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\ STRESS AND STRAIN DISTRIBUTIONS OF CYLINDRICAL SOIL SAMPLES by Fuli Dai M.S., Chorigqing Institute of architecture and Engineering, China, 1983 ,,._ Thesis submitted to the ..:..: Faculty of the Graduate School of the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Master of Science 1992
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This thesis for the Master of Science degree by Fuli Dai has been approved for the Department of Civil Engineering by Nien-Yin Chang Andreas S. Vlahinos date
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iii Dai, Fuli (M.S., Engineering) Stress and Strain Distributions of Cylindrical Soil Samples Thesis directed by Professor N.Y.Chang ABSTRACT The stress and strain distributions in cylindrical soil samples, including solid and hollow samples, have been investigated by previous researchers, either experimentally or numerically. The numerical study, so far, however, was limited to the effect of confining pressure and axial load. The effect of torsion, a very important concern, has seldom been studied. In this thesis, finite element analysis were performed to study the stress and strain distributions in a solid cylinder with aspect ratio equal to 2 and three hollow cylindrical samples with the height of 5, 10 and 15 inches respectively. Modified Cam-clay model was used for plastic simulation of the soil behavior. The hollow cylinders with height 10.0 and 15.0 inches provided good stress and strain distributions when subjected to torsional force. The solid cylinder under axial load can have enough
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lV room of end-effect free region, but the non-uniformity problem become serious when torsional force applied. This abstract accurately represents the content of the candidate's thesis. I recommend its publication. Signed N.Y.Chang
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v Acknowledgement I want to express my deepest appreciation and gratitude for the guidance and support of professor N.Y.Chang.
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vi CONTENTS List of Figures Xl Chapter 1. 2. Introduction 1.1 Background 1.2 Purpose of this work 1.3 Scope of this work Literature Review on Stress and Strain distribution of Laboratory Test Specimen 2.1 Simple Shear Test . 1 1 3 . . 4 5 6 2.2 Stress and Strain Condition ln Conventional triaxial Samples . 10 2.3 Stress and Strain distribution in Hollow Cylinder Specimens 24 2.4 Summary . . . 39 3. Calculation of Average Stresses and Strains in Hollow Cylinder . . . . . 43 3.1 Introduction 43 3.2 Radial and Circumferential Stresses 44
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3.2.1 Elastic Equal Effect o o o o o 3.2.2 Elastic Proportional Effect o 3. 2. 3 Linear Equal Effect 3.2.4 Linear Proportional Effect 3.3 Axial Stress 3.4 Shear Stress 3.5 Strain 3.5.1 Radial Strain 3. 5. 2 Circumferential Strain 3.5.3 Shear Strain 3.6 Principal Stresses and Strains 3.7 End Effect 4o Modified Cam-clay Model 0 0 0 4.1 Introduction 0 0 0 0 0 0 0 0 0 4.2 Governing Equations for Modified Cam-Clay Model 0 0 0 0 0 4.3 The Elasto-Plastic Stress Strain Relationship 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 vii 44 46 47 47 48 48 53 55 55 56 57 58 62 62 65 70 So Implementation of Modified Cam-Clay Model in the Finite Element Computer Code o o o o o 77 5.1 Introduction o o o 0 0 0 0 7 7 5.2 Basic Principles and Formulations 78
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Vlll 5.3 Development of Finite Element Computer Program CSSAP . . . . . . . 83 5.3.1 5. 3. 2 Formulation for Cylindrical Structures . . . . Boundary Condition 5.3.3 Strain Matrix and Shape Function 88 90 of Element . . . . . . 94 5.3.4 Load Force Distribution 97 5.3.4.1 Nodal Force Caused by Horizontal Pressures . . . . 99 5.3.4.2 Nodal Force Caused by Vertical Pressure .. 100 5.3.4.3 Equivalent Nodal Force From Torque ....... 101 5.3.5 Incremental Stiffness Method ... 102 5.3.6 Modified Cam-Clay Model and its Implementation ........... 105 5.4 Load Condition and Stress Strain Normalization 108 5.4.1 Load Condition and Stress State Estimate . . . 10 8 5.4.2 Vertical Deviator Load Alone . 110 5.4.3 Torque Alone . . .. 112 5.4.4 Vertical Deviator Load and Torsion 112 5.4.5 Stress and Strain Normalization . 113 5.5 The Presentation of Numerical Results . 118
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6. 7. Stress and Strain Distributions in a Solid Cylinder . . . . 6.1 Introduction 6.2 Under deviator Pressure Alone 6.2.1 Stress Distribution 6.2.2 Strain Distribution lX .. 119 .119 . .122 .122 .123 6.3 Under Deviator Pressure and Torsion .125 6. 3.1 Stress Distribution .125 6. 3. 2 Strain distribution .12 6 Stress and Strain Distributions in Hollow Cylinder . . . . . . .146 7.1 Introduction .146 7.2 Hollow Cylinder With H/2 = 2.5" .149 7.2.1 Under Deviator Pressure Alone .149 7. 2. 2 Under Deviator Pressure and Torsion .151 7.3 Hollow Cylinder With H/2 = 5.0" .. 152 7.3.1 Under Deviator Pressure Alone .152 7.3.2 Under Deviator Pressure and Torsion .. 154 7.4 Hollow Cylinder With H/2 = 7.5" .156 7.4.1 Under Deviator Pressure Alone .156 7.4.2 Under Deviator Pressure and Torsion .. 157 7.5 Effects of Torsion Resisting blades .157
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8. X Comparison of Analysis Results 19 7 8.1 8.2 8.3 8.4 Introduction .197 Under Deviator Pressure Alone .... 197 Under Deviator Pressure and Torsion .. 199 Torsion Resisting Blade Effect 8.5 Strain Distribution 2 01 .202 8.6 Recommendation of the Sample Dimensions .. 203 9. Summary and Discussions ............ 212 References . . 218
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xi List of Figures Chapter 2 Figure 2.1 Figure 2.2 Figure 2.3 Figure 2.4 Figure 2.5 Figure 2.6 Figure 2.7 Figure 2.8 Figure 2.9 Figure 2.10 Stress on the upper and lower faces of a simple shear sample . . . 8 Stress-strain distripution in square simple shear specimen . . 9 Stress-strain distribution in circular simple shear specimen . 9 Shear and normal stresses in sample in contacted with end cap ..... 11 Distribution of stress in triaxial specimen. .13 Distribution of strain in triaxial specimen . 14 Nonuniformity of sample and slip surface . . . . . . 16 Enlarged frictionless end platen .. 17 Final moisture content distributions . . 18 Stress contour throughout non-linear elastic unconfined compression specimen . . . . . . . 21
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Figure 2.11 Comparison of stress distribution by different investigations at the end surface for perfectly continued xii ends . . 22 Figure 2.12 Idealized stress conditions in a hollow cylindrical element ... 25 Figure 2.13 Stress components in Cartesian coordinates system for hollow cylinder specimen ....... 28 Figure 2.14 Normalized shear strain distribution at failure . . . . 28 Figure 2.15 Variation of ax 29 Figure 2.16 Shear stress-strain relation and calculated variation of ax 3 0 Figure 2.17 Effect of end restraint in torsion shear test on hollow cylinder Figure.2.18 specimen Definition of average stress and strain . . . . Figure 2.19 Variation in stress non-uniformity 30 34 with geometry and stress path .35 Figure 2.20 Variation in strain non-uniformity with geometry and stress path .35 Figure 2.21 Variation accuracy in strain with geometry and stress path . 36 Figure 2.22 Stress distribution across the wall of a hollow cylinder specimen 36 Figure 2.23 Linear elastic analysis of stress distribution in fixed end hollow cylindrical elements of different height . . . . . . 38
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xiii Figure 2.24 Influence of outer radius bon stress uniformity in hollow cylinder 111 wall thickness . . . . . 40 Figure 2.25 Stress strain attainable in the hollow cylinder apparatus with p0/Pi between Chapter 3 Figure 3.1 Figure 3.2 Figure 3.3 Figure 3.4 Figure 3.5 Chapter 4 Figure 4.1 Figure 4.2 Chapter 5 Figure 5.1 0 9 and 1 2 . . . . . . 41 The load condition of hollow cylinder . . . . . 45 (a) Derivation of shear stress tza 49 (b) Distribution and average tze 49 Strain components of a cylinder element . . . . 54 Principal stress rotation on an element . . . . . . 54 The top end stress condition of a strip of hollow cylinder wall 61 Yield surfaces in Cam-clay models .. 64 Some aspects of the modified Cam-clay model for triaxial condition . 66 A revolution shaft body twisted by couples applied at ends . . . 89
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Figure 5.2 Figure 5.3 Figure 5.4 Figure 5.5 Figure 5.6 Figure 5.7 Figure 5.8 Chapter 6 xiv The thickness of a cylindrical element . . . . . . 91 A slice of cylinder with center angle = 1 radian . . . . 91 Eight-node isoparameter element 98 Load distribution on a cylindrical element . . . . . 98 Increment stiffness approach to simulate the non-linear load-displacement relationship .. 104 The stress states of cylinders under different load .... 111 The stress components on an element . . . . . .114 Figure 6.1a One quarter structure of solid cylinder .............. 120 Figure 6.1b Finite element mesh of solid cylindrical structure ...... 121 Figure 6.2 Figure 6.3 Normalized stress distribution in solid cylinder under vertical deviator load alone (a) (O'z-O'o) /O'd (b) (O'r-O'o) /O'd (c) 'tzr/O'd (d) (O'e-O'o)/O'd . 128 . 129 130 . 131 Normalized strain distribution in solid cylinder under vertical deviator load alone (a) E z IE z. avg (b) Er/ Er.avg (c) Ee/Ee.avg 132 133 134
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Figure 6.4 Figure 6.5 Chapter 7 XV Normalized stress distribution in solid cylinder under vertical deviator load and torque (a) 0'1/0't.avg 135 (b) crdcr3.avg 136 (c) (crr-0'0 ) /O'd .137 (d) (O'e-O'o) /O'd 138 (e) tzaltze.avg 139 (f) a/aavg.. 140 Normalized strain distribution in solid cylinder under vertical deviator load and torque ( a ) E 1 I E 1. avg (b) E 3 / E 3. avg (c) Er/Er.avg (d) Ee/ Ee.avg (e) Eze/ Eze.avg " 141 142 143 144 145 Figure 7.1a One quarter structure of hollow cylinder ..... 147 Figure 7.1b Finite element mesh of hollow Figure 7.2 Figure 7.3 cylindrical structure . . .148 Normalized stress distribution in hollow cylinder (H/2=2.5") under vertical deviator load alone (a) (O'z-O'o) /O'd (b) (O'r-O'o) /O'd (c) (O'e-O'o) /O'd . 159 160 161 162 (d) t zr/ O'd Normalized stress distribution in hollow cylinder (H/2=2.511) under vertical deviator load and torque (a) crt I crl.avg 163 (b) 0'3/0'3.avg 164 (c) (O'r-0'0 ) /O'd 165 (d) tzaftza.avg 166 (e) a/ aavg 167
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Figure 7.4 Figure 7.5 7.6 Figure 7.7 Figure 7.8 Figure 7.9 Normalized stress distribution in hollow cylinder (HI2=5.0") under vertical deviator load alone XV1 (a) (O'z-O'o) IO'd (b) (O'r-O'o) IO'd 168 169 (c) (O'e-O'o) /O'd (d) tzriO'd Normalized stress distribution 1n hollow cylinder (HI2=5.0") under vertical deviator load and torque (a) 0' 1 I 0'1. avg . . . . . (b) 0'3IO'J.avg . . . . . (c) ( O'r-O'o) I O'd . (d) t zel t z9. avg . . . . . (e) alaavg . . . . . . Normalized strain distribution in hollow cylinder (HI2=5.0") under vertical deviator load alone (a) . (b) Er1Er.avg (c) EeiEe.avg Normalized strain distribution in hollow cylinder (H/2=5.0") under vertical deviator load and torque (a} E 1 I 1. avg (b) E3IEJ.avg (C) EzeiEze.avg Normalized stress distribution in hollow cylinder (HI2=7.5") under vertical deviator load alone 170 171 172 173 174 175 176 177 178 179 180 181 182 (a) ( O'z-O'o) IO'd 183 (b) (O'r-O'o) IO'd 184 (c) (O'e-O'o) lcrd 185 (d) tzriO'd 186 Normalized stress distribution in hollow cylinder (HI2=7.5") under
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Figure 7.10 Figure 7.11 Chapter 8 Figure 8.1 Figure 8.2 Figure 8.3 Figure 8.4 Figure 8.5 xvii vertical deviator load and torque (a) CJ1/0"1.avg 187 (b) cr3/cr3.avg 188 (c) (crr-0"0 ) /crd 189 (d) 'tze/'tza.avg 190 ( e) a/ O.avg 191 (Blade considered) normalized stress distribution in hollow cylinder (H/2=5.0") under vertical deviator load and torque (a) 0"1/crl.avg 192 (b) cr3/cr3.avg 193 (c) (crr-0"0 ) /O"d 194 (Blade considered) normalized strain distribution in hollow cylinder (g/2=5.0") under vertical deviator load and torque (a) E1/E1.avg (b) /.avg 195 196 Distribution of (crz-0"0)/crd in hollow cylinder under vertical deviator stress crd = 50 psi . . 204 Distribution of (crr-0"0)/crd in hollow cylinder under vertical deviator stress crd = 50 psi . . . . 205 Distribution-of (cr9-cr0 ) /crd in hollow cylinder under vertical deviator stress crd = 50 psi . . . . 206 Distribution of in hollow cylinder under-vertical deviator stress crd = 50 psi . . 207 Distribution of 0"1/0"t.avg in cylinder under vertical deviator stress
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Figure 8.6 Figure 8.7 Figure 8.8 XV111 crd = 50 psi, average 'tze = 27.2 psi 208 Distribution of cr3/cr3.avg in hollow cylinder under vertical deviator stress crd = 50 psi, average 'tze = 27.2 psi 209 Distribution of 0'1/0'l.avg and 0'3/0'3.avg in hollow cylinder (H/2=5.0") with or without blade under vertical deviator stress plus torque . 210 Distribution of a/a.avg in hollow cylinder under vertical deviator stress crd = 50 psi, average 'tze = 27.2 psi, flavg = 23.71 211
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1 1. Introduction 1.1 Background Cylindrical soil samples, such as solid cylinder and hollow cylinder, have been widely used in soil test to determine soil parameters, stress-path behavior and principal stress rotation effect, etc. A test may be static or dynamic, a load may be monotonic or cyclic. A solid cylindrical sample is used in a conventional triaxial test subjected to confining pressure and deviator stress. Principal stresses in a triaxial sample test cannot rotate gradually. They have a 90 jump rotation. When a hollow cylindrical sample is subjected to both an axial load and a torsion, its principal stress direction could rotate gradually, this provides a mechanism to simulate a more general field stress condition. As a cylindrical sample is placed between two platens of a test apparatus, end constraint effect occur. Besides, unequal internal and exterior pressures or torsion may also
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2 resulting nonuniform stress and strain distributions. The result of this non-uniformity is of major concern. Thus, the stress and strain distributions in both hollow and solid cylindrical samples, were investigated by researchers. The investigation approach can be grouped into experimental methods and numerical methods. A slenderness ratio or aspect ratio, 2, for solid cylinder, is adapted in this study as its stress and strain distributions have been studied extensively by previous researchers, either experimentally or numerically. As far as a hollow cylindrical specimen, some dimensions were recommended in previous studies: inner vs. outer radius ratio, wall thickness, sample height (Hight, 1983; Chen,l988). The numerical study, so far, however, was limited to the effect of confining pressure and axial load. The effect of torsion, which is very important has seldom been studied. In this study, assumed equal and radial pressures, both axial load inner and outer, and torsion are monotonically applied. Finite element analysis were performed to study the stress and strain distributions in cylindrical, solid and hollow, samples with or without.
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3 torsion. 1.2 Purpose of This Work Soils usually behave as an elasto-plastic material. Many constitutive models have been formulated to simulate their stress-strain behavior. The stress and strain distributions of an elastic material are somewhat different from those of an elasto-plastic material. Thus, in this finite element study both elastic and elasto-plastic models were used. The Modified Cam-clay model was used in simulating drained condition behavior of a specimen under axial load and/ or torsion. A computer program CSSAP (Cylinder Stress-Strain Analysis Program was developed for this investigation. Three hollow cylinders with different lengths were analyzed to investigate the influence of length on stress-strain distribution. Numerical results, presented in graphic form, were compared to see the influence of end restraint and cylinder wall curvature. Finally, the zone with an acceptable level of stress and strain distributions was identified, and proper specimen dimensions were recommended.
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4 1.3 Scope of This Work This study is divided into three parts: Literature review, development of a finite element computer code, and analysis. Part One 1s the literature review on stress and strain distributions in soil specimens, which covers three chapters. Previous research works on the stress-strain -distribution of a cylindrical sample was reviewed in Chapter 2. Chapter 3 listed the equations used 1n calculating the stress and strain distributions in cylindrical structure based on elastic and ideal plastic theories. Chapter 4 described the Modified Cam-clay model used in this numerical analysis. Part Two, Chapter 5, presented the implementation of the soil constitutive model in the finite element code. Part Three, contains Chapter 6, 7 and 8. Chapter 6 presented the stress-strain distribution in solid cylinder. Chapter 7 presented the stress and strain distributions in hollow cylinders of three different heights. Results were compared in Chapter 8. recommended. Sample dimensions were also Finally, summary and discussions were given in Chapter 9.
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5 2. Literature Review on Stress and Strain Distribution of Laboratory Test Specimen An ideal soil laboratory testing apparatus should be able to produce all responses of a given soil under all stress conditions similar to those encountered ln the field. This implies two points concernlng laboratory testing of soil. The first is a soil specimen should be assumed as a small element in a soil mass. The validity of this assumption relies on the uniformity of stress and .strain distribution of the soil specimen. The second is the test apparatus should be able to simulate all in-site stress condition. Unfortunately, it is difficult to design an apparatus capable of accomplishing all of these requirements. Different test apparatuses therefore are designed to simulate different stress-path or strain-path. Many attempts have been made by various researchers to design new laboratory test apparatus and to improve existing ones. Since the test purpose may vary from each other, test apparatuses are different from each other to
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6 provide the tool to get certain results. Types of tests most frequently used by researchers are: Simple shear test Conventional triaxial test True triaxial test Directional shear test, and Hollow cylinder test In the above tests, the simply shear test, directional shear test and hollow cylinder test could be used to simulate the stress condition when shear force is an important factor in the soil behavior. In theconventional triaxial test and the true triaxial test, the principal stress direction only can jump 9 oo instead of rotating gradually. In the directional shear test, both stress reversal and cyclic load cannot be applied, though this test can provide a gradually rotation of principal stresses and relatively uniform distribution of stress 2.1 Simple Shear Test Both circular and square samples are used in simple shear test apparatuses. It has been widely used in soil
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7 laboratory test due to its ease of operation. Stress and strain distributions in a simple shear specimen have been studied by many investigators. Roscoe(l953) analyzed the simple shear test with an assumption of plane strain condition and reached a conclusion that stress condition within the middle third of the sample would be more uniform than in the remainder of the sample. Duncon and Dunlop (1969) used FEM to analyze stress condition in Cambridge type simple shear on San Francisco Bay Mud. They found that stress conditions in a simple shear specimen were non-uniform. Prevost and Hoeg(1976) conducted an analysis similar to Roscoe(1953) and concluded that the stress distribution in a simple shear specimen is quite nonuniform. Fig.2.1 shows the stress condition on the upper and lower faces of a simple shear specimen. It can be seen that stress condition is quite nonuniform, and this is more serious with increase of degree of slippage. Wright et al. (1978) performed both theoretical and experimental studies on the simple shear test. Part of their result obtained from three dimensional frozen-stress photoelastic model test are shown in and Fig.2.3. It can be
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l.Z / y 1 I I i fl 0.8 o. ...__ 0.4 0.8 f--r-r--... \ ''" T = average shear stress T = shear stress xy o = normal stress y I I -\ r\\ !:.ill!iQ -l\.=0 --/1=0.2------A.= 0.5 ,-, I \ / "' 1-/ / l\ r---" A \ f-l\--..... .--:= 0.25 l 0.50 K/} 0.75 Fig.2.1 f j 1-Stresses on the upper and lower faces of a simple shear sample (Prevost and Hoeg, 1976) 8
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,--I 'I -II .. 1 Fig.2.2 Stress-strain distribution in square simple shear specimen (Wright et al., 1978) ,u ___. ___. . Fig.2.3 Shear stress distribution in circular simple shear specimen (Wright ewt al., 1978) 9
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10 seen from these figures that the stress ratio varies considerably from center to the edge of the specimen. The circular specimen shows the large stress variability. 2.2 Stress and Strain Conditions in Conventional Triaxial Samples The axial stress applied to the ends of the specimen and the confining pressure applied to the circumferential surface of a specimen ordinarily assumed to uniformly distributed. However, both theoretical and experimental investigations have shown that these stress distributions are not uniform. The non-uniformity is due to the effect of end platens. Balla(1960) calculated the stress distribution within a circular cylindrical specimen under compression. The solution, which took into account different degrees of end constraint, showed that the shearing stress existed at the perimeter of the end surfaces and also failed to satisfy the equilibrium conditions. One of his solutions in terms of normalized stress is shown in Fig.2.4. It can be seen that the shear stress distribution at both ends varies linearly, and the axial stress distribution at the edge of
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2.0 a .1.0 r Fig.2.4 Shear and normal stresses in sample in contacted with end cap (Balla, 1960) 11
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12 the specimen 1s much higher than that at the center. Shockley and Ahlvin (1960) the stress and strain non-uniformity conditions in a triaxial test. In testing dry sands, they found samples ranging from very loose to dense experienced the volume increase in their middle third at both ends. A similar change occurs in saturated sand subjected to axial strain under a constant volume condition. The test on clay, based on moisture change, showed a volume decrease in the zone of shear and a volume increase at the ends of specimens. Tests of large triaxial specimens(35.7 inches in diameter, 70 inches in height) of dry sand also showed the non-uniformity of stress and strain. Fig.2.5 shows the stress distribution in a large triaxial specimen. At the mid height of the specimen the stress at the center exceed those at the edge; whereas near the top, the edge stresses are larger. The strain distribution of this large triaxial specimen is shown in Fig.2.6. Strains are higher in the central 18-inch than at the edge. Since only one measurement of vertical strains at 53 inches and at 67 inches above the base of the specimen, the bottom two curves are drawn through the single point and followed a pattern indicated
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Specimen Diameter: 35.7 in. Hight: 70.0 in. ao I 1 ; : .. . I orrKT -s. ; :;to 0 JO ,. I I .. I I I ; I k: d 1.....::: It I-"' I I r-..... .-J-- I : I I 1--VtRTICAL STRtSS NCAR TOP ;rrTR-.:_ rl-1 or I ,_ .. Fig.2.5 Distribution of stress in triaxial specimen (Shockoley and Ahlvin, 1960) 13
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z.s .. II II u z.o .. .. I 0 .. .c .. .s c z c lr .... J c u ... ; lr II > o.s 0 I I Zl IN. ABOvE BASE r-...... I 1\ 1-.. IDHE:IGNT IJS '"' \ -"" ""' I U IN ABOVE BASil: r--.. ... ....... "" I IN. ABOVE 8ASC -f-CENT I.INE: Oil' I I I 1--,., :taovlt BASil: --i- I 10 0""111:T IN INCNI:S Fig.2.6 EOGI\1,.. II ,! \ It I '\ \ 11 "' 1 -'-ll 'I ....,.,. ,....-1: II .. zo Distribution of strain in triaxial specimen (Shockoley and Ahlvin, 1960) 14
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15 by the theoretical solution of Pickett(1944). The gross pattern indicates the nonuniform vertical strain throughout the specimen. Rowe and Barden(1964) also studied the non-uniformity in a triaxial sample. They found that the formation of dead z.ones at the ends of the sample, caused by the restraining effect of radial friction forces at the end platens, is one of the major causes of non-uniformity. There are large dead zones at the sample ends and the dilation zone in a conventional triaxial specimen as shown in Fig.2.7. The concentration of the dilation in narrow zones cause those zones to achieve the peak stress ratio ahead of the rest of the specimen. The soil in these zones is weaker and the stress ratio applied to the sample is decreased. Rowe and Barden(1964) suggested using the combination of rubber sheeting and silicone grease (Fig. 2. 8) to develop frictionless ends for triaxial compression test specimens. The results obtained from clay specimens tested with lubricated ends, shown in Fig.2.9, indicate that the use of lubricated end platens results in uniformity of This allows the moisture content throughout the test. sample to retain its cylindrical shape even at large
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\ \. _., DEAD ', ZONE . -v 'lr .. '-" .... .. / \ .. / \ v / I / I /\ ', / \ / I \ / \ / / I I '/ >< / / x I \ / / '/ \ /:.. \ ... ' /. ', / v '\ ' Fig.2.7 Nonuniformity of sample and slip surface (Rowe and Barden, 1964) 16
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0-ring !.e.,ls "" I I gr E'8St-d diam. porous disc mpmbrane I .. I ----:( araldil e juinl Fig.2.8 Enlarged frictionless end platen (Rowe and Barden, 1964) 17 ''-
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.... z .. ... . w 0\ z :. .. II II II .. ;c. (a) UNTESTED ... ...__;__ z .. ... a .. w ... ::! a I 0 wjr:.
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strains. Duncan and Dunlop (1968) both clay and sand. They 19 conducted triaxial tests on found that cap and base restraints result 1n a small increase in the modulus, a slight reduction 1n the strain at failure and a minor increase the angle of shearing resistance by as much as 3 for test in sand. They concluded that lubricated caps and base are required only when volumetric strain in sand must be measured. Lee (1978) reviewed most studies related to the influence of caps and base restraints in triaxial tests and concluded that there was a significant increase in static undrained strength with lubricated ends as compared with tests using regular ends for medium to dense sands. For drained tests on sand and undrained test on clays, the effect was found to be significantly greater than those observed in other studies. Kirkpatrick and Belshaw (1968) studied the effect of end restraints on the condition of strains in triaxial tests. X-ray technique was used for examination of radial and circumferential deformation of specimens tested with both rough or lubricated end platens. They found that
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20 relatively homogeneous strain conditions exist in specimens tested with lubricated end platens and the assumption of equality between the radial and the circumferential strain is valid for those tests with lubricated end platens. Nonuniform strain conditions occurred throughout the sample and for a large region of the specimen the radial and circumferential strains diverge widely when rough platens were employed. Perloff and Pombo (1969) analyzed the effects of the end restraint in the triaxial test using an assumed elastoplastic constitutive law. The distribution of stresses throughout a constrained unconfined specimen for which the slenderness ratio is equal to 2, is shown in Fig.2.10. Contours of vertical stress, crz, radial stress, crr, circumferential stress, Qa, and shear stress, are shown in respective quadrant of the schematic crosssections. Fig.2.10a and Fig.2.10b illustrate results for an average axial strain Ea=l. 0% and-2. 0%, respectively. The nonhomogeneity of stresses are apparent from these results; and it is more significant with increase of the axial strain. They suggested that the use of specimens with large slenderness ratio would assist in reducing the
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(a) Ez = 1% Fig.2.10 . I : 00.0 I v .1 r /o.o o. r-c..z "" eo (b) Cz = 2% J Stress contours throughout non-linear elastic unconfined compression specimen ( Perloff and Pombo, 1969) 21
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.. -!.-.. ,. co ---F'ilon --Prckefl -O'Appolonio -Newmork 1-----Boll.: --New solu1inn Oistonce Fig.2.11 Comparison of stress distribution by different investigations at the end surface for perfectly continued ends ( Al-Chalabi and Huang, 1974 ) 22
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23 end restraint effect on the observed results. Al-Chalabi and Huang (1974) studied the stress distribution within a homogeneous, isotropic and elastic circular cylinder in compression. A theoretical solution was derived for the stress and strain distribution within cylindrical specimens as a function of the interface friction between specimen and platens. The solution was compared with experimental results obtained from testing and epoxy cylindrical specimen with a strain gage embedded within and bonded to its surface. A comparison of this solution to other elastic solution is shown in Fig.2.11. The vertical stress, cr4 shear stress, circumferential stress, cr9 and radial stress I crr1 were normalized with average vertical stress 1 cr1 It can be seen that all solutions agree on the axial stress distribution, although the magnitude at the center and at the edge differ considerably. Only solution of Al-Chalabi and Huang (new solution in the figure) and that of Filon satisfied the boundary condition of no shearing stress at the cylindrical surface of the specimen. This is a large discrepancy in the circumferential and the radial stress distribution obtained by the different investigators.
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24 2.3 Stress and Stain Distributions in Hollow Cylindrical Specimens A hollow cylindrical specimen shown in Fig.2.12a is subjected to an axial load, w, which contributes to a vertical stress, crz; the difference between internal and external pressure, Pi and P0 establish a gradient of radial stress, crr, across the cylinder wall. The relationship between radial stress, crr, and the circumferential stress, cr9 is expressed by the equilibrium equation: da a8 -a = r--r r dr The stress condition in an element of a hollow cylinder is shown in Fig. 2 .12h. Both inner and outer pressure are applied on the membrane so there is no shear stress on the vertical boundaries. Without any end restraints, crr is always a principal stress because there are no shear stresses on circumferential surface throughout the wall. The magnitude and direction of principal stresses, shown in Fig.2.12c, are obtained by resolving stresses, crz, O'a, 'tza and 'taz. The most critical aspect of the use of hollow cyl1nder is the nonhomogeneity of stress and strain
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-t:Jo I 1 L I I ---1,' --'( .... ,l ', (al 11z Tz8 I I / (b) (C) Fig. 2.12 Idealized stress conditions in a hollow cylindrical element subject to axial load W, torque internal pressure, p1 and external, Po: (a) hollow cylinder sample; (b) stresses on an element in the wall; (c) Principal stresses on an element in the wall; (d) Mohr circle representation of stress in the wall ( Hight, et al, 1983 l 25
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26 distributions, developed in the wall of a specimen as a result of the wall curvature and end restraint. The curvature induced stress non-uniformity can be minimized by selecting an appropriate geometry of the specimen, such as recommended by Hight et al ( Hight, 1983 ) The influence of end restraint will vanish as one moves away from the end platens. Therefore, selecting dimensions of a specimen which can m1n1m1ze the non-uniformity of stress distribution is the most important task in the development of a hollow cylinder apparatus. Lade (1976) investigated the uniformities of stresses and strains in a hollow cylinder specimen. The height of the specimen was 5. 0 em, and the inside and outside diameters were 18.0 em and 22. em, respectively. The stress acting on an element of the specimen is shown in Fig.2.13. All stresses except crx were measured during the test. Since crx was unknown, it is determined by a trial -anderror process during the analysis. Therefore, the analysis of a torsional shear test is performed by specifying five independent stresses (cry I O'z I ax, 'tyz=O I 'tzx=O) and the shear strain ( "fx) The Lade's elastoplastic constitutive theory (Lade,
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27 1972) was adopted for the analysis. The normalized shear strain distribution along the wall of specimen is shown in Fig. 2 .14. The distribution of crx throughout a test was determined at discrete points ih the specimen. Fig.2.15 shows the variation of the crx determined for a range to shear stress-shear strain curve with a chamber pressure of 0. 60 kg/cm2 The variation of the crx depends on the developed shear strain. The distribution of crx, then, can be determined for a given value of shear stress 'txy. The calculated crx distributions for various values of 'txy 1s shown in Fig.2.16. The lowest values of tangential stress occur 1n the middle of the specimen, while the higher values of tangential stress occur at area close to cap and base platens. Analysis also showed that the highest stress levels occur in the middle of the specimen where the effect of end restraint was smallest, the values of crx are smallest, and the shear strain are greatest. From Fig.2.16, the middle half of the hollow cylindrical specimen seems to have a uniform stress distribution and the end effect would be small in this area. Based on the analysis, Lade suggested that taller
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I Fig.2.13 Stress components in Cartesian coordinate system for hollow cylinder specimen ( Lade, 1976 ) 1 2 'IIY"r-. Fig.2.14 Normalized shear strain distribution at failure (" Lade, 1976) 28
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O.Sl 1.110 i 1 .. ,.., llol l .e ,..,r&l Fig. 2.15 variation of crx determined for a range of shear stress -shear strain curves for test with crz = 0. 60 kg/cm2 and cry = crz/K0 on dense Monterey No. 0 sand ( Lade, 1976 29
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I ur'l 10 '"'''' ,_o " llrlt'c"'21 Fig.2.16 Shear stress -shear strain relation and calculated variations of crx for discrete values of applied shear stress in test with crz = 0. 60 kg/cm2 and cry = crz/K0 on dense Monterey No. 0 sand ( Lade, 1976 ) z.o 1-,.-,..----r----r------. -1.5-\ f Snu Monu Such $,,.,, """'\ .. r:J C, ZO 9& \ 10. ' 1.0 ... ; Yruu or Ocorr ''" lnchcd Ute O Pou A o ...... ,., 21 c"' Wll 1 em I.Dr.;t z.o I Fig.2.17 Effects of end restraint in torsion shear test on hollow cylinder specimen ( Lade, 1981 ) 30
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31 hollow cylinder specimens than those used in this study should be employed for improving the precision of the measurement. Lade (1981a) modified his earlier version of hollow cylinder apparatus by increasing the height of specimen from 5 em to 40 em, but kept the same average diameter and wall thickness. Using the new apparatus, he conducted testes on Santa Monica Beach sand with heights of 10 em and 40 em. Fig. 2 .17 shows normalized average tangential stress against specimen height for loose Santa Monica Beach sand. The result of a test on loose Monterey No. 0 sand performed with the same stress path as for other tests is also shown in Fig. 2 .17 to provide a data point corresponding to a specimen height of 5 em. It was concluded that the end effect can be neglected in a 40-cm tall specimen of loose sand, and the state of uniform stress might be achieved in specimen with height lower than 40 em. Wright et al. (1978) performed a study on the geometry of a hollow cylinder specimen. Based on the theory of elasticity, they evaluated the hollow cylinder specimens used by Lade (1976), Ishihada and Yasuda (1975), Saada and Zamani (1969), and Frydman et al. (1971) formulated an
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32 expression for the maximum relative difference in the shear stress under the assumption of uniform distribution as: ('t'ez) max ('t'ez) avg = ( 't'ez> where is the shear stress that resulted from torque applied on the specimen, n is the ratio between the inner and outer radii (r;/r0 ) The higher the n is, the more uniform the torsional stress distribution will be. By selecting Poisson ratio of 0.5 and the ratio of n of 0.65, the criteria for the length of a hollow cylinder specimen, L, was expressed as: These criteria have been followed by Saada and Samani (1969) in selecting the specimen dimensions of their hollow cylinders. Saada and his coworkers (Saada and Baar, 1967; Saada and Zamani, 1969; Saada and Ou, 1973; Saada and Bianchili, 1975; Saada and Shook, 1981; Saada and Townsend, 1981; Macky and Saada, 1984) developed a hollow cylinder test system and used it to study the effect of anisotropy on clays. They concluded that the thin hollow cylindrical
PAGE 51
33 specimens with the same internal and external pressures and subjected to the axial and/or torsional stresses, provide the best means for studying the mechanical properties of soils. The end effect can be minimized by using long specimens, and by using thin specimens, a uniform distribution of shear stress and shear strain can be reached. Hight, et al. (1983) developed a hollow cylinder apparatus. The expressions given in Fig. (2.18) were used to calculate the stress and strain in the test. To minimize the non-uniformities arising from curvature, Hight et al. performed a study for the geometry of a hollow cylinder specimen. By setting the bounds to the difference between the calculated and real averages, B1 and the level of non-uniformity, 83 they found that a ratio of inner to outer radius, a/b, of 0.8 was the most appropriate one for a hollow cylinder specimen under several selected stress paths. The variation in stress and strain nonuniformities with geometry and stress paths are shown in Figs.2.19 and 2.20. The variation of accuracy in strain with geometry and stress path is shown in Fig.2.21. Based on the ratio of inner and outer radii of 0.8, an example of
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Average venical stress i7, -__ w __ + 11102a2) Average radial stress ", (0 o J)a) 0 to.;. at (0 o2 -tJ. a2 ) 0 (02 -a2) (000-tl,a) Average c"cumterenhal stress;, 10 _a) JM, Average snear stress <,, -211103Average a..oal strain r 1 w -;;-(IJ -1.1) 0 Average radoal st raon r, -(O:a) tu0 v) Average corcumlerentoal strain r1 10 aJ 9 a angular circumlerenlial displacement (I) (2) (3) (4) (51 (6) 2Beo3 -a1 (81 Average snear stra i n 711 JHto2 a2) Fig. 2.18 Defination of average stress and strain (Hight, et al., 1983) 34
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0 /." ;.jdr J (0-3)0'L .:!13: ,,., "'lc I Nore lhal "ii.tti1 and I a/a, are 1nlercnangeaore I Ra11o or 1nner ro ourer radii a/0 Fig. 2.19 Variation in stress geometry and stress elastic analysis) non-uniformity with path (isotropic linear Hight et al., 1983 ) o fJ 3 ro-ar r L wnere 'L = 1 rr, 1 + 1 -r, I l/2 \J R.aho or 1nner ro outer raan a/tJ Fig. 2.20 --v = 0 --"= 0!5 Nole rnat if.J11, and are 1nlercnangeac1e Variation in strain non-uniformity with geometry and stress path (isotropic linear elastic analysis) ( Hight et al., 1983 ) 35
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JJ I !!:...:.L' wnere c'.. nr .. -1 r 1 1 12 \H)5 .., c: "' Ou4 -::.. -. _, OJ ---'-0 Note rnat lf"JIT, and IT/If. are ontercMangeaooe Aatoo ol onner 10 outer raau atb Fig. 2.21 Variation of accuracy in strain with geometry and stress path (isotropic linear elastic analysis) J 300 k!'; 200 --( Hight et al., 1983 ) ---Calculated averages p" r p. it' ---Calculated averages p-' Elastoc analysos Elastoptastic analysis Stress distributions at Q 3 Fig. 2.22 Stress distribution across the wall of a hollow cylinder specimen (Hight et al., 1983 36
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37 the stress distributions across the wall, the variations of crz, crr and cr9 found in both the linear elastic and elastoplastic analysis at the stress level, q, of two thirds the failure stress, qr, are plotted in Fig.2.22 To reduce the effect of end restraint, specimens with various heights were investigated. Fig.2.23 shows the results from a series of linear elastic analysis of fixed end specimens of different height and of inner and outer diameters of 7.87 inches and 9.84 inches, respectively. The applied stresses comprised an axial stress of 58.0 psi and inner and outer cell pressures of 50.8 and 58.0 psi, respectively. Fig.2.23 also shows that a height of 9.84 inches would provide a central length of specimen wall of 4.92 inches in which stress would differ by no more than 10% from those in an unrestrained linear elastic cylinder. This was considered to be an adequate central gauge length and so a specimen height of 10.0 inches was adopted. After selecting the ratio of inner and outer radii and the height of specimen, the inner and outer radii were determined by the thickness of the wall, and difference between tangential and radial stresses. A one-inch wall thickness was selected under consideration of sample
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1.d 200 mm o d 250 mm p0 400 kPa. C, 350 Pa. u: 400 kPa JJ mm H 100 'Tlm 90 /1 OS 75 110 fa! Con1ours ol a Ia : I e i,HI C All contours as ol stress oreOCied lor a 11near elaStiC e1ement wtnout eno restranl (bl Contours ol a.Ja,. .H 200 mm (c) Contours ol tt.frr, 01051 0 Fig. 2.23 105 110 -11:1" 125 120 1 "'t:. co-"'"' ... c j_ Linear elastic analysis of stress distribution in fixed-ended hollow cylindrical elements of different height (Poisson's ratio = 0.499) (Hight, et al., 1983) 38
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39 preparation and of instrumentation disturbance on the specimen. The difference between tangential and radial stresses developed by the difference between outer and inner pressures in a one-inch thick wall of different radii is shown in Fig.(2.24). From this figure it can be seen that a small improvement in stress uniformity is achieved for increases in outer radius above 125 mm (0.125 m). An outer radius of 5 inches was selected. Based on the value of 0.8 in ratio of inner and outer radius, an inner radius of 4 inches was adopted. For the selected dimension of specimen, stress status attainable with PoiPi between 0.9 and 1.2 were established for the cases of a of 0 and 90 and of a of 45, these are shown in Fig.2.25. 2.4 Summary From the review of literature, it can be concluded that no apparatus is perfect and is capable of simulating all situations occurring in the field. Each apparatus has its advantages and limitations. The stress-strain distribution in cylindrical specimen has been investigated by researchers via experimental and
PAGE 58
Oo !:: .. 0 0 0 b'-1 b"' 0 1 SQ 8 .. __________ !l:m Fig. 2.24 Influence of outer radius b on stress uniformity in hollow cylinder 1" wall thickness (Hight, et al., 1983) 40
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c:r,-cr2 c:r,-cr3 'r-G"r I 90 fir cr1 ., C1'2 ":-"3 ., o. go .. 45". 90" > 1 ., -0 "2 err- 1 "l "'I "ttt are un-oraerea principal stresses a,, az 03 are oraerea pnncil)al stresses Fig. 2.25 .i,ID1 < 0 .. -45 a, crl Stress strain attainable in the hollow cylinder apparaturs with PoiPi between 0.9 and 1.2 (Hight, et al., 1983) 41
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numerical methods. 42 It has been found the solid cylinder specimen could provide reasonable results when the slender ratio is near 2. For a hollow cylinder, it is found the ratio of internal radius and external radius equal 0.8 will lessen the curvature effect to a reasonable region. The ratio of cylinder height and the thickness of the wall has been investigated too, it is suggested that tall cylinder need to be used to lessen the end effect, but it is hard to glve a certain-value. For studying cyclic behavior of soils, hollow cylinder apparatus performed under equal external and internal pressures should be the best choice among these five apparatuses. However, the torsional force combined with vertical deviatoric pressure acting on cylinders haven't been studied by numerical method, although some experimental results indicated that the stress and strain are still distributed in a way that their non-uniformity is not so significant as to make test results unusable.
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3. Calculation of Average Stress and Strain in Hollow Cylinder 3.1 Introduction 43 Hollow cylinder apparatus is relatively a good one to complete the purpose of simple shear or the principal stress direction rotation in soil element. In general, stresses are not uniformly distributed but vary across the cylinder wall. To interpret the results of the hollow cylinder as an element, average stresses and strains are used. The calculation of average stress and strain in a hollow cylinder specimen is based on the applied externai forces and the deformation of the specimen. There are two sources which caused the stress and strain non-uniformity in the specimen: curvature of the cylinder and constraint at the ends. The effect of curvature will occur if internal pressure p1 and external pressure Po are not equal, and when if torsion is involved. The inner and outer radii
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44 ratio r1/r0 internal pressure and external pressure ratio p1/p0 and material's behavior decide the curvature effect. The load condition of a hollow cylinder is shown inFig.3 .1. 3.2 Radial and Circumferential Stresses The general solutions of radial and circumferential stresses of hollow cylinder at point r are: rl (pi -Po> r2 (pi -Po> r2 (3 .la) ( 3 .lb) There are four methods to determine the average radial and circumferential stresses. 3.2.1 Elastic Equal Effect In this method of averaging, it is assumed that the contribution of the stress on any fiber is independent of its radial distance. This equation appears to be the most
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45 w z r Figure 3.1 The load condition of hollow cylinder
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46 popular among researchers and has been used ln a number of publications, as Hight et al. (1983): (3.2a) (3.2b) 3.2.2 Elastic Proportional Effect This averaging assumes that the contribution of the radial stress crr on a fiber at a radial distance, r', is proportional to r. Momenzedeh(1985). ro2 r2 1 ro2 2 Ii This type of averaging was used by (3.3a) (3.3b)
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47 3.2.3 Linear Equal Effect These equations were used by Frydman(1973). Averaging without weighing gives circumferential stresses as: Po+ P1 a = r 2 the 3.2.4 Linear Proportional Effect average radial and (3. 4a) (3. 4b) Averaging with weighing was used in Miura(1986), and gives the following equations for the average stresses: (3.5a) (3.5b)
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48 3.3 Axial Stress The average axial stress is derived by dividing the resultant vertical force Pz by the area of cylinder A5 : ( 3. 6) 3.4 Shear Stress Refer to Fig. (3.2). Consider an element of thickness dr, at a radial distance of r, subtending an angle d9 at the center, subjected to a torsional stress of magnitude tz9 the torque on the element is given by: ( 3. 7) Integrating this over the specimen thickness from ri to r0 and over the whole circular surface: ra 21t T = J J "zer2d8dr ( 3 8) r1 o In order to compliment the integration of above equation,
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0 (a) (b) m -----( t'l!9) p Perfectly Plastic behaviour . . (II-' 1) ( t;el 1::::::::: 49 0 Linear Elastic behaviour 1 (11-2) o r, ro 0 r 0 Averaging by TorQue equilibrium Averaging by Shear force equilibrium Fig.(3.2) (a) Derivation of shear stress tri (b) Distributon and average of t'l!9
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50 need to know the For simplification, perfect plasticity and elasticity material can be assumed. When the material. behavior is assumed to be perfectly plastic, then the distribution is uniform and is a constant. For this condition: ro T = 2n't ze/ r2dr = n'tze r]) rJ. thus 3T ( 3 9) When it is assumed that the material behavior is linear elastic, then the shear stress varies linearly with radial distance. Given that is the shear stress at r=r0 the shear stress at any radial distance r is given by (3.10) The average shear stress that gives the same equivalent torque as the linearly distributed shear stress, may be obtained from the following equations:
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51 r., r., 'tze J I 2di = J 't';x I 3di = r1 r1 o but Tmax = 2TI0 n IJ) (3.11) found 3T 2n I]) (3.12) which is equal to the case when material behavior is perfect plasticity. It is worthwhile to look at the deviation of the average radial stress from the radial stress at the inner and outer boundaries. The average stresses which come from the Elastic Equal Effect were considered. oro = (1 + Rr) or 1 + Rrf?.r In below (3 .13a)
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52 a rJ. = ( 1 + Rr) or 1 + (3.13b) Similarly, for the circumferential stresses, the following equations are obtained: ( Oe) I'='I ___ ....;:;CI = and -(1 + R;) (1 + RI) (1 R_pRI) (3.14a) (3 .14b) For the shear stress, when the linear elastic distribution and average by torque equilibrium are assumed, then the maximum and minimum shear stresses at the r = ro and r = r1 respectively are given by: (3.15a)
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53 ( 1 -R3 ) 'tmin = 4 r R 't ze 3 ( 1 R:) l (3.15b) 3.5 Strain With reference to Fig. (3.3), which showing the strain components Ez, Er, E9 and Yza of an element in a cylinder. Defining extension strain is negative and compression is positive. In general condition the strain components of a cylinder are: au -; ar au = rae + ea=-(..!! I I 00 1 av ar v -; I au y rz = az e = z az aw + -; ar = av az (3.16) aw + raf (3.17) Consider the axisymrnetry of load on the cylinder, the displacements u,v and w must be independent of 8. Then for this special case, the strain components are: au -; ar u -; I (3.18)
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Fig. 3 3 Strain components of a cylinder element z"" Fig. 3.4 (a) Jnohal stress slate ICI Slress oncremenl state I r I u, ..p 6tl, .....-1"'---+-"7 (e) Final stress state IOJ lnohal stress state soecohed oy proncopal stress idl Stress oncrement slate specoloed oy oroncopal stress oncremenrs !0 Final stress state showong rotatoon of proncopal stress directoons Principal stress rotation on an element Hight, et al, 1983 54
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av = ar Yrz 3.5.1 Radial Strain = au + aw; az ar (3 .19) 55 It is obvious that the average radial strain is: (3.20) 3.5.2 Circumferential strain Since the circumferential strain varies across the wall of cylinder, the average strain is the one that will ensure equilibrium in circumferential displacement: ro ro J e8zdr = J (8 ) rdz but
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56 thus get (3.21) 3.5.3 Shear Strain The strain is the strain component of major interested in a hollow cylinder test. The circumferential displacement vat a point (r,z) in the cylinder is given by then v = H av = = az rd8 H averaging it over the area of specimen, .r, .r, y ze/ 2"'_rdr = J y ze2"'rd6dr .r1 .r1 (3.22) (3.23)
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57 get 2d8 Yze = 3H (3.24) 3.6 Principal Stress and Strain The principal stress and strain at a point of the cylinder structure are calculated by the following equations when assuming crr and Er are the secondary principal stress and strain if ignoring the end effect and based on the load condition of p1=Po (3.25a) (3.25b) The rotation of principal stress direction, see Fig. (3.4), is: 1 1( 2'tze ) Cl avg = -2 tanOz Oa (3. 25c)
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However, 58 the stress and strain are not uniformly distributed in the whole structure because of the curvature of cylinder and the end effect. Usually the average stress and strain are calculated using the above equations for the whole cylinder, in which, the stress and strain components are average values. In this thesis, therefore, the average stresses cr1j of the structure derived in previous sections that based on end restraint free situation were used to calculate the average principal stresses cr1 and cr3 The average strains 1j, which were used to calculate the principal strains were the average values of all the elements respectively. 3.7 End effect The end effect is another reason which caused the stress and strain non-uniformly in a specimen. The displacement,u along r direction will occur when a cylindrical specimen is subjected to vertical load or prescribed vertical displacement. Shear force will develop also between the platen of machine and the end of specimen if the platen was not fully lubricated. In many devices it
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59 is possible to reduce the frictional restraint by lubrication of the platens. Such a solution is very difficult in the hollow cylinder in where friction at the platens is necessary for transmission of torque to the sample. Restraint on radial displacement of the specimen ends is associated with the development of radial shear stress, 'tzr, and its complementary stresses, 'trz which decay with distance from the platen.-These shear stresses result predominantly in additional circumferential stress, 1n bending moment which affects the distribution of vertical stress and on rotation of principal stress out of the plane of the cylinder wall. The extent of the stress disturbance is clearly dependent on sample geometry, the specimen's constitutive law, and the applied pressure and load combinations. The displacement restraint of the r direction in turn affects the stress and strain conditions in circumferential direction, 0'9 and 9 which are the function of u. The deformation shape of specimen in r direction becomes a curve. A strip of hollow cylinder wall with the fixed end platen is isolated for investigating the stress condition
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60 when a specimen subjected to vertical load crz, see Fig.3.5. (1) At the end, the stiffness of the apparatus end platen is very large compared with that of specimen so the platen can be regarded as rigid, there would be no rotation at the end, thus a moment M is developed to keep the end of the specimen from rotating. (2) As there is no movement in the r direction at the top surface, shear force 'tzr was developed to keep the top surface from moving. (3) In the region very close to the top, strains er and e9 are almost equal to zero. There must have crr and cr9 acting there to keep the strain to zero. The total stresses acting on the end have been shown in the Fig.3.5. Furthermore, the non-uniformity of cr1 causes the non-uniform displacement w in z direction even on the same layer except the top and middle layers. When torsion 'is combined with vertical pressure, y9 will not be uniformly distributed in the specimen, since y9 = rd9/z, where z is the current height of elements which varies from element to element in a deformated sample, and r is different from element to element also.
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(Jl fixed to Figure 3.5 The top end stress condition of a strip of hollow cylinder wall 61
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62 4. Modified Cam-Clay Model 4.1 Introduction In this analysis the material is assumed following the modified Cam-clay model, a strain hardening constitutive law. The general concept of using a hardening plasticity model to describe the stress-strain behavior of soils was first proposed by Drucker et al. ( 1957) Essentially, Drucker et al. suggested putting a spherical 'cap' on the 'Drucker-Prager cone' The cap will enlarge by hydrostatic loading of the soil. Their paper speculates about what happens to the cap on elastic unloading during a triaxial test, but makes no firm proposals. The paper expresses doubts as to whether normality should be applied to the 'frictional' yielding on the cone. In constructing the critical state model, the Cambridge group picked up some of the proposal of Drucker et al., and discarded others. In doing so they managed to produce a model of soil behavior which is 'simple' in the sense that the model is derived
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63 from a small number of basic assumptions, yet the model manages to reproduce for the first time an appropriate description of volumetric response under shear. What really sets the critical state models apart from other attempted to formulate as elasto-plastic model for soil is the critical state line in the (P',V) plot,as shown in Fig.4.1. This allows a consistent and realistic treatment of both drained and undrianed tests. Although Cam-clay makes a significant step forward in the modelling of soil behavior, it is still deficient in some aspects of stressstrain modelling. Modified Cam-clay (Burland, 1965; Roscoe and Burland, 1968) addresses two particular dissatisfactions with the original Cam-clay model. First, based on experimental evidence, the shear strain predicted by the Cam-clay model is too high at low stress ratios, and the shape of the yield locus gives the stress increment that is incompatible with the condition when the yield locus intersect p' axis. Second, the predicted value of Ko(the coefficient of earth pressure at rest) is too high. The modified Cam-clay model is a modified version of the original Cam-clay model, which was extended to cover general three dimensional stress states by Roscoe and
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CSL Compression { I I ') 01 > 02 = 03 M. Yield Surfaces: Original Cam-Cay P1 = + EneDSion { < = CSL Figure 4.1 Yield surfaces in Cam-Oay models. (Ko and Sture, 1980) 64
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65 Burland (1968) One of the major differences between the two models lies in the shape of the yield surface. The modified model has an elliptical yield surface, as illustrated in Fig.4.2. It simulates the behavior of an isotropic, elastic-plastic strain hardening material. Only the volumetric strain is assumed to be partially recoverable, and the elastic distortional strain (shear strain) is assumed to be zero. Elastic volumetric strain is non-linearly dependent on hydrostatic stress. The total volumetric strain is composed of two parts: elastic strain and plastic strain. However, under an undrained condition, the plastic strain is equal to minus the elastic strain during plastic deformation. Although the Cam-clay model was originally proposed for stress ratios q/p' less than M, Schofield and Wroth(1968) extended the proposal for stress ratios greater than M. 4.2 Governing Equations for Modified Cam-clay Model (a) Volume -pressure relation: The equation of the isotropic Normally Consolidated Line (NCL) is: v = N + (A-le) Ln2-ALnp ( 4 .1)
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q Critical State Line Elliptical Yield Surface Elliptical 'Loading' Surface p' I I I Pc {a) Pc Figure 4. 2 Some aspects of the modified Cam-Clay model for triaxial conditions. 66
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67 where v is the specific volume, V=(l+e), e is void ratio,K and A are soil constants. The equation for the rebound curve is: viC = v + K Lnp 1 ( 4. 2) The equation for the yield cap is: I q2 f = (pI) 2 Po pI + M2 (4.3) where M is a material parameter, and is the slope of critical state line in the p'vs.q plot, Po is the original hardening parameter, which changes when stress state changes at plastic stage. (b) Critical state line: q = Mpl v = I' },lnp (4.4a) (4.4b) (c) Yield equation of the Stable State Boundary Surface:
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VA= r+(J.-lC) [Ln2-Ln(1+((.!l)jM)2 ] P' (d) Flow rule: (e) Yield locus: I 'I ( p -Po 2 ) 2 + ( q ) 2 = 1 M or I q2 f = (p1 ) 2 -Po P1 + M2 where q2 = a22> 2 + Ca22a33)2 + (a33an>2 + + 6 ( + + "";1) ] From Eq. (4.1), the volume change is: 68 (4.5) ( 4. 6) (4.7) (4.8a) (4.8a)
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= -A. 69 ( 4. 9) where Po' is the of yield locus with p' axis as shown in Fig.4.1. Also, the volume change ratio, or strain is: dev = v From Eq. (4.2), the elastic volumetric strain is: = l+e p' (4.10) (4.11) Combining Eqs. (4.10) and (4.11), give the plastic (or irrecoverable) component of the volumetric strain as: where = (A.-x) dp1 (l+e) p' (4.12) (4.13) From the yield locus, Eq. (4.3), noted the yield function f is and can be expanded or contracted as the hardening
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70 parameter P0 increases or reduces. The reduction of P0 corresponds to softening, and is not considered here. The increase of P0 corresponding to strain hardening. The increment of Po' is: (4.14) 4.3 The Elasto-Plastic Stress Strain Relationship In order to determine the elastic response, from Eq. ( 4 .11) have: K I da .lc1c3 ij 3 (1 + e)p (4.15) where O'kk = 0'1 + 0'2 + 0'3 and Bij is Kroneckel delta. If elastic distortional strain is identically zero, the elastic stress-strain rate equation is: From elasticity: = l+e p' where K' is the volume bulk modulus. (4.16) (4.17)
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71 Eq.(4.16) and Eq. (4.17) give: I Vi I K1 = ( 1 + e) = _:I!_ 1C 1C (4.18) obviously, K' is not a constant. It changes with the void ratio of material. As it 1s assumed no shear strain during elastic deformation, .the shear modulus G can be made quite large in computation. Starting from the associated flow rule after Drucker, we have (4.19) where f is the yield function: I q2 f = P12 Po P1 + Mz (4.20) Since The elastic stress-strain incremental relationship is: (4.21)
PAGE 90
72 in which D. is the elastic stress-strain matrix of the material. Now differentiating Eq. (4 .20): df Bf Bp1 da + Bf Bq da + Bf d P < ----ij aq aaij ij -,-eij = ap I aa ij Bpo (4.22) or (4.23) where (4.24) Substituting Eq. (4.21) into Eq. (4.23): (4.25) Thus (4.26) then dA can be found as:
PAGE 91
73 (4.27) Now consider Eq. (21): (4.28) The above equation is the incremental stress-strain relationship and D8P is the elasto-plastic compliance matrix. In which af 1 1 38 -(2p'-p ) +---.!.! aall 3 o Mz af l:. (2p'-p') + 3822 aa22 3 o Mz af l:. (2p'-p') + 3833 (__E_) aa33 3 o Mz (4.29) = = aaij 8f 6't12 8'tl2 M2 af 6't23 'tzJ M2 af 6't31 't31 M2
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74 and (4.30) D. is the elastic stress-strain compliance matrix, the 36 terms of it are: dll dl2 dl3 0 0 0 cizl d22 d23 0 0 0 D = d32 d33 0 0 0 Q 0 0 0 G 0 0 (4.31) 0 0 0 0 G 0 0 0 0 0 0 G where G is the shear modulus. In elastic stage G is assumed quite large while in plastic stage and 3K1(1-2v) G= 2 (l+v) (4.32)
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dll = (3K1+4G) I 4 ; dll = d22 = d33 d12 = (3K1-2G) /3 ; d12 = d13 = d21 = = d31 = d32 In Eq. (4.27), A is the strain hardening factor: as of = and from Eq. (4.14), get (4. 34) 75 (4.33a) (4.33b) (4.34) (4.35) (4.36) By multiplying Eq. (4.36) with Eq. (4.29), and consider (4.37) thus I (2 I I) A= -p p -p .A.-1e o (4.38)
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76 The modified Cam-clay model was developed for normally consolidated or lightly overconsolidated clay. The model has not been successfully applied to heavily overconsolidated clays with dilate, strain softening and produce nonuniform deformation during shearing. The Modified cam-clay model underestimates the shear strain at small stress levels. The model is an isotropic workhardening plasticity model which performs best for clays that have initially experienced isotropic consolidation. The assumption that plastic strains only occur when the stress path traverses the state boundary surface limits the prediction of stress hysteresis for unloading and reloading conditions under cyclic loading. However, the soil parameters used in this model are very easy to obtain. The prediction of stress-strain behavior of normally consolidated and lightly overconsolidated soil is very good under monotonic pressure loading.
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77 5. Implementation of Modified Cam-Clay Model in the Finite Element Computer Code 5.1 Introduction Finite element analyses were used to de .. ermine the stress and strain distributions of cylindrical specimens. A computer program CSSAP (Cylinder Stress-Strain Analysis Program} was developed to perform the numerical analysis. In this computer program, the Modified Cam-clay model was implemented to analyze the stress and strain distribution in drained condition. Eight-node isoparametric quadrilateral elements were used to discretize the structure, and incremental stiffness method was adopted in the analysis. In this chapter, the basic principle of Finite Element Method in the solid mechanics was described. The finite element discretation of cylindrical structure and the calculation of equivalent nodal force on the boundary were presented. The load condition applied on the structure was discussed. The definition of stress and strain
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78 normalization was given. 5.2 Basic Principle and Formulations The finite element method is a numerical procedure for solving a system of equations governing a physical problem. It has two major steps. First it utilizes discrete element to obtain the joint displacements. Second, its uses the continuum elements to obtain approximate stress and strain distributions in an element. The general procedure of using finite element method to solve solid mechanics problem can be described briefly as follow ( Zienkiewicz .O.C., 1977). A solid structure is discretized into the assemblage of elements. The connection among elements was realized by the common element nodes. First consider an element with n nodes, each node has three degrees of freedom. Then it is need to determine the: (a) strain: The displacements on any point in an element can be decided by the nodal displacements {a} of the element: tJ = [N] "{al" (5.1)
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79 where [N] is the shape function of the element, which 1s dependent on the element geometry. Knowing the displacement at a point, the strain at that point can be easily determined (initial strain was not considered} : {e} = [A] [N.J 9 {a} 8 (5.2) where [A] is an operator, or {e} = [B] {a} 8 ( 5. 3) in which [B] = [A]. [N] ( 5. 4} The operator [A] is determined by the mechanic problem, its expression for cylindrical structure will be give later. (b) stress: The stress crij at that point can be determined by the strain (the initial stress was not considered) : {a} = [D] {e} (5.5) where [D] is the elasto-plastic matrix of the element, which is the function of stress-strain history and material
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80 properties. For elastic problem, [D] is only the function of material properties. {c) equivalent nodal force: define the nodal force which are equivalent statically to the boundary stress and distributed load on the element is ( 5 6) the distributed load {b} are defined as those acting on a unit volume of material within the element with direction corresponding to those of displacement u at that point. On the boundary where subjected to a distributed loading, say {t} per unit area, a load term on the nodes of the element which has a boundary force A8 will has to be added. The equivalent nodal force q can be formulated by virtual displacement principle which is to impose an virtual nodal displacement and to equate the external and internal work done by the various forces and stresses during that displacement. Let such a virtual displacement be Ba at the nodes.
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81 This results, by Eqs. (5.1) and (5.2), in displacement and strains within the element equal to (5.7a) and &{e} 8 = [B] &{a} 8 (5.7b) The work done by the nodal forces is equal to the sum of the products of the individual force components and corresponding displacements: ( 5. 8) The internal work per unit volume done by the stresses, distributed forces and external load is: ( 5. 9) considering Eqs. (5.3) and (5.1), above equation becomes: (5.10) where {b} is the body force. Equating the external work with the total internal work
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82 obtained by integrating over the volume and boundary surface of the element,ve and A8 have: [B] T{a}dv-J [N] eT{b}dv-J [N] sT{t}) dA vs v A., ( 5 11 ) or qe = J [B] T{a}dv-J [N] eT{b}dv-J [N] eT{t}dA yB v A (5.12) Eqs. (5.3) and (5.5), above equation can be rewritten as: ( 5 .13) where K8 = f [B] T[D] [B] dv v (5.14a) and f 8 = -J [N] eT{b}dv-J [N] sT{t}dA (5.14b)
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83 The relationship between the displacements in a solid domain and the external forces can be derived also via virtual work principle. Let {u} stands for displacement vector in the solid domain, then {u} can be expressed as {u} = N{a} (5.15) in -which {a} is the all nodal displacement vector, and (5.16) is the shape function for each element. When the point of concern is within a particular element e and i is a point associated with that element. If point i does not fall within the element, N1=0. the structure have body force {b} on each element, and external force vector (or traction) acted on some boundary surfaces of the structure is {t}, then by virtual work principle, in similar way as derive for the element but integrate over the whole solid domain, get the follow equation:
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84 f[B] T[D] [B] dv-f[NJ T{b}dv-f[NJ T{t}) dA = 0 V V A (5.17) Similarly, Eq. (5.17) can be expressed in a system of ordinary equations: .Iqa} + f = o (5.18) where K = J [B] T[D] [B] dv v (5.19a) and f = -J [N] T{b}dv-J [N] T{t}dA V A (5.19b) The K in Eq. (5.19a) is the global stiffness matrix of the structure, which can be derived by assembling !r around each node for its related elements and Eq. (5 .19b) is a force vector. In the view of mathematic, Eq. (5.17) is the Galerkin's method of weighted residual if the shape function [N] is regarded as the weighted function. Those nodes with zero displacements due to the requirement of boundary condition will be eliminated from Eq. (5.19a), while some other nodal displacements and nodal
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85 forces may be given. The first condition is displacementcontrolled or strain-controlled while the second is loadcontrolled or stress-controlled. Although in displacementcontrolled problem, displacements.on some nodes are known, as prescribed displacements, the nodal displacements of all other nodes in the structure are still to be determined. This kind of problem can be solved by "stiff spring" method. The method is to add a "large" number to the leading diagonal term, say k11 of the stiffness matrix in the row in which the prescribed value is required. The term in the same row of the right hand side vector in Eq. (5.18) is then also set to a magnitude value by multiplying k11 by u1 The modified force {f} have the nature that will produce the same prescribed displacements on those nodes. Solving Eq. (5.18): {a} = K-1 {t} (5.20) the displacement of all nodes will be find out by this way. For a load controlled problem, the load displacements are easily to be found out by using Eq. (5.20).
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86 5.3 Development of Finite Element Computer Program CSSAP A finite element computer program was written to determine the stress and strain distribution in a cylindrical specimen of soil. The computer code is named CSSAP (Cylinder Stress-Strain Analysis Program) The program was developed with Fortran 77 on a VAX mainframe computer at CU-Denver. Its use on PC may be limited by the availability of memory and the processing power of the computer. The capability of the program is outlined as follows: (a) Types of analysis: undrained and drained behavior of solid and hollow cylindrical structures with axisymmetric load, (b) Soil models: Isotropic elasticity; modified Camclay. (c) Element types: Eight-node isoparametric quadrilateral element with stress and strain determined at the centroid of element. (d) Non-liner technique: Incremental (tangent) stiffness approach. Nodal coordinates are updated as the analysis progresses.
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87 (e) Boundary conditions: Free-end or fixed-end structures. Element sides on the boundary can be given prescribed increment values of displacements, applied load such as existing hydrostatic pressure, pressure on .element sides, and torque on the cylinder top. The following aspects of the program development are described in this chapter: The cylindrical structure simulation; The boundary condition considered in this program; The strain matrix and shape function which used in CSSAP; Nodal force distribution; Incremental stiffness method to simulate the nonlinear load-displacement relationship, and The implementation of modified Cam-clay model in CSSAP.
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88 5.3.1 Formulation for Cylindrical Structures A cylinder is Considered as a body of revolution, as shown in Fig.5.1 with its longitudinal axis as z axis and polar coordinates rand 8 for defining the position of an element in the plane of a cross section. The notations for stress components are crr, cr9 cr., 'trs' 'tra, ta., respectively. The components of displacements in the radial and circumferential directions denoted by u and v and the component in the z direction by w. A Cylinder is an axisymmetric structure, both boundary conditions as well as the region geometry must be independent of the circumferential direction 8. However, each node still has three degree of freedom. Eq. (5.14a) is the expression of element stiffness in three dimensions, by changing the Cartesian coordinate system into a cylindrical coordinate system, the stiffness matrix of an element becomes: [K] = J J J [B] T [D] [B] rdrd8dz (5.21) For a cylindrical structure subjected to an axisymmetric load, one piece of cylinder with center angle equals 1 radian (57 .29) can be used to simulate the whole structure.
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z Figure 5.1 A revolution shaft body twisted by couples applied at the ends (Timoshenko and Goodier, 1970) 89
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90 Then integrate Eq. (5. 8) over one radian, above equation becomes: [K1 e = J J [B] T [D] [B] rdrdz (5.22) and the problem becomes two dimensional problem. 5.3.2 Boundary Conditions Based on the simplification of the cylindrical structure in previous section, the structure to be analyzed is a slice of the cylinder wall. Further simplification can be made by assuming the identical boundary conditions for both the upper and lower ends. Then only one half height of the slice of cylinder needs to be analyzed, as shown in Fig.5 .3. The structure, with the coordinates system originated automatically, is then discretized by eight-node isoparametric elements into mesh. The bottom line of the wall is located at the middle height of the original structure, assuming where the restraint in z direction was fixed, while the restraint in r and 8 directions were free. On the top, the movements u (r direction) equal to zero, but v, w not equal to zero for fixed end problem, and
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91 Figure 5.2 The thickness of a cylindrical element Figure 5.3 A slice of cylinder with center angle = 1 radian
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u, v, w not equal to zero for free end problem 92 Under pressures (internal pressure, external pressure and axial pressure) the slice of cylinder will have deformations in r and z directions, in other words, u and w are not equal to zero. Under only torsion, u and w, corresponding to r and z directions, are zero but v is not. Since the structure was simplified as a two-dimensional one, it needs two different global stiffness matrices to determine the three nodal displacements, u, v, and w. This requires two different boundary conditions for the two global stiffness matrices: In one of the global stiffness matrix, when only pressure loads are applied, the v displacement (9 direction) is fixed, while in another one, when only torque is applied, the u and w displacements are 'fixed. When both pressure and torque are applied on the structure simultaneously, the two global stiffness matrices need to be assembled separately. By solving the first one with Eq. ( 5. 2 0) will produce the u and w displacement values; by solving the second one with Eq.(5.20), will produce the v displacement value. In practice test, the top and bottom end surfaces of specimen will keep to a plane parallel to that of the
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93 platens of the testing machine. In order to simulating the real situation, it requires the nodes of top layer of the discretized cylinder have the same vertical movement in the calculation. If in the calculation, the top boundary is not assumed fixed in both r and e directions, which corresponding to free end case, then all top boundary nodes have the same vertical displacement. This is not true for fixed end condition, if there is no an additional restraint to make the requirement be satisfied. In the fixed end situation, the vertical stress developed in the cylinder at any layer is not uniform even if the vertical load 1s uniformly distributed. This is resulted from a moment at the top end caused by the non-uniform displacement in r direction of the wall. The use of "stiff spring" method as described in section 5.1 could impose the top end boundary nodes of the cylinder.have the same vertical displacement value. The use of "stiff spring" method requires two steps calculation: first, solving the load control problem, find the mean vertical movement value that the structure will developed under that load, regard that value as the prescribed vertical movement. Second, utilize the displacement control method, solving the problem again use
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94 the mean vertical movement value as the known boundary displacement to get the new answer, which is the one that will satisfy the boundary condition. 5.3.3 Strain Matrix and Shape Function of Element The strain matrix of an element is the key function for determining the strain at any point, when the nodal displacements of the element were known. In an element, strains may different from one point to the other, since these points may have different displacements. Let the displacements at a point inside an element be u', v', and w', then the relationship of displacements at a point with the nodal displacements of the element is: = [M {a)' (5.23) where [N] is the shape function, and {a}8 is nodal displacement vector. Recalling Eq.(5.3), strains also are the function of nodal displacements. Let [B)= [A] [N] = [B1 B2 B3 B:1
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95 where nod is the number of nodes in an element. A typical submatrix of [B] for axisymmetric loading of cylinder is given by: aNj 0 0 ar aNJ Nj 0 I I 0 0 aNj az [Bj] = (5.24) aNj aNj az 0 a I 0 aNj NJ az r N:J ( aNj N 1 ) ar I 0 then the strains at a point can be written as the function of nodal displacements: {E} = [BJ {a} 6 ) (5.25a) or the function of displacement at that point: ( 5. 25b)
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96 Since there are now three displacements per node, there are six strains of any point taken in the order: a 0 0 ar r 1 1 a 0 I raf) e a {:J ez 0 0 az {} = = Yrz a a (5.26) az 0 ar ze a a 0 1 er az raf) 1 a a 1 0 raf) ---ar r The shape function of eight-node isoparametric element which used in this analysis is 1 4 ( ( 1-'11) ( -'11-1) _! (1-'112 ) 2 .! (1+'11) 4 _! (1+'11) 2 .! ( 1 ( 1 +'11) 4 .! (1-'112 ) 2 .! ( 1 ( 1-'11) ( -'11-1) 4 (5.27)
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97 in which, are the local coordinates in the element as shown in Fig.5.4. Fig.5.4a is an eight-node element with a center angle 9 equals 1 radian. Fig. 5. 4b is the corresponding isoparametric element with normalized local coordinates. At any point j, the shape function could be found by substituting the coordinate of point into above equation, in turn, the displacements at point j could be determined if the nodal displacements were known. 5.3.4 Nodal Force Distribution Since the structure was discretized by elements, the load acted on the cylinder surfaces need to be changed into equivalent nodal forces. The force vector can be derived by Eq. (5.14b) if not considering the body force: ... (5.28) where [N]1 is the shape function of node i located on the structural boundary, df is the surface area of the element where the load is applied and {P:I} = {Pr, P1 P9}T is the load components.
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(-1.1) 3. ( 1, 1) ,..., 7 (-1,-1) (1,-1) (a) (b) Figure 5.4 (a) (a) a cylindrical element (b) Eight-node isoparameter element in coordinate 3 4 (b) (c) Figure 5.5 Load distrubution on a cylindrical element 98 (a) horizantal load (b) vertical load (c) torsioanl laod
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99 For a cylindrical structure, the surface loads are horizontal pressure on the cylindrical surface, vertical load and torsional force at the upper end. The formulation for nodal forces caused by these loads are described in section 5.3.4.1 through 5.3.4.3. 5.3.4.1 Nodal Force Caused by Horizontal Pressure: This pressure is acted along the side of the cylinder wall, it could be internal pressure p1 or external pressure Po, as shown in the Fig.5.5. From Eq. (5.27), the shape function of node 5 ls: (5.29) At node 5, then (5.30) The surface r is proportional to r, then (5.31)
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100 where L is the length of element. Thus the nodal force at node 5 is: 1 (5.32) Fs J N5 PrrdTt = r -1 At node 6, = 1, N6 = (l-Tt2 ) F6 Lp r (5. 33a) = r 3 r (5.33b) At node 7 F7 = Fs = L r r 6Prr (5.34) 5.3.4.2 Nodal Force Caused by Vertical Pressure The vertical pressure Pz is uniformly distributed on the top surface of the element as shown in Fig.5.5b. The nodal force, however, is dependent on the distance of the node to the cylinder center, because the upper area has a fan shape, thus the nodal force is determined as: (5.35
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101 5.3.4.3 Equivalent Nodal Force From Torque Shear stress will arise, when torque T is applied on the top of cylinder. The shear stress at node 3 in Fig.S.Sc is: 't'3 = 2Tr3 (5.36) while at node 5 is: 2Tr5 1t (r: rt> (5.37) This situation is different from that of vertical pressure. The magnitude of nodal force not only dependent on the distance of the node to the center but also dependent on the shear stress distribution pattern. If the shear stress is linearly proportional to the radius, the nodal forces are:
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102 Express them in torque T, have Fi T (5.38a) F: = T (rs+rJ) 31t ( 5. 3 8b) Fi = T (r5-r3 ) 31t (r:-rt> (5.38c) 5.3.5 Incremental Stiffness Method Soils have been widely recognized as nonlinear materials. The causes of nonlinear response can be identified as being either geometric non-linearity or material non-linearity. Geometric non-linea.rity arises from large structure deformations. The equilibrium equations (based on the undeformed geometry) are no longer sufficiently accurate. Material non-linearity arises when the stress-strain relation of the material is nonlinear. In general, the non-linearity of a system may be due to geometric non-linearity, material non-linearity, or the combination of both. In the analysis of the cylindrical structure in this thesis, a small strain approach is used,
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103 and the geometric nonlinearly was not considered. The program, however, updates the coordinates of all nodal points as the analysis proceeds. The influence was found negligible when the material is in a hardening stage through the calculations in this thesis. There are a number of techniques for analyzing nonlinear problems using finite element. In this analysis the incremental or tangent stiffness approach was used 1n which the total load (or displacement) imposed on a structure is divided into a number of small increments and one incremental load (or displacement) was applied at a time. During each increment the stiffness properties at the previous stress levels are calculated and used in the current increment calculation. If only a few increments are used, this method produces a stiffer response for a strain hardening model and the displacements are always under-predicted as illustrated in Fig. 5. 6. The smaller is the increment, the better is the results. In practice, some accuracy of the calculation must be sacrificed to trade for the fewer calculation time. The plastic zone will spread to the whole cylindrical structure under the given load in the calculation, the use
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104 computed curve test curve Figure 5.6 Incremental stiffness approach to simulate the non-linear load-displacement relationship
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105 of incremental stiffness method guaranteed the convergence as long as the structure is not failed. 5.3.6 Modified Cam-clay Model and its Implementation This section, deals with the procedure for the calculation of elasto-plastic compliance matrix D.P for the modified Cam-clay model. All the equations associated with the elasto-plastic matrix, D.P. have been given in Chapter 4. The modified Cam-clay has the following material parameters as described in chapter 4: V=(l+e),M, A,K, p0 and Poisson's ratio v. The load (or displacement) is to be divided into n increments. For the ith increment, the calculation procedures for all elements are: (1) From the load (or displacement) increment, determine elastic stress increments dcrije using the equation: (5.39) The total stress at ith increment is then: 1-1 Vl:f + (5.40)
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106 Calculate p, q, from f(p,q,p0 ) and find: (5.41) if p' Po, the element is in plastic stage. (2) Calculate the elastic compliance matrix D. us1ng the updated stress cri/, the volume bulk modulus K' is: I K1 = ( 1 + e) 1!_ K (5.42) elastic compliance matrix D. is given in Eq. (4.31). ( 3) From the cri/-1 obtained in last step i-1, figure out each terms of < a > aaij A= I (2 I I) -p l-K .P Po (5.43a) (5.43b) ( 4) Calculate the elasto-plastic compliance matrix DP' the equation of D.P was given in Eq. (4 .28). (5) From the strain increment corresponding to a load (or displacement) increment and DP' calculate the stress increment dcr1 / and total stress a1/, where
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and .1-1 d .1 = Ojj + Ojj 107 (5.44) (5.45) (6) Calculate the specific volume for this increment using v1 = r+ (l.-lt) [Ln2-Ln(l+ ( /M) 2 ] P' and update the void ratio e1 = V1 -1 (5.46) (7) Calculate the change of Po' from the void ratio e1 and plastic strain increment d1l using the equation (5.47) add dpo, to Po, I get
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108 5.4. Load Condition and Stress-Strain Normalization In the numerical analysis, as in test, need to set the load conditions and load magnitude for the sample according to the purpose of analysis. As the material is elastoplastic, the stress. state which the sample will reached need to be estimated before the calculation is initiated. The numerical results, also, need to be normalized in order to present them in a meaningful way. 5.4.1. Load Condition and Stress State Estimate The loads applied to the cylindrical structure usually include internal pressure p1 external pressure Po, vertical pressure Pz and torque T. In a strain control problem, vertical or torsional displacement can be prescribed. In this thesis, the load increment was prescribed. To initialize the analysis using the modified Cam-clay model requires the application of a hydrostatic pressure p', before a deviator stress is applied The loads applied on the samples are: (a) Pressures: hydrostatic pressure p' = 100 psi, that means the
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internal and external pressure Pi=Po = 0.0; vertical deviator pressure Pz=SO psi. 109 Thus, the sample was subjected to 100 psi hydrostatic pressure and 50 psi vertical deviator pressure. (b) Torque: hollow cylinder: A torque of T = 3480 lb-in. was applied to hollow cylinder with external diameter D=10 inches. Under this load, yield average torsional shear stress 'tavg = 27.2 psi and 'tmax = 3 0 psi at the outside of the cylinder wall if it is elastic condition. solid cylinder: A torque ofT= 47.12 lb-in. was applied to solid cylinder with an external diameter of 2 inches. Under this load, yield average torsional shear stress 'tavg = 22.5 psi and 'tmax = 3 0 psi at the outside of the cylinder wall if it is elastic condition. The three hollow cylindrical specimens analyzed have inner and outer radius of 4.0" and 5.0", respectively, and a half height of 2.5", 5.0" and 7.5", respectively. The solid cylinder analyzed has a radius of 1.0" and a half height 2.0". Two different load conditions were analyzed:
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110 (1) vertical deviator load alone and (2) combination of vertical deviator load and torsional load. For both load conditions, the hydrostatic pressure p' = 100 psi is included. When the hydrostatic stress acted on the sample, specific volume V changes by following Eq. (4.1). Also, Be, the void ratio change caused by the change of hydrostatic stress was considered. But the original size of the cylinder is taken as the one after completion of consolidation The original hydrostatic stress is 100 psi. When it is applied, the soil reached plastic stage, at point I in Fig.5.7. This is corresponding to the stress condition of crr=cre=crz=100_ psi. 5.4.2 Vertical Deviator Load Alone: Initial stresses are 100 psi for crr, cr8 and crz. When Pz of 50 psi is applied, Ap = 16.67 psi and Aq = 50.0 psi because Aq/ Ap =3. This deviator load allows the stress state to reach to 0.716 strength (or failure load) at point J, in Fig.5.7. The yield locus with Po' = 143.9 psi.
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q 411 ZD 4() 6o 8o failure I j'/ !; -./of/ I I /' 14$-J 16 0 //,8 Figure 5.7 The stress states of cylinders under different loads Path I-J: vertical load acted on both solid and hollow cylinders; Path I-K: torsional force acted on solid cylinder; Path I-L: torsional force acted on hollow cylinder; Path I-M: combined load acted on solid cylinder; Path I-N: combined load acted on hollow cylinder. 111
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112 5.4.3 Torque Alone: When only a torque lS applied, the stress path lS a vertical line, with = 0.0 and = Mp' = 88.8 psi when the stress condition reached the critical state. If let 'tavg = 22.5 psi, as in solid cylinder case, then = 38.97 psi. It equals 44 percent of the failure load, at point Kin Fig.5.7. This corresponds the yield locus with Po'= psi. If let 'tavg = 27.2 psi, as in hollow cylinder case, then = 47.1 psi. It equals 53 percent of the failure load, at point L in Fig. (5.7). This corresponds the yield locus with Po'= 128.13 psi. 5.4.4 Vertical Deviator Load and Torsion: Under the combination of vertical deviator load and torsion the stress state and the corresponding yield locus can only be determined. (a) Solid cylinder: Pz =50 psi and 'tavg = 22.5 psi, thus = 16.67 psi, = 63.39 psi, The stress state is located at point M in Fig.5.7 on the corresponding yield locus with Po'=160 psi.
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113 (b) Hollow cylinder: Pz = 50 psi and 'tavg = 27.2 psi, thus = 16.67 psi, = 68.9 psi, The stress state is located at point N in Fig. (5.7) the corresponding yield locus with Po'=168 From Fig.5.7, it can be seen that all the load conditions discussed so far have not reached the critical state (or failure) condition. Thus, the soil sample behavior is in stable (or hardening) stage. 5.4.5 Stress and Strain Normalization It is necessary to define the stress and strain components. There are six stress or strain components on a cubic element. In a cylindrical cube, the stress and strain can be denoted as Fig.5.8. The nature of symmetry of deformation implies tre=tar=O. If in the principal stress space, there are only three stress components. The same situation is valid for strain components. The principal stress rotation direction a is defined ln z-9 plane, as shown in Fig.5.8. For better understanding of stress and strain distribution, all stresses and strains were normalized.
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114 Figure 5.8 The stress components on an element
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115 Two methods can be used to normalize stresses: one is to includes the hydrostatic pressure and another one not. In this thesis the stress (jz, (j1 (j3 and 'tze were normalized by their corresponding values of elastic stress without end effects, and the hydrostatic pressure is included in them. Thus, the normalized values are not only dependent on the applied vertical deviator pressure Pz but also on the hydrostatic pressure. The higher is the hydrostatic pressure the lower is the deviation of the normalized value. Therefore, the drawback of this normalization method is made the non-uniformity not so obvious. But this method can show the non-uniformity scale based on total stress. The stress components (jr, (j9 and 'tzr do not included the hydrostatic pressure and are normalized by vertical deviator pressure Pz because the non-uniformity of these stresses are proportional to the vertical deviator pressure. As for strains, however, only ez was normalized by a standard value. That is = H (5.48) where Bw is the displacement of the sample top in vertical
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116 direction and H is the height of the sample. As the radial displacements, ui and U0 vary along the axial direction, the normalized E9 and Er are relative values, so are E1 and E3 because of the lack of standard values. Let II I II denote the normalized components, the definition of stress and strain normalization as: (a) stress: (J' ,_ z O'z I (J'd (crd is the vertical deviator stress) (J''9 -( O'e-
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117 (b) strain: Ez -Ez/ Ez.avg Ee Ee/Ee.avg ( Ee is tension strain) Er ,_ Er/Er.avg (Er is tension strain) Ezr '= Ezr/Ez.avg El / El.avg / .avg ( lS tension strain) in which, E1.avg and E3.avg are given by Eq. (3.25b): and (Yze) avg is the mean of shear strain of all elements. (c) principal stresses direction normalized CX.' = CX./CX.avg CX.avg is calculated by eq.(3.25c): Where all stresses are the standard values. For solid sample, tz9=22 .5 psi, O'z=150 psi, 0'9=100 psi,
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Clavg=2 0 9 go 118 For hollow sample, 'tze=27. 2 I crz=150 psi I 5.5 Presentation of Numerical Results The stress and strain distributions are presented graphically in both 2-D and 3-D figures. The 2-D figure shows the contour map of stress and strain distributions. The 3-D figure shows the elevation map of stress and strain, the contour lines in this figure are corresponding to the lines in the 2-D contour map. The grid of x and y coordinates was developed automatically, the value of z at point (x,y) is estimated by Kriging (Krige, 1960) method. If no calculated data are available, the Kriging method will process the data surrounding and near that point at a certain distance to estimate the value z at that point. In the finite element computation, stress and strain are determined at the centroid of each element. Thus, the grid data between two element centriods and the grid data from the edge of the cylinder wall to its nearby element centroid are estimated by Kriging method.
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119 6. Distributions of Stress and Strain in Solid Cylinder 6.1 Introduction The solid cylindrical sample of soil has long been used in soil strength test. The stress and strain distribution in a sample has been investigated by researchers, as described in Chapter 2, but there are some discrepancies about the stress distribution among these works. The purpose of this chapter is to present the numerical results calculated by CSSAP and to compare it with other available results. In addition, a case with the simultaneous application of vertical load and torsion was studied. The aspect ratio of solid cylinder is 2. Considering the axisymmetric nature of the solid cylinder and assuming the top end and bottom end have the same boundary condition, only one half length of a slice cylinder was calculated, as shown in Fig. 6 .la. The structure was discretized into 5*10 elements of equal size, 5 elements along the r direction and 10 elements along the Z direction, as shown in Fig.6.lb. The stress and strain are determined at the centroid of each element.
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120 T H/2 1 Figure 6.1a. One quater of solid cylinder
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121 Hl2 I I I lL_/_ FIGURE 6.1 b Finite element mesh of solid cylindrical structure
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122 Fig.6.2 and Fig.6.3 are presented the stress and strain distribution, respectively, for solid cylinder under confining pressure and deviator pressure. Fig. 6. 4 and Fig.6.5 presented the stress and strain distributions of solid cylinder under pressure and torque. 6.2 Onder Deviator Pressure Alone The axial deviator pressure Poev. = 50 psi is applied on the structure. When no torsion is applied, it is assumed that the vertical stress crz is the major principal stress and the confining pressure is the minor principal stress. 6.2.1 Stress Distributions Fig.6.2 shows the normalized stresses distribution in the solid cylinder. It can be seen when ends are fixed, there is a stress concentration in region near the ends. The length of that region is about 20% of the total length of the cylinder. At the top end, crz is smaller near center and it increases with the increase in radius, and reaches a maximum value at the outer wall. The radial stress, crr, on the contrary, is higher when near the center. The shear
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123 stress, tzr increases from zero along the center line to a maxlmum near the outer side. There is the concentration of shear stress near the upper right hand corner and it diminished rapidly to zero at outer surface of the cylinder. The circumferential stress, cr9 is higher near the top end. It rapidly decreases toward the midpoint of a sample. Any non-zero normalized circumferential stress is resulted by the end-constraint The stress distribution patterns are similar to that of Perloff and Pornbe (1969, see Fig.2.10). In their work, stresses were not normalize as they were calculated by strain control method. It is difficult to discern the region which yields the stress close to a standard value. The shear tress distribution along top surface, tzr see Fig.6.1c, similar to that of AL-Chalabi and Hung (1974, see Fig.2.11). 6.2.2 Strain Distribution In the load control situation, the strain distribution usually cannot be used as a criterion for checking the validity of the result as the strain is dependent on material properties and constitutive law. Un;Like stresses
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124 they are independent of material properties but only static equilibrium. Although with the above mentioned drawback, the strain distribution is still used as an indicator of strain uniformity in a specimen. The vertical strain, E1 from Fig.6.3a, has the minimum value at -the top end, increase rapidly toward the midheight of the specimen, and reached its maximum at center line and midheight. The strain was normalized by the average strain value of all elements, which is slightly greater than the value from formula B1/(H/2), where B1 is the vertical displacement of the cylinder at top. By finite element method, the vertical strain is where n is the load iterative number, B1 and H1 are the vertical displacement and structure height at iteration step i, respectively. From the contour line of Fig.6.3a, the strain Ez in the uniformity area is higher than the average value. This means the vertical strain calculated in a test by B1/(H/2) was underestimated by a few percentage.
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Fig.6.3b shows the Er distribution. 125 It Jumps up from zero at the top to 1.0 at about one half inch away from the top surface. The gradient of Er is steeper at outer side than that at center. Fig.6.3c shows the Ee distribution. uniformly once out of the small concentration. 6.3 Under Deviator Pressure and Torsion It becomes quite area of stress A torsion was applied at the top of the cylinder. The torque T of 47.12 lb-inch yields psi at the outer surface of the wall. The deviator pressure Pdev of 50 psi is also applied. 6.3.1 Stress Distribution The high non-uniformity of principal stresses, cr1 and cr3 see Fig. 6. 4, is caused by shear stress (or the torque), see Fig.6.4d. The difference between maximum and minimum values is 15% when 45 and 44. The difference of principal stress rotation angle, is as high as 17, see Fig. 6. 4f. The angle is measured from vertical axis.
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126 By comparing Fig.6.2b with Fig.6.4c and Fig.6.2d with Fig.6.4e, is found the torsional force have influence on crr and cr9 as the material is plastic. The reason is the soil behavior is influenced by stress history. 6.3.2 Strain Distribution The strain distributions, as the stress distributions, are non-uniform when torsion applied. Fig.6.5a shows the distribution of 1/El.avg The major principal strain increases its value from center of the cylinder to the outer surface. Fig.6.5b shows the distribution of 3 / e3.avgr which is tension strain. The highly non-uniformity of 3 is caused by torsional shear strain. As to Er and 9 even in the end-effect free region, they are not so uniform as in the non-torsion case, refer to Fig.6.3b and Fig.6.3c. The distribution of shear strain, ez9 shown in Fig.6.5d, linearly from zero to its maximum value when one moves from cylinder center to the outer surface. The solid cylinder, when used in triaxial test, could give good results in stress non-concentration region. The. length of uniformity area can be reached up to 80% of the
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127 total length. However, when torsional force acting on solid cylinder, it will put the stresses and strains nonuniformity to the whole region of the cylinder. The solid cylindrical sample, therefore, 1.s not a good one when torsion involved in the soil behavior test.
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128 top end I.IQ ,.,.. I.ZD ., ... 0.10 ...... t/1 OAO til' middle line FIGURE 6.2 a Distribution of (crz-cr 0 )/crd in solid cylinder under vertical deviator stress crd = 50 psi
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129 top end I.IIQ '"" 1..010 1.20 o.aa ..... .... ... middle line FIGORB 6.2 b Distribution of (Or-00)/0d in solid cylinder under vertical deviator stress ad = 50 psi
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130 top end 1.60 1.20 LaO 0 .110 o.ao 0.40 0.40 middle line FIGURE 6.2 c Distribution of tzrlad in solid cylinder under vertical deviator stress ad = 50 psi
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131 top end 1.20 o.oo o.oo 0.60 0,60 middle line FIGURB 6.2 d Distribution of (a8-00)/0d in solid cylinder under vertical deviator stress ad = 50 psi
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132 040 0.40 o ""o ... .=-...._...._..__..__..__..__..__..&....-I,J1/Jom middle lir"e FIGURE 6.3 a Distribution of Ez/Ez.avg in solid cylinder under vertical deviator stress = 50 psi
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top end t.ea 1..30 0.80 0.60 middle line FIGURE 6.3 b Distribution of Er/Er.avg in solid cylinder under vertical deviator stress crd = 50 psi 133
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top end 0 .00 1 .00 O.J ..... 8 .. f\( 1.20 .... m iddle line FIGt:JRB 6.3 c Dis.tribution of 9/e9.avq in solid cylinder under vertical deviator stress crd = 50 psi 134 1 .00 1 .00 ..... 1 .20 O .ID ....
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top end 1.10 1.20 0.10 0.40 middle line FIGORB. 6.4 a Distribution of cr1/0't.avq in solid cylinder under vertical deviator stress crd = 50 psi, average t=9 = 22.5 psi. 135 1.10 1.20
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136 top end ,. ""' "" # .. ,, 7;,P' 1.10 _. (I I ; ( I ...:I # 1 { r 1 \ f\l o.ao [ 1 E l I 11M ll !I o.ooo.oo middle line PIGORB 6.4 b Distribution of 03/aJ.avq in solid cylinder under vertical deviator stress ad = 50 psi, average tze = 22.5 psi. ..... ,...., ...... 11M
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137 top end UD I.IID .. ,. 1..211 1.20 I. ., 0.811 \ O.IID t OAD 0.40 lf O.DO O.DO O.DO I.DO middle line FIGORB 6.4 c Distribution of (O'r-0'0 ) /O'd in solid cylinder under vertical deviator stress crd = 50 psi, average td = 22.5 psi.
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top end ,_ ...... 1'\," -.. middle FIGORB 6.4 d Distribution of (C19-C10 ) /C1d in solid cylinder under vertical deviator stress ad = 50 psi, average tze = 22.5 psi. \ \ I line 138 .... I.ZO .... 0.40 o.oo 1.00
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..... top end ..... T J I I 1.... .. ,..,. :1 a :1 :1 .a -0.411 0.00 ::z c : 0.1111 middle line PIGURB 6.4 e Distribution of tz91tavq in solid cylinder under vertical deviator stress ad = 50 psi, average = 22.5 psi. 139 '"" r :LaD I .. :! 1.lll ua ; 1J:JIIJD.r:JO
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140 to p end 1.111 1.2<1 o.aa middle line PIGORB 6.4 f Distribution of a!aavq in solid cylinder under vertical deviator stress crd = 50 psi, average 'tze = 22.5 psi. cx.vg = 20.99 l.lll o.aa 0.411
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141 top end .. t. ... <19
PAGE 160
top end middle line FIGURE 6.5 b Distribution of E3/E3.avv in solid cylinder under vertical deviator stress crd = 50 psi, average tri = 22.5 psi. 142 1AI
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143 8
PAGE 162
top end ta "" ... ___ ... LI!IIO ---.::-... ,.,. ..... I o.oo ( .. .. .., \ t:. ..... .... middle line 'ee FIGURE 6.5 d Distribution of E9/Ee.avq in solid cylinder under vertical deviator stress crd = 50 psi, average 'tze = 22.5 psi. 144 .oa . oc 1.00 1.20 o.oo ...... o.oc ,.co
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top end :1 I 1.110 I ,; !] a ;; 1.20 I I : ; !! c .. 0.80 \ I e !! a : 0.40 I :1 : : -;; -middle line FIGURE 6.5 e Distribution of Eze/Eze.avg in solid cylinder under vertical deviator stress crd = 50 psi, average = 22.5 psi. 145 1 ... :; :! 1.l0 :; 0.80 :! 0.40 .. .. ;.
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146 7. Distributions of Stress and Strain in Hollow Cylinders 7.1 Introduction Three hollow cylinders with height H/2=2.5", H/2=5.0" and H/2=7.5" were analyzed. They were subjected to deviator pressure, and deviator pressure combined with torsional force, respectively. Considering the axisymmetric nature of the hollow cylinder and assuming the top end and bottom end have the same boundary condition, only one half length of a slice of cylinder is needed in the analysis, as shown in Fig.7.1a. The structure has been divided into 5 by 10 elements of equal size, the stress and strain are determined at the centroid of each element, as shown in For the case of axial deviator pressure alone, the st-resses crz, crr, cr9 and tzr are presented; for the case of deviator pressure and torsional force cr1 cr3 and crr=cr2 and torsional shear stress tze are presented. The strain
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147 T H/2 1 Figure 7.1a.one quater of hollow cylinder
PAGE 166
148 -----T H/2 l FIGURE 7.1 b Finite element mesh of hollow cylindrical structure
PAGE 167
149 distributions, however, are only presented for the case of H/2=5.0". 7.2 Hollow Cylinder with H/2 = 2.511 This is the shortest cylinder used 1n the The end constraint effect is very obviously. 7.2.1 Under Deviator Pressure Alone Figs.7.2 show the distribution of normalized stresses under deviator pressure alone. Fig.7.2a shows the distribution of (crz-cr 0)/crd. The length of stress concentration region is smaller than the wall thickness. The stress distribution near top end is quite nonuniform. It increases from the inner wall toward the outer surface. In the region from approximately 0.5 inch from the top to the middle line, the stress distribution is quite uniform, although stresses at inner wall is greater than that at outer wall, and the difference is about %.
PAGE 168
150 Fig.7.2b shows the distribution of (crr-cr0)/crd. Stress Concentration region located at top, the stress caused by end restraint could be as high as 20% of the deviator vertical load. The contour line with 0.0 value stand for where crr equal to the hydrostatic stre'ss. The region from middle line to 2.00" height could be called as end-effect free region, where the stress caused by end restraint is negligible. Fig. 7. 2c shows the distribution of (cr9-cr0 ) /crd. The end effect has more disturbance on cr9 The stress induced by end restraint is as high as 30% of the vertical deviator load at the top of wall. There is no end-effect free. region in the whole cylinder. Fig. 7. 2d shows the distribution of 'tzrlcrd. The stress concentration region still is very large, only about one inch height area, from middle line to 1. 0 0" height, the stress is less than 5% of the vertical deviator load. The stress distribution at the top of the wall shaped like a hill with peak near outer side edge of the wall.
PAGE 169
151 7.2.2 Under Deviator Pressure and Torsion When torsional force acting, the stress distribution changed somehow, as shown in Figs.7.3. Fig. 7. 3a shows the distribution of cr1 /crl.avg. The stress distribution pattern is similar to that of the crz, but at the top of the wall, its deviation from standard value is much higher. In the region from middle line to 1.5" height, the stress could regarded as uniformly distributed, better than the case of crz in Fig. 7. 2a, because torsional shear stress offset the deviation. Fig. 7. 3b shows the distribution of cr3 /cr 3.avg. in the wall the stress is uniformly distributed. Nowhere This situation could attributed to the non-uniformity of cr9 and torsional shear stress. Fig.7.3c shows the distribution of (crr-cr 0)/crd. The existence of torsional force almost no influence on the distribution of crr, since the torsional force do not cause the structure deform in r direction. The same situation is apply to cr9 The influence of torsional force on crr and cr9 is the change of stress state in each element, which in turn, influenced the value of crr and cr9 only if the material is in plastic stage.
PAGE 170
* Fig. 7. 3d shows the distribution of 'tzal'tavg. 152 The difference of stress distribution between the inner wall and the outer surface is about 15%, The highest stress deviation is 8%. Fig.7.3e shows the distribution of a. The rotation angle was measured starting from vertical direction for all cases when torsion involved. The principal stress rotation angle, a, is influenced by both the end restraint and wall curvature. At top and inner corner of the wall, the angle have highest value while at middle line and inner corner of the wall have lowest value. In the low.er portion, from middle line to 1. 5" height, the difference of angle is less than 10% per cent. 7.3 Hollow Cylinder with H/2 = 5.0 .. The increase of cylinder length, from H/2=2. 5" to H/2=5.0", increased the end-effect free region and lessened the stress deviation even at a same distance from the top end of cylinder wall. 7.3.1 Under Deviator Pressure Alone
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153 Stress: Figs.7.4 show the distribution of normalized stresses. Fig.7.4a shows the distribution of (crz-cr 0 }/crd. The distribution of crz in the region from middle line to 4.0" height is quite uniform, the stress deviation in this area 1s only 1%. Fig.7.4b shows the distribution of (crr-cr0)/crd. More than 4 inches length of the cylinder start from the middle line of the wall is end-effect free region. Fig.7.4c shows the distribution of (cr9-cr0)/crd. The end effect still has more disturbance on the distribution of cr9 but in this case, there is a 2 inches length of region start from the middle line of the wall free from the end effect. Fig. 7. 4d shows the distribution of 'tzrlcrd. In the upper part of the cylinder, 'tzr' which caused by end restraint, is exist. From middle line to 3.0" height, the end effect is very small, but the shear stress changed sign and extended down to one more inch. Strain: Figs.7.6 show the distribution of normalized strains.
PAGE 172
* Fig. 7. 6a shows the distribution of tz/ ez.avg. 154 The distribution of vertical strain has the same form of the vertical stress crz. In the end-effect free area, its value is higher than the average one. Fig. 7. 6b shows the distribution of Er/ er.avg. It is tension strain. The highly non-uniformity of strain is located at the top end area, the length of that area is about one inch. The strain distribution is very uniform out of that top end area. Fig. 7. 6c shows the distribution of 9/Ee.avg This tension strain smoothly increases from zero at top of the wall to maximum value at center portion. 7.3.2. Under Deviator Pressure and Torsion Stress: Figs.7.5 shows the distribution of normalized stresses. Fig. 7. Sa shows the distribution of cr1/crl.avg. In the end-effect free region the deviation of cr1 from standard value is 1%, which is caused by torsional shear stress. case, Fig. 7. Sb shows the distribution of cr3/cr3.avg. In this the lower half area of the wall could be regarded as
PAGE 173
end-effect free region. 155 The deviation of stress in that area is 3%, which is caused by torsional shear stress. Fig.7.5c shows the distribution of 'tzel'tavg The difference of stresses distribution between the inner wall and the outer surface is about 15%, The highest stress deviation is 8%. The reason of contour lines are not straight, as they supposed should be, the stress condition 1.n the elements located on same r cannot be exactly the same due to end restraint and error arose from iterative calculation. Fig.7.5e shows the distribution of a. The rotation angle distribution also can be divided into two parts: (1) the stress concentration region, where the angle decreased when one moves away from the top end; (2) in the end-effect free region, the angle decreased when it moves from the outer surface to the inner of _wall, which is caused by torsional shear stress. The deviation value is less than 4%. Strain: Figs.7.7 show the distribution of normalized strains. Fig. 7. 7a shows the distribution of 1/Et.avg The end effect can be seen clearly from the upper part of the wall.
PAGE 174
156 The curvature effect also can be observed from the strain distribution: the value at inner wall is lower than that of outer surface. The deviation of strain is much notable in the end effect region than in the other regions. Fig. 7. 7b shows the distribution of E3/E3.avg It is strain. The deviation of strain distribution is higher than the situation of E1 Fig. 7. 7c shows the distribution of Ez9/Eze.avg The end constraint has no influence on the torsional shear strain. Its distribution, as expected, is inqreases from inner wall toward outer surface of the wall. The deviation of the normalized strain at inner or outer surface from the average strain is about 7%. 7.4 Hollow Cylinder with H/2 = 7.sn This cylinder is the longest one in the analysis. The end effect is not so obviously as in the two previous cases. 7.4.1 Onder Deviator Pressure Alone
PAGE 175
157 The stress distribution is like the case of H/2=5. 0" but have longer end-effect free region -as the result of length increased. See Figs.7.8. 7.4.2 Under Deviator Pressure and Torsion The stress distribution is like the case of.H/2=5.0" but have longer end-effect free region as the result of length increased. See Fig.7.9. 7.5 Effect of Torsion Resisting Blades The height of all blades are assumed equal to 0. 5", that is the height of one layer of element. The blades located at the top and bottom pressure plates of the machine and contacted with the specimen by wedging the blades into the soil sample so torsional force can be transferred from the machine to the specimen when the top plate turning horizontally to simulating torsional situation. In this computation, let the blades, as simulated by the top layer of elements, rotated in e direction rigidly, to let the sample been twisted on the end.
PAGE 176
158 The stress distributions cr1 cr3 crr are presented in Figs.7.10. The strain distributions 1 and 3 are presented in Figs.7.11. The effect of blade on the stress and strain distribution is local. In Fig. 7 .lOa and Fig(? .lOb the principal stresses, cr1 and cr3 in the blade area equal to O'z and cr9 respectively, because the torsional stress tze is zero since there is no torsional strain in the blade area. The blade no effect on the stress crr. In Fig.7.lla and Fig.7.llb the principal strains, 1 and 3 in the blade area equal to ez and e9 respectively, because the torsional strain Yze is zero. In other area, the stress and strain distributions are same as the case when no blade effect exist.
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159 top end l.QQ :Z.OO t.so /' 1.110 ., ( 1.00 ( 1.00 i I 1 o.so .. I o.oo o!r.._. 4.00 s.ooo.oo middle line PIGORB 7.2 a Distribution of (az-a0 ) Ja*
*
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160 top end 2.00 2.00 ,,,. 1.00 1.00 0.50 0.50 0.00 1....-i--1--i--.L-.l-.l--.l--.J.......J 0.00 4.00 5.00 middle line FIGUP.B 7.2 b Distribution of (ar-a0 ) /ad in hollow cylinder (H/2=2. 5") under vertical deviator stress ad = 50 psi
PAGE 179
JP ,. FIGURE 7.2 c 2.00 1..5a 1.00 0.:10 o.oa 4.00 161 top end .. .. Q.l .---2.00 .,:-"' 1.511 Q, 1.00 I / ... 0.511 Q' middle line Distribution of in hollow cylinder (H/2=2.5n) under vertical deviator stress ad = SO psi
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162 top end 2.00 l.QO 1.00 1.00 o.50 a.oo '---''---'---'---'---1.---I......J........:..--1 a. a a .ao 5.oo middle line FIGtJRB 7.2 d Distribution of t:rlad in hollow cylinder (H/2=2. 5 ) under vertical deviator stress ad = SO psi
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163 top end 2 .00 1.50 1.00 '\ -v 0 .50 13 middle line FIGORB 7.3 a Distribution of
PAGE 182
164 top end 2 .00 1.50 1.00 middle line FIGORB 7.3 b Distribution of rJ3/rJ3.avq in hollow cylinder (H/2=2. 5 n) under vertical deviator stress rJd = 50 psi, average tri = 27.2 psi. 1.50 1.00 0.50
PAGE 183
165 top end 1..511 0 II ,. 1.00 1.00 0.50 0.:50 0.00 0.00 4.00 5.00 middle line PIGURB 7.3 c Distribution of (Gr-G0)/Gd in hollow cylinder (H/2=2.5n) under vertical deviator stress Gd = 50 psi, average tri = 27.2 psi.
PAGE 184
166 tcp end :a.oo ..... oo o.ao middle line FIGURE 7.3 d Distribution of tzeltze.avg in hollow cylinder (H/2=2. 5") under vertical deviator stress ad = 50 psi, average tze = 27.2 psi. :a.oo ---Q ....
PAGE 185
167 top end middle line FIGURE 7.3 e Distribution of in hollow cylinder (H/2=2.5n) under vertical deviator stress aa = 50 psi, average t:e = 27.2 psi. aavq = 23.71 o.
PAGE 186
168 2.00 2.00 1.00 1.00 0.00 0.00 4.00 5.00 middle line .FIGOR.B 7 4 a Distribution of (az-00)/ad in hollow cylinder (H/2=5.0") under vertical deviator stress ad = 50 psi
PAGE 187
169 top end 4.00 5.00 ,.,...,....,lr"T17.,....,...,...,.,_, 4.00 4.00 J.OO J.OO 2.00 2.00 1.00 1.00 0.00 o.oo 4.00 5.00 middle line FIG'O'RB 7.4 b Distribution of in hollow cylinder (H/2=5. 0") under vertical deviator stress = 50 psi
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170 top end 4.00 5.00 J.CO 3.00 2.00 2.00 1.00 1.00 0.00 0.00 4.00 5.00 middle line FIGtJRB 7.4 c Distribution of (09-00)/0d in hollow cylinder under vertical deviator stress ad = 50 psi
PAGE 189
171 top end 4 .00 5.00 4.00 4 .00 3.00 J.OO 2.00 2.00 1.00 1.00 0.00 0.00 4.00 5.00 middle line FIGORB 7.4 d Distribution of in hollow cylinder {H/2=5. 0 n) under vertical deviator stress = 50 psi
PAGE 190
172 top end 4 co 5.00 5.00 5.00 2.00 2.00 1.00 1.00 0.00 0 00 4.00 5.00 middle l.ine PIGtJRB 7.5 a Distribution of G1/Gl.avg irt hollow cylinder (H/2=5. 0 n) under vertical deviator stress ad = 50 psi, average tze = 27.2 psi.
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173 PIGUl\B 7 5 b Distribution of a3/03. avQ in hollow cylinder (H/2=5. 0 ) under vertical deviator stress ad = 50 psi, average = 27.2 psi.
PAGE 192
174 4.00 4.00 3.00 3.00 2.00 2.00 1.00 1.00 0.00 0.00 4.00 s.oo middle line PIGORB 7.5 c Distribution of (ar-a0 }/Gd in hollow cylinder under vertical deviator stress ad = 50 psi, average = 27.2 psi.
PAGE 193
175 top end 4 .00 s. 00 4.00 4.00 J.OO 3.00 2.00 2 .00 1 .00 1.00 0.00 0.00 4 .00 5.00 middle line FIGORB 7.5 d Distribution of tz9/tza.avg in hollow cylinder {H/2=5. 0 n) under vertical deviator stress crd = 50 psi, average. tze = 27.2 psi.
PAGE 194
176 top end + .00 5.00 5.00 5.00 4 .00 4.00 J.OO 3.00 2.00 2.00 1.00 1.00 0.00 L.......lt-1.'-.&....ll....J.....u...I.-L.....J 0.00 +.00 5.00 middle line FIGURE 7.5 e Distribution of a/fLa.vq in hollow cylinder (H/2 =5. 0 n) under vertical deviator stress ad = 50 psi, average t:e = 27.2 psi. aavq = 23.71 o.
PAGE 195
177 top end s.oo 4.00 4.00 :l.OO J.OO 2.00 2.00 1.00 1.00 o.oo 0.00 4.00 5.00 middle line PIG'ORB 7.6 a Distribution of E:IE:.avq. in hollow cylinder (H/2=5. 0") under vertical deviator stress ad = 50 psi
PAGE 196
Fl:G'O'RB 7.6 b 4.00 J.OO 2.00 1.00 178 top end 5.00 5.00 4.00 J.OO 2.00 1.00 0.00 0.00 4.00 5.00 middle line Distribution of Er/Er.avq in hollow cylinder (H/2=5. 0 n) under vertical deviator stress crd = 50 psi
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179 top end 4.00 5.00 5.00 1 5.00 0.00 L.....J.....J......I....J.....L.....L...J.......I...J 0.00 4.00 5.00 middle line FIGO'RB 7.6 c Distribution of 9/Ee.avg in hollow cylinder (H/2=5. 0.") under vertical deviator stress crd = 50 psi
PAGE 198
180 2.00 2.00 1.00 1.00 0.00 0.00 4.00 5.00 middle line FIG'O"RB 7.7 a Distribution of 1 / Et.avq in hollow cylinder (H/2=5. 0 n) under vertical deviator stress Gd = 50 psi average = 27.2 psi.
PAGE 199
PIGORB 7.7 b 181 4.00 4.00 J.OO J.OO 2.00 2.00 1.00 1.00 0.00 0.00 4.00 s.oo middle line Distribution of /./lVq in hollow cylinder (H/2=5. 0 a) under vertical deviator stress crd = 50 psi average tri = 27.2 psi.
PAGE 200
ciP .... top 4.00 5.00 4.00 3.00 2.00 1.oo d 182 end 4.00 3.00 2.00 i ,.00 L......L. ........ ...................... o.oo 5.00 line FIGURE 7.7 c distribution of E::eiEze.avq in hollow cylinder (H/2=5. 0") under vertical deviator stress crd = 50 psi, average = 27.2 psi.
PAGE 201
183 0.00 0.00 4.00 5.00 middle line PIGURB 7.8 a Distribution of (
PAGE 202
184 e.oo e.oo .eo 4-.50 3.00 3.00 1.eo 1.50 0.00 0.00 4.00 s.oo middle line FIGORB 7.8 b Distribution of (crr-cr 0 )/crd in hollow cylinder (H/2=7.5") under vertical deviator stress crd = 50 psi
PAGE 203
185 o.co o .oo 4.00 s.oo middle line FIGORB 7.8 c Distribution of (cr8 -cr 0 )/crd in hollow cylinder (H/2=7.5) under vertical deviator stress crd = 50 psi
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186 top end 4.CO 5.00 7.50 7.50 5.00 6.00 I.
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187 s.oo 6.00 4 .50 4 .50 J.OO 3.00 1.50 1.!50 0.00 0.00 4.00 5.00 middle line FIG'O'U 7.9 a Distribution of (!1/at.avq in hollow cylinder (H/2=7. 5") under vertical deviator stress ad = 50 psi, average = 27.2 psi.
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188 top end 4 .00 5.00 7 .50 7 .50 6 .00 6 .00 4 .50 4.50 .J.OO .J.OO 1.50 1.50 0.00 0 .00 4 .00 5.00 middle line PIGORB 7.9 b Distribution of in hollow cylinder (H/2=7. 5 n) under vertical deviator stress = 50 psi, average t::e = 27.2 psi.
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189 .:s.oo 3.00 1.50 1.50 o.oo 0.00 +.00 5.00 middle line FIGURE 7.9 c Distribution of (ar-a0)/ad in hollow cylinder (H/2=7.5a) under vertical deviator stress ad = 50 psi, td = 27.2 psi.
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PIGtrRB 7.9 d 3.00 .oo 0.00 -4.00 r-r1iddle 190 3.00 .00 0.00 0.00 line Distribution of tz9/tze.avq in hollow cylinder {H/2=7. 5") under vertical deviator stress Gd = 50 psi, average tze = 27.2 psi.
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191 top end 4.00 5.00 7 .50 7.50 4.50 4.50 l)oo" 3 .00 3.00 "' Ill "" d 1 .50 1.50 0.00 LI..J"-L..I...u...L..L..J...J 0.00 4.00 5.00 middle line FIGURE 7.9 e Distribution of a/aavq in hollow cylinder (H/2 =7. 5") under vertical deviator stress ad = 50 psi, average tze = 27.2 psi. aavq = 23.71.
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192 top end 4 .00 5 .00 5.00 4.CO I I 4-.00 : "-o -j () J.OO J.CO 2.00 1.00 1.00 0.00 '--'--..1..-l...:..:.,_:.......u......_,_. 0.00 4.00 5.00 middle line FIGORB 7.10 a (blade considered) Distribution of 01/0t.avq in hollow cylinder (H/2=5. 0 n) under vertical deviator stress ad = 50 psi, average = 34.86 psi.
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193 top 5 4 ,00 4.00 3 .00 3 .00 2 .00 2.00 1.00 .00 0 .00 0.00 4.00 5 .00 middle line PIGORB 7.10 b (blade considered) Distribution of CJ3/CJ3. avq in hollow cylinder (H/2=5. 0") under vertical deviator stress ad = 50 psi, average tza = 34.86 psi.
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194 4.00 4.00 J.OO J.OO 2.00 2.00 1.00 1.00 0.00 0.00 4.00 5.00 middle line FIGORB 7.10 c (blade considered) Distribution of (ar-a0)/ad in hollow cylinder (H/2=5.0) under vertical deviator stress ad = 50 psi, average = 34.86 psi.
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195 4 .00 3 .00 J .OO 2 .00 2.00 1.00 1.00 0.00 0.00 4.00 5.00 middle line FIGURE 7.11 a (blade Distribution of E1/Et.avq in hollow cylinder (H/2=5. 0 n) under vertical deviator stress ad = 50 psi average tri = 34.86 psi.
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196 top end 4.00 5.00 5.00 f'Z , ::::::::::;' I 3 .. m
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197 8. Comparison of Analysis Results 8.1 Introduction The results from the numerical analysis are including solid and hollow cylinder's stress and strain distributions. The discussion about that of solid cylinder has been given 1n Chapter 6. In this chapter, only the results of hollow cylinders were compared, to investigating the suitable height of sample. The term 'suitable height' here means, with the height of sample, the end effect is nearly negligible in a certain part of the sample. That part provides enough room for instrumentation for measuring stresses and strains. Meanwhile, the total height of the sample should be optimized to. In the following discussion, different load conditions are considered separately. The normalized stress and strain values which defined in Section 5.3 are used here again for comparison purpose. 8.2 Under Deviator Pressure Alone: Fig.S.l shows the (az-a0)/ad contour line maps of the
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198 three hollow cylinders. In all the three cylinders, there are stress concentration region, start from top end of the wall to about 1.0 inch below, in other words, the length of nonuniform region equals to the wall thickness. The length of cylinder has influence on the stress distribution even beyond the stress concentration region. The cylinder with H/2=2.5 inch has more nonuniform area than the others. The cylinders with H/2=5. 0 inch and 7. 5 inch only have 1% deviation from the standard value in stress nonconcentration region. Fig. 8. 2 shows the contour line maps of (crr-0'0 ) /crd. The stress concentration regJ.on J.S same for the three cylinders. The stress distribution in non-concentration region is uniform. Fig.8.3 shows the contour line maps of (0'9-0'0)/crd. It doesn't like other stress distributions, the stress cr9 -induced by vertical load-decayed more slowly. It spreads about 3 inches down from the top end of the wall. The shortest cylinder shows greater value of induced stress than the longer one does. In region 2, the contour lines inclined down from outer surface to the inner wall, which indicate the deformation of cylinder wall shaped an outward
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199 curve if viewing in r-z plane. Fig. 8. 4 shows the contour line maps of tzr/crd. The shear stress developed in the cylinder stretched down from top end of the wall about 2 inches. Below that the stress changed sign and vanished in a small area. There are stress free region in the two longer cylinders while for H/2=2.5 inch shear stress covered whole cylinder. 8.3 Under Deviator Pressure and Torsion Principal stresses have been used here to present the distribution. Assuming O'r is the secondary principal stress. 0'1 and 0'3 were calculated by Eq. (3. 25a) Fig. 8. 5 shows the contour line maps of 0'1/0'1.avg First, study Fig.8.5(3), it is noted the contour line can be grouped into 3 regions: region 1 is the stress concentration area; in region 2, the stress is uniformly distributed and its value equal to that of the idealized situation. In region 3, the stress increased smoothly from the inner wall to outer surface of cylinder wall. In actually, the stress in region 3 reflected the end effect free condition, where 0'11 and 0'9 are also equal to the idealized condition stresses, that is end constraint free.
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200 Shear stress tse caused cr1 increase from the inner wall to outer surface. While in region 2, the non-uniformity of crz and cr9 offset the non-uniformity of tze' because in region 2, crz and cr9 are little higher near inner wall than that of outer surface, but tze have the contrary distribution. In Fig.8.5(1), there is almost no region 2. In Fig.8.5(3) region 3 is very short. Fig. 8. 6 shows the contour line maps of cr3/cr3.avg. The distribution of cr3 is not so uniform as cr1 but the difference of stresses distribution between the inner wall and outer surface is about 5% when one moves out of stress concentration region. The shortest cylinder, however, the value of cr3 is greater than the standard value in whole cylinder. The another two cylinders left some room for good values, in there the difference from standard value is very small. crr: The torsional force added on the structure doesn't change the distribution of crr. Its distribution is the same as axial deviator pressure acting alone. tzr= The end effect of the sample on the shear stress distribution caused by torsional force is relatively small if the cylinder's deformation in r direction is not so
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201 significant that let the geometric nonlinear problem occur. Refer to Fig.7.3d, 7.5d, and 7.9d. The influence of cylinder height on the shear stress distribution is also not so obvious. It is clearly showed for elastic-plastic material the distribution of torsional shear stress give a more favorite condition than elastic material does. For the given cylinders in this calculation, for instance, the shear stress at outer side of wall is 16% higher than that at inner side, while for elastic material, it will be 22% since from Eq. (3.15) where Rr = rJro = 0. 8 a: The principal stress direction varies from outer wall to inner wall because of the torsional shear stress The end effect region is the same for all the three cylinders but the longer is the cylinder the longer is the end-effect free region. 8.4 Torsional Resistant Blade Effect. The torsional resistant blade only have local effect on I
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202 the stress distributions. Fig.8.7 is the comparison of cr1 and cr3 distribution of two kind of condition. When blade effect is considered, only the stresses near top end of the wall, where blade located, are different with each other. 8.5 Strain Distributions Strain is highly material dependent, it is hard to find a standard value to check the distribution conditions, however, from the discussion about stress distribution, it can be concluded the region with spare contour lines could be considered as a strain uniformly distributed region. From Fig. 7. 6 and Fig. 7. 7, it can be seen the strain distribution similar to stress distribution, have strain concentration region and uniformity region. For the cylinder with height H/2=5.0 inch, about half length of it provided almost uniformly distributed strain region. 8.6 Recommendation of the Sample Dimensions Based on the above comparison, it can be seen that the length of cylinder is an important factor to minimize the end effect. The one with H/2=5.0 inch and H/2=7.5 inch could offer some room where the end effect is negligible.
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203 But the one with H/2=2.5 inch cannot, in most situations, offer enough end-effect free area. In test, the selection of cylinder length also dependent on other considerations besides the end effect. For instance, if wish to measure the vertical strain, horizontal displacement and pore pressure, the instruments need to be installed at inner side or outer side of the cylinder wall, or even into the wall. If the one with H/2=5.0 inch can give enough room for those instruments, therefore, it is. suggested here, don't use the longer one. Because the longer cylinder will increase the possibility of buckling. In addition, tall sample also will cause inconvenience of sample preparation. The thickness of cylinder wall in the analysis of this thesis is 1.00 inch, the ratio of inner radius and outer radius is 0.8, as suggested by Hight (Hight, 1983). It was found from the calculation that both the thickness of wall and the radii ratio are good under the load condition adapted in this thesis.
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204 top end 4.00 5 .00 7 .50 7 .50 6.00 6.00 4.50 4.50 3.00 3.00 2.00 2.00 .. 1.50 1.50 I .GO 1 .00 .. 0.00 0.00 o.oo o..oa 4.00 5.00 4.00 5.00 middle line .. middle line ... middle lin1 (a) H/2=2 .5 (b) H/2:5.0" (3) H/2:7.5' PIGtJ'lUI 8.1 Distribution of (az-ao) /ad in hollow cylinder under vertical deviator stress ac1 = 50 psi
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,. .. .. .. 205 top end 4.00 5.00 7.50 7.50 s.oo s.oo 4.50 4.50 J.OO J.OO 2.00 1.50 1.SO 1.110 .. 0.00 0.00 0.110 Q.OO 4.00 5.00 .ao .. middle line middle line .. middle (a) H/2=2.5" (b) H/2=5.0" (3) H/2:7.5 PIGORB 8.2 Distribution of (ar-a0)/ad in hollow cylinder under vertical deviator stress ad = SO psi
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tal H/2=2 .5 end s.aao.aa line (bl H/2:5.0" PIGORB 8.3 206 8.00 8.00 4-.SO 4-.50 J .OO J.OO 1.50 1.50 0.00 0.00 4.00 5.00 middle line (3) H/2:7.5 Distribution of (0'9-0'0 ) /ad in hollow cylinder under vertical deviator stress ad = 50 psi
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207 top end 4.00 5.00 7.50 7.50 e.oo e.oo 4.50 4.50 J.OO J.OO 2..1111 1.50 1.50 1..00 1..00 0.00 0.00 a.oo a.oo 4.00 S.CICI 4.00 5.00 .. .. -.. middle lin middle line middle line (a) H/2=2.5 Cbl H/2=5. 0 (3) H/2:7.5 PIGCJlUI 8., Distribution of 't:riO'i in hollow cylinder under vertical deviator stress O'd =50 psi.
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208 top end 4.00 5.00 7.50 t K 11 as"' "'"' 7.50 8.00 8.00 4.50 4.50 .ao ... OQ tapO'!CI .J.OO J.OO 3.00 3.00 I r t.._i 1M\ 1.50 11 1.50 . -J( .. .. i .. 1 r J -1 0.00 P'r I tr I II n 0.00 4,QG S.OQ 4.00 5.00 .. .. middle line middle line .. .. m iddle GM (al H/2=2.5" (b) H/2:5.0 (3) H/2:7.5 FIGtJRB 8.5 Distribution of r11/a1 ., in hollow cylinder under vertical deviator stress ad = SO psi, average = 2 7 2 psi
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.. .. .. .. .. middle Une top end 5.110 .oo. 2..00 1..00 a..aa 1.00 o..oa .00 !.OG middle line 209 1.50 1.50 0.00 0.00 4.00 5.00 middle line
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210 top end top end 4 .00 5.00 4.00 5.00 .00 s.oo s.oo s.oo .co 4,00 4.00 4 .00 4.00 .oo J.OO 3.00 3.00 3.00 3.00 3.00 .00 2.00 2.00 %.00 2.00 2.00 2.00 .00 1.00 1.00 1.00 1.00 1.00 1.00 ,,oo a.oo 4 00 5.00 0.00 0.00 4.00 5.00 o.oo 0.00 uo 5.00 0.00 0.00 4.00 5.00 middle line middle line middle line middle line :10 blade with blade no blade with blade (a) a1 PIGtJRB 8.7 Distribution of G1/GLavq and G3/Gl.avq in hollow cylinder (H/2=5.0), with or without blades. Veretical deviator stress plus torque.
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-.... 211 top end 4.00 5.00 7.50 7.50 8.00 8.00 4.50 4.50 4.00 4.1l0 top enc:l J.OO J.OO .1.00 ;s.ao ... 2.00 :Lao ... \ 1.50 1.50 \\ i .. \\\\\ 1.00 1.ao 0.00 0.00 I I --Q.OO o.ao 4.00 5.00 middle line .a a s.ao middle line middle line (a! H/2=2. 5 (bl H/2=5.0" (3! H/2=7.5" PIGURB 8.8 Distribution of in hollow cylinder under vertical deviator stress ad = 50 psi, average 't:e = 27.2 psi. a:svq = 23.71 o.
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212 9. Summary and Discussions The stress and strain distributions of cylindrical soil samples, a solid cylinder with aspect ratio equal of 2 and three hollow cylindrical samples with the height of 5,10 and 15 inches, were analyzed using Finite Element Method. Small strain assumption was followed, and drained condition of soil was used. Modified Cam-clay was adopted as the material's stressstrain model. Strain softening feature was not considered. The analysis is limited in the stable stage of stressstrain curve. The loads, including axial load and torsional force, are monotonically applied on the structure separately or combined. The stress and strain distributions of cylindrical sample were presented in 2-D and 3-D graphics.
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213 The FEM is a good tool to investigate the stress and strain distributions in the samples. The end effect, the curvature influence of the wall and the nonlinearity of material all have been displayed in this analysis, those are hardly been done so comprehensively neither by experimental nor analytical method. Based_on the results presented in Chapter 6 and Chapter 7, following conclusions can be drawn. 1. The stress and strain non-uniformity caused_ by two factors: end effect and wall curvature. 2. The end effect is a local problem, but its influence on stress cr9 is more obvious than on other stresses. The taller is the sample the less is the end effect. In the calculation, the end restraint is assumed fully affect. However, in practice test, the end restraint cannot be so idealized, there certainly will have some lateral displacements on the ends of sample. The end effect, therefore, is not so severe as in this analyses. The end effect diminish with distance, accord with the Saint-Venant's principle.
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214 3. The stress and strain non-uniformity caused by curvature of.wall is a global problem. The geometry of sample is very important to minimize that effect. The hollow cylinder with r1/r0=0. 8 and ro-r1=1. 0 inch suggested by Hight et al (Hight, 1983) provided good stress and strain distributions when subjected to torsional force. The hollow cylinder with H/2=2.5" is not suitable for test neither subjected to axial load nor combined load. 4. Plastic material can lessen the curvature effect compared with elastic material. When torsional force is applied on the sample, the upper bound and lower bound of the difference between outer side and inner side of the wall is easy to be calculated. If the material is linear elasticity, the ration of 'tmaxl'tmin = 1.25 for hollow cylinder in this thesis. It is the upper bound. If the material is assumed perfectly plasticity, the ratio of 'tmaxl'tmin = 1. It is the lower bound. The material considered in this thesis is strain hardening soil, the ratio of 'tmaxl'tmin: = 1.16.
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215 5. The solid cylinder under axial load can have enough room of end-effect free region, but the nonuniformity problem become serious when torsional force is applied. Although this analyses provided information about the stress and strain distributions in cylindrical sample, limitations of the results must be pointed out because of some artificial conditions added in the numerical analyses, those are: 1. Soil model:the soil used in the computation is a strain hardening material. In actually, the stress state at failure or post failure of sample is often investigated for some soils, for instance, the overconsolidated clay. The strain softening may occur under certain confining pressure for some soils, the stress and strain distributions would change in some way when the stress state transfer from hardening to softening. This situation will become more obvious when torsional force is a dominant load.
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216 2. Load condition: monotonic load following only one of the stress paths was adapted in the calculation. The loading-unloading and cyclic loading conditions are occur quite frequently in site and in laboratory test. Residual stress and strain would accumulated in the sample when plastic stage reached. In this case, the .distribution of stress and strain will differ from results calculated under monotonic load in some way. 3. Stress path: different stress paths are usually applied in laboratory test, for instance, axial load applied first,then the torsional force; or torsional force first then axial force; or axial load and torsional force applied together in a proportional way. Since stress history will determine the behavior of soil, different stress paths The certainly will yield different stress and strain distribution patterns. numerical method is an effective way of investigating the stress and strain distributions in a whole sample even there are some limitations on the
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217 generality of the results. In a test, the stress and the strain in a sample are hardly to be measured directly at many points. The numerical analysis could, in some degree, provides information of stress and strain distributions but one must beware of its limitation to avoid misleading by numerical analysis.
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218 REFERENCES Al-Chalabi,M., and Hung,C.L. (1974). Stress distribution within circular cylinder in compression." Int.J. of rock meckanics, Mineral Science, and Geomechanics Abstract, Vol. II, pp.45-56. Balla,A (1960) ."Stress conditions in triaxial compression." ASCE, Vol. 86, SM6. Burland,J.B. (1965). The yielding and dilating of clay", Geotechnique 15, pp. 211-214. Calladine, C.R., properties of (1971). "A microstructural view of the mechanical saturated clay", Chang,C.S. and Misra,A., "Deformation analysis of sands in cubical and hollow cylinder devices", Int .J. for numerical and analytical methods ln geomechanics, vol 13, 493-510 (1989). Chang,C.S., Misra A. and Weeraratne S.P., "A microstructural approach to constitutive modelling for granular at the 6th Specialty Conference of Engineering Mechanics Div.,ASCE,Buffalo, New York, May, 1987,pp.203. Chen,J.W.(l988), "Stress path effect on static and cyclic behavior of Monterey No. 0!30 sand", Ph.D thesis. Univ. of Colorado at Denver. Drucker,D.C., Gibson,R.E. & mechanics and work hardening ASCE,11,pp. 338-346. Henke 1, D J ( 19 57 ) so i 1 theories of plasticity", Duncan, J .M., and Dunlop, P. ( 1986) "The significance of cap and base restraint." J. of siol mechanics and foundation
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219 division, ASCE, Vol. 94, SMl, pp. 271-290. Frydman, S., Alpan, I., and Zeit len, I. G. ( 1971) "stress deformation behavior of sand studied in hollow cylinder and ttriaxial apparatus." Proc. 4th. Asian conference on soil mechanics and foundation engineering, Bankok, pp.17-23. Hight,D.W., Gens,A., and Symes,M.J. (1983). "The development of a new hollow cylinder apparatus for investigating the effects of principal stress rotation in soils." Geotechnique, Vol.33, No.4, pp.355-383. Ishihada,K., and Yasuda,S. (1975). "Sand liquefaction in hollow cylinder torsion under irregular excitation.", Soil and Foundation, Vol. 15, No.1, pp.45-59. Kirkpatrick,W.M., interpretation of 18, pp.336-350. and Belshaw,D.J. (1968) "On the triaxial test." Geotechnique, the Vol. Krige,D.G., (1960). "On the departure of ore value distributions from lognormal method in South Africa gold mines." J. South African Inst. Mining Metall., v.61, pp.231-244. Lade,P.V. (1972). "The stress-strain characteristics of cohesionless soils." University of California, Berkeley. and Ph.D. strength thesis, Lade,P.V. (1976). "Interpretation of torsion shear test on sand." Proc. of the 2nd int. conference on numerical method in geomechanics, Vol.l, pp.381-389. Lade,P.V. (1981a). "Torsion shear apparatus for testing." Proc. of ASTM symposium on laboratory strength of soil: ASTM STP740, pp.145-163. soil shear Lee,K.L. (1978). "End restraint effects on undrained static triaxial strength of sand." J. of geotechnical engineering division, ASCE, Vol. 104, GT6, pp.687-704. Macky,T.A., and Saada,A.s; (1974). "Dynamics of anisotropic
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220 clays under large strains." J. of geotechnical engineering division, ASCE, Vol.110, No.4, pp.487-504. Momenzadeh M. "Effects of principal and intermediate principal stress characteristics of sand", Dr. of Eng. Tohyo, 1985. stress axes rotation on teh deformatiom Thesis, University of Pande1G.N. 1 and SharmaiK.G., "A multi-laminate model of clays-a numerical study of the influence of rotation of the principal stress axes",Proc.Imp.of computer procedures and stress-strain laws in geotechnique engineering,C.S.Desai and R.H.Gallagher,Edss.John Wiley,New York,pp.45-59. Perloff,W.H., and Pombo,L.E. (1969). "End restraint effects in the triaxial test". Proc. of the 7th int. conference on soil mechanics and foundation engineering, Mexia, Vol. 1, pp.327-333. Pickett, G. (1944). "Application of the Fourier method to the solution of certain boundary problems in the theory of elasticity." J. of applied mechanics, ASCE, Vol. 11, pp.176-189. Roscoe,K.H.&: Burland,J.B. (1968), "On stress-strain behavior of 'wet clay' plasticity, Cambridge University Press. the generalised Engineering Rowe 1P.W., and Barden, L. (1964). "Importance of the free ends in triaxial testing." J. of soil mechanics and foundation Division, ASCE, Vol. 90, SM1, pp.1-27. Saada,A.S., and Baah,A.K. (1967). "Deformation and failure of a cross anisotropic clay under combined stress." Proc. of the 3rd Panamerican conference on soil mechanics and foundation engineering, Saada,A.S., and Bianchini,G.F. (1975). "Strength of one dimensionally consolidated clays." J. of geotechnical engineering division, ASCE, Vol. 101, GT11, pp.1151-1164. Saada1A.S. 1 and Ou1C.D. 1 (1973). "Strain-stress relations and failure of anisotropic clays." J. of soil mechanics
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221 and foundation divisioh, ASCE; Vol.99, SM12, pp.1091-1111. Saada,A.S., abd Shook.L. (1981). Behavior of clays subjected to slow cyclic loading and large strains." Int. conference on recent advance in geotechnical earthquack engineering and soil dynamics, Vol.1, St. Louis, pp.369-376. Saad,A.S., and Townsend,F.C., (1981). "State of the art: Laboratory strength testing of soils." Proc. of ASTM symposium on laboratory shear strength of soil, ASTM, STP740, pp.7-77. Saada,A.S., and Zamani,K.K. (1969). "The mechanical behavior of cross-anisotropic clays." Proc. of the 7th int. conference on soil mechanics and foundation engineering, Vol.1, Mexico, 1969. Schefield,A.N. & Wroth,C.P. (1968), "Criti8al state soil mechanics", Mcgrew-Hill, London. Shockoley, W .G., and Ahlvin, R.G. ( 1960) "Non-uniform conditions in the triaxil test specimen." Research conference on shear strength on cohesive soils, ASCE, Boulder, colorado, pp.341-357. Timoshenko and Goodier. (1970),"Theory of elasticity", Mcgrew-Hill, London. Wright,D.K., Gilbert,P.A.,and Saada,A.S. (1978). "Shear devices for determining dynamics soil properties." Proc. of the earthquake engineering and soil dynamics conference, Geotechnical engineering division, ASCE, Pasadena, Vol.II, pp.1056-1075. Zienkiewicz,O.C. (1977), "The Mcgrew-Hill, London. Finite Element Method",
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