





  

Eng.Mohammed Ashraf





       





       





 

             

    

      

   

     

    



        

     

Eng.Mohammed Ashraf





         

 

 

 

 



    

      



           



    

        

 

            

                  

               

  

Eng.Mohammed Ashraf





      

 







          

 

         

  



 



            



      

 

 

 

 

               



 

 

 

  

  

Eng.Mohammed Ashraf





    



   

 

 

  



    

              





  

 

 



  

 

        



 

         

  

Eng.Mohammed Ashraf





        





         

       

 

        



                



      

   

 

 

  

               

  

         

   

 

  

Eng.Mohammed Ashraf





     

    

       

    

    

     



 

   





  

  

         

   

   

  

Eng.Mohammed Ashraf





      

           

 

 

 

    

   



 



 

  



          



  



     





       

  



 

       



   

 



  

 

 



      

  

Eng.Mohammed Ashraf





        







    







       



    



     



    



      

    



  

     





 

 

 

    

 

 



  

  

Eng.Mohammed Ashraf





            



     

   















 











     



 

 

 

 

                      



                        



   

   

  

Eng.Mohammed Ashraf





      

     

         

     

     

 

     



           



    



       



        



        

 

 

  

Eng.Mohammed Ashraf





                               

 

  

Eng.Mohammed Ashraf



          







   

 

 





    

  

    

      

  

  



  

Eng.Mohammed Ashraf







                     



 

 





      





     

                                                                                                                                                                

Eng.Mohammed Ashraf







               

                

  

 

    

                                                





     

    

  

  

Eng.Mohammed Ashraf





   

   

   

   



 

      

  

Eng.Mohammed Ashraf





   

     

 

 

 

 



   

    

     

   



  



   



          

        

 

           

 

           

     

Eng.Mohammed Ashraf





        







          

 

        

  



   



         



       



 

 

 

   

               

 

 

  

   

    





 



  

      

 

    

         

  

  

Eng.Mohammed Ashraf



 



  





        



           

      



 

  

Eng.Mohammed Ashraf





   

  



 



   



 



      

 



  



  

  





















   

  

     

 

  

Eng.Mohammed Ashraf





                             

 

  

Eng.Mohammed Ashraf











         



        

 

               



           

 

 



   

 

        





  

Eng.Mohammed Ashraf





                       



 

                                                                                                                                                                                         

  

Eng.Mohammed Ashraf





                                      

  

  

 

                      

          

             





     

      

  

  

Eng.Mohammed Ashraf





   

     





 

         

  

Eng.Mohammed Ashraf





   

     

 

 

 

 



   

   

   

   



  



   



                             

        

 

           

       

Eng.Mohammed Ashraf





       









          

 

        

  



   



            



       



 

 

 

 

  

                 

 

   

  

  

Eng.Mohammed Ashraf





    



   

 

 

  



              



  

            



  

Eng.Mohammed Ashraf





          

        



         

 

         

  

       

         

     



 

 

 



 

      



                 



                 



 



  

Eng.Mohammed Ashraf





     

    

       

    

    

  

  

Eng.Mohammed Ashraf





                                     

 

  

Eng.Mohammed Ashraf









           

   



               

 

  

  



    



   

 

  

Eng.Mohammed Ashraf





       

  



  

 





 

        

 

    



 

 

    

   



  

 

 

 

 



 

 

     

  

Eng.Mohammed Ashraf





 

     

 

 

 

 



 

                     



       









             







          

 

        

 



 



         



       



 

 

 

 

                 

  

Eng.Mohammed Ashraf





   

         



  



 



    

               



   

  

Eng.Mohammed Ashraf





   





                                                

 

                     

   



   



   



     



 

 

       



    



  

Eng.Mohammed Ashraf





    

   

  

 

   

Eng.Mohammed Ashraf





       



  



        



 

 

      

  

Eng.Mohammed Ashraf





            



         





  



         

      

          

           

 



         

            

             

   

    

  

Eng.Mohammed Ashraf





                                       

  

Eng.Mohammed Ashraf





   

     

 

 

 

 





             



  



  



  



     



 





         



 



 



  







 

           

                







      

Eng.Mohammed Ashraf





 



         

       

      

   

  

  







 









                 

  

Eng.Mohammed Ashraf





       



       

 



   





  

 

  

  

Eng.Mohammed Ashraf





          

     

 

 

 

    

 

   

 

               





      

 





        



 

          

  



 



   



        









    



   

      



    





        

    



 

Eng.Mohammed Ashraf





 



     

 

 

 

   

 

 

    

 

  

Eng.Mohammed Ashraf





       

          





  

            

    

   

Eng.Mohammed Ashraf





  



  

Eng.Mohammed Ashraf





      

     

 

 

 

    

 

 

             





     







       

 

         

  



 



   

     







    



      

  





    





    



    



  

                                                      

  

Eng.Mohammed Ashraf





      





 

 

 

   

 

 



  

   



  

Eng.Mohammed Ashraf





      

         



     



     



 

 

 

 

 

                  

  

       

    

       

 

        

  

  



 







  

     

       





     

      





        

           



  

Eng.Mohammed Ashraf





        



  

Eng.Mohammed Ashraf





                                    

     

  

Eng.Mohammed Ashraf



             





     

                       

 

    

                      



  

 

  

Eng.Mohammed Ashraf





       





    

 



  

          

 

    

 

 

 

   

   

 

 

   

    

 

 

   

 

  

Eng.Mohammed Ashraf





 

     

 

 

 

 

 

 

                        

       



    



 



   







          

 

         

  



   



  

Eng.Mohammed Ashraf







                 





 

 

 

 

                 

 

 

  

  

Eng.Mohammed Ashraf





   

     

 

 

  

 

       

       

                



                 



           





  



   

 

         

 

       

  

Eng.Mohammed Ashraf







   

      

      

 

 

  

   

 

       



      

  

   

  

  

















  

Eng.Mohammed Ashraf





            

  

 

   

     



   

   

   

  

  

Eng.Mohammed Ashraf





    



             

          

 

 

 

    

 

 





  

           





       

 

       





 

        

  

Eng.Mohammed Ashraf









     





 

        







    



      



    



        









     

      



    

   

 

 

 

   

 



 

 

 



 



      



  

Eng.Mohammed Ashraf





     

   



     



             

 

 

 

  



  

 



  



              





  



 

                 

 

      

  

Eng.Mohammed Ashraf









      





 

        







    





     



    



       



    



   

        

 



     

 

 

 

    

 

  

 



 

    



  

    

Eng.Mohammed Ashraf





                                       



      

 

 

  



    

         



 

                



   



            

 



  

      

  

 

  

 

 



  

Eng.Mohammed Ashraf





     

       

        

  

  



 





 

     



       



    



       



         



        

 

       

  

Eng.Mohammed Ashraf









  

Eng.Mohammed Ashraf

 

Eng.Mohammed Ashraf

38800 Country Club Drive Farmington Hills, MI 48331 USA +1.248.848.3700 www.concrete.org

 

Eng.Mohammed Ashraf