Deep foundation behavior in non-linear materials

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(1)Deep Foundation Behavior in non-linear materials. by Eliana Carolina Amaya Mahecha. Advisor Prof. Arcesio Lizcano Ph.D.. Universidad de los Andes Faculty of Engineering Department of Civil and Environmental Engineering 2011.

(2) Contents I. Piles Desing. 4. 1. Single Plies. 5. 1.1. Ultimate Load Capacity of Single Piles . . . . . . . . . . . . . . . . . . . . . . . .. 5. 1.1.1. Ultimate shaft resistance . . . . . . . . . . . . . . . . . . . . . . . . . . .. 6. 1.1.2. Ultimate base resistance . . . . . . . . . . . . . . . . . . . . . . . . . . .. 6. 1.1.3. Cohesive soils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 13. 1.1.4. Granular soils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 16. 1.2. Lateral load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 17. 1.3. Negative Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 22. 1.4. Settlement analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 23. 2. Pile groups. 30. 2.1. Ultimate load capacity of pile groups . . . . . . . . . . . . . . . . . . . . . . . . .. 30. 2.1.1. Pile groups in cohesive soils . . . . . . . . . . . . . . . . . . . . . . . . .. 31. 2.1.1.1. Free standing gropus . . . . . . . . . . . . . . . . . . . . . . . .. 31. 2.1.1.2. Piled foundations . . . . . . . . . . . . . . . . . . . . . . . . .. 32. Pile groups in granular soils . . . . . . . . . . . . . . . . . . . . . . . . .. 33. 2.2. Settlement of pile groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 35. 2.3. Lateral Load of Pile Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 38. 2.4. Lateral Load of Pile Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 41. 2.1.2. II 3. Constitutive Models. 43. Elasticity. 44. 3.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 44. 3.2. Elasticity Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 44. ii.

(3) MIC 2011-I0-2B 4. 5. Elastoplasticity. 50. 4.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 50. 4.2. Elastoplastic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 50. 4.2.1. Yield criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 51. 4.2.2. Flow rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 52. 4.2.3. Kuhn-Tucker conditions . . . . . . . . . . . . . . . . . . . . . . . . . . .. 52. 4.2.4. Plasticity models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 53. 4.2.4.1. Perfect plasticity . . . . . . . . . . . . . . . . . . . . . . . . . .. 53. 4.2.4.2. Hardening plasticity . . . . . . . . . . . . . . . . . . . . . . . .. 54. Modified Cam Clay. 58. 5.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 58. 5.2. Critical State Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 58. 5.3. Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 62. 5.3.1. Yield criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 62. 5.3.2. Flow rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 63. 5.3.3. Hardening laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 64. 5.3.4. Consistent tangential moduli . . . . . . . . . . . . . . . . . . . . . . . . .. 65. Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 66. 5.4.1. Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 66. 5.4.2. Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 67. 5.4.3. Elastic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 67. 5.4.4. Yield condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 68. 5.4.5. Plastic corrector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 68. Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 72. 5.4. 5.5 6. Saniclay. 74. 6.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 74. 6.2. Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 74. 6.2.1. Anisotropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 74. 6.2.2. General Aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 75. Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 76. 6.3.1. Plastic Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 76. 6.3.2. Yield Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 78. 6.3.3. Hardening laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 78. 6.3.3.1. 79. 6.3. Preconsolidation mean stress ṗ0 . . . . . . . . . . . . . . . . . . iii.

(4) MIC 2011-I0-2B. 7. 8. 9. III. 6.3.3.2. Deviatoric stress-ratio tensor α̇i j . . . . . . . . . . . . . . . . .. 79. 6.3.3.3. Deviatoric stress-ratio tensor β̇i j. . . . . . . . . . . . . . . . . .. 80. 6.3.4. Plastic Multiplier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 80. 6.3.5. Consistent tangential moduli . . . . . . . . . . . . . . . . . . . . . . . . .. 81. 6.4. Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 81. 6.5. Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 87. Hypoplasticity. 90. 7.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 90. 7.2. Properties of granular soils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 90. 7.2.1. Barotropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 90. 7.2.2. Pycnotropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 91. 7.2.3. Critical Void Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 92. 7.3. Hypoplastic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 92. 7.4. Wolffersdorff hypoplasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 93. Visco-hypoplastic model. 96. 8.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 96. 8.2. Time-stress-strain behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 96. 8.3. Visco-hypoplastic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 98. 8.4. Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104. 8.5. Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107. Comparisson. 109. 9.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109. 9.2. Undrained triaxial tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110. 9.3. Drained triaxial tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113. Finite Element Model. 116. 10 Finite Element Model. 117. 10.1 Material Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 10.2 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 10.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 iv.

(5) MIC 2011-I0-2B A Tensor analysis A.1 Tensors . . . . . . . . . . A.2 Index notaion . . . . . . . A.3 Operations . . . . . . . . . A.3.1 Dot Product . . . . A.3.2 Dyadic product . . A.3.3 Double dot product A.3.4 Inverse . . . . . . A.4 Identity tensors . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. 123 123 124 125 125 126 127 128 130. B User Material Subroutines 134 B.1 Modified Cam Clay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 B.2 Saniclay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 B.3 Visco-hypoplasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158. v.

(6) List of Figures 1.1. Load transfer through a pile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 5. 1.2. Mohr-Coulomb failure envelope . . . . . . . . . . . . . . . . . . . . . . . . . . .. 7. 1.3. Stress field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 9. 1.4. Stress rotation effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 10. 1.5. Adhesion factor applied to granular soil overlying stiff soil . . . . . . . . . . . . .. 14. 1.6. Adhesion factor applied to weak soil underlain by stiff soil . . . . . . . . . . . . .. 14. 1.7. Adhesion factor applied to uniform layer of soil . . . . . . . . . . . . . . . . . . .. 15. 1.8. Pressure distribution in the soil under lateral load . . . . . . . . . . . . . . . . . .. 17. 1.9. Lateral Load. Poulos and Davis . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 18. 1.10 Broms Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 20. 1.11 Pressure distribution for Zhang et. all[43] . . . . . . . . . . . . . . . . . . . . . .. 21. 1.12 ξ values. Adapted from [27] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 24. 1.13 Interaction factors for floating piles L/d=10 . . . . . . . . . . . . . . . . . . . . .. 28. 1.14 Interaction factors for end-bearing piles L/d=10 . . . . . . . . . . . . . . . . . . .. 28. 1.15 Interaction factors for floating piles L/d=100 . . . . . . . . . . . . . . . . . . . . .. 29. 1.16 Interaction factors for end-bearing piles L/d=100 . . . . . . . . . . . . . . . . . .. 29. 2.1. Pile group dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 30. 2.2. Method 2:1 for settlement of pile groups . . . . . . . . . . . . . . . . . . . . . . .. 35. 2.3. Interaction factors for floating piles, L/d=25 . . . . . . . . . . . . . . . . . . . . .. 37. 2.4. Interaction factors for end bearing piles,L/d=10 . . . . . . . . . . . . . . . . . . .. 37. 2.5. Interaction factors for floating piles, L/d=100 . . . . . . . . . . . . . . . . . . . .. 38. 2.6. Interaction factors for end bearing piles,L/d=100 . . . . . . . . . . . . . . . . . .. 38. 2.7. Angle β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 40. 2.8. Angle β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 42. 4.1. Stress-Strain Curve 1D.[33] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 51. 4.2. Isotropic hardening: a) Load and unload path b) Yield surface . . . . . . . . . . .. 56. vi.

(7) MIC 2011-I0-2B 4.3. Kinematic hardening: a) Load and unload path b) Yield surface . . . . . . . . . . .. 56. 4.4. Combined hardening: a) Load and unload path b) Yield surface . . . . . . . . . . .. 57. 5.1. Critical State Line in V − p0 plane . . . . . . . . . . . . . . . . . . . . . . . . . .. 59. 5.2. Critical State Line and yield surface in q − p0 plane . . . . . . . . . . . . . . . . .. 60. 5.3. p0 ,V, q space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 62. 5.4. Undrained triaxial with K0 consolidation . . . . . . . . . . . . . . . . . . . . . . .. 72. 5.5. Undrained triaxial with isotropic consolidation . . . . . . . . . . . . . . . . . . .. 72. 5.6. Undrained triaxial with anisotropic consolidation . . . . . . . . . . . . . . . . . .. 72. 5.7. Drained triaxial with K0 consolidation . . . . . . . . . . . . . . . . . . . . . . . .. 73. 5.8. Drained triaxial with isotropic consolidation . . . . . . . . . . . . . . . . . . . . .. 73. 5.9. Drained triaxial with anisotropic consolidation . . . . . . . . . . . . . . . . . . . .. 73. 6.1. Surfaces in Saniclay[13] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 75. 6.2. Mapping definition of αibj [13] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 79. 6.3. Undrained triaxial with K0 consolidation . . . . . . . . . . . . . . . . . . . . . . .. 87. 6.4. Undrained triaxial with isotropic consolidation . . . . . . . . . . . . . . . . . . .. 87. 6.5. Undrained triaxial with anisotropic consolidation . . . . . . . . . . . . . . . . . .. 88. 6.6. Drained triaxial with K0 consolidation . . . . . . . . . . . . . . . . . . . . . . . .. 88. 6.7. Drained triaxial with isotropic consolidation . . . . . . . . . . . . . . . . . . . . .. 89. 6.8. Drained triaxial with anisotropic consolidation . . . . . . . . . . . . . . . . . . . .. 89. 7.1. Barotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 91. 7.2. Pycnotropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 91. 7.3. Critical Void Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 92. 7.4. Definition of θ and ψ angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 94. 8.1. Creep . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 97. 8.2. Isotach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 97. 8.3. Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 98. 8.4. Maxwell descomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 98. σi0j. 8.5. Shear stress τ under constant effective stress. . . . . . . . . . . . . . . . . . .. 99. 8.6. pe and p+ e location in p − q and p − q − e spaces. Adopted from Fuentes[15] . . . . 100. 8.7. Undrained triaxial with isotropic consolidation and velocity changes . . . . . . . . 107. 8.8. Isotropic consolidation with velocity changes . . . . . . . . . . . . . . . . . . . . 107. 8.9. Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108. 8.10 Creep simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 vii.

(8) MIC 2011-I0-2B 9.1 9.2 9.3 9.4 9.5 9.6 9.7. Undrained triaxial with K0 consolidation . . . . . . . . . . . . . . . . . Undrained triaxial with isotropic consolidation . . . . . . . . . . . . . Undrained triaxial with isotropic consolidation and changes of velocity . Undrained triaxial with anisotropic consolidation . . . . . . . . . . . . Drained triaxial with K0 consolidation . . . . . . . . . . . . . . . . . . Drained triaxial with isotropic consolidation . . . . . . . . . . . . . . . Drained triaxial with anisotropic consolidation . . . . . . . . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. 110 111 112 112 113 114 115. 10.1 10.2 10.3 10.4. Model . . . . . . . Model . . . . . . . Load-Displacement Pile Deflexion . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. 117 119 119 120. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. A.1 Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124. viii.

(9) MIC 2011-I0-2B. Acknowledgements. I would like to thank Professor Arcesio Lizcano for being part of this work, knowledge, time, and especially by the trust offered me, without he this work would not have been possible. To my family for their great support, my parents for believing in me and my brothers for the love and trust, without them I would not have got where I am. A Reibid Quiroga and his family for the love so great that have offered to me, for their prayers and encouragement. To the members of the Geotechnical Research Group, for their help and knowledge, especially Juan Daniel Moya and Luis Ángel Áviles. Finally, to my friends that was always present: Yessica, Erika, Sara, Paola, Felipe, Miguel Angel, Alejandro and Hernán.. 1.

(10) Introduction Common development of civilization goes by the hand of its infrastructure development, increasing the size and complexity of new structures in civil projects. At the same time, the safety requirements have lead to harder scenarios requesting better calculations during the design process. In traditional geotechnical engineering is usual to find empirical equations with a lack of mathematical tools, like the design of deep foundations, where the load capacity is calculated by MohrCoulomb’s theory and settlements with elastic theory, considering the soil as a linear-elastic material, valid only at small deformations (10−4 ). An alternative to enhance this kind of assumptions is the development of constitutive models, able to reproduce, in a realistic way, the soil behavior at some particular conditions, as is presented in the work of Roscoe and Burland in 1968[23] in elasto-plasticity and Kolymbas in 1977[12] in hypo-plasticity, among others. The purposes of this work is to analyze different constitutive models that represent in a realistic form the behavior of a cohesive soil, showing the applicability in the design of deep foundations under different loading conditions, and present a comparison between a soil-pile model and a real pile load test. This work is divided into three parts: the first part focuses on the design of piles, where Chapter 1 focuses on single piles, showing load capacity, lateral load, negative friction and settlements; and Chapter 2 shows the design of piles groups. Second part is focused on understanding the constitutive models; on chapters 3 and 4 is shown an overview of the Elasticity and Elastoplasticity constitutive models, as the basis for Modified Cam Clay and Saniclay models, in chapters 5 and 6 the last models are explained in detail, showing its implementation and results. In the following sections are explained some models which do not distinguish between plastic and elastic deformations. Chapter 7 presents the hypoplastic model developed by Kolymbas, Chapter 2.

(11) LIST OF FIGURES. MIC 2011-I0-2B. 8 presents the visco-hypoplasticity model developed by Niemunis, and in chapter 9 the Modified Cam-Clay, Saniclay and Visco-hypoplasticity models are compared with experimental results. The last part shows the implementation of a model in a finite element model which analyzes a pile under lateral load cycles, and is compared against a real pile load test. In Appendix A, is a detailed explanation of the tensor operations used in this work, similarly, the code to be developed in Fortran. In Appendix B, are the UMAT codes for the Niemunis incremental driver. These two appendices are an important part of the work, because one of the objectives is to develop a document in which a person with few bases can understand and develop each of the models. At the end of this document, the reader will be able to: • Understand the equations which are designed with deep foundations • Understand the basic concepts of elastoplasticity and hypoplasticity • Implement the three models studied • Know the advantages and disadvantages of the models, at the level of element test as in operation in a pile load test. 3.

(12) Part I Piles Desing. 4.

(13) Chapter 1 Single Plies 1.1. Ultimate Load Capacity of Single Piles. ∑ Fy = 0 Pb + Pf − Pu −W = 0 Z. Pf =. τdA. Pb = Ab · σvb Where: Pb : Ultimate base resistance Pf : Ultimate shaft resistance τ: Shear strength Ab : Area of pile base σvb : Normal stress in the base of the pile Pu = Pb + Pf −W. (1.1.1) 5. Figure 1.1: Load transfer through a pile.

(14) CHAPTER 1. SINGLE PLIES. 1.1.1. MIC 2011-I0-2B. Ultimate shaft resistance. The shear strength is given by Coulomb Theory: τ = ca + σh · tan δ Where: ca : Adhesion σh : Normal stress between pile and soil δ : angle of friction between pile and soil The shaft resistance is possible to find taking a differential of area dA: dA = p · dz Z. Pf =. {ca + σh · tan δ } pdz σh = Kσv. Where: p: pile perimeter K: lateral pressure coefficient σv : vertical stress. Z. Pf =. 1.1.2. {ca + Kσv · tan δ } pdz. Ultimate base resistance. Ultimate base resistance Pb is approach to the bearing capacity of shallow foundations[5]: In the case of an overconsolidate clay is known that: α=. π ϕ + 4 2 6. (1.1.2).

(15) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Figure 1.2: Mohr-Coulomb failure envelope. In the figure 1.1.2 could be seen the followings relations: 2α =. π +ϕ 2. sin 2α = sin. nπ. o. cos 2α = cos. nπ. +ϕ 2 π π sin 2α = sin cos ϕ + cos sin ϕ 2 2 sin 2α = cos ϕ. +ϕ. o. 2 π π cos 2α = cos cos ϕ − sin sin ϕ 2 2 cos 2α = − sin ϕ. 7.

(16) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. The values of shear stress τ and the normal stress σn0 could be determine: 1 τ = (σ10 − σ30 ) cos ϕ 0 2 1 0 1 0 0 σn = (σ1 + σ3 ) − (σ10 − σ30 ) sin ϕ 2 2 The relationship between the main effective stress σ10 y σ30 is get knowing the equation of the envelope of Mohr - Coulomb τ = c0 + σh0 tan ϕ 0 (1.1.3) 1 0 1 0 1 (σ1 − σ30 ) cos ϕ 0 = c0 + (σ1 + σ30 ) − (σ10 − σ30 ) sin ϕ 0 tan ϕ 0 2 2 2 (σ10 − σ30 ) cos ϕ 0 = 2c0 + (σ10 + σ30 ) − (σ10 − σ30 ) sin ϕ 0 tan ϕ 0 2 0 0 0 cos ϕ (σ1 − σ3 ) 0. sin ϕ. 0 cos ϕ. = 2c. 0. sin ϕ 0. + (σ10 + σ30 ) − (σ10 − σ30 ) sin ϕ 0. 2c0 cos ϕ 0 + (σ10 + σ30 ) sin ϕ 0 − (σ10 − σ30 ) sin2 ϕ 0 = sin ϕ sin ϕ. 2 0 0 0 cos ϕ (σ1 − σ3 ) 0. (σ10 − σ30 ) cos2 ϕ 0 = 2c0 cos ϕ 0 + (σ10 + σ30 ) sin ϕ 0 − (σ10 − σ30 ) sin2 ϕ 0 (σ10 − σ30 ) cos2 ϕ 0 + (σ10 − σ30 ) sin2 ϕ 0 = 2c0 cos ϕ 0 + (σ10 + σ30 ) sin ϕ 0 (σ10 − σ30 )(cos2 ϕ 0 + sin2 ϕ 0 ) = 2c0 cos ϕ 0 + (σ10 + σ30 ) sin ϕ 0 σ10 − σ30 = 2c0 cos ϕ 0 + (σ10 + σ30 ) sin ϕ 0 Equation 1.1.4 can be re-written as follows. 8. (1.1.4).

(17) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. σ10 − σ10 sin ϕ 0 = 2c0 cos ϕ 0 + σ30 + σ30 sin ϕ 0 σ10 (1 + cos 2α) = 2c0 sin 2α + σ30 (1 − cos2α) 1 − cos2α 0 1 + cos 2α σ1 ) = c0 sin 2α + σ30 ( 2 2 σ10 cos2 α = c0 (2 sin α cos α) + σ30 sin2 α σ10 = 2c0 tan α + σ30 tan2 α σ10 = 2c0 tan. nπ 4. +. π ϕ ϕo + σ30 tan2 ( + ) 2 4 2. (1.1.5). If an analysis is made for a normally consolidated soil, in which the cohesion is zero. The equation 1.1.5 would be:. σ10 = σ30 tan2. nπ 4. +. ϕo 2. Figure 1.3: Stress field. Using equation 1.1.6 is obtained 9. (1.1.6).

(18) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. σB0 = σ00 tan2. nπ 4. +. ϕo 2. But to calculate the stress at C, the stress rotate 90◦ or 180◦ in Mohr’s Circle, so should be used:. π ϕ qu = σ00 tan4 ( + ) 4 2. (1.1.7). However, the equation 1.1.7 can be improved by considering a stress field. To obtain a greater stress is made a small rotation dθ (see figure 1.1.2).. Figure 1.4: Stress rotation effects. The relation between d p and DB can find by a triangles ratio d p0 DB = sin 2dθ sin α. (1.1.8). Due to dθ is very small, the following approximations can be done: DA ≈ DB, sin α ≈ cos ϕ 0 y sin 2dθ ≈ 2dθ . In addition is known that DA = p0 sin ϕ 0 . So re-writing and re-arranging the equation 1.1.8: 10.

(19) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. d p0 = 2 tan ϕ 0 dθ p0. (1.1.9). The equation can be solved by integrating both sides, Z p0 0 2 dp p01. p0. Z π/2. =. 2 tan ϕ 0 dθ. 0. p02 = p01 e(π tan ϕ). (1.1.10). p01 correspond to the center of the first Mohr’s circle, for which σ0 is the lowest main stress. In the same way, p02 is the center of the second Mohr’s circle where qu correspond to the greatest main stress.. p01 − σ0 = r1 p01 − σ0 = p01 sin ϕ 0 p01 (1 − sin ϕ 0 ) = σ0 p02 − σ0 = r2 qu − p02 = p02 sin ϕ 0 p02 (1 + sin ϕ 0 ) = qu σ0 1 − sin ϕ 0 qu p02 = 1 + sin ϕ 0. p01 =. The equation 1.1.11 is introduced in the equation 1.1.10. 11. (1.1.11).

(20) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. 0 1 + sin ϕ 0 σ0 eπ tan ϕ 0 1 − sin ϕ 0 1 − cos 2α qu = σ0 eπ tan ϕ 1 + cos 2α 2 0 1 − cos α + sin α σ0 eπ tan ϕ qu = 2 1 + cos α − sin α 0 1 − cos2 α + 1 − cos2 α qu = σ0 eπ tan ϕ 2 2 1 + cos α − 1 + cos α 0 2 − 2 cos2 α qu = σ0 eπ tan ϕ 2 2 cos α 1 − cos2 α π tan ϕ 0 σ e qu = 0 cos2 α 0 sin2 α qu = σ0 eπ tan ϕ 2 cos α 0 qu = tan2 ασ0 eπ tan ϕ. qu =. 0 π ϕ0 qu = tan2 ( + )σ0 eπ tan ϕ 4 2. (1.1.12). Bolton suggests that the effective stress is given by a relation between the density γ, the diameter d and the friction angle ϕ:. 0. 0. σ = γ d tan ϕ. 0. (1.1.13) 0. 0. If the cohesion is considered, the normal stress increases a factor of c cot ϕ . So the equation of the load capacity 1.1.12 would be:. 0. 0. 0. 0. 2. qu + c cot ϕ = (σ0 + c cot ϕ ) tan. . π ϕ 0 π tan ϕ 0 ϕ 0 π tan ϕ 0 0 0 2 π + e + dγ tan ϕ [tan + e − 1] 4 2 4 2 (1.1.14). Some variables are introduced:. 2. Nq = tan. . π ϕ 0 π tan ϕ 0 + e 4 2. 12.

(21) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B Nγ = 2(Nq − 1) tan ϕ 0 Nc =. Nq + 1 tan ϕ 0. qu = c0 Nc + σ0 Nq +. γ 0d Nγ 2. (1.1.15). The ultimate stress qu is replaced by Pb /Ab and σ0 by the effective stress in the base σvq .. Pb = Ab(σvb Nq + 0, 5γdNγ + cNc ) The ultimate base resistance can be affected by the shape and depth. There are factors that represent these conditions, which have been determined empirically. Some authors express these conditions with an asterisk in the variables: Nq∗ ,Nγ∗ and Nc∗ . The typical values are: Nq∗ ≈ 102.7 tan ϕ and Nγ∗ ≈ 6 (for a homogeneous soil) The equation that determines the ultimate load capacity of a pile is:. Pu = Ab(σvb Nq∗ + 0, 5γdNγ∗ + cNc∗ ) +. Z. {ca + Kσv · tan δ } pdz −W. (1.1.16). The equation 1.1.16 is the more general equation for calculating the ultimate load capacity, however, it could be find specific equations for the case of cohesive soils and granular soils. 1.1.3. Cohesive soils. In a cohesive soil, the failure occurs in the short term, ie the soil is unable to drain pore pressure excess. In this case, is use the undrained soil cohesion Cu and therefore the friction angle ϕ will be zero.. Ultimate shaft resistance The adhesion between soil and pile is important in a short-term failure. Then the equation for shaft resistance would be: 13.

(22) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Z L. Pf =. pαCu 0. (1.1.17). Pf = AsαCu The adhesion factor decreases cohesion in some rate because only a fraction of this is achieved after disturbing the soil, this rate depends of soil stratigraphy and dimensions of the pile . Therefore, there is no physically based equations for its calculation, however, some authors have performed tests to obtain empirical values of this factor[5]. See the figures 1.5,1.6,1.7.. Figure 1.5: Adhesion factor applied to granular soil overlying stiff soil. Figure 1.6: Adhesion factor applied to weak soil underlain by stiff soil. 14.

(23) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Figure 1.7: Adhesion factor applied to uniform layer of soil. Ultimate base resistance Ultimate shaft resistance would be:. Pb = Ab(σvb Nq + 0, 5γdNγ +CuNc ) As ϕ = 0:. Nq = tan2. π 4. e0 = 1. Nγ = 2(1 − 1) tan ϕ 0 = 0 Then, the base resistance is defined by:. Pb = Ab(σvb +CuNc ). (1.1.18). The ultimate load of a pile in a cohesive soil is obtain approaching the weight of the pile and linking the equations 1.1.1, 1.1.17 and 1.1.18.. W ≈ AbγL Pu = Cu(αAs + Nc Ab) 15. (1.1.19).

(24) CHAPTER 1. SINGLE PLIES. 1.1.4. MIC 2011-I0-2B. Granular soils. For drained conditions, the failure may be occurs in long term, so the capacity equation for the pile shaft would be: Z L. Pf =. 0. pσv0 K tan δ dz. (1.1.20). Where the angle of friction between the soil pile and δ is found experimentally and is accepted using the following values: · Plain Steel: 0.5φ 0 ≤ δ ≤ 0.7φ 0 · Rugged Steel: 0.7φ 0 ≤ δ ≤ 0.9φ 0 · Precast Concrete: 0.8φ 0 ≤ δ ≤ 1φ 0 · Concrete cast in situ: δ = φ 0. The earth pressure coefficient K is affected depending on the constructive method: · High displacement: K0 ≤ K ≤ 2K0 · Low displacement: 0.75K0 ≤ K ≤ 1.75K0 · Pre-carved and cast in situ: 0.71K0 ≤ K ≤ K0 · Greenish Piles: 0.5K0 ≤ K ≤ 0.7K0. Similarly, the equation that governs the base resistance of the pile would be:. 0 Pb = Ab σvb Nq. (1.1.21). Where the friction angle should be calculated as follows: 0. · Driven piles: φ = φ +40 2 · Pre-carved piles: φ = φ 0 − 3·. Then the ultimate load of the pile for drained conditions is obtained from equations 1.1.1, 1.1.20 16.

(25) CHAPTER 1. SINGLE PLIES and 1.1.21:. Z L. Pu =. 1.2. MIC 2011-I0-2B. 0. 0 pσv0 K tan δ dz + Pb + Ab σvb Nq −W. (1.1.22). Lateral load. There are many theories proposed to calculate the final lateral load supported by a single pile: Brinch Hansen (1961), Broms (1964), Reese et al. (1974), Poulos and Davis (1980),Fleming et al. (1992). These differ considerably, making difficult choose one for the calculation. One of the major differences between theories is how they assume the pressure distribution in the soil. Some examples are shown in Figure 1.2. Figure 1.8: Pressure distribution in the soil under lateral load. • Poulos and Davis[31] proposed the stress distribution shown in Figure 1.9. 17.

(26) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Figure 1.9: Lateral Load. Poulos and Davis. Z Zr. Hu = Mu = Hu e = −. 0 Z Zr 0. Pu ddz −. Z L. Zr Z L. Pu dzdz +. Zr. Pu ddz (1.2.1). Pu dzdz. In the equation 1.2.1, Zr y Hu have different solutions, depending on the distribution of soil resistance. Uniform distribution:. Hu +L Pu d ( ) Mu Hu e 1 2Hu Hu 2 = = 1− − Pu dL2 Pu dL2 4 Pu dL Pu dL s 2Hu 2e 2 2e = 1+ +1− 1+ Pu dL L L 1 Zr = 2. . Linear distribution: 18. (1.2.2).

(27) CHAPTER 1. SINGLE PLIES. . Zr 4 L. MIC 2011-I0-2B. ( ) n o P0 12P0 e e Zr Zr 2 + + + 6 L L PL − P0 PL − P0 L L n eo P +P 2PL + P0 L 0 − =0 − 3 L PL − P0 PL − P0. (1.2.3). 2 Hu P0 Zr Zr 1 P0 P0 = 1− − 1+ + 2 PL dL PL L PL L 2 PL. (1.2.4). 3. • Brinch Hansen[8] proposes that ultimate lateral resistance is a function of the parameters of the soil c and φ , in the case of a cohesionless soil, the ultimate lateral resistance is: pu = Kq γzB Where: Kq is the Hansen earth pressure coefficient (Kq = f (φ )) and B is the diameter or width of the pile.. • Reese et all.[34] suggested that the value of pu can be determined by the lesser value given by the following equations: p p 1 0 + 1 tan φ sin β pu = γz B(K p − Ka ) + z(K p − K0 ) K p tan α + zK0 K p cos α pu = γzB(K p3 + K0 K p2 tan φ 0 − Ka ). • Broms (1964) suggested a pressure distribution simplification. (see figure 1.10) 19.

(28) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Figure 1.10: Broms Theory. Where:. f=. Hu 9cu d. Mmax = Hu (e + 1.5d + 0.5 f ). (1.2.5). Hu = 3K p γZd • Zhang et all.[43] suggests the following methodology for calculate the ultimate lateral resistance, based on a compilation of the different methods proposed: Considering the Smith propose, "the soil resistance to the lateral movement of the pile can be expressed in two components: the frontal normal reaction Q and the side friction reaction 20.

(29) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. F", then: pu = Q + F Q = ηPmax B F = ξ τmax B Where η is the shape factor to take account the nonuniform distribution of earth pressure in front of the pile and ξ is the shape factor to take account the nonuniform distribution of lateral shear drag Pmax = K p2 γz τmax = Kγz tan δ. Figure 1.11: Pressure distribution for Zhang et. all[43]. 21.

(30) CHAPTER 1. SINGLE PLIES. 1.3. MIC 2011-I0-2B. Negative Friction. The negative friction occurs when the soil settles more than the pile; this is normally due to consolidation processes in cases where a clay or granular fill is put over a clay layer. Another reason why this phenomenon occurs, is when there is a lowering of the water level.. Pf = kσv tan δ. Z h. ff =. kγ tan δ z(πd) = kγ tan δ (πd) 0. h2 2. Where h is the height of the layer that is generating the negative friction.. f f = kγ tan δ (πd). h2 2. (1.3.1). When there is negative fiction, has been found that this occurs from the surface to some depth, this depth is known as the neutral depth Ln . Bowles[7] suggested the following equation to calculate it:. Ln =. 2γh L − h L − h γh ( + )− Ln 2 γ γ Z Ln. ff =. πdk(γh + γz) tan δ dz 0. Then is obtaining the equation for the negative friction force in the case of a granular fill over clay. f f = (πdkγh tan δ )Ln + 22. Ln2 πkγ tan δ 2.

(31) CHAPTER 1. SINGLE PLIES. 1.4. MIC 2011-I0-2B. Settlement analysis. There are many theories about the calculation of settlements under a deep foundation. Most are based on the theory of elasticity. M. Das[27] proposes: Se = Se1 + Se2 + Se3. Where: • Se : Total settlement • Se1 : Elastic shortening of the pile Based on the theory of elasticity: σ = Ep · ε =. ε=. P Ap. P Se1 = A pE p L. P = Qwp + ξ Qws 23. (1.4.1).

(32) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Se1 =. (Qwp + ξ Qws )L A pE p. (1.4.2). Qwp : Pile base load Qws : Pile shaft load L : Pile Length A p : Pile cross-sectional area E p : Pile elasticity modulus ξ : Friction factor. The friction factor ξ depends on the distribution of frictional resistance along the shaft. Typical values can be seen in the figure 1.12:. Figure 1.12: ξ values. Adapted from [27]. • Se2 : Settlement in the base. Se2 =. qwp D (1 − µs2 )Iwp Es. qwp : Stress on pile base D: Width / diameter Es : Soil elasticity modulus under pile base 24. (1.4.3).

(33) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. µs Soil poisson ratio Iwp : Influence factor. • Se3 : Settlement produced by the load transmitted to the shaft. Se3 =. Qws pL. . D (1 − µs2 )Iws Es. (1.4.4). p: Pile perimeter L: Embedded pile length Iw s: Influence factor. The settlements Se2 and Se3 are based on the Timoshenko’s proposal[38]. Vesic[39] proposed a different calculation for settlements Se2 and Se3 :. Se2 =. QwpC p Dq p. (1.4.5). q p : Ultimate base resistance C p : Empirical coefficient Se3 =. QwsCs Lq p r. Iws = 2 + 0.35 Typical values of C p 25. (1.4.6) L D. (1.4.7).

(34) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Table 1.1: Typical values of C p according Vesic Soil Type Driven piles Drilled pile Sand (dense or loose) 0.02-0.04 0.09-0.18 Clay (hard or soft) 0.02-0.03 0.03-0.06 Slit(dense or loose) 0.03-0.05 0.09-0.12. ( Cs =. r. 0.93 + 0.16. L Cp D. ) (1.4.8). Poulos and Davis[31] make the analysis dividing the pile into segments and imposing displacement compatibility between the pile and its adjacent soil: • Soil displacement equation d sρi = Es. n. . db ∑ (Ii j p j )) + Es Iib pb j=1. (1.4.9). Where: sρi : soil displacement on the element i d: Pile diameter Es : Soil elasticity modulus Ii j : Vertical displacement factor for element i due to shear stress at element j p j :Stress on a element j Note: b means base. The equation 1.4.9 could be re-write like matrixes: {sρ } = • Pile displacement equation 26. d [Is ]{p} Es. (1.4.10).

(35) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. {p} =. d E p RA [I p ]{pρ } + {Y } 4δ. (1.4.11). Where: {p}: Shear stress vector {pρ }: Pile displacement vector [I p ]: pile action matrix {Y }:Constant vector, which contains the first component. P n and the rest of its elements πd 2 L/d. are 0 δ : Thickness of each element ap RA : Area ratio = πd 2 /4. • Displacement compatability. Consider that the soil displacements are equal to the pile n2 {p} = [I] − 4(L/d)2 . . . −1 K[I p ][Is ] {Y }. Where: [I]: Unit matrix K: Pile stiffness factor =. Ep RA Es. Some examples for the interaction factor are shown in the following figures[31]: 27. (1.4.12).

(36) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Figure 1.13: Interaction factors for floating piles L/d=10. Figure 1.14: Interaction factors for end-bearing piles L/d=10. 28.

(37) CHAPTER 1. SINGLE PLIES. MIC 2011-I0-2B. Figure 1.15: Interaction factors for floating piles L/d=100. Figure 1.16: Interaction factors for end-bearing piles L/d=100. 29.

(38) Chapter 2 Pile groups 2.1. Ultimate load capacity of pile groups. The determination of the load capacity of a pile group has not yet been fully resolved, because it has a very complex behavior. Depending on the diameter of the piles and the separation between them can each work individually or as a block . This is seen in what is known as efficiency η. The efficiency factor η is:. η=. Qgu ∑ Qu. Where, Qgu is the ultimate load capacity of group and Qu load capacity of individual piles.. Figure 2.1: Pile group dimensions. 30.

(39) CHAPTER 2. PILE GROUPS. 2.1.1. MIC 2011-I0-2B. Pile groups in cohesive soils. (a) Free standing gropus. 2.1.1.1. (b) Piled foundations. Free standing gropus. In this case, the pile cap is not in contact with the soil. There are many ways to find the efficiency for this case: • Converse-Labarre[6]: . (n1 − 1)n2 + (n2 − 1)n1 η = 1− θ 90 · n1 · n2 Where: θ : arctan. d s. • Das[27] η=. 2s(n1 + n2 − 2) + 4d n1 · n2. • Sayed and Baleer[35] η =2. [(n − 1)s + d] + [(n2 − 1)s + d] π · n1 · n2 · d 31.

(40) CHAPTER 2. PILE GROUPS. MIC 2011-I0-2B. However, Terzaghi and Peck method[37] is the most accepted, which take the group capacity like the lesser value between the ultimate load capacity of group and the sum of load capacity of individual piles. Ultimate load capacity of block. Pu = Pbb + Pf b −Wc −Ws − nWp. (2.1.1). Where: Pbb : Ultimate base resistance of block Pf b : Ultimate shaft resistance of block Wc : Cap weight Ws : Soil weight nWp : Piles weight. Pbb = 2(Bg + Lg )Lcu. (2.1.2). Pf b = Bg Lg cNc. (2.1.3). Ultimate load capacity of individual piles be calculated like in the Chapter 1. The ultimate load capacity of individual piles could. It is important to note the outcome of the investigation of Poulos and Davis[31] in which they noted that efficiency increases when the piles have small relation L/d, large areas or small number of piles in the group.. 2.1.1.2. Piled foundations. In this case, the pile cap is in contact with the soil. Like in free standing gropus, take the group capacity like the lesser value between the ultimate load capacity of group and the sum of load capacity of individual piles. Ultimate load capacity of a block 32.

(41) CHAPTER 2. PILE GROUPS. MIC 2011-I0-2B. This is the same as in free standing groups but considering the portion of the cap outside the perimeter of the block.[5]. Pu = Pbb + Pf b + (Ac − Bg Lg )cNc −Wc −Ws − nWp. (2.1.4). Where: Ac =pile cap area. Ultimate load capacity of individual piles. The sum of the ultimate load capacity of individual piles and the capacity of cap is giving for[31]:. Pu = n(ca AS + Ab cb Nc ) + Ncc cc (Bg Lg − nπd 2 /4). (2.1.5). Where: ca :Adhesion through pile cb : undrained cohesion at level of pile base cc : undrained cohesion beneath pile cap Ncc : bearing capacity factor for rectangular cap = 5.14(1 + 0.2Bg /Lg ). 2.1.2. Pile groups in granular soils. Some authors like Azizi calculate the efficiency of the group in sands in the same form as in cohesive soils, but considering drained conditions. However, others such as Poulos takes the test results, which are shown in the table 2.1 33.

(42) Soil Pile Lenght L Pile Diameter d L/d Group Spacing/d Group efficiency η Medium-grained-moist, 6-10 ft 5-6 in 12-20 2-8 Various >1 dense sand 23 ft 16 in 17 2 Various <1 Cambefort(1953) Humus 100in 2in 50 2-7 2 1.39 Stiff clay 3 1.64 Sand 5 1.17 Gravel 9 1.07 Kezdi(1957) Moist fine 80in 4in 20 4 2 2.1 sand (square) (In line) 3 1.8 4 1.5 6 1.05 4 2 2.1 (square) 3 2.0 4 1.75 6 1.1. Reference Press(1993). CHAPTER 2. PILE GROUPS MIC 2011-I0-2B. Table 2.1: Sumary of test data on large-scale pile groups in sand.[31]. 34.

(43) CHAPTER 2. PILE GROUPS. 2.2. MIC 2011-I0-2B. Settlement of pile groups. In section 1.4 it was show the settlement for a single pile and in the same way it was shown that there is no a single method that leads to answer closer to reality. Now, the settlement will be determined in a pile group, again using the elasticity theory, considering the effect caused by other piles to each of them. Bowles[7] used the method 2:1 for find the settlement of a pile group:. Figure 2.2: Method 2:1 for settlement of pile groups. It assumes that the load applied in the cap is transmitted to the ground starting at a depth of 2/3L from the top of the pile. This load is spread over a trapezoidal area in which the lateral lines have a slope 2:1. Given this, the induced stress is calculated, where only imports the layers below 2/3 of L and is a calculation of settlements as a shallow foundation. Poulos and Davis[31] was based on the principles of settlement of a single pile, but through a detailed analysis, included in the matrices of factors influence the contribution that make the other piles to each of them. This brings the solution for two cases: floating piles and end-bearing piles on rigid stratum. 35.

(44) CHAPTER 2. PILE GROUPS. MIC 2011-I0-2B. Poulos and Mattes[32] developed a method to analyze a group of 2 piles, where they find: {sρ } =. d I [1 + 2I ]{p} Es. (2.2.1). {p}=vector of soil displacement 1I + 2I = matrix of displacement influence factors, this contain the factors that consider the displacement in the elements of the pile 1 caused for the shear stress on the element j of the pile 1 and 2. The factors of interaction α are defined like[31]: α=. Additional settlement caused by ad jacent pile Settlement o f pile under its own load. (2.2.2). the equation of the settlement of a pile group in general can be find based on the above equations: n. ρk = ρ1. (Pj · αk j ) + ρ1 Pk. ∑. (2.2.3). j=1, j6=k. Where: ρ1 =Displacement of single pile under unit load Pj =load in pile j αk j =Interaction factor for spacing between piles k and j. Whit the above equations have a system of 2n unknown variables and n equations, so to make the system determined are formulated the following equations: For vertical load equilibrium: n. Pu =. ∑ Pj. (2.2.4). j=1. If the load is over a flexible pile cap, all the pile have equal load: P1 = P2 = ... = Pn. (2.2.5). If the load is over a rigid pile cap, all the piles have equal settlement: ρ1 = ρ2 = ... = ρn. 36. (2.2.6).

(45) CHAPTER 2. PILE GROUPS. MIC 2011-I0-2B. Here are some figures for calculating the influence factor, taken from Poulos and Davis[31]. Figure 2.3: Interaction factors for floating piles, L/d=25. Figure 2.4: Interaction factors for end bearing piles,L/d=10. 37.

(46) CHAPTER 2. PILE GROUPS. MIC 2011-I0-2B. Figure 2.5: Interaction factors for floating piles, L/d=100. Figure 2.6: Interaction factors for end bearing piles,L/d=100. 2.3. Lateral Load of Pile Groups. In the section 1.2 was shown how find the ultimate lateral resistance in a single pile. Similarly, in the section 2.2 it was shown how to find the settlement of pile groups considering the influence that have the other piles in each of them. In this section you will find how to calculated the ultimate lateral resistance in a pile group considering the influence factors, based on the theory of elasticity. 38.

(47) CHAPTER 2. PILE GROUPS. MIC 2011-I0-2B. The following method was developed by Poulos and Davis and is very similar to that shown in section 2.2 Making the same analysis that in the section 2.2 for two piles, the equation es practically the same that 2.2.1 d (2.3.1) {sρ } = [1I + 2I ]{p} Es But now [1I + 2I ] would be the matrix of displacement influence factors, this contain the factors for horizontal displacement that consider the displacement in the elements of the pile 1 caused for the stress on the element j of the pile 1 and 2. For this case, the influence factor would be for displacement and for rotation because the moment caused by the horizontal load.. αρ =. Additional displacement caused by ad jacent pile Displacement o f pile caused by its own loading. (2.3.2). Additional rotation caused by ad jacent pile Rotation o f pile caused by its own loading. (2.3.3). αθ =. Depending of the case of loading and head fixity Poulos and Davis gave different names to the influence factors:. αρH, αθ H: values for a free head pile subjected to horizontal load only αρM, αθ M: values for a free head pile subjected to moment only αρF:value for fixed head pile. The above factors depends on the stiffness modulus Kr and the β angle, this could be seen in the figure 2.8. 39.

(48) CHAPTER 2. PILE GROUPS. MIC 2011-I0-2B. Figure 2.7: Angle β. The displacement of a pile k in the group is find taken the equation 2.4.1 and making it general: n. ρk = ρH. ∑. (H j · αρHk j ) + ρH Hk. (2.3.4). j=1, j6=k. ρH =Displacement of single pile under unit horizontal load Pj =load in pile j αk j =Interaction factor for spacing between piles k and j. For the sum of horizontal forces:. n. HG =. ∑ Hj. j=1. Also is necessary considerer: • In the free head group each pile displaces equally • In the free head group each pile have equal horizontal load • In the fixed head group each pile displaces equally. 40. (2.3.5).

(49) CHAPTER 2. PILE GROUPS. 2.4. MIC 2011-I0-2B. Lateral Load of Pile Groups. In the Chapter 1.2 was shown how find the ultimate lateral resistance in a single pile. Similarly, in the Chapter 2.2 it was shown how to find the settlement of pile groups considering the influence that have the other piles in each of them. In this Chapter you will find how to calculated the Ultimate lateral resistance in a pile group considering the influence factors, based on the theory of elasticity. The following method was developed by Puolos and Davis and is very similar to that shown in Chapter 2.2 Making the same analysis that in the Chapter 2.2 for two piles, the equation es practically the same that 2.2.1 d (2.4.1) {sρ } = [1I + 2I ]{p} Es But now [1I + 2I ] would be the matrix of displacement influence factors, this contain the factors for horizontal displacement that consider the displacement in the elements of the pile 1 caused for the stress on the element j of the pile 1 and 2. For this case, the influence factor would be for displacement and for rotation because the moment caused by the horizontal load. αρ =. Additional displacement caused by ad jacent pile Displacement o f pile caused by its own loading. (2.4.2). Additional rotation caused by ad jacent pile Rotation o f pile caused by its own loading. (2.4.3). αθ =. Depending of the case of loading and head fixity Poulos and Davis gave different names to the influence factors: αρH, αθ H: values for a free head pile subjected to horizontal load only αρM, αθ M: values for a free head pile subjected to moment only αρF:value for fixed head pile. The above factors depends of the Stiffness Modulus Kr and the β angle, this could be seen in the figure 2.8 41.

(50) CHAPTER 2. PILE GROUPS. MIC 2011-I0-2B. Figure 2.8: Angle β. Taken the equation 2.4.1 and making it general, the displacement of a pile k in the group is: n. ρk = ρH. ∑. (H j · αρHk j ) + ρH Hk. (2.4.4). j=1, j6=k. ρH =Displacement of single pile under unit horizontal load Pj =load in pile j αk j =Interaction factor for spacing between piles k and j. For the sum of horizontal forces:. n. HG =. ∑ Hj. j=1. Also is necessary considerers: • In the free head group each pile displaces equally • In the free head group each pile have equal horizontal load • In the fixed head group each pile displaces equally. 42. (2.4.5).

(51) Part II Constitutive Models. 43.

(52) Chapter 3 Elasticity 3.1. Introduction. One of the most important aspects of a civil engineer is know the materials behavior. This requires realize many laboratory tests and analyze their results. With these tests were able to observe the relationship between the applied loads and material response (displacement). In a general way, the relationship between stresses and strains is found and the different ways in which a material can behave could be determined using this. For some materials have seen a range in which the unload does not produce residual deformation, which is called elasticity.. 3.2. Elasticity Model. For a lineal elastic material can be inferred the following[24]:. • The relationship between stress and strain is linear • The rate of load application does not affect the answer • In are no residual deformations • The strains are very small 44.

(53) CHAPTER 3. ELASTICITY. MIC 2011-I0-2B. The stress is a single function of the strain:. σi j = fi j (εi j ). (3.2.1). Where σi j is the CAUCHY stress tensor, ε the infinitesimal strain tensor and fi j (εi j ) is a second order tensor[16][18]. If fi j (εi j ) is linear in εi j then:. σi j = di jkl εkl. (3.2.2). di jkl is the fourth order stiffness tensor The stress could be written in terms of the potential strain energy f 1 σi j =. ∂ f (εi j ) ∂ εi j. (3.2.3). σ˙i j =. ∂ fi j ε˙kl ∂ εkl. (3.2.4). The stress ratio would be:. Normally is written in terms of a fourth order tensor, which is function of the strain or stresses, or rarely both ( fi jkl (εi j , σi j )). σ˙i j = fi jkl (εi j , σi j )ε˙kl. (3.2.5). If the material is consider as linear elastic, then: fi jkl (εi j , σi j ) = di jkl. σ˙i j = di jkl ε˙kl. 1 Note. that f is different of fi j. 45. (3.2.6).

(54) CHAPTER 3. ELASTICITY. MIC 2011-I0-2B. Also σ̇i j could be describe as a the differentiation of the energy:. ∂ 2 f (εi j ) σ˙i j = ε˙kl ∂ εi j ∂ εkl Then. (3.2.7). ∂ 2 f (εi j ) = di jkl ∂ εi j ∂ εkl. If an isotropic soil is consider, the mechanical behavior does not change if there is a change of direction. Hence the components of the stiffness tensor will be[16][24]:. 0. di jkl = di jkl. (3.2.8). The tensor I is the only isotropic second tensor, so using the components δi j could be write three independent isotropic fourth order tensors[24]:. Pi jk = δi j δkl Qi jk = δik δ jl Ri jk = δil δ jk And the tensor di jkl can be written as a linear combination of Pi jk , Qi jk y Ri jk :. di jkl = λ Pi jk + αQi jk + β Ri jk Where λ , α and β are scalar values. With the equations 3.2.9 and 3.2.2:. σi j = (λ Pi jk + αQi jk + β Ri jk )εi j σi j = λ Pi jk εi j + αQi jk εi j + β Ri jk εi j σi j = λ δi j δkl εi j + αδik δ jl εi j + β δil δ jk εi j. 46. (3.2.9).

(55) CHAPTER 3. ELASTICITY. MIC 2011-I0-2B. σi j = λ δkl ε j j + αεkl + β εlk. (3.2.10). Considering that εkl = εlk and defining 2µ = α + β . The CAUCHY tensor can be rewritten as:. σi j = λ eδi j + 2µεi j. (3.2.11). di jkl = λ δi j δkl + µ(δik δ jl + δil δ jk ). (3.2.12). Where e = ε j j . The tensor di jkl can be rewritten as:. The constants λ and µ are parameters of the material and are known as Lame’s constants. The Young’s Modulus and the Poisson’s Ratio can be write in function of these constants: With the equation 3.2.11 we can obtain εi j : 1 εi j = (σi j − λ εkk δi j ) 2. (3.2.13). The trace of σ :. Tkk = (3λ + 2µ)εkk. εkk =. σkk 3λ + 2µ. (3.2.14). Substituting the equation 3.2.14 in 3.2.13: 1 σkk εi j = σi j − λ δi j 2 3λ + 2µ 47. (3.2.15).

(56) CHAPTER 3. ELASTICITY. MIC 2011-I0-2B. The Young’s Modulus E and the Poisson’s ratio ν are defined. E=. σ11 ε11. (3.2.16). ε22 ε11. (3.2.17). ν =−. The equations 3.2.15 and 3.2.16 are used to obtain E: 1 1 ε11 = σ11 − λ 2µ (3λ + 2µ)σ11 δ11. ε11 =. 1 1 σ11 σ11 [1 − λ ]= 2µ (3λ + 2µ) E. E=. E=. 2µ (3λ +2µ)−λ (3λ +2µ). 2µ(3λ + 2µ) 2λ + 2µ. E=. µ(3λ +2µ) λ +µ. (3.2.18). And with the equations 3.2.15 and 3.2.17 its obtain ν:. ν=. λ 2(λ +µ). (3.2.19). In addition, the bulk modulus K and the shear modulus G could be introduce: The bulk modulus represents the resistance to change of the volume and is defined from a hydrostatic stress path[16][11]: σi j = Keδi j. 48.

(57) CHAPTER 3. ELASTICITY. MIC 2011-I0-2B. Using the equation 3.2.11:. K=. E 3(1 − 2ν). (3.2.20). The shear modulus is defined in a pure path as[11]:. G=. τ 2ε12. Defining τ and ε12 : τ = σ12. ε12 =. σ12 2µ. σ12 G= σ 12 2 2µ. G=µ. (3.2.21). dev Some authors express the stiffness tensor di jkl in function of the volumetric Iivol jkl and deviator Ii jkl :. dev di jkl = (3λ + 2µ)Iivol jkl + 2µIi jkl. Then the tensor di jkl. (3.2.22). 1 Iivol jkl = δi j ⊗ δi j 3 1 Iidev jkl = Ii jkl − δi j ⊗ δi j 3 could be rewritten as:. h i vol dev di jkl = 3KIi jkl + 2GIi jkl. 49. (3.2.23).

(58) Chapter 4 Elastoplasticity 4.1. Introduction. The classic material mechanics determined the behavior of a material depending of it stress-strain curve. When this curve show two different trends, where the first trend represent the elastic behavior and the second the plastic behavior, the material is considered as elastoplastic. When the stresses are in the elastic part, its behavior is described as in the Chapter 3, but when the stresses are in the plastic part the strains are going to be permanent. In the Figure 4.1 is shown a load unload path. The point y represents the elastic to plastic change and this is known as the yielding point. The R0 − R path represent the unload in the plastic part. Then the strains could be divided in two: permanent and recoverable. A elastoplastic model represent the yielding by a yield surface, so is necessary defined the boundary that divides the elastic of the plastic domain (Yield criterion). When the plastic state is reached and the stress state is increasing, the yielding surface also increase until the material fails. Then the model has to determined how change the direction of the plastic strain (Flow rule) and the yield surface (Hardening laws).. 4.2. Elastoplastic Model. The equation that govern this behavior would be a modification of the Elasticity equation (3.2.2). 50.

(59) CHAPTER 4. ELASTOPLASTICITY. MIC 2011-I0-2B. Figure 4.1: Stress-Strain Curve 1D.[33] The strain tensor is decompose by a elastic and a plastic part: εi j = εiej + εipj. (4.2.1). σi j = di jkl : (εkl − εklp ). (4.2.2). σ˙i j = di jkl : εkl − εklp. (4.2.3). The rate of the tensor σ is:. 4.2.1. . Yield criterion. In the figure 4.1 the point y show the boundary between the elastic and the plastic domain. This point is known as the yield stress. As is mentioned previously, in the elastoplastic models the yield stress becomes in a surface and is represented by a scalar function f . This function is "defined in 51.

(60) CHAPTER 4. ELASTOPLASTICITY. MIC 2011-I0-2B. terms of stresses σ and an internal set of hardening variables ξ , f = f (σ , ξ )"[16]. Depending of its value, the material would be in a different state: f (σ , ξ ) < 0 → Material behaves elastically. (4.2.4). f (σ , ξ ) = 0 → Material yields. (4.2.5). f (σ , ξ ) > 0 → Impossible stress state. (4.2.6). In the last case, f (σ , ξ ) > 0 the stress state is impossible because when the yielding surface is reached and the stresses increase, it will increase at the same time. Then, it would never presented stresses out of the yielding surface.. 4.2.2. Flow rule. The flow rule define the ratio between the plastic strain increments related to the stress state. If the model involves more than two dimensions of stresses this is defined by a plastic potential instead the flow rule. The plastic potential is the surface formed by the "initial points of the incremental strain vectors for each stress state"[40][18]. The plastic strains ε p appear when the yield surface is reached, then start to change whit a rate ε̇ p , ∂g and its magnitude is given for the rate of the this rate is proportional to a unit flow rule tensor ∂ σi j consistency parameter or plastic multiplier λ [18]. The unit flow rule tensor, like the yield surface, is function of the stresses and an internal set of hardening variables. In some models its talk about an associative flow rule, which means that the yield surface and the plastic potential describe the same surface.. ε̇ipj = λ. 4.2.3. ∂ g(σi j , ξ ) ∂ σi j. (4.2.7). Kuhn-Tucker conditions. The magnitude of the plastic strain increments is given by λ , if the stress state is in the elastic domain, this has to be zero (λ = 0) and if it is in the plastic domain,the plastic multiplier greater than zero (λ > 0). Therefore, it has to met the following condition: 52.

(61) CHAPTER 4. ELASTOPLASTICITY. MIC 2011-I0-2B. λ ≥0. (4.2.8). Besides, the yield criterion gives another condition. f ≤0. (4.2.9). Always has to be fulfilled the following, considering all the above information:. f < 0 ⇔λ = 0 ⇔ ε p = 0. (4.2.10). f = 0 ⇔λ > 0 ⇔ ε p 6= 0. (4.2.11) (4.2.12). The product of λ and f must to be zero to met the above conditions. λf =0. (4.2.13). The above are the Kuhn-Tucker conditions. These are using in the mathematical programming for restrictions of inequalities.(See reference [25]).. 4.2.4. Plasticity models. 4.2.4.1. Perfect plasticity. In the perfect plasticity the yield surface remains fixed in the stress space. Therefore the yield surface is only function of the stresses f (σi j = 0) because there is no hardening. ∂g (p) σ̇i j = di jkl εkl − εi j = di jkl εkl − λ ∂ σkl. (4.2.14). The yield surface is written in differential form, this is known as the consistency condition f˙: 53.

(62) CHAPTER 4. ELASTOPLASTICITY. MIC 2011-I0-2B. ∂f f˙ = σ˙i j = 0 ∂ σi j. (4.2.15). The plastic multipler λ is obtain combining the equation 4.2.14 and 4.2.15: ∂ f ∂ f ∂ g f˙ = σ˙i j = di jkl εkl − λ =0 ∂ σi j ∂ σi j ∂ σkl. ∂f di jkl ε̇kl ∂ σi j λ= ∂f ∂g d pqrs ∂ σ pq ∂ σrs. (4.2.16). The elastoplastic tangent moduli is obtain with the equations 4.2.14 and 4.2.19 (ep). σ̇i j = di jkl ε̇kl. (4.2.17). With (ep). di jkl = di jkl −. 4.2.4.2. ∂g ∂ f dmnkl ∂ σab ∂ σmn ∂g ∂f d pqrs ∂ σ pq ∂ σrs. di jab. (4.2.18). Hardening plasticity. When the plastic state is reached, the yield surface present changes. This changes are explained by the hardening laws. Depending on how it is that change, the hardening is isotropic, kinematic or combined. ∂f ∂f ˙ f˙ = σ˙i j + ξ ∂ σi j ∂ξ ∂ f ∂ g ∂f ˙ f˙ = di jkl εkl − λ + ξ =0 ∂ σi j ∂ σkl ∂ξ 54.

(63) CHAPTER 4. ELASTOPLASTICITY. λ=. MIC 2011-I0-2B. ∂f di jkl ε̇kl ∂ σi j ∂f ∂g ∂ f ξ˙ d pqrs − ∂ σ pq ∂ σrs ∂ ξ λ. (4.2.19). The tensor σ would be:. (ep). σ̇i j = di jkl ε̇kl. (4.2.20). ∂g ∂ f dmnkl ∂ σab ∂ σmn (ep) di jkl = di jkl − ∂g ∂ f ξ˙ ∂f d pqrs − ∂ σ pq ∂ σrs ∂ ξ λ. (4.2.21). With. di jab. Isotropic hardening The hardening is isotropic when the center of the elastic domain remains at the origin, i.e. it expands isotropically with respect to plastic strains, as is shown in the figure 4.2.4.2. Considering this, the internal hardening variable ξ iso and the plastic modulus K are introduce. K is a constant greater o equal than zero (if it is less than zero represent a strain-softening response).[19]. (p). ξ iso = −Kkεi j k. (4.2.22). f (σi j , ξ ) = kσi j k − (σiyj − ξ iso ). (4.2.23). The equation of the yield criterion is:. 55.

(64) CHAPTER 4. ELASTOPLASTICITY. MIC 2011-I0-2B. Figure 4.2: Isotropic hardening: a) Load and unload path b) Yield surface. Kinematic hardening The hardening is kinematic when the center of the yield surface moves in direction of the plastic flow and the size of the elastic domain remains constant. This behavior is shown in the figure 4.2.4.2. To model this motion is necessary introduce another internal hardening variable ξ kin known as the back stress and represent the location of the center of the yield surface.. (p). ξ kin = −Hεi j. (4.2.24). Where H is the slope of the path of loading and unloading when f = 0 and it is called the kinematic hardening modulus. Therefor, the equation for the yield criterion is:. f (σi j , ξ ) = kσi j + ξ kin k − σiyj. Figure 4.3: Kinematic hardening: a) Load and unload path b) Yield surface 56. (4.2.25).

(65) CHAPTER 4. ELASTOPLASTICITY. MIC 2011-I0-2B. The isotropic hardening is very simplify of the real hardening behavior in materials, then is more used the kinematic or the combined. The yield condition f for the last case is:. f (σi j , ξ ) = kσi j + ξ kin k − σiyj + ξ iso. Figure 4.4: Combined hardening: a) Load and unload path b) Yield surface. 57. (4.2.26).

(66) Chapter 5 Modified Cam Clay 5.1. Introduction. One of the most common elastoplastic models is the Modified Cam Clay. This model is based on the critical state theory and was developed by Roscoe and Burland[23] in 1968. As its name implies, this method is used to described the behavior of clays.. 5.2. Critical State Theory. Some parameters are define for describing the state in a triaxial test: the effective pressure p0 and the deviator stress q 1 p0 = tr(σi j ) (5.2.1) 3 r 3 q= kTi j k (5.2.2) 2 Where: 1 Ti j = σi j − tr(σi j )δi j (5.2.3) 3 In the triaxial test, the axial stress is σ11 and the radial stress is σ33 , the above equations could be re-written: 1 p0 = (σ11 + 2σ33 ) (5.2.4) 3 58.

(67) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B q = σ11 − σ 33. (5.2.5). With the specific volume V = 1 + e (where e is the void ratio) it is possible to observe the changes in the behavior of the soil sample using a three dimensional plot (p0 , V , q space). Nevertheless, will be observe different planes first.. Volume - Pressure plane. Figure 5.1: Critical State Line in V − p0 plane. In the figure 5.2.6 is shown the normal consolidation line (NCL) and the swelling/recompresion line (SL). In the first, the slope is λ and in the second, κ; these are parameters of the soil. Also appear the Critical State Line (CSL) witch is parallel to the normal consolidation line. The equation of the critical state line is: ln e = N − λ ln p0. (5.2.6). For an isotropic case, the point where the line of loading and unloading crosses whit the normal p0 consolidation line is the half of the preconsolidation stress 2c (see figure 5.2.6). Then: p0c 0 1 + ex − (1 + eo ) = −κ ln − ln po 2 59.

(68) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B. ex = eo + κ. Similarly:. ln p0o. !. p0c 2. p0c N − (1 + ex ) = −λ ln 1 − ln 2 p0 ex = N − 1 − λ ln c 2. (5.2.7). (5.2.8). The expression for the preconsolidation stress could be obtain combining equations 5.2.7 y 5.2.8: ln p0o. !. p0c eo + κ = N − 1 − λ ln p0c 2 2 N − κ ln p + 1 + e pc = exp λ −κ. Deviator - Pressure plane. Figure 5.2: Critical State Line and yield surface in q − p0 plane [26] 60. (5.2.9).

(69) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B. The behavior of a soil sample under a drained triaxial test is shown in the plane q − p0 (see figure 5.2). The effective stresses are equal to the total stresses, then the ESP have a slope of 3. The failure occurs when the Critical State Line intersects the ESP in the point F[?]. The equations of this point will be: q f = 3(p0f − p00 ). (5.2.10). q f = M p0f. (5.2.11). The coordinates of the point could be find using the above equations: 3(p0f − p00 ) = M p0f 3p0f − M p0f = 3p00 p0f (3 − M) = 3p00 3p00 (3 − M) 3M p00 qf = (3 − M) p0f =. (5.2.12) (5.2.13). In the equations 5.2.12 and 5.2.13 could be see that if M is less or equal to 3, the stresses will be infinite or negative, witch could be no possible. Then the slope M must be always greater or equal than 3.. Yield Surface. Also, the yield surface is shown in the figure 5.2. For soils, Wong and Mitchel in 1975 showed that a good approximation for the yield surface could be an ellipse. The equation is: q2 = M 2 p0 (p0c − p0 ). (5.2.14). Then all the combinations of q and p0 than lie within the ellipse are going to behave elastically, but if the combination lies on the surface, the soil stars to yield and the surface to expand. [26] 61.

(70) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B. Critical State Line The Critical State Line represents the final stress state for which it could not be possible to shear the sample of soil without produce a change of volume of the soil[29]. In the figure 5.3 is shown the critical state line in a three dimensional space.. Figure 5.3: p0 ,V, q space [29]. 5.3. Model. The constitutive equation that would described the behavior of the material its the seen in the previous Chapter, because this is an elastoplastic model:. σ˙i j = di jkl : ε̇kl − ε̇klp 5.3.1. . (5.3.1). Yield criterion. As its shown above, the yield surface is given for the equation 5.2.14. Rewriting its obtain f = p02 − p0 pc + 62. q2 M2. (5.3.2).

(71) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B. Note, the size of the initial yield surface depends on the preconsolidation mean effective stress.. 5.3.2. Flow rule. As is shown in the Chapter 4, the flow rule is defined by the following equation: ε̇ipj = λ. ∂g ∂ σi j. (5.3.3). Where g represents the plastic potential. Nevertheless, this model consider an associative flow rule, i.e, g = f . Then: ∂f (5.3.4) ε̇ipj = λ ∂ σi j Is necessary develop the above equation, considering the equations 5.2.1, 5.2.2, 5.2.3 and 5.3.2: ∂f ∂ f ∂ p0 ∂ f ∂ q ∂ f ∂ p00 = + + ∂ σi j ∂ p0 ∂ σi j ∂ q ∂ σi j ∂ p00 ∂ σi j ∂f =2p0 − p0c ∂ p0 ∂ f 2q = ∂ q M2 ∂f = − p0 ∂ p0c. The derivatives of. (5.3.5). (5.3.6) (5.3.7) (5.3.8). ∂ p0 ∂ q ∂ p0c , y are found using the differential of a function. ∂ σi j ∂ σi j ∂ σi j . 1 1 dP =d tr(σi j ) = d (δi j : σi j ) 3 3 1 1 =d δi j : σi j + δi j : dσi j 3 3 1 = δi j : dσi j 3 ∂ p0 1 = δi j ∂ σi j 3 63. (5.3.9).

(72) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B. dq =d. ! r 3 3 kTi j k = dkTi j k 2 2. dkTi j k =d. p 1 d(Ti j : Ti j ) Ti j : Ti j = p 2 Ti j : Ti j. r. d(Ti j : Ti j ) =dTi j : Ti j + Ti j : dTi j = 2(Ti j : dTi j ) Ti j is the deviatoric part of the stress tensor. Then could be written as Ti j = Idev i jkl : σmn , then: dTi j =Idev i jkl : dσi j Ti j : Idev i jkl : dσi j dkTi j k = p Ti j : Ti j Ti j. dkTi j k =. z }| { (Idev i jkl : Ti j ). kTi j k dkTi j k Ti j = ∂ σi j kTi j k ∂q = ∂ σi j. r. : dσi j. 3 Ti j 2 kTi j k. (5.3.10). ∂ p0c =0 ∂ σi j. (5.3.11). With the equations 5.3.5 to 5.3.11 is obtaine:. ". 1 2q ε̇ipj = λ (2p0 − p0c ) δi j + 2 3 M. 5.3.3. r. 3 Ti j 2 kTi j k. # (5.3.12). Hardening laws. The variable that governs the grown of the yield surface is pc . The evolution equation is: ṗc = pc ϑ ε̇vpi j 64. (5.3.13).

(73) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B. Where ε̇vpi j is the change of the plastic and volumetric strains, and is defined as: ε̇vpi j = tr(ε̇ipj ) ϑ=. 5.3.4. (5.3.14). V λ −κ. (5.3.15). Consistent tangential moduli. In the Chapter 4, was deducted the general equation for the consistent tangential moduli. The equation 4.2.21 could be re-written for a three-dimensional case and consider an associative flow rule:. (ep). (p). di jkl = di jkl − di jkl di jab :. (p). di jkl = χ=. ∂f ∂f ⊗ : dmnkl ∂ σab ∂ σmn χ. ∂f ∂f ∂ f ξ˙ : d pqrs : − ∂ σ pq ∂ σrs ∂ ξ λ dev di jkl = 3KIvol i jkl + 2GIi jkl. (5.3.16). (5.3.17) (5.3.18) (5.3.19). The function χ could be re-witting as: # " r h i ∂f h i T 2q 3 1 i j dev 0 0 vol dev δ + : 3KI + 2GI ṗc χ = 3KIvol + 2GI : (2p − p ) i j c i jkl i jkl − i jkl i jkl 3 M 2 2 kTi j k ∂ξ 2 ∂f 2q 2 − ṗc =K(2p − pc ) + 3G 2 M ∂ξ. 2q 2 ∂ f χ =K(2p − pc ) + 3G − ṗc M2 ∂ pc 2 2q ∂f ∂f 2 − pc ϑ =K(2p − pc ) + 3G 2 M ∂ pc ∂p 2. . . 2q χ = K(2p − pc ) + 3G M2 2. 65. 2 − ppc ϑ (2p − pc ). (5.3.20).

(74) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B. Develop the equation di jab :. ∂f = ∂ σab. ∂f : dmnkl ∂ σmn. " # r h i T 2q 3 i j dev 0 0 1 = 3KIvol i jkl + 2GIi jkl : (2p − pc ) δi j + 2 3 M 2 kTi j k r 1 2q 3 Ti j =3K(2p0 − p0c ) δi j + 2G 2 3 M 2 kTi j k (p). The numerator of di jkl would be: r r T 2q 3 1 2q 3 Ti j 1 i j ⊗ 3K(2p0 − p0c ) δi j + 2G 2 3K(2p0 − p0c ) δi j + 2G 2 3 M 2 kTi j k 3 M 2 kTi j k √ Ti j Ti j 24qG Ti j 0 0 2GKq 0 0 2 [K(2p − pc )] δi j ⊗ δi j + 6(2p − pc ) + ⊗ δi j + δi j ⊗ 2 M kTi j k kTi j k M 2 kTi j k. 5.4. Implementation. The following is the implementation of the model proposed by Borja. Which use and implicit integration, than means that the plastic multiplier is found using an iterative method.. 5.4.1. Parameters. Name Compression index Swelling index Initial preconsolidation mean stress Critical state slope in the space q vs. p Poisson ratio. Symbol λ κ Pc0 M ν. Table 5.1: Parameters Modified Cam Clay. 66.

(75) CHAPTER 5. MODIFIED CAM CLAY. MIC 2011-I0-2B. Name Initial Cauchy stress tensor Initial plastic strain rate tensor Void Ratio Preconsolidation mean stress. Symbol σi j ε̇i j e Pc. Table 5.2: Initial Conditions Modified Cam Clay. 5.4.2. Initial Conditions. 5.4.3. Elastic Relations. 1. Define material parameters and constants • Mean Stress p=. Tr(σi j ) 3. K=. (1 + e)p κ. • Bulk modulus. • Young modulus E = 3K(1 − 2ν) • Shear modulus G=. E 2(1 + ν). 2. Calculated the elastic tangent moduli E E ∗ Ivol ∗ Idev i jkl + 1−2∗ν 1 + ν i jkl. di jkl = 3. Trial Elastic Step. σitrial = σi j + di jkl : εi j j trial. Tr(σitrial j ). p. =. Titrial j. = σitrial − ptrial δi j j. 3 r. trial. q. = 67. 3 norm(Titrial j ) 2.

(76) CHAPTER 5. MODIFIED CAM CLAY. 5.4.4. MIC 2011-I0-2B. Yield condition 2. f. trial. qtrial + ptrial (ptrial − pc) = 2 M. • if f trial < 0 is elastic σi j = σitrial j p = ptrial q = qtrial e = e + Tr(εi j )(1 + e). • if f trial ≥ 0 → plastic corrector. 5.4.5. Plastic corrector. 1. Determinate the consistency parameter 1+e λ −κ ∆φ = 0 ϑ=. pc ini = pc Gtol = 10−7 Ftol = 10−7 68.