Numerical Modeling and Simulation of the Stability of ... · Figure 2.6: Forces considered in the...

1

Numerical Modeling and Simulation of the Stability of Earth Slopes

A Thesis Submitted to the Department of Nuclear Engineering

DEPARTMENT OF NUCLEAR ENGINEERING

SCHOOL OF NUCLEAR AND ALLIED SCIENCES

COLLEGE OF BASIC AND APPLIED SCIENCES

UNIVERSITY OF GHANA

BY

BRENDAN DAGEMANYIMA ATARIGIYA 10507155

BSc (KNUST Kumasi) 2012

In Partial Fulfilment of the Requirements for the Degree of

MASTER OF PHILOSOPY

IN

COMPUTATIONAL NUCLEAR SCIENCE AND ENGINEERING

July 2016

ii

DECLARATION

I hereby declare that with the exception of references to other peoplersquos work which

have duly been acknowledged this Thesis is the result of my own research work and

no part of it has been presented for another degree in this University and elsewhere

helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip helliphelliphelliphelliphelliphelliphelliphelliphelliphellip

Brendan Dagemanyima Atarigiya Date

(Candidate)

I hereby declare that the preparation of this project was supervised in accordance with

the guidelines of the supervision of Thesis work laid down by the University of Ghana

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip helliphelliphelliphelliphelliphelliphelliphelliphelliphellip

Dr Nii Kwashie Allotey Nana (Prof) A Ayensu Gyeabour I

(Principal Supervisor) (Co-Supervisor)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip

Date Date

iii

DEDICATION

I dedicate this work to God Almighty my family and friends

iv

ACKNOWLEDGMENT

Firstly I want to thank and praise God for the good health and strength given me during

all this period of schooling

My sincere and utmost gratitude goes to my Principal Supervisor Dr Nii Kwashie

Allotey of the Ghana Atomic Energy Commission (GAEC) and the University of Ghana

for his expertise in the field of research and exemplary guidance towards the progress

of this research I am also grateful to Prof (Nana) A Ayensu Gyeabour my Co-

Supervisor for his creative suggestions and motivation through this research project

My thanks also goes to Ms Rita Awura Abena Appiah of GAEC thank you for pushing

me to make this possible You helped me improve my programming skills a lot

To my big family the Atarigiya Aguyire and Allotey families thank you for the

support I could not have done it without you

To my amazing course mate Linda Sarpong thank you for your words of

encouragement during out period in school you are a strong woman Furthermore to

the nuclear engineering department (Samiru Efia Matilda Henryhellip) thanks for the

wonderful time we had together

Last and not the least to the woman who stood firmly behind me from day one from

when this journey began Mercy Selina Somhayin Namateng I LOVE YOU

GOD RICHLY BLESS YOU ALL

v

ABSTRACT

Ghana as most other countries has a considerable variation in its topography In an

attempt to build cheaper but yet the safe structures (ie roads apartments etc) we

are most often times faced with building on hill-sides and in valleys This then calls

for the need to correctly assess the stability of any adjacent slopes

In recent times due to the extensive need for stability analysis in engineering practice

slope stability analysis programs have been developed It is noted that these

commercial slope stability programs are used extensively in the industry but are very

expensive and require purchasing yearly licenses As a result of this slope stability

analysis is not routinely conducted in local geotechnical engineering practice The need

for cheaper more accessible options is thus considered needful

This research initiative uses MATLAB a commercially available user-friendly and

easy to access computing platform to develop a slope stability analysis program The

method used is the General Limit Equilibrium Method (GLE) with the adoption of the

Morgenstern-Price (M-P) factor of safety (FoS) approach to develop a cheap efficient

and yet effective model for slope stability analysis and design The results of the

program are validated by comparing with the results of SLOPEW a commercial slope

stability program

The results show four model outputs from the developed program and SLOPEW for a

homogeneous material Two different failure mechanisms are shown (ie toe and base

failures) It is noted that the percentage error in the M-P FoS is less than 5

It is anticipated that with the availability of this computer code Ghanaian Engineers

can more readily assess the safety of slopes in routine design works

vi

Contents

DECLARATION ii

DEDICATION iii

ACKNOWLEDGMENT iv

ABSTRACT v

LIST OF FIGURES ix

LIST OF TABLES xi

CHAPTER ONE INTRODUCTION 1

11 Background 1

12 Problem Statement 3

13 Relevance and Justification of Study 4

14 Research Goal 5

15 Research Objectives 5

16 Scope 5

17 Format of the Thesis 6

CHAPTER TWO LITERATURE REVIEW 7

21 Factors Causing Instability 7

22 Types of Slip Surfaces 8

23 Definition of FoS 10

24 Slope Stability Analysis Methods 12

241 Limit Analysis Method 12

242 Variational Calculus Method 13

243 Strength Reduction Method 14

244 General Discussion on Limit Equilibrium Method 15

25 General Limit Equilibrium Method of Slices (GLE method) 19

vii

26 The Ordinary or Fellenius Method (OMS) 20

27 Simplified Bishoprsquos Method 21

28 Janbursquos Simplified Method 22

29 Morgenstern-Price (M-P) Method 23

CHAPTER THREE RESEARCH METHODOLOGY AND SOFTWARE USED 27

31 Selection of Factor of safety method 27

32 Morgenstern-Price Method 28

33 Assumptions 29

34 Numerical Method Development 29

35 Derivation of Equations 30

36 Structured Program 32

37 Numerical Algorithm 33

38 Software and Programs Used 34

381 SLOPEW 34

382 MATLAB 36

CHAPTER FOUR RESULTS AND DISCUSSION 37

41 Introduction 37

42 Programme Test Examples 37

421 Toe Failure 37

422 Base Failure 42

43 Comparison with Cases from Literature 46

431 Case 1 46

432 Case 2 50

CHAPTER FIVE CONCLUSIONS 53

51 Conclusions 53

52 Recommendations 54

REFERENCES 55

viii

APPENDICES 60

Appendix A Structured Programme 60

Appendix B MATLAB Code for Solving FoS 63

ix

LIST OF FIGURES

Figure 11 Over 40 m high slope on the Ayi-Mensah-Aburi road 3

Figure 21 Types of circular slip failure surface 9

Figure 22 Typical non-circular slip surfaces 10

Figure 23 Various definitions for FoS 11

Figure 24 Swedish Slip Circle Method 16

Figure 25 Slice Discretization and Slice Forces in a Sliding Mass 17

Figure 26 Forces considered in the Ordinary Method of Slices 21


Figure 28 Forces considered in the M-P method 24

Figure 29 Inter-slice force function types 25

Figure 210 Variation of FoS with respect to Fm and Ff vs λ for the M-P method 26

Figure 31 Sketch of a Slope Section 30

Figure 32 Forces acting on a Single Slice from a Mass Slope 31

Figure 33 Algorithm flowchart for solving for the FoS 33

Figure 34 SLOPEW KeyIn Analyses Page 35

Figure 35 SLOPEW KeyIn Material Page 35

Figure 36 SLOPEW KeyIn Entry and Exit Range Page 36

Figure 41 SLOPEW Output of Toe Failure Case 1 38

Figure 42 MATLAB Output of Toe Failure Case 1 39

Figure 43 SLOPEW Output for Toe Failure Case 2 40

x

Figure 44 MATLAB Output for Toe Failure Case 2 41

Figure 45 SLOPEW Output for Base Failure Case 42

Figure 46 MATLAB Output for Base Failure Case 1 43

Figure 47 SLOPEW Output for Base Failure Case 2 44


Figure 49 Homogeneous Slope without Foundation 46

Figure 410 Analysis using SLOPEW - FoS = 1385 47

Figure411 FoS based on M-P approach for Toe failure - FoS = 1451 48

Figure 412 FoS based on M-P approach for Base Failure FoS = 1375 49

Figure 413 Slope Model Geometry from Slide 3 50

Figure 414 FoS based on M-P Approach for ACAD Problem 51

xi

LIST OF TABLES

Table 21 The Main Limit Equilibrium Methods 18

Table 31 Brief Comparison of Limit Equilibrium Methods 27

Table 41 Slope Dimensions and Material Properties for Toe Failure Case 1 38


Table 43 Slope Dimensions and Material Properties for Base Failure Case 1 42


Table 45 Slope dimensions and material properties 46

Table 46 Slope Dimensions and Material Properties for ACAD Problem 50

Table 51 Summary of FoS Outputs for all Case Studies 53

xii

LIST OF SYMBOLS AND ABBREVIATIONS

A cross-sectional area of slice

dx width of slice

l length of the bottom of the slice

c cohesion of soil

G total unit weight of soil

τf shear strength

τ shear stress

γw unit weight of water

angle of internal friction of soil

α inclination from horizontal of the bottom of the slice (degrees)

cal cos(α)

sal sin(α)

tph tan()

H Height of slope

N total normal force on the bottom of the slice

S shear force on the bottom of the slice

W weight of the slice

havg = average height of slice

u pore water pressure

xiii

∆X shear component of the inter-slice force

∆E inter-slice force on the downslope of the slice

ns number of slice faces

FoS factor of safety

F assumed factor of safety

Ff force factor of safety

Fm moment factor of safety

Xc x coordinate of centre of slip circle

Yc y coordinate of centre of slip circle

R Radius of slip circle

ytop Ground surface

ybot Slope surface

FSom Factor of Safety for Ordinary Method of Slices

FFm Moment Factor of Safety for Morgentern-Price Method of Slices

FFf Force Factor of Safety for Morgentern-Price Method of Slices

F(x) Interslice force function

λ Scale factor of the assumed f(x)

1

CHAPTER ONE INTRODUCTION

11 Background

Slope stability is the potential for ground slopes to resist movement [1] Slope

instability has been the subject of continued concern because of the tremendous loss of

life property and infrastructure caused annually in many places in the world [2] In the

field of construction slope instability can occur due to rainfall increasing the water

table and the change in stress conditions Similarly tracks of land that have been stable

for years may suddenly fail due to changes in the geometry external forces and loss of

shear strength [3]

Slope failures also called slides or landslides whether sudden or gradual are due to

the increased stress of slope materials or foundations compared with their mobilized

strength [3]

The majority of the slope stability analyses performed in practice still use traditional

limit equilibrium approaches involving the method of slices and has remained virtually

unchanged for decades

Analysis of the slope stability is carried out to assess the safety of artificial or natural

slopes (eg dams road cuts mining open pit excavations and landfills) For human

made slopes analysis of slope stability is used to evaluate various design options which

then provides a basis for a form of engineering design with associated costing

comparisons The efficient engineering of natural and artificial slopes has therefore

become a common challenge faced by both researchers and practitioners

Slope stability assessment mainly involves the use of the factor of safety (FoS) method

to determine how close a given slope is to the onset of instability or to what extent the

state of the slope is from failure

2

When this ratio is well above 1 the resistance to shear failure is generally higher than

the driving shear stress and the slope is considered stable When this ratio is near to 1

the shear strength is almost equal to the shear stress and the slope is close to failure If

the FoS is less than 1 the slope is considered to have failed or considered to be trigger-

point ready [4]

Ghana is not noted to be a frequent serious victim of mass movement (slope failures)

It is however noteworthy that Ghana has not been without slope failures In Ref [5]

reference is made of a slope failure which occurred in 1968 near Jamasi in the Sekyere

District involving about 1500 cubic meters of rock soil and vegetation The failure

blocked the main Kumasi-Mampong truck road for a total of ten days

Reference is also made in Refs [6 amp 7] on potential slope failures on the stretch of the

Accra-Aburi road when rocks began to fall unhindered onto the road in 2014

Furthermore Ref [8] notes that in 2013 after a heavy downpour of rain loose parts of

the Akoaso Mountain fell and blocked the Kasoa-Weija portions of the road resulting

in a significant traffic jam for hours

These recent records of slope instability in the country have served as a wakeup call for

researchers and practicing engineers to take a critical look at this issue

3

Figure 11 Over 40 m high slope on the Ayi-Mensah-Aburi road

12 Problem Statement

In an attempt to build cheap and yet the safe structures (ie roads living apartments

etc) for man-kind we are most times faced with building in valleys and on mountains

Either way we are faced with the problem of slope instability

In the past decades computer software for slope stability analysis and design have been

developed and marketed extensively These commercial software which have been

developed over many years are able to perform rigorous stability calculations and give

fast and accurate answers to complex slope stability problems These software have

become widely accepted in industry and are now part of most large design engineering

offices These software are however expensive and normally require the annual

renewal of licenses Notwithstanding their wide acceptance in industry most

Ghanaian geotechnical engineers still resort to the use of the rule of thumb slope

stability analysis methods and old charts for their daily slope stability analysis This is

4

due to the relatively high cost of these commercial software and the limited financial

capacity of local engineering firms

This has created a gap between local and international engineers and has resulted in

major geotechnical engineering projects involving complex slope instability problems

being awarded to international firms rather than local engineering firms

13 Relevance and Justification of Study

Landslides rock falls and mass movement of any kind are undoubtedly one of the

oldest natural disasters that have resulted in huge damages loss of lives and a great

deal of pain to mankind Like other mountainous countries Ghana has large variations

in its topography The impending threat of landslides in the case of [5 or 6] or rock

falls in the case of [7] is now accepted as life threating and the need for these slopes to

be properly engineered is critical

It has been already noted that the available commercial slope stability programs that

are extensive slope stability analysis are very expensive and require purchasing yearly

licenses This has necessitated the need to develop a simple yet efficient slope stability

program that can be easily accessed by local engineers to appraise local slope stability

problems using the most rigorous and accurate methods available

This research initiative uses MATLAB (a commercially available and easily

accessible computing platform with great user-friendly interface) using General Limit

Equilibrium method (GLE) and adopting the Morgenstern-Price FoS to develop a cheap

and efficient yet effective model for slope stability analysis and design

5

14 Research Goal

The goal of this thesis is to develop and implement a slope stability numerical model

that will aid local geotechnical engineers to readily appraise the stability of slopes in

local practice

15 Research Objectives

The objectives of the research are to

Develop a physical model to represent the problem

Develop mathematical equations to solve the problem

Develop a numerical algorithm and write a code to solve for the FoS of earth

slopes

Verify and validate the code using Geoslope Internationalrsquos SLOPEW

commercial slope stability programme

16 Scope

For the purpose of this study this thesis is limited to the development of a slope stability

programme for homogenous soil and rock media In this regard the goal of the study

is to develop the generic algorithm for slope stability analysis Furthermore similar to

the existing commercial programmes the study is limited to two-dimensional slope

stability problems

6

17 Format of the Thesis

Following the introduction to slope stability problems in Chapter 1 a detailed literature

review of the methods of slope stability analysis is provided in Chapter 2 Chapter 3

then presents the proposed solution method for calculating the factor of safety in which

the GLE method is explained and the solution algorithm developed

Chapter 4 presents sample results from the developed MATLAB code It presents

comparisons between the results of the developed code and results from the commercial

slope stability programme SLOPEW Chapter 5 finally presents the conclusions of

the study and also provides recommendations for further studies

7

CHAPTER TWO LITERATURE REVIEW

Analysis of slope stability problem is an important subject area in geotechnical

engineering The phenomenon of landslides and related slope instability is a problem

in many parts of the world Slope failure mechanisms and the geological history of a

slope can be very complicated problems and required complex forms of analyses

21 Factors Causing Instability

The failure of the slopes occurs when the downward movements of soil or rock material

because of gravity and other factors creates shear stresses that exceed the inherent shear

strength of the material Therefore factors that tend to increase the shear stress or

decrease the shear strength of a material increases the risk of the failure of a slope

Various processes can lead to a reduction of the shear strength of a soilrock mass

These include increased pore pressure cracking swelling decomposition of

argillaceous rock fills creep under sustained loads leaching softening weather and

cyclic loading among others

On the other hand shear stress within a rocksoil mass may increase due to additional

loads on top of the slope and increase in water pressure due to cracks at the top of the

slope an increase in the weight of soil due to increasing water content the excavation

of the base of the slope and seismic effects Furthermore additional factors that

contribute to the failure of a slope include the rocksoil mass properties slope geometry

state of stress temperature erosion etc

The presence of water is the most critical factor that affects the stability of slopes This

is because it increases both the driving shear stress and also decreases the soilrock

massrsquo shear strength The speed of sliding movement in a slope failure can vary from

8

a few millimeters per hour to very fast slides where great changes have occurred within

seconds Slow slides occur in soils with a plastic stress-strain characteristics where

there is no loss of strength with increased strain Fast slides occurs in situations where

there is a sudden loss of strength as in the liquefaction of sensitive clay or fine sand

Increase in shear stresses across the soil mass result in movement only when the shear

strength mobilized on given possible failure surface in the ground is less than the

driving shear stresses along that surface

22 Types of Slip Surfaces

To calculate the FoS of a slope it is always assumed that the slope is failing in some

shape normally in a circular or non-circular shape For computational simplicity the

slide surface is often seen as circular or composed of several straight lines [9] Different

sliding surfaces are normally assumed with the computation of a corresponding FoS

The sliding surface with the minimum FoS is then selected as the FoS of the slope in

question

A circular sliding surface like that shown in Figure 21 is often used because it is

suitable to sum the moments about a centre The use of a circle also simplifies the

calculations Wedge-like surfaces have their failure mechanisms defined by three or

more straight line segments defining an active area central block and the passive area

as shown in Figure 22 This type of sliding surface can be used for analysis of slopes

where the critical potential sliding surface comprises a relatively long linear sector

through low material bounded by a stronger material

As noted above the critical slip surface is the surface with the lowest factor of safety

The critical slip surface for a given problem analysed by a given method is found by a

9

systematic procedure to generate sliding test surfaces until the one with the minimum

safety factor is found [9]

Figure 21 Types of circular slip failure surface [3]

10

Figure 22 Typical non-circular slip surfaces [9]

23 Definition of FoS

FoS is usually defined as the ratio of the ultimate shear strength to the shear stress

mobilized at imminent failure There are several ways to formulate the FoS The most

common formulation assumes the safety factor to be constant along the sliding surface

and it is defined in relation to limit equilibrium force and moment equilibrium [4]

11

These definitions are given in Figure 23 below As will be developed further in the

limit equilibrium method the first definition is based on the shear strength which can

be obtained in two ways a total stress approach (su‐analysis) and an effective stress

approach (crsquo- φrsquo minusanalysis) The type of strength consideration depends on the soil

type the loading conditions and the time elapsed after excavation The total stress

strength method is used for shortndashterm conditions in cohesive soils whereas the

effective stress method is used in long- term conditions in all soil types or in short-term

conditions in cohesive soils where the pore pressure is known [3]

Figure 23 Various definitions for FoS [3]

12

24 Slope Stability Analysis Methods

Slope stability problems deal with the condition of ultimate failure of a soil or rock

mass Analyses of slope stability bearing capacity and earth pressure problems all fall

into this area The stability of a slope can be analysed by a number of methods among

others of which are the

Limit analysis method

Variational calculus method

Strength reduction method and

Limit equilibrium method

241 Limit Analysis Method

The limit analysis method theory is based on a rigid-perfectly plastic model material

Drucker and Prager [10] first formulated and introduced the upper and lower bound

plasticity theorems for soilrock masses

The general analysis process includes construction of a statically admissible stress field

for the lower-bound analysis or a kinematically admissible velocity field for the upper-

bound analysis

For both upper- and lower-bound analysis one of the following two conditions has to

be satisfied

Geometrical compatibility between internal and external displacements or

strains This is usually concerned with kinetic conditions ndash velocities must be

compatible to ensure no gain or loss of material at any point

Stress equilibrium ie the internal stress fields must balance the externally

applied stresses (forces)

13

The basis of limit analysis rests upon two theorems which can be proved

mathematically In simple terms these theorems are

Lower Bound any stress system in which the applied forces are just sufficient

to cause yielding

Upper Bound Any velocity field that can operate is associated with an upper

bound solution

The lower- bound approach has been used in 2D slope stability analysis by Zhang [11]

and Kim et al [12] The upper-bound approach was first used in 2D slope stability

analysis by Drucker and Prager [10] to determine the critical height of a slope

Subsequently Refs [13 14 amp 15] also applied and extended the upper-bound

approaches in 2D slope analysis Michalowski [16] proposed an upper-bound approach

based on a translational failure mechanism The vertical slice techniques which are

often used in traditional limit equilibrium approaches were employed to satisfy the

force equilibrium condition for all individual slices Two extreme kinematical solutions

neglecting the inter-slice strength or fully utilizing the inter-slice strength of the soil

were then obtained The traditional limit equilibrium solutions for slices with a proper

implicit assumption of failure mechanism can fall into the range of these two extremes

Donald and Chen [17] also presented an upper-bound method on the basis of a multi-

wedge failure mechanism in which the sliding body was divided into a small number

of discrete blocks

242 Variational Calculus Method

The variational calculus approach does not require assumptions on the inter-slice

forces It was first used for 2D stability analysis by Baker and Garber [18] This

14

approach was subsequently employed by Jong [19] for vertical-cut analysis in cohesive

directionless soils Cheng et al [20] also used it in their research where they developed

a numerical algorithm based on the extremum principle by Pan [21] The formulation

which relies on the use of a modern try and error optimization method and can be

viewed as an equivalent form of the variational method in a discretized form but is

applicable for a complicated real life problem

243 Strength Reduction Method

In recent decades there have been great developments in the area of the strength

reduction method (SRM) for slope stability analysis The general procedure of the

SRM analysis is the reduction of the strength parameters by the FoS while the body

forces (due to the weight of soil and other external loads) are applied until the system

cannot maintain a stable state This procedure can determine the FoS within a single

framework for both two and three-dimensional slopes

The main advantages of the SRM are as follows

The critical failure surface is automatically determined from the

application of gravity loads andor the reduction of shear strength

It requires no assumption about the distribution of the inter-slice shear

forces

It is applicable to many complex conditions and

It can give information such as stresses movements (deformations) and

pore pressures

One of the main disadvantages of the SRM is the long time required to develop the

computer model and to perform the analysis to arrive at a solution With the

15

development of computer hardware and software 2D -SRM can now be done in a

reasonable amount of time suitable for routine analysis and design This technique is

also adopted in several well-known commercially available geotechnical finite element

or finite difference programs In strength reduction analysis the convergence criterion

is the most critical factor in the assessment of the FoS

Investigation results show that the FoS obtained and the corresponding slip surface

determined by the SRM demonstrate good agreement with the results of the Limit

Equilibrium Method (LEM)

244 General Discussion on Limit Equilibrium Method

2441 Swedish Slip Circle Method of Analysis

The Swedish Slip Circle method assumes that the friction angle of the soil or rock is

zero In other words when angle of friction is considered to be zero the effective stress

term tends to zero which is therefore equivalent to the shear cohesion parameter of the

given soil The Swedish slip circle method assumes a circular failure interface and

analyses stress and strength parameters using circular geometry and statics as shown in

Figure 24 The moment caused by the internal driving forces of a slope is compared

with the moment caused by resisting forces in the sliding mass If forces resisting

movement are greater than the forces tending to cause movement then the slope is

assumed to be stable

16

Figure 24 Swedish Slip Circle Method [3]

2442 Method of Slices

Despite all the above methods limit equilibrium methods are by far the most used form

of analysis for slope stability studies They are the oldest best-known numerical

technique in geotechnical engineering These methods involve cutting the slope into

fine slices so that their base can be comparable with a straight line The governing

equilibrium equations equilibrium of the forces andor moments Figure 25 are then

developed According to the assumptions made on the efforts between the slices and

the equilibrium equations considered many alternatives have been proposed in Table

21 They give in most cases quite similar results The differences between the values

of the FoS obtained with the various methods are generally below 6 [22]

17

Figure 25 Slice Discretization and Slice Forces in a Sliding Mass [2324]

The idea of dividing a potential sliding mass into slices dates back to the early 1900rsquos

The first documented use of the method of slices is the analysis of the 1916 failure at

the Stigberg Quay in Gothenburg Sweden [25]

This Limit Equilibrium method is well known to be a statically indeterminate problem

and assumptions about the inter-slice shear forces are needed to make the problem

statically determinate On the basis of the assumptions on the internal forces and the

force andor moment equilibrium there are more than ten methods developed for

analysis of slope stability problems [26] Famous methods include those by Fellenius

[27] Bishop [28] Janbu [29 30] Spencer [31] Morgenstern-Price [32] and the General

Limit Equilibrium (GLE) [25]

Table 21 shows the differences between the various methods of stability analysis on

the basis of forces and moments equilibrium

18

Table 21 The Main Limit Equilibrium Methods [5]

No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function

1 Fellenius Swedish Circle or

Ordinary Method (1936) Yes No No No Yes No No

2 Bishop Simplified (1955) Yes No Yes No Yes No No

3 Janbu Simplified (1954) No Yes Yes No No Yes No

4 Spencer Method (1967) Yes Yes Yes Yes Yes Yes Constant

5 Morgenstern-Price Method

(1965) Yes Yes Yes Yes Yes Yes

Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19

All limit equilibrium methods utilise the Mohr‐Coulomb criterion to determine the

shear strength (τf) along the sliding surface The mobilised shear stress at which a soil

fails in shear is defined as the shear strength of the soil According to Janbu [29] a state

of limit equilibrium exists when the mobilised shear stress (τ) is expressed as a fraction

of the shear strength Nash [33] states that at the moment of failure the shear strength

is fully mobilised along the failure surface when the critical state conditions are

reached The shear strength is usually expressed by the Mohr‐ Coulomb linear

relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where

c is the soil cohesion and is the soil frictional angle

The strength of the soil available depends partially on the type of soil and the normal

stresses acting on it The mobilized shear strength on the other hand depends on the

external forces acting on the soil mass This defines the FoS as a ratio of the τf to τ in a

limit equilibrium analysis [30] as defined in Equation 22

25 General Limit Equilibrium Method of Slices (GLE method)

The GLE formulation was developed by Fredlund at the University of Saskatechewan

[34 35] This method encompasses the key elements of all the other methods of slices

(the Ordinary Bishoprsquos Janbu and M-P methods) The GLE formulation is based on

two factors of safety equations One equation gives the factor of safety with respect to

20

moment equilibrium Fm (Equation 23) while the other equation gives the factor of

safety with respect to horizontal force equilibrium Ff (Equation 24)

119865119898 = sum119888lowast119897lowast119877lowast119882lowastcos[120572119894]minusμl)lowast119877lowasttan[120601]

119882lowast119909 (23)

119865119891 = sum119888lowast119897lowastcos[120572119894]+(119873119894minusμ119897)lowasttan[120601]lowastcos[120572119894]

119873119894lowastsin[120572119894] (24)

Where

l is the length of the bottom of the slice R is the radius of the slip circle W is the weight

of slice Ni is the total normal force on the bottom of the slice c is the soil cohesion

microis the pore water pressure α is the inclination of the slip surface at the middle of the

slice

26 The Ordinary or Fellenius Method (OMS)

The ordinary method is considered the simplest of the methods of slices since it is the

only procedure that results in a linear factor of safety equation It is generally stated that

the inter-slice forces can be neglected because they are parallel to the base of each slice

[26] This notwithstanding the Newtons principle of action equals reaction is not

satisfied between slices The change in direction of the resultant inter-slice forces from

one slice to the next results in factor of safety errors that may be as much as 60 [37]

The normal force on the base of each slice is derived either from summation of forces

perpendicular to the base or from the summation of forces in the vertical and horizontal

directions The forces considered to act in this method are represented in Figure 26

below The FoS is based on moment equilibrium and computed as

21

Figure 26 Forces considered in the Ordinary Method of Slices

119865119898 =119888lowast119897+119873lowasttan()

119882lowastsin(120572) (25)

N = W lowast cos(α)ndash microl (26)

where

W is the weight of each slice

microis the pore water pressure

l is the base length of the slice

α is the inclination of the slip surface at the middle of the slice

27 Simplified Bishoprsquos Method

The simplified Bishop method neglects the inter-slice shear forces and thus assumes

that a normal or horizontal force adequately defines the inter-slice forces as shown in

Figure 27 below The normal force on the base of each slice is derived by summing

forces in a vertical direction The factor of safety is derived from the summation of

moments about a common point as in OMS since the inter-slice forces cancel out

Therefore the factor of safety equation is the same as for the ordinary method [28]

22

However the definition of the normal force is different The FoS is based on moment

equilibrium and computed as

Figure 27 Forces considered in the Simplified Bishoprsquos Method


119882lowastx (27)

119873119894 =119882minus[

119888lowast119897lowastsin[120572119894]minus119906lowast119897lowastsin[120572119894]lowasttan[120601]

119865119898]

Cos[120572119894]+sin[120572119894]lowasttan[120601]

119865119898

(28)

where lsquoxrsquo is the horizontal distance from the mid-base of the slice to the centre of

rotation

As can be seen in the equation above the expression is nonlinear due to the appearance

of Fm in both sides of the equation and will require an iterative procedure to reach a

solution

28 Janbursquos Simplified Method

In Janbus simplified method the normal force in each slice is derived from the

summation of vertical forces with the inter-slice shear forces ignored The horizontal

force equilibrium equation is used to derive the factor of safety The sum of the inter-

slice forces must cancel and FoS equation becomes

23

119873119894 =119882minus[


119865119891]


119865119891

(29)

119865119891 =sum119888lowast119897lowastcos(120572)+(119873119894minusμlowast119897)lowastcos(120572)lowasttan()

119873119894lowastsin(α) (210)

29 Morgenstern-Price (M-P) Method

This has become the most widely used method developed for analysing generalized

failure surfaces The method was initially described by Morgenstern and Price [36]

The Morgenstern‐Price method also satisfies both force and moment equilibriums and

the overall problem is made determinate by assuming a functional relationship between

the inter-slice shear force and the inter-slice normal force According to Morgentern

and price [32] the inter-slice force inclination can vary with an arbitrary function f(x)

as

119883119894 = 119864119894 lowast λ lowast f(x) (211)

where

f(x) = inter-slice force function that varies continuously along the slip surface and

λ = scale factor of the assumed function

The method suggests assuming any type of force function f(x) for example half‐sine

trapezoidal or user defined as shown in Figure 29 The relationships for the base normal

force (N) and inter-slice forces (E X) are the same as given in Janbursquos generalised

method The forces considered to act in this method are represented in figure 28 below

For a given force function the inter-slice forces and the Newton-Raphson numerical

technique can be used to solve the moment and force equations for the FoS and until

Ff is equals to Fm in equations (212) and (213) [33] below

24

Figure 28 Forces considered in the M-P method [24]


119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)

F is Fm or Ff depending on which equilibrium equation is being solved

and

∆119883119894 =∆119864119894 lowast λ lowast f(x) (215)

25

Figure 29 Inter-slice force function types

An alternative derivation for the Morgenstern-Price method was proposed by Fredlund

and Krahn in [34] It presents a complete description of the variation of the factor of

safety with respect to λ On the first iteration the vertical shear forces are set to zero

On subsequent iterations the horizontal inter-slice forces are first computed and then

the vertical shear forces are computed using an assumed λ value and side force function

Fm and Ff are solved for a range of λ values and a specified side force function These

26

FoS are plotted on a graph as shown in Figure 210 The FoS vs λ are fit by a second

order polynomial regression and the point of intersection satisfies both force and

moment equilibrium

Figure 210 Variation of FoS with respect to Fm and Ff vs λ for the M-P method [34]

27

CHAPTER THREE RESEARCH METHODOLOGY AND SOFTWARE

USED

31 Selection of Factor of safety method

There are different limit equilibrium methods having varying superiorities over each

other as discussed in Chapter 2 Methods are based on different assumptions on

equilibrium conditions to be satisfied Duncan and Wright [24] summarized some of

limit equilibrium methods with respect to their limitations assumptions and

equilibrium conditions to be satisfied as shown in Table 31

Table 31 Brief Comparison of Limit Equilibrium Methods [24]

Procedure Use

Swedish circle method Applicable to slopes where = 0

Ordinary method of slices

Applicable to non-homogeneous slopes soils where slip surface can be

approximated by a circle Very convenient for hand calculations In

accurate for effective stress analysis with high pore water pressure

Simplified Bishop method


approximated by a circle More accurate than ordinary method of slices

especially for analysis with high pore water pressures Calculations

feasible by hand or spreadsheet

Spencers method

An accurate procedure applicable to virtually all slope geometries and soil

profiles The simplest complete equilibrium procedure for computing

FoS

Morgenstern and Prices

method


profiles Rigorous well-established complete equilibrium procedure

Requires solution of nonlinear equations with an iterative procedure

The analysis and design of failing slopes requires an in-depth understanding of the

failure mechanism in order to choose the right slope stability analysis method

In this research as stated above the slice approach for GLE procedure is used and the

FoS according to the Morgenstern-Pricersquos procedure which satisfies all the

28

requirements for static equilibrium is adopted Regardless of whether equilibrium is

considered for a single free body or a series of individual vertical slices there are more

unknowns (forces locations of forces factor of safety etc) than the number of

equilibrium equations the problem of computing a FoS is thus statically indeterminate

Therefore assumptions must be made to achieve a balance of equations and unknowns

This method allows for analysis of any failure shape (circular non-circular or

compound) The solution for the (FoS) is derived from the summation of forces

tangential and normal to the base of a slice and the summation of moments about the

centre of the base of base slice The steps required to provide the input data for

performing the slope stability analysis include [37]

A survey of the elevation of the ground surface on a section perpendicular to

the slope

Estimation of ground stratigraphy from borehole logs and soilrock properties

from engineering soilrock tests

The determination of ground water level from piezometer readings to estimate

ground water pore-water pressures


Morgenstern and Price [32] developed the method for FoS similar to the Spencer

method [31] The method considers both normal inter slice force and shear forces

Therefore it satisfies both moment and force equilibrium

29

33 Assumptions

To begin with the generalized formulation the following assumptions are made for a

body of mass slope

1 The failure surface is assumed to be circular

2 The soilrock is a homogeneous

3 The soilrock is an isotropic material

4 The failure mass is a rigid body

5 The base normal force acts at the middle of each slice

6 The Mohr-Coulomb failure criterion is used

34 Numerical Method Development

In this work the geotechnical generalized method of slices approach which is a form

of the Finite Strip Method would be used for the computation of the Factor of Safety

(FoS) which is a measure of the degree of safety of the slope The Finite Strip Method

is a variant form of the known differential equation numerical discretization

approaches It discretizes the equilibrium differential equation (based on slices) and

imposes equilibrium boundary conditions at the ends to solve for the required

unknowns

The method of slices method is a numerical approach used to solve the equilibrium

differential equation at the limiting condition ie at the point of slope failure The

method consists of dividing the slope into a number of fine slices so that their bases can

be comparable with a straight line The equilibrium differential equations are then

developed for each slice and then the global force and moment equilibrium problem

solved numerically to obtain the FOS [24]

30

This study will make available a locally developed slope stability program that would

simulate the behaviour of a simple slope undergoing circular failure The results from

this work which will include both factor of safety for moment and force equilibrium

will be visualized in the command window of MATLAB

35 Derivation of Equations

Figure 31 is a sketch of a slope section and Figure 32 is a sketch of a slice of mass

from Figure 31 with the forces acting on it at the point of failure By taking moment

about the midpoint of the base of the slice in Figure 31

119864prime [(119910 minus 119910119905prime) minus (minus

dy

2)] minus (119864prime + dEprime) [119910 + dy minus 119910119905

prime minus dy119905prime + (

dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)

After simplification and proceeding to the limit as dxrarr0 it can be shown readily that

119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)

Figure 31 Sketch of a Slope Section [23]

31

Figure 32 Forces acting on a Single Slice from a Mass Slope [24]

For equilibrium in the N direction we have from Figure 32

dNprime = dWcos[120572] minus dXcos[120572] minus dEprimesin[120572] (33)

For equilibrium in the S direction we have from Figure 32

dS = dEprimecos[120572] minus dXsin[120572] + dWsin[120572] (34)

The Mohr-Coulomb failure criterion in terms of effective stresses may be expressed as

dS =1

119865119888primedxsec[120572] + (dNprime)tan[120601prime] (35)

Equation (35) defines the factor of safety in terms of shear strength

Eliminating dS from Equations (34) and (35)

1

119865[119888primedxsec[120572] + (dNprime)sin[120601prime]] = dEprimecos[120572] minus dXsin[120572] + dWsin[120572] (36)

By eliminating dNprime from Equations (33) and (36) and dividing by dxcos[120572] we have

119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)

In the specified co-ordinate system tan[120572]=-dy

dx and equation (37) becomes

32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)

Therefore the governing differential equation becomes using equations (32)

dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)

Ersquo and Ersquo+dErsquo are the horizontal inter-slice forces

X and X+dX are the vertical inter-slice forces

120601prime is the effective friction angle

W is the weight of ith slice

dN and dS are resultants of the normal and tangential forces on the slice base of length

li

α is the inclination of the slice with respect to the horizontal

F is the factor of safety

The governing differential equation is developed for each slice and summed up to

develop the equilibrium equation for entire mass Based on the assumption on the inter-

slice forces as provided in Chapter 2 the various expressions for Fm Ff N X amp E for

the different slices are computed

36 Structured Program

Below (Appendix A) is an algorithm developed to solve for the force and moment factor

of safetyrsquos (Ff and Fm) using the M-P method

33

37 Numerical Algorithm

Figure 33 shows the numerical algorithm used to solve the problem Equation numbers

referred to here are that from appendix A

Figure 33 Algorithm flowchart for solving for the FoS

No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi

Compute lb W and alpha-eq

10 7amp 6

Compute FSom-eq 15 amp16

Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol

Is FSf-FoS lttol else

NFSm=FSm NFSf=FSf

Increase

Compute N FSm FSf and

∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34

38 Software and Programs Used

381 SLOPEW

SLOPEW developed by GeoStudio International Canada is part of a professional

geotechnical software suite of programmes SLOPEW is the programme used for slope

stability analysis This software is based on the theories and principles of the LE

methods discussed in the previous sections To check the accuracy of the program

written in MATLAB and its output SLOPEW is employed For this study a full

licensed version of Geostudio 2007 has been used

For each model using the drawing tools the geometry of the slope is entered Then by

using the slip surface dialogue box the slip surface is specified by using a range

Then by using the ldquoMaterialsrdquo dialogue box soil parameters will be entered and the

selected soil will be assigned to the drawing in the software

After entering all of the input data into the software by hitting the ldquoStartrdquo button under

the ldquoSolve Managerrdquo the program starts to analyze the slope and find the minimum

factor of safety and its related failure surface Figures 34 to 36 show some of the stages

of developing the model using SLOPEW

35

Figure 34 SLOPEW KeyIn Analyses Page

Figure 35 SLOPEW KeyIn Material Page

36

Figure 36 SLOPEW KeyIn Entry and Exit Range Page

382 MATLAB

MATLAB (Matrix Laboratory) is a multi-paradigm numerical computing environment

and fourth-generation programming language A proprietary programming language

developed by MathWorks MATLAB allows matrix manipulations plotting of

functions and data implementation of algorithms creation of user interfaces and

interfacing with programs written in other languages including C C++ Java Fortran

and Python [38] The programme is also easily accessible and is available in most

design engineering offices These features of MATLAB make it suitable to be used for

this research work

37

CHAPTER FOUR RESULTS AND DISCUSSION

41 Introduction

This chapter presents outputs from the MATLAB code and SLOPEW programmes and

compares the results of the two First the analysis is run for assumed models (the same

geometry soil properties) of homogeneous soil mass and modelled for two failure

scenarios ie for toe and base failure conditions

The second set of comparison is done for existing case studies Case one is the study

from Griffiths and Lane [39] and Rocscience [40] and case two is from the study by

ACADS [41]

The MATLAB script is presented in appendix B

42 Programme Test Examples

The program was tested with two failure mechanisms toe and base failures In slope

stability design and analysis and for most design projects the worst case scenario is

always adopted The worst case when it comes to FoS in slope stability analysis is the

least FoS This is adopted in choosing the FoS in this MATLAB program in situations

when more than one FoS is outputted

421 Toe Failure

Two models were presented for the toe failure mechanism The slope angles for both

models are 45deg or 1H 1V (H horizontal distance V vertical distance) The rest of the

soil parameters and geometry are presented in Tables 41 and 42

38

Case 1

Table 41 Slope Dimensions and Material Properties for Toe Failure Case 1

cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5

Figure 41 SLOPEW Output of Toe Failure Case 1

39

Figure 42 MATLAB Output of Toe Failure Case 1

40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5

Figure 43 SLOPEW Output for Toe Failure Case 2

41

Figure 44 MATLAB Output for Toe Failure Case 2

42

It is noted from the outputs that the MATLAB code give reasonable outputs of FoS

compared to SLOPEW In both cases the percentage error is found to be 4 The

small error margin is an indication that the developed MATLAB code is working to an

acceptable degree

422 Base Failure

Two models were presented for the base failure mechanism The slope angles for both

models are 1H 1V The rest of the soil parameters and geometry are presented in Tables

43 and 44

Case 1

Table 43 Slope Dimensions and Material Properties for Base Failure Case 1

cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43

Figure 46 MATLAB Output for Base Failure Case 1

44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46

43 Comparison with Cases from Literature

431 Case 1

The geometry of a homogeneous slope without foundation is shown in Figure 49 This

case follows the analyses performed by [39] and [40] which were used as benchmark

cases to study the applicability of 3-D FE analyses to slope stability The slope angle is

2625deg or 2H 1V for this case According to Griffiths and Lane the adopted parameters

are based on cʹγH = 005 The height of the slope is assumed to be 40 m thus the

corresponding parameters are summarized in Table 45

Figure 49 Homogeneous Slope without Foundation

Table 45 Slope dimensions and material properties

cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40

The slope stability analyses using the computer program SLOPEW is shown in Figure

410 The FoS based on the GLE and the adopted M-P approach is conducted for both

toe and slope failures The FoS for toe and slope failures are found to be 1385 and

1373 respectively

47

The same analysis was conducted using the MATLAB code the FoS for toe and slope

failures are found to be 1431and 1375 respectively and shown in Figure 410-411

Figure 410 Analysis using SLOPEW - FoS = 1385

48

Figure411 FoS based on M-P approach for Toe failure - FoS = 1431

49

Figure 412 FoS based on M-P approach for Base Failure FoS = 1375

50

432 Case 2

In 1988 a set of 5 basic slope stability problems together with 5 variants was

distributed both in the Australian Geomechanics profession and overseas as part of a

survey sponsored by ACADS [41]

This problem is taken from the verification manual of Slide 30 (Verification 1) It was

distributed to the Australian Geomechanics profession and overseas in 1988 with some

other models to verify the efficiency and accuracy of Slide 30 The slope model

geometry is presented in Figure 413 The slope material properties are shown in Table

46

Figure 413 Slope Model Geometry from Slide 30 [41]

Table 46 Slope Dimensions and Material Properties for ACAD Problem

cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51

Figure 414 FoS based on M-P Approach for ACAD Problem

52

The FoS from this case study when modelled with Slide 30 and the GLE (M-P)

approach adopted was found to be 0986 Upon modelling the same problem with the

MATLAB code as shown in Figure 414 above the FoS is found to be 1035 Again

this can be said to be a reasonably accurate estimate of the FoS by the MATLAB code

The percentage error is 47

53

CHAPTER FIVE CONCLUSIONS

51 Conclusions

The primary focus of this research was to

Develop a MATLAB code for solving the FoS of homogeneous earth slopes

Verify and validate the code with SLOPEW models and also with cases from

literature

Seven cases were studied in this research as summarized below in Table 51

Table 51 Summary of FoS Outputs for all Case Studies

Case 1 Case 2

Percentage

Error ()

SLOPEW MATLAB SLOPEW MATLAB Case 1 Case 2

Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05

Comparison with Literature

Griffiths

and Lane MATLAB

Slide

(Rocscience) MATLAB Case 1 Case 2

Toe Failure 1385 1431 32

Base

Failure 1373 1375 0986 1035 01 47

From the Table 51 above it is seen that the MATLAB program gives a good estimate

of the FoS when compared with SLOPEW and some other problem evaluated with

different programs for a homogeneous soil material Two different failure mechanisms

are shown (ie toe and base failures) It is noted that the maximum percentage error

in the M-P FoS is 5



54

52 Recommendations

The current work could be extended in the future to include the following

The provision for a more user friendly interface and development of the

programme into a stand-alone interactive program

The direct extension of the model to cater for heterogeneous soil and rock

material

The direct extension of the model to include earthquake effect surcharge

loading and increase in pore water pressures

It is hoped that at the end of these improvements this work that has started during this

research study would be found in most geotechnical design offices in Ghana and other

less-developed countries who cannot afford the existing expensive commercial

software available

55

REFERENCES

[1] Wikepedia httpsenwikipediaorgwikiSlope_stability accessed June 2016

[2] Shioi Y and Sutoh S Collapse of high embankment in the 1994 far-off Sanriku

Earthquake Slope Stability Engineering Eds Yagi Yamagami amp Jiang Balkema

Rotterdam pp559-564 1999

[3] Abramson L W Lee T S Sharma S and Boyce G M Slope Stability and

Stabilisation Method 2nd edition John Wiley amp Sons New York 2002

[4] Rouaiguia A Dahim M A Numerical modeling of slope stability analysis

IJESIT Vol 2 No 3 May 2013

[5] Ayetey J K The extent and effects of mass wasting in Ghana Bulletin of the

International Association of Engineering Geology No 43 Paris 1991

[6] CitifmOnline httpcitifmonlinecom20140609death-trap-rocks-hang-

dangerously-along-ayi-mensah-peduase-road accessed May 2016

[7] GraphicOnline httpgraphiccomghnewsgeneral-news22493-danger-

looms-on-aburi-accra-road-as-rocks-fall-from-cliffhtml accessed May 2016

[8] GraphicOnline httpgraphiccomghnewsgeneral-news8384-landslide-can-

occur-around-akoasa-mountainhtml accessed May 2016

[9] DeNatale J S Rapid identification of critical slip surfaces structure Journal

of Geotechnical Engineering ASCE Vol 117 No 10 pp 1568-1589 1991

[10] Drucker D C and Prager W Soil mechanics and plastic analysis or limit

design Quarterly Journal of Applied Mathematics Vol 10 pp 157-165 1952

[11] Zhang X Slope stability analysis based on the rigid finite element method

Geotechnique Vol 49 No 5 pp 585-593 1999

56

[12] Kim J Salgado R and Lee J Stability analysis of complex soil slopes using

limit analysis Journal of Geotechnical and Geoenvironmental Engineering ASCE

Vol 128 No 7 pp 546-557 2002

[13] Chen W F Limit Analysis and Soil Plasticity Elsevier Scientific Publishing

Co New York 1975

[14] Karal K Energy method for soil stability analyses Journal of the Geotechnical

Engineering Division Vol 103 No 5 pp 431-445 1977

[15] Izbichki R J Limit plasticity approach to slope stability problems Journal of

the Geotechnical Engineering Division Vol 107 No 2 pp 228-233 1981

[16] Michalowski R L Slope stability analysis a kinematical approach


[17] Donald I and Chen Z Y Slope stability analysis by the upper bound

approach fundamentals and methods Canadian Geotechnical Journal 34 853-862

1997

[18] Baker G and Garber M Theoretical analysis of the stability of slopes


[19] Jong G D Application of the calculus of variations to the vertical cut off in

cohesive frictionless soil Geotechnique Vol 30 No 1 pp 1-16 1980

[20] Cheng Y M Zhao Z H and Wang J A Realization of Pan Jiazhengrsquos

extremum principle with optimization methods Chinese Journal of Rock Mechanics

and Engineering 27(4) pp 782-788 (in Chinese) 2008

[21] Pan J L Goh A T C Wong K S and Selby A R 2002 Three-

dimensional analysis of single pile response to lateral soil movements International

57

Journal for Numerical and Analytical Methods in Geomechanics 26(8) pp 747-758

2002

[22] Duncan JM State of art limit equilibrium and finite-element analysis of

slopes Journal of Geotechnical Engineering Vol 122 No 7 pp 557-596 1996

[23] Duncan JM and Wright SG Soil Strength and Slope Stability John Wiley

amp Sons Inc 2005

[24] Zhao Y Tong Z Luuml Q Slope Stability analysis using slice-wise factor of

safety Mathematical Problems in Engineering Volume 2014 Paper No 712146 6

pages 2014

[25] Petterson KE The early history of circular sliding surfaces Geotechnique

Vol 5pp 275-296 1955

[26] Krahn J Stability Modeling with SLOPEW An Engineering Methodology

July 2012 Edition Geo-Slope International 2012

[27] Fellenius W Calculation of stability of earth dams Proceedings of the 2nd

Congress Large Dams pp 445-463 Washington D C 1936

[28] Bishop A W The use of the slip circle in the stability analysis of slopes


[29] Janbu N Slope Stability Computations Embankment-Dam Engineering

Casagrande Volume Ed R C Hirschfield and S J Poulos John Wiley and Sons New

York pp 47-86 1973

[30] Janbu N Application of composite slip surface for stability analysis European

Conference on Stability of Earth Slopes Vol 3 pp39-43 Stockholm 1954

58

[31] Spencer E A method of analysis of embankments assuming parallel inter-slice

forces-Geotechnique Vol 17(1) pp 11-26 1967

[32] Morgenstern N R and Price V E The analysis of stability of general slip

surfaces Geotechnique Vol 15 No 1 pp 77-93 1965

[33] Nash D A comparative review of limit equilibrium methods of stability

analysis Slope Stability Ed M G Anderson and K S Richards pp 11 -75 1987

[34] Fredlund DG and Krahn J Comparison of slope stability methods of

analysis Canadian Geotechnical Journal Vol 14 No 3 pp 429-439 1977

[35] Fredlund DG Krahn J and Pufahl DE The relationship between limit

equilibrium slope stability methods Proceedings of the International Conference on

Soil Mechanics and Foundation Engineering Vol 3 pp409-416 1981

[36] Whitman R M and Bailey W A Use of computers for slope stability

analysis Journal of the Soil Mechanics and Foundation Division ASCE Vol 93 No

SM4 pp 475-498 1967

[37] Baba K Bahi L Ouadif L and Akhssas A Slope stability evaluations by

limit equilibrium and finite element methods applied to a railway in the Moroccan Rif

Open Journal of Civil Engineering Vol 2 pp 27-32 2012

[38] Matlab httpsenwikipediaorgwikiMATLAB accessed April 3 2016

[39] Griffiths DV and Lane PA Slope stability analysis by finite elements


[40] Rocscience Inc Application of the Finite Element Method to Slope Stability

Rocscience 2004

59

[41] ACADS 2D elasto-plastic finite element program for slope and excavation

stability analyses - Slope Stability Verification Manual Rocscience 2011

60

APPENDICES

Appendix A Structured Programme

Below is an algorithm developed to solve for the force and moment factor of safetyrsquos

(Ff and Fm) using the M-P method

1 By using the equation of a straight line define the slope surface

119910top = 119898 lowast 119909 + 119888 (1)

2 By using the equation of a circle define the slip surface

119910bot = 119910119888 minus radic(abs(1198772 minus (119909 minus 119909119888)2 (2)

3 By using the equation of a straight line define the piezometric line

119910w = 119898 lowast 119909 + 119888 (3)

4 Divide the slope into n number of slices by vertical lines

5 For each slice the width dx bottom inclination α and average height havg are

determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)

120572 = tanminus1((ℎ119894+1minusℎ119894)

dx) (6)

6 The area of the slice A is computed by multiplying the width of the slice (dx)

by the average height havg

7 The weight W of the slice is computed by multiplying the area of the slice by

the total unit weight of soil

8 W = γA (7)

9 If piezometric height is above slip surface continue with step 9 else skip to

step 17

61

10 The piezometric height is determined at the downslope boundary centre and

upslope boundary of each slice The piezometric height at the downslope and

upslope boundaries of the slice hb and ht respectively are used to compute the

forces from water pressures on the sides of the slice Here a triangular

hydrostatic distribution of pressures is assumed on the sides of the slice The

piezometric height at the centre of the slice hp represents the pressure head for

pore water pressures at the base of the slice

11 Hydrostatic forces from water pressures on the sides of the slice are computed

from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)

12 The pore water pressure is computed by multiplying the piezometric head at the

centre of the base of the slice by the unit weight of water micro = Ɣw hp

13 The length of the bottom of the slice l is determined the length can be

computed from the width dx and base inclination

119897 =119889119909

cosα (10)

14 A trial FoS (F) is assumed and a λ ranging from 0 to 1 set

15 Beginning with the first slice at the toe

∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)

16 The inter-slice shear force X for each slice is calculated from equationhellip

∆119883119894 = ∆119864119894λf(119909) (12)

62

Where λ is the percentage (in decimal form) of the function used and f(x) is

inter-slice force function representing the relative direction of the resultant

inter-slice force given by

119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)

17 The normal force N at the base of each slice is calculated from

119873119894 =119882minus(∆X)minus[


119865]

cos[120572119894]+sin[120572119894]lowasttan[120601]

119865

(14)

18 The new factor of safetyrsquos Ff and Fm are calculated from

Ff= sum119888lowast119897lowastcos[120572119894]+(119873119894minusμ119897)lowasttan[120601]lowastcos[120572119894]

119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)

19 Iterate for a number of slip surfaces and determine the slip surface with the least

FoS

63

Appendix B MATLAB Code for Solving FoS

this code calculates the factor of safety of a homogeneous soil mass

using the method of slices and the M-P interslice force function

Clear all close all clearvars clc

Slope and soil parameters

H = 5 inpu(Enter the height of the slope) height of

slope

Xc =1 inpu(Enter the x coordinate of center of slip circle Xc should be

gt0)

x coordinate of center of circle

Yc = 7 input (Enter the y coordinate of center of slip circle Yc should

be greater than H) y coordinate of center of circle

R = 8 5 Yc+10 radius of center of slip circle

fr = 45 slope angle in degrees

ns = 31 number of slice faces

G = 20 unit weigh of soil

c = 40 cohesion of soil

phi = 20 frictional angle of soil (degrees)

tph = tand(phi) tangent of the frictional angle of soil

tfr = tand(fr) slope angle in gradient

Slip surface generation

xmin = Xc - sqrt(abs(R^2 - Yc^2)) Exit point of slip circle

xmax = Xc + sqrt(abs(R^2 - (Yc-H)^2)) Entry point of slip circle

x = linspace (xminxmaxns) Positions where the forces

will be analyzed

ybot = Yc - sqrt(abs(R^2-(x-Xc)^2)) Equation for generating slip

circle

Slope surface generation

ytop = (xgt=0)(xlt(Htfr))(xtfr) + (xgt=(Htfr))H Slope surface

equation of a line is used

figure(1)

hold on

plot (xybot-rxytop-r)

xlabel (Distance (m))

ylabel (Elevation (m))

plot(reshape([xxx]1[])reshape([ytopybotytopNaN]1[]))

hold off

Other parameters

hs = ytop-ybot Height at each node

havg = (hs(1end-1)+hs(2end))2 5) Average height of each slice

xavg = abs(((x(1end-1)+x(2end))2)-Xc) midpoint distance of each

slice

dx = (xmax-xmin)(ns-1) Width of each slice

yb = (diff(ybot))

alpha = rad2deg(atan2(ybdx)) Slice angle With atan2 the sign

is clear

sal = sind(alpha)

cal = cosd(alpha)

A = dxhavg 6) Area of each slice

W = GA 7) Weight of each slice

s= Rsal offset f for non-circular slip surfaces

l = dx cal 12) Length of the bottom of the slice assuming

straight border

u = 0

factor of safety computation

upO = sum((cl) + (Wcaltph) - (ultph))

downO = sum(Wsal)

FSom = upO downO

OFSm = FSom

NFSm = 12FSom initial guessed F

OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1

f=sind (((xavg (d)-xmin) (xmax-xmin))180)

X=Elbdf

while abs(OFSm-NFSm)gttol

OFSm = NFSm

Nbu = ((W-X) - ((clsal) - (ultphsal)) OFSm)

Nbd = cal + (tphsal) OFSm

Nob = Nbu Nbd

upb = sum((clR) + (NobtphR) - (ultphR))

downb = sum(Wxavg)

FSm = upb downb

FSmm = ((FSm)) (ns-1))

NFSm = FSmm

FSM (i)=FSmm

end

while abs(OFSf-NFSf)gttol

OFSf = NFSf

Nju = ((W-X) - ((clsal) - (ultphsal)) OFSf)

Njd = cal + (tphsal) OFSf

Noj = Nju Njd

upj = sum( (clcal) + (tphcalNoj) - (ultphcal) )

downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end

E = (((cl-ultph)cal) OFSm) + (salNob)-((tphcal)NobOFSm)

E = (((cl-ultph)cal) OFSm) + (salNoj)-((tphcal)NojOFSm)

OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)

finding intersection point

[xi yi] = polyxpoly(y fSM y fSF)

xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)

mapshow(xiyiDisplayTypepointMarkero MarkerEdgeColork)

axis normal

strValues = strtrim(cellstr(num2str([xi yi](03f03f))))

text (xiyistrValuesVerticalAlignmentbottom)

[xi yi] Display intersection points

ii

DECLARATION

I hereby declare that with the exception of references to other peoplersquos work which

have duly been acknowledged this Thesis is the result of my own research work and

no part of it has been presented for another degree in this University and elsewhere

helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip helliphelliphelliphelliphelliphelliphelliphelliphelliphellip

Brendan Dagemanyima Atarigiya Date

(Candidate)

I hereby declare that the preparation of this project was supervised in accordance with

the guidelines of the supervision of Thesis work laid down by the University of Ghana

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip helliphelliphelliphelliphelliphelliphelliphelliphelliphellip

Dr Nii Kwashie Allotey Nana (Prof) A Ayensu Gyeabour I

(Principal Supervisor) (Co-Supervisor)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip

Date Date

iii

DEDICATION


iv

ACKNOWLEDGMENT



















v

ABSTRACT

















stability program






vi

Contents

DECLARATION ii

DEDICATION iii

ACKNOWLEDGMENT iv

ABSTRACT v

LIST OF FIGURES ix

LIST OF TABLES xi


11 Background 1



14 Research Goal 5


16 Scope 5












vii








33 Assumptions 29






381 SLOPEW 34

382 MATLAB 36


41 Introduction 37


421 Toe Failure 37

422 Base Failure 42


431 Case 1 46

432 Case 2 50


51 Conclusions 53


REFERENCES 55

viii

APPENDICES 60



ix

LIST OF FIGURES





















x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




iii

DEDICATION


iv

ACKNOWLEDGMENT



















v

ABSTRACT

















stability program






vi

Contents

DECLARATION ii

DEDICATION iii

ACKNOWLEDGMENT iv

ABSTRACT v

LIST OF FIGURES ix

LIST OF TABLES xi


11 Background 1



14 Research Goal 5


16 Scope 5












vii








33 Assumptions 29






381 SLOPEW 34

382 MATLAB 36


41 Introduction 37


421 Toe Failure 37

422 Base Failure 42


431 Case 1 46

432 Case 2 50


51 Conclusions 53


REFERENCES 55

viii

APPENDICES 60



ix

LIST OF FIGURES





















x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




iv

ACKNOWLEDGMENT



















v

ABSTRACT

















stability program






vi

Contents

DECLARATION ii

DEDICATION iii

ACKNOWLEDGMENT iv

ABSTRACT v

LIST OF FIGURES ix

LIST OF TABLES xi


11 Background 1



14 Research Goal 5


16 Scope 5












vii








33 Assumptions 29






381 SLOPEW 34

382 MATLAB 36


41 Introduction 37


421 Toe Failure 37

422 Base Failure 42


431 Case 1 46

432 Case 2 50


51 Conclusions 53


REFERENCES 55

viii

APPENDICES 60



ix

LIST OF FIGURES





















x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




v

ABSTRACT

















stability program






vi

Contents

DECLARATION ii

DEDICATION iii

ACKNOWLEDGMENT iv

ABSTRACT v

LIST OF FIGURES ix

LIST OF TABLES xi


11 Background 1



14 Research Goal 5


16 Scope 5












vii








33 Assumptions 29






381 SLOPEW 34

382 MATLAB 36


41 Introduction 37


421 Toe Failure 37

422 Base Failure 42


431 Case 1 46

432 Case 2 50


51 Conclusions 53


REFERENCES 55

viii

APPENDICES 60



ix

LIST OF FIGURES





















x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




vi

Contents

DECLARATION ii

DEDICATION iii

ACKNOWLEDGMENT iv

ABSTRACT v

LIST OF FIGURES ix

LIST OF TABLES xi


11 Background 1



14 Research Goal 5


16 Scope 5












vii








33 Assumptions 29






381 SLOPEW 34

382 MATLAB 36


41 Introduction 37


421 Toe Failure 37

422 Base Failure 42


431 Case 1 46

432 Case 2 50


51 Conclusions 53


REFERENCES 55

viii

APPENDICES 60



ix

LIST OF FIGURES





















x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




vii








33 Assumptions 29






381 SLOPEW 34

382 MATLAB 36


41 Introduction 37


421 Toe Failure 37

422 Base Failure 42


431 Case 1 46

432 Case 2 50


51 Conclusions 53


REFERENCES 55

viii

APPENDICES 60



ix

LIST OF FIGURES





















x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




viii

APPENDICES 60



ix

LIST OF FIGURES





















x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




ix

LIST OF FIGURES





















x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




x












xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




xi

LIST OF TABLES










xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




xii



dx width of slice


c cohesion of soil


τf shear strength

τ shear stress




cal cos(α)

sal sin(α)

tph tan()

H Height of slope






xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




xiii











ytop Ground surface

ybot Slope surface






1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




1


11 Background







shear strength [3]



strength [3]













2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




2





point ready [4]













3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




3















4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




4






















5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




5

14 Research Goal



local practice






slopes



16 Scope





stability problems

6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




6










7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




7
























8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




8














question










9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




9




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




10







11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




11










12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




12
















bound analysis


be satisfied






13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




13




to cause yielding


bound solution














of discrete blocks




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




14


















forces



pore pressures



15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




15




















16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




16












17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




17














18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




18


No Methods Moment

Equilibrium

Force

Equilibrium

Inter

Slice

Normal

Forces

Inter

Slice

Shear

Forces

Moment

Factor of

Safety

Force

Factor of

Safety

Inter Slice

Force Function








Constant Half-

Sine

Clipped-Sine

Trapezoidal

Specified

19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




19








relationship as

119891 = 119888 + tan() (21)

and

= 120591119891119865119900119878 (22)

where











20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




20




119882lowast119909 (23)


119873119894lowastsin[120572119894] (24)

Where




slice












21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




21



119882lowastsin(120572) (25)


where












22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




22





119882lowastx (27)

119873119894 =119882minus[


119865119898]


119865119898

(28)


rotation



solution






23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




23

119873119894 =119882minus[


119865119891]


119865119891

(29)


119873119894lowastsin(α) (210)








as


where










24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




24



119873119894lowastsin[120572119894] (212)


119882lowast119909 (213)

where

119873119894 =119882minus∆119883119894minus[


119865]


119865

(214)


and


25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




25








26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




26



moment equilibrium


27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




27


USED








Procedure Use











Spencers method



FoS


method








28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




28












the slope









29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




29

33 Assumptions


body of mass slope














unknowns







30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




30










dy



dy

2)] minus 119883 (

dx

2) minus (119883 +

dX) (dx

2) = 0 (31)


119883 =119889(119864prime119910119905

prime)

dxminus 119910(

dEprime

dx) (32)


31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




31







dS =1




1



119888prime

119865sec2[120572] +

tan[120601prime]

119865[dW

dxminus

dX

dxminus

dEprime

dxtan[120572]] =

dEprime

dxminus

dX

dx+

dW

dxtan[120572] (37)



32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




32

119888prime

119865[1 + (

dy

dx)2

] +tan[120601prime]

119865[dW

dxminus

dX

dx+

dEprime

dx(dy

dx)] =

dEprime

dxminus

dX

dxminus

dW

dx(dy

dx) (38)


dEprime

dx[1 minus

tan[120601prime]

119891(dy

dx)] +

dX

dx[tan[120601prime]

119865+

dy

dx] =

119888prime

119865[1 + (

dy

dx)2] +

dW

dx[tan[120601prime]

119865+

dy

dx](39)






li










33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




33





No

Yes

Input H

Xc Yc R

Generate ytop

and ybot eq

1amp2

Input c and phi


10 7amp 6


Set lambda and

initial FoS for the

first iteration

Is FSm-FoSlttol


NFSm=FSm NFSf=FSf

Increase


∆E-eq 17 15 16 amp11

Output

FSm(i)

FSf(i)

Output

FSm FSf

34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




34


381 SLOPEW















35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




35



36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




36


382 MATLAB








this research work

37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




37


41 Introduction







ACADS [41]








421 Toe Failure




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




38

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 15 38 18 5


39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




39


40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




40

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 5 28 16 5


41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




41


42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




42




acceptable degree

422 Base Failure



43 and 44

Case 1


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 1 40 20 20 5


43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




43


44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




44

Case 2


cacute rsquo γ H

(kPa) () (kNm3) (m)

Case 2 0 15 18 5


45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




45


46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




46


431 Case 1









cacute rsquo γ H

(kPa) () (kNm3) (m)

40 20 20 40




1373 respectively

47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




47




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




48


49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




49


50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




50

432 Case 2








46



cacute γ H

(kPa) () (kNm3) (m)

3 196 202 10

51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




51


52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




52






53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




53


51 Conclusions




literature



Case 1 Case 2

Percentage

Error ()


Toe Failure 5649 5406 3659 3514 4 4

Base

Failure 3152 3168 0611 0614 05 05


Griffiths

and Lane MATLAB

Slide



Base

Failure 1373 1375 0986 1035 01 47





in the M-P FoS is 5



54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




54

52 Recommendations





material






software available

55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




55

REFERENCES























56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




56



Vol 128 No 7 pp 546-557 2002


Co New York 1975









1997










57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




57


2002




amp Sons Inc 2005



pages 2014


Vol 5pp 275-296 1955









York pp 47-86 1973



58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




58














SM4 pp 475-498 1967








Rocscience 2004

59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




59



60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




60

APPENDICES





119910top = 119898 lowast 119909 + 119888 (1)




119910w = 119898 lowast 119909 + 119888 (3)



determined

ℎavg =ℎ119894+ℎ119894+1

2 (4)

dx =119883

119899119904minus1 (5)


dx) (6)





8 W = γA (7)


step 17

61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




61









from

micro119887 =ℽ119908lowastℎ119887

2

2 (8)

micro119905 =ℽ119908lowastℎ119905

2

2 (9)





119897 =119889119909

cosα (10)



∆119864119894 =119882[sin[120572119894]minus

tan[120601]cos[120572119894]

119865]minus

cl

119865

cos[120572119894]+tan[120601]sin[120572119894]

119865

(11)


∆119883119894 = ∆119864119894λf(119909) (12)

62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




62




119891(119909) = sin[(119883119894+1minus119883119894

119883119899minus119883119894) lowast 120587] (13)


119873119894 =119882minus(∆X)minus[


119865]


119865

(14)



119873119894lowastsin[120572119894] (15)


119882lowast119909 (16)


FoS

63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




63







slope


gt0)
















will be analyzed


circle




figure(1)

hold on





hold off

Other parameters




slice


yb = (diff(ybot))


is clear

sal = sind(alpha)

cal = cosd(alpha)





straight border

u = 0



downO = sum(Wsal)

FSom = upO downO

OFSm = FSom


OFSf = FSom

NFSf = 12FSom

Tol =0001

t =-1020

y=t

64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




64

i=1

k=1

n=1 ns-1

for lbd=t

E (1) =0

d=1 ns-1


X=Elbdf


OFSm = NFSm



Nob = Nbu Nbd


downb = sum(Wxavg)

FSm = upb downb


NFSm = FSmm

FSM (i)=FSmm

end


OFSf = NFSf



Noj = Nju Njd


downJ = sum(Nojsal)

FSf = (upj downJ)

FSf = (FSf) (ns-1))

NFSf = FSf

FSF (i)= FSf

end



OFSm = FSom

OFSf = FSom

i=i+1

end

fSM=FSM

fSF=FSF

figure(2)

plot(yfSM-ryfSF-b)



xlabel(lambda)

ylabel(FoS)

legend(FSm FSf)


axis normal




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Numerical Modeling and Simulation of the Stability of ... · Figure 2.6: Forces considered in the...

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