ACI Strut and Tie Model Examples-1

rn8 3 U rn Examples for the Design of Structural Concrete with Karl-Heinz Reineck ., Examples for the Design of Structural Concrete with Strut-and-Tie Models Prepared by members of ACI Subcommittee445-1, Strut and Tie Models, for sessions at the Fall Convention in Phoenix, October 27 to November 1,2002,and sponsored by Joint ACI-ASCE Committee 445, Shear and Torsion, and ACI Committee 318-E, Shear and Torsion Editor Karl-Heinz Reineck international SP-208 may exhibit structural damage or even brittle failure. Claudia M. shear walls with openings. nor intended to. This is the case in discontinuity regions (D-regions). are appreciated. former Chair of ACI Subcommittee 31 8E. In addition. including the making of copies by any photo process. supplant individual training. University of Stuttgart. The editor also acknowledges the efforts of several individuals at ILEK. and Jim Wight. Ramirez. University of Stuttgart. which if not properly detailed. Alcocer. and pile caps. or recording for sound or visual reproduction or for use in any knowledge or retrieval system or device. or by any electronic or mechanical device. Julio A. Sritharan. Reineck (chair). Barnes. Sri S. or judgment of the user. unless permission in writing is obtained from the copyright proprietors. Gary J. Daniel Kuchma. The papers of this special publication were reviewed in accordance with American Concrete Institute policies. bridge pier tables. Copyright 02002 AMERICAN CONCRETE INSTITUTE P. David Sanders. Hussein Coskun. Jay Holombo.4 and of this special publication. of the information presented. Michigan 48333-9094 All rights reserved. printed or written or oral. Karl H. the efforts of the reviewers and of the ACI Headquarters staff. The Institute is not responsible for the statements or opinions expressed in its publications. Schmick ACI Subcommittee 445-A Shear and Torsion: Strut and Tie Models Library of Congress catalog card number: 20021 12515 ISBN: 0-8703 1-086-0 Members: Sergio M. for their continuous support of the work of ACI Subcommittee 445. anchorage zones of prestressed members. Ali Daghigni. Closing date for submission of discussion is June 2003.DISCUSSION of individual papers in this symposium may be submitted in accordance with general requirements of the ACI Publication Policy to ACI headquarters at the address given below. Manager of Technical Documents. Germany Editorial production: Bonnie L. beams with indirect supports. responsibility. Todd Watson.2002. including Elfriede Schnee. The papers in this volume have been reviewed under Institute publication procedures by individuals expert in the subject areas of the papers. The contributions and the examples were prepared by members of ACI Subcommittee 445-A Shear and Torsion: Strut-and-Tie Models and were presented at two sessions at the ACI Fall Convention in Phoenix. Mario Rodriguez. co-sponsored by the technical committee ACI-ASCE Committee 445: Shear and Torsion and ACISubcommittee 318E: Shear and Torsion. and Angela Siller for their assistance in producing the document. Fernando Yanez . The editor finally wishes to thank Cathy French. Robert B. Anderson. All discussion approved by the Technical Activities Committee along with closing remarks by the authors will be published in the September/October 2003 issue of either ACI Structural Journal or ACI Materials Journal depending on the subject emphasis of the individual paper. Karl-Heinz Reineck Editor and Chair of ACI Subcommittee 445-A Shear and Torsion: Strut and Tie Models ILEK. including rights of reproduction and use in any form or by any means. deep beams with and without openings. Institute publications are not able to. Uribe. or the supplier. Chair of ACI-ASCE Committee 445. Printed in the United States of America Preface The new Appendix A of ACI 318-2002 on strut-and-tie models provides an excellent tool for the design of structural concrete where sectional design procedures for flexure and shear do not apply. The cooperation of the authors in the preparation of the manuscripts and their revision considering comments by reviewers is greatly appreciated.O. Denis Mitchell. dapped beam ends. especially Mr. Robert W. Klein. This special publication presents examples for the use of strut-and-tie models following Appendix A of ACI 3 18-2002 for designing D-regions such as corbels. October 27 November 1. Box 9094 Farrnington Hills. Heinz Reineck Part 2: Derivation of strut-and-tie models for the 2002 ACI Code James G. Lawrence C. Claudia M. Matamoros. Uribe and Sergio Alcocer Example lb: Alternative design for the non-slender beam (deep beam) Tjen N.Tjhin and Daniel A. Breen. MacGregor Part 3: Experimental verification of strut-and-tie models Denis Mitchell. Khaled A. Neil Wexler v Examples for the design of structural concrete with strut-and-tie models List of contents Corresponding members: John E. Adolfo B. Mary Beth D. James G. James K. and Sergio Alcocer Part 4: Examples Example 1a: Deep beam design in accordance with ACI 3 18-2002 Claudia M. MacGregor. James Lefter. Kuchma Example 3. Laura Lowes. H. Tjhin and Daniel A. Tjhin and Daniel A. James 0.List of contents Associate members: Dat Duthinh.. Jirsa. Nahlawi. Novak. Kuchma Example 2: Dapped-end T-beam supported by an inverted T-beam David Sanders Example 3. Cook. Kuchma 1 . William D. Uribe.1 : Corbel at column Tjen N. Wight Preface iii Part 1: Introduction Karl . Philip K.2: Double corbel Tjen N. Hueste. Tan. degrees from the University of Stuttgart.vi List of contents Example 4: Deep beam with opening Lawrence C. Barnes Example 9: Pile cap Gary J.1 "Practical Design". the design with strut-and-tie models and detailing of structural concrete and the design of highperformance concrete hot-water tanks. University of Stuttgart. Ramirez Introduction Example 7: Strut-and-tie model cable stayed bridge pier table Bob Anderson Karl . . and he is head of two research groups and managing director of the institute.-Ing. Novak and Heiko Sprenger Example 5: Beam with indirect support and loading Part 1 Wiryanto Dewobroto and Karl-Heinz Reineck Example 6: Prestressed beam Adolfo Matamoros and Julio A.-Ing. His research covers the shear design of structural concrete.Heinz Reineck Example 8: High wall with two openings Robert W. He is member of ASCE-ACI Committee 445 "Shear and Torsion" and chairs two subcommittees and also is a member of Jib Task Group 1. and Dr. Klein Part 5: Modeling structural concrete with strut-and-tie models summarizing discussion of the examples as per Appendix A of ACI 318-2002 Karl-Heinz Reineck Karl-Heinz Reineck received his Dip1. He is involved in both research and teaching at the Institute for Lightweight Structures Conceptual and Structural Design (ILEK). The development of strut-and-tie models has brought the unique chance toward gaining consistency in the design concept covering D-regions and B-regions with similar models. b)]. 1983) and Nielsen (1978. The Appendix A of ACI 3 18-2002 consequently reflects this international development in research and thus is in line with some codes like the CEB-FIP Model Codes 1990. This is especially true for discontinuity regions (D-regions). however. most codes still keep to the traditional concepts and only added a new chapter or appendix without integrating the new concept throughout the code. 1984) and their co-workers. Furthermore. EC 2.1922). as demonstrated by the use of truss models for the shear design e. Shortly thereafter the American Concrete Institute renamed the ACI 3 18 code accordingly. it is a major achievement for the engineers in practice and should trigger efforts to apply strut-and-tie models in daily practice. These recommendations are fully based on strut-and-tie models and show the direction for future developments. and separate checks are carried out for the different actions. The scope and aim of the Appendix A is described and extensive explanations are given in addition to those already presented in the Commentary of Appendix A. has been increasingly realized.g. which justify the use of strut-and-tie models for the design of structural cdncrete.g. One exception is the case of the shear design where a truss model has been used for steel contribution for many years. as well as the recent FIP Recommendations (1999) and the new German code DIN 1045-1 (2001-07). and one of its Working Groups developed the FIP Recommendations "Practical design of structural concrete" published in 1999 by fib. . are not dimensioned but left to be covered by detailing rules. (1987. The demand for developing clear models. Strut-and-tie models are a valuable design tool used since the beginning of reinforced concrete design. the application of strut-and-tie models emphasizes the essential role of detailing in design.2001). as pointed out at the IABSE Colloquium "Structural Concrete" April 1991 in Stuttgart [IABSE (1991 a. like strut-and-tie models. frame comers or corbels). This also formed the basis for strut-and-tie models after the works of Schlaich et al. 1912. even though improper design and detailing of D-regions led to structural damages including some failures [Breen (1991). The danger of a sectional design approach is that the overall flow of forces may be overlooked and that critical regions are not covered. which have been poorly addressed in codes. 1953) and others. was expressed by Schlaich (1 991) and Breen (1 991).Part 1: Introduction Part 1: Introduction 1 Historical remarks The last fifteen years has brought a major breakthrough in design methods for concrete structures which reflected in the terminology used. Podolny (1985)l. corbels and dapped beam ends. In addition. the dimensioning procedures and the checking procedures focus on sections. and they address the whole structure and not only sections when defining the requirements and principles for the design. the main objective of this Special Publication is to show with design examples the application of strut-and-tie models according to the Appendix A of ACI 3 18-2002. Part 3 presents a summary of important tests. Especially the regions with discontinuities due to the loading orland the geometry. After the introduction (Part I). Part 2 gives an insight into the development of Appendix A of ACI 3 18-2002 and of the discussions in ACI Committee 3 18 E "Shear and Torsion". apart from some special cases (e. Therefore. Rausch (1938. This Special Publication contains five parts. These differentiations were identified as artificial. Canadian Code and AASHTO. All these considerations triggered discussions at the IABSE Colloquium "Structural Concrete" in 1991 and to the conclusions published thereafter [IABSE (1991 a. leading to confusion in codes and in teaching as well as to unnecessary restrictions in practice. chaired by Julio Appleton. The limitations of purely empirical design procedures. prestressed concrete and partially prestressed concrete and even externally prestressed concrete or unreinforced concrete. Contrary to the principles. the D-regions. Even more. Morsch (1 909. the detailing rules given finally in codes are meant to ensure the overall safety of the structure. The term "structural concrete" was proposed as a unifying term for all kinds of applications of concrete and steel in order to overcome the traditional divisions between reinforced concrete. Among the tests are the classical examples for D-regions like deep beams. All this was pointed out in the State-of-the-Art Report on shear by the ASCE-ACI Committee 445 (1 998). b)]. such as moments and shear forces. by Ritter (1 899). The theory of plasticity was applied to the design of members under shear and torsion. Part 1: Introduction 2 Dimensioning procedures according to the present codes The principles for the design are clearly defined in most codes. 3 Aim and contents of this Special Publication The implementation of strut-and-tie models in ACI 3 18-2002 with Appendix A is an important step in direction towards a more consistent design concept. driving the demand for the development of clear design models. especially by Thurlimann (1975. However. Most of these ideas were taken up by the FIP Commission 3 "Practical Design". Part 1: Introduction Part 4 forms the major part of this Special Publication'presenting nine different examples designed with strut-and-tie models using Appendix A of ACI 3 18-2002. Most of these examples were taken from practice: - The Example 1 (deep beam), Example 2 (dapped beam end) and Example 3 (double corbel and corbel at column) are classical D-regions, which have been designed with strut-and-tie models since long, and for which even tests were carried out, as described in Part 3. - The Example 5 (beam with indirect supports) and Example 6 (prestessed beam) deal with well known D-regions of beams, which so far have mostly been dealt with in codes by rules for the shear design. - The Examples 7 (pier table) and Example 9 (pile cap) deal with D-regions of 3D-structures, for the design which most codes give only rare information. Some examples were selected to demonstrate the potential of strut-and-tie models to solve uncommon design problems, such as like Example 4 (deep beam with opening) and Example 8 (high wall with two openings). All examples show the approach of finding a model, which is the first and an important step in a design with strut-and-tie models. The examples also point out where problems in dimensioning or in detailing and anchoring the reinforcement occur and how the design could be improved. Part 5 gives a summary and discusses some issues which are either common for all examples or turned up in several examples. After a brief review of the procedures for finding a model, the uniqueness of a model is discussed and why different models may be selected by several engineers. The other issue addresses the transition between D- and B-regions of beams and is of general importance for many examples, because many D-regions are part of a larger structure and have to be "cut out" of it, i.e. the correct actions and forces have to be applied at the border of the D-region. Finally in Part 5 the importance of detailing is pointed out as it was demonstrated in several examples. Part 1: Introduction References AASHTO (1994): AASHTO LRFD Bridge design specifications, section 5 Concrete Structures. American Association of State Highway and Transportation Officials, Washington, D.C. 2001, 1994 ACI 3 18 (2002): Building Code Requirements for Structural Concrete and Commentary. American Concrete Institute, Farmington Hills ASCE-ACI 445 (1998): Recent approaches to shear design of structural concrete. Stateof-the-Art-Report by ASCE-ACI Committee 445 on Shear and Torsion. ASCE-Journ. of Struct. Eng. 124 (1998), No.12, 1375-1417 Breen, J.E. (1991):Why Structural Concrete? p. 15-26 in: IABSE Colloquium Structural Concrete, Stuttgart April 1991. IABSE Rep. V.62, 1991 CEB-FIP MC 90 (1993): Design of concrete structures. CEB-FIP-Model-Code 1990. Thomas Telford, 1993 CSA (1994): Design of Concrete Structures - Structures (Design). Canadian Standards Association (CAN3-A23.3-M84), 178 Rexdale Boulevard, Rexdale (Toronto), Ontario, December 1994 DIN 1045-1 (2001): Deutsche Norm: Tragwerke aus Beton, Stahlbeton und Spannbeton Teil 1: Bemessung und Konstruktion. S. 1 - 148. (Concrete, reinforced and prestressed concrete strutures - Part 1: Design). Normenausschuss Bauwesen (NABau) im DIN Deutsches Institut fur Normung e.V. Beuth Verl. Berlin, Juli (2001) EC 2 (1992): Eurocode 2: design of concrete structures - Part 1: General rules and rules for buildings. DD ENV 1992-1-1. BSI 1992 FIP Recommendations (1999): "Practical Design of Structural Concrete". FIP-Commission 3 "Practical Design", Sept. 1996.Publ.: SETO, London, Sept. 1999. (Distributed by:fib, Lausanne) IABSE (1991 a): IABSE-Colloquium Stuttgart 1991: Structural Concrete. IABSE-Report V.62 (1991 a), 1-872, Ziirich 1991 IABSE (1991 b): LABSE-Colloquium Stuttgart 1991: Structural Concrete - Summarizing statement. publ. in: - Structural Engineering International V.l (1991), No.3, 52-54 - Concrete International 13 (1991), No.10, Oct., 74-77 - PCI-Journal36 (1991), Nov.-Dec., 60-63 and: IVBH-Kolloquium "Konstruktionsbeton" - SchluBbericht. publ. in: - BuStb 86 (1991), H.9,228-230 - Bautechnik 68 (1991), H.9,3 18-320 - Schweizer Ingenieur und Architekt Nr.36,5. Sept. 1991 - Ze~nentund Beton 1991, H.4, 25-28 Morsch, E. (1909):Concrete Steel Construction. McGraw-Hill, New York, (1909), pp. 368 (English translation of "Der Eisenbetonbau", 1902) Morsch, E. (1912): Der Eisenbetonbau. 4. Aufl., K. Wittwer, Stuttgart, 1912 Part 2: Derivation of strut-and-tie models for the 2002 ACI Code Part 1: Introduction Morsch, E. (1922):Der Eisenbetonbau - Seine Theorie und Anwendung (Reinforced concrete Construction - Theory and Application). 5th Edition, Vol.1, Part 2, K. Wittwer, Stuttgart, 1922 Part 2 Nielsen, M.P.; Braestrup, M.W.; Jensen, B.C.; Bach, F. (1978): Concrete Plasticity: Beam Shear-Shear in Joints-Punching Shear. Special Publ. Danish Society for Structural Science and Engineering, Dec. 1978, 1-129 Derivation of strut-and-tie models for the 2002 ACI Code Nielsen, M.P. (1984): Limit State Analysis and Concrete Plasticity. Prentice-Hall, Englewood Cliffs, New Jersey, 1984, pp. 420 Podolny, W. (1985): The cause of cracking in past-tensioned concrete box girder bridges and retrofit procedures. PC1 Journal V.30 (1985), No.2, March-April, 82-139 Rausch, E. (1938): Berechnung des Eisenbetons gegen Verdrehung (Torsion) und Abscheren. 2. Aufl.. Springer Verlag, Berlin, 1938. pp. 92 James G. MacGregor Rausch, E. (1953): Drillung (Torsion), Schub und Scheren im Stahlbetonbau. 3. Aufl. Deutscher Ingenieur-Verlag, Dusseldorf, 1953 pp. 168 Synopsis Ritter, W. (1899): Die Bauweise Hennebique. Schweizerische Bauzeitung, Bd. XXXIII, Nr. 7., Jan., 1899 Schlaich, J.; Schafer, K; Jennewein, M. (1987): Toward a consistent design for structural concrete. PCI-Journ. V.32 (1987), No.3, 75-150 Schlaich, J.; Schafer, K. (2001): Konstruieren im Stahlbetonbau (Detailing of reinforced concrete). Betonkalender 90 (2001), Teil 11, 31 1 - 492. Emst & Sohn Verlag, Berlin 200 1 Schlaich, J. (1991): The need for consistent and translucent models. p. 169-184 in: IABSE Colloquium Structural Concrete, Stuttgart April 1991. IABSE Rep. V.62, 1991 - " Torsion., Bietzi~neund Schub in StahlThurlimann, B.; Grob, J.; Liichinger, P. (1975): ~ . , betontrtigern. ~ortbildun~skurs fiir Bauingenicure, Institut fiir Baustatik und Konstruktion, ETH Ziirich. April 1975. Thurlimann, B.; Marti, P.; Pralong, J.; Ritz, P.; Zimmerli, B. (1983): Anwendung der Plastizitatstheorie auf Stahlbeton. Fortbildungskurs fiir Bauingenieure, Institut fiir Baustatik und Konstruktion, ETH Zurich. April 1983 . This paper documents the decisions made by ACI Committee 3 18 to introduce strutand-tie models into the 2002 ACI Code. Sections 3 and 4 of this paper review code statements concerning the layout of strut-and-tie models for design. The format and values of the effective compression strength of struts are presented in Sec. 5. The first step was to derive an effective compression strength which gave the same crosssectional area and strength using Appendix A as required by another code for the same concrete strength and same unfactored loads. The final selection of design values of the effective compression strength considered test results, design values from the literature, values from other codes, and ACI Code design strengths for similar stress situations. A similar derivation of the effective compression strengths of nodal zones is summarized in Sec. 6 of the paper. The description of the geometry of nodal zones in code language proved difficult. The design of ties is described in Sec. 7 of this paper and requirements for nominal reinforcement are in Sec. 8. Nominal reinforcement is provided to add ductility, to improve the possibility of redistribution of internal forces, and to control cracks at service loads. James G. MacGregor received a B.Sc. in Civil Engineering from the University of Alberta, Canada in 1956, and a Ph.D. from the University of Illinois in 1960. He joined the Department of Civil Engineering at the University of Alberta in 1960, employed there until 1993. Dr. MacGregor serves on ACI technical committees on shear and torsion, columns, and the ACI Code committee. He chaired Canadian code committees on reinforced concrete design, and structural design. In 1992-93 he was president of ACI. Dr. MacGregor is an Honorary Member of ACI, a Fellow of the Royal Society of Canada, and a Fellow of the Canadian Academy of Engineering. In 1998 he received an honorary Doctor of Engineering degree from Lakehead University, and in 1999, an honorary Doctor of Science degree from the University of Alberta. Part 2: Derivation of strut-and-tie models for the 2002 ACI Code Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 1 Introduction The 2002 ACI Code (2002) includes a new Appendix A, "Strut-and-Tie Models" and changes to a number of code sections to allow the use of strut-and-tie models (STM) in design. In developing Appendix A, concepts were drawn from the AASHTO LRFD Specification (1998), the CEBJFIP Model Code (1993) as interpreted in the FIP Recommendations (1999), and the Canadian concrete design code, CSA A23.3-94 (1994). Research reports [ACI Committee 445 (1997)] also provided some of the basis for the appendix This paper, in combination with the ACI 3 18 Commentary for Appendix A [ACI (2002)l explains the decisions and assumptions made in the development of Appendix A. strut 2 Research significance This paper documents major decisions made in the development of Appendix A, "Strut-and-Tie Models" in the 2002 ACI Code. Fig. 1: A strut-and-tie model for a deep beam. 3 What are strut-and-tie models? 3.1 B-Regions and D-Regions 1 Concrete structures can be divided into beam-like regions where the assumption of the straight line strain distribution of flexure theory applies, and disturbed regions, adjacent to abrupt changes in loading at concentrated loads and reactions; or adjacent to abrupt changes in geometry such as holes or changes in cross section. In the latter regions the strain distributions are not linear. These different portions are referred to B-regions and D-regions, respectively. I The traditional theory of flexure for reinforced concrete, and the traditional (V, + V,) design approach for shear apply in B-regions. In D-regions, on the other hand, a major portion of the load is transferred directly to the supports by in-plane compressive forces in the concrete and tensile forces in reinforcement and a different design approach is needed. D-regions may be modeled using hypothetical trusses consisting of concrete struts stressed in compression, steel ties stressed in tension, joined together at joints referred to as nodes. These trusses are referred to as strut-and-tie models (STM1s).Thestrut and tie model of a single span deep beam shown in Fig. 1 is composed of two inclined struts, and a horizontal tie joined together at three nodes [ACI 3 18 (2002)l. The nodes are enclosed within nodal zones which transfer forces from the struts to the ties and reactions. Strut-and-tie models are assumed to fail due to yielding of the ties, crushing of the struts, failure of the nodal zones connecting the struts and ties, or anchorage failure of the ties. The struts and the nodal zones are assumed to reach their capacities when the compressive stresses acting on the ends of the struts or on the faces of the nodal zones, reach the appropriate efective compressive strength, f,,. d d .5".5 in Fig. Venant's principle and elastic stress analyses suggest that the localized effect of a concentrated load or a geometric discontinuity will die out about one member depth away from the load or discontinuity. D-region behavior controlled the strengths of beams with a/d ratios less than about 2. and anchorage of the ties.5. B-region behavior controlled the strengths of beams with d d greater than 2. For this reason.5 as shown by the steeply sloping line to the left of d d = 2. term in the traditional ACI shear strength equation is not included. rounded off to 25' (see ACI Sec. (b) what effective concrete strengths and 4 factors should be used. 2.) Figure 2.1.2 Decisions necessary to develop design rules for strut-and-tie models To codify strut-and-tie models for design. The words "about" and "approximately" are emphasized here because the extent of D-regions can vary from case to case (see ACI Sec. ACI Committee 3 18 limited the maximum lengths of isolated D-regions to d.). (e) the layout. 0 05 0 63 slrul and Ile model 0 0 secllonal1 1 model I 1 Fig. This establishes the smallest angle between a strut and a tie attached to one end of the strut as arctan (d / 2d) = 26. Three-dimensional strut-and-tie models are used for structures such as pile caps for two or more rows of piles. 2: Strengths of concrete beams failing in shear for various aid ratios.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code I Part 2: Derivation of strut-and-tie models for the 2002 ACI Code De St. Two-dimensional strut-and-tie models are used to represent planar structures such as deep beams. However. compares the experimental shear strengths of simply supported beams with various shear-span-to-depth ratios. If two D-regions each of length d or less. the major items to be defined and specified are: (a) the geometric layout of strut-and-tie models. reproduced from "Prestressed Concrete Structures" [Collins and Mitchell (1 991)]. from 1 to 7. (c) the shape and strength of the struts. D-regions are assumed to extend approximately one member depth from the load or discontinuity.5. A. they are considered in Appendix A to act as a combined D-region. A. . For a shear span in a deep beam the combined D-region has a depth of d and a length up to 2d one way or two ways from the disturbance.2.5 as shown by the approximately horizontal line to the right of d d = 2. Strut-and-tie models can also be used in the design of B-regions [Marti (1985)l. and to 2d for overlapping D-regions. come together and touch or overlap. (d) the arrangement and strength of the nodal zones. corbels and joints. the V. strength. and (f) the detailing requirements. 3. 1 in Appendix A [ACI(2002)] gives a step-by-step procedure for laying out a strut-and-tie model. Schlaich. .2. * On the other hand. and nodal zones rnust equal or exceed the forces in these members.4). The changes necessitated by the new load and resistance factors in the 2002 ACI Code are presented briefly. MacGregor (1 997)l. (ACI Sec. The widths of struts are chosen to carry the forces in the struts using the effective strength of the concrete in the struts. The effective concrete strength and the strength reduction factors in Appendix A were originally derived using the load and resistance factors in Chapter 9 of the 1999 ACI Code. The starting point is usually the computation of the reactions for the structure and loads. and the ties within 15" of the resultant tensile forces. Struts should not cross cracked regions. The selection of the model and the production of a drawing of the model are integral parts of Commentary RA. It therefore produces a statically adnissible force field.2 of the 2002 ACI presents several major requirements that must be satisfied by a strut-and-tie model: I . 2. the overlappingparts of the struts would be overstressed. Ifthe strength at every cross section equals or exceeds the strength required by the analysis in item I . the structure must have enough ductility to acconzmodate any needed redistribution offovces.. however. ties.2. the structure is said to have a safe distribution of strengths. and "Basic Tools ofReinforced Concrete Beam Design" [Marti (1985). 4. A.3). First andforemost. The calculation of the reactions and strut and tieforces satisfies statics.4) 6.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code The definitions of these items differ considerably in various codes and other design documents.6). The strengths of the struts.2. and the design of reinforcement in the ties are discussed. with examples. A. Schafer and Jennewein (1987)l recommend using a finite element analysis to determine the stress trajectories for a given load case. Generally speaking. the Canadian code [CSA(1994)] requires that finite element analyses be checked by independent analyses satisfying equilibrium. the location of the struts and ties can be arranged within the structure such that struts fall between cracks.5). A2.3 Geometric layout of strut-and-tie models A strut-and-tie model is a hypothetical truss that transmits forces from loading points to supports. For twodimensional structures some researchers [Schlaich. Appendix A is formulated on the assumption that strut-and-tie models will be used in design. The smallest angle between a strut and a tie that arejoined at a node is set at 25'. and nodal zones and support regions when laying out a strut-and-tie model (Sec. If pictures of the cracking pattern in similar structures are available. Schafer and Jennewein (1987)l "Prestressed Concrete Structures" [Collins and Mitchell (1991)]. in "Toward a Consistent Design of Structural Concrete" [Schlaich. It is generally necessary. The struts are then aligned within k 15" of the resultant compression forces from such an analysis. and suggests using strut-and-tie models for this check. A.2). Struts must not cross or overlap each other (Sec. It is not intended that design be reduced to equations for the shear resisted by struts and web reinforcement. In the early stages in the design o f a D-region it may be sufficient to consider only the axes of the struts and ties when laying out a strut-and-tie model. 3. (Sec. the strut-and-tie model must be in equilibrizrrn with the factored applied loads andfactored dead loads (ACI Sec. The selection of STM's. A structural design that is both staticallv admissible and satisfies the requirements o f a lower bound solution in the theory ofplasticity. a strut-and-tie model which minimizes the amount of reinforcement approaches the ideal model. the calculation of the forces in the struts and ties. Ties are permitted to cross struts or other ties. This implies that thefailure load computed by a strut-and-tie model underestimates the actualfailure load For this to be true. Ifstruts overlapped. to consider the widths of the struts.2. A2. 3. ties. 5. A. and [Reinforced Concrete: Mechanics and Design. Various authors have suggested methods of doing this [ACI (1999)].2. (Sec. Part 2: Derivation of strut-and-tie models for the 2002 ACI Code Section A. . Collins and Mitchell (1991). In developing a strut-and-tie model for a given application it is frequently useful to select initial trial locations for the nodes and use these in the initial cycle of calculations of the member forces. Schafer and Jennewein (1987) and MacGregor (1997). 8) on the prestressing force. and deep beams..7.. Iff.2 Design of struts Struts are designed to satisfy Eqs.2. F. If there is room for this spread to occur.14 Part 2: Derivation of strut-and-tie models for the 2002 ACI Code I> Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 4 Forces in struts and ties. In the 2002 Code.. A. the forces. is the nominal strength of the member. F. (If= 0. @.. is the effectiveness factor for a strut. @STM is the value of @ for struts.) In the 2002 ACI Code. and nodal zones are F. is the 0.. . and strength reduction factors. or nodal zone) due to the factored loads. 1 to 4. and nodal zones in strut-and-tie models. The term v is introduced as an intermediate step in the derivation of Eq.or y. the reactions to the applied loads and self weight loads are computed. The nominal strengths of struts. Fnt. In the CEBIFIP Model Code (1993) the Canadian Code [CSA(1994)]. respectively. material strength reduction factors. and y. Equation 1 includes the factored resistance (IF. a.85 for shear in beams. 5.1 and 0.13 on the design of post-tensioned tendon anchorage zones..2. 9.. is the end area of the strut acted on by f. 10. and @STMwere changed to 0. $. ties and nodal zones are then proportioned based on: where Fu is the force in the member (strut. is the effective compressive strength of the concrete in a strut.9 for flexure and = 0. Strut-and-tie models of prestress anchorage zones retain this @ factor and load factor because the tendon forces and the load factor on the tendon forces are unchanged. the load combinations and @ factors in Appendix C of ACI 3 18-99. the strut is idealized as being uniformly tapered. Most struts in two-dimensional strut-and-tie models will be bottle-shaped. were interchanged with those in Chapter 9 of ACI 3 18-99. The 1999 ACI Code used different strength reduction factors for each type of structural resistance.85 factor in ACI Sec. (Subscripts have been added to the @ factors in this paper to indicate the structural action corresponding to the various @ factors.75 for the design of strut-and-tie models using the load factors and strength reduction factors in Chapter 9 of the 2002 ACI Code. tie. 18.85 in the design of post-tensioned anchorage zones to be used with a load factor of 1. In strut-and-tie models. corbels.' and f. ACI 3 18-99 included a new Sec. the concrete in the beam webs adjacent to a strut is stressed by the lateral spread of the strut stresses into the adjacent concrete in the shear spans. That code specified @PA= 0. The struts. in all the struts. A reevaluation of the @ factor for flexure indicated @fcouldremain equal to 0. and qS. ties. The factored strength of a strut is computed as: Fns = fcuAc (2) where f. f.. are applied to the concrete and steel strengths. and Fnn. ties.2 (Sec. Once the reactions have been computed. is different at the two ends of a strut. and nodal zones loads are computed using truss analysis.1 Types of struts Struts vary in shape. the struts are said to be bottle-shaped. and the FIP Recommendations (1999).. 4 because different codes and researchers include different factors in their definitions of the effective compressive strength. In that strut-and-tie model. ties. taken equal to: or: @feu = @ v fc' = @STMa1 0 s fc' where v (nu) is called the effectiveness factor. they are generally idealized as prismatic or uniformly tapered members as shown by the straight sides of the idealized prismatic struts in the shear spans of the deep beam shown in Fig. based in part on strut-and-tie models. 1. 5. and (I is a strength reduction factor. Q After the initial strut-and-tie model has been selected.90. (1987) have analyzed this type of cracking and predict that it will occur when the compressive stress on the end of the strut exceeds approximately 0. these cracks may weaken the strut. or transverse cracks.2. The effective compressive strengths given in Appendix A and Sec. Struts frequently are wider at midlength than at their ends because the width of concrete that the strut stresses can spread into is greater at midlength than at the ends of the strut. Schlaich et al. differ from code to code. and a1 is the 0.3. 11 and 12 in Sec. solid-line boundaries of the struts in Fig. . 4 would be modified to agree. explained in various references as accounting for load duration effects. Schlaich et al.85 from ACI Sec. In the absence of reinforcement to confine this splitting. several relationships have been suggested as replacements for the a1 = 0. Schafer and Jennewein (1987)l. 3.1.5 of this paper reflect this concept.3. fcU 5. and Breen et al.7.1. bottle-shaped struts are idealized as the prismatic struts shown by the straight.2. In three-dimensional concrete structures like pile caps. Typically. (i) Bottle-shaued struts. The resulting reduction in the compressive strength of the struts is explained in the following paragraphs. 5. . 5. Depending on the emphasis placed on each factor when deriving values of the effectiveness factor.was taken equal to 0.85 factor defined in ACI Sec.'. The effective strength of the struts is given by Eqs.3 E f f e c t i v e c o m p r e s s i v e s t r e n g t h of s t r u t s . the a1 in Eq. Recently. In Eq. 3 have a slope of 1:2 as shown. (a) Load Duration Effects. Tests of uniformly-stressed.3. 1 and 4 where v = a1 p. (iii) Transverse tensile strains.7. In the event that one of these proposed revisions is accepted. square concrete panels by Vecchio and Collins (1982) have given rise to the values off. The strut may be crossed by cracks which will tend to weaken the strut [Schlaich. Part 2: Derivation of strut-and-tie models for the 2002 ACI Code strain in the strut due to the tie attached to one or other end of the strut. 10. or accounting for the vertical migration of bleed water.1 Factors affecting the effective concrete strength of struts The stress acting in a strut is assumed to be constant over the cross-sectional area at the end of the strut. dashed outlines of the struts in Fig.. This factor probably should be a function off. The divergence of the forces along the length of the strut tends to cause longitudinal splitting near the ends of the strut as shown in Fig. 4. Adebar and Zhou (1 993) have proposed equations for the effective compressive strength for use in designing pile caps. computed as a function of the angle between the axis of the strut and the axis of the tie. the compressive strength of a strut may be increased by the confinement resulting from the large volume of concrete around the strut.' increases [Ibrahim and MacGregor (1997)l. (1994) suggest that a the diverging struts in Fig. Tensile strains perpendicular to the axis of a cracked strut connected to a tie may reduce the compressive strength of the strut [Vecchio and Collins (1982)l.1. In design.. decreasing as f. In the Canadian Code [CSA(1994)] and the AASHTO Specifications (1998) it is assumed that the strength of a strut is a function of the transverse tensile Fig.7. or accounting for different stress regimes in the cylinders and the flexural stress blocks. 3: Splitting of a bottle-shaped strut. given by Eqs. diagonal. (b) Cracking of the struts.. The curved. 1. Three major factors that affect the effectiveness factor are given in the following paragraphs. 1 represent the effective boundaries of a typical strut.2.'. (ii) Cracked struts.85 in ACI Sec.Part 2: Derivation of strut-and-tiemodels for the 2002 ACI Code 5. The sub. 10. 10. the struts develop axial.55f. the values off. refers to strut. a. script "s" in p.3. (c) Confinement from surrounding concrete. Such a strut is said to be bottle-shaped. 2D + 1.Schlaich et al. Ps Initial values of $STMIXI corresponding to the 1999 ACI Code load factors and the ACI definition off. 9. In Sec. one. where U stands for ultimate load. QSTM = 0.OD). 5.2. f.2. - Ricketts (1985) reported ratios of test to calculated strengths of six two-span continuous deep beams. with L either 50 kips (0.55 to 0. The FIP Recommendations (1999) give the crosssectional area of a strut required for a given set of loads as: A ~ . Two concrete strengths were considered. 3000 and 6000 psi. .1 of the 2002 ACI Code.3 Effective concrete strength of struts-from other codes The values of 0.0.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 5. Concrete Strengths.6. from other codes cannot be used directly in Appendix A because other codes have different load factors. In FIP. (1991) related v and fcl with v ranging from 0. and F. When v was taken as 0.The cross-sectional areas of struts required by the FIP Recommendations. f. The basic load case considered is dead load plus live load (D + L).2 Effective compressive strength of struts-from tests and literature Various researchers have proposed values of the effectiveness factor v. 9. - Rogowsky (1983) stated that v = 1. ~ cisI the area of the strut or the compression stress block for a member designed using the ACI Code. and different ways of specifying the concrete strength.0 and 1.. Rogowsky and MacGregor (1986) proposed f.13.39 for f. p to A c . Appendix A: set equal to the corresponding areas.69 for concrete strengths ranging from 3000 psi to 6000 psi.85 and = 0. where a 1 is the factor from ACI Sec.'.8D (480 kips). v. U = 4. (1987) recommended values of v similar to those given in 19 Part 2: Derivation of strut-and-tie models for the 2002 ACI Code The values of $ S ~ Ma1 Ps needed in the ACI Code so that the strut area from the two codes givethe same load capacities were then computed using . For L = OSD. where D = 100 kips. For L = 0.and two-span deep beams. They observed that the selection of an appropriate truss was more important than the choice of v. and fcd. values of Ac.~lpis the cross-sectional area of the strut computed using the F P Recommendations.2.0 conservatively predicted the strengths of the corbels tested by Kriz and Raths (1965)..'.35D + 1.6.' = 0..3. different resistance factors. Many more references are listed in a Bibliography on Strut-and-Tie Models. between 1. were computed for the axial forces in a hypothetical strut due to assumed combinations of dead and live load.0 than to 0.6.4D +1.25D (225 kips) and for L = 2D. A further complication is introduced by the fact that the load and resistance factors in Chapter 9 of the 2002 ACI Code differ from those in Chapter 9 of the 1999 ACI Code. U = 1. ~ c lassuming . It is reasonable to assume that D and L are defined in a similar fashion in both American and European codes. ~ C Ifrom where A c .. = v f.~~ is the force in the strut due to ACI factored loads on the strut-and-tie model. U = 2. Virtually all of the corbel tests had effectiveness factors. gives: Section 5..5D) or 200 kips (2. - Ramirez and Breen (1991) proposed a relationship between v and ranging from 0. giving U = 200 kips and 440 kips for the two loading cases.7. - Marti (1985) suggested a constant value of v = 0. This indicates that v was closer to 1. the basic load combination changes to U = 1.5D. ~ equal same way in both the ACI and FIP.~lpwere computed using Eq.35D (435 kips).96 for v = 1. the basic load case is U = 1. - Based on tests of 24.A.effis the effective concrete strength from FIP. FIP Recommendaiions.5 of this paper.FIp is the strut force due to the sum of the factored loads acting on the strut. U = 2.6L.77 to 0. and rearranging.' of 3000 psi and 6000 psi. The failure loads predicted using strut-and-tie models. These were . F I=P FIIFIP (5) cdeff where Ac.1 OD (210 kips) and for L = 2D. the mean test to calculated strength ratio increased to 1. U = 4. for example.85f. A few comparisons are made here. Fu.85.1. - Bergmeister et al.3.1 of the 1999 ACI Code.' ranging from 3000 to 6000 psi.' were derived by calibration to the FIP Recommendations (1999).5 L.3. based on the load factors in the FIP Recommendations and for the corresponding concrete strengths. A c .7L. that the unfactored loads are defined in the Setting A c . In Sec. 10. 5 for specified concrete strengths. a with v . For unfactored strut loads of 100 kips dead load and 50 or 200 kips live load.3. Load Factors and Loadings. had a mean ratio of test / calculated strengths of 0. collected by the ACI-ASCE committee on Shear and Torsion [ACI 445(1997)]. For the 2002 Code the range is 0.85 is similar to the a. f..l(l . as e. which is a lower fractile than used in ACI to define f. FIP gives v2 = 0.645 x 0.3. 5.2f3) b) Struts with cracks parallel to the strut and bonded transverse reinforcement. = the smallest angle between the axis of the compressive strut and the axis of the tie attached to one end of the strut.25 = 65" relative to the tie. 10 using the following descriptions: 5. Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 21 5. 9 is applicable only in the compression zones for beams or axially loaded columns that are assumed to be uniformly stressed with fcd.3.= 0.1.39. The strength of concrete is defined in the same way as in ACI 3 18.85.5L.j.2I3) d) Struts transferring compression across large cracks. fck.3.85f. and a coefficient of variation = 0. The load factors in the Canadian Code are U = 1.5 to 25' and by implication to to an upper limit of 90 . 5.eff=v1 fled acting over the distance c from the neutral axis to the extreme compression fiber. Consider 6000 psi (41.002) Cot2 8. One definition of fcu is assumed to apply for all types of struts. The reduction in strut strength is due to the transverse tension and to the disturbances by the reinforcement and the irregular crack surfaces.86 . the range was 0.7. ranged from 0.effbased on v I will not be considered. e.64 to 0.2f3) c) Struts transferring compression across cracks with normal crack widths. For f ~=' 3000 psi: flcd = 0. In the strut-and-tie model in Fig. For the 2002 load and resistance factors psranged from 0. For the 1999 ACI load factors alv p. 1.2(3) a) Uncrackedstruts with uniform strain distribution.1 and Eq.8 MPa).' and a factored steel strength I$.60 and $s = 0. in members with axial tension or in tensionflanges. Consider 3000 psi (20.2.' 0. (I2) =tensile strain in the tie.OO x 1540 psi = 1540 psi and.' = 3000 + 1. These values are listed in Table 1 for comparison with values of Ps derived from other codes.85X2710 1. + 0.66.2. some FIP cases and ACI cases overlap and are listed two or more times. P.85. . FIP defines v2 in Eq. for @STM= 0.1.g. 4.1. rather than at a12 as assumed in the ACI Code rectangular stress block. b. A.3. Here. For 1999 Code load and I$ factors.5 = 1540 psi 6000 psi Concrete.45.33 x 600) . The effective compressive strength of the concrete in struts is: fcu = f: < 0.2. the range was 0.645 x 0.50.35 to 0. Because the verbal descriptions used in ACI Appendix A differ from those given in the FIP Recommendations. The FIP uses the 5th % fractile strength. From Eq.1.2. v2 = 1. and €1. For the 2002 load and resistance factors. For 3000 psi concrete.80 t1 7 0 ~ .7 MPa). FIP gives v2 = 0. the required mean strength.25D + 1.3. fck = 3600 (1 .10) = 5760 psi (39. FIP gives v2 = 0. in webs ofbeanzs.eff 'V 1 flcd or (10) fcd. values of fcd. For the 2002 load and @ factors.7 m a ) concrete with a standard deviation of 450 psi (3.fy where qC= 0.75. f.61 to 0. fc. and the values of p. the required mean strength.eff = V2 flcd where Eq.. @ = 0. ranged from 0. ranged from 0.63.eE is 1.4 MPa) concrete with a standard deviation of 600 psi (4.85.14 MPa). the strength of the concrete in a strut is taken as the smaller of (9) fcd.5.76 to 0.46 to 0. 5.37 for the 1999 Code. 0.80. The stirrups and minimum surface reinforcement are ignored when computing 0. The design strength for uniaxial compression from FIP is: where a = 0.46 to 0.00. equal to 1. 5.85 in ACI 10. 8 the design strength for uniaxial compression from FIP is fled = 3260 psi. the strut is anchored by the longitudinal tie and is crossed by stirrups or minimum reinforcement. From ACI Sec.1 m a ) .3. the corresponding values of b.' = (6000 + 2. In the FIF' Recommendations.10.83 for the 1999 load factors and $STM= 0.60.77 to 0. and yc is a resistance factor for concrete. . FIP defines concrete strength using the characteristic or 5th %ile strength. and those proposed for Appendix A.34 x 450 = 3600 psi. From ACI Sec. Because this locates the resultant compression force at c/2 from the extreme compression fiber.52. ranged from 0.15) = 2710 psi (18. For concrete with an ACI specified strength of 3000 psi.500 = 6900 psi.34 to 0. In the Canadian code [CSA(1994)] design is carried out using a factored concrete strength 9. f.15. fCk= fc.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 20 3000 psi Concrete.'. Canadian Concrete Design Code and AASHTO LRFD Specifcation. These two codes define fcu based on transverse tensile strains in the struts. and a coefficient of variation = 0. 8. (1 1) where E l = Es ES + (E.g. is taken as the angle between the axis of the strut and the tie which is limited in ACI Sec. For 3000 psi concrete. 1999 load factors Case Code Values 1 Chosen A...66 Struts transferring FIP 5.60 ACI App. . 1999 load factors Values (b) without reinforcement satisfying A.0 0.68. a. . f..7 = QSTMal P. = 0.60 A.60 h 0. = 0. vz = 0. and E. ZOO2 loads factors Values Chosen 1. Uncracked strut with a FIP 5. such as the compression zone of a beam.70 for compression in anchorage zones: - normal weight concrete .' p. .4--For all other cases 2002 loads factors Values Chosen 0. . .. For 0.3..39 0.85.40 FIP 5. . = E~ = 0.37 ACI App.34-0-0. for @ = 0.3.3. For 0. . fc. 2 1 of one and two span deep beams For 0. for0=0.= 0. . . Chosen Code Case p. a.2(3)b) 0.73f.0 cross section is the same as that at the nodal zones. .'.85.46-0.2.86 1 1 1 1 1 1.2. 0.40 0.61-0. = 0. A for+=0.5 1 .50 compression across cracks of normal width. Assuming 4 fCu= OSTM a1 fls f i . Eq.60 h 0. 0.80 lightweight concrete . . or the ACI App.2(3)~) 0. A 1.2. A Struts transferring FIP 5. E.0.3.75.0. .2. v2 = 0...00 Com~ressionzone in a beam Compression zone in a tied column A.. A tension flanges of members Struts transferring compression across large cracks A. the AASHTO L W D Specification [AASHT0(2002)] gives the following values of 0: for compression in strut-and-tie models . I MLE I-Lonrlnuea) s.-. 0.002.46-0.3. for these angles 0 are 0. = 30°. for Struts in Strut-and-Tie Models.2(3)a) 0.75.2(3)c) compression across cracks of normal width.06 ACI Chapter 10 1. p. v* = 1.35-0. for 4 = 0.83 uniform strain distribution.3.76-0.3.65 . = 0.52 0. v2 = 0.85.55f. = 0.3.85.002.3. . . = 0. 11 gives f.85. p.85 and $STM= 0.2.22 TABLE I-Values Part 2: Derivation of strut-and-tie models for the 2002 ACI Code of 0.2-Struts located such that the width of the mid-section of the strut is larger than the width at the nodal zones (Bottle-shaped struts) (a) with reinforcement satisfying A.. . . .60 0. . a.'. J( f. . = 0.85.1-Struts in which the area of the mid-strut ACI APP.3-Struts in tension members. . For strut-and-tie models.002. and 0.'.20 Part 2: Derivation of strut-and-tie models for the 2002 ACI Code - -.= 0.= 0. 11 from the Canadian Code [CSA(1994)] gives Eq. . A FIP 5. the values of P. a. . based on a.774.3.29 respectively. = 60" and f.3. .6 Struts in STM's of posttensioned anchorage zones Struts crossed by CSA 0 = 60' reinforcement at an angle 0 = 45' 0 with the axis of the strut 0 = 30" Back calculated from tests Ref.3 Struts with cracks and bonded transverse reinforcement. .85.3 ACI App. .31f. = 45' and E.2(3)d) 0.80 / ACI 10. The corresponding value of P.1 For a strut of uniform cross-sectional area over its length. except that.STM= @STMa 1 Ps fct a b Setting these equal and substituting @STM= 0.' = QSTM a 1 (3. Values of P.4 Effective Concrete Strength of Struts .. respectively.2.13 on tendon anchorage zones based in large part on strut-and-tie models. .90.' for the following three cases to minimize the differences at the interface between B-regions designed using the traditional flexure and shear design theory.6 $ f.' and 0. .. A.7 Part 2: Derivation of strut-and-tiemodels for the 2002 ACI Code The values of fcu presented in ACI Code Appendix A were chosen to satisfy four criteria: $ f. = 0. . (d) Compatibility with other codes or design recommendations. The first option depends on finding unambiguous descriptions of the state of cracking in the member. = gsTMa 1 Bsf. . Experiments gave P.85. 0. are given in the various codes cited: (a) FIP uses descriptions of the cracking of the struts to select the applicable values of v. .75. a = Ol.2.20. or (3. In making this choice. f.flex = $ f a 1 fcl a b If we assume the compressive force in a strut-and-tie model of the same beam is also equal to C.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 5. = 0..1 through A. verbal descriptions was adopted. C.7 $ f. The @ factor for tied columns is &. or if large inelastic rotations are expected.. For flexure by the 1999 ACI Code.06.82 for tied columns and 0. the factored effective compressive strength is limited to 0.7 h / a 1 = 0.' gives (3. to be computed. Values of (3. $f= 0. 11 and 12 which require E. .3. &. This value for the compression zone in a tied column is lower than the value of Ps for the compression zone in a beam because the ACI @ factors for columns were arbitrarily set below the @ for flexure to account for the more serious and more brittle nature of column failures. and D-regions designed using strutand-tie models. such as deep beams. .' and neglecting differences in the load factors gives Ps and 0.3 ..From Other Sections of ACI 318-99 Because these four criteria lead to different values of f.70 and for spiral columns $sc = 0. It would be desirable that @ v f. is 0. 5. . The italicized words in the following paragraphs are directly from ACI 3 18-02.7 hf. the first option. . = 0. depends on being able to compute a poorly defined strain in the web of the member.3. in areas where the concrete may be extensively cracked at ultimate due to other force effects.. Table 1 lists values of 0..85 we get 0. = 0. f. . A.85 has been proposed to correspond to the rectangular stress block for flexure. Ps = 0. . f.2 In 1999. For the same case using = Qf/ $STM = 1. for a given application.= 0. the flexural compressive force. = 0.82 (a) Simplicity in application. acting on the height. .20 for related cases.824.2. approaching 1. = I. ranging from 0.4 are the descriptions of the types of struts used in Appendix A of ACI 3 18-02. evidence of (3. .13.85 = 0. .'. Prestressed Tendon Anchorage Zones--ACI 318-99 Section 18.'. The Flexural Stress Block in Columns Failing in Compression... judgement was required in selecting the values off.: (14) CU. = @STMa1 (3. (b) Compatibility with tests of D-regions.'. In Appendix A.75 and @ f =0.90/0.76 to 1.3. ranging from 0. The effective concrete strength of a strut is given as the product $ f.71 for I$ f. For the 1999 ACI Code. v f. ST^ = 0. (c) Compatibility with other sections of ACI 3 18.7010.75 = 1.c.2 For struts located such that the width ofthe mid-section of the strut is larger than the width at the nodes (bottle-shapedstruts): (a) with reinforcement satisfjling A.0 in tests was also strongly weighted. dapped ends or corbels. .3. as high as 0..4. = 0. equal to @ a 1 P. (b) CSA and AASHTO base fcu on Eqs.2.3.88 for spiral columns. .=0.90 from the 2002 ACI Code.3. The bold headings numbered A. P. .85. are summarized in Table 1.75. of the rectangular stress block is: (13) Cu. for related cases from other documents are also listed in each section. The Flexural Stress Block in Beams. .85.0. This revision used f.6 @ f& Setting @ f. The second.3. .94. (3. Rewriting the expression for fcu as @ f. ACI 3 18 included a new Section 18. = 0.' and @ = 0.824 h.' agree with @STMa1 Bsf.5 Selection of fCu for Struts for Appendix A For strut-and-tie models of prestress anchorage zones AASHTO gives f. . . . was taken equal to 1. Two different methods of specifying f. and a. .46 to 0. The CSA code values are based on a different concept and will be disregarded. = 0. P. with a.. . Other codes give (3. . and nodal zones are the regions around the joint areas in which the members are connected. . . RA.37.: C3 acting on the sides. . .2. . .1 Classification of nodes and nodal zones It is desirable to distinguish between nodes and nodal zones. 6 Nodes and nodal zones This is illustrated in Fig. .2. The compression stresses assist in the transfer of forces from strut to strut. (a) the struts.. . . than was given to f. = wt cose + Cb sine (15) where w. . from other codes.2. wnl: w. 15 decreases as the stresses on the sides become more and more unequal. but not in the code itself. wt is the effective width of the tie. Extended Nodal Zones. this class of nodes was referred to as hydrostatic nodal zones. Derivation of strut-and-tie models for the 2002 ACI Code Hydrostatic nodal zones were extended to C-C-T or C-T-T nodes by assuming the ties extended through the nodal zones to be anchored on the far side by hooks or bond on the tie reinforcement beyond the nodal zone. . The strut width can be adjusted by changing wtor Cb. (b) the reactions.34 = 0. . . Appendix A uses extended nodal zones in place of hydrostatic nodal zones. Thus a C-C-C node anchors three struts. . (3. ties. Equations can be derived relating the widths of the struts. value for A. . The requirement for equal stresses on all faces of a hydrostatic nodal zone tends to complicate the use of such nodal zones. and bearing areas if it is assumed that the stresses are equal in all three members meeting at a C-C-T nodal zone. Nodes are classified by the types of forces that meet at the node. . t b is the length of the bearing plate. The area of the hypothetical anchor plate is chosen so that the bearing pressure on the plate was equal to the stresses acting on the other sides of the nodal zone. The similar case from FIP corresponds to p. 6.1. w. 6.3.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code (b) without reinforcement satisfving A. 4(a) where the darker shaded area is the hydrostatic nodal zone and the total shaded zone is the extended nodal zone.. . If the stresses were equal on all sides of the nodal zone. Appendix A assumes the faces of the nodal zone that are loaded in compression have the same width as the ends of the struts.60 A The term h for cracking of lightweight concrete was included in the !3.. = 0.4 For all other cases. (3.1. = 0. . . .3 For struts in tension members or tensionflanges of members. Once this has been done. a C-C-T node anchors two struts and one tie.40. .3.. it is necessary to check the stresses on all the faces of the nodal zone. A.2 Types of nodal zones and their use in strut-and-tie models The literature on nodes in strut-and-tie models is based on two quite different concepts. . . a C-T-T node anchors one strut and two ties. . Originally. Hydrostatic Nodal Zones.6). . In general. . Nodes are the points where the axial forces in the struts and ties intersect.3. These are nodal zones bounded by the outlines of the compressed zones at the intersection of: The selected values of (3. The effective area of the tie is the tie force divided by the permissible bearing stress for the struts meeting at a node. and 0 is the angle between the axis of the strut and the horizontal axis of the member. the ratio of the lengths of the sides of a hydrostatic nodal zone. = 0. there must be a minimum of three forces acting on the node in a planar structure like a deep beam. For vertical and horizontal equilibrium at a node. .60 Part 2. .. (3. are generally higher than those from other codes because more weight was given to values of fcu corresponding to related design cases in the ACI Code and values of fcu from tests. . Because the Mohr's circle for the in-plane stresses acting on such nodal zones plots as a point. . . Future code committees should consider adding such equations to the Commentary.3. C1: C. . . 4 (ACI Fig. 7. A. or strut to tie. . . to 0. The width of the faces anchoring ties will be discussed more fully in Sec.2(b) because the stabilizing effect of reinforcement transverse to the struts is not present and failure is assumed to occur shortly after cracking. . . . .~: wn3 are in the same proportions as the forces.3. . and (c) the assumed widths of ties including a prism of concrete concentric with the ties.one at a time. . . . This relationship is useful for adjusting the size of nodal zones in a strut-and-tie model. . The extended nodal zone falls within the area stressed in compression due to the reactions and struts. is the width of the strut. . This concept is represented using a hypothetical anchor plate behind the joint. The accuracy of Eq. . nodal zones were assumed to have equal stress on all in-plane sides. . Relationships between tlte dimensions of a nodal zone. Equation 15 was included in Fig. RA.zL I 1: - -.2. 6. The values are intended to reflect the weakening effect of the discontinuity of strains introduced when ties stressed in tension are anchored in a node stressed in compression. 4) fcu.60 in the Canadian code.91) triaxial stress state: up to 3.1 Effective compressive strength of nodal zones from tests Very few tests of the strength of nodal zones are available.3.02.3 Effective compressive strength of nodal zones Resolution of forces acting on nodal zones.80f.0 for comparison with the strength of test specimens and a1 = 0. (01 One bver of sreel biaxial stress state: up to 1. the nodes in the strutand-tie model acted on by more than three forces could be analyzed assuming that all the arut and tie forces act through the node.2 Canadian Code and AASHTO LRFD Specification The Canadian Code (1994) defines the effective compressive strength of nodal zones.' can be developed in such nodal zones if properly detailed For C-C-T nodal zones the average test / calculated strengths were 1.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code part 2: Derivation of strut-and-tie models for the 2002 ACI Code 6.85 this corresponds to & = 0.20 ficd (fin = 0.85 9. with the forces on one side of the nodal zone being resolved into a single resultant strut during the design of the nodal zone.71. = 0.94.3. f. The AASHTO LRFD Specification uses similar values of @ fcu. Assuming I$ = 1.17 with a standard deviation of 0. 8. This concept is illustrated in ACI Commentary Fig.2.85 and &rM = 0. a. .75 Oc fc' for nodal zones anchoring more than one tie: 0. f.2 Effective compressive strength of nodal zones from other codes 6. i i k (b) Dirlriburrd slerl Fig.80. For C-T-T nodal zones the mean test I calculated strength was 1. (1991) indicate that fCu= 0.' for nodal zones anchoring only one tie: 0.85. 0. equal to 0. where QC = 0. for biaxially. see A. If more than three forces act on a nodal zone in a two-dimensional structure.'.65 0. For a.14.70 and 0.2. a. this corresponds to = 0. Tests of ten isolated C-C-T and nine C-T-T nodal zones reported by Jirsa et al.85ficd where flcd is given by Eq.1 FIP Recommendations "Practical design of structural concrete" The FIP Recommendations limit fcu on nodal zones to the following values: for node regions anchoring one or more ties: 0.9) 6. 4: An extended nodal zone. 6. The values from the Canadian code are equivalent to 13. as: for nodal zones loaded by compressive struts and bearing areas: 0.65 to 0.3..3.88 fled (fin = 2.61 in ACI.3.4.3. Alternatively.or triaxially-loaded C-C-C nodal zones anchoring only struts: I- en. it is frequently necessav to resolve some of the forces to end up with three intersecting forces. 3 Selection of the Effective Compression Strength of Nodal Zones.2 requires the tie reinforcement to be distributed approximately uniformly over the width of the tie.03 C-C-T Nodal zones Nodal zones anchoring one or more ties. 4. will be based on the AASHTO and CSA values. .. ACI Eq.423.. Generally the major problem in the design of ties is the anchorage of a tie in a nodal zone.85 and 1999 load factors and @ = 0. load factors assume a. A non-prestressed tie is assumed to reach its capacity when the force in the tie reaches For the 1999 Code.2 of the Commentary to Appendix A. f.1 Ties in strut-and-tie models In strut-and-tie models based on hydrostatic nodal zones. modified to agree with the ACI load factors.A-6 assumes Afp equals 60 ksi.3 0. the tie reinforcement is spread over the height of the tie computed as The tie is assumed to consist of the reinforcement and a hypothetical prism of concrete concentric with the axis of the tensile force. The hypothetical concrete prism concentric with the tie does not resist any of the tie force. 4(a). the a1 factor. 7 Ties 7. If extended nodal zones are being used.61 0.85. leading to less deflection of the member.65-0. (See also Section RA.80 for C-C-T nodal zones anchoring only one tie for C-T-T or T-T-T nodal zones anchoring ties in more than one direction fin = for C-C-C nodal zones bounded by compressive struts and bearing areas 0.71 11P 0.85 and @sTM The second set of recommended values of fin are for use with the 2002 load factors assume a. values for each type of nodal zone. = 0. Thus.4. 4(a). the lower extreme value of the height of the tie corresponds to the steel being placed in one layer with wt taken equal to the diameter of the bars plus twice the cover to the bars as in Fig.65 . wt. Tests of Nodal zones Ref. 7. The effective compressive strength. 0.3.3 FIP zones anchoring only compression struts.0 fin = 0. 4(b).Nodal C-T-T Nodal zones Nodal zones anchoring one or . The following values were selected for inclusion in the 2002 ACI Code: fin = 1. In serviceability checks the reduced strain in the tie due to this concrete may reduce the elongation of the tie. @ = 0.94 The first set of recommended values of fin are for use with the 1999 = 0. feu. more ties. Notes: ' * 1 CSA . and @STM = 0.75 and 2002 load factors @fcu= @ Bn fc' Range of pn Code Case 2002 Load Factors C-C-C Nodal zones CSA A23.85 and @STM = 0.75.4. This is a reasonable approximation to the change in stress in the prestressed reinforcement as a member is loaded to failure.2 Strength of ties The values of feu for nodal zones from other codes are summarized in Table 2.60. rather than concentrating it near the tension face of the beam as shown in Fig. This may entail putting the reinforcement in several layers as shown in Fig.85.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 6. = 0. Table 2 shows these are a reasonable fit to the (3. 15 $I = 1. A.85 using Eq.0.73 . 0.0 for tests. ACI Sec. Table 2: Recommended values of fin for nodal zones in strut-and-tie models. the strength reduction factor for nodal zones was taken as @STM= 0.92-1.) A second term is added for prestressed ties. For alz between 0. A.60 h if this reinforcement is not provided. again without horizontal web reinforcement.1.0025. The direction of the bar is chosen so that a tensile bar force in the bar causes a compressive force in the concrete perpendicular to the crack. is the area of the bars in one direction and the angle y.acting on the crack resulting from a layer of confinement steel is 01 = 4. is the angle between the crack and the component of the force in the bar in question. A.3. FIP recommends that all the shear should be resisted by stirrups. the change in the tie forces at the truss node must occur within the width of the nodal zone. A d . 8.1 allows this requirement to be satisfied using layers of steel crossing the strut which satisfy 8. and 5 top and 2 bottom bars in the third. and as a result. . For shear spans with alz between 0. All three beams failed after both top and bottom steel yielded. 3(b). for a/z = 0. on the shear span be transferred to the supports by an inclined strut. V.0035. A.4. or. In these beams the relative amounts of top and bottom steel from an elastic analysis of a slender beam were 4 top bars and 3 bottom bars. and the rest.3. A.0. where z = jd is the internal lever arnl between the resultant tensile and compressive forces in flexure. The value of 0.5 and 2. is given by: 8.. This p.4.5. had varying distributions of flexural reinforcement.2. of the total shear. pv. This equation was derived assuming the normal stress. Thus.3. the full plastic capacity would be reached with a stirrup reinforcement ratio of 0.. drops to 0.5 and 2. ACI Sec.2 and A. A. of 0.3 Minimum shear reinforcement in deep beams There is a major discontinuity in the amount of shear reinforcement required at the limit between the deep beams and shallow beams in the 1999 and earlier codes.2. = V .3 allows the designer to compute the necessary steel either from an assumed localized strut-and-tie model of the strut as shown in Fig.8 with vertical stirrups giving a stirrup reinforcement ratio. 8 Reinforcing Requirements Part 2: Derivation of strut-and-tie models for the 2002 ACI Code ACI Sec. did not allow adequate moment redistribution.4. ACI Sec.3 Anchorage of ties ACI Secs. ACI Secs. A.3. A. A.3.2 Minimum reinforcement required in bottle-shaped struts.3. At the limits.3. should be considered by ACI Committee 3 18 Sub E as a possible addition to Appendix A.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 7. V2. be transferred by vertical web reinforcement. FIP recommends that all of the shear V be transferred by the inclined strut. Three. The FIP Recommendations (1998) provide a transition from a D-region to a Bregion for a/d ratios close to the limit of 2. or a similar provision. A. and V1 = V.3 require that the anchorage of the tie forces be achieved completely by the time the geometric centroid of the bars in a tie leaves the extended nodal zone. The trends from these tests and others with varying web reinforcement ratios suggested that in beams with vertical stirrups only.0015 and no horizontal web reinforcement. these two beams did not develop their full plastic load capacity.3.3. sin yl bsl where A.i to simplify the presentation.5 and 2. failed after the positive moment (bottom) reinforcement yielded but before the top steel did. ACI Section A. At nodal zones in beam-like structures where diagonal struts are anchored by stirrups. and to give some ductility to the struts. this equation gives V1 = 0.0.75 when computing the effective compression strength of bottle-shaped struts with reinforcement satisfying ACI Sec.1 Effect of minimum confinement reinforcement from tests Grids of reinforcement are desirable in the side faces of D-regions to restrain the splitting cracks near the ends of bottle-shaped struts.1 allows the results of the strut-and-tie model to be approximated using the empirical equation. with clear shear span to depth ratios of 0. This. V1.V2. 4. FIP requires that a portion. 4 top and 3 bottom bars in the second beam. A. but allows it to be placed in one layer in cases such as corbels.3. ACI Sec. ACI Sec. and for a/z = 2.3. Such a provision prevents strut-andtie models with struts which are at too small an angle with the longitudinal axis of the member. This point is shown in Fig.3. Eq.5 limits the angle to 25' for the same reason.2(a) and (b) allow the use of ps = 0.4. This required some moment redistribution. It has been written without the term f. for beams with a concrete strength of 6000 psi or less.1 states that this steel generally is placed in grids in two orthogonal directions in each face. Three similar beams with vertical stirrup ratios in the order of 0. a.3 itemizes other requirements for anchoring ties. two-span continuous deep beams tested by Rogowsky (1983) and Ricketts (1985). the amount V. The three beams varied the amounts of top and bottom bars from 2 top bars and 5 bottom bars.3 requires reinforcement transverse to the axis of the strut proportioned to resist the tensile force lost when the strut cracks due to the lateral spread of the strut forces. alz = 0.3. where Asi is the total area of reinforcement at a spacing si in a layer of bars at an angle yi to the axis of the strut. Hooks or mechanical anchorages may be required.3. 8.002 each way. The minimum steel was given in Eq. it is easier to place the web steel in one direction. to 1.4 and 11. with an overall mean of 1. with the vertical steel.i is the reinforcement ratio for web reinforcement at an angle yi to the axis of the strut. with a coefficient of variation of 0.4 Minimum web reinforcement from codes The 1994 CSA Code requires an orthogonal grid of reinforcing bars near each face with a ratio of total steel area to gross concrete area in each direction of at least 0. The Commentary to the AASHTO LRFD Specification states that. A.8. horizontal in the case of a corbel. In design this crack restraint is achieved by providing minimum web steel satisfying: Sections 11. This is referred to as crack control reinforcement. The critical web steel ratio taken perpendicular to the strut is p. the steel is placed in one direction at an angle of at least 40' with the axis of the strut. 1. These beams were relatively large.0 for beams with a/d = 1.3. The mean test/calculated ratios decreased from 2.03.0035.0 This is reasonable agreement for shear test data. The ratios of the lowest test/calculated shear strength ranged from 0. the full plastic capacity was obtained with a ratio of vertical steel equal to 0.001 in each direction. having overall depths of 500 to 1000 mm. and to provide ductility to the struts and nodal zones by confinement.2 allows uni-directional confining steel in corbels or similar cases.86 and 2. The two structural functions of minimum web steel in a D-region are to resist the transverse tension in the bottle-shaped areas near the ends of the strut after splitting cracks occur. such as those in a deep beam. and the steel area provided was equivalent to The FIP Recommendations (1999) suggest that deep beams should have a minimum reinforcement ratio of 0. while for thicker members it may be necessary to have multiple grids throughout the thickness. The strength of each shear span was taken to be the lowest of the five shear strengths for that shear span.6 Comparison of strut-and-tie models to tests of deep beams. These beams did not have horizontal web steel. ACI Sec. there should be a grid of bars at each face.003 in each direction. The portion of this reinforcement parallel to a tie and located within the assumed cross-sectional area of the tie can be considered as part of the tie reinforcement. A. It is accounted for by the increase in P. it is feasible to provide orthogonal grids of steel near the side faces of the deep beam.0025bws2.96 to 2.5 Selection of minimum web reinforcement. . For some D-regions. y was 55". where p. A-1 and A-2.001 Obws of vertical web reinforcement corresponds to a shear stress resistance. sin y. r Part 2 .3. C (pvi sinyi) ? 0. such as pile caps for more than two piles. Further checks against test data are presented by D. 8. In other cases.2 of this paper. such as corbels or dapped ends. Mitchell in another paper in this symposium.3. in each face.4 in terms of equivalent amounts perpendicular to the axis of the strut.003 (21) 8.3. Each 0. ranging from yield of the ties. 11-15. deep beams tested by Rogowsky and MacGregor (1986). In some structures. Each beam was assumed to have five failure limit states. A range of test results were considered in developing the ACI Code provisions for strut-and-tie models as reported in Section 5. A reduced strut strength is imposed by ACI Sec. In addition. respectively in deep beams.06. where y is the angle between the reinforcement and the axis of the strut.2.. for a total of not less than 0. it is frequently not feasible to place web steel in the three-dimensional strut-and-tie model.5 of ACI 3 18-99 require minimum ratios of vertical and horizontal web reinforcement of 0.14. In the Rogowsky and Ricketts tests quoted above.Part 2: Derivation of strut-and-tiemodels for the 2002 ACI Code 8.32 for beams with a/d = 2. The additional capacity provided by the vertical web reinforcement is not implicitly included in the strength calculation using ACI Eqs. to crushing of each end of each strut. in thinner members. These minimum amounts provide a considerable shear capacity. A.2(b) in such cases. and a / d ratios of 1.247.21.0015bws and 0. Vs conesponding to v = 60 psi.002. Derivation of strut-and-tie models for the 2002 ACI Code In the Rogowsky (1983) and Ricketts (1985) tests quoted earlier. AASHTO (2002) has similar requirements to CSA but the minimum total reinforcement ratio is set at 0. The strut-and-tie model of the failure region indicated that the critical strut had a slope of about 55". the final code statements were compared to tests of each end of six simple-span. Horizontal web reinforcement is much less efficient in transferring shear. calculated using ACI Eq.54. and spans of 2000 mm. and crushing of the nodal zones at each end of each strut. If in one layer. Jirsa.-7. 11.1.8. provided by horizontal stirrups dropped step-wise from 0. and the nominal strength of a tie. Appendix A was reviewed by ACI Committee 3 18 and approved over the course of four letter ballots. 1. In the same region. This is especially true for continuous deep beams where the moment approaches zero at the critical section defined in ACI Sec.99 to zero at En / d = 5.1 and 11. Strut-and-Tie Models. At tn/ d = 5.K. Subcommittee E.O Avfy ACI Sec. = area of vertical b. When strut-and-tie models were introduced. provided that enough bearing area is provided.1 mentions serviceability.1 and 11-8 The development of Appendix A of the 2002 ACI Code and the selection of the $ factors and effective concrete strengths is explained.2. Deflections can be estimated using an elastic frame analysis of the strut-and-tie model. / tuncr. Further study is needed in this area. who was succeeded as chair by J. the V. of vertical web reinforcement. and V. The 1999 Code Eqs. Wight during the period in which the Appendix was letter balloted. the nominal strength of a nodal zone.. tie. = specified eff = strength f. and the fraction having an uncracked stiffness respectively.)approach to shear design.Part 2: Derivation of strut-and-tie models for the 2002 ACI Code Part 2: Derivation of strut-and-tie models for the 2002 ACI Code 9 Other details 10 Summary 9. 9. web reinforcement within a spacing s. 11-29 for V. wn3 = lengths of the sides of a nodal zone. 10. approaches zero. F. 11.0..d end area of a strut. . but not requiring an exact match between the shape of the end of the strut and the corresponding shape of the face of the nodal zone acted on by the strut. 11. This allows the elongation of the tie to be modelled in a simple manner. it was necessary to change the definitions of deep beams in 1999 ACI Code Secs.. the second term on the right-hand side of ACI Eq. there was a major discontinuity in the magnitude of V. 10-7. lever arm between the resultant compressive and tensile forces in a beam. the V.01. In addition.1 Changes to 1999 ACI Code sections. the nominal strength of a strut. of the concrete in a strut-FIP. A. The axial stiffness of the ties can be modeled as cracked regions with axial stiffnesses of AsEs / E.7. Fu = force in a jd = internal S = spacing strut. The nodal zones are also three-dimensional with the struts from the various piles converging on the nodal zone at the base of the column.2 Pile caps Revisions were made to ACI Sec.O. compression strength of struts or nodal zones-ACI. Design checks and comparisons to tests were carried out by members of ACI 445. = effective fled = design Fn. 15.5.3 is intended to simplify the design of three-dimensional strut-and-tie models by requiring that the individual faces of the nodal zone have the areas computed from Eq. F. Extensive assistance and a number of widely differing design checks were provided by ACI Committee 445-A. Section A. 11 Acknowledgements ACI Code Appendix A was developed by ACI Committee 3 18.5 to 1. without replacements in Chap.2.3 Serviceability The last paragraph of RA. A. =width of the web of a beam. 12 Notation The following notation is used in this paper: wnl. from ACI Sec.where tcrand tun.30. concrete strength for uni-axial compression-FIP.5. 10.3 to allow the use of strut-and-tie models in the design of pile caps.8.A.8. were deleted. and uncracked regions with axial stiffnesses of ACE.. In a pile cap the strut-and-tie model is threedimensional. 9. = smaller A. 11 . "Deep Beams" was shortened by omitting the these sections and design was permitted either by non-linear analysis or by strut-and-tie models. chaired by K-H Reineck.5 Avfy at tn/ d = 4. As a result. Fnt = nominal strength. + V.5. or node due to the factored loads. The 1999 and earlier ACI Code design equations for deep beams were based on the (V. f. The layout of strut-and-tie models is reviewed. provided by vertical stirrups increased step-wise from 0. from 1999 ACI Eq.. because these equations did not reflect the true behavior of deep beams. 11-29 and 11-30 for V. Shear and Torsion.7. chaired by J. concrete compressive strength. w d .. A.are the portions of the length of tie which can be assumed to have a cracked section stiffness.' F. The new definitions are based on the definition of deep beams in ACI 2002 Sec. D. D. Transportation Research Board. Anderson. Qf. G: and Falconer. factor. (1997): ModiJication of the ACI Rectangular Stress Block for High-Strength Concrete.. and Jirsa. A =2 btvS 13 References. wt z = height = internal a1 = ratio Canadian Standards Association (CSA) (1994): A23. 90 (1993). Eng. C. ASCE-Journ.G. Vol. Sept. 1985. Vol. (1991): Experimental Studies of Nodes in Strut-and-Tie Models. B. (1985): Basic Tools of Reinforced Concrete Beam Design. London. 1375-1417 (see also ACI 445R-99. 88 (1991). Farmington Hills. ACI Structural Journal. 3.B.Part 2: Derivation of strut-and-tiemodels for the 2002 ACI Code Part 2: Derivation of strut-and-tie models for the 2002 ACI Code V. 1985 Rogowsky. pp 525-532 Kriz. 8 (1986). Thesis. ACI Journal. (1986): Design of Deep Reinforced Concrete Continuous Beams. 94 (1997). November 1983.D. 766 pp 4.. 5. Cornit6 EuroInternational du Beton. (Distributed by:fib.. = shear strength provided by stirrups. Englewood Cliffs. FIP-Commission 3 "Practical Design". tension strain in the concrete. 1216 pp. 2 0 0 2 . 4 4 3 ~ ~ . (1991): Dimensioning of Nodal zones and Anchorage of Reinforcement. L. (1985): Continuous Reinforced Concrete Deep Beams.3-94. K. ov. PC1 Journal. 16. Sept. Roberts. J. J.. 124 (1998). ACI Committee 445 (1997): Strut-and-Tie Bibliography. 10 (1965). 1998. (1991): Evaluation of a Modified Truss-Model Approach for Beams in Shear. ACI Structural Journal. and Bouadi.. Structural Concrete.Febr. (1997): Reinforced Concrete: Mechanics and Design.55 PP. and Raths. Ph. D. Hisham H. Bergmeister. pp 55 1-564.. Thomas Telford Services Ltd. 1991.12. Stuttgart. of a tie and its surrounding concrete. National Academy Press. University of Alberta. ratio for vertical web reinforcement Breen. p.. No. and MacGregor. London.. 82 (1985). Canadian Standards Association. J. Wollmann. Rexdale. Vol. State-of-the-Art-Report by ASCE-ACI Committee 445 on Shear and Torsion. 1993. Vol.Sc. Publ. AASHTO (1998): LRFD Bridge Specifications and Commentary.E. J.-Feb. May-June.50 pp. Thesis. D.E. 32 (1987). 178 pp Schlaich. American Association of Highway and Transportation Officials. 0 . Sanders. Breen. Collins. for shear. J. Denis (1991): Prestressed Concrete Structures. (1987): Toward a Consistent Design of Structural Concrete. 1. of the design strength of the concrete in a strut or node to the specified strength. Design of Concrete Structures. K. P. and Zhou.. K.O. J.M. J. (1993): Bearing Strength of Compressive Struts Confined by Plain Concrete. No. 1996. MI (1999). Barton. M. (1994): NCHRP Report 356. 74-150 . ACI Committee 3 18 (2002): Building Code Requirementsfor Structural Concrete (ACI 3 18-02) and Commentary (ACI 3 18R-02). 1.G. PC1 Journal. Farmington Hills. Lausanne) Ibrahim. R. 8. V. No.M. = shear strength provided by concrete.C. Jennewein.H. D. P. Waslungton. Edmonton.. 1991. 1994. Schafer. IABSE Colloquium Structural Concrete.H. pp 40-48 Jirsa. 199 pp p. Breen. strength reduction factors for flexure.. ACI Structural Journal Vol. Prentice Hall Inc. of the average stress in the rectangular stress block to the concrete strength. @STM= strength reduction factor. lever arm between the resultant compressive and tensile forces in a beam. (1965): Connections in Precast Concrete Structures: Strength of Corbels.437 pp ~1 = principal E:! = strain in a tie. 1999. 1997. 5. 1965. for strut-and-tie models. Vol. 1997. IABSE Colloquium Structural Concrete.E. and Breen. Bergmeister.A. D. Concrete International: Design and Construction. Prentice Hall.) Adebar. pp 49-58 Rogowsky. . ACI Bibliography No. FIP Recommendations (1999): Practical Design of Structural Concrete. No. J. pp 534-541. Stuttgart. Z. No. pp 46-56 Ramirez. y = angle h = correction v = effectiveness pv = reinforcement from axis of strut and axis of a layer of confining reinforcement. September-October 1991. Jan. factor for the strength of lightweight concrete. Dec. (1983): Shear Strength of Deep Reinforced Concrete Continuous Beams. pp 562-571 Ricketts.. M P and Mitchell. Pn = ratio CEB-FIP Model Code 1990 (1993): Design of concrete structures. Jan. J. MI. and MacGregor.O. J. American Concrete Institute. No. 3rd Edition. Structural Concrete. No. C. Washington. of Struct. Burdet. 1. 2nd Edition. 1991. ACI ASCE Committee 445 (1998): Recent approaches to shear design of structural concrete. No. Vol. University of Alberta.: SETO.E. August. American Concrete Institute. 1997. James G. 939 pp Marti. pp 16-47 MacGregor. H. M. SeptemberOctober 1993. 3 on the Design of Concrete Structures and the Canadian Highway Bridge Design Subcommittee on Seismic Design. She holds a B. ACI member.. as well as refined strut-and-tie models are presented. Standard Building Code. 3 18B. Both simple strut-and-tie models. William D. Alcocer Synopsis Results of experiments on a corbel. 332 pp Part 3: Experimental verification of strut-and-tie models Denis Mitchell. and Joint ACI-ASCE Committee 445. Cook. He received his Ph. degree from the National Autonomous University of Mexico (UNAM). Sergio M. Claudia M. FACI. is a professor in the Department of Civil Engineering at McGill University. His research interests include nonlinear analysis of reinforced concrete structures and the structural use of highstrength concrete. He chairs the Canadian Standard CSA A23. degree from McGill University in 1987. Publication 82-03. Medellin. Joints and Connections in Monolithic Concrete Structures. specializing in the behavior and design of regions near discontinuities in reinforced concrete members. Toronto. Uribe and Sergio M. March 1982. Performance Based Seismic Design of Concrete Buildings. Building Code Requirements. Shear and Torsion.D. degree from EAFIT. Alcocer. He is a member of ACI Committees 3 18. He is a member of ACI Committees 408.40 Part 2: Derivation of strut-and-tie models for the 2002 ACI Code Part 3: Experimental verification of strut-and-tie models Vecchio. Denis Mitchell. Collins. Colombia and a M. Bond and Development of Reinforcement. FACI.J. as well as a member of Joint ACI-ASCE Committee 352. Cook is a Research Engineer in the Department of Civil Engineering at McGill University. Claudia M. deep beams and a beam with dapped ends are presented to illustrate how strut-and-tie models are applied to these cases and to provide some experimental verification of the accuracy of the predictions. and 374. Reinforcement and Development. M.Sc. University of Toronto. .Sc. Department of Civil Engineering. William D. F. (1982): The Response of Reinforced Concrete to Inplane Shear and Normal Stresses.P. is a research professor at the Institute of Engineering of the UNAM and Director of Research at CENAPRED. Uribe is a former graduate research assistant in the Structural Engineering and Geotechnical Area at the National Center for Disaster Prevention (CENAPRED). 3. Figure 1 illustrates the strut-and-tie model for a double-sided corbel tested by Cook and Mitchell (1988). as well as in the 1984 and 1994 CSA Standards (1984. corbels and deep beams.Part 3: Experimental verification of strut-and-tie models Part 3: Experimental verification of strut-and-tie models 1 Introduction 2 Corbel Although truss models have been used since the turn of the century [Ritter (1 899) and Morsch (1909)]. This increased popularity is due to the fact that the designer can model the flow of the forces with struts and ties.63 in. The total yield force is Asfy= 4. the corbel was subjected to a vertically applied load.4 MPa (5860 psi). The testing of geometrically similar deep beams demonstrated that.54 in. Figure l(a) shows the details of the reinforcement and the dimensions of the test specimen.11. The concrete strength at the time of testing was 40. In selecting the experiments only well-instrumented. the FIP Recommendations (1999) provide guidance on suitable strut-and-tie models for different types of regions. 1994) and the FIP Recommendations (1999).7 mrn (0. This paper illustrates some simple strut-and-tie models and compares these predictions with experiments. the inclined struts are equilibrated by vertical struts with dimensions of a1 by b in the column and a horizontal strut with dimensions of a2 by b near the bottom of the corbel (see Fig. For a corbel.9. The corresponding predicted capacity. but also provides a clear understanding of the various resisting elements.)) weldable grade bars. V. l(c) are those resulting from yielding of the four No. 15 bars. In solving for the geometry. Therefore. which not only illustrates the flow of the forces.200 mm2'444 MPa = 355 kN (79. of 502 kN (1 13 kips).) and a:! was 21. (0. While the ACI Code and CSA Standard do not recommend strut-and-tie models for different cases. From equilibrium. full-size tests were chosen.2V represented the minimum horizontal design force required in the ACI Code (Clause 11. such as brackets.8 in. the response of smallscale specimens may not be representative of the response of actual structural elements. the designer can apply a strut-and-tie model. a uniform stress of 0.4).). This horizontal force of 0. a1 was calculated to be 39.2V. Walraven and Lehwalter (1 994) have summarized the importance of size effects not only for slender beams.. even for complex design situations. yet powerhl tool. where b is the width of the column and corbel. The reason for this increased popularity is that they provide the designer with a simple. compared to larger beams.85 in.) bearing plate was 25 mm (1 in. rather than using empirically based design approaches. but for deep beams as well. Empirically based approaches are not only limited in their applicability. V. smaller beams exhibit less severe crack propagation and as a result exhibit higher stresses for the crushing of the concrete near the bearing plates. 1(c)). and an outwards horizontal load that was kept equal to 0. The 50 mm. Design using strut-and-tie models provides an alternative to the empirically based code approaches for disturbed regions.) thick and was welded to four No. The forces shown in Fig. Design approaches using strutand-tie models have been codified in Appendix A of the 2002 ACI Code (2002). they have gained increased popularity in recent years.2 mm (1. Thus. An additional advantage of using strut-and-tie models is that sketching the flow of the forces within a member highlights the need for careful details of the reinforcement in key regions. In order to predict the capacity. . for the design of complex regions in reinforced concrete. unless special provisions are made to avoid tensile forces.85 fc' was assumed in the stress blocks. 15 (16 mm dia.300 mm (2 in. During testing. V. Figure l(c) shows the strut-and-tie model for this corbel. was 471 kN (106 kips) which is 94% of the measured failure load. but do not provide the designer with insight into the actual behavior.8 kips). it was necessary to find the geometry of the strut-and-tie model. This simplified model used the conservative assumption that only the main tie reinforcement contributed to the strength of the corbel. part 3: Experimental verification of strut-and-tte models 1lCa LLJ Part 3: Experimental verification of strut-and-tie models column and corbel width = 350 rnm (a) Details (b) Specimen after failure (a) Load. tested by Cook and Mitchell (1988). 2: Measured strains in reinforcing bars of double-sided corbel [Cook and Mitchell (1 988)l.ZV A . A 0 1 yield L strain = 0. H=O.0022 (b) Load. . V = 336 kN -* * no reading (c) Strut-and-tie model Fig. . 1: Double-sided corbel. V = 502 kN Fig. 15 bars making up the main tie. and then additional reinforcement is added to control cracking. with an area equal to 0. required to equilibrate the yield force of the tie was 76 mm (3. First yielding occurred in the main tie at the face of the column at a load.83f. Figure 2(b) shows the measured strains in the horizontal reinforcement just before failure. The area.4. The strut-and-tie model and the measured strains illustrate the need to adequately anchor the tie yield force at the support reaction areas.455 MPa = 546 kN (123 kips). The hooks at the ends of the tie reinforcement provided adequate anchorage of this steel. under the bearing plate was (471~1000)/(50~300) The 2002 ACI Code limits this stress to 0. The measurements were taken on these targets through small access holes in the concrete cover. was 1. .Part 3: Experimental verification of shut-and-tie models A small bearing plate was intentionally chosen for this test specimen to determine the crushing strength of the concrete under the bearing plate. Figure 3(d) shows the strains measured in the main tie reinforcement at two load levels. 3(a). The basic strut-and-tie model shown in Fig. for this compression-tension node.supported on rollers. a.68fc'. or 0. uniformly distributed within two-thirds of the effective depth adjacent to A. followed by crushing of the concrete under the bearing plate. It is clear that these limits are somewhat conservative for this particular case because a bearing stress of 0.0 in. The top two nodes were located a distance of d 2 below the top surface of the beam and in line with the resultant forces in the column (assumed to act at the quarter points of the column).) be provided by closed stirrups. (0. 3(c) neglected the presence of the additional stirrups on the left side of the beam. dd. the main tie is often selected using such a simplified strut-and-tie model. 3(a)). 2(a).75fc' and 0. the loading plates were welded to the four No.. Figure 2 shows the measured strains obtained from a mechanical extensometer.85Pnfc1= 0. On the left side of the beam. was reached at the failure load of 502 kN (1 13 kips). Part 3: Experimental verification of strut-and-tie models 3 Deep beam subjected to concentrated load Figure 3 illustrates the details and the strut-and-tie model for a deep beam tested by Rogowsky. As.4 MPa (6150 psi). Asfy= 6. As expected. respectively. At the predicted failure load of 471 kN (106 kips) the compressive stress at the node = 31. These experimental results emphasize the need to properly anchor themain tie at its ends. Failure occurred by yielding of the main tie. Clause 11. It is interesting to note that at failure the maximum strain in the main tie occurred close to the bearing plate. The equivalent rectangular stress block depth.) thick beam was supported on columns which were. V. l(a)). In design. of 336 kN (75. The reinforcement in the columns is not shown in Fig. 15 bars (16 mm dia. the member contained additional reinforcement consisting of five sets of 6 mm (0. The shear spanto-depth ratio. The two lower nodes of the truss were located at the intersections of the centerlines of the support reactions and the line of action of the main tie.9. failure took place on the side without the stirrups.)) having a total yield force. providing the necessary anchorage (see Fig. a larger bearing plate would typically be used.78fc'. The concrete strength at the time of testing was 42. measuring change of length on strain targets glued to the reinforcing bars. This 200 mm (7. MacGregor and Ong (1986). in turn.200 mm2.). is the total area of the tie and An is the area of tension reinforcement resisting the horizontal tension applied to the corbel. as well as yielding in the stirrups. As shown in Fig. In design. l(b) there was considerable spalling of the concrete in the region around the bearing plate. In this case.4 MPa (4550 psi).24 in. The 1994 CSA Standard and the FIP Recommendations (1999) limit the nodal zone stress to 0.) diameter closed stirrups (see Fig.9 in.63 in. The failure of the beam was governed by yielding of the main tie reinforcement.80f2.4 of the ACI Code requires that this additional reinforcement.5 (As-A. The main tie reinforcement consisted of six No. Figure 3(b) shows the crushing that took place in the concrete after yielding occurred in the main tie. At a load of 550 kN (124 kips) the tie had experienced yielding along most of its length.5 kips) as shown in Fig. The predicted capacity of 586 kN (132 kips) is 97% of the actual capacity of 606 kN (136 kips). 7 in. Following the FIP Recommendations (1 999). This 100 mm (3.6L = 864 rnm (34. Asfy= 4. was predicted to be 372 kN (83.32 in. The concrete strength (equivalent prism strength) at the time of testing was 30. the FIP Recommendations provide suggested geometries for some standard disturbed regions. tested by Leonhardt and Walther (1966) While the ACI Code and CSA Standard do not provide specific guidance on suitable strut-and-tie models for different cases. the distance fiom the centroid of the main tie to the centroid of compression was taken as Z = 0.). 3: Deep beam subjected to concentrated load. (c) Simplified strut-and-tie model (d) Refined strut-and-tie model Fig. The main tie reinforcement consisted of four 8 mm (0.26 mmz. This predicted value is considerably below the measured capacity of 1 172 kN (263 kips).) diameter reinforcing bars having a total yield force. tested by Leonhardt and Walther (1966).2 MPa (4380 psi). Figure 4(c) shows a simplified strut.). the capacity of the deep beam.) and the uniform loading applied to the top of the beam was located over the clear span distance of 1280 mm (50.6 kips).4 kips).50. 4 Deep beam subjected to uniform loading Figure 4 shows a deep beam.) thick beam was supported on 160 rnm (6. MacGregor and Ong (1986).) long bearing plates.0 in.9 in. tested by Rogowsky.1 kN (19. loaded uniformly on the top. The center-to-center span was 1440 mm (56. With the geometry of the strut-and-tie model established.and-tie model suitable for design. the uniform loading was replaced by two point loads at the quarter points of the clear span. In the strut-and-tie model.3 in.428 MPa = 86.4 in. 4: Deep beam subjected to uniform loading. in terms of the resultant of the uniformly distributed load. hoops1 1 Part 3: Experimental verification of shut-and-tie models 1300 (a) Details (b) Failure conditions (a) Details (b) Specimen after failure Load = 550 kN yield 80+1280/4 strain 925 mrn 150 925 (c) Strut-and-tie model I (d) Measured strains in tie reinforcement Fig.Part 3: Experimental verification of strut-and-tie models 48 5-6mm dia. . 5 = 732.21 .9 in.8 in.32 in.1000)/(2. The beam was supported on 160 mm (6. the distance.5 mm (28.) diameter reinforcing bars having a total yield force.3 in.) diameter horizontal closed stirrups.5 in. The ACI Code. the bearing stress was (1 172. From equilibrium of the strut-and-tie model shown in Fig. 5: Deep beam subjected to bottom loading. Part 3: Experimental verification of strut-and-tie models ! Figure 5(c) shows the simplified strut-and-tie model suitable for design.7 in. in terms of the total applied load. (d) Refined strut-and-tie model (c) Simplified strut-and-tie model Fig.100~160) = 36.9 kips). This bottom loading was achieved by hanging loads applied to a 400 mm (15.) long bearing plates and the thickness of the deep beam above the ledge was 100 mm (3. the CSA Standard. Although this provides a good prediction of the capacity it must be emphasized that considerable redistribution was assumed and extremely large strains would occur in the bottom horizontal tie. and the concrete. the FIP Recommendations all require additional.0 kN (18.7 in. This test demonstrates that high bearing stresses can be attained with the use of confinement in the bearing areas. This more refined model utilized the additional horizontal reinforcement that was spread uniformly over the beam depth and assumed that considerable redistribution of the tension and compressive resultants occurred.50. tested by Leonhardt and Walther (1966) . uniformly distributed reinforcement in the beam for crack control at service load levels. If the beam were designed using this more refined model then unacceptably wide cracks might occur at service load levels. that is 84% of the actual failure load. The two top nodes were located a distance of a12 down from the top face of the beam. was located at mid-depth of the beam and was assumed to yield with Asfy= 84.7 kips). 4(d). x. For this beam it was assumed that the bottom load was transferred to the beam by bond stresses between the vertical stirrups. It is noted that at the experimentally determined failure load of 1172 kN (263 kips). the capacity of the deep beam.428 MPa = 172 kN (38. t 5 Deep beam subjected to bottom loading 1600 (a) Details t (b) Specimen after failure Figure 5 shows a deep beam tested by Leonhardt and Walther (1966) that was bottom-loaded. The distance from the centroid of the main tie to the centroid of compression was taken as 800 .). 4(c) would give a conservative design with acceptable service load performance. = 8.:f This very high bearing stress was achieved by providing special confinement reinforcement directly above the bearing plates.). With the geometry of the strut-and-tie model established.2 MPa (4380 psi). The center-to-center span was 1440 mm (56. A.) and the simulated uniform loading was applied to the top of the bottom ledge over the clear span of 1280 mm (50. The main tie reinforcement consisted of eight 8 mm (0. 5(a). The concrete strength at the time of testing was 30. representing five 5 rnm (0.f.20 in. 5(c). The horizontal tie.6 MPa (53 10 psi) = 1.50 Part 3: Experimental verification of strut-and-tie models In order to provide a better estimate of the failure load the strut-and-tie model shown in Fig. A uniform stress transfer was assumed over the height of the beam and hence the deviation in the struts was assumed to occur at mid-height of the beam as shown in Fig.67. 4(d) was developed.) and the resulting predicted capacity was 979 kN (220 kips).4 in. was found to be 266 mm (10.26 mm2.) wide bottom ledge as shown in Fig. was calculated to be 630 kN (142 kips). This prediction is considerably below the measured capacity of 1102 kN (248 kips). The model illustrated in Fig.). that hang-up the load. ). In accordance with the FIP Recolnmendations (1999). Three additional shorter #8 bars were placed at the bottom to resist the forces required by the strut-and-tie model assumed. To assess the effect of confinement along the bar anchorage.1000/(210. From equilibrium of the truss. representing these two stirrups.) deep beam was supported on 400 mrn (1 5. At the bottom of the beam.9 in. having a total measured yield force of 5. 6(a)).5 in. From geometry. . A total of 12-#8 bars were placed at midspan (Fig.85 .2 MPa (4240 psi) or 0.52 Part 3: Experimental verification of strut-and-tie models The refined strut-and tie model for this deep beam is shown in Fig. ws = 150 cos63. for this case..60f: = 0. The FIP Recommendations (1999) limit the stress in the strut to 0. would be permitted. which is also a conservative prediction.97fc'.20 in. 5 mm (0. It was assumed that. as given by 113 (2 a/z . a/z. The 1994 CSA Standard limits the strut stress based on the amount of strain in the tie crossing the strut and on the angle between the strut and the tie.657 of the total load.60fc1 for the webs of beams. no stirrups were placed along the bearing region at one end of the beam (see Fig.51fc'. The center-to-center span was 3600 rnm (141.) and the resulting predicted capacity was 696 kN (156 kips). The actual yield stresses of the #4 stirrups and #8 longitudinal bars were 429 and 445 MPa (62.76f.) and the shear-span-to-effective depth ratio was 1. the distance. containing transverse reinforcement tested under monotonic loading by Uribe and Alcocer (2001). the refined strut-and-tie model predicted a force of 614 kN (138 kips) in the strut just above the reaction. part of the load is transferred from the loading plate directly to the support through an inclined strut (i:e.445 MPa = 1128 kN (254 kips).) long bearing plates. the concrete crushed in the 100 mm (3.8 by 47. The concrete cylinder compressive strength at the time of testing was 35 MPa (5075 psi).) apart. 5(d).) long plates centered 800 mm (3 1.8" + 160 sin63. The remainder was assumed to be carried by stirrups (tie) in a truss with two inclined struts at each beam end. with a = 1400 mm (55.).) and z = 942 mm (37.8" from the horizontal. 0. The top loading was applied through two 400 mm (1 5.100) = 29. direct strut mechanism). five bars were continuous over the span. giving a strut width.4 rnm dia.1 in.7 in. respectively. superimposed onto the direct strut mechanism.) spaced at 140 mm (5. consisted of #8 bars (25. the stirrups were predicted to carry 0.7 mm dia. At the failure load of the beam of 1102 kN (248 kips). the angle of this strut was found to be 63. 6 Deep beam with transverse reinforcement Figure 6(a) shows a deep beam.17.I).1 in.7 in. For this case a strut stress of 0.5 in. For the case with parallel diagonal cracks in a web the 2002 ACI Code limits the stress in the strut to 0. At this load level. which were.) thick portion just above the bottom flange.) in the clear shear span and the longitudinal reinforcement.1 in.).7 in.2 in. Figure 7 shows the strut-and-tie model developed in accordance with the FIP Recommendations (1999).) diameter. This model accounted for two of the horizontal.507 mm2. was located at their centroid. The additional horizontal tie. The total measured yield force of the 8-#8 bars was 1806 kN (406 kips). The transverse reinforcement consisted of #4 closed stirrups (12. the stress in the strut was 614. The 350 rnm thick by 1200 mm (13. the part of the load resisted by the stirrups depends on the shear span-to-internal lever arm ratio. Hence at failure.8" = 210 mm (8. x. It is evident that these code requirements are conservative in predicting the crushing of the struts. at each beam end.2 and 64. 6(a)). was found to be 23 1 mrn (9.85Psfcf= 0. closed stirrups that were considered to yield in the tension zone. The longitudinal reinforcement was anchored with 90' standard hooks following ACI 3 18-02. in turn.For this beam.5 ksi).3 in. Part 3: Experimental verificat~onof strut-and-tie models part 3 Expe~~mental ver~ficationof strut-and-tie models 3203 ~i anrianr lnmm (a) Strut and tie model A-A' B . 7: Strut and tie models for deep beam with transverse reinforcement... tested by Uribe and Alcocer (2001) Fig. 6: Deep beam with transverse reinforcement. (b) Direct strut mechanism (c) Truss mechanism (b) Beam after failure Fig.B' C-c Dimensions in mm (a) Details . tested by Uribe and Alcocer (2001) . .343 times the applied load) was a little larger....- . Figure 7(c) shows the predicted truss mechanism assuming that the seven stirrups yielded. I ~ I (a) Measured strains in stirrup reinforcement goo .. = 0.. ...-. I I I I 1 3 1 I TRANSVERSEREINFORCEMEM II ' I II II ' II .. Hence the predicted capacity of the truss mechanism is 762 kN. I . t -..5 kN) = 1702 kN (383 kips).----. .. For this beam.762 kN = 398 kN (89. third and fourth stirrups from the inner edge of the bearing plates.. . 8 5 ~ .. it was assumed that stirrup yielding that the controlled the failure mode.e. -. 0. .----.. This implies that the assumed contribution of the direct strut (0.657 times the applied load) was constant throughout the tests. . ..127 mm2. The failure of the beam was governed by yielding of the stirrup reinforcement.. .5 kips). .Part 3: Experimental verification of strut-and-tie models In predicting the strength of this beam..-----. 7(a)). by concrete crushing. Part 3: Experimental verification of strut-and-tie models I I I I I I I J I 1 t I I I I 1 ..-. . (b) Contribution of stirrup reinforcement to shear strength Fig. -. . 8: Stirrup contribution for deep beam tested by Uribe and Alcocer (2001) .. 646..." = 0 . Because both of these forces are less than the yield forces of the bars provided.. which were converted to stresses using the measured stress-Strain relation of the steel reinforcement. Figure 8(b) shows the predicted and measured stirrup forces as the applied load increased. The FIP Recommendations (1999) .2.5 kN = 1056 kN (237 kips) and in the 8-#8 bars is 409 kN + (2 . .I. .---......II I I 4 I I I I I I I I -----.-----. ..5 in. .. reauire * part of the total load carried by the stirrups be provided along the length a.I... ..94214 = 955 mm (37. 7absurned tie I I I Cenlroid of lhe assumed tie. it had smaller values. .. . The actual contribution was calculated from strains recorded in the test.. .85. In order to check the capacity of the longitudinal reinforcement it is necessary to combine the direct strut mechanism with the truss mechanism to give the complete strut and tie model (see Fig.6 in. extending from the outer edge of the loading plate to a region near the inner edge of the bearing plate.. I I ' 8 I 8 I 1 I I ' . 7(b)) is thus 1160 kN . . . This yield force is 7. .... The load carried by the direct strut mechanism (see Fig. . .. a. Yielding was recorded in nearly all of the stirrups. . . it was concluded that the correct yielding mechanism was chosen. .429 MPa = 762 kN (171 kips). + : I I I I . followed by wide cracking.1400 . --ldT --.) seven stirrups were considered effective.. I -I .. .I. ..Predicted cantribution Figure %(a)shows the strains measured in the stirrup reinforcement of the beam when the maximum load was attained.. The force required in the 5-#8 bars is 409 kN + 646..z/4.-. Oentroid af the . The additional 4-#8 bars in the midspan region were not included in the strut-and-tie model because they were not long enough to participate in the strut-and-tie model assumed. which is less than the measured strength of 1578 kN (355 kips). Although the stirrup contribution followed the trend of the prediction.-.-----. Therefore the beam capacity is predicted to be 762 kN10.I I. The largest strains were recorded in the second.657 = 1160 kN (261 kips)...) and hence with a stirrup spacing of 140 mm (5. as well as by local bending of the longitudinal reinforcement... .-----. ..---. . I.. . The prediction was calculated assuming that all seven stirrups participated and that the part of the load resisted by the stirrups (i. The strength of the deep beam was predicted to be 1160 kN (261 kips). . .. .. Figure 6(b) shows the specimen after failure.....--I . The yield force of the main vertical tie was ASSy= 4.0 kips).8 MPa (4320 psi). 4-No. .0 kips). The dapped ends resulted in a 250 mm (9. 3-No. It was assumed that the shear in the end projection flowed into the support reaction by means of the inclined concrete strut A-B.100 mm2.) steel angle to provide the necessary end anchorage. 2-No.) closed stirrups. 6mm (3. 10 closed stirrups 5-No.10 U bars Figure 9 shows a beam with dapped ends tested by Cook and Mitchell (1988).3 mm(0. tested by Cook and Mitchell (1988). At node A the horizontal tie.) between roller supports and was subjected to a concentrated load at midspan.) bottom bars (see Fig. 10(b) considerable yielding occurred in both the horizontal and vertical main ties.) deep projection at each end. In the full-depth portion of the beam. consisting of four No. was welded to a 7 5 ~ 7 5 . Horizontal tie A-D was required to provide equilibrium at the node just above the support reaction to equilibrate the outward thrust of strut A-B.9 mm (1.) dia. assuming that uniform bond stresses were developed. Node D represented the fanning compressive stresses resulting from bond stresses and was assumed to be at a distance equal to one-half of the development length beyond the main vertical tie.0.0 kips). Experimental verification of strut-and-tie models 7 Beam with dapped ends 2-No. 9(a)).5 kips).63 in. . The beam was designed using the strut-and-tie model shown in Fig.10 (c) Strut-and-tie model (forces in kN) Figure 9: Beam with dapped ends.2. 15 (16 mm (0. As can be seen from Fig.) dia) weldable grade bars.3. At the predicted failure load none of the other ties reached yield and no nodal zones or struts reached their limiting stresses. The struts represented the centerlines of the fanning compressive stresses.9 in.30 j I 1 37.25 in. Figure 10(a) and (b) show the strains in the reinforcement at first yielding and at the maximum load level of 307 kN (69. The predicted reaction at failure was 260 kN (58.445 MPa = 356 kN (80. The failure of the test specimen was predicted to occur by yielding of the main vertical tie B-C. The actual failure load was 307 kN (69.6 by 11.) beam spanned 3200 mm (126 in.) were provided.44 in. 10 U-stirrups spaced at 225 mm (8. It was assumed that the shear at the end of the beam collected at the bottom of the full-depth beam (node C) and then was lifted by the main vertical tie B-C to the top of the beam (node B). These bars were extended well into the full-depth portion of the beam to provide sufficient developn~entlength. The 600 mm deep by 300 mm wide (23. 9(c).10 U bars 4-No.5 100 1500 rnrn (a) Details (b) Specimen after failure The centroid of the top chord was assumed to be located at a distance of aJ2 down from the top surface and the bottom tie was located at the centroid of the tension reinforcement. No.10 U stirrups 225.8 in. 10 (1 1. Part 3.15 No. The significant tensile strains in the horizontal tie at the bottom of the end projection illustrate the need to anchor this tie at the bearing area.18 in. respectively.Part 3: Experimental verification of shut-and-tie models I . 10 U-bars at the end of the horizontal tension tie at the bottom of the full depth beam provided the additional force necessary to anchor the five No. The main vertical tie B-C consisted of four No. 30 (29.8 in.) dia. The addition of three No. The concrete strength at the time of testing was 29. Sept. D. London. American Concrete Institute. Canada. R. This paper presents the results of a number of experiments that have been carried out on disturbed regions. Morsch.: SETO. t I (a) Support reaction. CAN3-A23. 85. E. Other strut-and-tie models for the design of beams with dapped ends are given by Jirsa et al.Structural Concrete".. 368 pp.1000/(110. Breen J. Because of this detail.--yield strain = 0. Sept. Feb. pp. Cook.7 MPa (2130 psi) = 0. BulletinNo. The paper also discusses some important detailing considerations for disturbed regions. 1999. 199 pp. closed stirrups with 135'-bend anchorages were used for the main vertical tie. (1991): "Experimental Studies of Nodes in Strut-and Tie Models". In using the strut-and-tie model shown in Fig.3-94. (1 899): "The Hennebique Design Method (Die Bauweise Hennebique)".Part 3: Experimental verification of strut-and-tie models 60 I: It is noted that in checking the stresses at node B. K. The FIP Recommendations (1 999) provide useful guidance to the designer for the strut-and-tie geometries of a number of different standard cases such as deep beams and corbels. V. CSA Committee A23.3 (1984): "Design of Concrete Structures for Buildings".49f:. Experimental verification of shut-and-tie models 9 References ACI Committee 3 18 (2002): "Building Code Requirements for Structural Concrete (ACI 3 18-02) and Commentary ACI 3 18-R-02". Berlin. Jirsa. Rexdale. (1966): "Wandartiger Trager". ACI Structural Journal. 9(c) for design. 33. 281 pp. 443 pp. FIP-Commission 3 "PracticalDesign". FIP Recommendations (1999): Practical Design ofstructural Concrete. 178. pp. J. A companion specimen. Zurich. 525-532.. 1996. W. . provided that the horizontal tie extends beyond D' by a distance of at least one-half the development length of the bar. Deutscher Ausschuss fiir Stahlbeton.0022 U % I I CSA Committee A23. XABSE Colloquium Stuttgart 1991. 7. MI. Translation of the third German Edition by E.. McGraw-Hill Book Co. the spalling of the 40 mm concrete cover on the vertical tie reinforcement must be taken into account and hence the nodal zone stress is predicted to be 356. Canada. the CSA Standard (1984 and 1994) and the FIP Recommendations (1999) were used to predict the capacities of these test specimens. the nodal zone area at node B was effectively reduced and the specimen with open stirrups failed at a lower load by crushing of the compressive strut at this node. Wilhelm Ernst & Sohn. This emphasizes the need to provide closed stirrups for the main ties. (1991) and in the FIP Recommendations (1999). 10: Measured strains in reinforcing bars of beam with dapped ends [Cook and Mitchell (1988)l. Rexdale. The strut-and-tie design approach of the ACI Code (2002). D...3 (1994): "Design of Concrete Structures". No. (1909): "Concrete-Steel Construction (Der Eisenbetonbau)". Anderson. and Mitchell.O. The strut-and-tie model provides a conservative. Leonhardt F. W. the number of nodes may be reduced by shifting node D to location D'.. 206-216. and Walther R. Bergmeister. 2.E. Publ. Barton.P. (Distributed by:fib. yield strain = 0. Lausanne) (b) Support reaction. Schweizerische Bauzeitung (Ziirich). pp. with four No. CSA A23.220) = 14. and Bouadi. In order to provide adequate bearing at this nodal zone. Ritter. simple tool enabling the designer to visualize the flow of forces and to appreciate the need for careful detailing. H. . V. V = 173 kN Part 3. 59-61. V = 307 kN Fig. Canadian Standards Association. International Association of Bridge and Structural Engineering. The predicted capacities were compared with the measured failure loads and it is concluded that the strut-and-tie design approach for disturbed regions gives conservative predictions.0022 . 159 pp. (1988): "Studies of Disturbed Regions near Discontinuities in Reinforced Concrete Members". No. 10 open U-stirrups instead of closed stirrups was constructed and tested.3-M84.. Canadian Standards Association. Goodrich. Farmington Hills.D. New York. . 614-623. S. 585. No.M. 91.M. ISBN 970-628-607-1.. ACI Journal. pp. N. and Alcocer S. ACI Structural Journal..Y. and Lehwalter.. (1994): "Size Effects in Short Beams Loaded in Shear". Mexico. V. pp. No. 4. 5.G. D.Part 3: Experimental verification of strut-and-tie models Rogowsky. (2001): "Comportamiento de vigas peraltadas diseiiadas con el modelo de puntales y tensores". September-October. (In Spanish). and Ong. MacGregor. 83. Uribe C. J. (1986): "Tests of Reinforced Concrete Deep Beams". Centro Nacional de Prevencibn de Desastres. Part 4 Examples . J. July-August.593. Walraven.M. V. Uribe is a former graduate research assistant in the Structural Engineering and Geotechnical Area at the National Center for Disaster Prevention (CENAPRED). was designed in accordance with Appendix A of the ACI 318-2002. degree from EAFIT. Sergio M. . as well as a member of Joint ACI-ASCE Committee 352. Also. He is a member of ACI Committees 3 18. Claudia M. Colon~biaand a M. degree from the National Autonomous University of Mexico (UNAM). The analysis and design using the strut and tie model were performed in an efficient and straight forward manner.Sc. is a research professor at the Institute of Engineering of the UNAM and Director of Research at CENAPRED.Example 1 a: Deep beam design in accordance with ACI 3 18-2002 Example 1 a: Deep beam design in accordance with ACI 318-2002 Claudia M. She holds a B. supporting two concentrated loads at the top. Joints and Connections in Monolithic Concrete Structures. and 374. . He is Chainnan of the Committee on Masonry Standards and a member of the committee on Reinforced Concrete Standards of the Mexico City Building Code. Uribe Sergio M. His research interests include earthquake-resistant design of reinforced concrete and masonry structures. The strut and tie methodology provides a framework to understand and assess the flow of forces and the resisting mechanisms. Medellin. FACI. Alcocer Synopsis A deep beam. Alcocer. it is a valuable tool for achieving proper detailing of ductile concrete members. Her research interests include the seismic behavior of reinforced concrete structures. Building Code Requirements.Sc. Performance Based Seismic Design of Concrete Buildings. 6=214k(952kN) b = 14in. (406 mm) and the width is the same as the beam thickness.66 m). (1-1) in this example 0.6 MPa) fv = 60 000 psi (414 MPa) Fig.0 3400 (23. Struts located such that the width of the mid-section of the strut.6 MPa) and& = 60 000 psi (414 MPa). (Section A. with reinforcement 0.6) f: CCT 0.1: Member and loads Nodal zones Struts a. Neglect the selfweight. or bearing areas or both (type CCC) 16" (406mm) (406mm) factor to account for the effect of cracking and confining reinforcement on the effective compressive strength of a strut.80 0. the values of p2 in this example are 1:4 Section 16" P2 = 14" (356 m m ) f.75 satisfying A. Type Uncracked chord (prismatic) Inclined Stmt(bonle-shaped) P2=Ps feu.3. 134 k (596 kN) k (596kN) s e ~ i c eloads (1.3.75 2550 (17.5. (356 mrn) h = 48 in. larger than the width at the nodes. 14 in. The length of the bearing plate at each concentrated load location is 16 in.5. = 0.8 2720 (18. Nodal zones bounded by struts.00 0. 2.' = specified compressive strength of concrete f. 1. psi* (MPa) Type P2=Pn fop psi* (MPa) 1..66m) S. c. fin (Section A.3 (to be referred to as type c in this example) Nodal zones.2 of ACI 3 18-2002): e. psi (23. CTT 2040 (14.2 of ACI 3 18-2002): a. (356 mm). i..2 and A.6 .oo zone of a beam (to be referred to as type a in this example). (356 mm) and a 48 in.85P2 fi Usef.6 = 2 1 4 (952 ~ kN) each on a clear span of 12 ft (3. p.4) e. The beam has a thickness of 14 in.Example 1 a: Deep beam design in accordance with ACI 3 18-2002 1 Problem statement Example 1 a: Deep beam design in accordance with ACI 3 18-2002 2 Strut and tie model parameters Design the simply supported beam that carries two concentrated factored loads of 1 3 4 x~ 1. c. (1. Nodal zones anchoring a tie in one direction only g (type CCT) Nodal zones anchoring ties in more than one direction (type CTT) 1..8) g.e.3.22 m) Table 1-1: Strength of concrete in struts and nodal zones f.2 of ACI 318-2002) f. as shown in Fig. (1-1). Struts in uncracked zones and located such that the mid-strut cross section is the same as that at the nodes.85 ~ 4 0 0 0x P2 = 3400P2 .60 P = 134kx1.4P2.1) *from eq. CCC 1. P.' = 4 000 psi (27.22m) 12'-0" (3.' = 4 000 psi (27. such as the compression 1. or a nodal zone. (1. is or could be.1 Compressive strength of the concrete (Sections A.4) 0.O 3400 (23. MPa) (1-1) Struts.=0. b..22 m) overall depth. db NO 5 slrrrup .5) F.cover .= effective compressive strength A. = nominal strength of a tie A.. A-6) F.l3m) where s-.2.. (I354 kN m) Fig.6). = specified yield strength of nonprestressed tie Strength of nodal zones (Section A. = A.8 bars are provided for positive moment.. it is convenient to know the size (depth) of the compression block of concrete..2 Strength of struts. A-7) 69 F. = nominal f.2 of ACI 318-2002) 3.. (eq. nominal strength of the strut = @ = strength reduction factor (according to 9. the effective depth would be: d = h . = 3 Strut-and-tie model design procedure (Section A.1 of ACI 3 18-2002 Strength of ties (Section A. A-2) F.-0. A.Example 1 a: Deep beam design in accordance with ACI 3 18-2002 68 2.Oin. $ = 0...4) Mus@Mn Because the tie is made of nonprestressed reinforcement: F.3.A... ties and nodal zones The following shall be met: F 2F (eq. 1-2: Moment and shear diagrams (1-2) . is the vertical spacing between the two layers of No.S. = area of nonprestressed reinforcement in a tie f . = nominal compressive strength of the concrete in a strut f.625in.=44.1 Flexure design To develop the strut-and-tie model..4in. = effective compressive strength of a nodal zone A. Example 1 a: Deep beam design in accordance with ACI 3 18-2002 F.5in.-1. A-1) F..d b 8 -~ 1/2~S.. .(l. Assuming that two layers of No. or force acting on a node due to the factored loads F. for all truss elements. smallest effective cross-sectional area of the strut According to Section R9. (eq. =48in.75 Strength of struts (Section A.3) (eq..-%in. =&. = f .-l. = force in the strut or tie.. 8 bars. = area of the face of the nodal zone compressive strength of a nodal zone Mmax = 11 984 k in. in.70 Example 1 a: Deep beam design in accordance with ACI 3 18-2002 Example 1 a: Deep beam design in accordance with ACI 3 18-2002 where: 3.2 Strut-and-tie model From 3.-----. and the width of the loading plates and supports are the same (i.. with that required.5 in. for strut 3.5 in.in I solving for A. Therefore. For the strut and tie model try a depth of horizontal strut (a) equal to 10 in.2 (3 480 mm2). 1-3).. (216 mm) 3. Z17. (216 rnrn). The following is the strut-and-tie model proposed (Fig. (1 354 kN m). (450 mm) .7 in. calculated according to geometry. [454. w. Therefore. (254 rnm) (in the constant-moment region).4 mm) V = 214 000 Ib (952 kN) R = 214000lb(952W) fc' = f.9 (Section 9.cos 0= 16 sin 54. In this example.85~4000psi~14in. 1 A. w. I z = 17. or 356 mm)..1: c (distance from extreme compression fiber to neutral axis) = 8. 2 5.= 4 000 psi (27. will be calculated using eq. 4 = 0. the verification will be done by comparing the available strut or nodal width. ( 0.2. Fig. A. (254 mm) Fig. (1-4). (406 mm) n = 10 in. x 60 000 psi 2~0.6". are shown in Fig.9As ~ 6 0 0 0 psi 0 44. 1-3: Proposed strut-and-tie model ( z =39 in. struts - ties Dimensions in inches (1 in. because the thickness of the beam. see Figs. wpr0.4 mrn] )I a 21kk I z' = lb sin 0 + w.4 in.6 MPa) 60 000 psi (414 MPa) b = 14in. Therefore. 1-4: Forces in the truss 3.411984 k.89 in. = 11 984 k.7 in. Node 1 0node identification number truss element identification number .1) Forces in the truss.3 Truss solved 4 = strength reduction factor.(356mm) h = 48 in..e. (1-4).. = 25. (1.. (I-5a) to (1-5d).3" = 17. w. c = 8.3" D\ < /. 14 in.22 m) lb = 16 in.7" a= 26. 0= 54. for struts and nodes.3. M.3' + 8 cos 54.4 Verification of the strut and nodal zone strength The verification of the strut and nodal zone strengths should be done through comparing the available strut or nodal area with that required. [990 mm] ) / . = 17. For flexure. (254 rnm) can be proposed (w .3 in.. B .3" = 18. flexure design) I 14 6" = 10 in. Node 3 Table 1-2: Verification of strength of struts The width 18. . it is adequate. (1-3).00 w~~~~~ Adequacy Wreq 4 in.6 in. Fig. (371 mm) 82 = 1. From truss solved (Section 3.= 10 in.4 Fig.850. Fig.. (1-5b).88 in. 1-5b: Node 2 Strength check for other struts is shown in Table (1-2) and for nodes.0 (254) J 0 307323 (1367) 8. (1-5). from Table (1-1) b = thickness of member = 14 in.3" + 10 cos 54. 1-5c: Node 3 Node 4 214k z = 18. (449 mm) 153661 = 4. (mm) Proposed solution J:OK 153661 (684) 4. 10. in Table (13) and Fig. a width of . Therefore.'bW z1 = 17.3 and Fig.. ' mr / / I ' / . (1-4) in this example) Required width for struts (from eq. (mm) in. Therefore.69 in.0 (254) J 10.. strut Type a. According to the geometry of the nodal zone. for strut 4. shown in Fig. width of nodal zone = 14.8 in.f.8 in. (109 mm) 0. (478 mm) 'Provided width for struts. (254 mm) (from Wreq = FU - = Fu I$fcubw $0..0 in.. 8 2 = ps 16 sin 54.Example 1 a: Deep beam design in accordance with ACI 3 18-2002 72 Example 1 a: Deep beam design in accordance with ACI 3 18-2002 Strut 1(sample calculation) Node 2 For strut 1.85 x 1. Element 02' Node ~112 e2 (") Ib (kN) # i-i 1 2-4 1.8 in.O. 1-5d). (480 mm) I ' Notes: From Table 1-1 From STM proposed.0 (254) J +23. w. and bearing and support conditions. O X 4000 x 14 where. w.75 x 0. (1-4)) = 18. considering geometry. comes from solving Node 4 (Fig.3 (109) 10. it is an uncracked chord.6 (219) 10.2'.00 0 2 4-5 1.. 1-5d: Node 4 Therefore./ &z2' / Fig.). Provided width for nodes.1 in this example.1 MPa) (OK) 3.6 MPa) (OK) 17.0 (1372) 0.0 (1372) T3-1 153661 (684) 7. type CCC. OA 5 $0. it must be satisfied that o~ sf.. type CCT. headed bars. (1-3) --- ' in.8 (478) J 10.=F.96 (3200) 6 1-3 --. 5 stirmps @ 4 in. mechanical anchorages or straight bar development (Section A.5 (190) 54.2 (406 x 356 mm2) Therefore.78 MPa) l $0.= $2720 psi = 2040 psi (14. @ = 0. the amount of steel required in the ties is determined.75 Required width for nodes (from eq. = 3.4 (137) 0.3 (364) 8. X 16. 8 J A.7 (450) T3-2 214000 (952) 10 (254) 54. Fig..0 (406) J 1 0.. Verification of the bearing strength at loading and reaction points Tie 6 (from node 1to 3) From Fig. . /:OK 16.3 (312) 18.= $3400 psi = 2550 psi (17.8 CCT S2-1 263454 (1 172) 9. 8 bars (Asprov= 4.4 (187) S4-2 153661 (684) 4.8502 f. = 6 No. V = concentrated load From tniss solved (Section 3.2 [3060 mm2]) in two layers.. Node 0 2 ' ~~~e~~ o r c e ) .0 (254) J Distribution of steel Table 1-4: Steel requirements Element 0 F. 6 and No.2 (182) 8.8 (4390) Adequacy Proposed solution in.3 and Fig.0 (203) For node 1.4. (mm) 214000 (952) 7.2 (234) TI-3 153661 (684) 5..2 (234) 3 S2-4 153661 (684) 5. and bearing and support conditions. For node 4.Example 1 a: Deep beam design in accordance with ACI 3 18-2002 Table 1-3: Verification of strength of nodal zones Proposed Adequacy solution .0 (406) J 18.8 CCT R 2 S 1-2 263454 (1 172) 9. 8 bars were used instead of No..- Ib (kN) in. 0 2 = P. T = tie.4 (2190) 4.3 (109) According to Section 2.6 R = reaction. 10 bars to be able to distribute them. It is necessary to anchor the bars either by hooks.2 (4650) J 8 N o . (1-2).0 (203) 4 Example 1 a: Deep beam design in accordance with ACI 3 18-2002 1274 psi (8.J@h).0 CCCC V 214000 (952) 6. 8 and 2 No.0 7 3-6 As req 7.85P2f. No.' (mm2) As 153661 (684) 3.6 CTTT S3-4 263454 (1172) 12.74 (3060) 0 307323 (1367) 6. From STM proposed.74 in.2 (mm2) 90 214000 (952) 4. its area is AA = 16 x 14 in. 1274 psi (8.0 (152) S4-3 263454 (1 172) 7.5 Steel required in ties Once the strength of struts and nodes was verified.5 (190) in. that is.0 (203) T3-6 307323 (1367) 14. (1-4) in this example) ' Notes: Required area of ties (A. J A. '-1 (") Ib (kN) --- 5 2-3 ' Notes: From Table (1-1). . For tie 7.. the ultimate load and the reactions are From Table (1 -4) V = R = 2 1 4 k (952kN) Since the loading plate is 16x14 in.4 (137) 10.2 (2190 mm2): Use 6 NO. ..3 of ACI 318-2002).8 (3100) 4.78 MPa) 5 $0. = 8No.. 17.0 (254) T2-3 214000 (952) 7.4 in. Node if .85 P2hy. the compressive stress is equal to Asre.8 (478) 1. considering geometry. (1-4)) Provided area of ties.7 (450) 8. S = strut. since no more than 12 in. (305 mm) (controls) Use 8 No. = distance from critical section to edge of the beam .2 of ACI 318-2002 (1-6) 1. = where = . .3 Section 12. 1-6): Idh available 77 ldI. and 0. (305 mm) are cast below the bars.1. (226 mm) 12 in. x a = 1. 8 and 2 No.O. = ldh is modified by factor (a) of Section 12..3' . and a concrete cover 2 2 in.0. Tie 5 (from node 2 to 3) Fig.z. (864 mm) (to improve confinement in nodal zone). aval/ab/e = 16 + 4/tan 54. (64 mm) thick is provided in the perpendicular direction to the hook plane. h = 1. = 4. It is convenient to spread the reinforcement~miformlyover an area of concrete at least equal to the tension force in the tie.5.cover shear reinforcement. 5 closed stirrups (A.96 i n z [3200 mmz]).a: factor = 0. Fig.9 in.75 = 28.cover .E db = 19 in.z= 2. (483 mm) for No. spread the 8 No. an extra area of steel is needed and is calculated as 6.4 of ACI 3 18-2002 d 15 = 8. The straight No..3 in.5 in. Therefore. These stirrups shall have 135 deg bends alternatively around one or the other of these bars.OxO. since it is uncoated reinforcement. Space stirrups @ 4 in. 6 vertically over a height of 11 in. .2. h = 1.5 inside the top comers continuous along the beam. (5 1 mm) thick is provided in the direction of the bar extension.06 in. (861 mm) ldh ava~lable> Id (OK) Therefore. (338 mm) forNo. 8 bars spread (vertically) @ 3 in.9 in. (724 mm) 25- It was assumed that: Available anchorage is (Fig. Available anchorage is: ldh avarlable = length of the extended nodal zone .Oxl. ld. since it is normal-weight concrete.0. 8 bars from tie 6 is not enough. db = diameter of the bar (in this case. = 33.. .2 of ACI 3 18-2002. In this example.I Example 1 a: Deep beam design in accordance with ACI 3 18-2002 76 From Section 12..8 in. place No. the anchorage length of a 90-deg hook bar is 0. (279mm) From Table (1-4) From Section 11.=7.4 mm]) I.Oxl. (102 mm).7 = 13.3. Thus. (279 mm).8 bars. (425 mm) > Idh 60 OOOxl.7 because a concrete cover 2 2% in.5 in.5 . Use 2 No.8.4.8 and2No. (1-3).5. [25. (76 mm) and with Id = 34 in. divided by the applicable limiting compressive stress for the node.7 in.l Use 8No. p = 1... 6 bars are extended over the full span of the beam.= 19 in. To anchor the No. use the table from Section 12. w = 1 5 1 / b = 1 1 in.2in.z [4650mmZ]).db horizontal shear reinforcement Id. 8 bars Example 1 a: Deep beam design in accordance with ACI 3 18-2002 Thus.02ph f 1..1.0. d b horizontal (OK) ~dhavailable= 36 in.625 = 16. 6 bars. 1-6: Available anchorage length at the extended nodal zone in Node 1 Tie 7 (from node 3 to 6) From Table (1-4) It is clear that extending the 6 No.o. db=l in.74 in.O.625 in.5.6 (A. In this model. 1 in. part of the load is transferred from the loading plate directly to the support through an inclined strut.18 i n 2 (1 16 rnm2). . from Section 11. 1-8. Use No. (A-4) (eq. From eq.003 sin 54. The remainder is resisted by the stirrups. through a simple truss. it is convenient that the layer or grid of reinforcement be near each face. a. and is determined from equations that have been verified with test results. = 25.4 mm Fig.3.9 in. This model does not require vertical stirrups along the shear span (i. 1-8: Model in accordance with FIP 1999 10 No. (305 mm) Supposes = 7 in. 4 Arrangement of reinforcement Dimensions in inches. In the model assumed..Example 1 a: Deep beam design in accordance with ACI 3 18-2002 3. But. (203 mm). and because it required stirrups along the shear span. it is specified that layers or grids of reinforcement parallel to the plane of the member must cross struts 3 and 4.3" - a .e. large cracking at loads well below ultimate may be expected.5 of ACI 3 18-2002 d 15 = 8. su~dyi = 54.3.i -- b si 0. 1 in.. which was considered a safe detail. The distribution of load between these two mechanisms depends on the shear span-to-internal lever arm ratio. the load is transferred to the support by a single inclined strut (Fig. In Fig. i 2 0.8. similar to that assumed in the solution of this example. and therefore.0259 si Example 1 a: Deep beam design in accordance with ACI 3 18-2002 5 Optional models The strut-and-tie model solved (see Fig. 6 T'k Dimensions in inches. - Fig. 19). Asi = total area of reinforcement at spacing si in a layer of reinforcement with bars at an angle yi to the axis of the strut. (178 mm) in each face.32' 2A. because the web thickness is greater than 8 in. Also. direct strut and truss action) can be superimposed. 1-3) is one of several models that could have been selected. 1-7.4 bars spaced 7 in.6 Reinforcement for bottle-shaped struts 3 and 4 In Section A. 5 @ 4 (102mm) No. A-4) where.e. two other strut-and-tie models could have been selected. Optionally. (226 mm) (controls) 12 in. 4 @ 7 (178 mm) " NO. load transfer was considered to take place by the formation of a truss made of two inclined struts near each support. Suppose only horizontal reinforcement is provided. In the absence of stirrups. The FIP Recommendations consider that both load-carrying mechanisms (i. 5 @ 6 (152 z) \ 8 No. if minimum vertical stirrup reinforcement is provided. it can be argued about the suitability of a similar model. between the loading plate and the beam support) to maintain equilibrium. (178 mm). Nevertheless. = 25. the strut-and-tie model recommended in the 1999 FIP Recommendations is presented. is not recommended for design. 2 0.4 mm The final layout of the reinforcement is shown in Fig. 8 \2 No. This model was selected because of its simplicity. 1-7: Arrangement of reinforcement Another strut-and-tie model that could have been selected considers that at each beam end. His research interests include nonlinear analysis and design of concrete structures. The beam is 20 in.' . ACI Committee 3 18. . (508 mm) wide and 80 in.' -----------------i-. f. The reinforcement details are shown in Fig. V. FIP-Commission 3 "Practical Design". respectively. 1-9: Load transferred directly to the support Tjen N. it was demonstrated that it is a valuable tool for achieving proper detailing of ductile concrete members. The compressive strength of concrete. The anchorages of these bars are provided by means of standard 90' hooks. He is a member of ACI subcommittee 3 18E Shear and Torsion as well as joint ASCEIACI Committee 445 Shear and Torsion and its subcommittee 445-A Strut and Tie. f . Leonardo Flores in the preparation of the final version of the paper.A II < . 1996. A 20 ft (6. and the yield strength of the steel reinforcement. Lausanne) A simple strut-and-tie model shown in Fig..10 m) span deep beam was designed using the strut-and-tie method according to ACI 3 18-02 Appendix A. Farmington Hills.'. Tjen N. Bearing plates of 18 in... (1b-6).- Example lb: Alternative design for the non-slender beam (deep beam) . American Concrete Institute.'. 1 in. (2032 mm) deep and carries two concentrated factored loads.6 MPa) and 60 ksi (414 MPa). FIP Recommendations (1999): Practical Design of Structural Concrete. References Acknowledgements The authors gratefully acknowledge the participation and contribution of Mr. Michigan. Daniel A. . Sept. It was found that the strut and tie methodology provides a framework to understand and assess the flow of forces and the resisting mechanisms.: SETO.. 1999.ues IR Dimensions in inches. Also. *-*. Sept. Kuchrna Synopsis ACI 3 18-2002: Building Code Requirements for Reinforced Concrete and Commentary. Example 1b: Alternative design for the non-slender beam (deep beam) 81 II. (457 mm x 508 mm) are provided at all loading and support locations. are taken as 4 ksi (27. of 360 kips (1601 kN) each.. Publ. The self-weight was not considered in the design..Example 1 a: Deep beam design in accordance with ACI 3 18-2002 . Tjhin 6 Final comments The strut-and-tie model selected was readily analyzed. The provided reinforcement for the main tie is 2 layers of 5 #9 (#29 mm) bars. Kuchma is an Assistant Professor of Civil and Environmental Engineering at the University of Illinois at Urbana-Champaign. x 20 in. Daniel (Dan) A.-struts I . = 25. and the design and strength verification of its components was easily performed. (Distributed by:jb. Tjhin is a doctoral candidate in the Department of Civil and Environmental Engineering at the University of Illinois at Urbana-Champaign. London. 2002.4 mm Fig. (1b-2) is used in the design. The entire deep beam is a disturbed region because it is near statical discontinuities.80)(4000)= 2720 psi. The effective compressive strength of this node is f.5. A. it is only necessary to consider the left third of the structure to complete the design because the geometry and loading are symmetric about a vertical axis passing the midspan of the beam. the concentrated forces. The step-by-step design procedure is as follows: All Bearing Plates are 18" x 20" (457 mm x 508 mm) - Step 1: Check bearing capacity at loading and support locations. ( A 4 1 .8SP. 1b-1 : Structure and loading 3. The effective compressive strength of this node is limited to A. A. (1b-1).2 definition. [ACI Sec.' = 0.Step 6: Calculate the minimum reinforcement required for crack control. (A-8)] The nodal zone over the support locations is a compression-tension (CCT) node. = 360 kips Cross Sect~on V. = 18(20) = 360 in. The structure will be designed using the strut-and-tie method according to ACI 3 18-02 Appendix A.. = 360 kips - Step 7: Arrange the reinforcement.00)(4000)= 3400 psi.2 (232258 mm2). .85(0. However. . A. b = 20" (508 mm' V.2 eq.85(1. . The bearing stresses at points of loading and at supports are = 60 ksi (414 MPa) The nodal zone beneath the loading locations is an all-compression (CCC) node per ACI Sec.Step 2: Establish the strut-and-tie model and determine the required truss forces.8SP. [ACI Sec. = 4 ksi (27.5.1 Step 1: Check bearing capacity at loading and support locations Material strengths: f.Step 4: Design the nodal zones and check the anchorages - Step 5: Check the diagonal struts. within one section depth of the beam on either side of the discontinuity.82 Example 1b: Alternative design for the non-slender beam (deep beam) Example lb: Alternative design for the non-slender beam (deep beam) 1 Geometry and loads 2 Design procedure The structure and loading under consideration are shown in Fig.fc' = 0.f.. - Step 3: Select the tie reinforcement..5. i.e." = O.6 MPa) (normal-weight concrete) The area of bearing plate is A. = 0. f. 3 Design calculations Fig.2 eq. these strut and tie form a force couple. and therefore wt = 9. The horizontal position of nodes A and B is easy to define.85~.Bcjd = 0 V . and the width to anchor tie AD.. 2 5 and ~~ V.3. but the vertical position of these nodes must be estimated or determined.f.F. =80-1. wt must be minimum. = 360 kips (1b-4) Writing the moment equilibrium equation about point A as described in eq.~ ~ .. (1b-2) must reach the node capacity to anchor this To minimize wt tie force AD.(l b-3) Subtituting eqs. 3. the positions of these nodes have to be as close to the top and bottom of the beam. = 360 kips (1601 kN) ' Fig. which is defined in ACI Sec..75(2720) = 2040 psi (14. reach its capacity defined in ACI Sec. = 1. i.75(3400) = 2550 psi (17.5. (lb-1) gives wt = 1 . = 360 kips Fig. (80) . fi FU. lb-3: Free body diagram of the left third of the deep beam (1b-5) . V.. = 360 kips (1601 kN) f6"c V.= 0. / 2 . or Fu. V. w.. strut force BC.94 in. = @fCuA. As shown in Fig. )bw. = $(0.AD = = (Pfcu Ac = $(0. (lb-2) and (lb-3) into eq. A.e. the lever arm..58 MPa) at points of loading and @f. To minimize w. of the force couple must be set maximum. the area of bearing plates provided is adequate.95 in. jd.. (lb-3). = 0. where 0. = 360 kips jd = 8 0 . = 7. @f. In other words.BC = $F.')bw.84 Example 1b: Alternative design for the non-slender beam (deep beam) Because the bearing stresses are less than their corresponding limits.2. and this means that the width of strut BC. Strut BC and tie AD are required to equilibrate the truss.8 (CCT node).sc = F".2 Step 2: Establish the strut-and-tie model and determine the required truss forces ~ ~ . where P. F U . (1b-1) Fu.85~. (lb-2) is selected.0 (prismatic). FU. lb-2: Selected strut-and-tie model 1 (2032 mm) V . To fully utilize the beam. The truss consists of a direct strut AB (or strut CD) running from the applied load to the support.w 12 . = 0..2.07 MPa) at supports.125~~. = 360 kips 41 3 @ 80" = 240" (6096 rnrn) 6f(406 rn) V.AD.AD. (1b-5) and substituting eqs.. or A simple strut-and-tie model shown in Fig. must A. (lb-2) and (lb-4) into the equation give w. tie. AD = 360(80)171 = 406 kips (1 806 W). (102 mm) from the top of the beam and tie AD is located 1012 = 5 in. d = 80 -1012 = 75 in.. from bottom . ASt= 2(5)(1. (1806 kN) jd = 71" (1803 rnm) 3. represents the correction factor for excess of provided A. and F. : w 1 0 3 6.3 in. V.6" = 542 kips (241 1 kN). respectively. lb-5: Nodal zones A and B / w... As shown in Fig. The required anchorage length is F.2 layers of 5 #9 bars.3 layers of 6 #7 bars.1 layer of 6 # l 1 bars.2.. the stress in strut BC.812 -1012 = 71 in. = 360 kips 87 Example 1b: Alternative design for the non-slender beam (deep beam) Minimum tie reinforcement provided must satisfy @F. [ACI Secs.... (490 rnm).3. 1b-4: Strut-and-tie model dimensions and forces ..0 in. Because this is greater than 19.will ~ ~be anchored in just sufficient area. from bottom . @ 5 in. @ 2. (686 mm). This fixes the geometry of the truss and is illustrated in Fig. A.4.= 406 k required A.4 Step 4: Design the nodal zones and check the anchorages The 90' standard hook is used to anchor tie AD. ASt= 6(1. = 3601sin41. jd= 80 . the available development length is 27. (203 mm).. and w. (254 mm).9" (455 mm) . and wt just obtained are used for the dimensions of the struts and ties. ASt= 10. = 406 kips (1806 kN)..= @As. the anchorage length is adequate.00) = 10 in2. (lb-4). V. = 17.2. (lb-5) left. reinforcement. F u .36 in. ASt= 10 in.86 Example 1b: Alternative design for the non-slender beam (deep beam) If the values of w.2 requires that this development length start at the point where the centroid of the reinforcement in a tie leaves the extended nodal zone and enters the span... will be selected to be 10 in. w.6" 7 w = 10' L Nodal Zone A Fig. rnm) = 406 k For better steel distribution and for ease of anchorage length requirement.6" and F. In this design.. where h = < 7 '~ -1 1. and the force in tie AD.5 in... = 360 kips (1601 kN) Fig. @ 2. 5.5 and 7.56) = 9. and 8 in. ~will c be at its limit. Strut BC is located 812 = 4 in. will be selected to be 8 in.3 Step 3: Select the tie reinforcement ! Consider the following three steel arrangements: The angle and the force of diagonal strut AB are e = arctan(71/80)= 41.1 and A. ACI A.4. 3..Bc= F U.8 in.61 Thus. from bottom . F u . the required area of reinforcement for tie AD is .fy 2 F.. Therefore.2. choose 2 layers of 5 #9 (#29 mm) bars. (127 mm) from bottom of the beam.2 (6452 mm2). = 18" j-(457 mm) 8" (203 mrn) 8 = 41. 2 #9 Framing Bars Bearing Plates 18" x 20" (457 mm x 508 mm) (Typical) Horizontal Web Reinforcement #4 (#I3 mm) @ 12"(305 mm) Each Face (Typical) Vertical Web Reinforcement #5 (#I6 mm) @ 12" (305 mm) Each Face (Typical) . (493mm). 11.21 Because this is higher than the required force..3. = 0.6" =19. By assuming that sufficient crack control reinforcement is used to resist bursting force in the strut (P. and the force is [ACI Sec..equal to 0.0025.= 1. w. use #5 (#I6 mm) @ 12 in. minimum reinforcement provided must also satisfy 3. (305 mm) on each face over entire length.f.20)/20/12 = 0.41 and horizontal web reinforcement provided must be at least [ACI Sec.8.85)(0. 1b-6: Reinforcement details . For horizontal web reinforcement.6" +10cos41. = 0.6 and A.0026 > 0. 3.0015. the width at top of the strut is where yi is the angle between the axis of minimum reinforcement and the axis of strut.8.85P.. (305 mm) on each face over entire length.75 is used to calculate the strength of strut AB. J of .51 where s and s2cannot exceed dl5 or 12 in. .6" =17.1 eq. = 0. use #4 (#I3 mm) @ 12 in. = Q(0. strut AB (or CD) is adequate. A.5 Step 5: Check the diagonal struts From Sec. 11. r [ACI Secs.75) the capacity of strut AB is limited to QF".0015bs2.')bw.5 mm) [ACI Sec.5T (63..75(0.4in. = 542 kips (241 1 kN).6" +8cos41. A. and the width at bottom of the strut is w. For vertical web reinforcement. cos0 =18sin41. = I . 3. (A-4)] As shown in Fig.3.31)/20/12 = 0.(455 mm). (lb-6).5 #9 ~ (#29 mm) Bars (406 mm) (2032 mm) (1016 mm) Section A-A Vertical web reinforcement provided must be at least = 0.6: F.. Fig.9in.2.6 Step 6: Calculate the minimum reinforcement required for crack control A.~ 1 6 " ~ ~ 8 0 1 1 2~Layers .Example 1b: Alternative design for the non-slender bean1 (deep beam) 88 Example 1b: Alternative design for the non-slender beam (deep beam) 89 Because P. sin 0 + w. Based on the provided web reinforcement.75)(4)(20)(17. The reinforcement details are shown in Fig. AVh/bs2= 2(0. 5" (127 rnml 2. cos 0 =18sin41. the angle of strut AB (or CD) is 0 = 41. 3. sin 0 + w. A..0017 > 0.7 Step 7: Arrange the reinforcement Strut AB is expected to be a bottle-shaped strut.0025bs. A..2. (1b-5)./ bs = 2(0.i.9) = 685 kips (3047 kN).3.. Conventional B-regions in the dappedend beam are designed using conventional ACI beam design." The main steps in the design process of this deep beam involve defining the Dregion and the boundary forces acting on the region. Sanders received his Civil Engineering BS degree from Iowa State University and his MS and PhD degrees from the University of Texas at Austin.. for example parking structures.90 Example I b: Alternative design for the non-slender beam (deep beam) Example 2: Dapped-end T-beam supported by an inverted T-beam 91 4 Summary A design of a simply-supported deep beam under two point loads has been presented. He is ACI Fellow. Michigan. This design was completed using the provisions in ACI 3 18-02 Appendix A "Strut-and-Tie Models.e. David H.e.(2002): Building Code Requirements for Structural Concrete and Commentary. Sanders References ACI 3 18. the concentrated forces. within one section depth of the beam on either side of the discontinuity. In a number of applications. providing reinforcement to serve as the steel ties. Detroit. a D-region is formed. His research involves the behavior and design of concrete system with particular emphasis in bridges and seismic applications. The anchorage requirements were satisfied by the use of standard 90' hooks. Synopsis At the ends of dapped-end beams. ACI Committee 3 18. David H. i. dimensioning the struts and nodes. the transfer of load from the support into the beam is a D-region. visualizing a truss carrying the boundary forces in the D-region (i. The STM is excellent for modeling such a region. . A simple strut-and-tie model was employed in the design. This strut-and-tie model resulted in the use of 2 layers of 5 #9 (#29 mm) bars for the main tie.443 pp.. The following example used ACI 3 18-02 Appendix A to design both the end region of the dapped-end beam as well as the tie back and vertical reinforcement needed for each point load on the inverted T-beam. At each point where the load of the dappedend beam sets on the inverted T-beam. Example 2: Dapped-end T-beam supported by an inverted T-beam The entire deep beam is the D-region because it is near statical discontinuities. He is chair of ACI 341 Earthquake Resistant Concrete Bridges and a member of TAC. solving for the truss member forces. dapped-end beams are used in conjunction with inverted T-Beams. Particular attention was given to the anchorage of this main tie to ensure that it can carry the required force without having anchorage failure. American Concrete Institute. 2002. the strut-and-tie model). and providing distributed reinforcement for ductility. 2-1 : Geometry for T-beam ' Same Details and Dimensions The effective width be = b.2512 = 15..7 kN1m). 2-3. = 6. = 5. Step 2: B-region shear design.74" (19 mm) => Use 6 #9 (29 mm) bars. - The concrete strength is 5500 psi (38 MPa). 110 (2790 mm) Fig. The design positive moment (Mu) is 118 x 2.6 kN/m) based on 1. The following steps will be used.64 Wft (38. Therefore effective depth.000 psi (414 MPa).5 kN).5 .2 Dead Load + 1. The web is 10 inches (254 mm) wide and the section has an overall depth of 19 inches (483 m) (see Fig.6 kN/m) and the dead load including the superimposed dead load is 0. = 0. Each T-beam has a tributary width of 10 feet (3048 rnm). Step 4: Design and check capacities of struts. Step 6: Design and check capacities of struts. The T-beams are spaced at 10 feet (3048 mm) along the length of the inverted T-beam and span 32 feet (9900 mm). The concentrated support load will then be distributed into the inverted T-beam.1. 2-2).6 Live Load. The simply supported dapped-end T-beam will be designed for standard beam moment and shear. The inverted T-beam is also 19 inches (483 mm) deep and has a seat width of 7 inches (178 mm)(see Fig.375" (391 inm). The reinforcement has a yield stress of 60. d = 19 .9A. 3 7 5 (391 1" (25) 9" (229) mm) '3"'r 3. Step 3: Strut-and-tie model for dapped end layout. 13 ties (10 mm) 1.918 . Step 5: Strut-and-tie model for inverted T-beam. + 160) = 10 3. Therefore.1 Step 1: B-region flexural design The tension reinforcement was assumed to be 2 rows of #9 (29 mm) bars. The strut-and-tie model will be used to transition the forces from the beam section to the concentrated support load. 2-1). 16(5) = 90" (2286 mm) Setting Mu = 4183 in-kigs = (PM. Step 1: B-region flexural design.79 kN/mZ)and a superimposed dead load of 10 psf (0.92 Example 2: Dapped-end T-beam supported by an inverted T-beam Example 2: Dapped-end T-beam supported by an inverted T-beam 1 Geometry and loads 2 Design procedure A reinforced concrete dapped-end T-beam is supported by an inverted T-beam. ties and nodes 3 Design calculations 3.0 in2 (3871 mm2) Amount of tension reinforcement satisfies both minimum and maximum A. I #! 4-7' .1. The structure is a combination of D-Regions at the dapped end of the T-beam and near the load points of the inverted T-beam.625" (92 mm) (76) 1-32. A.48 kN/mZ). see Fig. The horizontal force at the support is taken as 10 kips (44.64 x 32S2 x 12 = 4183 in-kips (473 kN-m). There is a live load of 100 psf (4. The middle bars shown in Fig.87 k/ft (12.5 Fig.fy(d-a12) A. This produces a factored distributed load of 2. 2-2: Details of dapped end and inverted T-beam feet span (9900 mm) Fig. 1" (25) Cover= 1 5 (38 mm).318 . The remaining portions of the beams are B-Regions. ties and nodes.I !' 1 ! 1 5 .0 Wft (14. effective live load is 1.25" (32 mm) verllcal gap between I I I longitudinal bars.21 in2 (3363 rnrn ) and a = 0. 2-3: T-beam cross-section . 2-3 will be used as part of strut-andtie model in the dapped-end region and are not part of the standard flexural design. 94 Example 2: Dapped-end T-beam supported by an inverted T-beam Example 2: Dapped-end T-beam supported by an inverted T-beam 3. Tie EF was placed at two stirrup spacings away (12 in (305 mm)) from BC. OK Since s < dl2 = 7. The strut at the top of the section was assumed to be located at a depth equal to 10% of the overall depth. The distance between the centerlines of the top strut and the bottom tie is equal to (15.5 in (38 mm)) over the reinforcement.6 kips (194 kN) Horizontal Reaction = 10 kips (44.7) = 40. The distributed load is divided to the closest node by using the tributary width on each side of the node. The bearing surface is 4 in x 7 in (102 mm x 178 mm).22 in2 (142 mm2) and q = 6 in (152 mm). the tie at the bottom of the section is assumed to be located in the center of the longitudinal tension reinforcement. = 2. 1 (76) (102)(102) 1 1 (178) (mm or N) (127) Fig. On the right side of the strut-and-tie model. 2-4. Because of the very short development length.4 Step 4: Design and check capacities of struts. 2-4. A plate may not be necessary if a greater distance beyond the support or extra tie reinforcement was provided.9 kips (342 kN) Strut CD = 100 kips (445 kN) Tie CF= 64. See Fig.512 x 12 . From geometry I. The location of Tie AD permits sufficient room for an end plate and the cover requirements (1.2 Step 2: Shear design From statics ACI 3 18 (I) permits the design shear to be taken at a distance d from the face of the support.375 -0.9 kips (249 kN) V. This plate was also taken as 4 in x 7 in (102 mm x 178 mm).7 inches (1 96 mm) 3. StrutAB= 78. the design shear was determined at a distance d from the face of the short section (d = 7 inches (178 mm)).5 kN) If use #3 (I 0 mm) hoops. A.3 Step 3: Develop strut-and-tie model for the dapped-end beam The assumed strut-and-tie model is shown in Fig.9 kips (280 kN) Strut BE = 15. Tie BC will consist of several stirrups and therefore the centroid must be placed away from the end of the beam.64112 x (32.8 kips (101 kN) 95 Vertical Reaction = 43.1 Node A Node A. Current ACI does not allow a reduction in development length due to a high compression load on the bar.2 kips (67.2kips(348kN) Tie AD = 76.4 . 2-5). Because of the dapped end. In order to determine remaining geometry of the strut-and-tie model.48 in (342 mrn).3 kips (286 kN) Tie EF = 37. This could also be taken at the center of the compression zone as calculated in step 1. 2-4: Assumed strut-and-tie model 3. The potential development length for Tie AD extends to the point where Tie AD exits from Strut AB. d = 2 sqrt(5500) 10 x 15. Tie BC and Tie EF must be assumed. an end plate must be used on the end of the tie reinforcement to provide development.4.0 kips (164 kN) Strut DE= 55. ties and nodes 3. = 0.375/1000 = 22.1~19) 13.6 kN) Tie BC = 76. the location of Tie AD.5 kips (180 kN) VC=2 sqrt(f C) b . the connection of Strut AB and Tie AD at the end support.0 kips (338 kN) Strut BD = 62. must be checked (see Fig. . 5 (38) is required cover 2"(51) Fig.80 because there is one tie being anchored.(AD) => 75. The dimension given in Fig.8cos(36. use the web thickness for strut thickness (10 inches (25 mm)). This occurred because of the need to accommodate the reinforcement for Tie BC. p.. The bottom width of the node was set by the vertical reinforcement of Tie BC. The tie must be anchored at both the top and the bottom.003.84 in (123 mm) For Struts AB and BD.75 could be used if the #3 (1 0 mm) ties at 2 in (5 1 mm) were extended into the dapped end.69 in2 (1090 mm2) Use 3 #7 (22 mm) bars.8 x 5. P. The development length is measured from the far right boundary of the node towards the bearing plate.52 in (140 mrn) Assume that strut thickness equals beam width (10 inches (254 mm)).6 or 0.pC)60 => AS. 3.82 in (122 mm) Node Is smeared over all four #3 (10) stirrups.5) (4 x 7) = 79 kips (351 kN) > 75. . In Fig. AD = 1.75 (0.4.2 & 62.6 x 5. 2-5 from the bottom of the section to Tie CF is slightly different than the one in Fig.52 x10) = 116 kips ( 516 kN) >78.85 x 0. Struts AB and BD have 0.71 in2(1103 mm) Use 3 #5 (16 mm) stirrups bars AS.96 Example 2: Dapped-end T-beam supported by an inverted T-beam For the node.5) (4 x 7) = 79 kips (351 kN) > 43.3. The #5 (16 mm) stirrups have been placed 2 in (5 1 mm) apart creating a node face of 2 x 2 + 2 = 6 inches (152 mm). siny.BC= 1.9 kips = 0. In addition.3 Node C L 1 2"(51) L ! L 4"(102) L L L + 1 .75 (As. 0. the strut is headed towards Node D. Strut Capacity AB & BD = 0. Figure 2-5 shows the configuration of Node B.8 x 5.1 (CAsiIbsi siny.75 if sufficient stirrups are provided: CA.1.75 (0.85 x 0. cos0 = 74 sin(32.5) + 3.5) = 0.1) + 3. The ps for the struts is 0.75 (0. values of 0. this could be eliminated by moving the nodelcenter of the reinforcement to the right or by increasing the distance between the vertical reinforcement. the upper boundary of Strut BD is not show because the area is all compression. The plus 2 inches (5 1 mm) comes from allowing some (spacingl2) spreading of the tie width beyond the reinforcement.2 Node B For Node B. For Strut BE.6 kips (194 kN) OK Example 2: Dapped-end T-beam supported by an inverted T-beam 3.5) (5. Strut AB width = lbsin0 + h.80 in2(1 161 mrn2)(Could use more than one layer if wanted to shorten development length or increase node size. The BE strut will not control because the node has a lower than the strut. In Fig.80 because there is one tie being anchored.009 > > 0. ps can either be 0.). Tie BC = 76.9 kips (338 M\T) OK Tie Reinf. 2-5: Strut and tie details with proposed reinforcement ~nches(mm) For the node.86 in2(1200 mm) Tie Bearing Plate Capacity (A) = 0.85 x 0.13 and 7.AD)60 => AS.3. p.4. 6 inches (152 mm).75 (AS. cos9 = Strut AB width (perpendicular to line of action) = lb sinel j-ht cos0l Strut BD width = lb sin0 + h.AD= 1.75 by providing reinforcement that satisfies A3./bs. 6/2sin(32. As.003. ps is 1. Because the change was slight. is 0.1) = 4. So.5) = 5.8 x 10) = 101 kips (449 kN) > 78. which is a smeared node. Strut AB Capacity = 0.8 kips = 0.BC= 1. 6" (152) 3#5 (16) 1G +.6 x 5. 2 0.5) = 4.9 kips (348 kN & 280 kN) OK Development for Tie BC at Node B is governed by the stirrup development checks of ACI 3 18 12.221(10~2)x sin(90-32. The size of the struts is determined by the width of the Tie BC. is 0. is 0.5) (4.60 (bottled shaped) if reinforcement is neglected.003). 2-4 and used to calculate the angles.85 x 0. a bearing plate is shown. Support Capacity (A) = 0. Therefore ps could be increased to 0. 2 0.75 (0. = 0. AB=> 0.8cos(32. 2-5.60 because there are two ties being anchored. 2-5. 6/2sin(36.2 kips (348 kN) OK No strut reinforcement is needed.60 and therefore will control the design for compression. the angles and the forces were not recalculated. It will be shown that this is not necessary in order to provide capacity. For Strut AB.5) + kos(32. 5) (7.4.EF= 0.4. This is more consistent with the crack formation since it would allow the tension area to spread further into the beam.75 (0.0kips(338kN) Tie BC = 40.75 (As.9 inches (175 mm). This defines the left boundary of the node and the point from which Tie AD can be developed. 2-4).0 kips = 0. The 4 inches (102 mm) spacing provides a centroid at line EF and a distribution of stirrups between Node C and Node F. If all = 6. 99 tanB1 = 5. a strut (BD) crosses the crack. One such model is given in the FIP Recommendations "Practical design of structural concrete" (2) (see Fig. The width of the strut was assumed to be constant between Nodes C and D.EF)60 => AS. In this case. This is an acceptable strut-and-tie model solution.0 kips (169 kN) Tie CF = 33.4.85 x 0.0 in2(3871 mm2).3 in (719 mm) measured from left side of Node D.88 in2(567 mm2).1) = 6. (327 kN) > 64. In the first model (see Fig.EF= 0. The 6 #9 (29 mm) bars are extended into Node C then extra reinforcement will help with reducing the needed development length but it is still more than what is available.8 kips (1 50 kN) Vertical Reaction = 43.5) (5 x 7) = 73.3 and 3.6 kips (194 kN) Horizontal Reaction = 10 kips (44.CF= 1. the capacity checks at Node D are satisfied by the checks in 3.118 => el = 32.4. whereas in this model that compression area does not exist. In experiments.1" From statics Strut AB = 78. the value of Tie AD is the same as that shown in first model but the value of Tie BC and CF are much less in the FIP model. The distance available for developing Tie CF at the node is the distance from the concrete edge minus cover to where the tie leaves Strut CD: 6 .98 Example 2: Dapped-end T-beam supported by an inverted T-beam Tie CF Reinf. a crack forms in the comer to the right of the support.6 x 5. => 64.82 in2(529 mm2). In the alternative model. Use 4 #3 (10 mm) stirrups at 4 inches (102 mm) (AS.6 kN) .6.2 kips (286 kN) OK Strut CD width = lb sine t htcosO = 6sin(50. The struts coming into Node D are distributed (smeared). a 5-inch x 7-inch (127 mm x178 mm) bearing plate was added to the #9 bars (29 mm). Development length for AD => Id/db = 60000 (1)(1)(1)1(25 x sqrt 5500) => Id = 28.0 kips (338 kN) StrutBE= 76.6 Alternative models There are other potential models that have been developed for dapped-end beams.1) = 7.1)+4cos(50. The number of stirrups is based on the force in Tie EF.5 Nodes E and F Both nodes are smeared over multiple stirrups.85 x 0. 2-6: Alternative model From geometry 3. the only issue is the development of Tie AD. Therefore. Tie CF Bearing Plate Capacity = 0.4 Node D For Node D.5' tang2 = 8. 2-6). Strut CD Capacity = 0.75 (0.6 kips Example 2: Dapped-end T-beam supported by an inverted T-beam 3.2 x10) = 179 kips (796 kN) 100 kips (445 kN) OK 3.4 kips (180 kN) Strut CD = 52.6 x 5.5 + 2tan(50.2 in (1 82 mm) Assume that strut thickness equals beam width (10 in (254 mm)) and that P.2 kips = 0. = 0.3817 => O2 = 50.2 kips (348 kN) Tie AD = 76. P 1 1 1 3" 1 4" 1 4" 1 7" (mm or N) Fig. The width of a strut at Node D will be equal to or greater than the strut width at the other end (Node B and Node C). Tie EF = 37.7 kips (234 kN) Tie DF = 38.4.4.43 in2 (920 mm2).75 ( A s p ) 60 => AS.1. AB= 1.8 inches (96 mm).6 kips (296 W ) 56. . there is transverse tension that comes from the longitudinal bending of the inverted T-beam.61 x 10) = 118 kips (525 kN) > 71.75 (0. 2-8). normal beam theory will be used.4 x5. Use 4#5 (16 mm).5 kips (349 kN) > 43.5) (5. the strut thickness is the same as Strut AC (1 1.5) (4 x 7) = 78. Strut CD Capacity @ C = 0. therefore the height of the node is two times that or 3.60 because there would be two or more ties anchored in a three dimensional STM.4 because they are located in the tension zone of the member. 0.75 (0. The 3-inch (76 mm) spacing provides a distribution of stirrups over the T-beam seat width. 3. The bearing surface for the stem of the T-beam is 4 inches x 7 inches (127 mm x 178 mm). Fig.6 kips (194 kN) OK Tie Bearing Plate Capacity = 0. 0.3 kips (273 kN) > 56.75 (0.80 in2 (1 161 mm2) Strut AC width (perpendicular to line of action) = &sine + h. AS.6 kips (796 kN) OK Tie AB Reinf.85 x 0. The width was taken as the bar size plus 1 inch (25 mm) on either side.2 Node C For Node C. For Strut AC and CD.75 (AS& 60 => AS. ps can either be 0. This plate was taken as 4 inches x 7 inches (102 mm x 178 mm).1)/6.6 kips (252 kN) Example 2: Dapped-end T-beam supported by an inverted T-beam Support Capacity = 0. ps would be 0.6 kips = 0.48 in2. => 66.8 x 5.85 x 0.85 x 0. Because of thevery short development length.80 because there is one tie being anchored. 2-7: STM for inverted T-beam for struts and ties Fig.1 Node A The resultant of Strut AC and Tie AB at the inverted T-beam support must be checkedldesigned (see Fig.61 in (142 mm) Assume that strut thickness equals the section width 10 inches (254 mm).75 if sufficient stirrups are provided.100 Example 2: Dapped-end T-beam supported by an inverted T-beam 3.5)=74.CE= 1.5 kips (3 18 kN) OK 3.62) sin(37. The strut thickness is equal to the width of Tie CE reinforcement in the longitudinal axis of the inverted T-beam (9")plus half the spacing (1.5) (3. The location of Node C was taken at O. ties and nodes for an inverted T-beam. 2-8: STM details 3.6) + (3.8 x 11. Therefore P.5)(4. Strut AC Capacity @ A = 0.6 kips = 0.85 x 0.5 inches (292 mm)).6 Step 6: Design and check capacities of struts.8 x 5. Along the longitudinal axis of the inverted T-beam.8 inches (96 mm).6. Use 3 #7 (22 mm) bars.6 degrees From statics Strut AC = 71.AB)60 => AS.6) = 4.625) =z 0 = 37.4 x 5.5) = 61.5) (4 x 7) = 78. Distribute as shown in Figure 2-9. The width of the node is governed by the vertical reinforcement (Tie CE).6 kips (252 kN) OK The node does not control the design because of the larger P value in comparison to the struts.5 kips (318 kN) Tie CE = 43.5 Step 5: Strut-and tie model layout for inverted T-beam Fig. Strut AC Capacity @ A = 0. For Strut AB.61 in (1 17 mm).8)cos(37.75 (0. AS.lh above the bottom of the beam. Tie CE Reinf.5 kips(318 kN) OK For Strut CD.6 x 5.61 x 11. It is a common practice to use this type of plate and to weld the bars to the plate.6 or 0.5" (38 mm)) on each side of the reinforcement (1 1.97 in2. Currently this type of guidance is not given in the Appendix. => 43.4 kips(331 W ) > 71. 2-7 shows the assumed strut-and-tie model. CASi/bsisinyi.CE= 0.75 (0.6. For the node. is 0.6) + 4cos(37.003.5 inches (292 mm)) and the width is 1. The strut width of Strut AC is equal to lbsine + htcose 7 (2. is 0.5 kips (349 kN) > 66.85 x 0.AB= 1.cose = 4 sin(37. an end plate must be used.6) = 5.75 (AS. From geometry: tane = (5.6 kips (194 kN) Tie AB = Strut CD = 66. Bars will be at 3 inches (76 mm) on center in the longitudinal axis of the inverted T-beam.9 x 2 = 3.24 in2 (800 mm2). (Distributed by:jb. kips (kN) V. psi (MPa) effective height of a tie. kips (kN) V. in (mm) factored moment at a section. London. Sept. Publ. 2-9: Reinforcement details for inverted T-beam 4 Closure The example shows the ability of the provisions of Appendix A to model the flow of forces through a structure. in-kips (kN-m) s = spacing of stirrups along longitudinal axis of the member. degrees I I I ' Example 2: Dapped-end T-beam supported by an inverted T-beam 05aa (1 3) (76) (76) (76) (1 3) Fig. in2 (mm2) area of surface reinforcement in the ith layer crossing a strut area of reinforcement in Tie IJ. in (mm) area of nonprestressed tension reinforcement. 1996. The models provide an engineer with a rational design tool for portions of a structure where standard beam theory does not apply. = nominal shear strengh provided by concrete. = factored shear load. in (rnm) distance from extreme compression fiber to centroid of tension rebar (effective depth) . 1999. in (mm) specified compressive strength of concrete. psi (MPa) specified yield strength of nonprestressed reinforcement. kips (kN) p. in (mm) effective flange width of a T-beam based. = nominal shear strength at a section.5?+&-$3f' 103 SI = spacing of reinforcement in the ith layer adjacent to the surface of the member. = angle between the axis of a strut and the bars in the ith layer of reinforcement crossing that strut. = factor to account for the effect of the anchorage strength of a nodal zone ps = y. Sept. Lausa~me) . factor to account for the effect of cracking and confining inforcement on the effective compressive strength of the concrete in a strut 6 References 5 Notation depth of equivalent rectangular stress block.102 Example 2: Dapped-end T-beam supported by an inverted T-beam - Section A-A (I 6 ) af 3 I L " (76) A I o. FIP-Commission 3 "Practical Design".: SETO. in2 (mm2) width of a member. in (mm) V. in Imm) American Concrete Institute (2002): Building Code Requirements for Structural Concrete (ACI 3 18-02) and Commentary (ACI 3 18R-02). degrees @ = strength reduction factor B = angle between two struts or ties at a node. kips (IcN) V. in2 (mm2) area of shear reinforcement within a distance s. = nominal shear strength provided by shear reinforcement. Appendix A FIP Recommendations (1999): Practical Design of Structural Concrete. in (rnm) length of the bearing surface. 1-3) is selected to satisfy the code requirements. Normal-weight concrete with a specified compressive strength. of 1 1. (102 mm) from the face of the column. Kuchma is an Assistant Professor of Civil and Environmental Engineering at the University of Illinois at Urbana-Champaign.2 kips (49.Example 3. f .24. (3. A horizontal tensile force. Nu. The yield strength of reinforcement. (3. He is a member of ACI subcommittee 3 18E Shear and Torsion as well as joint ASCEiACI Committee 445 Shear and Torsion and its subcommittee 445-A Strut and Tie. These bars are welded to a structural steel angle of 3% in.12). Tjhin Daniel A. (356 mm) square column is designed using the strut-and-tie method according to ACI 3 18-02 Appendix A. Daniel (Dan) A.2 kips (250 kN) acting at 4 in. Tjen N. (89 mm x 89 mm x 13 mm). The main tie reinforcement provided is 5 #4 (#I3 rnm) bars. x 3% in. accounting for creep and shrinkage deformations. (3. of 56. aid. The corbel is to support a precast beam reaction force. A simple strut-and-tie mode1 shown in Fig. The structure and the loads are described in Fig. The corresponding shear span to depth ratio. (3. x % in. is 0.1-5). V. His research interests include nonlinear analysis and design of concrete structures. is taken as 60 ksi (414 MPa). The reinforcement details are shown in Fig.1: Corbel at column Example 3. .8 kN) is assumed to develop at the corbel top. Tjhin is a doctoral candidate in the Department of Civil and Environmental Engineering at the University of Illinois at Urbana-Champaign. The selected corbel dimensions including its bearing plate are shown in Fig. Kuchma I Synopsis A single corbel projecting from a 14 in. of 5 ksi (34.1: Corbel at column Tjen N.1-1). f.5 MPa) is assumed. The design is summarized as follows: . Section A . C (49..58 MPa).5 MPa) (normal-weight concrete) [ACI Sec. x 6 in. ACI Sec. .1 requires a span-to-depth ratio.2).Step 1: Determine the bearing plate dimensions. . - Step 6: Design the nodal zones and check the anchorages. the bearing size is adequate. .2 eq. Choose an overall corbel depth at column face of 18 in. No.2(1000)/72 = 781 psi (5.2 Step 2: Choose the corbel dimensions To be able to use ACI Appendix A.1-1). ACI Sec.Step 5: Select the tie reinforcement. (A-8)] (4 14 MPa) Choose a 12 in.9. Step 7: Check the struts. 3.9. (229 mm) at the free end of the corbel.5. The structure will be designed using the strut-and-tie method according to ACI 3 18-02 Appendix A. ald. = 0.1 Step 1: Determine the bearing plate dimensions The nodal zone beneath the bearing plate is a compression-tension (CCT) node. of less than 2. 1 1. In addition.1-2) summarizes the selected dimensions for the corbel.75(3400)= 2550 psi (17.1: Corbel at column 1 Geometry and loads 2 Design procedure The corbel to be designed and its loads are shown in Fig.Step 8: Calculate the minimum reinforcement required for crack control. The corresponding effective compressive strength is Material strengths: f. The bearing plate area is 12(6) = 72 in..1 : Corbel at column Example 3. of. Fig. (305 mm x 152 mm) bearing plate. A. Because this is less than the bearing stress limit.Example 3. To satisfy this requirement.2 (46452 mm.2 kips . (457 mm).1.8 kN) - Step 3: Establish the strut-and-tie model.38 MPa).' =5 ksi f.= 11. . +I 4"----4 (356 mm) Step 9: Arrange the reinforcement.A 3 Design calculations Fig.Step 4: Determine the required truss forces. The entire structure under consideration is a D-Region because it has abrupt changes in geometry and is in the vicinity of concentrated forces.e.2 requires that the depth at the outside of the bearing area is at least one-half of the depth at the column face.. (3. (3. 3. select a depth of 9 in. 11.1: Corbel geometry and loads 3.Step 2: Choose the corbel dimensions. and the bearing stress is 56. = 60 ksi (34. i. 1.DD~ into eq.8 -64.3 Step 3: Establish the strut-and-tie model To consider load eccentricities and erection tolerances.4 in.1 : Corbel at colunln Example 3. (79 mm). = 2550(14)w.2.1-1).1-1) and then solving it yield Fu.1-2: Selected corbel dimensions 3.6 in.Example 3.F .11 The horizontal tie DA is assumed to lie on the horizontal line passing through the sloping end of the corbel. A simple strut-and-tie model is selected. A. the position of V. (417 mm). Thus. (3. 3. Force (kips) CD CB BD BA DA DD' -60. toward the outer edge of the corbel from center of bearing plate. 3.7 +11. = 2550 psi per ACI Sec.. is the required compressive force in strut DD' and b is the out-of-plane dimension of the corbel. d = 18 . Fig. A negative sign indicates that the member is in compression.9. Thus..5. is shifted 1 in.1.. /I000 = 3 5 . 7 ~ ~ .. Thus.. of concrete cover.1 : Truss forces -- Member where F. The center of tie CB is assumed to be located 1. A positive sign indicates that the member is tension.DDs=I11 kips (494 kN) and w.2 +54.. which can be obtained by taking moments about node A as follows: Fig. and . from the top of the corbel. [ACI Sec. 10 in.1 : Corbel at column Substituting F. the new position from the face of column is 1 + 612 + 1 = 5 in. The position of strut DD' centerline is found by calculating the strut width w..1-3: Selected strut-and-tie model 3. node D is also a CCT node. Like the node beneath the bearing plate (node C). = @fc.bw.8 kN) (-494 kN) .=3.1-3). (3. its stress is limited to @f.6 = 16. considering one layer of steel bars and approximately 1 in. Table 3.. 11. This fixes the geometry of the truss.6 -1 11 (-271 W) (+I 55 kN) (-288 kN) (+243 kN) (i49. The geometry is given in Fig.9 +34.4 Step 4: Determine the required truss forces The required forces in all of the members of the truss are determined by statics and are shown in Table (3. . i. x 3% in. @fc.1-5). The required width for strut DD ' was determined in Sec.51 Choose 4 #4 (#I3 mm) bars. (A-2)] f.Thus. A. (79 mm). all the strut widths fit into the outline of the corbel region.. A. i.10 in. A. (3. this solution is accepted.3.1 eq. Therefore.. = 0. A. [ACI Sec.2. F.= 2(2)(0. To anchor tie CB.11) = 0.1-4).11 The nominal strength of strut CD is limited to The provided steel area must be at least [ACI Sec.97 in.Ac. 11. Other struts will be checked by computing the strut widths and checked whether they will fit within the space available. minimum reinforcement will be provided.9.b. = 14(2. 3.44i11. As shown in Fig. F.3 to satisfy the stress limit on the nodal zone.75. From Fig. QF. strut CD is adequate.2. Hence.~(284 mm2). x % in. Thus.20) = 0.2. [ACI Sec. the factored load of strut CD is 60.1 : Corbel at column Example 3. the calculations are presented in the next section. From Table 3. The details are shown in Fig.1-l). A.e. 3. only nodal zone C is checked in this section. (3.5.2. of nodal zone D was chosen in Sec. w. Because this is less than the limit. The area of reinforcement required for tie DA is [ACI Secs. tie BA has a larger tension than tie CB. the required width for strut BD is = 3 188 psi. must be at least equal to -c 34'8(1000) = 0. A. The effective compressive strength of strut BD is also limited to f.86) = 40.11 Choose 2 #3 (#lo mm) additional column ties at location DA. (A-3)] where = o.75(128) = 96 kips (427 kN). (5 1 mrn) on center. See Fig.04)/1000 = 128 kips. continue the 4 #4 bars down the column just to have a sufficient development length.. As indicated in Table (3.04in.e. [ACI Secs. weld the 4 #4 bars to a steel angle of 3% in.2 eq. These bars are spaced at 2 in.6 and A.3.5 Step 5: Select the tie reinforcement 3. the effective tie width. . (51 mm) width for strut BD...85(0. However.6 Step 6: Design the nodal zones and check the anchorages The width w. (25 mm). Therefore.2550(14) LAC1 Secs.Example 3.1-1. is the smaller area at the two ends of the strut. 3..3.7 Step 7: Check the struts The area of reinforcement required for tie CB is Strut CD will be checked based on the sizes determined by nodal zones C and D. Because @..z. = 4(0.2.6 and A.2 in (81 mm).. A . = 3 188(40.8sPsfc' = 0. To satisfy the stress limit of nodal zone C. [ACI Secs.' (516 mm2).6 and A. (3.1: Corbel at column 3.80 in.4...11 This limit is easily satisfied because the available tie width is 2(1. A. A.6) = 3. is assumed to be 0. = fc.9 kips (271 kN). 3.6 and A.75)(5000)= 3188 psi and A.. (3.11 Choose 2 in.1-4). (89 mm x 89 mm x 13 mm).4. this tie force should be resisted by column longitudinal reinforcement.1-4). According ACI Sec. 11. Based on the provided reinforcement and the angle of strut BD.3. (89 mm) spacing. (267 mm) from tie CB.1-4: Dimensions of strut-and-tie model components . Use 10.-A. provide 3 #3 ( # l o mm) closed stirrups at 3.with average spacing of Because 0.e. The area of these ties must exceed (81 mm) Nodal Zone C [ACI Sec.112 Example 3.4) = 10. i.5 in. bs.5(A3.41 A .0030. equal to 0. 5 ( ~.5 in. 11. 11. has to be greater than 40' because only horizontal reinforcement is provided.. of ACI Sec. = 3(2)(0. the minimum area required is A.4 requires closed stirrups or ties parallel to the reinforcement required for tie CB to be uniformly distributed within 213 of the effective depth adjacent to tie CB. and ASt A. t 0. crack control ACI Sec. = 0.3.513 = 3.3.66 in.5 in. (A-4)] where y.2. 3.) Try 3 #3 closed stirrups. is the area of reinforcement resisting the tensile force Nu.5 in.A . Hence.11)= 0. the smallest angle between strut and minimum reinforcement.=.e.9. where A .9.1 : Corbel at column Example 3. minimum reinforcement provided must also satisfy x L s i n y .8 Step 8: Calculate the minimum reinforcement required for. A. "). i. 213 (16. Because this amount of reinforcement satisfies both requirements.75 is used for the diagonal struts.9 in. Fig.3.. = 0 . y. 10.9.. [ACI Sec. A. distributed over a depth of 10.1 : Corbel at column 3. is the angle between the axis of minimum reinforcement and the axis of strut.1 eq. A. In this design. 3.. The design of a single corbel has been presented. The entire corbel is the D-region because there exist statical discontinuities. A simple strut-andtie model was employed in the design. and providing distributed reinforcement for crack control and ductility.9+. dimensioning the struts and nodes. Sufficient anchorage of this tie is required to ensure that it can carry the required force without having anchorage failure. This strut-and-tie model resulted in the use of one layer of 4 #4 (#I3 mm) bars for the main tie.9 Step 9: Arrange the reinforcement 4 Summary The reinforcement details are shown in Fig. This design was completed using ACI 3 18-02 Appendix A "Strut-and-Tie Models" and the ACI 3 18-02 Sec. Michigan. the entire corbel is the D-region. 1 1.1-5: Reinforcement details The main steps in this design involve defining the D-region and the boundary forces acting on the region.1: Corbel at column 3.1-5). providing reinforcement to serve as the steel ties. i. 4 ----14u ---- (229 mrn) (356 rnrn) L-910 J w 4 (229 rnm) (356 mrn) Section A -A Fig. Note: Column reinforcement is not shown Elevation View . This is achieved by welding all of the main bars to a structural steel angle. Detroit. and a simple strut-and-tie model was employed. ACI Committee 3 18. (3. .5" (89 mm) 3 #4 Framing Bars 2 #3 Hoops @ 2" b-.9 "Special Provisions for Brackets and Corbels". 2002.1: Corbel at column 4 #4 (#I3 mm) Main Bars References ACI 3 18-02: Building Code Requirementsfor Structural Concrete and Commentary. solving for the member forces of the strut-and-tie model. and geometrical discontinuities within one section flexural depth of the corbel on either side of the discontinuity. 12" x 6" (305 rnrn x 152 mm) Bearing P l a t e l 4 #4 (#I3 mrn) Main Bars Welded to Steel Angle 3 #3 (#I0 mrn) Hoops @ 3.- Example 3.443 pp.e. American Concrete Institute. the concentrated forces.Example 3. selecting a strut-and-tie model carrying the boundary forces in the D-region. ald.Example 3. Normalweight concrete is assumed.6 MPa) and 60 ksi (414 MPa).6 kN) is assumed to develop at each side of the corbel top. of 14.2: Double corbel Tjen N. The compressive strength of concrete. (356 rnm) square. Tjen N. (102 mm x 102 mm x 13 mm). (152 mm) from the face of the column at both ends. Kuchrna Synopsis A double corbel projecting from an interior colunln is designed using the strutand-tie method according to ACI 3 18-02 Appendix A. f. The column is 14 in.38. The selected dimensions including the bearing plates are shown in Fig. A simple strut-and-tie model shown in Fig. He is a member of ACI subcommittee 3 18E Shear and Torsion as well as joint ASCEIACI Committee 445 Shear and Torsion and its subcommittee 445-A Strut and Tie. (3. Tjhin Daniel A. . Tjhin is a doctoral candidate in the Department of Civil and Environmental Engineering at the University of Illinois at Urbana-Champaign. P. x 4 in.2-2). To account for beam creep and shrinkage deformations. Daniel (Dan) A. His research interests include nonlinear analysis and design of concrete structures. (3.3 kips (63.2: Double corbel Example 3. are taken as 4 ksi (27. of 275 kips (1223 kN). is 0. The corbel transfers precast beam reaction forces. a factored horizontal force. The reinforcement details are shown in Fig. x % in. The shear span to depth ratio. The anchorage of these bars is provided by welding each end of the bars to a structural steel angle of 4 in. respectively.8 kips (275 kN) acting at 6 in. f . The provided main tie reinforcement is 7 #4 (#I3 mm) bars. Kuchma is an Assistant Professor of Civil and Environmental Engineering at the University of Illinois at Urbana-Champaign. (3.2-5).2-3) was used for the design. Nu. and yield strength of steel reinforcement. Vu of 61. The upper column carries a factored compressive axial load.