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Shear design of pile caps and other members without transverse reinforcement Zhou, Zongyu 1994

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SHEAR DESIGN OF PILE CAPS AND OTHER MEMBERS WITHOUTTRANSVERSE REINFORCEMENTByZongyu ZhouB.Eng. Tongji University 1982, M.Eng. Tongji University 1987A ThESIS SUBMITTED IN PARTIAL FULFILLMENT OPTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinTHE FACULTY OF GRADUATE STUDIESDEPARTMENT OF CIVIL ENGINEERINGWe accept this thesis as conformingto the required standardTHE UNIVERSITY OF BRITISH COLUMBIAJanuary 1994© Zongyu Zhou, 1994In presenting this thesis in partial fulfilment of the requirements for an advanceddegree at the University of British Columbia, I agree that the Library shall make itfreely available for reference and study. I further agree that permission for extensivecopying of this thesis for scholarly purposes may be granted by the head of mydepartment or by his or her representatives. It is understood that copying orpublication of this thesis for financial gain shall not be allowed without my writtenpermission.(Signature)Department of V LThe University of British ColumbiaVancouver, CanadaDateDE-6 (2/88)AbstractThis thesis deals with the shear design of structural concrete members without transverse reinforcement. The three major parts of this study are the transverse splittingof compression struts confined by plain concrete, the development of a rational designprocedure for deep pile caps, as well as a general study of the shear transfer mechanismsof concrete beams.Three-dimensional compression struts that are unreinforced and confined by plainconcrete, as occur in deep pile caps, were studied both analytically and experimentally.Based on the study results, bearing stress limits are proposed to prevent compressionstruts from transverse splitting. The maximum bearing stress depends on the amount ofconfinement, as well as the aspect ratio (height to width) of the compression strut.The proposed bearing stress limit was incorporated into a strut-and-tie model todevelop a rational design procedure for deep pile caps. Two methods are proposed. Thefirst method is a direct extension of two-dimensional strut-and-tie models used for deepbeams. The second method is presented in a more traditional form in which “flexuraldesign” and “shear design” are separated. The shear design is accomplished by limitingthe bearing stress at the columns and the piles. The first method is more appropriate foranalysis, while the second method is more appropriate for design. The rationality andaccuracy of the proposed methods are demonstrated by the comparison with previoustest results.In the final part of this study, the influence of bond between concrete and longitudinalreinforcement upon the load transfer mechanism of both deep members and slendermembers without stirrups are investigated. An interpretation of an important shear11failure mechanism is presented.111Table of ContentsAbstract iiList of Tables viiiList of Figures ixAcknowledgement xvi1 Introduction 11.1 Background 11.2 Objectives and Outlines of Thesis 42 Analytical Study of Transverse Splitting 82.1 Finite Element Modelling 82.2 Internal Stress Distributions within Cylinders 92.3 Definitions of Geometrical Parameters and First Cracking 102.4 Location of First Cracking 102.5 Bearing Stress at First Cracking 112.6 Summary 123 Experimental Study of Transverse Splitting 173.1 Description of Test Program . . 173.2 Specimen Preparation 173.3 Instrumentation and Data Acquisition 18iv3.4 Test Observations3.4.1 Internal Cracking3.4.2 Influence of Concrete Confinement3.4.3 Failure Mode4 Bearing Strength of Concrete Compression Struts4.1 Comparison of Measured and Predicted Cracking Loads4.2 Comparative Study of Ultimate Bearing Strength4.2.1 Previous Studies of Bearing Strength4.2.2 Influence of Concrete Compressive Strength4.2.3 Influence of Size Effect4.2.4 Influence of Loading Geometry4.2.5 Summary4.3 Design Recommendations for Concrete Compression Struts191920• • • . 2029• • . . 2930• . .. 3032• . . . 333434• • . . 355 Shear Design of Deep Pile Caps5.1 Introduction5.2 ACI Code Approach for Pile Cap Design5.2.1 Code Procedure for Shear Design5.2.2 Inadequacies of ACI Code Procedure for Shear Design of Deep PileCaps5.2.3 Comparison of Different ACI Code Editions5.3 CRSI Approach for Shear Design of Deep Pile Caps5.4 Strut-and-Tie Model Approach5.4.1 CSA Approach for Deep Beams5.4.2 Proposed Design Method (1)5.4.3 Proposed Design Method (2)4545474749505153535457V5.5 Summary of Previous Pile Cap Tests5.6 Comparative Study of Different Design Procedures5.6.1 Comparison of Design Methods5.6.2 Comparison of Prediction Results5.7 Conclusions6 Shear Failure of Beams Without Stirrups6.1 Introduction6.2 Brief Review of the Literature6.2.1 Transition from Deep Beam to Slender Beam6.2.2 Behaviour of Slender Beams6.3 One Interpretation of Shear Failure of Beams without Stirrups6.4 Bond Effect in Longitudinally Reinforced Concrete Beams .6.4.1 Bond Influence in Uncracked Beams6.4.2 Bond Influence in Cracked Beams6.5 Shear Displacements along Cracks6.6 Load Transfer Mechanism6.7 Interpretation of Some Beam Test Results6.8 Conclusions7 Bond Splitting Failure7.17.27.37.47.57.658• . . . 636364669999100101102105108109112113114116117143143144146147149151IntroductionGeneral Bond ActionPrevious Experimental StudiesPrevious Studies of Ultimate Splitting FailurePrevious Studies of Splitting InitiationProposed Design Equation for Bond Splitting Initiationvi7.7 Some Comments and Conclusions.1538 Brief Summary and Further Research 168Bibliography 173Appendices 183A Measured Bearing Stresses and PUNDIT Readings 183B ACI Code and CRSI Handbook Predictions 193C Predictions from Proposed Design Method (1) 237D Predictions from Proposed Design Method (2) 250VIIList of Tables3.1 Summary of experimental results 223.1 (cont’d) Summary of experimental results 235.1 Summary of pile cap test results 685.1 (con’t) Summary of pile cap test results 695.2 Summary of ACT Building Code and CRSI Handbook predictions (flexureand bearing) 705.2 (con’t) Summary of ACT Building Code and CRST Handbook predictions(flexure and bearing) 715.3 Summary of ACT Building Code and CRSI Handbook predictions (shear). 725.3 (con’t) Summary of ACT Building Code and CRST Handbook predictions(shear) 735.4 Comparison of ACT Code and CRSI Handbook predictions: ratio of measured capacity to predicted capacity and failure mode 745.4 (con’t) Comparison of ACT Code and CRST Handbook predictions: ratioof measured capacity to predicted capacity and failure mode 755.5 Comparison of proposed design methods with experimental results. . . . 765.5 (con’t) Comparison of proposed design methods with experimental results. 77viiiList of Figures1.1 Strut-and-tie model for a deep beam or pile cap: (a) the idealized load-resisting truss; (b) linear elastic stress trajectories with transverse tensiondue to spreading of compression and (c) refined truss model with concretetension tie to resist transverse tension 61.2 Maximum bearing stress to cause transverse splitting in a biaxial stressfield, from Schlaich et al.[3} 72.1 Structural modelling of an idealized compression strut: (a) geometry; (b)finite element mesh; (c) isometric view showing stresses produced by axially symmetric loading and (d) an axially symmetric finite element ofrectangular cross section. Figs.(c) and (d) are from Ref.[15] 132.2 Typical internal stress distributions along the central Z axis within thecylinders: (a) tall cylinder and (b) short cylinder 142.3 Influence of D/d on the bearing stress to cause cracking for the case ofH/d=2 152.4 Analytical study of the bearing stress at first cracking: (a) influence ofconfinement and (b) influence of height 163.1 Schematic of test set-up 243.2 Photograph of test set-up 253.3 Typical relationships of load vs. PUNDIT readings (transit time increments) for the case of H/d = 2 263.4 Typical relationships of load vs. axial deformation for the case of H/d = 2 27ix3.5 Photograph of the typical failure mode of a concrete cylinder 284.1 Comparison of experimental results and analytical predictions: influenceof confinement for various heights 374.2 Comparison of experimental results and analytical predictions: influenceof height for various degrees of confinement 384.3 Comparison of ACT Building Code bearing strength prediction with experimental results: (a) present investigation and (b) previous tests. . . 394.4 Comparison of ACT Building Code bearing strength with Hawkins’s suggested equation for various concrete strengths 404.5 Influence of specimen height on ultimate bearing strength: (a) the forceflow in a single-punch test; (b) the force flow in a double-punch test and(c) the force flow in a tall double-punch test 414.6 Influence of height on accuracy of ACT Building Code bearing strengthprediction of double-punch tests 424.7 Comparison of suggested design equation with the analytical results. . . 434.8 Comparison of predictions from suggested design equation with experimental results 445.1 ACT Code specified critical sections for flexure and shear investigation ofpile cap 785.2 A deep two-pile cap 795.3 A deep three-pile cap 805.4 Comparison of two-way punching shear calculations: (a) the cap withsquare column; (b) the cap with circle column 81x5.5 CR51 approach for shear design of deep pile caps: (a) allowable shearstress, v, for one-way shear while w/d < 1.0; (b) allowable shear stress,v, for two-way shear while wfd < 0.5, from Ref.[36) 825.6 Flow chart for ACT and CRST design procedures for pile caps 835.7 Strut-and-tie model for deep beam, from Ref.[39] 845.8 Loading geometry of compression struts with linearly varying cross section 855.9 Various layouts of main reinforcing bars used by Blévot and Frémy, Ref.[32] 865.10 Various anchorage lengths used by Clarke, Ref.[44] 875.11 Comparison of one-way shear design methods for two-pile caps: (a) planview of pile cap; (b) to (d) influence of pile cap depth on column load forvarious pile cap widths 885.11 (cont’d) Comparison of one-way shear design methods for two-pile caps:(a) plan view of pile cap; (b) to (d) influence of pile cap depth on columnload for various pile cap widths 895.12 Comparison of two-way shear design methods for a typical four-pile cap:(a) plan view of pile cap; (b) influence of pile cap depth on column load 905.13 Comparison of ACT ‘77 predictions with experimental results 915.14 Comparison of ACT ‘83 predictions with experimental results 925.15 Comparison of ACT ‘[11.8] predictions with experimental results, clause11.8 considered 935.16 Comparison of CRST predictions with experimental results 945.17 Comparison of proposed method (1) predictions with experimental resultsof all specimens in Table 5.1 955.18 Comparison of proposed method (1) predictions with experimental resultsof specimens with bunched reinforcement in Table 5.1 96xi6.46.56.66.76.86.96.106.116.12123123124124• • • . 125• . • . 12512612712812997981201215.19 Comparison of proposed method (2) predictions with experimental resultsof all specimens in Table 5.15.20 Relationship of measured ultimate bearing stress and confinement on column zones of pile caps in Table 5.16.1 Distribution of transverse compressive stress for various shear span ratios,from Mau and Hsu, Ref. [52]6.2 Load near the support: transition from deep beam to slender beam, fromSchlaich et al., right side simple models; left side refined models, Ref.[3].6.3 Predictions of shear strength versus a/d ratio for tests reported by Kani[13J,from Collins and Mitchell, Ref.[39]. 122Truss model developed by Adebar, Ref.[54JTruss model developed by Reineck, Ref.[55jTruss models developed by Al-Nahlawi and Wight, Ref.[53].Structural model developed by Muttoni and Schwartz, Ref.[56].Structural model developed by Kotsovos, Ref.[57]Crack pattern of a beam tested by Kani, Ref.[13]Geometric relationship of a crack at reinforcement levelA simply supported and central loaded beamLinear elastic finite element analysis: modelling of (a) perfect bond and(b) no bond case6.13 Internal force flows in uncracked beams: (a) perfect bond case and (b) nobond case6.14 Modelling of bond effect upon transverse splitting of compression strutsin deep beam 130XII6.15 Bond influence upon transverse tensile stresses of compression struts indeep beam 1316.16 Internal stress distributions of cracked beams due to pure beam action,from Ref.[39J 1326.17 Shear stress distributions of cracked beams due to combined beam actionand arch action: (a) modelling of combined beam and arch actions; (b)tension force in longitudinal reinforcement; (c) shear stress distribution atsection n — n 1336.18 Shear displacement along cracks in shear span: (a) vertical cracks and (b)inclined cracks 1346.19 Shear transfer at cracks by aggregate interlock 1356.20 A beam with inclined cracks in shear span: (a) strut force due to arch action; (b) one assumed tension force in reinforcement; (c) another assumedtension force in reinforcement; (d) measured concrete strain and tensionforce in reinforcement from Ref.[71] 1366.21 Measured force in bar, bond stress and crack locations, from Ref.[59]. . 1376.22 A load transfer mechanism just before the occurrence of splitting alonglongitudinal reinforcement: (a) truss model and (b) tension force in reinforcement 1386.23 Design details of beams tested by Kotsovos: (a) a/d = 1.5 and (b) a/d>2.5, from Ref.[73j 1396.24 Load-deflection curves of beams shown in Figure 6.5: (a) a/d = 1.5; (b)aid = 3.3 and (c) a/d = 4.4, from Kotsovos, Ref.[73J 1406.25 Beam with internal stirrups tested by Chana, from Ref.[58] 1416.26 Beam with external stirrups tested by Kim et al., from Ref.[74] 1416.27 Beams tested by Muttoni et al., from Ref.[56j 142xlii6.28 Beams by Kuttab et al., from Ref.[75] 1427.1 Bond stress—slip relationship, from Gambarova et al., Ref.[80] 1557.2 Pullout tests: (a) concentric pullout-test specimen; (b) commonly used eccentric pullout-test specimen and (c) eccentric pullout-test specimen usedby Ferguson, Ref.[60] 1567.3 The University of Texas beam tests: (a) possible tension force distributionalong the top test bar after the occurrence of inclined crack; (b) side viewof specimen; (c) plain view of top reinforcement; and (d) moment diagram,adapted from Ferguson and Thompson, Ref.[61] 1577.4 Stub-beam or cantilever bond specimen, from Gergely, Ref.[77] 1587.5 Mechanism representation for bond, from Tepfers, Ref.[84] 1597.6 Bond Splitting Failure Patterns, from Oranguri et al., Ref.[82] 1607.7 Analysis of Bond Splitting Stresses: (a) bursting and bond stresses; (b)uncracked elastic state; (c) uncracked plastic state and (d) partly crackedelastic state. Adapted from Tepfers, Ref.[84} 1617.8 Comparison of test and prediction for bond splitting initiation, from Tepfers,Ref.[84] 1627.9 Comparison of Equation 7.7 with test data on cracking loads carried outby Kemp and Wilhelm Ref.[66] 1637.10 The relationship of bond splitting strength and ld/d proposed by Teng andYe Ref.[86] for concentric pullout tests 1647.11 Comparison of Equation 7.2 arid 7.10 with test results from Kemp andWilhelm Ref. [66] 1657.12 Comparison of Equation 7.10 with test results from Tepfers Ref.[84]. . . . 1667.13 Relationship of bond strength versus ld/db, presented by Eqs. 7.2 and 7.10. 167xiv8.1 Load transfer mechanisms of structural concrete members without transverse reinforcement: (a) very deep members (aid < 1) as well as moreslender members with no bond between concrete and reinforcement; (b)slender members with normal bond between concrete and reinforcement. 172xvAcknowledgementThe author would like to express his sincerest gratitude to his supervisor Dr. PerryAdebar for his constant guidance, criticism, and encouragement throughout the courseof research work and in the preparation of this thesis.Financial support to undertake this study was provided by the Natural Sciences andEngineering Research Council of Canada.Finally, this thesis is dedicated to the author’s family for their understanding andsupport, and especially to his wife, Jialing, and son, George, for their patience throughthe duration of this study.xviChapter 1Introduction1.1 BackgroundPlain concrete is strong and reasonably ductile in compression, but is weak and brittlein tension. Thus reinforcing steel bars are usually provided in concrete members to resistany tension force. In a reinforced concrete beam subjected to bending, for example,the flexural tension force is resisted by longitudinal reinforcing bars. For more complexloading conditions, such as combined shear and bending, structural concrete membersare usually reinforced with both longitudinal and transverse reinforcement.Structural concrete members reinforced with both longitudinal and transverse reinforcement can be designed using simple truss models or in the more general case strut-and-tie models [1-3]. It has been suggested by Schlaich et al. [3] that structures besubdivided into B-regions and D-regions for the purpose of design. B-regions are wherethe Bernoulli hypothesis of plane strain distribution is reasonable (B standing for beamor Bernoulli). Internal stresses in B-regions can be derived from the sectional forces without considering the details of how the forces are applied. Truss models for B-regions withstirrups involve compression and tension chords to model the flexural stresses, tensionties to model the stirrups and diagonal compression struts to model the cracked concreteweb. Truss models that have been developed for the shear design of beams with stirrupsare rational and work well [4-8].In D-regions (D standing for discontinuity or disturbance) the strain distribution is1Chapter 1. Introduction 2significantly nonlinear, and the details of how the forces are applied are very important.The more general form of truss models, strut-and-tie models can be used for both B- andD-regions. The theoretical justification for strut-and-tie (truss) models comes from thetheory of plasticity (limit analysis) [1]. Strut-and-tie models fall within the domain oflower-bound limit analysis. As strut-and-tie models allow designers to freely choose theload path, a sufficient amount of uniformly distributed reinforcement must be provided todistribute cracks, thereby ensuring adequate ductility (plastic deformation) for internalredistribution of stresses so that the assumed strut and tie system can eventually develop.Unfortunately, it is not always practical to provide the reinforcement needed for crackcontrol and ductility. Traditionally many structural concrete members are designed andbuilt without transverse reinforcement. In North America, both the ACT Building Code[9] and the Canadian Concrete Code [10] allow the following structural concrete membersto be designed without stirrups: slabs and footings; concrete joists; beams with a totaldepth less than 10 in. (254 mm); wide beams having a total depth less than the widthof the web; and tee beams having a total depth less than 2.5 times the thickness of theflange.Many of the current design procedures for members without stirrups are less thansatisfactory. For example, the design procedures currently specified in the ACT BuildingCode and the Canadian Concrete Code for deep pile caps (footings) are inappropriate.The ACT Building Code procedure for the shear design of footings supported on piles(pile caps) is the same sectional approach used for footings supported on soil and forslabs. The procedure involves determining the section thickness which gives a concretecontribution V greater than the shear force applied on the code defined critical section.While this approach is reasonable for slender footings supported on numerous pile, it isnot appropriate for deep pile caps which are entirely a D-region. On the other hand, applying strut-and-tie (truss) models to deep pile caps, which are usually without transverseChapter 1. Introduction 3reinforcement and have very limited ductility, is also questionable.The internal force flow within a deep pile cap can be represented by the simplestrut-and-tie model shown in Figure 1.1(a). The column load is transmitted directly tothe piles by compression struts. In order to prevent the piles from being spread apart,tension ties must be provided. The results of an experimental study on pile caps [11] hasdemonstrated that such a simple strut-and-tie model is a better model for deep pile capsthan the basis of the traditional design procedures for pile caps.A prismatic compression strut is a highly idealized representation of the stress statebetween a column and pile in a deep pile cap. In reality, the compressive stresses inthe compression strut spread out and transverse tensile stresses are created due to straincompatibility [see Figure 1.1(b)]. A refined strut-and-tie model for the compression strut,which includes a tension tie to model the transverse tension, is shown in Figure 1.1(c).As the compression struts which transfer load in deep pile caps are usually unreinforced,the tension tie must be provided by concrete tensile strength which is only about a tenthto a fifteenth of the concrete compressive strength. It is obvious that the failure ofunreinforced compression struts could be initiated by transverse splitting.Figure 1.2 from Schlaich et al. [3] shows the influence of the “amount of spreading”on the bearing stress to cause transverse splitting for a plane stress field (appropriate forthe case of a deep beam). Based on Figure 1.2 Schlaich et al. suggest that the concretecompressive stresses within an entire unreinforced D-region can be considered safe if themaximum bearing stress in all nodal zones is limited to O.6f (or in unusual cases O.4f)[3]. However what bearing stress limit is appropriate for compression struts confined bysurrounding plain concrete, such as occurs in deep pile caps, is not known. This hindersthe application of strut-and-tie models for the design of deep pile caps.While some truss models have been developed for slender members without transversereinforcement (i.e., B-regions), there is still no generally accepted interpretation of theChapter 1. Introduction 4shear failure mechanism for such members. Experimental investigations have shown thatthe bond between concrete and longitudinal reinforcement has a large influence uponthe shear capacity of reinforced concrete beams without transverse reinforcement [12-14].Truss models that have been developed for members without transverse reinforcementdo not adequately account for the influence of bond, particularly bond splitting failuresat the base of the critical diagonal crack.1.2 Objectives and Outlines of ThesisThis thesis deals with the shear design of structural concrete members without stirrups,and includes three major parts. The first part is a study of the transverse splitting ofcompression struts confined by plain concrete as occurs in deep pile caps. The secondpart involves the development of a rational design procedure for deep pile caps, whichmakes use of the bearing stress limit developed in the first part.In deep pile caps the column load is transmitted to the piles by direct compressionstruts. In more slender members a direct compression strut cannot form and the loadtransfer mechanism is influenced by the bond between concrete and reinforcing bars.A more general study of the load transfer mechanisms in members without transversereinforcement is undertaken in the third part of this thesis.In Chapter 2 three-dimensional compression struts confined by plain concrete arestudied analytically to investigate the internal stress distributions. The objective is todetermine the bearing stress at which first cracking occurs due to the transverse tension.The influence of various geometrical parameters upon the bearing stress at first crackingis presented.In Chapter 3 the transverse splitting of compression struts is investigated experimentally. The results from tests on plain concrete cylinders of various sizes loaded over aChapter 1. Introduction 5constant bearing area are presented.Design recommendations for the bearing stress limit of unreinforced concrete compression struts based on a combination of the analytical and experimental results areproposed in Chapter 4. Comparisons are also made with previous studies for ultimatebearing strength.In Chapter 5, the proposed bearing stress limit of compression struts is incorporatedinto a simple strut-and-tie model to develop a design procedure for deep pile caps. Twodesign procedures are proposed. The first design method is appropriate for a detailedanalysis, while the second design method is more suitable for design. A comparison ofpredictions from the proposed procedures, as well as the ACT Code and CRSI Handbookprocedures are carried out for available pile cap test results.In Chapter 6, a general study of the load transfer mechanisms in members withoutstirrups is carried out. Recently developed truss models are reviewed. The shear failuremechanism of beams without stirrups is studied from the compatibility of the criticaldiagonal cracking propagation. The effect of bond between concrete and reinforcing barsupon shear resistance mechanism are also studied. Based on the study in this chapter,some interesting conclusions are drawn. Finally in this chapter, experimental resultsfrom beam shear tests are interpreted.As a natural extension to Chapter 6, the bond splitting failure along reinforcing barsis investigated in Chapter 7. Based on previous studies on bond, an empirical equation isproposed to predict the tension force causing bond splitting in beams without transversereinforcemeilt.Finally, some major conclusions from this study and a few recommendations for further investigations are summarized in Chapter 8.Chapter 1. IntroductioncomPressio/,.,.N(a)Ti,I/ /J. N////Figure 1.1: Strut-and-tie model for a deep beam or pile cap: (a) the idealizedload-resisting truss; (b) linear elastic stress trajectories with transverse tension due tospreading of compression and (c) refined truss model with concrete tension tie to resisttransverse tension.6V Vnocal zoneLensiontie1T = V I tan eD=V/sine(b)a(c)Chapter 1. Introductionfbf,C10Figure 1.2: Maximum bearing stress to cause transverse splitting in a biaxial stress field.from Schlaich et al.[3J.FailureTransverse Cracking1 2 3 4 5 6 7 89ba1. bChapter 2Analytical Study of Transverse SplittingIn order to better understand the transverse splitting phenomenon in deep pile caps anddevelop an appropriate strength criterion for plain concrete compression struts, idealizedthree-dimensional compression struts are studied analytically. The study focuses on theinitial cracking within struts rather than the post-cracking stage, which is a much morecomplicated nonlinear problem.2.1 Finite Element ModellingThe compression struts are idealized as cylinders of exterior diameter D, and height H,subjected to concentric axial compression over a constant size circular bearing area ofdiameter d, both on top and bottom. See Figure 2.1(a). Linear elastic finite elementswere used to study the internal triaxial stress distributions, and to determine when andwhere first cracking occurs within the cylinders.For the problem at hand, the geometry, material properties and external loading areall axis-symmetric so that the problem is mathematically two-dimensional. The staticdisplacements and stresses are independent of angle 0, circumferential displacement vis zero and material points have only u(radial) and w(axial) displacement components.Non-zero stress components are O, o, o,. and Tzr. The stress o is a principal stressand o, r and Tzr are components of the two other principal stresses. In the analysis,axis-symmetric 4-node quadrilateral elements were used [15]. There are two integrationpoints in each one of radial, axial and circumferential directions. Stresses at the central8Chapter 2. Analytical Study of Transverse Splitting 9point of each element were calculated. See Figure 2.1(c) and (d).The finite element mesh used is shown in Figure 2.1(b), which also shows the geometry, grid layout and boundary conditions. Due to symmetry of the problem and thecharacteristic of the elements used, only a quarter of the cylinder is shown. The radialdisplacement u = 0 was prescribed at all nodes that lie on the Z axis and the rigidbody motion along Z direction was restrained by prescribing w = 0 at the mid-heightof the symmetrical cylinder. But no restrictions were imposed for vertical displacementon the Z axis and horizontal displacement along the radial direction with the exceptionof one location at the corner. By adding more elements either horizontally or vertically(i.e., increasing D or H), the influence of various loading geometry on the internal stressdistributions was investigated.2.2 Internal Stress Distributions within CylindersBased on the numerical analysis results the internal stress distributions along the centralZ axis are shown qualitatively in Figure 2.2. The vertical stress component r hasits maximum value beneath the loading plate and decreases gradually with increasingdistance from the loading plate. The stress o is distributed uniformly on the transversecross section at a certain distance away from the loading plate. The stress components 0and ,. on the Z axis are equal and may be negative (compressive), or positive (tensile),as well as zero at locations where the stress cr becomes uniformly distributed.The internal stress distribution within the cylinder can be divided into three zones[see Figure 2.2(a)]. A zone of triaxial compression exists immediately below the loadingplate. Further down from the loading plate is a zone with horizontal biaxial tension andvertical uniaxial compression. At the mid-height of relatively tall cylinders is a zone ofvertical uniaxial compression. In short cylinders, the third zone (of uniaxial compression)Cha0pter 2. Analytical Study of Transverse Splitting 10does not exist. See Figure 2.2(b).2.3 Definitions of Geometrical Parameters and First CrackingThe geometry of the problem can be summarized in terms of two parameters, namelythe ratio of the cylinder diameter to the loaded area diameter, D/d, and the ratio of thecylinder height to the loaded area diameter, H/d.The stress field within a cylinder can be expressed in terms of the four stress components oo, o, o,. and re,. or by three principal stresses and corresponding principal stressdirections. The stress O is always a principal stress. It was assumed that cracking willoccur when the maximum principal tensile stress reaches the concrete tensile strength.The location of the maximum principal tensile stress defines where the first cracking occurs. The influence of the other principal stress components upon the cracking strengthwas neglected for simplicity.2.4 Location of First CrackingAfter the internal stress distributions within the cylinders were examined, it was foundthat the maximum tensile stress would occur either on Z axis (at the centre of thecylinder) or near the exterior surface of the cylinder depending on the geometry. Hence,the location of first cracking could be located either in the inside of the cylinder or nearthe outside of the cylinder.The influence of loading geometry upon the location of first cracking is shown inFigure 2.3, where the predicted bearing stress at cracking versus the ratio of Did aresummarized for the case of H/ d = 2. There are two curves in the figure: one indicatingcracking inside, the other cracking outside. It can be seen that for small ratios of Did,first cracking occurs on the surface, while for D/d > 1.5 the first cracking occurs at theChapter 2. Analytical Study of Transverse Splitting 11centre of the cylinder.As cracking outside occurs only when there is very little confinement, which is not ofmuch practical interest, oniy cracking inside was considered below.2.5 Bearing Stress at First CrackingFigure 2.4 summarizes the analytical results regarding the the influence of D/ d ratio andH/d ratio upon the bearing stress at first cracking (inside). In order to plot the bearingstress as a function of the compressive strength of concrete, the concrete tensile strengthf’ was assumed to one-fifteenth of the concrete compressive strength f (i.e., f’ = f/l5).Figure 2.4(a) shows the relationship between bearing stress at first cracking and D/dfor specific H/d values. For small ratios of D/d (i.e., close to 1.0), the bearing stressat first cracking is independent of height and tends to be very large. In this rangethe uniaxial compression is more critical than the transverse tension. In the range ofD/d > 1.5 2.0, increasing the amount of confinement (i.e., increasing D/d further)will increase the bearing stress at first cracking, but not very much for small and mediumH/ d values. The significant transverse tension is introduced due to the spreading-out ofcompressive force. The transverse tensile stresses will reach concrete tensile strength atrelatively low external load in this range.Figure 2.4(b) shows the relationship between bearing stress at first cracking and H/dfor specific D/ d values. Figure 2.4(b) illustrates a similar trend to Figure 2.4(a). Inthe range of H/ d < 1, the bearing stress at first cracking is independent of confinementand tends to be very large. The localized uniaxial compressive stress field between twoloading plates is dominant within the compression strut. For H/d > 1.5, increasingheight will increase the bearing stress at first cracking. But increasing height further(i.e., Hid > 4 ‘-‘ 6) does not have significant influence upon the bearing stress at firstChapter 2. Analytical Study of Transverse Splitting 12cracking.It can be seen that the minimum bearing stress at first cracking is approximatelyO.6f at D/d = 2 and H/d = 1.5. More discussions about aspect ratio (H/d) uponthe bearing stresses at first cracking will be given in the comparative study of ultimatebearing strength in Chapter 4.2.6 SummaryThe numerical investigation indicates that the loading geometry has a significant influenceupon the internal stress distribution within a confined compression strut. See Figure 2.2.Qualitatively, the response of an idealized compression strut (cylinder) to external loadcan be described as follows. For a compression strut without confinement (D/d = 1),the strut is uniaxial stressed. The compressive stress is constant over the height andis equal to the bearing stress. However, if there is confinement outside the loaded area(D/d> 1), the additional concrete will be mobilized to sustain the external load, and theaxial compressive stress along the central axis of the strut will decrease continually withincreasing distance away from the loading plate. As expected, higher bearing stresses maybe possible due to the confinement. On the other hand, as a result of strain compatibilitythe transverse tensile stresses are introduced. Because concrete tensile strength f’ isabout a tenth to a fifteenth of concrete compressive strength f, internal cracking ofthe concrete compression strut (cylinder) may occur due to the weak tensile strength atrelatively low compression stresses.Chapter 2. Analytical Study of Transverse Splitting 132(C) (d)Figure 2.1: Structural modelling of an idealized compression strut: (a) geometry; (b)finite element mesh; (c) isometric view showing stresses produced by axially symmetricloading and (d) an axially symmetric finite element of rectangular cross section. Figs.(c)and (d) are from Ref.[15].d/2TTHJ(a)1cH12D12(b)z,w4 3x’Axis-\;—r,uChapter 2. Analytical Study of Transverse Splitting 14z5.rFy2— -bd H/2fbTension CompressionH (a)zII____D e.JZS(LTension Compression(b)Figure 2.2: Typical internal stress distributions along the central Z axis within thecylinders: (a) tall cylinder and (b) short cylinder.Chapter 2. Analytical Study of Transverse Splitting 15f /3- H/d=2 /fc’/ft’=152Cracking OutsideCracking Inside0 I I1 2 3 DdFigure 2.3: Influence of D/d on the bearing stress to cause cracking for the case ofH/d=2.Chapter 2. Analytical Study of Transverse Splitting 164H/d = 10HId =6H/d =43 H/d—8H/d =3H/d—2(a)0BD4Did =7D/d =5D/d—42D/d—6D/d =3D/d =2I(b)012HdFigure 2.4: Analytical study of the bearing stress at first cracking: (a) influence ofconfinement and (b) influence of height.Chapter 3Experimental Study of Transverse SplittingIn order to confirm the transverse splitting phenomenon within compression struts, andto investigate the influence of H/d and D/d ratios upon the bearing stress at first crackingexperimentally, a series of experiments were conducted on plain concrete cylinders.3.1 Description of Test ProgramTable 3.1 summarizes the properties of the 60 different specimens which were tested.The diameter of the concrete cylinders varied from 6 inches (152 mm) up to 24 inches(610 mm), while the diameter of the circular bearing plate was constant at 6 inches (152mm). The height of the cylinders varied from 9 inches (230 mm) up to 36 inches (914mm). The variation of geometrical parameters covers the range of 1 < Did 4 and1.5 Hid 6, which is believed to be the practical range. Note that D6-l2 in Table3.1 denotes a 6 inch (152 mm) diameter cylinder which is 12 inch (305 mm) high.3.2 Specimen PreparationThe specimens were constructed from two batches of concrete. Both batches were supplied by a local ready-mix supplier (25 MPa was specified) and had crushed stone asaggregate with a maximum size of 3/4 inch (19 mm). The specimens were cast in sonotube forms with plywood bases, which were stripped after about 7 days. The specimenscontinued to cure in the laboratory for an additional 50 to 60 days before being tested.17Chapter 3. Experimental Study of Transverse Splitting 18The specimens were always covered by plastic sheets during curing time. As it turnedout, Batch A concrete had a cylinder compressive strength of 30 MPa (4350 psi), whileBatch B concrete had a compressive strength of only 20 MPa (2900 psi) at testing.3.3 Instrumentation and Data AcquisitionThe test set-up used is shown in Figure 3.1 and 3.2. The specimen was placed on thetesting table of a 400 kip capacity Baldwin testing machine and loaded through 6 inch diameter, half-inch thick steel discs. Special care was taken to ensure the specimen was notloaded eccentrically. Two Linear Variable Differential Transducer (LVDT) displacementtransducers were mounted on two steel bars which were connected to the top loading(steel) disc. Two transducers (one transmitter and one receiver) of a Portable UltrasonicNon-Destructive Digital Indicating Tester (PUNDIT) [16] were put at the height wherethe maximum tensile stresses were expected based on the analytical results. Two smallwood pieces were glued to the side face of the specimen to act as supports for the twotransducers. An elastic rope was used to tighten the transducers to the specimen, andat same time to give as little transverse constraint to the specimen as possible.The axial deformations of specimens were measured by the two LVDTs. The axialdeformation versus the applied loads were plotted with the help of an X- Y plotter. Themeasured axial deformation included the deformations of the steel discs, plaster and theconcrete cylinder so the deformation of the steel discs and plaster had to be eliminatedbefore obtaining the true axial deformation of a concrete cylinder.An important piece of equipment used in the experimental investigation was thePUNDIT [16]. The PUNDIT measures the travelling time of an ultrasonic pulse whosevelocity is proportional to the density of the material. The transit time is displayedby three ‘in-line’ numerical indicator tubes. The transit time was recorded by hand atChapter 3. Experimental Study of Transverse Splitting 19predetermined load values. If the last digit either remained at a constant value or wouldoscillate between two adjacent-values with a bias for one of the values, this value wasrecorded. If the last digit oscillated between two adjacent-values evenly, the mean of thetwo values was recorded. In this way, an accuracy of 0.05 microsecond was possible.3.4 Test ObservationsLoad was applied at an approximate rate of 2.75 MPa every minute, continuously untilfailure. Major test observations are summarized below.3.4.1 Internal CrackingThe internal cracking within concrete cylinders could not been seen by eye so that thePUNDIT was used to detect it. By observing the change in ultrasonic pulse transit time,the internal cracking was determined indirectly.While load was being increased, the transit time of ultrasonic pulse passing throughthe specimens (i.e., from transmitter to receiver) also increased. The rate of changeof the ultrasonic pulse travel time varied at different loading stages. The transit timeincreased very little or remained almost stable at low load levels, and then increasedsteadily at medium load levels. Finally, the transit time increased very quickly justbefore failure. The quick increase of transit time was always a sign that failure wouldsoon occur. Typical load PUNDIT reading (transit time increments) relationships forspecimens with H/d = 2 are shown in Figure 3.3.The relationship between load and PUNDIT reading (transit time increment) is initially linear. As the velocity of ultrasonic pulses travelling in a solid material depends onthe density and elastic properties of the material, the transit time depends on both thepath length and the quality of concrete. In the linear stage, the increase in transmit timeChapter 3. Experimental Study of Transverse Splitting 20is believed to be a result of the transverse deformation of cylinders due to the Poisson’sratio effect (i.e., the small increase in path length). In the post-linear stage, both thequality of the concrete and the transverse deformation of concrete are believed to havean influence upon the transmit time increment. First cracking is believed to be indicatedby the onset of nonlinearity. The non-linearity in the PUNDIT readings for 6 inch (152mm) specimens is believed to be due to the micro-cracking that occurs when concrete issubjected to uniaxial compression.3.4.2 Influence of Concrete ConfinementIt is obvious that the confinement has an influence on the structural behaviour of concretecylinders subjected local compression on both the top and bottom surfaces. Typicalbearing stress versus axial deformation relationships for specimens with H/ d = 2 areshown in Figure 3.4, in which the deformations coming from the steel discs and plasterhave been eliminated and the applied loads have been converted to bearing stresses.It can be seen that increased confinement will make the axial stiffness of the specimens larger and allow higher ultimate bearing strengths to be achieved. In contrast,confinement has a minor influence on the bearing stress at first cracking. See both Figs.3.3 and 3.4.3.4.3 Failure ModeIn general, the failure of specimens occurred suddenly and with a loud noise. Typically,three or four radial cracks split the specimen into approximately equal segments. Concrete “cones” usually formed in the triaxial compressed zone below the loading plates.Figure 3.5 shows a typical concrete cylinder after failure.The 6 inch (152 mm) diameter and 8 inch (203 mm) diameter specimens were lessbrittle than other specimens. See Figure 3.4. In the 8 inch (203 mm) specimens surfaceChapter 3. Experimental Study of Transverse Splitting 21cracks were observed before failure. That is, the specimens resisted additional load afterexterior cracks were observed. Note that the 8 in (203 mm) specimens were predicted tocrack on the outside before the inside. See Figure 2.3. The 6 inch (152 mm) specimenswere standard cylinders that failed in compression.CD I I-s CD CI)CDC])I:I--iI--iI_,:I—::‘:i—.•.•i—,———;‘,—I.oo00000000)CDCCCI--sI--iI.-I.-.I—iI--iCCC.3CZC3I—IIçC.Cct’.)t.)t\)t\)ttt’.DsCAZL’3I—CL3I—C‘3ICt’I—i*C)oCD‘CD—‘I-st’.t3ISZt’.t’.ZIt\1t..t’.I-.)k1I—I—it’3tt.L2II-t.Z0000t’Z:tc‘.—I-”c-II-s CI),.—.‘.3CC’I——1).CI—iCCT©©—4©0000—1t’ZC—4CD O:sei F tN3Chapter 3. Experimental Study of Transverse Splitting 23Table 3.1: (cont’d) Summary of experimental results.Specimen Concrete Bearing Stress Max. BearingDesignation* Batch at Cracking Stress(MPa) (MPa)D14-9-1 B 15.9 32.5D14-9-2 B 14.6 31.7014-9-3 B 17.1 33.2014-18-1 B 26.8 48.1014-18-2 B 25.6 50.0D14-18-3 B 24.4 48.8D14-36-1 B 34.2 58.1014-36-2 B 28.1 53.2D14-36-3 B 29.3 51.7018-9-1 B 26.8 44.4018-9-2 B 19.5 42.2D18-9-3 B 20.7 44.4D18-12-1 A 26.8 40.2018-12-2 A 26.8 43.4D18-12-3 A 29.3 42.0D18-12-4 A 29.3 41.7D18-18-1 B 34.2 61.0018-18-2 B 31.7 55.6018-18-3 B 29.3 56.9D18-36-1 B 37.8 71.5D18-36-2 B 37.8 68.3018-36-3 B 39.1 68.3024-12-1 A 32.9 59.8D24-12-2 A 35.4 65.5024-12-3 A 34.1 62.0*D612 denotes 0 = 6 in; H = 12 in1 MPa = 145 psi1 in = 25.4 mmChapter 3. Experimental Study of Transverse Splitting 24d=6 in.Displacement1ransducerPlasterSpecimen H=VariesUNDIT PUNDITreceiverSteel discD =variesFigure 3.1: Schematic of test set-up.Chapter 3. Experimental Study of Transverse Splitting 25Figure 3.2: Photograph of test set-up.Chapter 3. Experimental Study of Transverse Splitting 268070• CrackingCl) 60-D24-12--G)-4-iC,)50 -D18-1240- / D12-1230 - D8-12D6-1220 -1io4 ( :0 I I0 1 2 3 4 5 6 7PUNDIT Transmit Time Increment(microseconds)Figure 3.3: Typical relationships of load vs. PUNDIT readings (transit time increments)for the case of H/d = 2.Chapter 3. Experimental Study of Transverse Splitting 278060- D24-12U)Cl) 50 -Dl 8-12ctI 40 -D12-12 D8-1230 -D6-1220 -10 -0 I I I0 0.5 1 1.5Displacement (mm)Figure 3.4: Typical relationships of load vs. axial deformation for the case of H/d = 2.Chapter 3. Experimental Study of Transverse Splitting 28IFigure 3.5: Photograph of the typical failure mode of a concrete cylinder.Chapter 4Bearing Strength of Concrete Compression Struts4.1 Comparison of Measured and Predicted Cracking LoadsFigure 4.1 and 4.2 compare the measured cracking loads with the linear elastic predictions. Figure 4.1 shows the influence of confinement (D/d) for various height, whileFigure 4.2 shows the influence of height (H/d) for various degrees of confinement. Thetensile strength used in the prediction, f = f/13, was chosen to give the best fit. Considering that only one parameter (f/f) is adjusted, there is a very good correlationbetween the analytical prediction and the experimental results regarding the influence ofDid and H/d on the bearing stress to cause first cracking.It should be pointed out that the large deviation of some experimental data pointsfrom analytical prediction in the case of H/cl = 2 of Figure 4.1 is not surprising. Asdiscussed in Section 2.4 (Figure 2.3) the analytical results predict that there is a rangeof first cracking inside as well as first cracking outside. The ratio of D/d = 1.33 forthe 8 inch (203 mm) diameter and 12 inch (305 mm) height specimens lies in the rangeof first cracking outside. Test observation also confirmed that some external crackingoccurred before the specimens failed. The early external cracking is believed to be themain reason for this deviation from the analytical prediction. It should be rememberedthat the predictions shown in Figure 4.1 represent that of first cracking inside.29Chapter 4. Bearing Strength of Concrete Compression Struts 304.2 Comparative Study of Ultimate Bearing StrengthWhile this study is concerned primarily with the bearing stress to cause initial transversecracking, the test results also give valuable information for ultimate bearing strength. Acomparison of the bearing stress which caused failure in the present tests and the bearingstrength predicted by the ACT Building Code [9] was therefore carried out. According tothe ACT Building Code the maximum bearing strength of concrete is 0.85f, except whenthe supporting surface area A2 is wider on all sides than the loaded area A1, the bearingstrength is multiplied by A2/A1,but not more than 2. Note that D/d = i.JA2/Ai.Surprisingly it was found that the ACT approach is unconservative in the range 1.5 <D/d < 3.5. See Figure 4.3(a). Thus, it seems that further study on ultimate bearingstrength is warranted.4.2.1 Previous Studies of Bearing StrengthAs bearing strength is important in the design of many concrete structures, a greatnumber of investigations have been done on the bearing strength of concrete.It is generally believed by most investigators that the principal variables influencingthe concrete bearing strength are:1. The geometry of loaded area and specimen;2. the nature of the supporting bed under the specimen;3. the supported area of specimen;4. the concrete strength, and5. the specimen size.The main conclusions based on the previous experimental work are that:Chapter 4. Bearing Strength of Concrete Compression Struts 311. The failure is due to the punching out of an inverted cone of concrete beneath theloaded area and the radial pressures exerted by this cone split the specimen;2. The ratio of bearing strength over concrete compressive strength increases continuously for increasing confinement, ie. increasing the ratio of D/d;3. There are cracking loads and failure loads which are sometimes different and dependupon the size of bearing plates and the height of specimens;4. The larger the loading plate size (smaller D/d) and smaller the relative height ofspecimen, the greater is the difference between failure load and external crackingload;5. Specimens supported on a compressible bed give less bearing strength than forsimilar specimens on a rigid support medium;6. Specimens subjected to localized forces from opposite ends exhibit lower bearingstrength compared to localized compression from one end only;7. The higher the compressive strength of concrete, the lower the ratio of ultimatebearing strength to concrete compressive strength;8. The ratio of bearing strength over concrete compressive strength decreases with theincrease in specimen size,Conclusions (3)—(8) are mainly from Niyogi’s tests [17].However, there are different opinions regarding the influence of specimen height. Someauthors considered that bearing strength over concrete compressive strength fb/f’ isindependent of specimen height [18-26]. But Niyogi [17] concluded that for double punchtests the ratio f&/f increases with the increase in the specimen height and supportingChapter 4. Bearing Strength of Concrete Compression Struts 32area. For single punch tests the bearing strength decreases with increasing specimenheight in some ranges of geometrical parameters, and in the reverse (the bearing strengthdecreases with decreasing specimen height) in other ranges of geometrical parameters.Double-punch tests involves the specimens being loaded symmetrically using bearingplates on both the top and bottom surfaces. In single-punch tests the base of the specimenis set directly on the lower platen of the testing machine and the upper platen is used toload the bearing plate.Figure 4.3(b) compares the ACT Building Code bearing strength prediction with previous tests conducted by Shelson [19], Au and Baird [20], Douglas and Trahair [23],Middendorf [24], Hawkins [26], Chen and Trumbauer [27], as well as Niyogi [17]. Asshown in Figure 4.3(a) for the present investigation, this figure also indicates that theACT prediction is unconservative for quite a number of data points. After the unconservative data points were carefully examined, it was found that the major reasons for thelower ratios of bearing strength to concrete compressive strength f are the influence ofconcrete compressive strength itself, size effect and loading geometry. These are discussedin detail below.4.2.2 Influence of Concrete Compressive StrengthThe Commentary to ACT 318-89 Section 10.15 makes reference to Hawkins [26] whosuggested the following expression for the bearing strength of concretefb = f +501J1(/A2/A— 1) (4.1)where f is in psi units. That is, the enhancement in bearing strength due to confiningconcrete is proportional to the tensile strength of concrete. In the ACT Building Code thisequation has been simplified so that the bearing strength enhancement is proportional toChapter 4. Bearing Strength of Concrete Compression Struts 33the concrete compressive strength. Figure 4.4 compares the ACT approach with Equation4.1 for different concrete strengths. Note that when the concrete compressive strength ishigh than 5,000 psi (34.5 MPa), the original expression suggested by Hawkins [26] giveslower bearing strengths than the ACT Building Code. The low strength reduction factorof 0.70 used in the ACT Building Code for bearing on concrete partly compensates forthis over-simplification.4.2.3 Influence of Size EffectA second factor which contributes to lower bearing strengths in Figure 4.3 is size effect.The ACT Building Code approach is based on tests of relatively small specimens, whiledata points shown in Figure 4.3(a) are from large specimens. Since bearing failuresinvolve fracture of concrete or crack propagation process due to indirect tension, there isa significant size effect involved.Based on the result of 42 double-punch tests, in which the specimens had the geometrical proportions D = H = 4d, Marti [28] concluded that the size effect was in goodagreement with Baant’s (nonlinear fracture mechanics) size effect relation [29, 30]. Thusthe bearing strength of concrete is proportional tofo cc (4.2))Dawhere D is the maximum aggregate size and .X is an empirical constant, which was foundto be 38.0 and 68.5 in Marti’s two series of tests [28]. For the geometrically identicalspecimens that were made with the same concrete, f = 3425 psi (23.6 MPa), and rangedin size from 8 to 128 times the maximum aggregate size (i.e., the size varied by a factorof 16), the bearing strength decreased by a factor of 1.6 from the smallest specimen tothe largest specimen.Chapter 4. Bearing Strength of Concrete Compression Struts 344.2.4 Influence of Loading GeometryThe most important factor to contribute to the actual bearing strength being lower thanthe ACT Building Code prediction is the geometry of loading.The majority of the test results, upon which the ACT Building Code procedure isbased, are from single-punch tests. In the present investigation, as well as some previous studies, the specimens were loaded by double-punch method. In single punch teststransverse tensile stresses are created at one end only and the lower platen of the testingmachine restrains the expansion of the specimen. In double-punch tests the compressionat both ends of the specimen results in higher tensile stresses at the mid-height of thespecimen as shown in Figure 4.5(b) and the restraint is eliminated. When the specimensare relatively tall [Figure 4.5(c)], the double-punch tests produce similar results as thesingle-punch tests except for the restraint issue [Figure 4.5(a)]. Figure 4.6 illustrates thatthe accuracy of the ACT bearing strength depends on the H/d ratio for double-punchtests.In addition to the influence of specimen height discussed above, the data points inFigure 4.3 also show that the effect of confinement is not as large as the ACT procedurepredicts in the range of small D/d ratios, where quite a number of test results either areunsafe or have a very low safety margin. Tt is interesting to compare Figures 4.3 and4.6 with Figure 2.4. Most lower bearing strength data points are located in the range1 < H/d < 3 and 1.5 < D/d < 3.5, exactly where the analytical results indicate theleast bearing stress to cause first cracking.4.2.5 SummaryOn the basis of present investigation, some conclusions which supplement the conclusionsof previous studies (Section 4.2.1) are that:Chapter 4. Bearing Strength of Concrete Compression Struts 351. Bearing failure is due to transverse splitting, provided that the supporting surfacearea is wider on all sides than the loaded area;2. The bearing stress at external visible cracking could be less than or equal to thebearing stress at failure depending upon whether first cracking occurs outside orinside;3. Bearing strength depends, to a major extent, upon the loading geometry, whichincludes the loading method, surrounding confinement (D/d) as well as specimenheights (H/d);4. Due to either higher concrete compressive strength or size effect or loading geometry, the bearing strength according to the ACT Code can be unconservative.4.3 Design Recommendations for Concrete Compression StrutsBased on the results of the analytical and experimental studies presented in Chapter 2and 3, it is suggested that when designing deep members (disturbed regions) withoutsufficient distributed reinforcement to ensure redistribution after initial cracking, themaximum bearing stress should be limited tofb O.6f(1 + 2a/3) (4.3)where= O.33(/ — 1) 1.00.33( — 1) 1.0The ratio h/b, which represents the aspect ratio (height/width) of the compressionstrut, should not be taken less than 1.0 (i.e., 3 0). The parameter a accounts forChapter 4. Bearing Strength of Concrete Compression Struts 36the amount of confinement, while the parameter j3 accounts for the geometry of thecompression stress field.The lower bearing stress limit of O.6f is appropriate if there is no confinement (i.e./A2/A1 1) regardless of the height of the compression strut, as well as when the compressive strut is relatively short (i.e. h/b 1) regardless of the amount of confinement.The upper limit of Equation 4.3, O.6f x 3 = l.8f, is chosen to correspond approximatelyto the upper limit of bearing strength given in the ACT Building Code. The interactionof confinement and geometry (aspect ratio) is chosen so as to give a reasonably simpleexpression and yet correspond well with the finite element predictions and the experimental results. Figure 4.7 compares Equation 4.3 with the finite element prediction,while Figure 4.8 compares predictions from Equation 4.3 with the experimental results.As mentioned previously, concrete bearing strength is actually proportional to theconcrete tensile strength. If the concrete compressive strength is significantly greaterthan 5,000 psi (34.5 MPa), a more appropriate limit for the bearing stress isfb 0.6f + 6aç8fl (4.4)where f in MPa units. If psi units are used, the 6cr6 in Equation 4.4 should be replacedby 72a/3.Chapter 4. Bearing Strength of Concrete Compression Struts 3780Experiment&:60 - • Cracking H/d=1.5o FailureCl)U)40Cl)20 I Predicted Cracking• ft’=1.52MPaQ) 02 3 480H/d=2.0s4-40_______Predicted Cracking20 ft’=2.3OMPa(Concrete Batch A)01 2 3 48060- o H/d=308Predicted Crackingft’=1.52 MPa01 2 3 4808H/d=660-40Predicted Cracking20 - ft’=1.52MPa0 2 3 DidFigure 4.1: Comparison of experimental results and analytical predictions: influence ofconfinement for various heights.Chapter 4. Bearing Strength of Concrete Compression Struts 3880Experimental:60 - • Cracking D/d=1.67Cl) o FailureCl)40 -Cl)20-edicted Crackingft=1.52MPa00 1 2 3 4 5 6 7 88060 - D/d=2.006 040 -20- “%-“edied Cracking1.52 MPa1 2 3 4 5 6 7 8808D/d =3.060- 040- 820- \_JedictedCracking1.52 MPa1 2 3 4 5 6 7 8H/dFigure 4.2: Comparison of experimental results and analytical predictions: influence ofheight for various degrees of confinement.Chapter 4. Bearing Strength of Concrete Compression Struts 39f,C65-4-.3- .III2- $ $ S11--, ACt Code01 I62 3 45543(a)f,C(b)Figure 4.3: Comparison of ACT Building Code bearing strength prediction with experimental results: (a) present investigation and (b) previous tests.2101 2 3A2/A1Chapter 4. Bearing Strength of Concrete Compression Struts 40f3-‘b/ABC2D- ACI codeSSSSSSSSF•1 SI ——FA = 2,500 psi (17 PAPa)B = 5,000 psi (34.5 PAPa)C = 7,500 psi (51.7 PAPa)D = 10,000 psI (69 MPa)01 2__3Figure 4.4: Comparison of ACT Building Code bearing strength with Hawkins’s suggestedequation for various concrete strengths.Chapter 4. Bearing Strength of Concrete Compression Struts 41——________________—_____________________II___(a) (b)(c)Figure 4.5: Influence of specimen height on ultimate bearing strength: (a) the force flowin a single-punch test; (b) the force flow in a double-punch test and (c) the force flow ina tall double-punch test.Chapter 4. Bearing Strength of Concrete Compression Struts 423b(exp)b(AcI)2-10 I I I I0 1 2 3 4 5 6 7 8 9 10H/dFigure 4.6: Influence of height on accuracy of ACT Building Code bearing strength prediction of double-punch tests.Chapter 4. Bearing Strength of Concrete Compression Struts 43—f;H/d =42-jH/d=3Hid = 1o I I I1 2 3 4D/d=1JA2/— —.2-,.-13Id4_-—-D/cI= 10I I I0 1 2 3 4H/dFigure 4.7: Comparison of suggested design equation with the analytical results.Chapter 4. Bearing Strength of Concrete Compression Struts 443(exp)(pred)2.• • ... I: • : :• ....4 • . .I SI01 2 3 4 5D/d=JA2/1Figure 4.8: Comparison of predictions from suggested design equation with experimentalresults.Chapter 5Shear Design of Deep Pile Caps5.1 IntroductionThis chapter deals with shear design of deep pile caps. The main objective is to develop a rational design procedure for deep pile caps based on the design recommendationproposed in Chapter 4 for concrete compression struts.The methods currently used in the design of reinforcement for deep pile caps aremainly classified as two categories. Some designers assume that a cap acts as a beamspanning between the piles and design the reinforcement on the basis of simple bendingtheory [9]. Others assume that the cap acts as a truss, the column load is transmittedto the piles through axial struts and the tension force required for equilibrium is takenby reinforcement [3], 32]. A literature survey for previous theoretical and experimentalstudies on pile caps can be found in Ref.[33].Bending theory and truss analogy may lead to similar amounts of steel, but suggestdifferent arrangements in plan: the former leads to a uniform grid whereas the latterfavours bands of steel running between the piles. In addition, the two methods havedifferent anchorage requirements. Nominal anchorage is required beyond the centre-linesof the piles for bending theory. Truss analogy, on the other hand, requires that fullanchorage lengths are provided beyond the centre-lines of the piles.A sectional method is usually used for the shear design of deep pile caps. The shearstresses on the code defined critical sections are limited to the shear strength contributed45Chapter 5. Shear Design of Deep Pile Caps 46by concrete. However a previous experimental study [11] revealed that a sectional methodis not appropriate for the shear design of deep pile caps. Further investigations on theuse of sectional methods for the design of deep pile caps are carried out in this chapter.This chapter is composed of six parts. In the first part, the ACT Code approachfor design of pile caps is summarized. Some practical design examples are presented,which show that the ACT Code approach is inadequate for shear design of deep pilecaps. The definition of deep pile caps is also given in this part. The background of theCR51 (Concrete Reinforcing Steel Institute) approach for shear design of deep pile capsis then discussed. The CRST suggested allowable concrete shear stresses for critical beamsections across the width of the pile cap at the face of the column and for two-way slabpunching shear on the periphery (faces) of the column are summarized.Thirdly, the Canadian Code (CSA A23.3-M84) approach of using strut-and-tie models to design deep members is summarized. However, there is actually no shear designprocedure for deep pile caps in the Canadian concrete code, i.e., there is no appropriatestrength criterion for compression struts confined by plain concrete. By applying the design equation developed in Chapter 4 for the bearing stress limit of concrete compressionstruts, two design methods are proposed for the shear design of deep pile caps.Previous experimental work for pile caps and some important conclusions suggestedby various investigators are summarized. A comparative study is carried out in the fifthpart of this chapter. The predictions using various design procedures (ACT Code, CRSIand proposed methods) are compared with the previous experimental results. Finally,some conclusions are drawn at the end of this chapter.Chapter 5. Shear Design of Deep Pile Caps 475.2 ACT Code Approach for Pile Cap DesignThe ACT Building Code (ACT 318) [9, 34-35] uses a sectional force approach for thedesign of pile caps regardless of the depth. The procedure involves three separate steps:(1) shear design, in which the depth of the slab or pile cap is chosen so that the concretecontribution to shear resistance is greater than the shear applied on the code definedcritical section; (2) fiexural design, in which the usual procedures for beams are used todetermine the required amount of longitudinal reinforcement at the critical section forflexure, and; (3) a check that the bearing stresses on the column and piles do not exceedthe bearing strength.The procedure to check bearing strength is to satisfy that bearing on concrete atcontact surface (e.g., the top of the pile cap under the column, the bottom of the pilecap above the pile) does not exceed the code bearing strength (ACT 318-89: 15.8.1.1) [9].This check is to ensure that the concrete does not crush at these locations.For moment and shear calculations the pile force may be considered concentrated atthe centre of the pile. Flexural design is straight forward. The critical moment is equalto the pile reaction times distance from pile centre to face of a concrete column (Clause15.4.2). For example, the moment at section m— m in Figure 5.1 is equal to the productof the reactions from three left side piles and the distance s. The required amount oflongitudinal reinforcement is provided to resist this moment, similar to the procedureused for the fiexural design of slabs or beams.5.2.1 Code Procedure for Shear DesignShear design involves checking one-way beam shear and two-way punching shear (seeFigure 5.1). For one way beam shear, the critical section is measured at a distance d fromthe face of supported member (column) (Clauses 15.5.2 and 11.12.1.1 of ACI 318-89). ForChapter 5. Shear Design of Deep Pile Caps 48two way punching shear, the critical section is taken at a distance d/2 out from perimeterof the column, or the pile (Clause 11.12.1.2), where d is effective height measured fromextreme compression fibre to centroid of longitudinal tension reinforcement. Any pilewhose centre is located d/2 or more outside the critical section shall be considered asproducing shear on that section, where d is diameter of pile. In Figure 5.1 the shear onthe section n-n is introduced by the reactions from three right side piles. All eight sidepiles produce shear on the critical section for two-way shear at d/2 from the perimeterof the column. Reaction from any pile whose centre is located d/2 or more inside thecritical section shall be considered as producing no shear on that section (Clause 15.5.3).Figure 5.2 shows a two pile cap, for which the pile produces no shear on the sectionn-n. For intermediate positions of pile centre, the pile reaction is assumed as distributedlinearly across the pile diameter d in the direction shear is accumulated (Clause 15.5.3.3).For pile caps where the pile reactions are located either under the column or outsidethe critical sections (d from the face of the column for one-way shear or d/2 for two-wayshear calculation), the code procedure for shear design is straight forward. It is proposedherein that such a pile cap, as shown in Figure 5.1, can be defined as a slender pile cap.However, a great number of pile caps used in practice are not slender pile caps. These pilecaps are herein called ‘deep’ pile caps. The effective height d of deep pile caps is equalto or greater than the distance from the centre line of the closet pile to the face of thesupported column. The ACT Code approach becomes less transparent and inadequatefor the shear design of deep pile caps. This inadequacy is illustrated in the followingexamples.Chapter 5. Shear Design of Deep Pile Caps 495.2.2 Inadequacies of ACI Code Procedure for Shear Design of Deep PileCapsFigure 5.2 shows a deep two-pile cap, for which the piles are located within the codedefined critical section from the face of the column (i.e., within the distance d from thecolumn face). There are no code provisions specifically for the shear design of such apile cap. Hence ACT 318-89 Commentary Section 15.5.3 says: “when piles are locatedinside the critical sections d or d/2 from face of column, analysis for shear in deep flexuralmembers in accordance with Section 11.8 needs to be considered.” In Clause 11.8 someguidance for deep beams is given. A critical section is established at 0.5 a from face ofsupport, where a is shear span and is defined as a distance from a concentrated load toface of support. It becomes hard to define what is the support (the pile or the column?).Based on the guidance for deep beams, it could be considered that piles are supports andcolumn load is concentrated load. In Figure 5.2, the critical section j-j is defined by thisinterpretation. The code equations 11-29 or 11-30 are applicable and thus shear strengthon that section can be calculated. However, if the cap depth is increased, this definedcritical section could eventually cut through the column so that again no code provisionscan be applied.Figure 5.3 shows a deep three-pile cap. If the piles are still considered as supports,it is difficult to determine the shear span a (i.e., where is the concentrated load ?). Itshould also be noted that there are no code provisions whatsoever applicable to performpunching shear calculations when the three piles are located within the square criticalsection shown in the figure.Figure 5.4 shows another example that demonstrates the inadequacy of the ACT Codeapproach. Two four-pile caps are identical except that one has a square column and theother has a circular colunm. The two columns have the same areas. For the pile capChapter 5. Shear Design of Deep Pile Caps 50that has the square column and four circular piles, as shown in Figure 5.4(a), the squarecritical section d/2 away from the perimeter of the column just has its corners at thecentres of piles. The ACT Code assumption that the pile reaction is distributed linearlyacross the depth, d (pile diameter) (Clause 15.5.3.3), can be utilized for punching shearcalculation. If the code is followed literally, half of all pile reactions will introduce shearon the critical section. On the other hand, if the square column is replaced by a circularcolumn with the same area and correspondingly the critical section is circular rather thansquare, [see Figure 5.4(b)], the pile reactions are now totally located outside the criticalsection so that the critical section is subjected to the shear from the full value of pilereactions. Consequently, the prediction for load capacity of the pile cap with the squarecolumn is as much as almost two times that of the pile cap with the circle column, basedon the two way punching shear calculation. The sensitivity of the predicted capacity tothe shape of the colunm is a demonstration that a sectional approach is not appropriatefor deep pile caps.5.2.3 Comparison of Different ACI Code EditionsIn the above section the 1989 ACT Code [9] procedures for shear design of pile caps havebeen summarized, and it has also been pointed out that these procedures are transparentand directly applicable to slender pile caps. From the 1977 edition to the 1989 edition,the ACT Code have made some changes in shear design clauses which do not affect theshear design of slender pile caps, but do affect the shear design of deep pile caps. Thesechanges are summarized below.In ACT 318-77 [34], Clause 11.1.3.1 stated that for nonprestressed members, sectionslocated less than a distance d from face of support may be designed for the same shearas that computed at a distance d. In the accompanying commentary to ACT 318-77, astatement is made that if the shear at sections between the support and a distance dChapter 5. Shear Design of Deep Pile Caps 51differs radically from the shear at distance ci, the shear at the face of the support shouldbe used.In ACT 318-83 [35], the contents of the cormnentary have been incorporated into theformal body of the code, which means that the requirement that the shear capacity atthe face of the support should be evaluated is imposed when there is a concentrated loadbetween the face of support and the code defined critical section.From ACT 318-83 to ACT 318-89 no additional changes have been made to Clause11.1.3. In comparison with the previous ACT Code editions, more explanations aboutClause 11.1.3 have been added to the Commentary (ACT 318R-89). Something new inthe ACT 318R-89, which is relevant to this study, is that the shear at the face of thecolumn should be investigated for footings supported on piles when the pile reaction islocated within a distance d from the face of the column.The ACT Building Code procedures for two-way shear have not been modified recently.The critical section remains at d/2 from the perimeter of the column regardless whetherthere is a concentrated load applied within the critical section.5.3 CRSI Approach for Shear Design of Deep Pile CapsAs there are actually no specific procedures in the ACT Code for the shear design of deeppile caps [36], formulas are presented in the CR51 Handbook [36, 37] for allowable shearon concrete as a function of the ratio w/d, where w is the horizontal distance from face ofcolumn to centre of pile reaction and ci is effective depth of pile cap. Based on recognitionof considerable reserve shear strength observed in deep beam tests [38], allowable shearstrength on concrete is assumed to increase rapidly from ‘diagonal tension’ to ‘pure shear’as the distance w varies from d to 0 for one-way shear and from d/2 to 0 for two-waypunching shear. In addition, the critical sections for shear calculation of deep pile capsChapter 5. Shear Design of Deep Pile Caps 52are specified.For one-way shear of deep pile caps, the critical section is at the face of the column.The formula for shear strength calculation, which is a modification of the ACT Equation11-30, isV = (--)(3.5 — 2.5)(1.9iTh+ 2500p )bd 10Jbd (5.1)where (w/d) < 1.0, 1.0 > M/Vd> 0 and f is in psi unit. In the CRSI handbook [37]the formula is further reduced for the case of p 0.002 and f’ = 3000psi toV = (-)(3.5 — 2.5#)(1.9.Th + O.l/)bd 1O/bd (5.2)which gives when the product of the last bracketed expressions = 2’J and(d/w) 1. Figure 5.5(a) summarizes the one-way shear calculations for deep pile caps.The CRSI critical section for two-way shear in deep pile caps is defined at the perimeter of the column face. The formula for the allowable shear stress at the critical section,which is a modification of the ACT Equation 11-36, isV.=(-)(1 + )(4/) <32./jb0d (5.3)where for a square column of dimension c, b0 equals 4 x c. When d/2w = 1, the CRSTformula leads to the same expression as the ACT approach where the critical section isat d/2 from the column perimeter (b0 equals 4 x (c + d)). Figure 5.5(b) summarizes theCRSI two-way shear calculations for deep pile caps.The flow chart summarizing ACT Code and CRSI approaches is presented in Figures5.6.Chapter 5. Shear Design of Deep Pile Caps 535.4 Strut-and-Tie Model ApproachAn alternative approach which can overcome the drawbacks of ACI approach for theshear design of deep pile caps is strut-and-tie models which consider the complete flow offorces within a pile cap rather than the forces at any one particular section. The internalforce flow indicates that the vertical column load is transmitted to the piles by inclinedcompression struts and in order to prevent the piles from spreading apart, tension ties(reinforcement) must be provided. Assuming reinforcing steel bars are properly anchoredin the nodal zone, it is believed that a “shear failure” of a deep pile cap will occur whena concrete compression strut fails prior to yielding of tension ties. It should be notedthat truss models previously proposed for the design of pile caps, as mentioned in theintroduction of this chapter, have been used only for “flexural design”.5.4.1 CSA Approach for Deep BeamsStrut-and-tie models have been adopted in the Canadian Concrete Code (CSA A23.3-M84 [10]) for the design of deep beams [39]. The internal flow of forces in a simplysupported deep beam can be approximated by the truss model shown in Figure 5.7.The truss is composed of several components. The zones of unidirectional compressivestress in the concrete are modelled as compression struts, while tension ties are used tomodel the principal reinforcement. The regions of concrete subjected to multi-directionalstresses, where the struts and the ties meet (the joints of the truss) are represented by‘nodal zones’.The Canadian Code [10] requires that the concrete compressive stresses in the nodalzones not exceed the following limits (Section 11.4.7.5):(a) 0.85f in nodal zones bounded by compressive struts and bearing areas;Chapter 5. Shear Design of Deep Pile Caps 54(b) O.75f in nodal zones anchoring only one tension tie; and(c) 0.60f in nodal zones anchoring tension ties in more than one direction,The tension tie reinforcement is distributed and anchored over an effective area ofconcrete at least equal to the tensile tie force divided by the stress limits given above(Sections 11.4.7.4 and 11.4.7.6).The key part of this approach is the establishment of stress limit for the compressionstruts based on the work by Vecchio and Collins [40]. The code requires that the concretecompressive stress f2 in the struts shall not exceed f2ma (Section 11.4.7.3), i.e.f2 f2rnax= 0.8 +170c(5.4)where f2ma shall not exceed f unless the concrete is triaxial confined and e is determinedby considering the strain conditions of the concrete and the reinforcement in the vicinityof the strut.This approach is rational for deep beams, however extending this approach to deep pilecaps seems questionable. In deep pile caps, compression struts are usually unreinforcedand confined by surrounding plain concrete. Thus the Canadian Code has no provisionsspecifically for the shear design of deep pile caps. In the Canadian Concrete DesignHandbook [41], Suter and Fenton have ignored shear calculations for deep pile caps andincorporated a more conservative flexural model to predict deep pile cap behaviour.5.4.2 Proposed Design Method (1)The stress limit for compression struts in the Canadian Code was established by testingreinforced concrete panels [40], which were subjected to in-plane stresses. The stresslimit is appropriate for planar reinforced concrete members and is not applicable to threedimensional pile caps where compression struts are confined by plain concrete.Chapter 5. Shear Design of Deep Pile Caps 55In Chapter 4 the design equations for the bearing stress limit of compression strutsconfined by surrounding plain concrete were proposed, based on the analytical and experimental investigation. For convenience, the equations are again written here:= 0.6f(1 + 2afl) (5.5)wherea = 0.33(J4- 1 1.0= O.33( — 1) 1.0If the concrete compressive strength is significantly greater than 5,000 psi (34.5 MPa),the limit for the bearing stress isfb 0.6f + 6a/3/ (5.6)where f is in MPa units. If psi units are used, the 6cq3 should be replaced by 72a.These equations are directly applicable to the idealized compression struts as studiedin Chapter 2 to Chapter 4. See Figure 5.8(a). However, the geometry of compressionstruts in deep pile caps may be different from that of the idealized struts. Some factorswhich may affect the determination of geometrical ratios of D/d and h/b have to be takeninto consideration.One situation that could occur in design of deep pile caps is shown in Figure 5.8(b)and (c). In addition to different loading areas on two ends, the compression strut doesnot have a constant cross-section along its length. Hence new ratios of both h/b and D/bare needed in using the design equations. Herein a weighted average method is proposedto calculate the new geometrical ratios. These areChapter 5. Shear Design of Deep Pile Caps 56D— (D1/b)2+ (D2/b) 5 7b— /b+Dh — (h/b1)2+ (h/b2) 5 8b— h/b+h/bwhere b1, D1, b2 and D2 are loading geometry of the compression struts. See Figure 5.8.The allowable bearing stress limit of the compression strut in Figure 5.8(a) is greaterthan that in Figure 5.8(b). And again the allowable bearing stress limit of the one inFigure 5.8(b) is less than that in Figure 5.8(c).It should be noted that the predicted bearing stress fb for each compression strut ison the end with smaller loaded area. As the force in a strut is constant, the bearingstress on the other end can be determined from equilibrium.By incorporating the developed bearing stress limit for compression struts confined byplain concrete into a simple three-dimensional strut-and-tie model, one design procedureis proposed.As strut-and-tie models suggested in the Canadian Concrete Code (CSA A23.3-M84)[39] for deep beam design, the first design procedure proposed here for deep pile caps doesnot split the procedure into “flexural design” and “shear design,” which are traditionallyused for the design of slender members. The procedure involves several steps to develop anequilibrium force system. The initial pile cap dimension can be chosen based on previousdesign experience or with the help of published design aids. The effective depth d of apile cap can be determined from the concrete cover requirement and then the locationof reinforcement can also be determined. The strut-and-tie model is drawn and thegeometric parameters of compression struts are then calculated. The bearing stress limiton the nodal zones and the corresponding force in the compression struts can be calculatedby using Equations 5.5 or 5.6, 5.7 and 5.8. The horizontal components of the inclinedcompression strut forces are resisted by providing properly anchored reinforcement. TheChapter 5. Shear Design of Deep Pile Caps 57sum of vertical components of the inclined compression strut forces is the calculated loadcapacity of the pile cap. Prediction examples can be found in Appendix C.5.4.3 Proposed Design Method (2)The second design method is a simplified procedure of the first design method and ispresented in a more traditional way with a separated “flexural design” and “shear design”.In the second design method, the calculation of the bearing stress limit is based onsome simple parameters such as the effective depth of the pile cap and the dimensionsof the column and the pile rather than more complicated geometric parameters of thecompression strut. As a result, this simplified procedure is more easy to apply in thedesign of deep pile caps.The traditional truss analogy principle is used for “flexural design”. The fiexuralcapacity depends strongly on the inclination of the compression strut, which is definedby the location of the nodal zones. The truss model used in this procedure is that thelower nodal zones are located at the center of the piles at the level of the longitudinalreinforcement, while the upper nodal zone is assumed to be at the top surface of the pilecap at the column quarter point.The “shear design” of a deep pile cap involves applying Equation 5.5 or 5.6 to limitthe concrete stresses in compression struts and nodal zones to ensure the tension tiereinforcement yields prior to any significant diagonal cracking in the plain concrete compression struts. The ratio A2/A1 is identical to that used in the ACT Code to calculatebearing strength. To calculate the maximum bearing stress for the nodal zone below acolumn, where two or more compression struts meet, the aspect ratio of the compressionstrut can be approximated as(5.9)Chapter 5. Shear Design of Deep Pile Caps 58where d is the effective depth of the pile cap and c is the dimension of a square column.For a round column, the diameter may be used in place of c. To calculate the maximumbearing stress for a nodal zone above a pile, where only one compression strut is anchored,the aspect ratio of the compression strut can be approximated as(5.10)where 4, is the diameter of a round pile.A general shear design procedure for deep pile caps can be accomplished as follows.First, the initial pile cap depth is chosen using the ACT Code one-way and two-way sheardesign procedures. In the case of one-way shear, the critical section should be taken atd from the column face, and any pile force within the critical section should be ignored(i.e., the ACT procedure prior to 1983). Secondly, the nodal zone bearing stress shouldbe checked using above described procedure. If necessary, the pile cap depth may beincreased (/3 increased), or the pile cap dimensions may be increased in order to increasethe confinement of the nodal zones (increase a), or else the bearing stresses may need tobe reduced by increasing the column or pile dimensions. Prediction examples showingthis method can be found in Appendix D.5.5 Summary of Previous Pile Cap TestsIn order to evaluate the various pile cap design procedures quantitatively, these procedures will be used to predict previously tested pile caps. In this section previousexperimental work on pile caps is summarized.Hobbs and Stein (1957) [42] tested 70 one-third scale models of two-pile caps to verifytheir design method, which is based on analytically determined bending stress distributions on vertical planes through two-pile caps. The specimens had various amounts ofChapter 5. Shear Design of Deep Pile Caps 59either straight or curved reinforcing bars, which were anchored by a number of differentmethods. During the test, the first cracks consistently appeared at or close to the pile capmid-span. Later, diagonal cracks, which originated at the top of the piles, propagated tothe loading plate and were usually followed directly by failure.Deutsch and Walker (1.963) [43] tested four two-pile caps. The pile caps were designedwith the same centre to centre pile spacing 42 inches (1067 mm), the same width 15 inches(381 mm), the same stub pile dimensions 10 x 10 in. (254 x 254 mm) and the same stubcolumn dimensions 6.5 x 6.5 in. (165 x 165mm). The depth of the pile caps and amount ofreinforcing steel were varied in an attempt to investigate the structural action of pile capsand to compare different design methods (sectional force methods and the truss analogywith compression struts inclined at 35 and 45 degrees). Owing to the limitation of theloading capacity of the testing apparatus, only two pile caps (No.3 and No.4) were testedto failure, which were designed on the truss analogy with compression struts inclined at45 and 35 degrees respectively. None of the pile caps were actually thought to have failedin shear. All pile caps behaved similarly with one main crack forming at the verticalcentre line, extending to within one inch from the top of the pile cap at failure.Blévot and Frémy (1967) [32] tested two series of pile caps. First series specimensincluded 51 four-pile caps, 37 three-pile caps and 6 two-pile caps and were all of abouthalf scale size. Second series specimens were of full scale size and had 8 four-pile caps, 8three-pile caps and 6 two-pile caps. The main objectives was to determine the influenceof different reinforcing steel layouts and to verify their proposed truss models, which wereused to design reinforcement. The height of the pile caps and the layouts of reinforcingsteel were varied in the study. Figure 5.9 shows the five different reinforcing steel layoutsinvestigated in the four-pile caps of first series specimens. These layouts can be namedbunched square, bunched diagonal, bunched hybrid, and grid. The three-pile caps of thefirst series specimens had similar layouts to those in the four-pile caps. In the secondChapter 5. Shear Design of Deep Pile Caps 60series specimens, three-pile caps and four-pile caps had two kinds of layouts, bunchedhybrid, and bunched square or triangle plus grid. Layouts in the two-pile caps of bothseries specimens are similar to that in simply supported deep beams.Clarke (1973) [44] tested 15 four-pile caps, with the objective to compare the strengthof caps with values predicted by various design methods. Moreover the effect of differentsteel layouts and different anchorage lengths upon the behaviour of pile caps at serviceloads and ultimate loads were also investigated. The specimens included two types ofpile caps, namely type A and type B. A pile diameter of 200 mm, a column dimensionof 200 x 200 mm and the total depth of 450 mm were used throughout, giving plandimensions of 950 mm square for the type A caps and 750 mm square for the typeB. Three different steel layouts (grid, bunched square and bunched diagonal) and fourdifferent anchorage lengths (nil, nominal, full and full-plus-bob) were considered. SeeFigure 5.10. Most pile caps failed in shear after reinforcement yielded. Four pile capswere thought to have failed in flexure.Clarke and Taylor (1974) [45] also tested a number of eight-pile caps at 1:4, 1:15 and1:38 scale to investigate the influence of pile stiffness upon distribution of the columnload in pile caps. The uncracked pile caps were found to distribute pile loads as predictedby an elastic solution. When the concrete cracked, load carried by the outer piles wasconsiderably decreased.In order to confirm their proposed shear design approach for deep pile caps, Sabnisand Gogate (1980, 1984) [46, 47] tested nine four-pile caps, all at a one-fifth scale. Thefirst three specimens, one of which was of plain concrete, served to finalize the test set-up.The remaining six specimens were used to study the effect of the amount of uniformlyplaced reinforcement upon the shear capacity of deep pile caps. All the tested pile capshad the same dimensions: 3 inch (76 mm) diameter piles and columns, 13 inch (330 mm)square caps and 6 inch (152 mm) total depth of caps. The reinforcement ratio was variedChapter 5. Shear Design of Deep Pile Caps 61between 0.0014 and 0.0079 for the main six specimens. In order to prevent anchoragefailure, the reinforcement was hooked and further extended vertically the full depth ofthe pile cap. Punching shear was thought to be the predominant failure mechanism forall the specimens tested.Adebar et al. (1990) [11] tested six full scale pile caps in an attempt to investigatethe applicability of strut-and-tie models (truss models) to the design of pile caps and toevaluate the validity of the current ACT design procedures for pile caps. Of six testedpile caps, four caps were of diamond shapes, one cap of rectangular shape and one capof cruciform shape. The pile caps all had an overall depth of 600 mm and were loadedthrough 300 mm square cast-in-place reinforced concrete columns. They were supportedby 200 mm diameter precast reinforced concrete piles. The layouts of reinforcing steelused were bunched bars, grid, and bunched plus grid. Full straight anchorage lengthswere provided for reinforcing bars passing over the piles. Of six pile caps, four are believedto have failed in shear while two are thought to have failed in flexure.From above mentioned experimental investigations for pile caps, some important conclusions obtained by the investigators are summarized in the following:1. Truss models are more appropriate for deep pile caps than simple bending theory.Blévot and Frémy [32] confirmed their truss models by carrying out a comprehensiveseries of tests. Best results were obtained with the imaginary struts inclined atbetween 45° and 55° to the horizontal. Clarke [44] concluded that the truss analogywas an adequate method of analyzing a four-pile cap in order to ascertain its flexuralcapacity and to determine the required amount of tensile reinforcement. Adebar etal. [11] concluded that the ACT Building Code (sectional method) failed to capturethe trend of the experimental results.2. Layouts of reinforcing steel have a great effect on the behaviour of deep pile capsChapter 5. Shear Design of Deep Pile Caps 62under service loading and on the ultimate capacity of deep pile caps. The mostdesirable behaviour was obtained by the layouts of the bunched hybrid or bunchedsteel plus a relative light grid of steel. Blévot and Frémy [32] observed that thebunched steel gave approximately 20% higher strength than the same weight of steelspread out in a grid pattern. Clarke [44] concluded, from his four-pile cap tests,that bunching the steel in the form of a square increases the load by about 14%,while bunching it along the diagonals only increases the strength by a negligibleamount. It was reported by both Blévot and Clarke that the crack control was verymuch improved by using bunched hybrid or bunched hybrid plus grid, instead ofusing bunched diagonal only.3. Only nominal anchorage of reinforcing steel is required. Deutsch and Walker [43]showed analytically and confirmed by their tests that only nominal anchorage wasrequired past the edges of the piles. Blévot and Frémy [32] recommended the use ofstandard hooks as regular requirement for anchorage based on their experimentalresults. Clarke [44] concluded from his four pile cap tests that caps with bunchedsquare steel with nil anchorage was the most efficient if the efficiency is defined asthe load carried by the cap divided by the total weight of steel used. The reasonthat the anchorage requirement can be reduced was, suggested by Clarke, thatreinforcing bars bunched over piles are subjected to high lateral bearing stresseswhich help to lock the bars into place. It was also shown in Clarke’s tests that thefull anchorage resulted in a 30% increase in load capacities. Clarke thought thatthis substantial increase was mostly due to the vertical portion of the reinforcementproviding reinforcing across shear cracks. These vertical portions of the reinforcingbars acted as stirrups. It should be noted that the reinforcing bars in the specimenstested by Sabnis and Gogate [47] were also full anchorage with the bars extendedChapter 5. Shear Design of Deep Pile Caps 63vertically the full depth of the pile cap.5.6 Comparative Study of Different Design Procedures5.6.1 Comparison of Design MethodsTo compare the one-way shear design procedures, Figure 5.11 summarizes the relationshipbetween the maximum column load and the width b and depth d of a two-pile cap. Whenthe width of the pile cap is the same as the column width (b = c), the pile cap is essentiallya deep beam. See Figure 5.11(b). When the width of the pile cap is increased, largershear forces can be resisted by the increased concrete area at the critical section, and thelimiting column load due to bearing strength is increased as a result of confinement, seeFigure 5.11(c) and (d).Three different ACT Code predictions for one-way shear are given in Figure 5.11. Theleast conservative prediction, entitled “ACI ‘77,” is what designers of pile caps would haveused prior to the 1983 edition of the ACT Building Code (any pile within d of the columnface is assumed to produce no shear on the critical section). The “ACT ‘83” prediction, inwhich the critical section is at the face of the column, gives very conservative predictions.The predicted column load based on one-way shear calculation at the critical section halfway (a/2) between the face of the column and the centre of the closest pile, “ACT [11.8],”gives an intermediate result. The CRST method, in which the critical section is also atthe face of the column, is much less conservative than “ACT ‘83” due to an enhancedconcrete contribution, but is more conservative than “ACT ‘77.”Note that all methods predict that as the pile cap becomes very deep, the maximumcolumn load is limited by bearing strength. When the pile cap is twice as wide as thecolumn (b 2c) the ACT Code predicts that the confinement is sufficient so that thebearing strength has reached the upper limit of 2 x O.85f.Chapter 5. Shear Design of Deep Pile Caps 64Figure 5.12 compares the influence of pile cap depth on two-way shear strength predictions for a typical four-pile cap. Although the CRSI Handbook method gives a concretecontribution which is significantly larger for deep pile caps, the maximum column loadis always smaller than the ACT method. This is because in the ACT method the criticalsection is at d/2 from the column face and any pile which intercepts the critical sectionis assumed to transmit part of the load directly to the column. For example, if a pileis centred on the critical section, only half of the pile reaction must be resisted by thecritical section.5.6.2 Comparison of Prediction ResultsTable 5.1 summarizes the 48 specimens which were chosen for the comparative study.These include two pile caps tested by Deutsch and Walker [43], eighteen pile caps testedby Blévot and Frémy [32] in their second series, fourteen pile caps tested by Clarke [44],eight pile caps tested by Sabnis and Gogate [47] as well as six pile caps tested by Adebaret at. [11]. Specimens not considered in the comparative study include the small widebeam models tested by Hobbs and Stein, the small-scale specimens (first series) testedby Blévot and Frémy, and the one specimen tested by Clarke and two specimens testedby Deutsch and Walker which did not fail.Table 5.2 and Table 5.3 give the ACT Code and CRST predictions. The predictedflexural strengths and bearing strengths are summarized in Table 5.2, while the predictedshear capacities are summarized in Table 5.3.In the case of one-way shear, three different predictions are given from the ACTBuilding Code: the 1977 edition of the ACT Building Code (critical section at d fromthe column face); the 1.983 ACT Building Code (critical section at the column face); andthe special provisions for deep flexural members (Clause 11.8, critical section at half waybetween the column face and the closest pile centre). Tn the case of two-way shear, ACTChapter 5. Shear Design of Deep Pile Caps 65Code procedures involve critical section at d/2 from the column face and the pile face.Details of the predictions can be found in Appendix B.Table 5.4 presents the ratio of measured pile cap capacity to predicted capacity forthe three ACT Code predictions as well as the CRSI prediction. As mentioned before,actually there are no shear design procedures for deep pile caps in ACT ‘77, therefore mostpile caps are predicted to fail in flexure. If Clause 11.8 is considered, the load capacityof some pile caps is limited by shear (especially specimens tested by Blévot and Frémy).Tf the 1983 ACT Building Code is applied, it can be seen that the load capacity of almostall pile caps is controlled by shear due to the lower predicted shear capacities.It is very interesting to examine the CRST predictions. The original objective of theCRSI equations is to reflect the considerable reserve strength observed in the tests of deepbeams. Thus the CRSI equations generally give higher shear capacities. This results inthe change of predicted failure modes for most of the pile caps to flexural failure as inthe ACT ‘77 predictions. The comparisons are illustrated graphically in Figure 5.13 to5.16.Table 5.5 summarizes the predictions from the proposed design methods and comparesthe predictions with the experimental results.The comparisons of measured pile cap capacities and the predictions from the proposed design method (1) are shown in Figure 5.17 and 5.18. The predictions for specimenslisted in Table 5.1 are shown in Figure 5.17, while the predictions for the pile caps withbunched reinforcement are shown in Figure 5.18. The comparison of measured test resultswith the predictions from the proposed design method (2) is shown in Figure 5.19.Chapter 5. Shear Design of Deep Pile Caps 665.7 ConclusionsSome conclusions can be arrived at from the study in this chapter. The ACT Codedesign procedures (especially shear design) are not adequate for deep pile caps. The ACTCode 1977 edition gives scattered and not conservative predictions. The ACT Code 1983edition, in contrast with 1977 edition, gives overly conservative predictions, obviouslydue to the equation used for shear calculation on the critical section of the face of thecolumn which underestimates the shear capacity of deep pile caps. Though this problemhas been realized in the development of CRST shear design method, the problem is nottotally solved. In addition, the ACT Code design procedures or CRSI procedures aresometimes very complex and tedious, involving many calculations in order to meet thecode requirement. For example, predicting the three-pile cap specimens tested by Blévotis difficult.The ACT Code flexural strength predictions are unconservative for deep pile caps.These fiexural strength procedures are meant for lightly reinforced beams which are ableto undergo extensive fiexural deformations (increased curvatures) after the reinforcementyields. As the curvature increases, the fiexural compression stresses concentrate near thecompression face of the member. Deep pile caps are too brittle to undergo such deformations, therefore, assuming the fiexural compression is concentrated near the compressionface is inappropriate. Assuming the flexural compression is uniform across the entirepile cap, which strain measurements have shown to be incorrect [11], leads to a furtheroverprediction of the flexural capacity.In Figure 5.20, the bearing capacities on column zones for previously tested pile capsare illustrated in terms of the amount of confinement at column zones. The dashed lineshown in the figure demonstrates a similar trend as explored analytically and experimentally on the double punch loaded concrete cylinders. This indirectly further confirmsChapter 5. Shear Design of Deep Pile Caps 67that the transverse splitting phenomenon is predominant for the shear failures of deeppile caps.Both proposed design methods herein capture the physical behaviour of deep pilecaps, and produce conservative and reasonably accurate predictions. The proposed design method (1) is a direct extension of the two-dimensional strut-and-tie model of theCanadian concrete code. The proposed design method (2) is a simplified method whichtreats “fiexural design” and “shear design” separately. This is the concept traditionallyaccepted by current engineering practice.Chapter 5. Shear Design of Deep Pile Caps 68Table 5.1: Summary of pile cap test results.Number d Pile Column f FailureSpecimen of Size Size Reinf. LoadPiles (mm) (mm) (mm) (MPa) Layouts (kN)Blévot & Frémy2N1 2 495 350 sq. 350 sq. 23.1 bunched 20592Nlb 2 495 350 sq. 350 sq. 43.2 bunched 31872N2 2 703 350 sq. 350 sq. 27.3 bunched 29422N2b 2 698 350 sq. 350 sq. 44.6 bunched 51002N3 2 894 350 sq. 350 sq. 32.1 bunched 44132N3b 2 892 350 sq. 350 sq. 46.1 bunched 58843N1 3 447 350 sq. 450 sq. 44.7 bunched 41193Nl 3 486 350 sq. 450 sq. 44.5 bunched 49043N3 3 702 350 sq. 450 sq. 45.4 bunched 60803N3 3 736 350 sq. 450 sq. 40.1 bunched 66694N1 4 674 350 sq. 500 sq. 36.5 b.& g. 68654N1 4 681 350 sq. 500 sq. 40.0 b.& g. 65714N2 4 660 350 sq. 500 sq. 36.4 bunched 64534N2b 4 670 350 sq. 500 sq. 33.5 bunched 72474N3 4 925 350 sq. 500 sq. 33.5 b.& g. 63754N3b 4 931 350 sq. 500 sq. 48.3 b.& g. 88264N4 4 920 350 sq. 500 sq. 34.7 bunched 73854N4b 4 926 350 sq. 500 sq. 41.5 bunched 8581Deutsch & WalkerNo.3 2 533 254 sq. 165 sq. 23.8 bunched 596No.4 2 373 254 sq. 165 sq. 23.6 bunched 289Chapter 5. Shear Design of Deep Pile Caps 69Table 5.1: (con’t) Summary of pile cap test results.Number d Pile Column f’ FailureSpecimen of Size Size Reinf. LoadPiles (mm) (mm) (mm) (MPa) Layouts (kN)ClarkeAl 4 400 200 rd. 200 sq. 20.9 grid 1110A2 4 400 200 rd. 200 sq. 27.5 bunched 1420A3 4 400 200 rd. 200 sq. 31.1 bunched 1340A4 4 400 200 rd. 200 sq. 20.9 grid 1230A5 4 400 200 rd. 200 sq. 26.9 bunched 1400A6 4 400 200 rd. 200 sq. 26.0 bunched 1230A7 4 400 200 rd. 200 sq. 24.2 grid 1640A8 4 400 200 rd. 200 sq. 27.5 bunched 1510A9 4 400 200 rd. 200 sq. 26.8 grid 1450AlO 4 400 200 rd. 200 sq. 18.2 grid 1520All 4 400 200 rd. 200 sq. 17.4 grid 1640A12 4 400 200 rd. 200 sq. 25.3 grid 1640Bl 4 400 200 rd. 200 sq. 26.9 grid 2080B3 4 400 200 rd. 200 sq. 36.3 grid 1770Sabnis & GogateSS1 4 111 76 rd. 76 rd. 31.3 grid 250SS2 4 112 76 rd. 76 rd. 31.3 grid 245SS3 4 111 76 rd. 76 rd. 31.3 grid 248SS4 4 112 76 rd. 76 rd. 31.3 grid 226SS5 4 109 76 rd. 76 rd. 41.0 grid 264SS6 4 109 76 rd. 76 rd. 41.0 grid 2805G2 4 117 76 rd. 76 rd. 17.9 grid 173SG3 4 117 76 rd. 76 rd. 17.9 grid 177Adebar, Kuchma and CollinsA 4 445 200 rd. 300 sq. 24.8 grid 1781B 4 397 200 rd. 300 sq. 24.8 bunched 2189C 6 395 200 rd. 300 sq. 27.1 bunched 2892D 4 390 200 rd. 300 sq. 30.3 bunched 3222E 4 410 200 rd. 300 sq. 41.1 bk g. 4709F 4 390 200 rd. 300 sq. 30.3 bunched 3026Chapter 5. Shear Design of Deep Pile Caps 70Table 5.2: Summary of ACT Building Code and CRSI Handbook predictions (flexure andbearing).Specimen Flexural Bearing CapacityName Capacity Column Pile(kN) (kN) (kN)2N1 2197 2746 54922Nlb 3756 5135 102702N2 3429 3244 64882N2b 5553 5310 106202N3 5413 3816 76322N3b 7259 5485 109703N1 3825 15378 238603Nlb 5290 15311 237573N3 6014 15631 242553N3b 7982 13808 214244N1 7969 15525 260834Nlb 8171 17005 285704N2 7812 15464 259774N2b 8546 14234 239134N3 8283 14234 239134N3b 10788 20549 345214N4 9864 14734 247534N4b 10866 17631 29621No.3 512 1100 3907No.4 270 1093 3881Chapter 5. Shear Design of Deep Pile Caps 71Table 5.2: (con’t) Summary of ACT Building Code and CRSI Handbook predictions(flexure and bearing).Specimen Flexural Bearing CapacityName Capacity Column Pile(kN) (kN) (kN)Al 1258 1421 3908A2 1266 1870 5140A3 1256 2115 5812A4 1258 1421 3908A5 1265 1829 5028A6 1252 1768 4860A7 1262 1646 4524A8 1266 1870 5140A9 1264 1822 5008AlO 1252 1238 3404All 1252 1183 3252A12 1262 1720 4728Bi 2022 1829 5028B3 1528 2468 6784SS1 133 242 808SS2 116 242 808553 194 242 808SS4 157 242 808SS5 317 318 1060SS6 455 318 1060SG2 302 139 463SG3 628 139 463A 2256 3794 5296B 2790 3794 5296C 4008 4146 8682D 5646 4636 6472E 7428 6288 8780F 5324 3091 6472Chapter 5. Shear Design of Deep Pile Caps 72Table 5.3: Summary of ACT Building Code and CRSI Handbook predictions (shear).Specimen One-Way Shear (kN) Two-Way Shear (kN)Name ACT CRSI Column Pile1977 1983 [11.8] ACT CRSI2N1 1053a 316 947 775 c c c2Nlb 1438a 432 1296 898 c c c2N2 b 488 1463 2438 c c c2N2b b 616 1846 2634 c c c2N3 b 673 2018 3364 c c c2N3b b 804 2414 4023 c c c3N1 2130a 1583a 4490 2130 3722a 6594 37893Nlb 2707a 1715a 4737 2639 4397a 8062 41573N3 b 2504a 7496 9321 b 21249 b3N3b b 2468a 7385 9871 b 22297 b4Nl b 2436 7308 12004 11831a d b4Nlb b 2575 7702 12114 12824a d b4N2 b 2377 7130 11319 10959a d b4N2b b 2318 6955 10672 11103a d b4N3 b 3201 9603 16005 56759a 13338 b4N3b b 3858 11286 19292 70248a 16111 b4N4 b 3236 9695 16178 54102a 13425 b4N4b b 3566 10437 17831 63794a 14870 bNo.3 1881a 329 925 558 c c cNo.4 285 230 504 d c c cChapter 5. Shear Design of Deep Pile Caps 73Table 5.3: (con’t) Summary of ACT Building Code and CRSJ Handbook predictions(shear).Specimen One-Way Shear (kN) Two-Way Shear (kN)Name ACI CRSI Column Pile1977 1983 [11.8] ACT CRSIAl b 578 1646 2714 2916a 1458 1996A2 b 662 1848 3078 3344a 1672 2288A3 b 704 1934 3246 3556a 1778 2432A4 b 578 1646 2714 2916a 1458 1996A5 b 654 1830 3046 3308a 1654 2264A6 b 644 1790 2988 3252a 1626 2224A7 b 620 1750 2902 3138a 1569 2148A8 b 662 1848 3078 3344a 1672 2288A9 b 654 1828 3042 3302a 1651 2260AlO b 538 1554 2550 2722a 1361 1860All b 526 1526 2498 2660a 1330 1820A12 b 634 1784 2962 3208a 1604 2196Bl b 516 2066a 2584 b 3308 bB3 b 600 2344a 3002 b 3843 bSS1 b 69 186 254 122 d 228SS2 b 68 178 249 122 d 228SS3 b 68 181 250 121 d 226SS4 b 71 190 258 122 d 228SS5 b 84 229 296 134 d 251SS6 b 89 229 315 134 d 251SG2 b 65 164 261 101 d 185SG3 b 84 164 273 101 d 185A 3248 2400 6056 6352 2360a d 6250B 3408 2084 5306 4278 1839 d 2764C 6304 1820 4938 3734 1899 d 2989D 3770 2432 6348 4726 1968 d 3107E 4478 3076 8140 7078 2475 d 3970F 1604 572 1224 1618 c c ca = Increased capacity since piles partially within critical section;b = Infinite capacity since piles totally within critical section;c = Procedure not applicable;d = CRSI prediction not applicable- use ACT.Chapter 5. Shear Design of Deep Pile Caps 74Table 5.4: Comparison of ACT Code and CRSI Handbook predictions: ratio of measuredcapacity to predicted capacity and failure mode.Specimen ReportedName ACT ‘77 ACT ‘83 ACT [11.8] CRSI FailureMode2N1 1.96 s1 6.52 s 2.17 i 2.66 s1 s2Nlb 2.22 s1 7.38 i 2.46 s1 3.55 s1 s2N2 0.91 b 6.03 s 2.01 s1 1.21 i s2N2b 0.96 b 8.28 i 2.76 Sj 1.94 i S2N3 1.16 b 6.56 .5 2.19 s1 1.31 i S2N3b 1.07 b 7.32 i 2.44 s 1.46 s1 s3N1 1.93 s 2.60 s 1.11 2 1.93 s1 s3Nlb 1.81 s 2.86 s 1.04 i 1.86 si s3N3 1.01 f 2.43 s 1.01 f 1.01 f s3N3b 0.84 f 2.70 s 0.90 s 0.84 f s4Nl 0.86 f 2.82 s1 0.94 s 0.86 f s4Nlb 0.80 f 2.55 s, 0.85 i 0.80 f s4N2 0.83 f 2.71 8L 0.91 i 0.83 f S4N2b 0.85 f 3.13 s1 1.04 si 0.85 f s4N3 0.77 f 1.99 s 0.77 f 0.77 f s4N3b 0.82 f 2.29 s 0.82 f 0.82 f s4N4 0.75 f 2.28 s 0.76 5j. 0.75 f 54N4b 0.79 f 2.41 i 0.82 s 0.79 f sNo.3 1.16 f 1.81 s 1.16 f 1.16 f fNo.4 1.07 f 1.26 s1 1.07 f 1.07 f fChapter 5. Shear Design of Deep Pile Caps 75Table 5.4: (con’t) Comparison of ACT Code and CRSI Handbook predictions: ratio ofmeasured capacity to predicted capacity and failure mode.Specimen ReportedName ACT ‘77 ACT ‘83 ACT [11.8] CRSI FailureModeAl 0.88 f 1.92 s1 0.88 f 0.88 f sA2 1.12 f 2.15 s1 1.12 f 1.12 f sA3 1.07 f 1.90 s1 1.07 f 1.07 f sA4 0.98 f 2.13 s 0.98 f 0.98 f sA5 1.11 f 2.14 s 1.11 f 1.11 f sA6 0.98 f 1.91 s1 0.98 f 0.98 f sA7 1.30 f 2.65 s 1.30 f 1.30 f sA8 1.19 f 2.28 s1 1.19 f 1.19 f sA9 1.15 f 2.31 s 1.15 f 1.15 f sAlO 1.23 b 2.83 s 1.23 b 1.23 b fAll 1.39 b 3.12 i 1.39 b 1.39 b fA12 1.30 f 2.59 s 1.30 f 1.30 f fBl 1.14 b 4.03 s1 1.14 b 1.14 b sB3 1.16 f 2.95 s1 1.16 f 1.16 f fSS1 2.05 2 3.62 i 2.05 s2 2.05 2 SSS2 2.11 f 3.60 i 2.11 f 2.11 f sSS3 2.05 s2 3.65 s 2.05 2 2.05 s2 sSS4 1.85 2 3.18 s1 1.85 s2 1.85 s2 sSS5 1.97 2 3.14 i 1.97 2 1.97 2 SSS6 2.09 s2 3.15 s1 2.09 2 2.09 s2 sSG2 1.71 2 2.66 s1 1.71 2 1.71 2 SSG3 1.75 2 2.11 s1 1.75 2 1.75 2A 0.79 f 0.79 f 0.79 f 0.79 f fB 1.19 2 1.19 2 1.19 s2 1.19 s2 sC 1.52 2 1.59 s1 1.52 1.52 2 SD 1.64 2 1.64 s2 1.64 2 1.64 2 SB 1.90 2 1.90 s2 1.90 2 1.90 2 SF 1.89 s 5.29 si 2.47 s 1.87 i sf=flexure; b=column bearing;1=one-way shear;s2=two-way shear; s=shear.Chapter 5. Shear Design of Deep Pile Caps 76Table 5.5: Comparison of proposed design methods with experimental results.Specimen Predicted Predicted (2) Experimental Exp. Exp.Name (1) Flexure Shear Pred.(1) Pred.(2)(kN) (kN) (kN) (kN)2N1 1663 2128 1053a 2059 1.24 1.96 s2Nlb 2921 3570 1438a 3187 1.09 2.22 s2N2 2581 3109 2148 2942 1.14 1.37 s2N2b 4186 5051 3504 5100 1.22 1.46 s2N3 3893 4835 2552 4413 1.13 1.73 s2N3b 5317 6443 3622 5884 1.11 1.62 s3N1 3207 3256 2l30a 4119 1.28 1.93 s3Nlb 4107 4531 2707a 4904 1.19 1.81 s3N3 5060 5070 7493 6080 1.20 1.20 f3N3b 6170 6767 6869 6669 1.08 0.99 f4N1 5025 6041 9037 6865 1.37 1.14 f4Nlb 5398 6178 9790 6571 1.22 1.06 f4N2 4876 5933 8868 6453 1.32 1.09 f4N2b 4864 6512 8393 7247 1.49 1.11 f4N3 6041 6208 10604 6375 1.06 1.03 f4N3b 7018 7010 13993 8826 1.26 1.26 f4N4 6704 7414 10825 7385 1.10 1.00 f4N4b 7704 8150 12450 8581 1.11 1.05 fNo.3 467 480 732 596 1.28 1.24 fNo.4 250 254 285 289 1.16 1.14 fChapter 5. Shear Design of Deep Pile Caps 77Table 5.5: (con’t) Comparison of proposed design methods with experimental results.Specimen Predicted Predicted (2) Experimental Exp. Exp.Name (1) Flexure Shear Pred.(l) Pred.(2)(kN) (kN) (kN) (kN)Al 792 1030 1420 1110 1.40 1.08 fA2 928 1031 1716 1420 1.53 1.38 fA3 992 1020 1868 1340 1.35 1.31 fA4 792 1030 1420 1230 1.55 1.19 fA5 916 1031 1688 1400 1.53 1.36 fA6 900 1020 1648 1230 1.37 1.21 fA7 868 1030 1572 1640 1.89 1.59 fA8 928 1031 1716 1510 1.63 1.47 fA9 916 1030 1684 1450 1.59 1.41 fAlO 696 1030 1296 1520 2.18 1.48 fAll 668 1030 1256 1640 2.46 1.59 fA12 888 1030 1620 1640 1.85 1.59 fBl 1008 1374 1592 2080 2.06 1.51 fB3 1028 1030 1972 1770 1.72 1.72 fSS1 98 97 122a 250 2.55 2.58 fSS2 84 85 122a 245 2.92 2.88 fSS3 116 144 121a 248 2.14 2.05 sSS4 112 116 122a 226 2.02 1.95 fSS5 149 238 134a 264 1.77 1.97 sSS6 149 347 134a 280 1.88 2.09 sSG2 71 231 lOla 173 2.44 1.71 sSG3 71 543 lOla 177 2.49 1.75 sA 1448 1445 1915 1781 1.23 1.23 fB 1659 1662 1696 2189 1.32 1.32 fC 1499 1502 1684 2892 1.93 1.93 fD 2320 3454 1968a 3222 1.39 1.64 sE 3505 5084 2475a 4709 1.34 1.90 sF 1774 3472 1303 3026 1.71 2.32 sACI ‘77 prediction critical; s=snear critical; f=flexure criticalChapter 5. Shear Design of Deep Pile Caps 78Flexural Calculations / d_________h LmI I / I II I j I I I I\ / \ J \Two WayPunching Shear0n Way Shear/ /\ \ II Ijl I I I I!‘/ HI I—d/2- IdP)/I I I I_ ____ ____ ____ __________________ ____ ____________________________________________L_______________________m n. . .Figure 5.1: ACT Code specified critical sections for fiexure and shear investigation of pilecap.Chapter 5. Shear Design of Deep Pile Caps 79Deep Beam Behaviour d HOne Way ShearN___,__/Idpi I)%_ “—a/2Figure 5.2: A deep two-pile capChapter 5. Shear Design of Deep Pile Caps 80-- One WayShear?—,--‘ I//r———--—-—-——-7—ITwo Wayd/2-÷Punching ShearI’/ /—,I—’Figure 5.3: A deep three-pile capChapter 5. Shear Design of Deep Pile Caps 81/ III / I /\ 4— _,Two Way_Hd12m__Punching Shear(a)// /N’ /Two Way___Punching Shear___________________(b)Figure 5.4: Comparison of two-way punching shear calculations: (a) the cap with squarecolumn; (b) the cap with circle column.Chapter 5. Shear Design of Deep File Caps 82k42IiU(a)(b)Figure 5.5: CRSI approach for shear design of deep pile caps: (a) allowable shear stress,v, for one-way shear while w/d < 1.0; (b) allowable shear stress, v, for two-way shearwhile w/d < 0.5, from Ref.[36}.0.3 0.1)[a.z.sU..400 0.Z 0.4 0. 0.8w-/d3 ——I‘——‘_‘•_Ic1‘1:C,.rutICkl }c)‘JcI÷d/cZr41—s {(I+d1c_..II-_w.0.1 0.Z 03 0.4 0.5 O.Chapter 5. Shear Design of Deep Pile CapsGiven:Column sizePile groupPile capacityfyC )83Yespile cap?NoSelect d:One-way shear(11.11.1.1, 11.1, 11.2, 11.3and 11.5)Two-way shear(11.11.1.2, 11.11.2 and 15.5)Minimum footing depth(15.7)(ACI 77)Have no appropriate designprocedure for shearSelect d:(ACI 83)One-way shear (11.1—3)Have no appropriate designprocedure for two-way shearSelect d: (CRSI)One-way shear (CRSI)Two-way shear (CRSI)Find As:Moment calculation(15.4.1 and 15.4.2)Detailing(15.4.3, 15.4.4 and 12.2)Compute cap thickness h:(7.7.1)Shear check:(for Individual pile)Perimeter shear(11.11.1.2 arid 11.11.2)Beam shear(11.11.1.1 and 11.3,1.1)Bearing check:(10.16)Figure 5.6: Flow chart for ACT and CRSI design procedures for pile capsf-Th C-) H U) C’, 0 0 CDcD(DCf)<— CD‘C)D (CD——:3-aCDCD.‘0 D-4. CD Cl) C— 9- -a ICD CD*-.QCD0(nCDD CDT1 0 0 —i,0 C) CD C’, 0 m D ci CDD 0 0. Q) N 0 D CD p co 01 C) 0-003c0 CD 0) Cl) 0 :301 0CD00 0)’ 10CDCDChapter 5. Shear Design of Deep Pile Caps 85D1Hh(a) (b) (c)Figure 5.8: Loading geometry of compression struts with linearly varying cross section.Chapter 5. Shear Design of Deep Pile Caps 86(b)(d)(e)(a)N:,(c)/ N--Figure 5.9: Various layouts of main reinforcing bars used by Blévot and Frémy, Ref.[32j.Chapter 5. Shear Design of Deep Pile Caps(1) NilL(2) Nominal—I7‘p87(3) Full (4) Full-plus-bobLFigure 5.10: Various anchorage lengths used by Clarke, Ref. [44].Chapter 5. Shear Design of Deep Pile Caps 8836”‘Id=1O”_____,-—/ I i(a)_‘i c=1O”b=c=1O(kip)40°300 lAd ‘77 CRSL—/ Proposed (2)200 -100- ACI [11.8](b)0——ZC18312 14 16 18 20 22 24 26 28 30d(in.)Figure 5.11: Comparison of one-way shear design methods for two-pile caps: (a) planview of pile cap; (b) to (d) influence of pile cap depth on column load for various pilecap widths.Chapter 5. Shear Design of Deep Pile Caps 89p 700(kip) 600500400300200100(c) 01d (in.)700(kip) 600500400300200100(d) 012 14 16 18 20d (in.)Figure 5.11: (cont’d) Comparison of one-way shear design methods for two-pile caps: (a)plan view of pile cap; (b) to (d) influence of pile cap depth on column load for variouspile cap widths.b=2c=20”b=4c=40” Proposed (2)Chapter 5. Shear Design of Deep Pile Caps 90(a)P 1800(kip)160014001200100080060040020015”36”15”12 16 20 24 28 32 36-d (in.)Figure 5.12: Comparison of two-way shear design methods for a typical four-pile cap: (a)plan view of pile cap; (b) influence of pile cap depth on column load./dp8/14’—-. f.. —I II15” 36” 15”—. .-4—(b)Chapter 5. Shear Design of Deep Pile Caps 914PexpACI ‘77pred.I I3 0 FlexureOne-way shearA Two-way shear2AA0*1 U0.0_0000 0 0Mean 1.31CCV 34.8%2000 4000 6000 8000 10000 12000Pexp (kN)Figure 5.13: Comparison of ACT ‘77 predictions with experimental results.Chapter 5. Shear Design of Deep Pile Caps 924Pexp ACI ‘83Ppred.• One-way shear2 : :A- Two-way shearA1CMean 3.09C0V55.1%2000 4000 6000 8000 10000 12000Pexp (kN)Figure 5.14: Comparison of ACI ‘83 predictions with experimental results.Chapter 5. Shear Design of Deep Pile Caps 93Pexp ACI [11.8]Ppred-• One-way shearA Two-way shear2*AA100 ,C*A AU0 10Mean 1.42CCV 38.9%2000 4000 6000 8000 10000 12000Pexp (kN)Figure 5.15: Comparison of ACT ‘[11.8] predictions with experimental results, clause 11.8considered.Chapter 5. Shear Design of Deep Pile Caps 944PexCRSIpred. Flexure• One-way shearA Two-way shearBearing.I •• •IA• A.0100 0&000 0Mean 1.40CCV 40.7%0 2000 4000 6000 8000 10000 12000Pexp. (kN)Figure 5.16: Comparison of CRSI predictions with experimental results.Chapter 5. Shear Design of Deep Pile Caps 954Pexp.Ppred. Proposed (1)3.3..2• $••.•..•••,•••• •.•••• • •• •• • ••• •Mean 1.57CCV 29.6%0 I0 2000 4000 6000 8000 10000 12000Pexp (kN)Figure 5.17: Comparison of proposed method (1) predictions with experimental resultsof all specimens in Table 5.1.Chapter 5. Shear Design of Deep Pile Caps 964Pexpp Proposed (1)pred.3 (bunched reinforcement)2.......•...Mean 1.32CCV 16.5%00 2000 4000 6000 8000 10000 12000Pexp (kN)Figure 5.18: Comparison of proposed method (1) predictions with experimental resultsof specimens with bunched reinforcement in Table 5.1.Chapter 5. Shear Design of Deep Pile Caps 974Pexp.Proposed (2)Ppred I3 ° “Flexure”o“Sheart0 L..S2.• 00• o .0.0 o 0 o01 00 0Mean 1.55• CCV 27.8%0 I Io 2000 4000 6000 8000 10000 12000Pexp. (kN)Figure 5.19: Comparison of proposed method (2) predictions with experimental resultsof all specimens in Table 5.1Chapter 5. Shear Design of Deep Pile Caps 983f b• Adebar et a— ° elévot and Frémy.c ‘ • ClarkeI Deutsch and WalkerSabns and Gogate 82.0...—0D00—00CC2 3 4 5Figure 5.20: Relationship of measured ultimate bearing stress and confinement on columnzones of pile caps in Table 5.1.Chapter 6Shear Failure of Beams Without Stirrups6.1 IntroductionIn the previous chapters the discussion focused on the shear resistance of deep pile caps.The model that was proposed assumes that the load is transmitted from the column tothe piles by direct compression struts, and that the tensile stresses which cause crackingof the compression struts is due only to the transverse spreading of compression stresses.As pile caps become more slender, other mechanism will be involved in creating tensilestresses which may lead to a shear (diagonal tension) failure. Note that this issue wasaccounted for in the proposed design method (Section 5.4) by suggesting that the ACTBuilding Code (empirical) method be used for slender pile caps.In order for this thesis to be a relatively comprehensive treatment of pile caps andother members without transverse reinforcement, it is necessary to consider the mechanisms involved in the shear failure of more slender members. Towards that end, a studywas made of the shear resisting mechanisms in slender beams. While this study is reallya separate topic, it is closely related to the first part of this thesis (transverse splittingin deep members). In fact, as will be shown in Chapter 8, the truss model which isassociated with the transverse splitting mechanism in deep members is in fact very similar to the truss model associated with what is believed to be an important shear failuremechanism in slender beams without transverse reinforcement.It should be pointed out however, that unlike the first part of this thesis, which was a99Chapter 6. Shear Failure of Beams Without Stirrups 100complete study of transverse splitting (it resulted in the development of a new improveddesign method for pile caps), this study on shear in slender beams is much more of apilot study. In this and the next chapter considerable information is given about theshear resisting mechanisms in slender beams, but additional research is needed beforethe concepts presented here can be implemented into a shear design procedure which canbe used by practising engineers.This chapter is divided into seven parts. A brief review of the literature is presentedand some important conclusions about the current knowledge of shear failures in beamswithout web reinforcement are summarized first. Then, an interpretation of an importantshear failure mechanism is presented based on the deformation compatibility of criticaldiagonal crack propagation. Next, bond influence upon internal stress distributions ofboth uncracked and cracked beams are investigated by using linear elastic finite elementsand the concepts of arch action and beam action. This is followed by a study on sheardisplacements along vertical cracks and inclined cracks in beams, and the presentationof a load transfer mechanism based on studies in this chapter and the experimentalmeasurements of previous work. As an application of the understanding obtained in thisstudy, the test results carried out by various researchers are explained in the sixth part.Finally, the conclusions arrived at are summarized.6.2 Brief Review of the LiteratureState-of-the-art summaries of the shear resistance of structural concrete members withouttransverse reinforcement have been reported previously in Ref. [38, 48-51]. The literaturereview presented here focuses on more recently developed models for shear resistance ofslender beams and understanding the transition from deep beams to slender beams.Chapter 6. Shear Failure of Beams Without Stirrups 1016.2.1 Transition from Deep Beam to Slender BeamThe distribution of the transverse compression stresses within the shear span estimatedby Mau and Hsu [52] is shown in Figure 6.1. When a/h = 0, transverse compression stressis maximum at the line of actions. While a/h = 0.25 and 0.5, the maximum values oftransverse compression stress still occur at the centre of the line in. connection with loadpoint and support point, but the magnitudes decrease with increasing a. While a/h = 1,the distribution of transverse stress shows the characteristics of two humps. These two-humps become more distinct and the transverse compressive stresses approach zero atthe centre of the shear span when a/h = 2. Hence, it is seen that direct compressivestress field between the load and the support is formed when a/h < 1.Following the internal force flow, Schlaich et al. [3] gave a description of the transitionfrom deep beams to slender beams using truss models for uncracked beams. See Figure6.2. If a single load is applied at a distance a < h near the support, the load is carrieddirectly to the support by a compressive stress field as simulated by a simple compressionstrut. A further examination indicates that the transverse tensile stresses will be introduced in the compression strut as shown by the refined truss model and as discussed inChapter 1. With increasing shear span a, the compressive member C1 joins the part ofthe tensile force T1 and simultaneously the transverse tensile ties blend into the verticalties of the truss model, which are now needed for hanging up the shear forces in slenderbeam. Thus the beam having a < h, where the load can be transmitted directly to thesupport, is treated as a deep beam.Collins and Mitchell [39] compared the shear strengths of a series of simply supportedbeams tested by Kani [13] with the predicted capacities from both sectional and strutand-tie analyses for assumed cracked beams. In Kani’s tests the shear span to depthratio a/d varied from 1 to 7 and no web reinforcement was provided. The comparisonChapter 6. Shear Failure of Beams Without Stirrups 102indicated that the shear resistance is governed by strut-and-tie action, which assumesthe loads are carried to the supports with direct compression struts. The failures aregoverned by a strut-and-tie model (crushing of the compression strut) at a/d less thanabout 2.5, but are governed by a sectional model for aid greater than 2.5. See Figure6.3. Their conclusions suggest that beams with aid < 2.5 can be treated as deep beams.The truss models developed by Al-Nahlawi and Wight [53] illustrate that when a/d>1, the shear failure of the beams without transverse reinforcement is characterized by thefailure of concrete tension ties. Their work suggested that the beams with a/d> 1 couldbe considered as slender beams. In the next section more details of Al-Nahlawi andWight’s truss models will be presented.6.2.2 Behaviour of Slender BeamsBased on the modified compression field theory which suggests that concrete tensile stressis present in cracked reinforced concrete members, a simple truss model with concretetension ties was developed by Adebar [54] for a diagonally cracked beam which is simplysupported, subjected to point loads and reinforced with only longitudinal reinforcement.See Figure 6.4. The inclination of compression struts in the web is equal to half of theinclination of the uniformly spaced diagonal cracks, and concrete tension ties are perpendicular to the compression struts. The shear capacity of the beam depends primarily onthe crack width, which is strongly influenced by the crack inclination.A mechanicaI model was developed by Reineck [55] based on a tooth model considering the shear carrying actions of friction along cracks, dowel force of longitudinalreinforcement, cantilevering action of tooth from the compression zone and shear forcecomponent in the compression chord. After a detailed analysis of the shear carryingactions involved, a stress field for a beam without transverse reinforcement was proposedand a truss model representing this stress field was developed. See Figure 6.5. This trussChapter 6. Shear Failure of Beams Without Stirrups 103model is similar to the one given by Adebar [54] except that the inclination of uniformlyspaced diagonal cracks are assumed to be 60 degrees. The failure is characterized by acrack further propagating into the compression zone and breaking off the tooth, and isdefined by the critical slip along the inclined crack which reaches a critical crack width.Based on an idea that a truss model must include a concrete tension member thatwill fail in tension before yielding of the flexural reinforcement, Al-Nahiawi and Wight[53] developed the truss models shown in Figure 6.6. The truss models have a 45-degcompression strut originating at the node under the applied load and a 35-deg compression strut originating at the support. The shear capacity of beams without stirrups iscontrolled by the failure of a concrete tension tie when its maximum tensile stress reachesthe tensile strength of concrete subjected to transverse compression. They consider thefailure of a concrete tensile tie as corresponding to the unchecked propagation of an inclined crack. One of conclusions from their models is that for beams with a/d> 1, shearfailure modes are similar and indicated by the failure of concrete tensile ties.Muttoni and Schwartz [56] developed a structural model for the loading stage whena typical crack pattern has formed. See Figure 6.7. They indicated that the form of thiscritical crack pattern is accompanied by a collapse of three shear carrying actions, whichare cantilever, interlocking and dowelling actions, and then the direct transfer of the loadto the support also becomes impossible because of the wider crack. Their model showsthat the compression strut turns in the central region of the beam and acts together witha concrete tie. Failure occurs either when the tensile strength in the region D is reachedor when the strength in zone E, which is subjected to both tension and compression, isexceeded.After carrying out tests designed to test the validity of current design concepts,Kotsovos [57] has concluded that aggregate interlock and dowel action make a negligible contribution to the load-carrying capacity and the strength of compressive zonesChapter 6. Shear Failure of Beams Without Stirrups 104increases due to triaxial stress/strain states, thus making a significant contribution toshear capacity. In an attempt to summarize the experimental information, the conceptof the ‘compressive force path’ has been developed by Kotsovos. The concept considersthat the load-carrying capacity of beams without transverse reinforcement is associatedwith the strength of uncracked concrete in the region of the paths along which compressive forces are transmitted to the supports. See Figure 6.8 (it should be noted that theoriginal model presented by Kotsovos in this figure does not satisfy equilibrium). Theshear failure is believed to be related to the development of tensile stresses mainly in theregion of the path. The tensile stresses may be developed due to changes in the pathdirection, the varying intensity of compression stress field along the path and bond failure at the level of the tension reinforcement between two consecutive fiexural or inclinedcracks, etc. More discussion about Kotsovos’ experimental work is given in Section 6.7.In summary, it is generally accepted that point loads can be transmitted directlyto the support in beams either uncracked or cracked while a < d. However, there aredifferent opinions about load transfer mechanisms for cracked beams when 1 <a/d < 2.5.Secondly, all the investigators, who tried to give rational interpretations of shear failure mechanism of reinforced concrete beams without transverse reinforcement, believethat the tensile strength of concrete plays an important role in the shear resistance forboth uncracked beams and cracked beams, which is illustrated by various truss models composed of tension ties. However the interpretations of where and how the tensilestrength of concrete is mobilized for shear resistance are different. Adebar [54], Reineck[55], A1-Nahlawi and Wight [53] considered that concrete tensile strength is needed forhanging up the shear force in the lower part of beams. Muttoni and Schwartz [56] considered that concrete tensile strength is needed in the upper part of beams to help thecompression strut to deviate from the critical diagonal crack. In Kotsovos’ compressiveforce path concept [57], concrete tensile strength mainly contributes to the change ofChapter 6. Shear Failure of Beams Without Stirrups 105the compressive force path direction. Thus it is evident that there is still no universally accepted shear failure mechanism for reinforced concrete beams without transversereinforcement.6.3 One Interpretation of Shear Failure of Beams without StirrupsIn order to investigate the shear failure mechanism, the behaviour of a slender, longitudinally reinforced concrete beam simply supported and subjected to two symmetricalpoint loads is examined herein. See Figure 6.9. The vertical flexural cracks first form inthe pure bending region and then the additional cracks form in the shear spans betweenthe concentrated loads and supports. The vertical cracks in the pure bending region keeppropagating upward vertically with increasing load. However the flexural cracks formedin the shear spans, after extending vertically to longitudinal reinforcement level, willbecome slightly inclined toward the load. Traditionally it is believed that these cracksbecome inclined because of shear. As the loading is further increased, a characteristicinclined crack, the so-called critical diagonal tension crack, forms. Its propagation intothe compression zone of the beam near the section of maximum moment is usually relatively sudden, splitting the beam into two pieces and causing collapse. Above shearfailure observations are well known and quite general.As reviewed briefly in Section 6.2, the various interpretations for shear failure mechanism given by previous studies are generally based on force transfer concept, whichemphasizes the various actions involved in shear carrying mechanism and on the satisfaction of equilibrium conditions. On the contrary, an alternate interpretation of shearfailure mechanism is based on the deformation compatibility of a critical diagonal crack.It is very reasonable to choose the crack opening of the critical diagonal crack atthe reinforcement level as an important geometric parameter to keep track of the crackChapter 6. Shear Failure of Beams Without Stirrups 106propagation. See Figure 6.10. If the crack opens horizontally only, the crack propagates vertically, which is the case in the pure bending region. If the crack opening hasboth horizontal and vertical components, the crack propagation will deviate from vertical direction and become inclined. This is the situation in the shear span range. Theorientation of stable crack propagation can be approximately determined by the ratioof horizontal and vertical opening components. Figure 6.10 shows this geometric relationship. Test observations indicate that there are several inclined cracks in the shearspan, all of which have the potential to become the critical diagonal crack. The shearfailure observations show that only one of the inclined cracks, the critical diagonal crack,penetrates into the compression zone and extends horizontally to load point before thebeam collapses. This means that of all inclined cracks, the critical diagonal crack thatdistinguishes itself from other inclined cracks must have a significant vertical openingcomponent, which is geometrically compatible with this crack propagation, at the reinforcement level or the other locations along the crack. Then it is logical to infer that theformation of a horizontal crack, which introduces the significant vertical opening component of the critical diagonal crack, is a critical stage in the shear failure of a reinforcedconcrete beam without transverse reinforcement.It is of interest that more information can be found for the phenomenon describedabove. The shear failure descriptions given in ACT 426 Committee report [38] are:Beams may exhibit a number of different modes of shear failure, the mostcommon of which is the crushing or shearing of the compression flange overthe inclined crack which is often accompanied or initiated by splitting alongthe tension reinforcement.It can be seen from this description that there are some uncertainties of which one occursfirst in this statement, i.e., the crushing of the compression flange or splitting along theChapter 6. Shear Failure of Beams Without Stirrups 107tension reinforcement.Fortunately, there is valuable experimental evidence that can help to clarify thispoint. The evidence comes from the experimental investigation carried out by Chana[58]. Chana used a high speed tape recorder in conjunction with electrical demountablestrain transducers to continuously monitor crack widths at critical locations while thetested beams were approaching failure. His test results clearly show that the beam shearfailure was preceded by splitting along the longitudinal reinforcing bars.Another factor which is of help to understand this mechanism is the crack controlefficiency of longitudinal reinforcement. The major function of longitudinal reinforcementin concrete beams is to compensate for the weakness of concrete low tensile strength andprovide enough tensile strength to have high concrete compressive strength fully utilized.This function is well obtained in the pure bending region of the beam. After the verticalcracks have occurred in this region, their propagation are most efficiently controlled bythe longitudinal reinforcement, which is at right angle to them. The crack opening atthe reinforcement level is proportional to the deformation of the reinforcement if thereis proper bond between concrete and reinforcement. Under a same load level a largerpercentage of reinforcing bars results in smaller stresses in the bars (or smaller crackopening). On the other hand, the force system in the pure bending region also favourschecking crack propagation. If more reinforcement is arranged at the bottom of the beam,more compressive force (or a larger compression zone) can be developed at the top of thebeam. This compressive force effectively restrains the crack from penetrating verticallyinto the compression zone.However in the shear spans of the beam, the cracks are diagonal rather than vertical.The reinforcement is skew to the cracks. In order to check the propagation of the diagonal cracks, not only axial action but also dowel action of the reinforcement should bemobilized. The crack control of longitudinal reinforcement to inclined cracks is much lessChapter 6. Shear Failure of Beams Without Stirrups 108efficient than to vertical cracks. The flatter the inclined cracks are, the less the crack control efficiency of longitudinal reinforcement. The crack control efficiency of longitudinalreinforcement is least when the cracks are horizontal.From the above discussion, an intuitive and rational conclusion about shear failuremechanism of slender, longitudinal reinforced concrete beams is that splitting along thelongitudinal reinforcement, which produces a considerable vertical opening component tothe critical diagonal crack, is the immediate cause of shear failure. In general, a distinctcharacteristic of slender beams without transverse reinforcement is lack of the abilityto control horizontal crack propagation and the brittleness of shear failure results fromthe occurrence of either flat inclined or horizontal cracks. Should the defect have beenovercome by any mechanism, the shear resistance behaviour will be greatly improved.Traditionally, the transverse reinforcement in conjunction with longitudinal reinforcementis provided for shear design of structural concrete members.6.4 Bond Effect in Longitudinally Reinforced Concrete BeamsIt was concluded above that the horizontal splitting along longitudinal reinforcement,which precedes the propagation of a critical diagonal crack into compression zone, indicates shear failure of a beam without transverse reinforcement. In this section the loadcarrying mechanism before the occurrence of this horizontal splitting and the cause ofthe horizontal splitting are investigated. The influence of bond between concrete andreinforcing bars upon the load transfer mechanism is focused on as so much informationfrom previous studies [13, 59-67] indicates that the bond has a significant effect, whichwas either ignored or considered to a less extent in previous studies of shear failure ofbeams without transverse reinforcement.Kani [13] tested a series of point loaded and simply supported beams without webChapter 6. Shear Failure of Beams Without Stirrups 109reinforcement to investigate the influence of various bond qualities on the shear failuremechanism. By introducing an intermediate layer of a vermiculite-cement mix withdifferent elastic properties and strengths between the reinforcement and the concrete inthe shear span, the varying bond quality (varying average ultimate bond stress) wasobtained. The tests showed that “the better the bond, the lower is the load capacity ofthe beam.” As the extreme case, the no bond beam presented flexural failure with nosingle crack occurring in the shear spans. Though the test results did not tell the wholestory about the shear failure mechanism of beams without transverse reinforcement, theydid reveal the fact that bond plays an important role in load sustaining mechanism.The second example showing the interaction of bond and shear failure is the bondtests done by Ferguson et al. [61, 62] and Kemp et al. [66]. Ferguson et al. tested beamsto investigate development length of reinforcing bars in bond. They found that therewere always the combinations of diagonal tension failure and bond splitting failure for thenarrow beams that were tested. Kemp et al. [64, 66] observed that the shear crack formedonly after a longitudinal bond crack and bond slip had occurred for the cantilever-typebond specimens tested (more discussions on these tests are given in the next chapter).The third example is the experimental work by Mains [59]. His measurement of bondstresses along reinforcing bars in beams indicated that the distribution was not uniformand that there was a significant localization after the occurrence of the critical diagonalcrack. This evidence conflicts with the basic assumption adopted by many authors thatthe stress variation of longitudinal reinforcement in cracked beams has a similar shapeas the moment diagram.6.4.1 Bond Influence in Uncracked BeamsIn order to investigate the effect of bond on the load transfer mechanism of reinforcedconcrete beams without web reinforcement, a simply supported and centrally loadedChapter 6. Shear Failure of Beams Without Stirrups 110beam was studied to find the internal stress distributions in the uncracked concrete.See Figure 6.11. Based on the work of Schlaich et al. [3], this beam can be dividedinto B-regions and D-regions, which are also shown in the figure. The prediction usingclassical beam theory, which assumes that cross sections remain plane during deformation(Bernouli assumption), is only valid for B-regions and not valid for D-regions so that thelinear elastic finite element method was used to investigate the effect of bond on theinternal stress distributions. Nine node Lagrange quadratic plane elements were used inanalysis. The biaxial stresses in x and y directions and principal stresses were calculatedat each node of the element.First, perfect bond and no bond cases for slender beams were studied. The beamwas assumed to already have a single flexural crack formed at mid-span. Because of thesymmetry of the problem, only half of the beam was modeled. See Figure 6.12. In Figure6.12(a) the force T modelling the tension force in the reinforcement is applied at the thefront of the beam, which is the simulation of the case with perfect bond between thelongitudinal reinforcement and concrete. In Figure 6.12(b), the corresponding force Tis applied at the end of the beam to represent the no bond case. Final analysis results,showing the internal stress distributions, are also given in Figure 6.12.The shear stresses, in both the perfect bond case and the no bond case, give similardistributions in the B-region. The shear stresses are confined within a certain part alongthe cross sections in the D-regions and distributed uniformly in the B-regions. The shearstress distributions demonstrate the shear transfer mechanism in the uricracked beams,but do not indicate the load transfer mechanism. The fundamental difference betweenthese two internal stress distributions is that the whole cross sections are subjected tocompression for the no bond case whereas the cross sections are subjected to tension onthe bottom and compression on the top for the perfect bond case.The load transfer mechanism can be found by looking at the internal force flows inChapter 6. Shear Failure of Beams Without Stirrups 111the beam. Figure 6.13 shows two principal stress trajectories for perfect bond and nobond situations respectively. In Figure 6.13(b), the load is transferred to the supportdirectly by a compression stress field (compression strut). The internal force flow isconsistent with the externally applied loads. The resultant of forces at the support andthe resultant of forces at the upper compression zone act along the same line but inopposite directions. In Figure 6.13(a), the force system is different from that in Figure6.13(b). As a requirement of static equilibrium, the action line of the resultant of forcesat the upper compression zone will still go through the point where the support reactionand the tension force in the reinforcement intersect. However the force flow in the beamis curved rather than straight so that the tensile strength is mobilized to make the forceflow change direction. Though the force flow in Figure 6.13(a) is different from that inFigure 6.13(b), this force flow is consistent with the corresponding external force system.As there is no horizontal force applied at the bottom corner of the beam, the forceflow originating from the support is almost vertical. Then the additional internal force(tension force) is required to make the force flow from the upper load point to the supportpossible.It has been mentioned before that the load can be directly transferred to the supportin deep beams (a/d < 1). The transverse splitting of compression struts dominates shearfailures of deep beams. In this section the influence of various bond stress distributionsupon the transverse splitting of compression struts is also investigated. Various bondstress distributions are shown in Figure 6.14. The analysis results are summarized inFigure 6.15. The bond influence on the transverse tension of the compression strut isconcentrated at the lower end, where the transverse tensile stresses are amplified due tobond. For the no bond case, the transverse tensile stress is almost uniformly distributedalong the inclined compression strut. The shear stress distributions along the crosssections for various bond cases are similar to those in Figure 6.12.Chapter 6. Shear Failure of Beams Without Stirrups 1126.4.2 Bond Influence in Cracked BeamsIn order to investigate the bond effect upon the load transfer mechanism of crackedbeams, the traditional concepts of arch action and beam action [68] are used in a simple equilibrium analysis procedure. It is believed that the bond effect can be betterunderstood by using these simple concepts.For a cracked beam in the shear span, the relationship between external moment andinternal moment of resistance, at a distance x from the support, can be approximated as[68]M = Tjd (6.1)where T = T(x) =tensile force resultant acting at the centroid of longitudinal reinforcement, j = j(x) =variable coefficient and d =effective depth measured from the extremecompression fibre to the centroid of longitudinal reinforcement. The shear force may beexpressed as V = dM/dx. Hence, by means of Equation 6.1 one obtainsV = jd + T- (6.2)where V is the constant shear force acting through the shear span. The first term atthe right-hand side of Equation 6.2 is usually termed the “beam action”, which reflectsthe change of the force in reinforcement, and the second term is called the “arch action”, which represents the inclination of the internal thrust force. The simultaneousoccurrence of both actions require the corresponding internal stress distribution and thecompatibility of the deformation within the beam.A half beam with idealized flexural cracks is shown in Figure 6.16. The shear resistance modes of the varying tension force in reinforcement with the constant lever armand the constant tension force in reinforcement with the varying lever arm are the twoChapter 6. Shear Failure of Beams Without Stirrups 113extreme cases, which could be termed the “pure beam action” and the “pure arch action” respectively. The internal stress distributions for the “pure beam action” are wellknown and first described by Mörsch [69]. See Figure 6.16. The shear stress is calculated by horizontal equilibrium of a piece of beam element and is found to be uniformlydistributed along the cross section. However, the shear stress for the “pure arch action”is concentrated as shown in Figure 6.12(b). The shear stresses combine with horizontalcompression stresses to produce a resultant which travels straight from the load point tothe support.The internal shear stress distribution, for combined “beam action” and “arch action,”is somewhat different from that of either one action. The same half beam, as shown inFigure 6.16, is again given in Figure 6.17, and a plot of the tension force in the longitudinalreinforcement is also given. It is assumed that the combined beam action and arch actionexists in the range of from the load point to section m — m with the resultant thrust forceshown in Figure 6.17(a). Correspondingly the variation of the force in the longitudinalreinforcement is shown in Figure 6.17(b). Then the shear stress distribution along sectionn. — n., coming from the contributions of both actions, can be determined. One part ofshear stresses is calculated by equilibrium consideration, in which the shear stress isrelated to the bond stress between the concrete and the reinforcement, as shown inFigure 6.16. Another part can be determined from the vertical component of inclinedthrust force. Summing up these two parts should be equal to external applied shear force.See Figure 6.17(c). Equation 6.2 is the mathematical expression of Figure 6.17.6.5 Shear Displacements along CracksAfter the shear stress distributions along the idealized vertical cracks have been examinedabove, the corresponding shear displacement along the cracks is investigated in thisChapter 6. Shear Failure of Beams Without Stirrups 114section. It is generally believed that there are two kinds of deformations in crackedbeams, which could introduce the relevant shear displacements along the cracks [70].One is the flexural rotation of the compression zone and another is the bending within aconcrete tooth caused by bond force LIT.For the idealized vertical flexural cracks in the shear span, the compatibility of deformations illustrates that only the bending of the concrete teeth causes the shear displacement along the cracks and thus develop uniform shear stress distributions along thecracks, which is consistent with the assumed beam action. See Figure 6.18(a). The rotation of the compression zone makes the vertical crack opening and does not introduce theshear displacement along the crack. This latter deformation for idealized vertical cracksis only consistent with the beam elements subjected to pure bending moment, where nobond force and no shear stresses will be developed.However, the existence of the shear force in the shear span makes the cracks inclinedrather than vertical so that the both deformations have contributions to the shear displacement. See Figure 6.18(b). The shear displacement will still be uniform along theinclined crack, if only the bending of the concrete teeth is considered. The shear displacement due to the rotation of the compression zone will not be uniform, with thelargest shear displacement occurring at the mouth of the crack. One of the importantconclusions about the shear displacement along the inclined cracks is that the flatter thecracks are, the more shear displacement will be introduced along the cracks for the samecompression zone deformation.6.6 Load Transfer MechanismBased on the previous detailed analysis, an insight into the load transfer mechanism canbe obtained by examining the shear failure process of a simply supported beam subjectedChapter 6. Shear Failure of Beams Without Stirrups 115to the point loading.When the beam is loaded to a certain stage, the obvious aggregate interlock at thecritical inclined crack is mobilized with the company of the occurrence of considerableshear displacement along the crack due to the increasing load and the reduced compression zone (i.e., the increasing bending of the tooth and the increasing rotation of thecompression zone). This can be related to the experimental observation that only thecritical crack keeps opening and the widths of other cracks remain almost unchangedunder increasing loading. In addition, the shear displacement along the crack is alwaysaccompanied by the crack opening for shear transfer by aggregate interlock action due tothe rough contacting surfaces. See Figure 6.19. At this stage, it can be considered thatboth beam action and arch action already exist for shear resistance, based on the previousanalysis. With the loading being further increased, in addition to the shear stresses, thecompression stresses along the crack will also be introduced as a result of the longitudinalreinforcement’s constraint to the crack opening. These shear stresses and compressionstresses acting on the crack are one of the direct reasons for the occurrence of the seconddiagonal crack, which is commonly observed in shear failures of longitudinal reinforcedconcrete beams. See Figure 6.20(a). As the loading is approaching this stage, just beforethe occurrence of the second diagonal crack, the full arch action can be approximatelyassumed in the range from the load point to the critical crack. See Figure 6.20. Thisassumption is justified by measurements of reinforcement strain and concrete strain inthe shear span [59, 71, 72], an example of which is also shown in Figure 6.20(d).If full arch action is assumed in the range from the load point to the critical crack,and full beam action with a reduced lever arm is assumed in the range from the criticalcrack to the support, the variation of tension force in reinforcement is shown in Figure6.20(b). Because the magnitude of bond stress is proportional to the gradient of thetension force in reinforcement, larger bond stresses will be developed near the support.Chapter 6. Shear Failure of Beams Without Stirrups 116If the variation of the tension force is further assumed as a solid line shown in Figure6.20(c), the bond stress will be amplified locally in the zone near the critical crack. Thisassumption can be fully confirmed by Mains’ measurements [59]. See Figure 6.21. It canbe concluded that the second diagonal crack or even splitting along reinforcement partlyresults from this high bond stress.A possible load transfer model is shown in Figure 6.22. The full arch action is assumedin the range from the load point to the cross section n — n where the critical crack occurs.Then the resultant compression force on the cross section n— n, F, should be as shownin the figure. Because the compression force flow N originating from the support is notalong the same line as F, the tension force T is required for equilibrium. The F. hastwo components, one meeting with the force N and another going downward to intersectthe longitudinal reinforcement.6.7 Interpretation of Some Beam Test ResultsWith the help of the model presented above, it is possible to give an alternative interpretation of the experimental information presented by Kotsovos in Ref.[57, 73] and otherresearchers in Ref.[56, 58, 74, 75].Kotsovos [73] tested a series of simply supported beams with various arrangements ofshear reinforcement, and subjected to two-point loading with various shear span to depthratios (aid). See Figure 6.23. The main test results are given in Figure 6.24 which showsthe load-deflection curves of the beams tested. Series C and D beams were found tohave a load-carrying capacity significantly higher than that of series A beams which hadno shear reinforcement throughout their span. Series D beams, in all cases, exhibited aductile behaviour, which is indicative of a flexural mode of failure, and their load-carryingcapacity was higher than that of series A beams by an amount varying from 40 to 100%Chapter 6. Shear Failure of Beams Without Stirrups 117depending on a/d. In addition, near the peak load the inclined crack of series D beamshad a width in excess of 2 mm. Kotsovos considers that the test results are conflict withthe concept of shear capacity of critical sections, the view that aggregate interlock makesa significant contribution to shear resistance, and even the truss analogy concept.Chana [58] tested beams with traditional internal stirrups, but locally arranged asshown in Figure 6.25. Kim et al. [74] tested beams with external stirrups. See Figure6.26. The prestressed external stirrups were provided at the outer third sections of beams.Muttoni [56] and Kuttab [75] tested beams with the arrangements of reinforcement shownin Figure 6.27 and 6.28 respectively. All beams failed in ductile modes and reached failureloads which are almost the full flexure capacities of the beams. The shear capacity ofthese beams was about double the capacity of similar beams without the nonconventionaltransverse reinforcement.The test results from Kotsovos or the other researchers are very interesting, but notthat surprising. The common point of these tests was that the beams were providedwith a mechanism that gave effective constraint to the development and propagation offlat (or horizontal) cracks in the critical zones, making the critical diagonal crack morestable. In reality, the arrangements of reinforcement used by these researchers can beconsidered as various modifications of the conventional form in which the stirrups areplaced throughout the whole beam and anchored in the compression zone.6.8 ConclusionsBased on the study in this chapter, some conclusions are arrived at:1. A distinct characteristic of slender beams without transverse reinforcement is thelack of ability to control horizontal crack propagation and the brittleness of shearfailure results from the occurrence of either flat inclined cracks or horizontal cracks.Chapter 6. Shear Failure of Beams Without Stirrups 1182. The shear stresses in both the perfect bond case and the no bond case have similardistributions along the cross sections in the B region of the beam. This suggeststhat simply focusing on on the shear stress distributions is not adequate for theinvestigation of shear failure mechanism. How a load transfers is much more important than shear stress distributions.3. As the entire cross section of a beam is subjected to compression stress and theshear stress in the no bond case, no inclined cracks will be developed in the shearspan. This was illustrated in the experimental work by Kani [13].4. Bond has an influence upon the transverse splitting of compression struts of verydeep beams (a/d < 1); however this influence can be ignored because the deteriorated bond in deep beams with inclined cracks will reduce this influence significantly.5. Generally, the shear stress distributions over the cross section of cracked beamsare not uniform. The localized shear stress distributions can be considered as thedirect result of arch action or vice versa.6. The ideal direct load transfer mechanism or “pure arch action” can only be realizedin the no bond condition, regardless of the shear span ratio a/d. The form of thedirect compression strut between the point load and the support depends on theresultant forces acting on the both ends of the strut. These resultant forces ontwo ends of the strut should act along the same line and in opposite directions.Consequently, the inclination of the strut originating from the support cannot bearbitrarily chosen when the truss models are developed.7. In addition to dowel action, which must be introduced due to the shear displacementalong the crack, the locally amplified bond stress at the zone near the critical crackChapter 6. Shear Failure of Beams Without Stirrups 119contributes to the occurrence of the second diagonal crack and the splitting parallelto the longitudinal reinforcement. This bond problem has also been recognized byother researchers [76] but from a different point of view.Chapter 6. Shear Failure of Beams Without Stirrupsa=h—, — ..,,,Z’’Cd) a/h 1/g h/2 ,I iI /‘ /(b) a/h = 0.25a=2h.-,,77,••,I FF‘ /.. —‘- —ICe) a/h = 2Figure 6.1: Distribution of transverse compressive stress for various shear span ratios.from Mau and Hsu, Ref.[52].120h/4—-S.‘I’/ I’hCT>I hiI iI II‘ I’‘1’—5---,— S.(a) a/h = 0— Distribution: /7 ot Transverse....7—-—j--------4.F CompressionI,1 h/2._._— Isostatic/ CompressiveI /i - CurveF‘ —-rCc) a/h = 0.5Chapter 6. Shear Failure of Beams Without Stirrups 121LrcF”jhf ‘I-*k--/,,I/\, ,Ij, z-v.Figure 6.2: Load near the support: transition from deep beam to slender beam, fromSchlaich et al., right side simple models; left side refined models, Ref.[3]..‘IS,,,Chapter 6. Shear Failure of Beams Without Stirrups 122• 6 x 6 x 1 in, (152 x 152 25 mm) plateo 25 -69 6 *9 x 2 in, (152 x 229 x 51 mm) plateo 6 3*0.38 in. (152* 76* 95mm)plate•67 V Via0.20}m)V 0.15 l 3940 psi (272 MPa)bdf max. aag.- 3/4 in. (19mm)d21.2in. (538mm)b6.1in. (155mm)As 3.53 in2 (2277 mm2)0.10 -• 72 f.=53.9 ksi (372 MPa)08165.76U.V.J 071 063 066strut and tie model sectionaI model0 I0 1 2 3 4 5 6 7aidFigure 6.3: Predictions of shear strength versus a/d ratio for tests reported by Kani[13j,from Collins and Mitchell, Ref.[39}.Chapter 6. Shear Failure of Beams Without Stirrups 123VvJ I(a) Geometry and Loadingr iiV1d,V4j f J2Ib) Truss Mode!Figure 6.4: Truss model developed by Adebar, Ref.[54}.F—/ \Nc —/_\_. —I’- 1Lf vFigure 6.5: Truss model developed by Reineck, Ref.{55}.Chapter 6. Shear Failure of Beams Without Stirrups 124hT* *O 450 450Figure 6.6: Truss models developed by Al-Nahiawi and Wight, Ref.{53j.Figure 6.7: Structural model developed by Muttoni and Schwartz, Ref.[56J.Chapter 6. Shear Failure of Beams Without Stirrups 125‘4I IFigure 6.8: Structural model developed by Kotsovos, Ref.[57J.24j Z/f/)fFigure 6.9: Crack pattern of a beam tested by Kani, Ref.[13}.Chapter 6. Shear Failure of Beams Without Stirrups 1261HHAFigure 6.10: Geometric relationship of a crack at reinforcement level.Chapter 6. Shear Failure of Beams Without Stirrups 127Figure 6.11: A simply supported and central loaded beam.cfIfjIL1LIwIUI[EW7 -uvi11Jijob-’.oCD CD C’, C, I-.c-l CD CD Iz 0 Cl) cD (I) Cl) 0 Cl) -‘ 0 C 0r----.----.-..III±:cD -S 0 ICD I I-’ L’311]y-rrrIrJnrrr1a-1ZQfffl1IIP.111Chapter 6. Shear Failure of Beams Without Stirrups 129I— — — — ———— -, ‘-— — .— .- — — — — — — —— /— —C. — —— C C c_ — —/ / ,/ / / 1 ,) / / K .‘ / /— ——,—— — — — — —— —- ///‘ / /7/ 7//il // / / / 1’! / / / i’, / 7 1 /I 7 / I I I I I I I I I I I I I--- ---------- ------ -1(a)------— — — — -, — -- —, — -— — — — — — — — — — — ——— .- .- ... ._ _ —, / — .-. — — — — — — ,, — — / /-— t.t. c .. —-. / / — — —. — — — — — — — ,, // ; /- — —— C C — — — —— .- .- — — — —— — ;./.H/— 2 — — — •_ — — _• — / /(b)Figure 6.13: Internal force flows in uncracked beams: (a) perfect bond case and (b) nobond case.Chapter 6. Shear Failure of Beams Without Stirrups 130--i———----—--q- - - - —— - — ——--(a)---- - ---- --—- - -LU(c)-1-- -- - —-- - - --- - — — —————- -(b)————--—1————- - -——---———- - - -——-LU(d)Figure 6.14: Modelling of bond effect upon transverse splitting of compression struts indeep beam.Chapter 6. Shear Failure of Beams Without Stirrups 131/1Tff/9.__/\/\____(a)(b)(c)(d)Figure 6.15: Bond influence upon transverse tensile stresses of compression struts in deepbeam.Chapter 6. Shear Failure of Beams Without Stirrups2T2b132—I—-;5z:---1-C+ACjd____— a___T T+ ATT ’1___Ax1T+ATshearstressesFigure 6.16: Internal stress distributions of cracked beams due to pure beam action, fromRef. [39].b idlongitudinal equilibrium longitudinalstressesChapter 6. Shear Failure of Beams Without Stirrups 133m m(a)_______pure arch action(b)+(c) beam arch combinedFigure 6.17: Shear stress distributions of cracked beams due to combined beam actionand arch action: (a) modelling of combined beam and arch actions; (b) tension force inlongitudinal reinforcement; (c) shear stress distribution at section n— n.Chapter 6. Shear Failure of Beams Without Stirrups(a) (b)Figure 6.18: Shear displacement along cracks in shear span: (a) vertical cracks and (b)inclined cracks.1347 7 / / /77Chapter 6. Shear Failure of Beams Without Stirrups 135V+dVVtOVV+dVFigure 6.19: Shear transfer at cracks by aggregate interlock.Chapter 6. Shear Failure of Beams Without Stirrupsnsecond C(a) lv(b)(C)njTCrackedConcreteSteel136Figure 6.20: A beam with inclined cracks in shear span: (a) strut force due to archaction; (b) one assumed tension force in reinforcement; (c) another assumed tensionforce in reinforcement; (d) measured concrete strain and tension force in reinforcementfrom Ref.[71].Cracked‘ 4.rr. NJChapter 6. Shear Failure of Beams Without Stirrups 137/Measured \steel tensionteéItensionCalculatedCritical flexural Icrack section 4(a)•lIRLI&ILP\JeL/I.I I(b)Figure 6.21: Measured force in bar, bond stress and crack locations, from Ref.[59].Chapter 6. Shear Failure of Beams Without Stirrups 138n-4(a)(b)Figure 6.22: A load transfer mechanism just before the occurrence of splitting alonglongitudinal reinforcement: (a) truss model and (b) tension force in reinforcement.JCDCDOc..cJ-CDCDTI_CDII.iii•ItIIIIIIIIi••IIII1WIIoIIII,-IybwIIIIICII‘I___I__I—.CDIIC’)III..IIIiII-.ICDiiiI_cJIIIIIIIIIWCDi‘I0Cl)0’(ID0 o C Cl)______p‘.‘Tn•1____I.I!0____—____I-’-90—iiooIH100WIIii0’LIaVa,CU’0 0W 0a, 1••UI 01Wo0DC)W0Chapter 6. Shear Failure of Beams Without Stirrups 140a) a)12 2I-\ZEx —ci) —C- 0 060/c5:—0S000? 002-0 0•I I I Ic 2 1. 6 8 10 l25 10 15 20 25 0.2 deflection- mmdeflection - mm(a) (b)ci) 12I-2 B-Dx0•0Ca.2Va)0.20-0V 00 2 L 5 8 10 12.2central deflection - mm(c)Figure 6.24: Load-deflection curves of beams shown in Figure 6.5: (a) a/d = 1.5; (b)a/d = 3.3 and (c) a/d = 4.4, from Kotsovos, Ref.[73].Chapter 6. Shear Failure of Beams Without Stirrups 1413530 Linknos32125 lWLnk 2/ Lnk IFailure of span without /20 links at 9SkN -15 /k310. 1/Failure load=158kN0 20 40 60 80 100 120 140 160TOTAL SHEAR FORCE - kNFigure 6.25: Beam with internal stirrups tested by Chana, from Ref.{58].+L4.J— L125-—r- f S I I PSIO7 I .iLIPS 24.1 *IPSI I I4.NUTSECT ION A—AFigure 6.26: Beam with external stirrups tested by Kim et al., from Ref.[74j.Chapter 6. Shear Failure of Beams Without Stirrups 142S—‘ ——— ,/. /(b)Figure 6.27: Beams tested by Muttoni et aL, from Ref.[56].Path of Compressive Force/ n-i 11Path of Compressive Forceaid = 2aid = 3.6‘[I_ujH1_rH I.1I I I I I I IFigure 6.28: Beams by Kuttab et al., from Ref.[75].Chapter 7Bond Splitting Failure7.1 IntroductionThe main objective of this chapter is to deal with the question of what causes the horizontal splitting along longitudinal reinforcement in a beam without transverse reinforcement.It is generally believed that splitting of concrete along reinforcing bars in a beam occursprimarily due to the combined effect of wedging action of bar deformations (bond) anddowel action of reinforcement. Ferguson et al. [61, 62j focused on bond splitting. Theynoticed that the existence of a diagonal crack resulted in lower bond strengths. Based onan extensive experimental study, Cergely [77] concluded that the dowel force is the mostimportant factor producing splitting in beams without stirrups, and that the dowel effectovershadows the pure bond effect in most situations, especially in beams with small barspacing. Jimenez et al. [78] found that bond strength and dowel capacity are independentof each other. Kemp et al. [66] observed from their experimental work that there is aweak interaction between dowel force and bond resistance until approximately 80 percentof the pure dowel capacity is reached at which time bond capacity decreases very rapidly.They suggested a 20 percent reduction in design ultimate bond to include the effect ofdowel action.The conclusions arrived at in Chapter 6 about the load transfer mechanism prior to theoccurrence of splitting, are consistent with above mentioned experimental observations.Dowel action and severity of bond stresses (i.e., amplification and localization) will both143Chapter 7. Bond Splitting Failure 144be introduced due to the occurrence of the critical diagonal crack. Consequently, dowelaction and bond action both will make contributions to splitting along reinforcement. Todate no satisfactory quantitative analysis results are available to describe the interactionof these two actions (bond and dowel). Traditionally bond and shear have been dealtwith separately so that previous investigations on bond splitting were mostly concernedwith ultimate bond splitting failure. However, the conclusions arrived at in Chapter6 make it evident that the occurrence of horizontal splitting along reinforcement in acritical zone indicates shear failure of beams without transverse reinforcement. Then itseems that the initiation of bond splitting could be the more appropriate criterion thanthe ultimate bond splitting strength for the study of shear capacity of beams withouttransverse reinforcement.In this chapter, the attention is focused on bond splitting with an objective to developa tentative design criterion for the initiation of longitudinal bond splitting. The chapteris divided into three parts. In the first part, previous studies are briefly reviewed. Thisincludes general bond actions, experimental studies, as well as bond splitting strength(ultimate and initiation). A proposed design equation for bond splitting initiation ispresented in the second part. Finally, some comments are given about Ferguson’s [61, 62]experimental work and a number of conclusions are drawn.7.2 General Bond ActionStudies of bonding forces for plain reinforcing bars and deformed bars by Lutz andGergely [79] showed that bond for plain bars is made up of three components: (1) chemicaladhesion, (2) friction and (3) mechanical interaction between concrete and steel. Whenplain bars without surface deformations are used, bond depends mainly upon chemicaladhesion, and after slip, upon friction. There is also some negligible mechanical actionChapter 7. Bond Splitting Failure 145due to the roughness of the bar surface. With use of deformed bars, the main reliance ischanged to bearing of lugs on concrete and to shear strength of concrete sections betweenlugs.Because slip of deformed bars can occur in two ways namely: (1) the lugs can splitthe concrete by wedging action and (2) the hugs can crush the concrete, two types ofbond failures can occur. If the surrounding concrete resistance is moderate, as it isfor ordinary concrete cover, the lugs of large steel bars can split the concrete withoutcrushing it. With small bars or with large cover over the bars, the lugs will shear theconcrete and pull out without splitting the concrete.Bond between concrete and a deformed reinforcing bar that is subjected to a pull-outforce as well as is with a moderate concrete cover, can be characterized by the relationshipbetween averaged bond stress along the embedment length and slip at the loaded end,with four different stages as shown in Figure 7.1 [80]. In Stage 1. (small values of thebond stress), bond is assured by chemical adhesion, and no bar slip occurs. In stage 2(larger bond stress values), the chemical adhesion breaks down and bonding is assuredby bearing action or wedging action of the bar lugs. In Stage 3 (still larger values ofbond stress), the first longitudinal cracks form as a result of the increasing wedge actionof lugs and more bond stress can be sustained by the interlock between concrete andreinforcement lugs. Once the longitudinal cracks break out through the whole cover,failure occurs abruptly in this stage if no transverse reinforcement is provided. If enoughtransverse reinforcement is provided, the confinement exerted by the reinforcement wouldallow the bond stress to reach a larger value in spite of concrete splitting. See Stage 4 inFigure 7.1.For bond related shear failures of structural concrete members without shear reinforcement, the confinement provided by transverse reinforcement is not available. Henceonly Stage 3 described above is relevant to the problem to be dealt with. The beginningChapter 7. Bond Splitting Failure 146and end of Stage 3 indicate splitting initiation and ultimate splitting failure. Both splitting initiation and splitting failure are investigated in the following. But it is believedthat the bond related shear failure is more relevant to splitting initiation than to ultimatesplitting failure, as dowel action, which can aggravate the bond splitting problem, willalways be introduced due to the occurrence of diagonal cracks in beams.7.3 Previous Experimental StudiesIn the history of bond research, the problem most considered has been bar developmentlength 1d, which is the embedment length necessary to assure that a bar can be stressedto its yield strength without failing in bond.In order to determine bond stresses or development length, a variety of test methodshave been used. Typically there are ordinary pull-out tests (also denoted as concentricpull-out test), eccentric pull-out tests, full-beam tests and semi-beam tests (often calledstub-beam or cantilever tests).In Figure 7.2 ordinary pull-out test specimens and eccentric pull-out test specimenscommonly used by researchers are shown. It was believed by Ferguson et al. [60] that theordinary pull-out test is not entirely realistic as the measure of bond strength in beams,because it carries no shear on the splitting plane. In a beam, a short 1x length of beam, asshown in Figure 6.16, transfers the change in bar tension by bond stress into a horizontalshearing stress, a considerable part of which acts on the section through the level of bars.Also the loaded end of a concentric pull-out specimen is in compression and is restraineddue to friction forces. Hence the concentric pull-out test gives higher bond strengths thanthose expected in beams, especially where splitting is an important factor. The eccentricpull-out test [Figure 7.2(b) and (c)] greatly improves the disadvantages mentioned forthe ordinary pull-out test and keeps the advantage of simplicity.Chapter 7. Bond Splitting Failure 147Full-beam tests are considered most reliable because the influences of both transverseshear stresses and flexural tension cracks are included. One kind of full-beam test used atthe University of Texas [61, 62] is shown in Figure 7.3. In order to eliminate the influenceof support reaction upon bond strength, the development length of the bar is placed ina negative moment region, where the L” shown in the figure is the development lengthinvestigated. However the full-beam tests are very expensive test procedures and difficultto generalize for different variables. Semi-beam specimens (or stub cantilever specimens),have been utilized at Cornell University [77] and West Virginia University [66], as wellas in Japan [81]. See Figure 7.4. This type of specimen still provides a realistic straingradient through the depth of the specimen as in beams, but the cost is much lower thanthat of full-beam specimen tests. In addition, it is quite versatile since ratios of bond,shear and flexure can be easily varied from one specimen to another.In summary, in order to get correct information about bond splitting in beams withouttransverse reinforcement, full-beam test procedures are most preferred. Semi-beam testsand eccentric pull-out tests are satisfactory substitutes for full-beam tests. The testresults from semi-beam specimens and eccentric pull-out specimens can be consideredapplicable to real beams.7.4 Previous Studies of Ultimate Splitting FailureThe generally accepted bond mechanism, when wedging action has been mobilized, isthat reinforcing bar force is transferred to surrounding concrete by inclined compressiveforces radiating out from lugs on reinforcing bars and making an angle with the baraxis. The inclined compressive forces can be decomposed into radial and tangentialcomponents. The radial components (bursting forces) are balanced by circumferentialtensile (ring) stress in the surrounding concrete. For the case of concern to this study,Chapter 7. Bond Splitting Failure 148longitudinal cracks (splitting) appear when the tensile rings, which are usually weakestin the thinnest concrete cover protecting the reinforcement, are stressed to the tensilestrength. See Figure 7.5. Several factors can affect bond splitting strength. These areconcrete cover or clear spacing of bars, concrete tensile strength (which is related to thecompressive strength f), embedment length and bar diameter etc.Orangun et al. [82] proposed an approach for determining bond strength or development length, that included all the variables mentioned above. The approach is based on aphysical bond model. The radial forces, generated between the lugs and the surroundingconcrete, can be regarded as water pressure acting against a thick—walled cylinder withan inner diameter equal to the bar diameter and a thickness c that is the smaller of theclear bottom or side cover cb or 1/2 the clear spacing c8 between adjacent bars. See Figure7.6. The capacity of the cylinder depends on the tensile strength of the concrete. Withcb greater than c3/2, a horizontal split develops at the level of the bars and is termed a“side split failure.” With c3/2 greater than cb, a “face-and-side split failure” forms withlongitudinal cracking through the cover followed by splitting through the plane of thebars. When c/2 is much greater than cb, a “V-notch failure” forms with longitudinalsplitting followed by inclined cracks that separate a V-shaped segment of cover from themember. From the results of 62 beam tests, an equation of the bond stress was developedby using a nonlinear regression analysis. That isu 3.23c 53db= 1.22 + db +(7.1)in which u is the ultimate bond strength; c is the smaller of c8 or c; c is the smaller ofone half of clear spacing or side cover; cb is the concrete cover; d6 is the bar diameter;and id is the development length or splice length (all units are in psi and inches).This equation was further modified by rounding the coefficients to obtain a somewhatChapter 7. Bond Splitting Failure 149more conservative value for u, denoted as uj(7.2)d, 1dOrangun et al. [82] compared the bond stresses calculated by Eq. 7.2 to test resultsobtained from a total of nine studies (over 500 tests) of splice and development strengthfor bars not confined by transverse reinforcement. The predicted strengths gave a closematch with the test results.According to the recent work by Darwin et al. [83], the dimensionless equations Eqs.7.1 and 7.2 for bond strength can better be expressed as the tension force in reinforcementin terms of the same variables. They believed that bond force provides a better measureof member response than bond stress, since bond strength can be considered as being astructural property rather than a material property. The recommended equation is= 3ld(c + 0.4db) + 200Ab (7.3)where Ab is bar area and f is steel stress at ultimate bond strength.7.5 Previous Studies of Splitting InitiationIt should be noted that the previous study results presented above are interesting butonly related to ultimate bond splitting strength. Equations 7.1 and 7.2 are valuablefor bond development design if shear and bond problems are dealt with separately. Asa matter of fact, shear and bond are always co-existing and interactive, therefore theinitiation of bond splitting is most concerned in this study. Unfortunately little workhas been carried out on bond splitting initiation in comparison with the work done onultimate bond splitting failure, especially experimental investigations.Chapter 7. Bond Splitting Failure 150Using the thick-walled pipe analogy, Tepfers [84] performed an analysis of bond cracking for the concrete in uncracked elastic state, plastic state and partly cracked elasticstate. See Figure 7.7. By setting the known maximum tensile hoop stress equal to material tensile strength and approximating the angle of inclined compression forces as 45degrees, expressions to predict splitting bond stress were developed for above mentioneddifferent states. These expressions are:i 12 u12-r) ) 7ft— ( + 1)2 + ()2(7.5)db= (0.3 + 0.6-) (7.6)where Equations 7.4, 7.5 and 7.6 are for the uncracked elastic state, uncracked plasticstate and partly cracked elastic state respectively, and tt is the bond stress indicatingcrack initiation, c is the thickness of cover, db is the diameter of a deformed reinforcingbar and f is concrete tensile strength.Figure 7.8 compares Tepfers’ [84] predictions with his eccentric pull-out test resultsand other test results [85]. It can be seen that the plastic prediction gives the upperbound solution, while the partly cracked elastic prediction gives a lower bound to theresults.Kemp and Wilhelm [66] conducted a linear regression analysis of their experimentaldata from semi-beam tests (see Figure 7.4) and suggested the following equation:= (2.64 + 2.37-) (7.7)Chapter 7. Bond Splitting Failure 151where c is concrete cover and db is diameter of test bars, all units in psi and inches. Acomparison of Equation 7.7 and test data on cracking loads is shown in Figure 7.9.7.6 Proposed Design Equation for Bond Splitting InitiationThere is a significant difference between the equations for bond cracking and ultimatebond splitting strength. The embedment length 1d of reinforcing bars is not included inthe Equations 7.4—7.7 for initial bond cracking in comparison with the Equations 7.1 and7.2 for ultimate bond strength. The uncertainties about the validity of Equations 7.4—7.7to predict bond splitting initiation strength in real beams exist due to lack of enoughtest data. The ld/d ratios of the specimens used by Kemp et al. and Tepfers were 11.34and 3.13 respectively. Because the bond stresses at which first concrete crack was visiblewere averaged along the embedment lengths in the studies of Kemp et al. and Tepfers,none of the equations developed by them can be readily applied to different beams withdifferent beam sizes.Teng and Ye [86] have carried out a series of concentric pull-out tests to study bondand slip relationships for deformed reinforcing bars. Both ultimate bond splitting loadsand bond splitting initiation loads were recorded during testing. Based on the statisticalregression analysis of test data, the following design equations were developed:= (1.106 + 1.3)- (7.8)= (1.162 + 1.8O2)- (7.9)where u, is bond cracking stresses and u is ultimate bond splitting strengths, all unitsin kg/cm2 and cm. These two Equations 7.8 and 7.9 are shown graphically in Figure7.10. It can be seen that the ultimate bond strength and cracking bond strength haveChapter 7. Bond Splitting Failure 152a similar trend. They both are a function of concrete cover, bar size, concrete strengthas well as embedment length. Although the two equations cannot be directly applied tobeams, as discussed in Section 7.2, the evidence revealed by the two equations can beused as a basic assumption to develop an empirical equation for the prediction of bondsplitting initiation strength in beams.The following equation is proposed for the bond splitting initiation strength——=1.2+3--+15--. (7.10)1dwhere all units of stress are in psi. This equation is actually a simple modification ofEquation 7.2, whose accuracy of predictions for ultimate bond splitting strength wasrecently confirmed by Darwin et al. [83]. Equations 7.2 and 7.10, as well as the experimental results measured by Kemp and Wilhelm [66], are shown in Figure 7.11. Equation7.10 results from Equation 7.2 by shifting a parallel displacement downwards for thetested idid ratio. The comparison of Equation 7.10 and experimental results measuredby Tepfers [84] is shown in Figure 7.12. The close agreement of the predictions with testresults is illustrated. Figure 7.13 demonstrates Equations 7.2 and 7.10 for the c/d& of0.5.Equations 7.10 can also be modified to express bar force at cracking normalized withrespect to as recommended by Darwin [83]. That equation is=37r1d(c + 0.4db) + 6OAb (7.11)where Ab is bar area and f3 is steel stress at cracking.Chapter 7. Bond Splitting Failure 1537.7 Some Comments and ConclusionsFrom Figure 7.13, it can be seen that the bond splitting initiation strength is much lowerthan ultimate bond splitting strength. The combination of this lower bond crackingstrength and dowel action can lead to lower shear strengths as well as lower ultimatebond splitting strengths.The University of Texas beam [61] is shown in Figure 7.3 with a potential diagonalcrack within development length L”. In experimental studies carried out by Ferguson etal. [61, 62], diagonal cracks always developed in narrow beams. These diagonal crackshastened the bond splitting process, particularly with longer L” values, and led to lowerbond strengths (splitting along whole development length). Sometimes, diagonal cracksdominated failures in shear with the ends of the bars still fully bonded near the inflectionpoint. In both cases, the occurrence of diagonal cracks led to lower shear capacities.Based on the study in Chapter 6, after the occurrence of a diagonal crack the variationof tension force in a reinforcing bar may be shown by the solid line in Figure 6.20(b). Asdiscussed before, dowel action will be introduced and bond stress will be locally amplifieddue to the development of a diagonal crack. Based on the study in this chapter, it canbe seen that the bond problem is further aggravated by the reduced development length<L” by referring to Equation 7.3 and lower allowable tension force in reinforcement bythe comparison of Equation 7.3 and Equation 7.11. See Figure 7.3. Then lower shearcapacity or lower bond splitting strength can be expected. When wider beams are usedfor testing, the unfavourable factors for bond splitting strength will be eliminated withthe disappearance of a critical diagonal crack. The higher shear capacities and the higherbond splitting strengths can be expected.Another issue is about whether dowel force or bond force dominates splitting alongreinforcing bars in beams without transverse reinforcement. If a single bar is put in a wideChapter 7. Bond Splitting Failure 154beam (small dowel force and large dowel capacity), bond splitting could be prominent asthe low steel percentage and the large steel stresses cause splitting on the bottom face,Ref. [62]. If a couple of bars are put in the same dimension beam with small spacing in onelayer (large dowel force and little dowel capacity), dowel action could dominate splittingas the splitting occurs on the sides under comparable load, Ref.[77]. Both situations,experimentally indicated, lead to lower shear capacities.The shear resistance mechanism and the bond strengths between concrete and longitudinal reinforcing bars are very complicated in a reinforced beam without stirrups. Theinteraction of shear and bond seems much more complicated. Another example showingthis complexity is Hall’s discussion [67] of the experimental work carried out by Baantand Kazemi [87] for the investigation of size effect on diagonal shear failure of beamswithout stirrups. Hall considered that the observed results are due to the location anddistribution of the reinforcement, which implies bond failure.In summary, the study in this chapter has led to a tentative equation for bond splittinginitiation and an improved understanding of the interaction of shear and bond in a beamwithout transverse reinforcement. Further work on this topic is necessary, however, theproblem is very complicated and many factors are involved. It is not possible to completethis topic as part of the present study. Suggestions for further study are given in thenext chapter.Chapter 7. Bond Splitting Failure 155Cl)C’)ci)Cl)-oCinadequate confinement0coBar SlipFigure 7.1: Bond stress—slip relationship, from Gambarova et aL, Ref.[80].Chapter 7. Bond Splitting Failure 156(a)(b)Reacrion(c)ReactionEach SideFigure 7.2: Pullout tests: (a) concentric pullout-test specimen; (b) commonly used eccentric pullout-test specimen and (c) eccentric pullout-test specimen used by Ferguson,Ref. [60].Chapter 7. Bond Splitting Failure 157(a)iFI I I I I I I I I I I I Potential inclined crack 1(b)(d)Figure 7.3: The University of Texas beam tests: (a) possible tension force distributionalong the top test bar after the occurrence of inclined crack; (b) side view of specimen;(c) plain view of top reinforcement; and (d) moment diagram, adapted from Fergusonand Thompson, Ref.[61].L” P.’.(c)0CD (I)F CJ( ccChapter 7. Bond Splitting Failure 159ZzrzEFigure 7.5: Mechanism representation for bond, from Tepfers, Ref.[84].Chapter 7. Bond Splitting Failure 160Failure planeC6>C/2. C=C,/2I?Side split failure Just before failureC >z-- -I At failure II ILzV-Notch failure Face-and-side split failureC,/2>>Cb C,/2>CbFigure 7.6: Bond Splitting Failure Patterns, from Orangun at aL, Ref.[82].Chapter 7. Bond Splitting Failure 161a1 / Or+dr(a) (b)___edb/2(c) (d)Figure 7.7: Analysis of Bond Splitting Stresses: (a) bursting and bond stresses; (b)uncracked elastic state; (c) uncracked plastic state and (d) partly cracked elastic state.Adapted from Tepfers, Ref.[84j.Chapter 7. Bond Splitting Failure 1626 .-—--________________________05—000/•/4: . ..Icbc 8° 0 •0c 00£ 0I Ct o • e0.%.• •2 -x °x1 - elastic stage (equabon 5)X • ordinary concrete0 Iightweght concreteI I I0 1 2 3 4 5 6c/dFigure 7.8: Comparison of test and prediction for bond splitting initiation, from Tepfers,Ref. [84].Chapter 7. Bond Splitting Failure 16315uc00 05 4 5Figure 7.9: Comparison of Equation 7.7 with test data on cracking loads carried out byKemp and Wilhelm Ref. [66].Chapter 7. Bond Splitting Failure 164Ufl 6 (Eq. 7.9)(Eq. 7.8)42=20db2- c—=1.5db• dbdb0 I0 5 10 15 20IddbFigure 7.10: The relationship of bond splitting strength and ld/d proposed by Teng andYe Ref.[86] for concentric pullout tests.Ultimate Strength— — —— Cracking StrengthChapter 7. Bond Splitting Failure 165_!_ . Crackingo Ultimate (1510- Equation 7.2:osFE7b0TFigure 7.11: Comparison of Equation 7.2 and 7.10 with test results from Kemp andWilhelm Ref.[66].Chapter 7. Bond Splitting Failure 166uc7r • Cracking15...Equation 7.10(---=3.13)0 I I0 0.5 1 1.5 2CdbFigure 7.12: Comparison of Equation 7.10 with test results from Tepfers Ref.[84j.Chapter 7. Bond Splitting FailureUfl125=0.516720 -15 -Ultimate StrengthCracking StrengthEquation 7.210 -5-dbEquation 7.100 I I I I0 5 10 15 20ddbFigure 7.13: Relationship of bond strength versus ld/db, presented by Eqs. 7.2 and 7.10.Chapter 8Brief Summary and Further ResearchThis study, on the shear design of structural concrete members without transverse reinforcement, includes three main topics: transverse splitting of compression struts, development of a rational design procedure for deep pile caps, as well as a more general studyof the load transfer mechanism in concrete beams without stirrups.For deep pile caps, the compression strut that transmits column load to a pile isusually unreinforced and confined by surrounding plain concrete. The compression in astrut will spread out thereby introducing transverse tension near mid-height of the strutdue to strain compatibility. As there is no reinforcement provided to resist this transversetension, the concrete tensile strength must be mobilized. A refined truss model includesa concrete tension tie to model the transverse tension. It is believed that the brittle shearfailure of deep pile caps is initiated by internal cracking due to this transverse tension.That is, the shear failure of a deep pile cap results from transverse splitting rather thancrushing of compression struts.In this study, the compression struts have been idealized as concrete cylinders ofvarious diameter D, and height H, subjected to concentric axial compression over aconstant size circular bearing area of diameter d. Linear elastic finite elements were usedto determine the triaxial stresses at first cracking within cylinders (Chapter 2). Thenumerical study has indicated that the bearing stress at first cracking within cylindersdepends on the amount of confinement, the aspect ratio (height/width), as well as theratio of concrete compressive strength to concrete tensile strength. In order to confirm the168Chapter 8. Brief Summary and Further Researth 169transverse splitting phenomenon within compression struts a series of experiments wereconducted on large size concrete cylinders (Chapter 3). A good correlation was foundbetween the analytical prediction and the experimental results regarding the influence ofD/d and H/d on the bearing stress to cause first cracking.Based on the results of the analytical and experimental studies, a bearing stress limitwas proposed in terms of the amount of confinement and the aspect ratio (height/width)of the compression strut, as well as concrete strength (Chapter 4). The proposed bearingstress limit is given for the maximum nodal zone bearing stress to prevent diagonal tension(shear) failures in deep pile caps with unreinforced compression struts. In contrast,the ACT Building Code bearing stress limit is intended to prevent crushing of concretein nodal zones and does not preclude a shear failure due to transverse splitting of acompression strut.By incorporating the proposed bearing stress limit into a strut-and-tie model thatemphasizes internal force flow rather than the “shear stress” on any prescribed section,two rational design methods for deep pile caps were proposed (Chapter 5). The twomethods are similar except for the details of how the bearing stress limit is applied.The first design method is a direct extension of the two dimensional strut-and-tie modelfor deep beams (CSA approach). The procedure involves defining an equilibrium forcesystem in a deep pile cap. The forces in the compression struts are calculated from theproposed bearing stress limit. The horizontal components of the compression strut forcesmust be equilibrated by tension forces in the provided reinforcement. The sum of thevertical components of the compression strut forces gives the designed load capacity ofthe deep pile cap.The second design method is presented in a more traditional way, in which “flexuraldesign” and “shear design” are separated. A truss model is used for “flexural design”(i.e., to calculate required longitudinal reinforcement). “Shear design” is accomplishedChapter 8. Brief Summary and Further Research 170by limiting the maximum bearing stress (between pile and cap or column and cap) belowthe proposed bearing stress limit. Although a similar force flow is implied in the secondmethod as in the first, the details of the strut geometry are not needed in the moresimplified second method.Comparison of predictions from the proposed design method with predictions fromACT Code procedure and CRST Handbook procedure for 48 previously tested pile capspecimens demonstrated that the proposed method is more rational and more accuratethan what is presently used to design deep pile caps (Chapter 5). Tn deep pile capsthe shear stress is concentrated in zones (compression struts) between the column andpiles, and is not uniform over the height making it difficult to calculate a meaningfulaverage shear stress. The sectional methods of the ACT Code and CRST Handbook arenot appropriate for the shear design of deep pile caps. For example, the one-way sheardesign provisions of the 1983 ACT Building Code (and subsequent edition) are excessivelyconservative for deep pile caps.It was found that the traditional ACT Building Code flexural design procedures areunconservative for deep pile caps. These flexural strength procedures are meant for lightlyreinforced beams which are able to undergo extensive flexural deformations (increasedcurvature) after the reinforcement yields. Deep pile caps are large blocks of plain concretewhich cannot undergo significant flexural deformations without triggering a brittle shearfailure.In the third part of this study, the load transfer mechanisms of beams, which unlikepile caps transmit the load in one direction, without stirrups were investigated. Additional considerations herein are the crack propagation of discrete diagonal cracks and theinfluence of bond between concrete and longitudinal reinforcing bars.Based on a study of the compatibility of the displacements of a critical inclined crack,an interpretation of an important shear failure mechanism of slender beams withoutChapter 8. Brief Summary and Further Research 171stirrups is presented (Chapter 6). The interpretation suggests it is the occurrence ofeither very flat or horizontal cracks near the critical inclined crack that is most indicativeof shear failure in beams. In order to get a better understanding of this interpretation,the influence of bond upon the load carrying mechanism of the beam was studied.In this study, two mechanisms of shear resistance in structural concrete memberswithout stirrups have been identified. In a very deep member (a/d < 1) or somewhatmore slender member with no bond between concrete and reinforcement, loads are transmitted directly to supports by compression struts [see Figure 8.1(a)]. The shear failureof such members is characterized by transverse splitting of compression struts. For moreslender members with normal bond between concrete and longitudinal reinforcement, theload transfer mechanism is as shown in Figure 8.1(b). A concrete tension tie is needed totransfer load to the support. In this case, the capacity of the concrete tension tie reliesupon the bond strength between concrete and reinforcement.An empirical equation for the strength of bond splitting initiation has been developed based on previous experimental results (Chapter 7). However, the shear resistancemechanisms in slender beams are very complicated, involving both shear and bond. Adesign procedure which can be used by practising engineers cannot be developed in thispilot study which is only a small part of this thesis. Additional concentrated researchis necessary. For example, more analytical and experimental research is required on thebond splitting initiation in beams without stirrups. The interaction of dowel action andbond splitting to cause cracking along reinforcement should be studied further both experimentally and analytically with the objective to develop a quantitative relationshipof this interaction. Based on a better understanding of the splitting phenomenon alonglongitudinal reinforcing bars, a rational design procedure for beams without stirrups canbe developed, in which bond splitting is avoided by limiting the maximum tension forcein longitudinal reinforcement.Chapter 8. Brief Summary and Further Research 172H**F__I \d(a)$1?(b)Figure 8.1: Load transfer mechanisms of structural concrete members without transversereinforcement: (a) very deep members (a/d < 1) as well as more slender members withno bond between concrete and reinforcement; (b) slender members with normal bondbetween concrete and reinforcement.Bibliography[1] Marti, P., “Basic Tools of Reinforced Concrete Beam Design,” ACT Journal,Proceedings, V. 82, No. 1, Jan.-Feb. 1985, pp. 46-56.[2] Collins, Michael P., and Mitchell, Denis, “Rational Approach to Shear Design-The 1984 Canadian Code Provisions,” ACT Journal, Proceedings, V. 83, No. 6,Nov.-Dec. 1986, pp. 925-933.[3] Schlaich, J., Schafer, K., and Jennewein, M., “Towards a Consistent Design ofStructural Concrete,” PCI Journal, V. 32, No. 3, May-June, 1987, pp. 74-150.[4] Ritter, W., “Die Bauweise Hennebique,” Schweizerische Bauzeitung (Zurich), V.33, No. 7, Feb. 1899, pp. 59-61.[5] MacGregor, James, “Dimensioning and Detailing,” IABSE Colloquium, “Structural Concrete,” Stuttgart, April 1991, pp. 391-409.[6] Ramirez, J.A., and Breen, J.E., “Evaluation of a Modified Truss-Model Approachfor Beams in Shear,” ACI Structural Journal, V. 88, No. 5, September-October1991, pp. 562-571.[7] Reineck, K.H., and Hardjasaputra, H., “Consideration of Strains in Shear Design ofReinforced Concrete and Prestressed Concrete Beams,” (in German), Bauingenieur,Vol. 65, February 1990, pp. 73-82.[8] Ramirez, J.A., “Truss Model Approaches for Shear Design in Beams,” State-ofthe-Art Report, ASCE-ACT Committee 445 Shear and Torsion,173Bibliography 174[9] ACT Committee 318, “Building Code Requirements for Reinforced Concrete andCommentary (ACT 318-89/ACT 318 R-89),” American Concrete Institute, Detroit,1989, 353 pp.[10] “Design of Concrete Structures for Buildings,” (CAN3 A23.3-M84), Canadian Standards Association, Rexdale, 1984, 281 pp.[11] Adebar, P., Kuchma, D., and Collins, M.P., “Strut-and-Tie Models for the Designof Pile Caps: An Experimental Study,” ACI Structural Journal, V. 87, No. 1,Jan.-Feb. 1990, pp. 81-92.[12] ACT-ASCE Committee 326, “Shear and Diagonal Tension,” ACT Journal,proceedings V. 59, Jan. and Feb., 1962, pp. 1-30, 277-333, and 353-395.[13] Kani, M.W., Huggins, M.W., and Wittkopp, R.R., “Kani on Shear in ReinforcedConcrete,” Department of Civil Engineering, University of Toronto, 1979, 225 pp.[14] Ferguson, P.M., Breen, J.E., and Jirsa, J.O., “Reinforced Concrete Fundamenatals,” 5th Edition, John Wiley & Sons, New York, 1988, 746 pp.[15] Cook, RD., Malkus, D.S., and Plesha, M.E., “Concepts and Applications of FiniteElement Analysis,” Third Edition, John Wiley & Sons, New York, 1989, 630 pp.[16] “PUNDIT Manual for Use with the Portable Ultrasonic Non-Destructive DigitalIndicating Tester,” C.N.S. 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No. 576, June1954, pp. 622-623.[32] Blévot, J., and Frémy, R., “Semelles sur pieux,” Annales, Institut Technique duBâtiment et des Travaux Publics(Paris), Vol. 20, No. 230, February 1967, pp. 223-295.[33] Adebar, P., “The Behaviour of Pile Caps: An Experimental Investigation,” MAScthesis, Department of Civil Engineering, University of Toronto, April 1989, 137 pp.[34] ACT Committee 318, “Building Code Requirements for Reinforced Concrete (ACT318-77),” American Concrete Institute, Detroit, 1977, 102 pp.[35] ACT Committee 318, “Building Code Requirements for Reinforced Concrete (ACT318-83),” American Concrete Institute, Detroit, 1983, 111 pp.[36] Rice, P.F., and Hoffman, E.S., “Pile Caps — Theory, Code, and Practice Gaps,”Structural Bulletin No.2, Concrete Reinforcing Steel Institute, Chicago, Feb. 1978,14 Pp.Bibliography 177[37] CRSI Handbook, Concrete Reinforcing Steel Institute, Chicago, 1984, 800 pp.[38] ACI-ASCE Committee 426, “Shear Strength of Reinforced Concrete 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Tests to Determine the Influence of Support Stiffness Upon the Distribution of Pile Loads on an Eight-Pile Cap,” Magazineof Concrete Research, Vol. 26, No. 86, March 1974, pp. 39-46.[46] Gogate, A.B., and Sabnis, G.M., “Design of Thick Pile Caps,” ACI Journal, proceedings, Vol. 77, No. 1, Jan.-Feb. 1980, pp. 18-22.Bibliography 178[47] Sabnis, G.M., and Gogate, A.B., “Investigation of Thick Slab (Pile Cap) Behaviour,” ACT Journal, Proceedings, Vol. 81, No. 1, Jan.-Feb. 1984, pp. 35-39.[48] ACT-AS CE Committee 326, “Shear and Diagonal Tension,” ACT Journal,proceedings V. 59, Jan. and Feb., 1962, pp. 1-30, 277-333, and 353-395.[49] Bresler, B., and MacGregor, J.G., “Review of Concrete Beams Failing in Shear,”proceedings, ASCE, V. 93, ST1, Feb. 1967, pp. 343-372.[50] Wairaven, J.C., “Shear in Elements without Shear Reinforcement,” Bulletind’Information 146, Comité Euro-Tnternational du Béton, Paris, Jan. 1982, pp. 9-41.[51] Reineck, K.-H., “Models for the Design of Reinforced and Prestressed ConcreteMembers,” Bulletin d’Information 146, Comité Euro-International du Béton, Paris,Jan. 1982, pp. 43-96.[52] Mau, S.T., and Hsu, T.T.C., “A formula for the shear strength of deep beams,”ACI Structural Journal, V. 86, No. 5, 1989, pp.[53] Al-Nahiawi, K.A., Wight, J.K., “Beam Analysis Using Concrete Tensile Strengthin Truss Models,” ACT Structural Journal, May-June 1992, pp. 284-289.[54] Adebar, P., Discussion of Paper by Peter Marti, “Design of Concrete Slabs forTransverse Shear,” ACI Structural Journal, September-October 1993.[55] Reineck, K.-H., “Ultimate Shear Force of Structural Concrete Members withoutTransverse Reinforcement Derived from a Mechanical Model,” ACT Structural Journal, V. 88, No. 5, Sept.-Oct. 1991, pp. 592-602.[56] Muttoni, A., and Schwartz, J., “Behaviour of Beams and Punching in Slabs withoutShear Reinforcement,” TABSE Colloquium Stuttgart 1991, IABSE Report, V. 62,Bibliography 179Zurich, 1991, pp. 703-708.[57] Kotsovos, M.D., “Compressive Force Path Concept: Basis for Reinforced ConcreteUltimate Limit State Design,” ACI Structural Journal, V. 85, No. 1, Jan.-Feb.1988, pp. 68-75.[58] Chana, P.S., “Investigation of the Mechanism of Shear Failure of Reinforced Concrete Beams,” Magazine of Concrete Research (Wexham Springs), V. 39, Dec. 1987,pp. 196-204.[59] Mains, R.M., “Measurement of the Distribution of Tensile and Bond Stresses AlongReinforcing Bars,” ACT Journal, Proceedings V. 48, No. 3, November 1951, pp. 225-252.[60] Ferguson, P.M., Turrin, R.D., and Thompson, J.N., “Minimum Bar Spacing as aFunction of Bond and Shear Strength,” ACT Journal, June 1954, pp. 869-887.[61] Ferguson P.R., and Thompson J.N., “Development Length of High Strength Reinforcing Bars in Bond,” ACT Journal, Proceedings V. 59, No. 7, July 1962, pp.887-922.[62] Ferguson P.R., and Thompson J.N., “Development Length for Large High StrengthReinforcing Bars,” ACT Journal, Proceedings V. 62, No.1, Jan. 1965, pp.71-94.[63] ACT Committee 408, “Bond Stress—The 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Bartos (Applied Science,London, 1982), pp. 331-341.Bibliography 182[82] Orangun, C. 0., Jirsa, J. 0., and Breen, J. E., “A Reevaluation of Test Data onDevelopment Length and Splices,” ACT Journal, Proceedings V. 74, No. 3, March1977, pp. 114- 122.[83] Darwin, D., McCabe, S.L., Idun, E.K., and Schoenekase, S.P., “DevelopmentLength Criteria: Bars Not Confined by Transverse Reinforcement,” ACT Structural Journal, V. 89, No. 6, November-December 1992, pp. 709-720.[84] Tepfers, Ralejs, “Cracking of concrete cover along anchored deformed reinforcingbars,” Magazine of Concrete Research, Vol. 31, No. 106, March 1979, pp. 3-12.[85] Tilantera, T., and Rechardt, T., “Bond of reinforcement in light-weight aggregateconcrete,” Otaniemi, Helsinki University of Technology, Division of Structural Engineering, 1977. Publication 17. pp. 1-36.[86] Teng, Z.M., and Ye, Z.M., “An experimental study of bond and slip relationship ofdeformed reinforcing bars,” Department of Civil Engineering, Tsinghua University,Beijing, China, 1984.[87] Baant, Z. P., and Kazemi, M. T., “Size Effect on Diagonal Shear Failure of Beamswithout Stirrups,” ACI Structural Journal, May-June 1991, pp. 268-276.Appendix AMeasured Bearing Stresses and PUNDIT ReadingsD6-12-3 D6-12-41 I -——(—15.9605040302010002 360-50Cl)Cl)a)30C605040302010004I2D8-12-1D6-1 2-5r605040 .30.. 24.42010r______2 3 4 5 0 2 3Transmit Time Increments (microseconds)4 5183Appendix A. Measured Bearing Stresses and PUNDIT Readings 184D8-1 2-2 D8-1 2-3605040302010rc 60-5oCl)040C)•1-(I)C1o006050403020100000( ,zz1 2 3 4 5D8-12-4r• 1’ 2 3 4 5Dl 0-9-13 4 5D8-1 2-5605040302010Dl 0-9-26050403020100041:IJ————5040n20102 4 5 Oo 1 2 3Transmit Time Increments (microseconds)/ 13.44 5Appendix A. Measured Bearing Stresses and PUNDIT Readings605040302010006050403020100060504030201000605040302010006050403020102 3 4 5Transmit Time Increments185010-9-30Cl,Cl,G).IC,)Cci)D10-18-160504030201000 15D10-18-2fZ.31 2 3 4 5D10-18-3(4 5Dl 0-36-1 Dl 0-36-2/c4T.......................................................................4 19.51 2 3(microseconds)Appendix A. Measured Bearing Stresses and PUNDIT Readings 1860U)U)ci)C)cim___ ___ ___010-36-3 Dl 2-9-16050403020100605040) 12 3 4 5,u204 1% ICl 2 34Dl 2-9-3I I12.25Dl 2-9-26050-403060D12-12-1460506001 23 4D12-12-26050—-.-—550403020102 3Transmit Time Increments (microseconds)4 - 5Appendix A. Measured Bearing Stresses and PUNDIT Readings 187012-12-3600-50Cl)cn4Oa)4-..C’)c,)Cci)D12-12-42 3D12-12-5 D12-12-6605040302010605040302010oc60504030201000/123.2) 1 3 4012-18-1122.0D12-18-2bU50 —. —4 7/30/‘ 24.420--0•n_________—I I2 3 4 5Transmit Time Increments2 3(microseconds)4 5Appendix A. Measured Bearing Stresses and PUNDIT Readings 188Cu0Cl)Cl)G)C’)cz)CI-CuCL)D12-18-3 Dl 2-36-1Dl 2-36-2 Dl 2-36-36050403020106050403020100019D14-9-1 Dl 4-9-26050403020102 3 4 5Transmit Time Increments1 2 3 4(microseconds)5Appendix A. Measured Bearing Stresses and PUNDIT Readings 189Dl 4-9-3 D14-18-1a50C’)C,)ci30c,)Co6050403020100/7’”26.8D14-18-2) 2 3D14-18-34 5605040302010/‘634Dl 4-36-10rz5Dl 4-36-260740 ——-3020102 3 4 5 00 1 2 3 4Transmit Time Increments (microseconds).1028.11-5Appendix A. Measured Bearing Stresses and PUNDIT Readings 190Dl 4-36-3 Dl 8-9-1Caa-50U)U)a)30c,)C60504030201026.8U01 2 34 54 29.31 2 3 4 5Dl 8-9-260—z——3 4D18-12-160—50403026.820 -0-n____Dl 8-9-260—50 -4030201020.7n4S ‘0 1 2 3D18-12-22 3 4 5Transmit Time Increments (microseconds)Appendix A. Measured Bearing Stresses and PUNDIT Readings6050403020100080706050403020102 3 4 5 00Transmit Time IncrementsD18-12-3191D18-12-4Cu0U)U)a)Ci)CC)m60504030201060504030201000I 2 3D18-18-14 12Th4D18-18-25.1060504030201000.K: 3D18-18-3554‘ 1 2 3 4Dl 8-36-14ZJ60504030201029.32 3(microseconds)4 5Appendix A. Measured Bearing Stresses and PUNDIT Readings 19280c3) 30C2O1oOüDl 8-36-2 Dl 8-36-3Kz2D24-1 2-14D24-1 2-260504030201000D24-1 2-32 3 4 5605040302010Transmit Time Increments (microseconds)Appendix BACT Code and CRSI Handbook Predictions?er/. I Coo. jyo),. c-f 2c’ J.ecre’€. le& To-hC- .ir&L — a o-O o1 P-Lo- -4CQ c Cc’: Zê’(’.€_ot or—- cc—J± +O—vj—Eedt-+ec& co&m. -&4e-oO€—v-&, SheCvrL3)—CIL4. c’c’2, O Q -&‘JCCdV&’. o-,Cp.U20 c’ +L ?kC/3. ts) 1r-he---i L4) —cJc,}-v c’v i_-C4LaL o* -cC_ Of ‘“ “.‘-- —‘-h€cy—.a—vJu-i -Th.e_j evy-C’2) — LcL’k , +o— €-.,--C r- C —frs —-cR —r-----pj93Appendix B. ACI Code and CRSI Handbook Predictions 194—t—- - vvJJOAVtA() ——Coi’-. ;OQ nco’’r. 14.’;L.-_w.&cit:.1=’4Ori€—’j c4—‘1J ZJf, cpa) — aij1Jt4f’D =—-<a.zS?J LH?’.)(oo3 ... o2J’r= c3_2. £IJ (?-)= (3—24 )C o43= (35—..5 )(o o3 4=-t-z5’-) to c)=—25 ) -- 7744C-) C3_2 )(°.°S (‘e?E)1)1:. rs) jj ( i L1?’)==(}‘3)-Appendix B. ACI Code and CRSI Handbook Predictions 195too 42.--Lone—-JJ.-___---1 I‘H-’L_2l.25 122Q—iQ heOX L.4):3/4;Q.€?(35 —25cQ. =az-c, 4cM__z3 /cAJ14-c3-L3I49.c —______= 112 L)=2 -rz z) 27 LkU)_-vj jLI)= o34J,’3 xos-4’f.S = 1.t t)== 3.7t)1caQ 2’iT4-—= le3 L4)øi—1wj skeL7.J:yo4J23.54s1’.1 (t))44Q==-2V’32.2.Lt) =beo-’- C3;10= /O32M(35—Z o.43) z.f 2.244.aLt) >3Lt) 1c4:’iN = 7&.6 it-) 47 L&)cñi’• 4’-):“%CQQ 0 •-“2OC)= 274i (43J)h€c srr& L2-):2 ‘•t-) —= 2 )=t Lt tS 1ftL&J4 <4±:L_J103’O—k)O. &le.&ri3)vc, = -.35 ( a.c39j -- t71 Q.O244 __7)4oe 49.;= 3q.Lt) . z.togS c4o4’,5’ O.L’f)\J3.5C-t) ZY7it) 7L’k$)Appendix B. ACI Code and CHSI Handbook Predictions i96}€tL ç4c4A=43.21e.-F= °/—= = Ct)c= 2? = 33 c-c) =- Qk)O—v)- SeocL-)Vt: = 0. 04 t4ts49.5=2t)_o Ct)44Ct)4J)fL)Yc= 3O4 -44’?.az’ (:)= V. = 73 ct))0$(3)p /Lk0)(4’5) = °2o3, =(-2. ,43= z.+3 . 25‘/= 24’( 5e3q, -i75.qo2o3)44>= zv= iaL&)=QP€ eL4):O.O203. %‘(3—2.-’)VC. 3(o.MJ= 4tt) . 44’? •=HO.l Ct)v=4u)beojn ii)=40c4 3Z 3(-)= 3L&)eo’r 4it2).— z.CCt)ca. == 04r1.1 Ct) a2Yjo c&’_)i(LAL—‘,it kec)A=42’crnt) tjl31 =7Q3cA r’e-;=Z7 Y/L______________________\— 4.74;3‘4=rtL743xOZ/= i74 ct)2iL 4 X)= 14 ck) 4Lt L)che3vc 3)=4273).OcZ,1Yz9o-t .OIr2 75.!,Q7- 2.Appendix B. ACI Code and CRSI Handbook Predictions 197cr. — (4J42.50t7/_(3—25o.. ) =;:; ÷i75.sIai72-)4ø73Ut) >V tt4.3Lt) ‘j.3c.= 2z4.3,(1iJ,)2ZfL ev+4.1.:Jo.4—4)=I2o3tt-r)= 3G/422tt)-)‘= 53e+)Ofl€LJ ‘e’.’’C)43.l/3)o.O14C3_1)(3) 2.737 2.= 2. (aof - oc4cL331.Lt) >.‘4=4iit)fr= .gç ,27g?c 33Ut)=3z44 c)k’ex’i c-he-, c)33Oä)=G€—)Rj k’L)nstOn.,—io C2-):Vc o.53o4-J4L3 3.4 it)— 1=b1.tt) ikt-’)p =aoi4, Ya=tU= 041(3. — 2.5LO)’tV j .3g(o 3J4;t; -I-= I4..3Ct)%4343Lb?, Q= 2Vzsi’4(4)Ltoi-j s4€.+kLIL..tc,L = Q.- )L?:SZ=1Ut)30rJ,= cO-== = Z2-7‘= 31/(c4 t173(t-m) ii’7.3 o>4i,= z’ L)Co- 2LI z.a)Appendix B. ACI Code and CRSI Handbook Predictions 198,oi:Or—”Wj c-L2..):+S4os-.4= 34.3 tt—= Lt) 93(kJone— skerc4)?.0L3, -‘-°4(3—2.5?L44)— 2.323(O3’ t$10.°I3)4ot4= 24 Lt..) 2 j7V71.5(.),=4O.24c4 $q=4q, dSI.tqirt= 4Jo)#é )= 7,— = 370.1 Ct-)W0t = 2)sipe t74Lt) 7251 ckiJ)ofl€—i.4 2J 4,‘O,‘4cr2(3.—2.5Q.24-)’ 2. )Z.5> xS’2 -V, = I Ct) zV = z44.’Z (t)‘ie- s4-H (..)5 -*41lo35 53It)= SLJ-)Ci-)r€—’i)P = OL93, -= oz4-(3.5 — 2.424.)= 2. >2.5(odj..t•>.‘/=azLt, kCD2205.th.2.L,ectr’r <.r•’4t)tsJCQ..Q5_ix33Pct) 3g (.)bem.r’j .1j-iL2)= 3?J et)lLQL r17.2C)= %32L4e-)@ne—’ seor-)to—’J-J chea,LZ)v= o53c4f 4o.2 =ALO 1t)= Ye...= 2IJp Szt)= 4-t-?..)arie—w e1Vr (4..)-‘==z.5Ct) )V 205iCt), NccZY4I0.2L4O23(J)Q.3tt)-Z(t’JAppendix B. ACI Code and CRSI Handbook Predictions 1993Ui-ttL__. L_J—As = 2.7cr.\1, 297 V/cm , 45ci.3 co334j.7t1f92— --A = 44)l2.LO 419qX2 —-.ç. _1,g=ac?&-7 I)(-t1t)• l.)L45I3’= I3°° 1*)29/-M= 3.13 =43.— -A 4’7.°’ e’?r IO.S297O+32252O —47 /= 1’°t’’ .f-& 2529x4547’lx57€= 57S (II4T1+ — 49 ) = 32 t-)= .ZO/3q= I4i. c-t)=4c4’iA.= !.3)O2/j-=22.Lt)I 37.Slo13sfLJ“ $.252c•ttj= fGq1= tE4 (L494— )= 3Ct)I= 3L3O.O qocL= 3Z5(L)Appendix B. ACI Code and CRSI Handbook Predictions 200I °Vt = S3a.J s%4474o5 (;t)==b—:43== I47.t)— 3I 3x2.4 tt)z13oL4i)Oi1.—Jvco3L441 = (t)12 —&;30 Ct), \/Zt4L+’6,3Lk)— 3 rJjI4 =eUc3):=— O.o1S_4= s—zo.)= 2.’L—z(o31fi -+ 75l:P +3/(i ‘447) =v=( 3S (-t)y=’.2a) , D2.L±) >(t +Ci&, iorkIS.75 .75Appendix B. AOl Code and CRSI Handbook Predictions 201o€—w k’r (4):b4T’S/çi.r I G, hOri ae—j’-” 72kt)—b icì. icrL,irore)= 3 zo k;C tjxi n-i)4 (4 + 447) 35 C.ftv=iJ;- x3 3• L)-.zi4?t=V==1V= tt) 372Ju-i)1jo—,Ja eCC2.)= =CW-, = Z.14/4v’ = o,o4-G,72.4 Ct) . 4jj?x4+!7J7.2tt)V= 72.+C) 44)shei’u.)— --7.4€ =. M’ii.°’7j-1a-3.i-Lt) 37 (-kiJ)1X2-i4- (z)L94e 3 tq-33p L+) z3&)Cf 37.Appendix B. ACI Code and CRSI Handbook Predictions 202A1:2ct144, - =43. c/‘---7 4OZ+!%3 ‘,—---444 i 3t—S4-.f44nA= A -t ‘—43(4—--)— 3L,Lt—) --4 3Lt)— 23) L), z-i - g4-x Z3C) f7t)= Lt) zc (:&eJ)-—M he—c’)vc=’*_______________________, *JLV— .3Appendix B. AOl Code and CHSI Handbook Predictions 203e—c h!G-)P 432/( o&4.I’) 24=Q3 (—25XO4) z.32.52.(O3Jç II.O t) r4t)‘4i6T.Q’i) -‘/ct—clAu/=••••) (3w— 2.’:2—’ 3.2,-aS‘4 .; (o3 -7cgLYøz44 -)ot=32.o t) >- goxs2. z4-)VL21%,+Lt), , Jz2cMt)bri- C i jrec) t’4q=3Mb1.o = 43L±=4737 &kiOne—’Jej heo’L4-)_____= )y — .(o3J --t7$ “c’L—37t) .O77J fD 31 4)°’4+ (3_2so7’fl 5S’-Z.5v (° 5°31 - 75.I i.’j Lt) cc—c. l2.I4/=o2 f3 4o.z/,=o4b, L3—2.5)=2.2.L2(asJ +jT d443X4b)2ZI.71t) > 2 J4t I Lt)v 4q4t) i-p’cV6 Z7L)-t’o—’i he)‘4 = .r2.ijTh *-4= 411 .o t)/\/ J — =I L_J \-/ r t-t= ,22.l(±)S2243Appendix B. ACT Code and CR51 Handbook Predictions 204O-’. ‘—r ()7I---4 = l2c5.t7cM, ‘4 J_ lk.t) == J1 4qL4) =F1T JfT’. 4S()=2 .L43S3’? o’Ct)= = 2.?t &)fl€-L &h-ø+K—= 73.m, ZPcS3a3=27$7t,‘2.43o COO = 3 cl . 44.3A.qs& - = z37= z4at1r4.73zS37(7o’z—t--) /47e=to44ct)A1=I7coO°= cA6 -= 2q13l/ =3 A= j4p9 cc’2&, ‘z4-cA4+4oco °=4m, -1’ 33k =W7ot3Z+331=t80X5SO.O23 (70.0— 94(.3 J ‘17.70 ‘j3c2€37= .‘T3 23?( 71 — Ct’) , JpIe= xlO/14 .%34tt)3iS’ 3zD44 £13.21t)= ,l4L-kL)C /A )Oi— WCM SieuC2):0—: V 304 ‘e,S(t)b— , V= 3’44C3 xt4-x-70 a.o(t)-c—cr i1S5.tLt)=3w =2)2O44f.N)Appendix B. ACI Code and CRSI Handbook Predictions 205O—vJQ S’.)Q +$/ rJ’OD, tJAj) z3A-’2c3 , (3—2T5?25— ) , )V=4Lt) ,C-—c-j v) 2.5V = 71 =zbI.ILt) > oQ2 4Z.Ct)V=432.!) ZJp€ = — 32L2.Lt)____Ct+4-L, orc&) aç4S = 744Ct) 74C-kJ)&e—vJc çhCOV((+)J— 4’,7g/T2’ L, p_4.=2E(5oj .jt75 _ _)3c7O.2 3,Lt) 4_44 O P /4r7o) °“‘4+,--i.i(o o3jZ- (cl ‘44 )M.-7o . Ct) .C—; 1A)/ jr P =43/LIsos 7Y2) 039 ,V 25(o 3J4 I75 ij4&39772IPtt)V. = 9z.3t) , = 4z5.t)— 3 33l = O.4-Ct 3 = 2! (k)“-): /A)LA)/ 2S4/9a7OI0n 4çJ 34.)SCL’) C/A )t’e&i r Z)°‘‘ iP_i€ =a.7 4.3>&31 34f.••= 3It) C7-Appendix B. AOl Code and CRSI Handbook Predictions 206-kLr&,L==2cAt, d.7lSe A2)c3C0S3/4J d,74.t±=4°’? A5.= - ==___$02 43U -zC3.C z7.3t.3±3o==‘74, A7e _ 4-34’ 4-3134c! )4-it &73. — i)= CVI) IlO.7?d/3.tz7-))=31J 3z7L3 $I%(t)- ‘7LL4d-)OflP_—OjerC4)O —VAJ t/L2.)oI-JiSZ73’’ o+.z(t)= J€k—i,: V= WP&.C—C: 3O4j3I’T 12i(t) , ct)== z5 Lt)1—vc .hoL3):)‘/2.(°34 c,ct73 z4Lt) -> x13’ 21-3;‘4=. 2.GA)=Q—w-—-U)= 42 2)Appendix B. ACI Code and CRSI Handbook Predictions 207‘! (øi -i-’, .s;c’o4.a 2.j3(r3.53‘‘c 3L1)= 1? ) 335.S.) -k)51,e3J<J) Ct—P’1 ecrC2)iJ/ 24/3 o= o—- o=4o.t; .- a+fox73=2Z7t)V c-) 4-L V z 73.,c.;= i2ZLkF)()(243 )(4? 140s Ott) t€ i )(43;2=7z(t)—4IJ 1:f.3c72.s/#14p7 cm’ 4= 23s/c4,o A______27,g+4Z7bZX.Lg+ i4o7.x Z3 )•‘ 594,cZ7b-:1-- = 272(47.4- —1—.= 4iZI = .-f)’ 3(&1)e71—uXAJ c-eci)o—)o’’-j heo-r2):v= o54-f xi’7.4- 244.C±)= Z.4W = Z43 (ki))Ofle—Jck1 h€o-’(3)P= 32>2.5V I7)OOO5)I(W7.4= 4z.3 L) > .s ;J ,ø7.4= 372. Ct)I I IL L.L1 LtJE1L_1 i__ITT Z rrj .G I 120 l°Appendix B. ACI Code and CHSI Handbook Predictions 208= 3(7,Ct)= aiJL :I?e=74a*)=73ofl1—W S€t2C(4-)=5/6ci-= O52 (3.’5—Z5)OS) 22V. — 2,p03J÷tiod74= ,IZ.O(t) ‘.‘Jt=V2PC±;)2IJ., cot 44J.1 I24(k04{)V 4-(sV 7.4) 94= 2O4Lt) 3( (-I&)-o---,JO.. 1ie?’-Cz)= c wi- acot)t,eOrift—ofZY32SS3.I L5ZtC441)etr4C):—-:QZ325=Lt)= 4t€’fLLVL= 4o.ocm 44’O‘8/ =e-f=4 2 t-F=5L7 d—4z.4t1,4’ = 4• 4443a&B5÷924.Sit7?cbb .__44455(—1k t-), 2Lw =‘-=44LL= 2()= 7!(&I)c€o’) C/Pt)fl€—VJ. A’L-):VL=4-ZØox’. j’31.3Lt), ztJJe = = Z24C+) 7kS)ce—-O- t’xL):/,a) 7/6 Zb, L322) 2.5 )‘2,5/= a.5ISj— Y3Q .. 34-o C&)Appendix B. ACT Code and CRSI Handbook Predictions 209= ct) = =‘7g5.4t) = 7732(&)OAi0j sie(C4)== (3—25XO22.’V 223(o.5° &ai=6i7kCt) 2. Ij4 v.1= 4t)= 67.1 ) = = = l2352CU z4Lk)v4= = = 37.)= z24&t.)4o-IMi-): ( 3/e& rio4_ioj erLS):Ci); 1?axfn1t=saxo5=I73+°t) c4os3 r(23tt)= 4= 2.I3.Z.Lt)2S57 ()+t-2795 S/- cL,Cm’, /k27 QS4Z7.7C?t’f?‘3‘Th /em? s -c”==7&G2x28b ) i3.4 (it)pZ2,65 L.)L19o37I=,•a±), ‘IotoJ-U): /A)one—t)Jc’ zHVL =53o4jo = i’Z z(tr) ziJr V. , cQ 2iJpe z+z.s1-ct 239j k.43)ofl€.—-j ieL3)=4AAvh&a L7/27 (39.—2,5)L027) 2.3’2.5{ 1O?(b5. 411.1-ct)> .5 jxjZ0Xc.9 = 3C35 c-t)vc.= 5(±) = J = ‘}27.Oct) 73C L-JJ)ore—vja OfC4)=.53, (3S_2.5)2.V 791 )iox6. = 77.l tk).Vc7JCt) 24L, 42Ct)lLJJ)Appendix B. ACI Code and CHSI Handbook Predictions 210—oj €&r)‘4 ô&7Jj 62’f.’ZL), Wt= = tlJ.c5(t)= 1ocC&t)_t14 /s.’i = o53 ‘ , nci aG—U ceckr(3):i,u—irw +‘4-,= izt37 x &=l576.(t)=4af{4r-Lk3oc, -F 4’so = c-*t Az d’4-cc=-34- /c4.42, A= +S.11cm> ..f —x4 +ibi4 4l((7—2.t--W)= 52.O/ =439 Ct)) Icøt 7i4C4. (j.)one—- heov)one — v’JL ¶4i€OYC?)o 3ô4j a7= fl2 Lt), zN,,t= ‘/. = 44= 23b.4 Ct) 23) (-?iJ)erC3)=4/rp °‘-°. /(/a) i75/,‘4= 2o5D5 + ,i3Ct)> 34(±)= 4-.1)= 1tJR. ) 4JpL ‘7.2 C1)or —icu.j 4ec’cL4);=o.i, 3-23-) z21(O5Of 4-1 t)L )(J=9c41Lt)< 2.!5i9 9o(t)V0 441() o44a,t)=jaZ)1e+Cz):4o= 4$Jpt 2L4.)Z17CkJ)-tuio—w&‘4=tC&77 bI4Lt),b-wj ?heo.rL7-)eA” ¶d7€+k(t)‘J0QQ4€X).JZLLO’t = 2M-3.4-’tt= zc1 r3 -&ii)Appendix B. ACI Code and CRSI Handbook Predictions 211iqi-:i= 0/cmj lj;’ z’i/-,c-= qsi f 4l5 = oCW(3! 6t l3,OX23lsO ——z 7 “Ltt 27o= 2W (2S — ) = 47. LtW)2j7L =- 147. 1Oz/ ‘4’223Lt) , 44L±) 2S3(k3)o—uX 4A.rC)_vo&Lj e.r(.Z):VO.3C4SL72 = 132 G) , Y = Zt’JpJ(l2.) ‘JeL = ç(±)32 4JJOflf—WO- €,A’():/cv..A) = = ojG., C3._2.5)cIV37,2.5‘4= 2S(0-qj.t7! oQo3’ --)!ox25 o.: t) - ,c(€o25 = !:t)V4-.tt) = 4\7’.2i) Q3(kFi)cfl’L-uJWJ hXC4);= —i z.O3J4.t5 +!72is°37 —)o25 !ze7 Lt) > O(72= &(*)V = J2=4—o—W.J hetML):Vt.7J4 -t-kD92-5= o3ft), 774.i)= (-JJ)-+1*30 — JCAV o4-(i-i 4X 923D(±) . 4-Jj 2Ox25=2o!.I (t)1JL-=ket) .j /f)__________________J,L 45l,4Lt)—c,L 24.3g.4ct) = a3’I3 (.J)Appendix B. AOl Code and CHSI Handbook Predictions 21243PS-çL Skv4k=2.L’-,-c=4s4/- =‘935 AI2.Gt 44/ =cWk3,I4’ ,;+Z /(A.t-, J ‘43 Ac. øCr (Zo+S4+ p24x414o)+ 3 ItjAr 94 °(‘1_ 43) 192. t4)rI‘-= = .ot-), =4J-j= 1IgCt-)= p7 (-?dJ)c’heo..rci): (/F)OhE_—v) t74XL2)v= j tV-2= i%.’7 Lt) , Jt= v’4 — ?j3.4(t)’ tkii)w&jc’= —,= + I752 ‘-2. ) Loj2. = 75.4-W) 5. I L)‘4= S7.4Lt) = L 1150.8(.t) —one_—W’- €-A4-= (32So.3)2.’ 2.Sv -= ‘ji ot,28 = 1371.9 ct) cga t et)V’3.&Lt) = 4JL= jq7.Lt) L9Z1Z(iJ)o—J& S1eosA’O:“4= i.o1o7j4oi.1z)Xc12. 24,4Lt) 7 .OCt) 9OZ4(4tJ)—bo — &tj ie12)LA)/cA ‘92 Q3vc _c.4-<j -÷ - 4xD)92= k2.8 34-17.0(t)il1-Z. L) =Y 4-ZCt,= (QJJ)-_J8vXC3): C’/A)e”s 4eM1Ii C7-)I’Jc.Q = OS 4I3xo 2D1S.3(t) O 4A3 X? 2kO(t)= oL(-kN) = = 320oCt)342 I (&)Appendix B. AOl Code and CHSI Handbook Predictions 213441+hec-tkPcs T.24 Z4 c )= 23cS’=27.7LC,72I4/eA1Jy=2J.jt ô/= tzLt) , L’ 4.”_________(‘/A)V=omo+Jtoc1;.L1= iS3t&) , k12 =vI3O.)2I’(k-I)t’ne—w15I1er(3):/j,J) = =9) L3S-2SX9)3O3>2,= 2.(3gJ +I75.oc44 )fQX 4j (t) > 443 ±)v = , —oie—v Me—rc4):.y __L o3J +i75flx&ôt4 j)o[ =jZ37,Lt)N/S4CtE) = — ‘f’7&)( 5)6.7Lt) io2 (*±J)o— ewrC)W/a 3/oVt. = (- e.4f4-3tt)tt. Vcr(3): /A)bexr 1-tJ seMHC,)t(35. 12t) 3 , Lt)= 1473+ (k.i) 1JJ.94 ‘25Z’ t)=A-73L&t)Appendix B. ACI Code and CRSI Handbook Predictions 214‘44*-€tt1 cm€’9i-fk-1/, =T4c13492.w, 1/CA%j344’4—---) = 3.1 -)4rk3 ‘93O1-= ;50Ct) , 1co 44( 1tO3pC)=)O/,CkiJ)e—Lc’j e-LU (/A)—j ç1iec’-U-)y=oo1-Jv’z.&=’ J.t) = 2-V 33.1t) 3,bc&i),‘he-rt3)O.OZ6 , 3.S-25XI)=3p3 25= - I o2.L, L = 32 L1) LR92. =a. L‘4 3jLt), kco’1cI42L) lo437L-ki)o—iioj iec+)= Z (PD1 PJ I’(5.& 021g) IC/J ) J.4L4)7 ‘—flV o9.ILt) , J€ = 2-V jI2Lt)= 73I(-&)o—u ectv- LU:I2,3t) lJ,L — = 4:9 Li)=’ J)Z.&rL2):-‘4 = o,3o4 — (I4 —T4-x T= lb3Lt) < 48i 23Z3Lt)‘6 13Lt) kcoL = II.3L) l’f7O (JJ)—je--W ie—[3); ‘-VA)he’f eM-rL2)=çI3)7q7.gIt)= 7’ -) R=4Ott)9&2-(42).)Appendix B. ACI Code and CRSI Handbook Predictions 215LLtJ S:AS2?.i 4iol,1:, = 2I ,fc A4 p4.%cYD=574= 1I,2= 5i2()4eoii)iL= a 1-2L9 p = iiCfr)kci;V=2js1—l’Lr 2V 740kr 32()an—i 4,e&fC3):‘)7/O4Z(35_2,XO12) = 4z.5-).2Z?r25C1J)a—vJ° eaA-c4-):= !7/2 -S) C3-Q.X°25) = 375i jgJ z?eq37 - ) iX1). O ?(52i#i k’i40t= 2V 54 =eu?r’ -iji-)=-Iz45 O=2.= 3’?O’It4JJ)o.3LZ1 U riiII2t”dLJ D (fi5JSQSi47#I_oq—w cheorcirjr1JcoQLr L.iJ)1o.1-+ku-t thep+ti:=a.92, =4,oD1z ‘=)47”, i1t=)= 3Z/=Appendix B. AOl Code and ORSI Handbook Predictions 216ene—Aj -heacL)= z5. .4 jc -LV — —øfl€— 4e2),= 23okkJ)—‘ h’3)GZ/ 5?(4vT) Ju+2 a4Z &7P=zJj3.44’(4&)on€_w&i ietrL4) _ +beoir ç-c) eoMYUj c.4--eL7)ki0Q &c(2 X34Q2 )L LS= 24 -r = g X34\O .43,,3 ep— 1clt 21J3kt)A1&A4A )L3i0 =5.4mr, jMoH1’si., =4oon)4ai______)= )2.7(—M.)2ipJ = 5r)LG/ =19 k) =4)=t2S8c4)( J7p)v= ‘=z9 =zV=7ØkJ)ofle—WR ie-r(’)‘°°7VfxoDZiL 3S92VQ—)O 1eo4)y=29o/ =o L3,5_1.)21SV= -= 9)4O(O M-3(&JJ200 ()—FI±Tf /L_F_HI 600 IL7Appendix B. ACI Code and CRSI Handbook Predictions 217Wco=‘€r-i) V - 33Z (2o+4o)4ov= I4sQJJ, zv—+-ja-4 ce(I) ‘1% •5.. V =G—1JO1 h-eAr(3): = ‘v3322J ki4i kt= 44=?q6(kM)‘9’:A & AfLQsr.L = - 410 , Iz66&iJ),33 &t-1C N/n)VOIJ5X944VJJ NcoL2V =2(jJ)________________z.87E,2S‘4= 25(57jE5 -1724,OPclZl 4ct,= 21--bJ . a-J = S‘424-J QVlC*)_v z.zc(I57Sf i724 >OOZ.1 97?c4OD j53’ ki)\I394VJ,____y=. Q33fl z4’o = = 2Y 3344 c-ku).lAJ/ = • V= 1j2-4&, LVc,= tCft3)__S2I mi’-j V 33zzg2.1 c72 J, 0Q= 4’4 z(J)A= 77SPrftf4Ih1&, 4’,-f3’-1 r4co41Jpl€)4= 774l’’— ‘2 z-{ 12O 4iJ= vj= 7o4c1)______ __?7/ct,4’a) = O2-5 (c-zsxo.zs)=2.75 ‘25V 2.i978JIt )95D?4O W7 * < 4fS3j D.x4o O5é -k= q c = 1 = ‘ Sk -k)‘4= Z5 (oI7J- I7z4 2O)ox4oO= -a3t) oj 7(frjj)‘/i3C) ‘c324(,’-kJ17eoJin $ +kL2)çexO)OQ-’AJ ?4L3)ane— -keoJ(4)—o—-. SoL):—)---Jo heoc).-Jo—1)1c.Jj 4At3).Aare—’ €O(U):OflPJ€X(3):o—Jaj ch4Appendix B. ACI Code and CRSI Handbook Predictions 218\= ‘3322j- 240D x4or I7f J, ‘icQ 2-V 5(J)_____________J/=o., v= 7-ii, c0iS4flIV = 332zSfl zc4o-= aiJ=JJ 4tJ 243zL-kJ)Jir L7)x c&z73413_M= 7g4)L4o (40o ccz) — ‘j= 35 J_____coQVfc= 2.5(PI57Ji + )5Ox4OD= 0411= ‘z-V,=3e’(IJ)= 225(o.IJ * 7.2cDPe2 )so4oo=1z3 &t QJIQ=37L)JI’JcoQ zV. = (4))c,&=2V33O&k)J)L4i7=o., =V=54t)____V.= Q.332ZJ 2I4’’ +—F) 4(mipa z,c-2L,i .2c’D= 2kJ)A17Smst2,{jZ6pK?ci. ,1r t2i-tt), ZkiL{_(/A )b5x4OO322_“4 = 2.5( 24J’ z,s4y=% jJVt95& co2V7(42J-)_ _17.z4 o)95rs4o= 4L4 g3oJ -ki’)=J cøt’ 2-Vt -— *‘eR’()—ha—bert S+-€AC1kC’):k0tQ(2;’-3LI 2C2I5C4&)one—wjseLtr3):one —ioj e’X(4)O—AWv ‘WLI)H-oio—woj c’1Z).3 —W ‘c3):,e.CLFin ske°14’(J):0—Wj e4rc2)o—1xi3)’3—o.%j SeiAL4):—-v.o—vi&q c*XC) Q32Z Zxi4 L48’zi)1 ct ZV 3QZL&)Appendix B. ACI Code and CRSI Handbook Predictionsj+{\; 7rn 24Zl4? ,—lZ2(’LtJ)4o’ —__--—-—-)=124.1J--nt) ]{ - (3 (k’i)—Nct= -V‘4= 724LO2 )5Dx4oo yt 43J zcV pL4i)= (JJ)Vt =_J2 5t3Jj*= = 2Y LO2.L)______________Jco.t 2V3i31J)_ _‘V’)/o.s,____332zj2C2\)t’O 537 Lk.i), 4lJLc z4U).ZG= 131 tCJ -z4(J)_9p& 22, 4t IZ,4LIJ)75-)(43 ) $J-), z’1= 12’O/,, 32 kiJ4-LktD__ ___ _v= z;(7SS 71 o’z1) rx4o q 4*) q3 ,cjc14pO=78Pr-)1’c 2-Vcv= I z I&-(kJ)V2.iL),,5 L(-ki), = 333Z L-k)219w—i- se&rc2); J/ =oE y= 2CJ)—-f.-L’) \4 (t), 14t= 4tJJ zztiC&k1)a -h’en -F+. ().Wco 0g 22&L2aO’ øZ-X2Oø 215(k) 4I(R)A1?o1€—W heci.€-r(2);e%C):O—IA3AJ sheoca)•-IiWQ- ]Cj ecaAL);s4Te1L1-;(J/)kli( C2).o-t.1C’Aj se-rL)—-VJ e4-ct) i/—O5)Appendix B. ACI Code and CRSI Handbook Predictions 220l€w h+; A 7yy1- -cL=\e.2 c4pt 125Z LJi)78;x4IMy 74ioC4— I22L41—ff)1 ZJ?& 52 4i)-; V= J \g=32 %.84OQS(*JJJ)€i9HiLt) -c s2O 2iJJtJ) oS ±!!x2,2 4x2oo l22(1Ji) 4pI 5D(4J)O,-WQ’M EOJL) --J -o€—c .he&rc3):flf—vJc- 1eL1-4C4*iO—iOA.i .1.eLkYL1)-Yio- icje’-’ fr+t)QO.W7Z2Y,2X2Ot )23(k1J)MIu-• /A)29L+)J), c.,L= 53C-))V = 2S Co.I7i l2O.21 )9 o4o= 77’7(-) .Wct=2V;= 4&)‘4=-2 =}75L4). 3J4oo 134L&bJ)cQ LkI-)i’4= 25Va(-kJ’))/= O. QV I3’ kiJV3t2-3j 4&)=+J, = b’ (-kg)12)A1LO ø cJ4)Jiie 4O4 (-)95m,7?7s4IO L4oo — ) = I ()J-fli-) zJt 25 .XI°= zJ £-kiJ)C1bI45X4 23(k)) cot - Z( (PJ)/. 2;(oi’37i4 - --°i-)9ox4= 73L-b). 33x4vqkJ)cZ’ z)4 — ia’ ki’i)= Lzç . o3o5)(cW)(4po= L3i6LJ)= i=ck)4c YcZ’ C’)O-Wo 4iecx-(e9(3)-OWOj i€/L4)-+Wo—W’ eo.tL1)—kJo—’?J-4 *E2-ecxrC3) I 2O?&)Appendix B. ACI Code and CHSI Handbook Predictions 221bead +flq+1L)1cot = 0,S5’c22 fl,3&fr1)(,3)o—iajMOV’O—-io—Mcj shea.rL2):to_V’.Oj ck€o-L3:eo”nrL2)xL74xX23(.i) lC= 4.=32Lk1J), 4CkJ)27S+.rX0D0ZI o,g3j 3s.9Z4ao =zL ki.3\/2’12-(k) lct iVC7g-(&1i)“4= C’57i - q4x=iIs)’3o5x4oo’’15tkJJ),V=°33flj 24odx=/,c14(J2P)1 bica — — 2O (ki))W/=oc,-1r“4 = 32aj &) = ztc1L cJ)c-høi- L2)CDr-- ozs!s4x2t= fl3zLkJ) I1=4l-i1Q= L+j.J)-çj=z,.9 HP144o)“ 101.tr s-410C400—l.9$2.tf)(75•D2Jhl1so3/ç’Ø0l1-kJJ, t”4Jp2(’ 2O22.,&)oe—9Aw”i4ct2Y iL L4’&)one—A3t- 4c 950S02) ;ga)=;; = O 125 (35_25XI2) )-Z‘4=.2(ol73j74,c Z1)0M0O31 (fJ’)> o4’3fj9974o= L41SJ)V’ L2JJ), Zp2o ?v 033(k1J), =4Ck-i)_____________z.31PoL Jt= 4$JQ ,2L+-)75x4l 0H 7x4ko(40o— 116.21’kii-P’9) l22X10/4ce =o.x ZxZ.32OJ= 14Q))(H-) (H-)-\ •/ \ •1 Poi -40O 1rI IAppendix B. ACI Code and CRSI Handbook Predictions 222o—u 4) (3S- z?02.7’2.____________________=25(o7SJ-i-i7z4-oo2I= lct 2V,’ 25.4L4&)—-uo-- -L): C/A)-4,—W- ct2XCZ-) i/ =O2 eoV — —i-- &&i i —i- x 37S 333g (-ji)- O25= V;c&) .-3o_vjcod(L2): /A)fmioi41LI): 9i1\(2).Ico-Q = ,cZOt 2.CF) L21 2O (27(b) =42N)A4”, co4L 12S)‘i3,c.95D 2Mp$( —Ofl€—”3Oj Shec) ( /A)onL2): ic,Q 2V./ UJ)ot€-wcj 7 ‘/c1.c>) O1é=v,= 25cI7w? - xOO , -— )T O-37L> e.4’3)c4oI25—V= 7’ ki4), ZIL - - 49t— a344.&iJ)Gflj h€L1-c(#) v=5(57sJ- ,4x°”L’ _L)r4tr 3,Ye IDI (kiJ), c24 3cn’2(kI) =±cio—vJaj SiecuU) (t’/A)*ecu) V 575Jgoox4ozvG = 5I24(R)ax(3) (7A)_ ___)J= oS .32,z1SL-kJJ), x33 xo2=ib% tkiJ) Jce474-LkI)Appendix B. ACI Code and CHSI Handbook Predictions5EZ—3223- 4E3 A12, c—-)=-si& i-41 PL7b0 ?-o.O77 / I‘779 97b(4.34— 4 3)23O /.4i’..20=32ZI3k-1p, t4ae2Lpkp= n,tLJ)-L43 ‘.12-—.4c t ) — . a3233(5+—— 25A r14’I077O(44_--)= 44 3 I-Iit4 B”+= DT’& k=- 427.=z z?63Q(4Z7S— )=8Q7 /4-, ZPJ$ 9o’35 èZp, 47L2k1I hd)-fD, A49’737i6’j =4Z75ItL, -P-= 2’o 1’“‘7i i243 (+.27V— ib -;. . 2iJJ = =i.i -lJpO2Z-k L-k)-2t’’ ?) As — o.33 &1, 4 joi- çpt,M °.334S J) 24’2 47 H-=z&’s- A.=°i’71, jOCO’ ?= J 4.lz’k = z3 L&)/ \-tAppendix B. ACI Code and CRSI Handbook Predictions 224kvf&) : /A).ii V,4.3T&= .c, p= 3.IiZ/ 4375 )V=(I .j I 35 13x47S344.c i1.—L)Yd/H=439445=l’7) f= or1r7/( 4.E;y: ‘.‘I412 1’. <3. J ,x434-= 3’-7 ik.3t Vc) Jc0Q 4413V/M=43$/2=I.74Iy=i_ -j- ao,7xi.74)3 (‘7Ot, fl. - 3.5S 374 IzL’v, L1I’Jr1.= i4&’()$4.: ‘4/= 43q4.7 4/3?(4)a.O2.5‘=--7’)34i4 793B j x34-34c=I1L I.9 7IL4)sc /K4’1 ‘= :142., I’ 3.5J J5MJ=V 4La= ‘ftJ)= = 737/xi..zY °1OZZ iL 35J’?( I3X47’ I6zk= 4, ILC.Q 44Q zock =5:!_ /o,x4,z) =q34 I. ‘2t4k= \4, 4 LQ47 4 = 5(-kN)‘Y= .3c-) p= )(42S) OOiZV= +z- oiz.) 3?c462 93 3i I2’4a-•= =Appendix B. ACT Code and CRSI Handbook Predictions 2253flij S4€OA”3):4375/. = 3, (3, — = 2.9V 25(9j, +2ODOZI 3YT) ia35= zo9Sfl,. I3xb3875= l,= ‘4,, = 4;J_t= ()= t4 c’kt)=4c/’,25=352) (3 25/) = 292c,36 ii,. ‘- 34sq4 Z3o4 ‘.2Q \/, Q 14tr)= 1Jg()Vd/ 43/ = (.3,5—278 ‘25’25(ç÷2v ko)97x34)13 (145 2a344 iI, - ( I3L13b5 2Z931 li’.2J V, UCG.9 444r140.’7 r43q4-5,/25 (352/) Z.7’T > 25V. =25( 1ij4s-k 2.)3430’3 2i/-1 b. . x3k.3’?+523ag ‘.1= Vc 4k=4z.S4v’ 1cbJ)4’3T5/j ,s_2s/3 ‘2 S2(rr7 -7 x3)(4Z= ‘2572.1 I1.2-(4tVc 2S2Ll , 4$L? 5I.-‘r (J),ss. vd/ 47/,zs=.2 2.5/.) = 2.y “zS’\4=2.c j+2vX3.Z 7)(34z1I5= c74r( ‘ G .f 34275 Z5’Z( ,2=V7’272) ( .1c49 51.4 1c. ‘r2. (i)S92: J/.j= 4.b t3c’25/)=2,72,SV xE5’c3.7) 3x4625 z’z2- i. ‘‘J (3 Ic’ 35 , = 4--‘ L4____v/= 4bzc%2,.=”3713Z5’7 I. > J(4i2= 3’5 17.2i’ V j )1cQ’ I1-(k)Appendix B. ACI Code and CRSI Handbook Predictions 226Seoc4);1J/ = 2S,4375 o57 i.° (—2’7)= 2.OSv.= -x2o9ic.78Z4 2.3.b1r, 4t=SSZ- )= = Z.O 0c 3414=ZWk=\4,j3 = 25/,7 .i . 3—l,7) 2.Dg‘/e’ 2pS77oo= 2) 4-f’ < e.2k:rJ= 44L= c2*fl25O(ktI)5’- 1J/4 z.,/4314 517 LO, (- 2s)-=2DSy,=- 2p3,r1c‘21.O Lf’ t-4-=L-.5o5=- z.O\/ _L2*ae 333z <oJ 142’(5 42. &?vs., i*t4L L4?’ 2’bLi4)ç<- ‘L= 3-’ <tGjj’4542..9 &,‘IL)= =e1E4 O) 2.4)2.I5T134- =tt=4=&’r=21-k’J)= 25/4= --G (,5—25O5k=Vc30.’74f?Appendix B. ACI Code and CRSI Handbook Predictions 227two—QvJ e.cct)SSI. r( 3+ 3S7) 23.Z , V 4j4.4 23.2 37=’ 27.4-kiit6= V =z7.4”=—zai., Vc=4 23.2s-+3q4 7544=i(43) Z3J .,IJcQVt 2l(&)J)SS+: 11(- 4s4.5)= 232 1*.. \4. 4j.i 7-Z3,2t 4s94-V — -ES- 22. ‘kr=<.s- b(34275)= ‘ •J ‘4= 3S(2 - S-i12) 240 jr., vtoX=• V’22Jt? o)a-JJ hc(a) (‘/ > o o+ -a-bt)—ie—WCj e%L):. (4.•375) = t) V4 43gt(5E—SS2: ba +149)t°Ij V 4i3 434S 4z.:)442L’= 4-V -I’k.? 2lJ)--(3- 4.’5)O.t V =4J (-kzp)=-J= 4V= *-‘P 2Z6 ke-)S2 2Z ‘kt)S-S; =s (3-4275)i, V4i o?27=4 ()4=4V 5&’A-’ =sikSG2SGi3: {3t4.625)= W°J y 4J t.ox4i o.4’k=— 4-’/ — =Appendix B. ACI Code and CRSI Handbook Predictions 2284-k ()S)S2= 3= -:WGQL-kT)= 242(4W)= 7i.(-ki)= 313 (-kiJ)cGrz= SS:t3S24= = I3 c-ki)I6O0j630d©0A:eir1g4-itzscs2=Ss=-1-5364r-Ss5 SS,:57J,(k)=Sz=2__fop (:&p)4)J.pLL= 4-r) ‘-‘ c&)i€wk:$44nn’,-L43-W. 47L1 141’o.H=9ox4J7(44o—I 234D)ki —-‘11-’.S) 7%.3 = z4.L1oJ))_ok a=o)z4r4r9r()= = 3J,. (.çJ3)41I)= ‘-‘/b3 21kF))iof__ 2(222561i)ne-W’- k€O.((-I)576 ÷(—. )Ciz4) I2-O ‘‘7ojOPOZ, —-= 25‘4 = (I51iJj -t-i724-x25) 2’4V4-S7 (*s)Tt-r tiu-*S)\4= 417c_ = fl3](-kN)1coZ 2( I3-1-4-) 324(-kJ)Appendix B. ACI Code and CHST Handbook Predictions 229o€-JjSotb’ 234Drrn)-) p -V(o.i7-H7z+ )Z3&V4O =9 Z’o44c o3C4}-1)heorC3) cj tp: r1,-crr’ -1-f e4)_______‘/= 4o_j- —Wcane—Joj $ke-):i@t= .3 -- —j) p—=(_34)52‘°‘ 2Z€-k-)Q4L3j 2344O2p c4)t4ØI -dXO1 eA1)‘4= z)20 =C-kN)Lt- 2qO- zI2O) boE(4eccL4-)ei:iSYo vZ-c=ca-)= l,3(-&)-z = z f(1ox+)4-*- 22 ---o44oj=V= °-33J* I31& 2J4L-kI)- =4 c> L,t-o), 3LJ)z6O3q6= . a. 5- -. - cykAppendix B. ACT Code and CHSI Handbook Predictions 230+_ WOj &FQ3);= 2(4W(I-)--= is= 33 24-?q2J33 ‘j- = 37 ji)24-7co-t LZ: L-ft -fr-#- -c- 2(3Z)j b2vt&1)dt-. 3Ornfl, zP=— ) = 22k U&-n., ,22q’ 7 (L)A=’l2, ---4’j ‘W1Z )(49 41i 1)- 6&22.C49j (4o45—z4S‘2 7÷6) Z7’o(k)3 --= 3bN0h-R((2): Ot2/3fOP v7(W) Q2O7J 23O?(3’TOj33Z (kJ)ti\—— zL)-Y v24L+-)-)“-‘Q): iot -(—:)vz1.= I.zTzmnl /az2i445)°r7-MtV=(ai7g ,LI’N) 737(k3)_______________çl.ot .irect Ph2oz;I) °‘= l/=33, C3.W—2,3S) ==2.(3 7’j - 7z4xi3 )23t3o=qi )o.4*3. sq&=z244(kJ.)1-eon;oe -c c4-) sI,ot €tü: =3373°i1) I 35 2. OT7)\/I4-t)’Appendix B. AOl Code and CRSI Handbook Prediciions 231Jo—J0’JV03322ç[ A B171ss9 kJ) 4cQY = kiJ3G—W l’et,%c7): (LJ/== > a.3)*i)o t4L hfs AArec1’ti b0 o--r(oo÷)=15Oc4 =2(4J 2(3)7+(kJ)on rectL-V 32ZJ2 x 2()= 2ll4&)jc& f.1Z =C:2600_______________I t f—.)o 0 0 ii- = o4-o NCQZ)(l0*&Z004fi5 tkiJ)crjQ 0Jk(b=z-4tjJ 4cZ 4oo CjJ)0—AcIjaom’n =°‘ =oooZj,ML4.‘Ic (P7iV7 4-)’7ci4o=4j-2(iii)= IVL =—I24 -Appendix B. AOl Code and CELSI Handbook Predictions 232oe_w&çcket1: Z6°)iV/ç 3Q/0,3V (O7 JoN)° ZOc’i3=3(-i4)=—tDAJ cJe&c (3):sko — 2omn , ?04_ /‘ =V = 2.(O.1578J 41’z4xo.,34-x Z. )2&t)(39O== zC’ LkJ)c’L iV 43k)Ofl€—j A&f)çhrr az: j .a L35_25,r77)V=.)(qIa= )7 L4jJ) <,Ic..f= vV394-f0J)-W SoW): bA z ft3’?o o)4+C4--3OV)CJ = oooo ht2\4=c22jxqq(kI4) jq&ktko—oj c11e(2):= nc GCøi4)Ol’J0J heic3:— OO--2lOO?(4O—co-r1i€r.:O = fleo-2l =&1,m’*., br(2b)4n2.303V,= q3 2j74?.37 = 127c-bi) J Ir..LL), 4)o7—- = 2.o).20i2+ — 2SV)7L4Qt)c2) m’t Vc 3322] L3< W71 Lki3) ‘l€ivce-4,1j+lZx3O+2W0x4 3b.4Y)ftt ,=--b.4-)= ;g73•w1.l2OD-i2’= (-- 4-)c)y3fl7fli< 17L mn1 v=43zJz7 ,74 i,9a)& o--2’3om, = 3omM-f,= 33.4&) t,i?o.4vc4. )=43.z,o(&4-m) cu)= 43)44(3)33(¼)2i+1Appendix B. ACI Code and CRSI Handbook Predictions 233Ofle—VJJ *e&r () 4ret.,v’: 042q7440)&34o/ 1’74\4=(oi 78!+L7 Fx0o3xI 74 2s7Jc1Oo220‘Jot ‘Ic — 4’tRs) 3 I20 L4()’)-i-;) 37r704p)ot reetoi b = 23O=--= = Q73S07q374oo27 )z3iG3-33z(bJ) o.z9or1J $23.)(3B0 1-J)Lo oAre-$ V = 3Z =tcQ 2134 + 33t) =Ofl—1XAJ ,IqeL-rL3) 23[.I, P .Y ==ZS3_________________I’ M j-o2.>2.SV=Zi57J -z4 27 3) z3I I $3o=zI72L q4?33J9L23IlX3&,=zd&)ree4-vt V= 2JZ ---‘ O2(&)J)22- OWt zCtc3Z+2(Z)=’heC) *rt rt.-: i:7 LS—2.%,1)=l.3V .27s S3Z—L-kM)—4)1017Lc. cret-Dv’. v= i7 74d,1kII)czt C’( 744) 47)-+?Jo—1.d €QvTt)) !bi = iC o%v my2\/o.3Z2j3o3.c1o7i,p bC) )4L= V.(1)/ A%so.71>0s-j 0I24—hLG A30&AJ 1L3): Shor& rt; Z)+60 ÷r1)oV 322..j3G? I2’4 c3S0 = LQ32?(O3-+ ZI0€ — 3107 LkiJ) co-h-Q22054oo 5&pn9 Ar€1Zc’i-.; b= 2X4Ot\/o32ZJ *1Z32J s.44’c eI (4L1)210Z’!(1 r )J9tk)101Appendix B. ACI Code and CRSI Handbook Predictions 234S1JC Short ritt;on: 33on’n 4?z4oo’43-i- o,c4’7q434#.i (411 — 4ML&4’9, k41,!,tar, 3O (Y’Y’t____________=‘______________-53X’S— I 9i”O, J 99= is177 1J)t=2-L2I37-N97)= 74zSL-ki.i)-frl€—WC i1eoc.4-u). an AIr€t..: 2i 57L,t f(4o?ci2l)VAd/14e7A/21Vc(0.’78iTh --.24.xooIeo ) 2.! -4’7.1 =kR7L&lJ)2qoJ4 x/238. b4o79 = t3Q43ritot Y=7 --=f1t)— 4+7S&ki))$hort O1I(€et Zøt = ‘?73v(o.Is7JI 1724 o34._L)2z,,(4i /P&&JJ) <Q29O7J .234’x4-J I )o (*MJLon3 oreet --= 474i)2_= 2(Jo7)a7Ji)O—WO *€cLF(3); sret 1?=23Iø.)mn, ‘ 1i4C3’94)22‘4 Z5(°.7kW +7.z 274)2316i 4ll344-(*jrtEoY‘4= 2.o3- =).JZ(J2Jo7±ZS3) l’fo (k)1)oe—J Iheut4): horétw. /444t/ i.37N4 = 1.37 247(&l) <lpnecto- VZ473’flbZ1)= 2C2.4-37 -r iloz.) = 9070 L4ü)*0—’j 4eocc 2 [ 441)4 300 -4o7q)4J =\4=33Zl4i.I Xl llTZS7 = c0L= V = 24-&)±0AIC’j JLZ). (-7I = 3/ 7S).o5 tpt)Appendix B. ACI Code and CHSI Handbook Predictions 235+Ie1S+(:reeefl: Aieomn 3BOl1Wt,-=44-1?24)377z(-kil-4v1)17 3a.3 x4L=A5=4emi4gooSb_Mr44,(4—1s7—(UJ-+)1\iL) = b’!.%5 = II53(h)t’4caQ ZCI 1153) =Gn-VJQLj gLl)j4ireci’ b=4”’, /(40,X)=q33- .74.I4’ 3OV(ol7S 23 &iJ-> oo4CO 25& i)NY2&L)) 1.J 22S&(1+ort ree+t4=-= i.szl.4A V‘4=(l7g.÷lz4 jxt.2)4ot)L3S2.k3Lk)J)Vt = 195 I ckw)I’icZ 2L--95)’ 57zLk)*(AJO_j 1..c-(3) 2O4.1’o•-t redt: 2A- tr-iT3- 9.w’= L3(d7&k)Ne 2(I37(1-1 )=qc€i) 35-QlJ3rL ,)rea b0= z4b+ir(w Z I G ‘I23t,2v)%MVt = 3322J I23.2.L4o7. = IO7+(,JJ)= 2o94(. I ) = 69e c-?J)I 6O___ColE4OOAppendix B. ACI Code and CRSI Handbook Predictions 236one—ij& 4hecA (3) odrL; = 4oc’mn,—23,C3_23> )25\425(o.I573ji-l- 24 253)4,Bo....L4)re: v=’tlcoL= q54)2-tk)SOr-tE recfii d/w3°/, S2)jt I9 — L)J) < C-kF4)1n4r’ect,1:zcs) I3C-kt)be.&r1 4 jq4-AL = o52 X 2Sx 23T+kJ)&.-1ct= O.)L2Xi7.! ]c.LE! (bi)beoc c-fr-ik ez)A: =NL4M32’ =:= 4,324 zC ()C,: tO,ZIz20 l4457C-ki)Ncd = i44J7 = £S2 L4ii)qSz3X2ottIui)‘J0L 4c b47Z ck)4kilL1c0= 4)(25J??.L= 53xvisek4-4-72kJ)Appendix CPredictions from Proposed’ Design Method (1)1. etZ x.m?i€I ° I t I ,= 4O.2 c, 4t, 3/2i E1 I =I I I I II I ml I I I the eJt & -rt- ‘netL_1 4- I I L__J I-i=.—tr-r’_____________________________tL__I__________0(2_________e=c(ç7)=3a2j oi(Q2=‘=— e—c=600—J’7.+ cf= 4o3S’vt2237Appendix C. Predictions from Proposed Design Method (1) 238-tue eii ‘1t cj -tt c esi?e. +mct okZ ?- i.34 Ø30= 30.27—5347 rA -—---.77--r Qf -*k co 6€ s-h-d cd c0lrrn’L 0n= ‘2?( Z. COSS.156h 53.b7 40 4-r3 -.= 22O±kQ. W€r9€ VOL ‘h — Z.I3’-+ T7r(LT — 2.l.77=—n\_ OJ9fl.9 i-u033(J43_l) OCI.ii--—t)=Lo44(S = Q.3L137— )=o ccl x1rt. 2elL’eFb’ 0kf(I = i-r 20 0.32) o.af(j€rw,’t)____CoS°.__ ________-c-‘_ ±._____— 2 cs4b.3° CS3’1 -= o.s4f 520A)2 caci toa. Qtve-co-t e-.d- of L’eI. -frut fm= C0s46.S9 I75 ? 35 = — 333 x440= ij Lt)o-%zL cao- o IicLn- -coc4- ._.2 — C) = ct)Appendix C. Predictions from Proposed Design Method (1) 239coL -LF V 2q7,Lt)- 2L1 -kt)2.kkt)= Scmt. c 4S,A ZCW2, 23/22,4-L,cmt, 1’2J°53.2.CMt b /= .7S,/=i92 (j4vt43.o47. =o.z bzffo.7fFv S4Stt) , F ii.aut) <-—— lb.L’U L,W3(1)1l11: C= Z7.tt) ‘9ZI(&)____!I=.3efrt, A= 25ck,2.3’74 q=24.33, o<=3°, f—iZ.5°b 32,OeM /=2.o4, (J)—’.*3= (J)1t.i4o= 1S3S3, b °ft, f O.iYfc’F= ‘3C)ia= ‘z.F=rl,ft, ?4O.2lC 47/z,4.2J .O’, I 23.9i, 23.74= = 33.7GeAt, h/= 97 .b2:—= ZIML, 4t/,=z.3. (f) =(=L,lz-3, c=oo47, f3c-E= Zr3.+Ct), F lqo,ct) =v i4—z,.%tt) 4k}.{= CAt, 4zcmt, 227/1, 3327 1r/.I7.T. ‘L b3.,4= 1=3.O2t7i,t.)j1fSAppendix C. Predictions from Proposed Design Method (1) 2402j4, J-=3 tx=oc47, t°-7F I,1t) F= 4I.oL1)<= 37.ott)= 3S3(-bi)• f47c/t L.ZCA?t 9= l5’ x— = —.= 7b.+e- , =S2.19cA’., /b’23, (J.)=4.3)yzLs J=ts / =o.o47 =o42,f =51fF =1-7t.1() E == 2f = = 37 CO)311-f emt, 4+7em., A-= - =z7 /c4.t2, .= I .4’2t, Q = 37.9 — 4°, 5.S O(- = = 4 .1=°•- b,z3.3ct, -k/E,=24 (J)=.7=j.9lct, (i*)22k/8) J4=ZP (‘O3,3 ob f’O.22fc’Fv loq.o’± i4Lt)= )t4h-’7ai4’JLt2‘icQ 3F7• 7’)____)1= 5’c.’t, c4I,I ott, 4C.øt?, 443 /i - i-’O- 9.imt, &= 7.1 33c,0 qi° ci 19.r, q= 3V°k=,bjøt, b=iz.1em , ,2=scA-i (J)=22/=7°, !-=.o f3ffv= 3.t) 1’?3Lt) 4.2(44= 794Lt)ISO(t =4ss, =zS 4s/=--O = =i7 )-CdQ ‘= =Appendix C. Predictions from Proposed Design Method (1) 241____=0c_1) 3h,cM,OI3e, =4’M°, c=,3a°, 7l0, zZ71’, l3.o= Z3 , h/b 3.b,l7I= z7.’, h/=.zc,h/b = 330, = Z,a2 O3 Q7p f2= =,=Z0.7Lt)3F,= 7,n c&= L4ce, Ac= Z7b f;7z.’/cS, O( 7.IJ CI 33.G, X i’7.7h=/e-’, b=43M h/,=L5= 1 (j3h/, ==a.71 Oo3—F o7b-f2’ f= fPj33LL)= 4Fv iZ.41t) o25kFi)Ac=*.ie 4S/1= &= 41S6, X 32’ a2S, c’22.7h=l7e4 =4j,et , ,i23, c])3h/tvJ4,r1 =-? , = =L79h, f=o4= I37 = 2-Lt) ‘- i54()kl. = 3cg kt*)4z; 7cøi, cL= •qe+tt, p= fL=37 S/CL=23.C41&, O(ZO.0°, O(-334° , ==i75k= 14jmt, ‘/b7b h/b LM-,J I4i çSfI ‘L+. 3 Ct) = 4o Ct) I-u, 54a L)4P = 4-7.2it)= 47’k1’))Appendix C. Predictions from Proposed Design Method (1) 2424: 9.oew. 34c-0.. =— 3L,.S, = — Of’- 4-3.4,_ko4r?4.t, b=4z4cwi K%=t52 (J)iiS3c’$t. h/= 3) (j)= 3o/b=L.44- J*=2.1 ) j=Oi4-h f=o.7I. ..= Fh=.Lt’-L= b.bLt)JeZ 4Fv 4%.0C±)4LAC-ktJ)43: let, 925O’).-, 43CAI 2,7cJ%q’-1.34-/c4&= 3.3° o(=’•9° LX’1°,(J). Lrpb= ‘• /b= 3.°4, (j)=3.,X=ki’7, =o33 ,-,=l.o2$L’, =1ft— I41) Fk J4.9Ct) (4()i’4P iLi)iO4L-kiJ)IOOCA) cA 14$1, As’ 2727, -4bZ& 0, +20 7.7x4b2oL)F k4-s$ lz9.c-tz=I7gtt)I1c- 4F 7iUz) 7aL-kJ)4k j=IOOeM., cJ.I.9e4fl1 47:3ekM2-,=ji’ ,=i7.9°, q=.z,.O° oizh4°, 42°‘‘ 9334cmt. = 4g.57ct, j-)1= i7I= -oent. (J)0.’3,..Q1lfL’-‘=o,aFPk= .°)= =l3tt)4F, b361t)704)=2flt As toe.4&1, /c/tj ob2 4=2o.7 22°k -A-tcn, k/= ()=%=2:17,Appendix C. Predictions from Proposed Design Method (1) 24324-3 (j-)= 2, =A13 °IJF= I&4-Lt), f= 5O.7Lt) <Lt)kQ 4Fv=Ct)z fflQ4d’J)121fl., 2wt. , 4!DJ&I.32i. 44ø, O(2-I.°, O&!Ll°1Zj°b°1/,= ro, (}&)c=O3L’7 =-‘=o[Llfc’= 25Lk) , FNL = Ioo L41’) = 4t7 (-k1)= I ., 4I?Q= O2ifrt, = 37.0° , , = q’= 33,’2.°, o=3S C? 41-, b= cj)=i,k/ qi,O=o257, (=.o-=‘fL(-r) F 37.3(p) ‘ 37.7‘21 = t.(p)= zt,(kii)Al &A4: = 4wt = ,k 4iv 4WMi’tc, -f c9‘ )4-it, Mis, O. 2r,k, 4 22., = t.c’, Q=4ic2m%, b1=2o,7rAm) h/03),, oL=o)7,4-, 44f:4R1 =z&€i)A2 & A = 4mM, 4o A= - = 4-to t4?’ -cL’ = 27. M= 7- e=.’ o(= 2.&°, I7.’) O(29o, L1j0=449,5bwt /=2t15 .z mn’t, h/1, 21), -=,3’L =764-, -=t.07f ,Appendix C. Predictions from Proposed Design Method (1) 244p = = 227, ((r)) ‘1 2V77(J)ls1= 4F— SckiJ))—4ionia =4’viflt 4Io4P,O=-477, X1’2° i97 1’32.k = 45 , = z , )=‘‘‘2o 39 nI P11 1/b2r (j=h4=319 )J.E=33 +‘!1U, t,zb- ,P= ‘24-iJ) F=.7Lk) -=D-.3L’kI’J)coL 4Fv TfZ L)l1=45or0Mi OmY0t, 1fm, -4o?-’W, f24’fr1L,t O4°, I?i3Oh 44mj b= Za4Sr11t, , J-21=t7—•?z= L,Qra1fl) h/zy, (J4,j-=3.32. X0’b4 4c4,-Ii,= c.36o-jFv -j) Fk= 7,z()27.7(k)4F qit4ci”)H=4°, 4oo1W1, A1T 21GM?c.= 9 (,vnt, 44-9° o = i-L°, q = ia’, i34l,h44.Z0t =2I.Mt’ h/k= 2,fl , (J)=k/1,=z7, )= 4’k/= 42 3Z, = 0.744, 49 -= io4f’, - = o3’7f= zVC4&) i-= -F = ‘o’(-kIJ)A7 450 rrtt1 d=4ooit, As -4io-1P, .L Z4’ç2MPOVLW(, U 430 c?—l97 (4° 4S-h= 443.Un, ,= tilo.44flt ., h/2.L} Lj)1=h(j)=4o=S- -R’=sf’Fv= 27 (L), F=7 )2277(kO1 s]coL4F/ 8(&tJ)Appendix C. Predictions from Proposed Design Method (1) 245*=4oiwl, d=4omo A42 c=41oHP, f2SMPa= ot, cp= az-’ 32., c=12k — 446P = ) .3 , = (j)l - ck k/= Z.7’(,, (7&+, 49 , -i.øE, FO3F= , F= 227.7LeJJ) 4F cc4J)H=.4om1rt a= 4cxD’niifl, A5n. ‘4-io HF, -=1ZHFci= zon’w, =4o.°, -•‘k 4OQfl1fl1 I2= 24P h/= t9q) (J)1=7c=$32..wm, /= , 4o=ou)cf, f=of =o46f’1e 4kJ) , F 2.3Lk)277&k))), k(1&)Aii )(=4omnt, c=4o-mit, f=-myvj, =4-i&-P -yc:= 4i1€O=4• O(,=2b, 4z3.I0, o=S1-.8’, q?2=—t°h = 4o8.o1ii b (j) =z43tn1M) ‘/,=I6S, (J)=4-h/b = !, = 3.3Z, O( 764 = O27 ?.R1= lVlLkrJ) )kiL 4Pv =4.qw1 = jJ)At2: 4omi11 =4niw Ach’f-&7fln) 2&°, 13J° 37.7°k= 44.4jPint = II3I, k,1,= .u ,7€,.511%t, /=2z, (f)?4-oh/33332 c=.7= zzC4iJ) , F4, =2cL?,L-ct)) j- zzY].7(-o)iL4Fv4-Z2t’ gc-ki)H=4omm, =4oo,rn1.,1= 2.CO1M-, Ø c%1=4%, -5°1:= 39.7Sn1fl b1= vzjo’ 1/,=Appendix C. Predictions from Proposed Design Method (1) 246‘2nt )a3.7c/bi37 , +°9°-, =33c:F,v V52L-kIJ) , = 78 8-Io)jt= 4=iooS(kJ)-(=4r3fl1M, t.=4ooYnnt1 4ç= 3331 3b3pU’= ‘Z P° i-L= A- 333.2 440 = 3k= 4J=.02.S (4I)£SI j= d=47t. A 4 ‘r3°? 43’Y-cu=o , AoPS4a43=.It2)P= Si)ct4t.2C4)• s 45 4., GOI f 97b7 $€a=o l-4tjDOt)(9,tJW1 = .31-kp)R’-LO= t&4I.g°= 47t2)-.kF=’ .4d1)=4ck)‘r1=.in, 436iv1., AOP7OS4) Z3OD1S, f’43o= ‘ Q35.S°, c=47Se, q= 67°. c°h= 4, . , h/ i.’v ,,(J)2=°.t-z, = , x = 0764., (3 o39 , 39 4-fL:) fc-z4fIS— b-?) , 9r)t’i= F’)= ckS4: -= t.. c As°9 ro99?, L:=436F_4,9D q,= c= 32°# t., I = 4.9 t, /= 3;- 147b (j4I=3 OO76+, -= a4 ,f’ow-F=Lk),Appendix C. Predictions from Proposed Design Method (1) 247____1t=, c&= A-.= O2lI 4-= 72O p-1,k= z.i4-’t., 3 = 331a = 4’T.°, 4= 7.9’. c. 0D f== =,h/b, t.- .h/=2 ()=+o/= L7 j=i c=o.7M-, 3’’ 2-8, -f=ciS4-f(’, ,“cz4fjR4-Pg = 33.& (ktp)’ (-ktJ)i-= = 4Z75 A=o,o13 2 9z43cB= 331 = 4o15lat., b1= Cj-)=irjk/.21 (J=4o1>_ i.73 3.31, O(=O.7LA-, o.S+ f=4-’=, F zSkj’) cki = 4Pj L&p) 14-9 C4J)-=k=4-snt.1 = j sz -rLQ.3’Z6 43=?z.5.2-°, q=i, I71 =—2.°1’ 393Ib= -71 fl•) h/,=/= I.75, =332, cc=o.7”r1 (3= 437 , -=o.S3- -= o.-g+’F 4(kF) = Yt-kp) < r L-&)JcJL 4P’-° 1Ick)J)= 71—t’ 44ovr&, Ont2 =471P =zr0.a.= 0, •= = Asfj = qo 47q == 4-3,.I4’1.6° a7c4vJ3JcoI=1tZct39Qr, 4’i9-,Appendix C. Predictions from Proposed Design Method (1) 248o.= a , — vz,vx47’F 57=7(ii.J)1=zq7 )= IQi’J);ht iret-o;-)= pii, ct= f: 4r7 f7 frP. -(,i°, A’c 2o41 5’4L?&)= ko-.t& =.$q7(I-i...hoi.tr a€Z-H=Ynnt ô3On1M-. Z1O)1VJ-1=H?R, fo.siPoa=iO’’, O=S7.2°i i7O) 11ab 3I2.O h/1h/.. cJ)=3S7h/bh92..J r3i4i,H4t42J1) 1 h io4+1kJ) j4= 2xTq4( —i- Z3’Lo Ck,)Skoit dreetvi-f=Ut’irnn1 -1riWt 14?,M?.j=4Zk°, c=41.T°, qg0 c4°, %= •h=4S.G4n11vt, b,zb.3rn1., ,z.nwt,1/=L4-I, (J)=3.s7,j=3 P O3i - ,-.°4fL’F1o(&t)<sc7(’kJ)± -)=&t)t=rIYtn, k=9Oflltt A Z4’Z 3-j?&—)fl1 O• —25,p)2 ,Appendix C. Predictions from Proposed Design Method (1) 249J4 (Q242, 7-PL’ c= 0tL&) Fk=WL vb44(i-- 2O) 1774 4eJ-L.)Appendix DPredictions from Proposed Design Method (2)1 Freet_____35L_1____t2tflA&’1C4t’, -= 1.’s /tf44o/c fr49.5t.1t“ ffL •‘;.4-&=-:;-=coV= zAPj±o4= 2 .4O.Zl (44 -=34)=‘ ka.r fr’+k”C.oLan: c=33(-L)047-o4o-tiqo 0b03J4443=275.31iLRL%) 275.3 )C 3i = 337.2 Lt) = 3337 L-)= o.047 33(—)o.(S7o4 x4f-ij-+ x o37= 5/c_i.2= 26.6Ofl.-4)0j,4ieo..fi) )4-33 () AiieJx .IeLt3= isf33(f&)‘a.I irr.j 600250Appendix D. Predictions from Proposed Design Method (2) 251=223/,== z4.Z52323. .Z /‘7tt) 2I2ki)A5=-4o.2its, = 4gg/- =4 ‘=40P- 2A-ci’8=i’.,3t7Lt)c ()2’Z A4O2Ie43, 473s/ctl, A1, 4.25&, d= .4tiM - 5L.2CA4= 2 z7 43 L±) = 43 kIJ)____A 4køtj0= zs41 7Lt) =4 the.- I c4e)= 4+ =c.c. =441c), of vttkMk Crcl—= o.Sciiic-e- t==7sr3/eA.,t.L 447ewt, +68jT— 32±) 325 (-k14)N;= x445 - 4”(*)= 1- €-_A=43 27/4, =7O2, J ,= z.S39 59 Ct) oO (ki;! A= 4.ld tj =6wcaL” 34o4-3L,Lt) Hlj= 7”’ /‘- - = Wf.+coL 4At47’ t:;= o(jJAppendix D. Predictions from Proposed Design Method (2) 252=3y-, k-ic2. 43.t7Z3 x 4Lt) 3(’k)4bZS: 090c41 Lh 7’42 ,4t) =43t =43.o3, =a7oS/21‘2S,KL= — -) Lr;2c.g iJ))•‘‘4M4: A47c Z91/44F732j4—-j !7c±)= ri4i4Lk1J)/‘ntj -tct= 4c3 =f=co)2I.2’7No.= 7 C-)A1’P4: 4iofr1? , -beat = 1o3aL4)A2AS: As=t’’i fj 4toMVt ,=4cDri =33’,toSi(k)=° Mo 4X ni, piiI =A4=A7=i=MO AW’AZ:=AA1;h1V, j=(2oo-)JZi2.lmfk.444o—7J)____A=Appendix D. Predictions from Proposed Design Method (2) 253= : = T2 ?- ‘ - i,= 4. og444- 7.. 72430 2! .S 14’) = ‘ 7 (&J)ft’, 9=43Q4L4=)23 1’2) l44Cd-)A— Th 4.4t 4 , 4A-1t= ?‘-°‘T s90’77o j.!t’) kJ)____= o.2fl , =7zr 4i4-Lo2flx92os.-[ 3.1(r)=3t)A33-?, 4c5.)39’3)L9ZOj Yaç J 42il-, 141i4.!iz-MoL 4?cz2 ,6c’ocD(sqS: Ao.-’ -L4l4,4o54-,o’eD 4cfl 122Lr)E4t.k1J)Ot3ThJ .ç= 41 fri?, 6{=44o’, _37m --440 5342. X9G) ‘ 49 - = !O ZL4) l4r.(L) !012..,L = 433 tl€ii)c=1O1I-rk33I4i-C)J395mM, ‘=----2acz Ic12GOThM.., jOT1”, h=37m’) =_fl 3=-. 32 ck1)Appendix D. Predictions from Proposed Design Method (2) 254di32O=OYt) ZA14°f 31]r5 234C&, ZrlJL) = 23C.4 LOO1JJ)=eCtO/=3,vrrn, 242441P 4-ltii 3flmrt’)330 = t’! LkI’), zJ1Jtt— 33DJ S3A=Z4’, 3,-I?€&, 375P.) ±0AL2OL41, 4LkJJ), 2,L) 234I’ic0L= 24i i)OS LJ)seJw ca____=23/i,CO1fl: , )=°“IJ_ 33(iY0P47 ,-f= o6x25S t DO4 O3j CILI) ki’1)&2.355 Pft47 J23 N,nz) 35O.9tY344otk)ane—ij he-orLI) NCOU3*lo3COi)=MoJ/nt, 45At, (37= oi37-S,1 =obx44 -\3 04 .bO3o •2’7:3 f/ 37.2;o7(kt)- =o,Arc - oi37 4-o 2Jb. WZL)= z. 3x2 31t)4)n—w&j netC()gt= ef= 1433(&N)Appendix D. Predictions from Proposed Design Method (2) 255L= 27g d=o33(--_-1)= o /,,;-)c6c3oT, =33(----l)o3,;— 31. 5=222.7L) 21 “)_ = Z7I X5= 429 (;t)= (ê)nei22 - = 4O3—I’7’7 =a.b4*o C4f’ j+5=.7o=o”47, —l) 323, 323j z71z i/oc) = 3z ct)o1o ziZ7. 4,t)7oeki)O)1— JAj sh4) nt i’ti irt f ,2J 3I5Zj- L-&J4)= 3a7 . 8.4eØ=bO47) ‘Q33(- t)>i.ô O327 212.4 /‘-ct—.O47 (-n.— 1)=o.i3, i-3 k,(W =295N c> 22.4 3’ zo,2.L) 22(&J), oi.°ti) = 4ii (k))Ofl-JOj ShQRC(i) ( ptLCLtbL) , N.t 1.L= z52 Ckt)4t7 k/=p4’7, H)11, q /2cottI); N7)= 2ZI.C Lt) 7oi4L.)ofl€-M he’O o- apt,b)f’= 45 l/ 44]e, C Zoitc Qf .cLftJ-E Iun=o.t)33(Ii)O’2cl, o4S-1- P’ 4!b2Ot o.33(-- —l).2) =33(4_4)0Q4.3 1’ )(OX4O43T-. flr?f Crt.& 3’1.541)3.,x4-.7=3.5Lt) = 22(kJ)) N(L2) = 3xO€—l-4 kc(i): C.L(3)O(ki’J), —‘J SJ’cLt)- NcL4)72Z(klJ)HC I.o1&)i)Appendix D. Predictions from Proposed Design Method (2) 256_____L= c4I, 33(— ) O3O 6s43. c4Oj4353233(-E—--D=op7, - 4S3. xoo7Io,zz’Ui)= 4-7 eiJ)1 —2774 ,?35 o2&(t) OD7cLJ)O—’ r’ -4j—Jo-4 eas’); 4.(4it)tL 27O7’)3N3; =7om=4i )=o2,-fb4’3- .O52j 44,3= 3773O.234, (- —I )oz, f=obs.43 + ?ictLu’ 3773 kS 7+.°Lt)7473(), L2) 3O2.b’3 I I ±) (fJ).cheo--u . €o4’t) 41o4L = nnn t— t74 1-ki.i)__4o = fl3,e(3=o33(-—)=ob2 OJ,4O &‘-j i/’.OzL3.3 33{lo25, - fthX4Oi -PP XOZ3,X 2T z7!.4=7004 t)SLkIi)) 97X2 (-)A)?)3J he’-rc.’)CrCO =e*-c o4 voJt y-= 3z.c = 4-ent,O33(1 )Q49,33_)oZ3, 33(-——1)’0233, _ =o)L37254 z3233j333’7tAtFi)1 4z4-3)L3I-k±)‘rt) N/ , o—’w e3rC) tJc.Lq.) 1k)g , ;f3’/:= 3 rTt; 2& ‘O) 23rAppendix D. Predictions from Proposed Design Method (2) 257)—)L s-e’-. ( /) 224= J) kM)42:=j) ?33(4—)= .44- , -- 3L7, 373 22 tb2’3 ‘5922x =2’23= = = 432.3(-ki)= 3b’(k’)112bS09 G33( )O# 3 &34i \ 5°4r’Th3 -‘z0Z3t,, jO.3(2- —O23 023IZ32Yr =g-34z= ;.i= ±22. 35= O9)tWO—i)) c9(4)= O(J)x=o.’’5S, 133t— )a.732 , x5 j=4• /ttZ3b, —I) 44--fb2— 3+IW- P> 23 443 I244 /el’icoLU ‘.tt’424 S14ct)5c15, ‘ie tobo4(kN)4’: c9a.Sehc=oZ1 33() Tb°’49‘.0 = I7Ø7 /t=O,23, 33(—l)=O.4k5E, =o4-t tlP 23o44f j/= I33()) = 4X-4OS 3” IbbT’5tt)=j353f-)3OV’1 €r1C): cc.2(i) 7o24C4zJ =____L ,-o353.4- o745r5 44 J/2CO23b, °.33 — lj=.’g, f =b 354-p 243g}.‘z.4.2‘ ccLL) 43 i22LLt) f75 (&‘))4-zLkN)1 := r)Appendix D. Predictions from Proposed Design Method (2) 25844: 2 42 F/f5L= 32=3(-j_1)=a444-,-,&4z--— 24jo9,5= 1t?) 4s24-.9 3= 44ct)=W{’)- c-tt4) 374(),—____-4p-k,O3fl3>ç}>.) =0ft°33f XIbXR363 z’23 ?$Zij- (&) 732(ki’i)1 z4okr)= 21(kIJ)&rt) 4c-fcy I31 L-kN) = 732(kI1)F=3+2.pci= .4;z1 ) 3kZ234?4 3g‘47o•3! —i)Q 2) p$NCØ-Lr 3ø’] I3.t4cp) 9z(-*jJ) , 2 -G’ 321k) vzLkt)Se4-(I) cett 2t-k)J), Nc= ftJcJ.A1&A1- -9=aQM?&, c=4oOmrto33(—>t, (,= o33(— L)O.3S ( ct nie,re vi€,,f=e•33( )2+3) -t)= 033‘= —j-(, $ 3SJi = 3.5 M?& b 2 G33o 4.3 M1’= 355I2O’ I’f 20 (4 t’ ct.) 4 c *3 K2= bo kij-—ijc shec(I)-—o- SeeL3): tq’ (-ku)= ii= I4’ZoC$)& A -(13S,=24.5 o33, -f=-o.x27-i- a24So33=p&rfr,()4) 24k)jo—Jc Sr(): 44Lkk1) . hAJc sbexc3) 2ZS(NJmJz= 7:(-kt)Appendix D. Predictions from Proposed Design Method (2) 259HP.Oi.°, , -.) =a,7c3I.L± 24G7933Jñ —4&, 2d— coL(ij = ZhLk)+J1”- , iO)’--i Sie(3) = Z432CkN).j,,= a.33S, -,= ob--&3j’2=o. f1 )c24.9 £z4xo:3j .9 MIicLLLY 422 ibSG4J) , 4cL2 I.4-2=(t)-kt*io-tj Seut3) Ns 2?44CdJ)= ‘rf1==‘2&OM?i, 43’-,o(i-°, X2-4j Jn 4i.2 I4?24Ø,‘+ xo2A 33jzh. = 11= 4i zøo=-S (-ki.3), = 4s i3.l x 2OI5 Z2YkUU) AS4.) V2±4}J) *30-WOj 5kcL2xt3). 22/ (e)CcJ lkL?N)tL + ==oz4z.-i- 3’.3 -i?o.035, rab4Z + oa433JtL)= 320 (ZC&4)) eJj) 4&9 ZJJj—-- r): tL4) 3t3(k)} j_3VJ z4Lk)=z-, =4e=I0,=38, =4z1M?=a.24 =O33-‘ ,c2,.S-+ 2433J =1 V?t’/ 4- TX2 233’L&).32(k •O-0-’ ((1.) ‘-AcJc)Appendix D. Predictions from Proposed Design Method (2) 260= 7r= 4g4A1O: -= 4Xi= IQ, =oS2+(O3, 4= o.2-t bNc) 32.4)c2’’ i29t&) , ‘4c’(1) 2frx4xL ,3+Ck)4GAL Sb_tLr(); L4)=ar22J.i) -) Sk€cx)= = (4v)All: -4}-, 4tG()0)-f,= 74--i-=o2J1S, (3= o.33,-f ou7 4-i- b$O.Z4 o3jE4-j2. 4?c2 (-1) 24-c= 7I (-k)Jt) c 4)Zki, e-): aC-ki)Th1- = (-kN)Al2; ‘z.Sl?&, =4etmrk-L-°,—= GX2,-S3SE 4o. fri.2=3) )(Z3 t 7 b‘ic-&L) 2DO lzoL&), — 17 —L2= zzt(ki-); NcJ4 ) -+UjO—&A3’ -r) e)=211l220(kJ)c=zTM =4on‘ =ck33 33(’)33’, =cb2ti - xooco38J7=.sL33Zl)) == qzci)= (S , , 3--(C 3’33= 44?2 “33, 4f7 HP=- =‘Appendix D. Predictions from Proposed Design Method (2) 261__________434y , o224.;r551 43gf11 aj,351=-,- PZ/.X. otpst) =é43+ 7ZX2Z(°I53j4 2t)oI 32= 4Io(k) 2(kL))J Nc-bo— ,hetct). lC-(4)= 22.LkJJ) O0V€ tfL): 1c.t1> ZZ()=434-?= L?) 2P)WLI) = 4-i. I tkp) 3(kJ). 4-x3 gJ.t-kL-) -—tuo-ij heof = raz ., -o—-w he’—t3‘i S Qi’22(tJJ)a33I)O0 (?b 1T7 ‘‘ 2J()0_eArL1) 4c.Jk) )2I&kJ) , te-W0j eL4-3) CI) Z2, L&J)Th= ZI4ei.))‘: c( .0 , O(t4.1i.,+?Z$1] SC-) 5V +2 2lFI4.5 = 374 LP)N L€fl) — =4tZLk?)2It4) ctzy4 ks’74I IoS(k)=ShW1): 4QI.I) JCAJ&J.5h1X(3) CL5)2I()SGz= s3= -= .2_,c7 p • y =,o, x= o-z= 4o& LP) q7j= 7o7 Pit)40X 2 (4p) -‘kN)-,.IckIz) rø 4Y7 42k)=€‘-ft l’5)c.-=Appendix D. Predictions from Proposed Design Method (2) 262oc dr3o; Lt+,=o4&,= —2S+kW 2),27(?)x2’== 3Lj()) iCO.e.L) = 34.(V± I’15 k)1Jc)-÷23 (-kM)Oe—ix eJ) MZL)= 32’fLk), —4A €X’) .J.j4) 26(kN)hearc3) (-kJ)= =2 ,o(=’ 31O= O43Z z÷432j 27.C M?)= o, = 23L1P&)= )zo t4vJ), L) a- )=a X ceQcz) IZ7(It 2Z.(4t)J)[3j i, 3kJL), -±wo-J&j shxi ) iLL+)——2,4(4UJ)i’v1l (&I3)SbGrt t: 2j M?m.. 9e ntcO332I)> o3(i)=o4z. f= Z7.1-i-xt42jj= LPR)sq -°,34-I c-ku), 34- (i-+ lg4-ck)= 2.I A2( f2ø= >. = A-° (I -‘- )=za (kiJ)Ofl—t’J 6h’erU): jy= 3o4jJ) ——j- ie.tr 9kJ)‘wo—t3J shzrL3) 3-±)=b1fl $JL3-IP,= 03Li)3ii2,c(= 1°, =o33(-j_l )=o2,= ;4,(-k), 4-C-)WD= 7(ki), c)T(—Appendix D. Predictions from Proposed Design Method (2) 263—) fLf:) c(3)37/€?J) , u—WM Sare).. c4c4-) L1OJJcL) 5r70 k=)11I” { Mc’k1r (kIJ)/=oc44, (iP)&z= t— 3° (1#)I -+- —‘= 4-I 3â= (4z)J) 27sqck)= 39.0 ) 2 ) *20 2339 c) NkL2-) -÷ = 347 (-k4)hEzVf(j) 4c.-t) -il)O_1Jj IAau.-to—iuj &1eaJ L3) ‘-co.L) 9cD ks))‘J--t ii’fJ z4L-ki’)rtt\ 03M’& O1Yt40= 33(_t) o, i)=0441, 33’13%.3_)2T) f3o.3- jj=’zI.4 (4&)Ncts •7s’= 827C-k3I), coW 37(I + I3ok1)4IJtc) = 2I42ZD r34(-k), M( + q75L-kJJ)&r€—IAIcj ec3rLI): -cC3) I104.(&J)0L iI4= ic-ki-

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