Abstract An experimental investigation into the effect of steel fibres on the shear strength of concrete beams and slabs containing no transverse reinforcement is presented. Eight slab elements were tested in the U B C Element Tester and eight beams were tested under third point loading. A number of combinations of concrete strength and fibre volumes were considered in the experimental program. The concrete strength was varied from 25 MPa to 95 MPa while the fibre content ranged from 0% to 1.5% by volume. The beams showed an increase in shear strength of up to 80 % when 1.5 % by volume of steel fibres were added to the concrete. Apart from considering the ultimate shear strength of the slabs and beams, the effect of steel fibres on crack widths and modes of crack opening was also considered. A method is presented to include the effect of steel fibres in predicting the flexural strength of steel fibre reinforced concrete sections. Previous results on the pull-out strength of individual fibres were used to define the tensile response of fibre reinforced concrete at a crack. With this proposed stress-strain relationship it is possible to predict the flexural response using a traditional strain compatibility based procedure. A tensile stress block is also derived to simplify flexural predictions. These methods are then compared to experimental results and show favourable agreement. ii Some possible additions to the Modified Compression Field Theory (MCFT) are suggested to account for the increase in shear strength due to steel fibres. To include the effect of steel fibres in a rationally based method for the shear strength of members containing no transverse reinforcement, the following areas are considered: the average post-cracking tensile response of concrete, the ability of a crack to transmit shear force by aggregate interlock and the crack spacing. The theoretical predictions are compared to the experimental results and some additional experimental work and show promise. iii Table of Contents Abstract i i Table of Contents iv List of Figures vi List of Tables vii i Chapter 1 - Introduction 1 Chapter 2 - Experimental Program 5 2.1 Outline 5"" 2.2 Loading Apparatus 5 2.3 Specimen Boundary Conditions for Slab Elements 11 2.4 Test Specimens 13 2.4.1 Slab Elements 13 2.4.2 Beams 15 2.5 Material Properties 16 2.5.1 Longitudinal Reinforcing Steel 16 2.5.2 Concrete 17 2.5.3 Steel Fibres 18 2.5.4 Polyolefin Fibres 19 2.6 Construction of Test Specimens 19 2.7 Instrumentation and Testing Procedure 23 2.7.1 Slabs 23 2.7.2 Beams 26 Chapter 3 - Summary of Experimental Results 28 3.1 Outline 28~ 3.2 Slab Element Test Results 28 3.3 Beam Test Results 38 3.4 Measured Crack Data for Slab Elements 43 3.5 Measured Crack Data for Beams 46 3.6 Summary of Collected Crack Data 51 Chapter 4 - Comparison of Predicted and Measured Strengths 52 4.1 Outline 52 4.2 Flexural Response of Sections Containing Steel Fibres 52 4.2.1 Theoretical Stress Distribution 52 4.2.2 Equivalent Stress Block Including the Effect of Steel Fibres 57 4.2.3 Predicting Flexural Capacities of Section Containing Steel Fibres 60 4.3 Flexural Capacities of Slab Elements 62 4.4 Flexural Capacities of Beams 67 4.5 Shear Capacities of Slab Elements 68 iv 4.6 Shear Capacities of Beams 72 4.6.1 Effect of Fibre Volumes on Shear Strength of Beams 73 Chapter 5 - Towards a Rational Design Method for FRC in Shear 75 5.1 Outline 15 5.2 M C F T for Members Containing Steel Fibres and No Transverse Reinforcing 75 5.2.1 Average Tensile Response of Cracked Concrete 76 5.2.2 Transfer of Shear Across Cracks 78 5.2.3 Crack Spacing Parameter 82 5.3 Shear Capacity of Fibre Reinforced Concrete Beams 83 Chapter 6 - Summary and Conclusion 88 Chapter 7 - Bibliography 93 APPENDICES 96 List of Figures 1.1 Cross sections of steel fibre reinforced beam and slab tests at U B C 2 2.1 (a) General test setup for beams 6 2.1 (b) Positioning of beam in Universal testing Machine 6 2.2 Overall test setup for beam specimens 7 2.3 Plan view of beam element tester 8 2.4 Partial elevation of beam element tester 8 2.5 Transverse shear application using three actuators and three rigid links 10 2.6 Application of transverse shear to slab elements 12 2.7 Slab element boundary conditions 13 2.8 Cross section of slab element 14 2.9 Cross section of beam 16 2.10 Stress-strain curves for samples of longitudinal reinforcing 17 2.11 Dramix®ZL 30/.50 steel fibre 18 2.12 Slab element ready for casting 22 2.13 Casting of beam and slab elements 22 2.14 Instrumentation for measuring deformation during testing of slab elements 24 2.15 Diagrammatic representation of displacements measured during the testing of the slab elements 24 3.1 Load-Displacement curves for SL1-SL3 (fc' = 50 MPa) 30 3.2 Load-Displacement curves for SL4-SL6 (fc' = 95 MPa) 30 3.3 Load-Displacement curves for SL7-SL8 (fc" = 25 MPa) 31 3.4 SL1 at failure (crack numbers marked on photograph) 32 3.5 SL4 at failure (crack numbers marked on photograph) 32 3.6 SL2 at failure (crack numbers marked on photograph) 33 3.7 SL5 at failure (crack numbers marked on photograph) 34 3.8 SL3 at failure (crack numbers marked on photograph) 35 3.9 SL6 at failure (crack numbers marked on photograph) 36 3.10 SL7 at failure (crack numbers marked on photograph) 37 3.11 SL8 at failure (crack numbers marked on photograph) 38 3.12 A typical failure in a beam containing no steel fibres 41 3.13 A typical failure in a beam containing steel fibre 41 3.14 Load-Displacement curves for beams 42 3.15 Calculating shear deformation of crack 46 4.1 Pull-out versus crack width for fibres inclined at various angles to a crack 54 4.2 Predicted shear versus curvature relationship for SL1-SL3 56 4.3 Predicted shear versus curvature relationship for SL4-SL6 56 4.4 Bi-linear representation of the pull-out force for a single fibre 57 vi 4.5 Calculation of stress block factors to include the tensile stresses caused by the steel fibres 59 4.6 Comparison between theoretical moment curvature response and that predicted using stress block factors 60 4.7 Section tested by Lim etal (1987) 62 4.8 Moment-curvature response for slab elements (V f = 0%) 63 4.9 Load-Displacement curves for SL1-SL3 with shear force that causes flexural yielding 65 4.10 Load-Displacement curves for SL4-SL6 with shear force that causes flexural yielding 65 4.11 Load-Displacement curves for SL7-SL8 with shear force that causes flexural yielding 66 4.12 Moment curvature response for beams (V f = 0%) 68 4.13 Calculation of lower bound of ((adopted from MacGregor, 1988) 70 4.14 Comparison of measured strengths with maximum and minimum predicted shear strengths for slab elements 71 4.15 Comparison of measured strengths with maximum and minimum predicted shear strengths for beams 73 4.16 The effect of increasing the fibre content on the shear strength of the beams (fc' kept constant) 74 5.1 Tensile response of concrete (f c '= 50 MPa) according to Tan et al. 77 5.2 Calculating shear contribution (V f i b ) due to presence of tension in steel fibres 81 5.3 Effect of fibre content (V f) on the shear strength of beams for different concrete strengths 86 vii List of Tables 2.1 Nominal characteristics of slab elements 14 2.2 Nominal characteristics of beams 15 2.3 Summary of tensile properties of the longitudinal reinforcing 16 2.4 Concrete cylinder strengths at time of testing 18 2.5 Actual fibre contents obtained from washout tests after casting of specimens 21 3.1 Test results and characteristics of elements SL1 to SL8 29 3.2 Test results and characteristics of B l to B8 38 3.3 Summary of maximum crack widths measured on slab elements 44 3.4 Comparison of crack widths for a similar diagonal shear crack for all slab elements at different load levels 45 3.5 Summary of crack widths along diagonal shear crack (B1-B3) 47 3.6 Summary of crack widths along diagonal shear crack (B4-B5) 48 3.7 Summary of crack widths along diagonal shear crack (B6-B7) 48 3.8 Summary of crack widths along diagonal shear crack (B8) 49 3.9 Summary of average diagonal crack widths at failure 49 3.10 Crack spacing at failure for beams 50 3.11 Average crack spacing for different fibre volumes 50 4.1 Comparison of predicted and experimental flexural capacities for sections tested by Lim etal (1987) 61 4.2 Predicted yield moments of slab elements 64 4.3 Comparison of predicted and measured flexural capacities for slab elements 66 4.4 Displacement ductilities for slab elements failing in flexure 67 4.5 Predicted ultimate flexural capacity of beams (V f = 0%) 68 4.6 Summary of measured and predicted shear strengths for slab elements that failed in shear 69 4.7 Summary of shear strength of beams using C S A 23.3-94 and ACI 318-89 (V f =0%) 72 5.1 Values of o t u for different fibre volumes 77 5.2 Summary of predicted and measured strengths of beams using a modified approach to predicting the shear strength of steel fibre reinforced beams (M/V = 918 mm) 84 5.3 Summary of predicted and measured strengths of beam elements (Adebar et a l , 1987) using the modified approach to predicting the shear strength of steel fibre reinforced beams (M/V = 753 mm) 85 5.4 Comparison between measured and predicted crack widths at failure using the modified approach 87 v i i i Chapter 1 Introduction The design of sections containing no transverse reinforcement has for many years been empirically based. Often the shear strength of such members is critical and the inclusion of transverse reinforcement is expensive or difficult. This is particularly true for the case of punching shear in slabs. Due to the difficulty in anchoring transverse reinforcement, designers are forced to either increase the slab thickness or the size of the supporting member, or provide special slab reinforcement such as stud rails. A l l these solutions are very expensive. Research has shown that adding steel fibres to concrete beams can significantly enhance their shear strength. Should this be true, steel fibres could be used to enhance the punching shear strength of concrete slabs. To increase the fundamental understanding of the effect of steel fibres on shear, it was decided to test slabs in one-way shear before considering the more complex two-way case. The slab elements that were tested in this study contained no transverse reinforcement and were designed to represent a typical floor slab in a high rise building. A series of eight slab elements was designed to gauge the effect of varying fibre content and concrete strength on the one-way shear strength of steel fibre reinforced slabs. The fibre content was varied between 0% and 1.5% by volume, while the concrete strength was varied from 25 MPa to 95 MPa. The slabs had an overall depth of 216 mm and were 610 mm wide (see Figure 1.1 (c)). Research into the effect of steel fibres on the shear capacity of beams containing no transverse reinforcement has been conducted over a long period of time. Few beams ranging in depth between 300 mm and 500 mm have been tested (Adebar et al, 1996), hence it was decided to test some beams within this range to fill the gap in existing knowledge. These tests would then complement existing test results in this field. The beam size decided upon was 200x400 (wide, deep). With the previous beam elements tested by Adebar et al. (1996), there will be results for a full range of sizes of beams/slabs tested in shear and containing steel fibres (see Figure 1.1). 150 CNJ CO 200 o CD in 610 CD CM CO (a) Adebar et al, 1996 (b) Beam Specimens (c) Slab Specimens Figure 1.1 Cross sections of steel fibre reinforced concrete beams and slabs tested at UBC: (a) beam elements tested by Adebar et al. (1996); (b) and (c) beam and slab specimens tested in the present investigation The fibre content and concrete strength of the beams was varied in a similarly to the slabs. That is the concrete strength varied from 25 MPa to 95 MPa and the fibre content varied from 0% to 1.5%. The beams were tested under third point loading using a universal testing machine. An additional beam was added to test the effect of a new polyolefin fibre. This study included the effect of different fibre types on the shear capacity of beams. In addition to the experimental work that was conducted, some analytical work was undertaken to try and better understand the effect of steel fibres on the flexural strength and the shear capacity of sections containing no transverse reinforcement. A method was developed that allows the affect of steel fibres to be included in calculating the flexural response of sections. A stress block is derived to simplify this method and aid in the prediction of the ultimate flexural capacity of sections containing steel fibres. The analytical work on shear was limited to modifying existing relationships in the Modified Compression Field Theory (MCFT) to account for steel fibres. Using the test results from the beams, a preliminary rational method for the analysis of such sections based on the M C F T is presented. The effect of steel fibres on some key relationships controlling the shear strength of sections is considered. The average post cracking tensile response of concrete is modified to account for the effect of fibres along with the effect of fibres on aggregate interlock and crack spacing. These changes are then incorporated into M C F T and the shear capacity of the beams is predicted and compared with the experimental results. The method presented is an initial attempt to quantify what is already known from much experimental data but predicted empirically. Chapter 2 outlines the experimental investigation considering both the methods of testing and detailed descriptions of the beam and slab elements and the relevant material 3 properties. The results of the experimental investigation are described in Chapter 3, with summaries of all the crack data that was collected. Particular attention was given to the crack data collected to investigate the role steel fibres play in reducing crack widths and spacing. Chapter 4 presents the method for incorporating the effect of steel fibres on the flexural capacities of the sections. The experimental and predicted flexural and shear capacities are then compared to the beam and slab elements. Possible changes to the M C F T are presented in Chapter 5. Chapter 6 presents a brief summary of the experimental work and the conclusions that can be drawn from the research conducted. 4 Chapter 2 Experimental Program 2.1 Outline This chapter describes the experimental program, which involved the testing of eight slab elements and eight beams in shear dominated loading. The loading apparatus that were used to load the slab elements and the beams are described in Section 2.2 while Section 2.3 describes the element boundary conditions for the slab elements. A complete description of the slab and beam specimens as well as the properties of the materials used to construct them are given in Sections 2.4 and 2.5, respectively. Section 2.6 describes the construction of the specimens and finally, Section 2.7 summarises the instrumentation and testing procedure. 2.2 Loading Apparatus Two separate test series were conducted. The first consisted of slab elements constructed specifically to be tested in the U.B.C. Element Tester (Adebar, 1994). The second consisted of a series of beams tested in a more traditional way. A Universal Testing Machine was used to test the 400 mm deep beams in third point loading as shown in Figure 2.1 (a). The tester had been modified to allow specimens to be loaded at a constant strain rate. The point loads were applied through a spreader beam 5 as shown in Figures 2.1 (a) and 2.1 (b). 19.05 mm x 65 mm steel plates were grouted to the beam at Load applied by testing apparatus Fixed Support Loading Beam 2754 Figure 2.1 (a) General test setup for beams 3 l 3000 1 . 1 . 918 1 " 918 918 Figure 2.1 (b) Positioning of beam in Universal Testing Machine -Specimen Roller Support i!=J 1 1 1 M fed I I I 90 65 | | 854 65 | 854 65 854 65 90 | | 123 I 918 918 918 123 I these points and placed at the supports to ensure that the load was applied evenly to the beam. One end of the beam was placed on a roller to prevent axial restraint forces 6 developing in the beam during loading, beams as well as the test machine used . Figure 2.2 shows the overall test setup for the Figure 2.2 Overall test set up for beam specimens The slab elements were tested using the U . B . C . Element Tester. The element tester was developed to apply axial load, bending moment and shear to short segments (elements) of structural concrete beams and columns. Figure 2.3 and Figure 2.4 illustrate the test set-up. The tester consists of four steel reaction frames, three hydraulic actuators, three rigid links and two loading yokes. 7 Reaction frames bolted to strong floor n II n n II n o; o; Rigid link Loading yoke T aid | Hj •o Structural concrete / specimen ^ — 1000 kN (225 kip) actuator Figure 2.3 Plan view of beam element tester (adapted from Adebar, 1994) Reaction frame Actuator Structural concrete specimen • r ~ 1 I > < L 7 " Strong floor Figure 2.4 Partial elevation of beam element tester (adapted from Adebar, 1994) 8 The four steel reaction frames are bolted to the strong floor in the structures laboratory and provide the required reactions for the hydraulic actuators and rigid links. The three actuators and three rigid links are connected to the two loading yokes which are in turn connected to the concrete element. Hence the loading yokes transmit the loads to the concrete element. The actuators provide active loads, while the three rigid links provide the equilibrating reactive forces. Each hydraulic actuator has a capacity of 1000 kN in compression and 800 kN in tension. A "load maintainer" allows the hydraulic pressures to be controlled and the rigid links have strain gauges so that they act as load cells. The maximum bending moment that can be applied (without axial compression) is 489 kNm, and the corresponding maximum transverse shear force for a 1.5m long span between rigid links is 533 kN. For the testing of the slab elements it was decided to set the distance between the rigid links to 1220 mm. This meant that the length of the specimen was 900 mm, and the maximum shear force that could be applied was 800 kN. As only one type of load (i.e. maximum shear force) was used during the testing of the slab elements, only this type of load application is described here. A description of how to apply other types of load combinations is given by Adebar (1994). Controlling the axial forces in the three hydraulic actuators is sufficient to control the axial force, bending moment and shear force applied to the specimen (see Figure 2.5). The rigid link (R6), which is parallel to the actuators, always applies a force which is equal and opposite to 9 the net force applied by the three actuators in order to satisfy longitudinal equilibrium. To satisfy moment equilibrium, the transverse rigid links (R4 and R5) must provide the equilibrating force couple (to satisfy transverse equilibrium R4 and R5 are always equal and opposite). To apply maximum transverse shear, the actuators are controlled so that equal moments (magnitude and direction) are applied to the two ends of the element. Thus in the slab element tests, equal and opposite bending moments were applied to the two ends of the element. The applied moment varied linearly (constant shear force) over the element such that the applied bending moment was zero at midspan. That is, the applied bending moment at any section along the element was equal to the applied shear force times the distance from midspan (see Figure 2.6). Figure 2.5 Transverse shear application using three actuators (A) and three rigid links (R) (adaptedfrom Adebar, 1994) 10 2.3 Specimen Boundary Conditions for Slab Elements The details of the connection between the test elements and the loading yokes is important as the boundary conditions need to be simulated correctly. The element-tester connection is shown in Figure 2.7. The slabs were built with a reinforced haunch to shift the critical zone (zone of maximum bending moment, see Figure 2.6) away from the element-tester boundary. The tensile and compressive reinforcement was welded to the element end plates. Although this could lead to an increased shear strength, the required forces in the reinforcement needed to be developed over a short distance. A further advantage was that specimens could be placed in the element tester without having specific tension and compression sides and could be reused i f needed (see Section 2.4). The shear force was transmitted via friction at the element-loading yoke boundary. The friction was ensured by securely bolting the element to the loading yoke. Shear studs were not welded to the end plates due to the high concentration of reinforcement in the haunches (see Figure 2.7). 11 V i -1220-V V ( V .916. ^ V V 0 V 0 V SFD BMD M = Va Figure 2.6 Application of transverse shear to slab elements 12 r- " ~i i i Loading yoke - r l •25.4 mm steel plate to distribute load (bolted to loading yoke) Haunch reinforcement welded to end plate (6-1OM @ 100 c/c) Flexural reinforcement (6-20M @ 100 c/c top and bottom) Slab element Flexural reinforcement welded to end plate End plate bolted to loading yoke Figure 2.7 Slab element boundary conditions A 25.4 mm thick steel plate was placed between the element and the loading yoke to distribute the moment evenly to the slab element since it extended above and below the loading yoke (see Figure 2.7). The slab element was bolted symmetrically to the loading yoke with equal numbers of bolts on the tension and compression sides. 2.4 Test Specimens 2.4.1 Slab Elements The slab elements were designed to represent a typical floor slab found in most high rise buildings. Eight elements were tested in three different series. A l l elements were identical in shape and amount of reinforcing, with the concrete strength and volume of steel fibres being the variables. Table 2.1 summarises the nominal characteristics of the slab elements. 13 A l l slab specimens had an overall depth of 216 mm and were 610 mm wide. The slabs were reinforced with equal amounts of top and bottom reinforcing steel. Each layer consisted of 6-20M at 100 mm centers. The cross-section is shown in Figure 2.8. Main compression and tension reinforcement 6 - 2 0 M @ 1 0 0 c / c 610 00 CM m CM CO o o o o o o o o o o o CM CD CM H 1 1 1 1 h— 55 100 100 100 100 100 55 Figure 2.8 Cross section of slab elements (all dimensions in mm) Table 2.1 Nominal characteristics of the slab elements (actual concrete strength and fibre volumes given in Section 2.5) Element Concrete Strength - fc' Volume of Fibres - V f (MPa) (%) SL1 50 0 SL2 50 0.75 SL3 50 1.5 SL4 95 0 SL5 95 0.75 SL6 95 1.5 SL7 25 0 SL8 25 0.75 14 In specimens SL1 and SL4 failure was concentrated in the slab concrete. The reinforced concrete haunches and longitudinal reinforcement in the slab was undamaged, thus these were reused in the construction of specimens SL7 and SL8. 2.4.2 Beams Seven beams were tested in three different series. A l l the beams were identical in size and amount of reinforcing with only the concrete strength and the quantity of hooked steel fibres being varied. An additional eighth beam was tested using poly olefin fibres. Table 2.2 summarises the characteristics of the beam specimens. The cross-section of the beam specimens is shown in Figure 2.9. There were two layers of 3-20M reinforcing bars on the tension side of the beam. No compression or transverse shear reinforcement was used in the beam. A l l beams were approximately 3.1m long. Table 2.2 Nominal characteristics of beams (actual concrete strengths andfibre volumes given in Section 2.5) Beam Concrete Strength - fc' Volume of Fibres - V f (MPa) (%) B l 95 0 B2 95 0.75 B3 95 1.5 B4 25 0 B5 25 0.75 B6 50 0 B7 50 0.75 B8 50 1.5 (Polyolefin Fibre) 15 200 o o o CO o CD o Longitudinal reinforcement two layers of 20M bars o CO I h + -I—I 40 60 60 40 Figure 2.9 Cross section of beam (all dimensions in mm) 2.5 Material Properties 2.5.1 Longitudinal Reinforcing Steel The longitudinal reinforcement used was Grade 400 weldable deformed steel bars. Tension tests were carried out on three samples. Each sample was 400 mm long and the strain was measured over 50 mm. These results are summarised in Table 2.3 and the resulting load deformation curves are shown in Figure 2.10. Table 2.3 Summary of tensile properties of the longitudinal reinforcing (nominal bar area was used to calculate stress from failure load) Sample fy(MPa) f„it (MPa) 1 410 588 2 415 590 2 408 586 16 600 500 400 "ST Q. — 300 in tn « 200 100 0 6 8 Strain (%) 10 12 14 Figure 2.10 Stress-strain curves from samples of longitudinal reinforcing 2.5.2 Concrete In order to measure the actual compressive strength of the concrete at the time of testing, three standard cylinders, with a diameter of 100 mm and height of 200 mm were cast along with each specimen. The cylinders were then field cured alongside each specimen and tested in the U B C laboratory. The cylinders were tested at the same time as each beam or slab was tested. The average of these results for each specimen is summarised in Table 2.4. A complete list of the individual cylinder strengths is given in Appendix A. 17 Table 2.4 Concrete cylinder strengths at time of testing Element Concrete Strength - fc' (MPa) SL1 50 SL2 50 SL3 52 SL4/B1 94 SL5/B2 96 SL6/B3 96 SL7/B4 24 SL8/B5 26 B6 49 B7 49 B8 49 2.5.3 Steel Fibres The Dramix Z L 30/.50 hooked steel fibre manufactured by Bekaert were used in all but one of the fibre reinforced specimens. The shape of the fibres is shown in Figure 2.11. The fibres are made from low carbon steel wire and have a diameter of 0.5 mm and overall length of 30 mm. The fibres have a minimum tensile strength of 1350 MPa and a length-to-diameter ratio of 60. Figure 2.11 Dramix ZL 30/. 50 steel fibre 18 2.5.4 Polyolefin Fibres Polyolefin fibres manufactured by 3M were used in Specimen B8. These fibres are straight with an elliptical cross section. The major axis is 1.00 mm while the minor axis is 0.68 mm and the total length is 50 mm. The ultimate tensile strength of these fibres range between 275 MPa to 300 MPa. 2.6 Construction of Test Specimens The construction of the slab elements involved fabrication of the steel end plates, welding of the steel reinforcement cage and the construction of the formwork and finally casting of the concrete. To ensure that each element could be bolted into the tester, a steel template was fabricated with a bolting pattern similar to that of the loading yoke. This template held the bolts in position while the end-plate were placed over the bolts and nuts were welded onto the end-plate. This process ensured that the nuts were correctly placed. The longitudinal reinforcement was welded to the end-plates once the nuts had been attached. The reinforcement in the haunch was then welded to the end-plate. Reusable wooden forms were constructed to hold the steel cages during casting of the concrete. Figure 2.12 shows a completed steel reinforcing cage in its form ready for casting. 19 Figure 2.12 Slab element ready for casting Reusable wooden forms were also constructed for the beams and the reinforcement was placed on plastic spacers in the forms to ensure the correct spacing and cover requirements. The concrete was obtained from a local ready-mix supplier. The steel fibres were placed directly into the mixing drum. To ensure that the correct amount of fibres were added to the concrete the volume of concrete was monitored. This was done by removing a set volume of concrete each time using a concrete bucket of known volume. 20 Initially plain concrete was removed from the truck to cast the specimens with no fibres. The concrete was placed into the forms from the bucket which was held in place by an overhead crane and vibrated using an immersion vibrator (see Figure 2.13). While these beams were being poured, super-plasticiser was added to the remaining concrete in the truck and then fibres were added to bring the volume of fibres up to 0.75%. Some of this concrete was then removed to cast the required slabs and beams. Super-plasticiser was added to the remaining concrete and the fibre content increased to 1.5% by volume and the final beams were cast. Figure 2.13 Casting of beams and slab elements 21 Washout tests were performed to measure the actual fibre contents obtained. These results are shown in Table 2.5. Table 2.5 Actual fibre contents obtainedfrom washout tests after casting of specimens Specimen Intended Fibre Content Actual Fibre Content vf(%) vf(%) SL1 0 0 SL2 0.75 0.65 SL3 1.5 1.37 SL4/B1 0 0 SL5/B2 0.75 0.58 SL6/B3 1.5 1.48 SL7/B4 0 0 SL8/B5 0.75 0.79 B7 0.75 0.65 B8 1.51 1.04' . Polyolefin fibre, all others hooked steel fibres The final series of beams (B6, B7, B8) were cast in a different way as two different fibres were used. A portable concrete mixer that could hold sufficient volume for one beam was used . A set volume of concrete was removed and placed in the mixer. The required amount of fibres was mixed into the concrete and then placed in the beams and vibrated as before. The concrete containing polyolefin fibres caused some problems during casting due to its high fibre content making the mix unworkable. The fibres were also longer than the cover provided which made placing the concrete difficult as it did not move freely between the reinforcing. A l l specimens were covered with plastic sheets and left to cure for seven days. They were then stripped and left in the laboratory until they were tested. 22 2.7 Instrumentation and Testing Procedure 2.7.1 Slabs The instrumentation for the slab elements included pressure transducers to measure the oil pressure in the three hydraulic actuators (active loads), load cells on the three rigid links to measure the corresponding reactive loads, three displacement transducers used to measure the overall displacement of the element, and photographs used to record the crack patterns and the general failure mode. Nine electronic signals (from the three pressure transducers, three load cells and three displacements) were converted from analog to digital before being recorded by a computer using the data acquisition software Labtech Notebook (manufactured by Laboratory Technologies Corporation). The overall displacements were measured in two directions (Figures 2.14 and 2.15). The axial displacement (AL) and transverse displacement (AT 2) were measured using two L V D T displacement transducers mounted on aluminium support arms which cantilevered from the end plate at one end. The displacements were measured along the centerline of the element. To verify the symmetrical nature of the deflections another support was cantilevered from the opposite endplate and an L V D T was attached to measure a 'transverse displacement' (AT^. 23 Figure 2.14 Instrumentation for measuring deformation during testing of slab elements L A L A T 2 A T , Figure 2.15 Diagrammatic representation of displacements measured during the testing of the slab elements 24 To record the crack patterns, the elements were painted with a thin coating of white latex paint. The cracks were marked with black markers on the paint alongside the actual crack. A camera, mounted on a specially made over-head tripod, was used to record the development of the crack pattern. Cracks less than 1 mm wide were measured using a special microscope while those greater than 1 mm wide were measured using a crack comparator. A label was placed next to the crack to indicate the width on the photograph and it was recorded in the experimental notes. To measure the crack inclinations and angles of opening, vertical lines were drawn on the slab before testing. A thin line was drawn on an overhead transparency and this was placed over the crack in such a way that it represented the average inclination of the crack. The inclination of this line to the vertical lines on the elements was measured to calculate the inclination of the cracks. The angle of opening was measured in a similar way accept that the line on the transparency was used to match what were initially the same points on the crack. By joining these two points with a straight line the angle that the crack was opening at could be measured with respect to the vertical lines on the element. The testing procedure was the same for all elements. In all tests the loads (the pressure in the jacks)'were slowly increased using the hydraulic load maintainer. A load stage was taken whenever a significant change occurred in the crack pattern. At this point, the 25 applied load was decreased slightly in order to maintain a constant deformation while the cracks were marked and measured, and photographs were taken. 2.7.2 Beams The instrumentation consisted of three displacement transducers to measure the load and the displacement of the beam and photographs to record the crack patterns and final failure mode. Three electronic signals were converted from analog to digital before being recorded by a computer using the data acquisition software Labtech Notebook. An L V D T displacement transducer was placed inside the testing machine to measure the displacement of the pendulum arm used to load the specimen. Using curve fitting parameters this displacement was converted to an applied load by the data acquisition software (This had been done prior to loading the beams and was calibrated on a load cell). The displacement of the specimen was measured by placing a L V D T displacement transducer under the center of the beam. Another displacement transducer was also used to measure the displacement of the loading beam. In order to record the crack patterns photographically the beam was painted with a thin coat of white latex paint. The cracks were marked with a black marker adjacent to the actual crack. The crack widths were measured using a microscope for cracks less than 1 mm wide and a crack comparator for cracks greater than 1 mm wide. Labels were placed next to the crack to indicate the widths. Photographs were then taken to record this information. 26 The testing procedure for all the beams was the same. The beam was placed in the testing machine as shown in Figure 2.1 (b). Loading plates were made and grouted to the beam to ensure an even load distribution at the loading points. A roller was placed under one : support to ensure that that the beam was not restrained during the test. The load was then gradually increased and a load stage was taken whenever a significant change occurred in the crack pattern or crack width data needed to be recorded. During the load stage, the load was maintained by the test machine and the rate of loading was set to zero. Figure 2.2 shows the overall test setup for the beams. 27 Chapter 3 Summary of Experimental Results 3.1 Outline In this chapter the results of the experimental work described in Chapter 2 are summarised. Brief descriptions of each test and the ultimate loads are presented in Sections 3.2 and 3.3 for the slab elements and beams respectively. Section 3.3 summarises the crack width data and crack opening characteristics collected from the slab element test series, while Section 3.4 summarises the crack width data collected from the beam tests. 3.2 Slab Element Test Results The maximum recorded shear forces and observed failure modes of all the slab elements are recorded in Table 3.1; also included are the volume of fibres and concrete strengths associated with each slab element. The failure types are initially characterised as either a pure shear failure or a flexural splitting type failure. These are described in more detail in the following paragraphs. Also included below is a description of what occurred during each test. The load-displacement curves for the different test series (concrete strength kept constant while fibre content was varied) are shown in Figure 3.1 to Figure 3.3. From these curves the flexural ductility of the elements is illustrated. Elements that failed predominantly in shear show little flexural yielding. The addition of the fibres increased the shear capacity 28 of the elements so that the flexural capacity was reached before the element failed in shear. It can also be noted that the addition of fibres increased the stiffness of the elements. Table 3.1 Test results and characteristics of elements SL1 to SL8 Element fc' (MPa) vf(%) V u (kN) Type of failure SL1 50 0 355 shear SL2 50 0.65 485 flexural splitting/shear SL3 52 1.37 528 flexural splitting/shear SL4 94 0 364 shear SL5 96 0.58 514 flexural splitting/shear SL6 96 1.48 480 flexural splitting/shear SL7 24 0 239 shear SL8 26 0.79 342 shear Specimen SL1 (V f = 0%) failed in shear. The failure was characterised by the initial formation of flexural cracks which became inclined as they propagated towards the compression face (flexural shear cracks). These cracks then moved diagonally down towards the tension face forming one predominant diagonal shear crack. This crack ran from the corner of the haunch on the compression face down to the tension face at an angle of approximately 45° (see Figure 3.4). The initial cracking of the element was symmetrical and hence a diagonal crack running from the compressive face to the tension face could be seen on each side of the specimen. Just prior to failure, a flat diagonal 29 600 S L 2 (485 KN) - V f = 0.75% S L 3 (528 KN) - V f = 1.5% SL1 (355 KN) - V f = 0% 10 15 D e f o r m a t i o n ( m m ) 20 25 Figure 3.1 Load-Displacement curves for SL1 to SL3 (fc'= 50 MPa) 600 SL6 (460 KN) - Vf = 1.5% SL5 (514 KN)-Vf = 0.75% SL4 (364 KN) - Vf - 0% H h 2 4 6 8 10 12 14 16 Displacement (mm) Figure 3.2 Load-Displacement curves for SL4 to SL6 (fc'~ 95 MPa) 30 500 400 1 f SL8 (342 KN) - V f = 0.75% o o i_ o 300 i S L 7 (239 KN) - V f = 0% 0 2 4 6 8 10 12 Deformation (mm) Figure 3.3 Load-Displacement curves for SL7 to SL8 (fc,z= 25 MPa) crack appeared and ran from the mid point of the dominant diagonal crack, down towards the opposite haunch (see Figure 3.4). This was the failure crack and loading could not be increased once this crack extended along the whole length of the element. The result of this crack was the spalling of the cover concrete from where the initial diagonal crack crossed the longitudinal tension steel to the opposite haunch. This observation was also made for SL4 (V f = 0%) (see Figure 3.5). Element SL2 (Vf = 0.75 %) exhibited significant flexural yielding before failing in a complex flexural splitting/shear type failure (see Figure 3.6). The failure was characterised by the initial formation of flexural shear cracks as described above. The flexural cracks began to become inclined at about half the overall depth of the element 31 Figure 3.4 SL1 at failure (Crack numbers marked on photograph) Figure 3.5 SL4 at failure (Crack numbers marked on photograph) (see Figure 3.6) as compared with SL1/SL4 where the inclination of the flexural crack started closer to the tension face. A stage was reached when the shear cracks stabilised and the deformations were concentrated at the haunches. This continued with the flexural cracks at the haunches opening up while the shear crack widths remained constant. Just 32 prior to failure, splitting cracks could be seen in the central portion of the specimen, and once these appeared the load could not be increased significantly. The failure crack was characterised by a large crack running diagonally across the length of the specimen as shown in Figure 3.6. A similar failure was witnessed with SL5 (see Figure 3.7). The angle of the failure crack though was steeper than that seen in SL2. Figure 3.6 5X2 at failure (Crack numbers marked on photograph) 33 Figure 3.7 SL5 at failure (Crack numbers marked on photograph) Element SL3 (V f = 1.5 %) exhibited significant flexural yielding before failing in a combination of a flexural splitting/shear type failure (see Figure 3.8). The failure was similar to that seen in SL2 with the difference being the formation of smaller flexural shear cracks. A zone of finer diagonal shear cracks was formed instead of a single dominant crack. The flexural shear cracks also began to become inclined closer to the compressive face than in SL2 or SL1. There were a number of flexural shear cracks developed along the length of the specimen, the further away they were from the haunch, the flatter they became. A l l the diagonal shear cracks eventually propagated towards the haunch on the compressive side. As was noted for specimen SL2 a point was reached in the testing of SL3 when the diagonal shear cracks stabilised and the deformations were concentrated at the haunches 34 in flexural hinging. The wide flexural cracks can be seen in Figure 3.8. At failure a diagonal crack ran from the haunch downwards along the flexural shear cracks (see Figure 3.8). The test was stopped at this point to prevent a sudden shear failure that would destroy the specimen completely, hence no definite failure plane could be seen. Figure 3.8 SL3 at failure (Crack numbers marked on photograph) SL6 (V f = 0%) failed in a similar manner to SL3 (see Figure 3.9). There were fewer flexural shear cracks and at failure there was crushing of the concrete at the compressive haunch. This is seen by the large splitting type cracks in the compressive side of the haunch in Figure 3.9. SL7 (V f = 0 %) failed in shear similar to SL1 (see Figure 3.10). There was the formation of a flexural shear crack but it was only seen on one side of the specimen. The growth of this crack was similar to the formation of the flexural shear cracks as described for 35 specimen SL1. On the opposite side only a single flexural crack at the haunch was seen. As the load was increased the single flexural shear crack began to open up. A l l the deformation seemed to be concentrated along this single crack. While the load was being held stable to measure crack widths and modes of opening, a flat diagonal crack suddenly appeared. This crack ran from the opposite compressive haunch across the specimen (see Figure 3.10). Upon reloading, the element failed along this crack. Figure 3.9 SL6 at failure (Crack numbers marked on photograph) 36 Figure 3.10 SL7 at failure (Crack numbers marked on photograph) SL8 (V f = 0.75 %) was the only slab element containing fibres to fail in a typical shear failure. Flexural shear cracks formed in a symmetrical manner as described for element SL1. The deformations, though, seemed to be concentrated on one side of the specimen. At failure, a flat diagonal crack similar to the one seen in SL7 appeared and the element was unable to carry an increased load (see Figure 3.11). A summary of the experimental notes for each test can be found in Appendix B. Appendix C contains a listing of the measured forces in the hydraulic actuators and the rigid links at failure. 37 Figure 3.11 SL8 at failure (Crack numbers marked on photograph) 3.3 Beam Test Results The maximum recorded shear forces and observed failure modes of all the beams are recorded in Table 3.2. Also included is a summary of the distinguishing characteristics of each beam. Table 3.2 Test results and characteristics of Bl to B8 Beam fc' (MPa) V f ( % ) V u ( k N ) Type of failure B l 94 0 130 shear B2 96 0.58 232 shear B3 96 1.48 260 shear B4 24 0 105 shear B5 26 0.79 156 shear B6 50 0 125 shear B7 50 0.65 227 shear B8 50 1.04 176 shear 38 A l l the beams failed in shear. The beams that contained no steel fibres all behaved in a similar manner as follows. There was an initial period of flexural cracking in the central span (i.e., between the point loads where the bending moment was constant). These cracks were vertical and as the load was increased new cracks developed between the point loads and the support. As these flexural cracks moved towards the compression face they began to become inclined and propagate towards the loading points. Once these cracks appeared the beam soon failed. The failure was usually very sudden (brittle), the only warning being the sudden appearance of a diagonal shear crack running from the loading point to the support. A stable crack pattern did not develop through the test and no crack seemed to dominate the deformations. Figure 3.12 shows a typical beam containing no steel fibres at failure. B6 (V f = 0%) underwent a slightly different shear failure as a definite diagonal shear crack was formed along which large deformations were measured. Since a pseudo displacement controlled testing machine was used and the load was not completely removed when crack width measurements were being made. Consequently during this time a large amount of creep was detected (see Figure 3.14). If the beam had been loaded continuously, crack widths of the magnitude detected would not have been formed as the beam would not have had time to redistribute the stresses across the crack and hence enhance its strength. 39 The beams containing steel fibres failed in a similar manner to those without fibres, the difference being that a dominant diagonal shear crack formed during loading. The deformation was concentrated at this crack while the other shear cracks remained stable. Also common to the failure of the beams containing steel fibres was the extension of the diagonal crack downwards to the support by a large number of finer cracks. The steel fibres seemed to induce a larger number of diagonal cracks before a single crack became dominant. This did not occur in the beams containing no steel fibres. Figure 3.13 shows a typical shear failure and crack pattern in a beam that contained steel fibres. B8 containing 1.04% polyolefin fibres failed in a similar manner as the beams containing steel fibres, though it was unable to reach the loads that the beams containing 0.75% steel fibres reached. The load-displacement curves for the different test series (concrete strength remains constant while fibre content is varied) are shown in Figure 3.14. The effect of the fibres in increasing the shear strength but not affecting the ductility can be seen. The fibres also do not effect the stiffness of the beams. Appendix D contains detailed descriptions of each test along with individual load displacement curves. 40 Figure 3.12 Typical shear failure in a beam containing no steel fibres Figure 3.13 Typical shear failure in a beam containing steel fibres 41 Deformation (mm) Figure 3.14 Load-displacement curves for the beams 42 3.4 Measured Crack Data for Slab Elements Throughout the testing of each slab element, detailed measurements of the crack widths and modes of opening were made. Appendix E lists all this data along with photographs illustrating where the crack width measurements were made. This section summarises the crack data collected and considers the effect of the fibres on crack widths. The presence of steel fibres in concrete is known to reduce crack widths and crack spacing in structural elements. A comparison of the crack width data collected is presented to try and aid the understanding of the effect of steel fibres on crack widths. Table 3.3 summarises the crack widths in each slab element at failure or close to failure. Since a large number of cracks were measured, the widths of the dominant diagonal shear cracks that were opening at failure are used. The cracks considered are listed in brackets next to the crack widths. From Table 3.3 it can be seen that by increasing the fibre content, the crack widths at failure are reduced. A comparison between SL7 and SL8, which both failed in a shear dominated failure, highlights this fact. At failure the crack width in SL7 (V f = 0 %) was 1.60 mm, while in SL8 (V f = 0.79 %) the width was 0.9 mm. This amounts to a 56 % reduction in crack width. 43 1 \ Table 3.3 Summary of maximum crack widths measured on slab elements (crack number in brackets) vf(%) Element # Crack widths 0.00 0.00 0.00 S L l SL4 SL7 1.40 (#C) 1.40 (#D) 1.80 (#B) 1.60 (#F) 2.00 (#G) 0.55 (#E) 0.90 (#B) 0.55 (#D) 1.60 (#D) 1.60 (#C) 0.65 0.58 0.79 SL2 SL5 SL8 0.40 (#C) 0.90 (#D) 0.35 (#E) 0.50 (#F) 0.25 (#E) 0.45 (#F) 0.20 (#D) 0.90 (#E) 0.65 (#G) 1.37 1.48 SL3 SL6 0.30 (#B) 0.15 (#D) 0.30 (#E) 0.20 (#F) 0.70 (#E) 0.20 (#G) 0.12 (#D) 0.20 (#C) Table 3.4 summarises the crack widths for a diagonal shear crack that was common to the elements. The actual crack used is listed next to the element number for reference purposes. The reduction in crack width due to the steel fibres can be seen for a given load. Considering elements SL7 and SL8 again, we see the crack width was reduced from 1.6 mm to 0.15 mm at the same load level. Similar conclusions can be drawn for the other elements. Since some of theses failures were dominated initially by flexural deformations the results may not hold entirely true for these elements. The shear deformation of the crack can be calculated as shown in Figure 3.15. A summary of the shear deformations of the cracks is listed in Appendix E. From the crack data, shown in Appendix E, it can be seen that as the load approaches failure, the shear deformation of the crack increases. A comparison is made between SL7 and SL8, which both failed in a shear dominated manner. A shear deformation of 0.71 mm and 1.34 mm was measured in SL7 just prior to failure. For the similar crack on SL8, a shear deformation of 0.24 mm was measured just prior to failure. This seems to suggest that 44 the presence of steel fibres reduces the shear deformation of the crack. Similar results are seen in the other elements although as mentioned the failures were not shear dominated. Table 3.4 Comparison of crack widths for a similar diagonal shear crack for all slab elements at different load levels (The crack for which the information is quoted is stated next to the specimen number) Load (kN) SL1 (#B) V f = 0 fc' = 50 SL2 (#E) V f=0.65 fc' = 50 SL3 (#D) V f = 1.37 fc' = 50 SL4 (#B) v f =o fc' = 95 SL5 (#F) V f=0.58 fc' =95 SL6 (#E) V f = 1.48 fc' = 95 SL7 (#D) V f = 0 fc' = 25 SL8 (#E) V f=0.79 fc' = 25 167 0.10 174 0.20 180 0.09 190 0.65 200 0.15 0.15 210 0.90 220 „ 0.40 1.00 233 0.40 240 0.40 1.60 0.15 250 0.50 0.60 260 0.60 0.20 275 0.90 280 0.70 290 0.70 0.45 300 0.12 0.90 0.05 310 0.95 0.55 330 1.00 0.10 340 1.50 0.90 350 1.80 0.20 370 0.15 0.05 0.25 400 0.25 0.10 0.08 0.30 420 0.35 0.15 440 0.10 450 0.35 0.10 0.25 0.7 470 0.15 0.35 500 0.15 0.40 510 0.45 45 crack edges a P Shear on crack ( A ) cos(a + P) Figure 3.15 Calculating the shear deformation of a crack 3.5 Measured Crack Data for Beams Throughout the testing program the crack widths in the beams were carefully monitored at various load levels. Appendix F summarises this data. The crack width data of interest are the width of the diagonal shear cracks. Tables 3.5 -3.8 summarise the widths of the diagonal shear cracks at different load levels for each test series. The widths were measured at various places along the crack starting at the tension face and moving towards the compression face. Table 3.5 shows the effect of steel fibres in preventing the formation of diagonal shear cracks. Considering the diagonal shear crack close to the tension face, a crack is first detected at 75 kN for B l (V f = 0%) while for B2 (V f = 0.75%) the first crack is detected 46 at 160 kN. A similar crack is detected on B3 (V f = 1.5%) at 140 kN. This suggests steel fibres inhibit the crack growth initially. Tables 3.5-3.8 serve to augment these results. Table 3.5 Summary of crack widths along diagonal shear crack for B1-B3 Specimen (V f) Shear (kN) Crack widths (mm) <- compression face tension face -> 75 0.05 100 0.05 0.1 B l 120 0.15 0.1 (0%) 128 0.2 130 0.2 0.3 160 0.1 180 0.2 B2 200 0.4 (0.58 %) 204 1.0 1.4 1.0 215 1.6 2.5 2.5 2.0 225 2.0 3.0 3.0 2.0 230 2.5 3.5 3.5 3.0 185 0.3 0.4 195 0.5 0.6 B3 200 0.7 0.95 (1.48%) 207 1.0 1.2 220 1.5 2.0 240 1.6 2.5 258 2.5 3.0 47 Table 3.6 Summary of crack widths along diagonal shear crack for B4-B5 Specimen Shear Crack widths (mm) (V f) (kN) <- compression face tension face -> B4 65 0.05 (0 %) 80 0.1 0.07 95 0.15 0.15 90 0.1 B5 105 0.15 (0.79 %) 125 0.15 0.18 0.25 140 0.35 0.5 0.5 150 0.5 0.8 0.8 Table 3.7 Summary of crack widths along diagonal shear crack for B6-B7 Specimen (V f) Shear (kN) Crack widths (mm) B6 1 70 0.08 0.05 (0%) 100 0.09 0.1 110 1.0 1.0 0.9 115 1.0 1.2 0.6 120 1.4 1.4 0.8 125 1.8 2.0 1.0 120 0.05 130 0.05 B7 150 0.1 (0.65 %) 170 0.18 0.2 185 0.1 0.3 0.5 200 0.5 1.0 0.9 215 0.7 1.6 1.2 200 0.8 1.7 1.3 220 0.9 1.8 1.4 1. The large crack widths measured are due to an error in the testing procedi that is explained in Section 3.2 48 Table 3.8 Summary of crack widths along diagonal shear crack for B8 (polyolefin fibres) Specimen (V f) Shear (kN) Crack widths (mm) <- compression face tension face -> B8 (1.04%) 120 0.4 0.6 0.5 130 0.8 1.0 0.5 140 1.0 1.2 0.6 154 1.2 1.6 0.8 165 1.6 1.6 0.9 Table 3.9 summarises the average diagonal crack width at failure for all the beams tested. From this we see that there is ah increase in the crack width at failure when steel fibres are added to the concrete. Table 3.9 Summary of average diagonal crack widths at failure V f ( % ) fc' (MPa) 0 0.75 1.5 95 0.25 3.1 2.8 25 0.15 0.7 -50 1.6 1.4 -From the data shown in Tables 3.5 to 3.9 the following conclusions can be drawn about the effect steel fibres have on the cracking of the concrete beams. Firstly the fibres inhibit the initial formation of shear cracks and secondly they allow greater crack widths at failure. The effect that steel fibres have on flexural and shear crack spacing is also important as this influences the relationship between strain and crack widths. Table 3.10 shows a brief analysis of the flexural crack spacing at failure for the beam elements. The crack spacing was calculated by counting the number of flexural and flexural shear cracks that could be seen at failure between the loading point and the support. This was done for both sides of the beam and then the average number of cracks was calculated. The crack spacing was then calculated by dividing the shear span (918 mm) by the number of cracks. Table 3.10 Flexural crack spacing at failure for beams Beam # vf(%) No of cracks No of cracks No of cracks Sfx (left) (right) (average) (mm) B l 0 6 6 6 153 B2 0.58 11 11 11 83 B3 1.48 8 12 10 92 B4 0 5 6 5.5 167 B5 0.79 10 7 8.5 108 B6 0 7 8 7.5 122 B7 0.65 10 9 9.5 97 B8 1.04 6 6 6 153 From Table 3.10 we can see the effect of fibres on reducing the flexural crack spacing in the beams. Table 3.11 summarises the average crack spacing for the different fibre contents. We can note that the presence of steel fibres reduces crack spacing. There is however only a small change in crack spacing once fibres have been added. Table 3.11 Average flexural crack spacing for different fibre volumes vf(%) S f x (mm) 0.0 147 0.75 96 1.5 92 50 3.6 Summary of Collected Crack Data The slab tests indicate a reduced crack width and an increased shear displacement on the crack at failure as the fibre content was increased. The presence of steel fibres also inhibited the formation of the first flexural and shear cracks. Few conclusions can be drawn on the effect of fibres on the crack widths and crack deformations from the slab elements as most of the slabs containing fibres did not fail in a shear dominated manner. The beam tests indicated that as the fibre content was increased, the diagonal shear crack width at failure also increased. The fibres also had the effect of reducing the crack spacing and allowing the formation of a larger number of smaller shear cracks. Steel fibres inhibited the formation of the first flexural cracks and first shear cracks in a similar manner as that observed in the slab elements. 51 Chapter 4 Comparison of Predicted and Measured Strengths 4.1 Outline The predicted and measured responses of the slab elements and the beams are compared in this chapter. In Section 4.2 a method is presented to predict the flexural response of sections containing steel fibres. This method is compared with the measured ultimate loads from the slab element tests in Section 4.3. Section 4.4 briefly considers the predicted flexural capacities of the beams and compares them to the experimental results. Sections 4.5 and 4.6 compare the predicted and measured shear strengths for the slab elements and beams, respectively. 4.2 Flexural Response of Sections Containing Steel Fibres The addition of steel fibres is known to increase the flexural strength of reinforced concrete sections. The steel fibres are able to carry a tensile force across the cracks and hence increase the tensile capacity of the section. By simply adding an additional tension stress block to account for this additional stress, the effect of the fibres can be included. The derivation of this stress block is outlined in the following sections. 4.2.1 Theoretical Stress Distribution A number of tests have been conducted on the pull-out of individual fibres inclined at different angles to a crack (Armelin and Banthia, 1996). A pull-out force per fibre, as a 52 function of crack width for each fibre inclination, can be defined with this information. These can be summed as proposed by Armelin and Banthia (1996) to calculate the average pull-out force per fibre (F p o) as a function of crack width. (This is explained in more detail in Section 5.2). The summation accounts for the random orientation of fibres and all possible inclinations and embedment lengths to give an average response of a single fibre. The individual responses of fibres inclined at various angles to a crack are shown along with the average pull-out force in Figure 4.1. The various inclinations referred to in Figure 4.1 are the inclination of a single fibre with respect to the crack plane. F h o o k refers to the pull-out strength of a fibre with just the hook embedded in the concrete. F p o(w) can be defined by a modified Ramberg-Osgood function f F, = 1.0677 x w\ -0.0208 + 1 + 0.0208 (4.1) pull-out V 1.7 J 53 0.2 22.5° 0.18 - * • " " " 0.16 . 0.14 . V - N . . * 1* - - . 45° Force (kt 0.12 . 0.1 . •+-» iv X Pull-oi 0.08 -0.06 . 0.04 . 0.02 . 0 ly / \ F ^ 67.5° 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 Crack width, w (mm) Figure 4.1 Pull-out force versus crack width for fibres inclined at various angles to a crack (FpQ is the average force pull-out force) Once the force in a single fibre is known it is possible to calculate the total force i f the number of fibres crossing a crack is known. Romualdi and Mandel (1964), have shown that the best representation of fibre spacing or number of fibres is given by N,=ax^- (4.2) Af where Nj = number of fibres per unit area Vf = Volume of fibres (expressed as a percentage of the total volume) Af = cross-sectional area of a single fibre of = 0.5 (for large structures) 54 By multiplying the average pull-out force per fibre (F p 0) by the number of fibres per unit area an equivalent stress can be calculated. This results in a tensile stress distribution through the tension zone of a section. It is assumed that the crack width varies linearly from zero at the neutral axis to a maximum at the bottom of the tension zone. The tensile force can then be calculated by multiplying by the area of concrete in tension. This tensile stress distribution was included in a computer program that uses a layer-by- . layer representation of a section to predict the moment curvature relationship for a given section. To estimate the crack width (w) at each layer, the following equation was used w = s*sx (4.3) Where w = crack width s = strain in the layer sx = crack spacing The flexural crack spacing (sx) was assumed to be uniform through the depth of the section. It was based on the maximum crack width proposed by the Gergeley-Lutz Expression. w m a x = 1 1 x P X fsc ^dJ- x 1 °~6 (units in MPa, mm) (4.4) where /?= factor accounting for strain gradient /3 = 1.0 for uniform strain gradients P = h2/h] for varying strains where hj is the distance from the tension steel to the neutral axis and /?2 is the distance from the extreme tension fibre to the neutral axis. f s c r = stress in reinforcement at crack location dc = distance from extreme tension fibre to the closest reinforcing bar A = effective area of concrete surrounding each bar 55 The resulting moment curvature relationships for the slab elements, SL1-SL6, are shown in Figures 4.2 and 4.3 (for comparative purposes the moment has been converted to a shear force so that the shapes of the curves can be compared with the actual load displacement curves of the slab elements, see Section 3 .2) . f; = 50 MPa SL3(V ,= 1.37%) SL2 (V,= 0.65%) SL1 (V = 0%) 0.05 0.1 Curvature (rad/m) 0.15 Figure 4.2 Predicted shear versus curvature relationship for SL1-SL3 700.0 600.0 -500.0 -z 400.0 -h. re o 300.0 -CO 200.0 -100.0 -0.0 -0 0.02 0.04 0.06 Curvature (rad/m) tV = 95 MPa SL6(V f =1.48%) SL5 (V f= 0.58%) SL4(V f =0.0%) 0.08 0.1 Figure 4.3 Predicted shear versus curvature relationship for SL4-SL6 5 6 4.2.2 Equivalent Stress Block Including the Effect of Steel Fibres The use of a complex stress strain-curve does not allow for hand calculations to be made. An equivalent stress block was derived that simplifies the tensile stress carried by the fibres to a rectangular stress block (see Figure 4.5 (c)). The function representing the average force per fibre is plotted in Figure 4.4. This can be converted to an average stress by multiplying the average force per fibre (F p o) by the number of fibres per unit area (Nj) as described in the previous section. This results in a complex curve that does not have a closed form solution to its integral. However this curve can be represented by a bi-linear curve as shown in Figure 4.4. The error caused by simplifying the curve to a bilinear one is small and was checked by comparing moment curvature responses using both the modified Ramberg-Osgood curve and the bilinear curve to define F p o(w). 0.16 -, . 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 Crack width, w (mm) Fpull-oul (modified Ramber-Osgood) Fpull-out (bi-linear) Figure 4.4 Bi-linear representation of the pull-out force for a single fibre 57 The bi-linear curve is described by the following equation w < 0.32 mm: Fpull-out = 490x w (N) w > 0.32 mm: ¥puU-0ut = 164 - 22.3xw (N) (4.5) The crack width can also be written in terms of y (depth below neutral axis) as follows w = sx x e sx eA (4.6) (h-c) Where sx = crack spacing s0 = tensile strain at bottom of section h = depth of section c - depth of neutral axis Figure 4.5 (a) represents Equation 4.6 graphically. Knowing how to define w in terms of y we can define the limiting crack width on the bi-linear curve (w = 0.32 mm ) in terms of y as follows 0.32 x (h-c) y,m = — 1 (4-7) Combining these equations we can define the stress distribution through the section as follows 491 x JV, xSxx ek y^yiim •• ffib=- J T ^ -*y (MPa) (4.8) 22.3xN, xSrx efc (h-c) y>yiim : fflb=\6AxN\- ^ ^ ^xy (MPa) (4.9) where y is measured from the neutral axis towards the tension side of the section. 58 The stress block factors are defined as shown in Figure 4.5 (c). They are calculated in the following manner. To ensure that the magnitude of the force is the same and the resultant of the force acts in the same place as that for the force derived from integrating the bi-linear distribution, the following integrals need to be solved h-c (1) \ffjb xbx(h-c)dy = af x TV, x f3f x(h-c)xb 0 h-c \ffih xbx(h-c)xydy (2) h-c \ffib xbx(h-c)dy = {h — c)x 1/ 2 (4.10) Once the integrals have been evaluated the following curves were fitted to the data to describe a f and p f ctf= -12.947*wb + 157.47 (4.11a) pj= 0.1578*ln(wb) + 0.8572 (4.11b) where w}j = maximum crack width (sx *£b) h-c Compression zone _0.32x(/z-c). P f * ( h - c ) N A ( h - c X l - P / 2 ) F f =a f *N,* pf*b*(h-c) a f * N , (a) Strain distribution (b) Bi-linear stress distribution (c) Assumed stress block Figure 4.5 Calculation of stress block factors to include the tensile stresses caused by the steel fibres 59 With the stress block factors it is possible to do a sectional analysis without calculating complex integrals. To evaluate the accuracy of the stress block it was compared with the computer prediction at a number of points on the moment curvature response for SL2 (see Figure 4.6). From Figure 4.6 we see that the use of the stress block accurately predicts the flexural strength for a given curvature when compared to a similar prediction made by a computer program using the actual curve representing the average pull-out force for a single fibre. 180 r 160 -Curvature (rad/m) ^ Calculated using stress block factors Figure 4.6 Comparison between theoretical moment curvature response and that predicted using stress block factors 4.2.3 Predicting Flexural Capacities of Sections Containing Steel Fibres The method can now be extended to predict the ultimate flexural capacities of sections containing steel fibres. Using the assumed concrete stress block and limiting the compressive strain to 0.0035 as stated in the Canadian concrete code (CSA A23.3-94), the flexural capacity can be calculated by varying the depth of the neutral axis until the section is in axial equilibrium. Once this has been done the moment capacity can be calculated by taking moments about any point in the section. This was done for a steel fibre reinforced section (shown in Figure 4.7) tested by Lim et al, (1987). The ultimate moments are predicted as outlined above and compared with the experimental results in Table 4.1. Since only the moment curvature results are included in the paper by Lim and a distinct yield plateau was exhibited in the testing, the average value of the yield plateau was used to define the experimental moment capacity listed in Table 4.1. Table 4.1 Comparison ofpredicted and experimental flexural capacities for sections tested by Lim et al (1987) Beam M e x D (kNm) M p r e d (kNm) M e x p / M p r e ( j 2/1.0/1.5-3.5 40.0 39.7 1.01 2/0.5/1.5-3.5 39.0 44.4 0.89 61 152 fc' = 27.2 MPa |3 V f= 0.00, 0.5, 1.0 As = 400 fy = 450 MPa Figure 4.7 Section tested by Lim et al (1987) Table 4.1 indicates that the prediction of the ultimate flexural strength of a steel fibre reinforced section is reasonably accurate when some simple assumptions are made regarding the strain distribution and the shape of the stress block. 4.3 Flexural Capacities of Slab Elements The flexural strength of the slab elements was predicted as outlined in Section 4.2. For each section, measured material strengths were used. These included actual concrete cylinder strengths (see Table 2.4) and reinforcing steel stress-strain curves (see Figure 2.9) . The stress-strain curve for the reinforcing steel was modeled using a tri-linear curve. The standard bilinear curve was used to fit the initial elastic-plastic behaviour and a cubic polynomial was fitted to the strain hardening part of the curve (see Section 2.5.1). This was done using a spreadsheet program. A stress strain curve for high strength concrete suggested by Collins et al, (1993) was used to estimate concrete stresses at varying strains. 62 The resulting moment curvature responses are shown in Figures 4.8 (for the slab elements containing no steel fibres). The predicted shear versus curvature relationships for the slab elements containing steel fibres are shown in Figures 4.2 and 4.3 . The effect of steel fibres on the flexural capacity of the sections is estimated using the method outlined in Section 4.2. Table 4.2 summarises the predicted yield moments. For simplicity the yield moment is defined as the average moment beyond the elastic range. It does not include the rising part of the moment curvature response after the initial elastic-plastic response . as this is due to the strain hardening of the reinforcement. 180 -, Curvature (rad/m) Figure 4.8 Moment-curvature response for slab elements (Vf — 0%) Figures 4.9 - 4.11 show the shear that causes flexural yielding plotted on the measured load displacement curves for the slab elements. From this we see that not all the elements failed in a shear dominated manner. Some had enough shear capacity to allow the yield moment to be reached. The yielding was detected experimentally when the deformations were concentrated in the flexural cracks at the intersection between the haunch and the slab. Table 4.2 Predicted yield moments of slab elements Element fc' (MPa) My 1 (kNm) V y 2 (kN) S L l 50 119 396 SL2 50 138 460 SL3 50 157 523 SL4 95 120 400 SL5 95 138 460 SL6 95 165 550 SL7 25 120 400 SL8 25 141 470 1. Yield Moment 2. Shear force associated with flexural yielding (M/V = 0.315) Table 4.3 compares the predicted and measured ultimate moments for those elements that failed in this flexural/splitting type failure. The observed failure seemed to indicate that as the section yielded, the depth of compression was reduced resulting in a highly concentrated compressive force. This force then caused a splitting type failure as it spread towards the tension side of the specimen. The appearance of splitting cracks just prior to failure seems to support this assumption. It is interesting to note that there is little difference in load when the elements began to yield for the different fibre contents (fc' remains constant). This would suggest that a section has a maximum flexural strength that cannot be exceeded even though the volume 64 of fibres might be increased. In other words, there is an optimum fibre content that allows a section to attain this maximum moment. Vf = 1.37% Deformation (mm) Figure 4.9 Load-displacement curves for SL1-SL3 (fc' = 50 MPa) with shear force that causes flexural yielding (dashed horizontal line) 600 500 400 300 Vf = 1.48% x SL5(514 KN) - Vf = 0.75% 1 SL6 (460 KN) - Vf = 1.5% SL4 (364 KN) - Vf - 0% 6 8 10 Displacement (mm) 12 14 16 Figure 4.10 Load-displacement curves for SL4-SL6 (fc' = 95 MPa) with shear force that causes flexural yielding (dashed horizontal line) 65 SL7 (239 KN) - Vf = 0% 6 8 Deformation (mm) 10 12 Figure 4.11 Load-displacement curves for SL7-SL8 (fc' = 25 MPa) with shear force that causes flexural yielding (dashed horizontal line) Table 4.3 Comparison ofpredicted and measured ultimate flexural capacities for slab elements Element fc' (MPa) v f (%) M p r e d (kNm) M e x p (kNm) M e Xp/Mp r e (j SL2 50 0.65 160 152.0 0.95 SL3 52 1.37 172 166.3 0.96 SL5 96 0.58 166 161.9 0.98 SL6 96 1.48 192 144.9 0.75 Fibres also affected the ductility of the elements. The elements that contained steel fibres exhibited a large amount of ductility for members that were predicted to fail in shear. Table 4.4 lists the displacement ductilties for those elements that underwent flexural yielding. 66 Table 4.4 Displacement ductilities for slab elements failing in flexure Element fc' (MPa) v f(%) P-A SL2 52 0.65 2.8 SL3 52 1.37 5.6 SL5 96 0.58 4.6 SL6 96 1.48 3.2 4.4 Flexural Capacities of Beams The flexural response of the beams was predicted using a computer program that uses a layer-by-layer representation of the section to predict moment curvature relationships as outlined in Section 4.2. The resulting moment-curvature responses are shown in Figure 4.12, while Table 4.5 summarises the predicted ultimate moment capacities of the beams. The only beam to have a shear capacity that exceeded the shear required to cause yielding of the tension reinforcement was B3. B3 had an ultimate shear capacity of 260 kN while a shear force of 241 kN would have resulted in a flexural failure. If the fibres are included in calculating the flexural resistance, its capacity is 285 kNm which is caused by a shear force of 261 kN (based on the method outlines in Section 4.2). Since no flexural yielding was detected, the failure was classified as shear dominated. As no flexural failures were detected it is not possible to make any comparisons between the predicted and experimental flexural capacities. 67 Table 4.5 Predicted ultimate flexural capacity of beams (Vf= 0%) Beam fc' (MPa) M v ' (kNm) V v 2 (kN) B1-B3 95 222 241 B4-B5 25 180 196 B6-B8 50 211 230 . Yield moment 2. Shear force that causes flexural yielding (M/V = 0.918 mm) 250 -r 1 1 i ! 0 0.005 0.01 0.015 0.02 curvature (rad/m) Figure 4.12 Moment-curvature response for beams (Vf - 0%) 4.5 Shear Capacities of Slab Elements The shear strength of the slab elements was predicted using two methods. The standard equation for the shear strength of members containing no transverse reinforcement has traditionally been that proposed by ACI 318-89. For members not containing any transverse reinforcement the Canadian concrete code provides an alternative, the General Method for shear design. This is an adaptation of Modified Compression Field Theory 68 (MCFT) developed by Vecchio and Collins (1986) and modified to account for beams containing no transverse reinforcement by Adebar and Collins (1996). A comparison of the two methods shows that the General Method predicts higher shear capacities for the slab elements than ACI 318-89. The elements that failed in shear ( S L l , SL4, SL7, SL8), however all failed at loads greater than the predicted capacities (see Table 4.6) Table 4.6 Summary of measured and predicted shear strengths for slab elements that failed in shear Predicted Shear Strength (kN) Element fc' (MPa) vf(%) ACI 318-89 CSA 23.3-94 Measured Shear Strength (kN) V y ' (kN) S L l 50 0 132 167 355 422 SL4 95 0 182 207 364 466 SL7 25 0 93 130 239 420 SL8 Z 25 0.79 - - 342 466 1. Shear force that causes flexural yielding 2. None of the above mentioned methods account for the affect of steel fibres The shear strength of members containing no transverse reinforcement can be written as VU=P^Fc*bd (MPa) (4.12) The well known lower bound is p= 2 (psi) ( P = 0.167 MPa). This was derived from early research into the shear strength of reinforced concrete sections and provides a conservative lower bound for most shear tests (see Figure 4.13). A well known upper bound is p= 0.4 (MPa). Using these values of P it is possible to create an envelope 69 between which the results for most shear tests should fall. Figure 4.14 shows these bounds with the results of those elements containing no steel fibres included. From Figure 4.14 it can be seen that although the shear strengths are higher than what would normally have been predicted, they do fall close to the upper limit. Some possible reasons for the high shear strengths are discussed in the following paragraph. 4 3 V e 2 i i i — f • • « • * * • m « • • • • • ft « • • • • • • • 1 t • • • 0 0 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 Aslb„d Figure 4.13 Calculation of lower bound of /? (from MacGregor, 1988) 70 0 20 40 60 80 100 W (Mpa) Figure 4.14 Comparison of measured strengths with expected range of shear strengths for slab elements without fibres Firstly, the size effect could have allowed for an increased shear capacity to be reached. It is well known that shallow beams have smaller crack widths than deep beams which could have greatly increased the shear strength of the sections. Secondly, the spacing of the longitudinal reinforcement could have reduced crack widths. It was well distributed through the width of the specimen, but was relatively close through the depth. This close spacing would result in effective crack control and hence enhance the shear capacity of the section. Thirdly, the haunches provided restraint due to reinforcement being welded to the endplates. This created a very stiff section in which dowel action could have significantly increased the shear strength of the section. 71 4.6 Shear Capacities of Beams The shear strength of the beams was calculated as outlined in Section 4.5. Table 4.7 summarises the results obtained using both CSA 23.3-94, ACI 318-89 along with the measured shear strengths. For the General Method the crack spacing was taken as 0.9d (300 mm). A comparison between the ultimate shear capacity of the beams and the shear force that would cause flexural yielding shows that B3 exceeded its moment capacity. The steel fibres though increased the flexural capacity of B3, preventing a flexural failure and allowing the beam to fail in shear. The General Method proves to be adequate in predicting the shear strength of members containing no steel fibres, but is very conservative when considering members containing steel fibres. Table 4.7 Summary of measured and predicted shear strengths of beams (Vf= 0%) Predicted Strength (kN) Element fc' (MPa) V f ( % ) CSA 29.3-94 ACI 318-89 Measured Strength (kN) B l 95 0 120 108 130 B4 25 0 75 56 105 B6 50 0 95 79 125 72 As outlined in Section 4.5 the shear strength of a section can be bounded by two values, an upper bound of p = 0.4 and a lower bound of P = 0.167, where p is defined in Equation 4.11. Figure 4.15 shows these bounds with the test results. 300 -250 . _ 200 -Z T 150 -O £ W 100 -50 -0 -0 20 40 60 80 100 fc' (MPa) Figure 4.15 Comparison of measured strengths with maximum and minimum predicted shear strengths of beams The test results for the beams containing no steel fibres are bounded by the two methods of prediction. The beams containing 0.75 % steel fibres fall above the upper bound. This suggests that some modifications have to be made to include the effect of steel fibres. 4.6.1 Effect of Fibre Volumes on Shear Strength of Beams From the test results we find that the shear strength does not increase proportionally to the fibre content, i.e., doubling the amount of steel fibres does not result in a doubling of 73 V , = 0.75 % the shear capacity of the beam. Figure 4.16 summarises the effect of increasing the fibre content for the different concrete strengths on the shear strength of the beams. 300 0 -I 1 1 1 1 1 1 0 0.25 0.5 0.75 1 1.25 1.5 vf (%) Figure 4.16 The effect of increasing the fibre content on the shear strength of beams {fc'kept constant) As the fibre content increases, the shear strength does not increase in direct proportion to the fibre content, instead it seems to approach a maximum value. This suggests that the beam has a maximum shear capacity no matter how large the fibre content. This implies that there could be an optimal fibre content for which a maximum increase in shear capacity is possible. 74 Chapter 5 Towards a Rational Method for FRC in Shear 5.1 Outline Many tests have been conducted to investigate the effect of steel fibres on the shear capacity of beams without transverse reinforcement. For a summary see Adebar et al. (1996). These tests show that steel fibres can enhance the shear capacity of such members. Little work, however, has been done to try to quantify this increase in shear strength using a rational model. With the introduction of the Modified Compression Field Theory (MCFT), the shear strength of members containing no shear reinforcement can be predicted using a rational method (Adebar and Collins, 1996). Some possible enhancements to this theory are presented here in an attempt to account for the presence of steel fibres. A complete solution is beyond the scope of this thesis. This chapter is limited to presenting some ideas on what modifications are required. Section 5.2 outlines some of these changes and some preliminary predictions are given in Section 5.3. 5.2 MCFT for Members Containing Steel Fibres and No Transverse Reinforcement A complete description of M C F T is not presented here (see Adebar and Collins, 1996), rather the influence of steel fibres on the variables used to control the shear capacity of sections is discussed. The three most important variables are: shear transferred by 75 aggregate interlock, a crack spacing parameter and the average tensile response of cracked concrete. It is proposed that steel fibres can enhance the ability of a crack to transmit shear forces. This is done in two ways. Firstly, steel fibres can directly carry a shear force across the crack. Secondly the fibres reduce the crack spacing and hence reduce the crack widths. This increases the ability of the crack to transmit a shear force through aggregate interlock. Steel fibres increase the ability of cracked concrete to carry a tensile force. This effects crack inclinations and the ability of cracked concrete to transmit a shear force. The following sections deal with the above mentioned modifications. 5.2.1 Average Tensile Response of Cracked Concrete The presence of steel fibres in concrete increases its ability to carry tensile stresses after cracking (Mitchell et al, 1996). An empirical formula (such as that used for concrete without fibres) has not been developed due to a lack of research. Tan et al (1992) have proposed an expression to define the average post cracking stress strain curve for concrete in tension: (5.1) where 76 Equation 5.1 is plotted in Figure 5.1. a t u is defined at the post tensile cracking resistance of fibre reinforced concrete and represents the stress that the concrete can transfer across a crack. The values for o~tu that correspond to the experimental work carried out in this thesis are shown in Table 5.1. (A full explanation of how a t u is derived is given in Appendix G). Table 5.1 Values of crtufor different fibre volumes v f ( % ) o-tu (MPa) 0 0.0 0.75 2.018 1.5 4.037 V,= 1.5 % V, = 0.75 % V, = 0 % •0.001 0.002 0.003 0.004 Strain 0.005 0.006 Figure 5.1 Tensile response of concrete (fc'= 50 MPa) according to Tan et al. (1992) 11 From Figure 5.1 we see that for large volumes of fibre, the average post tensile cracking strength surpasses the cracking strength of the concrete (fcr). Knowing the average tensile response of the concrete, the shear force carried by the section can be defined as V = fxco\e.bvdv (5.2) where f2 = the principle average tensile stress defined by the equation 5.1 9 = the inclination of the principle average tensile stress 5.2.2 Transfer of Shear Across Cracks The ability for a crack to transmit a shear force is critical to the strength of members without transverse reinforcement. This is what ultimately limits the shear strength of a section (Adebar and Collins, 1996). Again, little work has been done to consider the effect of steel fibres on the transfer of shear across cracks. Steel fibres reduce the crack width but no work has been done to quantify their effect on the ability to increase the shear transfer through aggregate interlock. One possible approach is to assume that by opening a crack there is a tensile force carried across the crack by the steel fibres. This implies that there is a compressive force on the crack surface and hence the shear can be transferred through friction. A shear friction approach can now be applied to the transfer of shear forces across cracks. 78 To apply a friction based approach, the compressive force on the crack or the tensile force carried by the steel fibres needs to be calculated. A large amount of work has been done at U B C on the pull-out of individual steel fibres lying at different angles to the crack (Armelin and Banthia, 1996). When a crack opens it is resisted by the steel fibres crossing it. By opening the crack, the fibres go into tension and hence a tensile force (FJ is transferred across the crack. Armelin and Banthia (1996) have proposed a number of expressions to describe the tensile force in a single fibre at a given inclination to the crack (fa) as a function of the crack opening (w). To calculate the average pull-out force for an individual fibre F p o(w), assuming a random orientation of the fibres and all possible inclination and embedment lengths, they propose the following equation: F(w) = — < p o \ / 2 + / , - ( » " ) (5-3) The total tensile force transferred across a crack by the fibres can therefore be calculated by multiplying the average force per fibre (Fp o(w)) by the total number of fibres (NJ crossing a crack. The total number of fibres per unit area (Nj) is given by 0.5. Vf Nx=—^ (5.4) Af where Vj= percentage volume of fibres Af= perimeter of fibre 79 Therefore the total number of fibres crossing a crack would be given by the number of fibres per unit area multiplied by the area of the crack. We can assume the area of the crack (A c r ) to equal A , = ^ (5.5) suit? where 6 = inclination of the crack The total number of fibres crossing a crack is thus given by N,=N,x Acr (5.6) With the tensile force carried across a crack (FJ, it is possible to calculate the shear carried across the crack by aggregate interlock (friction). Two approaches can be used. First, one could simply use a coefficient of friction to calculate the amount of shear carried across the crack, or the approach suggested by Vechio and Collins (1986) can be used. The use of a coefficient of friction proves to be conservative and the second approach which is based on experimental and analytical work is suggested. This is the bases for the aggregate interlock equation used in MCFT. Vechio and Collins (1986) proposed that the shear carried by aggregate interlock (v c i) can be expressed as follows: 80 v r f = 0.18v c / m a x +1.64/;, - ( 5 . 7 ) v . c/max where ' 24 V e , ^ 0.3 + \sin0 Va + 16v = compressive stress on crack a = aggregate size 6= inclination of crack sx = crack spacing in x direction (along length of specimen) Figure 5.2 illustrates how the tensile force in the fibres can be resolved onto a horizontal (Fh) and vertical (V f l b ) component. V f l b can be considered as a shear force that the fibres carry directly across the crack. It is defined as: ^ = ^ . 0 0 8 0 (5.8) where 0= inclination of the crack 81 F h can then be resolved to find its component that acts perpendicularly to the crack. This force can be defined as the additional compressive force on the crack. We can therefore define fci as follows ft=L±*t± (59) Acr Using this method the amount of shear a crack can transfer (V c r k ) through aggregate interlock and tensile forces in the steel fibres is Vcrk = Vflb + Vci where Vci =Acr*vci The shear capacity of a section is reached when V c r k is less than V (defined in Section 5.2.1) i.e. V = fxcoXa.bv.dv Element SL7: Failure shear = 239 kN Fibre content = 0.0 % Cylinder strength ( f c) = 24.0 MPa Date of casting: 27/11/1995 Date of testing: 17/01/1996 LOAD DEFORMATION CURVES (SL-7) 250 -. O t 1 , , , 1 , , 1 0 1 2 3 4 5 6 7 8 DISPLACEMENT (mm) Description of test: • We are reusing old specimens that have had the central slab portion removed and recast with a different concrete strength and fibre content. The haunches from the old specimens contain no fibres • 55 kN cracking along the interface of the old haunch/new slab seen • 140 kN something slips in test setup but it is not detected in the load deformation data • 167 kN large shear crack suddenly appears at #C • Shear cracking limited to one side of the slab element, a symmetrical crack patteren was not formed in this test • 174 kN shear crak propagates rapidly towrds the compressive zone in the haunch • 200 kN crack stabalises and begins to open up • the craccking down the side underside of the slab does not appear to be uniform • 239 kN ultimate shear reached • Later examination of the failures surface indicates that it did run vertically through the element 106 Element SL8: Failure shear = 342 kN Fibre content = 0.79% Cylinder strength ( f c) = 27.0 MPa Date of casting: 27/11/1995 Date of testing: 19/01/1996 LOAD DEFORMATION CURVES (SL-8) AXIAL DISPL T R A N S V E R S E DISPL (TOP) T R A N S V E R S E DISPL (BOT) 10 12 14 DISPLACEMENT (mm) Description of test: Note that we are reusing an element (see SL7) 70 kN initial flexural cracking detected 204 kN a flexural crack appears near #A but does not originate at intersection of slab and haunch 214 kN flexural shear crack develops at #H 225 kN flexural shear crack at #E develops 255 kN movement at shear cracks #H and #E-#G detected Uniform cracking down the slab detected Cracking along the longitudinal reinforcement on the sides seen (splitting cracks) 342 kN ultimate shear load reached While taking final measurements, the computer failed and a large part of the load deformation data is lost APPENDIX C FORCES IN HYDRAULIC ACTUATORS AND RIGID LINKS AT FAILURE OF SLAB ELEMENTS |o8 R l R 3 A 6 R 2 A 5 A 4 Figure C l Location of hydraulic actuators (A4-A6) and rigid links (R1-R3) Table C l Forces measured at failure in hydraulic actuators and rigid links Element # R l (kN) R2(kN) R3 (kN) A4 (kN) A5(kN) A6 (kN) SL1 355 -371 -352 376 -375 377 SL2 485 -498 -481 507 -506 509 SL3 528 -536 -520 556 -555 562 SL4 365 -365 -362 391 -387 394 SL5 514 -527 -509 545 -542 546 SL6 460 -472 -459 486 -484 487 SL7 238 -248 -238 256 -256 257 SL8 No data due to computer failure APPENDIX D SUMMARY OF BEAM TEST RESULTS II 0 Beam Bl : Failure shear =133 kN Fibre content (V f) = 0 % Cylinder strength (fc') = 94.0 MPa Date of casting: 14/07/1995 Date tested: 26/10/1995 Load Deformation Curve (B1) 140 0 1 2 3 4 5 6 7 8 Displacement (mm) Description of test:: • Initial flexural cracking at 33.6 kN • Notice a very even crack spacing for flexural cracks • 95 kN first shear cracks crack #F • 133 kN failed in shear on side that showed least cracking Beam B2: Failure shear = 232 kN Fibre content (V f) = 0.58 % Cylinder strength (fc') = 96.0 MPa Date of casting: 14/07/1995 Date tested: 28/10/1995 Load Deformation Curve (B2) 250 Displacement (mm) Description of test:: • 38 kN first flexural crack detected • Very even crack spacing noticed • 130 kN first diagonal shear cracks appear • 140 kN crack #J-#H (diagonal shear crack) appears. Most of the deformation to be concentrated here • 203 kN unloaded the beam to prevent a sudden shear failure (not completely unloaded) • 232 kN after reloading the beam it finally fails HZ Beam B3: Failure shear = 260 kN Fibre content (V f) =1.48 % Cylinder strength (fc') = 96.0 MPa Date of casting: 14/07/1995 Date tested: 24/10/1995 Load Deformation Curve (B3) 300 250 . (KN) 200 . • Force 150 . CO 0) 100 . CO 50 . 0 . 0 2 4 6 8 10 12 14 16 18 Displacement (mm) Description of test:: • Difficult to detect initial flexural cracks • 50 kN first flexural cracks thought to appear • 65 kN flexural cracking away from constant moment region • Notice that the paint is hiding some of the cracks as they can be seen on the unpainted side • 140 kN shear cracks begin to develop from the flexural cracks • 165 kN more shear cracks develop and begin to propagate diagonally down to the tension face of the beam • 190 kN the beam is unloaded to place rollers on one end to prevent any axial forces developing in the beam • 220 kN reach the capacity of the tester at the set scale, unload the beam to change scale on test machine • Reload the beam to failure • 260 kN beam fails with block of concrete failing off • A diagonal crack running from the loading point to just before the support defined a clean failure surface with all the fibres either breaking or pulling out 113 Beam B4: Failure shear = 105 kN Fibre content (V f) =0 % Cylinder strength (fc') = 24 MPa Date of casting: 27/11/1995 Date tested: 24/01/1996 Load Deformation Curves (B4) z 120 100 80 o 60 40 20 0 1 2 3 4 5 6 7 8 Displacement (mm) Description of test:: • 50 kN detect first flexural cracking • 65 kN first shear cracks begin to develop • 102 kN diagonal shear cracks suddenly appear near failure crack • 105 kN ultimate shear capacity reached, beam is unloaded to prevent sudden shear failure • Try to reload beam but it is unable to carry a larger load Beam B5: Failure shear = 156 kN Fibre content (V f) =0.79 % Cylinder strength (fc') = 26 MPa Date of casting: 27/11/1995 Date tested: 26/01/1996 Load Deformation Curves (B5) 0 2 4 6 8 10 12 14 Dis place m e nt (m m) Description of test:: • 40 kN first flexural cracks begin to develop • 90 kN first flexural shear cracks begin to develop • 105 kN diagonal shear cracks begin to appear • Deformation concentrated along a single diagonal shear crack (#A-#B) • 156 kN ultimate shear capacity reached Beam B6: Failure shear = 125 kN Fibre content (V f) =0 % Cylinder strength (fc') = 49 MPa Date of casting: 06/02/1996 Date tested: 07/03/1996 Load Deformation Curves (B6) 140 , , j ! ; T 0 2 4 6 8 10 12 Displacement (mm) Description of test:: • 58 kN first flexural cracks detected • 75 kN first flexural shear cracks begin to form • 110 kN sudden appearance of larger shear cracks on both sides of the beam • 118 kN beam seems to be about to fail so some load is removed, a large amount of creep is detected as the shear cracks open up • Able to slowly reapply the load and pass the previous level of 118 kN • 125 kN ultimate shear capacity reached • Uneven crack spacing noticed in constant moment region \\ Crack opening data: B2 Ultimate shear: 232 kN Crack pattern and location of measured cracks Crack # Load (kN) Width (mm) A 200 0.15 215 0.2 B 130 0.1 160 0.1 180 0.13 200 0.13 215 0.15 C 130 0.08 160 0.1 180 0.1 200 0.18 D 130 0.05 160 0.1 180 0.1 200 0.13 215 0.15 E 130 0.1 160 0.13 180 0.13 200 0.25 F 130 0.06 160 0.06 180 0.06 G 80 0.05 100 0.07 130 0.07 Crack # Load (kN) Width (mm) H 160 0.1 180 0.2 200 0.4 I 100 0.06 130 0.1 160 0.13 180 0.15 200 0.18 J 215 1.6 225 2.0 230 2.5 K 204 1.0 215 2.5 225 3.0 230 3.5 L 180 0.2 200 0.3 204 1.4 215 2.5 225 3.0 230 3.5 \'bO Crack # Load (kN) Width (mm) M 204 1.0 215 2.0 225 2.0 230 3.0 N 225 0.3 230 0.8 O 225 0.4 230 1.6 P 200 0.18 215 0.2 225 0.4 230 0.8 Q 225 0.3 230 0.3 131 Crack opening data: B3 Ultimate shear: 260 kN 1 3 ^ Crack opening data: B4 Ultimate shear: 105 kN f t Crack pattern and location of measured cracks Crack # Load (kN) Width (mm) A 65 0.05 80 0.08 95 0.1 B 65 0.05 80 0.07 95 0.1 C 65 0.05 80 0.1 95 0.15 D 80 0.07 95 0.15 Crack opening data: B5 Ultimate shear: 156 kN Crack pattern and location of measured cracks Crack # Load (kN) Width (mm) A 125 0.25 140 0.5 150 0.8 B 125 0.15 140 0.35 150 0.5 C 90 0.1 105 0.15 125 0.18 140 0.5 150 0.8 D 75 0.1 90 0.15 105 0.18 125 0.2 140 0.25 150 0.25 Crack # Load (kN) Width (mm) E 75 0.05 90 0.1 105 0.1 125 0.15 140 0.15 150 0.15 F 140 0.1 150 0.15 156 0.15 G 150 0.15 156 0.3 I3 l c 2 x Ij. where l c = critical length required to develop the ultimate fibre stress when a uniform ultimate bond stress, T u between the fibre and matrix is assumed. (The ultimate stress for steel fibres is assumed to be 1200 MPa). I3°l APPENDIX H LISTING OF FORTRAN PROGRAM TO CALCULATE SHEAR CAPACITY OF FIBRE REINFORCED CONCRETE BEAMS Listing of Fortran computer program to calculate shear capacity of steel fibre reinforced concrete beams C Calculation procedure for shear response of members without stirrups C Z Shear Span - M/V (mm) C bv, dv = Width and depth of shear area (mm) C Fy = yield stress of longitudenal steel (MPa) C As = Area of flexural rebar (mmA2) C Smx = Crack spacing in x direction (mm) C fc = compressive strength of concrete (MPa) C Agg = dimension of aggregate (mm) C Vf= percentage fibres C Lf= length of Fibre C Df = fibre diameter C BSTR = ultimate bond stress between concrete and fibre C FFy = ultimate strength of fibre C Vf = percentage fibres C Lf= length of Fibre C Df = fibre diameter C BSTR = ultimate bond stress between concrete and fibre C FFy = ultimate strength of fibre real NETN, MOMENT real Vf, Lf, Df, BSTR, FFy real NI, N2, NCALC, NCALCTOT C Input properties write (*,*)'' open (unit = 3, file = 'data.inp', status = 'unknown') open (unit = 4, file = 'data.out', status = 'unknown') read (3,*) NDB, TTI read (3,*) Z, REQN read (3,*) bv, d read (3,*) As, Fy, Smx read (3,*) fc read (3,*) Agg read (3,*) Vf, Lf, Df, BSTR, FFy write (*,*)'' 5000 dv = 0.9*d Vf= Vf/100 fc = abs(fc) Fcr = 0.33*sqrt(fc) Tol = 0.0001 *(As*Fy + fc*bv*dv) PI = 3.14159 Af=PI * (Df ** 2)/4 Pf=PI*Df Vcmax = 4.5 C Setting initial constants and initialising variables ALPHA = 0.5 NCALC = (ALPHA *Vf/Af) Epl = 0.00001 Eplmax = 0.01 Listing of Fortran computer program to calculate .shear capacity, ofsteel fibre reinforced concrete beams FLAG = 2 VI =0.0 C Calculating post cracking tensile resistance Lc = (FFy * Af) / (Pf * BSTR) if(Lf.lt.Lc)then NI =0.5 else NI = 1 - (Lc/(2*Lf)) endif N2 = ALPHA R = Af/Pf C PTENSCALC = post cracking tensile resisitance of fibre reinforced concrete PTENSCALC =( NI * N2 * Lf * Vf * BSTR) / (2 * R) C Writing general data to file write (*,*) do 2000 while (Epl.le.0.02 ) JJ= 1 THETA = 89*3.141592/180 XI = THETA goto 50 25 JJ = 2 THETA = TTI*3.141592/180 X2 = THETA 40 JJ =JJ + 1 X3 = X2 - ((X2 - X1)*F2/(F2 - Fl)) THETA = X3 50 continue C Calculate the concrete contribution Vc Epcr = 0.33/5000 if (Epl .It.Epcr) then if(Epl.lt.0)then Vc = 0 goto 35 else Ec = 5000 * sqrt(fc) fcl = Ec * Epl Vc = fcl * bv * dv / tan(THETA) goto 35 endif endif Listing of Fortran computer program .to calculate shear .capacity,of .steel -fibre .reinforced concrete .beams W = Ep 1 * Smx/sin(THETA) if(Epl.ltEpcr) W = 0 C Calculating BETA1 (shear carried by average tensile force in concrete) BETA = sqrt((Epl - Epcr)/0.005) Fcl = (Fcr + (BETA * PTENSCALC))/(1 + BETA) BETA1 =Fcl *bv*dv/tan(theta) if(Vf.eq.O)then BETA1 = (0.33*sqrt(fc)*bv*dv)/(tan(THETA)*(l+sqrt(500 + *Epl))) endif C Calculating the total number of fibres crossing a crack NCALCTOT = NCALC * bv*dv/sin(THETA) C Calculating the tensile force in an individual fibre Ft = 0.5*(((F0(W)/2 + F22_5(W) + F45(W) + F67_5(W))*0.25) + + FHOOK(W)) C Calculating the total tensile force carried across a crack FTOT = NCALCTOT * Ft * 1000 C Calculating the shear carried across a crack by the tensile force in the fibres Vfib = FTOT * cos(THETA) C Calculating Vcimax term used by Walraven Vci = sqrt(fc)/(0.3+(24*Epl*Smx/((Agg+16)*sin(THETA)))) C Calculating area of crack and compressive stress on crack Acr = bv*dv/sin(THETA) fci = FTOT*(sin(THETA))**2/Acr C Calculating shear transferred across a crack due to aggregate interlock if(Vf.eq.0)then Vcr = 0.18* Vci*bv*dv else Vcrl = (0.18 * Vci)+(1.64 * fci)-(0.82*(fci**2)/(Vci)) if (Vcrl.gt.Vcmax) Vcrl = Vcmax Vcr = Vcrl * bv * dv endif C Calculating BETA2 (total shear transferred at a crack) BETA2 = Vcr+Vfib Listing of Fortran computer program,to calculate shear capacity of steel fibre reinforced concrete beams C Calculating the shear carried by the section if (BETA l.lt.BETA2) then BETA = BETA 1 BFLAG = 1 else BETA = BETA2 BFLAG = 2 endif Vc = BETA if (NDB.eq.l) write (*,*) 'Vc = Vc/1000/kN' C Calculate f2 35 continue f2 = Vc/(bv*dv*tan(THETA)) if (NDB.eq.l) write (*,*) 'f2 = ', f2,'MPa' C Calculate f2max f2max = fc/(0.8 + 170*Epl) if (f2max.gt.fc) f2max = fc if (NDB.eq.l) write (*,*)'f2max = ', f2max,'MPa' C Calculate e2 if (f2/f2max.gt.l)then e2 = -0.002*(l+(f2max-l)**0.5) else e2 = -0.002*(l-(l-f2/f2max)**0.5) endif if (NDB.eq.l) write (*,*)'e2 = ', e2 C Calculate et et = (Epl + (e2*(tan(THETA))**2))/(l+(tan(THETA))**2) if (NDB.eq.l) write (*,*)'et= et C Calculate ex ex = Epl + e2 - et if (NDB.eq.l) write (*,*) 'ex = ',ex C Calculate the axial load corresponding to the required moment V = Vc MOMENT = V*Z RQDNP = 2*(ex*200000*As - MOMENT/dv) if (NDB.eq.l) write (*,*) 'M = ',MOMENT/1000000,'kN.m' if (NDB.eq.l) write (*,*) 'Np = ', RQDNP/ 1000,'kN' C Calculate net axial load RNV = 2*Vc/tan(2*THETA) if (NDB.eq.l) write (*,*) 'Nv = ', RNV/1000,'kN' NETN = RQDNP - RNV THETA = THETA * 180/3.141592 144 Listing of Fortran computer program to calculate shear capacity .oLsteeLfibre;reinforced.concrete beams if (NDB.eq.l)then write (*,*)' Epl THETA N V write (*,*) Epl,THETA, 'deg',NETN/1000,'kN',V/1000/kN' endif if (JJ.eq.l)then F1 = REQN * V - NETN goto 25 elseif (JJ.eq.2) then F2 = REQN * V - NETN FX12 = F1*F2 if(FX12.gt.0)then write (*,*) 'problem FX12 greater than 0' stop endif goto 40 endif F3 = REQN*V - NETN if(abs(F3).lt.Tol) then goto 100 endif if (JJ.gt.2000) then write (*,*) 'problem JJ greater then 2000' if(Vl.gt.V2)then write (*,*) 'Found Maximum Shear' else write (*,*) 'Did not find Maximum Shear' endif goto 1250 endif FX13=F1*F3 if(FX13.1t.0)then X2=X3 F2 = F3 else XI =X3 Fl =F3 endif goto 40 100 continue C Checks whether the shear capacity is incresing or decreasing incase a solution is not found, C therby telling us whether the maximum shear capacity has been exceeded if(V.gt.Vl)then V2=V1 VI = V Ep = Epl THMAX = THETA Listing of Fortran computer program to calculate shear.capacity .of.steel .fibre,reinforced concrete beams WMAX=W endif C Controlling increment of EP1 - depends on whether crack strain has been passed or not if (Ep 1 .le.Epcr) then Epl = Epl +0.00001 else Epl = Epl + 0.00005 endif 2000 continue C Writes maximum shear capcity to screen 1250 write (*,600) 'Vmax = '.Vl/lOOO/kN'/Epl = \Ep,Theta = ',THMAX +,'deg', 'w (max) = ',WMAX,'mm' 600 format(A,F5.1,lX,A,3X,A,F7.6,3X,A,F5.1,lX,A,3X,A,lX,F5.3,lX,A) write (4,800) fc, Vf* 100, Smx, Vl/1000, WMAX 800 format (lx,f5.2,5x,F6.3,5x,f5.1,5x,f5.1,5x,f6.3) read (3,*) fc,Vf,Smx if (fc.eq.999) stop goto 5000 close (unit = 4) close (unit = 3) end C Function to calculate F0 stress real function F0 ( W) real W real Ep, A, B, C Ep = 2.5 A = -0.015 B=12.1 C=1.2 F0 = Ep * W * (A + ((1 - A)/((l + (B * W)**C)**(1/C)))) return end C Function to calculate F22 5 stress real function F22_5 ( W ) real W real Ep, A, B, C Ep = 2.2 Listing of Fortran computer program to calculate, shear .capacity ,o£steel;.fibr,e reinforced concrete beams A = -0.017 B = 9.7 C = 1.1 F22_5 = Ep * W * (A + ((1 - A)/((l + (B * W)**C)**(1/C)))) return end C Function to calculate F45 stress real function F45 ( W ) real W real Ep, A, B, C Ep = 0.27 A = -0.1 B= 1.42 C = 4 F45 = Ep * W * (A + ((1 - A)/((l + (B * W)**C)**(1/C)))) return end C Function to calculate F67 5 stress real function F67_5 ( W ) real W real Ep, A, B, C Ep = 0.047 A = -0.2 B = 0.3 C = 4 F67_5 = Ep * W * (A + ((1 - A)/((l + (B * W)**C)**(1/C)))) return end C Function to calculate FHOOK real function FHOOK (W) real W if(W.le.0.7)then if(W.le.0.05)then FHOOK = 3.4 * W else FHOOK = -0.26 * W +0.182 endif else FHOOK = 0 endif return end