extension of mononobe-okabe approach

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EXTENSION OF MONONOBE-OKABE APPROACH TO UNSTABLE SLOPES by Sara Ebrahimi A thesis submitted to the Faculty of the University of Delaware in partial fulfillment of the requirements for the degree of Master of Civil Engineering Spring 2011 Copyright 2011 Sara Ebrahimi All Rights Reserved

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  • EXTENSION OF MONONOBE-OKABE

    APPROACH TO UNSTABLE SLOPES

    by

    Sara Ebrahimi

    A thesis submitted to the Faculty of the University of Delaware in partial fulfillment of the requirements for the degree of Master of Civil Engineering

    Spring 2011

    Copyright 2011 Sara Ebrahimi

    All Rights Reserved

  • EXTENSION OF MONONOBE-OKABE

    APPROACH TO UNSTABLE SLOPES

    by Sara Ebrahimi

    Approved: __________________________________________________________ Dov Leshchinsky, Ph.D. Professor in charge of thesis on behalf of the Advisory Committee Approved: __________________________________________________________ Harry W. Shenton III, Ph.D. Chair of the Department of Civil and Environmental Engineering Approved: __________________________________________________________ Michael J. Chajes, Ph.D. Dean of the College of Engineering Approved: __________________________________________________________ Charles G. Riordan, Ph.D.

    Vice Provost for Graduate and Professional Education

  • iii

    ACKNOWLEDGMENTS

    I owe my deepest gratitude and appreciation to my advisor Dr. Dov

    Leshchinsky for all his kind and unfailing support throughout the project. It was an

    honor for me to study under his supervision at the University of Delaware.

    I also would like to thank to Dr. Christopher L. Meehan and Dr. Victor N.

    Kaliakin, for their instructions and help during my graduate courses. I would also like

    to extend my gratitude to Mr. Fan Zhu for his assistance with the formulation and

    programming that was conducted during this project.

    Very special thanks to my husband and my best friend, Farshid. I would

    not have been able to complete my thesis without his encouragement, help and

    support.

    Finally, I like to gratefully thank my parents for their never-ending love

    and support.

  • iv

    TABLE OF CONTENTS

    LIST OF TABLES ......................................................................................................... vLIST OF FIGURES ....................................................................................................... viABSTRACT ................................................................................................................ xvi

    Chapter 1 INTRODUCTION .............................................................................................. 1 2 LITERATURE REVIEW ................................................................................... 4 3 PROBLEM DEFINITION AND FORMULATION ........................................ 10 4 RESULTS OF ANALYSIS .............................................................................. 26

    4.1. Kae-h Versus Batter Relationship ....................................................... 264.2. Effect of Vertical Seismic Coefficient ............................................. 664.3. Slip Surfaces ..................................................................................... 754.4. Studying Seismic-Induced Resultant Force ...................................... 88

    5 COMPARISON OF RESULTS ..................................................................... 100 6 CONCLUSION AND RECOMMENDATION ............................................. 108REFERENCES ........................................................................................................... 110

  • v

    LIST OF TABLES

    Table 5.1 Comparison of Kae-h (= 0, =0o, Kv=0) ............................................ 102Table 5.2 Comparison of Kae-h (= 30o, =0o, Kv=0) .......................................... 103Table 5.3 Comparison of Kae-h (= 45o, =0o, Kv=0) .......................................... 104Table 5.4 Comparison of Kae-h (= 0o, =0o, Kv=0) .......................................... 105Table 5.5 Comparison of Kae-h (= 30o, =0o, Kv=0) ........................................ 106Table 5.5 Comparison of Kae-h (= 30o, =0o, Kv=0) ........................................ 107

  • vi

    LIST OF FIGURES

    Figure 2.1 Notations and Conventions Used in the M-O Method ........................ 6

    Figure 3.1 Log-Spiral Failure Mechanism .......................................................... 11

    Figure 3.2 Notation and Rational Direction of Force Components Producing the Pseudostatic Resultant ................................................ 13

    Figure 3.3 Different Areas Used in Writing the Moment Equations Due to the Weight of Sliding Mass .............................................................. 15

    Figure 3.4 Direction of Resultant Force, Pae, Commonly Used in Classical Coulomb and M-O Analyses .............................................. 22

    Figure 4.1.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20, Backslope 1: , /=0 & 1/3) .................... 28

    Figure 4.1.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20, Backslope 1: , /=2/3 & 1) ................... 28

    Figure 4.2.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: , /=0 & 1/3) ................... 29

    Figure 4.2.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: , /=2/3 & 1 ) .................. 29

    Figure 4.3.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: , /=0 & 1/3) ................... 30

    Figure 4.3.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: , /=2/3 & 1) ................. 30

    Figure 4.4.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: , /=0 & 1/3) .................. 31

    Figure 4.4.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: , /=2/3 & 1) ................... 31

    Figure 4.5.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20, Backslope 1:10, /=0 & 1/3) ................. 32

    Figure 4.5.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20, Backslope 1: 10, /=2/3 & 1) ................ 32

  • vii

    Figure 4.6.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1:10, /=0 & 1/3) ................. 33

    Figure 4.6.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 10, /=2/3 & 1) ................. 33

    Figure 4.7.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: 10, /=0 & 1/3) .................. 34

    Figure 4.7.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: 10, /=2/3 & 1) .................. 34

    Figure 4.8.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: 10, /=0 & 1/3) .................. 35

    Figure 4.8.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: 10, /=2/3 & 1) .................. 35

    Figure 4.9.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20, Backslope 1: 5, /=0 & 1/3) .................... 36

    Figure 4.9.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20, Backslope 1: 5, /=2/3 & 1) .................... 36

    Figure 4.10.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 5, /=0 & 1/3) .................... 37

    Figure 4.10.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 5, /=2/3 & 1) .................... 37

    Figure 4.11.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: 5, /=0 & 1/3) .................... 38

    Figure 4.11.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: 5, /=2/3 & 1) .................... 38

    Figure 4.12.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: 5, /=0 & 1/3) .................... 39

    Figure 4.12.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: 5, /=2/3 & 1) .................... 39

  • viii

    Figure 4.13.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20, Backslope 1: 3, /=0 & 1/3) .................... 40

    Figure 4.13.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20, Backslope 1: 3, /=2/3 & 1) .................... 40

    Figure 4.14.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 3, /=0 & 1/3) .................... 41

    Figure 4.14.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 3, /=2/3 & 1) .................... 41

    Figure 4.15.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: 3, /=0 & 1/3) .................... 42

    Figure 4.15.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: 3, /=2/3 & 1) .................... 42

    Figure 4.16.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: 3, /=0 & 1/3) .................... 43

    Figure 4.16.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: 3, /=2/3 & 1) .................... 43

    Figure 4.17.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 2, /=0 & 1/3) .................... 44

    Figure 4.17.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 2, /=2/3 & 1) .................... 44

    Figure 4.18.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: 2, /=0 & 1/3) .................... 45

    Figure 4.18.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40, Backslope 1: 2, /=2/3 & 1) .................... 45

    Figure 4.19.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: 2, /=0 & 1/3) .................... 46

    Figure 4.19.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50, Backslope 1: 2, /=2/3 & 1) .................... 46

  • ix

    Figure 4.20.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20, Backslope 1:, /=0 & 1/3 )................... 47

    Figure 4.20.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20, Backslope 1: , /=2/3 & 1) ................... 47

    Figure 4.21.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:, /=0 & 1/3 )................... 48

    Figure 4.21.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: , /=2/3 & 1) ................... 48

    Figure 4.22.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1:, /=0 & 1/3 )................... 49

    Figure 4.22.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1: , /=2/3 & 1) ................... 49

    Figure 4.23.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1:, /=0 & 1/3 )................... 50

    Figure 4.23.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1: , /=2/3 & 1) ................... 50

    Figure 4.24.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20, Backslope 1:10, /=0 & 1/3 ) ................. 51

    Figure 4.24.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20, Backslope 1: 10, /=2/3 & 1) .................. 51

    Figure 4.25.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:10, /=0 & 1/3 ) ................. 52

    Figure 4.25.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: 10, /=2/3 & 1) .................. 52

    Figure 4.26.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1:10, /=0 & 1/3 ) ................. 53

    Figure 4.26.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1: 10, /=2/3 & 1) .................. 53

  • x

    Figure 4.27.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1:10, /=0 & 1/3 ) ................. 54

    Figure 4.27.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1: 10, /=2/3 & 1) .................. 54

    Figure 4.28.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20, Backslope 1:5, /=0 & 1/3 ) ................... 55

    Figure 4.28.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20, Backslope 1: 5, /=2/3 & 1) .................... 55

    Figure 4.29.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:5, /=0 & 1/3 ) ................... 56

    Figure 4.29.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: 5, /=2/3 & 1) .................... 56

    Figure 4.30.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1:5, /=0 & 1/3 ) ................... 57

    Figure 4.30.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1: 5, /=2/3 & 1) .................... 57

    Figure 4.31.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1:5, /=0 & 1/3 ) ................... 58

    Figure 4.31.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1: 5, /=2/3 & 1) .................... 58

    Figure 4.32.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20, Backslope 1:3, /=0 & 1/3 ) ................... 59

    Figure 4.32.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20, Backslope 1: 3, /=2/3 & 1) .................... 59

    Figure 4.33.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:3, /=0 & 1/3 ) ................... 60

    Figure 4.33.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: 3, /=2/3 & 1) .................... 60

  • xi

    Figure 4.34.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1:3, /=0 & 1/3 ) ................... 61

    Figure 4.34.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1: 3, /=2/3 & 1) .................... 61

    Figure 4.35.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1:3, /=0 & 1/3 ) ................... 62

    Figure 4.35.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1: 3, /=2/3 & 1) .................... 62

    Figure 4.36.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:2, /=0 & 1/3 ) ................... 63

    Figure 4.36.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: 2, /=2/3 & 1) .................... 63

    Figure 4.37.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1:2, /=0 & 1/3 ) ................... 64

    Figure 4.37.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40, Backslope 1: 2, /=2/3 & 1) .................... 64

    Figure 4.38.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1:2, /=0 & 1/3 ) ................... 65

    Figure 4.38.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50, Backslope 1: 2, /=2/3 & 1) .................... 65

    Figure 4.39.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1:, Kh=0.1, /=0 ) ............... 67

    Figure 4.39.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: , Kh=0.3, /=0) ................ 67

    Figure 4.40.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1:5, Kh=0.1, /=0 ) ................ 68

    Figure 4.40.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 5, Kh=0.3, /=0) ................ 68

  • xii

    Figure 4.41.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1:, Kh=0.1, /=2/3 ) ............ 69

    Figure 4.41.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: , Kh=0.3, /=2/3) ............ 69

    Figure 4.42.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1:5, Kh=0.1, /=2/3 ) ............. 70

    Figure 4.42.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30, Backslope 1: 5, Kh=0.3, /=2/3) ............. 70

    Figure 4.43.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:, Kh=0.1, /=0 ) ............... 71

    Figure 4.43.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: , Kh=0.3, /=0) ................ 71

    Figure 4.44.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:5, Kh=0.1, /=0 ) ................ 72

    Figure 4.44.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: 5, Kh=0.3, /=0) ................ 72

    Figure 4.45.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:, Kh=0.1, /=2/3 ) ............ 73

    Figure 4.45.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: , Kh=0.3, /=2/3) ............ 73

    Figure 4.46.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:5, Kh=0.1, /=2/3 ) ............. 74

    Figure 4.46.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1: 5, Kh=0.3, /=2/3) ............. 74

    Figure 4.47.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20, =0,Backslope 1: (Eq. 3-17) ........................................................................................... 76

    Figure 4.47.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20, =70,Backslope 1: (Eq. 3-17) ........................................................................................... 76

  • xiii

    Figure 4.48.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30, =0,Backslope 1: (Eq. 3-17) ........................................................................................... 77

    Figure 4.48.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30, =60,Backslope 1: (Eq. 3-17) ........................................................................................... 77

    Figure 4.49.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40,=0,Backslope 1: (Eq. 3-17) ........................................................................................... 78

    Figure 4.49.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =50,Backslope 1: (Eq. 3-17) ........................................................................................... 78

    Figure 4.50.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =50, =0,Backslope 1: (Eq. 3-17) ........................................................................................... 79

    Figure 4.50.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =50, =40,Backslope 1: (Eq. 3-17) ........................................................................................... 79

    Figure 4.51.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20, =0,Backslope 1: 5(Eq. 3-17) ........................................................................................... 80

    Figure 4.51.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20, =70,Backslope 1: 5 (Eq. 3-17) ........................................................................................... 80

    Figure 4.52.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =0,Backslope 1: 5 (Eq. 3-17) ........................................................................................... 81

    Figure 4.52.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =50,Backslope 1: 5 (Eq. 3-17) ........................................................................................... 81

    Figure 4.53.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20, =0,Backslope 1: (Eq. 3-22) ........................................................................................... 82

  • xiv

    Figure 4.53.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20, =70,Backslope 1: (Eq. 3-22) ........................................................................................... 82

    Figure 4.54.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30, =0,Backslope 1: (Eq. 3-22) ........................................................................................... 83

    Figure 4.54.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30, =60,Backslope 1: (Eq. 3-22) ........................................................................................... 83

    Figure 4.55.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =0,Backslope 1: (Eq. 3-22) ........................................................................................... 84

    Figure 4.55.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =50,Backslope 1: (Eq. 3-22) ........................................................................................... 84

    Figure 4.56.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =0,Backslope 1: (Eq. 3-22) ........................................................................................... 85

    Figure 4.56.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =50,Backslope 1: (Eq. 3-22) ........................................................................................... 85

    Figure 4.57.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20, =0,Backslope 1: 5(Eq. 3-22) ........................................................................................... 86

    Figure 4.57.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20, =70,Backslope 1: 5(Eq. 3-22) ........................................................................................... 86

    Figure 4.58.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =0,Backslope 1: 5(Eq. 3-22) ........................................................................................... 87

    Figure 4.58.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40, =50,Backslope 1: 5(Eq. 3-22) ........................................................................................... 87

  • xv

    Figure 4.59 Kae-h Versus Batter Using Eq. 3-17 for =20, Kv=0, and (a) =0; (b) =1/3; (c) =2/3; and (d) =1 ............................................. 91

    Figure 4.60 Kae-h Versus Batter Using Eq. 3-17 for =30, Kv=0, and (a) =0; (b) =1/3; (c) =2/3; and (d) =1 ............................................. 92

    Figure 4.61 Kae-h Versus BatterUsing Eq. 3-17 for =40, Kv=0, and (a) =0; (b) =1/3; (c) =2/3; and (d) =1 ............................................. 93

    Figure 4.62 Kae-h Versus Batter Using Eq. 3-17 for =50, Kv=0, and (a) =0; (b) =1/3; (c) =2/3; and (d) =1 ............................................. 94

    Figure 4.63 Kae-h Versus Batter Using Eq. 3-22 for =20, Kv=0, and (a) =0; (b) =1/3; (c) =2/3; and (d) =1 ............................................. 95

    Figure 4.64 Kae-h Versus Batter Using Eq. 3-22 for =30, Kv=0, and (a) =0; (b) =1/3; (c) =2/3; and (d) =1 ............................................. 96

    Figure 4.65 Kae-h Versus Batter Using Eq. 3-22 for =40, Kv=0, and (a) =0; (b) =1/3; (c) =2/3; and (d) =1 ............................................. 97

    Figure 4.66 Kae-h Versus Batter Using Eq. 3-22 for =50, Kv=0, and (a) =0; (b) =1/3; (c) =2/3; and (d) =1 ............................................. 98

    Figure 4.67 Impact of Vertical Seismicity ............................................................ 99

  • xvi

    ABSTRACT

    The resultant force of lateral earth pressures is commonly used in design

    of nearly vertical walls while flatter slopes are designed to be internally stable using a

    factor of safety approach. An unstable slope is considered to have unsatisfactory factor

    of safety unless supported by internal and/or external measures. However, from

    analytical viewpoint, the distinction between walls and unstable slopes is unnecessary.

    Using limit equilibrium analysis combined with a log spiral surface, a previous

    formulation is extended to deal with pseudostatic instability of simple, homogenous,

    cohesionless slopes. Hence, the original approach by Mononobe-Okabe (M-O) is

    extended to yield the resultant lateral force needed to stabilize an unstable slope.

    Given the slope angle, the design internal angle of friction, the backslope, the

    surcharge, the vertical and horizontal seismic acceleration, and the inclination of the

    resultant force, one can calculate the magnitude of this resultant. The approach allows

    for the selection of a rational inclination of the resultant for cases where soil-face

    interaction is likely to develop along vertical segments only. The approach generalizes

    the Coulomb (static) and the M-O (pseudostatic) methods as all are in the same

    framework of limit equilibrium. While all methods yield identical results for vertical

    slopes, where the critical slip surface defining the active wedge degenerates to the

    same planar surface, the presented approach becomes more critical for flatter unstable

    slopes where the active wedge is augmented by a curved surface. Hence, seamless

    extension of the M-O approach is produced.

  • 1

    Chapter 1

    INTRODUCTION

    The MononobeOkabe (M-O) method (Okabe 1926; Mononobe and

    Matsuo 1929) has been used in practice for decades to assess the resultant of lateral

    earth pressure acting on earth retaining structures. The formulation satisfies global

    force equilibrium for an active wedge leaning against the retaining wall. In assessing

    the stability of the wall, the location of the resultant force needs to be assumed as it is

    not part of the force equilibrium formulation. Current retrospective of the M-O

    approach is provided by Al Atik and Sitar (2010). In essence, the M-O formulation is

    an extension of the Coulomb formulation to include pseudostatic inertia force

    components due to ground acceleration. The seismic loading is momentary and not

    permanent as assumed in the pseudostatic approach. Hence, it is common in design

    guidelines to recommend using a fraction of the anticipated peak ground acceleration,

    PGA, typically 0.3 to 0.5 of PGA (e.g., Leshchinsky et al. 2009).

    Similar to Coulombs method, the M-O method becomes unconservative

    in the context of limit state formulation as the batter increases (e.g., Leshchinsky and

    Zhu, 2010). Simply, a planar mechanism for the active wedge assumed by the

    Coulomb method or the M-O method is less critical than a curved surface such as a

    log spiral. The objective of this work is to produce the corresponding lateral earth

    pressure coefficient which is compatible with the M-O concept but is theoretically

    valid for any batter representing unstable slopes. It uses the same framework of

    formulation as done by Leshchinsky and Zhu (2010), providing algorithms suitable for

  • 2

    pseudostatic loading combined with an active wedge defined by a log spiral as well as

    some practical results. The log spiral surface degenerates to a planar surface when the

    batter approaches zero and therefore, the present formulation provides seamless

    extension to the M-O formulation dealing with unstable slopes.

    Traditionally, geotechnical practice distinguishes between slopes and

    walls (e.g., FHWA 2009). Hence, seeking the resultant of lateral earth pressure in

    conjunction with slopes may appear awkward. However, unstable slopes (i.e., slopes

    for which the common factor of safety on shear strength is not in excess of one), need

    to be supported externally, internally, or both to ensure its long term performance.

    External support of unstable slopes can be achieved by using large concrete blocks,

    gabions, geocells infilled with soil, etc., stacked with a setback (batter), capable of

    sustaining the lateral pressures exerted by the slope. Design of such structures requires

    knowledge of the resultant force exerted by the slope so that sufficient resistance to

    sliding, overturning, and bearing failure can be provided. Internal support can be

    achieved using reinforcement (e.g., geosynthetics) connected to slender facing units.

    In this case, the sum of maximum tensile forces in all reinforcement layers, ignoring

    the impact of the bottom slim facing unit, is equal to the horizontal component of the

    resultant force. The National Concrete Masonry Association (NCMA 1997) requires

    that the sum of the connection forces (i.e., connection between the reinforcement and

    facings) should be equal to the horizontal component of the resultant force.

    Furthermore, the external stability of reinforced soil walls is assessed by considering

    the resultant force acting on the coherent reinforced mass. This force is calculated

    using the face batter as the inclination of the unstable slope retained by the reinforced

  • 3

    mass. Hence, extending the M-O concept to deal with unstable slopes is beyond just

    an academic interest; it has immediate design implications.

    The objective of this thesis is using limit equilibrium analysis combined

    with log spiral surface to extend the Mononobe-Okabe (M-O) formulation to deal with

    pseudostatic instability of simple, homogenous, cohesionless slopes. Chapter Two

    contains a literature review on the Mononobe-Okabe (M-O) method analysis. The

    formulation using a modified LE approach to find the lateral seismic earth pressure

    coefficient required to resist the soil is presented in Chapter Three. In Chapter Four,

    design charts of equivalent horizontal seismic lateral earth pressure coefficient and

    trace of critical slip surface are presented. Chapter Five provides a numerical

    comparison the results obtained from the M-O equation, the limit analysis (LA), and

    the equations developed by this work. Finally, in Chapter Six, conclusion of this work

    and recommendations for further study are presented.

  • 4

    Chapter 2

    LITERATURE REVIEW

    The dynamic analysis of earth retention systems is commonly simplified

    in practice by utilizing analytical methods which incorporate a pseudostatic force to

    simulate transient ground motion (e.g., Okabe 1926; Mononobe and Matsuo 1929;

    Seed and Whitman 1970; Steedman and Zeng 1990; Kim et al. 2010). Among several

    methods, the well-accepted M-O method (Okabe 1926; Mononobe and Matsuo 1929)

    is most widely utilized by current codes for the seismic design of earth retention

    systems (e.g., AASHTO 2007, Anderson et al. 2009). In these design codes, earth

    retention systems are designed in a way to counterbalance, by a satisfactory factor of

    safety, the resultant force of dynamic earth pressures that is determined using the M-O

    method.

    The development of the M-O method was motivated by catastrophic

    damages to many retaining structures observed after the 1923 Kanto earthquake,

    Japan. The method follows the limit-equilibrium approach and is basically a

    pseudostatic extension of the classical Coulomb (1776) lateral earth pressure theory.

    The M-O equation was derived for yielding retaining walls with cohesionless backfill

    materials and the equation does not take into account the cohesion of backfill soil, wall

    adhesion and external surcharge loading. The M-O method assumes that at the failure

    point, a planar failure surface is developed instantaneously behind the wall and the

    failure soil wedge behaves as a rigid body. The thrust point is assumed at 1/3 the

    height of the wall from the base. Figure 2.1 shows the forces acting on the active

  • 5

    wedge in the M-O analysis. Using the M-O approach, the active thrust under seismic

    conditions can be expresses as Equation (2-1):

    P HK1 K (2-1)

    Where is the unit weight of soil, H is the height of the earth structure, Kv is the vertical seismic coefficient, and Kae is the active seismic earth pressure

    coefficient. Kae is given in following equation:

    K cos

    cos cos cos 1 sin sin cos cos

    (2-2)

    Where: is the soil friction angle, is the batter, =tan-1(Kh/1-Kv), is the soil-facing interface friction angle and is the backslope.

  • 6

    Figure 2.1. Notations and Conventions Used in the M-O Method

    The performance of the M-O method and its underlying assumptions have

    been extensively evaluated through both numerical and experimental studies (e.g.;

    Seed and Whitman 1970; Sherif et al. 1982; Oritz et al. 1983; Ishibashi and Fang

    1987; Zeng and Steedman 1993; Veletsos and Younan 1994; Psarropoulos et al. 2005;

    Nakamura 2006; Al Atik and Sitar 2010). Several attempts have been made to modify

    the M-O method in order to improve the accuracy and address the limitations

    associated with the original M-O method (e.g., Seed and Whitman 1970; Fang and

    Chen 1995; Koseki et al. 1998; Kim et al., 2010).

    Similar to the Coulomb theory, the M-O theory assumes a planar slip

    surface for developing the failure wedge (Figure 2.1). However, experimental studies

    (e.g., Nakamura 2006) have shown that employing a planer surface cannot truly

    characterize the dynamic response of earth retention systems and the magnitude of

    seismic earth pressure is influenced by the assumed shape of failure surface. Based on

    these extensive investigations, it has been found that a curved failure surface or a two-

  • 7

    part surface (i.e., a curve in the lower part and a straight line in the upper part) is more

    compatible with the real failure surface formed in the backfill soil under dynamic

    loadings. The accurate equation for a curved failure surface is complex and yet has not

    been derived; instead, the curved surface is commonly represented by a circle or log

    spiral (e.g., Chang and Chen 1982; Fang and Chen 1995; Hazarika 2009)

    In the M-O method, the point of action of the resultant active thrust was

    taken at 1/3 the height of the wall above its base (Figure 2.1). However, this location

    has been extensively challenged in the literature (Kramer 1996). Seed and Whitman

    (1970) argued that the position of the resultant dynamic force varies in a range

    between 0.5H to 2H/3 depending on the magnitude of the earthquake ground

    acceleration. Within this range, Seed and Whitman (1970) recommended that D is

    equal to 0.6H be used as a rational value for design purposes. More recently,

    however, Al Atik and Sitar (2010) performed a set of experimental and numerical

    analyses and showed that D = H/3 is a more reasonable assumption for the position of

    the resultant.

    Koseki et al. (1998) modified the M-O method considering the

    progressive failure and strain localization phenomena. For this purpose, Koseki et al.

    (1998) proposed a graphical procedure to reduce the postpeak angle of friction in the

    backfill soil. In a similar attempt, Zhang and Li (2001) mathematically investigated

    the effect of strain localization on the M-O method.

    Since the M-O method is developed based upon the pseudostatic

    approach, there is an inherent over-conservatism associated with the method due to

    simulating transient ground motion by a constant force assumption. To overcome to

    this over-conservatism, the seismic coefficient is usually taken as a fraction of the

  • 8

    expected earthquake peak ground acceleration for design purposes (e.g., FHWA 1998;

    Leshchinsky et al. 2009; Anderson et al. 2009). For example, for the design of non-

    gravity cantilever walls, current AASHTO LRFD Bridge Design Specifications

    recommend a seismic design coefficient that corresponds to half of the earthquake

    peak ground acceleration (AASHTO 2007). Acknowledging the over-conservatism of

    current design methods, Al Atik and Sitar (2010) more recently suggested that seismic

    earth pressures on cantilever retaining walls can be neglected at peak ground

    accelerations below 0.4g for wall with horizontal backslope. In a similar fashion, Bray

    et al. (2010) recommended 0.3g as the boundary value below which there is no need

    for seismic design.

    The pseudostatic approach does not consider the amplification of the

    ground motion near the ground surface which will lead to a linear distribution of the

    resultant dynamic force along the wall height. To overcome to this drawback,

    Steedman and Zeng (1990) introduced a pseudo-dynamic approach which accounts for

    the time effect and phase change in shear and primary waves propagating in the

    backfill. The pseudo-dynamic method gives a non-linear seismic active earth pressure

    distribution behind the earth retention system and it has been further investigated and

    extended by others (e.g., Choudhury and Nimbalkar 2006; Ghosh 2008).

    In parallel to pseudostatic limit-equilibrium equations, several pseudo-

    static methods have been proposed using the limit analysis theory (e.g., Chang and

    Chen 1982; Chen and Liu 1990; Soubra and Macuh 2001). These methods are

    developed based on kinematically admissible failure mechanisms along with a yield

    criterion and a flow rule for the backfill soil, both of which are enforced along

    predefined slip surfaces (Mylonakis et al. 2007). Following this approach, seismic

  • 9

    earth pressure coefficients are derived by taking into account the equilibrium of

    external work and the internal energy dissipation.

    In this thesis an algorithm solving the moment equilibrium equation for a

    log-spiral slip surface is provided. Such a surface degenerates to the M-O planar

    surface when the slope face is near vertical. Hence, it provides a seamless extension

    to the M-O method dealing with unstable slopes. The formulation and solution scheme

    is basically a pseudo-static extension of the work presented by Leshchinsky and Zhu

    (2010). Implementing the presented algorithm in a computer code is simple as it

    represents a closed-form solution. Also, instructive charts, which constitute the critical

    solution to the log-spiral analysis, are presented in the familiar format. The

    formulation and results are limited to cohesionless soils.

  • 10

    Chapter 3

    PROBLEM DEFINITION AND FORMULATION

    Log-spiral slip surface has been used in the limit equilibrium (LE)

    analysis by Rendulic (1935) and Taylor (1937). Baker and Garber (1978)

    mathematically gained a log-spiral slip surface by using the variational limit

    equilibrium (LE) analysis with no prior assumption. The benefit of using the LE

    analysis of homogeneous soil is assessing particular problem without resorting to the

    static assumption. The moment equilibrium equation can be written for an assumed

    slip surface without explicit knowledge of the normal stress over that surface.

    Therefore, the LE moment equation can be solved iteratively until the critical log-

    spiral is found; i.e., the spiral that yields the lowest factor of safety is identified.

    Using notation proposed by Baker and Garber (1978), log-spiral geometry

    can be expressed as follows:

    R Ae (3-1)

    Where R is the radius of log-spiral, A is a constant (analogous to the

    radius of a circle), and = tan ()/ Fs, where is the internal angle of friction of soil and Fs is the safety factor for that log-spiral.

  • 11

    In this problem, a homogenous and cohesionless soil is considered. In

    design one will reduce the actual by a selected Fs value thus producing a design value of and subsequently that renders a system in a LE state. Figure 3.1 shows the failure mechanism expressed by Equation (3-1).

    Figure 3.1. Log-Spiral Failure Mechanism

    Figure 3.2 shows the notation and convention used in formulating the

    extended M-O problem. For the assumed direction of the resultant vector, reacting to

    the lateral earth pressure on line 1-3 (Figure 3.2), the classical expression for the

    horizontal component of this resultant force is:

  • 12

    P P 12 H1 KK cos 12 H

    1 KK (3-3)

    Where Pae_h is the horizontal component of the resultant Pae (i.e., Pae_h =

    Pae cos()); is the soil-facing interface friction angle; H is the height of the slope; is the unit weight of the soil; Kae is the pseudostatic lateral earth pressure coefficient

    assuming that the interface friction acts only along vertical surfaces (Leshchinsky and

    Zhu 2010); Kv is the vertical acceleration normalized relative to gravity g (fraction of

    PGA; note that the upward direction is positive); and Kae_h is a convenient parameter

    directly rendering the horizontal component of the resultant force.

    The resultant force is obtained from solving the moment equilibrium equation

    for an active wedge defined by a log spiral. The moment equilibrium equation written

    about the pole (Xc, Yc) of a log-spiral in a LE state, is independent of the normal stress

    distribution. Figure 3.2 shows forces acting on the log-spiral sliding mass. For

    cohesionless soils, the resultant of any elemental normal and shear stress at each point

    along the slip surface is passing through the pole thus the moment equilibrium

    equation can be presented as (Leshchinsky and Zhu 2010):

    Ph P tan h 1 KWh KWh+ (the moment due to the surcharge load)

    (3-4)

  • 13

    Where W represents the weight of sliding soil mass, and h1, h2, h3 and h4

    are moment leverage arms as illustrated in Figure 3.2. Also the moment induced by

    the surcharge load needs to be included to complete the equation.

    Figure 3.2. Notation and Rational Direction of Force Components Producing the Pseudostatic Resultant

    Considering the log-spiral geometry and Figure 3.2, it can be seen:

    h R cos D Ae cos D (3-5) h R sin D tan Ae sin D tan (3-6)

  • 14

    Where 1 and 2 are the polar coordinates of point 1 and 2 (See Figure 3.2; Point 1 is at the origin of the Cartesian coordinates where the slip surface emerges

    and Point 2 is the point where this surface starts).

    In the right hand side of the moment equilibrium equation (Equation 3-4)

    three terms needed to be defined.

    Calculation of Wh3:

    To define this term, it is needed to consider the moment generated by the

    weight of the sliding mass for a simple log-spiral with no backslope and a vertical

    facing (area (1-2-5) in Figure 3-3) as modified by Leshchinsky and San (1994) and

    subtracting areas of (1-3-6), (3-4-5-6) and (2-3-4) as shown in Figure 3.3.

    For a mass for a simple log-spiral with no backslope and a vertical facing

    (area (1-2-5) in Figure 3-3) Leshchinsky and San (1994) showed that the moment

    equilibrium equation can be written as:

    area1 2 5. h Ae

    cos A e cos Ae sin dx

    Aecos A e cos

    Ae sin Aecos sin d

    (3-7)

    Writing the moment equation for area (1-3-6) gives:

    H2 Htan R sin 13 H tan

    H2 tanAe

    sin H3 tan

    (3-8)

  • 15

    For area (3-4-5-6), writing the moment equation gives:

    H tanAe cos Ae cos HAe sin 12 H tan

    (3-9)

    In a similar way, writing the moment equation for area (2-3-4) gives:

    12 Ae sin Ae sin H tan Ae cos Ae cos H

    Ae sin H tan 13 Ae

    sin Ae sin H tan (3-10)

    Figure 3.3. Different Areas Used in Writing the Moment Equations Due to the Weight of Sliding Mass

  • 16

    Calculation of Wh4 :

    This term includes the horizontal seismic coefficient and renders a

    pseudostatic horizontal force due to an earthquake; customarily it is taken as a fraction

    of gravity and acts in the minus X-axis direction. It is suggested to use 0.3-0.5 times

    peak ground acceleration (PGA) for earth structures such as slopes, and 0.3-0.4 times

    PGA for geocell structures (Leshchinsky et al. 2009). The force due to Kh is assumed

    to act at the center of gravity of the sliding soil body. In the slice methods, this

    horizontal seismic force is acting at the center of each slice. Leshchinsky and San

    (1994) derived the moment equilibrium equation for seismic conditions and as same as

    the first term (moment due to the weight of the sliding mass) it is needed to subtract

    the moments induced by areas of (1-3-6), (3-4-5-6) and (2-3-4). The following

    equation is a modification and extension to the equation presented by Leshchinsky and

    San (1994).

    area1 2 5. h 12 Ae

    cos A e cos Aecos

    A e cos Ae cos sin d (3-11)

    Writing the moment for area (1-3-6) gives:

    H2 Htan Ae cos H H3

    (3-12)

  • 17

    Writing the moment for area (3-4-5-6) gives:

    H tanAe cos Ae cos HAe cos 12 Ae

    cos Ae cos H

    (3-13)

    And, the moment for area (2-3-4) can be calculated as:

    12 Ae sin Ae sin H tan w Ae cos Ae cos H

    Ae cos 13 Ae cos Ae cos H

    (3-14)

    Moment due to surcharge:

    Assuming q is perpendicular to the horizon and applied to Line 2-3 (as

    shown in Figure 3.2), the moment due to the surcharge can be calculated as:

    qR sin R sin H tan R sin H tan

    R sin R sin H tan2

    qAe sin Ae sin H tanAe

    sin Ae sin H tan 2

    (3-15)

    Substituting Equations (3-5) to (3-15) into the moment equilibrium

    equation (Equation (3-4)) gives:

  • 18

    PAecos D tanAesin Dtan 1 KAecos AecosAesinAecos

    sind H

    2 tan1 K H3 tan Ae

    sin 1 KHtanAecos Aecos H Aesin

    12 Htan 12 1 KAe

    sin Aesin HtanAecos Aecos H Aesin Htan 13 Ae

    sin Aesin Htan

    12 KAecos AecosAecos

    AecosAecos sind H

    2 Ktan Aecos H H3

    HKtanAecos Aecos H Aecos 12 Ae

    cos Aecos H 12 KAe

    cos Aecos HAesin Aesin Htan Aecos 13 Ae

    cos Aecos H

    qAesin Aesin H tan 12 Aesin

    Aesin H tan

    (3-16)

  • 19

    Rearranging the terms in Equation (3-16) in a form compatible with the

    M-O equation, one can get:

    K K cos 2H Ae

    cos AecosAesinAecos

    sind tan H3 tan Ae

    sin 2H tan Ae

    cos Aecos H Aesin H2 tan 1H Ae

    sin Aesin HtanAecos Aecos H Aesin Htan 13 Ae

    sin Aesin Htan

    1H K

    1 K Aecos Aecos Aecos

    sind K1 K tan Ae

    cos H H3 2H

    K1 K tanAe

    cos Aecos H Aecos 12 Ae

    cos Aecos H 1H

    K1 K Ae

    cos Aecos HAesin Aesin Htan Aecos 13 Ae

    cos Aecos H qH1 K Ae

    sin Aesin H tan

    /Aecos D tanAesin Dtan (3-17)

  • 20

    Equation 3-17 is developed based upon assuming a boundary condition

    where the critical slip surface, defining the active wedge, emerges at the toe. This

    boundary will naturally occur for cohesionless soils. Furthermore, it is a boundary for

    any backfill if one considers the physics of the problem. That is, while apparent

    critical surfaces may emerge away from the slope and the toe for large batters and

    homogenous cohesive soils, such surfaces are meaningless when considering the

    objective of the present problem. Such surfaces do not render the resultant of an

    active wedge on the face but rather signify a rotational slide that includes the facing as

    part of the sliding body. In fact, failure through the foundation is one of the design

    aspects of any retaining structure; however, it is not related to the resultant force

    sought in this work which, physically, can be produced only when the foundation soil

    is competent. Of surfaces intercepting the face, only the one emerging through the toe

    will yield the maximum resultant force. Hence, the toe is a natural boundary for the

    sought resultant of lateral earth pressure.

    One can adopt the common assumption associated with the resultant force

    of lateral earth pressure; i.e., the height at which Pae (or P) acts is usually taken at

    D=H/3. The homogenous problem formulated here has an infinite backslope.

    It can be verified that the trace of the log-spiral in Cartesian coordinates

    (shown in Figure 3.2) can be expressed by the following parametric equations:

    X X Ae sin (3-18) Y Y Ae cos (3-19)

  • 21

    Where Xc and Yc are the location of the pole of the log-spiral relative to

    the Cartesian coordinate system. Considering that Point 1 is at (0,0) and that Point 2

    must be on the crest (see Figure 3.2), manipulation of Equations 3-18 and 3-19 yields

    the following expression:

    A H 1 tan tan ecos sin tan ecos sin tan

    (3-20)

    Where is the backslope angle (See Figure 3.2) At this stage, Equation 3-17 can be solved via a simple maximization

    process (Leshchinsky et al. 2010):

    1. Assume values for 1 and 2 2. Solve Equation 3-20 to obtain the constant A of the log-spiral

    3. For assumed selected D (=H/3), solve Equation 3-17 to calculate Kae-h

    4. Considering all calculated values, is max(Kae-h) rendered? If yes, the complete

    critical solution Kae-h and its active wedge defined by the associated log-spiral

    (A, Xc, Yc) was found. If not, change 1 and 2 and go to Step 2. The process is repeated for all feasible values of 1 and 2 to ascertain that max(Kae_h) was indeed captured.

    5. Now Ph (Equation 3-3) can be calculated

    This numerical iterative process is analogous to finding the factor of

    safety in slope stability analysis that is associated with a circular slip surface. That is,

    in such analysis minimization of the safety factor is done by changing three

    parameters defining a circle: center (Xc, Yc) and radius (R). When circles that emerge

  • 22

    only at the toe are considered, the minimization is done with respect to two parameters

    only analogous to the log-spiral case here.

    To realize a formulation that is equivalent to the classical Coulomb and

    M-O methods in terms of inclination of interface friction, refer first to Figure 3.4. It is

    seen that the interface friction between the facing and the soil is along the slope.

    Hence, the horizontal component of the resultant force would be:

    P 12 H1 KK cos 12 H

    1 KK (3-21)

    Figure 3.4. Direction of Resultant Force, Pae, Commonly Used in Classical Coulomb and M-O Analyses

  • 23

    Similar to the manipulation used to derive Equation 3-17, one can

    assemble the following equation to yield Kae-h:

  • 24

    K K cos 2H Ae

    cos AecosAesinAecos

    sind tan H3 tan Ae

    sin 2H tan Ae

    cos Aecos H Aesin H2 tan 1H Ae

    sin Aesin HtanAecos Aecos H Aesin Htan 13 Ae

    sin Aesin Htan

    1H K

    1 K Aecos Aecos Aecos

    sind K1 K tan Ae

    cos H H3 2H

    K1 K tanAe

    cos Aecos H Aecos 12 Ae

    cos Aecos H 1H

    K1 K Ae

    cos Aecos HAesin Aesin Htan Aecos 13 Ae

    cos Aecos H qH1 K Ae

    sin Aesin H tan

    / cos cos cos Aecos D

    sin Aecossin D tan tanAesin Dtan sin tan Aecos sin D tan cos

    (3-22)

  • 25

    The right hand side of Equation 3-22 is different from Equation 3-17

    solely by the denominator. That is, the denominator represents the resisting moment

    generated by the resultant force that is needed to stabilize the mass augmented by a

    log-spiral defining the active wedge. Its value depends on the inclination of the

    resultant Pa. The solution process of Equation 3-22 is identical to that of Equation 3-

    17.

  • 26

    Chapter 4

    RESULTS OF ANALYSIS

    This chapter presents a series of design charts showing seismic lateral

    earth pressure coefficient (Kae-h) and also seismic component of the resultant force

    (Kae-h) in different states, combined with the trace of critical log spirals defining the active wedges for various seismic coefficients. These charts are developed utilizing

    Equations 3-17 and 3-22 and their associated solution schemes, as explained in

    Chapter 3. The proposed algorithms can be easily programmed and solved to produce

    results for possible cases which have not been presented in this chapter.

    4.1. Kae-h Versus Batter Relationship

    Figures 4.1 to 4.19 show the design charts using Equation 3-17 (i.e.,

    considering the modified direction of the resultant force as proposed by Leshchinsky

    et al. 2010), and Figures 4.20 to 4.38 show the design charts which are produced using

    Equation 3-22 (i.e., employing the conventional direction of the resultant force as used

    in the Coulomb and the M-O methods) . In these charts, the results are shown for

    different seismic coefficients, Kh, ranging from 0 to 0.5, batters varying from 0 to 90-, and values eqaul to 20, 30, 40 and 50. Also the charts are varied by the backslope angle from horizontal slope, 1V:10H, 1V:5H, 1V:3H and 1V:2H. In these

    charts, ratio is equal to 0, 1/3, 2/3 and 1. It should be noted that Figures 4.1 to 4.38

  • 27

    are procured for surcharge-free problems (i.e., q = 0) using Kv = 0. The effect of Kv

    will be investigated in the next section.

    Note that the higher soil strength leads to the smaller equivalent seismic

    lateral earth pressure coefficient and the value of Kae-h increases quickly as Kh

    increases. Utilization of these charts is straightforward: for a given , and selected batter for slope, one can determine the equivalent horizontal lateral earth pressure

    coefficient using the charts.

  • 28

    Figure 4.1.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=20, Backslope 1:, /=0 & 1/3)

    Figure 4.1.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=20, Backslope 1: , /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope 1:

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:

  • 29

    Figure 4.2.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1:, /=0 & 1/3)

    Figure 4.2.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: , /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    1.2

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:

  • 30

    Figure 4.3.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1:, /=0 & 1/3)

    Figure 4.3.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: , /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:

  • 31

    Figure 4.4.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: , /=0 & 1/3)

    Figure 4.4.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: , /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:

    0.0

    0.1

    0.1

    0.2

    0.2

    0.3

    0.3

    0.4

    0.4

    0.5

    0.5

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:

  • 32

    Figure 4.5.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=20, Backslope 1:10, /=0 & 1/3)

    Figure 4.5.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=20, Backslope 1: 10, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:10

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:10

  • 33

    Figure 4.6.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1:10, /=0 & 1/3)

    Figure 4.6.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 10, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:10

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:10

  • 34

    Figure 4.7.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: 10, /=0 & 1/3)

    Figure 4.7.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: 10, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:10

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:10

  • 35

    Figure 4.8.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: 10, /=0 & 1/3)

    Figure 4.8.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: 10, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:10

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:10

  • 36

    Figure 4.9.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=20, Backslope 1: 5, /=0 & 1/3)

    Figure 4.9.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=20, Backslope 1: 5, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:5

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:5

  • 37

    Figure 4.10.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 5, /=0 & 1/3)

    Figure 4.10.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 5, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:5

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:5

  • 38

    Figure 4.11.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: 5, /=0 & 1/3)

    Figure 4.11.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: 5, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:5

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    1.2

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:5

  • 39

    Figure 4.12.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: 5, /=0 & 1/3)

    Figure 4.12.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: 5, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:5

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:5

  • 40

    Figure 4.13.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=20, Backslope 1: 3, /=0 & 1/3)

    Figure 4.13.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=20, Backslope 1: 3, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20 , Backslope1:3=20o, Backslope1:3

  • 41

    Figure 4.14.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 3, /=0 & 1/3)

    Figure 4.14.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 3, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:3

  • 42

    Figure 4.15.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: 3, /=0 & 1/3)

    Figure 4.15.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: 3, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:3

  • 43

    Figure 4.16.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: 3, /=0 & 1/3)

    Figure 4.16.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: 3, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:3

  • 44

    Figure 4.17.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 2, /=0 & 1/3)

    Figure 4.17.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 2, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:2

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:2

  • 45

    Figure 4.18.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: 2, /=0 & 1/3)

    Figure 4.18.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=40, Backslope 1: 2, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:2

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:2

  • 46

    Figure 4.19.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: 2, /=0 & 1/3)

    Figure 4.19.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=50, Backslope 1: 2, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:2

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    1.1

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:2

  • 47

    Figure 4.20.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=20, Backslope 1:, /=0 & 1/3 )

    Figure 4.20.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=20, Backslope 1: , /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:

  • 48

    Figure 4.21.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1:, /=0 & 1/3 )

    Figure 4.21.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1: , /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    1.2

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:

  • 49

    Figure 4.22.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1:, /=0 & 1/3 )

    Figure 4.22.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1: , /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:

  • 50

    Figure 4.23.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1:, /=0 & 1/3 )

    Figure 4.23.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1: , /=2/3 & 1)

    0.0

    0.1

    0.1

    0.2

    0.2

    0.3

    0.3

    0.4

    0.4

    0.5

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:

  • 51

    Figure 4.24.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=20, Backslope 1:10, /=0 & 1/3 )

    Figure 4.24.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=20, Backslope 1: 10, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:10

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:10

  • 52

    Figure 4.25.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1:10, /=0 & 1/3 )

    Figure 4.25.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1: 10, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:10

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:10

  • 53

    Figure 4.26.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1:10, /=0 & 1/3 )

    Figure 4.26.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1: 10, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:10

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:10

  • 54

    Figure 4.27.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1:10, /=0 & 1/3 )

    Figure 4.27.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1: 10, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:10

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:10

  • 55

    Figure 4.28.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=20, Backslope 1:5, /=0 & 1/3 )

    Figure 4.28.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=20, Backslope 1: 5, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:5

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:5

  • 56

    Figure 4.29.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30, Backslope 1:5, /=0 & 1/3 )

    Figure 4.29.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1: 5, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:5

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:5

  • 57

    Figure 4.30.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1:5, /=0 & 1/3 )

    Figure 4.30.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1: 5, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:5

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:5

  • 58

    Figure 4.31.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1:5, /=0 & 1/3 )

    Figure 4.31.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1: 5, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:5

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:5

  • 59

    Figure 4.32.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=20, Backslope 1:3, /=0 & 1/3 )

    Figure 4.32.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=20, Backslope 1: 3, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60 70

    Kae

    _h

    Batter, (degrees)

    =20o, Backslope1:3

  • 60

    Figure 4.33.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1:3, /=0 & 1/3 )

    Figure 4.33.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1: 3, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30 , Backslope1:3=30o, Backslope1:3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:3

  • 61

    Figure 4.34.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1:3, /=0 & 1/3 )

    Figure 4.34.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1: 3, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:3

  • 62

    Figure 4.35.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1:3, /=0 & 1/3 )

    Figure 4.35.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1: 3, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:3

  • 63

    Figure 4.36.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1:2, /=0 & 1/3 )

    Figure 4.36.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1: 2, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:2

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40 50 60

    Kae

    _h

    Batter, (degrees)

    =30o, Backslope1:2

  • 64

    Figure 4.37.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1:2, /=0 & 1/3 )

    Figure 4.37.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=40, Backslope 1: 2, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:2

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0 10 20 30 40 50

    Kae

    _h

    Batter, (degrees)

    =40o, Backslope1:2

  • 65

    Figure 4.38.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1:2, /=0 & 1/3 )

    Figure 4.38.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=50, Backslope 1: 2, /=2/3 & 1)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    1.0

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:2

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    1.2

    0 10 20 30 40

    Kae

    _h

    Batter, (degrees)

    =50o, Backslope1:2

  • 66

    4.2. Effect of Vertical Seismic Coefficient

    Figures 4.39 to 4.46 illustrate the effect of vertical seismic coefficient on

    the resultant force for different conditions. Figures 4.38 to 4,42 are generated using

    Equation 3-17 (for the Pae inclination implied in Figure 3.2) and, Figures 4.43 to 4.46

    are produced using Equation 3-22 (for the Pae inclination implied in Figure 3.4). These

    figures demonstrate the impact of vertical acceleration as related to various batter

    angles. The results are shown for five different Kv/Kh ratios as, -1, -0.5, 0, 0.5, 1,

    when possible. For some cases (e.g., Figure 4.40.b), large Kv values with a positive

    sign have led to irrationally large Kae_h values and consequently, the results are not

    presented for such cases. The characteristic behavior exhibited at Kv=0 is retained

    whether Kv is greater or smaller than zero. For a rather large Kh Like 0.3, Kv may have

    significant impact on Kae-h compare to a small Kh like 0.1. Also increasing in

    backslope shows an increase in Kae-h and increasing in wall facing friction angle

    decreased the Kae-h.

  • 67

    Figure 4.39.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1:, Kh=0.1, /=0 )

    Figure 4.39.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: , Kh=0.3, /=0)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0Kv/Kh=0.5Kv/Kh=1

    =20 , Backslope1:,Kh=0.1=30o, Backslope1:,Kh=0.1,=0

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0Kv/Kh=0.5Kv/Kh=1

    =20 , Backslope1:,Kh=0.1=30o, Backslope1:,Kh=0.3,=0

  • 68

    Figure 4.40.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1:5, Kh=0.1, /=0 )

    Figure 4.40.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 5, Kh=0.3, /=0)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0Kv/Kh=0.5Kv/Kh=1

    =20 , Backslope1:,Kh=0.1=30o, Backslope1:5,Kh=0.1,=0

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0

    =20 , Backslope1:,Kh=0.1=30o, Backslope1:5,Kh=0.3,=0

  • 69

    Figure 4.41.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1:, Kh=0.1, /=2/3 )

    Figure 4.41.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: , Kh=0.3, /=2/3)

    0.0

    0.1

    0.2

    0.3

    0.4

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0Kv/Kh=0.5Kv/Kh=1

    =30o, Backslope1:,Kh=0.1,=2/3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0Kv/Kh=0.5Kv/Kh=1

    =20 , Backslope1:,Kh=0.1=30o, Backslope1:,Kh=0.3,=2/3

  • 70

    Figure 4.42.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1:5, Kh=0.1, /=2/3 )

    Figure 4.42.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-17 (=30, Backslope 1: 5, Kh=0.3, /=2/3)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0Kv/Kh=0.5Kv/Kh=1

    =20 , Backslope1:,Kh=0.1=30o, Backslope1:5,Kh=0.1,=2/3

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.9

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0

    =20 , Backslope1:,Kh=0.1=30o, Backslope1:5,Kh=0.3,=2/3

  • 71

    Figure 4.43.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq.

    3-22 (=30, Backslope 1:, Kh=0.1, /=0 )

    Figure 4.43.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1: , Kh=0.3, /=0)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0Kv/Kh=0.5Kv/Kh=1

    =20 , Backslope1:,Kh=0.1=30o, Backslope1:,Kh=0.1,=0

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0 10 20 30 40 50 60

    Kae

    -h

    Batter, (degrees)

    Kv/Kh=1Kv/Kh=0.5Kv/Kh=0

    Kv/Kh=0.5

    Kv/Kh=1

    =30o, Backslope1:,Kh=0.3,=0

  • 72

    Figure 4.44.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1:5, Kh=0.1, /=0 )

    Figure 4.44.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

    Eq. 3-22 (=30, Backslope 1: 5, Kh=0.3, /=0)

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0 10 20 30 40 50 60

    Ka