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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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Page 1: EXTENSION OF MONONOBE-OKABE APPROACH TO UNSTABLE …

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 ......................................................................................................... v LIST OF FIGURES ....................................................................................................... vi ABSTRACT ................................................................................................................ 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 ....................................................... 26 4.2. Effect of Vertical Seismic Coefficient ............................................. 66 4.3. Slip Surfaces ..................................................................................... 75 4.4. Studying Seismic-Induced Resultant Force ...................................... 88 

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

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LIST OF TABLES

Table 5.1 Comparison of Kae-h (= 0, =0o, Kv=0) ............................................ 102 

Table 5.2 Comparison of Kae-h (= 30o, =0 o, Kv=0) .......................................... 103 

Table 5.3 Comparison of Kae-h (= 45o, =0 o, Kv=0) .......................................... 104 

Table 5.4 Comparison of Kae-h (= 0o, =0 o, Kv=0) .......................................... 105 

Table 5.5 Comparison of Kae-h (= 30o, =0 o, Kv=0) ........................................ 106 

Table 5.5 Comparison of Kae-h (= 30o, =0 o, 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 (=20⁰, Backslope 1: , /=0 & 1/3) .................... 28

Figure 4.1.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20⁰, Backslope 1: , /=2/3 & 1) ................... 28

Figure 4.2.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: , /=0 & 1/3) ................... 29

Figure 4.2.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: , /=2/3 & 1 ) .................. 29

Figure 4.3.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: , /=0 & 1/3) ................... 30

Figure 4.3.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: , /=2/3 & 1  ) ................. 30 

Figure 4.4.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, Backslope 1: , /=0 & 1/3 ) .................. 31 

Figure 4.4.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, Backslope 1: , /=2/3 & 1 ) ................... 31 

Figure 4.5.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20⁰, Backslope 1: 10, /=0 & 1/3) ................. 32 

Figure 4.5.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20⁰, 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 (=30⁰, Backslope 1: 10, /=0 & 1/3) ................. 33 

Figure 4.6.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 10, /=2/3 & 1 ) ................. 33 

Figure 4.7.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: 10, /=0 & 1/3) .................. 34 

Figure 4.7.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: 10, /=2/3 & 1) .................. 34 

Figure 4.8.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, Backslope 1: 10, /=0 & 1/3) .................. 35 

Figure 4.8.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, Backslope 1: 10, /=2/3 & 1) .................. 35

Figure 4.9.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20⁰, Backslope 1: 5, /=0 & 1/3) .................... 36

Figure 4.9.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20⁰, Backslope 1: 5, /=2/3 & 1) .................... 36

Figure 4.10.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 5, /=0 & 1/3) .................... 37

Figure 4.10.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 5, /=2/3 & 1) .................... 37

Figure 4.11.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: 5, /=0 & 1/3) .................... 38

Figure 4.11.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: 5, /=2/3 & 1) .................... 38

Figure 4.12.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, Backslope 1: 5, /=0 & 1/3) .................... 39

Figure 4.12.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, 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 (=20⁰, Backslope 1: 3, /=0 & 1/3) .................... 40

Figure 4.13.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=20⁰, Backslope 1: 3, /=2/3 & 1) .................... 40

Figure 4.14.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 3, /=0 & 1/3) .................... 41

Figure 4.14.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 3, /=2/3 & 1) .................... 41

Figure 4.15.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: 3, /=0 & 1/3) .................... 42

Figure 4.15.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: 3, /=2/3 & 1) .................... 42

Figure 4.16.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, Backslope 1: 3, /=0 & 1/3) .................... 43

Figure 4.16.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, Backslope 1: 3, /=2/3 & 1) .................... 43

Figure 4.17.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 2, /=0 & 1/3) .................... 44

Figure 4.17.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 2, /=2/3 & 1) .................... 44

Figure 4.18.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: 2, /=0 & 1/3) .................... 45

Figure 4.18.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=40⁰, Backslope 1: 2, /=2/3 & 1) .................... 45

Figure 4.19.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, Backslope 1: 2, /=0 & 1/3) .................... 46

Figure 4.19.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=50⁰, 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 (=20⁰, Backslope 1: , /=0 & 1/3 )................... 47

Figure 4.20.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20⁰, Backslope 1: , /=2/3 & 1 ) ................... 47

Figure 4.21.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: , /=0 & 1/3 )................... 48

Figure 4.21.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: , /=2/3 & 1 ) ................... 48

Figure 4.22.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, Backslope 1: , /=0 & 1/3 )................... 49

Figure 4.22.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, Backslope 1: , /=2/3 & 1 ) ................... 49

Figure 4.23.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: , /=0 & 1/3 )................... 50

Figure 4.23.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: , /=2/3 & 1 ) ................... 50

Figure 4.24.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20⁰, Backslope 1: 10, /=0 & 1/3 ) ................. 51

Figure 4.24.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20⁰, Backslope 1: 10, /=2/3 & 1 ) .................. 51

Figure 4.25.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 10, /=0 & 1/3 ) ................. 52

Figure 4.25.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 10, /=2/3 & 1 ) .................. 52

Figure 4.26.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, Backslope 1: 10, /=0 & 1/3 ) ................. 53

Figure 4.26.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, 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 (=50⁰, Backslope 1: 10, /=0 & 1/3 ) ................. 54

Figure 4.27.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: 10, /=2/3 & 1 ) .................. 54

Figure 4.28.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20⁰, Backslope 1: 5, /=0 & 1/3 ) ................... 55

Figure 4.28.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20⁰, Backslope 1: 5, /=2/3 & 1 ) .................... 55

Figure 4.29.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 5, /=0 & 1/3 ) ................... 56

Figure 4.29.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 5, /=2/3 & 1 ) .................... 56

Figure 4.30.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, Backslope 1: 5, /=0 & 1/3 ) ................... 57

Figure 4.30.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, Backslope 1: 5, /=2/3 & 1 ) .................... 57

Figure 4.31.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: 5, /=0 & 1/3 ) ................... 58

Figure 4.31.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: 5, /=2/3 & 1 ) .................... 58

Figure 4.32.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20⁰, Backslope 1: 3, /=0 & 1/3 ) ................... 59

Figure 4.32.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=20⁰, Backslope 1: 3, /=2/3 & 1 ) .................... 59

Figure 4.33.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 3, /=0 & 1/3 ) ................... 60

Figure 4.33.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, 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 (=40⁰, Backslope 1: 3, /=0 & 1/3 ) ................... 61

Figure 4.34.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, Backslope 1: 3, /=2/3 & 1 ) .................... 61

Figure 4.35.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: 3, /=0 & 1/3 ) ................... 62

Figure 4.35.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: 3, /=2/3 & 1 ) .................... 62

Figure 4.36.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 2, /=0 & 1/3 ) ................... 63

Figure 4.36.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 2, /=2/3 & 1 ) .................... 63

Figure 4.37.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, Backslope 1: 2, /=0 & 1/3 ) ................... 64

Figure 4.37.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=40⁰, Backslope 1: 2, /=2/3 & 1 ) .................... 64

Figure 4.38.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: 2, /=0 & 1/3 ) ................... 65

Figure 4.38.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=50⁰, Backslope 1: 2, /=2/3 & 1 ) .................... 65

Figure 4.39.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: , Kh=0.1, /=0 ) ............... 67

Figure 4.39.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: , Kh=0.3, /=0 ) ................ 67

Figure 4.40.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 5, Kh=0.1, /=0 ) ................ 68

Figure 4.40.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, 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 (=30⁰, Backslope 1: , Kh=0.1, /=2/3 ) ............ 69

Figure 4.41.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: , Kh=0.3, /=2/3 ) ............ 69

Figure 4.42.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 5, Kh=0.1, /=2/3 ) ............. 70

Figure 4.42.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-17 (=30⁰, Backslope 1: 5, Kh=0.3, /=2/3 ) ............. 70

Figure 4.43.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: , Kh=0.1, /=0 ) ............... 71

Figure 4.43.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: , Kh=0.3, /=0 ) ................ 71

Figure 4.44.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 5, Kh=0.1, /=0 ) ................ 72

Figure 4.44.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 5, Kh=0.3, /=0 ) ................ 72

Figure 4.45.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: , Kh=0.1, /=2/3 ) ............ 73

Figure 4.45.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: , Kh=0.3, /=2/3 ) ............ 73

Figure 4.46.a Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 5, Kh=0.1, /=2/3 ) ............. 74

Figure 4.46.b Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, 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: =20⁰, ω=0⁰, Backslope 1:  (Eq. 3-17) ........................................................................................... 76

Figure 4.47.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=70⁰, 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: =30⁰, ω=0⁰, Backslope 1:  (Eq. 3-17) ........................................................................................... 77

Figure 4.48.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30⁰, ω=60⁰, Backslope 1:  (Eq. 3-17) ........................................................................................... 77

Figure 4.49.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1:  (Eq. 3-17) ........................................................................................... 78

Figure 4.49.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1: (Eq. 3-17) ........................................................................................... 78

Figure 4.50.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =50⁰, ω=0⁰, Backslope 1:  (Eq. 3-17) ........................................................................................... 79

Figure 4.50.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =50⁰, ω=40⁰, Backslope 1:  (Eq. 3-17) ........................................................................................... 79

Figure 4.51.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=0⁰, Backslope 1: 5 (Eq. 3-17) ........................................................................................... 80

Figure 4.51.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=70⁰, Backslope 1: 5 (Eq. 3-17) ........................................................................................... 80

Figure 4.52.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1: 5 (Eq. 3-17) ........................................................................................... 81

Figure 4.52.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1: 5 (Eq. 3-17) ........................................................................................... 81

Figure 4.53.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=0⁰, 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: =20⁰, ω=70⁰, Backslope 1:  (Eq. 3-22) ........................................................................................... 82

Figure 4.54.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30⁰, ω=0⁰, Backslope 1:  (Eq. 3-22) ........................................................................................... 83

Figure 4.54.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30⁰, ω=60⁰, Backslope 1:  (Eq. 3-22) ........................................................................................... 83

Figure 4.55.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1:  (Eq. 3-22) ........................................................................................... 84

Figure 4.55.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1:  (Eq. 3-22) ........................................................................................... 84

Figure 4.56.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1:  (Eq. 3-22) ........................................................................................... 85

Figure 4.56.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1:  (Eq. 3-22) ........................................................................................... 85

Figure 4.57.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=0⁰, Backslope 1: 5 (Eq. 3-22) ........................................................................................... 86

Figure 4.57.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=70⁰, Backslope 1: 5 (Eq. 3-22) ........................................................................................... 86

Figure 4.58.a Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1: 5 (Eq. 3-22) ........................................................................................... 87

Figure 4.58.b Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=50⁰, 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 Mononobe–Okabe (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 Coulomb’s 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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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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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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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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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 H K 1 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:

Kcos ω θ

cos θ cos ω cos δ ω θ 1sin δ sin α θcos δ ω θ cos α ω

(2-2)

Where: is the soil friction angle, is the batter, =tan-1(Kh/1-Kv), δ is

the soil-facing interface friction angle and is the backslope.

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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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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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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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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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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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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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P P12H 1 K K cos δ

12H 1 K K

(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):

P h P tan δ h 1 K Wh K Wh + (the moment due

to the surcharge load)

(3-4)

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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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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:

area 1 2 5 . h Ae cos β A e cos β Ae sin β dx

Ae cos β A e cos β Ae sin β Ae cos β sin β dβ

(3-7)

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

H2

Htan R sin 13

H tan H2

tan Ae sin H3

tan

(3-8)

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

H tan Ae cos Ae cos H Ae sin12

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 tan13

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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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).

area 1 2 5 . h

12

Ae cos β A e cos β Ae cos β

A e cos β Ae cos β sin β dβ

(3-11)

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

H2

Htan Ae cos β HH3

(3-12)

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

H tan Ae cos Ae cos H Ae cos β12

Ae cos β

Ae cos β 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 β H

Ae cos β13

Ae cos β Ae cos β 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:

q R sin R sin H tan

R sin H tanR sin R sin H tan

2

q Ae sin Ae sin

H tanAe sin β Ae sin β H tan ω

2

(3-15)

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

equation (Equation (3-4)) gives:

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P Ae cosβ D tanδ Ae sinβ Dtanω

1 K Ae cosβ Ae cosβ Ae sinβ Ae cosβ

ψ sinβ dβH2

tanω 1 KH3

tanω Ae sinβ

1 K Htanω Ae cosβ Ae cosβ H Ae sinβ

12

Htanω

12

1 K Ae sinβ Ae sinβ Htanω Ae cosβ

Ae cosβ H Ae sinβ Htanω

13

Ae sinβ Ae sinβ Htanω

12

K Ae cosβ Ae cosβ Ae cosβ

Ae cosβ Ae cosβ ψ sinβ dβH2

K tanω Ae cosβ HH3

HK tanω Ae cosβ Ae cosβ H Ae cosβ

12

Ae cosβ Ae cosβ H

12

K Ae cosβ Ae cosβ H Ae sinβ Ae sinβ

Htanω Ae cosβ

13

Ae cosβ Ae cosβ H

q Ae sin β Ae sin β H tan ω12

Ae sin β

Ae sin β H tan ω

(3-16)

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

M-O equation, one can get:

K K cos δ2

HAe cosβ Ae cosβ Ae sinβ Ae cosβ

ψ sinβ dβ tanωH3

tanω Ae sinβ

2H

tan ω Ae cosβ Ae cosβ H Ae sinβH2

tanω

1H

Ae sinβ Ae sinβ Htanω Ae cosβ Ae cosβ

H Ae sinβ Htanω

13

Ae sinβ Ae sinβ Htanω

1H

K1 K

Ae cosβ Ae cosβ Ae cosβ

ψ sinβ dβK

1 Ktanω Ae cosβ H

H3

2H

K1 K

tanω Ae cosβ Ae cosβ H Ae cosβ

12

Ae cosβ Ae cosβ H

1H

K1 K

Ae cosβ Ae cosβ H Ae sinβ

Ae sinβ Htanω Ae cosβ

13

Ae cosβ Ae cosβ H

qγH 1 K

Ae sin β Ae sin β H tan ω

/ Ae cosβ D tanδ Ae sinβ Dtanω

(3-17)

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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 β (3-18)

Y Y Ae cos β (3-19)

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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 H1 tan tan

e cos β sin β tan α e cos β sin β tan α

(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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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:

P12H 1 K K cos δ ω

12H 1 K K

(3-21)

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

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

assemble the following equation to yield Kae-h:

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K K cos δ ω2

HAe cosβ Ae cosβ Ae sinβ Ae cosβ

ψ sinβ dβ tanωH3

tanω Ae sinβ

2H

tan ω Ae cosβ Ae cosβ H Ae sinβH2

tanω

1H

Ae sinβ Ae sinβ Htanω Ae cosβ Ae cosβ

H Ae sinβ Htanω

13

Ae sinβ Ae sinβ Htanω

1H

K1 K

Ae cosβ Ae cosβ Ae cosβ

ψ sinβ dβK

1 Ktanω Ae cosβ H

H3

2H

K1 K

tanω Ae cosβ Ae cosβ H Ae cosβ

12

Ae cosβ Ae cosβ H

1H

K1 K

Ae cosβ Ae cosβ H Ae sinβ

Ae sinβ Htanω Ae cosβ

13

Ae cosβ Ae cosβ H

qγH 1 K

Ae sin β Ae sin β H tan ω

/cos δ

cos δ ωcos ω Ae cosβ D

sin ω Ae cosβ sin β D tan ω

tanδ Ae sinβ Dtanω sin ω

tan δ Ae cosβ sin β D tan ω cos ω

(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 20⁰, 30⁰, 40⁰ and 50⁰. 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 (=20⁰, Backslope 1: , /=0 & 1/3)

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

Eq. 3-17 (=20⁰, 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, Backslope 1: 

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29

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

Eq. 3-17 (=30⁰, Backslope 1: , /=0 & 1/3)

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

Eq. 3-17 (=30⁰, 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, Backslope 1: 

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, Backslope 1: 

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30

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

Eq. 3-17 (=40⁰, Backslope 1: , /=0 & 1/3)

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

Eq. 3-17 (=40⁰, 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, Backslope 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, Backslope 1: 

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31

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

Eq. 3-17 (=50⁰, Backslope 1: , /=0 & 1/3 )

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

Eq. 3-17 (=50⁰, 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, Backslope 1: 

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, Backslope 1: 

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32

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

Eq. 3-17 (=20⁰, Backslope 1: 10, /=0 & 1/3)

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

Eq. 3-17 (=20⁰, 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, Backslope 1: 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, Backslope 1: 10

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33

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

Eq. 3-17 (=30⁰, Backslope 1: 10, /=0 & 1/3)

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

Eq. 3-17 (=30⁰, 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, Backslope 1: 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, Backslope 1: 10

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34

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

Eq. 3-17 (=40⁰, Backslope 1: 10, /=0 & 1/3)

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

Eq. 3-17 (=40⁰, 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, Backslope 1: 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, Backslope 1: 10

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35

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

Eq. 3-17 (=50⁰, Backslope 1: 10, /=0 & 1/3)

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

Eq. 3-17 (=50⁰, 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, Backslope 1: 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, Backslope 1: 10

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36

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

Eq. 3-17 (=20⁰, Backslope 1: 5, /=0 & 1/3)

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

Eq. 3-17 (=20⁰, 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, Backslope 1: 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, Backslope 1: 5

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37

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

Eq. 3-17 (=30⁰, Backslope 1: 5, /=0 & 1/3)

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

Eq. 3-17 (=30⁰, 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, Backslope 1: 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, Backslope 1: 5

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38

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

Eq. 3-17 (=40⁰, Backslope 1: 5, /=0 & 1/3)

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

Eq. 3-17 (=40⁰, 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, Backslope 1: 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, Backslope 1: 5

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39

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

Eq. 3-17 (=50⁰, Backslope 1: 5, /=0 & 1/3)

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

Eq. 3-17 (=50⁰, 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, Backslope 1: 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, Backslope 1: 5

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40

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

Eq. 3-17 (=20⁰, Backslope 1: 3, /=0 & 1/3)

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

Eq. 3-17 (=20⁰, 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, Backslope 1: 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 , Backslope 1: 3=20o, Backslope 1: 3

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41

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

Eq. 3-17 (=30⁰, Backslope 1: 3, /=0 & 1/3)

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

Eq. 3-17 (=30⁰, 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, Backslope 1: 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, Backslope 1: 3

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42

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

Eq. 3-17 (=40⁰, Backslope 1: 3, /=0 & 1/3)

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

Eq. 3-17 (=40⁰, 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, Backslope 1: 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, Backslope 1: 3

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43

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

Eq. 3-17 (=50⁰, Backslope 1: 3, /=0 & 1/3)

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

Eq. 3-17 (=50⁰, 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, Backslope 1: 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, Backslope 1: 3

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44

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

Eq. 3-17 (=30⁰, Backslope 1: 2, /=0 & 1/3)

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

Eq. 3-17 (=30⁰, 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, Backslope 1: 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, Backslope 1: 2

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45

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

Eq. 3-17 (=40⁰, Backslope 1: 2, /=0 & 1/3)

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

Eq. 3-17 (=40⁰, 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, Backslope 1: 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, Backslope 1: 2

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46

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

Eq. 3-17 (=50⁰, Backslope 1: 2, /=0 & 1/3)

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

Eq. 3-17 (=50⁰, 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, Backslope 1: 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, Backslope 1: 2

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47

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

Eq. 3-22 (=20⁰, Backslope 1: , /=0 & 1/3 )

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

Eq. 3-22 (=20⁰, 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, Backslope 1: 

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48

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

Eq. 3-22 (=30⁰, Backslope 1: , /=0 & 1/3 )

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

Eq. 3-22 (=30⁰, 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, Backslope 1: 

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, Backslope 1: 

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49

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

Eq. 3-22 (=40⁰, Backslope 1: , /=0 & 1/3 )

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

Eq. 3-22 (=40⁰, 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, Backslope 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, Backslope 1: 

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50

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

Eq. 3-22 (=50⁰, Backslope 1: , /=0 & 1/3 )

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

Eq. 3-22 (=50⁰, 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, Backslope 1: 

0.0

0.1

0.2

0.3

0.4

0.5

0 10 20 30 40

Kae

_h

Batter, (degrees)

=50o, Backslope 1: 

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51

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

Eq. 3-22 (=20⁰, Backslope 1: 10, /=0 & 1/3 )

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

Eq. 3-22 (=20⁰, 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, Backslope 1: 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, Backslope 1: 10

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52

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

Eq. 3-22 (=30⁰, Backslope 1: 10, /=0 & 1/3 )

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

Eq. 3-22 (=30⁰, 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, Backslope 1: 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, Backslope 1: 10

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53

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

Eq. 3-22 (=40⁰, Backslope 1: 10, /=0 & 1/3 )

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

Eq. 3-22 (=40⁰, 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, Backslope 1: 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, Backslope 1: 10

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54

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

Eq. 3-22 (=50⁰, Backslope 1: 10, /=0 & 1/3 )

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

Eq. 3-22 (=50⁰, 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, Backslope 1: 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, Backslope 1: 10

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55

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

Eq. 3-22 (=20⁰, Backslope 1: 5, /=0 & 1/3 )

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

Eq. 3-22 (=20⁰, 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, Backslope 1: 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, Backslope 1: 5

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56

Figure 4.29.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter, Eq. 3-22 (=30⁰, Backslope 1: 5, /=0 & 1/3 )

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

Eq. 3-22 (=30⁰, 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, Backslope 1: 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, Backslope 1: 5

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57

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

Eq. 3-22 (=40⁰, Backslope 1: 5, /=0 & 1/3 )

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

Eq. 3-22 (=40⁰, 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, Backslope 1: 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, Backslope 1: 5

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58

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

Eq. 3-22 (=50⁰, Backslope 1: 5, /=0 & 1/3 )

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

Eq. 3-22 (=50⁰, 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, Backslope 1: 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, Backslope 1: 5

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59

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

Eq. 3-22 (=20⁰, Backslope 1: 3, /=0 & 1/3 )

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

Eq. 3-22 (=20⁰, 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, Backslope 1: 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, Backslope 1: 3

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60

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

Eq. 3-22 (=30⁰, Backslope 1: 3, /=0 & 1/3 )

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

Eq. 3-22 (=30⁰, 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 , Backslope 1: 3=30o, Backslope 1: 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, Backslope 1: 3

Page 77: EXTENSION OF MONONOBE-OKABE APPROACH TO UNSTABLE …

61

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

Eq. 3-22 (=40⁰, Backslope 1: 3, /=0 & 1/3 )

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

Eq. 3-22 (=40⁰, 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, Backslope 1: 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, Backslope 1: 3

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62

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

Eq. 3-22 (=50⁰, Backslope 1: 3, /=0 & 1/3 )

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

Eq. 3-22 (=50⁰, 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, Backslope 1: 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, Backslope 1: 3

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63

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

Eq. 3-22 (=30⁰, Backslope 1: 2, /=0 & 1/3 )

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

Eq. 3-22 (=30⁰, 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, Backslope 1: 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, Backslope 1: 2

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64

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

Eq. 3-22 (=40⁰, Backslope 1: 2, /=0 & 1/3 )

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

Eq. 3-22 (=40⁰, 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, Backslope 1: 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, Backslope 1: 2

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65

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

Eq. 3-22 (=50⁰, Backslope 1: 2, /=0 & 1/3 )

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

Eq. 3-22 (=50⁰, 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, Backslope 1: 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, Backslope 1: 2

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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.

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67

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

Eq. 3-17 (=30⁰, Backslope 1: , Kh=0.1, /=0 )

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

Eq. 3-17 (=30⁰, 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: , 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: , Kh=0.3, =0

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68

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

Eq. 3-17 (=30⁰, Backslope 1: 5, Kh=0.1, /=0 )

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

Eq. 3-17 (=30⁰, 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: 5, Kh=0.3, =0

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69

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

Eq. 3-17 (=30⁰, Backslope 1: , Kh=0.1, /=2/3 )

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

Eq. 3-17 (=30⁰, 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=‐1 Kv/Kh=‐0.5Kv/Kh=0Kv/Kh=0.5Kv/Kh=1

=30o, Backslope 1: , 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: , Kh=0.3, =2/3

Page 86: EXTENSION OF MONONOBE-OKABE APPROACH TO UNSTABLE …

70

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

Eq. 3-17 (=30⁰, Backslope 1: 5, Kh=0.1, /=2/3 )

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

Eq. 3-17 (=30⁰, 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: 5, Kh=0.3, =2/3

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71

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

3-22 (=30⁰, Backslope 1: , Kh=0.1, /=0 )

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

Eq. 3-22 (=30⁰, 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: , 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=30o, Backslope 1: , Kh=0.3, =0

Page 88: EXTENSION OF MONONOBE-OKABE APPROACH TO UNSTABLE …

72

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

Eq. 3-22 (=30⁰, Backslope 1: 5, Kh=0.1, /=0 )

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

Eq. 3-22 (=30⁰, 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=20 , Backslope 1: , Kh=0.1=20 , Backslope 1: 0.1, Kh=0.1, =0=20 , Backslope 1: , Kh=0.1=20 , Backslope 1: 0.1, Kh=0.1, =0=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

=20 , Backslope 1: , Kh=0.1=20 , Backslope 1: 0.1, Kh=0.1, =0=20 , Backslope 1: , Kh=0.1=20 , Backslope 1: 0.1, Kh=0.1, =0=20 , Backslope 1: , Kh=0.1=30o, Backslope 1: 5, Kh=0.3, =0

Page 89: EXTENSION OF MONONOBE-OKABE APPROACH TO UNSTABLE …

73

Figure 4.45.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

Eq. 3-22 (=30⁰, Backslope 1: , Kh=0.1, /=2/3 )

Figure 4.45.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

Eq. 3-22 (=30⁰, 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=30o, Backslope 1: , Kh=0.1, =2/3

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

Kae

-h

Batter, (degrees)

Kv/Kh=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=30o, Backslope 1: , Kh=0.3, =2/3

Page 90: EXTENSION OF MONONOBE-OKABE APPROACH TO UNSTABLE …

74

Figure 4.46.a. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

Eq. 3-22 (=30⁰, Backslope 1: 5, Kh=0.1, /=2/3 )

Figure 4.46.b. Equivalent Horizontal Lateral Earth Pressure Coefficient vs. Batter,

Eq. 3-22 (=30⁰, 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=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

Kv/Kh=0.5

Kv/Kh=1

=30o, Backslope 1: 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 70

Kae

-h

Batter, (degrees)

Kv/Kh=‐1 

Kv/Kh=‐0.5

Kv/Kh=0

=30o, Backslope 1: 5, Kh=0.3, =2/3

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75

4.3. Slip Surfaces

The traces of critical log spiral rendered by Equation 3-17 are presented in

Figures 4.47 to 4.52 and for Equation 3-22 are presented in Figures 4.53 to 4.58 with

the impact of seismic coefficient, Kh. The cases with horizontal backslope and

backslope 1V:5H considered to investigate.

Higher strength of soil results shallower slip surfaces. For low soil friction

angle, , flatter slope, or large friction at back of facing, the critical slip surfaces tend

to have larger curvature as well as becoming deeper. It means that increases the size

of the active wedge for near vertical slopes; its impact diminishes as the slope

becomes flatter. Also higher Seismic load leads to deeper failure.

For example, Figure 4.47b, which is for =20 and ω=70, a case where

the slope is statically just barely stable, shows that the sliding mass rapidly increases

with the seismic coefficient. Compared with Figure 4.47a which is for ω=0 , this

mass is actually larger, especially as Kh increases. Increasing larger mass means that

there is more rapid increase in Kae-h as the slope becomes flatter thus explaining the

behavior observed in Figures 4.59 and 4.63.

Comparing the failure surfaces generated by two different equations, it

can be seen that both of the equations have lead to identical slip surfaces.

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Figure 4.47.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=0⁰, Backslope 1: (Eq. 3-17)

Figure 4.47.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=70⁰, Backslope 1: (Eq. 3-17)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8

Y=

y/H

X=x/H

=20o, ω=0o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4 4.6 4.8 5 5.2 5.4

Y=

y/H

X=x/H

=20o, ω=70o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2

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Figure 4.48.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30⁰, ω=0⁰, Backslope 1:  (Eq. 3-17)

Figure 4.48.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30⁰, ω=60⁰,v Backslope 1:  (Eq. 3-17)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8

Y=

y/H

X=x/H

=30o, ω=0o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6

Y=

y/H

X=x/H

=30o, ω=60o, Backslope: Horizontal

Kh=0.0Kh=0.1 Kh=0.2 Kh=0.3

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Figure 4.49.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1: ,  (Eq. 3-17)

Figure 4.49.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1: (Eq. 3-17)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Y=

y/H

X=x/H

=40o, ω=0o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

Y=

y/H

X=x/H

=40o, ω=50o, Backslope: Horizontal

Kh=0.0Kh=0.1 Kh=0.2 Kh=0.3

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Figure 4.50.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =50⁰, ω=0⁰, Backslope 1:  (Eq. 3-17)

Figure 4.50.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =50⁰, ω=40⁰, Backslope 1:  (Eq. 3-17)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4

Y=

y/H

X=x/H

=50o, ω=0o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4

Y=

y/H

X=x/H

=50o, ω=40o, Backslope: Horizontal

Kh=0.0Kh=0.1 Kh=0.2 Kh=0.3

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Figure 4.51.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=0⁰, Backslope 1: 5 (Eq. 3-17)

Figure 4.51.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=70⁰, Backslope 1: 5 (Eq. 3-17)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8

Y=

y/H

X=x/H

=20o, ω=0o, Backslope: 1:5

Kh=0.0

Kh=0.1

-0.2

0.2

0.6

1.0

1.4

1.8

2.2

-0.2 0.2 0.6 1 1.4 1.8 2.2 2.6 3 3.4 3.8 4.2 4.6 5 5.4 5.8 6.2 6.6 7

Y=

y/H

X=x/H

=20o, ω=70o, Backslope: 1:5

Kh=0.0

Kh=0.1

Kh=0.0

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Figure 4.52.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1: 5 (Eq. 3-17)

Figure 4.52.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1: 5 (Eq. 3-17)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Y=y

/H

X=x/H

=40o, ω=0o, Backslope: 1:5

Kh=0.0 Kh=0.1 Kh=0.2Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Y=

y/H

X=x/H

=40o, ω=50o, Backslope: 1:5

Kh=0.0

Kh=0.1 Kh=0.2 Kh=0.3

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Figure 4.53.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=0⁰, Backslope 1:  (Eq. 3-22)

Figure 4.53.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=70⁰, Backslope 1:  (Eq. 3-22)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8

Y=

y/H

X=x/H

=20o, ω=0o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4 4.6 4.8 5

Y=

y/H

X=x/H

=20o, ω=70o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

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83

Figure 4.54.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30⁰, ω=0⁰, Backslope 1:  (Eq. 3-22)

Figure 4.54.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =30⁰, ω=60⁰, Backslope 1:  (Eq. 3-22)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Y=

y/H

X=x/H

=30o, ω=0o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8

Y=

y/H

X=x/H

=30o, ω=60o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

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84

Figure 4.55.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1:  (Eq. 3-22)

Figure 4.55.b. Traces of Critical Log Spirals Defining the Active Wedges for Various

Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1:  (Eq. 3-22)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4

Y=

y/H

X=x/H

=40o, ω=0o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Y=

y/H

X=x/H

=40o, ω=50o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

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85

Figure 4.56.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1:  (Eq. 3-22)

Figure 4.56.b. Traces of Critical Log Spirals Defining the Active Wedges for

Various Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1:  (Eq. 3-22)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2

Y=

y/H

X=x/H

=50o, ω=0o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Y=

y/H

X=x/H

=50o, ω=40o, Backslope: Horizontal

Kh=0.0 Kh=0.1 Kh=0.2 Kh=0.3

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86

Figure 4.57.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=0⁰, Backslope 1: 5 (Eq. 3-22)

Figure 4.57.b. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =20⁰, ω=70⁰, Backslope 1: 5 (Eq. 3-22)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8

Y=

y/H

X=x/H

=20o, ω=0o, Backslope: 1:5

Kh=0.0

Kh=0.1

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6

Y=

y/H

X=x/H

=20o, ω=70o, Backslope: 1:5

Kh=0.0 Kh=0.1

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87

Figure 4.58.a. Traces of Critical Log Spirals Defining the Active Wedges for Various Seismic Coefficients: =40⁰, ω=0⁰, Backslope 1: 5 (Eq. 3-22)

Figure 4.58.b. Traces of Critical Log Spirals Defining the Active Wedges for

Various Seismic Coefficients: =40⁰, ω=50⁰, Backslope 1: 5 (Eq. 3-22)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

Y=

y/H

X=x/H

=40o, ω=0o, Backslope: 1:5

Kh=0.0Kh=0.1 Kh=0.2

Kh=0.3

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

Y=

y/H

X=x/H

=40o, ω=50o, Backslope: 1:5

Kh=0.0Kh=0.1 Kh=0.2 Kh=0.3

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88

4.4. Studying Seismic-Induced Resultant Force

An effective presentation to showing the impact of seismicity is through the

difference between the seismic and static coefficients:

∆K K K , K K K K 0 (4-1)

The coefficients corresponding to static conditions, Kae-h (Kh=Kv=0), can

be determined from either Equation 3-17 or Equation 3-22. Alternatively, the charts

for its explicit values are presented in Figures 4.1 to 4.38 or by Leshchinsky and Zhu

(2010). Hence, combining Equation (4-1) with the known static coefficient produces

the seismic coefficient needed to calculate Pae-h.

Figures 4.59 to 4.62 show Kae-h considering the inclination of the

resultant implied in Figure 3.2 (i.e., Equation 3-17) while Figures 4.63 to 4.66

correspond to the inclination shown in Figure 3.4 (i.e., Equation 3-22). The vertical

seismic coefficient for Figures 4.59 to 4.66 was selected as zero. It is apparent that for

a strong seismicity and low friction angle (i.e., Kh=0.3 and =20), a small backslopes

will render the slope unstable regardless of the frontal support (i.e., the cohesionless

backslope will slide) or the Pae inclination.

Examining the results, especially Figures 4.59 and 4.63 for Kh=0.3 and

=20, one observes a seemingly irrational behavior. That is, generally Kae-h

increases when the batter gets larger (note that slope angle = 90 - ω). However, the

static load decreases with increase in batter angle thus implying that the combined

resultant of static and seismic loads could actually go down with the increase of batter.

As an example consider the following two cases: one with ω=0 and the second with

ω=70. For both cases take =20 , =0, and Kh equals to 0 (i.e., static), 0.1, 0.2, and

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89

0.3. Solving Equation 3-17 will produce for ω=0, Kae-h equals to 0.490, 0.569, 0.672,

and 0.830. For ω=70, Kae-h would be 0.000, 0.096, 0.266, 0.558. Hence, one realizes

that while Kae-h (Eq. 4.1) is larger for the flat slope (ω=70) than for the vertical

slope (ω=0), the absolute terms of Kae-h, and subsequently, as intuitively expected, the

horizontal force component of the resultant is larger for the steeper slope.

Review of either set of figures (i.e., Figures 4.59 - 4.62 or 4.63 - 4.66)

shows that the impact of vanishes as ω increases. Accordingly, is of practical

importance for, say, ω20. However, its value may be important for the entire range

of ω values should the resultant inclination be selected to follow Figure 3.4 (i.e., see

Equations 3-21 and 3-22).

Also with looking at Figure 4.47 it can be realized that increases the size

of the active wedge for near vertical slopes; its impact diminishes as the slope

becomes flatter. This increase in wedge size also results in increase of Kae-h for near

vertical slopes. It may appear that an increase of Kae-h with increase is

counterintuitive. However, looking at the absolute values of Kae-h for the active

wedges in Figure 4.47a, it can be verified by solving Equation 3-17 that for =0 and

Kh equal to 0, 0.1, 0.2, and 0.3, the corresponding Kae-h equals to 0.490, 0.569, 0.672,

and 0.830, respectively; for =2/3 and same Kh, Kae-h equals to 0.430, 0.511, 0.629,

and 0.824, respectively. That is, as expected, the values of Kae-h and hence the

horizontal force component decrease with increase of . While the presentation of

results using Kae-h rather than Kae-h is useful as it directly conveys the impact of

seismicity on horizontal force, it clearly might be confusing as it may give impression

opposite to the actual trend of results.

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90

Finally, refer to Figure 4.67 generated based on Equation 3-17 for the Pae

inclination implied in Figure 3.2. It demonstrates the Impact of vertical acceleration as

related to various batter angles. The characteristic behavior exhibited at Kv=0 is

retained whether Kv is greater or smaller than zero. For a rather large Kh (say, 0.3), Kv

may have significant impact on Kae_h.

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91

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

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92

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

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93

Figure 4.61. Kae-h Versus Batter Using Eq. 3-17 for =40, Kv=0, and (a) =0; (b)

=1/3; (c) =2/3; and (d) =1

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94

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

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95

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

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96

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

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97

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

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98

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

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Figure 4.67. Impact of Vertical Seismicity

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100

Chapter 5

COMPARISON OF RESULTS

Tables 5.1 to 5.3 provide a comparison of Kae-h values obtained from the

M-O equation, limit analysis (LA), Equation 3-17 and Equation 3-22 for ω=0, 30

and 45, respectively. To investigate the effect of seismicity Kae-h values resulted

from these four methods are also compared in Tables 5.4 to 5.6 for ω=0, 30 and 45,

respectively For brevity, the results are shown just for horizontal crest, Kv=0, and

various friction angles. The LA results are extracted from Chen and Liu (1990); it was

obtained using upper bound formulation in the framework of limit analysis of

plasticity using a composite log-spiral mechanism. Chen and Liu (1990) provide

results for ω 30.

As shown in Tables 5.4 to 5.6, for vertical slopes (ω=0), all four

approaches yield practically identical results as the critical wedge degenerates to

essentially the same plan. For ω=0 one may use the M-O equation as it is the simplest

to apply. For flatter slopes (e.g., ω=45), Kae-h is larger for the mechanism used in

Equation 3-17 (Figure 3-2). In fact, only looking at Kae-h could be somewhat

misleading as the static Kae-h is significantly larger for the log-spiral mechanism

compared with the M-O planar surface. Hence, as shown in Tables 5.1 to 5.3, the

combined static and seismic loading stemming from Equation 3-17 or Equation 3-22

are larger (i.e., more critical) than the M-O values. It indicates that the current limit

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101

equilibrium analysis provides a seamless extension to the M-O equation for ω>0,

especially when ω>>0. Generally, the results reported by Chen and Liu (1990) fall

between Equation 3-17 and Equation 3-22; in terms of Kae-h, the difference in the

results is not substantial. Since the limit analysis is fundamentally different from the

M-O's limit equilibrium, the LA method is not considered as a natural extension for

the M-O method.

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Table 5.1. Comparison of Kae-h (= 0, =0o, Kv=0)

/  Kh = 0o 

M‐O  LA*  Eq. 2  Eq. 6 

20o 

0  0.490  0.490  0.490  0.490 

0.1  0.569  0.570  0.569  0.569 

0.3  0.831  0.830  0.830  0.830 

2/3 

0  0.426  0.428  0.430  0.430 

0.1  0.511  0.516  0.511  0.511 

0.3  0.824  0.827  0.824  0.824 

30o 

0  0.333  0.330  0.333  0.333 

0.1  0.397  0.400  0.396  0.396 

0.3  0.569  0.570  0.569  0.569 

2/3 

0  0.279  0.282  0.282  0.282 

0.1  0.344  0.348  0.344  0.344 

0.3  0.537  0.536  0.537  0.537 

40o 

0  0.217  0.220  0.217  0.217 

0.1  0.268  0.270  0.268  0.268 

0.3  0.400  0.400  0.400  0.400 

2/3 

0  0.179  0.179  0.180  0.180 

0.1  0.228  0.232  0.228  0.228 

0.3  0.372  0.375  0.371  0.371 

50o 

0  0.132  0.130  0.132  0.132 

0.1  0.172  0.170  0.172  0.172 

0.3  0.275  0.280  0.275  0.275 

2/3 

0  0.108  0.109  0.109  0.109 

0.1  0.146  0.150  0.146  0.146 

0.3  0.257  0.259  0.257  0.257 

* Kinematic Limit Analysis (LA) method (Chen and Liu 1990)

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Table 5.2. Comparison of Kae-h (= 30o, =0 o, Kv=0)

/  Kh  = 30o 

M‐O  LA*  Eq. 2  Eq. 6 

20o 

0 0.283 0.294 0.285 0.308 0.1  0.370 0.381 0.365 0.383 0.3  0.651 0.650 0.684 0.651

2/3 

0  0.253 0.268 0.286 0.292 0.1  0.342 0.354 0.373 0.368 0.3  0.657 0.661 0.724 0.658

30o 

0  0.134 0.147 0.146 0.155 0.1  0.190 0.199 0.198 0.204 0.3  0.350 0.346 0.359 0.351

2/3 

0  0.120 0.128 0.144 0.148 0.1  0.175 0.187 0.200 0.198 0.3  0.346 0.355 0.381 0.354

40o 

0  0.051 0.061 0.064 0.066 0.1  0.085 0.087 0.097 0.097 0.3  0.184 0.182 0.196 0.187

2/3 

0  0.047 0.050 0.063 0.064 0.1  0.080 0.090 0.099 0.097 0.3  0.186 0.190 0.212 0.194

50o 

0  0.011 0.009 0.018 0.018 0.1  0.028 0.035 0.037 0.036 0.3  0.086 0.087 0.097 0.090

2/3 

0  0.011 0.010 0.018 0.018 0.1  0.027 0.030 0.039 0.038 0.3  0.090 0.090 0.110 0.098

* Kinematic Limit Analysis (LA) method (Chen and Liu 1990)

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Table 5.3. Comparison of Kae-h (= 45o, =0 o, Kv=0)

/  Kh  = 45o 

M‐O  LA*  Eq. 2  Eq. 6 

20o 

0 0.162 N/A 0.217 0.211 0.1  0.247 N/A 0.312 0.285 0.3  0.535 N/A 0.658 0.538

2/3 

0  0.147 N/A 0.225 0.209 0.1  0.232 N/A 0.328 0.288 0.3  0.551 N/A 0.712 0.566

30o 

0  0.046 N/A 0.076 0.074 0.1  0.088 N/A 0.129 0.118 0.3  0.226 N/A 0.293 0.243

2/3 

0  0.043 N/A 0.078 0.074 0.1  0.085 N/A 0.137 0.122 0.3  0.234 N/A 0.332 0.263

40o 

0  0.004 N/A 0.012 0.011 0.1  0.019 N/A 0.038 0.033 0.3  0.085 N/A 0.126 0.101

2/3 

0  0.004 N/A 0.012 0.011 0.1  0.019 N/A 0.041 0.036 0.3  0.092 N/A 0.149 0.116

50o 

0  0.004 N/A 0.000 0.000 0.1  0.000 N/A 0.001 0.000 0.3  0.021 N/A 0.039 0.030

2/3 

0  0.004 N/A 0.000 0.000 0.1  0.000 N/A 0.001 0.001 0.3  0.024 N/A 0.049 0.037

* Kinematic Limit Analysis (LA) method (Chen and Liu 1990)

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Table 5.4. Comparison of Kae-h (= 0o, =0 o, Kv=0)

/  Kh  = 0o 

M‐O  LA*  Eq. 2  Eq. 6 

20o 0

0.1 0.079 0.08 0.078 0.078 0.3  0.341 0.34 0.340 0.340

2/3 0.1  0.085 0.09 0.081 0.081 0.3  0.398 0.40 0.394 0.394

30o 0

0.1 0.063 0.07 0.063 0.063 0.3  0.236 0.24 0.235 0.235

2/3 0.1  0.064 0.07 0.062 0.062 0.3  0.258 0.25 0.255 0.255

40o 0

0.1 0.051 0.05 0.051 0.051 0.3  0.183 0.18 0.182 0.182

2/3 0.1  0.050 0.05 0.048 0.048 0.3  0.193 0.20 0.191 0.191

50o 0

0.1 0.040 0.04 0.040 0.040 0.3  0.143 0.15 0.142 0.142

2/3 0.1  0.038 0.04 0.037 0.037 0.3  0.149 0.15 0.148 0.148

* Kinematic Limit Analysis (LA) method (Chen and Liu 1990)

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Table 5.5. Comparison of Kae-h (= 30o, =0 o, Kv=0)

/  Kh  = 30o 

M‐O  LA*  Eq. 2  Eq. 6 

20o 0

0.1 0.087 0.09 0.080 0.074 0.3  0.368 0.36 0.399 0.343

2/3 0.1  0.089 0.09 0.087 0.076 0.3  0.404 0.39 0.438 0.366

30o 0

0.1 0.056 0.05 0.052 0.049 0.3  0.216 0.20 0.213 0.196

2/3 0.1  0.056 0.06 0.056 0.051 0.3  0.227 0.23 0.237 0.206

40o 0

0.1 0.034 0.03 0.033 0.031 0.3  0.133 0.12 0.132 0.121

2/3 0.1  0.033 0.04 0.036 0.033 0.3  0.139 0.14 0.149 0.130

50o 0

0.1 0.017 0.03 0.019 0.018 0.3  0.075 0.08 0.079 0.071

2/3 0.1  0.017 0.02 0.021 0.019 0.3  0.080 0.08 0.092 0.079

* Kinematic Limit Analysis (LA) method (Chen and Liu 1990)

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Table 5.6. Comparison of Kae-h (= 45o, =0 o, Kv=0)

/  Kh = 45o 

M‐O  LA*  Eq. 2  Eq. 6 

20o 0

0.1 0.085 N/A 0.095 0.074 0.3  0.373 N/A 0.441 0.327

2/3 0.1  0.085 N/A 0.103 0.079 0.3  0.404 N/A 0.487 0.357

30o 0

0.1 0.042 N/A 0.053 0.044 0.3  0.181 N/A 0.217 0.169

2/3 0.1  0.042 N/A 0.059 0.048 0.3  0.191 N/A 0.254 0.189

40o 0

0.1 0.015 N/A 0.026 0.022 0.3  0.081 N/A 0.114 0.090

2/3 0.1  0.015 N/A 0.029 0.025 0.3  0.088 N/A 0.137 0.105

50o 0

0.1 -0.003 N/A 0.001 0.000 0.3  0.017 N/A 0.039 0.030

2/3 0.1  -0.004 N/A 0.001 0.001 0.3  0.020 N/A 0.049 0.037

* Kinematic Limit Analysis (LA) method (Chen and Liu 1990)

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Chapter 6

CONCLUSION AND RECOMMENDATION

This thesis presented algorithms based on limit equilibrium which utilize

log spiral mechanism to find the resultant force needed to stabilize an unstable slope

subjected to pseudostatic loading. This resultant force can be reacted by facing units

such as masonry blocks or gabions or can mobilize the tensile resistance in the

embedded reinforcement. One algorithm considers the direction of the resultant to

result from normal and shear force acting on vertical facing segments only. The

second algorithm is as customarily assumed in the Coulomb method and the M-O

method; i.e., the normal and shear forces act along the average angle of the face of the

slope. The algorithms are valid for any inclination of seismically unstable,

homogenous, and simple slopes. The formulation provides a seamless extension to the

M-O pseudostatic formulation dealing with any slope angle or “wall” batter.

A limited number of stability charts has been presented. The charts

consider the slope angle, the horizontal design seismic coefficient, various soil friction

angles, and backslopes of the crest. The seismic-induced component of the resultant

force superimposed on the static case expressed by Kae-h can become larger as the

slope flattens, especially for small friction angles. Furthermore, the impact of the

interface friction, , on Kae-h also seems rather small. However, one should always

look at the total resultant force which is comprised of the static and the seismic-

induced superimposed load to assess the impact. That is, the static resultant is sensitive

to and it goes down dramatically with increase in the slope batter, ω.

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The comparison of the results shows practically small differences in Kae-

h resulting from different assumed resultant inclinations; however, the assumption that

the normal and shear force components of the resultant act on a vertical plane (Figure

3.2) lead to more critical results than the customary alternative. The comparison of the

M-O results with the limit analysis results shows that for vertical slopes all methods

practically yield the same results. However, even for flatter slopes the Kae-h is not

significantly different. Once again, the inclusion of the static load (as determined by

Leshchinsky and Zhu, 2010) to render a complete result will produce larger

differences in results as viewed through Kae-h alone. Leshchinsky and Zhu (2010) and

this work complement each other and, generally, will yield the most critical results.

The current work is considered as an extension of the M-O formulation and

Leshchinsky and Zhu (2010) as an extension of the Coulomb formulation. This is

because all of these methods are based on the same limit equilibrium principles,

yielding identical results when the critical mechanism becomes planar (i.e., when

ω=0). The current formation extends the classical work to deal more rationally with

ω>0 where the rotational active wedge is likely to become critical. Furthermore, it

deals with the reality that in some cases the soil-facing interface friction is physically

restricted to vertical plans only.

This research presented a robust analytical framework to extend the M-O

method to unstable slopes. The results were compared with the results given by the M-

O method and the LA method. For future investigations, the presented results can be

compared and verified versus numerical analysis results and/or experimental tests.

Moreover, the effect of progressive failure concept and strain localization can be

further investigated and if worthwhile, can be incorporated into the proposed method.

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