extension of mononobe-okabe approach
DESCRIPTION
MononobeTRANSCRIPT
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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
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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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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
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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.
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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
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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.
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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-
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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
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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
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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.
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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.
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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:
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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)
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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)
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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)
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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
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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)
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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:
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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)
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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)
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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)
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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
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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
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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)
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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.
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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
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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.
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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:
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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:
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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:
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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:
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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