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Seismic shear capacity of reinforced concrete elements Webster, Scott William Frederick 1995

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SEISMIC SHEAR CAPACITY OF REINFORCED CONRETE ELEMENTS by SCOTT WILLIAM FREDERICK WEBSTER B.A.Sc, The University of British Columbia, 1990 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF Master of Applied Science in The Faculty of Graduate Studies Department of Civil Engineering We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA October 1995 e Scott William Frederick Webster, 1995 In presenting this thesis in partial fulfilment for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the head of my department or by his or her representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my express written permission. Department of Civil Engineering The University of British Columbia 2324 Main Mall Vancouver, Canada V6T 1Z4 Date: fldt> /S'/fS" ABSTRACT The objective of this thesis was to develop a better understanding of reinforced concrete columns with minimal (widely spaced) transverse reinforcement subjected to reversed cyclic shear. This thesis involves three main parts which are described below. A rational model for the shear capacity of reinforced concrete is developed which takes into account the effect of previously existing diagonal cracks. The model is an extension of the modified compression field theory which is one of the most advanced shear design methods. The model was implemented into a computer program, and predictions based on this proposed method are compared to other shear design models. A copy of the program is included in the thesis. To forward the development of a seismic shear model, an analytic study was undertaken in which the magnitude of stresses (both compressive and shear) at an arbitrarily inclined crack were investigated. In addition, by means of a literature review, the degradation of aggregate interlock across a crack interface when subjected to reverse cyclic loads was studied. In addition to the analytical work a pilot experimental program was carried out, in which eight reinforced concrete bridge column elements were subjected to reversed cyclic loading. The columns were subjected to different load combinations (axial load, bending moment and shear), and had different distributions of longitudinal reinforcement. The results of the testing program is compared to predictions based on the analytical model presented earlier and other shear design models. An important part of this study was the construction and calibration of an element testing apparatus which was used to test the specimens. The experimental program also proved valuable as a pilot study, where much was learned about the operation of the tester and the most appropriate specimen connection details. The testing apparatus has subsequently been used to test a further 12 specimens as part of this ongoing study. ii TABLE OF CONTENTS ABSTRACT ii T A B L E OF CONTENTS iii LIST OF FIGURES v LIST OF TABLES viii ACKNOWLEDGMENT ix CHAPTER 1 Introduction 1 CHAPTER 2 Shear Strength of Members with Previously Existing Diagonal Cracks 4 2.1 Introduction 4 2.2 Background 4 2.3 Modified Compression Field Theory 7 2.4 Proposed Crack Check Model 10 2.4.1 Introduction 10 2.4.2 Compression on the Crack Interface 11 2.4.3 Calculation of Crack Width 12 2.4.4 Concrete Stress - Strain Relationship 14 2.4.5 Longitudinal Steel Yielding 15 2.5 Comparative Study 18 2.6 Concluding Remarks 19 CHAPTER 3 Towards a Rational Seismic Shear Model 26 3.1 Introduction 26 3.2 Influence of Crack Inclination on Interface Stresses 26 3.3 Effect of Cyclic Loading on Shear Transfer 30 3.4 Additional Considerations for Developing a Seismic Shear Model... 34 CHAPTER 4 Experimental Programme 44 4.1 Introduction 44 4.2 U . B . C . Beam Element Tester 44 4.3 Connection Detail 46 4.4 Test Specimens 47 4.5 Material Properties 49 4.6 Construction of Specimens 51 4.7 Testing Procedure 52 4.8 Load Control, Instrumentation and Data Acquisition 54 iii CHAPTER 5 Experimental Results 67 5.1 Introduction 67 5.2 Specimen CS1. 67 5.3 Specimen CS2 68 5.4 Specimen CS3 70 5.5 Specimen CS4 72 5.6 Specimen CS5 73 5.7 Specimen CS6 74 5.8 Specimen CS7 76 5.9 Specimen CS8 77 CHAPTER 6 Discussion of Results . . . 103 6.1 Introduction 103 6.2 Influence of Shear Span 103 6.2.1 Summary of Results 103 6.2.2 Comparison of Analytical Predictions and Experimental Results 105 6.3 Influence of Axial Load 109 6.3.1 Summary of Results 109 6.3.2 Comparison of Analytical Predictions and Experimental Results I l l 6.4 Influence of Distributed Reinforcement 114 6.4.1 Summary of Results 114 6.4.2 Comparison of Analytical Predictions and Experimental Results 116 6.5 Influence of Load History 118 6.6 Summary 118 CHAPTER 7 Summary and Recommendations 129 REFERENCES , 134 NOTATION 136 APPENDDX A Computer Program 138 APPENDDX B Beam Element Tester - Calibration Data 164 APPENDIX C Experimental Results - Peak Loads 192 iv LIST OF FIGURES Fig. 2.1 Comparison of average stresses and stresses developed locally at a crack (Collins and Mitchell, 1987) 21 Fig. 2.2 Comparison of concrete constitutive law for the MCFT and the proposed method 21 Fig. 2.3 Free body diagram of diagonal in a beam 22 Fig. 2.4 Longitudinal forces developed in top and bottom chord of beam subject to shear stresses (Collins and Mitchell, 1987) 22 Fig. 2.5 Comparison of predicted response for a beam with 0.40% transverse reinforcement 23 Fig. 2.6 Comparison of predicted response for a beam with 0.20% transverse reinforcement 23 Fig. 2.7 Comparison of predicted response for a beam with 0.05% transverse reinforcement 24 Fig. 2.8 Comparison of predicted and experimental observed shear strength 24 Fig. 2.9 Shear strength predicted without concrete tensile stresses 25 Fig. 3.1 a) Uniaxil compressive stress field inclined at 30 degrees b) Compressive stresses due to 45 degree "truss model" c) Stresses which must be resisted by shear at the crack face 36 Fig. 3.2 Portion of shear carried by Vc and Vs as a junction of stress field and crack inclination 37 Fig. 3.3 a) 30 degree stress field and a 30 degree crack b) 30 degree stress field and a 45 degree crack c) 30 degree stress field and a 60 degree crack 38 Fig. 3.4 Compressive and shear stresses on a crack interface as a junction of crack angle using a 30 degree stress field 39 Fig. 3.5 Compressive and shear stresses on a crack interface as a junction of crack angle using a 30 degree stress field 39 Fig. 3.6 Ratio of compressive stress to shear stress on a crack interface as a junction of deviant crack angle 40 Fig. 3.7 Typical specimen setup and test apparatus (White and Holley 1972) 41 Fig. 3.8 Typical test results (White and Holley 1972) 41 Fig. 3.9 Typical test setup (Paulay and Loeber 1974) 42 Fig. 3.10 Typical test result (Paulay and Loeber 1974) 42 Fig. 3.11 Typical test specimen and loading apparatus used by Mattock (1980) 43 Fig. 4.1 Plan view of Beam Element Tester 56 Fig. 4.2 Partial elevation of Beam Element Tester 57 Fig. 4.3 Method of load application using three actuators (A) and three rigid links (B) 58 Fig. 4.4 Tester capacity interaction diagrams 59 Fig. 4.5 Connection detail of specimen endplates 60 v Fig. 4.6 Inside face ofendplates showing shear stud configuration 61 Fig. 4.7 Configuration of distributed steel for Types A, B and C 61 Fig. 4.8 Typical stress-strain relationship for longitudinal (15M) and distributed reinforcement (10M) 62 Fig. 4.9 Sliding bar assembly 63 Fig. 4.10 Reinforcing cage in construction jig 63 Fig. 4.11 Completed reinforcing cage for Type A specimen 64 Fig. 4.12 Comparison of reinforcement for Types A, B and C 65 Fig. 4.13 Reinforcing cages placed in casting formwork 66 Fig. 4.14 Typical loading regime for test specimens 66 Fig. 5.1 Specimen CS1 after initial cycle of loading 79 Fig. 5.2 Hysteresis loop for Specimen CS1 80 Fig. 5.3 Specimen CS1 crack pattern at failure 81 Fig. 5.4 Specimen CS2 at ductility level 1.5 (corrected ductility of 2.3) 81 Fig. 5.5 Hysteresis loop for Specimen CS2 82 Fig. 5.6 Specimen CS2 at ductility level 3.0 (corrected ductility of 4.5) 83 Fig. 5.7 Specimen CS3 at ductility level 0.75 (corrected ductility of 0.96) 84 Fig. 5.8 Hysteresis loop for Specimen CS3 85 Fig. 5.9 Specimen CS3 at a ductility of 2.5 (corrected ductility of 3.2) 86 Fig. 5.10 Failure of stirrup anchorage in Specimen CS3 87 Fig. 5.11 Specimen CS4 at a ductility of 1.5 (corrected ductility of 1.65) 88 Fig. 5.12 Hysteresis loop for Specimen CS4 89 Fig. 5.13 Specimen CS4 crack pattern at failure 90 Fig. 5.14 Specimen CS5 at a ductility of 0.75 (corrected ductility of 0.97) 91 Fig. 5.15 Hysteresis loop for Specimen CS5 92 Fig. 5.16 Specimen CS5 at a ductility of 3.0 (corrected ductility of 3.87) 93 Fig. 5.17 Specimen CS5 crack pattern at failure 93 Fig. 5.18 Specimen CS6 at a ductility ofO. 75 (corrected ductility of 0.89).. 94 Fig. 5.19 Specimen CS6 at a ductility of 1.5 (corrected ductility ofl. 79) 94 Fig. 5.20 Hysteresis loop for Specimen CS6 95 Fig. 5.21 Specimen CS6 crack pattern at failure 96 Fig. 5.22 Specimen CS7 at a ductility of 1.5 (corrected ductility of 1.42) 96 Fig. 5.23 Hysteresis loop for Specimen CS7 97 Fig. 5.24 Specimen CS7 crack pattern at failure 98 Fig. 5.25 Damaged end of Specimen CS8 98 Fig. 5.26 Asymmetric crack pattern due to severe damage at opposite end of Specimen CS8 99 Fig. 5.27 Retrofit of damaged end of Specimen CS8 100 Fig. 5.28 Hysteresis loop for Specimen CS8 101 Fig. 5.29 Specimen CS8 crack pattern at failure 102 Fig. 6.1 Grouping of specimens into the four test series 120 Fig. 6.2 Comparison of Specimens CS1, CS2 and CS3 at failure 121 Fig. 6.3 Hysteresis loops for test series one 122 vi Fig. 6.4 Hysteresis loops for test series two., 123 Fig. 6.5 Comparison of Specimens CS3, CS4 and CSS at failure..... 124 Fig. 6.6 Hysteresis loops for test series three 125 Fig. 6.7 Comparison of Specimens CS1 and CS6 at failure. 126 Fig. 6.8 Hysteresis loops for test series four 127 Fig. 6.9 Comparison of Specimens CS3, CS7 and CS8 at failure 128 Fig. BI Calibration of Rigid Link Ul B4 Fig. B2 Calibration of Rigid Link »2 B5 Fig. B3 Calibration of Rigid Link #3 B6 Fig. B4 Calibration of Hydraulic Actuator M B7 Fig. B5 Calibration of Hydraulic Actuator #5.... B8 Fig. B6 Calibration of Hydraulic Actuator if6.... B9 Fig. B7 Plot showing inherent friction in Hydraulic Actuator #4...... B16 Fig. C I Labeling convention for rigid links and hydraulic actuators for the Beam Element Tester C I vii LIST OF TABLES Table 4.1 Summary of geometry type and loading condition 49 Table 4.2 Summary of steel reinforcing properties 50 Table 4.3 Specimen testing date 50 Table 4.4 Shear corresponding to predicted flexural strength 53 Table 6.1 Summary of Results - Test Series One 104 Table 6.2 Analytical Predictions and Experimental Results - Test Series One 107 Table 6.3 Sectional Shear Capacity and Shear Demand from Flexural Yielding Test Series One 109 Table 6.4 Summary of Results - Test Series Two 110 Table 6.5 Summary of Results - Test Series Three 110 Table 6.6 Analytical Predictions and Experimental Results - Test Series Two 112 Table 6.7 Analytical Predictions and Experimental Results - Test Series Three 112 Table 6.8 Sectional Shear Capacity and Shear Demand from Flexural Yielding Test Series Two 113 Table 6.9 Sectional Shear Capacity and Shear Demand from Flexural Yielding Test Series Three 114 Table 6.10 Summary of Results - Test Series Four. 115 Table 6.11 Analytical Predictions and Experimental Results - Test Series Four 116 Table 6.12 Sectional Shear Capacity and Shear Demand from Flexural Yielding Test Series Four 117 viii ACKNOWLEDGMENT First I would like to thank my supervisor D r . Perry Adebar for his expert guidance, criticism and encouragement throughout the course of the research work and in the preparation of this thesis. A special thanks to D r . Donald Anderson, my co-superyisor, for his valuble input throughout the duration of my studies. Financial support for this study was provided by the Bridge Engineering Branch of the Ministry of Transportation and Highways under the auspices of the Professional Partnership Program. Additional funding was provided by the Natural Science and Engineering Research Council of Canada. Additional thanks to the staff in the structures lab, and in particular, Paul Symons, for all their help and ingenuity throughout the course of the experimental program. This thesis is dedicated to my parents, whose constant love and support made everything possible. Scott Webster August, 1995 Vancouver, British Columbia ix Chapter 1 Introduction In the past, earthquakes have caused severe damage and destruction to reinforced concrete bridges. Brittle shear failure has been a major cause for the collapse of many of these structures. Dramatic examples have been seen in the Loma Prieta and Northridge earthquakes in California and in the more recent disaster in Kobe, Japan. In British Columbia, many bridges are thought to have inadequate seismic resistance. Although most bridge components have reasonable amounts of longitudinal reinforcement, they have very little transverse reinforcement. The concern is that if significant diagonal cracking occurs during the initial cycles of an earthquake, the bridge components (columns and pier-caps) may have very little shear resistance during subsequent cycles. Typically bridges constructed before the mid-seventies were designed for minimal seismic forces. A lateral force of about 4% of gravity was all that the structure was designed to resist. The idea of ductility was not understood and therefore, was never incorporated in the design of these structures. Consequently these older bridges have minimal transverse steel - usually #3 (71 mm2) at 12" (300 mm) on center. It should be noted that the transverse steel not only provides shear resistance, but also performs other roles - confining the concrete core in areas of plastic yielding and laterally supporting the longitudinal steel to prevent buckling (which increases the ductility of the member). Traditional monotonic shear design methods assume that the resistance of a reinforced concrete member is made up of a contribution from the transverse steel, Vs, and a concrete contribution, Vc. The current philosophy for seismic design is to neglect the concrete contribution and design the transverse steel to resist all the load. Considering the other beneficial effects of the additional transverse steel, this is a reasonable approach, albeit somewhat conservative. However, when an engineer is faced with evaluating the 1 Chapter 1 - An Introduction 2 seismic resistance of an existing structure, the cost of a retrofit warrants the use of a less conservative model. In this case the use of an empirical method is often used: the concrete contribution is assumed to degrade depending on the ductility demand. The objective of this thesis was to develop a better understanding of the response of reinforced concrete bridge elements subjected to reversed cyclic loading with emphasis on the degradation of the concrete contribution. The intent was to develop a rational model in which to capture this effect. The methodology for this model is consistent with the modified compression field theory (MCFT) which attributes the concrete contribution to the shear transmitted across diagonal cracks by aggregate interlock. The MCFT is the basis of a unified shear design method which has been incorporated into the new Canadian Bridge Design Code and the AASHTO Code To partially accomplish this objective, an analytical investigation of the effect of cyclic loads on reinforced concrete elements was undertaken. The initial investigation was to modify the MCFT in order to account for the fact that under seismic loads, the reinforced concrete element would likely suffer more damage than a similar element subjected to only monotonic loads. The inherent randomness of seismic loads could cause diagonal shear cracks to form at many mclinations as opposed to one or two discrete inchnations in the monotonic case. Consequently, the first task was to generalize the MCFT to account for the effect of these previously existing cracks. The second half of this analytical investigation was to study the effect that cyclic loads have on the transfer of stresses across cracks. This part of the study looked at two separate but related areas. The first was to study the magnitude of stresses (compressive normal stresses and shear stresses) developed at a crack relative to the applied shear stress. The second area, accomplished through a review of previous experimental work, was to look at the degradation of the ability to transfer these stresses across the crack interface. To complete the objectives of this thesis, a pilot experimental program was carried out, in which eight reinforced concrete elements were subjected to reversed cyclic loads Chapter 1 - An Introduction 3 under different loading conditions and specimen geometry. The results of this testing program is compared to predictions based on the analytical model presented earlier and from other shear design models. A major part of this pilot series was the construction and calibration of a testing apparatus which was used to test these specimens. Consequently, this experimental program was as valuable as a learning tool for working out the specimen connection details and operating the testing apparatus, as it was for the actual numerical results yielded from the experimental program. The testing apparatus has subsequently been used to test a further 12 specimens as part of this ongoing study. Chapter 2 presents the work done on modifying the MCFT to include previously existing diagonal cracks, which includes a comparative study of the new model and the MCFT. The analytical work examining the stresses at a crack is presented in Chapter 3. Chapter 4 describes the experimental program - including the details of the testing apparatus, the specimen geometry, testing procedure and instrumentation. Chapter 5 provides a summary of the events of each test for the individual specimens, while Chapter 6 presents a discussion of the results as they relate to the three parameters studied. This thesis concludes with Chapter 7 which summarizes the work presented in the preceding chapters and presents some recommendations for further study. Chapter 2 - Shear Strength of Members with Previously Existing Diagonal Cracks 2.1 - Introduction The purpose of this chapter is to present a shear design method which takes into account previously existing diagonal cracks, such as occurs in members subjected to reverse cyclic action. The Modified Compression Field Theory, (MCFT), considered as state-of-the-art in shear design, will be used as a basis for this new method. The heart of the new method is a modified crack check routine to account for previous (steeper) shear cracks. Section 2.2 gives a brief overview of the various methods of shear design, including the MCFT, and their application in current codes. Section 2.3 gives a general overview of the MCFT with a detailed discussion on the important aspects which play a key role in the proposed modifications. Section 2.4 presents the proposed method which examines the effects of previously existing cracks on the shear capacity of reinforced concrete members. Section 2.5 presents a comparative study which highlights the new method as compared to experimental results. This study also compares the new method to the MCFT and the General Method in the 1984 Canadian concrete code (1984). The final section presents some conclusions and summarizes the chapter. It should be noted that a computer program has been developed based on the proposed method and that a complete listing of both the algorithm and source code is provided in Appendix A. In addition, a worked example, complete with input and output files have been included in the appendix. 2.2 - Background The simplest method for shear design was presented in 1899 by Ritter, and was expanded and explained in more detail by Morsch in 1902. The method assumes a 45 degree truss model in which the stirrups resist the entire shear. The shear stresses in the 4 Chapter 2- Shear Strength of members with Previously . . . 5 concrete are carried by diagonal compressive stresses in struts while stirrups act as the vertical tension members of the truss. Equation 2.1 gives the sectional shear force resisted by the stirrups according to the 45 degree truss model. Av-f-dy 2.1 V, = 1— • s where Avfy is the yield capacity of a stirrup and d^/s (effective depth / stirrup spacing) is the number of stirrups crossing a 45 degree crack. This approach has proven to be to conservative by numerous experimental results. Consequently, the approach taken by the North American codes has been to add a concrete component, Vc, equal to the shear at initial cracking. The total shear resisted is then V = Vs + Vc. This procedure forms the basis for die ACI code (1983) approach and is used for the simplified method in both the 1984 and 1994 Canadian concrete codes (1984,1994). An alternative approach of improving the 45 degree truss model, is to account for the fact that the inclination of the diagonal cracks may be less than 45 degrees. The Europeans, who favour this approach, apply a lower hrnit to me inclination of the compressive struts of about 30 degrees. Equation 2.2 gives the shear capacity using this design procedure. • \ V s Av-fv-dv • 2.2 where dv/s tan 6 is the number of stirrups crossing a diagonal crack inclined at 0 from the horizontal. A variable angle truss model was proposed by Collins and Mitchell (1982), in which strain compatibility was used to establish the inclination of the compression struts. Chapter 2 -Shear Strength ofmembers with Previously ... 6 This method is called the compression field method. The General Method of shear design in the 1984 Canadian code is based on this approach. Rather than an arbitrary limit, the inclination of the compression struts is limited by the compressive strength of the concrete. Vecchio and Collins (1986) observed that reinforced concrete elements softened and weakened when subject to biaxial strains. They proposed the following expression: ci J c2max y \ECJ 2.3 where f c2n f. 0.8-0.34 <1 f ci max c2 S2 = uniaxial compressive strength = maximum compressive strength = average principal compressive stress = maximum compressive strain at maximum stress = average principal tensile strain = average principal compressive strain The inverse of this expression, solved in terms of principal compressive strain is: £ 2 - - 0 . 0 0 2 ( l - V l - / c 2 / / c 2 m a x ) Generally, as the shear force in the member increases, the strains increase which weakens the compressive resistance of the struts. If the longitudinal reinforcement is stiffer than the Chapter 2 - Shear Strength of members with Previously ... 7 transverse reinforcement (as is usually the case), the inclination of the struts decreases, which increases the diagonal compressive stresses. Ultimately, crushing of the concrete struts limit further redistribution of load, i.e., limits the inclination of the struts. Regardless of how flat the diagonal compression struts are permitted to go, a member with no transverse reinforcement is predicted to have no shear resistance if the concrete tensile stresses are totally neglected. The MCFT was put forward by Vecchio and Collins (1986) and it accounts for tensile stresses in the concrete after cracking. This method predicts that a member with no transverse reinforcement may still have considerable post cracking shear strength (especially if the crack control characteristics of the member are good). The details of this theory are given below in Section 2.3. Collins et al. (1991, 1995) have presented a general shear design method, based on the MCFT in which the shear resistance of the member is expressed in terms of a concrete contribution, Vc, and a stirrup contribution, V5. This design method is used in the 1994 Canadian concrete code (1994) and replaces the General Method of shear design in the 1984 code. It is also used in the Ontario Highway Bridge Code, the new Canadian Bridge Code and the AASHTO alternative design method. A more detailed explanation of the various shear design methods is given in the text by Collins and Mitchell (1987). 2.3 - Modified Compression Field Theory The MCFT, presented by Vecchio and Collins (1986), treats cracked reinforced concrete as a new material with its own stress-strain characteristics. Equilibrium, compatibility and constitutive relationships are formulated in terms of average strains and average stresses. The gauge length in which the strains are measured are large enough to span several cracks. Similar to the CFT, the MCFT limits the compressive stress on the diagonal compression struts to fC2max a s expressed in Eq. 2.3. The difference between the two theories is that the MCFT accounts for tension in the cracked concrete. Chapter 2 - Shear Strength of members with Previously ... 8 Based on experimental results, Vecchio and Collins (1986), proposed an empirical tension-stiffening relationship for reinforced concrete elements subject to shear. This relationship between the average principal tensile stress, fcj and the average principal strain, Sj shows a decreasing value of tensile stress with an increasing value of strain after cracking. The post-cracking relationship is: where fc and fcl are in MPa Since the MCFT accounts for tensile stress in cracked concrete, local stresses at a crack must be checked to see if the average tensile stress can be transferred across the crack. The MCFT employs a crack check procedure which considers only equilibrium of the forces at a crack. The method assumes a crack to have formed perpendicular to the direction of principal tensile average stress or average strain ( the principal directions of average stress and strain are assumed to coincide). At a crack, the tensile loads calculated in the reinforcement will be higher than those calculated in an average sense, as there can be no tensile stresses in the concrete normal to the crack interface. It is assumed that if there is reserve in both x and y steel, then there is no need for shear stress to be developed across the crack. In reality, this is a statically indeterminate problem depending on the relative stiffness of the longitudinal and transverse steel. However, since the reinforcement and the shear transfer mechanisms are ductile, compatibility is ignored locally in the MCFT, and a lower-bound equilibrium approach is used for simplicity. If, on the other hand, there is no reserve in one of either the longitudinal or transverse steels ( typically the transverse steel is weaker) then a shear stress must be mobilized across the crack face to maintain equilibrium. This stress is assumed to be 2.4 Chapter 2 -Shear Strength of' members with Previously ... 9 provided by the aggregate interlock of the two faces of the crack. This shear stress may be accompanied by a compressive stress on the crack face. Figure 2.1 compares equilibrium in the concrete between cracks and locally at a crack. Based on Walravens work (1981), Vecchio and Collins (1986) have suggested a relationship between the shear stress on the crack, vCJj and the compressive stress on the crack, f^. The relationship is expressed by: 0.82 f2. vc/ = 0.18Vc,max + 1.64/c/ ^ Vcimax 2.5 where vcimax is the maximum shear stress that can be resisted by a crack. Although initially included in their original paper, the compression on the crack face, fci, was later neglected in most applications of the MCFT. The added benefit of the compression is only significant at steep crack inclinations which are not important in many applications. Consequently, the maximum shear stress which can be transferred across a crack without compression was taken to be 0.18vcimax, which is a function of crack width. This function is expressed as 2.6 0.31 + — a + 16 where a is the maximum aggregate size (mm), w is the crack width (mm) zndfc is in MPa. The crack width is assumed to be equal to the product of the crack spacing (determined by the crack control characteristics of the element) and the average principal tensile strain. This is further explained in Section 2.4. Chapter 2 -ShearStrengthofmemberswith Previously... 10 More recently, Collins et al. (1991,1995), have presented a memod for shear design based on the MCFT; This general method expresses the shear force resisted by the tensile stresses in cracked concrete in the more familiar form of a concrete contribution V c given byEq. 2.7. : : Vc = Pifcbvdv 2.7 ^ _ 0.33 cot 0 where 1 + ^ 500^ 2,8 0.18 and ; V P± 24^" 2.9 . J 0.3 + - ^ -. ' a + 16 In Eq 2.8, /?is a factor which captures the tension-stiffening effect of the cracked concrete (based on Eq. 2.4), while Eq. 2.9 hmits the concrete component by checking equilibrium at a crack (derived from Eq. 2.6). In this method the compression on the crack face has been ignored. In summary, the MCFT includes a contribution from the tensile stresses developed in the concrete. This concrete component, Vc, is calculated, from an empirical tension stiffening relationship and the ability of the concrete to transmit shear stresses across the crack interface. The next section provides a generalization of these crack check equations. 2.4 - Proposed Crack Check Model 2.4.1 - Introduction In the MCFT the crack inclination at any load level depends on the relative stiffness of the longitudinal and the transverse reinforcement, and the relative stiffness of concrete in compression versus concrete in tension. For a member with little or no transverse Chapter 2 - Shear Strength of members with Previously ... 11 reinforcement, the crack inclination generally reduces with increasing loads, while the average tensile stresses reduce with increasing strains. In the crack check of the MCFT, the influence of previously existing cracks (i.e., from lower load levels) is ignored. Thus, for a member with little, or no transverse reinforcement, the crack check is done only for the present crack, which is the flattest, and the previously existing steeper cracks are ignored. In the following, the modifications that were made to the MCFT in order to check all previously existing cracks are summarized. 2.4.2 - Compression on the crack interface As mentioned earlier, compression may help in transferring shear across a crack interface. As the compression on the interface of a diagonal crack has a component transverse to the member (i.e., parallel to the applied shear), it is not a simple matter to determine when and how much compression is beneficial. A full discussion of this phenomenon was given by Adebar (1989), and a brief summary is given below. To find the maximum stress (expressed in terms of a principal average tensile stress normal to the crack direction) that can be transferred across a crack via shear and compression, some algebraic manipulation is needed. Assuming that the transverse reinforcement has yielded, equilibrium requires that: / c l = Vc,tan0-/c,. 2.10 Substituting Eq. 2.5 for in terms offd and equating the derivative offcl with respect to fd to zero, yields the following expression for compression on the crack face which yields the maximum principal tensile stress: f - U- 1 J ci ~ VcimaxI 1 1.64tant9y 2 11 Chapter 2 - Shear Strength of members with Previously ... 12 Substituting Eq. 2.11 into Eq. 2.5, results in the following expression for the shear stress on the crack face corresponding to maximum tensile stress in the concrete. - h - 1 Vci ~ Vci max * 3.28 tan2 0) 2.12 Finally substituting Eqs. 2.11 and 2.12 in Eq. 2.10 gives Eq. 2.13 for the maximum value of the principal average stress normal to the crack direction. ci max ' „ 1 ^ tant9-l+— -3.28tan6>y 2.13 It can be shown, that the steeper the crack, the more beneficial compression on the crack will be. This makes intuitive sense because as the inclination of the crack becomes steeper (relative to the longitudinal axis of the member), the vertical component of the compression decreases. Conversely, compression on the crack provides no benefit on cracks with an inclination of less than 31.4 degrees (Adebar, 1989). This provides good reason for ignoring the compression in the MCFT where only the flattest cracks are checked. 2.4.3 - Calculation of Crack Width In the MCFT the crack is assumed to be normal to the principal tensile average strain direction, and the crack width, w, is calculated as the product of the principal tensile average strain, e\, and a crack spacing parameter, smQ, which reflects the ability of the longitudinal and transverse reinforcement to control cracks of various inclinations. That is: Chapter 2 - Shear Strength of members with Previously ... 13 w = £l'Sm0 2.14 where ? a = — — S m 0 sin 0 t cosfl Smx Smy 2.15 and where s^ and smy are the crack spacings in the longitudinal and transverse directions respectively. This simple expression reflects the fact that the spacing of very steep cracks is controlled primarily by the longitudinal reinforcement and the spacing of very flat cracks is controlled primarily by the transverse reinforcement, i.e., = s^ for 0=90°and sm0 = smyfor 0=0°. In the generalized crack check procedure, the width of cracks is calculated from the average strains; however, the cracks may deviate considerably from the principal tensile average strain direction. The crack width is assumed to be equal to the product of the normal strain in the crack direction and the crack spacing, smQ, in the cracking direction, £2. w = £crk-smn 2.16 w h e r e cm 0= . _ — 2.17 * m n sin.Q cosQ + ..' Smx Smy The normal strain in the crack direction can be calculated from the following transformation of the average strain components £/, e2, 0, and-7. = gi + g2tan2(n-fl) and E c r k ~ l + tan2(---0) 2.18 Chapter 2 Shear Strength of members with Previously . . . 14 In these equations, Q, represents the angle of inclination of the crack to be checked and 6, represents the angle of inclination of the principal strain (and stress) field as it does in the MGFT. It can be seen that only the tensile strain at the crack is considered in calculating crack width. The shear strain parallel to the crack is ignored. 2.4.4 - Concrete Stress - Strain Relationship For a number of reasons the maximum principal compressive stress of concrete subjected to shear is considerably less than the uniaxial compressive strength. Collins (1986) suggested that one reason is that the compression field must be transmitted across previous diagonal cracks. Vecchio arid Collins (1986) conducted a large number of shear panel tests, and observed that the principal compressive response of concrete is softer and weaker as a result of transverse tensile strain. As mentioned earlier, they developed an empirical expression (Eq. 2.3) in terms of the principal tensile strain, which is largest when the difference in the stiffness or quantity of reinforcement in the two orthogonal directions is large. It is interesting to note that the amount of redistribution, i.e., the need to transfer stress across previously existing cracks increases when the transverse reinforcement is significantly less than the longitudinal reinforcement. As the proposed procedure involves checking all possible crack inclinations, it was thought to be inappropriate to include the empirical relationship for the weakening of the concrete due to transverse strain (the crack check was meant to capture this phenomenon). However, as the crack width is calculated from the average strains, the empMcal softening effect did need to be included. Thus, a new stress - strain relationship was developed so that the softening effect is very similar to the Vecchio and Collins relationship, but no weakening is included. That relationship is shown in Fig. 2.2 and is summarized below. Chapter 2 - Shear Strength of members with Previously . . . 15 £ceffj V 2.19 and where the inverse is given by: £2- e. 2.20 where \ 0.3675 J 2.21 1 >1.0 and 0.8 +170 ex 2.22 In developing the equations above, it was necessary to maintain the proper average stress - average strain response verified by the tests done by Vecchio and Collins. Therefore, the secant modulus of the cracked concrete was reduced in such a manner that when/c2 = 0.6fC2max, the compressive strain given by the proposed equation equaled the strain calculated by the MCFT equation. 2.4.5 - Longitudinal Steel Yielding The failure of a member subjected to shear and bending may be entirely flexure dominated, where the longitudinal reinforcement yields and the strains of the concrete are primarily uniaxial, or may be entirely shear dominated in which the longitudinal reinforcement does not yield and the largest concrete tensile strains are transverse to the longitudinal steel. On the other hand, the failure may also be somewhere between the two extremes, i.e., yielding of the longitudinal reinforcement may limit the "shear" capacity of a member. Thus, the stress increase in the longitudinal reinforcement at a crack must be Chapter 2 - Shear Strength of members with Previously ... 16 checked to ensure that it is not yielding. In this section, new equations are developed to account for the added demand on the longitudinal steel from compression on the crack. The free body diagram in Fig. 2.3 most effectively illustrates the demand on the longitudinal steel. By taking moments about the compression flange of the beam, the equation of equilibrium becomes the following: Nx-dv = V\x+-——1+-tmOJ 2sin2Q 2 tanQ 2.23 Dividing by dv and collecting terms gives: Nx=K+_!^+J^ + fcfbv-dv dv 2tanQ tanQ 2 sin >C1 2.24 The first term represents the demand on the longitudinal steel from the applied moment at the section. The remaining terms represent the axial force applied to the bottom flexural steel due to shear. However, it can be shown that these terms represent only half of the axial tension applied to the member from shear. Figure 2.4 illustrates the condition of equilibrium at a section of a beam subject to shear stresses where an axial tension at the top as well as at the bottom of the section is shown. Therefore, the total axial load due to shear on the element becomes: = Vs , 2Vc , fcibv-dv V c r k tanQ tanQ sin2^ 2 25 A detailed explanation of the first two terms is given by Adebar (1989). The third term was developed as part of the present study. Chapter 2 - Shear Strength of members with Previously 17 The task now is to calculate the compressive stresses on the crack interface corresponding to the concrete component, Vc. Assuming that the transverse steel has yielded, it is a simple calculation to find the proportion of shear carried by the steel, Vs, using the variable angle truss model equation. The shear carried by the concrete, Vc, is the difference between the applied shear and the shear carried by the transverse steel. Therefore, the first two terms in Eq. 2.25 are known. However, the third term is not as easy to calculate as Vc and fci are related. The minimum compressive stress at the crack interface must be calculated which can still provide the necessary Vc required for equilibrium. Noting from equilibrium at a crack interface Substituting Eq. 2.5 for vct- in Eq. 2.26 and collecting terms, gives Eq. 2.27 - a quadratic equation in terms of/c,-. 2.26 0.82 2.27 -0.18vc,max = 0 The lower of the two solutions to this equation is given by Eq. 2.28 -b-^b2-4ac 2a 2.28 where 0.82 a = b = 1.0 -1.64 tanQ c = -^—0.18Vc, bvdv Chapter 2 - Shear Strength of members with Previously ... 18 Having calculated the compressive stress on the crack face to satisfy equilibrium, Eq. 2.25 may be solved for the shear induced axial load. It should be noted that Eq. 2.25 gives the demand on the longitudinal steel at a crack from the applied shear, and any additional demand from axial load or moment must also be accounted for. An algorithm for the proposed procedure may be found in Appendix A. Some equations found in the appendix have been omitted from the body of this thesis as they are simple equilibrium or compatibility equations or have been derived previously (e.g., Collins and Mitchell, 1991). 2.5 - Comparative Study A study was undertaken to compare the predictions of the proposed method with predictions from the MCFT and predictions using the General Method of shear design presented in the 1984 Canadian concrete code (1984). The study utilized experimental results from a series of tests carried out by Adebar (1989). The beams tested were 290 x 310 mm in cross section, with 5 - 20M reinforcing bars top and bottom, and a varying amount of transverse shear reinforcement. The critical section for shear was assumed to correspond to a shear span of 375 mm (i.e., M/V = 375 mm). Complete details of the testing program may be found in Adebar (1989). Figure 2.5 compares the responses predicted by the modified compression field method and the proposed method, for a beam with 0.40% transverse shear reinforcement. The modified compression field method predicts that the initial crack occurs at an inclination of 70 degrees. After cracking, redistribution occurs (i.e., the compression struts become flatter). The stirrups yield when the compression strut is at an inclination of 40 degrees. After the stirrups yield, the stresses can still redistribute further. The shear force increases and the compression struts continue to get flatter. When the principal compression stress inclination falls to 34 degrees, the crack check indicates that a previous (steeper) crack inclined at 41 degrees can no longer resist Chapter 2 - Shear Strength of members with Previously . . . 19 the applied shear and the resistance of the beam decreases. If the previous steeper cracks are ignored, than the modified compression field method predicts that the principal compressive stress continues to redistribute further until the principal compressive stress reaches the diagonal crushing strength of the concrete at an inclination of 27 degrees. Figures 2.6 and 2.7, show the predicted responses for two other beams with less transverse reinforcement. For these beams, which have low amounts of transverse steel, the strength is dictated by slippage along the flattest cracks. The predicted shear strengths are compared with the experimentally measured capacities in Fig. 2.8. Note that for transverse reinforcement amounts less than about 0.25%, slipping along the flattest crack is critical. For larger amounts of stirrups, the previous (steeper) cracks are found to be more critical. The modified compression field method, in which the crack check is performed only on the flattest crack, gives slightly higher results for members with larger amounts of transverse steel. The strength predictions were repeated, but without a concrete contribution (the concrete tensile stresses were neglected). These predictions are compared with the experimental results in Fig. 2.9. It is interesting to note that the prediction which assumes no tensile stresses in the concrete and assumes that the inclination of the strut is limited by diagonal crushing (i.e., fc2 = fC2max) gives very good predictions for members with more than about 0.20% transverse reinforcement. This prediction is equivalent to the General Method in the Canadian concrete code (1984). Of course, the prediction is not appropriate for members with little or no transverse reinforcement since the concrete tensile stresses are not included. 2.6 - Concluding Remarks Different approaches can be used to limit the inclination of diagonal compression struts in truss models for cracked reinforced concrete subject to shear. For members with significant transverse reinforcement, one approach is to assume that the inclination can Chapter 2 - Shear Strength of members with Previously ... 20 reduce until the principal compressive stresses reach the diagonal crushing strength of the concrete. This approach is used in the General Method of shear design in the 1984 Canadian concrete code. In this chapter, an alternate approach has been proposed. The method involves checking the transfer of load across all possible crack inclinations, from the steepest, which is usually the initial crack, to the flattest, which is usually the most recent. The method used to check the transfer of load is a generalization of the modified compression field theory crack check procedure. It was found that for members with significant transverse shear reinforcement, the previous (steeper) cracks are generally most critical. For members with little or no transverse steel, the flattest cracks are always the most critical. In addition, regardless of the transverse steel ratio, the new procedure predicted the experimental data very well and was found to be consistent with the predictions of the modified compression field method. One of the main differences between members subjected to monotonic shear and members subjected to reverse cyclic shear is that the latter will have previously existing diagonal cracks due to previous load cycles. Any rational shear design method for seismic shear will need to account for this difference. One possibility is to develop empirical relationships for the softening and weakening of compression struts subjected to reverse cyclic action. Another possibility is to apply the method presented here in which the load transfer is checked across all possible crack inclinations. While the method of checking the transfer of load across all possible crack inclinations in a member subjected to reverse cyclic action would be more complicated to apply than an empirical equation, it would also be considerably more rational and would be more useful in trying to understand the phenomenon of reverse cyclic shear. Chapter! - Shear Strength of members with Previously ... 21 detail at crack Ca) Beam loaded in shear Cb) Calculated average (c) Local stresses at a crack stresses Figure 2.1 - Comparison of average stresses and stresses developed locally at a crack (Collins and Mitchell, 1987) Figure 2.2 - Comparison of concrete constitutive law for the MCFT and the proposed method Chapter 2 - Shear Strength of members with Previously .. Figure 2.4 - Longitudinal forces developed in top and bottom chord of beam subject to shear stresses (Collins and Mitchell, 1987) Chapter 2 - Shear Strength of members with Previously ... 23 350 FLATTEST CRACK 9 - 34" MCFT Proposed Method fc2 - fc2max 6 - 27* CRACK SLIDING e - *r 0.5 1 1.5 2 PRINCIPAL TENSILE STRAIN % 2.5 Figure 2.5 - Comparison of predicted response for a beam with 0.40% transverse reinforcement Figure 2.6- Comparison of predicted response for a beam with 0.20% transverse reinforcement Chapter 2 - Shear Strength of members with Previously 24 350 300 250 5" &. 8 200 p x CO 150 100 50 0 STIRRUPS YIELDING 6 -33* CRACKING e • r o 4 0.5 FLATTEST CRACK 8 - 3 2 * CRACK SLIDING G - 32* fc2 • fc2mu e - ir -Proposed Method 1.5 2 2.5 PRINCIPAL TENSILE STRAIN % 3.5 Figure 2.7 - Comparison of predicted response for a beam with 0.05% transverse reinforcement 400 300 z LU O rr O 200 u. tx x CO 100 MCFT fc2 » fc2max FLATTEST CRACK CRITICAL LONGITUDINAL REINFORCEMENT YIELDING CRACK SUDING 0.2 0.4 0.6 AMOUNT OF TRANSVERSE REINFORCEMENT (%) Figure 2.8- Comparison of predicted and experimental shear strength 0.8 Chapter 2 - Shear Strength of members with Previously ... 25 400 AMOUNT OF TRANSVERSE REINFORCEMENT (%) Figure 2.9 - Shear strength predicted without concrete tensile stresses Chapter 3 - Towards a Rational Seismic Shear Model 3.1 - Introduction Chapter two dealt with the issue of previously existing diagonal cracks on the shear capacity of concrete members. The equations used in that chapter were generalizations of the equations used for monotonic shear - no modification was made to include the effects of cyclic action. The intent of this chapter is to look at some of the important effects of cyclic loading. In particular, the magnitude of compressive and shear stresses at a crack interface will be studied. Also, previous work looking at the effect of cyclic loads on the shear transfer mechanism at a crack has been reviewed and some results will be presented. It was initially anticipated that this work would ultimately lead to the development of a new method which would be able to predict the seismic shear capacity of existing elements; but, this was simply not possible. However, this chapter presents some ideas which will help to contribute towards the development of a seismic shear model. Section 3.2 looks at the effect of crack inclination on the compressive and shear stresses on a crack interface, while Section 3.3 presents the results of the literature review with emphasis on previous work on the degradation of shear transfer through aggregate interlock. The final section presents some additional ideas and provides some recommendations for the further development of a seismic shear model. 3.2 - Influence of Crack Inclination on Interface Stresses The currently accepted method for the design of seismic shear in reinforced concrete elements is to neglect any contribution from tensile stresses in the cracked concrete, especially in areas of plastic yielding. In other words, the applied shear is assumed to be resisted solely by transverse steel and the concrete contribution is omitted. However, in members with previously existing diagonal cracks, such as members subject to cyclic loads, if the crack deviates from the assumed truss inclination, than concrete 26 Chapter 3 - Towards a Rational Seismic Shear Model 27 stresses must be mobilized at a crack interface and tensile stresses will be induced in the concrete between cracks. The following discussion illustrates this point. Suppose an unequally reinforced concrete element (such as a column or beam with light transverse reinforcement) is subject to pure shear in which the magnitude is such that the reinforcement in the weak direction has yielded. Figure 3.1a represents a possible stress state in the concrete under these conditions it is assumed that the concrete resists no tensile stresses and that the principal compression is inclined at 30 degrees to the longitudinal direction of the member (i.e., the x direction). The Mohr's circle representing the corresponding steel stresses has been omitted for simplicity. Now assume that there exists a 45 degree crack (caused by initial loading for example). Figure 3.1b represents the portion of concrete stresses needed to maintain equilibrium for a typical 45 degree truss model. The figure shows that there are no stresses normal to the crack interface and that the compressive stress in the transverse direction (i.e., the y direction) equilibrates the tension in the transverse steel, i.e., Avfv-Acfc = 0 3.1 fc = (Av/Ac)fv = pyfv This is the assumption of a 45 degree truss model. The third stress field, Fig. 3.1c, is the net difference between the previous two, and represents the portion of the total stress field that must be carried by a concrete contribution. Note that the resultants of this third stress field constitute a concrete truss with diagonal tension members normal to diagonal compression members. In the above discussion both the inclination of the total concrete stress field and the crack inclination were assumed. In reality these may vary over a considerable range. The procedure above was repeated for a range of crack inclinations from 0 degrees to 90 Chapter 3 - Towards a Rational Seismic Shear Model 28 degrees, and for total stress field inclinations of 20 degrees and 45 degrees in addition to 30 degrees. The results are summarized in Fig: 3.2 in terms of the total shear carried by the stirrup contribution and the portion carried by a concrete contribution. It can be seen that whenever the crack inclination is steeper than the assumed truss inclination (i.e., the inclination of the total stress field), a concrete contribution must be mobilized. For a very steep crack,, almost all the. shear must be carried by the concrete stresses at a crack interface. This makes intuitive sense since a vertical crack crosses no stirrups. It is apparent that even if the contribution from the tensile stresses developed in the concrete is neglected in the design of the member, the stress state at cracks which are steeper than the assumed truss inclination may be critical and should be checked. Since it is believed that seismic loading degrades the shear capacity of an element by reducing the Vc component (due to a reduction in aggregate interlock) it is important to understand the magnitude of these stresses and how they are transmitted across the crack interface. To look at the stresses which are developed on an interface of a crack which is not aligned with a principal stress direction, consider a non-isotropically reinforced concrete element subject to shear. For the purpose of this discussion, a 30 degree uniaxial compression stress field has been chosen to represents the stress state in the concrete member. In reality, there will be some principal tensile stresses developed in the concrete, but these have been ignored as they tend to complicate the analysis and do not significantly affect the trend of the results of the proceeding discussion. It should be noted however, that the member must have at least minimal transverse reinforcement otherwise no shear could becarried in the concrete under these assumptions; Figure 3.3a, b and c represent the casein which the stress field remains fixed (at 30 degrees), but the crack angle rotates from a low of 30 degrees to a high of 60 degrees. Mohr's circles indicate that as the crack angle initially starts to deviate from the assumed stress field inclination, stresses are mobilized along the crack interface. The shear stress along the crack interface increases rapidly while the corresponding rate of increase in the Chapter 3 - Towards a Rational Seismic Shear Model 29 compressive stress is much slower. At a crack angle of 45 degrees the ratio of shear to compressive stress is 3.73. However, as the crack angle continues to deviate, this ratio starts to decrease. At 60 degrees the ratio of shear to compressive stress drops to 1.73. Figure 3.4 plots the variation of both compressive and shear stress as a function of crack angle for a 30 degree uniaxial compressive stress field. It can be seen that there is a unique relationship between shear on a crack and compression on a crack as a function of crack angle. This analysis was repeated, but using a 45 degree stress field instead of the 30 degree stress field as before. The compressive and shear stresses, as a function of crack angle, are plotted in Fig. 3.5. This is almost identical to Fig. 3.4 except that it has been translated 15 degrees along the x-axis and that both the compressive and shear stresses, relative to the applied shear, have been scaled by the same magnitude from the previous case. Recognizing the similarity between Figs. 3.4 and 3.5, a third graph was produced in which the ratio of compressive stress to shear stress was plotted against the "deviant" angle, where the deviant angle is defined as the difference between the inclination of the stress field and the crack angle. As can be seen from Fig. 3.6, the results are identical for both inclinations of the initial stress field. This is an interesting observation because as has already been presented, the ability to transmit shear across a crack interface is a function of the compressive stress (Eq. 2.3). If a simplified model is considered, such as a shear-friction analogy, where the shear that can be resisted at an interface is directly proportional to the compressive stress (by a constant of proportionality, i.e., the friction coefficient) then it becomes clear that the ratio of compressive stress to shear stress is more important than the absolute amounts of stress. Subsequently, it can be seen from the figures that there is a range of deviant angles (less than 45 degrees) in which it is more likely that a shear failure will occur. Some important aspects come to light after studying these fundamentals of shear transfer. First, even though one can design a beam so that the transverse steel carries all Chapter 3 - Towards a Rational Seismic Shear Model 30 the shear, in all likelihood, cracks form randomly, and hence, aggregate interlock must be mobilized to pick up the shear which the transverse steel cannot. Second, the portion of the applied shear carried by the concrete increases rapidly with increasing crack inclination relative to the stress field inclination. Third, each crack interface has a unique ratio of shear to compressive stress induced on it depending on the "deviant" angle. The general trend is for the ratio of compression to shear to gradually increase as the deviant angle increases until it rises asymptotically to mfmity at an angle of 90 degrees. The next section considers the effect of cyclic action on the ability to transfer shear across a crack interface based on research that has been done in this area. Section 3.3 - Effect of Cyclic Loading on Shear Transfer By including the effect of previously existing diagonal cracks, Chapter two made a significant step forward in developing a rational model to predict the seismic shear strength of concrete members. However, as previously mentioned, no attempt was made to modify any of the empirical or analytical expressions to account for degradation of the shear carrying mechanisms in a member subject to cyclic loading. This section attempts to provide some ideas of how the degradation of the shear carrying mechanisms may be quantified and incorporated into the proposed method, presented in Chapter two, by looking at research already carried out. Many researchers have done work in the area of shear transfer by the mechanism of aggregate interlock under monotonic loading. Research into the effect of cyclic shear on aggregate interlock has not been so prolific. The main body of research into cyclic shear has been in one of two main categories, the first being in the area of fatigue. Fatigue is the phenomena related to the strength and stiffness degradation of concrete due to low intensity - high cycle loading. This research was carried out to better understand the effects of waves on off-shore structures. The second and more relevant category was to research the effect of high intensity - low cycle loading on the shear resistance of concrete elements. Chapter 3 - Towards a Rational Seismic Shear Model 31 However, the aim of almost all of this research was to establish the cyclic stress-strain relationship under this loading history. This body of experiments was carried out to study the effects of earthquake loads on pre-cracked nuclear containment vessels ( the vessels being cracked from internal pressure). The shear strains looked at in these studies were quite small compared to the shear strains induced by seismic events on normal structures. Only a couple of researchers have studied the effect of high intensity - low cycle cyclic loading on the ultimate shear carrying capacity of reinforced concrete members. These studies were very fundamental in nature, isolating the mechanism of aggregate interlock under cyclic loading. Of these studies, only one gives conclusions and recommendations on the degradation of aggregate interlock at ultimate conditions. The following summarizes the work done by the researchers in the second category and a discussion of the findings of the latter study. The bulk of the research done on the cyclic effects of shear transfer across a crack interface comes from research carried out at Cornell University. A group of researchers have worked together on preparing numerous papers and reports on the behavior of pre-cracked nuclear containment vessels subject to reversing seismic loads (White and Holley 1972; Jimenez, Perdikaris, Gergeley, & White, 1976; Jimenez, White, & Gergeley 1982). It should be noted that dowel action was also studied in some of these reports. However, it has been concluded by Mattock (1981) that for ultimate strengths, where the transverse reinforcement yields, dowel action plays a negligent role. The stress and strain levels considered in this research was far below ultimate values and consequently dowel action contributed significantly to the load deformation response of these specimens. A quick summary of the research by White and Holley (1972), which gives a good representation of the work done by these researchers, is described below. The specimens tested were cast monolithically and subsequently pre cracked along a pre-formed plane of weakness. Figure 3.7 shows a typical test specimen configuration. It should be noted the use of external restraints rather than embedded reinforcement in the Chapter 3 - Towards a Rational Seismic Shear Model 32 test set-up. This was done for ease in pre-setting crack widths and also to eliminate any contribution from dowel action. The test specimens were subject to fully reversed cyclic loads - typically 25 but up to 55 for some specimens. The shear stresses applied to the specimens were between 0.84 and 1.10 MPa. At the end of loading, one final cycle was applied to each specimen to a much higher, albeit still moderate, shear stress of 2.07 MPa. A typical plot taken from these reports is shown in Fig. 3.8. This particular specimen was subject to a constant shearing stress of 1.10 MPa. The initial crack width was set at .02 inches. It was then loaded for 25 cycles. The crack width was then increased to .03 inches and again loaded another 25 times. Finally, on one side the crack width was set at .015 inches and on the other by .03 inches. It was cycled an additional 5 times. In summary, the main conclusions drawn from this body of work relevant to this thesis study are threefold: First, slip in the shear plane and the width of the crack at the plane both increased under cyclic shear loading; however, the increase in crack width, with the exception of only two (out of sixteen), was less than 0.254 mm from an initial crack width of 0.762 mm. Second, the force in the clamping bars remained low during cycling (about 30 - 50% of applied shear), but grew considerably larger during the last cycle (approximately equal to the applied shear). Third, the significant damage to the integrity of the shearing surfaces occurred during the final load application and not during the cyclic loading which was done at a low stress level. Paulay and Loeber (1974) also carried out monotonic and cyclic shear tests on specimens similar to the specimens tested by the former researchers (Fig. 3.9) In total 44 specimens were tested. Thirty-five were tested under monotonic loading and the remaining 9 were subject to repeated shearing stresses of about 5.6 MPa. Of the 35 monotonic experiments, 27 were carried out at a constant crack width of either .127, .254 or .508 mm. The other 8 specimens had crack widths adjusted linearly with shear stress (0.689 MPa/0.0508 mm). All monotonically loaded specimens were taken up to failure. The Chapter 3 - Towards a Rational Seismic Shear Model 33 cyclically loaded specimens were subject to 33 repeated loads of approximately 5.6 MPa each and then up to failure. As before, the main body of this report was concerned with the shear stress - shear strain relationship of the concrete. Figure 3.10 shows a typical plot of shear stress - shear displacement for a specimen with an initial crack width of 0.254 mm. A conclusion drawn from this research is the fact that the largest single factor affecting shear displacement is the width of the crack across which shear stresses are to be transferred. Within ranges considered, the shear displacement was found to be approximately proportional to the selected crack width. Mattock (1981), performed a series of tests to gather information on the effect of fully reversing cyclic loading on the mechanism of interface shear transfer. Eleven pairs of specimens were tested. One specimen from each pair was tested monotonically to failure and the other was subjected to a regimen of gradually increasing cyclic loading. The loading regime consisted of 10 cycles at 50% of the ultimate shear strength predicted by the ACI 318-77 concrete code. After these ten cycles, the specimen was subject to five additional cycles at an increased shear load of 58% of predicted ultimate (an additional 8%). After each set of five cycles, the load was increased by the same amount and loaded for five complete cycles. This was continued until the specimen failed. Test specimen details and the testing apparatus are shown in Fig. 3.11. Some interesting conclusions reported by Mattock were that at ultimate loads, dowel action will not be significant in resisting shear. This hypothesis is supported by the following: "... When an external compression force is applied to a shear plane, the shear transfer strength is increased by the same amount as would result from providing additional reinforcement with a yield strength equal to the external compression force (Mattock & Hawkins 1972). If significant dowel action Chapter 3 - Towards a Rational Seismic Shear Model 34 occurred after bar yielding one would expect a greater increase in shear strength from provision of the extra reinforcement than from the action of the equivalent compression force." Another interesting point is to do with the loading history. The effect of randomly varying loading history was not considered in these tests. However, because of the stable behavior of the specimens until just below the failure load, it is believed that the strength under randomly varying shearing stresses would not deviate significantly from the results obtained in this study. One of the main differences between Mattock and the other researchers is that Mattock allowed the crack width to dilate or contract as it wanted. The transverse steel cast into the specimens was the only means of applying compression to the crack interface. This is more representative of what happens in reinforced concrete. Another large difference was that, unlike the other experimental tests where the constitutive relationship of cyclic loading was sought, Mattock was only interested in the ultimate conditions. In other words, how does fully reversed cyclic loading effect the ultimate shear strength of reinforced concrete? In general, it was found that cyclic loading on the specimens degraded the ultimate shear stress transfer across an interface by only 20% of the monotonic case. Some specimens were cast in two sections fonning a cold joint which subsequently formed the crack interface. A bond-breaking agent was applied to the crack interface prior to casting the second half of the specimen. In this case, the strength degradation between ultimate monotonic and ultimate cyclic shear strength was approximately 40%. Section 3.4 - Additional Considerations for Developing a Seismic Shear Model An intrinsic quality of seismic induced displacements is that they are completely random. They do not gradually increase from small amplitudes to large and they do not Chapter 3 - Towards a Rational Seismic Shear Model 35 necessarily fully reverse. It would be wise then to establish the effect that the load history has on the ultimate load carrying capacity of ah element. Mattock has suggested that because of the stable nature of the hysteresis loops for his specimens until just below failure that the load history may not have a very significant effect upon the ultimate cyclic shear strength of an element. Also, from the experimental results presented from White and Holley (1972), the significant damage occurred on the last load cycle in which the specimen was taken up to its ultimate shear capacity. However, it is known that cycling a specimen has a strong influence on its stiffness, and as such, directly influences the shear and flexural forces induced in the specimen. Another characteristic of seismic events or of the interaction between seismic events and structures is that the displacements of a structure is essentially the same whether it behaves linearly or bi-linearly. This introduces the idea of ductility. The National Building Code of Canada (1991), allows a reduction in design force as long as the elements are detailed such that they have adequate ductility to safely carry the loads through the imposed displacements. This is the major difference for the design of concrete elements subject to monotonic loads and seismic loads. Consequently, a model predicting the seismic shear strength of a reinforced concrete element, should also address the issue of ductility. . Chapter 3 - Towards a Rational Seismic Shear Model 36 Figure 3.1a - Uniaxial concrete compressive stress field inclined at 30 degrees 45 degree crack Figure 3.1b - Concrete compressive stresses due to 45 degree "truss model" 45 degree crack Figure 3.1c - Concrete stresses which must be resisted by shear at the crack face Chapter 3 - Towards a Rational Seismic Shear Model 37 Crack Angle Figure 3.2 - Portion of shear carried by Vc and Vs as a function of stress field and crack inclination Chapter 3 - Towards a Rational Seismic Shear Model Figure 3.3b- 30 degree stress field and a 60 degree crack Chapter 3 - Towards a Rational Seismic Shear Model 39 Figure 3.5- Compressive and shear stresses on a crack interface as a function of crack angle using a 45 degree stress field Chapter 3 - Towards a Rational Seismic Shear Model 40 Figure 3.6 - Ratio of compressive stress to shear stress on a crack interface as a function of the deviant crack angle Chapter 3 - Towards a Rational Seismic Shear Model Restraint booms bolted to concroto 2 , ^-Rostralnt rods bottod to restraint booms + Figure 3.7 - Typical specimen and test apparatus (White and Holley 1972) Figure 3.8 - Typical test results (White and Holley 1972) Chapter 3 - Towards a Rational Seismic Shear Model -2-% bars Tension crack across shear plant 3* „ 3--it Stirrups Loading direction jQp 4j(11.4) BOTTOM '-'(0.64) 1 7 j (19.04) 1(0.64) -\'(1.9) Figure 3.9 - Typical test specimen (Paulay and Loeber 1974) Figure 3.10 - Typical test results (Paulay and Loeber 1974) Chapter 3 - Towards a Rational Seismic Shear Model 43 Shear Plane —..-4 J I I C U I I l \ _ QU- I0x5in. Closed stirrup Shear Transfer Reinforcement T Secondary ^ ( reinforcement, / l ^ d'o- holes #3bq Shear Transfer Reinforcement Section Faces "A" ihear plane-spherical seat ft f specimen m .load cell -pull-rod -60 K center -hole rams .glmbal support for rams Section A - A gripping plates A i _ r gripping rods glmbal support for rams Section B-B Figure 3.11 - Typical test specimen and loading apparatus used by Mattock (1981) Chapter 4 - Experimental Program 4.1 - Introduction This and the following two chapters describe the experimental study undertaken as part of this thesis. Eight reinforced concrete columns were tested to investigate the effect of cyclic loading on shear strength. Three parameters were studied during this program. They were the shear span, the axial load level and the amount of distributed longitudinal reinforcement. This chapter describes the testing apparatus, connection details, physical properties of the specimens, construction of specimens, testing procedure, data acquisition and instrumentation. Chapter 5 describes the test results for each specimen, including a brief synopsis of events during the test. Chapter 6 contains a discussion of the test results including the effect of particular parameters on the ductility and strength of the specimens. 4.2 - UBC Beam Element Tester The specimens which were tested as part of this study were loaded using a specially developed testing apparatus called the UBC Beam Element Tester. As the characteristics of the tester influenced the experimental program, the details of the tester are presented first. The tester was designed by Adebar (1993) so that it could be used for a wide variety of beam and column elements. The tester was constructed (assembled) and calibrated as part of the present study and, therefore, these details are described below. The apparatus is mounted horizontally on the strong floor of the structures laboratory. It consists of four separate floor mounts as shown in Fig. 4.1. The hydraulic actuators are attached to the two large end mounts. These actuators apply axial load and end moments to the specimen. The two smaller floor mounts, seen at the side of the drawing are used to attach two rigid links which resist the applied shear. Because only 44 Chapter 4 - Experimental Program 45 three actuators are needed for equihbrium, a third rigid link is used instead of a "fourth" actuator on the bottom end mount. The jacks and rigid links are attached to loading yokes - one on each end of the tester. Figure 4.2 shows a partial elevation of the Beam Element Tester. The jacks and rigid links are attached to both the floor mounts and the loading yokes by means of large steel pins. Aluminium bushings are inserted to keep the line of action of the actuators and the rigid links through the centre of the specimen so as to avoid eccentric loading. The floor mounts are attached to the strong floor by high strength threaded rods. Both end mounts have 6 rods, while the shear mounts have only 4 rods. The rods have a nominal diameter of 2 inches and are prestressed to approximately 890 kN. This provided in the case of the end mounts, approximately 5340 kN of vertical force and in the shear mounts approximately 3560 kN. This was enough to maintain a high enough slip load to prevent sliding of the floor mounts during the experiment. The unique aspect of this apparatus is that the loading applied at the ends of the specimen by the two yokes can be controlled individually, so that any combination of moment, shear, and axial load can be applied to the specimen. For example, if equal end moments are applied (single curvature), no shear is required for equilibrium, while equal and opposite end moments (double curvature) require a shear force of 2M/L. Consequently, a specimen tested in the Beam Element Tester can be fairly short even when large shear spans are desired. Also, by superimposing a constant axial load, either compression or tension, any combination of the three sectional forces may be applied. Figure 4.3 illustrates the flexibility with which the Beam Element Tester can apply different loads. Figure 4.4 summarizes the capacity interaction diagrams for the tester. As the hydraulic actuators have higher capacity in compression than tension, the interaction diagrams are asymmetric. Since the end reaction frames are independent of each other, they are free to be moved to any position to which they may be bolted to the strong floor Chapter 4 - Experimental Program 46 (a 600 mm by 600 mm grid). The maximum shear force applied to a specimen depends on its length, as shown in the shear-axial interaction diagram. Another unique characteristic of this tester is the manner in which the shear stress is applied to the specimen. Rather than apply the shear through a bearing plate on the compression face, the shear stress is distributed along the ends of the specimen in a reasonably uniform fashion. This is important because it reduces the chance that the shear will be transmitted directly to the support by a direct compression strut. 4.3 - Connection Detail Specimens are connected to the tester by end plates which are cast integrally with the specimen. The end plates have holes drilled through them so that bolts can be used to fasten the specimen to the face plate of the loading yoke. The face plate of the loading yoke has a grid of 5/8 inch diameter holes spaced 6 inches vertically and 3 inches laterally. The end plates of the specimens must have a corresponding network of holes. Figure 4.5 illustrates the bolting detail of the specimens. Typically, a specimen which is to be loaded in the Beam Element Tester has the longitudinal steel welded onto the end plates. Any intermediate or skin reinforcement may or may not be welded to the end plate. However, due to concerns of artificial restraint in the specimens which contained very little transverse reinforcement, the connection of the flexural steel to the end plate was modified. From previous experiments (Adebar 1989) it is known that if the longitudinal reinforcement is welded directly on to the end plate, there is opportunity for considerable compressive strut action and significant dowel action of the flexural reinforcement may be mobilised. In order to eliminate the restraint, it was decided to modify the generic end plate connection described earlier. The end plate was still cast integrally with the concrete specimen, however, the flexural steel passed through slots milled into the end plate. The rebar was welded to a flat bar, which could slide up and down on the outside of the end plate. \ Chapter 4 - Experimental Program 47 To facilitate easy connection to the loading yoke of the Beam Element Tester, 6 -Grade 5 bolts were welded to the end plate. These bolts were situated such that the head of the bolt was on the inside of the end plate and the shaft passed through the end plate to leave the exposed threaded shank on the outside. When placed in the tester, these exposed bolts would pass through the grid of holes in the loading yoke and could then be easily fastened. It can be seen ( Fig. 4.5) that there were two rows of three holes at the top and bottom of the end plate. These were for 6 - Grade 8 bolts which could be saved and reused after each specimen was tested. Placing of these bolts posed no problem as the concrete did not extend far enough to cover these holes. In order to apply the shear force to the concrete, 12 - 16 mm diameter by 116 mm long shear studs were welded to the inside of the end plate. To allow free movement of the sliding plates, spacer blocks were placed between the end plate of the specimen and the loading yoke of the Beam Element Tester. The photograph in Fig. 4.6 shows the inside of the end plate and the configuration of the shear studs. During testing of the second specimen, it was discovered that this elaborate end connection detail was not needed. Due to the placement of a stirrup close to the end plate, there was very little movement of the longitudinal reinforcement during the test. Consequently, artificial restraint was not a problem. In addition, it was observed during the testing that the longitudinal steel was not resisting any load on the compression cycle as the sliding bar welded to the longitudinal steel simply moved away from the end plate. It had been hoped that the bond stresses between the concrete and the longitudinal steel would be sufficient to yield the steel in compression. In light of the fact that artificial restraint was not a problem and because the longitudinal steel did not yield in compression, it was determined to weld the sliding bar to the specimen end plate. 4.4 - Test Specimens Eight reinforced concrete specimens were tested as part of this study. The overall Chapter 4 - Experimental Program 48 dimensions of all specimens was 405 x 405 mm x 1372 mm. The specimens all contained the same transverse reinforcement and were cast from the same batch of concrete. Six specimens had identical longitudinal reinforcement, while the remaining two had different amounts of distributed longitudinal reinforcement. The main flexural reinforcement consisted of 5 - 15M bars on both top and bottom faces (i.e., 10 bars in total) in all specimens. The first six specimens had 2 - 10M on both sides of the specimen (i.e., 4 bars in total). This arrangement of longitudinal steel is referred to as type "A". The seventh specimen had no 10M bars, while the eighth specimen had 4 - 10M bars on both sides (i.e., a total of 8 bars). For future reference, the configuration of specimen CS7 and CS8 will be referred to as type "B" and "C" respectively. Figure 4.7 distinguishes between the three different arrangements of longitudinal reinforcement. All eight specimens had the same amount of transverse reinforcement Each specimen contained 4 - 10M stirrups spaced at 305 mm (12 inch) intervals. The stirrups were hoops that completely encircled the longitudinal steel and terminated in 135 degree hooks which were anchored 60 mm into the concrete core. In addition to the arrangement of longitudinal reinforcing, two other parameters were studied during this experimental program. First, the influence of maximum shear span was studied. It should be noted that the term "shear span" in the context of this experiment, simply means the ratio of applied moment to shear at a point along the beam. In these specimens, where the shear was constant and the moment was linear, the maximum shear span was the distance from the point of inflection to the section of maximum moment - in this case - the end plate. The maximum shear span ranged from a minimum of 0.686 m to a maximum of 1.600 m. Second, the effect of axial load was investigated. One specimen was subject to axial compression, while two other specimens were subject to axial tension. The remaining specimens were not subject to any axial load Bridge columns are generally Chapter 4 - Experimental Program 49 under very little axial load, and as such, it is possible that seismic displacements could induce tensile forces in the columns (i.e., bridge bents must resist overturning forces where one column may go into tension). Therefore, the influence of this parameter on ductility was considered important. The axial load was applied as a ratio of the shear as this was thought to be most representative of the actual case. The following table (Table 4.1) summarizes the basic experimental parameters and loading conditions for each specimen Table 4.1: Summary of Geometry Type and Loading Condition Specimen Longitudinal Reinforcement Arrangement Shear Span (m) Axial Load N/V CS1 A 0.686 0 CS2 A 1.600 0 CS3 A 1.029 o CS4 A 1.029 -1.87 CS5 A 1.029 + 1.64 CS6 A 0.686 + 1.10 CS7 B 1.029 0 CS8 C 1.029 0 4.5 - Material Properties As mentioned in the previous section, the main flexural reinforcement were 15M bars, while the stirrups and distributed reinforcement were 10M bars. However, another difference between the two was that the 15M longitudinal steel was type G30.16, while the hoops and distributed reinforcement were the more common G30.12. The plots below in Figs. 4.8(a) and 4.8(b), show the measured stress - strain curves/for the different Chapter 4 - Experimental Program 50 reinforcement. Table 4.2 summarizes the properties of the two different types of reinforcing bars. The yield and ultimate strengths were calculated by dividing the measured loads (yield and ultimate) by the nominal areas. The concrete was obtained from a local ready-mix company. The concrete had a specified 28 day strength of 25 MPa. The maximum aggregate size was 19 mm and the specified slump was 75 mm. All specimens and cylinders were cast on November 12, 1992. Two sets of three cylinders were tested for compressive strength. The first set was tested December 16, 1992 (30 days old) and had an average compressive strength of 25.5 MPa. The second set was tested February 24, 1993 (104 days old) and averaged 31.1 MPa. Table 4.3 summarizes the date of testing of each specimen. Table 4.2: Summary of Steel Reinforcing Properties Size Specified yield strength f v (MPa) Measured yield strength fv (MPa) Ultimate measured strength f u (MPa) Nominal Area (mm^) 15M 400 485 640 200 10M 400 440 690 100 Table 4.3: Specimen Testing Date Specimen Time of Testing Age (Days) CS1 January 27, 1993 86 CS2 January 28, 1993 87 CS3 February 10, 1993 100 CS4 February 17, 1993 107 CSS February 26, 1993 116 CS6 March 5, 1993 123 CS7 March 11, 1993 129 CS8 March 17, 1993 135 Chapter 4 - Experimental Program 51 4.6 - Construction of Specimens The construction of the steel reinforcing cage was complicated by the fact that the flexural reinforcement was not welded directly to the end plate. A jig was constructed which allowed the 5 - 15M bars to be welded to the sliding bar. However, since this bar was to slide along the outside of the end plate, the weld had to be made on the outside of the sliding bar. Consequently, the sliding bar had 5 holes drilled into it, through which the rebar passed. The rebar was then welded on the outside. The photograph ( Fig. 4.9) shows the rebar and sliding bar assembly during construction. Once both top and bottom longitudinal reinforcement had been welded (one side only)j the free ends of the rebar were threaded through the milled slots in the end plate. Four stirrups were placed over the two sets of bars. The second end plate was then threaded over the reinforcement. This whole assembly was placed in a second jig which kept the end plates parallel and spaced at the proper length. Finally, the longitudinal steel, which protruded through the second end plate, had the second sliding bar placed over them and welded in place. The photograph (Fig. 4.10) shows the reinforcing cage in the jig. Once air the welding was completed, the intermediate skin reinforcement was placed into the core of the cage. The stirrups were then spaced at the appropriate intervals and tied at all four corners to the longitudinal steel. Once the stirrups were in place, the skin reinforcement was tied to the stirrups. The skin reinforcement, although not welded to the end plate, were affixed with small plates which could develop the yield strength of the bars. The space between these small plates and the specimen end plate was approximately 10 mm. Figures 4.11 and 4.12 show the completed cage for a type "A" specimen and a comparison between the three different types of specimens - type "A", "B" and "CV The formwork was designed so that the lateral pressure from the wet concrete would be self-equilibrating. The outer walls were reinforced with bracing. There were two separate forms which held four specimens each. Figure 4.13 illustrates the forms and Chapter 4 - Experimental Program 52 how the reinforcing cages fit into them. As the rebar fit through milled slots in the end plate which were larger than the rebar, an opening was left, through which upon casting, the concrete could flow. To prevent this concrete from blocking the sliding mechanism, rigid styrofoam plugs were inserted into the openings. Wet concrete was placed in the form in approximately three levels and vibrated at each level. The cylinders for testing were also cast at the same time. Once the specimens were finished, they were covered by polypropylene and allowed to cure for the next four days. Subsequently, the forms were removed and the specimens continued curing with all four sides exposed to the air. 4.7 - Testing Procedure The loading regime was consistent for all specimens in terms of numbers of cycles and ductilities with the exception of specimen CS3 (the details are given in Section 5.3 of the next chapter). Each specimen was cycled in sets of five fully reversed cycles at predetermined ductility levels. The specimens were subjected to increasing ductilities until failure. The displacement ductility was measured as a ratio of the displacement of the inflection point of the specimen at any given cycle to the displacement of the inflection point at flexural yield. In order to establish the yield displacement, the moment capacity of the section was calculated, using the ACI equivalent rectangular compression stress block factors and the actual material strengths. This moment will be referred to as the ideal moment, Mi. The shear corresponding to this ideal moment could then be calculated by dividing the moment by the shear span. It should be noted that the ideal moment capacity was a function of the specimen type (A, B or C) and the axial load level. In short, each specimen had a different shear level corresponding to its ideal moment. Table 4.4 summarizes the predicted flexural capacities and the corresponding shears. All specimens were first loaded up to a shear corresponding to 75% of the ideal Chapter 4 - Experimental Program 53 moment. The deflection measured at this load level was taken as the 0.75 displacement ductility level. The remaining four cycles at this ductility level were displacement controlled as were all subsequent cycles at increased ductility levels. The next ductility level was taken at a displacement ductility level of 1.5. That is, the specimen was loaded to a displacement equal to twice the displacement measured at a displacement ductility of 0.75. From then on the ductility levels were increased by 0.5 on each successive set of cycles. Figure 4.14 illustrates a typical loading history for a specimen. Table 4.4: Predicted Flexural Strengths and Corresponding Shears Shear Ideal moment capacity M i (kN*m) Shear span M / V (m) Corresponding shear M i / ( M / V ) (kN) CS1 195 0.686 285 CS2 195 1.600 122 CS3 195 1.029 190 CS4 265 1.029 258 CS5 150 1.029 146 CS6 150 0.686 164 CS7 163 1.029 158 CS8 224 1.029 164 If the actual maximum load level was considerably less than predicted, corrected ductility levels were calculated. This was done using the following procedure. From the hysteresis loop, the maximum shear capacity was recorded and this was multiplied by 0.75. The displacement corresponding to this reduced shear load was found on the initial loading curve (i.e., cycle 1A of the hysteresis loop). This displacement was taken as the corrected Chapter 4 - Experimental Program 54 0.75 ductility displacement. Using this procedure all the hysteresis loops were labelled with the corrected ductilities. It should be noted that this method is very sensitive to the shape of the initial loading curve. Since the specimens exhibited significant non - linear behaviour (relative to a perfectly elastic - plastic material) before yielding, the corrected ductilities are quite different from the initially assumed ductilities. 4.8 - Load Control, Instrumentation, and Data Acquisition The actuators were controlled by a multiple pressure hydraulic load maintainer made by Edison systems. The unit is hand operated and utilizes a cantilevered weight system to vary the pressure in each channel. The main hydraulic line entered the hydraulic load maintainer and split into six separate channels. These channels attached to six independent proportional valves and then were connected to the hydraulic actuators. The proportional valves, which controlled the pressure in the out going lines, were adjusted by a set of six cantilevered weights. The magnitude of the weight for each individual proportional valve could be adjusted, so that the pressure to each channel could be controlled. A mechanically operated crank advanced the weights along the cantilever, which allowed the individual proportional valves to increase the pressure to each channel. All channels increased by the same proportion, although depending on the individual weight, not to the same magnitude. Because the actuators generally required different pressures in tension than in compression (different areas), six different channels had to be used for the three jacks - three in tension and three in compression. Therefore, a hydraulic switchboard was used to facilitate the operation of the actuators. Connected to each actuator was a pressure transducer which measured the pressure of the hydraulic oil. The rigid links acted as load cells as they were fitted with strain gauges. A multi-channeled amplification/power supply unit was used to collect the signals from the pressure transducers and strain gauges. Displacement transducers were used to measure the axial and transverse Chapter 4-Experimental Program 55 displacement of the specimen; These two displacement transducers were fixed to a linear tracking system which itself was attached to the end plate of the specimen. The displacement transducers were then attached to a small steel pin which was epoxyed to the inflection point of all specimens except CS2. The inflection point of CS2 actually fell 225 mm off the end of the specimen. Therefore, rather than trying to measure the strains at the inflection point for this specimen, it was decided to use the opposite end plate as the reference point. The effect of this on the results are negligible (this will be discussed in more detail in Section 5.3). Originally, the tracking arm was attached to the end of the 1 inch thick end plate of the specimen. As it turned out, however, the end plate underwent significant rotational displacement, which was discovered when testing specimen GS2. This rotation, multiplied by the length of the tracking arm, gave displacements too large in one direction, whiie too short in the opposite direction. The tracking arm was then moved to the 50 mm thick end plate of the yoke, which was stiff enough to ehminafe the problem. However, there was still a chance that the end plate of the specimen and the yoke could experience a relative slip. Therefore, the tracking arm was moved again and welded to the center of the end plate of the specimen, which due to the nature of the connection, was restrained against rotation. The signals once amplified by the power supply, were sent into a computer equipped with a standard analog to digital board. The software program "Notebook" was used to collect and plot the data from the eight different channels of instrumentation. Chapter 4 - Experimental Program 56 Figure 4.1 - Plan view of Beam Element Tester Chapter 4 - Experimental Program Figure 4.2 - Partial elevation of Beam Element Tester Chapter 4 - Experimental Program Figure 4.3 - Method of load application using three actuators (A) and three rigid links (R) Chapter 4 - Experimental Program Axial force (kN) Bending moment compression + ^ tension Axial force (kN) Figure 4.4 - Tester capacity interaction diagrams Chapter 4 - Experimental Program 60 Figure 4.5 - Connection detail of specimen endplates Chapter 4 - Experimental Program 61 Figure 4.6 - Inside face of endplates showing shear stud configuration Type A Type B Type C Figure 4.7 - Configuration of distributed steel for Types A, B and C Chapter 4 - Experimental Program 15M Weldable Reinforcement kN 10M Distributed Reinforcement % strain Figure 4.8a and 4.8b - Typical stress-strain relationship for longitudinal (15M) and distributed reinforcement (10M) Chapter 4 - Experimental Program Figure 4.9 - Sliding bar assembly Figure 4.10 - Reinforcing cage in construction jig Chapter 4 - Experimental Program 64 Figure 4.11 - Completed reinforcing cage for Type A specimen Chapter 4 - Experimental Program 65 Figure 4.12 - Comparison of reinforcement for Types B, A and C Chapter 4 - Experimental Program 66 Figure 4.14 - Typical loading regime for test specimens Chapter 5 - Experimental Results 5.1 - Introduction This chapter describes the details of the individual tests for each specimen. In all cases, the characteristics of the specimen and the loading regime are first summarized. Subsequently, a brief overview of the events of the test are presented. As previously mentioned the predicted ductility of each specimen was corrected using the procedure outlined in Section A.1. These corrected values have been presented (in brackets) along with the predicted ductility in the following discussion. The hysteresis loops, presented in following figures, have also been labelled with the corrected ductilities. 5.2 - Specimen CS1 Specimen CS1 had the standard cross section - type "A". The ultimate moment capacity of the section, calculated using the ACI compression block factors and using experimentally determined material strengths with material factors taken as unity, was 195 kN*m. The specimen was loaded in double curvature without any axial load. The point of inflection was at the center of the specimen and consequently the moment to shear ratio at the end of the specimen (location of maximum moment) was 0.686 m. Thus, the shear force applied to the specimen corresponding to the ideal moment was 285 kN. The specimen was loaded up to a force level corresponding to 75% of this shear (i.e., 214 kN). This shear force corresponded to ah inflection point displacement of 10.12 mm. As shown in the photographs (Fig. 5.1) cracks initiated as flexural cracks and turned into diagonal shear cracks at approximately 90 kN. As the load increased, more shear cracks formed. At a shear force of 214 kN there was in total 4 diagonal cracks on south end of specimen (left side of photograph) and three diagonal cracks on the north end. As a matter of convention, the initial direction of loading will be referred to as cycle "A" while loading in the opposite direction will be denoted as cycle "B". For example, cycle 4A at 67 Chapter 5 - Experimental Results 68 ductility 1.5, means the fourth cycle in the initial direction at a ductility level of 1.5. The specimen was unloaded and then was loaded in the reverse direction (cycle "IB"). The crack pattern was similar to that exhibited in the first direction; however, as the load increased, it became apparent that the second flexural-shear crack on the south end of the specimen was becoming critical. In fact, the specimen failed in shear at a load of 197 kN. As the specimen had not reached the ideal flexural capacity of the section, the failure exhibited little ductility. Figure 5.2 shows the hysteresis loop for this specimen. The critical shear crack formed between the first and second stirrup. Therefore, the stirrups probably did not yield. The photograph in Fig. 5.3 clearly shows the failure mechanism. On the top face of the specimen, the concrete spalled off between the second and third stirrup. It can be seen that the longitudinal reinforcement has come away from the adjacent concrete due to the shear deformation. This displacement was the cause of the spalling of the concrete between the stirrups. 5.3 - Specimen CS2 Specimen CS2, which was identical to CS1* was tested with the inflection point 1600 mm away from the heavily loaded end . As the specimen was only 1372 mm long, the inflection point was located 228 mm off the far end. Rather than try to rig a device to measure the displacements at the inflection point, the reference point was taken at the far end of the specimen (i.e., 1372 mm away from the heavily loaded end). The error in using this point to measure ductility demand is believed to be small. The reason being that the initial cycle to determine the yield displacement was load controlled. To take the specimen to say twice this initial ductility (as was done in this test) it was a simple matter of inducing twice the deformation at the reference point. The error results in the fact that deformations are made up of two components - namely flexure and shear. If the ratio of these deformations stay fixed, then there is no error in this procedure. However, if the proportion of deformation from shear increases, than measuring the strains away from the Chapter 5 - Experimental Results 69 inflection point becomes an approximation. However, since the specimen was completely dominated by flexure (and the difference in location of the reference point was small) the error was considered negligible. This specimen was subject to no axial load. The ideal moment for this specimen, as for CS1, was 195 kN*m. With this configuration, the applied shear force corresponding to the ideal moment was 122 kN. During the first ductility level, the specimen was loaded to a shear force of 107 kN (the shear span was mistakenly taken to be the length of the specimen. The specimen should have been loaded to only 92 kN.). The development of the crack pattern was, as expected, dominated by flexure. However, as discussed in detail in Chapter 4, problems arose with the tracking arm during the testing of this specimen. As such, the typical loading of this specimen during the first ductility level had to be altered. After the modifications were completed, it was determined to subject the specimen to only as many load cycles as necessary to establish a relatively stable hysteresis loop. It was found that two complete cycles were adequate. The displacement corresponding to a ductility level of 0.75 (a corrected ductility of 1.15) was 12.5 mm. The specimen was then subjected to five fully reversed cycles at a ductility of 1.5 (corrected ductility of 2.3) After the first couple of cycles, large flexural cracks formed near the end plate. The longitudinal steel buckled upon compression which caused the concrete to spall off in the area of buckling. Although, the picture (Fig. 5.4) shows the cover concrete still intact in the damaged area, it is not contributing to the strength of the member. The hysteresis loop for these five cycles were quite stable and showed little deterioration of the peak shear load. The specimen was then subject to five cycles at a displacement ductility of 2.0 (corrected ductility of 3.1). The concrete had now completely spalled off the west side of specimen from the end plate up to the first stirrup. At this ductility level, the buckled longitudinal steel had forced the first stirrup to bow out. As can be seen from the Chapter 5 - Experimental Results 70 hysteresis loop (Fig. 5.5) the peak shear load, especially on the "A" side, has dropped considerably from approximately 117 kN to 80 kN. This drop, being greater than 20% of the initial shear strength, constitutes failure. However, as the reverse direction did not exhibit such a sharp decrease in strength, the specimen was subject to further deformations. The specimen was finally subject to five fully reversed cycles at a displacement ductility of 3.0 (corrected ductility of 4.5). Concrete spalled off completely on both sides of the specimen up to the second stirrup. The longitudinal steel continued to buckle and the lateral deflection of the stirrups increased (i.e. bowing due to the buckling of the flexural steel). The degree of damage of the member can be seen in the photograph (Fig. 5.6). The hysteresis loop also exhibits the extent of degradation of the strength of the specimen. In both directions of loading the peak shear load dropped dramatically on each successive cycle. Testing was terminated after the fifth full cycle. 5.4 - Specimen CS3 This specimen was identical to the first two. The ideal moment was 195 kN*m and again, the specimen was subject to no axial load. However, the shear span was changed to 1.029 m. Consequently, the shear associated with the ideal yield moment was 190 kN. Since the previous two specimens were tested with inflection points at 0.686 m and at 1.600 mm, and had failed in shear and flexure respectively, this test represented a middle ground between the two. Consequently, it was still unknown whether the third specimen would fail in shear or flexure. For this reason it was decided to decrease the jump in ductility levels between successive sets of cycles. Therefore, this specimen was tested at ductilities that increased by 0.25 in each successive series. The specimen was first subject to five cycles at a ductility level of 0.75 (corrected ductility of 0.96). The crack pattern can be seen in Fig. 5.7. No degradation of the peak shear load was recorded for the cycles at this ductility level. In addition, for the next two ductility levels, namely 1.0 and 1.25, (corrected ductilities of 1.28 and 1.60 respectively), Chapter 5 - Experimental Results 71 no appreciable decrease in the peak shear load was found. At a ductility level of 1.75 (corrected ductility of 2.23), the specimen started to exhibit a loss in peak shear load in the "B" direction of loading. On the last cycle in the initial loading direction (i.e. direction "A") there was a large loss in stiffness of the member. The stiffness loss was accompanied by a large flexure - shear crack at the end plate. However, the stiffness (on the last cycle) was unaltered in the "B" direction. It was decided to load the specimen for a sixth cycle at this ductility level. Upon loading for the "A" side, the stiffness greatly increased from the previous cycle. The reverse direction exhibited the same stiffness as the previous cycle. One can see from the plot of the hysteresis loop (Fig. 5.8) the extent of the degradation of stiffness and the subsequent increase in stiffness in the sixth cycle. The reason for this apparent anomaly is unknown. At a ductility level of 2.0 (corrected ductility of 2.55), degradation of the peak shear load from one cycle to the next was exhibited in both directions of loading. The loss of strength was approximately 20%. It was decided that as the important aspects of the specimen behaviour had been observed, (i.e. flexure or shear dominated) that an increase in ductility from 2.0 to 2.5 (corrected ductility of 3.2) was justified. The specimen was subject to two and a half cycles of load before it failed in shear. The shear crack which formed on the first cycle to a ductility of 0.75 (corrected ductility of 0.96) remained critical throughout the whole test (Fig. 5.9). The shear failure was initiated because the cyclic load tended to crush the concrete underneath and adjacent to the outer most flexural steel. As this area degraded, the flexural steel deformed under the load from the stirrups. This deformation allowed the stirrups to become rounded at the corners. As a consequence, the cracks were able to widen without much resistance from the stirrups. Subsequently, the shear resistance degraded rapidly as well. As a result, the failure was relatively sudden with the second stirrup pulling out of the concrete core due to the excessive deflections and cracking (Fig. 5.10). Chapter 5 -. Experimental Results 72 5.5 - Specimen CS4 Specimen CS4 was loaded similar to specimen three in the sense that the point of inflection was located at a shear span of 1.029 m. However, in addition to the applied shear and moment, an axial compressive force was also applied. This axial load was applied proportionally to the applied moment. When the moment was zero, the axial load was zero and when the moment reached the ideal yield moment, the axial load was at 10% of the gross axial compressive strength (i.e., 'Q.\*Ag*f£. As the specimen was subject to an axial load, the ideal moment for the section changed, even though the specimen was identical to the previous three specimens. The ideal moment was calculated to be 265 kN. The shear load corresponding to this ideal moment was 258 kN. As was seen from the previous specimen, it was determined that a 0.5 jump in ductility was small enough to give the full behaviour of the specimen. Therefore, I specimen 4 was subject to a load history at ductility levels of .75, 1.5, 2.0 and 2.5. The response of the specimen at a ductility level of 0.75 (corrected ductility of 0.82), was quite stiff compared to specimen CS3. It subsequently lost some stiffness but remained very stable from the second cycle to the fifth. At a ductility level of 1.5 (corrected ductility of 1.65), a large shear crack formed at a relatively flat angle (due to the presence of axial load). This crack can be seen in the photograph (Fig. 5.11). This crack seemed to have degraded the response of the specimen (on the "A" side) throughout the rest of the load cycles. It can be seen from the hysteresis loop (Fig. 5.12) that side "A" suffered more than side "B". In fact, at this ductility level side "A" was very close to its limit of a 20% reduction in its flexural capacity. The specimen was taken up to a ductility level of 2.0 (corrected ductility of 2.19). From the hyteresis loops it can be seen that the specimen exhibited a large increase in stiffness in the first cycle in the "B" direction. It is thought that the increased compression force in the previous cycle (in the "A" direction) re-established significant aggregate interlock causing the increase in stiffness. On subsequent cycles at this ductility, the specimen showed a relatively stable Chapter 5 - Experimental Results 73 behaviour. Upon loading to a ductility of 2.5 (corrected ductility of 2.74), the specimen failed in the first "A" cycle. Failure was along the major shear crack that developed in the first cycle of 0.75 displacement ductility. Figure 5.13 clearly shows the failure of the specimen along the crack. It should be noted the magnitude of buckling of the longitudinal steel and the extent to which this buckling has caused the stirrups at either end of the crack to bulge in the center. Once again, as in specimen CS3, the cycling has enabled the stirrups to become rounded in the corners, which allowed the crack to open without a lot of resistance from the stirrups. 5.6 - Specimen CS5 Specimen CS5 was subject to an axial tensile load in addition to the applied moment and shear. In order to keep the ideal moment capacity high enough to induce a reasonably large shear, it was necessary to use a reduced axial load relative to the compressive load applied to CS4. As was done with the previous specimen, the axial load was loaded proportionately to the applied moment, such that, when the ideal moment was reached, the specimen was subject to an axial load of 0.05*Ag*fc (as opposed to 0AQ*Ag*fc for Specimen CS4). During the application of the first of the cycles at ductility level 0.75 (corrected ductility of 0.97), the effect of axial tension on the specimen became apparent. As shown by the photograph (Fig. 5.14), the specimen exhibited multiple cracks perpendicular to the longitudinal axis at the location of the first 3 stirrups. Only one significant diagonal shear crack formed in the initial cycles. By the end of the fifth cycle at a ductility level of 1.5 (corrected ductility of 1.93), a significant diagonal shear crack pattern had developed. The hysteresis loop was very stable and exhibited little loss in the peak shear load (Fig. 5.15). However, flexural cracks had developed at the end plate which widened upon their respective tension cycle to approximately 4 or 5 mm. These large cracks translated into large rotations with relatively little resistance. This explains the high degree of pinching in Chapter 5 - Experimental Results 7 A the hysteresis loop. At a ductility of 2.0 (corrected ductility of 2.58), extensive bond splitting had occurred along both the top and bottom of the specimen (i.e. the location of longitudinal steel), and flexural cracks increased in width. The shear cracks remained very stable. Upon unloading the diagonal shear cracks closed up. Very little degradation of peak shear load occoured. At a ductility level of 2.5 (corrected ductility of 3.22), the hysteresis plots began to show a reduction in the peak shear load with increasing number of cycles. At ductility 3.0 (corrected ductility of 3.87), the reduction in the peak shear load had reduced by about 20% after the fifth cycle. Figure 5.16 clearly shows the extent of cracking of the specimen at this level of ductility, as well as the spalling that occurred at the corners of the specimen. The specimen was finally loaded to a ductility level of 3.5 (corrected ductility of 4.51). The specimen lasted for the full five cycles, however, the peak shear load degraded constantly until, on the last cycle, it was only about 55% of the peak shear on the first cycle. Figure 5.17 shows the extent of damage of the specimen. The concrete had spalled off on the east side from the end plate to the second stirrup. Conspicuously absent from this specimen, as compared to the last specimen, was the large degree of buckling of the longitudinal steel. Since there was a large reduction in the extent of buckling of the flexural steel (due to the tensile axial load) there was a corresponding reduction in the lateral distortion of the stirrups. This may contribute to why this specimen failed in flexure while the previous specimen, under axial compression failed in shear, albeit at a much higher shear load. 5.7 - Specimen CS6 Specimen CS6 was loaded under the same axial tensile force as specimen CS5. However, the shear span was reduced to 0.686 m to try and induce a shear failure. Since this specimen was the same type as specimen CS5 and was subject to the same axial force, Chapter 5 - Experimental Results 75 the ideal moment was also the same - namely 150 kN*m. However, the reduction in shear span increased the corresponding shear to 164 kN. It should be noted that this test was similar to the first test with the exception of the axial tensile force. The first five cycles of loading at a ductility of 0.75 (corrected ductility of 0.89), the specimen exhibited quite a stiff response as expected (short shear span). The specimen developed significant diagonal cracks in both directions of loading (at both ends). The pictures clearly show the symmetric pattern of cracking (Fig. 5.18). The specimen was then subject to five more cycles at a ductility level of 1.5 (corrected ductility of 1.79). At this ductility, large shear cracks opened. At the worst spot, the shear crack had opened to approximately 5 mm. The stirrups still helped control the width of the shear cracks within the region of the stirrups. The peak shear load had dropped from approximately 190 kN down to 105 kN on cycle "B". However a large percentage of the drop had occurred between the first and second cycle. Figure 5.19 reveals the extent of bond splitting. In fact, the whole east side of the specimen had developed a bond split running from end to end. The specimen was subject to a further five cycles at ductility 2.0 (corrected ductility of 2.38). The hysteresis loop (Fig. 5.20) was quite pinched; however, the degradation of the peak shear load was not nearly as evident as it was for the previous ductility. The specimen developed more shear cracks and the bond tension cracks opened wider. The concrete on the east of the specimen had all but spalled off; however, the specimen still exhibited considerable strength. It was decided to take the specimen up to a displacement ductility of 2.5 (corrected ductility of 2.98). The specimen was loaded, but before the displacement ductility could be reached, the specimen started to fail in a combination of shear and bond failure. The concrete spalled completely off the east side of the specimen. The pictures show the extent of damage to the specimen (Fig. 5.21). Chapter 5 - Experimental Results 76 5.8 - Specimen CS7 Specimen CS7 had type "B" reinforcement which did not have skin reinforcement. The only longitudinal steel was the five 15M bars on the top and bottom of the specimen. This specimen was not subject to axial load. The ideal moment capacity was calculated to be 163 kN*m. The shear span of this specimen was 1.029 m. The shear corresponding to the ideal moment was 158 kN. The specimen was subject to the typical loading history of five cycles at 0.75, 1.5 and 2.0 displacement ductilities (corrected ductility of 0.71, 1.42, and 1.89 respectively). During the cycles the specimen behaved very well. There was little degradation of the peak shear load for all three ductility levels. The most significant observation of these cycles was the relatively large branching of cracking at the level of the longitudinal steel compared to the interior of the beam where there was no reinforcement. During the cycles at a ductility level of 1.5, the flexure - shear cracks were opening to a width of approximately 1.5 mm in the worst place and between 0.3 and 0.5 mm in the location where a stirrup intercepted the diagonal crack (Fig. 5.22). The specimen was then subject to five more cycles at a ductility of 2.5 (corrected ductility of 2.36). As can be seen from the hysteresis loop (Fig. 5.23), the specimen had basically failed because the peak shear load had dropped far below 80% of the peak. The average width of the shear cracks had increased to approximately 5 mm and up to 7 or 8 mm in spots. The shear slip represented approximately half of the transverse displacement of the specimen. It was decided, however, to cycle the specimen at a ductility level of 3.0. On the first cycle at this ductility, the specimen failed in shear (Fig. 5.24). The critical shear crack had formed on the first cycle at a ductility of 0.75 and had remained critical throughout the test. The failure of this specimen in shear was rather ductile as the stirrup never yielded but rounded and deformed with the ultimate sudden shear failure occurring with the stirrup pulling out of the core concrete. Chapter 5 - Experimental Results 11 5.9 - Specimen CS8 Specimen CS8 had type "C" reinforcing. Instead of two 10M bars, this specimen had four 10M bars on each side. The ideal flexural moment calculated for this specimen which was not subject to any axial load, was 224 kN*m. The shear span, as in the previous test was 1.029 m. The shear corresponding to this ideal moment was 164 kN. Although these bars were included to study the effect of crack control on the shear resistance of the specimen, it also enhanced the moment capacity which was reflected in the larger ideal moment. However, as the skin reinforcement was not welded to the end plate (to prevent artificial restraint), the skin reinforcement did not pick up as much of the applied moment as first thought. Not until sufficient deformations were encountered did the skin reinforcement start adding to the moment resistance of the section. As a result, on calculating the shear load level for 0.75 displacement ductility, the calculations erred on the high side. Consequently on loading die specimen, it started to yield just as it reached the calculated 0.75 ductility shear load. This posed no real problem as the true ductility level was determined after the experiment had been completed. After the specimen was subject to five complete cycles at a ductility level of 0.75 (corrected ductility of 1.0), it was subject to another five cycles at a ductility level of 1.5 (corrected ductility of 2.0). However, on the "B" side of loading, during the first cycle, a manual error caused the hydraulics to increase rather than decrease the pressure upon unloading. Subsequently, before the situation was brought under control, the specimen was subject to deformations corresponding to a ductility level of approximately 4.5 (corrected ductility of 6.0). Needless to say, the end of the specimen was damaged too severely to continue testing on this end. The photograph (Fig. 5.25) shows the damaged east side of the specimen due to the high loading. Fortunately, the inflection point of this specimen was at the three quarter point which left the north end of the specimen undamaged (this was similar to specimen CS2). Therefore, the specimen was taken out, turned around and replaced in the Beam Element Tester. Chapter 5 - Experimental Results 78 The specimen was once again loaded up to a displacement ductility of 0.75. However, it become obvious after the first complete cycle that the previously damaged end would cause problem at higher ductility levels. Figure 5.26 illustrates the asymmetric crack pattern which developed after the first full cycle. The specimen was removed and any loose concrete was chipped out and a steel collar was fit around the base of the specimen and grout placed between the collar and the specimen. Figure 5.27 shows the specimen after the retrofit. It was decided that since the specimen had already been subject to one cycle at 0.75 ductility, that only four more cycles should be applied. The subsequent crack pattern was quite symmetric as was the hysteresis loop (Fig. 5.28). The specimen was then subject to five cycles at a ductility of 1.5. The specimen was fairly stable with only about 11% loss in peak shear load. Shear cracks were fairly well controlled with widths being on average 0.5 to 0.6 mm and in the worst spots 0.9 mm. (This should be compared to the 1.5 mm seen in specimen CS7). The specimen was loaded for another five cycles at a ductility of 2.0 (corrected ductility of 2.65). The crack pattern had developed into a myriad of fine cracks which enabled the main diagonal shear cracks to remain fairly stable. By the fourth full cycle it had become evident that the specimen would not fail in shear and that flexural failure would govern. As can be seen by the hysteresis loop, the degradation of the peak shear load had dropped by half. Significant bond splitting had occurred on both faces of the specimen. The specimen was subject to two full cycles at a ductility of 2.5 (corrected ductility of 3.32). As can be seen from the hysteresis loops, the flexural capacity had degraded so much that it could not induce a large enough shear to cause a shear failure. Even one stirrup could resist the shear force induced by the final cycle at this ductility. Figure 5.29 illustrates the crack pattern for the final testing stages of specimen CS8. Chapter 5 - Experimental Results 79 Figure 5.1- Specimen CS1 after initial cycle of loading Chapter 5 - Experimental Results 80 CS1 HYSTERESIS PLOTS Figure 5.2 - Hysteresis loop for Specimen CS1 Chapter 5 - Experimental Results 81 Figure 5.4 - Specimen CS2 at ductility level 1.5 (corrected ductility of 2.3) Chapter 5 - Experimental Results 82 CS2 HYSTERESIS PLOTS REFERENCE POINT DEFLECTION [mm] Figure 5.5 - Hysteresis loop for Specimen CS2 Chapter 5 - Experimental Results 83 Figure 5.6 - Specimen CS2 at ductility level 3.0 (corrected ductility of 4.5) Figure 5.7 - Specimen CS3 at ductility level 0.75 (corrected ductility of 0.96) Chapter 5 - Experimental Results 85 CS3 HYSTERESIS PLOTS 200 -i 1 1 INFLECTION POINT DEFLECTION [mm] Figure 5.8 - Hysteresis loop for Specimen CS3 Figure 5.9 - Specimen CS3 at ductility level 2.5 (corrected ductility of 3.2) Chapter 5 - Experimental Results 87 Figure 5.10 - Failure of stirrup anchorage in Specimen CS3 Figure 5.11 - Specimen CS4 at a ductility of 1.5 (corrected ductility of 1.65) Chapter 5 - Experimental Results 89 CS4 HYSTERESIS PLOTS 300-1 — i — r— -300 -i 1 1 1 i 1 r 1 r-—I T r — i i 1 i r 1 — f 1 -18 -14 -10 -6 -2 2 6 10 14 18 I N F L E C T I O N P O I N T D E F L E C T I O N [ m m ] Figure 5.12 - Hysteresis loop for Specimen CS4 Chapter 5 - Experimental Results Figure 5.13 - Specimen CS4 crack pattern at failure Figure 5.14 - Specimen CSS at a ductility of 0.75 (corrected ductility of 0.97) Chapter 5 - Experimental Results 92 CS5 HYSTERESIS PLOTS H 1 1 —i 1 1 h l i i '—I -25 -15 -5 5 15 25 INFLECTION POINT DEFLECTION [mm] Figure 5.15 - Hysteresis loop for Specimen CS5 Chapter 5 - Experimental Results Figure 5.17 - Specimen CS5 crack pattern at failure Chapter 5 - Experimental Results 94 Figure 5.18 - Specimen CS6 at a ductility of 0.75 (corrected ductility of 0.89) UNIVERSITY B C WFBSTEH M»R I '33, Figure 5.19 - Specimen CS6 at a ductility of 1.5 (corrected ductility of 1.79) Chapter 5 - Experimental Results 95 CS6 HYSTERESIS PLOTS -200 H — i — i — i — i — i — i — i — r — i — I — i — i — i ' i i—i—H 1 1 1 -10 -8 -6 -4 -2 0 2 4 6 8 10 INFLECTION POINT DEFLECTION [mm] Figure 5.20 - Hysteresis loop for Specimen CS6 Chapter 5 - Experimental Results 96 Figure 5.21 - Specimen CS6 crack pattern at failure S P E C I M E N DUCT 11_ 1T < C 1 C L E SHEAR UNIVERSITY B C | ; T f D MHRJ0IS2 Figure 5.22 - Specimen CS7 at a ductility of 1.5 (corrected ductility of 1.42) Chapter 5 - Experimental Results 97 CS7 HYSTERESIS PLOTS INFLECTION POINT DEFLECTION [mm] Figure 5.23 - Hysteresis loop for Specimen CS7 Chapter 5 - Experimental Results Figure 5.25 - Damaged end of Specimen CS8 Chapter 5 - Experimental Results Figure 5.26 - Asymmetric crack pattern due to severe damage at oppositeend of Specimen CS8 Chapter 5 - Experimental Results 100 Figure 5.27 - Retrofit of damaged end of Specimen CS8 Chapter 5 - Experimental Results 101 Figure 5.28 - Hysteresis loop for Specimen CS8 Chapter 5 - Experimental Results Figure 5.29 - Specimen CS8 crack pattern at failure Chapter 6 - Discussion of Experimental Results 6.1 - Introduction The eight specimens can be divided into four separate series to study the effect of three variables. The three variables are shear span, axial load level and longitudinal reinforcement distribution. Figure 6.1 summarizes how the specimens can be grouped into the four different series. For each test series, a brief overview is given, discussing the results in terms of the variable being studied. A second section, comparing the analytical results from the proposed method (computer program) and the ACI code is presented. Results from the computer program and the ACI code are discussed with consideration of the mode of failure as well as ductility levels. 6.2 - Influence of Shear Span 6.2.1 - Summary of Results Specimens CS1, CS2 and CS3 make up a series that can be used to study the effect of shear span, which is the ratio of the maximum bending moment in the element divided by the shear force. That is, the maximum shear span is equal to the distance from the point of inflection (zero moment) to location of maximum moment. Table 6.1 gives a brief summary of the results for three specimens with different maximum shear spans (M/V)^. Specimen CS1, with the shortest shear span, failed suddenly in shear. Specimen CS2, with the longest shear span, was very ductile and failed in flexure. Specimen CS3, with a shear span between the other two, exhibited considerable flexural ductility, but ultimately failed in shear. Figure 6.2 compares the three specimens after failure and Fig. 6.3 shows the hysteresis loops for the three specimens. 103 Chapter 6 - Discussion of Experimental Results 104 Table 6.1: Summary of Results - Test Series One Specimen <M/V)max [ml Axial Load N/fcAg Section Type Maximum Ductility Failure Mode CS1 0.350 0 A < 0.75 Brittle shear failure CS2 0.700 0 A 3.0 Ductile flexural failure- no indication of any shear problem. Lost considerable stiffness. CS3 1.270 0 A 2.6-3.3 Flexure dominated but ultimately failed in shear. Subject to more load increments than normal. Shear span effects the ductility of a member indirectly by determining the governing mode of failure of the element. It is obvious that if a member fails in brittle shear (no flexural yielding) the member does not even reach a ductility level of one. However, if the shear span is long enough to produce a flexural failure or even just to delay a shear failure (as occurred in CS2 and CS3 respectively), it seems that a ductility level of 2.0 can be justified by the experimental results. Consequently, when trying to predict the ductility of a reinforced concrete element, it is essential to have accurate models to predict the ultimate sectional shear capacity and flexural capacity of the element. The ability to predict the potential shear demand on an element versus the ability to resist this demand is key in predicting the seismic shear response of the element. Although Specimen CS1 failed after the first cycle it still provides valuable information. From the photograph in Fig. 6.3, it can be seen that the crack formed between the first and second stirrup. Although dowel action was mobilized in this failure, the transverse steel did not act as modelled in a typical truss analogy. Consequently, when dealing with small amounts of transverse steel, especially when the spacing is relatively large compared to the depth of the member, the typical truss equations for V s may not be realistic. This situation is probably worsened in the case of cyclic shear as successive cycles may allow the crack to work its way between stirrups. The spacing of transverse Chapter 6 - Discussion of Experimental Results 105 steel according to the ACI code (as well as the Canadian code) should be limited to d/2. However, reviewing the typical reinforcement of bridge piers, it was found that most columns had stirrups at 12 inches on center, regardless of the dimensions of the piers. Therefore, the specimens which were considered full scale, also had stirrups spaced at 12 inches. However, the test specimens were quite small in comparison to existing sections, and therefore, in future experimental tests, consideration should be given to scaling down the specimens and using closer spaced ties. 6.2.2 - Comparison of Analytical Predictions and Experimental Results When using the ACI code equations, it is understood that the maximum shear span is to be used for calculating Vc (i.e., the shear span measured at the face of the end plate). This is not the case with the proposed model (see Chapter 2 for details on the proposed model). The model accounts for the influence of moment on the shear strength of an element by calculating the effect that the moment has on crack widths. The proposed model uses crack width to calculate the ability of the specimen to transmit shear by means of aggregate interlock. Therefore, for the proposed model, it makes practical sense to measure the critical shear span at the point on the diagonal shear crack which was initiated by flexure. This gives the moment which must be resisted by the flexural steel for this diagonal crack. Since it is not known prior to testing where the critical crack will be initiated, it is assumed, and verified by experiment, that at a point "d" from the end plate is a reasonable choice in which to calculate the shear span. Consequently, calculations were done at a shear span of 350 mm, 1270 mm and 700 mm for Specimens CS1, CS2 and CS3 respectively. The ACI code proposes several equations for predicting the shear strength of reinforced concrete members; three of which were used in this comparison. The three equations are 11-4, 11-6, and 11-9 in the ACI 318-83 code and are presented below. ACI Eq. 11-6 suggests a dependence on the maximum shear span for the V c component. ACI Chapter 6 - Discussion of Experimental Results 106 Eq. 11-4 is used for members subject to axial compression, while ACI Eq. 11-9 is used for members subject to axial tension. ACI Eq. 11-4 ACI Eq. 11-6 ACI Eq. 11-9 1+ Nu 2000 Aj Jfcbwd Vc = [l.9^/c-r2500pw^jbwd Vc = 2\ 1+-500A Wcbwd 6.1 6.2 6.3 where AL M„ = factored axial load normal to cross section occurring simultaneously with Vu lbs (positive for compression) = factored moment at section, lb-in = factored shear force at section, lbs = gross area of section, sq. in = web width, in = distance from extreme compression fiber to centroid of tension steel = As/byjd Table 6.2 below tabulates the predictions from the ACI 318 - 83 Building Code, the proposed method and the experimental values for the first three specimens. As can be seen, two predictions were made using the proposed model. The first prediction (Vp r e d ultimate), assumed that the longitudinal reinforcement did not yield. This was accomplished by effectively raising the yield strength of the longitudinal reinforcement. It was done this way so as not to effect the crack width calculations which, as previously discussed, are used to determine the ability of the concrete to transmit shear across the crack. This enabled a comparison of the proposed model with the ACI shear equations, Chapter 6 - Discussion of Experimental Results 107 regardless of the flexural capacity of the section. A second prediction using the proposed method (Vp r e d yield), was made allowing the longitudinal steel to yield (if the strain in the reinforcement reached the yield strain). Table 6.2: Analytical Predictions and Experimental Results - Test Series One Specimen M/V Mi Mj/(M/V) V ACI Eq. VACI Eq. V pred V pred V exp [m] [kN*in] 11-6 ultimate yield CS1 0.686 195 285 255 241 273 254 215 CS2 1.600 195 122 243 241 226 123 116 CS3 1.029 195 190 248 241 249 187 165 Note: All values are in kN unless noted otherwise It will be shown that the ratio of the shear calculated from the proposed model (without yielding of the flexural steel) to the shear associated with flexural yielding gives a good indication of the mode of failure of the specimen. If for instance this ratio is large, than there is ample reserve in the member to resist the shear induced by flexural yielding of the section. Consequently, the member is likely to fail in flexure - even if there is some degradation of the shear transfer mechanism due to cyclic loads. On the other hand, if this ratio is low, than a brittle shear failure is probable. It is apparent from the table, that as the shear span increases (M/V), the shear force at failure decreases. This trend is consistent throughout the table with the exception of the simplified ACI prediction which is not dependent upon the shear span. As can be seen for specimens CS2 and CS3, the flexural calculations based on a stress block approach give almost identical solutions to the proposed model (Vp r e d yield). This close correlation is expected as specimens CS2 and CS3 were dominated by flexure. The proposed method (Vpted yield) overestimated the shear strength of Specimens Chapter 6 - Discussion of Experimental Results 108 CS1, CS2 and CS3 by 18%, 6%, and 13% respectively. A large part of the disparity between the predicted and actual shear capacity of Specimen CS1 can be accounted for by the fact that the actual shear crack formed between stirrups, and therefore, never fully realized the contribution from the transverse steel. The predictions using the proposed model (Vp r e d ultimate) can be compared to the ACI code shear predictions as they both neglect the flexural capacity of the section. The prediction for shear using Equation 11.6 and the proposed model (Vp r e d ultimate) are within 7%,8% and 0.5% for specimens CS1, CS2, and CS3 respectively. It is interesting to note that the proposed method predicts that shear span has a larger effect than does the ACI code equations. The main point of interest in this test ( and the following tests) is how well the ductilities achieved by an element can be predicted by the ratio of the sectional shear capacity of an element to the shear associated with flexural yielding. Table 6.3 tabulates the ACI code and the proposed methods prediction (Vp r e d ultimate) for the sectional shear strength of these three specimens. For completeness, the shear calculated by the proposed method (Vp r e d yield) is also included. The ratio of sectional shear strength to the shear associated with flexural yielding is calculated in three different ways. The first ratio is calculated using the ACI code prediction for shear (ACI Eq. 11-6) and the shear associated with the ideal moment capacity, Af,-. The second ratio substitutes the ACI shear prediction with the shear calculated from the proposed model, (Vp r e d ultimate). The third ratio is calculated using the prediction from the proposed model for both terms -i.e., (Vp r e d ultimate)/(Vpred yield). It can be seen that the magnitude of these ratios generally correlate well with the type of failure experienced by the specimen. The lower the ratio, the more likely the specimen will fail in shear. Conversely, as the ratio increases, the specimen tends to be governed by flexure. Knowing the mode of failure gives a good indication of the expected ductility of the specimen. Chapter 6 - Discussion of Experimental Results 109 Table 6.3: Sectional Shear Capacity and Shear Demand from Flexural Yielding Test Series One Specimen M;/ (M/V)max (A) V ACI (B) Vpred (yield) (Q (ultimate) (D) (B/A) (D/A) (D/C) Failure Mode CS1 285 255 254 273 0.89 0.96 1.075 Brittle shear failure. CS2 122 243 123 226 1.99 1.85 1.837 Ductile flexural failure - no indication of any shear problem. Lost considerable stiffness. CS3 190 248 187 249 1.31 1.31 1.332 Flexure dominated but ultimately failed in shear. Subject to more load increments than normal. Note: All values are in kN except for dimensionless ratios As the ultimate shear capacities predicted by both models are relatively close, the ratios are also fairly consistent. These ratios do not always correlate as well as will be seen in the following sections. 6.3 - Influence of Axial Load 6.3.1 - Summary of Results There are two separate test series which were used to investigate the effect of axial load on the shear strength and ductility of reinforced concrete members. The first test series involved specimens CS3, CS4 and CS5. Specimen CS1 and CS6 composed the second series. The only difference between the two was that the specimens in the first series were all tested with a maximum shear span of 1.029m while the second set were tested at a smaller shear span of 0.686m. Tables 6.4 and 6.5 below summarize the important properties and parameters of both sets of test specimens. As can be seen, the specimens within each separate test series differed only by the axial load applied to the Chapter 6 - Discussion of Experimental Results 110 specimens during testing. As explained in Chapter 5, the axial load was applied proportionately - i.e., the axial load reached its specified level when the specimen was subject to its ideal moment. The hysteresis loops (Fig. 6.4) illustrate the trend of increasing shear strength with axial compression with ultimate shear loads of 130 kN, 165 kN, and 240 kN for specimens CS5, CS3, and CS4 respectively. Figure 6.5 compares the three specimens after testing. Table 6.4: Summary of Results - Test Series Two Specimen [ml Axial Load N/r c A S Section Type Corrected Ductility Failure Mode CS3 1.029 0 A 2.6-3.3 Flexure dominated but ultimately failed in shear. Subject to more load increments than normal CS4 1.029 -0.10 A 2.2 Similar to CS3. Barrelling of plastic hinge region due to compressive load. Shear failure. CSS 1.029 +0.05 A 4.0 Ductile flexural failure - no indication of any shear problem. Hysteresis was quite pinched. Table 6.5: Summary of Results - Test Series Three Specimen <M/V)raax[m] Axial Load N/r c A g Section Type Corrected Ductility Failure Mode CS1 0.686 0 A < 0.75 Brittle shear failure CS6 0.686 +0.05 A 1.0-1.85 Shear failure. Large loss of strength from 1st to 2nd cycle. Regained large portion of strength on subsequent ductility level. More ductile than CS1. The interesting trend of this test series is how the axial load level influenced the Chapter 6 - Discussion of Experimental Results 111 ductility of the specimens. It can be seen from the hysteresis plots that specimen CS3 and CS4 failed in shear, while CS5 was governed by flexure. It can also be seen that, although the strength has been diminished, the ductility of the specimen has been enhanced from a low of approximately 2.2 (specimen GS4) to a ductility of approximately 4.0 (specimen CS5). Specimen CS3 falls in the middle at approximately 3.0. Figure 6.6 compares the hysteresis loop for specimen CS1 and CS6. Specimen CS6 reveals a dramatic increase in ductility over specimen CS1. It should be noted that the only difference between these two specimens was that specimen CS6 was subject to a small axial tensile load (maximum 5% offfAg). Figure 6.7 compares these two specimens after failure. These test series confirm the expectation that compressive stress increases the shear strength of reinforced concrete elements. However, it also illustrates the adverse effect of axial compression on the ductility of the member. It becomes apparent that special consideration should be made when evaluating the strength of elements which are subject to axial compression. On the other hand, although axial tension diminishes the shear strength of an element, it significantly enhances the ductility of the specimen. 6.3.2 Comparison of Analytical Prediction and Experimental Results The ACI code presents two simple equations for the prediction of V c for concrete elements subject to axial loads. Equation 11-9 predicts the shear strength of the member when subject to tension, Equation 11-4 predicts the strength for a member under compression. Equation 11-6 may also be modified to predict the shear capacity of a member subject to axial compression. As before, these equations are not dependent on the flexural capacity of the section. Table 6.6 and 6.7 tabulate the analytical predictions from the ACI code, the proposed method and the experimental results for the two test series. For test series two, it can be seen that the proposed method (Vp r e d yield) predicts a 13%, 1% and 16% increase in shear strength from what was seen in the experimental Chapter 6 - Discussion of Experimental Results 111 values for specimens CS3, CS4 and CS5 respectively. It is believed that some of this experimental decrease represents the degradation of the shear transfer mechanism when subject to cyclic loads. Table 6.6: Analytical Predictions and Experimental Results - Test Series Two Specimen M/V [m] Mi [kN*ra] Mi /(M/V) V ACI Eq. 11-6 VACI Eq. 11-4,9 V pred ultimate V pred yield V exp CS3 1.029 195 190 248 241 249 187 165 CS4 1.029 265 258 248 275 270 247 245 CSS 1.029 150 146 248 168 236 149 129 Note: All values are in kN unless noted otherwise Table 6.7: Analytical Predictions and Experimental Results - Test Series Three Specimen M/V [ml M; [kN*m] Mj/(M7V) V ACI Eq. 11-6 VACI Eq. 11-4,9 V pred ultimate V pred yield V exp CS1 0.686 195 285 255 241 273 254 215 CS6 0.686 150 219 255 174 255 204 191 Note: All values are in kN unless noted otherwise For test series three, the proposed method (Vp r e d yield) over estimated the shear strength for specimens CS1 and CS6 by 18% and 8%. As discussed in section 6.2.2., the large stirrup spacing, which allowed the diagonal crack to form between the transverse steel, partially accounts for the large discrepancy between the predicted shear strength and the experimental result for CS1. Table 6.8 below summarizes the shear capacity and associated shear demand from flexural yielding for the specimens involved in test series two. It can be seen that the ratio Chapter 6 - Discussion of Experimental Results 113 of shear capacity predicted by the proposed method (Vp r e d ultimate) to the shear demand predicts quite well the relative ductility of each specimen. Specimen CS4, which failed in a relatively brittle fashion, has a low ratio of 1.05. Specimen CS5, on the other hand, which did not fail in shear, but exhibited large ductilities due to yielding of the longitudinal steel, has a large value of 1.62. Specimen CS3, which was subject to no axial force, has a ratio of 1.31 which fell in the middle of the other two specimens. This is consistent with the observed behaviour of Specimen CS3, which was not as brittle as CS4 but not as ductile as CSS. Use of the ACI equations in the prediction does not work well as the ACI code is conservative in predicting the shear strength of members in tension. Table 6.9 below summarizes the same results for test series three. Again, the proposed model predicts a more ductile failure for the specimen subject to higher tensile loads. As can be seen, specimen CS6 has a shear capacity to demand ratio of 1.16 compared to a ratio of 0.96 for CS1. The reduction in ultimate shear capacity predicted by the proposed method (V d ultimate), is not as great as the reduction in shear associated Table 6.8: Sectional Shear Capacity and Shear Demand from Flexural Yielding Test Series Two Specimen Mj/ (M/V)inax (A) V ACI (B) V pred (yield) (C) V pred (ultimate) (D) (B/A) (D/A) (D/C) Failure Mode CS3 190 248 187 249 1.31 1.31 1.332 Flexure dominated but ultimately failed in shear. Subject to more load CS4 258 275 247 270 1.07 1.05 1.093 increments than normal. Similar to CS3. Barrelling of plastic hinge region due to compressive load. Shear CS5 146 168 149 236 1.15 1.62 1.584 failure. Ductile flexural failure - no indication of any shear problem. Hysteresis was quite pinched. Note: All values are in kN except for dimensionless ratios Chapter 6 - Discussion of Experimental Results 114 Table 6.9: Sectional Shear Capacity and Shear Demand from Flexural Yielding Test Series Three Specimen ivy (M/V)inax (A) V ACI (B) V pred (yield) (C) Vpred (ultimate) (D) (B/A) (D/A) (PIC) Failure Mode CS1 285 255 254 273 0.89 0.96 1.075 Brittle shear failure. CS6 219 174 204 255 0.79 1.16 1.250 Shear/bond failure. Large loss of strength from 1st to 2nd cycle. Regained large portion of strength on subsequent ductility level. More ductile than CS1. Note: All values are in kN except for dimensionless ratios with flexural yielding. However, the predictions using the ACI code equations suggest the exact opposite. The reduction in predicted ultimate shear capacity for Specimen CS6 is greater than the reduction in flexural capacity - hence the ratio drops. Lxwking at the shear capacity predictions from the ACI code equations and the proposed method (Vp r e d ultimate), it can be seen while the proposed method predicts only a 7% decrease in the ultimate shear capacity of the section, the ACI code equations predict almost a 30% loss in strength. These observations are consistent with what was seen in the test series involving Specimens CS3, CS4 and CS5 and reinforce the fact that the ACI code equations are conservative when dealing with combined shear and tension. 6.4. - Influence of Distributed Reinforcement 6.4.1 - Summary of Results The final series, consisting of specimens CS3, CS7 and CS8 investigated the influence of longitudinal steel placement on the shear strength of reinforced concrete elements. Table 6.10 summarizes the properties and parameters for this test series. Other than the difference in distributed steel, the three specimens were identical and subject to no axial load. Chapter 6 - Discussion of Experimental Results 115 Table 6.10: Summary of Results - Test Series Four Specimen [m] Axial load N/rcAg Section Type Corrected Ductility Failure Mode CS3 1.029 0 A 2.6-3.3 Flexure dominated but .ultimately failed in shear. Subject to more load increments than normal. CS7 1.029 0 B 1.9-2.4 Dominated by flexure but ultimately failed in shear. CS8 1.029 0 c 2.0-2.6 Similar to CS7 but did not fail in shear. The specimen exhibited smaller diagonal crack than CS7. The actual results from this series were not significantly different to draw any quantitative conclusion. The ductility levels achieved for all three specimens were comparable. The main point of interest is that while Specimen CS3 and CS7 ultimately failed in shear, Specimen CS8 did not exhibit any indication of an imminent shear failure. The hysteresis loops for the three specimens are shown in Fig. 6.8. It can be seen that Specimen CS7 is significantly different from the other two. The difference between CS7 and CS3 is much more prominent than is the difference between CS3 and CS8. One possibility may be the fact that the crack control is significantly enhanced by the initial addition of intermediate steel, while any additional steel provided after that provides a diminishing effect on crack control. This is verified by the proposed computer model prediction of the strength of the three specimens in which the predicted shear was 173, 187 and 190 kN for Specimen CS7, CS3 and CS8 respectively. Figure 6.9 compares the crack pattern for these three specimens at failure. It can be seen from the photographs the enhanced crack control properties of Specimen CS8. A fine network of diagonal cracks have formed in the plastic hinge region of the section. Specimen CS7 lacks this extensive crack pattern. Chapter 6 - Discussion of Experimental Results 116 6.4.2 - Comparison of Analytical Predictions and Experimental Results The predictions from the ACI code and the proposed computer program are presented with the experimental results in Table 6.11. While the predicted shear strength, using the proposed model (Vp r e d yield) for specimens CS3 and CS8 are quite similar to one another at 187 kN and 190 kN respectively, the predicted shear strength for specimen CS7 Table 6.11: Analytical Predictions and Experimental Results - Test Series Four Specimen M/V [m] Mi [kN*mJ Mi (M/V) V ACI Eq. 11-6 VACI Eq. 11-4,9 V pred ultimate V pred yield V exp CS3 1.029 195 190 248 241 249 187 165 CS7 1.029 163 158 246 241 222 173 162 CS8 1.029 224 218 251 241 262 190 182 Note: All values are in kN unless noted otherwise is quite low at 173 kN. These predictions show analytically the dependence of shear strength on the crack control properties of the section. The distribution of longitudinal reinforcement can influence the shear capacity of a section by controlling the formation of cracks and hmiting the crack widths. This allows more aggregate interlock which increases the shear capacity. Although the experimental program results reveal that CS3 is stronger than CS7, and CS8 is stronger than CS3,- the results do not show as much increase with distributed steel as the proposed method predicts. However, it is believed that the load history of specimen CS3 caused a higher decay in flexural stiffness, and hence, a lower shear demand, than for either specimen CS7 or CS8. If specimen CS3 had been subject to the identical loading regime as the other two specimens, it is believed a strength increase close to that predicted by the proposed method would have been seen. The proposed method predicted a 13%, 7% and 4% strength increase over the experimental values for Chapter 6 - Discussion ofExperimental Results 117 specimens CS3, CS7 and CS8 respectively. Table 6.12 summarizes the shear capacity predictions from the ACI code and the proposed model. It can be seen from the shear capacity/demand ratios, that both the ACI code equations (B/A) and the proposed method (D/A), rank the specimens in the same order of ductility with Specimen CS8 being the most brittle, CS3 being less brittle and CS7 being essentially ductile. The experimental results, however, were just the opposite. It is thought that the distributed steel did not develop its full tensile capacity and consequently, the specimens never reached their theoretical moment capacities. Therefore, the specimens Table 6.12: Sectional Shear Capacity and Shear Demand from Flexural Yielding Test Series Four Specimen Mj/ (M/V)max (A) v ACI Vpred (yield) (C) V pred (ultimate) (D) (B/A) (D/A) (D/C) Failure Mode CS3 190 248 187 249 1.31 1.31 1.332 Flexure dominated but ultimately failed in shear. Subject to more load increments than normal. CS7 158 246 173 222 1.56 1.41 1.283 Dominated by flexure but ultimately failed in shear. CS8 218 251 190 262 1.15 1.20 1.379 Similar to CS7 but did not fail in shear. The specimen exhibited smaller diagonal crack than CS7. Note: All forces in kN unless noted otherwise benefited from the enhanced shear capacity provided by the extra reinforcement, while the shear demand, associated with flexural yielding, was reduced. On the other hand, the ratio using the proposed method for both the capacity and demand (Vp r e d ultimate/Vpred yield) suggests that even if the longitudinal steel was welded to the endplate, (increasing the shear demand on the specimen), the ductility of the member would have increased. This is because the prediction from the proposed model, V p r e d Chapter 6 - Discussion of Experimental Results 118 yield, accounts for the demand on the longitudinal steel from shear, and as such, reduces the flexural capacity of the section relative to the ACI prediction. 6.5 - Influence of Load History It is believed that by subjecting the specimens to fully reversed repeated loads, the shear transfer mechanism was degraded. In these specimens, were the transverse steel provides a significant portion of the shear resistance, this degradation is believed to be in the order of 10% (of the total shear resistance). This might explain the consistent higher strength predicted by the proposed method (Vp r e d ultimate) and the ACI code (with the exception of CS7 for the ACI prediction), where neither prediction accounts for the loss in shear strength due to cyclic loads It was felt that this effect was most pronounced in Specimen CS3, as this specimen was subject to cyclic loads at 0.25 ductility increments rather than the standard 0.5 increments. The best illustration of this effect was demonstrated in the last test series to study crack control. The proposed program predicts that specimen CS3 and CS8 would have relatively similar shear capacities while CS3 and CS7 should have a significant difference. However, the experimental program did not verify this prediction. 6.6 - Summary The first series yielded the conclusion that the larger the shear span the more likely the failure of the specimen will be dominated by flexure, and hence, a more ductile failure mechanism. The next two test series established the important fact that axial compression increases the flexural capacity faster than increasing the shear capacity. Consequently, the mode of failure of the specimen tends to go from that of flexure to one of shear and results in a element with less ductility. The important issue is that tensile loads enhance the ductility of specimens subject to seismic loads. The final test series confirmed the trend that better crack control can improve the shear strength of a reinforced concrete element by Chapter 6 - Discussion of Experimental Results 119 maintaining aggregate interlock by ensuring small crack widths. One issue that is important in a experimental program is the fact that the crack control parameter should be controlled in such a manner as to not increase the flexural capacity of the section. It is questionable to what extent this experimental program succeeded in that endeavour, however. These conclusions lead to a singular observation. All three parameters which were investigated in the experimental program can be thought of as how they affected the ratio of shear capacity of the section to the shear associated with yielding of the tension reinforcement. For instance, increasing the shear span, decreases the shear demand, while maintaining the sectional shear strength of the section. Increasing the axial compression increases both the flexural capacity ( and hence shear demand) and the sectional shear strength. The last series predicts an increase in sectional shear strength with increasing crack control, while having less affect on the shear associated with flexure. The four test series have shown that it is important to use a rational theory to predict the monotonic shear capacity of an element under all types of loading, to give meaningful predictions to the strength and ductilities of reinforced concrete elements subject to cyclic loads. It is proposed that using the ratio of the ultimate shear capacity calculated from the proposed method to the shear associated to flexural yielding gives a good indication of the mode of failure of an element subject to bending and shear forces. It is suggested that a ratio greater than 1.6 would suggest that a flexural failure is probable with expected ductilities around 3.0. Likewise, a ratio less than approximately 1.2 implies a brittle shear failure (i.e., no ductility). Between these two values would be all failures predominately governed by flexure but failing in shear at ductilities between 1.0 and 3.0. Chapter 6 - Discussion of Experimental Results Figure 6.1 - Grouping of specimens into the four test series Chapter 6 - Discussion of Experimental Results 121 Figure 6.2 - Comparison of Specimens CS1, CS2 and CS3 at failure SHEAR [kN] 300 200 100 0 -100 -200 -300 Chapter 6 - Discussion of Experimental Results 123 - CS4 -30 -10 10 30 SHEAR [kN] 300 200 100 0 -100 -200 -300 - CSS -30 •10 10 30 Chapter 6 - Discussion of Experimental Results 1 2 4 Figure 6.5 - Comparison of Specimens CS3, CS4 and CS5 at failure Chapter 6 - Discussion of Experimental Results 125 SHEAR [kN] 300 200 100 0 -100 -200 -300 - CS1 -10 -8 -6 -4 -2 0 2 4 6 8 10 300 200 SHEAR [kN] -200 -300 • CS6 DISPLACEMENT [mm] Figure 6.6 - Hysteresis loops for test series three Chapter 6 - Discussion of Experimental Results Figure 6.7 - Comparison of Specimens CS1 and CS6 at failure Figure 6.8 - Hysteresis loops for test series four Chapter 6 - Discussion of Experimental Results Figure 6.9 - Comparison of Specimens CS3, CS7 and CS8 at failure Chapter 7 - Conclusions and Recommendations 7.1 - Conclusions The objective of this thesis was to develop a better understanding of the response of reinforced concrete bridge elements subjected to reversed cyclic loading with emphasis on the degradation of the concrete contribution. An analytical investigation was carried out in which the effect of previously existing diagonal cracks on the ultimate shear strength of reinforced concrete elements were studied. The modified compression field theory was used as a basis for the new method. The crack check routine in the MCFT was generalized in order to account for the older steeper cracks. A significant part of the new crack check procedure was the inclusion of compressive stresses on the crack interface. In addition, the method included a new constitutive relation for the concrete in compression in which, unlike the MCFT, did not include any weakening of the concrete when subject to transverse tensile strains. The secant modulus was reduced in order to maintain the softening that had been observed experimentally. In general, it was found that for members with significant transverse steel, the previous (steeper) cracks are the most critical. For members with little or no transverse reinforcement, the flattest cracks are always the most critical. In addition, regardless of the transverse steel ratio, the new procedure predicted the experimental data very well and was found to be consistent with the predictions of the modified compression field theory. It was shown that when a shear design model was used, which did not allow for tensile stresses in the cracked concrete (e.g., the 1984 Canadian Concrete Code general shear design method), a portion of the shear must still be carried by the concrete whenever the crack deviated from the assumed (or calculated) truss inclination. The 129 Chapter 7 - Conclusions and Recommendations 130 magnitude of this shear increases as the crack becomes steeper until all the shear must be carried by the concrete (aggregate interlock) at a 90 degree crack Further analytical work was conducted in which the magnitude of stresses were studied when the crack angle and the principal stress field did not coincide. It was found that regardless of the absolute inclination of the principal stress field, the ratio of compressive stresses to shear stresses was a function of only the difference in the inclination of the crack and the stress field (termed the "deviant angle"). It was also proposed that there was a range of deviant angles in which shear transfer across the crack face would be critical The effect of cyclic loads on the ability of a reinforced concrete member to transfer shear stresses across a crack interface was studied. A literature review was completed and the results from numerous researchers were presented. Although most research was done on the cyclic shear-slip relationship of reinforced concrete (and usually at loads considerably less than ultimate), one researcher who had performed tests at ultimate loads suggested that the aggregate interlock mechanism degraded to approximately 80% of that for the monotonic case. To complete the objectives of this thesis, a pilot experimental program was carried out, in which eight reinforced concrete elements were subjected to reversed cyclic loads. Three parameters were studied; axial load level (both tension and compression), distributed (crack control) longitudinal reinforcement and the maximum shear span. It was found that axial tension increases the ductility of reinforced concrete members subject to cyclic shear. Similarly, members with larger shear spans exhibited higher ductilities than identical members with smaller shear spans. Also, the experimental program confirmed the expectation that better crack control enhances the shear transfer mechanism by controlling crack widths. Chapter 7 - Conclusions and Recommendations 131 Another conclusion drawn from the experimental work is that the ratio of shear capacity to shear demand (calculated indirectly from the plastic moment capacity of the section) is a good indicator of mode of failure and hence available ductility. This supports the work done by other researchers, e.g., Ghee,Preistley and Paulay (1989) and the method proposed in ATC - 6 (1983). It is very important to use an appropriate shear design method to predict the shear capacity of an element. For example, traditional shear methods predict tensile load reduces shear capacity faster than it reduces the flexural capacity, and as such would predict a more brittle failure. However, this research has shown, both experimentally and analytically, that the flexural capacity reduces faster than the shear capacity of a section when subject to tensile forces. Consequentiy, axial tension produces a more ductile failure than that predicted using the traditional ACI shear design method. Based on the present experimental results, it is suggested that a ratio of monotonic shear capacity (predicted using a shear model such as the one presented in this thesis) to the shear demand greater than 1.6 gives a good indication that the mode of failure of the specimen will be by yielding of the flexural steel and will reach displacement ductilities greater than 3.0. Ratios less than about 1.2, suggest that the specimen will probably fail in a brittle manner with ductilities less than 1.0. Ratios between 1.2 and 1.6 indicate a mode of failure predominately governed by flexure but the element will ultimately fail in shear with ductilities between 1.0 and 3.0. 7.2 - Recommendations As this is the first phase of an ongoing study, some recommendations are given below on the detailing of the specimens and their connections. In addition, recommendations for further areas of study are suggested. Chapter 7 - Conclusions and Recommendations 132 7.2.1 - Specimens and Connections 1) The longitudinal bars should be welded to the end plate so that they may reach their yield stress in compression when subject to fully reversed load applications. 2) Since the longitudinal steel is welded to the end plate, the specimens should have both ends strengthened in order to move the plane of failure away from the end plate. This will accomplish two things: first, it will prevent formation of compression struts carrying load to the endplates and the anchored reinforcement, thereby, ehrrunating any artificial increase in shear strength. Second, this allows the longitudinal steel a chance to debond in the plastic hinge region, and simulates the conditions in an actual concrete element away from any supports. 3) Small scale specimens should be tested by using smaller gauge stirrups at closer spacing (i.e., the percentage of transverse reinforcement should remain similar). As mentioned previously (Section 5.1), the shear cracks can work their way around transverse reinforcement if the spacing of this steel is too wide. 7.3.2 - Further testing This thesis proposes a method of predicting the ductility of a reinforced concrete element subjected to reverse cyclic loading by looking at the ratio of predicted shear strength versus the shear demand. Both the predicted shear strength and the shear demand (calculated from the flexural capacity and the maximum shear span) have been based on the monotonic capacity. It would be valuable to establish the relative degradation of the shear mechanism and the flexural mechanism under different parameters such as shear span and axial load levels at all ductility levels. It is assumed in the ATC - 6 model that the flexural capacity and hence the shear demand, remains constant over all the ductility levels -which is not the case. The flexural mechanism degrades under cyclic load as the longitudinal reinforcement debonds. Chapter 7 •• Conclusions and Recommendations 133 As seismic ground motion is very hard to predict, the emphasis on design (or retrofit) should concentrate the same, if not more, on mode of failure and ductility as it does on strength. Structures can not economically or feasibly be expected to withstand all earthquakes without damage or failure; however, a better understanding of the interaction between strength and ductility, and the effect different parameters have on this relationship, will help designers engineer safer and more efficient structures. REFERENCES ACI Committee 318, "Building Code Requirements for Reinforced Concrete (ACI 318 - 83)," American Concrete Institute, Detroit, 1983, 111 pp. Adebar, P.E., "Shear Design of Concrete Offshore Structures," Ph.D. Thesis, Dept. of Civil Eng., University of Toronto, 1989, 413 pp. Collins, M.P., and Mitchell, D., "Prestressed Concrete Basics," Canadian Prestressed Concrete Institute, Ottawa, Canada, Edition 1, 1987, 614 pp. Collins, M.P., and Mitchell, D., "Prestressed Concrete Structures," Prentice Hall, Englewood Cliffs, New Jersey, 1991, 766 pp. Collins, M.P., Mitchell, D., Adebar, P.E., and Vecchio, F.J., "A General Shear Design Method," ACI Structural Journal (in press - to be published early 1996). CSA Standard, "Design of Concrete Structures for Buildings (CAN3 A23.3 - M84)," Canadian Standards Association, Rexdale, Canada 1984, 281 pp. CSA Standard, "Design of Concrete Structures (A23.3 - 94)," Canadian Standards Association, Rexdale, Canada 1994 199 pp. Jimenez, R., Perdikaris, P., Gergely, P., and White, R.N., "Interface Shear Transfer and Dowel Action in Cracked Reinforced Concrete Subject to Cyclic Shear," Methods of Structural Analysis Proceedings, ASCE, Madison, Wis., Aug., 1976, pp. 457 -475. Jimenez, R., White, R.N., and Gergely, P., "Cyclic Shear and Dowel Action Models in Reinforced Concrete," Journal of the Structural Division, ASCE, May 1982, p. 1106. Mattock, A.H., "Cyclic Shear Transfer and Type of Interface," Journal of the Structural Division, ASCE, Oct., 1981, p. 1945. Paulay, T., and Loeber, P.J., "Shear Transfer by Aggregate Interlock," Shear in Reinforced Concrete, SP-42 American Concrete Institute, Detroit, Mich., 1974. Vecchio, F.J., and Collins, M.P., "Modified Compression Field Theory for Reinforced Concrete Elements Subjected to Shear," ACI Structural Journal, Proceedings, V. 83, No 2, Mar. - Apr., 1986, pp. 219 - 231. 134 References 135 Vecchio, F.J., and Collins, M.P., "Response of Reinforced Concrete to In-Plane Shear and Normal Stresses," Publication No. 82-03, Department of Civil Engineering, University of Toronto, Mar. 1982, 332 pp. White, R.N., and Holley, M J . , Jr., "Experimental Studies of Membrane Shear Transfer," Journal of the Structural Division, ASCE, August 1972, p. 1835. N O T A T I O N a = maximum aggregate size Ac = area of concrete Ag = gross area of section, sq. in Av = area of transverse reinforcement bv = minimum effective web width within depth dv bw = web width, in d = distance from extreme compression fiber'to centroid of tension steel dv = distance from the resultants of the tensile and compressive forces due to flexure f'c = specified compressive strength of concrete fc = concrete stress fcl = average principal tensile stress fc2 = average principal compressive stress fc2max = crushing strength of cracked concrete subject to tensile strains fcl = crack interface compressive stresses normal to crack plane fY = stress in the transverse steel fy = yield strength of reinforcement Mu = factored moment at section, lb-in Nu = factored axial load normal to cross section occurring simultaneously with V„ lbs Nvcrk = force in longitudinal steel due to shear at a crack Nx — force in longitudinal steel due to flexure and shear at a crack s = spacing of transverse reinforcement smO = crack spacing parameter at mclination 0 to horizontal smx = crack spacing parameter (cracks perpendicular to longitudinal reinforcement) smy = crack spacing parameter (cracks parallel to longitudinal reinforcement) V = total applied shear Vc = shear resistance attributed to the concrete vcf = crack interface shear stresses vcimax = maximum shear stress resisted across a crack without compressive stresses Vj = shear resistance provided by transverse reinforcement Vu = factored shear force at section, lbs w = crack width (mm) 0 = angle of inclination of principal stress field P = factor depending on the average tensile stresses in the cracked concrete £2 = angle of mclination of the crack interface 136 Notation factor depending on the average principal strains in the cracked concrete concrete strain corresponding to f'c modified concrete strain corresponding to f'c for softening model average principal tensile strain average principal compressive strain average strain in the cracked concrete normal to crack interface Aslbyji ratio of area of transverse steel to area of concrete APPENDIX A - Computer Program -A computer program based on the preceding algorithm has been written. A complete listing of the source code has been included in this Appendix. Along with the source code, a sample problem has been run and the input and output files are included (the sample problem is actually the prediction for specimen CS8 with the longitudinal steel check included). Information about the physical characteristics of the element and the specific loading conditions is supplied by the user through a simple input file. The following is a list of the specific information, and the order in which it is to be supplied in the file. (See the sample input for specimen CS8). First line consists of: CONAXL Applied constant axial load (N) RN Applied axial load ratio (decimal) Z Shear span i.e. M/V (mm) BV,DV Width and Depth of shear area (mm) FC Cylinder compression strength (MPa) AGG Maximum aggregate size (mm) ALPHA Cracking stress reduction factor NFLAG Toggles between proposed method (1) and MCFT (0) Second line consists of: AV FYY S CV DBV Area per stirrup (mm2) Yield stress of longitudinal (x) reinforcement (MPa) (Spacing of stirrups mm) Maximum distance to stirrups to point of cracking (mm) Diameter of stirrups (mm) 138 Appendix A - computer program 139 Third line consists of: ASX FYX ES CL SL DBL Flexural tension leinforcement area (mm2) YieldI .stress of transyerse (y) reinforcement (MPa) Young's modulus for rruld reinforceraent (MPa) Spacing of longitudinal reinfor^ment (mm) Diameter of longitudinal reinforcement (mm) Fourth line consists of: APS EPS FSE Area of prestressing strands (mm*2) Young's modulus for prestressing strands (MPa) Initial prestressing stress (MPa) Most of the input is self explanatory however there are a few exceptions that will be defined here: ALPHA allows the user to set the stress at which the concrete will crack. As a default, it is set at 0.33^f/e. A value of zero, gives the concrete no cracking strength, while a value of 2 will double the tensile stress at cracking. This variable has no effect on the tension-stiffening equations used after initial cracking. NFLAG allows the user to toggle between predictions based on the MCFT and the proposed method. Setting the value of NFLAG to zero gives the prediction based on the MCFT, while setting the value to 1 (or any other number) uses the proposed method for the calculations, (refer to sections 2.3 and 2.4 for a discussion of these two different methods). S,CV,CL and SL are used in the crack spacing parameters (S is also used elsewhere). The crack spacing equation used in this model comes from the paper by Bhide and Collins (1989). Figure A l best illustrates the definition of these variables. Once the input file has been filled in, the program may now be run. To start the program simply type "main" ( without the quotations) at the dos prompt. The program will start and prompt the user for one more input: Appendix A - computer program 1 4 0 "Enter 0 to emit longitudinal steel check" If the user wants to emit the longitudinal steel check then enter zero at this prompt. If the user wishes to include this check, then enter any other number. The program completes the calculations after this has been entered. The program presents the results in the form of three output files. All three output files are included in this appendix. The output files are quite self explanatory, but a brief summary of each is given below. The first file echos the input data and then prints out information at each strain state. Although this file includes the crack check routine values, the value of the shear, V, is based on the MCFT. The second file outputs all the average strain values with the corresponding inclintion of the principal stresses. Also included is the shear resisted at this strain state (using the proposed model if NFLAG = 1 or the MCFT if set to zero). This file is set up so that the load - deformation relationship can be graphed easily. The third file allows the user to track both the average stress - strain state and the crack check for all previous angles at each particular strain state. At each strain state the minimum shear is printed and the user is able to see which crack angle is becoming critical. This file is the most transparent in dealing with the crack check procedure. Al) A2) A3) A4) Appendix A - computer program A. Equilibrium Routine Choose a value of the principal tensile strain ei. Estimate the inclination of the average stress field 0. Estimate the stirrup stress fv. Calculate the stirrup contribution V s < 141 ' s-tanO (2.1) A5) Calculate V c (2.2) where on average _( 0-33^ ' 1+^ 500-ff, (2.4) A6) Calculate the average principal compressive stress fc2. f - V . + V c 4V,-tang c 2 tan0-bv-dv b v -d v A7) Calculate the average principal compressive strain ei *-v[ww-'/;] (2.15) where . f 0.3675 e =£ • , e" c 1^1-^1-0.6^ Appendix A - computer program 142 and „_ 1 0.8 + 170£i A8) Calculate the transverse reinforcement strain e, _ e}+£2 tan2 6 ' ~ l+tan0 A9) Calculate fv / v = 200000.*, <Lfy If the calculated stirrup stress does not equal the assumed stirrup stress, than go to A3 until the specified tolerance is achieved. A10) Calculate the longitudinal strain ex All) Calculate axial force corresponding to longitudinal strain A12) Calculate the average axial load due to shear V 2V tan0 tan20 A13) Calculate the net axial load N=N,-NV Appendix A - computer program If the net axial load N does not equal the actual applied load, than go to A2 and make a new guess for 0. A14) Call subroutine to check load transfer across previous cracks. A15) Call subroutine to check yielding of the longitudinal reinforcement. A16) Return to A l and increment ex. B. Crack Check Subroutine BI) Choose the crack angle £1 B2) Calculate the maximum shear that can be resisted by the stirrups crossing the crack, (i.e. assume yielding of the stirrups) B3) Calculate the crack spacing parameters. Appendix A - computer program 1 4 4 B4) Calculate the crack width (2.14) where £crt = g,+ e2tm2(Cl-6) l+tan2(fi-0) and smo = B5) sinfl [ cosfl Calculate the maximum shear which can be transmitted across the crack vcimax 47. \ 0.3 + 2Aw a + l6j (2.6) B6) Calculate the compressive stress fc{ which gives the maximum V c fc [ 1 -(l.64.tann)} a max (2.11) BT) Calculate the interface shear stress Vc[ which gives the maximum V c ^ " f 1 (3.28.Ln ln)]' dmax (2.12) B8) Calculate the maximum Vc which the present crack can resist v - [ v — - d Appendix A - computer program 145 B9) Calculate the maximum shear which can be transmitted across the crack BIO) Compare the maximum shear which can be transmitted across the crack with the calculated average shear. If larger than the crack is not critical, if it is smaller than the crack is failing. C. Longitudinal Steel Subroutine CI) Choose the crack angle Q C2) Calculate maximum shear which can be transmitted by stirrups V s w s- tanQ C3) Calculate the shear that must be transmitted by aggregate interlock Vc C4) Calculate the interface compressive stress f ci corresponding to Vcneed« ( 2 2 1 ) 2a _ 0.82 where a - ' v d n u b = ——1.64 tanQ V V Appendix A - computer program C5) Calculate the yield force of the longitudinal reinforcement. C6) Calculate the axial force required in the longitudinal reinforcement for shear. N„.=—^+ E-+JEL. (2.18) Vt* tanQ tanQ sinJQ CT) Calculate the total axial force on the longimdinal reinforcement >R N ^ M HERK=—+—ST-+— 2 2 rf. C8) Compare N x y with Nc rk- If N x y is greater than Ncrfc, the longitudinal steel has yielded. Appendix A - computer program Main program C CALCULATE THE RESPONSE OF A BEAM SUBJECTED C TO AXIAL LOAD, BENDING MOMENT C AND SHEAR FORCE C C INCLUDES A "CONCRETB CONTRIBUTION* TO SHEAR RESPONSE C C LIST OF INPUT VARIABLES C C RN Applied axial load ratio C CONAXL Applied constant axial load C Z Shear span ie. M/V (mm) C BV, DV Width and Depth of shear area (mm) C CV Maximum distance to stirrups (mm) C CL Maximum distance to longitudinal reinforcement (mm) C SL Spacing of longitudinal reinforcement (mm) . C DBL Diameter of longitudinal reinforcement (mm) C DBV Diameter of stirrups (mm) C FC Cylinder compression strength (MPa) C ALPHA Cracking stress reduction factor C AGG Maximum aggregate size (mm) C AV Area per stirrup (mm2) C S Spacing of stirrups (mm) C FYX, FYY Yield stress of longitudinal (x) and C transverse (y) reinforcement (MPa) C ASX Flexural tension reinforcement area (mm2) C SMX Crack spacing in x direction (mm) C EP Young's modulus for longitudinal C APS Area of prestressing strands (mm*2) C EPS Young's modulus for prestressing C strands (MPa) C and transverse reinforcement (MPa) C FSE Initial prestressing stress (MPa) IMPLICIT REAL(A-H, O-Z) COMMON /MAT/ BV,DV,AV,S,FYY,ASX,FYX,SMX,FC,AGGrRN,Z, 1 ES,EPS,AI>S,FSE,ALPHA,CL,SL,CV,DBL,DBV,CONAXL,NFLAG CHARACTER*20 FIN1 .DEFAULT.OUT1 .OUT2.0UT3 DO 5 1=1, 30 WRITER,*) ' ' 5 CONTINUE WRITE (•,10) 10 FORMAT (//' Enter data file'/' (eg. XYZ.dat)1) DO 111=1,20 WRITER,*) ' ' 11 CONTINUE READ (M5) FIN1 15 FORMAT(A20) DO 16 1-1,30 WRITER,*) 1 1 16 CONTINUE WRITE (•,20) 20 FORMAT (//' Enter 0 to omit longitudinal steel check') DO 211=1,20 WRITER,*) ' 1 Appendix A - computer program 21 CONTINUE READ (•,•) STLCHK DO 221=1, 30 WRirE(V) • ' 22 CONTINUE WRITE (•,*) THINKING. DO 23 1=1,20 WR1TE(V) ' • 23 CONTINUE OPEN (UNIT - 5,FILB=FIN1,STATUS - 'OLD") OPEN (UNIT = 7,Fn-B=,0UTl •.STATUS - 'UNKNOWN") OPEN (UNIT - 8,FILB='OUT2'fSTATUS - 'UNKNOWN') OPEN (UNIT = 9,FILE=,OUT3',STATUS - 'UNKNOWN') READ (S,*) CONAXL,RN,Z,BV,DV,FC,AGG,ALPHA,NFLAG READ (5,») AV,FYY,S,CV,DBV READ (5,») ASX,FYX,ES,CL,SL,DBL READ (5,*) APS.EPS.FSE IF (S .EQ. 0.0 .OR. AV J2Q. 0.0)THEN SMV = 0.0 ELSE SMV = 2.0*(CV+S/10.0)+0.1*DBV/2.0/AV/BV/S END IF SML = 2.0*(CL+SL/10.0)+0.1*DBL/2.0/ASX/BV/DV WRITE (7,200) 200 FORMAT (/'SECTIONAL AND MATERIAL PROPERTIES:'/) WRITE (7,205) 205 FORMAT ('MEMBER DIMENSIONS:*) WRITE (7,210)BV 210 FORMAT C BV *,F8.1,' mm") WRITE (7,215)DV 215 FORMAT f DV '.F8.1,' mm') WRITE (7,220) 220 FORMAT (/'LOADING CONDITIONS:") WRITE (7,222)CONAXL 222 FORMAT C CONAXL (CONSTANT AXIAL LOAD) *,F8.1,) WRITE (7,225)RN 225 FORMAT C RN (APPLIED AXIAL LOAD RATIO) ',F8.1.) WRITE (7,230)Z 230 FORMAT C Z (SHEAR SPAN ie. M/V) • J8.1,' mm*) WRITE (7.235) 235 FORMAT (/'CONCRETE PROPERTIES:') WRITE (7,240)FC 240 FORMAT C FC (CYLINDER STRENGHT) • jn.l,' MPa*) WRITE (7,245)ALPHA 245 FORMAT C ALPHA (CRACKING REDUCTION FACTOR)....'.F8.1,) WRITE (7,250)AGG 250 FORMAT (' AGG (MAXIMUM AGGREGATE SIZE) '.F8.1,' mm') WRITE (7,255) 255 FORMAT (/'STEEL PROPERTIES:') WRITE (7,260)AV 260 FORMAT C AV (AREA OF STIRRUPS) \ra.l,* mm*2') WRITE (7,265)FYY 265 FORMAT C FYY (YIELD STRESS OF STIRRUPS) 'JPi.l,' MPa") WRITE (7,270)S Appendix A - computer program 270 FORMAT (' S (SPACING OF STIRRUPS) 'fS-V mm') WRITE (7,271)CV 271 FORMAT (* CV (MAX. DISTANCE FROM STIRRUPS).... '.FS.l,' mm') WRITE (7,272)DBV 272 FORMAT f DBV (DIAMETER OF STIRRUPS) '.F8.1,' mm") WRITE (7.275)ASX 275 FORMAT C ASX (AREA OF TENSION STEEL) \F8.1,' mm*2') WRITE (7,280)FYX 280 FORMAT C FYX (YIELD STRESS OF STEEL) *,F8.1,' MP**) WRITE (7,285)CL 285 FORMAT (' CL (MAX. DISTANCE FROM X-STEEL) \F8.1,* mm') WRITE (7,286)SL 286 FORMAT (' SL (SPACING OF LONGITUDINAL STEEL).. *,F8.1,' mm") WRITE (7,287)CL 287 FORMAT C DBL (DIAMETER OF X-STEEL) \F8.1,' mm") WRTTH (7,290)ES 290 FORMAT C ES (YOUNGS MODULUS) '.F8.1,' MP«') WRITE (7,295) 295 FORMAT (/'PRESTRESSING STEEL PROPERTIES:*) WRITE (7,300)APS 300 FORMAT C APS (AREA OF PRESTRESSING STEEL).... '.F8.1,' mm'2') WRITE (7.305)EPS 305 FORMAT C EPS (YOUNGS MODULUS) '.F8.1,' MP»') WRITE (7,310)FSE 310 FORMAT C FSE (INITIAL PRESTRESSING STRESS)... '.F8.1,' MP»') WRITE (7,315) WRrTE (9,315) 315 FORMAT (/'CALCULATED PROPERTIES:') IF (S .EQ. 0.0)THEN rhoY •= 0.0 ELSE fhoY " AV/S/BV*100 ENDIF WRITE (7.320)rhoY WRITE (9,320)*oY 320 FORMAT C mo Y (PERCENT TRANSVERSE STEEL).... ',F8.2,' %") WRITE (7,325)ASX/BV/DV* 100 WRITE (9,325)ASX/BV/DV* 100 325 FORMAT C mo X (PERCENT LONGITUDINAL STEEL).. \F8.2,' *•) WRITE (7,330)SML WRITE (9,330)SML 330 FORMAT (' SMX (CRACK SPACING [X] ) 'JF8.1,' mm') WRITE <7,335)SMV WRITE (9,335)SMV 335 FORMAT C SMY (CRACK SPACING m ) mm*) WRITE (7,50) WRITE (7r52) IF(NFLAG .EQ. 0)THEN WRrTE (8,55) ELSE WRITE (8,56) ENDIF ULTM = ASX*FYX+BV*DV*FC TOL «= ULTM/100.0 FV = FYY VMAX » 0.0 BETA1 -0 .0 BETA2 - 0.0 FCR - ALPHA*0J3»FC**0J Appendix A - computer program ECR = ALPHA*0 J3/5000. GAMMA •= 0.0 THETA= 80.0*3.14159/180.0 DELTA = ECR/5.0 IF (DELTA XT. .000001) THEN NSTART - INT(J3/5000.»1E+09) NSTOP- INT(J3/5000*lE+09) NSTEP- INT(J3/5000.*1E+09) ELSE STRNIN = DELTA STRNFN - ECR NSTART = INT(STRNIN*lE+09) NSTOP = INT(STRNFN*lE+09) NSTEP - INT(DELTA*lE+09) ENDIF DO 100 NEP1 - NSTART.NSTOP.NSTEP EP1 « FLOAT(NEP1)/1E+09 CALL SUB1 (THETA,TOL,VS,VC,FC2(FC2MAX,FV, 1 E2,ET,EX,EP1,BETA1,FCR) IF (NEP1 .EQ. NSTOP) THEN GAMMA «» THETA THEDA •= THETA»180./3.14159 ENDIF THEDA = THETA*180./3.14159 V = VS + VC WRITE (7,62)m,THEDA,FV,FC2/FC2MAX,VS/1000., 1 BETA1,BETA2,VC/1000.,V/1000. WRITE (8.65)EP1,E2,THEDA,EX,ET 100 CONTINUE NSTART = INT(.0001'1E+09) NSTOP •= INT(.0500*1E+09) NSTEP « INT(.00005»1E+09) DO 101 NEP1 - NSTART.NSTOP.NSTEP EP1 = FLOAT(NEP1)/1E+09 CALL SUB1 (THETA,TOL,VS,VCJ'C2,FC2MAX,FV, 1 E2,ET,EX,EP1 .BETAl.FCR) THEDA - THETA*180./3.14159 V - VC+VS yfgTTEP »•)•••••*»•**•••*••**•+•••***••— CALL SUB3 (THETA.GAMMA.EPl ,E,FV,V,BETA2,VMINpVMCFT,STLCHIQ$dcbug WR^^E(9,•), • WRTTEvVSO) WRITE(9>60)EP1 ,THEDA,FV,FC2/FC2MAX,VS/1000., 1 BETA1,VC/1000.,V/1000. THEDA - THETA*180./3.14159 WRITE (7,62)EP1,THEDA,FV,FC2/FC2MAX,VS/1000., 1 BETA1,BETA2,VC/1000.,V/1000. IF(NFLAG .EQ. 0)THEN WRITE (8,65)EP1,E2,THEDA,EX,ET,VMCFT/1000 Appendix A - computer program ELSE WRITE (8,65)EP1,E2,THEDA,EX,ET,VMIN/1000 IF (VMIN .GT. VMAX) THEN VMAX - VMIN ELSEIF (VMIN .LT. 0.7*VMAX) THEN WRITE (•,*) 'CAPACITY OF SECTION HAS DEGRADED TO 70* OF MAX.' STOP ENDIF ENDIF 101 CONTINUE 50 FORMAT(//' epl Tbctt fv tell V i BeU IBeU Vc V ) 52 FORMATC fc2msx (avj) (c lrack)'/) 55 FORMAT(//' epl ep2 Theta ex et 1 VMCFT'/) 56 FORMAT(//' epl ep2 Theta ex et 1 VMIN'/) 60FORMAT(F8.6,F8.1,F8.1,F8J,F8.1,F9J,F11J,F9JI) 62FORMAT(F8.6,F8.1Jf8.1,F8J,F8.1,2F9J,Fll^,F9^) 65 FORMAT(2F10J,F10.2,2F10J,F10^) 450 FORMATC epl TheU fv fc2/max Vs BeU(sv) Vc I V ) STOP END Subroutine One SUBROUTINE SUB1 CrHETA,TOL,VS,VC,FC2,FC2MAX,FV,E2,ET, 1 EX,EP1,BETA1,FCR) COMMON /MAT/ BV,DV,AV,S,FYY,ASX,FYX,SMX,FC,AGG,RN,Z, 1 ES,EPS,APS,FSE,ALPHA,CL,SL,CV>DBL,DBV,CONAXL,NFLAG 5 JJ « 0.0 SIGN - 1.0 XI - THETA CALL SUB2 (THETA>TOL,RNETVSVC (FC2,FC2MAX,FV,E2,ET, 1 EX,EP1,BETA1,FCR) F l = CONAXL+RN*(VS+VC)-RNET 10 THETA - THETA - 5.<W.14159/180.0*SIGN X 2 - T H E T A CALL SUB2 (THETA^TOL,RNETVSVC,FC2(FC2MAX,FV,E2,ET, 1 EX,EP1,BETA1,FCR) F2 - CONAXL+RN*(VS + VC) - RNET F12 « F1«F2 IF(F12.GT.0.0)THEN IF (ABS(F1) .GT. ABS(F2)) THEN Fl - F2 XI » X2 ELSE SIGN = -1.0 T H E T A - X I ENDIF GOTO 10 ENDIF 40 JJ=JJ+1 Appendix A - computer program X 3 =X2-((X2-X1)*F2/(F2-F1)) T H E T A = X 3 IF ( T H E T A X T . 0.0 .OR. T H E T A .GT. 90.0*3.14159/180.0)THEN T H E T A = 10.0*3.14159/180.0 G O T O 5 E N D I F C A L L SUB2 ( T H E T A , T O L , R N E T , V S , V C , F C 2 , F C 2 M A X , F V , E 2 , E T , 1 E X . E P 1 ,BETA1 ,FCR) F 3 = C O N A X L + R N * ( V S + V C ) - R N E T IF(ABS(F3) X T . T O L ) T H E N G O T O 100 E N D I F IF(JJ . G T . 50 )THEN W R I T E R , * ) •mmwmmim P R O B L E M W . G T . 50 tmmmsw WRITE(*,*) WRITE(*,*) EP1 STOP E N D I F F13=F1*F3 IF(F13 X T . 0 . ) T H E N X 2 = X 3 F2-=F3 E L S E X1«=X3 F1=F3 E N D I F G O T O 40 100 IF ( N F L A G .EQ. 0) T H E N IF (FC2 . G T . F C 2 M A X ) T H E N STOP E N D I F E L S E IF (FC2 . G T . F C ) T H E N STOP E N D I F E N D I F R E T U R N E N D Subroutine Two SUBROUTINE SUB2 CIWTA,TOIJRNET>VS,VC>FC2,FC2MAX,FV,E2,ET, 1 EX.EPl.BETAl.FCR) COMMON /MAT/ BV,DV,AV,S,FYYASX,FYX,SMX,FC,AGG^N^., 1 ES,EPS^.FSE,ALPHA,CL^L,CVJ>BL,DBV,CONAXl nNFLAG C CALCULATE V S 10 IF (S .EQ. 0.0 .OR. A V .EQ. 0.0)THEN V S - 0.0 ELSE VS=AV*FV*DV/(S*TAN(rHETA)) ENDIF C CALCULATE V C EC-5000 .0*SQRT(FC) Appendix A - computer program ECR-FCR/EC IF(EP1 .LT. ECR)THEN BETA1 -EC*EPl/fTANCrHETA)*FC**0 S) ELSE BETA1 = ALPHA'O J3/(TANCTHETA)*(1+(500*EP1)»*0^)) ENDIF VC=BETA 1 »SQRT(FC)*BV*DV C CALCULATE FC2 FC2= ((VS+VQ»(l/TANCrHETA))+(VS)*CrANCTHBTA)))/(BV*DV) C CALCULATEFC2MAX FC2MAX=FC/(0.8 + 170*EP1) IF(FC2MAX .CT. FC) FC2MAX-FC PSI - FC2MAX/FC C CALCULATE E2 IF (NFLAG .EQ. 0)THEN IF(FC2/FC2MAX .GT. 1.0)THEN E2=-.002»(l+(FC2iTC2MAX-l)'*0 S) ELSE E2=-.002'(H1-FC2/FC2MAX)**0.5) ENDIF ELSE ECC = OJ675*.002/(l.O-(l.O-0.6»PSD*»0J) IF(FC2 .GT. FQTHEN E2 »=-ECC»(l + (FC2/FC-l)->*0.5) ELSE E2 »=-ECC*(l-(l-FC2/FQ»'0.5) ENDIF ENDIF C CALCULATE ET ET«= (EP1+E2*(rAN(rHETA))**2)/(l+(TAN(rHETA))**2) C CALCULATE FV FVA=200000*ET IF(FVA .GT. FYY)FVA°FYY FYT »= ABS (FV-FV A) IF(FYT .GT. 1)THEN FV=FVA GOTO 10 ENDIF C CALCULATE EX EX-EP1+E2-ET C CALCULATE THE INTERGRATED AXIAL FORCE ON THE SECTION V-VS+VC RRM=V«Z RNP«2.*(EX«(ES*ASX+EK*APS)+APS*FSB-RRM/DV) C CALCULATE THE NET AXIAL LOAD RNV-VS/TAN(THETA)+2.()*VCn,AN(2.()<THETA) RNET «= RNP-RNV RETURN END Subroutine Three Appendix A - computer program SUBROUTINE SUB3 CTHETA.GAMMA.EPl ,E2,W.V.BETAa.VMD^.VMCFT.STLCHK) COMMON /MAT/ BV,DV,AV,S,FYY,ASX,FYX,SMX,FC,AGC,RN,Z, 1 ES,EPS,APS,FSE,AU'HA,CL,SL,CV,DBI^DBV,CONAX1,NFLAO WRITE (9,400) VMIN = V VMCFT - V N ST ART •= INT(THETA*>lE+09) NSTOP = INT(GAMMA*IB+09) NSTEP = INT((GAMMA-THETA)/5.0*lB+09) DO 10 NOMEGA •> NSTART,NSTOP,NSTEP OMEGA " FLOAT(NOMEGA)/1E+09 IF (S .EQ. 0.0 .OR. AV .EQ. 0.0)THEN VS « 0.0 ELSE VS=AV*FYY*DV/(S*TAN(OMEGA)) ENDIF SML • 2.0->(CL+SL/10.0)+0.1*DBL/2.0/ASX/BV/DV IF (S .EQ. 0.0 .OR. AV .EQ. 0.0)THEN SCRK •= SML/SDM(OMEGA) ELSE SMV » 2.0*(CV+S/10.0)+0.1*DBV/2.0/AV/BV/S SCRK - 1.0/(SIN(OMEGA)/SML+COS(OMEGA)/SMV) ENDIF ECRK => (EPl+E2»frAN(OMEGA-THETA)),>^)/(l+CrAN(OMEGA-THETA))*,>2) W «= ECRK*SCRK VCiMAX = SQRT(FQ/(OJ + P4.0»W)/(AGG+16.0)) FCi - (1.0-1.0/(1,64*TAN(OMEG A)))*VCiMAX IF (FCi XT. 0.0) THEN FCi = 0.0 ENDIF VCi = (l.-l./(3.28*TAN(OMEGA),>*2))*VCiMAX IF(VCi XT. 0.18*VCiMAX)THEN VCi - 0.18*VCiMAX ENDIF VC - (VCi - FCi/TAN(OMEGA))*BV**DV BETA=Va(SQRT(FC)*BV>DV) IF (NFLAG .NE. 0) THEN V^UTE(9,20)OMEGA*18073.14159,ECRK,SCRK,Va,FCi,BETApVS/10C)0.( 1VC/1000.,(VC+VS)/1000. ENDIF IF (NOMEGA .EQ. NSTART)THEN V»TUTE(9,20)OMEGAM80./3.14159,ECRK,SC'RKtV^ ivc/iooo.,(vc+vsyiooo. BETA2 » BETA IF(VC+VS .LT. VMCFDTHEN VMCFT «= VC+VS ENDIF ENDIF IF(VC+VS XT. VMIN)THEN Appendix A - computer program VMIN - VC+VS ENDIF 10 CONTINUE IF (NFLAG .EQ. 0)THEN WRITE (9,26)VMCFT/1000 ELSE WRITE <9,25)VMIN/1000 ENDIF IF (STLCHK .EQ. 1)THEN CALL SUB4 (THETA,GAMMA,EP1 ,E2,FV,V,BETA2,VCiMAX,VMIN,VMCFT) ENDIF 20 FORMAT (F8.1,F8.4,F8.1,2F8.2,F8.3,3F8.2) 25 FORMAT (/' Minimum ihearii 'JJ8^, ,kN') 26 FORMAT (/' Minimum MCFT (bear U',F8.2,'kN') 400 FORMAT C Theta ecrk ac* Vci Fci Bela(crk) Va 1 Vc V ) RETURN END Subroutine Four SUBROUTINE SUB4 (THETA.GAMMA.EPl >E2,FV,V,BETA2,VCiMAX,VMIN>VMCFT) COMMON /MAT/ BV,DV,AV,S.FYY,ASX,FYX,SMX,FC,AGGJIN.Z, 1 ES,EK,APS,FSE,AU>HA,CL,SL,CV,DBL,DBV,CONAXL,NFLAO NSTART «= INT(THETA*lE+09) NSTOP - INT(GAMMA*lE+09) NSTEP = INT((GAMMA-THETA)/5.0*1E+09) IF (NFLAG .EQ. 0)THEN NSTOP = NSTART ENDIF DO 10 NOMEGA •= NSTART,NSTOP,NSTEP OMEGA - FLOAT(NOMEGA)/1E+09 IF (S .EQ. 0.0 .OR. AV .EQ. 0.0)THEN VS = 0.0 ELSE VS=AV*FYY*DV/(S*TAN(OMEGA)) ENDIF IF (NFLAG .EQ. 0)THEN VC - VMCFT - VS ELSE VC = VMIN - VS ENDIF A = 0.82/VCiMAX B - 1.0/TAN(OMEGA)-1.64 C - VC/BV/DV-0.18*VCiMAX RAD - B**2-4*A»C IF (B .GT. 0 .OR. C XT. 0)THEN FCi - 0.0 ELSEIF (RAD XT. 0.0)THEN WRrTE(*,*)'RAD IS NEGATIVE' RAD - 0.0 ELSE FCi - (-B-SQRT(RAD))/(2*A) Appendix A - computer program ENDIF F0RCE1 = ASX"FYX+APS*1600 RNVCRK = VS/rAN(OMEGA)+2.0*VC/TAN(OMEGA) l+FCi*BV*DV/SIN(OMEGA)**2.0 FORCE2 = CONAXX/2+RN*(VS+VC^+RNVCRKy2+Z*(VS+VC)/DV IF (FORCE1 XT. FORCE2)THEN WR1TE(9,18) OMEGA*180.0/3.14159 18 FORMATC Cannot maintain equilibrium at*,F6.2,' degrees.*) WRTTE<9,19) 19 FORMATC Yielding of longitudinal reinforcement*) STOP ENDIF 10 CONTINUE RETURN END Appendix A - computer program Input File - IN 0 0 707 400 332 30 20 1 1 CONAXL^lN,Z,BV,DV (FC^GO,ALPHA,NFLAO 200 450 305 180 10 AV,FYY,S,CV,DBV 1120 485 200000 30 80 15 ASX,FYX,ES,CL,SL,DBL 0 200000 0 APS,EPS,FSE Output File - OUT! Appendix A - computer program 158 SECTIONAL AND MATERIAL PROPERTIES: MEMBER DIMENSIONS: BV 400.0 mm DV 332.0 mm LOADING CONDITIONS: CONAXL (CONSTANT AXIAL LOAD) 0 RN (APPLIED AXIAL LOAD RATIO) .0 Z (SHEAR SPAN ie. M/V) 707.0 mm CONCRETE PROPERTIES: FC (CYLINDER STRENGHT) 30.0 MP» ALPHA (CRACKING REDUCTION FACTOR)... 1.0 AGG (MAXIMUM AGGREGATE SIZE) 20.0 mm STEEL PROPERTIES: AV (AREA OF STIRRUPS) 200.0 mm"2 FYY (YIELD STRESS OF STIRRUPS) 450.0 MP» S (SPACING OF STIRRUPS) 305.0 mm CV (MAX. DISTANCE FROM STIRRUPS).... 180.0 mm DBV (DIAMETER OF STIRRUPS) 10.0 mm ASX (AREA OF TENSION STEEL) 1120.0 mm*2 FYX (YIELD STRESS OF STEEL) 485.0 MP* CL (MAX. DISTANCE FROM X-STEEL) 30.0 mm SL (SPACING OF LONGITUDINAL STEEL).. 80.0 mm DBL (DIAMETER OF X-STEEL) 30.0 mm ES (YOUNGS MODULUS).. 200000.0 MPa PRESTRESSING STEEL PROPERTIES: APS (AREA OF PRESTRESSING STEEL) 0 mm*2 EPS (YOUNGS MODULUS) 200000.0 MPa FSE (INITIAL PRESTRESSING STRESS)... .0 MPa CALCULATED PROPERTIES: rho Y (PERCENT TRANSVERSE STEEL) 16 % rho X (PERCENT LONGITUDINAL STEEL).. .84 % SMX (CRACK SPACING PC]) 76.0 mm SMY (CRACK SPACING fY] ) 421.0 mm epl Theta fv fcV V i fc2max .000013 76.1 .1 .001 .0 .000026 76.1 .1 .001 .0 .000040 76.1 .1 .002 .0 .000053 76.1 .1 .003 .0 .000066 76.1 .1 .004 .0 .000100 75.1 1.1 .004 .1 .000150 73.8 2.4 .004 2 .000200 72.6 2.7 .005 2 .000250 71J 4.1 .005 3 .000300 70.2 5.8 .006 JS .000350 69.0 7.8 .007 .6 .000400 67.9 10.0 .007 .9 .000450 66.9 12.4 .008 12 .000500 65.9 15.2 .009 13 .000550 64.9 18.1 .010 1.8 .000600 64.0 21J .011 23 .000650 63.1 24.6 .012 2.7 .000700 62.3 28.2 .012 3.2 .000750 61.5 31.9 .013 3.8 .000800 60.8 35.8 .014 4.4 Beta Beta Vc V (•vg) (crack) .016 .000 11.85 11.85 .033 .000 23.76 23.77 .049 .000 35.65 35.65 .065 .000 47.54 4734 .082 .000 59.42 59.42 .072 2.474 52.38 52.44 .075 2.391 54.73 54.88 .079 2.309 57.27 57.46 .082 2.230 59.91 60.21 .086 2.154 62.43 62.89 .089 2.081 64.91 65.56 .093 2.011 67 JO 68.18 .096 1.945 69 31 70.73 .099 1.882 71.72 73.20 .101 1.823 73.75 75.60 .104 1.766 75.67 77.93 .106 1.712 77.47 80.18 .109 1.661 79.15 82.37 .111 1.613 80.73 8430 .113 1.567 82.21 8637 Appendix A - computer program 159 .000850 60.1 39.9 .015 5.0 .115 1.523 83.59 8839 .000900 59.4 44.1 .016 5.7 .117 1.482 84.89 9036 .000950 58.8 48.4 .017 6.4 .118 1.443 86.09 92.48 .001000 58.2 52.9 .018 7.1 .120 1.405 87.22 9436 .001050 57.6 57.4 .018 7.9 .121 1.370 88.27 96.20 .001100 57.1 62.1 .019 8.8 .123 1.336 89.26 98.01 .001150 56.6 66.9 .020 9.6 .124 1.304 90.18 99.79 .001200 56.1 71.7 .021 10.5 .125 1.273 91.03 10133 .001250 55.6 76.6 .022 11.4 .126 1.244 91.83 103.25 .001300 55.1 81.5 .023 12.4 .127 1.216 92.57 104.94 .001350 54.7 86.6 .024 133 .128 1.189 93.26 106.60 .001400 54.3 91.7 .025 143 .129 1.163 93.91 108.25 .001450 53.9 96.8 .027 15.4 .130 1.139 9431 109.88 .001500 53 J 102.0 .028 16.4 .131 1.115 95.07 111.49 .001550 53.2 107.3 .029 17.5 .131 1.092 95.59 113.08 .001600 52.8 112.6 .030 18.6 .132 1.071 96.07 114.66 .001650 52.5 117.9 .031 19.7 .133 1.050 9632 116.22 .001700 52.2 123.3 .032 20.8 .133 1.030 96.93 117.77 .001750 51.9 128.7 .034 22.0 .134 1.011 9732 11931 .001800 51.6 134.2 .035 23.2 .134 .992 97.67 120.84 .001850 51.3 139.7 .036 24.4 .135 .975 98.00 12236 .001900 51.0 145.2 .037 25.6 .135 .957 98.31 123.87 .001950 50.8 150.7 .038 26.8 .136 .941 9839 12537 .002000 50.5 156.3 .040 28.0 .136 .925 98.84 126.87 .002050 50.3 161.9 .041 29.3 .136 .910 99.08 12835 .002100 50.1 167.5 .042 30.5 .137 .895 99.29 129.83 .002150 49.8 173.1 .043 31.8 .137 .880 99.49 13131 .002200 49.6 178.8 .045 33.1 .137 .867 99.67 132.78 .002250 49.4 184.4 .046 34.4 .137 .853 99.83 134.24 .002300 49.2 190.1 .047 35.7 .137 .840 99.97 135.70 .002350 49.0 195.8 .049 37.1 .138 .828 100.10 137.16 .002400 48.8 201.5 .050 38.4 .138 .816 100.22 138.61 .002450 48.6 207.3 .051 39.7 .138 .804 100.32 140.05 .002500 483 213.0 .053 41.1 .138 .792 100.41 14130 .002550 48.3 218.8 .054 42.4 .138 .781 100.49 142.94 .002600 48.1 224J .056 43.8 .138 .771 10035 14438 .002650 48.0 230.3 .057 45.2 .138 .760 100.61 145.81 .002700 47.8 236.1 .058 46.6 .138 .750 100.65 147.24 .002750 47.7 241.8 .060 48.0 .138 .740 100.69 148.67 .002800 47.5 247.6 .061 49.4 .138 .731 100.71 150.10 .002850 47.4 253.4 .063 50.8 .138 .721 100.73 15133 .002900 47.2 259.3 .064 52.2 .138 .712 100.74 152.95 .002950 47.1 265.1 .066 53.6 .138 .703 100.74 15437 .003000 47.0 270.9 .067 55.1 .138 .695 100.73 155.79 .003050 46.8 276.7 .069 56.5 .138 .687 100.72 157.21 .003100 46.7 282.5 .070 57.9 .138 .678 100.70 158.63 .003150 46.6 288.4 .072 59.4 .138 .670 100.67 160.05 .003200 46.5 294.2 .073 60.8 .138 .663 100.64 161.46 .003250 46.4 300.1 .075 623 .138 .655 100.60 162.88 .003300 46.3 305.9 .076 63.7 .138 .648 10035 164.29 .003350 46.2 311.8 .078 65 .2 .138 .641 10030 165.70 .003400 46.0 317.6 .080 66.7 .138 .634 100.45 167.11 .003450 45.9 323.5 .081 68.1 .138 .627 10039 16833 .003500 45.8 329.3 .083 69.6 .138 .620 10033 169.94 .003550 45.7 336.2 .085 71.3 .138 .614 100.26 17137 .003600 45.7 342.1 .086 72.8 .138 .607 100.19 172.98 .003650 45.6 348.0 .088 74.3 .138 .601 100.11 174.40 .003700 45.5 353.9 .090 75.8 .138 .595 100.03 175.81 .003750 45.4 359.8 .091 77.3 .137 .589 99.95 177.23 .003800 45 J 365.7 .093 78.8 .137 383 99.86 178.64 .003850 45.2 371.6 .095 80.3 .137 .578 99.78 180.05 .003900 45.1 377.5 .096 81.8 .137 .572 99.68 181.46 .003950 45.1 383.4 .098 83.3 .137 .567 9939 182.88 .004000 45.0 389.3 .100 84.8 .137 .561 9930 18430 .004050 44.9 395.2 .102 86.3 .137 .556 99.40 185.73 .004100 44.8 400.2 .103 87.6 .137 .551 9930 186.95 .004150 44.8 406.2 .105 89.2 .136 .546 99.20 188.40 Appendix A - computer program 160 Output File - OUT2 epl ep2 .00001 .00000 .00003 .00000 .00004 .00000 .00005 .00000 .00007 .00000 .00010 .00000 .00015 .00000 .00020 .00000 .00025 -.00001 .00030 -.00001 .00035 -.00001 .00040 -.00001 .00045 -.00001 .00050 -.00001 .00055 -.00001 .00060 -.00001 .00065 -.00001 .00070 -.00001 .00075 -.00001 .00080 -.00001 .00085 -.00002 .00090 -.00002 .00095 -.00002 .00100 -.00002 .00105 -.00002 .00110 -.00002 .00115 -.00002 .00120 -.00002 .00125 -.00002 .00130 -.00002 .00135 -.00002 .00140 -.00003 .00145 -.00003 .00150 -.00003 .00155 -.00003 .00160 -.00003 .00165 -.00003 .00170 -.00003 .00175 -.00003 .00180 -.00004 .00185 -.00004 .00190 -.00004 .00195 -.00004 .00200 -.00004 .00205 -.00004 .00210 -.00004 .00215 -.00005 .00220 -.00005 .00225 -.00005 .00230 -.00005 .00235 -.00005 .00240 -.00005 .00245 -.00005 .00250 -.00006 .00255 -.00006 .00260 -.00006 .00265 -.00006 .00270 -.00006 .00275 -.00006 .00280 -.00007 .00285 -.00007 .00290 -.00007 Theta ex 76.13 .00001 76.10 .00002 76.10 .00004 76.10 .00005 76.10 .00006 75.05 .00009 73.81 .00014 7236 .00018 71.33 .00022 70.16 .00026 69.01 .00030 67.92 .00034 66.86 .00038 65.86 .00041 64.90 .00045 63.99 .00048 63.13 .00051 62.31 .00055 6133 .00058 60.79 .00061 60.09 .00063 59.42 .00066 58.79 .00069 58.19 .00072 57.62 .00074 57.07 .00077 5635 .00079 56.06 .00082 55.59 .00084 55.14 .00087 54.71 .00089 5430 .00091 53.91 .00094 5333 .00096 53.17 .00098 52.83 .00100 5230 .00103 52.18 .00105 51.88 .00107 5139 .00109 5131 .00111 51.04 .00113 50.78 .00115 5033 .00117 50.29 .00120 50.05 .00122 49.83 .00124 49.61 .00126 49.40 .00128 49.20 .00130 49.01 .00132 48.82 .00134 48.64 .00136 48.46 .00138 48.29 .00140 48.12 .00141 47.96 .00143 47.81 .00145 47.66 .00147 4731 .00149 4737 .00151 47.23 .00153 et VMIN .00000 .00000 .00000 .00000 .00000 .00000 52.44 .00001 54.88 .00001 57.46 .00002 60.21 .00003 62.89 .00004 6536 .00005 68.18 .00006 70.73 .00008 73OO .00009 75.60 .00011 77.93 .00012 80.18 .00014 8237 .00016 8430 .00018 8637 .00020 8839 .00022 9036 .00024 92.48 .00027 9436 .00029 96.20 .00031 98.01 .00034 99.79 .00036 10133 .00038 103.25 .00041 104.94 .00043 106.60 .00046 108.25 .00049 109.88 .00051 111.49 .00054 113.08 .00056 114.66 .00059 116.22 .00062 117.77 .00065 11931 .00067 120.84 .00070 12236 .00073 123.87 .00076 12537 .00078 126.87 .00081 12835 .00084 129.83 .00087 13131 .00090 132.78 .00092 134.24 .00095 135.70 .00098 137.16 .00101 138.61 .00104 140.05 .00107 14130 .00110 142.94 .00113 14438 .00115 145.81 .00118 147.24 .00121 148.67 .00124 150.10 .00127 15133 .00130 152.95 Appendix A - computer program .00295 -.00007 .00300 -.00007 .00305 -.00007 .00310 -.00008 .00315 -.00008 .00320 -.00008 .00325 -.00008 .00330 -.00008 .00335 -.00008 .00340 -.00009 .00345 -.00009 .00350 -.00009 .00355 -.00009 .00360 -.00009 .00365 -.00010 .00370 -.00010 .00375 -.00010 .00380 -.00010 .00385 -.00010 .00390 -.00011 .00395 -.00011 .00400 -.00011 .00405 -.00011 .00410 -.00011 .00415 -.00012 47.10 .00155 46.97 .00157 46.84 .00159 46.72 .00161 46.60 .00163 46.48 .00164 46.37 .00166 46.26 .00168 46.15 .00170 46.05 .00172 45.95 .00174 45.85 .00176 45.75 .00178 45.66 .00180 4536 .00181 45.47 .00183 4539 .00185 4530 .00187 45.22 .00189 45.14 .00191 45.06 .00192 44.98 .00194 44.90 .00196 44.83 .00198 44.75 .00200 .00133 15437 .00136 155.79 .00139 157.21 .00142 158.63 .00145 160.05 .00148 161.46 .00150 162.88 .00153 164.29 .00156 165.70 .00159 167.11 .00162 16833 .00165 169.94 .00168 17137 .00171 172.98 .00174 174.40 .00177 175.81 .00180 177.23 .00183 178.64 .00186 180.05 .00189 181.46 .00192 182.88 .00195 18430 .00198 185.73 .00201 186.95 .00204 188.40 Appendix A - computer program 162 Output File O U T 3 CALCULATED PROPERTIES: mo Y (PERCENT TRANSVERSE STEEL) 16 * rho X (PERCENT LONGITUDINAL STEEL).. .84 * SMX (CRACK SPACING [X] ) 76.0 mm SMY (CRACK SPACING fY] ) 421.0 mm TheU ecrk •crk Vci Fci Beta(crk) V i Vc V 75.1 .0001 75.0 1737 15.03 2.474 26.16 1799.87 1826.03 75.1 .0001 75.0 1737 15.03 2.474 26.16 1799.87 1826.03 75 J .0001 75.0 1738 15.08 2.485 25.78 1807.70 1833.47 75.5 .0001 75.0 1739 15.12 2.496 2539 181533 1840.92 75.7 .0001 75.0 17.60 15.16 2307 25.01 182336 184837 75.9 .0001 75.0 17.61 15.21 2318 24.63 1831.21 1855.84 76.1 .0001 74.9 17.62 15.25 2328 24.25 1839.06 186331 Minimum shear is 52.44kN epl Theta fv fc2/max Vs Beta(av) Vc V .000100 75.1 1.1 .004 .1 .072 5238 52.44 Theta 73.8 73.8 74.3 74.7 75.2 75.6 76.1 ecrk .0001 .0001 .0001 .0001 .0001 .0001 .0001 scrk 75.2 75.2 75.1 75.1 75.0 75.0 74.9 Vci 1735 1735 1738 17.41 17.43 17.46 17.48 Fci 14.66 14.66 14.75 14.84 14.94 15.03 15.12 BeU(crk) Vs 2391 28.45 2391 2.414 2.438 2.461 2.484 2308 28.45 27.61 26.76 25.92 25.08 24.25 Vc 1739.17 1739.17 1756.10 1773.06 1790.06 1807.10 1824.18 V 1767.62 1767.62 1783.70 1799.82 1815.98 1832.19 1848.43 Minimum shear is 54.88kN epl Theta fv fc2/max Vs Beta(av) Vc V .000150 73.8 2.4 .004 .2 .075 54.73 54.88 Theta ecrk scrk Vci Fci Beu(crk) Vs Vc V 72.6 .0002 75.4 17.13 14.28 2309 30.77 1679.75 171032 72.6 .0002 75.4 17.13 14.28 2309 30.77 1679.75 171032 733 .0002 753 17.18 14.43 2345 29.45 170536 1735.01 74.0 .0002 75.2 17.22 1437 2380 28.13 1731.45 175938 74.7 .0002 75.1 17.26 14.72 2.416 26.83 1757.42 1784.25 75.4 .0002 75.0 1730 14.86 2.452 2534 1783.48 1809.02 76.1 .0002 74.9 1734 15.00 2.488 24.25 1809.64 1833.89 Minimum shear is 57.46kN epl Theta .000200 72.6 fv fc2/max 2.7 .005 Vs a. Beta(av) Vc .079 57.27 V 57.46 Data deleted in middle of output file OUT3 Theta ecrk scrk Vci Fci Beta(crk) V i Vc V 44.8 .0041 91.2 6.89 3.85 351 9836 40035 499.11 44.8 .0041 91.2 6.89 3.85 351 9836 40035 499.11 51.1 .0040 853 8.28 5.24 .738 79.10 536.87 615.98 573 .0039 80.9 9 3 8 633 .948 62.81 689.89 752.70 63.6 .0037 77.9 103 3 7.79 1.180 48.66 858.43 907.09 69.8 .0033 75.9 11.19 9.06 1.436 35.96 1044.44 1080.40 76.1 .0030 74.9 11.99 1038 1.721 24.25 1251.69 1275.94 Minimum shear is 186.95kN epl Theta fv &2/max Vs Beta(av) .004100 44.8 400.2 .103 87.6 .137 Appendix A - computer program Vc v 99.30 186.95 Theta ecrk acrk Vci Fci Beta(crk) Vs Vc V 44.8 .0041 913 6.84 3.82 346 98.81 396.88 495.69 44.8 .0041 91.3 6.84 3.82 346 98.81 396.88 495.69 51.0 .0041 853 8.22 5.21 .732 79 XI 532.76 612.03 573 .0039 80.9 933 6.49 .942 62.92 685.48 748.40 63.6 .0037 77.9 10.28 7.75 1.174 48.72 853.87 90239 69.8 .0034 75.9 11.14 9.02 1.430 35.99 1039.88 1075.87 76.1 .0030 74.9 11.95 1034 1.715 24.25 124736 1271.61 Minimum shear is 188.40kN epl Theta fv fc2/max Vs Beta(av) Vc V .004150 44.8 406.2 .105 89.2 .136 99.20 188.40 Theta ecrk acrk Vci Fci Beta(crk) Vs Vc V 44.7 .0042 91.4 6.78 3.78 341 99.06 393.27 49233 44.7 .0042 91.4 6.78 3.78 341 99.06 393.27 49233 51.0 .0041 853 8.17 5.17 .727 79.43 528.71 608.14 57.2 .0040 81.0 9.28 6.46 .936 63.02 681.13 744.15 633 .0037 77.9 10.23 7.71 1.168 48.78 8493 6 898.14 69.8 .0034 75.9 11.10 8.98 1.423 36.02 103538 1071.40 76.1 .0030 74.9 11.91 1031 1.709 24.25 1243.07 126732 Minimum shear is 189.84kN Cannot maintain equilibrium at 44.68 degrees. Yielding of longitudinal reinforcement APPENDIX B - Beam Element Tester-Calibration of rigid links: The rigid links are equipped with four strain gauges which are mounted on a machined portion of the rigid link. The rigid links were placed in the Baldwin testing machine located in the structures lab at the University of British Columbia. The rigid links were cycled approximately ten times to shakedown the strain gauges and ensure that they were solidly connected to the links. They were then loaded in the testing machine in compression while the output voltage from the strain gauges were recorded. After the three rigid links were tested in compression, the Baldwin testing machine was re configured and used to test the rigid links in tension. Once again the output voltage was recorded from the strain gauges while loading the rigid links. Graphs (Figures 1 through 3) and the raw data collected from these tests are presented following this discussion. A linear regression was performed on this data and the rigid link and strain gauge assembly was found to be very linear with identical calibration factors for both tension and compression. Calibration of Hydraulic Actuators: The hydraulic actuators were also tested in the Baldwin testing machine. Each jack had a separate pressure transducer which measures the magnitude of pressure seen by each jack (one transducer per actuator). Under compression the jacks have a larger capacity than they do under tension becuase of the reduction in area on the tension side of the jack (the actuators are double acting). After fixing the actuator in the testing machine, the pressure in the jack was increased by 200 psi and then the load on the Baldwin machine and the corresponding output voltage from the pressure transducer was recorded. In effect, the Baldwin machine was used as a load cell. All three jacks were tested in compression 164 Appendix B - Beam Element Tester 165 and then in tension. Again the results from these tests are presented subsequently (Figures 4 through 6). In addition to the above, tests were performed to get an idea of the inherent friction in the jacks. This was performed as follows: As the pressure in the hydraulic actuators was increased in the compression side (i.e. the pressure was increased on the side of the piston which caused elongation of the jack), the load was recorded from the Baldwin testing machine. This load is the difference between the product of the recorded pressure and the area of the compression side of the jack and the inherent friction force which opposes the motion of the actuator. The pressure was subsequently increased and the relevant data was recorded. This was repeated five times until reaching the capacity of the jack. Now the pressure in the actuator was slowly released until it reached the corresponding pressure on the inital loading cycle (i.e. the second last point recorded on the initial cycle). Now the force read by the Baldwin testing machine represented the sum of the product of pressure and area and the friction force. Again the pressure was reduced until it hit the next lower point. This was repeated until the pressure in the jack was reduced to zero. By calculating the difference in force for a given pressure in the jack, the friction could be calculated as one half of this difference. Figure B7 shows the results for hydraulic actuators #4. It can be seen the inherent friction is relatively constant throughout the entire range of the hydraulic actuator and very small relative to the ultimate capacity of the jack (approximately 1 %). Also it is to be noted that there is negligible difference in the friction force between different jacks. A brief summary of the calibration factors for both the rigid links and the hyraulic actuators derived from the linear regression of the test results is presented in Table B I . Appendix B - Beam Element Tester 166 Summary of Linear Regression Analysis of Rigid Links and Hydraulic Actuators: Tension Compression Rigid Link #1 28446 lbs/v 28657 lbs/v Rigid Link #2 29037 lbs/v 29163 lbs/v Rigid Link #3 28749 lbs/v 28820 lbs/v Hydraulic Actuator #4 17207 lbs/v 23354 lbs/v Hydraulic Actuator #5 17250 lbs/v 23206 lbs/v Hydraulic Actuator #6 17376 lbs/v 23371 lbs/v Table BI - Summary of Calibration Results Appendix B - Beam Element Tester Calibration of Rigid Link #1 In compression 250 | " " 8 Voltage [volts] Calibration of Rigid Link #1 In tension 250 T ~ ~ 8 Voltage [volts] Figure BI - Calibration of Rigid Link #1 Appendix B - Beam Element Tester Calibration of Rigid Link #2 In c o m p r e s s i o n 250 i 0 2 4 6 8 Voltage [volts] Calibration of Rigid Link #2 In t e n s i o n 250 | • : 1 0 2 4 6 8 Voltage [volts] Figure B2 - Calibration of Rigid Link to Appendix B - Beam Element Tester Calibration of Rigid Link #3 In compression 250 | Voltage [volts] Calibration of Rigid Link #3 In compruslon 250 T Voltage [volte] Figure B3 - Calibration of Rigid Link $3 Appendix B - Beam Element Tester 170 Calibration of Hydraulic Jack #4 In compression 250 "I Voltage [volts] Calibration of Hydraulic Jack #4 In tension 200 | — 2 4 6 8 10 Voltage [volts] Figure B4 - Calibration of Hydraulic Actuator U4 Appendix B - Beam Element Tester Calibration of Hydraulic Jack #5 In compression 250 i — Voltage [volts] Calibration of Hydraulic Jack #5 In tension 200 i • Voltage [volts] Figure B5 - Calibration of Hydraulic Actuator #5 Appendix B - Beam Element Tester Calibration of Hydraulic Jack #6 In compression 250 T Voltage [volts] Calibration of Hydraulic Jack #6 In tension 200 | — Voltage [volts] Figure B6 - Calibration of Hydraulic Actuator if 6 Appendix B - Beam Element Tester 173 Rigid Link 1 tension Load output Kips v o l t s 0 0 Regression Output: 20 0.691 Constant 0.564380 40 1.386 Std Err of Y Est 0.397110 60 2.085 R Squared 0.999967 80 2.77 No. of Observations 11 100 3.49 Degrees of Freedom 9 120 4.19 140 4.89 X Coefficient(s) 28.44564 160 5.6 Std E r r of Coef. 0.053852 180 6.32 200 7.03 Rigid Link 1 comp Load output Kips v o l t s 0 0.245 Regression Output: 20 0.948 Constant -6.90933 40 1.638 Std Err of Y Est 0.214333 60 2.334 R Squared 0.999990 80 3.022 No. of Observations 11 100 3.73 Degrees of Freedom 9 120 4.42 140 5.12 X Coefficient(s) 28.65713 160 5.82 Std Err of Coef. 0.029281 180 6.53 200 7.23 Appendix B - Beam Element Tester I 7 4 Rigid Link 2 t e n s i o n Load Kips 0 20 40 60 80 100 120 140 160 180 200 output v o l t s 0 0.692 1.382 2.06 2.743 3.45 4.12 4.82 5.51 6.2 6.89 Regression Output: Constant 0.039043 Std Err of Y Est 0.193609 R Squared 0.999992 No. of Observations 11 Degrees of Freedom 9 X Coefficient(s) Std Err of Coef. 29.03769 0.026801 Rigid Link 2 Comp Load Kips 0 20 40 60 80 100 120 140 160 180 200 output v o l t s 0 0.689 1.374 2.054 2.738 3.43 12 ,81 ,48 ,17 ,86 Regression Output: Constant -0.01813 Std Err of Y Est 0.143364 R Squared 0.999995 No. of Observations 11 Degrees of Freedom 9 X Coefficient(s) Std Err of Coef. 29.16367 0.019932 Appendix B - Beam Element Tester Rigid Link 3 tension Load output Kips v o l t s 0 Regression Output: 20 0.7 Constant 0 .102949 40 1.388 Std Err of Y Est 0.226578 60 2.083 R Squared 0 .999987 8 0 2 .765 No. of Observations 10 100 3.48 Degrees of Freedom 8 120 4 .17 140 4 .86 X Coefficient(s) 28 .74929 160 5 .56 Std Err of Coef. 0.035858 180 6.27 200 6.95 R i g i d Link 3 Comp Load output Kips v o l t s 0 - 0 . 0 6 3 Regression Output: 20 0.758 Constant 4 0 1.45 Std Err of Y Est 60 2.142 R Squared 80 2.832 No. of Observations 100 3.54 Degrees of Freedom 120 4.23 140 4.93 X C o e f f i c i e n t ( s ) 28.82022 160 5.62 Std Err of Coef. 0.028618 180 6.3 200 7 Appendix B - Beam Element Tester Hydraulic Jack tA - Tension 0 0 Regression Output: 1000 0.167 Constant 0 10500 0.728 Std Err of Y Est 1261.636 22000 1.406 R Squared 0.999531 34000 2.064 No. of Observations 17 45500 2.727 Degrees of Freedom 16 58000 3.43 69500 4.1 X Coefficient(s) 17207.46 82000 4.82 Std Err of Coef. 52.67253 93500 5.49 105000 6.15 118500 6.9 129500 7.5 140500 8.11 154000 8.93 167500 9.65 170500 9.86 Hydraulic Jack »*4 - Compression 0 0.011 Regression Output: 16250 0.71 Constant 0 32000 1.404 Std Err of Y Est 469.6635 54750 2.364 R Squared 0.999947 71000 3.048 No. of Observations 14 79250 3.4 Degrees of Freedom 13 94000 4.05 111000 4.77 X Coefficient(s) 23354.24 126500 5.43 Std Err of Coef. 24.19720 141000 6 157000 6.7 173000 7.42 189000 8.09 206250 8.82 Appendix B - Beam Element Tester 177 Hydraulic Jack /5 - Tension 0 0 Regression Output: 12500 0.736 Constant 0 24000 1.43 Std Err of Y Est 543.4189 35000 2.088 R Squared 0.999902 47000 2.761 No. of Observations 16 59000 3.45 Degrees of Freedom 15 70000 4.09 81500 4.76 X Coefficient(s) 17249.88 93000 5.43 Std Err of Coef. 22.91796 105000 6.11 116500 6.75 128000 7.41 140000 8.11 152500 8.82 164000 9.47 169500 9.79 Hydraulic Jack /5 - Compression 0 0.044 15000 0.693 30500 1.384 46000 2.04 62000 2.702 77500 3.38 93000 4.05 109000 4.72 125000 5.4 140500 6.06 158000 6.78 171500 7.37 188000 8.07 203750 8.75 Regression Output: Constant 0 Std Err of Y Est 921.4949 R Squared 0.999803 No. of Observations 14 Degrees of Freedom 13 X Coefficient(s) Std Err of Coef. 23206.41 47.78874 Appendix B - Beam Element Tester Hydraulic Jack /6 - Tension 0 0 Regression Output: 9500 0.656 Constant 0 21500 1.325 Std Err of Y Est 866.3540 33250 1.977 R Squared 0.999754 44750 2.622 No. of Observations 16 56250 3.27 Degrees of Freedom 15 69000 4 79500 4.63 X Coefficient(s) 17376.10 91000 5.26 Std Err of Coef. 37.23093 102000 5.9 115000 6.614 127000 7.3 139000 7.99 152000 8.69 163000 9.35 169000 9.7 Hydraulic Jack #6 - Compression 0 0.066 Regression Output: 15000 0.722 Constant 0 31000 1.401 Std Err of Y Est 1081.666 46250 2.044 R Squared 0.999730 62000 2.696 No. of Observations 14 78250 3.38 Degrees of Freedom 13 93500 4.04 109500 4.7 X Coefficient(s) 23371.07 125500 5.37 Std Err of Coef. 56.37663 140500 6.01 156500 6.69 173000 7.38 188500 8.03 204000 8.7 Appendix B - Beam Element Tester 179 9 Appendix B - Beam Element Tester -ryes- ****J. hi/fl-CALIBRATION DATA $HEET fc/f rigid link #j hydraulic Jack # Q /tbppsl trial 1 trial 2 trial 3 comments V_v asm 1 C.C14 1 |.*?3 2^«S" t-?70 1*L0 0Trt> 4.W +.10 4.«9 T.»*0 iso cm Iffb att •3-<ro > 168 0Tt> Appendix B - Beam Element Tester 181 f CALIBRATION DATA SHEET - \y rigid link #f hydraulic Jack * Q lbs psl trial 1 trial 2 trial 3 comments 0 - ft!4t a. t4r "Ltrtnn /). 94X irCTWD t>0~7rrrT) 2-.*44 1 o-tnm 4.1-f 4.4t- 4.42-fill- r i T -1 bOVTT) s' »4 fT.es 6.49 6.C4 *.l-4 Appendix B - Beam Element Tester CALIBRATION DATA SHEET OZi rigid 1tnk#^ hydraulic Jack # Q /lb? psl trial 1 trial 2 trial 3 comments — — — — — — 0 a. coo o.ooo | idotrO B.6&I I *.?M OTTb •Z_C?e 1_0*O *-.-»4^ \Ob OTro 3.46 5.^ 110 OTTO 4.(4 4.1* i4o cm 4.e>^  \U> (rtrv \SrO OTT& ICO olro V Appendix B - Beam Element Tester 3~Ar7 boJlV v CALIBRATION D A T A S H E E T [j/j rigid llnk#£ hydraulic Jack # Q lbs psl trial 1 trial 2 trial 3 comments 0 O 0 o rt. I l.tt* crfrro $0 D /; h V *<T I A* +.11-AfrOTT) A.si +:«) %#~trcrb r.+9 r.4y *.l? 1.21 > Appendix B - Beam Element Tester / CAL1BRATJ0N DATA SHEET V riflld «nk #3 hydraulic Jack # Q lbs psl trial 1 trial 2 trial 3 comments (>•% s/ .61*1 .Ten I .5M /.•?&« i lanrnr-r. M«f *.4* ^ errVb 4-.it 4. ft 4.1 q-sA-nrnro IJro <rv\ •Urn frTn 6.n Appendix B - Beam Element Tester /CALIBRATION DATA SHEET - \ / rig'd«nk#3 hydraulic Jack # Q lbs psl trial 1 trial 2 trial 3 comments 0 - 0 0-24 -o.etfS vo<m> /.-4C7D toovO 5o <m> \y6 crtrt 4. to 4-.V2-\*rfi cnr-f. * . » 0 4-. ? 2 -\iO rtvt S.S1 s.cz-1 bO 6m> C?>o 140 err* LOT) Appendix B - Beam Element Tester CALIBRATION DATA SHEET / f~j rigid link # hydraulic Jack #4. y i Ibs/psI trial 1 trial 2 trial 3 comments ofo atoo ium> 14-irb 1.406 4<TS7/T. /i<n ffcvTr* /ion & K'Q-b 1 \"urt> A- in Slxnrv /1-4^0 cn /{<>(*> rl iji-lostvh /['ten / tern Mn ZrrD / lA(rb A.1% it?<Sirtl l&UT) <? t<r 9. xt ( Appendix B - Beam Element Tester CALIBRATION DATA SHEET / [""*] rigid link # hydraulic jack *4 fpf Ibs/psl trial 1 trial 2 trials comments 1-1 ;— 6/0 - D.Ol) to»tM£ZStoti —^= - cV3/t - l.4o+ a&B? / On -?.t?4? -J.40 4.0-s IK em HJSSVTI / Cn Iricrirb 1"UUrh -9- 4-v Itrttnn fiArt) lO(,l&)/UCTt m. i Ut / Iflirb M-<r<rrt / \<C~b Appendix B - Beam Element Tester CALIBRATION DATA SHEET f \ | rigid Jink # hydraulic Jack #5 t/ IbSy'psI trial 1 trial 2 comments 0/0 a 000 1 ICrvTO /urx> 170VT> A<*> M4S oi* A Men pdeasote <x<rri/in er*<cn A<rtn> %.44 s 6Qcnn> /1101 4.10 4.o<3 ttref \Atrv 4 . t » 4-96 ii^ov / i(.ffi> 9i<rn> <.|f loStnrh hty uccm /twi f>t<rn 7.4/ \*,VS<n(lAen *-l6 JMl \Arm M l >49 i&sn [lev* Uis<n> Appendix B - Beam Element Tester Pen. 1/12-CALIBRATION DATA SHEET J \ \ rigid link # hydraulic Jack # 5 -lbs psl trial 1 trial 2 trial 3 comments e/o \<(nn /un <U13 * » 5 n / 4<rt> 1^040 lot-Wh /to* <norb Aufo mm, ft*™ llSatn 1 Herb S.40 1.61 ittrm l-um CA? Wren / W e V>\?ST> /urrf. ^ it, Mm \-<krt tlUJD 1M !T.ro l-am 1 Appendix B - Beam Element Tester CALIBRATION DATA SHEET f f~| rigid link* hydraulic lackey; lbs psl triad trial 2 trial 3 comments 0 lo a.666 f (Zfanf urb 6. 1.46) (LCrb ClMrT> /gtrj l-Mf, wisij /knn ?.?S •J?«rr> / i t * * 4v* 4-.40 nST/Tb /lC(fO <.tf 6-6\ l.W mr/rf /iA(rt l.ol VA(nrT> futrt P. SO Ul L A W Am, i.a«tf i— f-t-—•— /Urvb Appendix B - Beam Element Tester 191 CALIBRATION DATA SHEET f \ \ rigid link* " hydraulic Jack tf£ ] / lbs psl trial 1 trial 2 trial 3 comments wvsi /(cn i. m 44?S"0 /X<T1 sstsfc /lavT> C^cm /iijrh 4x0*0 34<n> /\Atn Slit (aim/ t&rf S".90 HiTfrrf, litm <.CI 1.S6 Micnrt fi4(rt) 7.<n (Finn, 1 um Idem / xxvs MS* APPENDIX C Numerical Experimental Results The following presents the numerical results from the experimental program in a condensed format. For every specimen, the peak loads on every cycle for each ductility level is given. Since there was only one pressure transducer for each hydraulic actuator, the data acquisition equipment was set up to record two forces (compression and tension) from the one pressure reading. Since the pressure only acts on one side of the actuator at a time, the other reading is extraneous. The values have been left in the tables but have been shaded to avoid confusion. Also included is the transducer readings for the axial and transverse displacements. Figure CI shows the numbering of the rigid links and hydraulic actuators of the Beam Element Tester which corresponds to the headings in the tables. Ten (Comp) 3 Mm •a 3 Link 2 Q Ten (Comp) 2 Ten (Comp) 1 Figure CI - Labeling convention for rigid links and hydraulic actuators for the Beam Element Tester 192 Appendix C - Numerical Experimental Results c3 Bo I *3 6 10 p CN 19 C H '9 194 Appendix C - Numerical Experimental Results 5 w s O 195 Appendix C - Numerical Experimental Results u s D 1 a a 196 Appendix C - Numerical Experimental Results a o E H a . a ••e w B ft a B ft Appendix C - Numerical Experimental Results 197 IT) 9 Q a 9 198 Appendix C - Numerical Experimental Results vs u 9 fi 5 CJ 9 fi Appendix C - Numerical Experimental Results Trans ll m C N i m C N m C N i <n C N C N i 1 Axial 1 co co Comp 2 || Comp 3 S C N oo C N i f O N C O C N Comp 2 || Comp 3 o to m. 1 00 N O C N i l l ©: sg © C N Comp 1 H r~--en m C N m o p ro e H m ro CN 2 O O i r -C N S i r f C N Ten 2 fft OS-a i n O N C N roN co m C N oo t>-a 1 C O to Ov O N IS & mi i n © to •a m ro i VO C O i n co s 1 | Link 2 C N O i co m 00 i 00 C N o i I Cycle H Link 1 I O N C O o C O 1 CQ © C N C N © i : vo o 2 / Appendix C - Numerical Experimental Results & S u 201 Appendix C - Numerical Experimental Results u 9 o B<o Appendix C- Numerical Experimental Results in 3 u s fi Appendix C - Numerical Experimental Results © <N 3 9 a (J 9 o Appendix C - Numerical Experimental Results u s fi 9 o Appendix C - Numerical Experimental Results E D a CA I/) a •c fa) if) 9 Appendix C - Numerical Experimental Results w D o g Appendix C - Numerical Experimental Results Appendix C - Numerical Experimental Results I 9 a Appendix C - Numerical Experimental Results 209 rs H<N| 18 s •3 IK I O N I I in ON Appendix C - Numerical Experimental Results 210 o 3 9 o Appendix C - Numerical Experimental Results 211 o B 

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