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53244919 Flexural Comparison of the ACI 318 08 and AASHTO LRFD Structural | Strength Of Materials | Bending
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Flexural comparison of the ACI 318-08 and AASHTO LRFD structural concrete codes
Nathan Jeffrey Dorsey
Dorsey, Nathan Jeffrey, "Flexural comparison of the ACI 318-08 and AASHTO LRFD structural concrete codes" (2008). Master's Theses. Paper 3562. http://scholarworks.sjsu.edu/etd_theses/3562
A Thesis Presented to The Faculty of the Department of Civil Engineering San Jose State University
by Nathan Jeffrey Dorsey May 2008
UMI Number: 1458121 Copyright 2008 by Dorsey, Nathan Jeffrey
UMI Microform 1458121 Copyright 2008 by ProQuest LLC. All rights reserved. This microform edition is protected against unauthorized copying under Title 17, United States Code. ProQuest LLC 789 E. Eisenhower Parkway PO Box 1346 Ann Arbor, Ml 48106-1346
©2008 Nathan Jeffrey Dorsey ALL RIGHTS RESERVED .
APPROVED FOR THE DEPARTMENT OF CIVIL ENGINEERING j/jfl^LDr. McMullin 1 Mr. Daniel Merrick APPROVED FOR THE UNIVERSITY £<? / Au0Lr~**~~ Vi/zo/os' . Akthem Al-Manaseer Dr. Kurt M.
a safe working design of a structural concrete section. however there are fundamental differences between the codes regarding the calculation of member properties. Parametric studies were conducted using software developed by the author specifically for this purpose.ABSTRACT FLEXURAL COMPARISON OF THE ACI 318-08 AND AASHTO LRFD STRUCTURAL CONCRETE CODES by Nathan Jeffrey Dorsey There are two prevailing codes utilized during the design of structural concrete members in North America. . This study focused on the code's influence during the classical approach to analysis of a shallow reinforced T-beam and the strut and tie model method as applied to a series of deep beams with openings. It was concluded that both codes provided similar and safe results regardless of the very different approaches to solution taken by each. The purpose of this paper was to investigate these differences between codes encountered during the calculation of flexural strength or moment capacity of a section. ACI 318-08 and AASHTO LRFD. Each takes a unique approach to achieve the same result.
Al-Manaseer gave me full academic freedom during the course of this research and by doing so provided me with a tremendous learning experience that went far beyond reinforced concrete. And I would especially like to thank my lovely wife Jaya for always believing in me and for making sure that I always believed in myself.ACKNOWLEDGEMENTS I would like to thank Dr. Dr. v . Akthem Al-Manaseer for his patience and support during this process.
0 LITERATURE REVIEW 2.1 Flexure 5.5 Node Effectiveness Factors f3„ 5.2.2.0 INTRODUCTION 1.1.2 STM 5.1.1 Stress Block Depth Factor^ 4.3 Deep Beams 3.2.2 Scope 1.2 Hand Calculations 4.3 Strength Reduction Factor ^ 4.3 Location of Neutral Axis c.2 ACI 318-08 4.1.4 Strut Effectiveness Factors f3s 4.0 THEORY OF FLEXURE 3. ACI 318-08 versus ASHTO LRFD 5.2.2.2 ACI 318-08 vs.1.1 Introduction 2.2.1.1 Asmin m&Asmax.2.1. and Depth of Compressive Block a 4.1.2 Location of Neutral Axis c.3.1 General 1.Table of Contents 1.2 Location of Neutral Axis c and Depth of Compressive Block a 4.2.2.1 Shallow Beams 4.3 AASHTO LRFD 4.1. AASHTO LRFD 4.0 RESULTS and DISCUSSION 5.2 As versus Mu 5.2. and Depth of Compressive Block a 5.3 Strength Reduction Factor <f> 4.1.1.3.1.2.1.1 Introduction 4.3.1.STM versus FEA 1 1 1 1 2 3 3 4 24 37 37 39 43 48 48 50 51 51 52 54 59 59 60 62 66 66 69 70 70 72 72 73 73 76 76 77 83 90 92 98 99 vi .0 ANALYTICAL PROCEDURE 4.1.2 Theory of Flexure in Shallow Reinforced Concrete Sections 3.1 Limits of Reinforcement 4.3 Excel Spreadsheets 4.1 Maximum Allowable Concentrated Load .1 Graphical Solution 4.2 Shallow Beams 2.3 Objective 1.2 STM 4.3 Theory of Flexure in Deep Beams 4.4 Strength Reduction Factor ^ 5.3 Strength Reduction Factor <j> 4.4 Outline 2.1 Background 3.1 Limits of Reinforcement 4.2.1.1.
2 vii .2.2.2 As versus Mu for Flexure 104 6.Maximum Allowable Concentrated Load without $.2.5 Maximum Load Capacityfor STM 105 Works Cited 106 Appendix A .2 Conclusion 104 6.STM versus FEA 100 6.1 Asmin and Asmax for Flexure 104 6.0 SUMMARY and CONCLUSIONS 103 6.4 Strength Reduction Factor <f> for Flexure 105 6.1 Summary 103 6.2.Deep Beam STM Models 108 Appendix B .2.Deep Beam STM Truss Models 113 Appendix C .2.3 Depth of Neutral Axis c for Flexure 104 6.STM Truss Free Body Diagrams & Solutions 119 5.
2 . maximum test load comparisons 102 viii .1. RA. R9. RA.D-regions and discontinuities 45 Figure 3-4 ACI 318-08 Fig. 2002 ACI codes and AASHTO LRFD and Naaman recommendation 18 Figure 2-3 ACI 318-05 and AASHTO LRFD limits of reinforcement 22 Figure 3-1 Beam in flexure with saddling effect 38 Figure 3-2 Graphic representation of forces within a reinforced concrete section 42 Figure 3-3 ACI 318-08 Fig.1 (a and b) .List of Figures Figure 2-1 AASHTO LRFD Procedure for calculation of sx 7 Figure 2-2 Summary of reinforcement limits by 1999.Strength reduction factor 54 Figure 4-4 ACI 318-08 calculation of strength reduction factor ^ 55 Figure 4-5 Sample excel spreadsheet analysis 56 Figure 4-6 Sample excel spreadsheet showing preliminary ACI 318-08 calculations 57 Figure 4-7 ACI 318-08 excel program logic flowchart 58 Figure 4-8 Determination of AASHTO LRSYi Asmca 62 Figure 4-9 AASHTO LRFD and ACI 318-08 limits of reinforcement 63 Figure 4-10 Sample excel spreadsheet for AASHTO LRFD analysis 64 Figure 4-11 AASHTO LRFD flowchart 65 Figure 4-12 Generic beam with and without opening 67 Figure 4-13 Strut and tie modeling procedure 75 Figure 5-1 Graphic representation of section geometry 76 Figure 5-2 Asmax andAsmin as a function of b 79 Figure 5-3 Asmax andAsmin as a function of bw 80 Figure 5-4 Asmax and Asmi„ as a function of d 81 Figure 5-5 Asmax and Asmin as a function of tf 82 Figure 5-6 As versus Mu for variable b 86 Figure 5-7 As versus Mu for variable bw 87 Figure 5-8 As versus Mu for variable d 88 Figure 5-9 As versus Mu for variable tf 89 Figure 5-10 As versus c for variable b 94 Figure 5-11 4* versus c for variable bw 95 Figure 5-12 As versus c for variable d 96 Figure 5-13 As versus c for variable tf 97 Figure 5-14 Generic beam with and without opening 98 Figure 5-15 Predicted load vs. maximum test load comparisons 102 Figure 5-16 Predicted load without ^ vs.Hydrostatic nodes 47 Figure 4-1 Section geometry 49 Figure 4-2 IF logic statement for computation of stress block depth factor /?/ 50 Figure 4-3 ACI 318-08 Fig.1.5 (a and b) .3.
Table 5-3 Results for location of neutral axis Table 5-4 Cutout locations from lower left corner in inches Table 5-5 Predicted load versus maximum test load comparisons Table 5-6 Predicted load without ^versus maximum test load comparisons 49 67 74 77 83 91 98 99 101 ix .List of Tables Table 4-1 Values for variable and fixed geometry for each case Table 4-2 Cutout locations from lower left corner in inches Table 4-3 ACI 318-08/AASHTO LRFD effective strength coefficients Table 5-1 Rates of change for ACI 318-08 and AASHTO LRFD Asmin and Asmax Table 5-2 Ratio of AASHTO LRFD to ACI 318-08 for Mu per unit A.
positive if tensile.9d modulus of elasticity of the material.9d depth from the extreme compressive fiber to the centroid of the tensile force in the prestressed reinforcement depth from the extreme compressive fiber to the centroid of the tensile force in the non-prestressed reinforcement. can be taken as 0. i. taken as a positive quantity not less than Vud„ efficiency factor factored axial force.List of Variables a a' Ac' Aps As As As As' Av b bv bw c d D dc de d„ dp ds dv E Ec Ep Es f*c f'c fps fpu fsy fy kz Mu n Nu effective depth of the compressive block distance from center of concentrated load to edge of support area of concrete on flexural compression side of member area of prestressed longitudinal reinforcement placed on the flexural tension side of the section area of flexural reinforcement area of non-prestressed longitudinal reinforcement placed on the flexural tension side of the section area of conventional non-prestressed tensile reinforcing steel area of conventional non-prestressed compressive reinforcing steel cross sectional area of provided stirrup reinforcement within distance s section width effective web width web width distance from extreme compressive fiber to neutral axis distance from the extreme compressive fiber to the centroid of the flexural reinforcement overall member depth strut width depth from the extreme compressive fiber to the centroid of the tensile force in the non-prestressed reinforcement. negative if compressive x .e.e. reinforcing bars also called the effective depth of main flexural reinforcement effective shear depth. taken as 0. reinforcing bars effective shear depth. commonly referred to as Young's Modulus elastic modulus of concrete modulus of elasticity of prestressed reinforcement modulus of elasticity of non-prestressed reinforcement effective concrete strength compressive concrete strength stress in the prestressing steel at nominal bending resistance tensile strength of prestressing steel yield strength of main longitudinal reinforcing steel yield strength of conventional non-prestressed reinforcing steel ratio of the in situ strength to the cylinder strength factored moment. i.
parallel to strut strain in the horizontal direction equivalent strut width over which ties contribute angle of strut to horizontal angle of strut to longitudinal axis a stress xi .s tf Tu V Vc Vs Vu w z s Si £2 Ex Q 9 6 6 stirrup spacing flange thickness factored torsion shear force shear strength contribution from concrete shear strength contribution from reinforcing steel factored shear node width over which shear force is applied distance between node centers strain major principal strain. normal to strut minor principal strain.
2 Scope The crux of this thesis is the comparison of the two prevailing concrete design codes regarding the design and detailing of concrete beams in pure flexure with no other loading present. ACI 318-02 and AASHTO LRFD 3 rd Ed.3 Objective The objective of this thesis is to clarify the differences between the two prevailing concrete design codes. no laboratory experiments were carried out. To accomplish this a series of Excel spreadsheets were created to ensure the accuracy and consistency of all calculations performed.1 1.0 INTRODUCTION 1. Comparisons were based on but not limited to maximum predicted allowable load. Both sections focus solely on analytical results. vet and utilize these tools.1 General The purpose of this thesis is to compare and contrast the ACI 318-08 and the AASHTO LRFD 3rd Ed. 1. however as programming is not the point of this research several simplifying assumptions were made to reduce the time required to create. 1. The discussion on shallow beams used a series of twelve flanged beams as its focus while the deep beam discussion focused on a series of five deep beams and the Strut-and-Tie Model (STM) method. code provisions and design philosophies. and categorize them as .
minor. Chapter 6 provides the concluding remarks on the findings of this study. Chapter 3 provides a brief introduction and review of the design philosophy of both flexural analysis for shallow reinforced concrete beams and the STM method of design for reinforced concrete deep beams. . A comprehensive literature review providing coverage of examples illustrating additional differences found between the ACI 318 and AASHTO LRFD codes beyond pure flexure and deep beams is included. is referenced.4 Outline Chapter 2 contains a literature review of relevant academic and industry articles regarding a specific structural member type. a specific concrete design code or a comparative study of both codes. In cases where other editions are referenced. shall be omitted. the edition under discussion shall be noted.2 major. Chapter 4 details the analytical procedure as described by each code and what methods were implemented by the software tools developed. For simplicity when the AASHTO LRFD 3r Ed. 1. An additional comparison will be made between the results produced by the two codes when using the method of STM and the actual experimental test loads used during experimental work carried out by Ha in 2002. the 3rd Ed. or insignificant. Chapter 5 contains the results obtained from this study.
3 2. However many of the conclusions presented by the reviewed articles followed the same general trend. This flexural resistance is only one of a myriad offerees that needs to be considered when designing or evaluating a structural member. older articles on these subjects were available but were not considered for review due to the significant revisions made to the codes since the time of publication. .1 Introduction The flexural resistance or moment capacity of a structural member is a fundamental part of the overall analysis required when designing or evaluating an assembly of structural concrete sections. Articles detailing differences in method and results when using the strut-and-tie model (STM) method were rather numerous and articles reviewed or discussed in this paper date back as far as 1996.0 LITERATURE REVIEW 2. torsional and shear forces must not be overlooked. Of the many comparisons between the American Concrete Institute (ACI) 318 Building Code Requirements for Structural Concrete and the American Association of State Highway Transport Officials Load and Resistance Factor Design (AASHTO LRJFD) Bridge Design Specifications for non-prestressed concrete members reviewed only one article directly addressed differences when analyzing or designing a structure with respect to the flexural resistance. Other forces due to direct loading or reactions such as axial.
V„=K+VS (2-1) The traditional 45 degree truss equation is defined as VS=^IA s (2-2) The code also allowed for a simplified conservative calculation to be used for Vc although the detailed approach provided for more accurate and less conservative results. Vs. To aid in their determination of the safety of these traditional methods of analysis and design 24 reinforced concrete elements having concrete compressive strengths ranging from 4280 to 12. The results from these tests were used to evaluate the design provisions of the ACI 318-99 and AASHTO LRFD "Bridge Design Specifications and Commentary " 2" Ed. 1991.600 psi were loaded under a variety of shear and compressive axial load combinations.2 Shallow Beams Gupta and Collins (537-47) performed a study that primarily focused on the question of the safety of using traditional shear design procedures based on the failure of the Sleipner offshore platform on 23 August. The ACI 318-99 code provisions made a general assumption that the shear stress of a member V„. Both equations are shown below. and the provided shear capacity of any stirrups present using the traditional 45 degree truss equation.4 2. could be defined as the sum of the shear load at which diagonal cracks form Vc. The simplified and conservative approach is detailed in .
. N. Mm=Mu-Nu 'Ah-d^ 8 (2-5) Regardless of the method utilized for calculation of the shear load at which diagonal cracks form the ACI 318-99 code placed a restriction on the maximum value for Vc as defined by Vc<3. rxd (2-3) 1. These detailed equations for Vc were derived by ACI-ASCE Committee 326 in 1962 and were based on the principal stress as found at the location of the diagonal tension cracking. rather it relied on the more involved modified compression field theory (MCFT) which in turn uses relationships between equilibrium.4. The AASHTO LRFD shear design procedure did not use the general assumptions utilized by the ACI 318-99 provisions.9^77 + 2 5 0 0 ^ .3 while the detailed approach is shown in Equation 2.'„ = 1 + 2 20004. Both equations require the use of English units.54JrXdfi + N„ (2-6) 5004. .^ V * (2-4) Nu in Equation 2-3 represents the load due to axial compression and Mm in Equation 2-4 represents the modified moment as defined by the relationship shown in Equation 2-5.5 Equation 2. V.
V„. taken as 0.6 compatibility and stress and strain to predict the shear capacity of cracked concrete section/elements.^ d v s The variables were defined as: Av = cross sectional area of provided stirrup reinforcement within distance s bv = effective web width dv = effective shear depth.9d f'c = compressive strength of concrete fy = yield stress of steel reinforcement s = stirrup spacing cot0 (MPa) (2-7) Values for /?and 6 were derived from calculating the stresses transmitted across diagonal cracked concrete sections which contained no less than the minimum required .0830Jf\b v d v + . was more involved and incorporated several different sub-equations. In addition to the obvious difference between the backgrounds of the two code provisions note that the AASHTO LRFD used SI units of MPa whereas the ACI 318-99 provided solutions in English units of psi. AASHTO LRFD defined shear resistance as Vn = 0. Hence the AASHTO LRFD expression for shear resistance of a section.
and the longitudinal mid-depth strain of the section.5Vcoi6 Flanged Forces. was defined and calculated by AASHTO LRFD as <mi„ = 0.sc (2-9) Where the terms s.5F. A's \ A< Flanged Compression -^ +0. V„ Ac / •-*_ A<i Vcoid 0. . Web Forces and Section Forces dv Longitudinal Strains f \ Idealized Section Flanged Compression Section Figure 2-1 AASHTO LRFD Procedure for calculation of ex Mathematically the longitudinal mid-depth strain of this section is defined by the relationship shown in Equation 2-9.57V -0. v. £•„ = e. sx.cot^ \ \ » • / Section %\ M ^ d.7 transverse reinforcement for crack control. N. This minimum amount of transverse reinforcement Av>mi„. Figure 2-1 details the idealized section used by AASHTO LRFD in the calculation of sx for this procedure. represented the longitudinal tensile strain in the flexural tension flange and ec represented the longitudinal compressive strain in the flexural compressive flange.083V77^ (MPa) J y (2-8) The values for /? and 9 were dependant on the shear stress.5N+Q.
v = ^~ (2-12) The result of the relationships defined above was that for an increase in axial compression the variables Nu. In order to obtain empirical data to corroborate the analytical predictions from each code's procedure a series of 24 specimens were built and tested in the University of Toronto's shell element testing apparatus. This contributed to an increased shear capacity for any given section.10) ^± + Q. ^*—0.5F 7 cot<9 U U £ i = _ l (2. Specimens were designed to represent the .5N+0. cote 7 U U £ = Ji ( 2 .ii) EA + EA' The newly introduced variables were defined as: Ac' = area of concrete on flexural compression side of member Ec = elastic modulus of concrete Es .8 Mathematically st and sc were calculated as shown in Equations 2-10 and 2-11.5N -0. sx and # decreased while /? would increase.5V.elastic modulus of steel And Kwas defined by the relationship shown in Equation 2-12.
A compressive stress of up to 9400 psi was applied along with equal but opposite moments applied to each end of the specimen bending it in double curvature. the concrete compressive strengthf'c. shear and moments were increased proportionally until the specimens reached failure. were measured from locations directly adjacent to the steel end plates. Strains from the flexural tension side of the member ext and the flexural compression side of the member sxc. Loads were induced by five series of six jacks that applied pressure onto steel transfer beams located on the top and bottom faces of the specimens. The parameters that were variable were the compression to shear ratio N/V. In each test case the loads from axial.9 sections of a structural wall that was simultaneously subjected to high axial compression and high out of plane shear loads. . total percentage of longitudinal reinforcement px. Tangential deformation was used as the means of quantifying the total deformation of each specimen. Strain for the shear reinforcement was measured directly from the Theaded bars used to provide shear reinforcement and reinforcement across the member width. specimen width b and the shear reinforcement provided r:. and total percentage of transverse reinforcement across the specimen width py were not varied during the course of investigation. overall section depth h. Specimen length L.
It was demonstrated that it was possible to properly design a section using the detailed method described by this procedure and yet only provide a factor of safety as low as 1. The AASHTO LRFD procedure for calculating Vc provided far more accurate and consistent results for shear failure and the upper limit of shear capacity as induced by an increase in axial compression was correctly predicted.10. whereas when using the more detailed method of calculating Vc was used. the mode of failure can be brittle shear and occur at loads significantly less than those predicted. In the six flexural failures it was determined that the longitudinal reinforcement yielded and the applied moment ME. on average 68% less. This experiment demonstrated that when the ACI 318-99 conservative method for calculating Vc was utilized the shear strength and mode of failure for a reinforced specimen loaded under high axial compression can be consistently predicted. or flexural F.10 Failures were classified as shear S. Of the 24 specimens tested 18 were classified as failing in shear while the remaining 6 were classified as flexural failures. approximately equaled or exceeded the predicted moment at failure M0. based on the observed strains in the Ext and sxc directions at the ends of the specimen when compared to the magnitude of the applied moment at the specimen ends ME and the shear strains yx:. .
When Rahal and Collins' report was published the current version of each code contained torsional design provisions that were similar other than the method used to determine the angle 6. It was concluded that there was a lack of data available that directly correlated the torsional provisions of each code. 2nd Edition. Hence Rahal and Collins ran four large scale experiments to compare the results of the calculated torsion-shear interaction diagrams obtained from both the AASHTO LRFD and ACI 318-02 provisions. The ACI 318-02 code described the basic truss equation relating the provided hoop reinforcement to torsional strength as follows: . rectangular sections 340mm wide and 640mm deep reinforced with non-prestressed longitudinal bars. Rahal and Collins noted that the AASHTO LRFD provisions had been extensively checked for shear cases whereas the ACI 318-02 provisions had been extensively checked for pure torsion as well as combined torsion and bending. The beams tested were solid.11 Two major recommendations arose from these results: • • The detailed expression for Vc be removed from the ACI 318 code The term for axial compression NJAS in Equations 2-3 and 2-6 not be taken greater than 3000 psi Rahal and Collins (277-82) undertook research to provide an evaluation of the design provisions for combined shear and torsion load cases as described by both the ACI 318-02 and the AASHTO LRFD Bridge Design Specifications.
^ cot 9 (SI Units) s (2-13) Where A0 represented the area enclosed by the shear flow path and was permitted to be taken as 0. see Equation 2-14.3. related the torsional strength to the quantity of longitudinal reinforcement provided.6 the angle 6shall not be less than 30 degrees but it is suggested that <9= 45 degrees for non-prestressed members and 6= 37. 0 represented the angle of inclination of the compressive diagonals and was noted to have rather ambivalent instructions to its suggested versus its analyzed values.tan 3 (SI Units) Ph (2-14) Rahal and Collins observed that when comparing equations 2-13 and 2-14 the equivalent torsional strengths could be obtained by using less hoop reinforcement but increasing the longitudinal reinforcement.5 degrees for members that are prestressed.6. As described in ACI 318-02 §11. fyv was the yield strength of the hoop reinforcement and At represented the cross sectional area of one leg of the transverse reinforcement. and where A0h represented the area enclosed by the outermost transverse torsional reinforcement. . A similar truss equation. Af T„=2A0 -!+*.85*Aoh.12 T„ = 2A0 .
K = K + K= 0A66j]\bwd The variables were defined as: As = area of flexural reinforcement bw . .web width d = distance from the extreme compressive fiber to the centroid of the flexural reinforcement f'c = compressive concrete strength fy = yield strength of reinforcing steel Vc .13 When these two equations were set equal to each other the required area of transverse reinforcement could be solved for as shown by Equation 2-15.shear strength contribution from concrete Vs = shear strength contribution from reinforcing steel + .^ d (MPa) (2-16) Additionally the ACI 318-02 code required that the nominal shear stress for solid sections be limited to avoid concrete crush prior to reinforcement yield. Equation 2-17 describes this limitation. A f Al=^-Ph^cot20 s fyt (SI Units) (2-15) The ACI 318-02 equation for a non-prestressed section that described the relationship between transverse reinforcement and shear strength is shown in Equation 216.
to relate the area of one leg of the transverse reinforcement^. Rahal and Collins noted that when /?was set equal to 2.083£Jf~cbvdv +—< s cot 9 (SI Units) (2-18) The variable bv represented the web width.^ r < 0.. For sections containing stirrups the values for J3 and # depended on the nominal shear stress v. Also when examining nonprestressed sections sx could be taken as 1. however dv was defined as the effective shear depth which can be taken as 0.00x10" which provided a value of 36 degrees for 0. and when the AASHTO LRFD value for 6 equaled 36 degrees the results for Vs were 24% higher than those obtained from the ACI 318-02 procedure. Equation 2-13.22 the AASHTO LRFD value matched the ACI 318-02 value for Vc exactly. . However the AASHTO LRFD relationship between the minimum required shear strength V„ and the required area of transverse reinforcement Av was noticeably different from that shown for ACI 318-02 in Equation 2-18. to the required torsional strength T„. V = Vc + Vs = 0.14 v^2 b f . and the mid-depth longitudinal strain ex.9*d.83SK (SI Units) (2-17) Jj The AASHTO LRFD code used the same basic truss equation as the ACI 318-02.
at the mid-plane could be taken as l.5N„ + 0.9d Es = modulus of elasticity of non-prestressed reinforcement Ep = modulus of elasticity of prestressed reinforcement fpu = tensile strength of prestressing steel Mu = factored moment.25/'c to avoid failures due to concrete crushing.15 Equation 2-19 details the nominal shear stress v.5 cot 6JV„ +0-Wu = -1 2(EsAs + EpAps) ' v 2 f° J "0. !L £x 2 + 0.area of prestressed longitudinal reinforcement placed on the flexural tension side of the section dv= effective shear depth.OOxlO'3 or it could be calculated using the relationship shown in Equation 2-20.25/' c (MPa) J (2-19) The longitudinal stress sx.7/aA (2-20) Variables in Equation 2-20 were defined as: As = area of non-prestressed longitudinal reinforcement placed on the flexural tension side of the section Aps . can be taken as 0. taken as a positive quantity not less than Vudv . M. 2 (T PA j A \Aoh 2 <0. for a solid section as calculated under the AASHTO LRFD provisions for a solid section under combined shear and torsion which was required to be no greater than 0.
<i> \\<t> ) \ <t>1Ao j (2-21) The test variable in the series was the torsion to shear ratio which varied from zero to 1.216m. The AASHTO LRFD provisions required that the tensile capacity of the longitudinal reinforcement on the flexural tension side of the section be no less than the force TE. Conversely when the minimum value for #of 30 . positive if tensile.V M. Rahal and Collins determined that use of the ACI 318-02 provisions produced very conservative results when the maximum value of 45 degrees for the angle of the compression diagonals 6. TE was calculated as shown in Equation 2-21. was used. Specimen failures were attributed to excessive yielding of the closed stirrups as well as spalling and crushing of the concrete in the test region. to prevent premature failure of the longitudinal reinforcement.16 Nu = factored axial force. negative if compressive Tu = factored torsion Vu = factored shear Rahal and Collins noted that for prestressed sections sx would often approach zero. In these cases the value of 6 would vary from 22 to 30 degrees depending on the level of shear stress present. fir WJMM V / / W C T .
Use of this value provided results of a consistent and reasonable nature that closely replicated the observed crack patterns. Naaman noted that the changes made from the ACI 318-99 to the ACI 318-02 codes relocated the limits for tension and compression controlled sections and added the transition region between the two. in the transition zone. i.e. These flaws were not directly correlated to the corresponding AASHTO LRFD provisions but they did reflect similar results for other ACI provisions as described by several other publications where solutions appeared conservative but in reality did not provide adequate factors of safety. The various regions for reinforcement limits and the definitions and from different codes are shown in Figure 2-2. . the flaw lie in this definition for these regional boundaries. Naaman (209-18) investigated the differences between the ACI 318-02 and AASHTO LRFD codes regarding sections that were classified as being between tension controlled and compression controlled. Results obtained from use of the AASHTO LRFD provisions provided a consistent value of approximately 36 degrees for 6.17 degrees was used the ACI 318-02 provisions demonstrated less consistent results and provided failure loads for high torsion-to-shear ratios much higher than those observed. Naaman found and described several examples where the ACI 318-02 provisions regarding the limits of reinforcement for flexural members lead to unintended erroneous results that brought the validity of the provision into question.
42 Under-reinforced (t for bending O I ± -m.375 _J_ c 0.0038 0.75ft.60 a) O < o> i i 1 1 l for bending 1 036A 1 C 1 1 1 .002 Q LU CO o C) +for bending Minimum 0.j for compression (no transition) c Figure 2-2 Summary of reinforcement limits by 1999. o OH -L I AASHTO LRFD t Tension controlled (under-reinforced) m I< |" Transition ^ I ^ compression controlled "• | (over-reinforced) Minimum 0.600 CM O O CM 0.<>for compression (no transition) ^ factor LU Q 0. A 1 LU Q O O | Under-reinforced i 1 1 0..18 Ptmm 0. 2002 ACI codes and AASHTO LRFD and Naaman recommendation .44 ft factor Transition -ft for compression 0.002 b) O O O < +tor bending ft factor Transition -(t for compression 0.60 Q.002 r i 0.0038 1 0.005 0.» .44 1 0.
— < 0. This allowed for values of if) found using the ACI 318-02 to be different from those obtained using the AASHTO LRFD for identical sections because the definition of dt and its corresponding st was defined as the distance to the centroid of the extreme tensile reinforcement only.19 The ACI 318-99 code used the ratio of c/ds where c represented the depth of the compression block and ds represented the depth from the extreme compressive fiber to the centroid of the tensile force in the non-prestressed reinforcing bars.42 de (2-22) . This did not take into account tensile resistance provided by a multi-layered arrangement and implied that a section which contained more than one layer of reinforcement or a section having any combination of plain reinforced. While the ACI 318-02 code offered this somewhat conflicting definition for the limits of reinforcement between editions the corresponding AASHTO LRFD provision detailed that for all cases the maximum reinforcement was bound by the relationship detailed in Equation 2-22. In the 2002 edition of the ACI 318 code the ratio was changed to c/dt where dt was defined as the depth from the extreme compressive fiber to the centroid of the extreme layer of the non-prestressed reinforcement. partially prestressed or fully prestressed steel would be controlled by the extreme layer of reinforcement exclusively.
It also made the type of reinforcement present irrelevant since the tensile force T must equal the compression force C for all cases. independent of whether the section was plain reinforced.e. reinforcing bars (2-23) Naaman noted that when the quantity de was used to calculate the limits of reinforcement for a section the results were guaranteed to be the same. This consistency was attributed to the fact that the equation guaranteed simultaneous equilibrium of forces as well as strain compatibility in any case. i. Reference Equation 2-23 de = AJpAp + Afyds Apsfps + Asfy The variables used were defined as: As = area of non-prestressed reinforcement Aps = area of prestressed reinforcement fps = stress in the prestressing steel at nominal bending resistance fy = yield strength of conventional reinforcing steel dp .depth from the extreme compressive fiber to the centroid of the tensile force in the prestressed reinforcement d„ = depth from the extreme compressive fiber to the centroid of the tensile force in the non-prestressed reinforcement.20 Where c again represented the depth of the compression block and de was calculated as the weighted sum assuming yield of the steel reinforcement provided. partially prestressed or fully prestressed. .
and the upper limit of the transition region remained 0. It should be noted that the ACI 318-05 provisions did include some but not all of the recommendations proposed by Naaman.60 consistent with Naaman's recommendation. The relationship used to define the regions was changed to use the quantity de rather than the less accurate dt. Reference Figure 2-3 to compare the new limits of the ACI 318-05 to those of the AASHTO LRFD.375 rather than the 0.21 More specifically Naaman described that because the assumed failure strain in the concrete ecu. . was taken as a constant in equations 2-24 and 2-27 there was a direct relationship between c/de and the tensile strain in the concrete at the centroid of the tensile force ste that was unique. however the lower limit of the transition region remained set at 0. (2-24) de=dsfor Aps=0 de=dpfor As=0 d-c (2-25) (2-26) (2-27) vA j The use of de was proposed by Naaman to replace dt when determining the ratio c/dt to avoid erroneous reinforcement level classification of a section.44 that Naaman had proposed.
HSC sections require larger amounts of transverse reinforcement due to their behavior of cracking at much higher shear stresses than conventionally reinforced concrete sections. Under-reinforced . Their study focused on high-strength concrete (HSC) based on its increased use in construction. Canadian Standards Association (CSA) A23. ~ „ — Minimum AASHTO l UWD . crack widths at the estimated service load and at the post-cracking reserve strength were used to evaluate the performance of each specimen. The observed cracking patterns.3 and AASHTO LRFD Specifications 2 nd Ed.tion ' ' • I .22 0. . ^ Factor 1 L ACI 318-05 Minimum 1 1 Tension controlled (under-reinforced) i p Transition • " 1 ^ ( • Compression controlled f (over-reinforced) .002 -i ~4 I f for bending > 1 —J I __ Trans.005 0. ^ for c o m p r e s s i o n 4. unaer-reinrorceq I (j> for bending 0-42 . ._J •• „ ' ^L£_ ^forcoinpres$jol](ngtransjtioD) • J—*• Over-reinforced Figure 2-3 ACI318-05 and AASHTO LRFD limits of reinforcement Rahal and Al-Shaleh (872-78) conducted a study to examine the differing requirements for minimum transverse reinforcement as specified by the ACI 318-02 Code.
which has effects on both concrete and steel reinforcement contributions Vc and Vs respectively. . The average concrete cylinder strength^ was 75% of the concrete cube strength fcu and split tensile strength/^ was approximately 0. on the effect of the longitudinal steel as well as the stress resultants for bending moment and axial load. addressing this influence from longitudinal reinforcement was the objective of Rahal and Al-Shaleh's test program. the ACI 318-02.74 Jf~. The CSA A23.) wide.23 The ACI 318-02 used detailed equations that accounted for the contribution from the concrete Vc. torsional and axial loading in the calculation of the strain indicator sx.87 in. flexural.) deep and 2750 mm (108 in. and the AASHTO LRFD were not united in their approached to longitudinal reinforcement provisions and were noted to have differed significantly.3. Beam dimensions for all specimens were 200 mm (7. Regardless of the difference in the approach to provided longitudinal reinforcement none of the aforementioned codes accounted for the influence of the longitudinal reinforcement.) long with a shear span of 900 mm.57 in. The AASHTO LRFD Specifications accounted for the influence of longitudinal reinforcement. Eleven four point load shear tests were performed on 65 MPa (9500 psi) beams that had minimal transverse reinforcement and two levels of longitudinal reinforcement. 370 mm (14.
S5/3J<c (2-28) . and the strut efficiency factor fis. fcu=0. Sankovich.3 provided adequate performance for members containing large amounts of longitudinal reinforcement • No evidence was found in performance between beams designed with the maximum stirrup spacing as defined by each code • Shear capacity equations in the ACI 318-02 and AASHTO LRFD 2nd Ed were conservative 2. Bayrak and Jirsa (348-55) completed a study of the behavior and efficiency factors assigned to bottle shaped struts when used in calculations as described by the method of STM.3 Deep Beams Brown. as shown in Equation 2-28. ACI 318-05 defined^..24 Their study produced the following conclusions: • Behavior of members containing large amounts of longitudinal steel was far superior when compared to members that contained very little or no longitudinal steel • The ACI 318-02 and CSA A23. A fundamental difference found between the two provisions was in the calculation of the strength of a strut based on specified concrete strength as determined by a cylinder test/' c . Historically these efficiency factors were assigned based on good practice rather than actual results from experimentation.
3. is based on the type of strut under consideration and the amount of transverse reinforcement present. Y ^ s i n r . . i.3-1 / " where el=e.e. It was noted that many engineers have difficulty choosing an appropriate tensile concrete strain to be used during design and have therefore expressed reservations about using the MCFT based ASSHTO LRFD provisions. + (e. When ACI 318-05 is used in cases of a bottle shaped strut that was crossed by adequate transverse reinforcement as defined by ACI 318-05 Equation A-4 (Equation 2-29) the strut will control the strut-node interface in all cases other than CTT. an interface node having one strut and two ties. + 0.8 + 170s.002) cot2 as (2-31) = Ls. 0. < 0.003 bsi (2-29) The AASHTO LRFD utilized the MCFT to define fcu and therefore the definition is much more involved than that described by the ACI 318-05 method.6. Equation 2-30 repeats the AASHTO LRFD Equation 5.3.25 The strut efficiency factor J3S.85/'c (2-30) as was defined as the smallest angle between the compressive struts and the adjoining tie. < 0.
Each specimen had a unique amount and placement of reinforcing steel and in each isolated strut test the same mode of failure was observed. Of the 26 test specimens 20 of the AASHTO LRFD determined efficiency factors yielded results for the isolated struts that were less conservative but more consistent with .68. rather it would change direction. When the average value for the experimental efficiency factor was divided by the predicted ACI 318-05 efficiency the result was 1. Failure was first indicated by a vertical crack which would form in the center of each panel. (305 x 152 x 51 mm) steel bearing plate. (914 x 914 x 152 mm) were loaded using a 12 x 6 x 2 in. Failure was described as crushing and spalling of the concrete near but not adjacent to the loading plate. Of these 25 specimens it was noted that the results were conservative but erratic when compared to the test data.26 To more directly measure the effect these two approaches had on modeling struts 26 concrete panels measuring 36 x 36 x 6 in. Of the 26 specimens tested the efficiency factors presented in ACI 318-05 provided a safe estimate of the isolated strut capacity. The same failure mode was observed in every test regardless of the boundary conditions present. One test used a different panel thickness and bearing plate to observe and examine the effect of specimen geometry on the efficiency factor. That crack then propagated from panel midheight to the loading points but would not intersect them.
85/c as described by AASHTO LRFD Equation 5. historically these two strength factors were combined into a single parameter.3. The plastic truss model has two possible failure modes. They defined the clear span a. A third failure mode was proposed by in 1998 by Foster where splitting or bursting of the strut should be considered. To simplify their discussion regarding deep beams and corbels Foster and Malik standardized the nomenclature used in their study.6. Due to the well known behavioral and material properties of reinforcing steel tension failures can be predicted with a high degree of confidence and therefore this mode is not discussed.27 the test data. All relevant equations were then recalculated to incorporate this split between variables. . concrete crush in the struts and yielding of the ties.3.3-1 (Equation 2-30). The STM model that was most extensively investigated was the plastic truss model where all truss members enter the nodal zones at 90°. Foster and Malik (569-77) reviewed a comprehensive set of test data on deep beams and corbels and compared them to the proposed efficiency factors. as the distance from the centers of the strut nodes and they also split up the in situ strength factor ks and the strut efficiency factor v. All of the AASHTO LRFD data was governed by the limitation placed on the maximum strut strength of 0.
geometry and strength of a member in equilibrium are shown below.cbdc (2-32) (2-33) (2-34) fc=k.28 The fundamental plastic truss model equations that relate material properties. Material: T = Afssy sJ C = f.yf\ Geometry: a = a +w Q (2-35) =d+2 m0 = (2-36) w z a m (SI) (2-37) n = d-Jd2-2aw<2(D-d) d=Strength: V = min w sin^ (2-38) (2-39) a (2-40) The variables were defined as: a = shear span a' = distance from center of concentrated load to edge of support .
As = area of tensile reinforcement b = section width d = effective depth of main flexural reinforcement dc = strut width D = overall member depth f'c = compressive concrete strength fc — effective concrete strength fay = yield strength of main longitudinal reinforcing steel ks = ratio of the in situ strength to the cylinder strength V= shear force w = node width over which shear force is applied z = distance between node centers v= efficiency factor 6 . V V ~Kf'cbw .equivalent strut width over which ties contribute Using these relationships Foster and Malik calculated the efficiency factor shown in Equation 2-41.angle of strut to longitudinal axis Q .
0. whereas another undertaken in 1997 suggested v= 0.for /?<0. A 1986 study suggested v= 0. In 1998 Chen revised these factors for deep beams and proposed the following relationship v.6(1 . a/D<0.85 and placed greater emphasis on the selection of an appropriate truss model.4-) v= P= &. In 1978 Nielsen introduced an efficiency factor to calibrate the concrete plasticity models they had developed for members in shear.6.25Z))(100/? + 2)(2 . In 1986 the MCFT was proposed to describe and define that concrete is not perfectly plastic and the corresponding loss of strut capacity due to transverse tension fields. 1990 and 1997 also used efficiency factors that were functions of strut or node location and the degree of disturbance these struts or nodes experienced.0.0 Where p was defined as the reinforcement ratio for main longitudinal steel (SI) (2-42) . Foster and Malik also investigated the controversy and debate surrounding the assignment and values used for efficiency factors in this same report. ASCE-ACI Committee 445 offered a comprehensive review of this work.25 0.30 This efficiency factor was used to reduce predicted member capacity to compensate for the fact that concrete is not a perfectly plastic but rather a brittle material.02 If. The greater this disturbance the lower the efficiency factor assigned. Models created in 1987. Z)<1.
Based on their parametric study Bachelor and Campbell proposed the following equation be used to define the efficiency factor.1 + 0.31 In 1986 Batchelor and Campbell proposed a reduction of the effective compressive strength based on their theory that the diagonal struts are in a state of biaxial tension which in turn reduced the strength of the web concrete.18 500 and v = 0. . From this Warwick and Foster concluded that concrete strength and the ratio of the shear span to member depth were the main contributors that affected the efficiency factor. and the quantities of horizontal and vertical reinforcement as the variable parameters and by comparing a series of experimental data with non-linear finite element analyses.53-•£-*. 2 5 — ^ .471 d) V b) (SI) (2-44) Warwick and Foster investigated the effects of concrete strength on the efficiency factor using a range of 20 to lOOMPa and proposed + A v = 1 .342-0. ^ ( d\ In v— = 3.1991 •7.for aA>2 (SI) 500 /d (f) ° {f) -Xf°ra/d-2 (SI) (2_45) (2-46) Equations 2-44 and 2-45 were developed in parametric studies using the concrete compressive strength f'c.0 . 7 2 [ .
angle of strut to horizontal Foster and Gilbert demonstrated in 1996 that the relationship defined by Collins and Mitchell could be modified as a function of/'c and the ratio a Yd. parallel to strut Sx = strain in the horizontal direction 6 .14+ 0.64+ l (SI) " 470 A z) J c (2-49) Foster and Gilbert further simplified this equation. . In doing so the strut angle was approximated as tan9~ d/a'. k3v = 1. This produced the result shown in Equation 2-49. normal to strut £2 = minor principal strain. Collins and Mitchell proposed Equation 2-47. termed the modified Collins and Mitchell relationship. where e-sx 1 + tan # 2 tezZhl (2-48) Variables were defined as: £•/ = major principal strain.8 + 170*. Lv = 3 (2-47) 0.32 Based on panel testing performed by Vecchio and Collins (1986). based on the insensitivity of vtof'c to produce Equation 2-50. more precisely tan6~ z/a.
75(- (2-50) u For special situations where a/z = 0. si = 0 the MCFT infers that v= 1 and thus the relationship is further modified to yield Equation 2-51. 1 1 + KcKf (2-54) where . v= l - j- (2-51) (a\ 1.55 + .14 + 0.88. i.H L (MPa) (2-53) The variable vi in Equation 2-52 was defined as a factor dependant on the potential of damage to the strut(s) under consideration. Vecchio and Collins later revised their efficiency factor based on newly available data of the time to produce Equation 2-54.e. + 0.66 \z) It must be noted that when Equation 2-51 is used ks = 0.33 V = T-77 (SI) 1. MacGregor proposed that the efficiency factor be defined by Equation 2-52 k3v = vlv2 where (2-52) v2 = 0.
80 (2-58) For cases where a/z = 0 the MCFT implied an efficiency factor value of v.1*25Jf'e >1.28 J (2-55) V c2 kf =0.8 a ( A 0.1 and kf= 1. ) / ' / (2-60) .35 0.0025 and the relationship shown in Equation 2-57 £. =£X + 1 k-g 2 ) .80 k = 0.52 + 1.0 and substituting Equation 2-58 into Equation 2-54 with these special case values produced Equation 2-59.83 + £c£7 (2-59) Based on the development of a reinforced concrete cracked membrane model for plane stress elements the efficiency factor define by Equation 2-60 was adopted 1 v= (0.0 (M/ty (2-56) Using an assumed value for £2 of-0.34 /- \0. tan 2 # (2-57) The following is obtained for kc K = 0.4 + 3 0 * . v = • 1 0.35 and ^--0.
Two parameters were used to define the efficiency factors. 2-59 and 2-61 were compared against the experimental data and models proposed by the AS3600 model with cutoff and models proposed by Batchelor and Campbell. Equations 2-51.35 When the major principal strain si. is treated as a function of both sx and l/tan2 6 then Equation 2-60 takes the form of Equation 2-61. K= 1 . multi-parameter model predictions and MCFT. MacGregor and Warwick and Foster.22-0..24 when compared to the experimental data ./'c and a/z and the following observations were made: • Poor correlation with high degrees of variability existed between experimental data and efficiency factor models based solely on concrete strengthf'c • Multi-parameter efficiency factor models also exhibited poor correlation with high degrees of variability between experimental data and predicted behavior • MCFT models provided coefficients of variation from 0. 135 specimens that had been determined to have failed in compression were analyzed and efficiency factors assigned based on concrete strengthf'c.—jT-<1 cl+c2(a/zY\f</i l (2-61) where Ci and C2 are empirically derived constants From their review of sixteen previous studies. Chen.
provided the best predictions for efficiency factors Most of the reviewed articles addressed topics other than the same two specific areas of pure flexure and deep beams covered by this thesis. this was attributable to the paucity of relevant available literature. . However their inclusion was justified by the wide array of examples contained within these articles.36 • Boundary conditions play are a significant factor in obtaining data that is reliable and has a low degree of scatter • Strut angle was the most significant factor and models based on the shear span to depth ratio a/z. The examples illustrated that there has always been division between the codes on approach to analysis and design as well as the actual formulae to be utilized when designing or analyzing a reinforced concrete section. member or structure.
1 Background In the classical approach to solving for the flexural strength of a section generally there are two assumptions required in order to provide a simplified method to the solution of the required calculations. These two assumptions form the backbone of the elastic case for flexural theory that states that normal stresses within a beam due to bending vary linearly with the distance from the neutral axis. Flexural theory is based upon the relative distances between sections and because it has been proven that warpage does not violate .0 T H E O R Y O F F L E X U R E 3. However adjoining planes are also similarly warped and therefore the distance between any two points on adjoining sections for all practical purposes remains constant whether or not warpage is considered.37 3. 2) Hooke's Law can be applied to the individual fibers within the beam section. There are two components to the first assumption as discussed in §6.4 of Mechanics of Materials: The first component is based on rigorous mathematical solutions from the theory of elasticity that demonstrate some warpage does actually occur along plane sections and that this warpage is greatest when shear is applied along with a moment. The assumptions are: 1) Plane sections remain plane.
Likewise the negative strains along the outermost compressive surface are accompanied by positive transverse strains. this type of curvature is classified as anti-synclastic curvature. See Figure 3-1. Remain Straight Figure 3-1 Beam in flexure with saddling effect . The second part of this assumption is that when a beam is subjected to pure bending the positive strains on the outermost tensile surface are accompanied by negative transverse strains.38 this relationship between plane sections and the assumption that plane sections remain plane remains valid. this curvature is classified as anticlastic curvature. The classical approach to solutions of reinforced concrete sections ignore this behavior.
2 Theory of Flexure in Shallow Reinforced Concrete Sections In the study of reinforced concrete additional assumptions are made in series with the first two presented. commonly referred to as Young's Modulus a= stress 3. assumptions 4. 3) The strain in the reinforcing steel is the same as the surrounding concrete prior to cracking of the concrete or yielding of the steel . Whitney but they were obtained from "Design of Reinforced Concrete ACI 318-08 Code Edition" for this thesis.this is a continuation of the second assumption 4) The tensile strength of concrete is negligible and assumed to be zero 5) The stress-strain curve of the steel is elastically perfectly plastic 6) The total force in the compression zone can be approximated by a uniform stress block with magnitude equivalent to 0. based on Hooke's law. e=E Where the variables are defined as follows: s = strain E = modulus of elasticity of the material.85/' c multiplied by a depth of a 7) The maximum allowable strain of concrete is 0.39 The second assumption. states that individual fiber strains can be used to calculate individual fiber stresses and visa-versa. 6 and 7 are generally attributed to Charles S. Equation 3-1.003 (3-1) .
Rather the stress distribution is in the form of a parabola as shown in Figure 3-2 c.40 The general arguments that form the basis of Hooke's Law remain as valid for reinforced concrete sections as they do for isotropic homogenous material sections when subjected to small strains. When loaded the tension zone of a concrete section will begin to crack under very light loads destroying the continuity of the section and any tensile reinforcement will be forced to carry the tensile load in its entirety. This block has . In effect. The stress distribution in the compressive region does not maintain a linear relationship with respect to distance from the neutral axis due to the nature of the constituent materials used to manufacture concrete. Whitney developed an equivalent rectangular stress block that provides results of equal accuracy for the compressive strength of a concrete section that avoids the rigorous mathematical calculations required to compute the area of a parabola. The assumption that the stress-strain curve of steel is elastically perfectly plastic implies that the ultimate strength of steel is equivalent to its yield strength. this results in an underestimation of the overall ultimate strength of a given section due to the reinforcing steel but it produces a more predictable mode of member failure. In general the tensile strength of concrete is around 10% of the compressive strength.
41 depth. counteracting force required for a beam to remain in equilibrium while under the application of an external force. is multiplied by a modification coefficient Pi yielding the result shown in Equation 3-2.85-0. The value of 0. or limit strain. a = pxc The coefficient /?/ varies as summarized in Equation 3-3.003 in the extreme compressive fiber as the assumed maximum allowable strain.65 (3-2) A (3-3) The ACI has adopted a strain of 0. Compared with results determined from extensive empirical data this value represents the lower bound of the limit strain. a. To calculate the depth of the compression block the distance from the extreme compressive fiber c. 0.4 0 0 0 j 1000 J for fc> &000psi 0.85/' c .85 far f'c<4000psi 4000 < / ' c < 8000 mz 0.05 / ' c . and an average compressive strength equivalent to 0. for a concrete section. This can be . Using these seven basic assumptions a series of equations can be derived that provide quantitative values for the equivalent.85 was derived from extensive laboratory testing of core test results of concrete in structures where the concrete was a minimum of 28 days old.
c6 where a = (5xc :. equal in magnitude and opposite in direction.85/'c/?.85fcab a -PlC o (a) (b) (c) T=Afy (d) Figure 3-2 Graphic representation of forces within a reinforced concrete section . the tensile forces provided by the reinforcement present in a section are opposite and equal to the sum of the compressive force provided by the concrete. acting at any two locations along the centerline of the cross section. Summing the forces in the axial direction T = C. reference Equations 3-4 through 3-8.42 explained by visualizing the two forces as vectors.c-al[5x T= C=zAtfy=0. i.Z5fcab (3-6) (3-7) (3-8) (3-4) (3-5) •C=0. T = A.fy C = 0.e.
3 = 0.85/>Z>(V£) (3-10) Theory of Flexure in Deep Beams Deep beams are a common structural member for which an accurate solution cannot be reached using the aforementioned techniques. a= Af '-^— 0.43 Variables are defined as: a = effective depth of the compressive block As = area of non-prestressed steel fy = yield strength of non-prestressed steel f'c = compressive strength of the concrete b = the width of the section c = distance from extreme compressive fiber to neutral axis Using the relationships show in Equations 3-4 through 3-8 and solving for a yields the result shown in Equation 3-9. Deep beams are structural members defined by the relationship between beam width bw. beam depth h and clear . shown in Equation 3-10. Mn -AJ^d-^j 3.85 fcb (3-9) Solving for the internal resisting couple between the tensile and compressive forces yields the nominal strength or moment resistance of the section M„.
The overarching motivation of STM is weighted toward conservatism in the solutions it provides and to transfer as much of the applied loading as possible into compression using the fewest number of members. Therefore other more advanced methods must be employed. This invalidates the first and most fundamental assumption discussed in this paper for the solution of reinforced sections. or.7. Per ACI 318-08 §10. equal to or less than four time the overall member depth. One of these is the method of Strutand-Tie Model (STM) an inherently conservative method for solving the forces within these member's sections. deep beams are members loaded on one face and supported on the opposite face so that compression struts can develop between the loads and the supports and have either: a) clear spans /„. . and plane sections are no longer assumed to remain plane. The foundations of STM are attributed to the truss method work done by Ritter in 1899 developed as a means to explain the dowel action of stirrup reinforcement. However it was not until the 1980s that the truss method transformed into STM and was used to find solutions for the discontinuous regions within deep beams. where Pu represents the concentrated load.1.44 span /„. b) regions with concentrated loads within twice the member depth from the face of the support Deep beams or deep beam regions begin to show crack propagation at loads in the range of xhPu to xliPu.
1 (a and b) . RA. or D-regions. h. dapped-end beams and the discontinuous regions. it also has valid applications in corbels. Mathematically a D-region is defined as a region located within a distance equal to the member depth h.D-regions and discontinuities (Reprinted with permission from the American Concrete Institute) A D-region within a beam is defined as an inter-beam span where traditional moment and shear strength theory no longer applies as based on the assumption that plane sections remain plane. h h 4)1*.1.The method of strut and tie models is not exclusively limited to deep beams. reference Figure 33 for examples. 'TO O (tf 2h 1)1' lt)l" (HE JJ \>>t -4 ~A~ Ay? ttmmtmt fW Loading and geometric discontinuities (a) Geometric discontinuities Figure 3-3 ACI318-08 Fig. from the beam/support interface or a region located a distance h from each side of a concentrated load. •KH h. as such these D-regions can be isolated and viewed as a . within shear spans. and loads are not reacted by beam action but rather they are reacted primarily by arch action.
only equilibrium and yield conditions need be satisfied. only gives an estimation of member strength. At its core STM disregards kinematic restraints. a truss model is chosen to represent the flow of forces within that member. The intersection of any two or more members is designated a node. The solutions for the capacity of members found using the application of this lower bound theory are estimates that will provide member capacities which will be less than or equal to the load required to fail the member. and how the truss members are chosen. The two greatest similarities between the ACI 318-08 and AASHTO LRFD code provisions are the basic set up of the solution. .46 deep beam. After the global forces have been determined for a particular member. the method of sections. In these regions the ratio of shear deformations to flexural deformations can no longer be considered negligible. and it conforms to the lower boundary of theory of plasticity. thus the inherently conservative nature of STM solutions. The goal for development of any truss model is to use the lowest number of members required to satisfy equilibrium and safely transmit the forces into the supports.e. Members in compression are designated struts and members in tension are designated ties. a combination of both or using a CAD program. After a truss model has been developed the forces within that truss are analyzed by using the method of joints. i. as shown in Figure 3-4.
The basic premise for the implementation of the method using either code is the same but the way that these forces are resolved within the individual strut and nodal members is the area of greatest divergence between the two codes.1. RA.r— (a) CCC Node (bJCCTNode Figure 3-4 ACI318-08 Fig.Hydrostatic nodes (Reprinted with permission from the American Concrete Institute) There are few similarities that exist between the ACI 318-08 and AASHTO LRFD codes with respect to STM efficiency factors. . and § 5.5 (a and b) .47 T-c.2 provides a comprehensive explanation of the differences in the STM between the ACI 318-08 and the AASHTO LRFD 2nd Ed. § 4.2 discusses the differences in the results obtained from each code.
4.0 ANALYTICAL PROCEDURE 4.1 Shallow Beams In order to simplify the programming phase of the project, an analytical approach was used rather than a design approach. Doing so reduced the required number of IF logic statements, shortened the overall length of the EXCEL program, simplified the program used and focused the project on the analysis of results rather than programming. All spreadsheets used in the shallow beam analysis were vetted via direct comparison against example problems 7.4 and 7.5 from the Notes on ACI 318-08 Building code Requirements for Structural Concrete. Results from the spreadsheets matched those in the example problems exactly.
Each series of analytical calculations were performed using sections having similar geometric and identical material properties. The yield strength of the reinforcing steely, was set at 60ksi and the crushing strength of the concrete f'C) was set at 4ksi. Ten different geometries were analyzed in a series of four different calculations with each calculation using three different values for the following variables; flange width b, web width bw, flange depth tf, and the depth of reinforcement d. Ten arbitrary sections with the following geometries, graphically described in Figure 4-1, were chosen for analysis.
-fig w
bw —*•
Figure 4-1 Section geometry The values used for each variable's analytical case are listed below in Table 4-1.
Table 4-1 Values for variable and fixed geometry for each case
VARIABLE VALUES 36in. b 42in. 54in. 14in. bw 18in. 21 in. 24in. 30in. d 36in. 2in. 6in. tf 10in. FIXED GEOMETRY 4in. tf 14in. bw d 24in. b 36in. 4in. tf d 24in. 36in. b 14in. bw 4in. tf 36in. b bw 14in. 24in. d
For each case the assumed area of provided reinforcing steel was varied from zero to an arbitrary value of 20 square inches. This was accomplished in incremental steps of 0.50 square inches. However only data within and inclusive of the boundaries determined by Asmi„ and Asmax would be considered for final analysis.
Because each code specifies a unique analytical procedure, the focal crux of this paper, separate programming approaches were required for both the ACI 318-08 and the AASHTO LRFD in order to reach solutions. Each procedure is described in detail in the following sections.
Stress Block Depth Factor fit The sole quantity that was independent of the code provision used and could be
calculated simultaneously was the stress block depth factor, /?/. Used in computation of the location of the neutral axis the method for computing the value of /?/, is identical for both ACI 318-08 and AASHTO LRFD codes and was computed using a simple nested IF statement as described in Figure 4-2 below.
fc < 4.0fc/
Pl = 0.85
4.0 < f\ < 8Msi No f\ > 8.0*5/ >
A=0.85-(0.05*(/' c -4))
» px = 0.65
Figure 4-2 IF logic statement for computation of stress block depth factor /?/
\ J yJ ((6-0/+0. as a fraction of the balanced area of steel for a section. .1. is the maximum value of the following two equations. and the crushing strength of concrete f'c.63375 A . A.1000 (4-2) j ACI 318-08 defines the maximum area of reinforcement Asmax. =0. smax. the yield strength of reinforcement and the crushing strength of concrete.51 4. both of which are functions of section geometry. stress block depth factor ft.2 ACI318-08 4. yield strength of reinforcement fy. 4„=0.1. sbal (4-3) Where the balanced area of reinforcement ASM.1 Limits of Reinforcement As defined by ACI 318-08 the minimum area of reinforcement Asmin. Asbai is also a function of section geometry.2.375 A M ) (4-4) . . A.85 i f . is computed as shown in Equation 4-4. = Maximum] s mm 3V/'C1000 V (( Aiooo ->y Kd 200 hd v/.
1. i. a'=1. a' c'=— (4-7) A Both of these preliminary values were used only for the evaluation of section behavior via comparison to the flange depth tf.e.52 4. based on the ratio of reinforcement and computed as shown in Equation 4-5. as shown in Equation 4-6. and the working value for . denoted in this paper as a' for clarity.^ = p^>F bdfc fc (4-5) crwas used to evaluate the preliminary value for the depth of the compressive section.2.18fflr/ (4-6) This preliminary value a' was used in the calculation of the preliminary depth of the neutral axis which has been denoted in this paper for clarity as c' shown in Equation 4-7. Af f m = . These preliminary results for a' and c' determined the method used to calculate the working value for the depth of the compressive section a.2 Location of Neutral Axis c and Depth of Compressive Block a The ACI 318-08 utilizes a variable called the reinforcement index zu. the determination as to whether the section was in rectangular action or flanged action and therefore if the web carried any compressive load.
This in turn provided the appropriate formula to solve for the factored moment capacity for the section. had been established.e. by means of IF logic statements. an IF statement was used to determine whether the section behaved under T-section or rectangular action. See Figure 4-6.53 the depth of the neutral axis c. had been obtained and the depth of the neutral axis c. . i. The grayed out cells on the left hand side of the flow chart in Figure 4-6 offers a comprehensive view of this procedure. if the depth of the compressive section was greater than or equal to the flange depth a < t/. After a value for the depth of the compressive section a.
005 i = 0.65 <p = 0.65 + (£.3 Strength Reduction Factor (j> The depth of the neutral axis was also used in the determination of the strength reduction factor <f> based on ACI 318-08 reproduced below in Figure 4-3.1.7 + (£.0. The moment capacity could then be computed using the given values for A . = 0.54 4. fy. R9.002) (22S) Other Compression! controlled e.0. /?/. = 0. section geometry. and the calculated values for c.002) (2QQ 0.2.fc.3.Strength reduction factor (Reprinted with permission from the American Concrete Institute) Another nested IF statement was used within the Excel program to determine this value as shown in Figure 4-4. .70 0.600 Transition Tension controlled €.90 0. Mnl and Mn2 respectively. and if required the web and flange moment capacities..002 § = 0. <p = 0.375 Figure 4-3 ACI 318-08 Fig.2 . . A sample Excel worksheet is shown in Figure 4-5.
these cells are represented by lightly shaded columns. Figure 4-4 ACI318-08 calculation of strength reduction factor ^ . This procedure is illustrated by the flowchart in Figure 4-7.55 Figure 4-6 shows the preliminary calculations which were contained in hidden cells.
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No o=- .tf.=^/.) <0.inittri 1 Is ^ < a' .ttfoji'iAKJ ^^^MS'i •i.S5feb) a c=- Is a>t / •No Yes M„=^(M„ 1 + M„ 2 ) a M.f'c.'.bw.d.fyand As ' * ^ .58 0«pdi«« Given b. Is ^ < c' No Yes tf = ((4-A/)/.• * * • ^*Mmm i-iSA-'^Jj^A^ffl fcftSI &:* •W».85/' e 6J (4/.) (0.|rf-Figure 4-7 ACI318-08 excel program logic flowchart .Yes.%Mtaa.
3.3 4.3.1 Maximum Reinforcement The maximum amount of prestressed and nonprestressed reinforcement shall be such that: c <0.1 defines the limitation for maximum reinforcement as follows: 5.3.3.1.3.3.7.7.3.7. (4-8) However. A. . .1-1) d„ in which: A fd +A f d (5.1 AASHTO LRFD Limits of Reinforcement The procedure for determining Asmin as specified by AASHTO LRFD Equation 5.3.) If Equation 1 is not satisfied the section shall be considered to be overreinforced.2-1 did not require the use of any IF logic statements and was determined as shown in Equation 4-8.7bd s mm w f .) de = the corresponding effective depth from the extreme compressive fiber to the centroid of the tensile force in the tensile reinforcement (in.3.42 (5. determination of the AASHTO LRFD value for Asmax did require an IF statement that was dependent on the value calculated for the maximum depth of neutral axis c. AASHTO LRFD § 5.1-2) d^=_sJ_y^ PsJps P + Afy y dps fpi sJ psJ ps where: c = the distance from the extreme compressive fiber to the neutral axis (in.7.3. = 0.3.59 4.7.1.
For clarity within this paper this preliminary value for c will be denoted maxc. the conservative approach and the refined approach.3.42(i (4-10) Using one of the two approaches described in detail in §4.60 The final statement leads to the conclusion that is summed up mathematically in Equation 4-9. c — > 0.2 Location of Neutral Axis c. as shown in Equation 4-10.3.1. and Depth of Compressive Block a The AASHTO LRFD code provides two methods of determining the depth of the neutral axis.2 and solving for c one can readily determine the limiting area of reinforcement allowed. also reference Figure 4-9.3.1.7. maxc = 0. .1-1 equivalent to Equation 4-9 and using the known depth for the centroid of reinforcement the maximum value for the depth of the neutral axis for a section while remaining underreinforced can be found simply.42 -> Overreinforced d (4-9) Setting AASHTO LRFD Equation 5.3. 4.
c=Afy-A\fy\ All AASHTO LRFD analysis performed and discussed in this paper utilized the refined approach during the calculation of the depth of the neutral axis and an EXCEL IF statement was programmed using the value for maxc from Equation 4-10 as shown in Figure 4-8.61 When using the refined approach the method used to determine the depth of the neutral axis is based on the action experienced by the section. i.85/cM. whether or not the beam is under rectangular action or T-section action.e. .S5f'cft(b-bw)hf l n 0. AJy-As\fy\-0. If the beam is under T-section action then c is computed as shown in Equation 411. reference Equation 4-12. The gray box in Figure 4-11 highlights this process within the overall AASHTO LRFD analytical procedure. c= (4-11) Otherwise c is computed by the method described as the conservative approach which is identical to the method used by ACI 318-08.
3 Strength Reduction Factor ^ The strength reduction factor </).9. as defined by AASHTO LRFD also uses the limit of Asmax to determine its value.4. otherwise ^ = 0.£>„ + fy Figure 4-8 Determination of AASHTO LRFD.85/'c/?.85fcj3l(b-bjtf p. 4max=maxc0. 4. If the area of reinforcement is less than Asmax then ^ = 0. .3.7.62 Is maxc>tf } Yes 0.1. See Figure 4-9.
60 ste # Factor 1 i \ A - t | ' — Transition f J • Compression controlled ue " *f (over-reinforced) *r. and strength reduction factor 0.\ * j — * • f for compression (no transition) I i j—+• Over-reinforced Figure 4-9 AASHTO LRFD and ACI 318-08 limits of reinforcement The AASHTO LRFD moment capacity was computed using the identical values forAs. reinforcement yield strength^. „ . . section geometry.002 1 1 J » <* for compression I 0. . as for the ACI 318-08 test cases along with the AASHTO LRFD specific calculated values for depth of neutral axis c.375 Tension controlled (under-reinforced) 0.„ Minimum AASHTO I LRFD i-.005 1 < ACI318-05 Minimum (c/djmi„ ^forbending 1 I Transition 0.BuinT.42 I ^i£_ j ^ .Under-reinforced * ^(f> for bending 0. and concrete crushing strength/' c .„nmMmaBiMS /„„ tm. A sample EXCEL worksheet is shown in Figure 4-10 and Figure 4-11 illustrates this procedure by means of a flowchart. .63 0.
084 0.000 6.9 * &*«m ! ^hflSBiSc.319 7.000 1.153 1.9 0.9 0.002 6.7 As 0.061 7.0 390.9 0.9 0.171 4>.080 4.481 c/d e 0.2 5564.108 0.000 0.892 8.0 480.9 4373.294 0.144 0.9 0.471 1.060 0.167 0.1 11372.490 0. filly l JflHB^^b jHH^S^H^i c>=t.0 600.716 1.0 240. 56 8 0.9 0.0 360.0 630.000 AASHTO .371 6.9 0.595 2.172 3.288 60.096 0.8 1914.000 8.000 2.0 540.9 0.3 3768.012 0.961 2.024 0.285 10.000 4.018 2.865 1.802 8.696 2.9 0.111 4.9 0.036 450.9 0.^ w l f l * JMPMr'-i *• ••MMSht'.0 120.000 5.0 570.9 0.095 a 0.7 1282.64 Moment k*in.000 10.418 0.0 0.741 5.000 3.0 4. Bending 0.0 Figure 4-10 Sample excel spreadsheet for AASHTO LRFD analysis .0 210.8 8283.442 1.263 0.632 7.387 0.9 0.500 8.120 0.4feUws 0.9 Mi 8900.523 a.4 7.000 9.027 10.225 1.0 300.749 4.420 0.0 0.500 6.2 2539.0 416.9 0.9 0.9 4972.7 10912.9 0.206 2.500 10.9 0.500 9.LRFD C kip c 0.156 0.7 9940.356 0.500 11.0 150.000 0.451 2.'.000 11.202 0.500 1.000 0.7 8354.980 1.5 7864.0 12242.036 0.9 0.""i'^roWksa • 'lfflWHIrvi •««3^Wtes £.0 6149.0 0.500 5.5 11816.6 7299.072 0.936 7.500 4.000 30.186 3.9 0.460 3.9 0.132 0.0 90.577 0.2 10434.325 0.730 2.500 2.735 0.0 330.544 9.048 0.1 6727.307 2.9 0.9 0.1 3157.0 510.000 4.941 3.0 270.0 644.500 6.884 3.2 9428.836 5.500 3. fc a>at f • .2 420.578 6.400 3.245 0. :14M: i {.0 180. 0.0 660.262 7.232 0.
v — Figure 4-11 AASHTO LRFD flowchart .bw.S5 feftbw) (4/.S5PJ'c(b-bw)tJ a.85/VA^) a = /^c J Is c<^/ No Yes Mu=tAsffy\d--} a Mu =Msffy(d~yO.fc.d.85/'cfl&J No (0.fvm& As Is tf> AAy sJ 0.tf.65 Given b.85/' c fl(ft-ft i r )f / ) c=- (0.-0.
the lowest value used by Ha. Concrete strength was assumed to be 7ksi.2 4. x 36in. This spreadsheet then converted the unit load into an equivalently proportioned maximum allowable service load which led to the final solution to the problem. code provisions. Geometric section properties were matched to those used by Ha with identical beam sections measuring 44in. then the truss geometries from the first step were solved for using a unit load and classical hand calculations. The solutions for this series of beams were worked in three phases. and reinforcing steel strength was assumed to be 60ksi.1 STM Introduction Each series of deep beams was analyzed using the STM method as described by both the ACI 318-08 and AASHTO LRFD 2nd Ed.2. Results were then compared to the work performed by Ha on identical sections and compared as a percentage of the maximum allowable load to the test load. finally an EXCEL spreadsheet created specifically to solve for the STM model element capacities (struts and ties) based on the previously defined geometry from step 1 was utilized.66 4. first a graphical STM solution was solved to determine the maximum allowable strut and tie member sizes. x 4in. The location of the openings was specific to each series and constituted the only variable .
12in.67 within this problem the locations of which matched those used by Ha. x 12in. Reference Figure 4-12 and Table 4-2. 36in. Figure 4-12 Generic beam with and without opening Table 4.2. -12in- 7\ 4in. OPEN i TJ 4in. 44in.1-1 lists the values used for the x and y coordinates of the lower left corner of the 12in. Table 4-2 Cutout locations from lower left corner in inches Opening Location ""• „ ^ _ X SERIES 1 SERIES 2 SERIES 3 SERIES 4 SERIES 5 N/A 16 16 16 20 y N/A 12 8 16 12 . 40in. cutout.
the value oifcu recommended by ACI in Equation (15.68 The method of STM analysis as employed for purposes of this thesis were for all intents and purposes identical. outside of the values for strength reduction factors was identical." For completeness Equation 15. = smaller of 0.. ..7 from the Naaman text reads: fa.1 is appropriate enough. the strength reduction factors remained unique to each code.85/?/'c (4-13) This upper limit of the effective strength as defined by the AAHTO LRFD of 0. when these assumptions were used the method of STM as described by both ACI 318-08 and AASHTO LRFD 2nd Ed.85/y c was used for all AASHTO LRFD based STM calculations performed in this paper.7) and Table 15. given the uncertainty on the evaluation of the state of strain in the struts. This assumptions greatly reduced the complexity of many of the iterative steps required during the analytical process. Use was made of Naaman's proposal regarding the compressive strength of concrete that ".85/y c or 0.
Figures A-l through A-5. See Appendix A .1 Graphical Solution Pro/E 3-D design software was utilized to provide true scale models and drawings of each deep beam section to ensure that the largest possible strut and tie members had been placed in the section. that is they do not satisfy all of the necessary requirements for a stable and determinate structure.2.1. Geometric model constraints were used to ensure that no strut possessed cross sectional area greater than that of any adjacent node and that all STM members remained prismatic. It should be noted that Figures A-2 through A-5 are unstable trusses. The convention for calculating the determinacy and stability of a structure is shown below.69 4.Deep Beam STM Models. r + b = 2n Stable and Determinate r + b > 2n Stable and Indeterminate D -r + b -2n Degree of Indeterminacy r + b<2n Unstable (4-14) (4-15) (4-16) (4-17) With variables defined as: r = number of reactions b = number of beams/truss members n = number of nodes .
reinforcement yield strength^. Due to the symmetry of the truss models unique solutions for each member were possible. See Appendix C . and beam member . Results from these calculations were used as the input into the EXCEL Spreadsheets. the number of reactions is 3 and the number of nodes is 7.3 Excel Spreadsheets A series of Excel spreadsheets was developed that used concrete strength/'c.1. Therefore r + b = 3 + 9< 2*7 demonstrating that the structure is unstable.STM Truss Free Body Diagrams & Solutions Series 1 through Series 5. basic equations of equilibrium and the method of joints. strength reduction factor(s) ^. the individual truss member forces were solved for using a unit load.70 Using Figure A-2 as an example one can see that the number of members is 9. A thorough examination of "Examples for the Design of Structural Concrete with Strut and Tie Models" reveals that many effective STM models are not considered stable truss structures when evaluated using Equation 4-17.2.2 Hand Calculations After the truss geometry had been determined as shown in Appendix B . for the asymmetric truss model shown in Figure B-6 this is not the case and unique solutions for each member were not found. However. 4.Deep Beam STM Truss Models Figures B-l through B-5. 4. but they are symmetric trusses and provide satisfactory support because the members are confined by the surrounding concrete.2.1.
geometry as inputs to calculate the maximum concentrated load that could be applied. An iterative process that determined STM applicability was used, i.e. after all maximum equivalent forces had been determined and converted into a single point load placed at the midsection of the beam the validity of Vu < </>V„ was checked. If the criteria of Vu < <f>Vn was met then the magnitude of the maximum point load was considered acceptable.
Spreadsheet logic flow followed the procedure shown in Figure 4-12 and verification of the strut capacity at midlength and nodal interface was computed using the code specific relationship details described in Table 4-3.
For all cases documented for this thesis all struts were treated as bottle shaped with the cross sectional area at strut midlength assumed to be greater than the maximum area available at the strut-node interface.
Unit load values found during the hand analysis portion were input into the spreadsheet and the minimum required member width was determined using Pr = 0Pn and Pn = fcuAcs, where Pr is the calculated force within the strut member, Acs represents the Since member depth was
cross sectional area of the strut and/CM = 0.&5f'cmin(fln,/3s).
held constant at 4.0 inches to remain consistent with Ha's experimental beams the specific version of the equation used was,
£ 4.0/«.0.85/'cmin(/fr,/fo)
The minimum required "unit width" calculated from Equation 4-18 was then divided into the maximum attainable widths as determined using Pro/E to obtain the percentage increase available. The limiting case was selected and used to calculate the individual member forces. This value was then used with the previously worked hand solutions and converted into the maximum applied concentrated load, to be compared against the test loads used by Ha in his work. These comparisons formed the basis of this portion of this thesis.
ACI318-08 vs. AASHTO LRFD As shown in Table 4-3 the effective concrete compressive strength^, is
calculated the same regardless of which code is used; however the member effectiveness factors and strength reduction factors used by the two codes were very different.
Strength Reduction Factor ^ ACI 318-08 provision used 0.75 for all STM cases whereas the AASHTO LRFD
provided two distinct strength reduction factors based on the member type; <f> = 0.7 for struts and <j> = 0.9 for ties. Reference Table 4-3 for application.
Strut Effectiveness Factors fis ACI 318-08 contained five distinct strut effectiveness factors J3S, based on strut
geometry, provided strut reinforcement and weight of concrete used. In cases where the provided strut reinforcement does not meet the requirements of ACI 318-08 section A.3.3 the weight of concrete is used to determine the strut effectiveness factor via the modification factor X which was related to the unit weight of concrete; X = 1.0 for normal-weight concrete, X = 0.85 for sand-lightweight concrete and X = 0.75 for alllightweight concrete. However for purposes of simplicity the requirements of ACI 31808 A.3.3 were always assumed satisfied and therefore fis = 0.75. The AASHTO LRFD 2n Ed. code only provided two strut effectiveness factors based solely on strut geometry, i.e. if they were prismatic or bottle shaped struts.
Node Effectiveness Factors /& The number of node effectiveness factors /%, defined by each code was equal at
three; however the two values provided by the AASHTO LRFD 2nd Ed. for C-C-C and CC-T nodes were conservative when compared against their counterparts in the ACI 31808. See Table 4-3 for exact values.
65 /:T=fA .LRFD 0.75 0.3.85p.3 p s (ACI 318-08 Only) Struts in Tension ps (ACI 318-08 Only) All other Cases ps (ACI 318-08 only) Strength of Nodes C-C-C Nodes pn C-C-T Nodes pn C-T-T.75 0 60' 0.3 ps (ACI 318-08 Only) w/o Reinforcement Satisfying A. T-T-T Nodes Pn Strength of Ties 0.3.85 0.85psf'c 1. ' n = 'cu"n ' 'i fcu=0.. : .8 0.40 0.0 0.74 Table 4-3 ACI 318-08/AASHTO LRFD effective strength coefficients ACI 318-08 Strength Reduction Factor (p Strength of Struts Uniform Cross Section p s w/Reinforcement Satisfying A.r .fc 0.9 Tension f =f A fcu=0.85psfc 1.6 fT=fyAS for0.60 ' n — ' cu"n AASHTO .7 Compression 0.0 0.75 f =f A fcu=0..-.0 ' s ' cu"s ' ! • .85psf'c 1.
75 bVuz<pV„=t(ioJfcbM Yes Select Strut & Tie Truss Model to Carry Loads/Reactions No Model Truss Member Dimensions Using Maximum Available Area Solve Truss Model Using Unit Load Determine Equivalent Maximum Member Forces Verify Capacity of Struts at Midlength and Nodal Interface Detail D-Region Verifying Minimum Reinforcement Requirements have been Met Figure 4-13 Strut and tie modeling procedure .
1 Flexure Twelve beam sections were analyzed using both the ACI 318-08 and AASHTO LRFD code provisions. The section geometry and variables used were: 'W Figure 5-1 Graphic representation of section geometry Where the fixed values used for these dimensions were as follows: Flange width Flange depth Web depth Depth to centroid of reinforcement b = 36in. reinforcement limits were held respective to the provisions of each code. bw= 14in. The reinforcement provided was varied over the range and inclusive of the limits Asmi„ to Asmax.0 RESULTS and DISCUSSION 5.76 5. d = 24in. tf = 4in. .
The difference between results for Asmin and Asmax as calculated using the ACI 318-08 and AASHTO LRFD varied greatly depending on the variable in question.375.283 1.1.000 0. Table 5-1 highlights these differences via comparison of the slope of the lines generated when the limits Asmin and Asmax were calculated using different section variables.63 89.09 From Table 5-1 several things become obvious: section flange depth.29 83.028 0.253 0. whereas flange depth will have the greatest impact upon the difference between the slope of the lines generated when determining the maximum allowable area of reinforcement as provided by the ACI 318-08 and the AASHTO LRFD.193 0.227 0.207 0.880 0.048 0. regardless of the code used.000 0. ACI318-08 versus ASHTO LRFD There were no cases where the ACI 318-08 calculated value forAsmax would allow the ratio of y.00 63.00 117.25 63.25 0. to be greater than 0. b and //-have no impact on the minimum amount of reinforcement allowed.000 0. The percentage difference between these results is .044 0. Table 5-1 Rates of change for ACI 318-08 and AASHTO LRFD Asmin and A.060 "smax 0.1 Asmin and Asmax.293 0.65 70.77 5.000 0. a change in web width will have the greatest impact upon the difference between the slope of the lines generated when determining the minimum allowable area of reinforcement between the ACI 318-08 and the AASHTO LRFD. Slope ACI-381-05 Variable b bw d tf Slope AASHTO LRFD A A % ACI-AASHTO "smin "smax A "smax 0.076 0.
It is of interest to note that the difference between the codes for minimum reinforcement when based on section flange width and section flange depth. b and tf respectively. was consistent between the codes while the difference between the codes for maximum reinforcement was always less when using the ACI 318-08 than the AASHTO LRFD except when comparing results found using the flange width b as the variable. Reference Figures 5-2 through 5-5.65% faster than for those produced when using the AASHTO LRFD provisions. . These percentages were calculated by dividing the values for the slopes of the results found using the ACI 318-08 by the slopes of the results found using the AASHTO LRFD. In this case the slope of the results found using ACI 318-08 increased by a rate of 17.78 tabulated in the far right columns of Table 5-1. The slopes of the results found using ACI 318-08 for all other variables were an average 81% less than the corresponding values calculated using the AASHTO LRFD.
L f 26in *BB> I HH> •t 14in. — 36 42 44 46 48 38 40 Flange Width.A A S H T O .• . b (in.) Figure 5-2 Asmax and Asmin as a function of b . . — b t 24in.-ACI318-08 Asmin -AASHTO-LRFDAsmin -ACI318-08 Asmax .LRFD Asmax 4in.
oo CO o I C O * i 26i c X (0 _ .3 o a o u a .- o .80 c E <n < Q 14- X E U) < _J 4in. £ < E < Q LL 01 18-0 18-0 oo CO 36in.a I a U4 J2 1 -SO 1 s W) <b6-4? ('Uj) 'xeuusv ^ ujuisv . rl < * s { I C O + I < s '' _c ~ i 1 _ r«fK> • MM *o o CM .
1 0 >< re •—-~i jj E r 8 08 c 1 6 v> < A A .16 -ACI 318-08 Asmin ^ •AASHTO-LRFDAsmin •ACI 318-08 Asmax -AASHTO-LRFDAsmax 14 12 c ••=.) Figure 5-4 Asmax and Asmin as a function of d . -i 24 ? — --6—1 i32 o 34 S 36 14in 26 30 28 Steel Depth. d (in.
t.) Figure 5-5 ^4jmax and Asnun as a function of tf oo to .18 -ACI318-08Asmin -AASHTO-LRFDAsmin •ACI 318-08 Asmax -AASHTO . zd_L 26in. 1 24in. (in. O < 6 ft ? — [==t ? — t=jf ? — ^ = * 7 5 6 ^ _ A 10 14in.LRFD Asmax 16 14 ~ . 12 _ c X g 10 < tf eg c 8 36in. - Flange Depth.
09 99.17 % 99.01 1189.10 1178.85 98. Table 5-2 Ratio of AASHTO LRFD to ACI 318-08 for M„ per unit .04 1171.07 99.56 1201. the average difference was 0.04 99.83 5.93%. All results presented in this section in the form of tables or percentages represent a ratio between the results found for the ACI 318-08 and AASHTO LRFD code requirements for determining flexural capacity.67 1495.79 1196.79 1481.20 1193.01 19.53 99.63 99.4.13 99. reference Equations 5-1 through 5-3.43 1197.04 1167.30 1178.13 98.97 13.34 bw 12.54 99.31 %AVG 99.13 tf 11.69 98.10 16.69 .13 d 13.66 99.27 1186.11 1185.34 99.34 1176.2 As versus Mu During the determination of the flexural strength of a section based on the area of reinforcing steel by either ACI 318-08 or AASHTO LRFD code.39 1163.61 1164.78 1782. certain geometric dimensions were found to have greater influence than others on the amount of reinforcement required to resist a given moment.05 1152.49 1186.22 A MAX 10.80 1133. VARIABLE b VALUE 36 42 54 14 18 21 24 30 36 2 6 10 ACI 318-05 AASHTO LRFD 1189.58 1798. results are summarized in Table 5-2 and Figures 5-6 through 5-9.22 98.11 1178.35 99. It was observed that sections analyzed using AASHTO LRFD code consistently required a larger area of steel to support the same moment than that required for an otherwise equivalent ACI 318-08 section.79 1189. the difference in corresponding moment capacity was never greater than 1.1.4%.31 % MAX 99. For all cases analyzed.30 AAVG 9.92 99.13 99.19 99.
M„ = 0MU. the range Asmin to Asmax as defined by each code.84 The percentage difference ratio as shown in Table 5-2 is the sum of the applied factored moments divided by the sum of the areas of reinforcement required for a section analyzed using the AAASHTO LRFD to carry that moment over the entire viable range of reinforcement.e. A% = where (AASHTO)Mu:As (ACI)MU:A •* 100 J (5-1) Asmax AASHTO. divided by the area of reinforcement required for the ACI section.e. i. The applied moment is a representation of the applied loads only however it is used in the determination of required reinforcement during design or analysis of a section via the calculated resistive moment of the section. This was considered a convenient method of quantifying the difference in results produced by each code for the area of reinforcement required to resist an applied moment on a per unit basis. Therefore in essence this percentage represents a comparison between the ACI 318-08 and the AASHTO LRFD of the allowable applied load as defined by each code for based on each cross section's geometry.MU:AS=^^Asmin (5-2) and . applied moment per unit area of reinforcement. Equations 5-1 through 5-3 describe the formulae used in the calculation of this percentage more concisely. i.
85 As max v4C/. i. although this amount averaged only 1.M « • yl = As max ^sas— •smin s IK is max (5-3) /f^min 1-4. the AASHTO LRFD will provide a more conservative solution for the resistive moment per unit of reinforcing steel. Examination of the data in Table 1 highlights two consistent traits between the codes: a) a section designed using the AASHTO LRFD code will always require a larger area of reinforcing steel than an equivalent section designed using the ACI 318-08 code. has the greatest effect on the ratio of steel to flexural strength for any arbitrary T or L flanged section.40% for the section geometries studied. . b) web width bw.e.
Mu (k*in.) o b M O O O -N O O CD 00 o o o o o o o o o o o to -N o o o o o o o> o o o 00 o o o to b CO b b b (D = • OS n> o CO b oo b n s o g to 1 < 65 b »' cr E|P| • SHT o CT II * • a • SHT X CO 1 O • > 318 • 00 CO 1 1HS > > > > > > > > > > > > > o > o > O m » M jo. o o > > > a a HUH q O ait en i q O cr II *.Resistive Moment. p' •c 3 <o to 3" ? p a> p a> 5' 98 . cr CO en en A. M q O cr II co O) II ui *• M <) > W 00 3 <n •b- > V) X o <) > (A o cII co > < > / X o () > w X n j^ u co <" > (> / X o —t o oII CT " ^ cr -fc.
0 11. ACI 318-08 Asmax for bw= 18 in. 0) 0.0 13. AASHTO Asmax for bw= 18 in. As (in. ACI 318-08 Asmax for bw= 21 in.0 2.0 12.0 2 7. AASHTO Asmax for bw= 14 in.0 Area of Steel.0 6. » .0 10. 4in.14000 - -ACI bw=14 -ACI bw=18 -ACI bw=21 -AASHTObw=14 -AASHTObw=18 -AASHTO bw=21 ACI 318-08 Asmax for bw= 14 in.0 8.0 4.0 3.0 1.0 5.0 9.0 14. ) Figure 5-7 As versus M„ for variable Z. AASHTO Asmax for bw= 21 in.
0 10.0 15.0 12. AASHTO Asmax for d= 24 in.0 6. A ACI 318-08 Asmax for d= 24 in.i (A d t 26in I 1.0 13. 36in.-ACI d=24in.0 5.2) Figure 5-8 As versus M„ for variable d oo oo . • • £ o •*m— 4in. -AASHTO d=24in. -AASHTO d=30in. AASHTO Asmax for d= 36 in. • O c • 3 15000 - ACI 318-08 Asmax for d= 36 in.0 14. -AASHTO d=36in.0 3. ACI 318-08 Asmax for d= 30 in.0 7.0 4.0 11. -ACI d=30in. As (in.0 8.0 14in.0 2. AASHTO Asmax for d= 30 in. i > \ .0 9. -® '' 0) 0.0 Area of Steel. -ACI d=36in.
0 14in.0 Area of Steel. 6000 24in. • D • ACI 318-08 Asmax for tf= 10 in.0 16. AASHTO Asmax for tf= 6 in. T 26in. ' -AASHTOtf=6 -AASHTO tf=10 A • 0 ACI 318-08 Asmax for tf= 2 in.0 10.0 6. AASHTO Asmax for tf= 2 in.0 2.-ACI tf=2 -ACI tf=6 -ACI tf=10 -AASHTO tf=2 .0 18." 12. ACI 318-08 Asmax for tf= 6 in. AASHTO Asmax for tf= 10 in.2) Figure 5-9 As versus Mu for variable tf 00 . A s (in.0 14. 10000 - 8000- 36in.0 4.0 8. 01 0.
Because the value for Asmax that was produced by the ACI 318-08 was always less .3. or whether the section is in rectangular or T-section behavior when using ACI or AASHTO LRFD Simplified Conservative Approach. c=A.85/' c fl6 c AJy-A\fy-^5fcP. 4Jy-A'sfy 0.2. which was the method used for all AASHTO LRFD based calculations in this paper.fy-A.85/' e /?A. that is rectangular or flanged. The formulae are repeated below for convenience. differs when sections exhibit flanged behavior as described earlier in §4.250 square inches increase in the area of reinforcement provided. and Depth of Compressive Block a The depth of the neutral axis has been conventionally defined as described by Equation 5-4 which is used by both the ACI 318-08 and AASHTO LRFD.fy 0.3 Location of Neutral Axis c.1.85/^6 (5-4) The AASHTO LRFD refined approach.{b-bJhf 0. This equation applies regardless of section geometry. this range was inclusive of the lower limit of Asmin and the upper limit of Asmax as found by the ACI 31808.90 5. The results for the depth of the neutral axis using Equation 5-6 were computed for every 0.1.
681 2.409 0.409 97.234 2 2.27 0. From these values the arithmetic mean was calculated.130 100. The AASHTO LRFD provided the option to neglect any additional compressive strength contributions that the flanged area.580 21 2.06 6 2.539 2.056 30 2. would provide.539 2.839 24 2.143 101.95 42 2.31 .073 18 2.140 93. minus the web width.540 2.43 d 0.83 bw 0. When the option to use the AASHTO LRFD refined approach was not taken both codes would produce identical values for the depth of the neutral axis.602 3.750 36 3. The results are listed in Table 5-3. Table 5-3 Results for location of neutral axis Depth of Neutral Axis ACI vs.181 14 2.167 A %max 94.086 97.102 54 2. both in the approach used in and the results obtained.827 2.874 10 3.130 2.158 0.02 0.409 98.04 tf 0. AASHTO AVE[RAGEc A ACI AASHTO VALUE AVG A %AVG 36 2. This option to choose the level of fidelity in the calculation of the depth of the neutral axis was a significant difference between the two codes.130 102.695 2.732 2.745 VARIABLE b Amax 0.014 2.91 than that found by using the AASHTO LRFD it was used as the limiting value for this comparison.224 2.141 3.
that is when the section is no longer in rectangular action and the refined approach described by Equation 5-6 was utilized. C = 0. artificial or otherwise.375 for all ACI 318-08 analyses and y. was always less than or equal to 0.85 f'c f5xcb .e. or 18 inches for the sections analyzed. Reference Figures 5-10 through 5-13. This abrupt change in slope is due to the direct proportionality of the tensile force Tto the compressive force C. T = C. i. i.e. this provided a larger compressive area and hence required a larger area of tensile reinforcement to obtain force equilibrium.92 As shown in Table 5-3 the results using the refined approach of Equations 5-5 and 5-6 produced a greater depth for the location of the neutral axis.9 for every test point. Indirectly this states that the ratio y. of the section. a balanced section. was equal to 0. This effect was maximized when bw was equal to one-half the value of flange width b. The point at which the slope change occurs is when the depth of the neutral axis exceeds the flange depth tf. T = C . was always less than or equal . had the greatest impact on the difference between results obtained from ACI 318-08 and AASHTO LRFD codes. Notice should be made of the abrupt change in slope that is attributable to the effect of provided reinforcement in the calculation of the location of the neutral axis. 5. Web width bw. the strength reduction factor </>.4 Strength Reduction Factor <f> Due to limits placed on the sections analyzed in this paper on the allowable amount of reinforcement.1. and the role the location of the neutral axis has in the calculation of the compressive force as shown in Equation 3-5.
Because the value of <f> never differed between the ACI 318-08 and the AASHTO LRFD for the shallow beam flexural analyses performed in this paper no additional comparisons on this particular code provision were made.93 to 0. .42 for all AASHTO LRFD analyses.
= —•—AASHTO b = 42 in. 3 0) £ & 4.0 —A— ACI b = 36 10.0 ^ 1 —o-ACI b = 54 O 8. As (in.0 14.0 2.( 4in.0 Area of Reinforcement Steel.0 10. = < £ 6.0 0.* — AASHTO b = 54 in.) Figure 5-10 As versus c for variable b .( —*-AASHTO b = 36 in.0 16.0 8.I Q 2.0 4.12.0 12.0 6.0 0. = .
12.000 -*— ACI bw=14 -*—ACI bw=18 -o—ACI bw=21 -*— AASHTO bw=14 -+—AASHTO bw=18 -I—AASHTO bw=21
4ia 36 in.
24 in. 26in
a. « Q
0.000 2.0 6.0 4.0 8.0 10.0 12.0 14.0
Area of Reinforcement Steel, As (in. )
Figure 5-11 As versus c for variable £„
-4—ACI d=24in. -»—ACI d=30in. -*— ACI d=36in. -*—AASHTO d=24in. -I—AASHTOd=30in. -K—AASHTOd=36in.
14 n -
>f Neut ral Axis>,c (in.
—-. to n _
0.0 3 0.0 2.0 4.0 6.0 8.0 10.0
Figure 5-12 As versus c for variable d
ACI tf=2 ACI tf=6 ACI tf=10 AASHTO tf=2 MSHTOtf=6 AASHTOtf=10
Area of Reinforcement Steel, As (in.)
Figure 5-13 As versus c for variable tf
A 4in. 36in. Section geometry and variables are shown in Figure 5-14 and Table 5. Cutout Location ^-^^ X y SERIES 1 N/A N/A SERIES 2 12 16 SERIES 3 16 8 SERIES 4 16 16 SERIES 5 20 12 .2 STM Five deep beams previously tested by Ha were reevaluated using the Strut and Tie Method (STM) as described by both the ACI 318-08 and AASHTO LRFD code provisions.4. Figure 5-14 Generic beam with and without opening Table 5-4 Cutout locations from lower left corner in inches 12in. l-^— 40in. x 12in. 12in. 44in.98 5. OPEN -12ia4in.
2.00 100. Table 5-5 Predicted load versus maximum test load comparisons SERIES 1 2 3 4 5 Ha 175.17 98.89 A% 4.87 4. this was consistent with expectations given the conservative nature of the STM model process.04 68.90 99.89 % 68.12 119. The gray columns in Table 5-5 represent the STM method predicted loads as a percentage of Ha's test loads.73 7.67 80.00 125.04%.68 106.35 73.99 5.57 4. The predicted values for all other cases analyzed were well below the experimental test loads used by Ha.67 % AASHTO . All design loads and their corresponding ACI 318-08 and AASHTO LRFD counterparts are summarized in Table 5-5 and are shown graphically in Figure 5-15.00 5.78 Of the five comparisons made only series 4 produced predicted load values that exceeded the test load used by Ha.00 ACI 318-08 127.53 86.63 86.25 63.STM versus FEA In the work performed by Ha the maximum test loads applied to the sections varied from 70kips to 175kip depending on the beam geometry. this was only true when the ACI 318-08 code provisions were utilized and this exceedance was 5.04 80.04% of the actual test load value used by Ha of 70 ksi.97 70. In the instance of series 4 several factors can explain the fact that the predicted safe design load inclusive of safety factors was greater than the load actually reached .00 70.LRFD 73.43 68.1 Maximum Allowable Concentrated Load .26 105. The magnitude of the predicted loads calculated using the AASHTO LRFD in series 4 were 98.54 79.98 66.00 150.96 85.
. 0. When a comparison was made between the area of reinforcing steel required by each code the AAHSTO-LRFD code requirement was 77. J ^C/318-08 _ „ .3 = 77. .9 0/ (5-8) 5.STM versus FEA A second comparison was made between the experimental design loads used by Ha and the analytical results derived in this paper without application of the strength reduction factor ^.75 ___ %Loaa— =>93.3% of that analyzed using ACI 318-08. These percentages were directly attributable to the different values specified for the strength reduction factors by each code. . %Load Mcnn-m </>AASHTO-LRFD . The load predicted as analyzed using AASHTO LRFD was always 93.2 Maximum Allowable Concentrated Load without </> . Analytical and predictive calculations make several assumptions any of which can be the source of this discrepancy: • • • • compressive strength of concrete is consistent throughout the section placement of the concrete is without voids or inconsistencies area and depth of reinforcement are exact cross section geometry and member length are exact In every series the maximum load the beam was predicted to be capable of carrying was greater when the ACI 318-08 code was used.8% of that of the ACI 318-08 sections.2.8 0AASHTO-LRFD 0. Reference Equations 5-7 and 5-8. All design loads and their corresponding ACI 318-08 and AASHTO .100 during Ha's laboratory testing.
Table 5-6 Predicted load without # versus maximum test load comparisons SERIES 1 2 3 4 5 T.l A% 0.00 0.101 LRFD counterparts without the application of </> are summarized in Table 5-6 and are shown graphically in Figure 5-16.00 ACI 318-08 170. 175.62 114. .00 0.00 125.04 115. This finding clearly demonstrates the importance of using the strength reduction factor to account for any imperfections during the mixing and placement of concrete as well as the location of reinforcing steel.H.79 98.00 In the cases of both series 4 and 5 it can be observed that the predicted strengths have exceeded the actual test loads as determined by Ha.00 0.00 0. The strength reduction factors assigned by the ACI 318-08 and AASHTO LRFD codes are very different from each other and therefore would never produce identical results for the same section same regardless of truss or member geometry.39 S4<53 140.00 100.79 98.00 70.56 % 97.04 115.56 AASHTO LRFD 170.50 91.24 141.06 118.00 150. The removal of the strength reduction factors also illustrated the significance of largest difference between the ACI 318-08 and AASHTO LRFD when using the method of strut and tie models as shown in Table 5-6. Again the gray columns in Table 5-5 represent the STM method predicted loads as a percentage of Ha's test loads.56 m ' mmi .24 141.62 114. only by ignoring the contribution of the strength reduction factors altogether could identical results be achieved for each code.
maximum test load comparisons .00 IT.00 Load Values 80.00 Figure 5-15 Predicted load vs. maximum test load comparisons Load Comparison Without 4 T.00 120. Ha's Loads ACI 318-08 Loads AASHTO .LRFO Loads Figure 5-16 Predicted load without i vs. Ha's Loads IACI 318-08 Loads IAASHTO-LRFD Loads 0.Load Comparison 140.
Many other differences existed but would only be noticed if prestressed reinforcement was incorporated into the design. flange depth t/. Maximum load capacity was predicted for a concentrated load located at the center on the top face of the beam. x 12in. opening.1 Summary Twelve different flanged shallow beam sections were designed and analyzed for flexural resistance by applying both the ACI 318-08 and the AASHTO LRFD code provisions using hand calculations and analytical software developed by the author.„ for each section was calculated for reinforcing steel amounts that were increased by increments of 0. Differences found between the code provisions included but were not limited to the method prescribed for finding the allowable limits of reinforcement or the method used for determining the location of the neutral axis. each category represented a geometric feature to be varied. hand calculations and analytical software developed by the author. flange width b. The methods employed to reach solutions for both .0 SUMMARY and CONCLUSIONS 6. web width bw. Flexural shallow beam sections were divided into four categories. The deep beam sections investigated were identical.103 6.50 square inches. An additional five rectangular deep beam sections were analyzed using the method of strut and tie models by applying both the ACI 318-08 and AASHTO LRFD code guidelines. apart from the inclusion and location of a 12in. utilizing Pro/E CAD software. this fell beyond the scope of this study. and the depth of the centroid of reinforcement dt. Factored moment capacity M.
2 As versus Mu for Flexure Web width bw. had the greatest impact upon the minimum allowable reinforcement for a given section.4% of each other.104 sets of problems could vary greatly between the two codes however there were no significant differences in the end results derived from using either set of provisions.2. and the flange depth t/.3 Depth of Neutral Axis c for Flexure The method for calculating the depth of the neutral axis c.2.2.2 6. Significant differences were encountered between the ACI 318-08 and AASHTO LRFD provisions when analyzing or designing sections for moment capacity in regards to the method prescribed for finding the allowable limits of reinforcement.1 Conclusion Asmin and Asmax for Flexure Web width bw. had the greatest effect on a given section's moment capacity regardless of what code was utilized. whereas the minimum allowable reinforcement Asmin. 6. was the most noticeable difference between the ACI 318-08 and AASHTO LRFD code provisions. 6. However the results calculated for sections from either code discussed in this paper were . 6. In spite of these differences the results obtained from both codes for all cases analyzed within this paper were always within 1. had the greatest influence upon the maximum allowable reinforcement Asmax for a section.
These differences in maximum predicted load capacity ranged from 4.105 never more than 9% apart. was always equal to 0. Results using non-reinforced sections were never considered for this thesis. the strength reduction factor.9 and therefore <j) did not have a differing affect on the results obtained from either code.2. plain reinforced sections were analyzed. <f>.00%.2. 6.4 Strength Reduction Factor $ for Flexure Due to the limits of reinforcement placed each section by both codes. 6. .5 Maximum Load Capacityfor STM Maximum predicted load capacity for any of the beam series studied was always greater when analyzed using the ACI 318-08 Appendix A STM method provisions than when using the AASHTO LRFD provisions. This was attributed to the fact that only non-prestressed.57% to 7. Inherent differences in member effectiveness factors and the strength reduction factors were the source of this variance.
and Michael P. Pawan R. and Adrian R. 2005. Sun Thuc. American Concrete Institute. San Jose State University. AASHTO LRFD. Notes on ACI 318-05 Building Code Requirements for Structural Concrete with Design Applications 9th ed.4 (2001): 537-47." ACI Structural Journal 100. Malik. Portland Cement Association. Washington DC. Stephen J." ACI Structural Journal 103.C. American Association of State Highways and Transportation Officials. Farmington Hills. "Toward a Unified Nomenclature for Reinforced-Concrete Theory. Ha. R.5 (2002): 569-77. IL." Journal of Structural Engineering 122. Gupta. 2004. "Behavior and Efficiency of Bottle Shaped Struts. 2005. Thomas T. 3rd ed. 6th ed.. Notes on ACI 318-02 Building Code Requirements for Structural Concrete with Design Applications.. IL. Design of Concrete Deep Beams with Openings and Repair Using Carbon Fiber Laminates. McGraw-Hill. Pearson. Mechanics of Materials. Hsu. and Chia-Ming Uang.. Mahmoud E. Prentice Hall. AASHTO LRFD Bridge Design Specifications. Fundamentals of Structural Analysis. Skokie. I .3 (1996): 275-83. Collins. Skokie. ACI Committee 318. Sankovich et al. and Commentary (318R-08). Kenneth M. 2002. Kamara. 2005. 2002. American Concrete Institute. Leet. Building Code Requirements for Structural Concrete 318-08." Journal of Structural Engineering 128. San Jose. Farmington Hills. 2002. Attila. 1st ed. Brown. 2002. Rabbat. CA. Beres. "Evaluation of Efficiency Factor Models used in Strut-and-Tie Modeling of Nonflexural Members. Portland Cement Association.. Building Code Requirements for Structural Concrete 318-02. Foster.3 (2006): 348-55. Upper Saddle River. New York.106 Works Cited ACI Committee 318. NJ. "Evaluation of Shear Design Procedures for Reinforced Concrete Members under Axial Compression. and Basile G. Hibbeler. 8th ed. Michael D. and Commentary (318R-02).. C.
. and Michael P. 2" ed. Khaldoun N. Naaman. MI. Antoine E.. Antoine E. Collins. 2006. Techno Press 3000. Huang.4 (2003): 277-82. Daia. McCormac. "Strut-and-Tie Design Methodology for ThreeDimensional Reinforced Concrete Structures. American Concrete Institute. Karl-Heinz. 2004. Englewood Cliffs. et al. 2003. and Gustavo J. Al-Shaleh. Parra-Montesinos. Tan K. NJ. Naaman. Strut and Tie Model for Deep Beam Design . Prestressed Concrete Analysis and Design Fundamentals.107 Leu. Cheng. James T. Hoboken. Wiley. and Khaled S.H. and James K.6 (2004): 872-78.. Ann Arbor. Design of Reinforced Concrete ACI 318-05 Code Edition.. James G. Wright. Pretence Hall. MI. 7th ed. Jack C. 1997. Reineck." Journal of Structural Engineering 132." ACI Structural Journal 98. Limits of Reinforcement in 2002 ACI Code: Transition.A Practical Exercise using Appendix A of the 2002 ACI Building Code. Liang-Jeng. Rahal. 3rd ed. Farmington Hills. NJ. and G.5 (2006): 673-85. ACI Structural Journal 101. MacGregor. 2002. and Solution. "Size Effect on Shear Strength of Deep Beams: Investigating with Strut-and-Tie Model. Flaws. . "Minimum Transverse Reinforcement in MPa Concrete Beams.5 (2006): 686-93.Mechanics and Design." Journal of Structural Engineering 132. Rahal. Nelson. "Experimental Evaluation of ACI and AASHTO LRFD Design Provisions for Combined Shear and Torsion. "Critical Inclination of Compression Struts in Concrete Beams. Reinforced Concrete . H. and Thomas Vogel. Examples for the Design of Structural Concrete with Strut and Tie Models.2 (2004): 209-18." ACI Structural Journal 101. Concrete International.6 (2006): 929-38. Khaldoun N. Zwicky." Journal of Structural Engineering 132.
1 Deep beam series 1 STM model o .41 5.41 Figure A .Deep Beam STM Models SERIES 1 MEMBER MINIMUM WIDTH 4.Appendix A .011 1 2 i 5.
64 5.00 5.00 5.Load P MEMBER SERIES 2 MINIMUM UNIHH 5 62 1 4.64 2 3 4 5 6 7 4.00 4.00 Figure A .2 Deep beam series 2 STM model o v© I .62 4.00 « 9 4.
3 Deep beam series 3 STM model .00 4.&0 1 Z 3 4 5 6 7 9 i 4.64 4.00 Figure A .00 4.84 5.SERIES J MEMBER MINIMUM WIDTH 5.00 5. so 4.00 5.
00 5. U Z 3 4 5 6 7 9 5.4 Deep beam series 4 STM model .59 Figure A .00 4.00 5.MEMBER SERIES 4 MINIMUM KIHTH 1 5.53 4.24 4.00 $ 3.63 4.
O0 5.62 4.5 Deep beam series 5 STM model K> .63 5.S3 e 3 4.47 4.62 3.38 2 3 4 5 6 1 3.Load MEMBER SERIES 5 MINIMUM WIDTH 1 5.38 5.00 Figure A .
Deep Beam STM Truss Models Load P 3 6 .Appendix B . CO O T Figure B .1 Deep beam series 1 STM truss model .
Load P 5 OPEN 12.COIO.05- Figure B .2 Deep beam series 2 STM truss model .DO- 2.
OPEN Figure B .3 Deep beam series 3 STM truss model .
00 - 2.go- Figure B . 00 14.4 Deep beam series 4 STM truss model CTs .00.Load P s OPEN 16.
00- Figure B .89- 26.00 24.76- .5 Deep beam series 5 STM truss model -~j .008.00- OPEN 12.Load 34.
6 Deep beam series 5a Alternate STM truss model . C O 26.83 Figure B .36.
Appendix C .0 Element Force 0.468 From symmetry Ft = F3.566 C 0.500 9a a 0.STM Truss Free Body Diagrams & Solutions Deep beam series 1 free body diagram and solution Truss Element 1 2 3 34.566 -> +SFX = 0 => F2 -0.Fx sin 62.10° = 0 Fl = 0.-1 = 0 RR= 0.500 .10° 18 t +2F r = 0 => 0.500 t +IFr = 0 => RA + 0.566 .566 C ^ + IMA =0=>RB* 36m.500 JL F2 34 0= tan"1 — =>0 =62.".10° = 0 F2 = 0.464 T 0. F3 = 0.566 cos 62.1 = 0 RA =0.500 .
0 18.Deep beam series 2 free body diagram and solution 33.14 24.500 C 0.0 36.625 T 0.664 C 0.500-1 = 0 RA = 0.0 Truss Element 1 2 3 4 5 6 7 8 9 Element Force 0.500 C 0.664 C 0.0 0.0 10.437 T £+2MA = 0 => RB*36in.625 C 0.500 t + L F r = 0 => ^ +0.-l = 0 RB= 0.800 C 0.0 26.800 C 0.500 .
F8 = 0 FQ = 0.81° 7 8.625 t +SF7 = 0 => 0.81° = 0 F6 = 0.437 From symmetry F.800cos38.800 -> +EFX = 0 => F2 -0.81° = 0 F9 = 0. = F3. = tan-1 — =>0ff =48. F4 = F1 & F5 = F6 .6 = tan"1 — z*9 =38.66°.800 cos 38. sin 38.66° .500 -F6 sin 48.66° = 0 Fx = 0.66° = 0 F2 = 0.500 -> +1LFX = 0 => 0.66° 10 t + S F r = 0=> 0.800sin38.664 cos 48.F7 = 0 F1 = 0.664 -> +HFX = 0 => F9 -0.625 0.0 t +2F r = 0=> 0.500-F.
378 T 0.0 ^ + TMA = 0 => RB * 36m.710 C 0.1 * 18w.0 32. .86 Truss Element 1 2 3 4 5 6 7 8 9 Element Force 0.1 = 0 ^=0.627 C 0.504 T •0. = 0 RB = 0.02 18.500 .Deep beam series 3 free body diagram and solution 32.378 C 0.0 RA 0.500 C 0.710 C 0.98 36.500 t +ZFY = 0 => RA + 0.500 C 0.0 RB 3.627 C 0.500 .
6n = tan-1 — =>0a. = 44.F sin 52.504 From symmetry Fl = F3.627cos52.Fn = 0 F7 .378 F8 eff = tan-1 ^ ^ =* 0.500 .0. F4 = F1 & F5 = F6 .378 t +EFr = 0 => 0.710cos44.98 t +Siv = 0 => 0.500 .627 sin 52.77° = 0 Fg = 0.52.02 t +ZFr = 0 => 0.627 cos 52.95° .95° = 0 ^ F2 = 0.95° .0.710 -» +EFX = 0 => F9 -0.77° = 0 F6= 0.500 ->+EF x = 0^0.95° 3.95° = 0 Y Fx = 0.F6 sin 44.627 -» +SFX = 0 => F2 .F 8 = 0 Fa = 0.77° r 14.
77 27.813 T 26.0 36.500 t +ZFr = 0 => ^ + 0.MA =0=>RB* 36m.500-1 = 0 RA = 0.651 C 0.-1 = 0 RB = 0.124 Deep beam series 4 free body diagram and solution 32.500 C 0.85 12.0 Truss Element 1 2 3 4 5 6 7 8 9 Element Force 0.417 T 0.651 C 0.954 C 0.500 C 0.417 C 0.0 ^ + T.954 C 0.500 .
F6 sin 31.954 cos 31.651sin50.19° = 0 F 2 = 0.0 .0 t +SFK = 0 z=> 0.19°-F7=0 F7 = 0.=0=>0.F.500 ->+EF A . 4 92 0t7= tan-1 . =0.59° = 0 -> +SF X = 0 => F9 .813 From symmetry Ft = F3.59° = 0 F9 = 0.417 .651cos50.500 .651 ->+LFy = 0 ^ F 2 . =0. sin 50. F4 = F7 & F5 = F6 .500. 6 5 lcos50.19° 10 t +EF r =0=> 0.19° = 0 F.417 t+SFr=0^0.19°-F g =0 F.0.954 8.59° .^ = > 0 7 =31. F6 = 0.0= tan-1 — =$0= 50.
500 t + S F r =0=>RA +0.640 C 0. = 0.-1 = 0 RB = 0.500-1 = 0 R.831 C 0.640 C 0.500 C 0.664 T 18.831 C 0.0 ^ + ZMA = 0=>RB* 36/n.Deep beam series 5 free body diagram and solution Truss Element 1 2 3 4 5 6 7 8 9 Element Force 0.399 C 0.500 C 0.399 T 0.0 29.81 36.500 .
Fg = 0 Fa = 0. = 0 => 0. = 36.399 0.42° .F6 sin 36.97° = 0 F6 =0.^ .500 . = F3.0. = 0.831 cos 36.664 o on From symmetry F.e = tan"1 — ^0= 51.42° = 0 F.42° 6.0.640 cos 51.97° f y 11.sin51.500-F.500 -> + I F .399 f +SFr = 0 => 0.F4 = F7 & F5 = F6 .81 t +ZFr = 0 => 0. = tan"1 .831 -> +1FX = 0 => F9 .=> 0.42° .640 sin 51.640 -> +EFy = 0 => F2 .19 t + S F r =0=>.640 cos 51.97° = 0 F9 = 0.F7 = 0 F7 = 0.42° = 0 F2 = 0.
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