aspects of design and analysis of reinforced soil dams(thesis)

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1994

Aspects of design and analysis of reinforced soildamsMohammad Reza MagharehUniversity of Wollongong

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Recommended CitationMaghareh, Mohammad Reza, Aspects of design and analysis of reinforced soil dams, Doctor of Philosophy thesis, Department of Civiland Mining Engineering, University of Wollongong, 1994. http://ro.uow.edu.au/theses/1227

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ASPECTS OF DESIGN AND ANALYSIS OF

REINFORCED SOIL DAMS

A thesis submitted in fulfilment of the requirements for the award of the degree

DOCTOR OF PHILOSOPHY

from

THE UNIVERSITY OF WOLLONGONG

by

MOHAMMAD REZA MAGHAREH, BSc, MSc.

UNIVERSITY OF

DEPARTMENT OF CIVIL AND MINING ENGINEERING

July 1994

"To my late father who really was a Father and died when 1 was investigating this thesis

at the University of Wollongong. I was notable to participate in his occasional passing

ceremonies, or on the first anniversary of his passing."

"Godbless him."

STATEMENT

I here certify that the work presented in this thesis has not been submitted for a degree

to any other university or similar institution.

MOHAMMAD REZA MAGHAREH

ABSTRACT

This thesis was concerned with the design, analysis and geometrical optimisation of

reinforced soil dams (RSDs, singular RSD) and aimed to develop a computer program

for these tasks. In order to achieve this objective, the following tasks were carried out

as parts of this thesis.

(a) Comprehensive literature survey. This part included an overview of the history

of reinforced soil, its application, material components, design considerations,

construction methods, and economic considerations of reinforced soil. The

fundamentals of design and analysis of conventional earth dams were also considered.

This included the history, classification, factors governing the choice of dam type,

stability analysis, design criteria, and limitations of conventional earth dams. This

project also gave a detailed evaluation and design criteria of a number of existing

RSDs. A summary of recent investigations on the behaviour of RSDs was outlined.

This continued by considering the classification of RSDs and the forces acting on them.

(b) Stability analysis of RSDs. This part was focused on some formulae in order to

optimise the cross sectional area of RSDs. The external stability analysis of RSD was

evaluated as a whole structure based on analytical approach. Some design formulae

were given for RSD optimisation concerning the minimum base length required against

sliding, overturning, overstressing, bond failure, and rupture failure. The internal

stability of RSD was also taken into account based on a semi-empirical method. Some

empirical relationships were proposed to eliminate the tangent discontinuity which exists

in the Coherent Gravity Method formulae. These relationships reflect the non-linearity

indicated by the field data and eliminate unknown parameters existing in the formulae of

Modified Coherent Gravity Method. They also offer a better fit with the available field

observations. Relationships between the lateral earth pressure and the apparent friction

factor with the fill depth were proposed. The apparent friction factor versus the

reinforcement length were also undertaken.

i

(c) Analysis of the behaviour under seismic loads. Although many researchers have

investigated the effects of earthquakes on soil dams, many problems are still unsolved,

specially for RSDs. A comparison between the natural frequency of conventional earth

dams and RSDs were considered in this project. The practice of inserting

reinforcement into the earth dam material allows reduction in fill volume and reduction

in displacement. However, this also leads to an increase in the natural frequency of

such structures compared with conventional earth dams. This may increase the

possibility of failure. The natural frequency of RSD is increased because of its

geometry and its overall stiffness. In this project, the increases in natural frequency of

RSDs due to these two major factors were separately discussed. Formulae concerning

the magnification of the natural frequency of the structure due to reinforcement

insertion were derived, and in some cases tabulated and plotted.

(d) Development of a computer program. A computer program was developed for

geometrical optimisation and stress-strain analysis of RSD. The outcomes of the

program are (a) geometrical optimisation of RSD based on analytical and semi-

empirical formulae, and (b) stress-strain analysis of the optimised RSD based on the

finite element method.

(e) Analysis of models of RSDs. Six models of RSDs were analysed for various

heights and safety factors to find the optimum geometry. A 30m high RSD was also

analysed considering the following configurations: (a) without reinforcements, (b) with

the assumed increased stiffness of the soil fill, (c) with horizontal reinforcements, and

(d) with inclined reinforcements to evaluate the variation of stresses versus the

direction of reinforcements. It was concluded that putting reinforcements in soil dams

decreases displacements and stresses values.

ii

ACKNOWLEDGMENT M y special thanks go to the Iranian Ministry of Culture and Higher Education, because

this thesis was made possible by its scholarship.

I am also grateful to the Department of Civil Engineering, University of Wollongong

N S W , Australia for encouragement and facilities for research. I wish to record my

utmost gratitude to Dr. R. M . Arenicz for his supervision of this thesis. Also, the

comments and suggestions of Professor R. N. Chowdhury are acknowledged and

appreciated.

M y gratitude goes to m y wife, Mrs. Z. G. Haghighi for her assistance during the past

three years. I also thank Reinforced Earth Company Pty Ltd in Gosford for allowing

access to its library and sending some brochures during the initial stage of this project.

iii

TABLE OF CONTENTS

ABSTRACT i

ACKNOWLEDGMENT iii

TABLE OF CONTENTS iv

LIST OF FIGURES ix

LIST OF TABLES xv

CHAPTER ONE: INTRODUCTION, AIMS AND SCOPE 1

1.1 INTRODUCTION 1

1.2 AIMS AND SCOPE OF THE THESIS 2

1.3 THESIS OVERVIEW 3

CHAPTER TWO: PRINCIPLES OF REINFORCED EARTH 6

2.1 INTRODUCTION 6

2.2 HISTORY 7

2.3 APPLICATIONS 8

2.4 MATERIALS 14

2.4.1 Soil '. 15

2.4.2 Facing 17

2.4.3 Reinforcement 18

2.5 SOIL-REINFORCEMENT INTERACTIONS 22

2.5.1 Development of theory 22

2.5.2 Stability considerations 36

2.6 CONSTRUCTION METHODS AND STRUCTURAL SYSTEMS 39

2.7 DURABILITY 42

2.8 COSTS AND ECONOMICS 46

iv

2.9 CONCLUSIONS 49

CHAPTER THREE: EVALUATION OF SOIL D A M S 51

3.1 INTRODUCTION 51

3.2 CONVENTIONAL EARTH D A M S 52

3.2.1 History 52

3.2.2 Classification 53

3.2.3 Associated facilities 56

3.2.4 Factors governing selection of a type 56

3.2.5 Materials 57

3.2.6 Design procedure 57

3.2.7 Sections of earth dams 58

3.2.8 Limitations of conventional earth dams 62

3.3 R E I N F O R C E D SOIL D A M S 64

3.3.1 History of reinforced soil dams 64

3.3.2 Other investigations 69

3.3.3 Classification of reinforced soil dams 94

3.4 F O R C E S A C T I N G O N SOIL D A M S 98

3.4.1 External water pressure 98

3.4.2 Internal water pressure and seepage gradients 100

3.4.3 Uplift pressure 106

3.4.4 Ice pressure 107

3.4.5 Silt pressure 108

3.4.6 Weight of structure 109

3.4.7 Earthquake force 110

3.4.8 Reaction of foundation 112

3.4.9 Load combinations 114

3.5 C O N C L U S I O N S 116

v

CHAPTER FOUR: STABILITY ANALYSIS OF REINFORCED SOIL D A M S 118

4.1 INTRODUCTION 118

4.2 EXTERNAL STABILITY 119

4.2.1 Sliding 121

4.2.2 Overturning 125

4.2.3 Overstressing 130

4.3 INTERNAL STABILITY 137

4.3.1 Coefficient of lateral earth pressure 138

4.3.2 Apparent friction factor 142

4.3.3 Extension of failure zone 149

4.3.4 Reinforcement effect 150

4.3.5 Design equations 153

4.3.6 Internal erosion and piping failure 155

4.3.7 Hydraulic fracture failure 159

4.3.8 Distortional settlement 161

4.4 CONCLUSIONS 163

CHAPTER FIVE: BEHAVIOUR OF SOIL DAMS UNDER SEISMIC LOAD 165


5.2 FREE H A R M O N I C VIBRATION 167

5.3 FORCED H A R M O N I C VIBRATIONS 169

5.4 DAMPING 171

5.5 N A T U R A L FREQUENCY 172

5.5.1 Stiffness function 174

5.5.2 Shape function 176

5.6 EXAMPLE 180

5.7 CONCLUSIONS 184

vi

CHAPTER SIX: COMPUTER PROGRAM 185


6.2 FINITE ELEMENT FORMULAE 185

6.2.1 Elastic behaviour of soil 186

6.2.2 Inelastic behaviour of soil 188

6.2.3 Soil-reinforcement interaction 190

6.3 RSDAM COMPUTER PROGRAM 195

6.3.1 Purpose 196

6.3.2 Input data 196

6.3.3 Program operation 197

6.3.4 Output data 201

6.4 CONCLUSION 202

CHAPTER SEVEN: ANALYSIS 203


7.2 GEOMETRICAL OPTIMISATION 204

7.3 NUMERICAL ANALYSIS 210

7.3.1 Loading steps 210

7.3.2 Mesh information 211

7.3.3 Material property 213

7.3.4 Stages of analysis 214

7.3.5 Displacement variation 214

7.3.6 Stress variation 215

7.3.7 Variation of the vertical facing displacement 216

7.4 CONCLUSIONS 221

CHAPTER EIGHT: CONCLUSIONS AND RECOMMENDATIONS 224


8.2 PART A- THEORETICAL OPTIMISATION AND ANALYSIS 224

vii

8.2.1 Geometrical optimisation 224

8.2.2 Semi-empirical relationships 225

8.2.2 Natural frequency 227

8.3 PART B-NUMERICAL ANALYSIS 228

8.3.1 Computer program 229

8.3.1 Results of analysis 230

8.4 RECOMMENDATIONS 232

8.4.1. Reinforced soil arch dams 232

8.4.2. Cross sectional optimisation 232

8.4.3. Behaviour of reinforcement 233

8.4.4 Reinforcement width 233

8.4.5 Natural frequency 233

8.4.6 Seismic load based on dynamic analysis 234

8.4.7 Stress concentration 234

REFERENCES: Rl

APPENDICES Al

APPENDIX A- EARTH DAM FAILURES A2

APPENDIX B- TYPICAL TYPES OF DAM'S SOIL A5

APPENDIX C-ICE PRESSURE TABLES A8

APPENDIX D- BOND AND BREAK FAILURES EQUATIONS A9

APPENDIX E- RSDAM PROGRAM FLOWCHART Al 1

APPENDIX F- RUNNING THE RSDAM PROGRAM ,. A17

INTRODUCTION A17

INPUT DATA A17

OUTPUT DATA A24

EXAMPLE A30

APPENDIX G- RSDAM PROGRAM LISTING A77

viii

LIST OF FIGURES

Fig. 2.1.1 Reinforced earth components 7

Fig. 2.3.1 Reinforced soil arch 9

Fig. 2.3.2 The slot storage system 10

Fig. 2.3.3 Stepped highway structures 10

Fig. 2.3.4 Rock crushing plant 11

Fig. 2.3.5 Typical section of sea wall 11

Fig. 2.3.6 The sea wall using Z-shaped panels 11

Fig. 2.3.7 Modes of embankment reinforcing 12

Fig. 2.3.8 Critical embankment orientations 13

Fig. 2.3.9 Typical cross-section of a RSD compared with a conventional earth

dam 14

Fig. 2.3.10 Cross-section of Vallon des Bimes dam 14

Fig. 2.4.2.1 Typical examples of shapes of facing panels 18

Fig. 2.4.3.1 Typical shapes of reinforcements 19

Fig. 2.4.3.2 A reinforcement system connected to a facing panel 20

Fig. 2.4.3.3 Facing panels and reinforcement systems of various techniques 23

Fig. 2.5.1.1 Improvement in strength due to reinforcement 24

Fig. 2.5.1.2 Increase in brittleness due to reinforcement 25

Fig. 2.5.1.3 Increase in G\ due to reinforcement 25

Fig. 2.5.1.4 Coulomb analysis 26

Fig. 2.5.1.5 Comparison of theoretical and experimental results 28

Fig. 2.5.1.6 Failure condition for constant o'r 29

Fig. 2.5.1.7 Friction angle <j>r as a function of <t>and F 30

Fig. 2.5.1.8 Increase in the friction angle of soil because of reinforcement 31

Fig. 2.5.1.9 Composite Mohr envelope 31

Fig. 2.5.1.10 The LCPC interpretation 32

Fig. 2.5.1.11 The Aa'3 interpretation 33

ix

Fig. 2.5.1.12 Variation of strength with aspect ratio 34

Fig. 2.5.1.13 Variation of Aa'3 with a'3 34

Fig. 2.5.2.1 The forms of failures 37

Fig. 2.5.2.2 a) Trapezoidal distribution pressure and b) Meyerhof distribution

pressure 38

Fig. 2.5.2.3 Comparison between maximum tension stress inside reinforcement 39

Fig. 2.6.1 The cross section of Concertina Method 41

Fig. 2.6.2 A typical section of facing of a Telescope Method 42

Fig. 2.6.3 A typical section of Sliding Method 42

Fig. 2.7.1 Development of corrosion 44

Fig. 2.7.2 Loss of thickness during time for galvanised steel and unprotected

steel 45

Fig. 2.8.1 Comparison between the height of the reinforced soil structures and

the percentage of the costs of reinforced soil relative to the costs of reinforced

concrete cantilever walls 47

Fig. 2.8.2 Economy versus height of structure 47

Fig. 2.8.3 Variation of percentage of total material cost with height of structure 48

Fig. 2.8.4 Comparison between the costs of reinforced soil structures and

reinforced concrete structures 49

Fig. 3.2.2.1 General classification of dams 55

Fig. 3.2.4.1 A general view of a composite dam 57

Fig. 3.2.7.1 Cross-section of a thin core earth dam 58

Fig. 3.2.7.2 Typical sections of impervious foundation of earth dams 60

Fig. 3.2.7.3 Typical sections for shallow pervious foundation of earth dams 61

Fig. 3.2.7.4 Typical sections for deep pervious foundation 63

Fig. 3.3.1.1 Vallon des Bimes dam 65

Fig. 3.3.1.2 L'Estella D a m 65

Fig. 3.3.1.3 A general view of Taylor Draw D a m 66

x

Fig. 3.3.1.4 Front face elevation and cross-section of Taylor Draw D a m 66

Fig. 3.3.1.5 a) The cross-section of Bridle Drift dam b)Downstream elevation

after the flood 67

Fig. 3.3.1.6 a) The cross-section of the Xonxa Dam, b) Reinforcing system

designed 68

Fig. 3.3.1.7 N e w section of earth dam at Lake Sherburne 68

Fig. 3.3.1.8 N e w section of Jamesville, N e w York dam 69

Fig. 3.3.2.1 Standard sections of reinforced embankments 70

Fig. 3.3.2.2 Relationship between embankment deformation, the strain

distribution of grid and saturation degree 71

Fig. 3.3.2.3 Embankment section used in stability analysis 72

Fig. 3.3.2.4 Embankment with geocell 75

Fig. 3.3.2.5 The observed and the predicted by F E M values of stress on steel

bars... 82

Fig. 3.3.2.6 The bearing forces applied to the plates 83

Fig. 3.3.2.7 Embankment 84

Fig. 3.3.2.8 Settlement along a horizontal section in the subsoil at the ground

level 84

Fig. 3.3.2.9 Settlement profile along a vertical section 85

Fig. 3.3.2.10. Vertical and principal stress distribution 85

Fig. 3.3.2.11 Stress distribution of reinforcement 86

Fig. 3.3.2.12 The geometry and finite element mesh of the embankment 87

Fig. 3.3.2.13 Stress and strain profiles 88

Fig. 3.3.2.14 Ground surface settlement 89

Fig. 3.3.2.15 Surface horizontal displacement 89

Fig. 3.3.2.16 Reinforcement strains and forces 90

Fig. 3.3.2.17 Stranstead Abbotts Embankment 91

Fig. 3.3.2.18 Displacement distribution along ground surface 93

xi

Fig. 3.3.2.19 Pore water pressure at point B varying with time 93

Fig. 3.3.2.20 Tension distribution in the grid 94

Fig. 3.3.3.1 Cross-sections of a homogeneous fill RSD and a zoned RSD 95

Fig. 3.3.3.2 A typical cross-section of an impervious upstream shell dam 95

Fig. 3.3.3.3 A central core RSD compared to an inclined core RSD 96

Fig. 3.3.3.4 A general classification of RSDs based on material used and cross-

section shape 96

Fig. 3.3.3.5 A classification of RSDs based on their foundations 97

Fig. 3.3.3.7 A possible classification of RSDs 97

Fig. 3.3.3.6 Cross-section of an imaginary reinforced soil arch dam 98

Fig. 3.4.1.1 External water pressure acting on an earth dam 99

Fig. 3.4.1.2 External water pressure acting on a vertical downstream face RSD 99

Fig. 3.4.2.1 Seepage lines through (a) a homogeneous earth dam without

blanket (b) a homogeneous earth dam with a drainage blanket (c) a non-

homogeneous earth dam 101

Fig. 3.4.2.2 Seepage lines through: (a) a RSD without blanket (b) a RSD with a

drainage blanket (c) a zoned RSD 101

Fig. 3.4.2.3 The seepage line through a RSD compared with the seepage line

through a conventional earth dam with the same height 103

Fig. 3.4.2.4 Seepage line through the foundation of a conventional earth dam 104

Fig. 3.4.2.5 Seepage line through the foundation of a RSD 104

Fig. 4.3.5.1 A comparison between the path of water under a conventional

earth dam and a RSD with the same height 105

Fig. 3.4.3.1 Uplift pressure acting on an impervious rigid foundation dam 106

Fig. 3.4.3.2 Uplift water pressure acting on a pervious foundation dam 107

Fig. 3.4.4.1 Location of ice pressure acting on a dam 108

Fig. 3.4.5.1 Silt pressure 109

Fig. 3.4.6.1 Zoned RSD 110

xii

Fig. 3.4.7.1 The horizontal earthquake force due to water slashing 112

Fig. 3.4.7.2 The value of Cp 113

Fig. 3.4.8.1 Trapezoidal reaction of foundation 113

Fig. 3.4.8.2 Possible non-linear reaction of foundation 114

Fig. 4.1.1 Stability analysis of RSDs 118

Fig. 4.2.1 The cross section of a parametric RSD with imaginary horizontal

layers 119

Fig. 4.2.2 Forces acting on a RSD 120

Fig. 4.2.3.1 Reactions of foundation 130

Fig. 4.3.1.1 Coefficient of lateral earth pressure 139

Fig. 4.3.1.2 Comparison between the formula (for <j)=45) and the results of

observed experiments 139

Fig. 4.3.1.3 Comparison between the field data and experimental formulae 141

Fig. 4.3.2.1 Apparent friction factor 143

Fig. 4.3.2.2 Comparison between theoretical and typical values of apparent

friction factor for smooth strips 145

Fig. 4.3.2.3 Comparison between theoretical and typical values of apparent

friction factor for ribbed strips 145

Fig. 4.3.2.4 The results of pull-out tests 147

Fig. 4.3.2.5 Linear relationship between f* and the ratio H/L 148

Fig. 4.3.3.1 Effective length of reinforcing strip 150

Fig. 4.3.6.1 Piping through a homogeneous fill RSD without drainage blanket 156

Fig. 4.3.6.2 Piping through a homogeneous fill RSD with a horizontal drainage

blanket 156

Fig. 4.3.6.3 Piping through a zoned RSD 157

Fig. 4.3.6.4 Piping under a RSD 158

Fig. 4.3.6.5 Use of heavy stones in downstream side for preventing piping 158

Fig. 4.3.7.1. Idealised flow gradients in the upstream part of RSD 160

xiii

Fig. 4.3.7.2 Seepage line through a homogeneous fill RSD without drainage

blanket 160

Fig. 4.3.8.1 The distortion settlement of RSD 162

Fig. 5.2.1 a) A typical RSD divided into several imaginary layers b)the first

and the second blocks of the RSD 167

Fig. 5.3.1 The relation between N and n 171

Fig. 5.5.1 a) A typical conventional earth dam and b) a typical RSD with

vertical downstream facing 173

Fig. 5.5.1.1 Comparison between reinforced and unreinforced soil elements 175

Fig. 5.5.1.2 Variation of *F versus p for yvlys=2>.9, M=2.63 176

Fig. 5.5.2.1 Minimum slope ranges of a conventional earth dam compared to

an equivalent RSD 177

Fig. 5.5.2.2 Maximum slope ranges of a conventional earth dam compared to

an equivalent RSD 178

Fig. 5.5.2.3 Variation of (D versus Wt/H 179

Fig. 5.6.1 The illustrative example of a reinforced and a conventional earth

dam 180

Fig. 5.6.2 The illustrative example of the conventional earth dam 181

Fig. 5.6.3 Pseudo acceleration verses period, T, for various values of damping

coefficients based on four major earthquakes happened in U S A 182

Fig. 6.2.1.1 A n elastic stress-strain curve 186

Fig. 6.2.2.1 Possible elasto-plastic stress-strain curves for an element of soil

under unload-reload condition 189

Fig. 6.2.3.1 Reinforcement elements within a. RSD 190

Fig. 6.2.3.2 A typical reinforcement carrying the horizontal forces induced in

the nodal points of reinforcement 192

Fig. 6.3.3.1 Abbreviated flowchart 198

Fig. 6.3.3.2 A general view of a typical RSD showing subdivisions 200

xiv

Fig. 7.2.1 Minimum required base length versus height for a 20m high dam 206



Fig. 7.3.1 The 30m high vertical downstream earth dam 210

Fig. 7.3.1.1 Variations of seepage lines 211

Fig. 7.3.2.1 A general view of the RSD showing nodal points 212

Fig. 7.3.2.2 Positions of horizontal reinforcements 212

Fig. 7.3.2.3 Positions of inclined reinforcements 212

Fig. 7.3.5.1 The dam before loading 215

Fig. 7.3.5.2 Displacement result of the dam 217

Fig. 7.3.6.1 Variations of principal stresses acting on the elements 218

Fig. 7.3.6.2 Variations of horizontal stresses acting on the elements 219

Fig. 7.3.7.1 Variations of vertical and horizontal movements of the vertical

facing based on -0.08m base displacement 220

Fig. 7.3.7.2 Variations of vertical and horizontal movements of the vertical

facings based on 0.15m base displacement 220

Fig. IF The explanation of elements A20

Fig. 2F The consequence of the nodal points A21

Fig. 3F The cross section of a parametric RSD with imaginary horizontal layers A26

Fig. 4F The consequence of the nodal points A32

xv

LIST OF TABLES

Table 2.4.1.1 Grading restriction for non-cohesion material 16

Table 2.4.1.2 Grading restrictions for cohesive frictional material 17

Table 2.4.3.1 Degradation resistance of various synthetic fibres 19

Table 2.4.3.2 Properties of typical sheet and strip material 20

Table 2.4.3.3 Frictional properties for various strip material 20

Table 2.4.3.4 A comparison between the properties of general polymers 21

Table 2.5.2.1 Equations for calculation of the maximum tension in the

reinforcements of a vertical reinforced soil structure 38

Table 2.7.1 Corrosion allowance for metallic components exposed to various

environment 45

Table 3.2.2.1 Classification of dams based on storage and height 53

Table 3.3.2.1 Test cases 70

Table 3.3.2.2 The property of fill material... 71

Table 3.3.2.3 Analysis for the integration effect at collapse without

reinforcement 74

Table 3.3.2.4 Properties of reinforcing elements 76

Table 3.3.2.5 The summarised condition of the parameters used 77

Table 3.3.2.6 The property of foundation and embankment soil 79

Table 3.3.2.7 The constants used for finite element analysis..... 79

Table 3.3.2.8 The methods used for analysing maximum tensile force and their

formulae 80

Table 3.3.2.10 The properties of soil used in embankment 87

Table 3.3.2.11 The properties of soil used in foundation 87

Table 3.3.2.11 The results of analysis 88

Table 3.3.2.12 Summary of predictions and observations 90

Table 3.3.2.13 Physical and mechanical parameters of soils 91

Table 3.3.2.14 Computation parameters of foundation soils 92

xvi

Table 3.4.7.1 Earthquake acceleration HI

Table 3.4.9.1 Cases of load combinations 115

Table 4.2.1 Summary of the forces in sliding and overturning states 121

Table 4.2.1.1 Results of driving and resistance forces acting on RSD in sliding

situation 122

Table 4.2.2.1 Results of driving and resistance moments acting on the dam in

overturning situation 126

Table 4.2.3.1 Summary of the forces used in analysis of soil bearing capacity 131

Table 4.3.5.1 Factors of safety formulae against both break and bond failures

based on Proposed Method 154

Table 7.2.1 Final widths of the dam computed by the program 204

Table 7.2.2 Assumptions accepted during the analysis of the models 205

Table 7.3.3.1 Assumed soil properties 213

Table 7.3.3.2 Assumed concrete facing properties 213

Table 7.3.3.3 Assumed interface element properties 213

Table 7.3.3.4 Assumed reinforcement properties 214

Table L A Earth dam failures due to hydraulic problems A 2

Table 2.A Earth dam failures due to structural failures A3

Table 3.A Earth dam failures due to seepage failures A 4

Table L B Typical types of soil in or under dams A5

Table 2.B Typical types of soil in or under dams A 6

Table 3.B Soil performance in or under dams A 7

Table l.C Ice pressure A 8

Table l.D Factors of safety formulae against both break and bond failures

based on C G M A 9

Table 2.D Factors of safety formulae against both break and bond failures

based on M C G M A10

xvii

INTRODUCTION, AIMS AND SCOPE CHAPTER ONE

CHAPTER ONE

INTRODUCTION, AIMS AND SCOPE

1.1 INTRODUCTION

A significant part of the cost of any dam is associated with its design and construction.

This indicates that there is a need for a careful assessment of the cost involved. A right

type of structure with a suitable shape would reduce the cost considerably. The use of

an earth dam instead of some other types would usually reduce the cost. For example,

an earth dam can be constructed at less than half the cost of a concrete dam with equal

capacity and height. However, the use of an earth dam is restricted by its geometrical

area, weir restriction, height limitation and the availability of a sufficient amount of

earth material. These restrictions can be alleviated by the use of a reinforced soil dam

(RSD, plural RSDs) with an additional reduction in material cost. For example, at least

two RSDs may be constructed with the material needed for one earth dam.

Soil reinforcement is a reliable and suitable method for augmenting strength and

solidity of soil. Reinforced soil can be substituted for concrete and soil in the

construction. In the current form of reinforced soil, which was introduced by H. Vidal

in the 1960s, the soil is reinforced by strips located in particular directions regular in a

pattern. The concept of reinforced soil is based on making a composite structure by

frictional action between the soil and the reinforcements.

Although many researchers have been investigating the behaviour of reinforced soil,

there are still many unsolved problems in the analysis and design of RSDs. Shape,

surface properties, dimensions, strength and stiffness of the reinforcement are the main

parameters that affect the performance and behaviour of a RSD. The location,

orientation and spacing of reinforcement affects the soil reinforcement interaction.

1


Grading, particle size, mineral content, index properties, degree of saturation, density,

overburden pressure and state of stress are other parameters that change the behaviour

of the soil used for RSDs. The stability of RSDs under some loads such as dead load,

uplift pressure, hydrostatic pressure and, particularly, earthquake have not yet been

fully investigated. The seepage effects and the piping phenomenon in RSDs should

also be investigated. These problems clearly show the need for further research in this

area.

RSDs, based on their shape, can be classified into four groups; vertical downstream

face, vertical upstream face, inclined downstream face, and inclined upstream face.

The design and analysis of RSDs are affected by the type of dam foundation, material

homogeneity, type and shape of reinforcements, and shape and position of the core in

the zoned type. In RSDs, with a vertical downstream side, the material costs can be

reduced by eliminating the downstream material, and allowing for the construction of a

spillway on the top.

1.2 AIMS AND SCOPE OF THE THESIS

This thesis is concerned with the investigation of design and analysis of RSDs. The first

goal of the study is to contribute a better understanding for this analysis and design.

Some analytical formulae and some semi-empirical formulae are derived in order to

achieve this objective. Detailed tasks of this project may be undertaken in as follows:

(1) Literature review concerning the concept of reinforced soil and its

behaviour in RSDs.

(2) Literature review concerning the evaluation of conventional earth

dams and RSDs.

(3) Literature review concerning the soil reinforcement interaction.

(4) Literature review concerning the structural stability analysis of

RSDs.

2


(5) Study of the forces acting on soil dams and the behaviour of RSDs

under the forces.

(6) Classification of RSDs.

(7) Comparison of the behaviour of conventional earth dams and RSDs

under seismic loads.

(8) Consideration of the minimum required base length of RSD (against

sliding, overturning, overstressing, bond failure and break failure)

required for its geometrical optimisation.

(9) Analysis of semi-empirical relationships needed for internal stability

of reinforced soil structures.

(10) Development of a computer program (called RSDAM) for

geometrical optimisation and stress-strain analysis of RSDs.

(11) Analysis of models of RSDs using the computer program.

A major part of the project was concerned with the development of the computer

program using: (a) the analytical approach for geometrical optimisation of RSDs, and (b)

the finite element method to model the behaviour of the dam. Two-dimensional

quadrilateral elements and a general stress-strain curve are assumed in the program to

simulate the behaviour of the soil. A non-linear hyperbolic stress-strain curve is used to

represent the primary loading, while a linear response is assumed for the unloading or

reloading behaviour of the soil. The interface elements are used in the program to

permit relative movement between the soil and the concrete facing panels.

1.3 THESIS OVERVIEW

Chapter 2 gives a comprehensive literature survey on the mechanics of reinforced soil.

A n overview of the history of reinforced earth, its application, material components,

fundamental behaviour, design considerations, construction methods, and construction

cost of reinforced soil are included in this chapter.

3


Chapter 3 presents fundamentals of design and analysis of conventional earth dams. It

includes the history, classification, factors governing the choice of dam type, stability

analysis, design criteria, and limitations of conventional earth dams. This chapter also

considers detailed evaluation and design criteria of a number of existing RSDs. An

historical perspective, stability analysis, and a summary of recent investigations into the

behaviour of RSDs are investigated. This chapter continues by considering the

classification of RSDs and the forces acting on them. In reality, there are no major

differences between the forces acting on a RSD and the forces acting on other types of

dams. However, the behaviour of RSD and other dams is different in withstanding the

forces. The forces resulting from the weight of structure, the pressures of water, silt,

ice, seepage and earthquake are considered here.

Chapter 4 presents a stability analysis of the RSD to optimise the cross sectional area.

This includes the external stability analysis of the dam as a whole structure based on

analytical approach. Some proposed formulae are given for earth dam optimisation

concerning the minimum base length of the dams required against sliding, overturning,

overstressing, bond failure, rupture failure, hydraulic fracture failure. The semi-

empirical relationships of Coherent Gravity Method (CGM) and Modified Coherent

Gravity Method (MCGM) are taken into account. The relationships between the lateral

earth pressure and the apparent friction factor with fill depth are proposed to eliminate

the tangent discontinuity which exists in the CGM formulae and the unknown

parameters which exists in MCGM formulae. These relationships reflect the non-

linearity indicated by the field data and offer a better fit with the available field

observations The apparent friction factor versus the reinforcement length are also

undertaken in this chapter.

Chapter 5 considers the behaviour of dams under seismic loads. Although many

researchers have investigated the effects of earthquakes on dams, many problems

remain unsolved, specially for RSDs. A comparison, between the natural frequency of

conventional earth dams and RSDs, is considered in this chapter. It is shown that the

4


practice of inserting reinforcement into the earth dam material leads to an increase in

the natural frequency of such structures compared with conventional earth dams. This

may increase the possibility of failure. In this chapter, the increase in natural frequency

of RSD due to its geometry and its overall stiffness are discussed. Formulae concerning

the magnification of the natural frequency of the structure due to reinforcement

insertion are proposed.

Chapter 6 is concerned with the development of a computer program for optimisation

of RSDs and for stress-strain analysis based on the finite element method. The

purposes of the program are (a) the geometrical optimisation of RSDs based on

analytical and semi-empirical formulae, and (b) the stress analysis of RSDs using the

finite element method. In the finite element section, the quadrilateral elements are

assumed to model the elements of the soil and the one-dimensional bar elements are

considered to model the behaviour of reinforcements. At the beginning of this chapter,

the formulation of soil reinforcement interaction is presented.

Chapter 7 considers the analyses of six models of RSDs using the computer program for

various heights and safety factors to find the optimum base length. A 30m high RSD is

also analysed considering the following four configurations: (a) without

reinforcements, (b) with the assumed increased stiffness of the soil fill, (c) with

horizontal reinforcements, and (d) with inclined reinforcements to evaluate the

variation of stress and displacement. It is concluded that putting reinforcements within

the soil dams can decrease displacement and stress values.

Finally, Chapter 8, which contains two parts, represents a summary of main findings of

this thesis. The first part, summarises the results of the field data analysis and the

second part summarises the findings from the developed computer program.

5

PRINCIPLES OF REINFORCED EARTH CHAPTER TWO

CHAPTER TWO

PRINCIPLES OF REINFORCED EARTH

2.1 INTRODUCTION

Soil reinforcement is a modern technique for improving the mechanical properties of

soil, using the concept of frictional interaction between the soil and the reinforcement.

In the composite material consisting of soil and reinforcement, the generation of the

frictional forces between the soil and reinforcement is fundamental to its behaviour. In

these structures, the compressive and tensile stresses are borne, respectively, by the soil

and reinforcement. In fact, the contribution of reinforcement, in the reinforced earth

structure, is to unify a mass of soil by preventing its lateral displacement.

Reinforced earth is a general concept which has many applications in construction of

bridge abutments, foundations, sea walls, and dams. In some countries, e.g. United

States of America and the United Kingdom, Reinforced Earth is a trade mark and refers

to a special structure which was invented and developed by a French architect, H. Vidal,

in the early 1960s. In comparison with similar techniques, reinforced earth has many

advantages e.g. reduction in cost and ease of construction. These advantages have

caused reinforced earth to be accepted as a suitable substitution for reinforced concrete

in some structures such as, sea walls, bridge abutments and dams.

Reinforced earth is formed from two basic components, fill and reinforcements. The

reinforcement material can be wood, steel, geotextile or other materials such as

polymers. It can be used in different forms such as bar, strip, grid and sheet. Either

cohesive or non-cohesive soil can be used as the back-fill material. However, the non-

cohesive soil is preferred because of its higher internal friction angle. In a vertical

reinforced earth structure, besides the above components, another feature is necessary,

6


to prevent the erosion of the soil at its vertical face. This additional component, called

'facing', is usually provided by precast concrete panels, arched or plain steel sheets, or

timber. Fig. 2.1.1 shows the main components of a reinforced earth structure.

The most important considerations in the analysis and design of the reinforced earth

retaining structures are the internal stability of the composite material and the external

stability of the structure. The latter is necessary for a gravity retaining structure by a

conventional design method. In the following section of this chapter, the reinforced

earth history, its application, material components, fundamental behaviour, design

considerations, construction methods, and economy will be discussed.

Facing units Reinforcing strips

"777

Selected till

Fig. 2.1.1 Reinforced earth components

2.2 HISTORY

Although the modem technique of soil reinforcement has been developed scientifically

since the 1960s, its original concept is not new and goes back thousands of years

(Ingold, 1982). The earliest remaining structure of soil reinforcement is Al-Zigurate in

the ancient city of Ur in Iraq (1500 BC). The great wall of China, which dates back to

the third century BC, is another example of a man-made reinforced earth structure (Al-

Ashou, 1990).

7


After 1966, intensive research on reinforced earth began in countries such as France, the

United State of America and the United Kingdom. The first fundamental research on

behaviour, analysis and design procedures of reinforced earth wall was undertaken at the

LCPC (Laboratoir Central des Ponts et Chaussees) in 1967 (Ingold, 1982). At the same

time, similar research was continuing in the United State of America by A S C E

(American Society of Civil Engineers) and the United Kingdom. The first reinforced

earth structure of this period was built to the north of Los Angeles by the California

Department of Transportation (Caltrans) in 1972 (Hausmann, 1990). It was constructed

on a landslide area and its facing was of the sheet steel type (Chang, 1974). The large

number of international symposia and conferences, held in different parts of the world

such as USA, U K , France, Australia, Japan and India, clearly shows the universal rapid

growth of the reinforced earth technique during the past thirty years.

2.3 APPLICATIONS

Reinforced earth is a technique which can be used as a method for designing different

types of structures such as; bridge abutments; arches; tunnels; slabs; foundations;

retaining walls; sea walls; embankments and dams. Each of the above structures may

have various engineering applications in: industry; military use; housing; highway

making; railway construction and coastal protection. In the next section of this chapter,

some applications of reinforced earth will be discussed.

A successful application of a reinforced earth slab was made on State Route (SR200)

near Norristown, Pennsylvania. This slab was designed by the Reinforced Earth

Company to cover a collapsed section of foundation soil under the embankment of a

highway. The slab was constructed in the form of a low wall lm high with semi-

elliptical steel facing units forming its perimeter. In comparison with a reinforced

concrete slab, the reinforced earth slab was 25 percent cheaper.(Steiner, 1975)

8


Another application of the soil reinforcement is the improvement of the characteristics of

the soil under the foundation. In such an application, the reinforcement is used to ensure

stability, reduce settlement and increase the bearing capacity of the foundation. In

comparison with other soil reinforcing applications, only a very small amount of research

has been done on this application of the reinforced earth in foundation engineering. This

is because the reinforced earth foundations are not economically superior to the other

soil reinforcing techniques such as, lime piles or vibroflotation (Jones, 1985).

Some laboratory and analytically investigations have been carried out by Andrawes et al

(1978) to determine the increment of the bearing characteristics resulting from the use of

reinforcement. As the result of these investigations show, the maximum bearing capacity

ratio (q/qo) was found to occur at the depth ratio (d/B, when d is the depth of the top

layer and B is the width of the footing.) of 0.4. At a depth ratio between 0.8 and 1.8,

the smooth steel is found to give a reduced bearing capacity ratio. A similar research has

been done by Bassett & Last (1978) who advocated the use of discrete reinforcements

installed at various inclinations. This system has the great advantage that it can be

installed beneath new or existing foundations, without the need for excavation.

Reinforced earth technique can be used for underground arches and tunnels. Models of

the arch and tunnel have been successfully tested. Fig. 2.3.1 shows the plane-strain arch

studied by Behnia (1972)

Fig. 2.3.1 Reinforced soil arch (after Behnia, 1972)

9


The largest proportion of application of reinforced earth structures are reinforced earth

walls. According to Ingold (1982), "at the end of 1978 Vidal's licences had completed

in excess of 2000 projects involving 1.3 million square meter of facing". Typical cross

sections of reinforced earth wall, shown in Figs. 2.3.2 to 2.3.4, illustrate the application

of reinforced earth in retaining walls for different structures.

Reinforced earth volume L- -X Reinforced earth volume

Fig. 2.3.2 The slot storage system (after Ingold, 1982)

Fig. 2.3.3 Stepped highway structures (after Vidal, 1970)

Reinforced earth retaining walls can also be used in marine structures. In such cases, the

structure should resist wave forces, tidal conditions and corrosion. Figs. 2.3.5 and 2.3.6

10


illustrate two different cross-sections of sea walls (Reinforced Earth Company

Brochures).

-777— j 1

k.-

^ = = = "

' n '1 r*i " ' Vr^ ///

"

. v** w

Fig. 2.3.4 Rock crushing plant

Tetrapods ^£3$?

§§§}§§&£>'

, , , ^ . jCi

Fig. 2.3.5 Typical section of sea wall (Reinforced Earth Company Brochures)

5m to

6m

T V 2m\

2m

f ^ = i/n\y

Fig. 2.3.6 The sea wall using Z-shapedpanels (Reinforced Earth Company Brochures)

11


Fig. 2.3.7 shows three different purposes for using reinforcement in embankments

(Iwasaki & Watanabe, 1978). In Fig. 2.3.7a, the contribution of reinforcement is edge

stiffening and superficial slope reinforcement. Such reinforcement gives resistance to

seismic erosion and seismic shock as well as permitting heavy compaction plant to

operate close to the shoulder of embankment, hence effecting good compaction in this

sensitive area. In Fig. 2.3.7b, the main body of the embankment is reinforced by a

geogrid net. This type of reinforcement can improve the seismic stability and static

stability, especially against lateral spread of the embankment, during compaction

operation. Reinforcement of weak embankment foundation represents another

application of reinforced earth (Fig. 2.3.7c). Forsyth (1978) used similar techniques to

improve the resistance of an embankment using car tyre.

777" (a) Superficial embankment reinforcement

777 ' (b) Major embankment reinforcement

*77

77\ V7

(c) Embankment foundation reinforcement

Fig. 2.3.7 Modes of embankment reinforcing (Iwasaki & Watanabe, 1978)

The ideal and most efficient orientation for placing the reinforcement is in the

embankment along the axis of principal strain (Sims & Jones, 1979). At this orientation

12


a considerable increase in strength will be obtained. If, however, the reinforcement is

placed parallel with the failure plane, the strength of the embankment may be decreased.

In fact, it depends on the friction angle between the soil and the reinforcement. Fig.

2.3.8 shows these critical orientations for embankments.

^ ^

**^^ ^ ' " ^

(a) Approximate tensile strain orientation

^ «r

(b) Approximatefailure surface orientation

Fig. 2.3.8 Critical embankment orientations (Sims & Jones, 1979)

Reinforced earth can also be used in earth dam construction. The use of soil

reinforcement in the construction of earth dams allows the reduction or elimination of

the downstream slope of the structure resulting in a considerable reduction in the fill

volume. It also allows for a dam spillway to be built at the crest of the structure. In the

event of high water level during construction, it is possible to allow a portion of the flow

to spill over the unfinished dam. In comparison with the other types, RSDs also have

many other advantages e.g. structural flexibility on moderately compact foundation soils,

an increase in the speed of construction, and the integration of embankment work with

construction of reinforced earth spillway (Reinforced Earth Company Brochures). Fig.

2.3.9 compares a RSD with a conventional earth dam.

The first RSD was constructed in the Bimes Valley situated in the south of France

(Ingold, 1982; Taylor and Drioux, 1979; Cassard et al. 1979). The dam was constructed

13

with 9 m height and vertical downstream face using precast concrete facing units. Fig.

2.3.10 shows the cross section and the view of the Vallon des Bimes dam. More details

about RSDs will be discussed in Chapter Three.

Upstram water table -.^ ^ ^

f * ^r *

Jf j£_

Jf ^T~ S^ f ^r

f ^T ^T f

f

\ i •

i *

^ ( Downstram water table -

/// ///

Fig. 2.3.9 Typical cross-section of a RSD compared with a conventional earth dam

->^ ! w

jS??: ': ^f^.:',<:-.-

y*?Xv: ••):••:•

sS^' ••'•'•'.'•'••

^f----^S^ .*<" T •

s^^ ^ r " ? •

^S^yy.yy

9m

Fig. 2.3.10 Cross-section of Vallon des Bimes dam

2.4 MATERIALS

Recognition of the material components needed for the construction of reinforced earth

structures is necessary for prediction of the behaviour and mechanics of the reinforced

soil. The availability of the material components is another major factor for constructing

reinforced earth structures. Selection of the material components depends on type of

14


structure and the cost. Technical requirements of the structure and the basic economics

relate to the selection of material components (Jones, 1985).

Soil, reinforcement and facing are three major components of any reinforced soil

structure. However, other features may also be required for special reinforcement

structures. For example, joining elements and capping units may be necessary as barriers

and facing in some cases (Jones, 1985).

The type of structure has an important role in the selection of material. Some materials

may be suitable for use as components of some reinforced earth structures, but may not

be suitable to be used as components of other reinforced earth structures. For example,

a 'marginal material' may be used to construct reinforced embankments. However, it

may not be suitable for the use in construction of reinforced soil walls (Jones, 1985).

2.4.1 Soil

Soil forms the major part of reinforced soil structures and it usually occupies the largest

volume within the reinforced earth structure. The increase in internal friction of soil can

normally results in the reduction of the stress and strain within the reinforced soil

structure. This may also result in reduction of the amount of soil needed for

construction. According to Jones (1985) only few types of soil can be recommended in

constructing reinforced earth structures. Generally, the soil used for the filling may be

classified into four groups: non-cohesion (or granular) material; cohesive frictional

material; cohesive fill material and waste material (Jones, 1985).

Non-cohesion materials are usually well grained material which have special properties

and granularity. They should normally be used in constructing important reinforced

earth structures with long term use, because of their high internal friction coefficients,

free drainage, less reinforcement corrosion problems and cost considerations. All non-

cohesion materials, which are suitable to be used as fill, should pass through a sieve size

125. More than ninety percent of them should not pass through the sieve size 63 mp..

15


Recommended restrictions in the grading of non-cohesion soil are shown in Table

2.4.1.1. Density, uniformity coefficient, the friction coefficient between reinforcements

and non-cohesion material should be specified for selecting this type of material.

Table 2.4.1.1 Grading restriction for non-cohesion material (Jones, 1985)

Sieve size

125

90

10

600 tim

63 urn

2iim

Passing percentage

100

85-100

25-100

10-65

0-10

0-10

Cohesive frictional materials may also be used in the construction of some reinforced

structures. M o r e than ten percent of this type of soil should pass through the sieve size

63 mu.. Recommended restrictions in the grading of frictional cohesive soil is shown in

Table 2.4.1.2. Apart from grading, other parameters such as: uniformity coefficient;

resistivity; internal friction angle; the friction coefficient between soil and reinforcements

(the adhesion between reinforcement and fill material); soil cohesion; the index of

plasticity and liquid limit should also be considered in the design. The design life period

of the structure (long versus short term) will affect the required properties and type of

soil fill.

Cohesive soil m a y also be used as a material for reinforced soil structures, however, this

type is not suitable for long term structures, especially in the case of wet conditions.

The problem of corrosion is greater in this category of soil. Therefore, to use cohesive

soil, it is necessary to select a reinforcement with low susceptibility for corrosion. Long

term deformation is also a problem, particularly affecting construction of vertically

facing structures. Nevertheless, cohesive materials are used in the construction of some

reinforced structures, the main reason is the availability of the cohesive material on or

16


near to the construction site. This may reduce the costs of the reinforced earth structure.

Therefore, if the use of this type of material is more economical, it can be used provided

the material requirements are met.

Table 2.4.1.2 Grading restrictions for cohesive frictional material (Jones, 1985)

Sieve size

125

90 mm

10 mm

600 Ltm

63 Ltm

2 pm

Passing percentage

100

85-100

25-100

11-100

11-100

0-10

Waste materials may be used as a fill material in the construction of some reinforced

earth structures. For example, some industrial waste materials may be suitable for the

use as filler in the construction of non-important structures e.g. embankments. This may

help in reduction of environment problems and may be economical. For example,

pulverised fuel ash (as a light weight fill) has been used in the construction of

embankments (Jones, 1985).

2.4.2 Facing

Surface erosion of the reinforced soil structures is usually prevented by facing panels,

especially in vertical structures. The use of the facing panels can provide an attractive

architectural facing. The panels may be made of concrete, steel, timber, plastic or from

other materials. Form, size, shape and material are significant parameters which should

be considered for the designing of suitable facing panels. Examples of several shapes of

facings panels are shown in Fig. 2.4.2.1.

17


D O ^ Square facing panel Hexagonal facing panel Flexible facing panel Fig. 2.4.2.1 Typical examples of shapes of facing panels

2.4.3 Reinforcement

The types of materials which may be used as reinforced soil are very different. Steel,

aluminium, wood, rubber, fibre glass, concrete, some kinds of polymers, or plastics may

be used. In a general classification, the reinforcements may be divided into metallic

reinforcements and non-metallic reinforcements. Metallic reinforcements are usually

stronger than none-metallic reinforcements however, the second type is cheaper and

more flexible than the first. Non-metallic reinforcements may be made of one polymer

or combination a of polymers. The degradation resistance of various synthetic fibres is

shown in Table 2.4.3.1 (Cannon, 1976).

Shapes and properties of reinforcements vary. Strips, planks, grids, geogrids, sheets and

anchors may be used. They may be combined to create other types. Typical shapes of

reinforcements are shown in Fig. 2.4.3.1, while a system of reinforcement is shown in

Fig. 2.4.3.2.

The friction coefficient between reinforcement and soil, and the durability of

reinforcement against corrosion should be considered in selecting the reinforcement

The durability of the chosen reinforcement should be compared with the required length

of life of the reinforced structures. Some properties of strip and sheet materials are

presented in Table 2.4.3.2 and the frictional properties of various strip materials are

shown in Table 2.4.3.3.

18


Table 2.4.3.1 Degradation resistance of

Resistance to

attack by

Fungus

Insects

Vermin

Mineral acids

Alkalis

Dry heat

Moist heat

Oxidising agents

Abrasion

Ultraviolet light

various synthetic fibres (Cannon, 1976)

Types of synthetic fibres

Polyester

Poor

Fair

Fair

Good

Fair

Good

Fair

Good

Excellent

Excellent

Polyamide

Good

Fair

Fair

Fair

Good

Fair

Good

Fair

Excellent

Good

Polyethylene

Excellent

Excellent

Excellent

Excellent

Excellent

Fair

Fair

Poor

Good

Poor

Polypropylene

Good

Fair

Fair

Excellent

Excellent

Fair

Fair

Good

Good

Good

PVC

Good

Good

Good

Good

Good

Fair

Fair

-

Excellent

Excellent

Steel bar _ J 1 " 1

Steel bar

Anchor plate

Key

Webbing Tyre

Fig. 2.4.3.1 Typical shapes of reinforcements

19

Longitudinal reinforcements

Tranverse members

\ Facing panel

Fig. 2.4.3.2 A reinforcement system connected to a facing panel

Table 2.4.3.2 Properties of typical sheet and strip material (Jones, 1985)

Materia]

Aluminium alloy

Copper

Carbon steel (galvanised)

Stainless steel

M a x i m u m thickness to which stresses

apply (mm)

6

10

10

6-10

Basic permissible stresses

Axial

tension

mm

120

108

120 - 192

126 - 220

Shear

< " 2 >

mm 72

65

72-115

Bearing

< \ >

mm 180

163

200 -350

75-132 1 210-360

Table 2.4.3.3 Frictional properties for various strip material (Boden etal., 1978)

Effective stress

range 0-40 kPa

Effective stress

range 0 - 100 kPa

Angle of friction

of soil without

reinforcement

(<*>')

37

37

Coefficient of friction between fill and

reinforcement (p)

A

0.38

0.36

A = Galvanised mild steel

B = Stainless steel

C = Glassfibre reinforced plastic

B

0.40

0.39

C

0.53 to

0.64

0.53 to

0.64

D

0.51 to

0.58

0.51 to

0.58

E

0.36

0.37

F

0.42

0.40

D = Aluminium coated mild steel

E = Plastic coated mild steel

F = Polyester filaments in polyethylene

20


Types of non-metallic reinforcements are usually provided by polymers. Although the

strength of the non-metallic reinforcements is lower than that for the metallic, the

applications of the non-metallic group has increased during recent years. The most

important reasons for the increased use of non-metallic reinforcements are their

avauability and low cost (John, 1987). In particular, polyamide, polyester,

polypropylene and polyethylene may be used as soil reinforcements. A comparison

between the properties of non-metallic materials is shown in Table 2.4.3.4.

Table 2.4.3.4 A comparison between the properties of general polymers (John, 1987)

Comparative properties

Strength

Elastic modulus

Strain at failure

Creep

Unit weight

Cost

RESISTANCE TO:

Stabilised U. V. light

Unestablished U. V. light

Alkalies

Fungus, vermin, insects

Fuel

Detergents

Polyester

high

high

medium

low

high

high

Polyamide

medium

medium

medium

medium

medium

medium

Polypropylene

low

low

high

high

low

low

Polyethylene

low

low

high

high

low

low

high

high

low

medium

medium

high

medium

medium

high

medium

medium

high

high

low

high

medium

low

high

high

low

high

high

low

high

The reinforcement system and the face panel of the first bar mesh reinforced wall are

shown in Fig. 2.4.3.3a This wall was constructed by Caltrans near Dunsmuir in

California in 1975 (Hausmann, 1990). The facing elements of the wall were of

concrete type and beam shaped panels. This technique was designed as Mechanically

Stabilised Embankment (Hausmann, 1990).

21


The Hilfiker Reinforced Soil Embankment, the Vorspann System Losinger Retained

Earth and the Georgia Stabilised Embankment are other techniques of reinforced earth

embankment with different facing panels, bar mesh geometry and construction details

(Hausmann, 1990). In the Hilfiker Reinforced Soil Embankment, which was

introduced in 1983, the precast facing panel and the reinforcements are formed by a

beam shaped and cold-drawn wire mesh, respectively. In this system, the bar mesh is

connected to the facing panels. The shape of the panel and mesh of the Hilfiker

Reinforced Soil Embankment are shown in Fig. 2.4.3.3b.

The Vorspann System Losinger Reinforced Earth represents another welded wire mesh

system with precast concrete facing. The first wall using this system was built in

California in 1981. During the period time, 1981 to 1984, about 100 walls were

constructed by using this system (Hausmann, 1990). The shape of its facing and its

wire mesh are shown in Fig. 2.4.3.3c. Also, the panel and reinforcement system of the

Georgia Stabilised Embankment system are shown in Fig. 2.4.3.3d. This system was

introduced by the Georgia Department of Transportation.

2.5 SOIL-REINFORCEMENT INTERACTIONS

2.5.1 Development of Theory

From 1966 until the present (1994), numerous tests have been undertaken to understand

the behaviour of reinforced earth and various theories have been presented to describe

reinforced soil behaviour in analytical terms. Extensive researches in this area has been

pursued by Laboratoir Central des Ponts et Chaussees (LCPC), N S W Institute of

Technology, The University of California- Los Angeles (UCLA) amongst other

researchers. Some of these works will be discussed in the following sections.

22

V. V, p

i ~""C

•U—

V

•K

s o ! ; o | : •"> -.

V

1 s vo

—1

V r V

S

7

oo

"a

+

+ +

4 <*>

--

— —

--

--

_ _

--

1 1 1 1

S

d

"«

"1

s: + + + +

+ + + +

s •»»•»

e -fc

to

a

•2

5J to

fc

"a e a to

s; o< oo fc

In 1966, Vidal found that if a horizontal reinforcement strip was put within the loaded

soil, the friction between the soil and reinforcement raises the lateral stress from c'3 to

tf'3 + Aa'3 in failure condition. The increase in the lateral stress (A03) increases the

bearing vertical stress of the soil from c'\ up to (a'i)r. Figure 2.5.1.1 shows the new

situation of the stresses in the Mohr's circle based on Vidal's (1966, 1969) theory, due to

reinforcement insertion within the soil.

t

Ogl y°\ y0 g+AO 3

^ C <t>

J(0\)T

- ^

Fig. 2.5.1.1 Improvement in strength due to reinforcement (based on Vidal, 1966,

1969)

Other experimental and theoretical research was conducted by Long et al. in 1972.

Figure 2.5.1.2, a plot of the deviator stress versus axial strain, shows that the reinforced

samples are brittle. The researchers concluded that the failure envelopes of reinforced

and unreinforced samples have the same angle of friction. The results from a series of

triaxial tests, carried on 100 mm diameter samples of special sand with D50= 0.15 mm

and a mean dry density of 1.67 g/cm3, are illustrated in the figure. The figure shows

that the amount of axial strain increases for unreinforced soil with same deviator stress.

Also, the additional strength of the reinforced samples results from apparent cohesion,

c'.

24


*-> -* VI vi

£ o

Of

2500

2000

1500

1000

500

0 0 2 4 6 8 10

Axial Strain - %

Fig. 2.5.1.2 Increase in brittleness due to reinforcement (based on Long et al, 1972)

In the LCPC cohesion theory, presented by Schlosser and Long in 1973, it was

suggested that the value of o'l is equal to the sum of passive earth pressure (Kpo'3) and

Aa'i. This means that:

oJ1 = £poJ3+Ao

J1 (2-D

25

Vertical

stress °\ 2^

(100 kN/m2) 15

10

5

0

™ M '

M

/

/

f .' , Reinforced

7 / Unreinforced

) 2 4 6 8 10

Confining pressure o y 700 kN/m )

Fig. 2.5.1.3 Increase in Cj due to reinforcement ( based on Schlosser & Long, 1973)

Reinforced

Unreinforced

25


They then compared this equation to the Rankine-Bell Equation for c'-(t>' soil and

concluded that:

Ao1 c =• l-rP

(2.2)

Schlosser and Long (1978) also presented a theoretical procedure to calculate c'. Fig.

2.5.1.4 shows the element which was assumed for calculating the amount of c'. In

regard to the figure and using equilibrium it can be concluded that:

F + G ' ~ A tana = c', Atan(a-0') (2.3)

where F is the sum of tensile forces induced in the reinforcements, A is the cross section

of the sample, a is the angle of failure plane, c\ and 03 are the vertical and lateral

stresses respectively, and 0' is the angle of friction of soil.

Fig. 2.5.1.4 Coulomb analysis (Schlosser & Long, 1973)

O n the other hand, because F is the sum of tensile forces in reinforcements, it can be

written as:

26

F=ATjma (24)

h

where h is the vertical space between reinforcements and T is the tensile force in each

reinforcement. By substituting Eq. 2.4 in Eq. 2.3 and by differentiating, the maximum

value of a'l may be given as:

K T G\=K o' +-£— (2.5) 1 P 3 h

By comparing this equation with Eq. 2.1, the Aa'i can be obtained as:

, K T A C T 1 = - £ - (2.6)

1 h

By substituting the Aa'l in Eq. 2.2, the value of c' is found as follows:

c =

T IK L. (2.7)

2h

This equation was found to be in close agreement with the experimental results which

were undertaken by Long in 1972. The comparison of the theoretical and experimental

results is shown in Fig. 2.5.1.5.

A modified version of Eq. 2.8 is:

Tr \K~

c<= V P 2h (2.8)

where r is the ratio of the plane area of the reinforcing ring to the cross sectional area of

the sample.

27


Cohesion

(kN/m )

300

200

100-

0

experimental

theoretical

0 50 100 150 200 250

Ratio T/h (kN/m2)

Fig. 2.5.1.5 Comparison of theoretical and experimental results (Schlosser & Long,

1973)

Hausmann (1976) worked on two models, called Sigma and Tau, both considered tensile

and bond failure. The results from the two models were similar. In the Sigma model, it

was assumed that reinforcements prevented lateral expansion and, in the second model,

it was assumed that horizontal and vertical shear stresses were induced by reinforcement

into the soil. The Sigma model was analysed in two situations; when the failure happens

because of the rupture of reinforcement and, when the failure occurs due to slippage.

In former situation, it was assumed that the sum of lateral stress a'3 and the value of a'r

is equal to Ka multiplied by vertical stress a'l in failure condition. This means that:

1 r a 1 (2.9)

or

c' =K a' +K a' 1 p 3 p r

(2.10)

A comparison between this equation and Rankine-Bell's equation yields:

28

r 2 (2.11)

By substituting the value of G'T = aA/dy dz, Hausmann (1976) obtained that:

c ^CA^P Cr Id d

y z

(2.12)

where O" is the stress in the reinforcement, A is the cross section of reinforcement, Kp is

the coefficient of soil in passive condition and dy and dz are the dimensions of the soil

element. Fig. 2.5.1.6 shows failure condition in Sigma model in the case of constant c'r.

x • Reinforced Unreinforced

<t>'

G3 ka°l

Fig. 2.5.1.6 Failure condition for constant G'r (Hausmann, 1976)

Hausmann also considered the Sigma model when the failure happens because of lack of

bonding between soil and reinforcement. It was assumed that the friction along the

reinforcement is in linear proportion to vertical stress. This means that:

rj' = F o \ r 1

(2.13)

Substituting the o*r from Eq. 2.13 to Eq. 2.9 results in:

29

-^-+F = K

On the other hand, it is known that:

CT' l-sin(j>' 3__ T r

G1. 1 + sin <>' 1 T r

Substituting the CT'3/a'i from Eq. 2.15 to Eq. 2.14 results in:

(2.14)

(2.15)

(K -F-l) sin(<j>' ) = —^

(F-K -1) a

(2.16)

A series of laboratory experiments, undertaken by Hausmann, to prove his theory in

bond failure condition, indicated a suitable agreement with the LCPC theory. However,

the failures which occurred due to rupture, did not correspond very well with this

theory. Therefore, it was concluded that, at high stress level, the failure of soil is the

result of rupture of reinforcement and at low stress condition the failure may occur

because of slippage. Fig. 2.5.1.7 shows the variation of §\ due to F.

Fig. 2.5.1.7 Friction angle §r as a function ofty and F (Hausmann, 1976)

30

The increase in the friction angle of the soil due to failure by slippage between the soil

and reinforcement is shown in Fig. 2.5.1.8. The failure of soil because of reinforcement

rupture at high stress level, and the failure of soil because of slippage at low stress level

are shown in Fig. 2.5.1.9. The Mohr stress circle for a reinforced and unreinforced

samples, with the same lateral pressure, a'3 is shown in Fig. 2.5.1.10. This figure shows

that the effect of reinforcement is an increase from a'l to (a'i)r in the vertical stress or

the increase in the inducing cohesion, cr, of soil.

X ' . Reinforced

A—$r

03 ka G'I

> < * ,

Unreinforced

a> t?

Fig. 2.5.1.8 Increase in the friction angle of soil because of reinforcement (Hausmann,

1976)

Reinforced. x 4

J

Unreinforced

03 G\ 03 G\ a

Fig. 2.5.1.9 Composite Mohr envelope (Hausmann, 1976)

31


Fig. 2.5.1.10 The LCPC interpretation

In 1972, Chapuis rejected the assumption that in a mass of soil with horizontal

reinforcement, the principal stresses were vertical and horizontal. It was assumed that

the horizontal and vertical planes were not able to be principal planes. Chapuis(1972)

considered that the main principal stress (a'3)r is higher than a'3 and the term, Aa'3, is

approximately equal to:

Ao%= — = — (2.17) 3 BH H

The second side of this equation is the same as that presented by Hausmann

(G'T=GA/BH). However, Aa'3 is a stress increment, whereas a'r is a stress decrement.

Chapuis also found that cohesion relates to the distribution of stress along the

reinforcement. The Aa'3 interpretation is shown in Fig. 2. 5.1.11.

32

X <

°3

Reinforced/?

(a3 ) r CT1

Unreinforced

1

(^l)r

Fig. 2.5.1.11 The AG'S interpretation (Chapuis, 1972)

Yang (1972) undertook the same experiments using triaxial tests on sand by using

samples with 71 mm diameter and height variations between 20 mm and 162 mm. In a

series of experiments, he investigated the reasons for the failure using strong rigid

reinforcement. It was found that the compressive strength of the samples increased

while the space between reinforcements decreased believing the samples failed at

constant effective stress ratio. It was concluded that any increase in a'l at failure

condition in the reinforced samples was due to a modified confining pressure, Aa'3, as

follows:

G. - K G~+K Aa~ 1 p 3 p 3

(2.18)

or,

Aa~ = K G.-G~ 3 a 1 3

(2.19)

Fig. 2.5.1.12 shows that the equivalent confining pressure per initial confining pressure

decreases when the aspect ratio (height / diameter) increases based on Yaung (1972).

Fig. 2.5.1.13 illustrates that the variation of the confining pressure, Aa'3, increases

linearly with the applied confining pressure, a'3. Therefore, it was concluded that the

33


value of Aa'3 would be constant, the equivalent of the Eq. 2.17. However, according to

Ingold (1982), there was a poor agreement between predicted and increased values.

20 Height

Diameter

10

0 0

8 12 16

Equivalent confining pressure_ 3 3

Initial confining pressure

Fig. 2.5.1.12 Variation of strength with aspect ratio (after Yang, 1972)

100

75

Increase in confining 50

pressure

Ac3(psi) 25

a

d

-

psi -= 6.9 kN/m ) /

h/d=0.57

0 10 20 30 40 50

Applied confining pressure G ? (psi)

Fig. 2.5.1.13 Variation ofAG'3 with G'3 (after Yang, 1972)

34

In 1985, Jones presented a theory based on principles of soil mechanic. In an element of

unreinforced soil, the value of vertical stress and lateral stress may be given as follows:

Gx=yh (2.20)

G3=KQyh (2.21)

where, y is the unit weight of soil, h is the soil depth where element is located, and Ko is

the coefficient of lateral earth pressure at the rest condition. When the element of soil

starts for expanding laterally, the coefficient of lateral earth pressure reduces from Ko to

Ka where:

K = 2 (2.22) a (l+sin<j))

Jones (1985) argued that in a compacted reinforced soil, the reinforcement doesn't

permit the soil to expand because of the friction between soil and reinforcements. This

results in creation of tensile stress and strain in any units of the reinforcement, as

follows:

(2.23)

(2.24)

(Jrp '

5 = r

*0al a r

Grp

E r

or,

5 = * 0 a l (2.25) r a E

r r

where, ar and Er are, respectively, cross sectional area and elastic modulus of

reinforcement, GT is the tensile stress in the reinforcement, and 5r is the strain of

35


reinforcement due to Gj. When the effective stiffness of reinforcement, ar Er, increases,

the strain in the reinforcement decreases. It was also assumed that the values of strains

in soil (er) and reinforcement (5r) are equal.

er=8r (2.26)

Thus, it was concluded that the lateral strain in the soil, er, reduces to zero when

effective stiffness of the reinforcement (ar Er) is high, and the lateral strain increases

when the effective stiffness decreases. However, the coefficient of lateral earth pressure

of the soil, Ko, decreases to Ka in the second situation (Jones, 1985).

2.5.2 Stability Considerations

Studies of the relationships between stress and strain within reinforced earth structures,

and studies of sliding, bearing, slip, tear and tension failures should be considered during

the stability considerations of reinforced soil structures. The consequential reinforced

earth structures such as walls, abutments, and dams are involved this problem. Transfer

of stress from soil to a single strip, as tensile stress, should be considered here. Two

modes of failure may occur in the reinforcements: the breaking of the reinforcements due

to tensile stresses, and the failure due to pull-out of reinforcements. Force equilibrium

and moment equihbrium have been used to calculate tension in the reinforcements. Fig.

2.5.2.1 shows the possible forms of failures in a reinforced soil structure. These will be

discussed during the following sections.

Based on the equilibrium of a reinforced soil element, the tensile force in the

reinforcement, T, is usually calculated as:

T = KGSS, (2.27) a v v h

36


where, Sv is the vertical space between reinforcements, Sn is the horizontal space

between them, av is the vertical stress over the soil elements and Ka is the coefficient of

lateral earth pressure in active condition.

Fig. 2.5.2.1 The forms of failures (Jones, 1985)

On the basis of Coulomb's wedge theory, the tension, Tf, in the ith layer of a vertical

reinforced soil structure is usually calculated as:

T.=—^—KyHAH (2.28) 1 (n + 1) a

where n is the number of reinforcement layers, y is unit weight of soil, Tf is the depth of

structure and AH is the vertical space between reinforcements.

Trapezoidal distribution and Meyerhof s distribution of pressure under the base of

vertical structures are shown in Figures 2.5.2.2a and 2.5.2.2b, respectively. Several

equations have been presented so far for calculation of the maximum tension in the

reinforcements of a vertical reinforced soil structure as shown in Table 2.5.2.1.

37


H

e = H/3

iiikkkikikki

Fig. 2.5.2.2 a) Trapezoidal distribution pressure and b) Meyerhof distribution pressure

In Table 2.5.2.1, L is the length of reinforcements, y is unit weight of soil, H is the depth

of the structure, AH is the vertical spacing between reinforcements and Ka is the

coefficient of lateral earth pressure in active condition.

Table 2.5.2.1 Equations for calculation of the maximum tension in the reinforcements

of a vertical reinforced soil structure

Rankine Eq.

Trapezoidal Distribution Eq.

Meyerhof Distribution Eq.

Coulomb Moment Balance Eq.

Elastic Analysis Eq.

Equation

T max

T max

T max

T max

T max

= K yHAH

= K yHAH(l + K A 2 ) a u j_.

K yHAH a'

(1-0.3* (^)2) a L,

n2K yHAH a1 (n2-D

= 0.35yHAH

Eq. no.

(2.29)

(2.30)

(2.31)

(2.32)

(2.33)

38


A comparison between Eq. 2.29 to Eq. 2.33 are shown in Fig. 2.5.2.3. The figure is

plotted based on areas of one row of reinforcement per metre width, when H is assumed

to be 5 metres. It shows that the numerical differences between these equations increase

when the vertical spacing between reinforcements rises .

Vertical spacing (mm)

1000

800

600

400

200

0 0 40 80 120 m

Area of reinforcement (mm/metre width)

Fig. 2.5.2.3 Comparison between maximum tension stress inside reinforcement (Jones,

1985)

2.6 CONSTRUCTION METHODS AND STRUCTURAL SYSTEMS

Construction of a structure is the final stage of a project. In order to reduce the

associated costs, the construction processes should be simplified as much as possible,

leading to a short construction time. For this, a number of factors ought to be

considered. Theses will be discussed in the next two sections.

According to Hambley (1979), the following considerations should be taken into

account during construction of reinforced soil structures including (a) the use of

materials obtainable and easy to work with, (b) the use of simple shape foundations, (c)

39

a= Trapezodial distribution b= Elastic analsis c= Meyerhof distribution d= Coulomb moment balance e— Rankine theory /= Coulomb wedge


the use of horizontal or vertical surfaces, (d) fixing reinforcement first then placing soil,

(e) avoiding small sections, (f) avoiding massive large elements, (g) considering the

stability of detailed elements of reinforced soil structure during the stages of

construction, and (h) considering the requisite distance between reinforcements.

Differential vertical settlement is another important feature which may adversely affect

the construction process. Construction methods, reinforced systems, labour and plant,

rate of construction, compaction, damage and corrosion, distortion, logistics and

constructor's construction sequences have to be considered to optimise the construction

procedure (Hambley, 1979).

There are three major methods used in the construction of reinforced soil structures: the

Concertina Method, the Telescope Method and the Sliding Method. These methods will

be discussed in the following paragraphs.

The Concertina Method was developed by Vidal (1966). The largest reinforced soil

structures have been constructed after the development of this method (Jones, 1985). In

this method, the structure of the reinforced earth wall is formed from reinforced soil with

metallic flexible faces and reinforcing material. The face of the structure is formed from

semi-elliptical cross section facing units. Each 2 5 0 m m high facing unit is typically

connected to the reinforcements by bolts which pass through the strips and the edges of

the facing units. The weight of each unit is usually 1\5kg and its length is typically up to

10m. The thickness of the unit is about 1.5 to 3mm. The facing may be settled

proportional to the soil settlement. Therefore, the settlement does not destroy the facing

units. It means that the facing will be deformed without any destruction during the

internal settlement. A cross section of this method is shown in Figure 2.6.1.

The Telescope Method (Fig. 2.6.2), was also developed by Vidal (1978). In this method

the face of the structure is made of concrete panels instead of flexible face units. The

weight of a standard concrete panel is about lOOOfcg, the sizes of panel is about

1.5mxl.5m and its thickness is 18cm. To get rid of the panels rapidly, there are 4 lugs in

40


each panel for connecting the fourth edges of each panel and they are rebated. The

vertical distance between luges is 75cm centre to centre and the horizontal distance

between them is lm. Settlement is achieved by horizontal gaps between the facing

panels which will be filled after gravitational settlement of the layers. Therefore, in this

method the facing panels will be fixed after the internal settlement of the soil.

Fig. 2.6.1 The cross section of Concertina Method (Vidal, 1966)

The Sliding Method (Fig. 2.6.3), was developed by Jones (1985). In this method, the

facing is formed by light weight glass reinforced cement. The weight of each facing is

only \%kg. The shape of the cross section of each facing is a hexagonal- based pyramid

22.5cm deep and 60cm across the flats. Vertical movement is provided by installation of

two rods. In this method, the differential settlement may be achieved by vertical sliding

of the facing panels. W h e n the soil is settled, the end of reinforcement may simply slide

down because of its vertical pole. This facing panel has two roles: that of protection of

soil from erosion and as a structural element A typical section of sliding method facing

is shown in Fig. 2.6.3.

41


Fig. 2.6.2 A typical section of Telescope Method (Vidal, 1978)

Fig. 2.6.3 A typical section of Sliding Method (Jones, 1985)

2.7 D U R A B I L I T Y

The required durability of reinforced earth structures relates directly to their design life

span. The corrosion of reinforcement strips may adversely affect their durability and

hence lessens the length of life of the reinforced earth structures. Most reinforced soil

structures, their reinforcements in particular, are susceptible to corrosion. If the

42


question of durability is not addressed, then the structure may fail. Therefore, durability

should be considered as a function of design life.

Corrosion is a major problem which effects the durability of metallic reinforcement. The

corrosion happens under ground, its problem is not seen until the failure occurs.

Material deterioration may occur because of electrochemical, bacterial or physical

corrosion problems (Jones, 1985). All metallic reinforcements should be protected

against electrochemical corrosion. This can be done by the use of cathodic protection

systems or through electrical compatibility (Jones, 1985). Cohesive fill material is more

corrosive especially in the case of metallic reinforcement, hence it reduces the durability

of reinforcement. Therefore, non-cohesion fill material is preferable to the other types,

particularly for using in the permanent structures.

O n the basis of life span, reinforced earth structures may be classified into three

categories: temporary structures, short life structures and permanent structures. The

first category includes structures with the life span of less than 100 weeks. Durability is

not considered to be a problem for these structures. The second category are structures

with a life span of between 2 and 20 years. Durability in this category should be

considered as a minor problem. The permanent structures, with a life span of between

60 and 120 years, form the third category, and durability is a major problem for this type

of structure (Jones, 1985). Most dams are categorised in this third category because

their life span is usually more than 20 years.

Corrosion will develop in the metallic reinforcement during this period, however the rate

of corrosion development will decrease in time (Romanoff, 1959). This reduction in the

rate may be the result of the creation of an external corroded layer on the surface of the

reinforcement, acting as a protection for the underlying material. In some cases, the

corroded layer can reduce the penetration of corrosion. Therefore the actual cross

section of metallic reinforcements should be the sum of the net cross section area (to

carry the expected level of stress) and a portion of cross section area (to be sacrificed

43


because of corrosion). A diagram which indicates the development of corrosion versus

time is shown in Fig. 2.7.1.

Corrosion

^^^ X=Ktn

LJ > 10 years t

Fig. 2.7.1 Development of corrosion (after Romanoff, 1959)

The use of other types of metallic reinforcements may reduce corrosion. For example,

the use of galvanised steel instead of unprotected steel can decrease corrosion. A

comparison between galvanised steel and unprotected steel is shown in Fig. 2.7.2. In the

case of galvanised steel, corrosion may only happen after destruction of the protected

surface of the reinforcement. Therefore, protection of the external layer of the

reinforcement strips can increase the length of life of the reinforced soil structures.

The P H of soil, the water content, the redox potential and the soil resistivity may also

affect corrosion. In some cases, it may be desirable to use materials which are not

susceptible to corrosion or more durable than metallic reinforcements. For example, the

degradation problems of polyamide, polyester, polypropylene and polyethylene

reinforcements appears to be less extensive than the corrosion of metallic reinforcements

(Jones, 1985). Typical corrosion allowance for metallic reinforcements are shown in

Table 2.7.1.

44


Fig. 2.7.2 Loss of thickness during time for galvanised steel and unprotected steel

(Jones, 1985)

Table 2.7.1 Corrosion allowance for metallic components exposed to various

environment (Department of Transport BE, 1978)

Aluminium alloy

Cooper

Galvanised steel

Stainless steel

Sacrificial thickness to be allowed for each surface exposed

to corrosion (mm)

Atmospheric environment

Urban, industrial,

industrial costal

0.85

0

Other

.

0.3

0

Buried in fill

Frictional

fill

0.15

0.15

0.75

0.1

Cohesive

frictional fill

0.3

0.3

1.25

0.2

45


2.8 COSTS AND ECONOMICS

Even one percent reduction in the construction costs of big projects can save millions of

dollars. The greatest disadvantage in the use of concrete in the construction of marine

structures, is its cost. For example, construction of a reinforced soil standard bridge

abutment, instead of the conventional piled standard bridge abutment, may reduce the

costs of the project by fifty percent (Jones, 1985).

A group of researchers in the U K have undertaken a comparison between the height of

the reinforced soil structures and the percentage of the costs of reinforced soil relative to

the costs of reinforced concrete cantilever walls. This comparison is shown in Figure

2.8.1. The figure shows that the cost of a reinforced soil structure with the height of

10m may be about 30 percent of the costs of the same structure which has been

constructed of reinforced concrete. Although the costs of reinforced earth structures

vary with the type of material, the costs of reinforced concrete structures are normally

higher than those of the reinforced earth structures. This is a conclusion which can be

drawn from Fig. 2.8.1.

The variable percentage of construction costs to the reinforced soil cost; relative to the

reinforced soil cost, versus the height of reinforced soil structure is shown in Fig. 2.8.2.

The figure shows that the costs of a reinforced soil structure with the height of 20m may

be about half the cost of a conventional structure. The figure also indicates that use of

reinforced soil is more economical in the case of high structures.

The distribution of material costs of reinforced soil structures has three major

components: the cost of reinforcement, the cost of facing and the cost of soil fill. Fig.

2.8.3 shows the variation of percentage of the total material cost with the height of

structure. This figure shows that the cost of facing reduces with the increasing the

height of structure but, the cost of soil fill and the cost of reinforcement both rise. All

three material cost components tend to stabilise with increasing the height of structure.

46


Percentage

cost

100 -

75 ~

50

25 ~

0 T 1

15 0 5 10

Height of structure (m)

~T~

20

Fig. 2.8.1 Comparison between the height of reinforced soil structures and the

percentage of the costs of reinforced soil walls relative to the costs of reinforced

concrete walls (Jones, 1985)

100 -

Economy

0 5 10 15


r 20

Fig. 2.8.2 Economy versus height of structure (Jones, 1985)

47


Prcentage

of total cost

60_

50 _

40~

30~

20~

10

0

Reinforcement

*N*. Facing elments

Soilfi.il

1 1 1 1 0 5 10 15 20


Fig. 2.8.3 Variation of percentage of total material cost with height of structure (Jones,

1985)

Another diagram which shows the other comparison between the costs of reinforced soil

structures and reinforced concrete structures is shown in Fig. 2.8.4. This comparison is

undertaken for the construction of one (6 m high) reinforced soil wall and one (6 m high)

reinforced concrete wall. This diagram shows the cost of energy content of construction

material, process water used in the manufactured materials, the labour for manufactured

material, material transport and construction in the case of reinforced soil and reinforced

concrete structures. The process water used in the case of reinforced soil is the only

item when cost is exceeded (Jones, 1985).

48

http://Soilfi.il


120

100

80

X 60

40

20

0

reinforced soil wall X = — • xlOO

reinforced concrete wall

a f 8 h

a

d

f g h

energy content of construction material process water used in manufacture of materials despoiling of land in production of materials S(?2 - emission dust - emission labour - manufacture of materials

labour - material transport labour - construction

Fig. 2.8.4 Comparison between the costs of reinforced soil structures and reinforced concrete structures

2.9 C O N C L U S I O N S

One way to improve mechanical behaviour of soil is to use reinforcement in the soil.

The low tensile strength of the soil can be increased by reinforcement, hence the

combination of soil and reinforcement results in a new stronger material which can

withstand loads higher than the soil without reinforcement. Prevention of lateral

expansion, which is the main role of reinforcement, can decrease the lateral displacement

of soil and this increases the lateral stress. From the theoretical research and

observations concluded so far, it is obvious that the use of reinforcement in soil increases

49


the strength of soil mass. Some researchers claim that reinforcement increases the

cohesion of soil. Others believe that the reinforcement can increase the frictional angle of

soil. A general conclusion may be made that the effect of reinforcement in the soil is an

increase in the angle of friction in low stress levels and an increase in cohesion of soil in

high stress levels.

50

EVALUATION OF SOIL DAMS CHAPTER THREE

CHAPTER THREE

EVALUATION OF SOIL DAMS

3.1 INTRODUCTION

Earth-fill and rock-fill dams have a greater role than that of concrete dams in water

collection. According to Wolff (1985), about three - fourths of all large dams are

constructed of earth and rock-fill. The earth dam is the most important structure among

water resource structures, because it is the most economical. N o earth dam, which has

been built based on modern soil mechanic concepts, has failed. In recent years, the earth

dams are considered to be as safe as concrete dams (Singh, 1976).

The use of reinforcement in earth dams allows the reduction in displacement, stress

level, fill volume, and at the same time, increases the safety factor of the slope of the

dam. Other advantages of RSDs are: speed of construction; the flexibility of the

structure; the possibility of spillway construction in the crest of dam; and the possibility

of spilling a portion of flow over the unfinished dam. By the use of reinforcement, it is

possible to eliminate the downstream slope and reduce the upstream slope of the dam.

This results in a considerable reduction in the fill volume and the costs.

The state of stresses in RSD, the application of loads acting on dam, the lateral stresses

acting on the facing panels, and the assessment of shear stresses along the

reinforcements are not yet completely understood. The best method to be used to

analyse the internal stability of RSDs, h o w the influence of construction stages should

be simulated in the design of RSDs, and how the state of stresses should be stabilised at

the end of construction have not yet been fully answered. These should be considered

in the design criteria. In this chapter, history, classification, forces evaluation, stability

51


analysis, and foundation behaviour of conventional earth dams and RSDs will be

presented and discussed.

3.2 CONVENTIONAL EARTH DAMS

3.2.1 History

The date of construction of the oldest dams is not known for certain but the oldest

known earth dams were constructed about 500 B C in India (Singh, 1976). However,

Smith (1971) claims that the Sadd-el Kafara, D a m of the Pagana, which was discovered

in 1885, was built sometimes between 2950 and 2750 B C . The oldest known arch dam

was constructed in Iran (Smith, 1971). Today, there are many large rigid arch dams,

gravity dams and buttress dams in the world.

During ancient times, earth dam construction was improved. Construction

improvements were mostly undertaken by architects (Smith, 1971). By 1900 there were

less than 10 earth dams over 30m in heights (Singh, 1976). N o dam exceeding 40m in

height had been constructed until 1925 (Singh, 1976). Since 1925, the increase in the

ability of engineers to build safe and economical earth dams has led to the construction

of a greater number. From this date, the number of earth dam constructions has been

greater than in all previous history (Sherard, 1976). Causes of soil dam failures based on

Sowers (1961) is tabulated in Appendix A.

In reality, the improvements of large earth dams started after the improvement of soil

mechanics. For example, the 111m high Aswan dam with a capacity gross storage 156.2

milliard cubic meters, the 235 high Oroville dam with the gross storage 4.3 milliard cubic

meters, the 300m high Nurek dam in Russia, were all built after soil mechanic

improvements. The Nurek dam has created a reservoir with total storage 10.5 milliard

cubic meters and generation capacity of 2100MW (Singh, 1976).

52

Some other important earth and rock-fill dams with their capacities and heights are: the

Djatiluhur dam in Indonesia with a height of 91.5m and 7.1 million cubic meters

capacity, Beam dam with a 134m height and 32.5 million cubic meters capacity, Mica

dam with a 244m height and 32.1 million cubic meters capacity, and the Portage

Mountain dam with a 138m height and 70 milliard cubic meters capacity (Singh, 1976).

3.2.2 Classification

Dams may be classified based on: construction material; rigidity; use; structure; and

hydraulic design. A general classification of dams is shown in Fig. 3.2.2.1. O n the basis

of rigidity, dams are classified into two major categories: rigid and non-rigid. In both

categories, further classification is made with respect to construction material.

Rigid dams may be constructed from concrete, masonry, timber and even steel. The

latter two are not particularly common at the present time. Based on their structures,

types of rigid dams are arch dams, gravity dams, buttress dams or a composite of all

these.

Non-rigid dams are usually of the gravity type and made of earth or rock-fill materials.

This category is classified into two groups: earth dams and rock-fill dams. According to

U. S. Army Corps of Engineering (1982), dams may be classified based on height and

capacity storage as follows:

Table 3.2.2.1 Classification of dams based on storage and height (U. S. Army Corps of

Engineering, 1982)

Category

Small

Intermediate

Large

Storage (acre-feet)

50 < volume < 1000

1000 < volume <50,000

50,000 < volume

Height (feet)

25 < height < 40

40 < height < 100

100< height

53


In rigid dams the technique of construction is usually very complicated. For example, in

concrete arch dams special casings with particular arch should be used. Special casings

need professional workers, special machines and equipment for construction and

installation. These requirements cause the costs of dam construction to rise. Most parts

of the materials of a concrete dam are transported from factories to the location of the

dam. For example, reinforcement and cement are transported to the location of a

concrete dam from factories, increasing the costs of dam construction. Therefore, the

project may become un-economical.

The stages of construction in non-rigid earth dams are adaptable to the local area.

Techniques of construction are not very complex, in comparison with the concrete type.

Embankment dams do not need any casing. The need of special machines and specialist

workers is very low for the earth dam in comparison with that of the concrete dam. This

leads to a decrease of cost of dam construction. Provision of the earth dam material is

much easier than that of a concrete dam. Local materials are usually used for earth dam

or rock-fill dam construction. Therefore, the cost of the earth dam construction per unit

length (of the dam) is generally less than that of the concrete dam. For example, the cost

of a concrete work per unit volume in a concrete dam may sometimes be 20 times more

expensive than an earth work per unit volume in an earth dam (Singh, 1976).

A gravity concrete dam with the length of 500m and height of 100m may need at least 3

million cubic meters concrete and this volume of concrete needs about 1 rnillion ton

cement, which is usually an expensive material. Therefore, it is better to built an earth

dam instead of a concrete one because it is usually more none-economical and the

concrete can be used in the construction of other structures such as bridges, hospitals,

airports and buildings instead of using in dams. In conclusion, the readily availability of

the materials needed around the actual location of the earth dam, the compatibility of

earth dam with the environment, and the need for only simple technology have all gives a

better role for earth dams to be used and assisted in a reduction in the costs of earth dam

construction.

54


CLASSIFICATION OF DAMS

BASED ON CONSTRUCTION MATERIAL

BASED ON FLEXIBILITY

BASED ON USE

BASED ON STRUCTURE

BASED ON HYDRAULIC DESIGN _ ^ ^ _ _ _ _ _

CONCRETE DAMS

MASONRY DAMS

EARTH-FILL DAMS

ROCK-FILL DAMS

TIMBER DAMS

STEEL DAMS

MIXED DAMS

RIGID DAMS

NON-RIGID DAMS

STORAGE DAMS

FLOOD CONTROL DAMS

POWER

NAVIGATION

MULTI DAMS

PURPOSE

ARCH DAMS 1 BUTTRESS DAMS I

GRAVITY DAMS

COMPOSITE DAMS 1

OVER FLOW DAMS |

NON-OVERFLOW DAMS

MIXED DAMS Z3 Fig. 3.2.2.1 General classification of dams

55


3.2.3 Associated Facilities

Locks, power stations, spillways, fish ladders, baffle piers, side channels and outlet

works are the associated facilities which are normally necessary in the site plan of dams.

The costs of the associated facilities of an earth dam may be equal, or even more than,

the constructional cost of the structure. The type, shape and size of a dam influences the

location and position of power station. For example, a power station may be easily

constructed within the concrete dam, however, the construction of a power station

within an earth dam is usually a costly project.

The construction of spillway is necessary to control and regulate the outflow from the

reservoir. There may be for example free fall, side channel chute, tunnel or a glory

spillway. The constructions of spillways such as sharp crested, broad crested, or ogee

shaped on the top of concrete dams are usually recommended. These are impossible to

construct on the top of earth dams. Therefore, to discharge out-flow from the reservoir

of an earth dam, the construction of a separate spillway such as glory spillway may be

recommended.

3.2.4 Factors governing selection of a type

The shape of valley, the geological condition, the topography, the spillway location, the

foundation condition, the earthquake situation, the material availability, and, finally, the

comparative costs are factors dictating the type of dam to be constructed. For example,

the shape of dam is a function of the length and the height of the valley. If the height of

valley is more than 3 times of the length of valley and both abutments are formed from

rock or other high strength material, an arch dam may be suitable and economical, while

other types may not be economical and their constructions may even be impossible.

Gravity dams and buttresses dams are usually used in average valleys. Embankment

dams are usually suitable and economical for wide valleys with deep over-burden. A

type of composite section may be used in irregular valleys as shown in Fig. 3.2.4.1.

56


Fig. 3.2.4.1 A general view of a composite dam

3.2.5 Materials

Typical types of soil in or under dams and their properties (including: permeability, shear

strength, compressibility, workability, and sensivity to seepage and piping), based on

United States Bureau of Reclamation, 1974, are shown in Appendix B.

3.2.6 Design procedure

The purpose of dam construction, the location and type of dam, the necessary types of

material for dam construction and environmental considerations are all concerns in the

initial study of the dam. Factors governing design including availability of materials, the

diversion of river, the shape of valley and the characteristics of foundation should also be

considered at this stage. The design of associated facilities, the details of construction

stages and cost calculations should be evaluated in other stages of design.

Design consideration needs evaluation and investigation about the design parameters

involved. In earth dams, design considerations are divided into three groups: factors

influencing design, factors relating to the type of earth dam, and factors affecting design

details. Factors influencing design are: availability of materials for embankment

construction; characteristic of foundation; shape and size of valley; river diversion;

location of spillway; situation of spillway; probable wave action, earthquake activity and

availability of time for construction. Factors relating to the type of earth dam are: types

of alternative earth dams; shape and size of shells and core; downstream drains and

alternative sections. Factors affecting design details include: embankment side slopes;

57


internal and external stability; filter zones; embankment freeboard; crest width and

chamber.

3.2.7 Sections of Earth Dams

The sections of non-homogeneous earth dams are generally formed from core and shells.

Each non-homogeneous earth dam has an impervious zone called core within its body.

This plays an important role in preventing water leakage. Non-homogeneous earth dams

include the central core types or inclined core types. The central core types are usually

suitable for both large and small earth dams, however, inclined core types are usually

suitable for low earth types. The core is constructed using clay, silt, concrete or

asphaltic materials. Sometimes, the designers choose a thin core type because of

economical considerations and the availability of materials. In this type, the thickness of

the core is less than the others. A cross-section of a thin core type is shown in Fig.

3.2.7.1.

2.5

lv^' ^t^""^ Upstream shell

10

Thin core

1P* 2

f 1 \ ^\l 7 ; 1 \ Transitions^**^

\ Downstream shell ^ v .

Fig. 3.2.7.1 Cross-section of a thin core earth dam

The significant role of shells is as a protection to both sides of the core. The shells are

normally provided from local materials. The upstream materials should be provided

from the pervious material, because the water within the upstream shell should be

58


followed rapidly when rapid drawdown occurs. Otherwise, the upstream shell is in

danger of cylindrical or inclined sliding. Pervious material, semi-pervious material, semi-

impervious material or even random material is usually used as external shell material. It

is necessary to place a transition layer between the core and the shell when the shell

material is obtained from coarse materials. The location of a transition layer is illustrated

in Fig. 3.2.7.1.

In homogeneous earth dams, the role of core and shell are provided by the body of dam.

In this case, the material of dam is normally chosen from impervious or semi-impervious

types. Other types are not permitted. Therefore, the use of homogeneous earth dam is

usually un-economical, unless the particular homogeneous needed material is available.

Sections of earth dams are usually chosen based on foundation type and dam height

Based on the type of foundations, dams are divided into: impervious foundations,

shallow impervious foundations and deep impervious foundations.

Typical sections for impervious foundations, according to U. S. Army Corps of

Engineers, are shown in Fig. 3.2.7.2. A central core dam with two zoned shells is

suitable in large earth dams with impervious foundation. In this case, upstream and

downstream shells are usually formed from two zones. The internal shell zones may be

chosen from random materials. The central core type, suitable for high and moderately

high dams is shown in Fig. 3.2.7.2a.

The inclined core type can provide wider area in core. This type is usually useful in

small earth dams. The inclined core type, which provides a wider working zone in the

core for low dams, is illustrated in Fig. 3.2.7.2b. A homogeneous type, requires

relatively flat slopes. This is illustrated in Fig. 3.2.7c.

59


Fig. 3.2.7.2 Typical sections of impervious foundation of earth dams

In the shallow pervious foundations, the shape of the core is not different from that of

the impervious foundation. However, the core should cut the impervious foundation

layer. Cutting the impervious foundation layer is not necessary for the shells in these

cases. Three types of shallow pervious foundation dams, according to U. S. Army

60


Corps of Engineers, are shown in Fig. 3.2.7.3. A central core type, suitable to high and

moderately high dams, is shown in Fig. 3.2.7.3a. The inclined core type, providing a

wider working zone, in the core for low dams is illustrated in Fig. 3.2.7.3b. A modified

homogeneous type, is illustrated in Fig. 3.2.7.3c.

X P/R M\R \ p X.

Cut-off trench \ / Pervious stratum

(a)

/ P / Mf SM, SP OR P Nv


(b)

y^ M OR SM ^ v


(c)

M= impervious; P= pervious; SP= semipervious; SM= semi-impervious; R= random

Fig. 3.2.7.3 Typical sections for shallow pervious foundation of earth dams

61


In the deep pervious foundation dam, the core is connected to the impervious foundation

layer by a curtain wall. The curtain wall is usually chosen from clay, concrete or

asphaltic material. Typical sections for deep pervious foundations, according to U. S.

Army Corps of Engineers, are shown in Fig. 3.2.7.4. A central core type, an inclined

core type and a modified homogeneous type are shown, respectively, in Figures 3.2.7.4a

to 3.2.7.4c.

The range of slopes for the two shells of earth dams is determined, based on internal

friction coefficients of soil, types of materials and their unit weight, the plane zones of

sliding in the shells and a safety factor. The range of upstream slope is usually about

2.5-3 horizontal per 1 vertical. This range is usually about 2-2.5 horizontal per 1 vertical

in the downstream slope.

3.2.8 limitations of Conventional Earth Dams

Although, there are many reasons for preferring earth dams, the use of earth dams is

limited because of some restrictions including: weir limitation; spillway limitation; power

house limitation; outlet restriction and the large amount of material needed for

construction of conventional earth dams.

The crest of concrete dam is usually used as spillway in overflow conditions, however

the use of the crest on the top of an earth dam as spillway is impossible. The

construction of conventional earth dam is economical only if there is a suitable hill, to be

used as spillway, near the dam's location. Otherwise, a costly spillway arrangement is

needed to be built for the earth dam. This limits the use of earth dam in any location.

The construction of a power house in the body of concrete dams is possible, however

this is impossible in the body of earth dams. The power house needs to be located

independently. The separate construction of the power house increases the cost of dam

project.

62


Curtain wall Deep pervious

foundation

(a)

(b)

MORSM

Curtain wall Deep pervious

foundation

(c)

M= impervious; P= pervious;

SP= semipervious;

SM- semi-impervious;

R= random

Fig. 3.2.7.4 Typical sections for deep pervious foundation

63


Finally, another important problem connected with the construction of the earth dam is a

great amount of material needed. A concrete dam with the length of 1000m and height

of 100m needs about one million cubic metres concrete. However, the volume of a

conventional earth dam needs at least 22 million cubic metres of soil for the same

construction. This means that the volume of work in the construction of the earth dam

may be 20 times that of the construction of concrete dam. Therefore, the volume of

earth material in the conventional earth dams should be reduced.

3.3 REINFORCED SOIL DAMS

3.3.1 History of reinforced soil dams

As shown in Chapter two, the first RSD was constructed in the Bimes Valley near

Hyeres situated in the south of France. The dam was constructed with 9 m high vertical

downstream face using precast concrete facing units. A general view of the Vallon des

Bimes dam is shown in Fig. 3.3.1.1. A cross-section of the dam has been shown in

Chapter two (Fig. 2.3.7.2). Much larger dam, called the L'Estella Dam, was

constructed in Estelle with a maximum height of 29.5m as shown in Fig. 3.3.1.2.

The most complex dam built of reinforced earth is the Taylor Draw Dam, on the White

River in Colorado (USA), which is 380m in length, and with a flowrate of its spillway

which can reach 1850 m3/s (Reinforced Earth Company Brochures). The vertical

downstream side of dam has allowed construction of spillway on the top of its central

section which has been formed by reinforced earth. A core of impervious material was

used in the foundation to control the water penetration of foundation. T w o vertical

drainage zones, upstream and downstream drainage zones, were used in its reinforced

earth area to control the water penetration in its body. A thin layer of impervious soil

and a reinforced concrete slab topped the reinforced earth zone, to prevent water

penetration from the crest. It has been estimated that the construction of this dam by

reinforced earth could save about 1.5 million dollars (Reinforced Earth Company

64


Brochures). The general view, front face elevation and typical cross section of Taylor

Draw D a m are shown in Figures 3.3.1.3. and 3.3.1.4a & b, respectively.

Fig. 3.3.1.1 Vallon des Bimes dam (Reinforced Earth Company Brochures)

Filter

Impervious zone

i;ni;iiimninnimnn!

ft'*^y*YO':'i.i.i.i.i.i.:

29.5 m

^v

Fig. 3.3.1.2 L'Estella Dam (after Taylor & Drioux, 1979)

65


Fig. 3.3.1.3 A general view of Taylor Draw Dam (Reinforced Earth Company

Brochures)

1616.5 m 1616.8 m 1620 m

M -7—r-

(a)

16m 22.5m

Fig. 3.3.1.4 Front face elevation and cross-section of Taylor Draw Dam (after

Reinforced Earth Company Brochures)

Pells (1977) has reported that the techniques of using downstream zones of reinforced

rock-fill in the completion of three embankment dams (Bridle Drift, Xonxa and Lesapi

Dams) in South Africa were not successful. In the case of Xonxa D a m a major failure

developed, while at the Bridle Drift and Lesapi dams minor failures occurred when

66


floods passed over the partially constructed dams (Pells, 1977). The cross section of

Bridle Drift dam is shown in Fig. 3.3.1.5.a, its downstream elevation after the flood is

shown in Fig. 3.3.1.5.b. The cross section of the Xonxa dam and the reinforcing

system designed for the rock-fill at the Xonxa dam are shown in Figs. 3.3.1.6a & b,

respectively.

Fig. 3.3.1.5 a) The cross-section of Bridle Drift dam b)Downstream elevation after the

flood (after Pells, 1977)

Reinforced earth can also be used for increasing the height of existing dams using a

double-faced structure. A good example of this application is the earth dam at Lake

Sherburne in Montana (USA). This dam is 60 years old and rises to a height of 26m. In

1983 it was topped with a double-faced reinforced earth wall 7.3m wide and 350m

long, increasing the reservoir holding capacity to approximately 200 million cubic

metres. The reinforced earth solution was 3 5 % less expensive than the other methods

67


for raising the dam (Reinforced Earth Company Brochures). The cross section of dam

is shown in Fig. 3.3.1.7.

Fig. 3.3.1.6 a) The cross-section of the Xonxa Dam, b) Reinforcing system designed

(Pells, 1977)

Fig. 3.3.1.7 New section of earth dam at Lake Sherburne (Reinforced Earth Company,

1988)

68


The use of reinforced earth is also possible for earth dam restoration. According to the

Reinforced Earth Company, the uncertain condition of the Jamesville (a 100 years old

dam) in N e w York, was repaired by reinforced soil. The stability of this dam, was

improved because reinforced earth zone was added to the old dam, which had a height

of 15m and the capacity of 8,000,000, cubic metre in water retaining. The resulting

cross section of the dam is shown in Fig. 3.3.1.8 (Reinforced Earth Company

Brochures).

Reinforced concrete spillway cap

Heavy stone filling^5== Existing

dam

//>/// ////////////////

Bed rock

Fig. 3.3.1.8 New section of Jamesville, New York dam (Reinforced Earth Company

Brochures)

3.3.2 Other Investigations

According to Miki et al. (1988), a sequence of experiments were directed on test

embankments in order to establish a suitable design method for this type of structure.

The embankments were 3 m high with 1:0.7 slope, variable length and spacing of grid

laying as shown in Table 3.3.2.1 and Fig. 3.3.2.1. They were subjected to a severe test

of 15 mm/hr rain.

Three types of grid laying and three types of grid layers were used. The surface and

internal displacement were measured, using inclinometers and displacement gauges.

The foil strain gauges attached to the grids measured the strains on polymer grids. The

degree of saturation was determined from the moisture distribution inside the

embankment. The manometers inserted in the embankment measured the depth of

69


ground water. The properties of soil fill used within the reinforced embankment are


Table 3.3.2.1 Test cases (after Miki et al; 1988)

Test case

Case 0.0

Case 1.3

Case 2.3

Case 3.3

Case 3.1

Case 3.2

Case 3.3

Grid laying

length L

0

1

2

3

3

3

3

Grid laying

layer N

0

3

3

3

1

2

3

8m

2 m 2.1 m 3.64 m

3 m

\j s \> s

Foil strain gauge

s s s (a) Case 3.1

3 m

1 m

i 1 m /" i i i i i

Polymer grid i i i i

1 m

(b) Case 3.2

2.6 m

2.5 m

Fig. 3.3.2.1 Standard sections of reinforced embankments (after Miki et al; 1988)

The grids (Table 3.3.2.1 Cases 3.1, 3.2 and 3.3) have a length of 3m and different

number of layers: 1, 2 or 3, called Case 3.1, 3.2 and 3.3, were tested. The relationships

70


between embankment deformation, the strain distribution of grid and the degree of

saturation are illustrated in Fig. 3.3.2.2.

Table 3.3.2.2 The property offi.il material (after Miki et al; 1988)

Natural water content

Specific gravity

Gravel fraction

Sand fraction

Silt fraction

Clay fraction

Maximum grain size

Uniformity coefficient

Optimum moisture content

Maximum dry density

Permeability

22.4 - 24.3 (%)

2.7

1 - 2 (%)

70 - 74 (%)

12 - 20 (%)

9 - 12 (%)

4.76 (mm)

5.5 -15.9

17 - 18.6 (%)

1.64- 1.70 (t/m3)

1.5 -1.6 X 10 -4 (cm/s)

Case 1.3

Case 2.3

Case 3.3

Sr(%)

Sr(%)

Sr(%)

Case 3.1

Case 3.2

Case 3.3

Sr(%) - Degree of saturation

Sr(%)

Sr(%)

Sr(%)

Fig. 3.3.2.2 Relationship between embankment deformation, the strain distribution of

grid and saturation degree (after Miki et al; 1988)

71

http://offi.il


In the case of 1 layer used, the horizontal displacement of the top slope increased

quickly when the accumulated rainfall reached 110mm. The slope was eroded for a

depth of 5 0 0 m m at the time of 2 1 0 m m rainfall. In the case of both 2 and 3 layers, there

was only surface erosion without any sliding, the accumulative reached the final

rainfall of test. The basic equations that were used to evaluate the internal stability of

the reinforced embankment were:

M +AM FS = -* r-

M,

(3.1)

and

AM =Y(7\y.) r *->K riJiJ

(3.2)

where Mr and Md are, respectively, the resisting and driving moments of soil mass,

AMr is the resisting moment due to grid reinforcements, Tri is the pull-out resisting

forces due to /th layer of grid reinforcement, and y/ is the vertical distance of the ith

layer of grid to the centre of slip circle as shown in Fig. 3.3.2.3.

y. y

.o

y

O'

r

Fig. 3.3.2.3 Embankment section used in stability analysis (after Miki et al; 1988)

72


Tn is calculated by the smaller value of either the allowable tensile strength, Ta, of

grid, or the pull-out grid resistance Tpi which can be equated as:

r?.=2a/tan(J,L. {33)

where, ai is the vertical stress on the z'th layer grid, L; is the boarding length of /th layer

grid, and the constant 2 represents both sides of the grid.

The researchers were concerned that the safety factor, FS* (obtained by substituting the

value of Tji in Equation 3.22 with the tension, T, which is equal to multiplying grid

strain e; by stiffness J) is found to be smaller than the FS (obtained by Equation 3.21

with pull-out resistance force Tn). It was concluded that the grid and earth integrated

into a rigid body with a decreased deformation of the grid reinforcement. The

differences of both factors of safety can be calculated as:

AFS = FS - FS* =*ACIL (3.4) Md

where, L L and R are the length and radius of slip circle, Md is the driving moment of

soil mass, and A C is the increase of an apparent cohesion.

The rate of increase of apparent cohesion, Rc, was determined by the following

equation:

R =£±AC c C

where, C is the cohesion coefficient, and A C is the increase of apparent cohesion. The

results of computation for FS*, FS, and Rc for different types of reinforcement are


73

°0 Os h-1

•oo

Q to

-it!

to

K

B o So

#

Is. to

s

>w tn

1 to Q to

« is. 60

S K "•O

to

is.

•s. <>>

*->

-O

g

en cs

en en

cs CO

,-H

en

co

O t-» </3

en cs • * '

1-H

OO

in r—(

s

t->

c s o

H

en cs ,-H

oo

en i-H

Is

* - >

c 6 o

e 60 S3 • IH

* - > CO • -H CO .3

H

in 1-; in

in

cs

in

in

>n cs

in cs

in in

in

in

60

3

i

o

§ cs

1 cs

o o oo T—1

o O oo

© O NO

O O

o i-H

o o oo

o o VO

cs

o r-VO

to

VO

o ,-H c

•a s-t-> CO

T3 3 to

• *

oo ©

en r-©

VO in

©

"3-en ©

oo in

©

T_( en d

cs en ©

in

oo

d

ON

I—1

en

cs

en cs

i-H

/—*\

s ts» ^^ i.

• ^

3 °IH

60 O t->

<D 3 *J

3

E o 6 60 C t-> C/3

«iH CO

»H

o H

VO

en i-H d

es ON

es d

vo CS

d in

en d

en cs en d en oo cs d vo en cs d

vo en d

ON

VO CS

d

1"

s ea CO C/3

<D u, t-> 3 O

OH

in

d

sF §? 5 •S 60

1 «3

c:

cd

o 1—1

in i—H

OO

q

OO

q i—i

OO

q

c

3 tz:

3 CO

C3 D s c o CO

J3 *

cs 1—1

en cs 1—*

1-H

O N

q

cs en c o t->

3 cr >->

vo en d

o

d

en

en d

vo en d

^

vj

<

+ vj 3 O «.H CO

<U J-3 O

o 3

2 ea

<

es

q

cs cs

r-

q ,-H

vj

G <

+ II VJ

It was also concluded that in the testing, with the grids laying 0.75m vertically spaced,

Case 1.3, with the rates of laying length to embankment height L/H<0.33, the external

stability governs, and for Case 2.3 where UH>0.67, the internal stability governs the

overall stability of the embankment. As a result of interaction between the grids and

fill material, the deformation of the reinforced zone is decreased when the grids are laid

horizontally in several layers. Also, the analysis of the embankment as an elasto-plastic

body was done by finite element method analysis, and the results were in close

agreement with those of the tests. Therefore, it was concluded that the finite element

method is suitable for analysing the reinforcing mechanism in the embankment (Miki et

al., 1988).

Dean and Lothian (1990) used a geocell mattress, illustrated in Fig. 3.3.2.4, to

overcome problems encountered in the construction of a 9 m embankment over an area

of variable soft deposits. It was expected that the underlying soft layers would reached

plastic failure mode under the pressure of the embankment constructed without

reinforcement. In this case the embankment would not be able to bear the internal

strain and would fail in the centre. The application of geocell mattress was expected to

prevent the failure by reducing the settlement and internal stresses.

Fig. 3.3.2.4 Embankment with geocell (after Dean and Lothian; 1990)

75


Preventing slip failure and transverse rupture were also expected to be other benefits of

reinforcing the embankment base. However, the mattress did not behave as predicted.

It is still considered that the use of a geocell mattress is economical compared to other

solutions and it can reduce the time of construction (Dean and Lothian, 1990).

Koga et al. (1988a) used non-woven fabric nets and steel bars in 14 cases of model

shaking tests of embankments to investigate the seismic resistance of an embankment

constructed on an inclined ground. A steel box of 2 m high, 8 m long and lm wide was

used for the model of a bed slope. The properties of reinforcement used are shown in

Table 3.3.2.4 and the summarised condition of the parameters used are shown in Table

3.3.2.5.

Table 3.3.2.4 Properties of reinforcing elements (Koga et al, 1988a)

Type

Non woven fabrics

Plastic net

Steel bar

Properties

Nylon 70%, Polyester 30%, Thickness .2 m m

Polyethylene 100%, Grid 2.5 #2.5 m m

Pianowire, diameter 3.5 m m

The kind of reinforcement, the spacing between them, the slope surface gradient, and

the existence of benches on a bed slope were varied during the experiments. It was

assumed that Poisson's ratio v is 0 for reinforcements. Also, the reinforcement ratio, R,

which represents the ratio of strength increase of a reinforced soil to an unreinforced

one at a specified reference strain, was defined as:

R= —^ (3.6) a3QAH

where, £ 3 ^ is average horizontal tensile strain of the reinforced soil, E is the Young

modulus of reinforcements, t is the thickness of reinforcements, a30 is the horizontal

confining pressure, and A H is the spacing between the reinforcements.

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The tests were conducted under sinusoidal wave loading, starting from 4 Hz and 210

sec, while the acceleration was increased step by step. After each step, the tensile

force in reinforcement and the acceleration were recorded.

During the tests, the embankment model was sliding along a slip surface and its crest

was settling under the large acceleration. The deformation of reinforced embankments

were less than the unreinforced ones for some slope gradients and the value of crest

settlement became less when the spacing of non-woven fabrics became smaller. The

settlement of the embankment with plastic nets as reinforcement, which has larger

tensile stiffness, was less than that where non-woven fabrics in the same spacing was

used. The deformation of embankments also decreased when the reinforcements were

overlapped on their slope surface. The embankment settlement and deformation also

became less when the reinforcements were fixed to the bed slope. The deformation of

the embankments became larger when the slope became steeper (Koga et al., 1988a).

Fukuoka and Goto (1988) had investigated design and analysis of steel bars with anchor

plates used to strengthen the high embankment on soft foundation. A n embankment

was constructed (10m in thickness) on soft ground, mainly used for rice fields. The

steel bar reinforcing method was used to reduce deformation at the ground surface and

to strengthen the embankment.

The preconsolidation pressure a'p and effective overburden pressure a'v both increase

with depth. The dimensions of bearing plates used as anchors were 250x300x9 mm.

The diameter of the steel bars were 22 mm, placed at 500 mm horizontally and 600 mm

vertically. The properties of soil used in the embankment and foundation are shown in

Table 3.3.2.6, while the constants used for finite element method (FEM) analysis are

shown in the Table 3.3.2.7. The formulae and the resulting tensile force in the

reinforcements are shown in Table 3.3.2.8. Fig. 3.3.2.5 shows the observed values, and

the predicted values by F E M of stresses on steel bars. Fig. 3.3.2.6 shows the bearing

78

forces applied to the plates. These forces were calculated from the tensile forces on the

tensile bars.

Table 3.3.2.6 The property of foundation and embankment soil (Fukuoka and Goto,

1988)

Soft clay

Embankment

1

m

17

20

Unconfined compressive

kN strength qu (—~-)

mz

20 for 0 < Z <3m

20 + 6.6 (Z-3) for 3m < Z

_

Coefficient of

consolidation

(Cv'xl07)

3.3(m2/s)

_

C

kN

m

-

10

<t>*

-

30

Table 3.3.2.7 The constants used for finite element analysis (Fukuoka and Goto, 1988)

Unit weight of submerged soil (kN/m*)

Poisson's ratio v

Coefficient of earth pressure K

Coefficient of deformation E (MN/m2)

Embankment

20

0.3

0.43

10

As

10

0.3

0.43

Ac

.33

0.5

Therefore, based on experiments done by Fukuoka and Goto (1988), reinforced steel

bars with gravel compaction piles can be used to strengthen high embankments on soft

foundation and to reduce their displacement. The largest tensile force in the bars occur

at the lower layer and its ratio to that in the middle layer was about 2. The analysis of

reinforced earth embankment done by F E M was in good agreement with the results

from the field experiments (Fukuoka and Goto, 1988).

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Upper bar

Middle bar

Lower bar

7.0 m

Stress 10 r~

(kN/m2) 5

0

-5 -10

Upper bar

Stress 5 (kN/m 2) 0

-10

-20

-30

A——A

Middle bar

Stress Q -

(kN/m2) .10

-20

-30

-40

-50

Yield point (-35)

Lower bar

O Bank height 4.2m 122 days after

• Bank height 4.2m (Predicted by FEM)

A Bank height 7.0m 193 days after

A Bank height 7.0m (Predicted by FEM)

Fig. 3.3.2.5 The observed and the predicted by FEM values of stress on steel bars (after

Fukuoka and Goto, 1988)

82


7.0 m

Middle bar

Lower bar 4.2 m

Plate

Stress on

plate (kN)

10

0

-10

-20

-30

-40

-50

-60

~Q — >

—

-

¥ \

\

N. i

\

o- -

\'

.o'\

O"

1 JF

, Ny

\

^o~*

Middle

o—

Lower

Fig. 3.3.2.6 The bearing forces applied to the plates (after Fukuoka and Goto, 1988)

Koga et al. (1988b) used the finite element method to study the behaviour of reinforced

soil by a geogrid system with particular reference to an embankment on soft soil. The

geometry of the embankment and its material properties are shown in Fig. 3.3.2.7. The

use of geogrid within the embankment allows higher compaction to be achieved, hence

a reduction in the width of the embankment, and economical construction are the

results. The stiffness of the reinforced soil embankment may be increased and the

amount of settlement may be reduced by the use of a geogrid mattress in the

embankment.

The behaviour of the individual elements containing soil, reinforcement and interface

elements were analysed and the results were as shown in Figures 3.3.2.8 and 3.3.2.9.

Fig. 3.3.2.8 shows the settlement profile along a horizontal section in the subsoil at the

ground level. Fig. 3.3.2.9 shows the settlement profile along a vertical section at the

centre of the embankment. The vertical stress (Cy) and the maximum principal stress

83


distribution along the horizontal section at the top level of subsoil, below the

reinforcement, are plotted in the Figures 3.3.2.10a and b, respectively. As shown in the

Fig. 3.3.2.10a, the distributions of vertical stress for the three cases are the same. Also,

based on the results shown in Fig. 3.3.2.10b, the use of grid reinforcement reduces the

maximum tensile stress up to about 50%. The stress distribution in the reinforcement is

shown in Fig. 3.3.2.11. Therefore, provision of geogrid reinforcement reduces the

settlement profile.

15m

5m Clay E=300 t/m2

, 14m

1 E=10 t/m2

v=0.4 7

¥=1.8 t/m

v=0.45

82m

:L

2

Y=1.6 t/m

Gravel 9 E=15 t/m v=0.3 9

T=0.8 t/m

Fig. 3.3.2.7 Embankment (after Koga et al, 1988b)

10

Settlement 0

Distance from centre line

20 t 3p ty40(m) 12-^ Grid 2

Grid 1 No grid

Fig. 3.3.2.8 Settlement along a horizontal section in the subsoil at the ground level

(after Koga etal 1988b)

84

0 Settlement

0

Depth 10 _.

20

(m)

<m%

No grid

Fig. 3.3.2.9 Settlement profile along a vertical section (after Koga et al, 1988b)

Distance from centre line

10 20 30 40 (m)

0

Stresses

(t/m )

10 -

20

30 J

j . i

Gridl

No grid

(a) Vertical stress

Tension 10 -•

Stresses

(t/m ) 0

10 ~t

No grid

Grid 2

(m)

Gridl

Compression

(b) Principal stress

Fig. 3.3.2.10. Vertical and principal stress distribution (after Koga etal; 1988b)

85

EVALUATION OF SOIL DAMS

Tension

4000 -

Stresses

m2, °

4000 ~

Compression

" »--__Hjj^--s_

• — ^ o ^ m >

Lower 1 1

10

Middle

s W ^ _ y

i 20

Upper

^ T M ^ \ ^

"T -^^p— 30 (m) j

Fig. 3.3.2.11 Stress distribution of reinforcement (after Koga et al; 1988b)

The behaviour of a reinforced embankment on soft ground was investigated by Hird et.

al (1990). The numerical modelling of the embankment was evaluated using the

computer program CRISP. The geometry and finite element mesh of the embankment

is shown in Fig. 3.3.2.12.

Constant pore suction was assumed to exist within the embankment. A summary of

input parameters, including the property of the foundation and embankment soils, is

shown in Tables 3.3.2.10 and 3.3.2.11. Fig. 3.3.2.13 shows the effective vertical stress,

preconsolidation pressure, over-consolidation ratio, effective horizontal stress and

undrained shear strength within the foundation. As shown in Fig. 3.3.13.d, the

computed shear stress under undrained conditions is seen after the predicted strength.

The tensile modulus of the reinforcement, which was assumed to be linear elastic

material, was taken as 450 kN/m. The results of six analyses are shown in Table

3.3.2.11 and the distribution of the settlement of the original ground surface and surface

displacement are plotted in Figs. 3.3.2.14 and 3.3.2.15, respectively. At least 7 0 % of

the settlement occurred in the clay layer. Fig. 3.3.2.16 shows the distribution of

reinforcement strains and forces. It can be seen in the figure that the results of Analysis

No. 4 was in good agreement with pattern.

86

Fig. 3.3.2.12 The geometry and finite element mesh of the embankment (Hird et al,

1990)

Table 3.3.2.10 The properties of soil used in embankment (Hird et al, 1990)

Embankment

Fill

E' (kN/m2)

15000

•

V

0.3

C*

0

<&•

30

y(kN/m2)

20

Table 3.3.2.11 The properties of soil used in foundation (Hird et al, 1990)

Foundation

Peat

Clay

X

2.8

0.25

K

0.56

0.05

M

1.7

1.2

r

16.5

3.2

<

V

0.14

0.3

y(kN/m2)

11.4

16.2

87


Effective vertical stress (kN/m2) 0 W 20 30 40 50 60 0Y~

Depth (m) 6

8'

10-

Overconsolidation ratio 0 2 4 6 8 10 12

Preconsolidation Pressure

Beneath working 'lateform Depth

Outside working plate form

10-

Beneath working ' plateform

Outside working plateform

0

2

4

Depth (m)

6

8

10-

Effective horizontal stress (kN/m2) 0 10 20 30 40 50 60

1—T

Beneath working plateform

Depth


8

id-

i—r

Undrained shear strength (kN/m2) 0 5 10 15 20 25 30 0~

2

4 (m) 6


T

Beneath working plateform

Shear stresses computed by

CRISP

Fig. 3.3.2.13 Stress and strain profiles (Hird et al, 1990)

Table 3.3.2.11 The results of analysis (Hird et al, 1990)

Analysis

No

1

2

3

4

5

6

Embankment

Representation

Equivalent vertical loading

Equivalent vertical loading

Elements added in layers




Embankment

Suction

Not applicable

Not applicable

High

High

Low

Low

Foundation

Clay

Drained

Undrained

Drained

Undrained

Drained

Undrained

88


Toe Distance from embankment toe (m)

2.5 -Settlement (m)

No.l

Fig. 3.3.2.14 Ground surface settlement (Hird et al, 1990)

Distance from embankment toe (m)

0.5-

Horizontal displacement (m)

Fig. 3.3.2.15 Surface horizontal displacement (Hird et al, 1990)

The prediction of the results and the observations are summarised in Table 3.3.2.12.

Based on these results, the reinforcements appear to play a minor role in increasing the

stability of the embankment. Another analysis, conducted by later authors on the effect

of lack of the reinforcement, showed that although the horizontal displacement will be

increased up to 15%, the vertical displacement would remain constant (Hird, 1990).

89


Fig. 3.3.2.16 Reinforcement strains and forces (Hird et al, 1990)

Table 3.3.2.12 Summary of predictions and observations (Hird et. al, 1990)

Item

Settlement (mm) on centre-line

Maximum horizontal displacement of Inclinometer A (mm).

Depth (m)

Maximum horizontal displacement of Inclinometer B (mm).

Depth (m)

Maximum Reinforcement tension (kN/m)

and Strain (%).

Distance from centre-line.

Maximum excess pore water pressure (kN/m2) in peat (0-3.5m

depth)

Maximum excess pore water pressure (kN/m2) in clay (3.5-

10m depth)

Predicted

Value

2000 + 350

430 + 90

1.8 + 0.5

500 + 70

2.5 + 0.5

12.2 + 2

2.7 + 0.4

10 + 0.5

Assumed

zero

80 + 80

Observed

Value

2250

270

2.9

300

3

2.3

8

20

90

The effect of reinforcement in an embankment was predicted and analysed using the

finite element method and Biot's consolidation theory, and compared with the measured

one by Yin Zong Ze (1990). The cross section of the embankment, called Stranstead

Abbotts Embankment, is shown in Fig. 3.3.2.17, while the physical and mechanical

parameters of soils are shown in Table 3.3.2.13.

90


Grey clay,

k

Brown

Peat

Gravel

1m

Fig. 3.3.2.17 Stranstead Abbotts Embankment (Yin Zong Ze, 1990)

Table 3.3.2.13 Physical and mechanical parameters of soils (Yin Zong Ze, 1990)

Soils

1-Fill

2- Sand

3- Brown clay

4- Peat

5- Grey clay

7

kN

m 3

19

19

16

10.5

16

C'

kN

m2

20

0

0

0

0

0>

20

35

33

25

33

Cu

kN

m2

15

14

40

W

(%)

64-127

400 - 605

35-37

my

2

m . ^MN

0.24 - 0.87

1.08 - 6.33

0.38 - 1.27

Cy

MN

1.7 - 3.7

2.7-31.4

8.4 - 22

The elliptical and parabolical yield surfaces, based on stress-strain relationship, were

developed and the following equations were presented.

F+-hep,P vl a

(3.13) M, {{P+Pr) i-«j; vl

aq

G M2(P + Pr) -q)

(3.14)

where, P, q and G are defined as follows:

91

p_al + a2 + <T3

3

q = [(a, - a 0 ) + (a0 -a,) + (CT3-a1) '1 " 2 / , v " 2 "3-

(3.75)

(3.76)

G=KGPa^n

a

(3.17)

in which, Pa is the atmospheric pressure, KQ, n, h, t, a, Mj, M2 and Pr are parameters

determined from triaxial drain tests and for the embankment foundation soils. These

parameters and AT are shown in Table 3.3.2.14.

Table 3.3.2.14 Computation parameters of foundation soils (Yin Zong Ze, 1990)

Soils

Brown clay

Peat

Grey clay

KG 20

10

20

n

0.7

1

0.7

Pr 0

0

0

M]

1.5

1.6

1.5

M2 1.3

1.1

1.3

h

6

5

6

t

0.5

1

0.5

a

0.1 0.1

0.1

K .004

.01 .004

The thickness, the average Young modulus and the Poison's ratio of the grid was

assumed to be 3 mm, 150 MPa, and 0.3, respectively. The comparison of computation

results with the data measured on site, including vertical and horizontal displacement

distribution along ground surface, pore water pressure at point B, and tension

distribution in the grid are shown in Fig. 3.3.2.18 to 3.3.2.20, respectively.

As shown in Fig. 3.3.2.18, the computed and measured vertical and horizontal

displacements were close to each other. Fig. 3.3.2.19 shows that the results of the

computed pore water pressure were not in close agreement with the measured ones.

Fig. 3.3.2.20 shows that the measured tensile distribution in the grid was lower than the

computed one (Yin Zong Ze, 1990).

92


(a) vertical

-©-

(b) horizontal

End of construction

18 months after

computed measured

Fig. 3.3.2.18 Displacement distribution along ground surface (Yin Zong Ze, 1990)

u(m

40^

30-

20~

0 1

0

fm2)

f t

f i S * f t i *

J > ) t / t

J *

f * It It

\ Computed

i \

\\ Measured

1 S \ ^ S

V. ** x

1 1

20 40

i

60 t(day)

Fig. 3.3.2.19 Pore water pressure at point B varying with time (Yin Zong Ze, 1990)

93


Tension (kN/m)

20

10 — ^y^7 - • "

0

End of construction

18 months after

• o o

computed

° W\ o v\

measured ©

Fig. 3.3.2.20 Tension distribution in the grid (Yin Zong Ze, 1990)

3.3.3 Classification of Reinforced soil dams

An understanding of the general behaviour of RSD is necessary for the selection of a

suitable RSD for a particular site. Categorising RSDs allows for the classification and

recognition of their general behaviour. General and possible classifications of RSDs,

based on their construction will be considered in the following paragraphs.

3.3.3.1 General Classification

Based on the material used, RSDs can be classified into two main groups: homogeneous

fill types and zoned types. Typical cross-sections of a homogeneous fill RSD and a

zoned RSD are shown in Fig. 3.3.3.1. The components of RSDs are soil,

reinforcements and facing panels.

Zoned RSDs can be further divided into impervious upstream shell types and central

impervious core types. In the later, shown in Fig. 3.3.3.1a, the central core should

made from impervious materials such as clay, clayey silt or clay mixtures for water

retardation. Fig. 3.3.3.2 shows the cross-section of an impervious upstream shell dam.

94


Unreinforced shell

Impervious core

Filter

Reinforced shell

Filter

(a) Zoned RSD (b) Homogeneous fill RSD

Fig. 3.3.3.1 Cross-sections of a homogeneous fill and a zoned RSDs

Impervious upstream shell

Facing panels

Downstream reinforced soil shell

Fig. 3.3.3.2 A typical cross-section of an impervious upstream shell dam

Another classification can be made with respect to the form of the core, including

central core RSDs and inclined core RSDs. A central core RSD is illustrated in Fig.

3.3.3.3a, while a RSD with inclined core is shown in Fig. 3.3.3.3b. All these types can

be further divided into four groups which are, vertical upstream face, vertical

downstream face, vertical both sides and inclined both sides as shown in Fig. 3.3.3.4.

In the cases of the vertical face of dams, the use of facing panels is necessary to prevent

erosion of soil and to facilitate the connection of reinforcements.

95


Upstream shell

Reinforced earth shell Reinforced earth shell

Upstream shell

Inclined impervious core

(a)

Central impervious core

(b)

Fig. 3.3.3.3 A central core RSD compared to an inclined core RSD

CLASSIFICATION OF RSDs BASED O N THEIR

SECTIONS

H O M O G E N E O U S FILL TYPES

Vertical

upstream

face

Vertical

downstr

eam face

Vertical

both

faces

Inclined

both

faces

ZONED TYPES

IMPERVIOUS UPSTREAM SHELL TYPES

Vertical

upstream

face

Vertical

downstr

eam face

Vertical

both

faces

Inclined

both

faces

IMPERVIOUS INTERNAL C O R E TYPES

CENTRAL C O R E TYPES

Vertical

upstream

face

Vertical

downstr

eam face

Vertical

both

faces

ZL INCLINED C O R E TYPES

Inclined

both

faces

Vertical

upstream

face

Vertical

downstr

eam face

Vertical

both

faces

Inclined

both

faces

Fig. 3.3.3.4 A general classification of RSDs based on material used and cross-section

shape.

96


The foundations of RSDs can be classified as soft foundation and rigid foundation. The

behaviour of RSDs is different in these two situations. Based on the type of foundation

soil, RSDs are classified as shown in Fig. 3.3.3.5. Other classifications may also be

considered based on hydraulic design and use. For example, based on hydraulic design,

RSDs may be classified into over-flow types and non over-flow types similar to

conventional earth dams (for more detail see Chapter 1).

CLASSIFICATION OF RSDs BASED O N FOUNDATION SOIL

SOFT FOUNDATION TYPES RIGID FOUNDATION TYPES

S H A L L O W SOFT FOUNDATION

TYPES

DEEP SOFT FOUNDATION

TYPES

Fig. 3.3.3.5 A classification of RSDs based on their foundations

3.3.3.2 Possible Classification

Although most RSDs are of the gravity type, there is no reason to claim that reinforced

soil arch and buttress dams can not be built in the future. As an illustrated example, the

cross section of an imaginary reinforced earth arch dam is shown in Fig. 3.3.3.6. In

this case, reinforcement may stabilise the structure by increasing the strength of the soil

and connect the facing panels of two sides. Therefore, the RSDs can be, potentially,

classified as gravity, arch or buttress types as shown in Fig. 3.3.3.7.

CLASSIFICATION OF RSDs

GRAVITY TYPES

BUTTRESS TYPES

ARCH TYPES

Fig. 3.3.3.7 A possible classification of RSDs

97


*r —

TA

/r^i

m

H — y.~

Fig. 3.3.3.6 Cross-section of an imaginary reinforced soil arch dam

3.4 FORCES ACTING ON SOIL DAMS

A n understanding of the forces acting on soil dams is essential for the analysis and

design of the structures. Actually, there are no main differences between the forces

acting on RSDs and the forces acting on other types of dams. However, the behaviour

of RSDs and other dams in withstanding forces is different. The forces resulting from

water pressure, silt pressure, ice pressure, earthquake pressure, foundation reaction,

seepage and the weight of structure all act on a RSD. In subsequent sections these

forces will be discussed separately, and the combinations of the loads (including usual

loading, unusual loading and critical loading) will be described.

3.4.1 External water pressure

Upstream and downstream hydrostatic pressures (Vj and V-j) and the weight of water

on the upstream and downstream sides (Wj and W2) of soil dams are external water

pressures which act on the dam. A schematic diagram illustrating the external water

pressures acting on an earth dam is shown in Fig. 3.4.1.1. The weight of water on the

upstream (or downstream) side is zero when the upstream (or downstream) face is

vertical as shown in Fig. 3.4.1.2. The value of the downstream water pressure is low,

consequently the effect of downstream water pressure can be ignored during the

analysis of soil dams.

98


Fig. 3.4.1.1 External water pressure acting on an earth dam

Fig. 3.4.1.2 External water pressure acting on a vertical downstream face RSD

Referring to Fig. 3.4.1.1, the external water pressures are calculated as follows:

r2

V, Y #i

_ 'w 1 (3.18)

Y #o _ ' w 2

(3.19)

W, Y Hf

_ 'w 1 1 2tan0

(3.20)

1

Y #o 2 2tan0^

(3.21)

where H 7 and //2 are the depths of water on upstream and downstream sides,

respectively, 01 and 92 are the angles of upstream and downstream side slopes and yw

is the unit weight of water. Locations of Vl and Vl are respectively at Hj/3 and #2/3

99

from the bottom of the reservoir, and the locations of W] and W2 from the upstream

and downstream toes of dam (Xj and X2) are respectively calculated as follows:

H, _2_

3tan6, X2=TZ^~ <3-23)

3.4.2 Internal water pressure and seepage gradients

Seepage is the gradual motion of water through the soil causing driving force acting on

the particles of soil dams. If the value of seepage force (transmitted to a soil particle) is

greater than the resultant of the resistance force, an unstable condition occurs for the

particle. In conventional homogeneous earth dams, seepage appears on the downstream

slope regardless of the impermeability of the soil. The downstream slope is finally

affected by seepage to a height of approximately one third of water depth in the

reservoir (USBR, 1977).

The upper limits of seepage through two conventional homogeneous earth dams (one

with a horizontal drainage blanket and one without drainage blanket) and a non-

homogeneous earth dam are shown in Figs. 3.4.2.1a to 3.4.2.1c, respectively.

Similar to conventional earth dams, seepage occurs through the RSDs. The upper limit

of seepage lines through two homogeneous fill RSDs (one with a horizontal drainage

blanket, and one without drainage blanket) and a zoned RSD are shown in Figs. 3.4.2.2a

to 3.4.2.2c, respectively.

100


^

Seepage line

H2=Hj/3

(a)

Seepage line

Horizontal drainage blanket

Seepage line

Filter

H;

H2

(c)

Fig. 3.4.2.1 Seepage lines through (a) a homogeneous earth dam without any blanket (b) a homogeneous earth dam with a drainage blanket (c) a non-homogeneous earth

dam

Fig. 3.4.2.2 Seepage lines through: (a) a RSD without any blanket (b) a RSD with a

drainage blanket (c) a zoned RSD

101


On the basis of Darcy's law, the discharge of seepage flow in unit of time is proportional

to (a) the coefficient of permeabihty for the soil, k, (b) the hydraulic gradient, i, or the

rate of loss of head, dh/dl, and (c) the gross area of soil which the flow takes place. The

seepage flow, Q, is usually given as follows:

Q = kiA (3.24)

On this basis, the velocity of seepage flow, V, in unit of time is proportional to the

coefficient of permeabihty for soil and the hydraulic gradient as follows:

V = -ki (3.25)

The combination of Darcy's law with the continuity equation leads to:

—+— = 0 (3.26) dx dy

where u and v are the velocity components in both x and y directions. The Laplacian

equation of seepage for steady condition is usually formulated as follows:

___|+___| = 0 (3.27) dx2 dy2

where, ty=-kh is the flow potential.

It should be noted that the slope of seepage line through a RSD is normally steeper than

the slope of seepage line through a conventional earth dam (with the same condition),

because the base length of RSD is normally less than the base length of the conventional

earth dam as compared in Fig. 3.4.2.3. This causes an increase in the rate of loss of

head, dh/dl Hence, the hydraulic gradient, i, increases, because i=dh/dl. The increase

102


in the hydraulic gradient causes an increase in the seepage flow and seepage velocity,

because both are linearly proportional to the hydraulic gradient (See Eqs. 3.24 and 3.25).

A n increase in the seepage velocity through the RSD fill causes an increase in the driving

force acting on the particles of dams. For any particle of soil under the upper limit of

seepage line, the value of seepage force should be less than the value of resistance force

divided by an appropriate safety factor to avoid the occurrence of unstable conditions for

the soil particle.

Fig. 3.4.2.3 The seepage line through a RSD compared with the seepage line through a

conventional earth dam with the same height

Seepage may also emerge through the foundation of the conventional earth dams as

shown in Fig. 3.4.2.4. The motion of water through the foundation soil causes a force

which acts on the soil particles. Referring to Fig. 3.4.2.4, if the value of seepage force,

F3, (transmitted to the soil particle Q is greater than the weight, W, of particle C, an

unstable condition occurs for the particle C.

Similarly, seepage may also emerge through the RSD foundation as shown in Fig.

3.4.2.5. If the value of F3 is greater than the particle weight, W, an unstable condition

occurs in the downstream side.

103


H2 3

%

W Pervious foundation

Seepage line

~~ I B h - > — - -

(a)

Fig. 3.4.2.4 Seepage line through the foundation of a conventional earth dam

0

A

W 1

Pervious foundation Sgepage ^

(a) **

H2 rF3

c\-w

H2 l i s Seepage line- /W

(b) *lw

Fig. 3.4.2.5 Seepage line through the foundation of a RSD

104


Cutoff trenches are normally used to reduce the seepage force under the dams as seen in

Figs. 3.4.2.4b and 3.4.2.5b. The use of cutoff trench causes an increase in the length of

seepage line. The increase in the seepage line causes a decrease in the hydraulic gradient

i causing a reduction in the seepage flow and seepage velocity. Reduction of seepage

velocity causes a reduction in the driving force transmitted to the particle.

A comparison between the path of water under a conventional earth dam and under a

RSD with the same heights is shown in Fig. 3.4.2.6. Since the length of water path under

the RSD is normally less than the length of water path under the conventional earth dam,

as shown in Fig. 3.4.2.6, the hydraulic gradient (i-dh/dl) under the RSD is more than the

hydraulic gradient under the conventional earth dam. Hence, the seepage force under a

RSD is more than the seepage force under a conventional earth dam with the same

condition.

_..-_.. Reinforced soil dam

Hy

\Seepage line under reinforced soil dam / Vs. \ /

Conventional earth dam

Ho

Seepage line under conventional earth dam

Fig. 4.3.5.1 A comparison between the path of water under a conventional earth dam

and a RSD with the same height

Similarly, an increase of seepage velocity through the foundation of RSD causes an

increase in the driving force acting on the foundation particles. For any particle of soil

under the dam, the value of seepage force should be less than the value of resistance

force divided by an appropriate safety factor to avoid the occurrence of unstable

conditions for the particle. More detail will be presented in Sec. 4.3.5.

105

3.4.3 Uplift pressure

Uplift pressure is an internal water pressure which should be discussed here. Based on

the types of foundations, the magnitude of the uplift pressure may vary. The

commonly assumed distribution of the uplift force, acting under a dam, is shown in Fig.

3.4.3.1. The value of uplift water pressure and its distance to the toe of dam will be

equal to:

U = JUa + UhWh

2

X = _W*(2Ua+Uh)

6U

(3.28)

(3.29)

If the dam is built on impervious rigid foundation, as shown in Fig. 3.4.3.1, the values

of U a and U^, are calculated as follows:

a 'w 1 (3.30)

Ub = V * 2 <3-31>

in which W D is the width of dam at the base, and y w is the unit weight of water.

Fig. 3.4.3.1 Uplift pressure acting on an impervious rigid foundation dam

106

If the dam is built on pervious foundation, as shown in Fig. 3.4.3.2, the values of Ua

and Ub are calculated as follows:

U = a

H l-< gl- H2>*V,

ab

g, -(//, -H2)Lh b L

r w ab

L U = L +L\ + WU ab a b b

(3.31)

(3.32)

(3.33)

where, L a is the weighted distance from the beginning of upstream apron to the

upstream face of dam, Lb is the weighted distance from the beginning of downstream

apron to the downstream face of dam, Lab is the weighted length of path from the

beginning of upstream apron to the end of downstream apron, and y w is the unit weight

of water.

Hi ^^^m

Ua

1 h- 1 Vl 'h

~R2

ub x -

I I 1 1 1 Lfth 1 I 1

Fig. 3.4.3.2 Uplift water pressure acting on a pervious foundation dam.

3.4.4 Ice pressure

The magnitude of ice pressure can be calculated based on solar energy, ice thickness

and earth temperature. The table which shows the value of ice pressure, based on these

107


three factors, is illustrated in Appendix C. The position of ice pressure / acting on a

dam is shown in Fig. 3.4.4.1.

Fig. 3.4.4.1 Location of ice pressure acting on a dam

3.4.5 Silt pressure

According to the United States Bureau of Reclamation, the horizontal component of silt

pressure acting on a dam is assumed to be equivalent to that of liquids weighting 35

percent of the hydrostatic pressure, while the vertical pressure is equivalent to 90

percent. However, in silt retention dams, both components are calculated by Rankine's

formulas as follows:

V = s

y #2tan2 (45-|) 's s 2 (3.34)

y H2

' s s W =• s tan 9,

(3.35)

where Vs and Ws are horizontal and vertical components of silt pressure respectively, as

shown in Fig. 3.4.5.1, Hs is the depth of silt, $ is the angle of internal friction of silt

material, ys is the submerged unit weight of silt, and 01 is the angle of upstream side

slope.

108


Fig. 3.4.5.1 Silt pressure

3.4.6 Weight of structure

In homogeneous fill dams, the weight of structure per unit length of dam is calculated

by the multiplication of cross sectional area of dam by the unit weight of material as

follows:

W= Ay m

(3.36)

where, Wis weight, A is the cross sectional area of dam and ym is its unit weight.

In zoned types (Fig. 3.4.6.1), the weight of each area should be calculated separately.

The sum of the weights results in the total weight of the structure per unit length as

follows:

W. = A. y . i i ' mi

(3.37)

W= IW. (3.38)

where, W is the total weight of unit length of the dam, W( is the weight of each part of

dam cross section, and ym; is its unit weight. The location of W should be obtained

based on static equilibrium of the cross section.

109


•S^^H, y J—1

/^ Wi A JcV

—

Fig. 3.4.6.1 Zoned RSD

3.4.7 Earthquake force

The forces due to earthquakes are classified as direct and indirect (slashing). Direct

force is the result of inertia force of an earthquake acting on the body of dam, while

indirect force represents hydrodynamic forces acting on the upstream side of dam.

These two forces are evaluated in the following sections, while a comparison between

the natural frequency of RSDs and conventional soil dams will be presented in Chapter

Five.

3.4.7.1 Direct force

The static method and response expectra method are two methods for analysing the

direct effect of an earthquake acting on soil dams. In static analysis, it is assumed that

a horizontal force, equal to a portion of the acceleration due to gravity, acts on the

centre of gravity of dam. This force is calculated as follows:

F = (—)(ag) = Wa (3J9> 8

where W is the weight of dam, g is the acceleration due to gravity, and a is the

earthquake acceleration specified for the dam site and the surrounding area. The force

should be calculated separately for each zone of dam, when the cross section of dam

contains several zones. This force should also be checked separately for the horizontal

layers of dam. Generally in this method, the acceleration considered does not indicate

110


the duration of shaking (or the frequency of earthquake), which is usually necessary for

determination of the acceleration period and the natural frequency of dam.

In the response expectra method, the magnitude of earthquake acceleration, a, is

calculated with reference to the acceleration, frequency and duration of forces acting on

the dam. Based on the work done by Schnable and Seed (1983) and accepted by the

United States Bureau of Reclamation, the value of a is calculated for two situations: the

maximum credible earthquake (MCE) and the operating basis earthquake (OBE). Both

M C E and O B E are considered in the design of dams. The O B E is obtained based on

probabilistic and statistic approaches. There should be no permanent damage under the

OB E , and the dam should be able to resume operation after an earthquake. Under the

M C E , it should not cause the release of water from the reservoir (National Research

Council (U.S.) Panel on Regional Networks, 1990).

According to the National Research Council, the earthquake record for the region, the

length and the depth of all major faults, the types of foundation material (soil or rock)

and the distance of dam from the faults are parameters to be considered at the first stage

of the determination of maximum credible acceleration. The amount of earthquake

force, the maximum stress in the dam and the material strength required to resist these

stresses are considered as the second stage of this method. The magnitude of

earthquake acceleration, a, in the M C E and O B E are shown in Table 3.4.7.1 (National

Research Council (U.S.) Panel on Regional Networks, 1990). The numbers shown in

the first column of this table are usually represented on seismic zone maps.

Table 3.4.7.1 Earthquake acceleration

Region

0

1

2

3

MCE

o* o.i/? 0-2*

0-3*

O B E

o* .05*

o-i* 0.2*

Note: * = acceleration due to gravity

111


3.4.7.2 Indirect force

Referring to Fig. 3.4.7.1, the magnitude of horizontal earthquake force due to water

slashing is normally calculated from the following equation.

V =0.726 C Xy Y e p 'wJ

(3.40)

where, Ve is the magnitude of horizontal forces above the elevation considered, A, is the

earthquake intensity, y is the vertical distance from the reservoir water table to the

elevation considered, Cp is determined from the curve shown in Fig. 3.4.7.2 by

reference the values of y/h and a, which is the angle between the upstream slope and

vertical line shown in this figure, and the height of water in the upstream side of the

dam, h. The position of Ve is at 0.41y above the elevation considered.

Fig. 3.4.7.1 The horizontal earthquake force due to water slashing

3.4.8 Reaction of foundation

The linear reaction of foundation is usually assumed to have a trapezoidal distribution

as shown in Fig. 3.4.8.1. The upstream and downstream sides of the trapezoidal

pressure are calculated as follows:

/, o-iUL (1..5L) a W7 W,

(3.41)

112


b wb wb (3.42)

where, Z F V is the sum of vertical forces acting on the dam without uplift force, 7LFn is

the sum of horizontal forces acting on the dam, WD is the width of dam in the top level

of foundation, and e is eccentricity, which is a function of ZFy/ZFh and its location.

Fig. 3.4.7.2 The value ofCp (US Bureau of Reclamation, 1977)

Fig. 3.4.8.1 Trapezoidal reaction of foundation

113


Several non-linear foundation reactions may also be assumed for the distribution of

foundation reaction of RSDs. The exact reaction of foundation is not fully understood.

A possible non-linear reaction of foundation is shown in Fig. 3.4.8.2.

R a

Rb

Fig. 3.4.8.2 Possible non-linear reaction of foundation

3.4.9 Load Combinations

The combination of forces acting on a RSD should be considered in order to understand

the critical state of loading combinations. According to United State Bureau of

Reclamation (USBR), the three following combinations should be considered for

analysing the critical state of stresses and strains due to the forces acting on dams;

including usual loading, unusual loading and critical loading.

In usual loading, forces due to upstream and downstream hydrostatic pressure, ice and

silt pressure and the weight of dam are considered for purpose of analysing the

behaviour of dam. In this case, the level of water should be considered in both normal

and maximum situations. In unusual loading analysis, the maximum upstream

hydrostatic force, downstream hydrostatic force, weight of dam and the silt force are

considered. In critical loading analysis, usual and unusual load combinations are

analysed in consideration of the maximum force due to an earthquake. The cases of

load combinations m a y be summarised as shown in Table 3.4.9.1.

114

e o a •S sS

s o "a «

to

a *»i

Oo

Extreme loading

(case 2)

Extreme loading

(case 1)

Unusual

loading

em B

es

"3 S

D

• •

o • * *

s •a CA

O

x Cd

o

x cd

X es

UH"

o

X ed

(D Vo

3 co co

a o .—* Cst H—»

CO

O J-c

T3 > .3

s Co H-»

CO

a

CO

<D >*

CO

co 0H>

CO

<D

3-i

-3 CO CO

e a o cS H—»

co

o UH

"O Pi .3

s cd CO

3 O

Q

co

CO

CO

CO

e cd

o H—»

J3

X cd

so

o

X cd

X ed

UH'

O

x' ed

<D

•u

s ed O U H S J CO

D. 3 3 O UH

<D s—»

cd

o H—»

at

CO

><

eo

CO

><

CO

><

<t> -rH CO

e cd (D

£S CO

C O 3

o UH

<D H-*

ed SSH

O H-»

,3 M »»o

O

CO

><

O

CO

><

a 3 CO CO

a o o

CO

CO

CD ><

CO

<D ><

CO

(D UH

3 CO CO

K H->

CO

co <D

© So

o Us

o5 H->

I-"

&

ed 3 CT

3 B3

CO

CO

<D

O

o

eg o

• 1-4

T3 3

<D O UH

M ed 3 CT UH

ed

w



The construction costs of earth dams are much lower than these of concrete types such

as arch dams, buttress dams and gravity dams. However, the impossibility of spillway

construction on the crest of earth dams, the great amount of material needed for

constructing an earth dam, and the high costs of incorporating outlets and power

stations into the body of earth dams are restrictions which should be considered in the

selection of earth dams.

The insertion of reinforcement within earth dams reduces the restriction on the great

amount of earth work needed for construction of the earth dam. At least two RSDs may

be constructed using the material of one conventional earth dam with the same heights.

Reductions in fill volume, stress level, and displacement result from the use of

reinforcement inside earth dams. The structural flexibility, increase of safety factor,

elimination of downstream zone, reduction in upstream slope, and decrease in the time

needed for RSD construction should be considered as advantages of using

reinforcement in earth dam construction. The stiffness of RSD fill material is increased

due to the presence of reinforcement. The use of geogrid in earth dam construction

allows higher compaction to be achieved, resulting in a reduction in the width of earth

dam, and a more economic construction.

The design and analysis of reinforced earth embankments and soil dams (based on the

researches done so far) by the finite element method seems to be in a close agreement

with the observed behaviour. It is, therefore, an useful tool for further research in this

area.

The selection of an appropriate RSD for a particular site can be assisted by the use of a

classification system based on consideration of c o m m o n behaviour of dams. In this

chapter general and possible classifications of RSDs based on their components and

types have been considered in regards to external stability and internal stability

analysis. RSDs, based on the material used, have been classified into two basic groups:

116


homogeneous fill types and zoned types. Components, properties and types of both

groups have also been considered.

The identification of forces acting on RSD is fundamental to a study of the behaviour of

RSD. In reality, there are no differences between the forces acting on a RSD and the

forces acting on other types of dams. However, the behaviour of RSD and other dams

in withstanding the forces are different. The forces due to water pressure, silt pressure,

ice pressure, earthquake pressure, foundation reaction, seepage and the weight of

structure are the main forces acting on a RSD.

117

STABILITY ANALYSIS OF REINFORCED SOIL DAMS CHAPTER FOUR

CHAPTER FOUR

STABILITY ANALYSIS OF REINFORCED SOIL DAMS

4.1 INTRODUCTION

The stability analysis of RSDs, which may be classified as shown in Fig. 4.1.1, should

be accurately evaluated regarding its two main parts: internal stability and external

stability. Sliding, overturning and over-stressing should be carefully thought about in

external stability analysis. The failures due to reinforcement failure, and lack of bond

between the soil and reinforcement should be considered in the internal stability

evaluation. The minimum required base length for a no failure state due to sliding,

overturning, over-stressing should also be considered in order to optimise the geometry

of dam. In this chapter, the external stability of RSDs based on an analytical approach,

and the internal stability analysis based on semi-empirical methods will be evaluated.

It is assumed that the whole reinforced soil structure acts as a unit in external stability

analysis.

S T A B I L I T Y A N A L Y S I S O F RSDs

SLI]

EXTERNAL STABILITY OF RSDs

DING OVERTURNING OVER-STRESSING OF SOIL

INTERNAL STABILITY OF RSDs

REINFORCEMENT STABILITY

BOND STABILITY

Fig. 4.1.1 Stability analysis of RSDs

118


4.2 EXTERNAL STABILITY

Sliding, overturning, and overstressing should be considered in the external stability

analysis of RSDs. These should also be evaluated for each layer of the dam. For

evaluation, the cross-section of a parametric RSD with vertical downstream facing is

assumed to be as shown in Fig. 4.2.1, illustrating its cross section with imaginary

horizontal layers. During analysis, it is assumed that the height of the dam is a constant

parameter fixed with respect to the flow-rate of water in the river. The ratio of the crest

width to base width of the dam is assumed to be expressed by a constant parameter

h=Wt/Wb. For simplification, the average unit weight of the dam is assumed to be

another constant parameter called ys. Fig. 4.2.2 shows the forces, their locations and

their directions, acting on the RSD.

w t

M 1

AT 1

AT

/ 1

r

i

M 1

^ •^ ..

—

wb

\Hi H2 V

. A

Hi

r

3

H

*

Fig. 4.2.1 The cross section of a parametric RSD with imaginary horizontal layers

119


Fig. 4.2.2 Forces acting on a RSD

In the cases of sliding and overturning, the weight of the water, and the weight of silt,

both acting on the upstream side of dam (or a layer of dam as shown in Fig. 4.1), and

the weight of dam (or the layer), should be considered as resistance forces. The

upstream hydrostatic force, ice force, direct and indirect forces of earthquake and the

horizontal force of silt pressure should be considered as driving forces acting on dam

(or the layer). Downstream hydrostatic forces can be neglected in the stability analysis

of dam, while the weight of water acting on the vertical downstream side of dam (or the

layer) is zero. For sliding and overturning evaluations, the summarised states of the

forces under the four cases of load combinations including usual loading, unusual

loading, and the two cases of extreme loading (Case 1 and Case 2) are shown in Table

4.2.1.

120


Table 4.2.1 Summary of the forces in sliding and overturning states

Forces:

1-Upstream hydrostatic force

2-Downstream hydrostatic force

3-Uplift pressure

4-Weight of water acting on

upstream side of dam

5-Weight of dam

6-Ice pressure • *

7- Weight of silt

8-Silt pressure

9-Earthquake force (direct force)

10-Earthquake force (indirect force)

Neg. = Negligible

Vl

V2

U

Wj

W

Vl

Ws

Vx E

Ve

Usual

loading

Drv.

Neg.

Drv.

Res.

Res.

Drv.

Res.

Drv.

-

.

Unusual

loading

Drv.

Neg.

Drv.

Res.

Res.

_

Res.

Drv.

-

_

Drv. = Driving Force

Extreme

loading

(case 1)

Drv.

Neg.

Drv.

Res.

Res.

Drv.

Res.

Drv.

Drv.

Drv.

Extreme

loading

(case 2)

Drv.

Neg.

Drv.

Res.

Res.

-

Res.

Drv.

Drv.

Drv.

Res. = Resistance Force

4.2.1 Sliding

Failure due to sliding will occur, if:

where XZ>/ and IRi are, respectively, the sum of driving and resisting forces acting on

RSD dam, and p is the coefficient of friction between its layers, between the RSD dam

and its foundation, or between the layers of the foundation. Actually, u can be

expressed as:

a) p= tan fa between the layers of RSD dam

b) p= tan 5 between the dam and its foundation

c) p= tan fa between the layers of foundation

121

where <t>r and fa are, respectively, angles of internal friction of soil within the dam and

within its foundation, and tan8 is angle of friction between the dam and its foundation.

The RSD dam will fail theoretically if:

2ZD. ^ L > - ^ - (42)

£/?. SF ( ' i s

where SFS is a greater than 1 safety factor against sliding failure. Referring to Fig.

4.2.2, the sum of driving forces, £Df, the sum of resistance forces, ZJ?/, and the results

of the sum of driving forces divided by the sum of resistance forces, HDf/LRi, for the

four cases of loading, can be illustrated as shown in Table 4.2.1.1.

Table 4.2.1.1 Results of driving and resistance forces acting on RSD in sliding situation

•LD;

ZRi

x*.

Usual loading

Vl + Vl+Vs

W+Wi+Wg-U

V+VT + V l i s W + W.+W -U 1 s

Unusual

loading

Vi + Vs

W+Wi+Ws-U

V +v

W + W.+W -U 1 s

Extreme loading

(Case 1)

Vl+Vl+Vs + E+Ve

W+ Wi + Ws -U

V+VT+V +E+V l i s e W + W.+W -U

1 s

Extreme loading

(Case 2)

Vj+ Vs+ E+Ve

W+ W] + Ws -U

V.+V +E + V Is e W + W.+W -U

1 s

As can be seen from Table 4.2.1.1, the critical loading at the time of sliding, is Case 1

of the extreme loading because, in this state, the sum of the driving forces is maximum.

The sum of resistance forces in the all states is equivalent. This clearly shows that Case

1 of the extreme loading is the critical state of loading. Therefore, by replacing the left

side of Eq. 4.2 with the extreme loading, the following result (for no base sliding) will

be expressed:

122


W+W.+W • 1 s

-U

V+VT + V +E+V l i s e

> SF s V

(4.3)

in which SFS is a greater than 1 safety factor for no sliding, p is the coefficient of

friction expressed previously, and W, W], WSf U, V]f Vjt Vs, E and Ve are the forces

shown in Fig. 4.2.2 6.1b. Separation of the variables, which are a function of the width

of the crest, Wt, and/or the width of the base, WD, in the left side of the equation, results

in the following expression for a no sliding situation.

(W+W.+W -U)\i-SFE 1 £ *—>V+V +VT + V (4.4)

SF I s l e s

or

(W+W.+W -U)\i-SF E 1 S. _ §— > v (4.5)

SF s

where Vis the sum of all the horizontal forces acting on the dam except the direct force of earthquake. By factoring W from the left side of Eq. 4.5, the following expression

can be written.

(l + P+X-a>)p-SFa W s— > V (4.6)

SF s

where, E

W (4.7)

p . V f o - ^ w (4.8) W H2(\+%)y

123


Ws "s^-^sub X = W = H2{1 + ™

y} = ILjhw + hw2)'Yw (4J0) W H(l + )ys

( ;

where, W, is the weight of RSD dam (or the layer), Wj and Ws are, respectively, the

weight of water and silt on its upstream side (or the layer), E is the direct force of

earthquake acting on the dam (or the layer), U is the uplift pressure acting on the dam

(or the layer), ys is the average unit weight of the dam (or the layer), yw and ysub are,

respectively, the unit weights of the water and the silt on the upstream side of the dam

(or the layer) and, finally, H, hw and Hs are, respectively, the height of dam (or the

layer), the height of water and the height of the silt on the upstream side of dam (or the

layer).

Assume;

(l + P + X-x>)v.-SFa m = — (4.11)

SF s

Substituting Eq. 4.11 in 4.6, yields the following:

mW>V (4.12)

Since,

(Wt + Wb)Hys Wh(l + Wls (4J3) 2 2

substituting W from Eq. 4.13 to Eq. 4.12 and solving for Wb results in the following

condition for a no sliding failure state:

V i b Hy

(4.14)

124

where £l\ is a factor expressed as the following equation:

^ ((l + ^+X-v)[i-SFa)(l + ^) Ql= ISF— (4J5)

s

where p, %, D, p, SFs, £ are factors expressed before. Since, these are dimensionless

parameters, Q.\ is a dimensionless factor in conventional static analysis (because

earthquake acceleration, a, is a dimensionless factor in such analysis), while in

dynamic analysis, €1\ linearly depends on the earthquake acceleration, a.

4.2.2 Overturning

Referring to Fig. 4.2.2, in stability analysis of RSD against overturning, the sum of

driving moments and resistance moments in the critical state of loading should be

calculated. In each layer, the point of rotation is the common point of the base line of

the layer with the vertical line of downstream facing. The sum of the driving moments

divided by the sum of the resisting moments, in the four cases of loadings, are

calculated and compared in Table 4.2.2.1.

Failure of RSD dam due to overturning occurs, if:

ZD.h. l-±>l (4.16)

ZRx. i i

Regarding to the safety factor for no overturning failure, the sum of overturning

moments divided by the sum of resistance moments should be less than the inverse of

the safety factor.

X D.h. i ±—LL<-±- (4.17)

ZR.x. SF i i o

where SF0 is a greater than 1 safety factor for no overturning failure.

125


Table 4.2.2.1 Results of driving and resistance moments acting on the dam in

overturning situation

Loading case

Usual loading

Unusual loading

Extreme loading (Case 1)

Extreme loading (Case 2)

The sum of driving moments divided by the sum of

resistance moments

(iD.h.) ^ i i

V i i)

V fu + V7 hT + V h + Ux 1 1 II s s u Wx + W.x, + W x

11 s s Vh, + V h + Ux 1 1 s s u Wx + W.x, + W x

11 s s V h, +VThT + V h +Ehr, + V h +Ux 1 1. I I s s E e e u Wx + W.x,+W x

11 s s Vh,+V h +Ehr7 + V h +Ux 1 L s s E e e u Wx + W. x.+W x

1 1 s s

As can be seen from Table 4.2.2.1, Case 1 of the extreme loading is the critical state of

the load combinations, because in this state, the sum of driving moments is maximum

while the sum of resistance moments is constant. Therefore, for no overturning failure

the following expression should be met.

Wx + W.x.+W x 11 s s

Vh,+VThT + Vh +Ehr + Vh+Uxit 11 II s s E e e u

>SF (4.18)

Separation of the variables, which are functions of the width of crest and / or base of

the dam in the left side of the above, results in the following condition for no

overturning failure.

Wx + Wxxx+Wsxs -(EhE +Uxu)SFo>SFoMh (4.19)

126

in which,

Mh -<v 1vv JVv.VW (4-20)

Since the safety factor against overturning failure, SFD, is a positive number, the

following expression should be met for no overturning failure.

Wx W U , W J C . JE%„ Hr [ 1 + — ^ + - £ - £ - ( — E - + — ^ ) S F ]>M, (4.21)

SF Wx Wx Wx Wx o h o

Referring to Fig. 4.2.2, the horizontal distance from the centre of gravity of the dam (or

the layer) to its toe, x, and the horizontal distances from the centre of gravity of the

water and the silt, both on the upstream side of the dam (or the layer) to the toe of dam

(or the layer), xj and xs, can be calculated as follows:

x = W2+WWU+W

2 WAl+^+Z,2) - t t b b _ b ^ T (4 22)

3(Wf + Wb) 3(1 + $)

(l-c> x,=WAl- w) (4.23) 1 bK riff

3J7

(l-x)h x =WA1- -&•) (4.24) s bK 2>H

The distance of the result of uplift pressure, xu, to the toe of the dam is shown in Fig.

4.2.2. The ratio of the horizontal distance of the centre of gravity of the water on

upstream side, the ratio of the horizontal distance of the centre of gravity of the silt, ,

and the ratio of the horizontal distance of the result of uplift pressure, all to the

horizontal distances of the centre of gravity of dam, (Pi, X\ and v\) can be calculated

as follows:

x, (l + %)QH-hw+h^) (425)

1 x (l + ^+%2)H

x (l + x)(3H-h +h$) tJ^ y = s ___K s s*J (4.26) 1 x (\+%+%2)H

127


X (l + %)(2h +h )

x a+$+e)(hw+hw2)

Substituting the values of x, pi, X\ and -oi from Equations 4.22, 4.25, 4.26 and 4.27 in

Equation 4.21 results in the following expression for no overturning failure state.

WAl + %+^2)(l+W +XX -vx>.SF -ahFSFnlx) W-£ i -1 1—Q £—Q > M, (4.28)

3(1 + |)SF h

where hE is the vertical distance of the centre of gravity of dam to the base level, x, x],

xs, are horizontal distances shown in Fig. 6.2, and a, 0 and % are the parameters

calculated from Equations 4.7 to 4.9. Substituting the value of W from Eq. 4.13 in 4.28

results in the following condition for no overturning failure state:

r2, W, £(l + l; + ¥)(l + Wl + XX1-vv1SFo-ahESFo/x) Mh (4.29)

6SF Hy o s

Referring to Fig. 6.2, the value of hE can be calculated as follows:

2HW+HWU H(2c\ + \)

h 1 b_=H(li; + l) Q) 3(Wt + Wb) 3(^ + 1)

Substituting the values of x from Eq. 4.22 and hE from Eq. 4.30 in 4.29, and solving

for W&, results in:

m^W2 + m2Wb + m^ > 0 (4.31)

where,

(l+jc+x2)(l + j3p1 +%%i -x>v, SF ) tr,=: - — M H 1 A A 1 I—QL (4.32) 1 6SF

o

128


aH(l + 2t) m2=

y — — (4.33)

Mu m3=--^

tL- (4.34) Hy s

in which Mh is calculated from Eq. 4.20, m$ is a negative value depending on the

moments due to the forces acting on dam, Mh, the dam height, H, and the unit weight,

ys of the dam. Since £;, P, Pi, x, Xh v, t)l and SFo are dimensionless positive values as

explained before, mj should be a dimensionless value. Finally, based on Eq. 4.33, m2

is a negative value, because a, H and % are positive values. Since, H was assumed to

be a constant parameter for a particular site and £, is a dimensionless constant

parameter, m2 is a parameter which linearly depends on earthquake acceleration, a.

Therefore, m2 is a constant factor in conventional static analysis, while in dynamic

analysis W 2 depends linearly on the earthquake acceleration, a.

Solution of 4.31 for W b results in the following equations:

-mn +J(mn -4m.m~) W = 2 V 2 1_3_ (435)

bl 2mx

~ m 2 ~\^m2 ~^m\m'x) W = ___________ ______ (4.36) b\ 2nu

Since m2 and m j are negative values, Wbl is always a positive value and WD2 is

always a negative value when ml is a positive value. Therefore, the correct solution of

4.31 is only Wbl calculated from Eq. 4.35, because Wb should always be a positive

value. Therefore, for no overturning failure state, the following condition should be

met.

- m 0 +J(m~ -4m,m-.) W =

2 V 2 1 _ _ (4J7) b 2mx

129


4.2.3 Overstressing

Based on the type of soil used, the reaction of the foundation may change. Actually,

the form of foundation reaction force is not exactly understood. Linear distribution and

several types of non-linear distribution may be assumed. The linear reaction of the

foundation of a RSD is shown in Fig. 4.2.3.1a. T w o simple forms of non-linear

reactions of foundation are shown in Fig. 4.2.3.1b & c.

miuun. j.muuu-*a

Wi

R

(a) (b) (c)

Fig. 4.2.3.1 Reactions of foundation

The following horizontal and vertical forces are considered in overstressing analysis

including: upstream hydrostatic force, force due to silt pressure, force due to ice

pressure, the two forces of earthquake (direct force and indirect force), weight of dam,

weight of water on the upstream side in normal and maximum situations, and weight of

silt on the upstream side. These should be considered in the four loading cases (usual,

unusual and two cases of extreme loadings). The role of these forces acting on a dam

for overstressing analysis is shown in Table 4.2.3.1, while direction and location of the

forces were shown in Fig. 4.2.2.

130


Table 4.2.3.1 Summary of the forces used in analysis of soil bearing capacity

Forces:

1-Upstream hydrostatic force

2-Downstream hydrostatic force

3-Weightofdam

4-Weight of water acting on upstream side of dam

5- Weight of water acting on downstream side of dam

6-Ice pressure

7-Weight of silt

8-Horizontal force due to silt pressure

9-Earthquake force (direct force)

10-Earthquake force (indirect force)

Vl

V2

W

Wi

W2

VI

Ws

Vs

E

Ve

Usual loading

Driv.

Neg.

Driv.

Driv.

0

Driv.

Driv.

Driv.

-

-

Unusual loading

Driv.

Neg.

Driv.

Driv.

0

-

Driv.

Driv.

-

-

Extreme

loading

(case 1)

Driv.

Neg.

Driv.

Driv.

0

Driv.

Driv.

Driv.

Driv.

Driv.

Extreme

loading

(case 2)

Driv.

Neg.

Driv.

Driv.

0

-

Driv.

Driv.

Driv.

Driv.

Neg. = Negligible Drv. = Driving Force

4.2.3.1 Linear Reaction

Referring to Fig. 4.2.3.1a, and based on the equilibrium of the forces, the two following

equations can be written:

wh

(R+Ru)-?- = ZR, a V 2 R W£ (R,-R )WU2 ___£__£_+_____£—aL-b_+1D h =2-R x

2 6 l l l l

(4.38)

(4.39)

where Ra and Rb are, respectively, the upstream and downstream values of foundation

reaction and £_?i, L/?f, xi, and ID; hi are the sum of vertical loads, the sum of

resistance moments and the sum of driving moments, respectively. The solution of

131

Eqs. 4.38 and Eq. 4.39, yields Ra and Rb being the upstream and downstream sides of

the foundation reaction as following equations:

(6ZR.x.-2Wh2ZR.-6ZD.h.) R = LJ OIL Li_ (4A0) a Wb2

(-6ZR.x.+4Wuj:R.+61D.h.) R = LJ ____! L±- (4A1) a Wb2

For no overstressing failure, the values of Ra and Rb should both be a positive number

smaller than the ratio of allowable bearing capacity, R*, over the factor of safety, SFb-

Therefore the following expressions should be met.

0<R <— (4.42) a SFb

0 <RU<— (4-43) b SFb

Substitutions of Ra and Rb from 4.40 and 4.41 to 4.42 and 4.43, respectively, yields the

following expressions:

-2 l— ZR.x. + ZD. h.<0 (4.44) 3 — i i — i i

* WUZR. W2R ,, _ - b l+Y,R.x.-<ZD.h.—2 <0 (4.45)

3 i i ' * 6SFb

b±-L+2^R.x..^D.h. <0 (4.46) 3 — i i — i i

2W,_/.. whR* ,AA^ h l-ZR.x.+lD.h.--b < 0 (4-47) 3 i i * i i 6SFb

132

Solution of the set of conditions (Eqs. 4.44 to 4.47) results in the following expressions

for no overstressing failure:

W2R* —- > 0 (4.48) 6SFb

WUJ,R. —- l- > 0 (4.49)

3

W.^R. 2W2R* --2 l- - < Q (4.50)

3 6 ^

Conditions 4.48 and 4.49 are always met because all terms (Wb, R* and -_J?j) of the

equations are positive. However, for fulfilling condition 4.50, the following condition

should be met.

W, > b l (4.51) b /?*

Substitution of the sum of vertical loads acting on the dam, YRi, from Table 4.2.1.1 to

Eq. 4.51 results in:

(W+W.+W)SFU W > : 1 _____ (4.52) b /?*

or,

yq+p+x)-. (4J3) b R*

where p and % are dimensionless factors, respectively, shown in Equations 4.8 and 4.9.

Substituting Whom Eq. 4.13 in 4.53 results in the following equation for no failure.

Wb(l-p)>W{p (4-^4)

where

133

HySF(l+$+x) p = —s__o (455) V 2R* ' '

4.2.3.2 Non-linear Reaction

Similar to the above procedure (linear reaction), other procedures are evaluated based

on the non-linear reaction of foundation. Referring to Fig. 4.3b and 4.3c, the

foundation reaction is assumed to be a parametric polynomial curve of two degree as

follows:

R = ax2 +bx + c (4.56)

where R is the function of reaction, and x is the distance from the left side of the dam at

base level as shown in Fig. 4.3b and c. After calculation, it can be found that the

following equations should be met for fulfilling the vertical equilibrium equation.

-6HR. 3R 3RU a = —^r

±+—4-+—%- (4.57)

w? wr w£ b b b

61/?. AR 2RU b= l &- &• (4.58)

W2 WL W, wb b b c = R (4.59) a

where Ra and Rb are, respectively, the upstream and downstream value of foundation

reaction, and X/?/, IRj x(, and XD; hi were shown in Tables 4.2 and 4.3.

The following equation should be met for the fulfilment of moment equilibrium

equation around the toe.

aW* bwl cW2 tA</.. — - £ - + _ _ _ - + _ k - ^ R . x . - Z D . h . (4.60) 12 6 2 l l l l

There are five unknown variables (a, b, c, Ra and Rb) in four equations (Eqs., 4.57 to

4.60). Another equation is needed for the solution of the set of equations. The fifth

134

equation can be written to describe the shape of the reaction of foundation. For

example, the following conditions can be written for Figs 4.3b, 4.3c, respectively.

dR For x = 0: — = 0 -» fo = 0 (4.61a)

dx For x = Wu: — = 0 -> b = -2aWu (4.61b)

b dx b

The solution of the set of equations (Eqs. 4.57 to 4.60, and 4.61a) leads to the

following results.

-TRi 4(2ZR-x.-lD.h.) \ = ^ + ll

2 ll (4.62a)

a Wb Wb

5Y Ri S(2ZR.x.-2ZD.h.) R = ______ J___I_L^1____ (4.62b) b Wu w2

b wb Also, the solution of the set of equations (Eqs. 4.57 to 4.60 and 4.61b) leads to the

following results.

-4YJ?,' 8(X#.x.-_D.fc.) R = * ± R i + ^ i i ^ n> (463a)

a Wb W2

».2__.____(_^5l_2 (4.63b) b 2Wb W

2

Substituting the values of Ra and Rb, Equations 4.62a and b, in Equations 4.42 and

4.43 results in the following conditions for no failure:

-YRi ACZR:X.-2ZD.h) R* 0<R _ _ _ _ _ + i-l L _ _ < _ — (4.64)

a Wu Wn SF, b b2 b

0 </.,= — LJ--5 —- <-—- (4.65) b wt rf SFb

135

Solution of 4.64 and 4.65 leads to:

CLR:X.-lDh) R* 0< L L , — L J - < T T — (4-66)

W2 2SFb

X*, R* 0< L< (4.67)

W b SFb

The left side of 4.67 is always met, and for meeting the right side, the following

condition should be considered:

SFUZR-Wu >—-

l- (4.68) b R*

For complying with 4.68, which is similar as 4.51, the following expression should be

met:

Wb(l-p)>Wfp (4.69)

where p has been defined by Eq. 4.55. Also, for meeting Eq. 4.66, the two following

conditions should be concerned for no failure:

(__/?.*.-XD./j.) 0< l l , l l (4.70)

W£ b

(lR.x.-TD.h) R* i i < _ L ; — (4.71) W2 25F, wb b

ZR.x. For meeting 4.70, the ratio of resistance moments over driving moments, — ^ should

i i

be greater than 1, which has been fulfilled for no overturning failure (See Eq. 4.16).

Also, for fulfilling 4.71, the following condition should be met:

\Yl > LJ i i b (4.72) b R*

136


4.3 INTERNAL STABILITY

The internal stability of reinforced earth structures can be analysed using methods based

on conventional principles of soil mechanics (Rankine and Coulomb-type of analysis). It

has become known, however, that certain theoretical assumptions of these methods have

not been supported by observations. In particular, since reinforcement changes the state

of stress within soil, the direction of principal stresses are no longer vertical and

horizontal, and the ratio of the vertical stress to the horizontal stress is not constant

(Arenicz & Chowdhury, 1987). This, together with other field data regarding the bond

between soil and embedded reinforcement, has led to the development of semi-empirical

methods, with one of them (proposed by McKittrick and Schlosser in 1978) being

adopted as a recommended design method by the Reinforced Earth Company. The

method, the Coherent Gravity Method (CGM), was structured around a set of bi-linear

functions, which represent and interpret field data in a simplified manner. Some

modifications to this method were suggested by Arenicz and Chowdhury in 1987,

Modified Coherent Gravity Method (MCGM), to reflect field observations more closely.

Factors of safety against tensile and bond failure of reinforcements are needed for design

purposes. The apparent friction factor, the coefficients of lateral earth pressure, and the

maximum tension line are needed to establish the safety factor formulae. The field data

on which both the CGM and the MCGM are based, is re-examined and analysed in order

to assess and reduce discrepancies between methods; their assumptions and field

observations. Therefore, the apparent friction factor, coefficient of lateral earth pressure

and the maximum tension line will be considered here. O n this basis, new empirical

formulae will be proposed for design purposes.

The assumptions accepted in CGM are that,

(a) the failure surface of the reinforced earth model is of a bi-linear shape

originating at the toe of the wall,

137

(b) the maximum force in the reinforcement occurs at some distance from the

facing panels,

(c) the coefficient of earth pressure varies linearly between Ko at the top of wall

to Ka at the depth of 6m, and

(d) the friction factor between the soil and reinforcements varies between fo near

the top to/* at 6m depth.

4.3.1 Coefficient of Lateral Earth Pressure

In the CGM, the coefficient of lateral stress is assumed to be a bi-linear function of the

fill depth, as shown in Fig. 4.3.1.1. It can be seen that, for y<6m, the lateral earth

pressure coefficient varies linearly from Ko to Ka and, for y>6m, it remains constant and

equal to Ka. This has been formulated as Eq. 4.73a & b, suggested by Schlosser (1978).

K = \ ' '' +Ka for, y<6m 6 ' " 0 JU" ^yj"y (4.73)

K for, y > 6m a J J

where K is the coefficient of the lateral earth pressure, Ka and Ko are the coefficients of

the lateral earth pressure in active and at rest conditions, respectively, and y is the depth

of soil fill above the level considered. A comparison between Eq. 4.73 (for § = 45) and

the field data of the experiments is shown in Fig. 4.3.1.2.

The main problem regarding the Schlosser equation is that the results of observations

indicate a non-linear change of K with fill depth (Baquelin, 1978, Arenicz & Chowdhury,

1987). The tangent discontinuity of Eq. 4.73 at y=6m can not be justified in terms of

stress state in a non-stratified soil fill, nor can it be supported by the field data.

138


T y = 6m \

y

Ka K0 ^ K 1

/

Fig. 4.3.1.1 Coefficient of lateral earth pressure

2.5-

2.0-

1.5-

K/K a

i.o-

0.5-

o.o -\ t

•.

N

)

Schlosser's Formula

. [ •

10

Depth (m)

t

20

• Asahigaoka

• Granton

• Gringy

* Lille |

• Silvermine

• (7c/a j

* Vicksbourg

Fig. 4.3.1.2 Comparison between the formula (for §=45) and the results of observed

experiments

To eliminate discontinuity and allow for a non-linear change of K, as observed, the

following function was proposed by Arenicz and Chowdhury (1987):

K = dlKa + $y(d2K0-dlKa) (4.74)

139

where, P is a constant equal to 0.75, d} and d2 are dimensionless coefficients that

change depending on boundary conditions. The previous equation was derived assuming

a non-linear decrease of K from KQ for y=0 to Ka for y = °°, with the lower limit

effectively reached for y=7m (Arenicz & Chowdhury, 1987).

There is, however, a practical problem associated with this proposal. In order to be used

for design purposes, the two unknown parameters it contains (dj and d_) have to be

determined. Based on the comparison between the theoretical and observed variation of

K/Ka with the fill depth, illustrated in Fig. 4.3.3.1, both dj and d2 have been suggested

to have a value of 1. On the other hand, different values have been suggested to

conform to the measurements taken in Lille abutment (d]=0.25 & d2=1.92) and

Dunkerque wall (dj=0.6 & d2=2.8). Therefore, Eq. 4.74 cannot be used for design

calculations since the actual determination of the two parameters for such a purpose

have not been addressed.

The analysis of the results of the field investigations suggests that, in order to eliminate

some of the problems described above, Eq. 4.74 should be altered to:

Ka+$yd1(K0-Ka) for, y<6m

„=J ' (4.75)

K for, y>6m a

v.

where,

<L --i-I+l (4.76) 1 36 3

and, p is the constant equal to 1.2.

The assumptions accepted in the new formulae are the same as these accepted in the

MCGM, except that for any depth of backfill exceeding 6m (instead of 7m) the value of

K remains constant. This gives a better agreement between the proposed function and

the interpolated average of observed data than in the MCGM, as shown in Fig. 4.3.1.3.

140

2.0-

KJKa ' i

1.6-

1.2-

f) S-

• The average of interpolated data of the experiments

Schlosser, 1978

- - - - Arenicz & Chowdhury, 1987

K<+ """ """ Proposed Equation

V'"-o.

t/.o~. ' 1 ' 1 ' 1 ' 1 • 1 • I • 1

0 2 4 6 8 10 12 14 Depth (m)

Fig. 4.3.1.3 Comparison between the field data and experimental formulae.

It should be noted that there is no tangent discontinuity in the proposed equation

because:

dK(y) = dK(y)

dy dy

y -> 6 + y->6~

= K(6) (4.77)

A comparison between Eq. 4.75, Schlosser (1978) formula, the formula of Arenicz and

Chowdhury (1987), and the average of the field data is shown in Fig. 4.3.3. It can be

seen from Fig. 4.3.1.3 that Eq. 4.75 offers a better fit with the average of observed data

than the alternative formulae.

141

4.3.2 Apparent Friction Factor

It has been known that both the length of reinforcement and the overburden pressure can

affect the apparent coefficient of friction between soil and reinforcement. Although

some formulae have been proposed to describe these effects, there are still problems

regarding their application in design. In the following paragraphs, frictional formulae

reflecting the influence of vertical pressure and strip length will be presented and

analysed. On this basis, some new formulae are proposed.

4.3.2.1 Vertical Pressure Effect

According to Schlosser and Segrestin (1979), the apparent friction factor for smooth and

ribbed strips, its variation with depth shown in Fig. 4.3.2.1, can be calculated as follows:

I) For smooth strips:

/* = 0.4 (4.78)

U) For ribbed strips:

* / =

y(tan(])-/0) +

— ~ + /0 for, y<6m (4.79)

tantj) for, y>6m

in which,

/0*=1.2 1ogCM (4.80)

and Cu is uniformity coefficient, fo* is the apparent friction factor at the top of

reinforced earth zone, (j> is internal angle of friction of the soil, and y is the depth of soil.

The figure shows clearly that Eq. 4.79 has a tangent discontinuity at y=6m, which

cannot be justified in terms of physical interaction between the soil and reinforcement

Also, Figures 4.3.2.2 and 4.3.2.3 shown a significant disparity (on the conservative side)

between Eq. 4.79 and the field data.

142

t y = 6m

>

/• /„

1

y

X i

Ribbed strip

Smooth strip

f

Fig. 4.3.2.1 Apparent friction factor

To address these problems, Arenicz and Chowdhury (1987) suggested the following:


/* = tan\|/+ct-),(n-tan\|/) (4.81)

H) For ribbed strips:

/* = tan<|>+ay(m/0-tan(|>) (4.82)

where f* is the apparent friction coefficient between soil and reinforcement, \\f is the

angle of friction between soil and reinforcement measured in direct shear box, y is the

depth, fo* is calculated from Eq. 4.80, a is an empirical coefficient suggested to be 0.6,

n and m are maximum value factors, respectively, for smooth and ribbed strips.

As shown in Figures 4.3.2.2 and 4.3.2.3, their formulae (Equations 4.81 & 4.82) have

eliminated the tangential discontinuity but remains conservative. Although these

formulae are in closer agreement with the experimental results in comparison with

143

Schlosser's proposals, a significant disparity still remains. Hence, the following

modifications in the calculation of the apparent friction factor are proposed.


/ =

tan\|/ + sl(2.4-tan\|/) for, y<6m

tan\|/ for, y>6m

(4.83)

II) For ribbed strips

* tan(j) +0.9^^(3.85^-tan(j)) for, y<6m

tan<|) for, y > 6m

(4.84)

where „; is calculated from Eq. 4.76. It should be noted that there is no tangent

discontinuity in Equations 4.83 and 4.84 because:

3/*(y) _ff*(y) By dy

y -» 6 + y -> 6~

= f*(6) (4.85)

A comparison between the proposed Equations 4.83 and 4.84, Schlosser and Segrestin's

formulae (1979), Arenicz and Chowdhury's formulae (1987), and typical values of the

apparent friction factor based on observations is shown in Figures 4.3.2.2 and 4.3.2.3.

The figures illustrate that Equations 4.83 and 4.84 eliminates the problem of tangential

144


discontinuity, reflects the non-linearity suggested by the field data, and offers a closer

agreement with the observations.

8-1

f* •

6-

4-

2-

A Observed (After Schlosser & Elias, 1978)

Schlosser and Segreston, 1979

Arenicz & Chowdhury, 1987 — - Proposed formula (Eq. 4.83)

U 1 • 1 • 1 • ! • 1 • 1 ' 1

0 2 4 6 8 10 12

Depth (y) 'm'

Fig. 4.3.2.2 Comparison between theoretical and typical values of apparent friction

factor for smooth strips

f* 8-i

2-

0 0

A Observed (After Schlosser & Elias, 1978)

Schlosser & Segrestin, 1979

Arenicz & Chowdhury, 1987 — - Proposed Formula (Eq. 4.84)

~r 2

-r 4

T-6 8

T— 10

Depth (y) 'm'

12

Fig. 4.3.2.3 Comparison between theoretical and typical values of apparent friction

factor for ribbed strips

145


4.3.2.2 Strips length effects

The length of strips appears to have an important effect on the value of apparent friction

coefficient as found by several tests carried out in the U S A and France. Based on the

results of pull-out tests in Satolas in France (Alimi et al, 1973) and in California (Chang

& Forsyth, 1977) for smooth strips, there appears to be a non-linear relationship

between the strip length and the apparent friction coefficient. The result of the tests are

shown in Fig. 4.3.2.4.

Arenicz and Chowdhury (1987), have suggested the calculation of the effect of length of

reinforcement strips as follows:

/*=/*(l-aL) (4.86)

where / is the apparent friction coefficient between soil and reinforcements, fc is the

maximum value of /*, L is the length of reinforcement strips and a is a constant

suggested to be 0.72.

There are two points that ought to be made regarding this equation. Firstly, it does not

relate the value of/* to the height (H) of reinforcement earth wall, even though the field

tests results (Fig. 4.3.2.4) indicate that/* depends on H. Secondly, the value of f* is

unknown and appears to change with the depth of fill. Since no method of detennining

the value of fc* has been proposed by Arenicz and Chowdhury (1987), the equation

cannot effectively be used for design purposes.

An analysis of the results of the field tests, shown in Fig. 4.3.2.4, indicates that although

the relationship between the apparent friction factor/* and the length of reinforcement L

is non-linear - as confirmed by others (Alimi et. al, 1973; Chang and Forsyth, 1977;

Arenicz and Chowdhury, 1987) - there is a linear relationship between/* and the ratio of

H/L, as illustrated in Fig. 4.3.2.5 based on the same field data.

146


Results of pull-out tests at Satolas (France)

2.2 ~\

f* 1.8 -

1.4 -

1.0 -

0.6 -

0.2 H=2m

1——T 6 7

~i——

8 L 9

Results of pull-out tests at Highway 39 Wall (USA)

1.5 -i

f* 1.2-

0.9-

0.6-

0.3-

0.0 — T 2 L 5

Fig. 4.3.2.4 The results of pull-out tests

From Fig. 4.3.2.5, the apparent friction coefficient/* decreases linearly with the increase

in ratio H/L. This relationship can be written as:

* H

f* = f -m(—) JC >

(4.87)

where m is the tangent of the lines.

147

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4.3.3 Extension of failure zone

Observations and measurements taken from the existing structures, which remain in a

state of safe equilibrium, can only identify the position of a potential failure surface. This

surface is commonly assumed to coincide with the maximum tension line, ie. the line

passing through the loci of maximum tension in reinforcement strips.

For a number of existing reinforced earth structures with vertical facings, the shape and

position of the maximum tension line has been determined through field observations. In

general, they are curvilinear and located between 0.1H to 0.3H away from the face of

structure. To facilitate empirical design, Schlosser (1978) suggested a simple bi-linear

function (Fig. 4.3.3.1), with a tangential discontinuity at 0.5H, to represent the observed

maximum tension lines. Therefore, the required length of the strips, L, can be calculated

as follows.

L = L +0.3H for y<— (4.88) e 2

L = L +0.6#-0.6y for y > — (4.89) e J J J 2

where, H is the height of the reinforced earth structure, Le is the effective length of the

reinforcing strips and, y is the depth of soil from the top.

Later, Arenicz and Chowdhury (1987), suggested a modification to the Schlosser

proposal, which eliminated tangent discontinuity and reflected field observations more

closely. This was achieved by proposing the following function:

_L = \r2-(±)2-a (4.90)

H V H

where r is radius of cylindrical failure surface, x is distance from facing panels to the

point of maximum tension in reinforcements and y is the thickness of the soil fill above

149


the strip considered. Based on Eq. 4.90, therefore, the required length of the strips, L,

can be calculated as follows:

L=L +x e

(4.91)

where Le is calculated (as will be discussed in the following section) and:

x = H(^r2-(jj)2-a) (4.92)

H

r

H/2

0.3H '

L

Le

Fig. 4.3.3.1 Effective length of reinforcing strip

4.3.4 Reinforcement Effect

4.3.4.1 Bond Failure

The friction force between soil and reinforcements,//, can be calculated as follows:

/. =2B.L f a Jf i eJ v

(4.93)

150

where 5/ is the total width of reinforcements at level i, Le is the effective length of the

reinforcing strips, /* is the friction factor between soil and reinforcement, and o*v is the

vertical stress acting on the reinforcements. On the other hand, the pull out force, fa can

be calculated as follows:

/ = KSVTSTJG Ja V H v

(4.94)

where K is the coefficient of lateral earth pressure, Sy and Sjj are vertical and horizontal

spacings between reinforcements, respectively, and GV is the vertical stress. At the time

of bond failure, the following equation can be written considering the safety factor:

FS, a (4.95)

where FS§ is the safety factor against bond failure. Substituting ff and fa from

Equations 4.93 and 4.94 in Eq. 4.95 and solving for Le results in the following equation:

L = e

KSySHFS^

IB.f* r

(4.96)

Substituting K from Eq. 4.75, /* from Eq. 4.83 and 4.84 to Eq. 4.96, results in the

following:

I) for smooth strips:

L =< e

[*«+I-2H<WvV5* 2B. [tan y + d^ (2.4 - tan \|/)]

K S-,,STjFS. a V H § 2_5.tan\|/

i Y

for, y < 6m

for, y>6m

(4.97)

II) for ribbed strips:

151

e

»y [ * a + 1 . 2 ^ l ( „ Q - * a ) ] V ^

2B. [tan<|> + 0.9y dl (3.85 - tan <)))]

K S,rSrjFS± a V H §

2i5.tan(l) i Y

/or, y < 6 m

/or, y < 6 m

(4.98)

where L e is the effective length of the reinforcements, Sy and S H are the vertical and

horizontal spacings between reinforcements, respectively, Ka and Ko are the coefficients

of lateral earth pressure in active and at rest conditions, respectively, FS§ is the safety

factor against bond failure, B{ is the total width of reinforcements at level i, d\ is

calculated from Eq. 4.76, y is the depth of soil, \|/ is the angle of friction between soil and

reinforcement measured in the direct shear box, and § is the internal angle of friction of

soil.

4.3.4.2 Reinforcement Failure

At the time of failure due to the rupture of reinforcing strips, the following equation can

be written:

A/ ________

FS y

f. a (4.99)

where, fa is the force given by Eq. 4.94, FSy is the safety factor against rupture failure of

reinforcement, As is the cross sectional area of the reinforcement, and/y is the ultimate

tension in a unit area of reinforcement. Substituting fa from Eq. 4.94 for Eq. 4.99

provides a solution for the cross section area, As, as follows:

A -s

KS SrjO SF v H v s

/_.

(4.100)

Substituting the value of K from Eq. 4.75 to Eq. 4.100 leads to the following equations

for no failure due to the rupture of the reinforcement:

152

A = \ s

[K +1.2ydAK -K )]S„SrjO FS 1 a 1 o a V H v y

— for, y < 6m

K SvS„c FS <4M1>

a V H v y , — — for, y<6m U

where S y and Sfj are vertical and horizontal spacings between reinforcements,

respectively, Ka and K o are coefficients of lateral earth pressure in active and at the rest

conditions, respectively, FSy is the safety factor against rupture failure,/y is the ultimate

tension in a unit area of reinforcement, d\ is calculated from Eq. 4.76, c v is vertical

stress on the reinforcement, y is the thickness of soil in the level considered.

4.3.5 Design Equations

In the design of reinforced earth structures, factors of safety against both break and bond

failures have to be calculated. In current practice, this is usually carried out through

design formulae derived from the static equilibrium, which incorporates the empirical

functions discussed in the previous sections of this chapter. The sets of formulae derived

by Arenicz and Chowdhury (1987) for the C G M and the M C G M are given in Appendix

D for comparison with the alternative design formulae based on the semi-empirical

functions proposed in this chapter (given in Table 4.3.5.1).

In table 3.4.5.1, A/ is the total cross-section of reinforcement at level i, ot is the

allowable tensile stress for the reinforcement, y is the thickness of soil fill above the strip

considered, Lt is the length of reinforcement strip at level i, y is the total unit weight of

reinforced soil fill, Ka is the coefficient of active earth pressure, K o is the coefficient of

lateral earth pressure at rest, dj is calculated from Eq. 4.76, Sv and SH are, respectively,

the vertical and horizontal spacing of the reinforcement, 5; is the total width of

reinforcement at level i, H is the height of reinforced earth structure, <|> is the angle of

internal friction of soil fill, and y is the angle of friction between soil and reinforcement.

153

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4.3.6 Internal erosion and piping failure

Piping is the removal of soil particles by percolating water causing creation of channels

through the soil. As mentioned in Sec. 3.4.2, seepage flow through the soil exerts force

on the soil particles. If this force is greater than the resultant of resistance forces, the

particles start to move. The resistance forces acting on the last particle (which is at the

end of a flow path) are minimum. Therefore the removal of particle firstly occurs there.

After removing the last particle, this process will be repeated for the next particles.

Continuation of this process leads to creation of a channel through the soil. This process

is normally accelerated for the next particles, because the seepage force is increased due

to decrease of the seepage path length.

The type of RSD is an important factor in the prevention of internal erosion and

occurrence of piping failure. For example, the piping may affect the stability of

downstream facing panels in homogeneous fill RSDs without drainage blanket. However,

piping does not directly affect the downstream facing panels in zoned RSDs.

Construction of drainage blanket has an important role in the prevention of piping. In

the following sections, these will be further discussed as: (a) piping in homogeneous fill

RSD without drainage blanket, (b) piping in homogeneous fill RSD with drainage

blanket, (c) piping in zoned RSD, and (d) piping under RSD.

(a) Piping in homogeneous fill RSD without drainage blanket. In homogeneous

fill RSD without drainage blanket, the effect of seepage force on the last particle is nearly

a horizontal force, Fp, at the downstream side of dam. This force acts on the facing

panels, which are below the upper line of seepage, toward the downstream side as

shown in Fig. 4.3.6.1. If the upper seepage limit is assumed to act at about one third of

the height, this force m a y be represented as a push-out force and is given as:

V2

F = f 3 C , p A — (4.107) p h dv 2

155


where V is the seepage velocity calculated from Eq. 3.25, C j is the drag shape

coefficient relating to the shape of facing panel suggested to be 2 for the square shapes

(Streeter and Wylie, 1979), p is the density of water, and A is the cross sectional area of

the facing panels. This force should be considered in calculation of cross sectional area

of reinforcement needed against break failure.

Hi y

y^' yS \ -

S? S\f y)*h i#\_

't' Particle C

Hj/3

Fig. 4.3.6.1 Piping through a homogeneous fill RSD without drainage blanket

(b) Piping in homogeneous fill RSD with drainage blanket. Referring to Fig.

4.3.6.2, in homogeneous fill RSD with a horizontal drainage blanket, the seepage force,

Fp, does not act on the facing panels. This force may cause water to percolate through

the drainage blanket. This should be considered in the design of drainage blanket which

is beyond the scope of this project.

Draniage blanket

particle C

Fig. 4.3.6.2 Piping through a homogeneous fill RSD with a horizontal drainage blanket

156


(c) Piping in zoned RSD. Seepage lines through a zoned RSD is shown in Fig

4.3.6.3. Similar to the Case (b), the piping force, Fp, does not act on the facing panels.

However, this force should be considered as external force acting on the reinforced soil

zone when there is no filter between the reinforced zone and the core. The stability of

reinforced soil zone should be checked against this force. If a filter is constructed

between the core and reinforced soil zone, this force may be ignored.

(d) Piping under RSD As the seepage force percolates upward under RSD at the

downstream side, it tends to uplift the soil particle (See Fig. 4.3.6.4). This force reduces

the effective weight of particle C. W h e n the seepage force exceeds the weight of

particle, then the piping failure starts and particle C floats out. If particle C floats out,

the length of water path is reduced. Reduction of the path length increases the driving

force, Fp. This causes the floating out of the next particle. Continuation of this process

leads to rapid creation of a channel under the dam. This should be considered in the

design of foundation of RSDs which is beyond the scope of this project.

H,

Hn

(a)

(b)

Reinforced soil zone

Filter

Draniage blanket

Reinforced soil zone

Draniage blanket

Fig. 4.3.6.3 Piping through a zoned RSD

157


Fig. 4.3.6.4 Piping under a RSD

The use of cutoff trenches under the RSD and the use of heavy stones on downstream

side may prevent the piping failure through the foundation (see Fig. 4.3.6.5). Using

heavy stones on the downstream side causes an increase in the vertical stress acting on

the particles. This prevents piping. Using cutoff trenches increases the length of seepage

line, this means prevention of piping. Detailed evaluation of these solutions are beyond

the scope of this thesis.

Reinforced soil dam

Heavy stones

H-

Seepage line under reinforced soil dam

Fig. 4.3.6.5 Use of heavy stones in downstream side for preventing piping

158


4.3.7 Hydraulic fracture failure

The shearing strength of fill material is reduced when pore water pressure occurs in the

soil mass. The increase in pore water pressure may lead to hydraulic fracture failure.

According to Singh (1975), pore water pressure may occur in earth dams under three

stages: (a) during the earth dam construction, (b) under steady seepage, and (c) during

or after a quick drawdown. The first one is the result of weak compaction of fill

material. The second one occurs after the reservoir is full of water for a certain time.

The last one occurs when the reservoir is emptied rapidly.

Three flow gradients may occur in the upstream side of RSD as shown in Fig. 4.3.7.1.

After a quick drawdown, the seepage reverses and flows towards the upstream side.

Reversed seepage lines through a RSD at rapid drawdown situation are shown in Fig.

4.3.7.2.

The total pore water pressure, u, is normally calculated as follows:

u = B [a3 +A (cx -CT3)] (4.108)

where A and B are Skempton's pore pressure coefficients, determined based on triaxial

test, and "1-0*3 is the deviator stress difference. When the pore pressure is more than

minimum principal stress, the soil mass is increased in volume and may be floated out as

a dense liquid with a unit weight more than the submerged unit weight of soil.

Therefore, the minimum principal stress should always exceed the pore water pressure at

any infinitesimal element of the soil to prevent hydraulic fracture failure.

According to Mitchell (1983), upward gradient can be expected when lateral flow

gradients conform to the slopes. In some cases (Case c in Fig. 4.3.7.2), this gradient

reaches to a critical hydraulic gradient, ic which is the hydraulic gradient at a point

where water exits out of the soil fill. The hydraulic gradient is normally given as:

i --^-cose. (4.109) C y 1

'w

159

where y is the submerged unit weight of soil mass, y w is the unit weight of water, and

61 is the upstream slope angle as shown in Figs. 4.3.7.2.

(a) Hydrostatic pressure

(b) Drawdown

(c) Artesian pressure

u

u

u-yh 'w

u < y h 'w

u> y h 'w

Fig. 4.3.7.1. Idealised flow gradients in the upstream part of RSD

Upstream water table (level 1)

Reversed seepage lines

H

Upstream water table (level

\

Seepage line

H,

Fig. 4.3.7.2 Seepage line through a homogeneous fill RSD without drainage blanket

160


When the hydraulic gradient reaches its critical value, a continued erosion of the slope

may occur leading to hydraulic fracture failure. This phenomenon may be seen in the

case of rapid drawdown. Therefore, the upstream slope of dam should be as flat as

practicable to prevent hydraulic fracture failure in rapid drawdown. This slope may be

given as:

8. -Arc cos Y * 'w c

y SF (4.110)

where SF is the safety factor against hydraulic fracture failure. Since,

tan6- = H

wb-wt (4.111)

Therefore, for no hydraulic fracture failure, the minimum required base length of dam

may be calculated as:

wb>-H

tan /

Arc cos Y i 'w c Y SF

T + W > (4.112)

4.3.8 Distortional settlement

Distortional settlement, which usually causes the appearance of cracks in conventional

earth dams, is also a major problem for RSDs. Foundation character and construction

method affect the distortional settlement. The foundation may suffer under the forces

acting on dam, causing the creation of cracks. The construction method has also an

important role in the reduction of settlements in RSD. Maximum compaction should be

obtained when the dam is being constructed. Using thinner layers during the compaction

causes better compaction, which leads to a reduction of settlement. A rapid construction

of earth dams may cause post-distortional settlement after construction. T w o main

161


reasons for considering the distortional settlement are: to ensure that the distortional

strains are sufficiently low to prevent internal cracking and, the compressive stresses are

greater than the water pressure at any location to prevent hydraulic fracture failure.

Original reinforced soil dam ^>*

Deformed reinforced soil dam ^ , - '

^ g g j g l l l l ^ ^

5

Bd

-'-- — •-- - h j

H

W

Fig. 4.3.8.1 The distortion settlement of RSD

Referring to Fig. 4.3.8.1, the distortional settlement, 5, of RSD may be calculated as:

6 = 5 _ + 5 / (4.113)

where 67 is the foundation settlement and 8j is the dam compression which may be

calculated as:

§j=\r\m ACT =]/?m zyd =m y — „ JU v v Jv v ' z v ' 9

(4.114) z v • 2

in which m v is the compression modulus of the compacted reinforced soil fill. Hence,

the vertical strain may be estimated as:

162


8 , m yH E =-^= —* (4.115) v // 2

For average settlements, the base extensional strain may be approximated as:

2J(W2+b2) E -i JL__ ____ C4.-Z76;

Therefore, the distortional strain under plain strain is:

_ e ___._____!_______! v h H H

To prevent distortional cracks, the distortional strain should be less than the maximum

distortional strain, ^max, which causes rupture in reinforced soil sample, tested in

laboratory condition.


The stability analysis of RSD should be accurately addressed from the point of view of

internal and external stability. The external stability of RSD has been evaluated, based

on an analytical approach. In external stability analysis, it has been assumed that the

whole reinforced soil structure acts as a unit.

To optimise the geometry of these dams, the functions of minimum required base

length for no failure state due to sliding, overturning, and overstressing have been

presented in the first part of this chapter.

Soil reinforcement is a modern technique for improving the mechanical properties of

soil, using the concept of frictional interaction between soil and reinforcement. In the

composite material (consisting of soil and reinforcement) the generation of frictional

forces between soil and reinforcement is fundamental to its behaviour. The mechanisms

163


involved in this process are, however, not yet fully understood. Various analytical

theories developed so far are still not in satisfactory agreement with the observed

behaviour of reinforced earth structures; which necessitates the use of empirical

relationships in current design practice.

Some of the theories developed so far, their relationships and the field data on which

they were based, have been analysed in the second part of this chapter and, subsequently,

modified. Empirical formulae reflecting the observed behaviour of reinforced earth

structures were suggested.

The semi-empirical relationships suggested in this chapter have eliminated the tangential

discontinuity existing in the formulae of the CGM. They have reflected the non-linearity

indicated by the field data, have eliminated unknown parameters existing in formulae of

the MCGM, and have offered a closer agreement with available field observations.

In regards to the apparent friction factor, (which is fundamental to design of reinforced

earth structures), a linear relationship between the factor and the ratio of fill depth to

strip length has been discovered in the analysis of the field data.

On the basis of the coefficient of lateral earth pressure, apparent friction factor, and the

maximum tension line, the formulae for calculation of the factor of safety against tensile

failure FSy, and the bond failure of reinforcements FS(j), have been proposed.

164

BEHAVIOUR OF SOIL DAMS UNDER SEISMIC LOADS CHAPTER FIVE

CHAPTER FIVE

BEHAVIOUR OF SOIL DAMS UNDER SEISMIC LOAD

5.1 INTRODUCTION

Earthquakes, a world wide problem, act on structures as a kind of dynamic force. In

the past, many dams have been damaged by earthquakes e.g. the Sheffield dam in the

U S A failed as a resulted of the Santa Barbara earthquake in 1925. About nine cases of

damage and/or even failure were reported from 1930 to 1946 (Ambraseys 1960). The

range of side slopes of these nine earth dams were from 2 to 3.5 horizontal to 1 vertical.

The failure of the Hebgen dam in Montana in the U S A was the result of another

earthquake which occurred in 1959. Slope sliding, settlement, slumping, longitudinal

cracks and even the complete wash-out of earth dams were some of the results of the

earthquakes (Singh, 1976). The 1939 Ojika Earthquake in Japan resulted in 12

complete dam failures. More than half of these failures occurred during the 24 hour

period after the earthquake. The sandy soil embankments suffered the greatest damage.

However, there were no total failures in clay soils embankments. It is well known that

crest settlement and formation of cracks are the most frequent types of damage to dams

as a result of earthquake. Cracks may cause damage to outlets of tunnels resulting in

leakages. Blockage of the outlets, piping and even overtopping, may also appear after

an earthquake.

The failure of the earth dam may be the result of relative dam displacement caused by

major fault movement in the foundation soil, loss of freeboard due to differential

tectonic ground movement, slope failures induced by ground motions. Other factors

deserving consideration are the sliding of the dam on weak foundation materials, piping

165


failure through cracks induced by ground motions, overtopping of dam due to slides or

rock-falls into the reservoir and failure of the spillway or outlet works (Seed, 1983).

In an earthquake, the earth moves in an approximately random manner in both

horizontal and vertical directions. Variation of acceleration due to earthquake is a

function of time (Newmark, 1965). The velocity and displacement caused by the

earthquake can be calculated by integration from the acceleration-time function.

At least, two relationships have been formulated by Richter (1958) and Bath (1966)

between the magnitude of earthquake, M on the Richter Scale, and the energy released

from the earthquake shock, E in Ergs, as follows:

log E= 11.4 +1.5 M (5.1)

log E = 12.24 + 1.44 M (5.2)

Regarding the second equation, the energy released by an earthquake of the magnitude

7 on the Richter Scale is equal to 7.2 x 1 0 2 0 Ergs or 9 times the energy released by the

Hiroshima atom bomb. The energy released by an earthquake of the magnitude 8 on the

Richter Scale is equal to 241 times as this bomb. The El-Centro earthquake, which

occurred in California, on the M a y 1940 is one of the strongest earthquakes recorded.

It is known that the maximum ground acceleration of the earthquake was 0.32g, the

maximum ground velocity was about 0.35 m/s, and the maximum ground displacement

was 0.21 m (Newmark, 1965).

Although many researchers have investigated the behaviour of conventional earth dam

under seismic load, many problems still remained unknown in this regard (Wahlstorm

1974; Wolff 1985;). The behaviour of RSD under seismic loads is another problem

which should be studied. Seismic resistance of reinforced soil dams and embankments

may be tested by the shaking table tests (Koga et al. 1988a), or may be modelled

numerically by the finite element methods. However, evaluating the natural frequency of

166


the RSD is a very simple method for addressing the maximum safe proportion of

reinforcement needed for a RSD. Comparison between the natural frequency of

conventional earth dam and RSD is the aim of this chapter.

5.2 FREE HARMONIC VIBRATION

Fig. 5.2.1a shows a typical RSD divided into several imaginary layers. Each layer may

be considered to act as a block for the purpose of dynamic analysis: the first and the

second blocks of the dam are shown in Fig. 5.2.1b.

w t

•4 • m y e * g \ x x

g X X X

g • / / g X X X

m * * *

- • - ^

\ X X

g X X _x

wb r

(a)

in? 1 ?2

i

i -

Hi

r

^^ N. l

First layer Second Layer

W W • i i

i

a / — •* \ X X \ X

X X X X \

1 1

iff. / 1 fe2 /, . ,—-J-' N' S' -' -' \ S

' /

model of internal resistance

(b)

Fig. 5.2.1 a) A typical RSD divided into several imaginary layers b)the first and the

second blocks of the RSD

It is assumed that a load Fi is acting on a block and then is suddenly removed. The

motion of the block, neglecting damping effects of soil under the load, can be

expressed by the following equation:

m4-£+F.=0 dt1 V

(5.3)

167


in which y is the vertical displacement of the block at time t, m is mass, m = Wl g, and

¥\ is the force necessary to keep the block in its place. Substitution of F, = K y in the 1 sJ

above equation, in which Ks is the spring constant of soil, results in the following:

dt

The general solution of the above expression is usually written as follows:

y = cx s i n ( r ^ ) + c2 cos(r^J-) (5.5)

in which c\ and C2 are constants. Since, y=0 when t=0, C2 should also be zero and the

solution of Eq. 5.5 is:

K y = c1sin(?AM-) (5.6)

1 V m

The dimension of the term IK lm is 1/sec, because the term tjK lm has no

dimension. Substitution of cj and IK lm with a and co, respectively, results in the

following:

y = asm((at) (5.7)

in which a is the amplitude of a sinusoidal harmonic vibration and co is the angular

velocity which can be represented by a vector of length a which rotates with a constant

angular velocity around the equilibrium position of the centre of gravity of the block.

In fact, the vibrations are damped because of the internal resistance of the soil.

Therefore, the amplitude will decrease over time until Vibration stops completely. Thus

Eq. 5.3 may be modified as follows:

d2y m—%-+F, + F,=0 (5.8)

dt2 d 2

168

in which Fj is the damping force, which could be described in terms of c^, the

damping coefficient, as follows:

Fd=

c4 (5-9> Substitution of 5.9 in 5.8 yields;

dt2 d dt s

or;

^+2A.-^ + co2y = 0 (5.11) dt1 dt

in which X, the damping ratio, is related to c^ as follows:

\ = ^L (5.12) 2m

In this case, the amplitude of sequential cycles have the ratio:

a n + 1 _ e-<K a n

(5.13)

in which x is the period of the vibration which may be related to the circular frequency,

co, as follows:

T = 2TC

__. (5.14)

5.3 FORCED HARMONIC VIBRATIONS

Impulses causing vibrations are repeated frequently. In this case, the sinusoidal

periodic impulse and the associated equation of motion are represented as follows:

F= FjSint.Gy) (5.15)

169


— ^ + ( 0 02 y = a1_0

2sin(co10 (5.16) dt

where COQ is the natural frequency of the system consisting of a block and spring and aj

is a ratio of Fj and Ks:

F. 0.--1- (5.17)

1 K s

The solution of Eq. 5.16 is:

y = N [sin(co10-«cos(C000] (5.18)

in which the amplification factor, is given by:

_V=- —5L- f5.i9j d-nl)

where n is the ratio between the frequency of the periodic impulse to the natural

frequency of the block, coi/coo- The relationship between N and n can be plotted as

illustrated in Fig. 5.3.1. The figure shows clearly that if the frequency of the periodic

impulse is equal to the natural frequency of the block (n=l), resonance of vibrations

occurs and the magnification factor reaches infinite. The first term of Eq. 5.18

represents a forced vibration with an amplitude N and a circular frequency a>\. The

second term of the equation is a forced vibration with circular frequency COQ and

amplitude -nN.

170


N

al

0

HAZARD ZONE

Fig. 5.3.1 The relation between N and n

5.4 D A M P I N G

The infinite value of magnification factor N at n=l is theoretically correct for the ideal

case of an undamped system. However, since damping occurs, N is not infinite even for

n=l. With the inclusion of damping, the equation of motion can be modified as

follows:

1 a v av ? 2 —^-+2X,^-+coty = fl1cotsin(co10 dt1 dt U i U 1

(5.20)

where X is the damping ratio given by Eq. 5.12. In this case, the amplitude is

maximum when:

1 a = a l2_.f_W

%i 4 coQ

(5.21)

and the magnitude of circular frequency associated with this amplitude at resonance is

given by:

171

<» = c o n l 1 " ^ — > 2 (5-22)

res 0A/ 2 fi>0

5.5 NATURAL FREQUENCY

It is of considerable interest to estimate the natural frequency of any structure which may

be subjected to dynamic forces such as those due to an earthquake. The analysis and

design of such a structure must recognise the possibility of resonance during an

earthquake. This will require selection of the appropriate design earthquake and a

comparison of the natural frequency of the structure with the frequency characteristics of

the design earthquake.

This chapter is concerned with the development of a simple approach for the estimation

of the natural frequency of a RSD. The method is developed by considering the overall

stiffness of such a composite structure in terms of the stiffness of the unreinforced mass

and that of the reinforcing elements.

A conventional earth dam and a RSD will be shown schematically in Fig. 5.5.1a and b.

Each of these may by subdivided into imaginary layers which are horizontal. It is

assumed that any such layer acts as an individual block. It is also assumed that the

natural frequency of a system of soil layers is:

0 K w

where, k is the stiffness of the spring assumed to be in the system, W is the weight of

the system of the structure assumed, g is the gravity acceleration and COQ is the natural

frequency of the system.

172


mi y 1

Ai i i

^

WM (a)

\ni 1

p\ I 1

ft

m? / 1

/ • > ^

I 1 **

H

(b)

Fig. 5.5.1 a) A typical conventional earth dam and b) a typical RSD with vertical

downstream facing

Referring to Fig. 5.5.1, the weight of the conventional earth dam, Wj, and the weight

of the RSD, W2, may be calculated as follows:

Wl = ^+wh^Hh

W, _(^2_^_2____

(5.24)

(5.25)

in which h is dam height, wtj and w g are the crest widths of both dams, wbJ and wjj2

are the base widths of both dams as shown in Fig. 5.5.1, and y\ and 72 we,

respectively, the average unit weights of the soil in the conventional earth dam and in

IheRSD.

The unit weight of the soil material used within the RSD, Y2, is calculated as follows:

y2=|3Yr+(l-p)Y1 (5.26)

173


in which yr is the unit weight of reinforcement and P is the proportion by weight of the

reinforcement used within the RSD to the total weight of the dam. Substitution of

Equations 5.24 to 5.26 in 5.23, and finding the ratio of natural frequency of RSD per

natural frequency of conventional earth dam gives:

co

co ^ - - c p x ¥ (5.27)

01

where

2— L L + m1 +«H

H 1 1

I 2-^ + m, H

(5.28)

¥ =

[p-_. + (l-P)]

h (5.29)

[P__ + (1_P)]

in which COQI and COQ2 are, respectively, the natural frequencies of both conventional

and reinforced soil dams, m\ and nj are respectively the slopes of upstream and

downstream of the conventional dam as shown in Fig. 5.5.1, mj is the upstream slope

of the RSD, kj and ki are, respectively, the spring constants for the elastic support in

both conventional and RSDs. In reality, (p is a function of geometry of dams, and V is

a function of the overall stiffness of materials of dams.

5.5.1 Stiffness Function

In order to find ¥, it is necessary to find the proportion of spring constant of reinforced

soil to the spring constant of the soil, fc?A/- Referring to Fig. 5.5.1.1, the ratio of the

174

spring constant of the reinforced soil, > to the spring constant of the unreinforced soil,

k], can be expressed as follows:

b, (ErAr+EsAs)/0-5(wt2+^ kl <VVV°-5(",i+"M)

(5.30)

or,

(1+ p—^)M K _

s

U + P)

(5.31)

where Er is the elastic modulus of the reinforcements strips, Es is the elastic modulus

of soil, P is the ratio between Ar and As which are, respectively, the cross-section area

of reinforcement strips and the cross-section area of soil element as shown in Fig.

5.5.1.1, and Mis a dimensionless factor equal to (w . + w,,) / (w ~ + Wuy )•

Reinforcement strip

d

EA / r

W////////////K

EA s s

a- Reinforced soil element

Soil elements

b- Unreinforced soil element

Fig. 5.5.1.1 Comparison between reinforced and unreinforced soil elements

Substitution of Eq. 5.31 in Eq. 5.29 gives:

¥ =

[p(l + p-f)Af + (l-pz)] E s

(5.32)

[p(l + P ) - ^ + (l-pZ)]

175


Eq. 5.32 shows that ¥ is a function of (a) the ratio of the reinforcement used within the

soil, P, (b) the ratio of elastic modulus of reinforcements strips to elastic modulus of

soil E/Es, (c) the ratio of the unit weight of reinforcement to the unit weight of soil

Yj/yj, and (d) the ratio of the middle width of conventional earth dam to the middle

width of RSD. The variation of *F versus p for Yr/Ys=3-9, M=2.63 and the various

Ej/Es is shown in Fig. 5.5.1.2. This figure clearly represents that by increase the

proportion of the reinforcement used within the RSD leads to an increase in the

function of material, T.

Fig. 5.5.1.2 Variation of*¥ versus >fory/is-3.9, M=2.63

5.5.2 Shape Function

Inclination of upstream and downstream slopes of conventional earth dams depends on

the internal friction angle of soil, unit weight of material used, plane zones of sliding in

the slopes, and the safety factor (Sherard, 1963; Janbu, 1973; Singh, 1976). However,

the minimum value for small earth dams, shown in Fig. 5.5.2.1, is about 2:1 in both

176

upstream face and down stream face (United States Bureau of Reclamation, 1977).

Assuming the upstream slope inclination of the equivalent RSD with vertical

downstream facing is 1.5:1 as shown in Figures 5.5.2.2 and 5.5.2.3, the substitution of

the minimum values in the function of geometry, Eq. 5.28, gives:

<P mm

2-*-+4 H

2-*-+1.5 H

(5.33)

- '

I* r

l

2

Upstream

Wt i * — *

, V

1

• ' /

/Reinforced

/ soil dam

wb

- 3

2

1

Downstream

M

1

7

H

Fig. 5.5.2.1 Minimum slope ranges of a conventional earth dam compared to an

equivalent RSD

The maximum slope inclination of the earth dams, shown in Fig. 5.5.2.1, is about 4:1 in

upstream face and about 2.5:1 in downstream face (United States Bureau of

Reclamation, 1977). Therefore, substitution of the maximum values in the function of

geometry, Eq. 5.28, gives:

w 2 — + 6.5 H

max

n (5.34)

+ 1.5

177

where, <pmi'n and (Pmax are, respectively, the minimum and the maximum values of the

function of geometry.

,_ *

IH_

r

i

-'" i

4

Upstream

Wt v*-*

, -jr

i

1 lr /Reinforced

soil dam

wb

t

* • s 2.5

>. 1

Downstream ' - ,

fo|

1

H

_

Fig. 5.5.2.2 Maximum slope ranges of a conventional earth dam compared to an

equivalent RSD

Therefore, the value of cOQ2/tO0b Eq- 5.21, can be expressed as:

co Tmax Q)

02

01

<cp . V ^min

(5.35)

Equations 5.34 and 5.35 show that the minimum and maximum values of the function of

geometry, cp-^ and cpm/„, are functions of the ratio of the crest width of dam to dam

height, W/H. Variations of (pmin and (praax versus W/H is shown in Fig. 5.5.2.3. This

figure illustrates that the maximum value of cpmin and (pmax happens when Wt/H is

zero. This means that both cpm/n and (p,^ are maximum when H is maximum, or the

maximum values of both <?min and (p,-^ happen when the RSD is compared with the

total conventional earth dam.

The values of shape functions, (p, are calculated and shown in Table 5.1 for three (20m,

25m and 30m high) dams with a crest width 5m and various side slopes to find the

effect of dam shape on natural frequency. Table 5.1 shows clearly that replacing a

178

conventional dam by a RSD causes increase in the value of shape function. The

increase in the shape function leads to the increase of natural frequency of the structure.

Fig. 5.5.2.3 Variation o/cp versus Wt/H

Table 5.1 The values of shape functions, q>, for various side slopes

Conventional

Earth Dams

US-DS

2:1-2:1

2.5:1-2:1

2.5:1 - 2.5:1

3:1 -2:1

3:1-2.5:1

3.5:1-2:1

3.5:1-2.5:1

4:1-2:1

4:1-2.5:1

Reinforced soil

dam

US-DS

1.5:1-0:1

1.5:1-0:1

1.5:1-0:1

1.5:1-0:1

1.5:1-0:1

1.5:1-0:1

1.5:1-0:1

1.5:1-0:1

1.5:1-0:1

Shape

functions, <p

(H=20 m )

1.500 (cpm7-~)

1.581

1.658

1.658

1.732

1.732

1.803

1.803

1.871 (cpw/7r)

Shape

functions, cp

(H=25 m)

1.522 (cpm,-„)

1.606

1.686

1.686

1.762

1.762

1.835

1.835

1.906 (qw)

Shape

functions, cp

(H=30 m )

1.537 (cpm/„)

1.624

1.706

1.706

1.784

1.784

1.859

1.859

1.931 (qw)

US = Upstream slope DS = Downstream slope

179

5.6 E X AMPLE

As an illustrative example, the natural frequency of the RSD, shown in Fig. 5.6.1, is

compared to the natural frequency of the conventional earth dam, shown in the same

figure. Assume the elastic modulus of reinforcement, Er, is 2xl08 kN/m2; the elastic

modulus of dense soil, Es, is 50000 kN/m2; the unit weight of reinforcement, yr, is 78

kN/m3; the unit weight of soil, ys, is 20 kN/m3 and the ratio of reinforcement weight

used within the dam to that of soil, p, is 0.02.

1*

1

2.5

1 Jpstreaml S

tJ»L

1 X

r Reinforced soil dam

i 1 50m

140m

2

Downstream

j

1

tl rl

r

30

<

Fig. 5.6.1 The illustrative example of a reinforced and a conventional earth dam

Based on Fig. 5.5.1.2, the value of *F is 2.21, because E/Es=4000, Yr/Ys=3-9 and

M=[0.5x(140+5)]/[0.5x(50+5)]=2.63. Regarding Table 5.1, the values of (p is 1.624

(because the upstream and downstream slopes are, respectively, 2.5:1 and 2:1 for the

conventional earth dam, and are 1.5:1 and 0:1, respectively, for the RSD). Replacing

these values in 5.27 yields the following:

—^ = 2.21 x 1.624 = 3.59 (5.35)

180


From Eq. 5.35 it appears that the value of natural frequency of the RSD is more than 3.5

times the value of natural frequency of the conventional earth dam. The natural

frequency of the conventional earth dam is calculated from Eq. 5.23.

Referring to Fig. 5.6.2, the stiffness of the soil within the conventional earth dam, k, is

assumed to be calculate as follows:

E A pH E AHx 1 fc = — £ — = f - s

L Jo A L

(5.36)

Fig. 5.6.2 The illustrative example of the conventional earth dam

For the conventional earth dam shown in Fig. 5.6.2, k=50000xdH/dL=20690 kN/m.

Substituting the values of (a) weight of the conventional dam, W, (b) stiffness of the soil

within the conventional earth dam, k, and (c) acceleration due to gravity, g=9.8 m/s2, in

Eq. 5.21 results in.

CO 20690 [kNImlm] 9.8 [mis1] ^ 2 1 6 §ec-i

01 -'(5 + l40x30) [ m2 ]x20 [kNIm3]

(5.37)

181


From Eq. 5.35, the natural frequency of the RSD can be expressed as follows:

coQ2 =3.59x2.16 = 7.75 Sec"1 (5.38)

Pseudo acceleration verses period, T, for various values of damping coefficients based

on four major earthquakes which occurred in the U S A is shown in Fig. 5.6.3. It is

assumed that such earthquakes acts on the conventional earth dam as shown in Fig.

5.6.1. Since the earthquake frequency is 20.94 sec-1 for maximum acceleration of such

earthquakes, the value of ngj can be calculated as:

co "01 =

01 _ 2.16

co 0 20.94

= 0.1 (5.39)

\ccelaration

4

3

2

0

iffs*0

__ 0.05

S s^ Damping 7 t

^l_d • --S_3 a—.

0 10 ~„ Period (sec) ™

Fig. 5.6.3 Pseudo acceleration verses period, T,for various values of damping

coefficients based on four major earthquakes happened in USA (After Adely, 1987)

Regarding Eq. 5.38, the ratio of the natural frequency of the RSD to earthquake

frequency, nQ2, is calculated as follows:

182


10 CY) 7 75 nm = —^- = — — = 0.37 (5.40) 02 „ 0 20.94

As seen from Fig. 5.6.3, the ratio between the natural frequency of the RSD to the

earthquake frequency, COQ2/COO, is less than 0.5. Referring to Fig. 5.3.1, this does not

cause the phenomenon of resonance in the structure which could lead to a significant

damage of the structure. However, if the value of P is increased to 0.05, then \\f is 4.77

(See Fig. 5.5.1.2). O n this basis, the ratio of natural frequency of the RSD to that of

conventional earth dam is:

—^- = 4.77x1.624 = 7.75 (5.41) ffl01

Therefore,

co02 = 7.75 x 2.16 = 16.73 Sec"1 (5.42)

In this situation the value of nryi will be:

%l = ii73_080 (5A3) 02 „ 0 20.94

which is very close to resonance situation regarding Fig. 5.3.1. To prevent the

resonance phenomenon in this example, the proportion of the reinforcement used in the

above example should be kept below 3% .

Again, if the value of p is increased to 0.11, then Eq. 5.32 yields \jr=9.45. On this basis,

the ratio of natural frequency of the RSD to that of the conventional earth dam is:

—Q2- = 9.45 x 1.624 = 15.35 (5.44)

%1

Therefore,

coQ2 = 15.35 x 2.16 = 33.16 Sec"1 (5-45)

In this situation the value of nrj2 will be:

183


C002 33.16 _ co R n - = - ^ = = 1.58 (546) 0 2 co0 20.94 (J' '

which is far enough from the resonance situation (See Fig. 5.3.1). Again, to prevent the

resonance phenomenon, the proportion of reinforcement used in the dam should be kept

higher than 11% (ignoring cost). Therefore, to prevent resonance in the RSD, the

proportion of reinforcements used within the dam should be sufficient to result in a

considerable reduction or increase in the value of cp (Less than 4% or more than 10% in

the above example). Using reinforcements with low stiffness such as polymers results in

a considerable decrease in the values of cp

5.7 CONCLUSIONS

Construction of a RSD would normally lead to a considerable cost savings. However, it

is necessary to calculate the natural frequency of such a dam to find its behaviour under

earthquake force. Knowledge of the natural frequency of the structures can assist

designers in assessing the potential for the resonance phenomenon in the structure,

which may result in its total destruction. The practice of inserting reinforcement into

the earth dam material allows reduction in fill volume, reduction in displacement, and

increases the safety factor. However, this also leads to an increase in the natural

frequency of such structures compared with conventional earth dams which may

increase the possibility of failure.

In reality, the natural frequency of RSD is increased because of its geometry and its

overall stiffness. In this chapter, the increases in natural frequency of RSD due to these

two major factors have been separately discussed. Formulae concerning the

magnification of the natural frequency of the structure due to reinforcement insertion

have been derived, and in some cases tabulated and plotted.

184

COMPUTER PROGRAM CHAPTER SDl

CHAPTER SIX

COMPUTER PROGRAM

6.1 INTRODUCTION

A computer program is developed as part of thesis based on the calculation of the

forces acting on RSD (Chapter 3), the equations of stability analysis (Chapter 4), and

the formulae of soil-reinforcement interaction which will be explained in this chapter.

The purpose of the program is to assist a designer in geometrical optimisation and

stress-strain analysis of RSDs.

6.2 FINITE ELEMENT FORMULAE

The finite element method is now widely used as a numerical solution method for the

systems of partial differential equations describing the mechanical behaviour of

material. The deformation of soil, the flow of fluids and the natural behaviour of

metals are some of the fields in which mechanical effects can be simulated by this

method. A RSD, which covers the three materials, can also be analysed by this method.

The following section presents equations representing the elastic and/or elasto-plastic

behaviour of such a system.

The state of stress inside soil normally varies from point to point. Although the stresses

within the soil are not necessary elastic, it is helpful to describe the elastic behaviour of

the soil before any explanation of the plastic response or elasto-plastic behaviour of a

soil mass. Such an analysis is normally based on consideration of the internal forces

acting on an infinitesimal element of soil mass. These will be further explained in the

following sections.

185

COMPUTER PROGRAM CHAPTER SIX

6.2.1 Elastic behaviour of soil

Fig. 6.2.1.1 shows an elastic stress-strain curve. Initially the relationship between

stress and strain is linear (OA in this figure), but become non-linear after point A.

When the soil is unloaded, the stress-strain relationship follows the same path but in the

reverse direction to the origin even though the path is not linear.

Fig. 6.2.1.1 An elastic stress-strain curve

The relationship between the stress components, ox, cy, xxy, acting on a soil element,

based on elastic deformation in a two dimensional form, are usually formulated as

follows:

3c? <K -*+__-_/r

dy

da dx *y _ = F

(6.1)

(6.2) d y a x y

where Fx and Fy are, respectively, the body forces per unit volume in directions x and

y. The direct strains of the element, EX, £ V and y^, are usually calculates as follows:

Bd

x ax (6.3)

186


5 d

y a y

dd dd Y = — ^ + — * • r*y a JC a y

(6.4)

(6.5)

The relationship between the strains and the normal stresses are normally represents as:

a vo\, (6.6)

(6.7)

£ = X

c — y

Y

£ £

x , y E

X 2(1 + v)

E (6.8)

These equations can be shown in matrix form as follows:

y

'xy

1

F

1

-V

0

-v 0

1 0

0 2(1+ v)

X

a x

c y X

xy

(6.9)

or,

xy (1-v2)

1 v 0

v 1 0

0 0 0.5(l-v)

which can be written as:

x

Y xy

(6.10)

to-MM (6.11)

In the case of saturated soil, the effective stresses, a*x, a*y, are defined instead of the

stresses, ax, oy as follows:

187


CT = G -u (6.12) A. ^

CT = CT - M (6.75J y y

where « is pore water pressure which discussed before (Sec. 4.3.6). This is known as

Terzaghi's principle of effective stress which states that the change in soil stress is due

to change in effective stress. In this case, Eq. 6.11 changes to:

{a'}-[A'] {e} (6.14)

in which [A'] includes elastic moduli E' and v' rather than E and v.

6.2.2 Inelastic behaviour of soil

Familiarity with plasticity theory is necessary to find how soil deformations can be

modelled. The shape of stress-strain curves is an important factor for predicting the

deformation of the soil. M a n y investigations are being undertaken to understand how

the deformation of a soil mass can be predicted. A possible stress-strain curve for an

element of soil under an unload-reload condition is shown in Fig. 6.2.2.1a, while

Figures 6.2.2.1b to 6.2.2. Id show the simplify models of Fig. 6.2.2.1a.

188

K .© Sst

©

"© © © Ss

©

s <s)

S3

>!

h ••si

SO

SO

-S

SO

©

•—-a

-©

©

a, CN

CN VO


From Fig. 6.2.2.1a, it can be seen that the relationship between stress and strain is not

elastic. If the element is unloaded beyond point B, the unload path is not reversible but

the path BC will be followed. If the element is loaded again, the path CD will be

followed. Since the reverse path beyond B, BC, is nearly parallel to the origin path of

primary loading path, OA, as shown in Fig. 6.2.2.1a, the figure may be simplified to

Fig. 6.2.2.1b to Fig. 6.2.2. Id. In these figures it is assumed that the unload-reload

paths are the same and linear.

6.2.3 Soil-reinforcement interaction

In a RSD, shown in Fig. 6.2.3.1, it is assumed that compressive forces are induced in

the soil mass as active forces while, tensile forces are induced in the reinforcements due

to the frictional bond between the soil and the reinforcements as reaction forces. The

finite element method will be used to model the soil deformation, to find the tensile

stress within the reinforcements, and to predict the level of bond between the soil and

reinforcements. To simplify the problem, it is assumed that the loading generates a

group of nodal forces at the contact points between the soil and reinforcement. These

forces cause some deflections within the soil and reinforcements.

Fig. 6.2.3.1 Reinforcement elements within a RSD

190


For a no bond failure state, the soil deflections should be compatible with the

deflections of the nodal points. In this condition, the reinforcement and soil would

require to be combined by joining or spring elements, modelling the slip behaviour

between soil and reinforcement (Goodman et al, 1968). In the following sections, the

soil reinforcement interaction will be discussed for a layer of reinforced soil.

Following this, the interaction will be extended to the whole structure.

Each reinforcement, Fig. 6.2.3.2, carries the horizontal forces induced in the nodal

points of the reinforcement. The displacement relative to the first node, AT is

calculated as shown in the following equation assuming constant cross-section area of

reinforcement, Ar, and constant stiffness of the reinforcement, Er:

n 2 F*Lr.

• 1 ' l

A r=-—i (6.15) 1 ErAr

in which LF.r is the sum of the nodal points, and ll. is the length of reinforcement

from the first nodal point to the assumed nodal point. Therefore, the nodal

displacements relative to the first node, p, are calculated as follows:

p.-0

= A. -A. (6.16)

P2= A2" A1 p3=A3-A1

The relationship representing the displacement of the soil nodes, As., which are parallel

to the reinforcement nodes, being caused by the active forces, may be calculated as

follows:

n _ F?LS.

i EsAs

191


?S • in which XF. is the sum of the active forces at the nodes, L. is the length of soil

elements, Es is the Young modulus of the soil elements, and As is the cross section area

of soil elements.

Fig. 6.2.3.2 A typical reinforcement carrying the horizontal forces induced in the nodal

points of reinforcement

For no bond failure, the difference between the displacement of the nodal points of soil

elements and the displacement of the nodal points of reinforcement elements should be

zero. The difference can be expressed as:

-r Tr ?S rS ^F!L. J^FfU.

5. = A r _ A * ______ L_L i i i £rAr

(6.18) s As _ M

Since lengths of reinforcement elements are equal to lengths of soil elements, the above

equation yields to:

8. =Ar-A* = ^ - L L_L (6.19)

ErAr S AS EM

192

This assumes that the force on reinforcement nodes should be equal and opposite to the

force acting on the corresponding soil nodes and results in:

Substituting this equation to Eq. 6.19 yields:

8. = < — r - r + — r - r ) £ ^ ^ ErAr ESAS i i

(6.21)

This can be written as follows:

{8/} = A{IF.L.} (6.22)

in which;

A = (ESAS -ErAr)

ErArEsAs (6.23)

Eq. 6.23 can be given in matrix form as follows:

= A

0 0 0 0 0

0 1 1 1 1 0 1 2 2 2

0 1 2 3 3 0 1 2 3 4

Flh F 2 L2 F3L3

FALA F5 L5 (6.24)

From equilibrium:

F 1 L 1 + F 2 L 2 + F 3 L 3 + F 4 L 4 + F 5 L 5 = ° C6.25)

Addition of this equation into the above matrix results in:

193


0

= A

0 0 0 0

0

1

0 1 1

1

1

1

0 1 2

2

2

1

0 1 2

3

3

1

0 . . 1 . . 2 . .

3 . . 4 . .

1 1 1 0

F.L 1 1

FiLi F3 L3 F4L4

F5L5

- A l

(6.26)

or; F1L1 F 2 L 2 F 3 L 3

F 5 L 5

- A l

1

A

0 0 0 0 0 0 1 1 1 1 0 1 2 2 2

0 1 2 3 3 0 1 2 3 4

1 1 1 1 1 1 1 0 0

(6.27)

The solution of this matrix yields a group of nodal forces from a given set of

differential displacements together with the chosen distances between the nodal points.

The set of soil displacement can be obtained from the solution of the soil stiffness

matrix under the external forces as:

[*]{As} = {F) (6.28)

in which [K\ is the stiffness matrix of soil mass, {As} is the vector of nodal

displacement in soil, and {F} is the external force acting on whole elements.

For a number of reinforcements, if the stiffness of reinforcements is assumed to be

constant, the displacements of the nodes can be formulated as:

Ar. U

n r r

_ FT.lL. . 1 U lJ i = l J

ErAr

(6.29)

194


in which i is the number of nodal points assumed in a reinforcement and j is the number

of reinforcements assumed in the whole structure. The nodal displacements of the soil

elements are calculated as follows:

n 2 F?.LS..

Af. = __L__! (6.30) ij EsAs

In this case, the difference between the displacement of the nodal points of soil

elements and the displacement of nodal points of reinforcement elements can be

expressed as:

8..= (—-—+ ——)_.F..L.. (6.31) V ErAr ESAS lJ lJ

On the other hand, from equilibrium;

2^2 =° (6.32) SF3I3-O

Therefore, for each;", the equation 6.32 can be repeated and solved.

6.3 RSDAM COMPUTER PROGRAM

The program, R S D A M , has been compiled using Fortran 77 and contains two main

sub-programs. The first includes 15 subroutines and optimises the geometry of a RSD.

Then, the dam is divided into several incremental elements in order to perform the

analysis by the second part, which includes 13 subroutines and computes the stresses

and displacements within the elements of the dam, based on two dimensional finite

element formulation.

195


6.3.1 Purpose

The purposes of the program are: (a) to optimise the geometry of RSD and (b) to find

the stresses and strains inside the dam. Firstly, the program allows for the geometrical

optimisation of RSD and the necessitate reinforcement used within the dams based on

the formulae of external and internal stability analysis presented in Chapter 4.

Secondly, it allows for the calculation of the stresses and strains within the dam, based

on plain strain analysis. It should be noted that although the program is particularly

adapted to RSDs, it may also be used for a variety of reinforced earth walls and

embankments with only a small change in the configuration of the program.

6.3.2 Input Data

In the program, the information regarding dam geometry, loading, safety factors, fill

material, reinforcement, facing panel, and foundation material are used as input data.

Geometrical data covers the dam height, the initial width of crest and the initial width

of base. Final widths are calculated by the program. Loading data covers the height of

water acting on the upstream side of the dam, the height and unit weight of silt settled

on the upstream side of dam, the height of water acting on the downstream side of dam,

a possible ice force, and the coefficient of earthquake acceleration. The safety factor

data includes those against sliding, overturning, over-stressing, bond failure and rupture

failure.

Fill material data contains unit weight, angle of internal friction, elastic modulus,

Poisson's ratio, unload-reload coefficient, coefficient of uniformity of fill materials

used within the dam and the frictional coefficient between the soil and the

reinforcement. Reinforcement data covers width and the admissible tension of

reinforcements used. Facing panel data contains the width and length of facing panels,

and the number of reinforcements connected to each facing panel. Foundation material

data covers allowable bearing capacity of foundation soil. The other input data are as

196


follows: the method of internal stability analysis, the number of nodal points in x-

direction, the number of fixed nodal points in y-direction, and possible displacements

of base nodal points. More detail together with the input data of an example of a 20m

high RSD is shown in Appendix F.

6.3.3 Program Operation

The program written by the author contains two main sub-programs which are further

described below. A n abbreviated flowchart for the program is shown in Fig. 6.3.3.1,

with more details included in Appendix E. A guide for running the program is

presented in Appendix F, while the listing of program is included in Appendix G.

6.3.3.1 First main sub-program

The first main sub-program serves to control calling subroutines (INPUTD, O U T P U T ,

H O R F O R C E , V E R F O R C E , DIST, B E A O P T M , SLIDOPTM, O V T U O P T M ,

O V S T O P T M , C G M , M C G M , N C G M , NOFAIL, REINAREA, M E S H ) , the processes

of iterations of calculation, preparing output data for graphical figures, and preparing

material properties. These will be explained in the subsequent paragraphs.

During the execution of the first sub-program, the dam is divided into several layers.

The number of layers is equal to the ratio of dam height to panels height. Every layer

is taken from the top of dam to a specified layer depth.

The horizontal forces acting on each layer of the RSD, including upstream hydrostatic

force, downstream hydrostatic force, the horizontal force due to silt pressure, and the

direct and indirect forces of earthquake are calculated in Subroutine H O R F O R C E .

Layer weight, uplift force, and the weights of the water and silt both acting on the

upstream side of layer are calculated in Subroutine V E R F O R C E .

197


START THE FIRST MAIN SUB-PROGRAM (SOB

I CALL INPUTDATA

I r, f

LOOP -m~ S M _ 1 T Q NUMBER OF LAYERS-.

T PROCESS OF THE FRST MAIN

SUB-PROGRAM (SOB)

CALL MESH

T END THE FIRST MAIN SUB-PROGRAM (SOB)

I START THE SECOND MAIN SUB-PROGRAM (MAINI!

J LOOP

N=l TO NUMBER OF ITERATION FOR LODING STEP

T PROCESS OF THE SECOND MAIN

SUB-PROGRAM (MAINI)

T -.--__-.

END THE SECOND MAIN SUB-PROGRAM (MAINI)

Fig. 6.3.3.1 Abbreviated flowchart

198


On the basis of the forces acting on each layer, the layer is analysed by the program.

The effective distances from the forces to the point of rotation including the horizontal

distances for the vertical forces, and the vertical distances for the horizontal forces

acting on the layers, are calculated in Subroutine DIST.

Regarding the external stability analysis of the layers, the minimum required base

width of the layer is checked in Subroutines B E A O P T M , SLIDOPTM, O V T U O P T M ,

and O V S T O P T M against sliding, overturning, bearing capacity, and overstressing

failure states, respectively.

Internal stability analysis of the layers can be analysed based on CGM, MCGM, or N e w

Coherent Gravity Method included in Subroutines C G M , M C G M , or N C G M ,

respectively (for detailed methods see Chapter 4). The choice of the method which will

be used is optional.

Subroutine R E I N A R E A computes the minimum required cross-sectional area of the

reinforcements. The area should be designed considering the rupture failure and talcing

into account to the methods of internal stability analysis. Minimum required

reinforcement lengths within the layers of the dam against bond failure are calculated in

this subroutine based on equations of the three methods mentioned above. The

optimum net weights of reinforcements within the dam at different levels and the

optimum net total weights of the reinforcements are calculated based on the methods in

this subroutine. In Subroutine O P T M , the base widths are compared and the minimum

required to prevent failure is determined.

Subroutine M E S H serves to subdivide the optimum geometry of the dam into

incremental four-node elements, and to prepare input data for the second main sub

program. The number of nodal points, the number of elements including interface

elements, the coordination of nodal points, and the slopes of the dam facings are

calculated at this stage. More detailed properties of the material used within the dam,

the position of forces acting on the nodal points, the locations of reinforcements, and

199


the boundary condition, required as input data for the second main program, are

prepared at this stage. A general view showing the subdivisions of a typical RSD is

illustrated in Fig. 6.3.3.2.

Fig. 6.3.3.2 A general view of a typical RSD showing subdivisions

6.3.3.2 Second main sub-program

The second main sub-program serves to control the calling subroutines (NDF,

E B T E D A , T S S M , SSMILV, T A N E S H , E S M , SBE, SIE, VSE, STIE, S E E P A G E ,

PSTMS), and the processes of iterations of calculation. This is further explained in the

subsequent section.

Subroutine N D F computes the structure of the stiffness matrix. The initial stresses

within the dam, due to its weight is calculated by Subroutine E B T E D A . Subroutine

T S S M assembles the results in the stiffness matrix. The equations representing the

stiffness matrix and loading vectors are solved in Subroutine SSMILV. The stresses

and strains of two dimensional elements are computed in Subroutine T A N E S H .

Subroutine E S M computes the stress-strain matrix. Principal stresses and maximum

shear stresses for material elements are calculated in Subroutine P S T M S . The stiffness

of reinforcement are computed in Subroutine SBE. The stiffness of interface elements

are computed in Subroutine SIE. Subroutine STIE calculates the stresses in interface

200


elements. Subroutine S E E P A G E computes the equivalent seepage level due to pore

water pressure changes within the dam. Ultimately, Subroutine V S E models the

material properties used within the dam.

A two-dimensional quadrilateral element has been used in the program to represent the

soil behaviour, while a general stress-strain curve is assumed in order to model the

behaviour of the soil within the dam. A non-linear hyperbolic stress-strain curve is

used in the program to represent the primary loading, while a linear response is

assumed for the unloading or reloading behaviour of the soil.

One dimensional interface elements have been used in the program to permit relative

movement between the soil and the concrete facings. Interface elements, which have

no thickness, have been defined by four nodes, with each of two pairs having the same

coordinates. The interface elements response has been modelled in the program

considering linear or hyperbolical variation of the shear stress with shear displacement

until a specified shear strain is reached.

6.3.4 Output Data

Output data contains two main parts, called D A M L O U T and D A M 2 . 0 U T , which will

be explained in the following two sections:

6.3.4.1 First Part

The output data consists of the minimum required base width of the cross-sectional area

of RSDs to prevent the failures due to sliding, overturning, over-stressing, insufficient

bond, and rupture. The optimised base width is printed out based on the minimum

required base length for no failures.

In this part, the results of analysis are printed out after each iteration. The output data

includes (a) the numbers and the heights of layers, (b) the depths of the water and the

silt acting on the upstream side of the layers, (c) the values of the base width of the

201


layers, (d) the weight of layers, (e) the weight of water and silt acting on the upstream

side of layers, (f) the values of hydrostatic force, (g) the silt force, (h) the ice force, and

(i) the direct and indirect forces of earthquake acting on the layers. The optimum

required crest and base widths of the dam are computed by the program. The main

output data in this part represents the minimum required base width of the layers versus

sliding, overturning, overstressing, rupture failure and lack of bond. If any of these

failures is likely to occur, the program will stop and a massage describing the mode of

failure will be shown in the output data file. The detailed output for the example of

RSD with the height of 20m is shown in Appendix F.

6.3.4.2 Second Part

In the second part, the nodal point data and element data are printed out in

D A M 2 . 0 U T . The results of analysis include (a) the horizontal and vertical

displacements of nodal points, (b) the horizontal, vertical and principal stresses within

the elements, and (c) the maximum shear stresses for material elements. These are

printed out after each iteration. The values of stresses in the reinforcements are

included in this part, too. The output for the second part of example of the 20m high

RSD is shown in Appendix F.

6.4 CONCLUSION

In this chapter, the RSD dam has been divided into several layers which have been

separately analysed based on the proposed equations governing sliding, overturning,

and overstressing as the equations of external stability analysis, and the equations

governing bond failure and rupture failure as internal stability analysis equations of the

dam. The formulae of soil-reinforcement interaction has also been presented in this

chapter. A computer program has been developed based on these equations for: (a)

optimisation of a parametric RSD, and (b) the analysis of the optimised RSD based on

the finite element method.

202

ANALYSIS CHAPTER SEVEN

CHAPTER SEVEN

ANALYSIS

7.1 INTRODUCTION

Six models of RSDs, with the heights of 20m, 25m, and 30m have been analysed for

safety factors equal to 1 and 1.5. The purpose of analysis of these models was to find

the variation of the minimum required base width of dam (or the layers) versus the dam

height for various safety factors to find the geometrical optimisation of these dams. It

was found that increases in the safety factors cause a non-linear increase in the

minimum required base length of RSD (or the layers of dam). Also, the increase in the

height of dam leads to a non-linear increase in the minimum required base length of

dam (or the layers of dam) to maintain stability. Although these effects are small when

the safety factors are equal to 1, they increase greatly when the safety factors increase

from 1 to 1.5. This will be discussed in greater detail in Sec. 7.2.

In addition, the 30m high RSD has been analysed by the program in order to find the

variation of stresses and deformations. The dam was analysed under plane strain

conditions in the following four configurations: (a) without reinforcements, (b) with an

assumed increased stiffness of the soil fill (due to the presence of reinforcement), (c)

with horizontal reinforcement, and (d) with inclined reinforcements. It was found that

placing reinforcement within the dams, can reduce the displacement and stress values in

the dam fill. Changing the direction of reinforcement results in further reduction of

these values. Also, it was found that the analyses of dams based on the soil stiffness

increase is much less effective than the analyses which include the existence of

reinforcement in soil fill. This is given in greater detail in Sec. 7.3.

203


7.2 G EOMETRICAL OPTIMISATION

Initially, a 20m high RSD was analysed using the RSDAM Program under the two

following conditions. Firstly, it was assumed that the safety factors against internal and

external modes of failures were equal to 1, and then, in second part, it was assumed that

the safety factors were equal to 1.5.

For both these safety factors, the following data were assumed: (a) the levels of water

in upstream side and in downstream side of the dam of, respectively, 20m and 2m; (b)

the height of silt acting on the upstream of 6m; (c) the unit weight of the silt of 15

kN/m?; (d) the unit weight of the reinforced earth soil of 20 kN/m?; (e) the coefficient

of earthquake acceleration of 0.15; (f) the allowable bearing capacity of foundation soil

of 700 kN/m2; (g) the internal angle of friction of soil of 35 degree, and (h) the

coefficient of uniformity of soil of 200.

Each 60 mm wide reinforcement was also assumed to be connected to one lxlm facing

panel, and the allowable tension of reinforcements was assumed to be 240 MN/m2 in

both conditions. Initial widths of the crest and the base of the dam were assumed to be

4m and 10m, respectively. The final widths of the dam base and crest, computed by the

program, are shown in Table 7.2.1. The minimum required base width to prevent bond

failure was also assumed to be calculated based on the New Coherent Gravity Method

(See Chapter 4).

Table 7.2.1 Final widths of the dam computed by the program

Final crest width

Final base width

Safety factors = 1

6m

17.4m

Safety factors = 1.5

6m

32m

The minimum required base widths of the layers versus height of the 20m high dam

(for safety factors equal to 1 and 1.5) are shown in Fig. 7.2.1. This figure shows that

204


the increase in the safety factor leads to an increase in the minimum required base

width of the dam (or the layers). This figure also shows that the minimum required

base width of the layers appears to be governed by:

(a) the sliding failure in about the bottom half of dam when safety

factor is 1.

(b) the bond failure in about the top half of dam when safety factor is 1.

(c) the sliding failure in about the bottom two third of dam when safety

factor is 1.5.

(d) the bond failure in about the top one third of dam when safety

factor is 1.5.

In a similar way, four other RSDs with the heights of 25m and 30m, and safety factors 1

and 1.5 have been analysed in order to find the effect of dam height versus the

optimum required base width. Different assumptions made during the analysis of these

dams are shown in Table 7.2.2. Also, it has been assumed that the levels of upstream

water were equal to the heights of dams.

Table 7.2.2 Assumptions accepted during the analysis of the models

Model No.

Height (m)

Safety factor against sliding

Safety factor against overturning

Safety factor against over-stressing

Safety factor against bond failure

Safety factor against rupture failure

1

25

2

25

1.5

1.5

1.5

1.5

1.5

3

30

1

1

1

1

1

4

30

1.5

1.5

1.5

1.5

1.5

205


Height (m)

30 -

20 -

10 -i

0 -c

Q ^

)

Height (m)

30 -

20 -

20 -

rt .

0

5F=i

____ Bond failure

^^- Overstressing

_ _ Overturning

!L . Sliding

20 40 60

Minimum required base width (m)

SF=1.5

Bond failure

"•L Overstressing

"*L "L , Overturning

\ ta Sliding

20 40 60 Minimum required base width (m)

80

1

80

Fig. 7.2.1 Minimum required base length versus height for the 20 m high dam

206


The minimum required base widths of the layers versus height for 25m and 30m high

dams are shown in Figs. 7.2.2 and 7.2.3, respectively. As indicated by these figures,

the increase in the height of dam leads to an increase of the minimum required base

width of the dam (or the layers). The changes are small when safety factors are equal

to 1, while the changes increase greatly when the safety factors increase to 1.5.

The minimum required base width of the layers appears to be governed by:

(a) the sliding failure in about the bottom half of dam when safety

factor is 1.

(b) the bond failure in about the top half of dam when safety factor is 1.

(c) the sliding failure in about the bottom two third of dam when safety

factor is 1.5.

(d) the bond failure in about the top one third of dam when safety

factor is 1.5.

(e) the overstressing failure in the base when safety factor is 1.5 only

for 30m high dam.

As a result, for no failure (due to sliding, overturning, overstressing and bond failure),

the minimum required base widths of the layers of dams should be checked against the

minimum required base width for no bond failure in about the top one third of the dam

when safety factor is 1.5. These should also be checked against the required base

length for no sliding failure in the remaining part of the dam when safety factor is 1.5.

Over-stressing failure needs to be considered when the height of dam reaches 30m and

factor of safety is 1.5.

207


Height (m)

30 -

20 ~l\ V

-

10 -

n •=

0

Height (m)

30 ~

20 -

10 -

o

0

SF=1

___ Bond failure

^^ Overstressing

L ___ Overturning

\ J L Sliding

20 40 60


SF=1.5

Bondfailure

±M ^^Overstressing

it Tl ^^.Overturning

4* Ti Sliding

20 40 60


- -+ 80

i \

80

Fig. 7.2.2 Minimum required base length versus height for a 25 m high dam

208


SF=1

Bondfailure

Overstressing

verturning

Sliding

20 40 60


80

0

SF=1.5

Bondfailure

Overstressing

Overturning

Sliding

20 40 60


80

Fig. 7.2.3 Minimum required base length versus height for a 30 m high dam

209


7.3 NUMERICAL ANALYSIS

It was assumed that the 30m high earth dam with vertical downstream face was built on

a rigid foundation, with the base width of the dam to be 40.4m. Two 0.2m thick

concrete facings are assumed to be on both side slopes of the dam, as shown in Fig.

7.3.1.

5.4m

30m

Concrete facings

Rigid Foundation

40.4m

Fig. 7.3.1 The 30m high vertical downstream earth dam

For the static analysis, it was assumed that the dam was constructed on rigid foundation

hence there was no horizontal nor vertical movements at the dam base level. In time-

history analysis it was, however, assumed that the dam base had moved proportional to

0.15m and -0.08m base displacements.

7.3.1 Loading steps

The dam has been analysed under loadings which included the weight, hydrostatic

pressure, seepage, and earthquake. The horizontal and the vertical forces, due to

upstream hydrostatic pressure, were calculated and applied to the upstream nodal points

210


(Nodal points 155 to 165 in Fig. 7.3.2.1), while the downstream hydrostatic pressure

was assumed to be zero. The height of its water was assumed to be 30m in maximum

condition. The variations of top seepage lines were assumed to be as illustrated in Fig.

7.3.1.1. The coefficient of earthquake acceleration, based on static method, was taken

to be 0.2g in the calculation of the earthquake force acting on the nodal points. T w o

increments of earthquake displacement acting on the base of dam were applied to the

dam using time-history analysis.

Fig. 7.3.1.1 Variations of seepage lines

7.3.2 Mesh information

Number of four-external nodal elements including interface elements, but excluding

reinforcement elements was 140 as shown in Fig. 7.3.2.1. The total number of nodes

considered in the analyses was 165, and the number of interface elements was 20. The

positions of horizontal and the inclined reinforcements used are shown in Figs. 7.3.2.2

and 7.3.2.3, respectively.

211


Fig. 7.3.2.1 A general view of the RSD showing nodal points

• i" ,• _ Horizontal reinforcements

i * >

i \

Fig. 7.3.2.2 Positions of horizontal reinforcements

-, U . - V-X-^S-S S S. Inclined reinforcements

Fig. 7.3.2.3 Positions of inclined reinforcements

212


7.3.3 Material property

The assumed properties of the soil, concrete facings, interface elements, and

reinforcements are shown in Table 7.3.3.1 to Table 7.3.3.4, respectively.

Table 7.3.3.1 Assumed soil properties

Unit weight

Angle of internal friction

Cohesion of the soil

Initial tangent exponent

Initial unload-reload exponent

Loading coefficient

Unloading coefficient

Ratio of measured strength at failure to ultimate

strength

Minimum initial tangent modulus

Bulk modulus exponent

Bulk modulus coefficient

Tangent modulus at failure

16 kN/m3

35 degrees

0

0.5

0.5

300

500

0.8

' 1 MN/m2

0.2

250

50 MN/m2

Table 7.3.3.2 Assumed concrete facing properties

Unit weight

Young modulus

Poisson's ratio

24 kN/m3

25 GN/m2

0.2

Table 7.3.3.3 Assumed interface element properties

Interface cohesion

Interface friction angle between soil and concrete

Initial shear stiffness

Failure shear stiffness

Initial normal stiffness

Failure normal stiffness

0

25 degree

40 kN/m2

lkN/m2

1 GN/m2

1 kN/m2

213


Table 7.3.3.4 Assumed reinforcement properties

Unit weight

Young modulus

Poisson's ratio

78 kN/m3

250 GN/m2

0.2

7.3.4 Stages of analysis

The 30m high RSD was analysed by the program in the following stages: (a) without

reinforcements, (b) assuming increased stiffness of the soil due to the presence of

reinforcements, (c) with horizontal reinforcement strips, and (d) with inclined

reinforcements. The results are given in the following sections.

7.3.5 Displacement variation

In the first stage, the dam was analysed without reinforcements. After the analysis, it

was found that significant horizontal displacements appeared in the nodal points of the

dam due to the action of forces such as weight, hydrostatic force, seepage force and

earthquake. Locations of the nodal points of the dam before loading and after loading

are shown, respectively, in Figures 7.3.5.1 and 7.3.5.2a.

In the second stage, the dam was re-analysed under conditions assuming increased

stiffness of the soil (used within the dam) due to the presence of reinforcements.

Similarly to the first stage, the soil and the concrete facings formed the only dam

materials. It was assumed that the effect of the reinforcement insertion increases the

stiffness of the soil material proportional to the ratio of the cross-sectional area of

reinforcements to unit area of soil. It was found that although the displacements values

were reduced, a considerable horizontal displacement still appeared in the dam, due to

the forces acting on the dam as shown in Fig. 7.3.5.2b.

In the third stage, the dam was re-analysed again with horizontal reinforcements as

illustrated in Fig. 7.3.5.2a. The result of the deformed dam was plotted in Fig.

214


7.3.5.2c. W h e n this figure is compared with Figures 7.3.5.2b and 7.3.5.2a, it is shown

clearly that the horizontal displacements of the nodal points are reduced. Therefore,

horizontal displacements can be decreased by inserting horizontal reinforcements

within the dam.

As the final stage of displacement analyses, the dam was re-analysed once more with

inclined reinforcements as shown in Fig. 7.3.5.2.b. It was found that inserting inclined

reinforcements still decreases the displacements even more than the other stages. Fig.

1.3.5.26. shows the deformation of the dam (with inclined reinforcements) after the

analysis.

Fig. 7.3.5.1 The dam before loading

7.3.6 Stress variation

Stress variation should be considered in the analysis of reinforced earth structures to

locate the high stress levels. The variation of principal stresses, due, to the forces

acting on the dam in the four stages of analysis has been contoured and plotted in

Figures 7.3.6.1a to d. These clearly show that using reinforcement within the earth

dam with vertical downstream face reduces the maximum principal stress acting on the

elements to more than half of the maximum principal stress of the dam without

215


reinforcement. Moreover, the use of inclined reinforcements reduces the value of the

maximum principal stress.

The changes to the horizontal stresses due to the forces acting on the dam for the four

stages of analyses are also pictured in Figures 7.3.6.2a to 7.3.6.2d supporting the

conclusion that the use of reinforcement leads to a reduction of horizontal stress level

after using reinforcement within the dam.

7.3.7 Variation of the vertical facing displacement

Vertical and horizontal movements of the vertical facing (nodal points 1 to 12 shown in

Fig. 7.3.2.1) affect stresses within reinforcements. Such displacements are plotted in

Fig. 7.3.7.1 and Fig. 7.3.7.2 regarding the four steps of analysis (a) without

reinforcements, (b) with increased stiffness of the soil fill, (c) with horizontal

reinforcement strips, and (d) with inclined reinforcements strips. Figure 7.3.7.1 shows

the variations of the facing movements based on -0.08 m base displacement, while

Figure 7.3.7.2 shows the variations based on 0.15 m base displacement.

These figures clearly show that the horizontal movement of the vertical facing is

maximum in Case a (dam without reinforcement) while the magnitude of horizontal

displacement is minimum in Case d (dam with inclined reinforcements). As shown in

Fig. 7.3.7.1, the maximum value of horizontal movement is about 0.63m in Case a at a

height equal to about 27m above the base. It is only about 0.28m in Case d and about

0.36m in Case c (dam with horizontal reinforcements) near to the dam crest. Similar

conclusions can be made by referring to Fig. 7.3.7.2.

The minimum value of vertical displacement is associated with Case a, while the

maximum value is with Case b (increased stiffness of the soil fill). Fig. 7.3.7.1b

indicates that the maximum value of vertical displacement, which is about 0.20 m,

happens at height 24m from the base for Case b, while it is 0.15m for Case c.

Maximum value of vertical displacement, which is about 0.10m, occurs near the crest.

216

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30 A. m

20..

Height

10..

-0.6 -0.4 -0.2 0

Horizontal Displacement

-0.2 -0.1 0 0.1

Vertical Displacement

Fig. 7.3.7.1 Variations of vertical and horizontal movements of the vertical facing based on -0.08 m base displacement

-0.5 -0.3 -0.1 +0.1 +0.3

Horizontal Displacement

-0.2 -0.1 0 0.1

Vertical Displacement

Fig. 7.3.7.2 Variations of vertical and horizontal movements of the vertical facings

based on 0.15m base displacement

220


7.4 CONCLUSIONS

Results of the analysis of the six models of RSDs indicate that an increase in the safety

factor leads to an increase in the minimum required base length of the RSD (or the

layers of the dam). Also, an increase in the height of dam leads to an increase in the

minimum required base length of dam (or the layers of dam). The changes are small

when the safety factor is 1, but they increase considerably when the safety factor

increases to 1.5. This is most noticeable in the minimum required base length for no

overstressing failure when the safety factor is equal to 1.5.

The presence of reinforcement also leads to a reduction in displacement and stress

level. A 30m high RSD with vertical downstream facing was analysed assuming the

following four stages: (a) without reinforcements, (b) with increased stiffness of the

soil fill, (c) with horizontal reinforcement embedded within the fill, and (d) with

inclined reinforcement within the fill. It was found that inclusion of reinforcement

within the dam material could reduce the vertical and horizontal displacements of the

dam. Changing the direction of reinforcements could also result in further reduction in

the displacements of the nodal points of the dam.

In the first stage, the dam was analysed without reinforcements within the dam. After

the analysis, it was found that significant horizontal displacements appeared in the

nodal points due to the action of forces such as weight, hydrostatic force, seepage force

and earthquake. Locations of the nodal points of before loading and after loading were

shown in this chapter.


stiffness of the soil used within the dam due to the presence of reinforcements.

Similarly to the first stage, the soil and the concrete facings formed the only dam

materials. It was assumed that the effect of reinforcement insertion increases the

stiffness of soil material proportional to the ratio of the cross-sectional area of

reinforcements to unit area of soil. It was found that although the displacements values

221


were reduced, a considerable horizontal displacement still appeared in the dam, due to

the forces acting on it.

In the third stage, the dam was re-analysed again with horizontal reinforcements. It

was shown that the horizontal displacements of the nodal points are considerably

reduced.

In the fourth stage, the dam was re-analysed once more with inclined reinforcements.

It was found that inserting inclined reinforcements still decreases the displacements

even more than the other stages. The deformations of dams after the analyses were

shown in this chapter.

The variation of maximum principal stress, due to the forces acting on the dam, in the

four stages of analysis has also been contoured and plotted in this chapter. These show

that using reinforcement within the earth dam with vertical downstream face reduces

the maximum principal stress acting on the elements to more than half of the maximum

principal stress of dam without reinforcement. The use of inclined reinforcements still

reduces the value of the maximum principal stress.

The changes to the horizontal stresses due to the forces acting on the dam for the four

stages of analyses have also been pictured in this chapter supporting the conclusion that

the use of reinforcement leads to a reduction of horizontal stress level by using

reinforcement within the dam.

Vertical and horizontal displacements of the vertical facing have been plotted in this

chapter regarding the four steps of analysis. The variations of facing movements based

on -0.08m and 0.15m base displacements are shown in this chapter (Sec. 7.3.7). It was

concluded that the horizontal displacement of the vertical facing is maximum in Case a

(dam without reinforcement) while the magnitude of horizontal displacement is

minimum in Case d (dam with inclined reinforcements). The maximum value of

horizontal movement is about 0.63m in Case a at a height equal to about 27m above the

222


base. It is only about 0.28m in Case d and about 0.36m in Case c (dam with horizontal

reinforcements) near to the dam crest.

The minimum value of vertical displacement is associated with Case a, while the

maximum value is with Case b (increased stiffness of the soil fill). It has been shown

that the maximum value of vertical displacement, which is about 0.20m, happens at

height 24m from the base for Case b, while it is 0.15m for Case c. Maximum value of

vertical displacement, which is about 0.10m, happens near the crest.

223

CONCLUSIONS AND RECOMMENDATIONS CHAPTER EIGHT

CHAPTER EIGHT

CONCLUSIONS AND RECOMMENDATIONS

8.1 INTRODUCTION

Conclusions and implications of the results of the investigations on the design on RSDs

are presented in this chapter. Part A deals with (a) the semi-empirical formulae obtained

from analysing the field data, (b) the theoretical formulae obtained from the analytical

investigation, and (c) the formulae of the natural frequencies of RSDs. Part B outlines

the main features of the computer program based on these formulae and its application in

the design of RSDs.

8.2 PART A- THEORETICAL OPTIMISATION AND ANALYSIS

The work presented in this part can be classified into three categories:

- Theoretical formulae in geometrical optimisation of RSDs.

- Semi-empirical formulae in the design of reinforced soil structures.

- The formulae of the natural frequency of RSDs.

Several investigations have been performed in each category. The results from the

developed computer program concern the overall behaviour of RSDs.

8.2.1 Geometrical optimisation

The aim of research was to find the theoretical formulae in the geometrical optimisation

of RSDs. Initially, in this thesis (Chapter Three), general and possible classifications of

RSDs based on their types and components have been considered. RSDs, based on the

material used, can be classified into homogeneous fill types and zoned types. The

components, properties and types of homogenous fill and zoned RSDs have all been

considered.

224


The identification of the forces acting on RSD is also fundamental to study this

behaviour. Although there are no major differences between the forces acting on RSD

and the forces acting on other types of dams, the behaviour of RSD and other dams are

different in withstanding the forces. The main forces assumed to act on a RSD are those

due to water pressure, silt pressure, ice pressure, earthquake pressure, foundation

reaction, seepage and the weight of the structure. In Chapter Three, the forces acting on

a RSD were individually discussed, and, at the end, the combination of the loads

(including usual loading, unusual loading and critical loading) were defined.

This was followed by the stability analysis of RSDs which was addressed from the point

of view of both internal and external stabilities. In Chapter Four, the external stability of

RSDs was evaluated based on the analytical approach. Sliding, overturning, and

overstressing were considered in the external stability analysis. In the external stability

analysis, it was assumed that the whole reinforced soil structure acts as a unit. To

optimise the geometry, the formulae of minimum required base length for no failure due

to sliding, overturning, and overstressing was proposed and evaluated separately for the

dam and its layers.

8.2.2 Semi-empirical relationships

The research undertaken in this area was concerned with the analysis of the field data

using the concept of frictional interaction between soil and reinforcement. Various

analytical theories developed so far are not in sufficient conformity with the observed

behaviour of reinforced earth structures. This necessitated the use of empirical

relationships in the current design practice. Some of these relationships and the field

data, on which the previous experimental formulae were based, were re-analysed in this

thesis. N e w proposed semi-empirical formulae reflecting the observed behaviour of

reinforced earth structures have been suggested. The proposed formulae make

adjustments to exiting formulae based on the field observations. The findings,

corresponding to the investigations performed, are n o w summarised.

225


The internal stability of the reinforced earth structures can be analysed using methods

based on conventional principles of soil mechanics. It has been known, however, that

certain theoretical assumptions, accepted in these methods, were not supported by the

observations. In particular, since reinforcement changes the state of stress within soil,

the directions of principal stresses are no longer vertical and horizontal, and the ratio of

the vertical stress to the horizontal stress is not constant. This, together with other field

data regarding the bond between soil and embedded reinforcement, have led to the

development of the semi-empirical methods. One of the methods termed CGM,

proposed by McKittrick and Schlosser in 1978, was adopted as a recommended design

method by Reinforced Earth Company and the French Code. This method has been

structured around a set of bi-linear functions representing and interpreting the field data

in a somewhat simplified way. Some modifications to the method were later suggested

by Arenicz and Chowdhury in 1987, in order to reflect the field observation more

closely.

The proposed semi-empirical relationships suggested in this thesis (Chapter Four) have

eliminated the tangent discontinuity existing in the formulae of the CGM, and have

reflected the non-linearity indicated by the field data. Also, the proposed relationships

have eliminated unknown parameters existing in the formulae of the MCGM, and have

offered a better fit with the available field observations.

A linear relationship between the apparent friction factor and the ratio of fill depth to

strip length was discovered by analysing the field data. Formulae for calculation of the

safety factor against the tensile failure and bond failure of reinforcements have been

proposed regarding the formulae of lateral earth pressure coefficient, apparent friction

factor, and maximum tension line.

8.2.2 Natural frequency

The resonance phenomenon in the structure can be prevented by the designer's

familiarity of the natural frequency of the structure. Although the practice of inserting

226


reinforcement within earth dams allows reduction in fill volume, displacement, and stress

level, this causes an increase in the natural frequency of RSDs compared with the

conventional earth dams. This leads to an increase of the natural frequency of the

structure which may result in the possibility of total destruction of such dams.

Therefore, the calculation of the natural frequency of RSDs is necessary to find their

behaviour under earthquake forces.

Chapter Five discussed that the natural frequency of a RSD may be more critical than the

natural frequency of a corresponding conventional earth dam. This is because of the two

major functions: (a) the geometrical function concerning the change in dam geometry

and (b) the overall stiffness function concerning the change in the dam's flexibility.

Formulae concerning the major functions were derived and in some cases plotted. The

following general suggestions have been proposed in order to prevent resonance in

RSDs:

a) The volume of reinforcements used within the dams should be calculated

based on the formulae of the natural frequency of RSDs. Any additional increase in the

volume of reinforcement may result in the extreme situation of resonance in the structure

under an earthquake condition.

b) Using reinforcements with low stiffness such as polymers can yield a

considerable decrease in the value of the geometry function of the dams compared with

reinforcements with high stiffness

c) The effect of geometry in increasing or decreasing the natural frequency has

been found as shape coefficient and has been tabulated in Chapter Five. As a result,

inserting reinforcement within earth dams causes a decrease of the dam width and at the

same time, causes an increase in the natural frequency of the dam.

8.3 PART B NUMERICAL ANALYSIS

The state of stresses inside a soil mass normally varies from point to point. In reality, the

situation of stresses within the soil is not elastic. However, it is helpful to describe the

227


elastic behaviour of the soil before giving the explanation of the plastic response or

elasto-plastic behaviour of a soil mass. The equations representing the elastic and/or the

plastic behaviours of the soil were modelled in the program. The deformation of the

soil, the concrete facing panels, and the natural behaviour of the reinforcements were

simulated by the finite element program.

In the body of RSDs it is assumed that some forces (normally compressive forces) are

induced in the soil mass as acting forces, while some forces (normally tensile forces) are

induced in the reinforcements considering the frictional bond between the reinforcements

and the soil as reaction forces. The finite element method has been used to model the

soil deformation, to find the tensile stress within the reinforcements, and to predict the

behaviour of the bond between the soil and reinforcements. It is assumed that the

loadings cause a group of nodal forces in contact point between the soil and

reinforcements. The forces cause some deflections within the soil and reinforcements.

For a no bond failure state, the soil deflections should be compatible with the deflections

of nodal points. In this condition, the reinforcement and soil would need to be combined

by joining or spring elements modelling the slip behaviour between soil and

reinforcement. Each reinforcement carries the horizontal forces induced in the nodal

points of reinforcements.

For no bond failure, the difference between the displacement of nodal points of soil

elements and the displacement of the nodal points of reinforcement elements should be

zero. It is assumed that the displacements, due to force on nodal points within

reinforcements, should be equal and opposite to the displacements due to the forces

acting on the corresponding soil nodes. The set of soil displacement equations has been

met by a solution of the soil stiffness matrix under the external forces assuming the

stiffness of reinforcements are constant. The difference between the displacements of

nodal points of soil elements and the displacements of nodal points of reinforcement

elements were formulated in Chapter Six.

228


8.3.1 Computer Program

The program, RSDAM, written by the author as a part of the investigations, has two

main sub-programs. The purpose of the program is (a) geometrical optimisation of

RSDs and (b) evaluation of stresses and strains inside the dam increments. The

optimisation includes the fill volume optimisation of material and the reinforcement

volume optimisation. It should be noted that the program is particularly adapted to

RSDs, however, it may also be used for a variety of reinforced earth walls and

embankments.

The first sub-programs including 15 subroutines optimises the geometry of RSDs based

on semi-experimental formulae. Also, the dam is subdivided into several increments in

order to be prepared for the analysis by the second part. The second, including 13

subroutines, computes the stresses and displacements within the elements of the dam

based on the two dimensional finite element formulation.

8.3.1 Results of Analysis

Results of the analyses of six RSD models, presented in Chapter Seven, show that an

increase in the safety factor leads to an increase in the minimum required base length of

these RSDs (or the layers of dams). Also, an increase in the height of these dams leads

to an increase of the base length of the dams (or the layers of the dams). In this case, the

changes are small when the safety factor is equal to 1, but the changes increase

considerably, when the safety factor increases from 1 to 1.5. The greatest increase is

seen in the minimum required base length for a no overstressing failure state.

A 30m high RSD with vertical downstream facing was analysed by the program under

the four following stages: (a) without reinforcements; (b) assuming increased stiffness of

the soil due to inserting reinforcements; (c) with horizontal reinforcements; and (d) with

inclined reinforcements. It was found that the addition of reinforcements within the

dams can reduce the vertical and horizontal displacements. Changing the direction of

229


reinforcements may result in further reduction in the displacements of the nodal points of

dam. In some cases, the changes of the stress levels are more than 50%.

In the first stage, the dam was analysed without any reinforcements within the dam.

After the analysis, it was found that significant horizontal displacements appeared in

the nodal points of dam due to the action of forces such as weight, hydrostatic force,

seepage force and earthquake.


stiffness of the soil used within the dam due to the presence of reinforcements. It was

assumed that the effect of reinforcement insertion increases the stiffness of the soil

material proportional to the ratio of the cross-sectional area of reinforcements to the

unit area of soil. It was found that although the displacements values were reduced, a

considerable horizontal displacement still appeared in the dam, due to the forces acting

on dam.

In the third stage, the dam was re-analysed again with actual horizontal reinforcements.

It was shown that the horizontal displacements of the nodal points were considerably

reduced.

As the final stage of displacement analyses, the dam was re-analysed once more with

inclined reinforcements within the dam. It was found that inserting inclined

reinforcements still decreases the displacements even more than the other stages.

Locations of the nodal points of the dam before loading and after loadings were shown

in Chapter Seven.

The variation of principal stresses, due to the forces acting on the dam in the four stages

of analysis were contoured and plotted in Chapter Seven. These showed that using

reinforcement within the dam face reduced the maximum principal stress acting on the

elements of dam to more than half of the maximum principal stress of the dam without

reinforcement. The use of inclined reinforcements reduced the value of the maximum

principal stress more than the other cases.

230


The changes to the horizontal stresses due to the forces for the four stages of analyses

were also pictured in the chapter supporting the conclusion that the use of

reinforcement leads to a reduction of horizontal stress level within the dam.

Vertical and horizontal displacements of the vertical facing were also plotted in Chapter

Seven regarding the four steps of analysis. The variations of the facing movements

based on -0.08m and 15m base displacements were shown in the chapter.

It was shown that the horizontal displacement of the vertical facing is maximum in the

case dam without reinforcement, while the magnitude of horizontal displacement was

minimum in the case dam with inclined reinforcements. The maximum value of

horizontal movement was about 0.63m in the case dam without reinforcement at a

height equal to about 27m above the base. It was only about 0.28m in case dam with

inclined reinforcements and about 0.36m in case dam with horizontal reinforcements

near to the dam crest.

The minimum value of vertical displacement is associated with case dam without

reinforcement, while the maximum one is with Case b (increased stiffness of the soil

fill). It was shown that the maximum value of vertical displacement, which is about

0.20m, occurs at height 24m from the base for this case, while it is 0.15m for the case

dam with horizontal reinforcements. The maximum value of vertical displacement,

which is about 0.10m, occurs near the crest.

8.4 RECOMMENDATIONS

8.4.1. Reinforced soil arch dams

Although most RSDs are of the gravity type, there is no reason to claim that the

reinforced soil arch dam and the reinforced soil buttress dam can not be built in the

future. In these cases, the reinforcement may stabilise the structure by increasing the

strength of the soil and by connecting the facing panels of two sides. Therefore, it

would be of considerable interest to investigate the possibility of construction of arch

231


and buttress RSDs and, to construct experimental and numerical models of them to find

their behaviour under the forces acting on dams.

8.4.2. Cross sectional optimisation

A stability analysis of the RSD based on an analytical approach and a semi empirical

method was presented in Chapter 4 to optimise the cross sectional area. Some proposed

formulae were given for optimisation of the RSD to minimise the base length against

sliding, overturning, overstressing, bond failure, and rupture failure. Since the main

aim of this thesis was to develop a computer program for the design and analysis of

RSD, it is recommended that some hydraulic (experimental) models of the RSD be

constructed to compare the results of the experimental models to the results of the

analysis based on this computer program.

8.4.3. Behaviour of reinforcement

There is a difference between the behaviour of reinforcements used within the reinforced

soil walls and within the RSD. As indicated in Chapter 4, most reinforced soil theories

(eg. Vidal's Theory (1966), CGM (1978), and MCGM 1987)) claim that inserting

reinforcement within the soil induces a tension force in the reinforcement. The

experiments done so far (explained in Chapter 4) support this theory. It should be noted,

however, that the forces acting on the reinforced soil walls are usually perpendicular to

the reinforcement directions. While, the directions of forces acting on RSD are not

perpendicular to the reinforcement direction. Therefore, forces acting on the

reinforcement seem not to be pure tension. This needs further investigation in the

future.

8.4.4 Reinforcement width

The width of reinforcement has an important role in apparent friction factor. The

optimum width of reinforcement in RSD, and the relationship between the reinforcement

232


and the height of structure are questions which are not fully answered yet. Therefore, it

is recommended that a numerical and/or an experimental model be used to conduct

further study of these aspects.

8.4.5 Natural frequency

In Chapter Five, the natural frequency of RSD and that of conventional earth dam were

theoretically calculated and compared. It was concluded that inserting reinforcement

increases the natural frequency of the structure which may increase the possibility of

failure. It was also concluded that the natural frequency of RSD is increased because of

two major factors: (a) its geometry and (b) its overall stiffness. Comparison between

both natural frequencies using some experimental models to compare the results of

experiments with the results of theory presented in Chapter Five might lead to new

findings.

8.4.6 Seismic load based on dynamic analysis

The computer program presented in this project can analyse RSD based on static analysis

or time history analysis. However, it is an alternative to model the RSD based on

dynamic analysis. The program can be modified based on dynamic analysis.

8.4.7 Stress concentration

In Chapter Seven, it was shown that there is a stress concentration at the toe of a 30

high RSD with vertical downstream side. A n increase in the RSD height increases the

rate of stress concentration at the toe of RSD. This may be the reason why the

maximum heights of RSDs constructed so far are not more than 30m. The effect of

height increase on the stress concentration of RSD needs more investigation. It seems,

for example, that using buttresses at the downstream side of RSD may reduce the stress

concentration at the toe. Therefore, it is recommended that the stress concentration of

233


RSD is taken into account to find the relationship between height and stress

concentration at the toe of RSD.

234

REFERENCES

REFERENCES:

Al-Ashou, M. O. and Hanna T. H., (1990), "Deterioration of Reinforced Earth Elements

Under Cyclic Loading", Proceedings of International Conference on Performance of

Reinforced Soil Structures, London, British Geotechnical Society, pp. 303-307.

Alimi, I., Bacot, J., Lareal, p., Long, N. T., Schlosser, F., (1973), "Etude de .'adherence

sol-armatures", Proceedings of the 9th International Conference on Soil Mechanics and

Foundation Engineering, Moscow, Vol. 1, pp. 11-14.

Alimi, I., (1978), "Critere de Choix des Materiux de la Terre Armee - Etude de L'

adherence" Terre-armature, Thesis, LCPC.

Ambraseys, N. N., (1960), "On the seismic behaviour of earth dams", Proceedings of

the 2th World Conference on Earthquake Engineering, Tokyo, 1-35.

Andrawes, K. Z., McGown, A. and Al-Hussaini, M. M., (1978) "Alteration of soil

behaviour by the inclusion of materials with different properties" Ground Engineering,

Vol. 11, no. 6, pp. 35-42.

Arenicz, R. M., and Chowdhury, R. N., (1987), "Empirical formula in reinforced earth

design", Journal of Australian Civil Engineering Transactions, Vol. C E 29, N o 3, pp.

198-179.

Arenicz R. M., and Chowdhury, R. N., (1988), "Observed and theoretical failure

surfaces in reinforced earth backfill and their design implications", Research Report No.

S088/1, Department of Civil Engineering, University of Wollongong, Australia.

Bannerjee, P. K., (1975) "Principles of analysis and design of reinforced earth retaining

walls" J. Inst. Highway Engineering, 22, No. 1, 13-18.

Rl

REFERENCES

Baquelin, F., (1978), "Construction and instrumentation of reinforced earth wall in

French Highway Administration", Proceedings of the Symposium on Earth

Reinforcement, Pittsburgh, pp. 186-201.

Barry Cook J. and J. L. Sherard, (1985), "Concrete Face Rockfill Dams - Design,

Construction and Performance", ASCE, N e w York, pp 98 -120.

Bassett R. H., and Last, N. C, (1978), "Reinforcing earth below footings and

embankments" Proceedings of A S C E Symposium on Earth Reinforcement, Pittsburgh,

pp 202-231.

Behnia C, (1972), E'tude des routes en terre arme'e" Ing. Thesis-Paris University.

Boden, J. B., Irwin, M. J., and Pocock, R. G. (1978), " Construction of experimental

reinforced earth walls at the TRRL" Ground Engineering Vol. 11, no. 7 pp 28-37.

Cassard, G., Kern, F. and Mathieu, H. G., (1979), "Utilisation des techniques de

renforcementdans les barrages en terre", Proceedings of the International Conference on

Soil Reinforcement, Paris, Vol. 1, pp. 229-233.

Chang, J. C, Forsyth, R. A., (1977), "Design and field behaviour of reinforced earth

wall", Journal of the Geotechnical Engineering Division, ASCE, July, pp. 677-692.

Chang, J. C, (1974), "Earth reinforcement techniques", Final Report CA-DOT-TL-

2115-9-74-37, Dept. of Transport, California.

Chapuis, R., (1972), "Rapport de recherche de DEA". Institut de Mecanique de

Grenoble (unpublished internal report).

R2

REFERENCES

Das, B. M., (1990), "Principles of Geotechnical Engineering", Second Edition, P W S -

K E N T Publishing Company, Boston.

Das, B. M., (1983), "Fundamentals of Soil Dynamics", New York, Elsevier.

Dean, R. and E. Lothian, (1990), "Embankment construction problems over deep

variable soft deposits using a geocell mattress", Performance of reinforced soil

structures, British Geotechnical Society, pp 443 - 447.

Department of Transport (1978), "Reinforced earth retaining walls and bridge abutments

for embankments" Tech. M e m o , BE3/78.

Forsyth R. A., (1978), "Alternative earth reinforcements", Proceedings of ASCE

Symposium on Earth Reinforcement, Pittsburgh, pp. 358-370.

Fukuoka, M., and M. Goto, (1988), "Design and construction of steel bars with anchor

plates applying to the high embankment on soft ground", International Geotech.

Symposium on Theory and Practice of Earth Reinforcement; Rotterdam, Oct., pp. 389-

394.

Geylord H., and C. N. Geylord, (1979), "Structural engineering handbook", New York,

McGraw Hill Book Company.

Hall, C. D., (1985), "Reinforced soil structures in coastal protection", Proceedings of

Australian Conference on Coastal and Ocean Engineering, Dec. 1985, Volume 2, pp.

247-254.

Hambley, E. C, (1979), "Bridge foundation and structures", Building Research

Establishment Report, Department of the Environment.

R3

REFERENCES

Hausmann, M., (1976), "Strength of reinforced soil" Proceedings of the 8th Aust. Road

Research Conference, Vol. 8, sect. 13, pp. 1-8.

Hausmann, M., (1990), "Engineering principles of ground modification", New York,

McGraw-Hill Publishing Company.

Hird, C. C, and I. C. Pyrah, (1990), "Predictions of the behaviour of a reinforced

embankment on soft ground", Performance of reinforced soil structures, British

Geotechnical Society, pp 409 - 414.

Ingold, T. S., (1982), "Reinforced Earth", Thomas Telford Ltd, London.

Ingold, T. S., (1988), "Some factor in the design of geotextile reinforced embankments",

International Geotechnical Symposium on Theory and Practise of Earth Reinforcement;

Rotterdam, pp 413-418.

Iwasaki, K. and Watanabe, S., (1978), "Reinforcement of railway embankments in

Japan" Proceedings of the A S C E Symposium on Earth Reinforcement, Pittsburgh, pp

473-500.

Janbu, (1973)., N., "Embankment-Dam Engineering", John Wiley and Sons, New York.

John N. W. M., (1987), "Geotextiles", Blackie, Glascow, Champman and Hall, New

York.

Jones J. F. P., (1985), "Earth reinforcement and soil structures", Butterworths, London,

Boston, Sydney.

R4

REFERENCES

Koga, K., G. Aramaki, and S. Valliappan, (1988b), "Finite element analysis of grid

reinforcement", International Geotech. Symposium on Theory and Practice of Earth

Reinforcement; Rotterdam, pp 407- 411.

Koga, K., Y. Ito, S. Washida and T. Shimazu, (1988a), "Seismic resistance of reinforced

embankment by model shaking table tests", International Geotechnical Symposium on

Theory and Practise of Earth Reinforcement; Rotterdam, pp 413-418.

Lee, K. L., Adams, B. D.,& Vagnernon, J. J., (1972), Reinforced earth walls" Rep. No.

UCLA-ENG-7233.

Londe P., (1980), "Lessons from earth dams failures" Symposium on Problems and

Practice of D a m Engineering, S. N., Thailand, pp 65 - 92

Long , N. T, Guegan, Y. 8c Legeay, G, (1972), "Etude de la terre armee a l'appariel

triaxial", Rapp. de Recheche, No. 17, LCPC.

McKittrick, D. P., & M., Durbin, (1979), "World-wide development and use of

reinforced earth structures", Ground Engineering, 12, No. 2, pp. 15-21.

McKittrick, D. P., (1978), "Reinforced earth: Application of theory and research to

practice", Proceedings of the Symposium on Soil Reinforcing and Stabilising

Techniques, Sydney, pp. 1-44.

Miki, H. and K. Kutara, T. Minami, J. Nishimura, N. Fukuda (1988), "Experimental

studies on the performance of polymer grid reinforced", International Geotech.

Symposium on Theory and Practice of Earth Reinforcement; Rotterdam, Oct. 1988, pp.

431- 436.

R5

REFERENCES

Mitchell, R. J. (1983), "Earth structures engineering", Goerge Allen & Unwin Ltd.

London.

National Research Council (U.S.) Panel on Regional Networks, (1990), "Assessing the

nations of earthquakes: the health and future of regional seismograph networks",

Washington D C, National Academy Press.

Newmark, N. M., Rosenbleuth, E., (1971), "Fundamentals of earthquake engineering",

Prentice-Hall.

Newmark N. M., (1965), "Effects of earthquakes on dams and embankments",

Geotechnique, Vol. 15 (2), pp. 139-160

Pells P. J. N., (1977) "Reinforced rockfill for construction flood control", University of

Wollongong.

Reinforced Earth Company, (1985), "Steel strip durability, technical information sheet-

design", No. 2, Internal Brochure.

Reinforced Earth Company, (1988), "Reinforced earth marine and dam structures",

Internal Brochure.

Reinforced Earth Company, (1990), "The advanced retaining wall construction

technology", Internal Brochure.

Richter, C. F„ (1958), "Elementary seismology", W. H. Freeman and Co.

Romanoff M., (1959), "Underground corrosion" US National Bureau of Standards,

Circular 579.

R6

REFERENCES

Schlosser, F., and Long, N. T., (1973), "Etude du comportment du materiau terre

armee" Annies de l'inst Techq. du Batiment et des Trav. Publ. Suppl. No. 304. Se'r.

Mater. No. 45.

Schlosser , F. & N. T., Long, (1974), "Recent results in French research on reinforced

earth", Journal of Const. Div., A S C E , 100, No. C03, pp. 223-237.

Schlosser, F., Elias, V., (1978), "Friction in reinforced earth", Proceedings of the

Symposium on Earth Reinforcement, Pittsburgh, pp. 735 - 763.

Schlosser, F., (1978), "History, current and future developments of reinforced earth",

Proceedings of the Symposium on Soil Reinforcing and Stabilising Techniques, Sydney,

pp. 5-28.

Schlosser, F., Segrestin, P., (1979), "Local stability analysis method of design of

reinforced earth structures", Proceedings of the International Conference on Soil

Reinforcement, Paris, Vol. 1, pp. 157 - 162.

Schlosser, F., and P. D. Buhan, (1990), "Theory and design related to the performance

of reinforced soil structures", Proceedings of the International Conference on

Performance of Reinforced Soil Structures, London, British Geotechnical Society.

Sherard J. L., R. J. Woodward, S. F. Gizienski and W. A. Clevenger, (1963), "Earth and

earth rock dams", Wiley, N e w York.

Shercliff D. A., (1990) "Reinforced embankment theory and practice", Thomas Telford,

London.

R7

REFERENCES

Sims, F. A, and Jones C. J. F. P., (1979), "The use of soil reinforcement in highway

schems" Proceedings of the International Conference on Soil Reinforcement, Paris, vol.

2, pp 367-372.

Singh B., (1976), "Earth and rockfill dams", Nauchandi, Meerut: Sarita Prakashan.

Smith, A. K. C. S, & P. L., Bransby, (1976), "The failure of reinforced earth wall by

overturning", Geotechnique 26, No. 2, pp. 376-381.

Smith N., (1971), "A history of dams", P. Davis, London.

Sowers G. B. & Sowers G. F., (1970), "Introductory soil mechanics and foundations",

New York, Macmillan.

Sowers G. F., (1979), "Introductory soil mechanics and foundations: Geotechnical

engineering", New York, Macmillan.

Steiner, R. S., (1975), "Reinforced earth bridges highway sinkhole" Civil Engineering,

ASCE, July, pp. 54-56.

Streeter V. L., and Wylie E. B. (1979), "Fluid mechanics", McGraw-Hill, USA.

Taylor, J. P. and Drioux, J. C, (1979), "Utilisation de la terra arme'e dans le domaine

des barrages", Proceedings of the International Conference on Soil Reinforcement; Paris,

Vol. 2, pp 373 - 378.

Terre Armee International, (1987), "Quay walls built underwater", Australian and

Canadian Prototypes, Technical Report, N o M 6 .

R8

REFERENCES

United State Committee on Large Dams and Committee on Failures and Accidences to

Large Dams, (1975), " Lessons from dam incidents, USA", A S C E and U S C O L D , N e w

York.

United States Bureau of Reclamation, (1977), "Design of small dams", United States

Department of the Interior, Bureau of Reclamation, 2nd edition, Washington, D. C ,

U.S. Government Printing Office.

Vidal, H., (1966), "Diffiusion restpeinte de la terre armee" Institute Technique du

Batement et des Trovause Publics, pp 888-939 No. 223-4.

Vidal, H., (1969), "The principle of reinforced earth", Highway Res. Rec, No. 282, pp.

1-16.

Vidal, H., (1978), "The development and future of reinforced earth" Proceedings of the

Symposium on Earth Reinforcement, Pittsburgh, Pennsylvania, pp. 1-61.

Vidal H., (1986), "A brief history of terre armee (Reinforced Earth)", Reinforced Earth

Company, Technical Report No. 635.

Wahlstorm, E., (1974), "Dams, dams foundations, and reservoir sites", Elsevier

Scientific Publishing Co.

Wolff T. F., (1985), "Analysis and design of embankment dam slopes: A probabilistic

approach", Ann Arbor, Michigan.

Wu P. and R. J. H. Smith, (1990), "Reinforced earth marine wall experienced in Canada

and United Kingdom", Proceedings of the International Conference On Performance of

Reinforced Soil Structures, British Geotechnical Society.

R9

REFERENCES

Yamanouchi, T., (1970), "Experimental study on the improvement of the bearing

capacity of soft ground by laying a resinous net", Proceedings of the Symposium on

Foundations on Interbended Sands, Australia, Commonwealth Scien. & Indus. Res.

Orgn. pp. 144-150.

Yang, Z, (1972), "Strength and deformation characteristics of reinforced sand" Ph.D.

Thesis, U C L A .

Yin Zong Ze, (1990), "Effect of reinforcement in embankment", Performance of

reinforced soil structures, British Geotechnical Society.

RIO

APPENDICES

Al

EARTH DAM FAILURES APPENDIX A

APPENDIX A- EARTH DAM FAILURES

Table 1A- Earth dam failures due to hydraulic problems (Sowers, 1961)

Form

Overtopping

Wave erosion

Toe erosion

Gulling

General characteristics

Flow over embankment, washing out dam

Nothing of upstream face by waves, currents

Erosion of toe by outlet

discharge

Rainfall erosion of dam face

Causes

Inadequate spillway capacity

Clogging of spillway with debris

Insufficient freeboard due to settlement, skimpy design

Lack of riprap, too small riprap

Spillway too close to dam

lack of sod or poor surface drainage

Preventive measures

Spillway designed for maximum flow

Maintenance, trash booms, clean design

Allowance for

freeboard and settlement in design; increase crest height or add flood parapet

Properly design riprap

Training wall

sod, fine riprap; surface drains

A2


Table 2.A- Earth dam failures due to structural failures (Sowers, 1961)

Form

Found ation slide

Upstre am slope

Downs tream slope

Flow

side


Sliding of entire dam, one face or both faces in opposite directors, with bulging of foundation

Slide in upstream face with little or no bulging in foundation below toe

Slide in downstream face

Collapse and flow of soil

in either upstream or downstream direction

Causes

Soft or weak

foundation

Excess water pressure in confined sand or silt seams

Steep slope

Weak embankment

soil

Sudden drawdown of

pond

Steep slope

Weak slope

Loss of soil strength by seepage pressure or saturation by

seepage or rainfall

Loose embankment

soil at low cohesion, triggered by shock,

vibration seepage, or foundation movements

Preventive measures

Flatten slope; employ broad berms; remove weak material; stabilise

soil

drainage by deep drain trenches with protective filters; relief wells

Flatten slope or employ berm at toe

Increase compaction;

better soil

Flatten slope, rock berms; operating rales

flatten slope or employ berm at toe

Increase compaction;

better soil

core internal drainage with protective filters; surface drainage

Adequate compaction

A3


Table 3. Earth dam failures due to seepage failures (Sowers, 1961)

Form

Loss of water

Seepage erosion or

piping


Excessive loss of water from reservoir /or occasionally increased

seepage or increased groundwater levels near reservoir.

Progressive internal erosion of soil from downstream side of dam or foundation

toward the upstream side to

form an open conduit or

pipe.

Often leads to a washout of

a section of the dam.

Causes

Pervious reservoir rim or bottom.

Pervious dam foundation.

Pervious dam

Leaking conduits.

Settlement cracks in dam.

Shrinkage cracks in dam.

Settlement cracks in

dam.

Shrinkage cracks in dam

Pervious seams in

foundation

Pervious seams, roots,

etc., in dam.

Preventive measures

Banked reservoir with

compacted clay or chemical admix: grout

seams, cavities.

Use foundation cutoff; grout; upstream blanket

Impervious core.

Watertight joints; water stops; grouting.

Remove compressible foundation, avoid sharp changes in abutment

slope, compact soil at high moisture.

Use low plasticity clays for core, adequate

compaction.

Remove compressible foundation, avoid sharp

changes, internal

drainage with protective

filters.

L o w plasticity soil;

adequate compaction;

internal drainage with

protective filters.

Foundation relief drain

with filter; cutoff.

A4

TYPICAL TYPES OF DAM"S SOIL APPENDIX B

APPENDIX B- TYPICAL TYPES OF DAM'S SOIL

Table LB Typical types of soil in or under dams (U. S. Bureau of Reclamation, 1974) Typical names of soil groups

Well graded gravels, gravel sand mixture, little or no fines

Poorly graded gravels, gravel sand mixtures, little

or no fines

Silty gravels, poorly graded gravel - sand - silt mixture s

Clayey gravels, poorly

graded gravel, clay mixtures

Well graded sands, gravel sands, little or no fines

Poorly graded sands,

gravelly sands, little or no fines

Silty sands, poorly graded sand - silt mixtures

Clayey sands, poorly graded sand- clay mixtures

Inorganic silts and very fine sands, rock flour, silty or clayey fine sands with slight plasticity

Inorganic clay of low to

medium plasticity, gravelly clays, sandy clays, silty

clays, lean clays

Organic silts and organic silt

- clays of flow plasticity

Inorganic silts, micaceous or

diatom aceous fine sandy or

silty soils, elastic silts

Inorganic clay of high plasticity, fat clays

Organic clays of medium to

high plasticity

Group

symbols

G W

GP

G M

GC

sw

SP

SM

SC

ML

CL

OL

M H

CH

OH

Seepage important

1

2

3

4

6

5

7

8

9

10

Seepage not

important

1

3

4

6

2

5

7

8

9

10

11

12

13

14

Permanent

reservoir

Positive cutoff or blanket

Positive cutoff or blanket

Core trench to none

None

Positive cutoff or upstream blanket and toe drains

Positive cutoff or

upstream blanket and toe

drains

Upstream blanket and toe drains

None

Positive cutoff or

upstream blanket and toe drains

None

None

None

None

None

Flood water retarding

Control only within volume acceptable plus pressure relief if required


None

None



Sufficient control to

prevent dangerous seepage piping

None

Sufficient control to prevent dangerous seepage piping

None

None

None

None

None

Note: No. 1 is considered the best

A5

TYPICAL TYPES OF DAM'S SOIL APPENDIX B

Table 2.B Typical types of soil in or under dams (U. S. Bureau of Reclamation, 1974)

Group

symbols

GW

GP

GM

GC

SW

SP

SM

sc

ML

CL

OL

MH

CH

OH

Homogeneous

embankment

-

-

2

1

-

-

4

3

6

5

8

9

7

10

Core

-

-

4

1

-

-

5

2

6

3

8

9

7

10

Shell

1

2

-

-

3 if gravelly

4 if gravelly

-

-

-

-

-

-

-

-

Resistance to piping

Good

Good

Poor

Good

Fair

Fair to poor

Poor to very poor

Good

Poor to very poor

Good to fair

Good to poor

Good to poor

Excellent

Good to poor

Note: No. 1 is considered the best

A6

TYPICAL TYPES OF DAM'S SOIL APPENDIX B

Table 3.B Soil performance in or under dams (Bureau of Reclamation, 1974)

Group

symbols

GW

GP

GM

GC

SW

SP

SM

sc

ML

CL

OL

MH

CH

OH

Permeability

when compacted

Pervious

Very pervious

Semipervious to

impervious

Impervious

Pervious

Pervious

Semipervious to

impervious

Impervious

Semipervious to

impervious

Impervious

Semipervious to

impervious

Semipervious to

impervious

Impervious

Impervious

Shear strength

when compacted

and saturated

Excellent

Good

Good

Good to fair

Excellent

Good

Good

Good to fair

Fair

Fair

Poor

Fair to poor

Poor

Poor

Compressibility

when compacted

and saturated

Negligible

Negligible

Negligible

Very low

Negligible

Very low

Low

Low

Medium

Medium

Medium

High

High

High

Workability as

a construction

material

Excellent

Good

Good

Good

Excellent

Fair

Fair

Good

Fair

Good to Fair

Fair

Poor

Poor

Poor

A7

ICE PRESSURE TABLES APPENDIX C

APPENDIX C- ICE PRESSURE TABLES

Table l.C Ice pressure (kN/m) (

Vertical shores, solar energy considered

Ice

thickness

m

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

Air temperature increase( C)

9 18 27

0

17

31

43

54

64

74

84

94

104

114

124

134

0

35

58

64

86

96

106

116

126

136

146

158

166

0

60

90

115

130

140

150

160

170

180

190

200

210

(US Bureau of Reclamation, 1977)

Vertical shores, solar energy considered

Ice

thickness

m

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2


9 18 27

0

26

50

62

82

100

115

130

145

160

175

190

205

0

50

83

112

129

146

163

180

197

214

231

246

265

0

110

160

180

202

218

234

250

266

282

296

314

330

Vertical shores, solar energy neglected

Ice

thickness

m

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2


9 18 27

0

10

20

30

40

50

60

70

80

90

100

110

120

0

16

29

41

52

63

74

85

96

107

118

129

140

0

30

55

75

80

103

114

125

136

147

158

169

180

Vertical shores, solar energy neglected

Ice thickness

m

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2


9 18 27

0

19

37

54

70

85

100

115

130

145

160

175

190

0

20

39

67

84

101

118

135

152

169

186

203

220

0

60

95

120

140

163

180

197

214

231

248

265

282

A8

BOND AND BREAK FAILURES EQUATIONS APPENDIX D

APPENDIX D- BOND AND BREAK FAILURES EQUA TIONS

Table l.D Factors of safety formulae against both break and bond failures based on

CGM DESIGN

EQUATIONS

FSy

(for both smooth and ribbed strips)

FS(|)

(for smooth strips)

FS<>

(for ribbed strips)

C G M (McKITTRICK, 1978)

yiK0+(Ka-K0)i]ysvsH

i yK yS Sv ' aJ v H O.SB.(L.-0.3H)

i i '

[Kn+(K -Kn)^-]S ST, 0 a 0y6J v H

O.%B.(L.-0.3H)

K S Srr a v H O.SB.[L.-0.6(H-y)]

[K~+(K -Kn)2-]S Srr 0 a 0 g v H

0.8_S.[L.-O.6(ff-y)]

K S SJJ a v H

2B.(L.-0.3H)[f*(l-^)+^ tan(|>]

[Kn+(K -Kn)^]S Srr L 0 a 0 6 v H

2_S.(L.-0.3//)tan<|>

K S Srr a v H

2B.[L.-0.6(H-v)][/J(l-^) + tan<j>]

[Kn+(K -Kn)?-]S Srr 1 0 a 0^6J v H

2B.[L.-0.6(H-y)] tan(J)

[K^+iK -K~)-]S Srr L 0 a 0 g v H

CONDITION

for (0 < y < 6m)

for (y < 6m)

for (0 < y < 6m & 0 < y < 0.5H)

for (y > 6m & 0 < y < 0.5H)

for (0 < y < 6m & 0.5H < y < H)

for (y > 6m &

0.5H < y < H)

for (0 < y < 6m & 0 < y < 0.5H)

for (y > 6m & 0 < y < 0.5H)

for (0 < y < 6m &

0.5H<y<H)

for (y > 6m & 0.5H<y<H)

A9

BOND AND BREAK FAILURES EQUATIONS APPENDIX D

Table 2.D Factors of safety formulae against both break and bond failures based on

MCGM DESIGN

EQUATIONS FSy (for both smooth and ribbed strips)

FS<> (for smooth strips)

FS<b (for ribbed strips)

MCGM (ARENICZ & CHOWDHURY, 1987)

Ai°t«-Ka^ i

j[Ka+o.6y(K0 -Ka)]ySvSH

2B. [L. -j(2.6H)2 -y2 + 2.3H][tan\\r + 0.6 (1.5- tan\|/)]

y[K +o.6y(K„-K )] S 5„ 'a 0 a v H

2B. [L. -•y/(2.6//)2-y2 + 2.3H][ton$ + 0.6y(l.lf*- tan (J))]

[K +o.6y(Kn-K )]S Srr a 0 a v H

A10

RSDAM PROGRAM FLOWCHART APPENDIX E

APPENDIX E- RSDAM PROGRAM FLOWCHART

c START SOB \

CALL INPUDATA

__ 1 . S^* LOOP

- - " -8^C^N=1 TO NUMBER OF

1 LA1

CALL VERFORCE

1 CALL HORFORCE

'

\ CALL DIST

\

fERsJ)

CALL BEAOPTM

• CALL SLIDOPTM

• CALL OVERTOPTM

1 3ALL OVERSTOPTM

/ \. Y \ K K = 1 / " ^ ~

I N

/ \.Y \ KK=2 /*""

I N

CALL NCGM • 1—z

CALL CGM

CALL MCGM

X All


? CALL OPTM

• CALL NOFAIL

r { CALL REINAREA

} r

^END LOOP) S X

r" - CALL MESH

1

- IDAM1.UUT1

^ ~S

S^—>. C^JEIN0 SOB^>

( DAM.IN ] • V J C3TART FEM >

v_ C

;SS 8 Br^^=l TO NUMBEF

\ "START MAINT_>

VI " •

CALL NDF

• CALL EBTEDA

1 LOOP

. OF ITERATIC

/ X Y ^ K C = 3 y~*

NY-_«

—© N FOR LODING SI

*"" CALL SEEPAGE

'EP^

OT>

A12

i ? \ Y

<KC=5y>~*~

Nl

CALL TSSM

1

CALL NDF

— r a

CALL SSMILV |

• CALL TANESH

• C^END LOOP

1

-©

-- 1 DAM2.UUTI

A13


CALL TSSM

I CALL SSMILV

I CALL TANESH

T LOOP

•j8***C N=l TO NUMBER OF ELEMENTS

T CALL PSTMS

\

CALL VSE

T ^JDLOOp

LOOP N=l TO NUMBER OF REINFORCEMENTS

T CALL SBE

A14


? ([^START TSSM

1 z> • >v^J=l TO NUMBER OF ELEMENTS^

\

CALL SIE

\

(END LOOP)

' -®*K^N^2- TO NUMBER OF REINFORCEMENTS^

1 CALL SBE

(END LOOP)

(RETURN)

A15


CALL ESM

T CALL PSTMS

T CALL PSTMS |

CALL VSE

1 END LOOP.

N=l TO NUMBER OF INTERFACE ELEMENTS

T CALL STIE

A16

RUNNING THE RSDAM PROGRAM APPENDIX F

APPENDIX F- RUNNING THE RSDAM PROGRAM

INTRODUCTION

It is possible to find the minimum base length required for a RSD to prevent the

following modes of failures: sliding, overturning, overstressing, bond failure, and rupture

failure. A computer program, called RSDAM, has been developed based on the

calculations of the forces acting on RSD (presented in Chapter 3), the equations of

stability analysis of RSD (presented in Chapter 4), and the formulae of soil-reinforcement

interaction (presented in Chapter 6). The purpose of the program is to assist a designer

in geometrical optimisation of RSDs and their analysis. This program has been compiled

using Fortran 77 and contains two main sub-programs.

The first main sub-program includes 15 subroutines and optimises the geometry of RSDs.

At the end of this main sub-program, a dam is divided into several incremental elements

in order to perform the analysis by the second main sub-program. The second main sub

program includes 13 subroutines and computes the stresses and displacements within the

elements of the dam based on two dimensional finite element formulation. It should be

noted that although the program is particularly adapted to RSDs, it may also be used for

a variety of reinforced earth walls and embankments with a small change in the

configuration of the program. A guide to this program will be presented here and, as

illustrative example, a model of a RSD with a height of 2 0 m will be analysed.

INPUT DATA

In the program, the information regarding dam geometry, loading, safety factor, fill

material, reinforcement, facing panel, and foundation material are used as input data.

This information, which is asked by the program at the time of running, will be explained

in the following stages:

A17


a) First stage

The first stage covers the dam height, the upstream and downstream water tables, the

upstream silt height and, the initial widths of the crest and base as follows:

HEIGHT OF DAM =? UPSTREAM WATER TABLE =? DOWNSTREAM WATER TABLE =? HEIGHT OF SILT =? INITIAL TOP WIDTH OF DAM =? INITIAL BASE WIDTH OF DAM =?

FOR CHANGING DATA TYPE 1 FOR CONTINUE TYPE 2

(m) (in) (m) (m) (m) (m)

b) Second stage

The second stage covers the unit weight of silt, the average unit weight of dam, and the

safety factors against the modes of failures (sliding, overturning, overstressing, bond

failure, and rupture failure) as follows:

UNIT WEIGHT OF SILT =? (KN/m3) AVERAGE UNIT WEIGHT OF DAM =? (KN/m3) ***************************************************** SAFETY FACTOR AGAINST SLIDING =? SAFETY FACTOR AGAINST OVERTURNING =? SAFETY FACTOR AGAINST BOND FAILURE =? SAFETY FACTOR AGAINST OVER-STRESSING =? SAFETY FACTOR AGAINST RUPTURE FAILURE =? FOR CHANGING DATA TYPE 1 FOR CONTINUE TYPE 2

c) Third stage

The third stage covers the ice force (which should be obtained by referring to Table CI

presented in Appendix C), and the coefficients of direct and indirect forces of earthquake

acceleration (see Chapter Three) as follows:

A18


ICE FORCE =? (KN) INITIAL COEFFICIENT OF EARTHQUAKE ACCELERATION =? COEFFICIENT OF INDIRECT FORCE OF EARTHQUAKE =?


d) Fourth stage

The fourth stage covers the width, height and thickness of facing panel as follows:

WIDTH OF FACING PANEL =? HEIGHT OF FACING PANEL =? THICKNESS OF FACING PANEL =?


(m) (m) (m)

e) Fifth stage

The fifth stage covers the width, unit weight, allowable tension and the number of

reinforcements connected to a facing panel as follows:

WIDTH OF REINFORCEMENTS =? (m) UNIT WEIGHT OF REINFORCEMENTS =? (KN/m3) ALLOWABLE TENSION OF REINFORCEMENTS =? (KN/m2) NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL =?

FOR CHANGING DATA TYPE 1 FOR CONTINUE TYPE 2 .

f) Sixth stage

The sixth stage covers the allowable bearing capacity of foundation soil, internal friction

angle, and uniformity coefficient of the dam soil as follows:

ALLOWABLE BEARING CAPACITY OF FOUNDATION SOIL =? (KN/m2) ANGLE OF INTERNAL FRICTION OF DAM SOIL =? (DEGREE) COEFFICIENT OF UNIFORMITY OF DAM SOIL =?

FOR CHANGING DATA TYPE 1 FOR CONTINUE TYPE 2 .

A19


g) Seventh stage

The seventh stage covers the selection of the method, which the internal stability analysis

of the dam is based, as follows:

1- INTERNAL STABILITY ANALYSIS BASED ON COHERENT GRAVITY METHOD 2- INTERNAL STABILITY ANALYSIS BASED ON MODIFIED COHERENT GRAVITY METHOD 3- INTERNAL STABILITY ANALYSIS BASED ON NEW COHERENT GRAVITY METHOD


h) Eighth stage

The eighth stage covers a question for mesh generation of dam. The number of nodal

points in x-direction of dam should be determined here. The number of the nodal points

in y-direction is calculated by the program and equal to the ratio of dam height per facing

panel height.

NUMBER OF

1-2-

STATIC

NODAL POINTS

ANALYSIS =? TIME HISTORY

IN

ANALYSIS =

X-

-?

-DIRECTION =?

The explanation of elements and the consequence of nodal points are shown in Figures

IF and 2F respectively.

Interface elements Thickness = 0~

Facing Panels

m*, Interface elements Thickness = 0

Facing Panels

Fig. IF The explanation of elements

A20


2n

n

4

3

mn

n+1 2n+l 3n+l 4n+l 5n+l 6n+l 7n+l

(m-l)n+l

Fig. 2F The consequence of the nodal points

i) Ninth stage

The ninth stage covers the facing panel properties as follows:

UNIT WEIGHT OF FACING PANELS =? YOUNG'S MODULUS OF FACING PANELS =? POISSON'S RATIO OF FACING PANELS =?


(KN/m3) (KN/m2)

j) Tenth stage

The tenth stage covers the soil properties for finite element analysis as follows:

UNIT WEIGHT OF THE MATERIAL =? (KN/m3) COHESION OF THE MATERIAL =? (KN/m2) FRICTION ANGLE =? (DEGREE) LATERAL EARTH PRESSURE COEFFICIENT AT REST =? INITIAL TANGENT MODULUS EXPONENT =? INITIAL TANGENT MODULUS COEFFICIENT =? FOR CHANGING DATA TYPE 1 FOR CONTINUE TYPE 2 **************************************************** UNLOAD-RELOAD MODULUS COEFFICIENT =? MIN. INITIAL TANGENT MODULUS FOR NON-ELASTIC MATERIALS =? BULK MODULUS EXPONENT =? BULK MODULUS COEFFICIENT =? YOUNG'S MODULUS =? POISSON'S RATIO =? FOR CHANGING DATA TYPE 1 FOR CONTINUE TYPE 2

(KN/m2)

(KN/m2)

A21


k) Eleventh stage

The eleventh stage covers the number of fixed nodes in y-direction, x-direction, both x

and y directions and z-rotation as follows:

NUMBER NUMBER NUMBER NUMBER

OF OF OF OF

NODAL NODAL NODAL NODAL

FIXED FIXED FIXED FIXED

POINTS POINTS POINTS POINTS


IN IN IN

Y-X-X

-DIRECTION =? -DIRECTION =? AND Y DIRECTIONS =?

AGAINST ROTATING =?

I) Twelfth stage

The twelfth stage covers the numbers of fixed nodes in y-direction, if any, as follows:

NODAL NUMBERS AGAINST Y-MOVEMENT =?


m) Thirteenth stage

The thirteenth stage covers the number of fixed nodes in x-direction, if any, as follows:

NODAL NUMBERS AGAINST X-MOVEMENT =?


n) Fourteenth stage

The fourteenth stage covers the numbers of fixed nodes in both y and x-direction, if any,

as follows:

NODAL NUMBERS AGAINST BOTH Y AND X-MOVEMENT =?


A22


o) Fifteenth stage

The fifteenth stage covers the nodal numbers of fixed nodes against rotation, if

follows:

NODAL NUMBERS AGAINST ROTATIONS =?


p) Sixteenth stage

The sixteenth stage covers the numbers and the elastic modulus of reinforcements

installed within the dam as follows:

NUMBER OF REINFORCEMENTS =? ELASTIC MODULUS OF THE REINFORCEMENTS =? (KN/m2)


q) Seventeenth stage

The seventeenth stage covers the nodal numbers, and cross-section area of these

reinforcements together with the angle between reinforcements and a horizontal line as

follows:

NODAL ANGLE CROSS-

NUMBERS OF THE BETWEEN THE

Nth REINFORCEMENT = Nth REINFORCEMENT AND

-SECTIONAL AREA

FOR CHANGING DATA FOR CONTINUE TYPE

= ? HORIZONTAL

OF THE Nth REINFORCEMENT =?

TYPE 1 2

LINE =?

r) Eighteenth stage

The eighteenth stage covers the displacement of the base nodes, if time history analysis

has been chosen in the eightieth stage, as follows:

A23


DISPLACEMENTS OF BASE NODAL POINTS =? DELTA(X) AND DELTA(Y) OF NODE J =? DELTA(X) AND DELTA(Y) OF NODE J+NMP =?


s) Nineteenth stage

The nineteenth stage covers the phreatic surface at the present and at the new levels as

follows:

NUMBER OF PHREATIC SURFACE END POINTS =? X-COORDINATE OF NODE J =? PRESENT LEVEL (Y-COORDINATE) OF THE PHREATIC SURFACE AT NODE J =? NEW LEVEL (Y-COORDINATE) OF THE PHREATIC SURFACE AT NODE J =? FOR CHANGING DATA TYPE 1 FOR CONTINUE TYPE 2

OUTPUT DATA

Output data contains two files, called D A M L O U T and D A M 2 . 0 U T , which will be

explained in the following two sections:

a) First Section

Initially the values of the input data will be printed out in the D A M L O U T file for

checking the input data as follows:

A24


HEIGHT OF DAM = m

UPSTREAM WATER TABLE = m DOWNSTREAM WATER TABLE = m HEIGHT OF SILT = m TOP WIDTH OF DAM = m BOTTOM WIDTH OF DAM = m

UNIT WEIGHT OF WATER = KN/m3 UNIT WEIGHT OF SILT = KN/m3 AVERAGE UNIT WEIGHT OF DAM = KN/m3

SAFETY FACTOR AGAINST SLIDING = SAFETY FACTOR AGAINST SLIDING = SAFETY FACTOR AGAINST BOND FAILURE = SAFETY FACTOR AGAINST OVER-STRESSING = SAFETY FACTOR AGAINST RUPTURE FAILURE =

ICE FORCE =

COEFFICIENT OF EARTHQUAKE ACCELERATION =

KN

COEFFICIENT OF INDIRECT FORCE OF EARTHQUAKE =

WIDTH OF FACINGS = HEIGHT OF FACINGS = WIDTH OF REINFORCEMENTS = UNIT WEIGHT OF REINFORCEMENTS =

NUMBER OF REINFORCEMENTS CONNECTED TO A

ALLOWABLE BEARING CAPACITY OF SOIL = ANGLE OF INTERNAL FRICTION OF SOIL =

ALLOWABLE TENSION OF REINFORCEMENTS = COEFFICIENT OF UNIFORMITY OF SOIL =

m m m KN/m3

FACING PANEL =

KN/m2 DEGREE

KN/m2

THE NAME OF THE METHOD WHICH THE INTERNAL STABILITY ANALYSIS IS BASED

Then, the output results of analysis for optimisation of the RSD are printed out in

D A M L O U T . In this stage the dam has been divided into several layers. Each layer is

taken from the crest to a specified depth as shown in Fig. 3F. The first layer is

considered as the whole dam and the second layer means the dam from its crest to the

depth of the first facing panel near the base.

A25

w ,

/ '" "

r

<

_

wb

#2 ! ^1

T

Fig. 3F The cross section of a parametric RSD with imaginary horizontal layers

The output data includes:

- the height of the layer,

- the depth and weight of water and silt acting on the upstream side of

the layer,

- the base width of the layer,

- the weight of the layer,

- the hydrostatic force acting on the layer,

- the silt force acting on the layer,

- the ice force acting on the layer,

- the direct and indirect forces of earthquake acting on the layer,

- the rrtinimum required base width of dam for no sliding, overturning,

overstressing, rupture failure and lack of bond,

- the optimum required base width of the dam

- the minimum required length of reinforcement

- the minimum net thickness of reinforcement needed for each layer

- the minimum cross-section area of each reinforcement

- the minimum net volume of each reinforcement

- and the minimum net weight of the reinforcements

If any failure in the above output data occurs, the program will stop and a massage

describing the mode of failure will be shown in output data file. These stages are iterated

A26


in the first layer (whole dam) analysis until the base width of the RSD is optimised.

Then, the above operation is repeated and printed out for other layers of the dam. In

these layers, the base length of the layer should be compared with the rninimum required

base lengths of the layer. The above output-data are printed out as follows:

**************************************************** LAYER NO. = ****************************************************

HEIGHT OF LAYER = UPSTREAM WATER TABLE = DOWNSTREAM WATER TABLE = HEIGHT OF SILT = TOP WIDTH OF LAYER = BOTTOM WIDTH OF LAYER = RATIO OF TOP WIDTH TO BOTTOM WIDTH =

WEIGHT OF LAYER = WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER = WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER = UPLIFT PRESSURE ACTING ON THE LAYER = HYDROSTATIC FORCE ACTING ON LAYER = ICE FORCE ACTING ON LAYER = SILT FORCE ACTING ON LAYER = INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER = BEARING CAPACITY FAILURE WILL NOT HAPPEN

rn m rn rn m m

=

MIN. REQUIRED BASE LENGTH FOR NO SLIDING = MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING = MIN. REQUIRED BASE LENGTH FOR NO OVER-STRESSING = MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE =

MIN. REQUIRED BASE LENGTH FOR NO FAILURE =

NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL

MIN. REQUIRED LENGTH OF REINFORCEMENT = MIN. NET THICKNESS OF REINFORCEMENT = WIDTH OF REINFORCEMENT =

MIN. CROSS SEC. AREA OF REINFORCEMENT = MIN. NET VOLUME OF REINFORCEMENT = MIN. NET WEIGHT OF REINFORCEMENT =

*

*

KN KN KN KN KN KN KN KN KN

m m m m

m

1

m mm cm

cm2/m2 AREA m3/m2 AREA Kq/m2 AREA

b) Second Section

In this section, the numbers of nodal points, soil elements, interface elements,

reinforcements, and loading steps are printed out in DAM2.0UT as follows:

A27


NUMBER NUMBER NUMBER NUMBER NUMBER

OF OF OF OF OF

NODAL POINTS = ELEMENTS = REINFORCEMENTS = INTERFACE ELEMENTS = LOADING STEPS =

This is followed by the reinforcement installation and the loading steps as follows:

STAGE No.

* *

NO. OF ITERATION

* *

CONSTRUCTION TYPE

* *

The material properties are printed out as follows:

MATERIAL

* *

GAMMA

* *

COHESION

* *

PHI

* *

TENS. STRENGTH

* *

KO

* *

In addition, the coordinates of the nodal points are printed out as follows:

COORDINATES OF NODAL POINTS

NODAL POINT X-COORDINATE

* * * *

Y-COORDINATE

* *

The boundary conditions of dam is also printed out in D A M 2 . 0 U T as follows:

NODES WITH BOUNDARY RESTRAINTS

NO x-MOVEMENT =

NO Y-MOVEMENT =

NO X OR Y MOVEMENT = * * *

* *

* * *

A28


This is followed by the data representing dam geometry, including the number of soil

element, followed by the four numbers representing the number of nodes of this element.

A number representing the type of element materiel is also printed out after the numbers

of nodes as follows:

ELEMENT DATA ELEMENT No. I J K L MATERIAL

* * * * * * * * * * * *

The specification of the reinforcement instalation and/or loading step is printed out in the

next stage as follows:

**************************** STAGE NUMBER 1

******************** THE FOLLOWING

REINFORCEMENT No.

* *

* ********

*************************

******* REINFORCEMENTS

I

* *

J

* *

*********** ARE ADDED

DISP. TC

******* HEREIN

ACTIVATE

* *

or

***************************************************** STAGE NUMBER 2

*****************************************************

FORCE AND/OR DISPLACEMENT LOADING IS SPECIFIED FOR THIS INCREMENT

NODE X-LOAD Y-LOAD NODE X-LOAD Y-LOAD

* * * * * * * * * * * *

The results of analysis, including the coordinates of the nodal points, and the horizontal

and vertical displacements of nodal points are printed out here as follows:

A29


DISPLACEMENT

NODAL POINT

* *

X

* *

RESULTS

Y

* *

FOR STAGE 1

TOTAL UX

* *

TOTAL UY

* *

PORE PRESS

* *

The coordinates of the middle of soil element, the horizontal stress, the vertical stress

and the principal stresses within the elements, and the maximum shear stresses for soil

elements are printed out as follows:

ELEM NO

* *

STRESSES VALUES

X Y

* * * *

FOR STAGE

SIGMA X

* *

1

SIGMA Y

* *

TAU XY

* *

SIGMA 1

* *

SIGMA 3

* *

Then the interface element results are printed out as follows:

INTERFACE ELEMENT RESULTS FOR STAGE 1

ELEM NO X Y NORMAL STRESS SHEAR STRESS NORMAL STIFF SHEAR STIFF

* * * * * * * * * * * * * *

Finally, the value of tension in the reinforcements are printed out as follows:

REINFORCEMENT RESULTS FOR STAGE 1

REIN. NUM. I J TYPE COMPR FORCE INCR COMPR STIFFNESS COSA

**** * * ** * * * * * * * *

EXAMPLE

The input data and output data for the example of the 20m high RSD is presented here.

A30


Input Data

The input data of a RSD with 20 m height are as follows:

HEIGHT OF DAM =20 UPSTREAM WATER TABLE =20 DOWNSTREAM WATER TABLE = 2 HEIGHT OF SILT = 6 INITIAL TOP WIDTH OF DAM = 2 INITIAL BASE WIDTH OF DAM =10

(m) (m) (m) (m) (m) (m)

UNIT WEIGHT OF SILT = 18 AVERAGE UNIT WEIGHT OF DAM =20

**************************************

SAFETY FACTOR AGAINST SLIDING = 2 SAFETY FACTOR AGAINST OVERTURNING = 2 SAFETY FACTOR AGAINST BOND FAILURE =3 SAFETY FACTOR AGAINST OVER-STRESSING = SAFETY FACTOR AGAINST RUPTURE FAILURE

(KN/m3) (KN/m3)

************

2 = 3

ICE FORCE = 0 (KN) INITIAL COEFFICIENT OF EARTHQUAKE ACCELERATION =0.2 COEFFICIENT OF INDIRECT FORCE OF EARTHQUAKE =0.2

WIDTH OF FACING PANELS = 1 HEIGHT OF FACING PANELS = 2 THICKNESS OF FACING PANELS = 0.2

(m) (m) (m)

WIDTH OF REINFORCEMENTS =0.08 (m) UNIT WEIGHT OF REINFORCEMENTS = 78 (KN/m3) ALLOWABLE TENSION OF REINFORCEMENTS = 24 0000 (KN/m2) NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL = 1

ALLOWABLE BEARING CAPACITY OF FOUNDATION SOIL = 900 (KN/m2) ANGLE OF INTERNAL FRICTION OF DAM SOIL =35 (DEGREE) COEFFICIENT OF UNIFORMITY OF DAM SOIL = 150

1-2-3-

INTERNAL INTERNAL INTERNAL

STABILITY STABILITY STABILITY

ANALYSIS ANALYSIS ANALYSIS

BASED BASED BASED

ON ON ON

COHERENT GRAVITY METHOD MODIFIED COHERENT GRAVITY METHOD NEW COHERENT GRAVITY METHOD = 3

NUMBER OF NODAL POINTS IN X-DIRECTION 11

A31


Therefore, the consequence of the nodal points of this example is as shown in Fig. 4F.

ll

10

9

8

4

3

2

33 55 77 99 22, 44

i l l \

34 12 23

\ \ \ \ \ w \ 7 7_r

\ \ \ W ^ x \ \ \ \ \ ^

-, 113

M v 117 \ \ \ \ \ ^

45 56 67 78 89 • ;;;

100

Fig. 4F- The consequence of the nodal points

UNIT WEIGHT OF FACING PANELS =24 (KN/m3) YOUNG'S MODULUS OF FACING PANELS = 2500000 (KN/m2) POISSON'S RATIO OF FACING PANELS =0.2

UNIT WEIGHT OF THE MATERIAL =20 (KN/m3) COHESION OF THE MATERIAL = 0 (KN/m2) FRICTION ANGLE =35 (DEGREE) LATERAL EARTH PRESSURE COEFFICIENT AT REST =0.5 INITIAL TANGENT MODULUS EXPONENT =0.5 INITIAL TANGENT MODULUS COEFFICIENT = 3 00 **************************************************** UNLOAD-RELOAD MODULUS COEFFICIENT = 5 00 MIN. INITIAL TANGENT MODULUS FOR NON-ELASTIC MATERIALS = 1000 (KN/m2) BULK MODULUS EXPONENT =0.2 BULK MODULUS COEFFICIENT =250 YOUNG'S MODULUS = 50 00 (KN/m2) POISSON'S RATIO =0.1


OF OF OF OF

NODAL NODAL NODAL NODAL

FIXED FIXED FIXED FIXED

POINTS POINTS POINTS POINTS

IN Y-DIRECTION = 0 IN X-DIRECTION = 0 IN X AND Y DIRECTIONS = AGAINST ROTATING = 0

= 11

NODAL NUMBERS AGAINST BOTH Y AND X-MOVEMENT = 1 12 23 34 45 56 67 78 89 100 111

A32


NUMBER OF REINFORCEMENTS =10 ELASTIC MODULUS OF THE REINFORCEMENTS = 250000000 (KN/m2)

Since the number of reinforcements are specified to be 10, the input data in regard to the

specifications of the reinforcements are asked 10 times as follows:

NODAL NUMBERS OF THE 1th REINFORCEMENT = 1 111 ANGLE BETWEEN THE Nth REINFORCEMENT AND HORIZONTAL LINE = 0 CROSS-SECTIONAL AREA OF THE Nth REINFORCEMENT =

NODAL NUMBERS OF THE 2th REINFORCEMENT = 2 112 ANGLE BETWEEN THE Nth REINFORCEMENT AND HORIZONTAL LINE = 0 CROSS-SECTIONAL AREA OF THE Nth REINFORCEMENT = **************************************************** NODAL NUMBERS OF THE 3th REINFORCEMENT = 3 113 ANGLE BETWEEN THE Nth REINFORCEMENT AND HORIZONTAL LINE = 0 CROSS-SECTIONAL AREA OF THE Nth REINFORCEMENT =




NODAL NUMBERS OF THE 10th REINFORCEMENT = 10 120 ANGLE BETWEEN THE Nth REINFORCEMENT AND HORIZONTAL LINE = 0 CROSS-SECTIONAL AREA OF THE Nth REINFORCEMENT = ****************************************************

Then the input data in regard to the displacements of the base nods are asked as follows:

A33


DISPLACEMENTS OF BASE DELTA(X) AND DELTA(Y) DELTA(X) AND DELTA(Y)

DISPLACEMENTS OF BASE DELTA(X) AND DELTA(Y)

DELTA(X) AND DELTA(Y)

NODAL POINTS OF NODE 1 = OF NODE 12 =

NODAL POINTS OF NODE 23 =

OF NODE 34 = **********************************.

DISPLACEMENTS OF BASE

DELTA(X) AND DELTA(Y) DELTA(X) AND DELTA(Y)


DELTA(X) AND DELTA(Y) *********************

DISPLACEMENTS OF BASE






3. 0

0 0

V *

0 0

0 0

I (

.1

1

1 t *

1 1

1 1

*****************


OF NODE 100 =


0 1 : 0.1

= 0.]

)

0

0 0

0 0

0 0

0

0

0

Since the number of phreatic surface end points are specified to be 2, the input data in

this regard are asked 2 times as follows:

NUMBER OF PHREATIC SURFACE END POINTS = 2

X-COORDINATE OF NODE J = 0

PRESENT LEVEL (Y-COORDINATE) OF THE PHREATIC SURFACE AT NODE

NEW LEVEL (Y-COORDINATE) OF THE PHREATIC SURFACE AT NODE J =

NUMBER OF PHREATIC SURFACE END POINTS = 2

X-COORDINATE OF NODE J = 6 PRESENT LEVEL (Y-COORDINATE) OF THE PHREATIC SURFACE AT NODE

NEW LEVEL (Y-COORDINATE) OF THE PHREATIC SURFACE AT NODE J =

J =

0

J = 0

2

20

Output Data

The output data contains two files, called D A M l . O U T and DAM2.0UT, which will be

explained in the following two sections:

A34


a) First Section (Dam 1.out)

Initially the input data of this example (printed out in the D A M L O U T file for checking)

are presented as follows:

***************************************************** * * * * INPUT DATA *

* * *

* * *****************************************************

HEIGHT OF DAM = 20.0000 m UPSTREAM WATER TABLE = 20.0000 m DOWNSTREAM WATER TABLE = 2.00000 m HEIGHT OF SILT = 6.00000 m TOP WIDTH OF DAM = 2.00000 m BOTTOM WIDTH OF DAM = 20.0000 m

UNIT WEIGHT OF WATER = UNIT WEIGHT OF SILT = AVERAGE UNIT WEIGHT OF DAM =

SAFETY FACTOR AGAINST SLIDING = SAFETY FACTOR AGAINST SLIDING = SAFETY FACTOR AGAINST BOND FAILURE = SAFETY FACTOR AGAINST OVER-STRESSING = SAFETY FACTOR AGAINST RUPTURE FAILURE =

ICE FORCE =

10.0000 KN/m3 18.0000 KN/m3 20.0000 KN/m3

2.00000 2.00000 3.00000 2.00000 3.00000

0.000000

COEFFICIENT OF EARTHQUAKE ACCELERATION = 0.200000 COEFFICIENT OF INDIRECT FORCE OF EARTHQUAKE = 0.200000

WIDTH OF FACINGS = HEIGHT OF FACINGS = WIDTH OF REINFORCEMENTS = UNIT WEIGHT OF REINFORCEMENTS =

NUMBER OF REINFORCEMENTS CONNECTED TO A

ALLOWABLE BEARING CAPACITY OF SOIL = ANGLE OF INTERNAL FRICTION OF SOIL =

ALLOWABLE TENSION OF REINFORCEMENTS = COEFFICIENT OF UNIFORMITY OF SOIL =

INTERNAL STABILITY ANALYSIS IS BASED ON

NUMBER OF LAYERS =

1.00000 2.00000 0.800000E-01 78.0000

FACING PANEL =

900.000 35.0000

240000. 150.000

KN

m m m KN/m3

1

KN/m2 DEGREE

KN/m2

NEW COHERENT GRAVITY METHOD

10

A35


Then the output data of this example for stability analysis and optimisation of the whole

dam (printed out in the DAM LOUT file) are presented as follows:

***************************************************** * * * * * OUTPUT

*****************************************************

LAYER NO.= 1 ITERATION NO.= 1

HEIGHT OF LAYER= 20.0000 m UPSTREAM WATER TABLE= 2 0.0000 m DOWNSTREAM WATER TABLE= 2.0000 0 m HEIGHT OF SILT= 6.00000 m TOP WIDTH OF LAYER= 6.00000 m BOTTOM WIDTH OF LAYER= 2 0.0000 m

RATIO OF TOP WIDTH TO BOTTOM WIDTH= 0.300000

WEIGHT OF LAYER= 5200.00 KN WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= 1400.00 KN WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= 22 6.80 0 KN UPLIFT PRESSURE ACTING ON THE LAYER= 2200.00 KN

HYDROSTATIC FORCE ACTING ON LAYER= 2000.00 KN ICE FORCE ACTING ON LAYER= 0.000 00 0 KN SILT FORCE ACTING ON LAYER= 87.8 0 07 KN INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 116.160 KN DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 1040.00 KN

BEARING CAPACITY FAILURE WILL NOT HAPPEN

MIN. REQUIRED BASE LENGTH FOR NO SLIDING= 76.1570 MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= 50.402 0 MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING= 42.3742 MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= 16.5226

MIN. REQUIRED BASE LENGTH FOR NO FAILURE= 76.1570 m FOR NO FAILURE BASE LENGTH SHOULD BE INCREASED

m m m m

A36


gggggggg@ggggg@@ggggg@@g@g@gg@g@gggg@@gg@gg@gg@g(a@g@g *****************************************************

OUTPUT

***************************************************** ggggggggggggggggggggggggggggggggggggggggggggggggggggg

LAYER NO.= ITERATION NO.

HEIGHT OF LAYER= UPSTREAM WATER TABLE= DOWNSTREAM WATER TABLE= HEIGHT OF SILT= TOP WIDTH OF LAYER= BOTTOM WIDTH OF LAYER=

RATIO OF TOP WIDTH TO BOTTOM WIDTH=

1 2

20.0000 20.0000 2.00000 6.00000 6.00000 76.1570

m m m m m m

0.787847E-01

WEIGHT OF LAYER= WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= UPLIFT PRESSURE ACTING ON THE LAYER=

HYDROSTATIC FORCE ACTING ON LAYER= ICE FORCE ACTING ON LAYER= SILT FORCE ACTING ON LAYER= INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER=


MIN. REQUIRED BASE LENGTH FOR NO SLIDING= MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING= MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE=

MIN. REQUIRED BASE LENGTH FOR NO FAILURE=

16431.4 7015.70 1136.54 8377.27

2000.00 0.000000 87.8007 116.160 3286.28

KN KN KN KN

KN KN KN KN KN

70.4086 43.4634 34.3741 16.5226

m m m m

70.4086 m

A37


ggggggg@ggggg@@ggggggggg@ggggggggggggggggggg@@ggggg@g ***************************************************** * * * * * OUTPUT * * * * * ***************************************************** ggggggggggggggggggggggggggggggggggggggggggggggggggggg

LAYER NO.= ITERATION NO.

1 3

20.0000 20.0000 2.00000 6.00000 6.00000 73.2828

m m m m m m


RATIO OF TOP WIDTH TO BOTTOM WIDTH= 0.818746E-01

WEIGHT OF LAYER= 1585 6.6 KN WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= 6728.28 KN WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= 1089.98 KN UPLIFT PRESSURE ACTING ON THE LAYER= 8061.11 KN



MIN. REQUIRED BASE LENGTH FOR NO SLIDING= 70.4829 m MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= 43.5895 m MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING= 34.4680 m MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= 16.5226 m

MIN. REQUIRED BASE LENGTH FOR NO FAILURE^ 70.4829 m

2000.00 0.000000 87.8007 116.160 3171.31

KN KN KN KN KN

A38


gggggggggggggggggggg@@g@ggggg@gg@ggggggg@ggg@@ggggggg *****************************************************

* OUTPUT * * * * * ***************************************************** ggggggggggggggggggggggggggggggggggggggggggggggggggggg

LAYER NO.= ITERATION NO.:



1 4

20.0000 20.0000 2.00000 6.00000 6.00000 71.8829

m m m m m m

0.834691E-01




MIN. REQUIRED BASE LENGTH FOR NO SLIDING= MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING--MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE=


15576.6 6588.29 1067.30 7907.11

2000.00 0.000000 87.8007 116.160 3115.31

KN KN KN KN

KN KN KN KN KN

70.5213 43.6543 34.5166 16.5226

70.5213

m in m m

m

A39


ggggggggggggggggggggggggggggggggggggggggggggggggggggg ***************************************************** * * * * * OUTPUT *


LAYER NO.= 1 ITERATION NO.= 5

HEIGHT OF LAYER= 2 0.0000 m UPSTREAM WATER TABLE= 2 0.0000 ro DOWNSTREAM WATER TABLE= 2.00000 m HEIGHT OF SILT= 6.0000 0 m TOP WIDTH OF LAYER= 6.00000 m BOTTOM WIDTH OF LAYER= 71.2021 m


WEIGHT OF LAYER= 1544 0.4 KN WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= 6520.21 KN WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= 1056.27 KN UPLIFT PRESSURE ACTING ON THE LAYER= 7832.23 KN

HYDROSTATIC FORCE ACTING ON LAYER= 2 000.00 KN ICE FORCE ACTING ON LAYER= 0.000000 KN SILT FORCE ACTING ON LAYER= 87.8007 KN INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 116.160 KN DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 3 088.08 KN


MIN. REQUIRED BASE LENGTH FOR NO SLIDING= 7 0.5406 MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= 43.68 67 MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING= 34.5410 MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= 16.5226

m m m m

MIN. REQUIRED BASE LENGTH FOR NO FAILURE= 70.5406 m

A40


ggggggggggggggggggggggggggggggggggggggggggggggggggggg *******************************************************

* OUTPUT * * * _ ******************************,*****.**************,*, ggggggggggggggggggggggggggggggggggggggggggggggggggggg

LAYER NO.= ITERATION NO.=

HEIGHT OF LAYER= UPSTREAM WATER TABLE= DOWNSTREAM WATER TABLE: HEIGHT OF SILT= TOP WIDTH OF LAYER= BOTTOM WIDTH OF LAYER=

20.0000 20.0000 2.00000 6.00000 6.00000 70.8713

m m m m m m


WEIGHT OF LAYER= 15374.3 KN WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= 6487.13 KN WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= 1050.92 KN UPLIFT PRESSURE ACTING ON THE LAYER= 7795.85 KN

HYDROSTATIC FORCE ACTING ON LAYER= 2 000.00 KN ICE FORCE ACTING ON LAYER= 0.000000 KN SILT FORCE ACTING ON LAYER= 87.8007 KN INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 116.160 KN DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 3074.85 KN


MIN. REQUIRED BASE LENGTH FOR NO SLIDING= 70.5501 m MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= 43.7027 m MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING= 34.5530 rn MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= 16.5226 m

MIN. REQUIRED BASE LENGTH FOR NO FAILURE= 70.5501

A41


ggggggggggggggggggggggggggggggggggggggggggggggggggggg *****************************************************

OUTPUT


LAYER NO.= ITERATION NO.:


1 7

20.0000 20.0000 2.00000 6.00000 6.00000 70.7107

m m m m m m


WEIGHT OF LAYER= 15342.1 KN WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= 6471.07 KN WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= 1048.31 KN UPLIFT PRESSURE ACTING ON THE LAYER= 7778.18 KN



2000.00 0.000000 87.8007 116.160 3068.43

KN KN KN KN KN

MIN. REQUIRED BASE LENGTH FOR NO SLIDING= MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= 43.7105 MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING^ 34.5589 MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= 16.5226

70.5547 m m m m

MIN. REQUIRED BASE LENGTH FOR NO FAILURE: 70.5547 m

A42



* OUTPUT * *

* * ggggggggggggggggggggggggggggggggggggggggggggggggggggg

LAYER NO.= ITERATION NO.=


20.0000 20.0000 2.00000 6.00000 6.00000 70.6327

m m m m m m


WEIGHT OF LAYER= 15326.5 WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= 64 63.27 WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= 1047.05 UPLIFT PRESSURE ACTING ON THE LAYER= 7769.60

HYDROSTATIC FORCE ACTING ON LAYER= 2 000.00 ICE FORCE ACTING ON LAYER= 0.00 00 00 SILT FORCE ACTING ON LAYER= 87.8007 INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 116.16 0 DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 3 065.31


MIN. REQUIRED BASE LENGTH FOR NO SLIDING= 70.5570 MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= 43.7143 MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING= 34.5618 MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= 16.5226

MIN. REQUIRED BASE LENGTH FOR NO FAILURE= 7 0.557 0

KN KN KN KN

KN KN KN KN KN

m

NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL= MIN. REQUIRED LENGTH OF REINFORCEMENT^ MIN. NET THICKNESS OF REINFORCEMENT= WIDTH OF REINFORCEMENT=

m m m m

16.5226 33.8737 8.00000

m mm cm

MIN. CROSS SEC. AREA OF REINFORCEMENT= MIN. NET VOLUME OF REINFORCEMENT= MIN. NET WEIGHT OF REINFORCEMENT=

13.5495 0.223872E-01 1.74620

cm2/m2 AREA m3/m2 AREA KN/m2 AREA

After calculation of the stability analysis and optimisation of the RSD, the output data of

this example for the other layers of the dam (printed out in the DAMLOUT file) are

presented as follows:

A43


g@g@@g@gggggggg@g@gggg@@ggggg@ggg@ggggggg@g@ggg@g@gg@ *****************************************************


LAYER NO.



18.0000 m 18.0000 m 0.000000 m 4.00000 m 6.00000 m 64.1694 m

0.935025E-01

WEIGHT OF LAYER= 12 63 0.5 WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= 5235.25 WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= 465.355 UPLIFT PRESSURE ACTING ON THE LAYER= 5775.25

HYDROSTATIC FORCE ACTING ON LAYER= 1620.00 ICE FORCE ACTING ON LAYER= 0.000000 SILT FORCE ACTING ON LAYER= 3 9.0225 INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 94.0896 DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= 2 526.10




NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL= MIN. REQUIRED LENGTH OF REINFORCEMENT= MIN. NET THICKNESS OF REINFORCEMENT= WIDTH OF REINFORCEMENT=

KN KN KN KN

KN KN KN KN KN

60.2596 39.2878 24.7662 17.3078

60.2596 rn

m m m m

1 17.3078 30.4864 8.00000

m mm cm

MIN. CROSS SEC. AREA OF REINFORCEMENT: MIN. NET VOLUME OF REINFORCEMENT= MIN. NET WEIGHT OF REINFORCEMENT=

12.1945 0.211061E-01 1.64627


A44


ggggggggggggggggggggggggggggggggggggggggggggggggggggg • i * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * ^

* OUTPUT * *

* * *****************************************************

LAYER NO.= -3



16.0000 16.0000 0.000000 2.00000 6.00000 57.7062

0.103975

m m m m m m

WEIGHT OF LAYER= WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER: WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= UPLIFT PRESSURE ACTING ON THE LAYER=

HYDROSTATIC FORCE ACTING ON LAYER= ICE FORCE ACTING ON LAYER= SILT FORCE ACTING ON LAYER= INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER: DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER=


MIN. REQUIRED BASE LENGTH FOR NO SLIDING= MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= MIN. REQUIRED BASE LENGTH FOR NO OVERSTRESSING: MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE=

MIN. REQUIRED BASE LENGTH FOR NO FAILURE:

10193.0 4136.49 116.339 4616.49

1280.00 0.000000 9.75563 74.3424 2038.60

KN KN KN KN

KN KN KN KN KN

56.2094 40.1675 18.2379 17.9998

m m m m

56.2094 m

NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL= MIN. REQUIRED LENGTH OF REINFORCEMENT= MIN. NET THICKNESS OF REINFORCEMENT WIDTH OF REINFORCEMENT=


10.8396 0.195111E-01 1.52187

17 27 8.

.9998

.0990 00000

cm2/m2 m3/m2 KN/m2

m mm cm

AREA AREA AREA

A45




LAYER NO.



14.0000 14.0000

0.000000 0.000000 6.00000 51.2429

0.117089

m m m m m m

8014.00 3167.00 0.000000 3587.00

980.000 0.000000 0.000000 56.9184 1602.80







KN KN KN KN

KN KN KN KN KN

50

50.4058 38.1480 13.5319 18.6025

.4058 m

m m m m

18.6025 23.7116 8.00000

m mm cm


9.48464 0.176438E-01 1.37622


A46




LAYER NO.


12.0000 12.0000 0.000000 0.000000 6.00000 44.7796







m m m m m m

6093.56 2326.78

0.000000 2686.78

720.000 0.000000 0.000000 41.8176 1218.71

KN KN KN KN

KN KN KN KN KN

43.3298 32.9942 9.90900 19.1190

m m m m

43.3298


ro

19.1190 20.3242 8.00000

m mm cm

MIN. CROSS SEC. AREA OF REINFORCEMENT--MIN. NET VOLUME OF REINFORCEMENT= MIN. NET WEIGHT OF REINFORCEMENT=

8.12969 0.155432E-01 1.21237


A47




LAYER NO.:



10.0000 10.0000 0.000000 0.000000 6.00000 38.3164

0.156591






m m m m m m

4431.64 1615.82

0.000000 1915.82

500.000 0.000000 0.000000 29.0400 886.327

KN KN KN KN

KN KN KN KN KN

36.2484 27.7815 7.01611 19.5520

m m m m

36.2484

NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL: MIN. REQUIRED LENGTH OF REINFORCEMENT= MIN. NET THICKNESS OF REINFORCEMENT= WIDTH OF REINFORCEMENT=

m

19.5520 16.9369 !. 00000

m mm cm


6.77475 0.132460E-01 1.03319


A48



* OUTPUT * *

* *

* ***************************************************** gggggggggggggggggggggggggggg@@@g@@@@g@g@g@@ggg@gggggg LAYER NO.= n



8.00000 8.00000 0.000000 0.000000 6.00000 31.8531

0.188365

m m m m m m






3028.25 1034.12

0. 000000 1274.12

320.000 0.000000 0.000000 18.5856 605.649

KN KN KN KN

KN KN KN KN KN

29.1578 22.4771 4.70376 19.9035

m m m m

29.1578 m


1 19.9035 13.5495 8.00000

m mm cm


5.41980 0.107873E-01 0.841409


A49




LAYER NO.



6.00000 6.00000

0.000000 0.000000 6.00000 25.3898

0.236315

m m m m m m

1883.39 581.694 0.000000 761.694

180.000 0.000000 0.000000 10.4544 376.678






NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL: MIN. REQUIRED LENGTH OF REINFORCEMENT= MIN. NET THICKNESS OF REINFORCEMENT= WIDTH OF REINFORCEMENT=

KN KN KN KN

KN KN KN KN KN

22

22.0509 17.0239 2.87675 20.1753

.0509 m

m m m m

20.1753 10.1621 8.00000

m mm cm


4.06485 0.820093E-02 0.639673


A50


ggggggggggggggggggggggggggggggggggggggggggggggggggggg ***************************************************** * * * OUTPUT *

***************************************************** gggggggggggggggggggggggggggggggggggggggg@@@@@gg@g@ggg

LAYER NO.=


4.00000 4.00000 0.000000 0.000000 6.00000 18.9265







NUMBER OF REINFORCEMENTS CONNECTED TO A FACING MIN. REQUIRED LENGTH OF REINFORCEMENT= MIN. NET THICKNESS OF REINFORCEMENT= WIDTH OF REINFORCEMENT=

m m m m m m

997.062 258.531 0.000000 378.531

80.0000 0.000000 0.000000 4.64640 199.412

KN KN KN KN

KN KN KN KN KN

14.9101 11.3263 1.47904 12.2295

14.9101 m

m m m m

PANEL= 12.2295 5.87985 8.00000

m mm cm


2.35194 0.287630E-02 0.224351


A51



* OUTPUT * * * ***************************************************** ggggggggggggggggggggggggggggggggggggggggg@@g@@@@g@@@g

LAYER NO.= 10

HEIGHT OF LAYER= 2.00000 UPSTREAM WATER TABLE= 2.00 000 DOWNSTREAM WATER TABLE= 0.0000 00 HEIGHT OF SILT= 0.0 00 000 TOP WIDTH OF LAYER= 6.00000 BOTTOM WIDTH OF LAYER= 12.4633







m m m m m m

369.265 64.6327

0. 000000 124.633

20.0000 0.000000 0.000000 1.16160 73.8531

KN KN KN KN

KN KN KN KN KN

7.67795 5.32100 0.493391 7.54333

7.67795

m m m m

NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL= MIN. REQUIRED LENGTH OF REINFORCEMENT^ MIN. NET THICKNESS OF REINFORCEMENT= WIDTH OF REINFORCEMENT=

7.54333 2.14445 8.00000

m mm cm


0.857782 0.647053E-03 0.504701E-01


b) Second Section (Dam2.out)

Then the values of the analysis by the finite element method (printed out in the

DAM2.0UT file) are presented as follows:

A52


REINFORCED EARTH DAM ANALYSIS

STAGE

1 2 3 4 5


NO.

MATERIAL

1 2

OF OF OF OF

OF

NODAL POINTS = 121 ELEMENTS = 100 INTERFACE ELEMENTS = 2 0 LOADING STEPS = 5

ITERATION CONSTRUCTION TYPE

1 1 1 1 1

GAMMA

24.00 20.00

REINFORCEMENT INSTALLATION HYDROSTATIC FORCE

SILT FORCE EARTHQUAKE FORCE OR DISPLACEME

SEEPAGE LINE VARIATION

COHESION PI TEN. STRGTH

0.00 0.00 0.00 0.00 35.00 0.00

K0

0.000 0.500

COORDINATES OF NODAL POINTS

NODAL POINT X-COORDINATE Y-COORDINATE

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.200 0.2O0 0.200 0.200 0.200 0.200 0.200 0.200

11.910 10.832 9.755 8.677 7.599 6.522 5.444 4.366 3.288 2.211

0.000 2.000 4 .000 6.000 8.000

10.000 12.000 14.000 16.000 18.000 20.000 0.000 2.000 4.000 6.000 8.000

10.000 12.000 14 .000 16.000 18.000 20.000 0.000 2.000 4.000 6.000 8.000 10.000 12.000 14.000 16.000 18.000 20.000 0.000 2.000 4.000 6.000 8.000 10.000 12.000 14.000 16.000 18.000

A53


44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114

1.133 23.610 21.456 19.301 17.147 14.993 12.838 10.684 8.530 6.376 4.221 2.067 35.320 32.088 28.856 25.624 22.392 19.160 15.928 12.696 9.464 6.232 3.000 47.020 42.711 38.403 34.094 29.785 25.476 21.168 16.859 12.550 8.242 3.933 58.730 53.344 47.957 42.571 37.185 31.799 26.412 21.026 15.640 10.253 4.867 70.430 63.967 57.504 51.041 44.578 38.115 31.652 25.189 18.726 12.263 5.800 70.430 63.967 57.504 51.041 44.578 38.115 31.652 25.189 18.726 12.263 5.800 70.630 64.167 57.704 51.241

20.000 0.000 2.000 4.000 6.000 8.000

10.000 12.000 14.000 16.000 18.000 20.000 0.000 2.000 4.000 6.000 8.000 10.000 12.000 14.000 16.000 18.000 20.000 0.000 2.000 4.000 6.000 8.000

10.000 12.000 14.000 16.000 18.000 20.000 0.000 2.000 4.000 6.000 8.000 10.000 12.000 14 .000 16.000 18.000 20.000 0.000 2.000 4.000 6.000 8.000 10.000 12.000 14.000 16.000 18.000 20.000 0.000 2.000 4 .000 6.000 8.000

10.000 12.000 14 .000 16.000 18.000 20.000 0.000 2.000 4.000 6.000

A54


115 116 117 118 119 120 121

NO X OR Y MOVEMENT 111

44.778 38.315 31.852 25.389 18.926 12.463 6.000

1 1 23

8.000 10.000 12.000 14.000 16.000 18.000 20.000

34 45 56 67 78 89 100

ELEMENT DATA ELEMENT

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

I

23 24 25 26 27 28 29 30 31 32 100 101 102 103 104 105 106 107 108 109 12 13 14 15 16 17 18 19 20 21 34 35 36 37 38 39 40 41 42 43 45 46 47 48 49 50 51 52 53 54 56 57 58

J

24 25 26 27 28 29 30 31 32 33

101 102 103 104 105 106 107 108 109 110 13 14 15 16 17 18 19 20 21 22 35 36 37 38 39 40 41 42 43 44 46 47 48 49 50 51 52 53 54 55 57 58 59

K

13 14 15 16 17 18 19 20 21 22 90 91 92 93 94 95 96 97 98 99 2 3 4 5 6 7 8 9 10 11 24 25 26 27 28 29 30 31 32 33 35 36 37 38 39 40 41 42 43 44 46 47 48

L

12 13 14 15 16 17 18 19 20 21 89 90 91 92 93 94 95 96 97 98 1 2 3 4 5 6 7 8 9 10 23 24 25 26 27 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 45 46 47

MATERIAL

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

A55


54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

100

59 60 61 62 63 64 65 67 68 69 70 71 72 73 74 75 76 78 79 80 81 82 83 84 85 86 87 89 90 91 92 93 94 95 96 97 98 111 112 113 114 115 116 117 118 119 120

60 61 62 63 64 65 66 68 69 70 71 72 73 74 75 76 77 79 80 81 82 83 84 85 86 87 88 90 91 92 93 94 95 96 97 98 99 112 113 114 115 116 117 118 119 120 121

49 50 51 52 53 54 55 57 58 59 60 61 62 63 64 65 66 68 69 70 71 72 73 74 75 76 77 79 80 81 82 83 84 85 86 87 88

101 102 103 104 105 106 107 108 109 110

48 49 50 51 52 53 54 56 57 58 59 60 61 62 63 64 65 67 68 69 70 71 72 73 74 75 76 78 79 80 81 82 83 84 85 86 87

100 101 102 103 104 105 106 107 108 109

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1

*************************************************** STAGE NUMBER 1

THE FOLLOWING 10 REINFORCEMENTS ARE ADDED HEREIN

REINFORCEMENT NUMBER I J

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9

10

111 112 113 114 115 116 117 118 119 120

A56


STRESSES VALUES FOR STAGE 1

ELEM X Y SIGMA SIGMA TAU SIGMA SIGMA NO X Y XY 1 3

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

0 0 0 0 0 0 0 0 0 0 5 5 4 4 3 3 2 2 1 0

16 15 13 12 10 8 7 5 4 2

28 25 22 20 17 14 11 9 6 3

39 35 31 27 24. 20. 16, 12. 9. 5.

50. 45. 40. 35. 31. 26. 21. 16. 11. 6.

61. 55. 49. 43. 37. 31.

.10

.10

.10

.10

.10

.10

.10

.10

.10

.10

.79

.25

.71

.17

.63

.09

.55

.01

.47

.94

.95

.34

.72

.10

.49

.87

.26

.64

.02

.41

.12

.43

.73

.04

.35

.65

.96

.27

.57

.88

.28

.51

.74

.97 ,20 ,43 ,66 ,89 ,12 ,35 ,45 ,60 76 .91 06 21 37 52 67 82 62 69 77 84 92 99

1. 3, 5, 7, 9.

11, 13. 15. 17. 19. 1. 3. 5. 7 9

11 13 15 17 19 1 3 5 7 9

11 13 15 17 19 1. 3 5 7 9

11 13, 15 17, 19, 1. 3. 5, 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11.

.00 ,00 .00 ,00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 ,00 ,00 ,00 ,00 ,00 ,00 ,00 ,00 00 ,00 00 .00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 8 7 5 3 1 1 1 1 1 9 8 6 4 3 1 1 1 9 8 7 6 5 3 2 1 8 7 6 6 5 4 3. 2 1 9 5. 4 4. 3, 3, 2. 2. 1, 1. 3, 1. 1. 1. 1. 1. 8.

.000E+00

.OOOE+00

.000E+00

.000E+00

.OOOE+00

.000E+00

.000E+00

.000E+00

.OOOE+00

.000E+00 •913E+02 .678E+02 .463E+02 .272E+02 .072E+02 .925E+01 .059E+01 .298E+01 .366E+01 .280E+01 .533E+02 .456E+02 .309E+02 .154E+02 .914E+01 .153E+01 .423E+01 .709E+01 .158E+01 .623E+01 .180E+02 .064E+02 .877E+01 .894E+01 .865E+01 .725E+01 .364E+01 .918E+01 .523E+01 .463E+01 .733E+01 .758E+01 .788E+01 .059E+01 .337E+01 .698E+01 .984E+01 .979E+01 .849E+01 .250E+00 .405E+01 .861E+01 .273E+01 .650E+01 .103E+01 .605E+01 .276E+01 .911E+01 209E+01 822E+00 888E+01 697E+01 514E+01 316E+01 079E+01 369E+00

0 0 0 0 0 0 0 0 0 0 3 3 2 2 2 1 1 1 6 2 3 2 2 2 1 1 1 9 6 3 2 2 1 1 1 1 1 7 5 2 1 1 1 1 1. 9 7. 5 3, 1 1. 9 8, 7. 6, 5 4, 3, 2. 7, 3. 3, 3. 2, 2. 1.

.OOOE+00

.000E+00

.000E+00

.000E+00

.OOOE+00

.000E+00

.OOOE+00

.000E+00

.OOOE+00

.000E+00

.826E+02

.356E+02

.926E+02

.543E+02

.145E+02

.785E+02

.412E+02

.060E+02

.733E+01

.560E+01

.065E+02

.913E+02

.618E+02

.307E+02

.983E+02

.631E+02

.285E+02

.417E+01

.315E+01

.245E+01

.359E+02

.128E+02

.975E+02

.779E+02

.573E+02

.345E+02

.073E+02

.836E+01

.045E+01

.926E+01

.747E+02

.552E+02

.358E+02

.212E+02 , 067E+02 .396E+01 ,969E+01 .957E+01 .697E+01 .850E+01 ,081E+02 .722E+01 .547E+01 ,301E+01 .207E+01 .211E+01 .552E+01 .822E+01 417E+01 ,644E+00 776E+01 395E+01 029E+01 632E+01 158E+01 674E+01

0 0 0 0 0. 0 0 0. 0. 0, 0, 0, 0. 0. 0, 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0 0. 0. 0. 0. 0. 0. 0 0. 0. 0. 0. 0 0, 0. 0 0, 0 0, 0, 0. 0. 0. 0. 0 0, 0, 0, 0. 0, 0 0, 0 0, 0, 0, 0,

•O00E+00 •000E+00 .000E+00 .000E+00 .000E+00 .000E+00 .000E+00 .OOOE+00 ,000E+00 .000E+00 .000E+00 .000E+00 ,000E+00 .000E+00 .000E+00 .000E+00 .000E+00 .000E+00 .OOOE+00 .000E+00 ,000E+00 .000E+00 ,000E+00 .000E+00 .000E+00 .000E+00 .000E+00 .000E+00 .OOOE+00 .000E+00 .000E+00 •000E+00 .000E+00 •000E+00 •000E+00 .OOOE+00 •000E+0O .OOOE+00 .OOOE+00 •000E+00 •000E+00 .OOOE+00 •OO0E+00 •000E+00 •000E+00 •OOOE+00 .OOOE+00 •OOOE+OO •000E+00 •000E+00 •000E+00 •000E+00 •000E+00 .OOOE+00 •000E+00 .OOOE+00 .000E+00 .OOOE+00 •000E+00 •000E+O0 .000E+00 .000E+00 OOOE+00 .000E+00 .000E+00 .000E+00

0 0 0 0 0. 0 0. 0 0 0 3 3. 2. 2 2. 1 1. 1 6 2 3 2 2. 2 1. 1 1. 9. 6. 3 2 2 1 1. 1. 1 1. 7 5. 2 1. 1 1 1 1. 9 7 5 3 1 1 9 8 7 6 5 4 3 2 7 3 3 3 2 2 1

.000E+00

.000E+00

.000E+00

.000E+00

.000E+00

.000E+00

.000E+00 •OOOE+00 •000E+O0 •OOOE+00 •826E+02 •356E+02 •926E+02 •543E+02 •145E+02 •785E+02 •412E+02 •060E+02 •733E+01 •560E+01 •065E+02 .913E+02 •618E+02 •307E+02 •983E+02 .631E+02 .285E+02 •417E+01 •315E+01 .245E+01 •359E+02 .128E+02 •975E+02 .779E+02 .573E+02 •345E+02 •073E+02 .836E+01 .045E+01 .926E+01 .747E+02 •552E+02 .358E+02 .212E+02 .067E+02 .396E+01 .969E+01 .957E+01 .697E+01 .850E+01 .081E+02 .722E+01 .547E+01 . 301E+01 .207E+01 .211E+01 .552E+01 .822E+01 .417E+01 .644E+00 .776E+01 .395E+01 .029E+01 .632E+01 .158E+01 .674E+01

0 0 0 0 0. 0 0. 0, 0. 0. 1. 1. 1. 1. 1. 8. 7. 5, 3. 1, 1. 1. 1, 1. 9. 8, 6. 4. 3. 1, 1. 1. 9, 8, 7, 6 5. 3 2. 1 8. 7. 6. 6. 5 4 3 2 1 9 5 4 4 3 3 2 2 1 1 3 1 1 1 1 1 8

•000E+00 •000E+00 •000E+00 .000E+00 •000E+00 •000E+00 .000E+00 •000E+00 .000E+00 •000E+00 .913E+02 •678E+02 463E+02 •272E+02 .072E+02 .925E+01 059E+01 .298E+01 366E+01 .280E+01 533E+02 .456E+02 309E+02 .154E+02 914E+01 .153E+01 .423E+01 •709E+01 158E+01 •623E+01 •180E+02 .064E+02 •877E+01 •894E+01 .865E+01 •725E+01 •364E+01 .918E+01 •523E+01 •463E+01 •733E+01 .758E+01 •788E+01 .059E+01 .337E+01 .698E+01 .984E+01 .979E+01 .849E+01 .250E+00 .405E+01 . 861E+01 .273E+01 .650E+01 .103E+01 .605E+01 .276E+01 .911E+01 .209E+01 .822E+00 .888E+01 .697E+01 .514E+01 .316E+01 .079E+01 .369E+00

A57


87 88 89 90 91 92 93 94 95 96 97 98 99 100

26. 20. 14, 8

67, 60, 54 47 41 34 28 22 15 9

,07 .15 .22 .30 .30 .84 .37 .91 .45 .98 .52 .06 .59 .13

13. 15. 17. 19. 1. 3. 5. 7. 9

11. 13 15 17 19

.00 00 .00 ,00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00

6. 6. 6.

-1. 0. 0 0 0. 0 0 0 0 0 0

108E+00 077E+00 .795E+00 .516E+O0 •OOOE+00 •OOOE+00 •OOOE+00 •000E+00 .OOOE+00 .OOOE+00 .000E+00 .000E+00 .OOOE+00 .OOOE+00

1, 1. 1. -3 0 0 0 0. 0 0 0 0 0 0

222E+01 215E+01 •359E+01 •032E+00 •000E+00 •OOOE+00 .000E+00 .000E+00 .OOOE+00 .000E+00 .000E+00 .OOOE+00 .OOOE+00 .000E+00

0. 0. 0, 0. 0. 0. 0 0 0 0. 0 0 0 0

.000E+00 000E+00 .000E+00 , OOOE+00 .OOOE+00 .000E+00 .OOOE+00 .OOOE+00 .000E+00 .OOOE+00 .OOOE+00 .OOOE+OO •000E+00 .OOOE+00

1 1 1 -1 0 0. 0 0 0 0 0 0 0 0

.222E+01

.215E+01 •359E+01 •516E+O0 •000E+00 .OOOE+00 .000E+00 .OOOE+00 .OOOE+00 .OOOE+00 .OOOE+00 .000E+00 .OOOE+00 .OOOE+00

6, 6. 6,

-3. 0. 0. 0 0. 0 0. 0 0 0 0

108E+00 077E+00 795E+00 •032E+00 .OOOE+00 •OOOE+00 .OOOE+00 •000E+00 .OOOE+00 •000E+00 .OOOE+00 .OOOE+00 .000E+00 .OOOE+00

INTERFACE ELEMENT RESULTS FOR STAGE

EM

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

NO X

0. 0. 0. 0. 0. 0. 0. 0 0. 0

67 60 54 47 41 34 28 21 15 9

.20

.20

.20 ,20 .20 .20 .20 .20 .20 .20 .20 .74 .27 .81 .35 .88 .42 .96 .49 .03

Y

1. 3. 5. 7. 9,

11. 13. 15, 17, 19, 1 3, 5 7 9 11 13 15 17 19

NORMAL STRESS SHEAR

00 00 ,00 .00 .00 .00 ,00 ,00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00

7, 0, 0 2 , 0 1 0 4 0. 4 8 1 1 1 1 1 1 1 7 1

823E+00 .000E+00 .OOOE+00 •049E+00 •000E+00 .257E+00 .OOOE+00 .657E-01 .000E+00 .657E-02 .513E-01 .524E+00 .243E+00 .478E+00 .350E+00 .187E+00 .676E+00 .676E+00 .451E-01 .304E+00

0. 0. 0, 0 . 0, 0. 0. 0. 0. 0 0 0. 0

o. 0 0 0 0 0 0

STRESS

000E+00 000E+00 .OOOE+00 000E+00 .000E+00 .OOOE+00 .OOOE+00 •000E+00 •OOOE+00 .000E+00 •OOOE+00 •000E+00 .OOOE+00 •OOOE+00 .OOOE+00 .OOOE+00 .000E+00 .OOOE+00 .OOOE+00 .OOOE+00

NORMAL STIFF

1. 1. 1. 1. 1. 1. 1 1. 1. 1 1 1 1 1 1 1 1 1 1 1

000E+08 000E+02 .000E+02 OOOE+08 .000E+02 .000E+08 .000E+02 .OOOE+08 •000E+02 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .000E+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08

SHEAR STIFF

4. 1, 1. 4. 1 4 1 4 1. 4 4 4 4 4 4 4 4 4 4 4

OOOE+03 .OOOE+02 .000E+02 .000E+03 .000E+02 .000E+03 .000E+02 .000E+03 .000E+02 .000E+03 .000E+03 .000E+03 .00OE+O3 .000E+03 .OOOE+03 .000E+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03


REIN. NUM.

1 2 3 4 5 6 7 8 9

10

I

1 2 3 4 5 6 7 8 9

10

J

111 112 113 114 115 116 117 118 119 120

TYPE COMPR FORCE

1 0.000000E+0Q 1 0.O0O0OOE+OO 1 0.000000E+00 1 0.OOO0OOE+O0 1 O.OOOOOOE+00 1 0.0O000OE+O0 1 0.000000E+00 1 0.OOOOO0E+00 1 0.00O0O0E+O0 1 O.OOOOOOE+00

INCR COMPR

0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

OOE+00 00E+00 OOE+00 OOE+00 00E+00 00E+00 OOE+00 00E+00 OOE+00 OOE+00

STIFFNESS

6.7500 6.7500 6.7500 6.7500 6.7500 6.7500 6.7500 6.7500 6.7500 6.7500

00E+05 OOE+05 00E+05 OOE+05 OOE+05 00E+05 00E+05 OOE+05 OOE+05 OOE+05

A58


*************************************************** STAGE NUMBER 2

FORCE AND/OR DISPLACEMENT LOADING IS SPECIFIED FOR THIS STAGE

NODE

111 113 115 117 119 121

-2. -3. -2 -1 -8 0

X-LOAD

. 00000E + 02

.20000E+02

.40000E+02

.60000E+02

.00000E+01

.OOOOOE+OO

-6. -1. -7. -5 -2 0

Y-LOAD

. 50000E+02

.00000E+03

.80000E+02

.20000E+02

.60000E+02

.00000E+00

NODE

112 114 116 118 120

0

-3. -2. -2. -1 -4 0.

X-LOAD

.60000E+02

.80000E+02

.0OO00E+02

.20000E+02

.OOO00E+01

.OOOOOE+00

-1 -9 -6. -3 -1 0

Y-LOAD

.20000E+03

.00000E+02

.50000E+02

.90000E+02

.30000E+02

.00000E+00

DISPLACEMENT RESULTS FOR STAGE 2

NODAL POINT

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

X

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20

11.91 10.83 9.75 8.68 7.60 6.52 5.44 4.37 3.29 2.21 1.13

23.61 21.46 19.30 17.15 14.99 12.84

Y

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00

TOTAL UX

0.00000E+00 -9.66641E-03 -2.25790E-02 -3.06625E-02 -3.78562E-02 -4.33966E-02 -4.72485E-02 -4.93257E-02 -4.98057E-02 -4.87903E-02 -4.86997E-02 0.00000E+00 -9.65705E-03 -2.25741E-02 -3.06547E-02 -3.78522E-02 -4.33936E-02 -4.72473E-02 -4.93254E-02 -4.98057E-02 -4.87909E-02 -4.86996E-02 0.00000E+00

-9.65728E-03 -1.97643E-02 -3.06544E-02 -3.78521E-02 -4.33935E-02 -4.72472E-02 -4.93253E-02 -4.98057E-02 -4.87910E-02 -4.86996E-02 0.OOOOOE+00

-9.95573E-03 -2.02009E-02 -2.90837E-02 -3.77028E-02 -4.41208E-02 -4.85027E-02 -5.08358E-02 -5.11112E-02 -4.98042E-02 -4.84247E-02 0.OOOOOE+00

-1.25726E-02 -2.21393E-02 -3.09764E-02 -3.84480E-02 -4.51733E-02

TOTAL UY

0.OOOOOE+00 -1.22706E-03 -9.08989E-04 -1.24448E-03 -1.27741E-03 -1.38915E-03 -1.30659E-03 -1.30794E-03 -1.17737E-03 -1.15954E-03 -1.25621E-03 0.00OOOE+00 6.55410E-04

-1.85596E-04 -3 .63533E-04 -7.14513E-04 -8.49476E-04 -1.07506E-03 -1.12469E-03 -1.26667E-03 -1.26971E-03 -1.16573E-03 0.00000E+00 -2.39635E-03 -4.06288E-03 -4.74494E-03 -4.89627E-03 -4.46839E-03 -3.63578E-03 -2.72890E-03 -1.69778E-03 -9.93194E-04 -9.81562E-04 0.OOOOOE+00 9.01700E-06 -2.03284E-04 -9.70777E-04 -1.91997E-03 -2.62305E-03 -2.94632E-03 -2.71885E-03 -2.08639E-03 -1.03563E-03 -3.70832E-04 0.O0O0OE+O0 -2.76861E-03 -4.24053E-03 -4.75471E-03 -4.87349E-03 -4.89617E-03

PORE PRESS 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

A59


51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121

10. 8. 6. 4. 2.

35. 32. 28. 25. 22. 19. 15. 12. 9. 6. 3 .

47. 42, 38. 34. 29. 25. 21. 16. 12 8 3 58 53 47 42 37 31 26 21 15 10 4

70 63 57 51 44 38 31 25 18 12 5

70 63 57 51 44 38 31 25 18 12 5

70 64 57 51 44 38 31 25 18 12 6

68 53 38 22 07 32 09 86 62 39 16 93 70 46 23 00 02 .71 .40 ,09 ,79 ,48 .17 ,86 .55 .24 .93 .73 .34 .96 .57 .18 .80 .41 .03 .64 .25 .87 .43 .97 .50 .04 .58 .11 .65 .19 .73 .26 .80 .43 .97 .50 .04 .58 .11 .65 .19 .73 .26 .80 .63 .17 .70 .24 .78 .31 .85 .39 .93 .46 .00

12. 14. 16. 18. 20. 0. 2. 4. 6. 8.

10. 12. 14. 16. 18. 20. 0. 2. 4. 6. 8.

10. 12. 14. 16. 18, 20. 0. 2. 4. 6. 8.

10. 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8

10 12 14 16 18 20 0 2 4 6 8

10 12 14 16 18 20

00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 .00 00 .00 .00 ,00 .00 .00 .00 .00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00

-4. -5. -5. -5. -4. 0.

-1. -2. -3. -4. -4. -5. -5. -5. -5. -4. 0.

-1. -2. -3. -4, -4 -5. -5. -5 -5 -4 0

-9 -2 -3 -4 -4 -5 -5 -5 -5 -4 0

-8 -2 -3 -3 -4 -4 -5 -5 -4 -4 0

-1 -2 -3 -3 -4 -4 -4 -4 -4 -4 0

-9 -2 -3 -3 -4 -4 -4 -4 -4 -4

99906E-02 25863E-02 28276E-02 09024E-02 83737E-02 00000E+00 28875E-02 45714E-02 33215E-02 05170E-02 61660E-02 08548E-02 36319E-02 38874E-02 16189E-02 84485E-02 00O00E+O0 •30506E-02 •44252E-02 •45416E-02 •19100E-02 •72777E-02 •09291E-02 .35281E-02 .40132E-02 .15180E-02 •84739E-02 •00000E+00 .19290E-03 •18989E-02 .29591E-02 .16002E-02 .71757E-02 .05278E-02 .22017E-02 .29021E-02 .07735E-02 .81074E-02 .OOOOOE+00 .61350E-03 .20175E-02 .11377E-02 .85978E-02 .43546E-02 .81436E-02 .01491E-02 .06764E-02 .99556E-02 .72268E-02 .OOOOOE+00 .00216E-02 .26582E-02 .08441E-02 .80129E-02 .35467E-02 .73717E-02 .94188E-02 .98763E-02 .88108E-02 .62745E-02 .OOOOOE+00 .70039E-03 .25526E-02 .07110E-02 .78746E-02 .34164E-02 .72559E-02 .93277E-02 .98066E-02 .87788E-02 .62688E-02

-4. -3. -2. -1. -2. 0.

-4. -8. -1. -1. -1. -9. -7. -5. -2. -2. 0.

-7. -1. -1. -1. -1. -1. -1. -9, -4. -4. 0.

-1, -2. -2 . -2 -2. -2 -2 -1 -7 -1 0

-1 -2 -3 -4 -4 -3 -3 -2 -1 -2 0

-1 -2 -3 -4 -4 -3 -3 -2 -1 -2 0

-9 -2 -3 -4 -4 -3 -3 -2 -1 -2

53903E-03 82648E-03 64485E-03 27354E-03 72025E-04 0O000E+00 86885E-03 54078E-03 06540E-02 12751E-02 07864E-02 69127E-03 69882E-03 14939E-03 33852E-03 31698E-04 00000E+00 67479E-03 35124E-02 74231E-02 91577E-02 88400E-02 68891E-02 39573E-02 .53121E-03 62503E-03 .66477E-04 .OOOOOE+00 .13967E-02 .03076E-02 •60613E-02 .88770E-02 .88405E-02 .62180E-02 .16085E-02 .57274E-02 .90649E-03 .18001E-03 .OOOOOE+00 .08538E-02 .57712E-02 .64069E-02 .17358E-02 .21101E-02 .84662E-02 .16737E-02 .27137E-02 .25233E-02 .11478E-03 .00000E+00 .04194E-02 .55744E-02 .64992E-02 .19180E-02 .23611E-02 .87059E-02 .19003E-02 .29617E-02 .28778E-02 .40937E-03 .OOOOOE+00 .95582E-03 .51590E-02 .62534E-02 .18383E-02 .24221E-02 .88787E-02 .21539E-02 .32645E-02 .32006E-02 .73496E-03

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 .00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

A60



3LEM NO

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

0 0 0 0 0 0 0 0 0 0 5 5 4 4 3 3 2 2 1 0

16 15 13 12 10 8 7 5 4 2

28 25 22 20 17 14 11. 9 6. 3.

39. 35. 31. 27. 24, 20. 16. 12. 9. 5.

50. 45. 40. 35. 31. 26. 21. 16. 11. 6.

61. 55. 49. 43. 37. 31.

X

.10

.10

.10

.10

.10

.10

.10

.10

.10

.10

.79

.25

.71

.17

.63

.09

.55

.01

.47

.94

.95

.34

.72

.10

.49

.87

.26

.64

.02

.41

.12

.43

.73

.04

.35

.65

.96

.27

.57

.88

.28

.51

.74

.97 ,20 .43 ,66 ,89 ,12 35 45 60 76 91 06 21 37 52 67 82 62 69 77 84 92 99

1 3 5 7 9

11 13 15 17 19 1 3 5 7 9

11 13 15 17 19 1 3 5 7 9

11 13 15 17 19 1 3 5 7 9

11 13, 15 17, 19 1. 3 5, 7. 9,

11. 13, 15. 17. 19. 1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11.

Y

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00 ,00 .00 ,00 ,00 .00 .00 .00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

3 -8 1

-1 -5 -4 -2 -3 1 1 2 1 1 1 1 9 7 5 3 1 1 1 1 1 1 9 7 5 3 1 1 1 1 1 1 9 7 5 3 1 1 1 1. 1. 9. 8, 6. 4, 2, 9. 1. 1. 1. 1. 9. 7. 6. 4. 2. 4. 1. 1. 1. 1. 8. 7.

SIGMA X

.425E+01

.259E+00

.207E+00

.516E+01

.698E+00

.660E+00

.515E+00

.660E-01

.669E+00

.983E+00

.057E+02

.784E+02

.501E+02

.274E+02

.080E+02

.027E+01

.243E+01

.504E+01

.655E+01

.412E+01

.760E+02

.696E+02

.534E+02

.345E+02

.140E+02

.354E+01

.357E+01

.391E+01

.545E+01

.708E+01

.649E+02

.491E+02

.383E+02

.240E+02

.084E+02

.091E+01

.128E+01

.096E+01

.098E+01

.493E+01

.609E+02

.438E+02 •266E+02 •127E+02 •800E+01 •343E+01 .774E+01 .840E+01 .754E+01 .495E+00 534E+02 .382E+02 238E+02 077E+02 234E+01 659E+01 189E+01 663E+01 542E+01 747E+00 102E+02 260E+02 204E+02 062E+02 818E+01 046E+01

3 3 3 2 1 9 3 7 -9 -4 4 3 2 2 2 1 1 1 6 2 3 3 2 2 2 1 1 9 6 3 3 2 2 2 1 1 1 8 5. 2 2 2 2, 1, 1, 1, 1, 8. 4. 1. 2. 2, 1. 1. 1. 1. 9. 7. 4. 1. 1. 1. 1. 1. 1. 9.

SIGMA Y

.807E+02

.384E+02

.346E+02

.462E+02

.592E+02

.194E+01

.256E+01

.324E+00

.207E+00

.274E+00

.035E+02

.500E+02

.995E+02

.571E+02

.146E+02

.767E+02

.389E+02

.033E+02

.606E+01

.529E+01

.363E+02

.200E+02

.887E+02

.539E+02

.161E+02

.754E+02

.355E+02

.678E+01

.228E+01

.078E+01

.012E+02

.690E+02

.466E+02

.204E+02

.934E+02

.633E+02 •274E+02 •994E+01 •435E+01 .757E+01 .744E+02 .441E+02 •127E+02 •872E+02 •623E+02 •396E+02 •152E+02 •275E+01 •790E+01 900E+01 399E+02 .171E+02 919E+02 643E+02 390E+02 150E+02 506E+01 425E+01 332E+01 185E+01 486E+02 660E+02 589E+02 424E+02 202E+02 760E+01

1 -5 3 -1 1

-1 1 5 -9 3 2 2 2 1 1 7 3 1 -8 -8 3 2 2 2 1 1 5 2 -4 -1 2 2. 2 1. 1 1 7. 3 2. -1 2. 2. 1. 1. 1 1. 7 . 3 . 2.

-1. 1, 1. 1, 1. 9. 6. 4. 2 .

-3, -1. 7. 1, 9, 7, 6. 4.

TAU XY

.299E+02

.963E+01

.144E+01

.228E+01

.458E+01

.673E+00

.971E+00

.418E+00

.227E+00

.925E+00

.799E+01

.608E+01

.285E+01

.651E+01

.145E+01

.410E+00

.856E+00

.038E+00

.889E-01

.505E-01

.041E+01

.771E+01

.412E+01

.055E+01 •565E+01 .054E+01 .962E+00 .107E+00 .740E-01 .106E+00 .940E+01 .501E+01 .202E+01 .889E+01 .579E+01 .205E+01 •786E+00 .587E+00 .479E-01 .231E+00 •574E+01 •249E+01 .870E+01 .569E+01 •285E+01 •016E+01 •187E+00 .709E+00 •223E-01 .288E+00 .611E+01 •660E+01 506E+01 .208E+01 .363E+00 .926E+00 731E+00 479E+00 .099E-01 716E+00 708E+00 154E+01 719E+00 909E+00 009E+00 600E+00

4 3 3 2 1 9 3 1 6 3 4 3 3 2 2 1 1 1 6 2. 3 3. 2 2. 2. 1 1 9 6. 3 3 2. 2. 2 1 1 1. 9 5 2 2 2 2 1. 1. 1. 1 8 4 1 2. 2 1. 1 1 1. 9 7. 4 1 1 1 1. 1 1 9

SIGMA 1

.240E+02

.484E+02

.375E+02

.467E+02

. 605E+02

.197E+01

.268E+01

.012E+01

.942E+00

.874E+00

.074E+02

.539E+02

.029E+02

.591E+02

.158E+02

.773E+02

.391E+02

.033E+02

.609E+01 •536E+01 .419E+02 .250E+02 •929E+02 .573E+02 •184E+02 .767E+02 •361E+02 •689E+01 •229E+01 .087E+01 .072E+02 .740E+02 •509E+02 .239E+02 .962E+02 .653E+02 .285E+02 •027E+01 .435E+01 .768E+01 .799E+02 .489E+02 .166E+02 .904E+02 .647E+02 •414E+02 •163E+02 •315E+01 •790E+01 •917E+01 •428E+02 .204E+02 •951E+02 •668E+02 .408E+02 •162E+02 •573E+01 •447E+01 .333E+01 •225E+01 . 501E + 02 .691E+02 •612E+02 .440E+02 .213E+02 .836E+01

-9 -1 -1 -1 -6. -4 -2 -3. -1 -6. 2 1. 1. 1 1 8 7 5. 3 1, 1. 1. 1 1. 1. 9, 7. 5, 3. 1, 1. 1. 1. 1. 1. 8. 7, 5. 3, 1. 1, 1. 1. 1. 9. 8. 6. 4. 2 9 1. 1 1 1 9 7 6 4 2 4 1 1 1 1 8 6

SIGMA 3

.026E+00

.823E+01

.733E+00

.573E+01

.978E+00

.689E+00

.626E+00

.164E+00

.448E+01

.165E+00

.019E+02 •745E+02 •466E+02 .253E+02 067E+02 •964E+01 •221E+01 •501E+01 •652E+01 .406E+01 .705E+02 .647E+02 .493E+02 311E+02 116E+02 .220E+01 .301E+01 .381E+01 .545E+01 .699E+01 •588E+02 441E+02 •340E+02 •204E+02 •055E+02 •896E+01 022E+01 •063E+01 098E+01 •481E+01 •553E+02 •390E+02 •227E+02 •096E+02 •553E+01 •165E+01 •667E+01 •801E+01 .754E+01 •324E+00 •505E+02 .348E+02 .206E+02 .052E+02 .053E+01 .538E+01 .123E+01 .641E+01 •542E+01 .354E+00 .087E+02 .229E+02 .181E+02 .046E+02 .709E+01 .970E+01

A61


87 88 89 90 91 92 93 94 95 96 97 98 99

100

26 20 14 8. 67 60 54 47 41 34 28 22 15 9

.07 ,15 .22 ,30 .30 .84 .37 ,91 .45 .98 .52 .06 .59 .13

13 15, 17 19. 1 3. 5 7. 9

11. 13. 15 17. 19

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

5 3 2

-2 -1 -2 -1 -1 -1 -1 -1 -1 -6 -2

.317E+01

.903E+01

.684E+01

.82OE+O0 •294E+03 .624E+03 •543E+03 .800E+03 .792E+03 .647E+03 .394E+03 .090E+03 .941E+02 •581E+02

7 5 4 3 3 6 4

-1 -1 -1 -1 -1 -1 -5

.541E+01

.845E+01

.347E+01

.446E+00

.188E+03

.170E+02

.195E+01

.323E+02

.734E+02

.802E+02

.858E+02

.733E+02

.289E+02

.211E+01

3 1

-3 -2 1 7 5 6 5 5 4 3 2 5

.286E+00

.579E+00

.548E-01

.423E+00

.225E+03

.571E+02

.262E+02

.011E+02

.752E+02

.339E+02

.421E+02

.398E+02

.285E+02

.756E+01

7 5 4 4 3 7 2 6 1

-6 -4 -6 -4 -3

.589E+01

.858E+01

.348E+01

.274E+00

.502E+03

.851E+02

.007E+02 •176E+01 .016E+01 .449E+00 .129E+01 .104E+01 .807E+01 .712E+01

5 3 2

-3 -1 -2. -1 -1 -1. -1 -1. -1 -7. -2.

•27OE+01 .890E+01 •683E+01 . 647E+00 •607E+03 .793E+03 .702E+03 .994E+03 .976E+03 .821E+03 .538E+03 .202E+03 .749E+02 •731E+02



1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20

0 0. 0 0. 0 0

o. o. 0 0 67 60. 54 47 41. 34 28. 21. 15. 9.

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.74

.27

.81

.35

.88

.42

.96

.49

.03

1 3 5 7 9

11. 13 15. 17 19 1. 3. 5 7 9. 11 13 15 17, 19,

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

1 0 0 0 0 0 0 2. 7 2 6 1. 1 1 1 9 7. 5 3. 8.

•914E+01 •000E+00 .000E+00 .000E+00 .OOOE+00 •000E+00 .000E+00 .794E-01 .264E-06 .841E+00 .376E+01 .348E+02 .395E+02 .287E+02 .109E+02 .078E+01 .134E+01 .234E+01 .343E+01 .941E+00

-6 0 0 0 0 0 0

-4 -7 9 2 4 7

-1 -2 -3 -3 -3 -4 -4

.104E+00 •000E+00 .000E+00 .000E+00 .000E+00 •000E+00 .OOOE+00 •071E+00 .729E-03 .214E-01 .947E+00 .288E+00 .248E-01 .841E+00 .917E+00 .309E+00 •146E+00 .205E+00 .072E+00 .390E+00

1 1. 1 1. 1 1 1 1. 1 1. 1 1 1 1 1 1 1 1 1 1

.000E+08 •000E+02 •000E+02 .000E+02 .000E+02 •000E+02 .000E+02 .OOOE+08 .OOOE+08 .000E+08 •OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .000E+08 .000E+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08

4. 1. 1 1 1 1. 1 1. 1 4 4 4 4 4 4 4 4 4 4 4

.OOOE+03

.000E+02 •000E+02 .000E+02 . 000E + 02 •00OE+02 •000E+02 •OOOE+02 •000E+02 •000E+03 .OOOE+03 •000E+03 .OOOE+03 .000E+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03


REIN. NUM. I J TYPE COMPR FORCE INCR COMPR STIFFNESS

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9

10

111 112 113 114 115 116 117 118 119 120

1 1 1 1 1 1 1 1 1 1

0 2 -1 3 1 1 4 1 6

-7

. OOOOOOE + 00

.293917E+01

.781573E+01

.272714E+01

.238929E+01

.338003E+01

.986394E+00

.327693E+00

.361865E-01

.737335E+00

0 3. -2 4 1 1 7 1 9

-1

.000000E+00

.398396E-05

. 639368E-05

.848465E-05

.835451E-05

.982227E-05

.387251E-06 •966953E-06 .424984E-07 •146272E-05

6 6 6 6 6 6 6 6 6 6

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

A62


*************************************************** STAGE NUMBER 3



111 113 115 117 119 121

-2. -2. 0 0, 0. 0.

•90000E+01 •00000E+01 .00000E+00 •00000E+00 •OOOOOE+00 .00000E+00

-3. -2 0. 0 0 0

•50000E+02 .30000E+02 .00000E+00 •OOOOOE+00 .00000E+00 .OOOOOE+00

112 114 116 118 120 0

-3. 0 0. 0. 0 0.

•90000E+01 .00000E+00 •OOOOOE+00 •00000E+00 •00000E+00 •00000E+00

-4. 0 0 0. 0 0.

.70000E+02

.00O00E+O0

.OOOOOE+00 •OOOOOE+00 .OOOOOE+00 •O0000E+O0


NODAL POINT

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

X

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 o. o. o.

11. 10, 9. 8. 7. 6. 5. 4. 3, 2. 1.

23. 21. 19. 17, 14. 12.

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20

.20 ,20 .20 .91 .83 .75 ,68 .60 ,52 .44 .37 .29 .21 .13 .61 .46 30 .15 99 84

Y

0 2 4 6 8

10 12 14 16 18 20 0 2 4 6 8

10 12. 14 16 18. 20 0. 2. 4 6. 8,

10. 12, 14. 16. 18, 20, 0, 2. 4. 6. 8.

10. 12. 14. 16. 18. 20. 0. 2. 4. 6. 8.

10.

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00 ,00 ,00 .00 ,00 .00 ,00 .00 ,00 .00 ,00 ,00 00 .00 .00 00 .00 00 00 .00 00 00 00 00 00 .00 00 00

0 -9 -2 -3 -3 -4 -4 -4 -4 -4 -4 0

-9 -2 -3 -3 -4 -4 -4 -4 -4 -4 0

-9 -1 -3. -3 -4 -4 -4 -4. -4. -4 0.

-1. -2 -2. -3. -4. -4, -5. -5, -4, -4. 0,

-1, -2. -3, -3. -4.

TOTAL UX

.00000E+00

.29390E-03

.19444E-02

.01371E-02

.74785E-02

.31643E-02

.71013E-02

.92201E-02

.97055E-02

.86902E-02 •86042E-02 .00OOOE+00 •28491E-03 •19394E-02 .01294E-02 .74745E-02 •31613E-02 .71000E-02 .92197E-02 .97056E-02 .86908E-02 .86040E-02 .OOOOOE+00 .28519E-03 .96293E-02 .05217E-02 .77371E-02 .32834E-02 .71404E-02 .92197E-02 .97056E-02 ,86909E-02 .86040E-02 .0OOO0E+O0 .00224E-02 ,00712E-02 ,89665E-02 .75792E-02 .40031E-02 83894E-02 07271E-02 .10059E-02 97033E-02 83290E-02 OOOOOE+00 24844E-02 20844E-02 08603E-02 83266E-02 50499E-02

0 -1 -9 -1 -1 -1 -1 -1 -1. -1 -1. 0, 6

-1. -3. -7 -8 -1. -1. -1. -1. -1 0

-2. -4 -4. -4. -4 -3. -2. -1. -9. -9 0 2. -1 -9 -1. -2 -2 -2 -2 -1 -3, 0

-2, -4, -4 -4. -4.

TOTAL UY

.OOOOOE+00

.18995E-03 •15243E-04 .24498E-03 .28584E-03 .38940E-03 . 30786E-03 .30355E-03 .17423E-03 .15420E-03 •25308E-03 .00000E+00 .21295E-04 •74961E-04 •58022E-04 .00258E-04 .42770E-04 .06666E-03 .12138E-03 .26153E-03 .26626E-03 .15984E-03 •OOOOOE+00 .36644E-03 .00443E-03 •69558E-03 •85242E-03 •43172E-03 .60508E-03 •70312E-03 .67655E-03 •77095E-04 •71149E-04 .OOOOOE+00 .10586E-05 .70920E-04 .31043E-04 •88057E-03 . 58690E-03 .91427E-03 .69032E-03 .06131E-03 •01416E-03 53036E-04 •OOOOOE+00 •77175E-03 .24640E-03 •74626E-03 •85546E-03 .87224E-03

PORE PRESS 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

A63


51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121

10. 8. 6. 4. 2.

35. 32. 28. 25. 22. 19. 15. 12. 9. 6. 3.

47. 42. 38. 34. 29. 25. 21. 16. 12. 8. 3.

58, 53 47. 42 37. 31 26 21 15 10 4

70 63 57 51 44 38 31 25 18 12 5

70 63 57 51 44 38 31 25 18 12 5

70 64 57 51 44 38 31 25 18 12 6

68 53 38 22 07 32 09 86 62 39 16 93 70 46 23 00 02 71 40 09 79 48 .17 .86 ,55 ,24 ,93 ,73 .34 .96 .57 .18 .80 .41 .03 .64 .25 .87 .43 .97 .50 .04 .58 .11 .65 .19 .73 .26 .80 .43 .97 .50 .04 .58 .11 .65 .19 .73 .26 .80 .63 .17 .70 .24 .78 .31 .85 .39 .93 .46 .00

12. 14. 16. 18. 20. 0. 2. 4. 6. 8.

10. 12. 14. 16. 18. 20. 0. 2. 4. 6. 8.

10, 12. 14, 16, 18, 20, 0. 2. 4. 6 8 10 12. 14 16 18 20 0 2 4 6 8

10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8

10 12 14 16 18 20

00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 .00 ,00 .00 ,00 ,00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00

-4.98724E-02 -5.24739E-02 -5.27205E-02 -5.08004E-02 -4.82780E-02 0.OOOOOE+00 -1.28039E-02 -2.44456E-02 -3.32206E-02 -4.03993E-02 -4.60460E-02 -5.07354E-02 -5.35174E-02 -5.37782E-02 -5.15151E-02 -4.83529E-02 0.00000E+00 -1.31414E-02 -2.44302E-02 -3.44561E-02 -4.18091E-02 -4.71638E-02 -5.08136E-02 -5.34147E-02 -5.39039E-02 -5.14126E-02 -4.83777E-02 0.OOOOOE+00 -9.51986E-03 -2.20710E-02 -3.29632E-02 -4.15022E-02 -4.70628E-02 -5.04151E-02 -5.20924E-02 -5.27945E-02 -5.06686E-02 -4.80091E-02 0.OOOOOE+00 -9.02932E-03 -2.24031E-02 -3.10573E-02 -3.83935E-02 -4.41852E-02 -4.80221E-02 -5.00446E-02 -5.05730E-02 -4.98524E-02 -4.71282E-02 0.OOOOOE+00 -9.61286E-03 -2.20318E-02 -3.03283E-02 -3.76442E-02 -4.33172E-02 -4.72262E-02 -4.93128E-02 -4.97761E-02 -4.87106E-02 -4.61747E-02 0.OOOOOE+00 -9.31813E-03 -2.19202E-02 -3.01840E-02 -3.74979E-02 -4.31833E-02 -4.71092E-02 -4.92217E-02 -4.97068E-02 -4.86786E-02 -4.61691E-02

-4.51294E-03 -3.80039E-03 -2.61947E-03 -1.24981E-03 -2.51793E-04 0.00000E+00 -4.82731E-03 -8.49374E-03 -1.06176E-02 -1.12435E-02 -1.07579E-02 -9.66440E-03 -7.67385E-03 -5.12595E-03 -2.31563E-03 -2.10179E-04 0.OOOOOE+00 -7.57060E-03 -1.33447E-02 -1.72698E-02 -1.90489E-02 -1.87690E-02 -1.68428E-02 -1.39250E-02 -9 .50689E-03 -4 . 60414E-03 -4.45818E-04 0.OOOOOE+00 -1.18685E-02 -2 .04748E-02 -2.59546E-02 -2.87019E-02 -2.86998E-02 -2.61296E-02 -2 .15580E-02 -1.56973E-02 -7.88600E-03 -1.16161E-03 0.OOOOOE+00 -1.20427E-02 -2.72880E-02 -3.71329E-02 -4.18080E-02 -4.19582E-02 -3.83186E-02 -3.15888E-02 -2.26721E-02 -1.25009E-02 -2.10052E-03 0.OOOOOE+00 -1.18640E-O2 -2.74047E-02 -3.73599E-02 -4.20412E-02 -4 .22279E-02 -3.85657E-02 -3.18159E-02 -2.29191E-02 -1.28545E-02 -2 .39550E-03 0.OOOOOE + 00 -1.13540E-02 -2.70067E-02 -3.71466E-02 -4.19766E-02 -4.22920E-02 -3.87371E-02 -3.20679E-02 -2.32210E-02 -1.31769E-02 -2.72084E-03

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

A64



ELEM NO

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 5. 5. 4. 4. 3 , 3. 2. 2. 1, 0.

16. 15. 13. 12. 10 8 7 5 4 2

28 25 22 20 17 14 11 9 6 3

39 35 31 27 24 20 16 12 9 5

50 45 40 35 31 26 21 16 11 6 61 55 49 43 37 31

X

10 10 10 10 10 10 10 10 10 10 79 25 71 17 .63 .09 .55 .01 .47 .94 .95 .34 .72 .10 .49 .87 .26 .64 .02 .41 .12 .43 .73 .04 .35 .65 .96 .27 .57 .88 .28 .51 .74 .97 .20 .43 .66 .89 .12 .35 .45 .60 .76 .91 .06 .21 .37 .52 .67 .82 .62 .69 .77 .84 .92 .99

1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11. 13, 15, 17, 19. 1 3. 5 7 9

11 13. 15 17 19 1 3 5 7 9

11 13 15 17 19 1 3 5 7 9 11 13 15 17 19 1 3 5 7 9

11

Y

00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 .00 00 .00 .00 .00 ,00 ,00 ,00 .00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00

3. -6. 7.

-1. -5. -4. -2. -3. 1. 2. 2. 1. 1. 1. 1. 9. 7. 5, 3. 1. 1. 1. 1, 1. 1. 9. 7. 5 3 . 1. 1 1 1 1 1 9 7 5 3 1 1 1 1 1 9 8 6 4 2 9 1 1 1 1 9 7 6 4 2 4 1 1 1 1 8 7

SIGMA X

627E+01 916E+00 708E-01 493E+01 784E+00 589E+00 554E+00 439E-01 614E+00 015E+00 062E+02 789E+02 501E+02 274E+02 080E+02 029E+01 244E+01 503E+01 653E+01 .413E+01 758E+02 •694E+02 .535E+02 .345E+02 •140E+02 .354E+01 •357E+01 •391E+01 •545E+01 .708E+01 .646E+02 •488E+02 .380E+02 .239E+02 .083E+02 .090E+01 .127E+01 .096E+01 .097E+01 .493E+01 .604E+02 .438E+02 .266E+02 . 126E + 02 .795E+01 .340E+01 .773E+01 .841E+01 .755E+01 .515E+00 .589E+02 .399E+02 .243E+02 .078E+02 .220E+01 .651E+01 .187E+01 .665E+01 .544E+01 .771E+00 .257E+02 .369E+02 .238E+02 .060E+02 .754E+01 .005E+01

3. 3. 3. 2. 1. 9. 3. 6.

-9. -4. 4. 3. 2. 2. 2. 1. 1. 1. 6. 2. 3. 3. 2. 2. 2. 1. 1. 9. 6. 3. 3. 2. 2. 2 1 1 1 8 5 2 2 2 2 1 1 1 1 8 4 1 2 2 1 1 1 1 9 7 4 1 1 1 1 1 1 9

SIGMA Y

793E+02 378E+02 340E+02 457E+02 587E+02 153E+01 218E+01 958E+00 557E+00 419E+00 036E+02 501E+02 996E+02 571E+02 147E+02 768E+02 389E+02 033E+02 610E+01 532E+01 361E+02 198E+02 888E+02 539E+02 161E+02 754E+02 .355E+02 .682E+01 .230E+01 .079E+01 .007E+02 .687E+02 •463E+02 •202E+02 .933E+02 .633E+02 .274E+02 .997E+01 .438E+01 .758E+01 .727E+02 .429E+02 .120E+02 .868E+02 .620E+02 .394E+02 .152E+02 .276E+01 .793E+01 .903E+01 .514E+02 .187E+02 .914E+02 .639E+02 .386E+02 .147E+02 .494E+01 .422E+01 .335E+01 .189E+01 .845E+02 .895E+02 .659E+02 .435E+02 .201E+02 .725E+01

1. -5. 2.

-1. 1.

-6. 1. 5.

-9. 4. 2 . 2. 2. 1. 1. 7. 3 . 1.

-8. -8. 3 . 2 , 2, 2 . 1. 1. 5 . 2

-4 . -1 2 2 2 1 1 1 7 3 2 -1 2 2 1 1 1 1 7 3 2

-1 1 1 1 1 9 6 4 2

-2 -1 1 1 9 7 5 4

TAU XY

214E+02 475E+01 871E+01 014E+01 314E+01 399E-01 227E+00 859E+00 519E+00 012E+00 758E+01 618E+01 288E+01 653E+01 146E+01 418E+00 861E+00 047E+00 979E-01 613E-01 036E+01 748E+01 .409E+01 056E+01 .567E+01 .057E+01 .987E+00 •127E+00 •591E-01 •100E+00 •914E+01 .492E+01 .186E+01 .882E+01 .577E+01 .206E+01 .807E+00 .611E+00 .681E-01 .219E+00 .589E+01 .234E+01 .852E+01 .546E+01 .271E+01 .011E+01 .186E+00 .724E+00 .376E-01 .281E+00 .936E+01 .720E+01 .490E+01 .177E+01 .066E+00 .772E+00 .682E+00 .480E+00 .985E-01 .710E+00 .232E+01 .369E+01 .381E+00 .436E+00 .778E+00 .427E+00

4 . 3. 3 . 2. 1. 9. 3. 1. 7. 3. 4. 3. 3. 2. 2. 1 . 1. 1. 6. 2. 3. 3. 2. 2. 2. 1. 1. 9. 6. 3. 3. 2, 2, 2. 1. 1 1 9 5 2 2 2 2 1 1 1 1 8 4 1 2 2 1 1 1 1 9 7 4 1 1 1 1 1 1 9

SIGMA 1

179E+02 463E+02 365E+02 461E+02 598E+02 153E+01 222E+01 021E+01 066E+00 940E+00 074E+02 540E+02 030E+02 592E+02 159E+02 774E+02 391E+02 033E+02 613E+01 538E+01 416E+02 247E+02 929E+02 573E+02 185E+02 768E+02 361E+02 692E+01 231E+01 087E+01 .067E+02 •737E+02 .505E+02 .237E+02 •961E+02 .653E+02 .285E+02 .030E+01 .439E+01 .770E+01 .784E+02 .477E+02 .159E+02 .899E+02 .644E+02 .412E+02 .162E+02 .316E+01 .794E+01 .920E+01 .553E+02 .223E+02 .946E+02 .663E+02 .403E+02 .159E+02 .559E+01 .444E+01 .335E+01 .228E+01 .870E+02 .929E+02 .679E+02 .449E+02 .211E+02 .795E+01

-2. -1. -1. -1. -6. -4. -2. -3. -1. -6. 2. 1. 1. 1. 1. 8. 7. 5. 3. 1. 1. 1. 1. 1. 1. 9. 7. 5. 3. 1. 1, 1. 1. 1. 1. 8 7 5 3 1 1 1 1 1 9 8 6 4 2 9 1 1 1 1 9 7 6 4 2 4 1 1 1 1 8 6

SIGMA 3

375E+00 540E+01 684E+00 532E+01 828E+00 593E+00 597E+00 597E+00 501E+01 344E+00 024E+02 750E+02 467E+02 253E+02 068E+02 966E+01 221E+01 500E+01 651E+01 406E+01 702E+02 646E+02 494E+02 311E+02 117E+02 220E+01 300E+01 380E+01 544E+01 .699E+01 .586E+02 .438E+02 •338E+02 .204E+02 .055E+02 .894E+01 •021E+01 •063E+01 •097E+01 .481E+01 .547E+02 .390E+02 .227E+02 .095E+02 .552E+01 .163E+01 .667E+01 .801E+01 .755E+01 .345E+00 .551E+02 .363E+02 .211E+02 .054E+02 .049E+01 .534E+01 .122E+01 .643E+01 .544E+01 .382E+00 .232E+02 .336E+02 .218E+02 .046E+02 .654E+01 .935E+01

A65


87 88 89 90 91 92 93 94 95 96 97 98 99 100

26. 20. 14. 8

67. 60 54. 47 41 34 28 22 15 9

,07 ,15 ,22 .30 .30 ,84 .37 .91 .45 .98 .52 .06 .59 .13

13, 15, 17, 19, 1. 3. 5, 7. 9. 11 13. 15 17 19

.00 ,00 .00 ,00 ,00 ,00 ,00 ,00 .00 .00 .00 .00 .00 .00

5. 3 2. -2 -8 -2 -1 -1 -1 -1 -1 -1 -6 -2

.307E+01 •903E+01 .686E+01 •802E+00 •771E+02 .501E+03 .718E+03 .926E+03 .861E+03 .674E+03 .400E+03 .088E+03 .921E+02 .575E+02

7 5 4 3 4 5 -2 -1 -1 -1 -1 -1 -1 -5

•520E+01 •837E+01 •347E+01 •480E+00 •167E+03 •799E+02 .082E+02 .359E+02 .483E+02 .609E+02 .788E+02 .731E+02 .306E+02 .304E+01

3 1

-3 -2 1 7 5 6 5 5 4 3 2 5

.202E+00 •560E+00 •449E-01 •413E+00 .122E+03 .315E+02 •669E+02 .409E+02 .988E+02 .425E+02 •447E+02 .385E+02 .280E+02 .727E+01

7 5 4 4 4 7

-1. 6 4 1

-3 -6 -4 -3

•565E+01 •850E+01 .348E+01 .300E+00 .405E+03 •447E+02 .907E+01 •995E+01 •033E+01 •356E+01 •398E+01 •149E+01 .962E+01 .809E+01

5 3 2 -3 -1 -2 -1. -2 -2 -1 -1 -1 -7. -2

•261E+01 •891E+01 .686E+01 .622E+00 .115E+03 .666E+03 .907E+03 .131E+03 . 050E+03 .848E+03 •545E+03 .200E+03 •731E+02 .724E+02



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0. 0. 0. 0. 0. 0 0 0. 0 0

67 60 54 47 41 34 28 21 15 9

• 20 .20 .20 .20 .20 .20 .20 .20 .20 .20 .20 .74 .27 .81 .35 .88 .42 .96 .49 .03

1. 3 . 5. 7 9

11. 13 15 17 19 1 3 5 7. 9 11 13 15 17 19

.00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00

2. 2 4 3 1. 8 2 2 0 2 9 1 1 1 1 9 7 5 3 8

.206E+01 •499E-02 •462E-02 •277E-02 .925E-02 •132E-03 •021E-03 .703E-01 .OOOE+00 .854E+00 .059E+01 .754E+02 .550E+02 .306E+02 .108E+02 .037E+01 .100E+01 .222E+01 .343E+01 .992E+00

-5. 5. 4. 3. 2. 2. 2.

-4.

o. 9 1 4

-2 -3 -3 -3 -3 -3 -4 -4

975E+00 •592E-03 .583E-03 .673E-03 .978E-03 .613E-03 •238E-03 •069E+00 .000E+00 .557E-01 .221E+00 .422E-01 •306E+00 •097E+00 •387E+00 .485E+00 •199E+00 .201E+00 .059E+00 .387E+00

1. 1. 1. 1. 1. 1. 1. 1. 1. 1 1 1 1 1 1 1 1 1 1 1

000E+08 OOOE+08 .OOOE+08 .OOOE+08 OOOE+08 •OOOE+08 .OOOE+08 •OOOE+08 .000E+02 .OOOE+08 •OOOE+08 .OOOE+08 •OOOE+08 •OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08

4 4 4 4 4 4 1 1 1 4 4 4 4 4 4 4 4 4 4 4

•OOOE+03 .OOOE+03 •000E+03 .OOOE+03 •000E+03 .OOOE+03 .000E+02 .000E+02 .OOOE+02 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03


REIN. NUM. I J TYPE COMPR FORCE INCR COMPR STIFFNESS

1 2 3 4 5 6 7 8 9

10

1 2 3 4 5 6 7 8 9

10

111 112 113 114 115 116 117 118 119 120

1 1 1 1 1 1 1 1 1 1

0 1.

-1 3. 1 1 5 1 8

-7.

.000000E+00

.635870E+01

.637198E+01

.167185E+01

.305019E+01

.285954E+01

.331912E+00

.083230E+00

.405101E-01

.811441E+00

0 -9 2

-1. 9 -7 5

-3 3 -1

.000000E+00

.748852E-06

.138899E-06

.563400E-06

.791111E-07

.710914E-07

.118782E-07

.621681E-07

.027017E-07

.097869E-07

6. 6. 6. 6 6 6 6 6 6 6

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

.750000E+05

A66


*************************************************** STAGE NUMBER 4



1 23 45 67 89 111

1. 1. 1. 1. 1. 1.

•0OOOOE-•00000E-•00000E-•00000E-•00000E-.00000E-

-01 -01 -01 -01 •01 -01

0. 0. 0 0. 0 0.

•OOOOOE+00 •00000E+00 .O00O0E+O0 .00000E+00 .00000E+00 •OOOOOE+00

12 34 56 78 100

0

1 1. 1 1. 1 0.

.00000E-01

.00000E-01

.00000E-01 •00000E-01 .00000E-01 .00000E+00

0 0 0 0 0 0,

•00000E+00 •00000E+00 •OOOOOE+00 •OOOOOE+00 .00000E+00 •00000E+00


NODAL POINT

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

X

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20

11.91 10.83 9.75 8.68 7.60 6.52 5.44 4.37 3.29 2.21 1.13

23.61 21.46 19.30 17.15 14.99 12.84

Y

0. 2. 4 6. 8

10, 12 14. 16 18 20 0 2 4 6 8

10 12 14 16. 18 20 0 2 4 6 8

10 12. 14 16 18 20. 0. 2. 4, 6. 8.

10. 12. 14, 16. 18. 20. 0. 2. 4. 6. 8.

10.

.00 ,00 .00 ,00 .00 ,00 .00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 ,00 .00 ,00 .00 ,00 ,00 ,00 ,00 ,00 ,00 ,00 ,00 00 ,00 00 00 00 00

TOTAL

ux 1.00000E-01 9.07302E-02 7.81036E-02 6.99333E-02 6.25979E-02 5.68958E-02 5.29159E-02 5.07664E-02 5.02681E-02 5.12356E-02 5.11215E-02 1.00000E-01 9.07393E-02 7.81087E-02 6.99409E-02 6.26020E-02 5.68987E-02 5.29171E-02 5.07667E-02 5.02681E-02 5.12350E-02 5.11216E-02 1.00000E-01 9.07390E-02 8.04188E-02 6.95487E-02 6.23394E-02 5.67766E-02 5.28766E-02 5.07668E-02 5.02681E-02 5.12349E-02 5.11216E-02 1.00000E-01 8.99889E-02 7.99503E-02 7.10626E-02 6.24508E-02 5.60168E-02 5.16156E-02 4.92572E-02 4.89513E-02 5.02114E-02 5.14681E-02 1.00000E-01 8.75187E-02 7.79215E-02 6.91459E-02 6.16811E-02 5.49549E-02

TOTAL UY

0.OOOOOE+00 -1.18993E-03 -9.18942E-04 -1.24981E-03 -1.29476E-03 -1.40099E-03 -1.32273E-03 -1.31741E-03 -1.19000E-03 -1.17508E-03 -1.28636E-03 0.00000E+00 6.17085E-04

-1.79756E-04 -3.65719E-04 -7.07965E-04 -8.51873E-04 -1.07585E-03 -1.13445E-03 -1.27581E-03 -1.27872E-03 -1.16176E-03 0.00000E+00

-2.36892E-03 -4.00997E-03 -4.70419E-03 -4.86381E-03 -4.44613E-03 -3.62526E-03 -2.73185E-03 -1.71259E-03 -1.02000E-03 -1.05233E-03 0.00000E+00 1.85673E-05

-1.73438E-04 -9.31039E-04 -1.87593E-03 -2.57767E-03 -2.90473E-03 -2.68546E-03 -2.06130E-03 -1.01557E-03 -3.86103E-04 0.OOOOOE+00

-2.76985E-03 -4.24341E-03 -4.74190E-03 -4.84963E-03 -4.86557E-03

PORE PRESS 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 o.oo 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

A67


51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121

10.68 8.53 6.38 4.22 2.07

35.32 32.09 28.86 25.62 22.39 19.16 15.93 12.70 9.46 6.23 3.00

47.02 42.71 38.40 34.09 29.79 25.48 21.17 16.86 12.55 8.24 3.93 58.73 53.34 47.96 42.57 37.18 31.80 26.41 21.03 15.64 10.25 4.87 70.43 63.97 57.50 51.04 44.58 38.11 31.65 25.19 18.73 12.26 5.80

70.43 63.97 57.50 51.04 44.58 38.11 31.65 25.19 18.73 12.26 5.80

70.63 64.17 57.70 51.24 44.78 38.31 31.85 25.39 18.93 12.46 6.00

12.00 14.00 16.00 18.00 20.00 0 .00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00

5.01235E-02 4.75069E-02 4.72379E-02 4.91217E-02 5.15530E-02 1.00000E-01 8.71994E-02 7.55594E-02 6.67868E-02 5.96053E-02 5.39565E-02 4.92608E-02 4.64668E-02 4.61870E-02 4.84186E-02 5.14975E-02 1.00000E-01 8.68657E-02 7.55828E-02 6.55567E-02 5.82047E-02 5.28439E-02 4.91871E-02 4.65741E-02 4.60669E-02 4.85297E-02 5.14833E-02 1.00000E-01 9.04917E-02 7.79473E-02 6.7062OE-02 5.85198E-02 5.29576E-02 4.95947E-02 4.79039E-02 4.71816E-02 4.92753E-02 5.18608E-02 1.00000E-01 9.09838E-02 7.76183E-02 6.89717E-02 6.16478E-02 5.58468E-02 5.20025E-02 4.99615E-02 4.94093E-02 5.00919E-02 5.27413E-02 1.00000E-01 9.04120E-02 7.80181E-02 6.97451E-02 6.24321E-02 5.67430E-02 5.27929E-02 5.06762E-02 5.01977E-02 5.12136E-02 5.36933E-02 1.00000E-01 9.07058E-02 7.81291E-02 6.98888E-02 6.25787E-02 5.68778E-02 5.29112E-02 5.07676E-02 5.02677E-02 5.12471E-02 5.36999E-02

-4.50502E-03 -3.79170E-03 -2.61203E-03 -1.24678E-03 -2.64942E-04 0.O00O0E+O0

-4.82784E-03 -8.49295E-03 -1.06162E-02 -1.12404E-02 -1.07522E-02 -9.65906E-03 -7.66991E-03 -5.12343E-03 -2.31428E-03 -2.20945E-04 0.00000E+00 -7.57676E-03 -1.33525E-02 -1.72744E-02 -1.90520E-02 -1.87697E-02 -1.68383E-02 -1.39208E-02 -9.50649E-03 -4.60827E-03 -4.57878E-04 0.00000E+00

. -1.18765E-02 -2.04895E-02 -2.59725E-02 -2.87127E-02 -2.87074E-02 -2 .61335E-02 -2.15537E-02 -1.56950E-02 -7.89169E-03 -1.17720E-03 0.00000E+00 -1.20472E-02 -2.73009E-02 -3.71541E-02 -4.18358E-02 -4.19748E-02 -3.83296E-02 -3.15938E-02 -2.26665E-02 -1.25058E-02 -2.12310E-03 0.00000E+00

-1.18721E-02 -2.74264E-02 -3.73949E-02 -4.20797E-02 -4.22532E-02 -3.85750E-02 -3.18156E-02 -2.29109E-02 -1.28532E-02 -2.41761E-03 0.OOOOOE+00 -1.13617E-02 -2.70279E-02 -3.71813E-02 -4.20153E-02 -4.23179E-02 -3.87469E-02 -3.20678E-02 -2.32129E-02 -1.31752E-02 -2.74217E-03

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 o.oo 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0. 00 0 . 00 0 .00 0 .00 0 . 00 0. 00 0.00 0. 00 0 .00 0.00 0.00 0. 00 0 .00 0 . 00 0.00 0. 00 0.00

A68



ELEM

NO

21

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

0

0 0 0 0 0 0 0 0 0 5 5 4 4 3 3 2 2 1 0

16 15 13 12 10 8 7 5 4 2

28 25 22 20 17 14 11. 9. 6. 3,

39. 35. 31. 27. 24. 20. 16. 12. 9. 5.

50. 45. 40. 35. 31. 26. 21. 16. 11. 6.

61. 55. 49. 43. 37. 31.

X

.10

.10

.10

.10

.10

.10

.10

.10

.10

.10

.79

.25

.71

.17

.63

.09

.55

.01

.47

.94

.95

.34

.72

.10

.49

.87

.26

.64

.02

.41

.12

.43

.73

.04

.35

.65

.96

.27

.57

.88 ,28 .51 .74 .97 20 43 66 89 12 35 45 60 76 91 06 21 37 52 67 82 62 69 77 84 92 99

1 3 5 7 9

11 13 15 17 19 1 3 5 7 9

11 13 15 17 19 1 3 5 7 9

11 13 15 17 19 1 3 5 7 9

11 13. 15. 17. 19, 1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11.

Y

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00 ,00 ,00 ,00 ,00 ,00 .00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

3 -6 1

-1 -5 -3 -1 1 1 2 2 1 1 1 1 9 7 5 3 1 1 1 1 1 1 9 7 5 3 1 1 1 1 1 1 9. 7. 5. 3. 1. 1. 1. 1. 1. 9. 8. 6. 4. 2 . 9. 1. 1. 1. 1. 9. 7. 6. 4. 2. 4. 1. 1. 1. 1. 8. 6.

SIGMA

X

.635E+01

.936E+00 •582E+00 .454E+01 .736E+00 .396E+00 .291E+00 .773E-01 .758E+00 .040E+00 .063E+02 .790E+02 .502E+02 .276E+02 .082E+02 .043E+01 .251E+01 .512E+01 .668E+01 .376E+01 .758E+02 .695E+02 .536E+02 .346E+02 .141E+02 .361E+01 .363E+01 .392E+01 .538E+01 .679E+01 .646E+02 .488E+02 .380E+02 .239E+02 .083E+02 .091E+01 .128E+01 .095E+01 ,090E+01 .478E+01 •604E+02 438E+02 266E+02 126E+02 791E+01 336E+01 770E+01 839E+01 753E+01 532E+00 590E+02 400E+02 243E+02 078E+02 213E+01 643E+01 180E+01 659E+01 545E+01 821E+00 258E+02 370E+02 239E+02 060E+02 745E+01 997E+01

3 3 3 2 1 9 3 9

-7 -3 4 3 2 2 2 1 1 1 6 2 3 3 2 2 2 1 1 9 6 3 3 2 2 2 1 1 1. 9. 5. 2. 2. 2. 2. 1. 1. 1. 1. 8. 4. 1. 2. 2. 1. 1. 1. 1. 9. 7. 4. 1. 1. 1. 1. 1. 1. 9.

SIGMA Y

.820E+02

.406E+02

.369E+02 •484E+02 .614E+02 .400E+01 .437E+01 .113E+00 .383E+00 .176E+00 .037E+02 .502E+02 .997E+02 .572E+02 .148E+02 .769E+02 .390E+02 .034E+02 .611E+01 .527E+01 .361E+02 .198E+02 .888E+02 .539E+02 .162E+02 .755E+02 .356E+02 .683E+01 .226E+01 .073E+01 .007E+02 .687E+02 .463E+02 .202E+02 .933E+02 •633E+02 •275E+02 .004E+01 •443E+01 •759E+01 .728E+02 .430E+02 .121E+02 868E+02 620E+02 394E+02 152E+02 281E+01 802E+01 911E+01 517E+02 189E+02 915E+02 639E+02 387E+02 148E+02 493E+01 423E+01 341E+01 198E+01 846E+02 898E+02 661E+02 437E+02 202E+02 730E+01

1 -5 2 -8 1 4 3 6

-1 4 2 2 2 1 1 7 3 1

-8 -6 3 2 2 2 1 1 6 2 -2 -9 2 2 2 1 1 1 7, 3 3.

-1, 2, 2. 1. 1. 1. 1. 7. 3. 3 .

-1. 1. 1. 1. 1. 9. 6. 4. 2.

-2. -1. 1. 1. 9. 7. 5. 4.

TAU XY

.199E+02

.338E+01

.744E+01

.992E+00

.206E+01

.208E-01

.729E-01

.599E+00

.040E+01

.642E+00

.753E+01

.614E+01

.284E+01

.652E+01

.148E+01

.473E+00

.907E+00

.055E+00

.202E-01

.257E-01

.034E+01

.746E+01

.409E+01

.056E+01

.570E+01

.062E+01

.077E+00

.246E+00

.896E-01

.092E-01 •913E+01 •492E+01 •185E+01 •883E+01 •579E+01 .210E+01 •872E+00 •710E+00 •913E-01 •090E+00 •586E+01 •231E+01 •850E+01 545E+01 273E+01 014E+01 234E+00 796E+00 271E-01 227E+00 927E+01 713E+01 483E+01 174E+01 053E+00 793E+00 729E+00 548E+00 187E-01 653E+00 220E+01 358E+01 281E+00 332E+00 738E+00 410E+00

4 3 3 2 1 9 3 1 8 4 4 3 3 2 2 1 1 1 6 2 3 3 2 2 2 1 1 9 6 3 3. 2 2. 2. 1. 1. 1. 9, 5, 2, 2, 2. 2. 1. 1. 1. 1. 8. 4. 1. 2. 2. 1. 1. 1. 1. 9. 7. 4. 1. 1. 1. 1. 1. 1. 9.

SIGMA 1

.196E+02

.487E+02

.391E+02

.487E+02

.622E+02

.400E+01

.438E+01

.261E+01

.547E+00

.757E+00

.075E+02

.541E+02

.031E+02

.593E+02

.160E+02

.775E+02

.392E+02

.034E+02

.614E+01

.530E+01

.416E+02

.247E+02

.929E+02

.573E+02

.185E+02

.768E+02 •362E+02 •695E+01 •227E+01 •079E+01 •067E+02 •737E+02 •505E+02 •237E+02 •961E+02 .653E+02 .285E+02 039E+01 444E+01 768E+01 .785E+02 477E+02 159E+02 899E+02 644E+02 412E+02 162E+02 322E+01 803E+01 926E+01 555E+02 224E+02 947E+02 663E+02 404E+02 159E+02 559E+01 446E+01 342E+01 235E+01 871E+02 930E+02 681E+02 451E+02 212E+02 800E+01

-1 -1 -6 -1 -6 -3 -1 -3 -1 -5 2 1 1 1 1 8 7 5 3 1 1 1 1 1 1 9 7 5 3 1 1 1 1. 1. 1. 8. 7. 5. 3. 1, 1. 1, 1, 1, 9. 8, 6. 4. 2. 9, 1. 1. 1. 1. 9. 7. 6. 4. 2. 4. 1. 1. 1. 1, 8, 6.

SIGMA 3

.162E+00

.495E+01

.481E-01

.485E+01

. 602E + 00

.398E+00

.295E+00

.324E+00

.417E+01

.892E+00

.025E+02

.751E+02

.468E+02

.255E+02

.069E+02

.979E+01

.228E+01

.510E+01

.665E+01

.372E+01

.702E+02

.646E+02

.494E+02

.311E+02

.117E+02 •225E+01 •304E+01 •381E+01 •538E+01 •674E+01 •586E+02 •438E+02 •338E+02 •204E+02 •055E+02 •894E+01 •020E+01 •060E+01 •089E+01 •469E+01 •548E+02 •390E+02 •227E+02 •095E+02 •548E+01 •159E+01 •663E+01 •798E+01 •752E+01 •378E+00 •552E+02 364E+02 212E+02 054E+02 043E+01 526E+01 114E+01 636E+01 545E+01 458E+00 234E+02 337E+02 219E+02 046E+02 647E+01 927E+01

A69


87 88 89 90 91 92 93 94 95 96 97 98 99

100

26. 20. 14. 8.

67. 60, 54 47 41 34 28 22 15 9

07 .15 ,22 ,30 .30 .84 .37 .91 .45 .98 .52 .06 .59 .13

13. 15. 17. 19. 1. 3. 5 7. 9. 11 13 15 17 19

00 .00 .00 .00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00

5. 3, 2,

-2. -8 -2 -1 -1 -1 -1 -1 -1 -7 -2

.297E+01

.896E+01

.686E+01

.765E+00

.698E+02 •490E+03 .710E+03 •923E+03 .869E+03 .689E+03 .410E+03 .095E+03 .070E+02 .736E+02

7. 5. 4 3. 4 5 -2 -1 -1 -1 -1 -1 -1 -5

•523E+01 •835E+01 •352E+01 •568E+00 •170E+03 .811E+02 .062E+02 .341E+02 .513E+02 .639E+02 .781E+02 .763E+02 .323E+02 .227E+01

3. 1

-2 -2 1 7 5 6 6 5 4 3 2 6

.231E+00

.615E+00 •622E-01 •338E+00 •116E+03 .301E+02 .629E+02 .415E+02 .000E+02 .480E+02 .487E+02 .392E+02 .343E+02 .237E+01

7 5 4 4 4 7 -1 7 3 1

-3 -6 -4 -3

•569E+01 •849E+01 •352E+01 •338E+00 .406E+03 .459E+02 .887E+01 .213E+01 .753E+01 .264E+01 .204E+01 .470E+01 .894E+01 .591E+01

5. 3. 2

-3 -1 -2 -1 -2 -2 -1 -1 -1 -7 -2

•252E+01 .883E+01 .686E+01 .534E+00 .106E+03 .654E+03 .897E+03 .129E+03 .058E+03 .865E+03 .556E+03 .207E+03 .904E+02 .900E+02



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0, 0. 0. 0, 0 0. 0 0. 0. 0

67 60 54 47 41 34 28 21 15 9

.20

.20

.20 ,20 .20 .20 .20 .20 .20 .20 .20 .74 .27 .81 .35 .88 .42 .96 .49 .03

1. 3. 5, 7, 9

11. 13. 15 17. 19 1 3 5 7 9 11 13 15 17 19

.00

.00 ,00 .00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00

2 . 2. 0, 0, 3. 1. 2, 1. 3. 2 9 1 1 1 1 9 7 5 3 9

169E+01 499E-02 .000E+00 .0O0E+00 •918E-01 .498E+00 .237E+00 •015E+00 •725E-07 •481E+00 •040E+01 •757E+02 •554E+02 •305E+02 .108E+02 •037E+01 •100E+01 .204E+01 •315E+01 •085E+00

-5, 7. 0. 0.

-1. -2. 9 -4 -2 7 1 3 -2 -3 -3 -3 -3 -3 -3 -4

.972E+00

.556E-03 •000E+00 •OOOE+00 •501E-02 •998E-02 •060E-04 •071E+00 •611E-03 .363E-01 •196E+00 •579E-01 •458E+00 •263E+00 •520E+00 .532E+00 •152E+00 .147E+00 .999E+00 .342E+00

1. 1. 1. 1. 1. 1 , 1. 1, 1. 1. 1. 1. 1 1 1 1 1 1 1 1

OOOE+08 OOOE+08 000E+02 000E+02 .OOOE+08 OOOE+08 OOOE+08 OOOE+08 OOOE+08 .000E+08 •000E+08 •000E+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08

4 • 4. 1. 1, 4 4 4. 1 1. 4 4 4 4 4 4 4 4 4 4 4

OOOE+03 OOOE+03 000E+02 000E+02 •OOOE+03 •000E+03 •000E+03 .000E+02 •000E+02 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .000E+03 .000E+03 .OOOE+03 .OOOE+03


NUM.

1 2 3 4 5 6 7 8 9

10

I

1 2 3 4 5 6 7 8 9

10

J

111 112 113 114 115 116 117 118 119 120

TYPE

1 1 1 1 1 1 1 1 1 1

COMPR FORCE

O.OOOOOOE+00 1.646934E+01 -1.719676E+01 3.000217E+01 1.297475E+01 1.215547E+01 3.149264E+00 -8.278439E-01 2.521005E-01 -7.751091E+00

0, 1. -1 -2. -1 -1 -3 -2 -8 8

INCR COMPR

•O0OOOOE+00 .639128E-07 •221895E-06 .473593E-06 .117587E-07 .043081E-06 .233552E-06 .831221E-06 .717179E-07 .940697E-08

6. 6, 6. 6. 6 6 6 6 6 6

STIFFNESS

•750000E+05 •750000E+05 .750000E+05 .750000E+05 .750000E+05 .750000E+05 .750000E+05 .750000E+05 .750000E+05 .750000E+05

A70

RUNNING THE RSDAM PROGRAM

:***************,*******«****»*,*,*********** STAGE NUMBER

SEEPAGE LOADING IS SPECIFIED FOR THIS INCREMENT

NUMBER OF WATER LEVEL CHANGES SPECIFIED

X-COORD OF BOUNDARY

PRESENT LEVEL

NEW LEVEL

0.00 6.00

2.00 20.00

O.OO 0.00


NODAL POINT

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

X

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20

11.91 10.83 9.75 8.68 7.60 6.52 5.44 4.37 3.29 2.21 1.13

23.61 21.46 19.30 17.15 14.99 12.84

y

0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00

10.00

TOTAL UX

1.00000E-01 9.18048E-02 8.12350E-02 7.55360E-02 7.13068E-02 6.85893E-02 6.67098E-02 6.54552E-02 6.49130E-02 6.56933E-02 6.68319E-02 1.00000E-01 9.18134E-02 8.12406E-02 7.55434E-02 7.13113E-02 6.85931E-02 6.67121E-02 6.54564E-02 6.49136E-02 6.56939E-02 6.68319E-02 1.00000E-01 9.18131E-02 8.35508E-02 7.62569E-02 7.10488E-02 6.84710E-02 6.66717E-02 6.54565E-02 6.49135E-02 6.56938E-02 6.68320E-02 1.00000E-01 9.32471E-02 8.60010E-02 7.91981E-02 7.30601E-02 6.82883E-02 6.52204E-02 6.35424E-02 6.35009E-02 6.49777E-02 6.68722E-02 1.00000E-01 9.44769E-02 8.83178E-02 8.10120E-02 7.41486E-02 6.78647E-02

TOTAL UY

0.OOOOOE+00 -1.07595E-03 -8.07301E-04 -1.08613E-03 -1.11719E-03 -1.25857E-03 -1.28305E-03 -1.36982E-03 -1.31898E-03 -1.30157E-03 -1.31771E-03 0.OOOOOE+00 5.17785E-04

-2.65887E-04 -4.99654E-04 -8.52247E-04 -9.85142E-04 -1.17807E-03 -1.22645E-03 -1.35443E-03 -1.42151E-03 -1.42605E-03 0.00000E+00

-2.47771E-03 -3.78615E-03 -4.14875E-03 -4.19436E-03 -3.65968E-03 -3.44588E-03 -3.83093E-03 -3.36798E-03 -2.60641E-03 -1.82659E-03 0.00000E+00 -1.20963E-03 -2.14154E-03 -3.28552E-03 -4.36388E-03 -5.08759E-03 -6.10823E-03 -6.42548E-03 -5.07781E-03 -3.53701E-03 -2.55833E-03 0.00000E+00

-4.20731E-03 -6.58483E-03 -7.48384E-03 -7.68276E-03 -7.58165E-03

PORE PRESS

-2.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

-2.60 -0.60 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

-2.60 -0.60 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

-2.60 -0.60 0.00 0.00 0.00 0.00

-6.33 -1.10 0.00 0.00 0.00

-5.40 -3.40 -1.40 0.00 0.00 0.00

A71


51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121

10, 8. 6, 4. 2

35, 32 28. 25, 22 19, 15. 12. 9. 6 3. 47 42 38 34 29 25 21 16 12 8 3 58 53 47 42 37 31 26 21 15 10 4

70 63 57 51 44 38 31 25 18 12 5

70 63 57 51. 44 38. 31 25. 18. 12, 5.

70, 64. 57. 51, 44, 38, 31, 25. 18, 12. 6,

.68 ,53 ,38 .22 .07 ,32 .09 .86 .62 .39 .16 .93 .70 .46 .23 .00 .02 .71 .40 .09 .79 .48 .17 .86 .55 .24 .93 .73 .34 .96 .57 .18 .80 .41 .03 .64 .25 .87 .43 .97 .50 .04 .58 .11 .65 .19 .73 .26 .80 .43 .97 .50 .04 .58 ,11 .65 .19 ,73 ,26 ,80 ,63 ,17 .70 ,24 ,78 ,31 .85 ,39 ,93 ,46 ,00

12. 14, 16, 18. 20. 0. 2. 4. 6. 8.

10. 12 14. 16. 18 20. 0 2 4. 6 8

10 12 14 16 18 20 0 2 4 6 8

10 12 14 16 18 20 0 2 4 6 8 10 12. 14 16 18 20 0. 2 4 6. 8

10, 12 14 16, 18, 20, 0. 2. 4. 6. 8.

10. 12. 14. 16. 18. 20,

.00 ,00 ,00 ,00 .00 ,00 ,00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 ,00 .00 .00 ,00 ,00 ,00 .00 ,00 ,00 .00 ,00 ,00 ,00 .00 ,00 ,00 .00

6. 6 6. 6 6 1. 9 8. 8. 7 6. 6 6 6. 6 6 1 9 8 7 7 6 6 6 6 6 6 1 9 8 7 7 6 6 6 6 6 6 1 9 8 7 7 6 6 6 6 6 6 1. 9 8 7 7 6 6 6 6. 6 6 1 9. 8. 7. 7. 6. 6. 6. 6. 6. 6

.32008E-02

.11789E-02

.15591E-02

.39281E-02

.68473E-02

.00000E-01

.40924E-02

.67067E-02

.01941E-02

.39547E-02

.84105E-02

.36379E-02

.05436E-02

.02732E-02

.30577E-02 •67719E-02 .00000E-01 .06950E-02 .39196E-02 .74659E-02 .23426E-02 .79900E-02 .43003E-02 .13024E-02 .02685E-02 .28223E-02 .67176E-02 .00000E-01 .27749E-02 .29353E-02 .55651E-02 .04660E-02 .73166E-02 .49633E-02 .30257E-02 .17194E-02 .33985E-02 .68180E-02 .00000E-01 .17702E-02 .01520E-02 .41461E-02 .03833E-02 .80989E-02 .65959E-02 .51982E-02 .41095E-02 .43814E-02 .71369E-02 .00000E-01 .15420E-02 .11914E-02 .54173E-02 •12121E-02 .84923E-02 .66270E-02 .53769E-02 •48494E-02 .56618E-02 •79471E-02 .00000E-01 •17823E-02 .12486E-02 ,54891E-02 .12791E-02 .85623E-02 ,67052E-02 54558E-02 •49162E-02 .56908E-02 •79513E-02

-7 -7 -6. -4 -2 0 -5 -9 -1. -1. -1. -1 -1. -8. -6 -3. 0

-8. -1. -1 -1. -1. -1 -1 -1 -8 -3 0

-1 -2 -2 -2 -2 -2 -2 -1 -1 -4 0

-1 -2 -3 -4 -4 -3 -3 -2 -1 -5 0

-1 -2 -3 -4 -4 -3 -3 -2 -1 -6 0 -1 -2 -3 -4 -4 -3 _3 -2 -1 -6

.58374E-03

.58667E-03

.46377E-03

.52293E-03

.78183E-03

.OOOOOE+00

.76020E-03 •99426E-03 •24827E-02 •33147E-02 •30489E-02 •21143E-02 •05080E-02 .66246E-03 •06968E-03 •18198E-03 .OOOOOE+00 .20939E-03 •41376E-02 •81046E-02 •99713E-02 •98823E-02 •84056E-02 •59691E-02 .23020E-02 .20758E-03 .88243E-03 .OOOOOE+00 •23045E-02 •11158E-02 •65622E-02 .90729E-02 .90176E-02 .66936E-02 •27845E-02 .77785E-02 .10669E-02 .79952E-03 .00000E+00 .21430E-02 .75558E-02 .74787E-02 .20374E-02 .20004E-02 .83609E-02 .21034E-02 .41502E-02 .51544E-02 .88470E-03 .OOOOOE+00 .20742E-02 .78792E-02 .78735E-02 .22951E-02 .21232E-02 .83714E-02 .21593E-02 .43796E-02 .55508E-02 .13539E-03 .00000E+00 .15520E-02 .74729E-02 .76603E-02 .22377E-02 .21933E-02 .85353E-02 .23838E-02 .46424E-02 .58374E-02 .42996E-03

0. 0. 0. 0. 0,

-8. -6, -4, -2. -0. 0. 0. 0. 0 , 0. 0,

-11. -9, -7. -5. -3. -1. 0, 0. 0. 0. 0,

-13, -11. -9. -7, -5. -3. -1, 0, 0. 0. 0

-16. -14. -12 -10. -8 -6 -4 -2 -0 0 0

-19 -17 -15 -13 -11 -9 -7 -5 -3 -1 0

-19 -17 -15 -13 -11 -9 -7 -5 -3 -1 0

00 .00 00 .00 .00 20 .20 .20 .20 .20 00 .00 00 00 .00 00 ,00 .00 00 .00 00 00 .00 00 .00 .00 00 ,80 ,80 ,80 ,80 80 ,80 .80 ,00 ,00 ,00 .00 ,60 .60 .60 .60 .60 .60 .60 .60 .60 .00 .00 .40 .40 .40 .40 .40 .40 .40 .40 .40 .40 .00 .40 .40 .40 .40 .40 .40 .40 .40 .40 .40 .00

A72



SIGMA SIGMA TAU SIGMA SIGMA X Y XY 1 3

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

0 0 0 0 0 0 0 0 0 0 5 5 4 4 3 3 2 2 1 0

16 15 13 12 10 8 7 5 4 2 28 25 22 20 17 14 11 9 6 3.

39. 35, 31, 27, 24. 20. 16. 12. 9. 5.

50. 45. 40. 35. 31. 26. 21. 16. 11. 6.

61. 55. 49. 43. 37. 31.

.10

.10

.10

.10

.10

.10

.10

.10

.10

.10

.79

.25

.71

.17

.63

.09

.55

.01

.47

.94

.95

.34

.72

.10

.49

.87

.26

.64

.02

.41

.12

.43

.73

.04

.35

.65

.96

.27

.57

.88 ,28 ,51 ,74 ,97 20 43 66 89 12 35 45 60 76 91 06 21 37 52 67 82 62 69 77 84 92 99

1 3 5 7 9

11 13 15 17 19 1 3 5 7 9

11 13 15 17 19 1 3 5 7 9

11 13 15 17 19 1 3 5 7. 9

11. 13. 15. 17. 19. 1, 3. 5. 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 1. 3. 5. 7. 9.

11.

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00

.00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

3 -9 -8 -1 -9 -4 -8 7 3 -9 2 1 1 1 1 9 8 5 3 1 1 1 1 1 1 9 8 5 3 1 1 1 1 1 9 8. 6. 4. 3. 1. 1. 1. 1. 1. 9. 7. 6. 4. 2. 1. 1. 1. 1. 1. 1. 7. 5. 3. 2. 5. 1. 1. 1. 1. 1. 7.

•678E+01 .314E+00 •181E-01 .614E+01 .962E+00 .140E+00 .637E-01 .760E-01 .721E-01 .046E-01 .112E+02 .735E+02 .443E+02 .234E+02 .049E+02 .951E+01 .634E+01 .919E+01 .504E+01 .269E+01 .862E+02 .658E+02 .433E+02 .249E+02 .083E+02 .862E+01 .330E+01 .720E+01 .644E+01 .772E+01 .779E+02 .537E+02 .344E+02 •143E+02 .724E+01 .223E+01 .565E+01 .942E+01 ,278E+01 .670E+01 .734E+02 580E+02 357E+02 145E+02 310E+01 542E+01 020E+01 402E+01 778E+01 123E+01 698E+02 567E+02 443E+02 255E+02 008E+02 605E+01 658E+01 996E+01 358E+01 799E+00 320E+02 517E+02 446E+02 294E+02 056E+02 985E+01

3 3 3 2 1 1 8 5 3 1 4 3 2 2 2 1 1 1 6 2 3 3 2 2 2 1 1 9 6 3 3 2 2 2 1 1. 1. 8. 5. 2. 2. 2. 2. 1. 1. 1. 1. 7. 4. 2. 2. 2. 1. 1. 1. 1. 8. 6. 4. 1. 1. 1. 1. 1. 1. 9.

.726E+02

.330E+02 •335E+02 .457E+02 .760E+02 .405E+02 .779E+01 .039E+01 .243E+01 .325E+01 .142E+02 .493E+02 .969E+02 .558E+02 .139E+02 •912E+02 .607E+02 .081E+02 .488E+01 .350E+01 .557E+02 .256E+02 .864E+02 .490E+02 .125E+02 .845E+02 .487E+02 .616E+01 .222E+01 .166E+01 .206E+02 .782E+02 .462E+02 .135E+02 .852E+02 ,591E+02 ,261E+02 .806E+01 .406E+01 .926E+01 913E+02 520E+02 126E+02 801E+02 518E+02 299E+02 083E+02 987E+01 635E+01 019E+01 686E+02 283E+02 942E+02 603E+02 308E+02 054E+02 607E+01 733E+01 109E+01 194E+01 916E+02 951E+02 691E+02 443E+02 176E+02 363E+01

1 -5 3

-1 1

-6 6 1

-1 1 2 1 1 1 8 5 5 5 3 8 1 1 1 1 1 7 3 2 1

-3 8 1 1 1 1 8 3

-2. -6. -5. -3. 1. 3. 4. 5. 5. 3. 8.

-1. -1. -5. -5. -3. -1. 6. 1. 1. 2.

-1. -1. 2. 9.

-3. -3. -1. 1.

.158E+02

.438E+01

.051E+01

.377E+01

.357E+01

.178E+00

.962E+00

.491E+00

.746E+00

.584E+00

.084E+01

.976E+01

.715E+01

.304E+01

.303E+00

.988E+00

.742E+00

.271E+00

.092E+00

.482E-01

.573E+01

.704E+01

.702E+01

.512E+01

.223E+01

.916E+00

.986E+00

.119E+00

.457E+00

.176E-01

.818E+00

.000E+01

.070E+01

.119E+01

.063E+01

.251E+00

.372E+00

.987E-02

.962E-01

.989E-01

.088E+00

.099E+00 583E+00 .758E+00 166E+00 054E+00 990E+00 930E-01 637E+00 300E+00 231E+00 490E+00 237E+00 169E+00 314E-01 605E+00 455E+00 733E-01 700E+00 942E+00 169E+00 481E-01 967E+00 625E+00 292E+00 498E+00

4 3 3 2 1 1 8 5 3 1 4 3 2 2 2 1 1 1 6 2 3 3 2 2 2 1 1 9 6 3. 3 2. 2. 2 1, 1. 1, 8, 5, 2. 2, 2. 2, 1. 1. 1. 1. 7. 4. 2. 2. 2. 1. 1. 1. 1. 8. 6. 4. 1. 1. 1. 1. 1. 1. 9.

.086E+02

.414E+02

.363E+02

.464E+02

.770E+02

.407E+02

.833E+01

.043E+01

.253E+01

.342E+01

.163E+02

.515E+02

.988E+02

.571E+02

.145E+02

.916E+02

.611E+02

.086E+02

.519E+01

.357E+01

.572E+02

.274E+02

.884E+02

.508E+02

.139E+02

.852E+02

.489E+02

.627E+01

.230E+01

.167E+01

.211E+02

.790E+02 ,472E+02 ,148E+02 .864E+02 .599E+02 .263E+02 .806E+01 409E+01 929E+01 913E+02 520E+02 127E+02 804E+02 523E+02 303E+02 086E+02 989E+01 650E+01 037E+01 689E+02 287E+02 944E+02 603E+02 308E+02 055E+02 614E+01 733E+01 126E+01 251E+01 916E+02 952E+02 698E+02 451E+02 178E+02 379E+01

7 -1 -3 -1 -1 -4 -1 7 2 -1 2 1 1 1 1 9 8 5 3 1 1 1 1 1 1 9 8 5 3, 1, 1, 1. 1, 1, 9, 8, 6, 4, 3 . 1. 1. 1, 1. 1. 9. 7. 5. 4. 2. 1. 1. 1. 1. 1. 1. 7. 5. 3 . 2 . 5. 1. 1. 1. 1. 1. 7.

.444E-01

.774E+01

.580E+00

.686E+01

.095E+01

.404E+00

.407E+00

.312E-01

.773E-01

.080E+00

.091E+02

.713E+02

.424E+02

.221E+02

.042E+02

.912E+01

.590E+01

.863E+01

.473E+01

.263E+01

.848E+02

.640E+02

.413E+02

.231E+02

.069E+02

.790E+01

.305E+01

.708E+01 ,636E+01 .771E+01 ,774E+02 ,529E+02 ,334E+02 ,130E+02 .597E+01 .135E+01 547E+01 942E+01 275E+01 667E+01 734E+02 580E+02 356E+02 142E+02 265E+01 495E+01 987E+01 400E+01 764E+01 105E+01 695E+02 563E+02 441E+02 255E+02 008E+02 596E+01 651E+01 996E+01 341E+01 237E+00 319E+02 517E+02 440E+02 286E+02 055E+02 969E+01

A73


87 88 89 90 91 92 93 94 95 96 97 98 99 100

26. 20. 14. 8.

67. 60. 54. 47 41 34 28 22 15 9

07 15 .22 ,30 ,30 ,84 .37 .91 .45 .98 .52 .06 .59 .13

13. 15. 17. 19. 1. 3 5. 7 9 11 13 15 17 19

00 00 .00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00

5. 3. 2.

-1. -4. -1, -8. -9, -9 -1 -1 -9 -6 -2

457E+01 650E+01 372E+01 .863E+00 .967E+02 .768E+03 .685E+02 .199E+02 .173E+02 .002E+03 .058E+03 .727E+02 .438E+02 .378E+02

7 , 5. 4. 3 . 4. 6

-1 -6 -1 -1 -1 -9 -8 -8

169E+01 495E+01 004E+01 .953E+00 .272E+03 .784E+02 .142E+02 .521E+01 .009E+02 .485E+02 .426E+02 .708E+01 .094E+01 .607E+01

2. 1. -1 -2. 9 5 3 3 2 3 3 3 2 5

.883E+00

.455E+00

.570E+00

.278E+00

.604E+02

.167E+02

.041E+02

.186E+02

.961E+02

.320E+02

.407E+02 •210E+02 •117E+02 . 012E+01

7 5, 4 4 4 7 -6 4 -4 -3 -2 8 -1 -7

.216E+01

.507E+01 ,019E+01 .739E+00 •458E+03 .831E+02 .889E+00 •047E+01 .823E+00 .455E+01 •973E+01 .001E+00 .022E+01 .101E+01

5, 3. 2,

-2. -6, -1 -9. -1 -1 -1 -1 -1 -7 -2

410E+01 638E+01 357E+01 649E+00 .828E+02 ,872E+03 .758E+02 .026E+03 .013E+03 .116E+03 .171E+03 .078E+03 .146E+02 .529E+02


ELEM

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20

NO X

0. 0. 0. 0. 0. 0. 0. 0, 0. 0,

67 60 54 47 41 34 28 21 15 9

20 20 20 20 20 .20 ,20 .20 ,20 .20 .20 .74 .27 .81 .35 .88 .42 .96 .49 .03

Y

1. 3. 5. 7. 9.

11. 13. 15. 17. 19 1 3 5 7 9

11 13 15 17 19

NORMAL STRESS SHEAR

00 00 00 .00 ,00 ,00 ,00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00 .00

2. 0. 0. 0. 0, 1. 2, 2 0, 4 9 1 1 1 1 9 7 5 3 9

389E+01 000E+00 OOOE+00 OOOE+00 OOOE+00 .478E-01 .097E+00 .133E+00 .000E+00 .322E-01 .006E+01 .750E+02 .541E+02 .295E+02 .101E+02 .076E+01 .328E+01 .314E+01 .020E+01 .297E+00

-5. 0. 0. 0. 0. 2. -1 -4 0

-3 4

-1 -4 -4 -2 -8 -4 -1 -4 -4

STRESS

991E+00 000E+00 OOOE+00 .000E+00 .000E+00 .373E+00 .450E+00 .200E+00 .000E+00 .171E+00 .767E-01 .700E+00 .839E+00 .398E+00 .560E+00 .897E-01 .402E-01 .924E+00 .230E+00 .377E+00

NORMAL STIFF

1. 1. 1. 1. 1. 1. 1. 1. 1, 1 1 1 1 1 1 1 1 1 1 1

OOOE+08 000E+02 OOOE+02 OOOE+02 000E+02 OOOE+08 OOOE+08 .000E+08 OOOE+02 .000E+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .OOOE+08 .000E+08 .OOOE+08 .OOOE+08

SHEAR STIFF

4. 1. 1. 1. 1. 1. 4. 1. 1. 1. 4. 4 4 4 4 4 4 4 4 4

OOOE+03 000E+02 000E+02 OOOE+02 000E+02 000E+02 000E+03 000E+02 OOOE+02 .OOOE+02 .OOOE+03 .000E+03 .OOOE+03 .000E+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03 .OOOE+03


NUM.

1 2 3 4 5 6 7 8 9 10

I

1 2 3 4 5 6 7 8 9

10

J

111 112 113 114 115 116 117 118 119 120

TYPE

1 1 1 1 1 1 1 1 1 1

COMPR FORCE

0.000000E+00 1.516247E+01 -9.161447E+00 3.167562E+01 1.866586E+01 1.821873E+01 3.136692E+00 -4.022528E-01 -2 .177604E+00 1.640831E+00

0. -1, 1 2. 8 8

-1 6

-3 1

INCR COMPR

000000E+00 936103E-06 .190417E-05 .479181E-06 .431263E-06 .982606E-06 .862645E-08 .305054E-07 .599562E-06 .391396E-05

6. 6. 6. 6. 6. 6 6 6 6 6

STIFFNESS

750000E+05 750000E+05 .750000E+05 .750000E+05 .750000E+05 •750000E+05 .750000E+05 .750000E+05 .750000E+05 .750000E+05

A74

RSDAM PROGRAM LISTING APPENDIX G

APPENDIX G- RSDAM PROGRAM LISTING

Q ******************************************************************

C C * PROGRAM FOR C * C * (A) GEOMETRICAL OPTIMISATION OF REINFORCED EARTH DAMS C * C * AND * C C * (B) STRESS-STRAIN ANALYSIS WITHIN REINFORCED EARTH DAMS * C Q ******************************************************************

COMMON /MESH1/ PJ(99),KS(15,3),NIT(15),X11(1000),Y11(1000),NUS(15) &,TEJ(99),IB(35,3),JDN(99),STF(99),X(999),Y(999),MOD(99,15),IC(200) &,ALPHA(99),EZ(99),AO(99),FR(99),GAM(99)

C COMMON /MESH2/ XPB(99),BC(99),PHI(99),XXP(99),COHE(99),EIMN(99), & TN(99),HCF(99),ULF(99),IDN(99),GUE(99),REDJ(99) ,BR(35,9) /COJ(99) , & FRJ(99),XI(10 00),Y1(10 00),FX1(100)

C COMMON /MESH3/ IDT(99),STI(99),STS(99),STN(99),FX(10 0),FY(100),NMP & ,HIZ,HWIZ/GSUB,HSIZ,AKA,ALFA,CP,WT,WBIZ/TF

C DIMENSION Vl(lOO) ,VS(100) ,VE(100) ,V(100) ,E(100) ,W1(100) ,AKISI(100) DIMENSION H1(100),HS(100),RU(100),ANU(100),AMH(100),U(100),XK(100) DIMENSION WB(IOO),ANU1(100),SV(100),WR(100),AN(100),HE(100),Z(100) DIMENSION WS(100),W(100),H(100),HW(100),HW2(100),HI(10 0),DOV(100) DIMENSION AM(IOO),AM1(100),AM2(100),AM3(100),DELTA(100),D1(100) DIMENSION XX{100),XXI(100),XX2(100),XX3(100),XX4(100),XX5(100) DIMENSION XX6(100),XX7(100),AK1(100),AK2(100),AK3(100),VR(100) DIMENSION CMM(IOO),FST(100),AS(100),WEIGG(100),CC(100),ZZ(100) DIMENSION BET(IOO),BET1(100),SIG(100),SIGl(100),DD(100)

OPEN (UNIT=5,FILE='DAM1.OUT',STATUS='OLD') OPEN (UNIT=9,FILE='DAM.IN',STATUS='NEW')

_, ****************************************************************** C * MAIN PROGRAM c ******************************************************************

CALL INPUDATA(H(1),HW(1),HW2{1),HS(1),WT,WB(1),GW,GSUB,GS,SFS,SFO k ,SFB/SFOS,SFY,VI,ALFA,CP/Bl,B2,B,UR,N,RSTAR/FEE/FY,CU,KK,P,TF) FEEl=FEE/57.3248 EE=(45-FEE/2) EEl=EE/57.2958 AKA=(tan(EEl))**2 AK0=AKA*(1+SIN(FEE1)) AK=SFS/(tan (FEED ) XJ=H(1)/B2 J=XJ IF (XJ.GT.J) J=J+1 WRITE (5,19) WRITE (5 , *) 'NUMBER OF LAYERS= ' -J

WRITE (5,19) NMP=J+1 HIZ=H{1) HWIZ=HW(1) HSIZ=HS(1) IF (H(l).LE.15) THEN WT=H(1)/5+3 ELSE

WT=6 END IF 1 = 1

C WRITE (6,19) C WRITE (8,19) A75


50 ZZ(1)=0 AKISI(I)=WT/WB(I) AN(I)=1+AKISI(I) CALL OUTPUT(I,H(I),HW(I),HW2(I),HS (I) ,WT,WB(I),AKISI(I)) CALL VERFORCE(Wl(I) ,WB(I) ,WT,HW(I),HW2(I),U(I),GW, H(I) ,WS(I) ,HS(I) 5c ,GSUB,W(I) ,GS,BET(I) ,SIG(I) , ANU (I) ) CALL HORFORCE(VI(I),GW,HW(I),VS(I),GSUB,HS(I),EE1,VE(I),CP,ALFA, & E(I),W(I),V(I),VI) CALL DIST(HKI) , HW( I) , HI (I) , HS (I) , HE (I) ,AMH(I) , VI (I) , VI, VS (I) , & VE(I)) CALL BEAOPTM(DOV(I),RSTAR,H(I),GS,BET(I),SIG(I),AKISI(I)) CALL SLIDOPTM(AMd) , BET (I) ,SIG(I) ,ANU(I) , AK, ALFA, XX (I) ,V(I) ,AN(I) , Sc H(I) ,GS) CALL OVTUOPTM(AKISI(I),H(I),BET1(I),HI(I),SIG1(I),HS(I),CMM(I), & BET(I),SIG(I),AM1(I),AM2(I),AM3(I),SFO,ALFA,AMH(I),GS,DELTA(I), & XXI(I),XX2(I),HW(I),HW2(I),ANU(I),ANU1(I)) CALL OVSTOPTM(AKKI) , RSTAR, SFOS, AN (I) ,H(I) ,GS,BET(I) ,SIG(I) , CC (I) , & AKISI(I) ,AK2(I),ALFA,AMH(I),AK3(I),DD(I),XX4(I) ,XX5(I),RU(I) , & XX7(I),WT) IF(ZZ(I).LT.XX(I)) ZZ(I)=XX(I) IF(ZZ(I) .LT.XXKI) ) ZZ(I)=XX1(I) IF(ZZ(I).LT.XX2(I)) ZZ(I)=XX2(I) IF(ZZ(I).LT.XX7(I)) ZZ(I)=XX7(I)

Q ****************************************************************** C * CHECK FOR NO BOND FAILURE OF REINFORCEMENTS WITHIN A * C * LAYER OF A REINFORCED EARTH DAM Q ******************************************************************

Z(I)=H(I) F0ST=1.2*ALOG10(CU) IF (KK.EQ.l) THEN GO TO 12 ELSE IF (KK.EQ.2) THEN GO TO 13 ELSE GO TO 14 END IF END IF

12 CALL CGM(Z(I),FST(I),FEE1,FOST,XK(I),AKA,AK0,XX3(I),Bl,B2,SFB,N,B, &H(D) IF(ZZ(I).LT.XX3(I)) ZZ(I)=XX3(I) GO TO 15

13 CALL MCGM(FSTd) ,Z(I) , FOST, FEE1, XK (I) ,AKA,AK0,XX3 (I) , Bl, B2 , SFB, N, B &,H(D) IF(ZZ(I).LT.XX3(I)) ZZ(I)=XX3(I) GO TO 15

14 CALL NCGM(Z(I),D1(I),FST(I),FEE1,FOST,XK(I),AKA,AK0,XX3(I),B1,B2, &SFB,N,B,H(1))

15 IF(ZZ(I).LT.XX3(I)) ZZ(I)=XX3(I) CALL NOFAIL(ZZ(I),WB(I)) IF (I.EQ.l) THEN IF (WB(1).LT.ZZ(l)) THEN WB(1)=ZZ(1) GO TO 50 END IF IF (WB(l)-ZZ(l).GT.0.1) THEN WB(1)=(WB(1)+ZZ(1))/2 GO TO 50 END IF END IF CALL REINAREA(SV(I) ,Z(I),GS,AS(I),XK(I) ,SFY,Bl,B2,N,FY,XX6(I) ,B, &XX3(I),VR(I),WR(I) ,UR,H(I)) WEIGG(1)=0 WEIGG(I+1)=WEIGG(I)+WR(I) IF(I.LT.XJ)THEN 1=1 + 1 H(I)=H(1)-(I-1)*H(1)/XJ HW(I)=HW(1)-(I-1)*H(1)/XJ HW2(I)=HW2(1)-(I-1)*H(1)/XJ HS(I)=HS(1)-(I-1)*H(1)/XJ WB(I)=WT+(WB(1)-WT)*H(I)/H(1)

A76


ELSE C WRITE (8,*) 'TOTAL WEIGHT OF REINFORCEMENTS=',WEIGG(I+l)

GO TO 999 END IF WBIZ=WB(1) IF(O.GT.HSd) ) HS(I)=0 IF(O.GT.HWd) ) HW(I)=0 IF(0.GT.HW2(I)) HW2(I)=0 GO TO 50

999 CALL MESH 121 FORMAT (7 F 10.2) 19 FORMAT (/)

STOP END

p ******************************************************************

C * SUBROUTINE FOR INPUT DATA C ******************************************************************

SUBROUTINE INPUDATA(H,HW,HW2,HS,WT,WB, GW, GSUB, GS,SFS,SFO,SFB,SFOS &/SFY,VI,ALFA,CP,Bl,B2,B,UR,N,RSTAR,FEE,FY,CU,KK,P,TF) WRITE (*,11) WRITE (* * ) • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * ' WRITE (*,*)'HEIGHT OF DAM=? (m)• READ (*,*) H WRITE (*,*)'UPSTREAM WATER TABLE=? (m)' READ (*,*) HW WRITE (*,*)"DOWNSTREAM WATER TABLE=? (m)' READ (*,*) HW2 WRITE (*,*)'HEIGHT OF SILT=? (ra) ' READ (*,*) HS WRITE (*,*)'INITIAL TOP WIDTH OF DAM=? (m) ' READ (*,*) WT WRITE (*,*)'INITIAL BASE WIDTH OF DAM=? (m)' READ (*,*) WB GW=10. WRITE ( *,*) ' FOR CHANGING DATA TYPE 1 WRITE ( * , *) 'FOR CONTINUE TYPE 2 READ ( * , *) P IF (P.EQ.l) GO TO 1 WRITE (*,11) WRITE (* *) •*****************************************************' WRITE (*',*)'UNIT WEIGHT OF SILT=? (KN/m3)' READ (*,*) GSUB WRITE (*,*)"AVERAGE UNIT WEIGHT OF DAM=? (KN/m3 READ (*,*) GS WRITE (* * ) * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * WRITE {*',*) 'SAFETY FACTOR AGAINST SLIDING=? READ (*,*) SFS WRITE (*,*)'SAFETY FACTOR AGAINST OVERTURNING=? READ (*,*) SFO WRITE (*,*)'SAFETY FACTOR AGAINST BOND FAILURE=? READ ( *,*) SFB WRITE (*,*)'SAFETY FACTOR AGAINST OVER-STRESSING=? READ ( *,*) SFOS WRITE (*,*)'SAFETY FACTOR AGAINST RUPTURE FAILURE=? READ (*,*) SFY WRITE (*,*)"FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ (*,*) P IF (P.EQ.l) GO TO 2 WRITE (*,11) WRITE (* *) •**************************************************' WRITE (*',*) 'ICE FORCE=? 'KN> READ (*,*) VI WRITE (*,*)'INITIAL COEFFICIENT OF EARTHQUAKE ACCELARATION=? READ (*,*) ALFA WRITE (*,*) 'COEFFICIENT OF INDIRECT FORCE OF EARTHQUAKE=? READ (*,*) CP WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ ( *,*) P IF (P.EQ.l) GO TO 3

********* '

A77


WRITE (*,11) WRITE (*,*)"**************************,*»,***»****»**************,

4 WRITE (*,*)"WIDTH OF FACING PANELS=? lm) READ (*,*) Bl WRITE (*,*)'HEIGHT OF FACING PANELS=? (m) READ ( *,*) B2 WRITE (*,*) 'THICKNESS OF FACING PANELS=? (m) READ (*,*) TF WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ ( *,*) P IF (P.EQ.l) GO TO 4 WRITE (*,11) WRITE (*,*)'*****************************************************,

5 WRITE (*,*)"WIDTH OF REINFORCEMENTS=? (m) READ ( *,*) B WRITE (*,*) 'UNIT WEIGHT OF REINFORCEMENTS=? (KN/m3) READ (*,*) UR WRITE (*,*)'ALLOWABLE TENSION OF REINFORCEMENTS=? (KN/m2) READ ( *,*) FY WRITE (*,*)'NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL=? & READ ( *,*) N WRITE ( * , *) " FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ ( *,*) P IF (P.EQ.l) GO TO 5 WRITE (*,11) WRITE (*,*) ' *****************************************************•

6 WRITE (*,*)'ALLOWABLE BEARING CAPACITY OF FOUNDATION SOIL=? (KN/m2 &) * READ (*,*) RSTAR WRITE (*,*)'ANGLE OF INTERNAL FRICTION OF SOIL=? (DEGREE) READ ( *,*) FEE WRITE (*,*)'COEFFICIENT OF UNIFORMITY OF SOIL=? READ ( *,*) CU WRITE ( *,*) ' FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ ( *,*) P IF (P.EQ.l) GO TO 6 WRITE (*,11) WRITE (* * ) • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * '

7 WRITE (*,*) '1- INTERNAL STABILITY ANALYSIS BASED ON COHERENT GRAVI &TY METHOD WRITE (*,*) '2- INTERNAL STABILITY ANALYSIS BASED ON MODIFIED COHER &ENT GRAVITY METHOD' WRITE (*,*) '3- INTERNAL STABILITY ANALYSIS BASED ON NEW COHERENT G &RAVITY METHOD READ (*,*) KK WRITE (* * ) ' * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * ' WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (* , *) 'FOR CONTINUE TYPE 2 READ ( *,*) P IF (P.EQ.l) GO TO 7 WRITE (*,11)

11 FORMAT (/1111 /7 / / / /1111111111•/) 19 FORMAT (/)

WRITE (5,*)'ggggggggggggggggggggggggggggggggggggggggggggggggggggg WRITE (5 * ) * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * WRITE (5,*)'* * WRITE (5,*) '* WRITE (5,*)'* INPUT DATA * WRITE (5,*)"* WRITE (5,*)'* WRITE (5 *) '***************************************************** WRITE (5,*)"@g@gggggggggggggggggggggggggggggggggggggggggggggggggg

WRITE (5,*)'HEIGHT OF DAM= WRITE (5,*)'UPSTREAM WATER TABLE= WRITE (5,*) "DOWNSTREAM WATER TABLE= WRITE (5,*)'HEIGHT OF SILT= WRITE (5,*) 'TOP WIDTH OF DAM=

,H, 'm ' ,HW,' m' ,HW2,' m' ,HS,' m' ,WT,' m'

A78


WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE WRITE &,N WRITE WRITE WRITE WRITE WRITE WRITE WRITE

5,*)'BOTTOM WIDTH OF DAM= 5,19) 5,*)"UNIT WEIGHT OF WATER= 5,*)"UNIT WEIGHT OF SILT= 5,*)'AVERAGE UNIT WEIGHT OF DAM= 5,19) 5,*)'SAFETY FACTOR AGAINST SLIDING= 5,*)'SAFETY FACTOR AGAINST SLIDING= 5,*)"SAFETY FACTOR AGAINST BOND FAILURE= 5,*)'SAFETY FACTOR AGAINST OVER-STRESSING= 5,*)'SAFETY FACTOR AGAINST RUPTURE FAILURE 5,19) 5,*) 'ICE FORCE=

',WB,' m'

',GW,' KN/m3' ',GSUB, ' KN/m3' ',GS,' KN/m3'

' ,SFS ' ,SFS ' ,SFB ',SFOS ' ,SFY

19) • VI, KN'

*)'COEFFICIENT OF EARTHQUAKE ACCELARATION= ",ALFA *)"COEFFICIENT OF INDIRECT FORCE OF EARTHQUAKE=',CP 19) *) 'WIDTH OF FACINGS= • , Bl,

5,*)'HEIGHT OF FACINGS= ',B2, 5,*) 'WIDTH OF REINFORCEMENTS= ',B, ' 5,*) 'UNIT WEIGHT OF REINFORCEMENTS= ' ,UR,

m' m' n" KN/m3'

m

19) *)"NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL=

19) *)'ALLOWABLE BEARING CAPACITY OF SOIL= *)"ANGLE OF INTERNAL FRICTION OF SOIL= 19) *)"ALLOWABLE TENSION OF REINFORCEMENTS= *)'COEFFICIENT OF UNIFORMITY OF SOIL=

RSTAR," KN/m2 FEE,' DEGREE'

",FY,' KN/m2 " ,CU

(5,19) IF (KK.EQ.l) WRITE (5,*)"INTERNAL STABILITY ANALYSIS IS BASED ON C

COHERENT GRAVITY METHOD" IF (KK.EQ.2) WRITE (5,*)'INTERNAL STABILITY ANALYSIS IS BASED ON M MODIFIED COHERENT GRAVITY METHOD' IF (KK.EQ.3) WRITE (5,*)'INTERNAL STABILITY ANALYSIS IS BASED ON N &EW COHERENT GRAVITY METHOD' RETURN END

19

C C C C

SUBROUTINE OUTPUT(I,H,HW,HW2,HS,WT,WB,AKISI) WRITE (5,*) 'ggggggggggggggggggggggggggggggggggggggggggggggggggggg WRITE (5 *) ****************************************************** WRITE (5,*)'* * WRITE ( 5 , * ) '* * WRITE (5,*) '* OUTPUT * WRITE (5,*)'* * WRITE (5,*) "* * WRITE (5 *) ****************************************************** WRITE (5,*) 'ggggggggggggggg@@g@ggggggggggggggggggggggggggggggggg@ WRITE (5,19) WRITE (5,*) 'LAYER NO.= ',I WRITE (5,19) WRITE (5,*) 'HEIGHT OF LAYER= ',H,' m* WRITE (5,*)'UPSTREAM WATER TABLE= ',HW,' m' WRITE (5,*) "DOWNSTREAM WATER TABLE= ',HW2, ' m' WRITE (5,*)'HEIGHT OF SILT= ',HS,' m' WRITE (5,*) "TOP WIDTH OF LAYER= ",WT, " m' WRITE (5,*)'BOTTOM WIDTH OF LAYER= ',WB,' m' WRITE (5,19) WRITE (5,*)"RATIO OF TOP WIDTH TO BOTTOM WIDTH=',AKISI WRITE (5,19) FORMAT (/) RETURN END ****************************************************************** * SUBROUTINE FOR CALCULATION OF HORIZONTAL FORCES ACTING ON * * A REINFORCED EARTH DAM * ****************************************************************** SUBROUTINE HORFORCE(VI,GW,HW,VS,GSUB, HS,EE1,VE,CP,ALFA,E,W,V,VI) Vl=GW*HW**2/2

A79


c

c

C

VS=GSUB*HS**2*(TAN(EEl))**2/2 VE=0.726*CP*ALFA*GW*HW**2 E=ALFA*W V=V1+VI+VS+VE WRITE (5,19) WRITE (5,*)'HYDROSTATIC FORCE ACTING ON LAYER= \V1,'KN' WRITE (5,*) "ICE FORCE ACTING ON LAYER= ',VI,'KN' WRITE (5,*)"SILT FORCE ACTING ON LAYER= ',VS,'KN' WRITE (5, *) 'INDIRECT FORCE OF EARTHQUAKE ACTING ON LAYER=' ,VE, 'KN' WRITE (5,*)'DIRECT FORCE OF EARTHQUAKE ACTING ON LAYER= ',E,'KN'

C WRITE (5,19) C WRITE (5,*)'SUM OF HORIZENTAL FORCES EXCEPT DIRECT FORCE OF EARTHQ C &UAKE=',V,' KN'

WRITE (5,19) 19 FORMAT (/)

RETURN END

n ****************************************************************** C * SUBROUTINE FOR CALCULATION OF VERTICAL FORCES ACTING ON A C * REINFORCED EARTH DAM *

****************************************************************** SUBROUTINE VERFORCE(Wl,WB,WT,HW, HW2,U,GW,H,WS,HS,GSUB,W,GS,BET,SIG Sc ,ANU) W1=(WB-WT)*HW**2*GW/(2*H) WS=(WB-WT)*HS**2*GSUB/(2*H) W=(WT+WB)*H*GS/2 U=GW*(HW+HW2)*WB/2 WRITE (5,*) 'WEIGHT OF LAYER= ',W, 'KN' WRITE (5,*)'WEIGHT OF WATER ON UPSTREAM SIDE OF LAYER= ',Wl,'KN' WRITE (5,*)'WEIGHT OF SILT ON UPSTREAM SIDE OF LAYER= ',WS,'KN' WRITE (5,*)'UPLIFT PRESSURE ACTING ON THE LAYER= ',U,"KN' BET=W1/W S1G=WS/W ANU=U/W RETURN END ******************************************************************

SUBROUTINE FOR CALCULATION OF THE DISTANCES OF THE C * FORCES ACTING ON A LAYER OF A C * REINFORCED EARTH DAM * Q ******************************************************************

SUBROUTINE DIST(HI,HW,HI,HS,HE,AMH,VI,VI,VS,VE) Hl=HW/3 HI=HW HE=0.4*HW AMH=Vl*Hl+VI*HI+VS*HS/3+VE*HE

C WRITE (5,19) C WRITE (5,*)'SUM OF DRIVING MOMENTS= ',AMH,' KN-m' C WRITE (5,19) 19 FORMAT (/)

RETURN END

Q ****************************************************************** C * SUBROUTINE FOR NO BEARING CAPACITY FAILURE STATE Q ******************************************************************

SUBROUTINE BEAOPTM(DOV,RSTAR,H,GS, BET,SIG,AKISI) DOV=2*RSTAR/(H*GS*(1+BET+SIG)) IF(AKISI.GT.DOV)THEN WRITE (5,*)'BEARING CAPACITY FAILURE WILL HAPPEN' ELSE WRITE (5,*)'BEARING CAPACITY FAILURE WILL NOT HAPPEN' END IF WRITE (5,19)

19 FORMAT (/) RETURN END n ****************************************************************** C * SUBROUTINE FOR NO SLIDING FAILURE STATE WITHIN A C * LAYER OF REINFORCED EARTH DAM ****************************************************************** SUBROUTINE SLIDOPTM(AM,BET,SIG, ANU, AK,ALFA,XX,V,AN,H,GS)

A80

AM=(1+BET+SIG-ANU-(AK*ALFA))/AK IF(AM.LE.O) THEN WRITE (5,*)'WARNING; SLIDING MAY HAPPEN' WRITE (5,19) END IF XX=2*V/(AM*AN*H*GS) WRITE (5,*)'MIN. REQUIRED BASE LENGTH FOR NO SLIDING= ",XX," & m"

19 FORMAT (/) RETURN END

Q ****************************************************************** C SUBROUTINE FOR NO OVERTURNING FAILURE STATE WITHIN * c * A LAYER OF REINFORCED EARTH DAM * Q ******************************************************************

SUBROUTINE OVTUOPTM(AKISI,H,BET1,HI,SIG1,HS,CMM,BET,SIG,AMI,AM2, & AM3 , SFO, ALFA, AMH, GS, DELTA, XXI, XX2 , HW, HW2 , ANU, ANU1) BET1=(1+AKISI)*(3*H-H1+H1*AKISI)/((1+AKISI+AKISI**2)*H) SIG1=(1+AKISI)*(3*H-HS/3+HS*AKISI/3)/((1+AKISI+AKISI**2)*H) IF (HW.EQ.0.AND.HW2.EQ.0) THEN ANU1=0 ELSE ANU1=(1+AKISI)*(2*HW+HW2)/((1+AKISI+AKISI**2)*(HW+HW2)) END IF CMM=(1+BET*BET1+SIG*SIG1-ANU*ANU1*SF0) AM1=(1+AKISI+AKISI**2)*CMM/(6*SFO) AM2=(-1)*ALFA*H*(1+2*AKISI)/6 AM3=(-1)*AMH/(H*GS) DELTA=AM2 **2-4*AMl*AM3 IF (DELTA.LT.O) THEN IF (AMI.LT.O) THEN WRITE (5,19) WRITE (5,*) "NO ANSWER FOR THE EQUATION OF OVERTURNING FAILURE' WRITE (5,*) "FOR NO OVERTURNING FAILURE, MINIMUM BASE LENGTH SHOUL &D BE INCREASED" WRITE (5,*) END IF ELSE XX1=(AM2 +DELTA* * 0.5)/(-2 *AM1) XX2=(-1*AM2+DELTA**0.5)/(2*AM1) IF (AMI.LT.O)THEN WRITE (5,*) XX1,'<MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING<' , & XX2,' m' ELSE WRITE (5,*)"MIN. REQUIRED BASE LENGTH FOR NO OVERTURNING= ",XX2 &, " m' END IF END IF


Q ****************************************************************** C * SUBROUTINE FOR NO OVER-STRESSING FAILURE STATE WITHIN * C * A LAYER OF A REINFORCED EARTH DAM * Q ******************************************************************

SUBROUTINE OVSTOPTM(AK1,RSTAR,SFOS,AN,H,GS,BET,SIG,CC,AKISI,AK2, & ALFA,AMH,AK3,DD,XX4,XX5,RU,XX7,WT) AKl=RSTAR/SFOS-AN*H*GS*(1+BET+SIG)12 CC=H*(2*AKISI+1)/(3*(AKISI+1)) AK2 = -3 *AN*H*GS*ALFA*CC AK3=-6*AMH DD=AK2**2-4*AK1*AK3 IF (DD.LT.O) THEN WRITE (5,*) "NO ANSWER FOR THE EQUATION OF OVER-STRESSING FAILURE' ELSE IF (AK1.EQ.0) THEN XX5=-AK3/AK2 WRITE (5,*)'MIN. REQUIRED BASE LENGTH FOR NO OVER-STRESSING=',XX5, &' m' GO TO 99 END IF

A81

XX4=(AK2+DD**0.5)/(-2*AKl) XX5=(-1*AK2+DD**0.5)/(2*AK1) IF (AK1.LT.O)THEN WRITE (5,*) XX4,'<MIN. REQUIRED BASE LENGTH FOR NO OVER-STRESSING< &', XX5,' rn' ELSE WRITE (5,*)'MIN. REQUIRED BASE LENGTH FOR NO OVER-STRESSING=',XX5, &' m' END IF

99 END IF RU=H*GS*SFOS*(1+BET+SIG)/(2*RSTAR) XX7=WT*RU/(1-RU) IF (XX7.LT.0) THEN WRITE (5,19) WRITE (5,*)'OVER-STRESSING FAILURE WILL HAPPEN' PRINT *,'ERROR: BEARING CAPACITY OF FOUNDATION SOIL IS VERY LOW' WRITE (5,*)'BEARING CAPACITY OF FOUNDATION SOIL IS VERY LOW' WRITE (5,19) STOP ELSE WRITE (5,*)'MIN. REQUIRED BASE LENGTH FOR NO OVER-STRESSING=',XX7, &' m' END IF


Q ****************************************************************** C * SUBROUTINE FOR NO BOND FAILURE OF REINFORCEMENTS WITHIN A * C * LAYER OF A REINFORCED EARTH DAM * C * BASED ON COHERENT GRAVITY METHOD * Q ******************************************************************

SUBROUTINE CGM(Z,FST,FEE1,FOST,XK,AKA,AKO,XX3,Bl,B2,SFB,N,B,H) IF (Z.LE.6)THEN FST=Z*(tan(FEEl)-FOST)/6+FOST XK=Z*(AKA-AKO)/6+AKO ELSE FST=tan(FEEl) XK=AKA END IF XX3=XK*B1*B2*SFB/(2*N*FST*B) IF (Z.LE.H/2)THEN XX3=XX3+0.3*H ELSE XX3=XX3+0.6*(H-Z) END IF WRITE (5,*)'MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= ',XX3, &' m' WRITE (5,19)


C ****************************************************************** C * SUBROUTINE FOR NO BOND FAILURE OF REINFORCEMENTS * C * WITHIN A LAYER OF A REINFORCED EARTH DAM * C * BASED ON MODIFIED COHERENT GRAVITY METHOD * Q ******************************************************************

SUBROUTINE MCGM(FST,Z,FOST,FEE1,XK,AKA,AKO,XX3,Bl,B2,SFB,N,B,H) FST=((0.6)**Z)*(1.7*F0ST-tan(FEEl))+tan(FEED XK=((0.75)**Z)*(AKA-AKO)+AKA XX3=XK*B1*B2*SFB/(2*N*FST*B) XX3=XX3+H*((6.76-(Z/H)**2)**0.5-2.3) WRITE (5,*)'MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= ',XX3, &' m' WRITE (5,19)

19 FORMAT (/) RETURN END Q ****************************************************************** C * SUBROUTINE FOR NO BOND FAILURE OF REINFORCEMENTS * C * WITHIN A LAYER OF A REINFORCED EARTH DAM C * BASED ON NEW COHERENT GRAVITY METHOD *

A82


Q ****************************************************************** SUBROUTINE NCGM(Z,Dl,FST,FEE1,FOST,XK,AKA,AKO,XX3,Bl,B2,SFB,N,B,H) IF (Z.LE.6)THEN Dl=Z**2/36-Z/3+l FST=tan(FEEl)+0.9**Z*D1* (3 . 85*F0ST-tan(FEED) XK=1.2**Z*D1*(AKA-AKO)+AKA ELSE FST=tan(FEEl) XK=AKA END IF XX3=XK*B1*B2*SFB/(2*N*FST*B) XX3=XX3+H*((6.76-(Z/H)**2)**0.5-2.3) WRITE (5,*) 'MIN. REQUIRED BASE LENGTH FOR NO BOND FAILURE= ',XX3, &' m' WRITE (5,19)


r» * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

C * SUBROUTINE FOR CHECK OF MINIMUM REQUIRED BASE LENGTH C * FOR NO FAILURE Q ******************************************************************

SUBROUTINE NOFAIL(ZZ,WB) IF (ZZ.LE.WB) THEN WRITE (5,*) 'MIN. REQUIRED BASE LENGTH FOR NO FAILURE= ',ZZ,' m' ELSE WRITE (5,*)'MIN. REQUIRED BASE LENGTH FOR NO FAILURE= ",ZZ,' m' WRITE (5, * ) ' FOR NO FAILURE BASE LENGTH SHOULD BE INCREASED' WRITE (5,19) END IF


£, ****************************************************************** C * SUBROUTINE FOR CALCULATION OF THE CROSS SECTION AREA OF C * REINFORCEMENTS WITHIN A LAYER OF C * A REINFORCED EARTH DAM

****************************************************************** SUBROUTINE REINAREA(SV,Z,GS,AS,XK,SFY,Bl,B2,N,FY, XX6, B,XX3,VR,WR,

ScUR,H) SV=Z*GS AS=XK*SV*SFY*B1*B2*10000/(N*FY) XX6=AS/(B*100) VR=AS*XX3*N/(10000*B1*B2) WR=VR*UR WRITE (5,19) WRITE (5,*) 'NUMBER OF REINFORCEMENTS CONNECTED TO A FACING PANEL=' Sc, N WRITE (5,*) 'MIN. REQUIRED LENGTH OF REINFORCEMENT= &,XX3,' m' WRITE (5,*)'MIN. NET THICKNESS OF REINFORCEMENT= Sc,XX6*10, " mm" WRITE (5,*) 'WIDTH OF REINFORCEMENT= &,B*100,' cm" WRTTF (5 19) WRITE (5!*)'MIN. CROSS SEC. AREA OF REINFORCEMENT= ',AS*N/(B1*B2) Sc, ' cm2/m2 AREA' WRITE (5,*) "MIN. NET VOLUME OF REINFORCEMENT= ',VR,' & m3/m2 AREA' WRITE (5,*)'MIN. NET WEIGHT OF REINFORCEMENT= ',WR,' & KN/m2 AREA'


c ****************************************************************** C * SUBROUTINE FOR MESH GENERATION r ******************************************************************

C

SUBROUTINE MESH COMMON /MESH1/ PJ(99),KS(15,3),NIT(15),Xll(1000),Yll(1000),NUS(15) &,TEJ(99),IB(35,3),JDN(99),STF(99),X(999),Y(999),MOD(99,15),IC(200) &,ALPHA(99),EZ(99),AO(99),FR(99),GAM(99)

A83


COMMON /MESH2/ XPB(99),BC(99),PHI(99),XXP(99),COHE(99),EIMN(99), & TN(99),HCF(99),ULF(99),IDN(99),GUE(99),REDJ(99),BR(35,9),COJ(99), & FRJ(99),X1(1000),Y1(1000),FX1(100) COMMON /MESH3/ IDT(99),STI(99),STS(99),STN(99),FX(IOO),FY(100),NMP &. ,HIZ,HWIZ, GSUB, HSIZ, AKA, ALFA, CP, WT, WBIZ , TF WRITE (*, *) ***************************************************** ' WRITE (9,*)'REINFORCED EARTH DAM ANALYSIS' WRITE (*,*) 'NUMBER OF NODAL POINTS IN X-DIRECTION=? READ (*,*) NUP NNP=NUP*NMP NUMEL=(NUP-1)*(NMP-1) NUMJT=2*(NMP-1) NUMBA=0 NBEAM=0 NBTYPE=0 NMOD=0 NC=5 INIT=1 IHORIZ=0 ITRD=2 ILIST=1 IOUT=0 WRITE(9,5010) NNP,NUMEL,NUMJT,NUMBA,NBEAM,NBTYPE,NC,NMOD,INIT, & IHORIZ,ITRD,ILIST,IOUT GW=10 PATM=100 WRITE(9,5015) GW,PATM NMAT=3 NNNNN=1 NUMSOL=NMAT-NNNNN IATYP=0 INOSLIP=0 WRITE(9,502 0) NMAT,NUMSOL,0,1,0,0,0,0,IATYP,INOSLIP DO 2 0 N=1,NC+1 KS(N,2)=0 KS(N,3)=0 MOD(1,N)=0 KS(1,1)=5 NIT(1)=1 NIT(N)=1 NUS(N)=1 IF (N .EQ. 1) WRITE(9,5040) KS(N,1),KS(N,2),KS(N,3),NIT(N),NUS(N), & MOD(1,N),'REINFORCEMENT INSTALATION' IF (N .EQ. 2) THEN KS(N,1)=9 END IF IF (N -EQ. 3) THEN KS(N,1)=9 WRITE(9,5040) KS(N,1),KS(N,2),KS(N,3),NIT(N),NUS(N),MOD(l,N), Sc 'HYDROSTATIC FORCE' ENDIF IF (N .EQ. 4) THEN KS(N,1)=9 WRITE(9,5040) KS(N,1),KS(N,2),KS(N,3),NIT(N),NUS(N),MOD(1,N), & 'SILT FORCE' ENDIF IF (N .EQ. 5) THEN WRITE(*,*) '1- STATIC ANALYSIS=? & 2- TIME HISTORY ANALYSIS=? READ (*,*) ZEL IF (ZEL .EQ. 1) THEN KS(N,D=9 WRITE(9,504 0) KS(N,D,KS(N,2),KS(N,3),NIT(N),NUS(N),MOD(l,N) , & 'EARTHQUAKE FORCE' END IF IF (ZEL .EQ. 2) THEN KS(N,D=8 WRITE(9,5040) KS(N,1),KS(N,2),KS(N,3),NIT(N),NUS(N),MOD(l,N), & 'EARTHQUAKE FORCE OR DISPLACEMENT' END IF ENDIF

A84


IF (N .EQ. 6) THEN KS(N,1)=3 WRITE(9,504 0) KS(N,1),KS(N,2),KS(N,3),NIT(N),NUS(N),MOD(l,N), & 'SEEPAGE LINE VARIATION' ENDIF IF (N .EQ. NC+1) GO TO 100

2 0 CONTINUE 100 WRITE (*,*) •***************FACING PANELS PROPERTY***************'

DO 160 N=l,NUMSOL IF (N .EQ. 1) THEN ZTZ=0 WRITE(*,*) 'UNIT WEIGHT OF FACING PANELS=? (KN/m3) READ(*,*) GAM(N) COHE(N)=0 PHI(N)=0 TN(N)=0 AO(N)=0 XXP(N)=0 HCF(N)=0 ULF(N)=0 FR(N)=0 EIMN(N)=0 XPB(N)=0 BC(N)=0

2 WRITE (*,*) 'YOUNG,S MODULUS OF FACING PANELS= (KN/m2) READ (*,*) EZ(N) WRITE (*,*) 'POISSON,S RATIO OF FACING PANELS READ (*,*) GUE(N) WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ (*,*) P IF (P.EQ.l) GO TO 2 WRITE (*,11) ELSE WRITE (*,*) *********************SOIL PROPERTY*******************'

1 WRITE {*',*) 'IS THE MATERIAL DRAINED? WRITE (*,*) '0-NO

& 1-YES READ(*,*)ZTZ IF (ZTZ.EQ.l) IDN(N)=1 IF (ZTZ.EQ.0) IDN(N)=0 , WRITE(*,*) "UNIT WEIGHT OF THE MATERIAL=? (KN/m3) READ(*,*) GAM(N) WRITE '(*,*) 'COHESION OF THE MATERIAL=? (KN/m2) READ(*,*) COHE(N) mPPRFF)' WRITE (*,*) "FRICTION ANGLE=? (Ub^Ktuj READ (* *) PHI(N)

c WRITE '(*,*) 'MIN. ALLOWABLE VALUE OF MINOR PRINCIPAL STRESS=? c READ (*,*} TN(N)

TN(N)=0 WRITE (*,*) "LATERAL EARTH PRESSURE COFFICIENT AT REST=? READ (*,*) AO(N) WRITE (*,*) 'INITIAL TANGENT MODULUS EXPONENT=? READ (*,*) XXP(N) WRITE (*,*) 'INITIAL TANGENT MODULUS COEFFICIENT? READ (*,*) HCF(N) WRITE (*,*) "FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ ( * , *) P IF (P.EQ.l) GO TO 1 WRITE (*,11) ENDIF WRITE(9,5080) IDN(N),GAM(N),COHE(N),PHI(N),TN(N),AO(N),XXP(N), & HCF(N)

SITE if;)1!™^********************************** — **; 21 WRITE (*,*) 'UNLOAD-RELOAD MODULUS COFFICIENT=?

READ (*,*) ULF(N) PR (jj) -l WRITE"(*,*)'MIN. INITIAL TANGENT MODULUS FOR NON-ELASTIC MATERIALS Sc =? (KN/m2) '

A85


READ (*,*) EIMN(N) WRITE (*,*) "BULK MODULUS EXPONENT=? READ (*,*) XPB(N) WRITE (*,*) 'BULK MODULUS COEFFICIENT? READ (*,*) BC(N) WRITE (*,*) 'YOUNG,S MODULUS=? (KN/m2) ' READ (*,*) EZ(N) WRITE (*,*) 'POISSON,S RATIO=? READ (*,*) GUE(N) WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE ( *,* ) 'FOR CONTINUE TYPE 2 READ ( *,*) P IF (P.EQ.l) GO TO 21 WRITE (*,11) ENDIF ALPHA(N)=0 WRITE(9,5100) ULF(N),FR(N),EIMN(N),XPB(N),BC(N),EZ(N),GUE(N), Sc ALPHA (N)

160 CONTINUE DO 170 N=NUMSOL+l,NMAT JDN(N)=0 COJ(N)=0 WRITE (*,*)"FRICTION COFICIENT BETWEEN FACING PANELS AND SOIL=? READ (*,*) PJ(N) TEJ(N)=0 IDT(N)=0 WRITE(9,5085) JDN(N),COJ(N),PJ(N),TEJ(N),IDT(N) STI(N)=4000 STS(N)=100 STN(N)=100000000 STF(N)=100 FRJ(N)=0 REDJ(N)=0 WRITE(9,5090) STI(N),STS(N),STN(N),STF(N),FRJ(N),REDJ(N)

17 0 CONTINUE X(1)=0 X(NUP)=WBIZ X(NMP)=0 X(NNP)=WT Y(1)=0 Y(NUP)=0 Y(NMP)=HIZ Y(NNP)=HIZ K=l DO 190 N=1,NNP,NMP IF(N .EQ. 1) THEN WRITE(9,5120) N,0,0 WRITE(9,5120) N+NMP-1,0,HIZ GO TO 190 ENDIF IF(N .EQ. NMP+1 .OR. N .EQ. 2*NMP+D THEN WRITE(9,5120) N,TF,0 WRITE(9,5120) N+NMP-1,TF,HIZ GO TO 190 ENDIF IF(N .EQ. NNP-2*NMP+1 .OR. N .EQ. NNP-3*NMP+1) THEN WRITE(9,5120) N,WBIZ-TF,0 WRITE(9,512 0) N+NMP-1,WT-TF,HIZ GO TO 190 ENDIF IF(N .EQ. NNP-NMP+1) THEN WRITE(9,5120) N,WBIZ,0 WRITE(9,5120) N+NMP-1,WT,HIZ GO TO 190 ENDIF WRITE(9,5120) N,TF+(WBIZ-2*TF)*K/(NUP-5),0 WRITE(9,5120) N+NMP-1,TF+(WT-2*TF)*K/(NUP-5),HIZ K=K+1 19 0 CONTINUE 3 4 0 WRITE(*,*)•******************************************************' 3 WRITE (*,*) "NUMBER OF NODAL FIXED POINTS IN Y-DIRECTION=?

A86

RSDAM PROGRAM USTING APPENDIX G

READ (* , *) NOY WRITE (*,*) 'NUMBER OF NODAL FIXED POINTS IN X-DIRECTION=? READ (*,*) NOX WRITE (*,*) "NUMBER OF NODAL FIXED POINTS IN X AND Y DIRECTIONS=?' READ (*,*) NOXY WRITE (*,*) 'NUMBER OF NODAL FIXED POINTS AGAINST ROTATING=? READ (* , * ) NOROT WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE ( *,*) 'FOR CONTINUE TYPE 2 READ (*,*) P IF (P.EQ.l) GO TO 3 WRITE (*,11) WRITE(9,5020) NOY,NOX,NOXY,NOROT WRITE ( * * ) •******************************************************' IF(NOY .EQ. 0) GO TO 4 00

4 WRITE (*,*) 'NODAL NUMBERS AGAINST Y-MOVEMENT=? READ (*,*) (IC(N),N=l,NOY) WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ (*,*) P IF (P.EQ.l) GO TO 4 WRITE (*,11) WRITE(9,5020) (IC(N),N=l,NOY)

400 IF(NOX .EQ. 0) GO TO 460 5 WRITE (*,*) 'NODAL NUMBERS AGAINST X-MOVEMENT=?

READ (*,*) (IC(N),N=l,NOX) WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ (*,* ) P IF (P.EQ.l) GO TO 5 WRITE (*,11) WRITE(9,5020) (IC(N),N=l,NOX)

460 IF(NOXY .EQ. 0) GO TO 510 6 WRITE (*,*) 'NODAL NUMBERS AGAINST BOTH X- AND Y- MOVEMENTS=?

READ (*,*) (IC(N),N=l,NOXY) WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ (*,*) P IF (P.EQ.l) GO TO 6 WRITE (*,11) WRITE(9,5020) (IC(N),N=1,NOXY)

510 IF(NOROT.EQ.0) GO TO 520 7 WRITE (*,*) 'NODAL NUMBERS AGAINST ROTATIONS=?

READ (*,*) (IC(N),N=1,NOROT) WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ ( * , *) P IF (P.EQ.l) GO TO 7 WRITE (*,1D WRITE(9,5020) (IC(N),N=1,NOROT)

C 520 IF(NUMEL.EQ.O) GO TO 660

DO 570 N=1,NUMEL IF (N .EQ. 1) THEN WRITE(9,5020) N,(2*NMP+D,(2*NMP+2),(NMP+2),(NMP+1),NMAT WRITE(9,5020) N+NMP-2,3*NMP-1,3*NMP,2*NMP,2*NMP-1,NMAT ENDIF IF (N .EQ. NMP) THEN WR1TE(9,5020) N,(NUP-2)*NMP+1,(NUP-2)*NMP+2,(NUP-3)*NMP+2, &(NUP-3)*NMP+1,NMAT WRITE(9,5020) N+NMP-2,NNP-NMP-1,NNP-NMP,NNP-2*NMP,NNP-2*NMP-1,NMAT ENDIF IF (N -EQ. 2*NMP-1) THEN WRITE(9,5020) N,NMP+1,NMP+2,2,1,1 WRITE(9,502 0) N+NMP-2,2*NMP-1,2*NMP,NMP,NMP-1,1 ENDIF K=2 DO 571 II=2,NUP-4 IF(N .EQ. (K+1)*NMP-K) THEN WRITE(9,5020)N,(K+1)*NMP+1,(K+l)*NMP+2,K*NMP+2,K*NMP+1,2 WRITE (9, 5020 )N+NMP-2, (K + 2)*NMP-1, (K + 2)*NMP, (K+l) *NMP, (K+D*NMP-1,2

A87


ENDIF K=K+1

571 CONTINUE IF(N .EQ. (NUP-2)* (NMP-D+1) THEN WRITE(9,5020) N,(NUP-1)*NMP+1,(NUP-1)*NMP+2,(NUP-2)*NMP+2, &(NUP-2)*NMP+1,1 WRITE(9,5020) N+NMP-2,NNP-1,NNP,NNP-NMP,NNP-NMP-1 1 ENDIF

570 CONTINUE C C REINFORCEMENTS INSTALATION C 660 WRITE (*,*) '**************************************************** 8 WRITE (*,*) 'NUMBER OF REINFORCEMENTS=?

READ ( *,*) NUMBAR IF(NUMBAR .EQ. 0) GO TO 960 WRITE (*,*) 'ELASTIC MODULUS OF THE REINFORCEMENTS=? (KN/m2) READ (*,*) HJ WRITE (*,*) 'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ ( *,* ) P IF (P.EQ.l) GO TO 8 WRITE (9,514 0) NUMBAR WRITE (*,11) WRITE (*,*) ***************************************************** DO 68 0 N=l,NUMBAR

9 WRITE (*,*) 'NODAL NUMBERS OF THE ', N,'th REINFORCEMENT=? READ (*,*) IB(M,1),IB(M,2) WRITE (*,*) 'ANGLE BETWEEN REINFORCEMENT AND HORIZONTAL LINE=? READ (*,*) ZAVIEH WRITE (*,*) 'CROSS-SECTIONAL AREA OF THE ',N,'th REINFORCEMENT=? READ ( *,*) HK WRITE (*,*)'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ (*,*) P IF (P.EQ.l) GO TO 9 WRITE (*,11) IB(M,3)=1 ZAV=ZAVIEH/57.2958 WRITE(9,5140) N,(IB(M,I),1=1,3),COS(ZAV),SIN(ZAV),0,HJ*HK,0

680 CONTINUE C C CALCULATION OF GRAVITY FORCE C 960 IF(NUMBAR .EQ. 0) WRITE (9,5140) NUMBAR

WRITE f * *) ******************************************************' DO 918 1=0,NUP-1 DO 919 J=1,NMP Xl(i*NMP+j)=0 Y(j)=HIZ*(j-l)/(NMP-l) X(I*NMP+J)=X(I*NMP+1)-(X(I*NMP+1)-X((I+1)*NMP))*Y(J)/HIZ

919 CONTINUE 918 CONTINUE

DO 980 1=0,NUP-1 DO 982 J=1,NMP Xl(i*NMP+j)=0 IFU.EQ.O .OR. I.EQ.l .OR. I.EQ.NUP-1 .OR. I.EQ.NUP-2) THEN Yl(I*NMP+j)=-l*TF*HIZ*GAM(l)/(NMP-1) IF(J .EQ. 1 .OR. J .EQ. NMP) Yl(I*NMP+j)=Y1(I*NMP+j)12 GO TO 982 ENDIF IF(I .gt. 1 .AND. I. It. NUP-2) THEN IF(J .EQ. 1) THEN Yl(I*NMP+j)= -1*(X( (I + 1)*NMP+J+D-X( (1-1) *NMP+J + 1) +X ( (I + l)*

Sc NMP+J)-X( (I-1)*NMP+J) ) *HIZ*GAM(2)/ (8* (NMP-1) ) GO TO 982 ENDIF IF(J .EQ. NMP) THEN Yl(I*NMP+j)= -1*(X((I+1)*NMP+J)-X((1-1)*NMP+J)+X((I+l)*NMP+J-1

&)-X((I-1)*NMP+J-1))*HIZ*GAM(2)/(8*(NMP-1)) GO TO 982

A88


ELSE Yl(I*NMP+j)= -1*(X((I+1)*NMP+J+1)-X((1-1)*NMP+J+1)+X((I+1)*NMP &+J-1)-X((1-1)*NMP+J-1))*HIZ*GAM(2)/(4*(NMP-1))

ENDIF ENDIF

982 CONTINUE 980 CONTINUE 979 WRITE(*,*)'*****************************************************

DO 1915 I=1,NNP,2 IF (I .EQ. NNP) THEN GO TO 850 ENDIF

1915 CONTINUE C C CALCULATION OF HYDROSTATIC FORCE C 850 WRITE(9,5010) NMP

DO 290 1=1,NMP HWW=HWIZ-(i-1)*HIZ/(NMP-1) FX(I)=-GW*HWW*HIZ/(NMP-1) IF (I .EQ. 1 .OR. I .EQ. NMP) FX(I)=FX(I)12 IF (FX(I) .GT. 0) FX(I)=0 XWW=(WBIZ-WT)/(NMP-1) FY(I)=-1*GW*HWW*XWW IF (I .EQ. 1 .OR. I .EQ. NMP) FY(I)=FY(I)/2 IF (FY(I) .GT. 0) FY(I)=0

290 CONTINUE DO 291 1=1,NMP,2 IF (I .EQ. NMP) THEN WRITE(9,5160) NNP-NMP+I,FX(I),FY(I) GO TO 857 ENDIF WRITE (9,5160) NNP-NMP+I,FX(I),FY(I),NNP-NMP+I+1,FX(I+l),FY(I+D

291 CONTINUE C C CALCULATION OF SILT FORCE C 857 WRITE(9,5010) NMP

DO 390 1=1,NMP HSS=HSIZ-(i-1)*HIZ/(NMP-1) FX(I)=-GSUB*HSS*HIZ*AKA/(NMP-1) IF (I .EQ. 1 .or. I .EQ. NMP) FX(I)=FX(I)12 IF (FX(I) .GT. 0) FX(I)=0 XWW=(WBIZ-WT)/(NMP-1) FY(I)=-1*GSUB*HSS*XWW IF (I .EQ. 1 .OR. I .EQ. NMP) FY(I)=FY(I)12 IF (FY(I) .GT. 0) FY(I)=0

39 0 CONTINUE DO 391 I=1,NMP,2 IF (I .EQ. NMP) THEN WRITE(9,5160) NNP-NMP+I,FX(I),FY(I) GO TO 858 ENDIF WRITE (9,5160) NNP-NMP+I,FX(I),FY(I),NNP-NMP+I+1,FX(I+l),FY(I+l)

391 CONTINUE C C EARTHQUAKE FORCE OR DISPLACEMENT C 858 IF (ZEL .EQ. 2) GO TO 1117

DO 299 I=NNP-NMP+1,NNP HWW=HWIZ-(1-1)*HIZ/(NMP-1) FX1(I)=-0.72 6*CP*ALFA*GW*HWW*HWW*HIZ/(NMP-1) IF (I .EQ. NNP-NMP+1 .OR. I .EQ. NNP) FX1(I)=FX1(I)12 IF (FX1(I) .GT. 0) FX1(I)=0

299 CONTINUE 820 WRITE(9,5010) NNP

DO 1919 1=1,NNP,2 X11(I)=ALFA*Y1(I) Y11(I)=ALFA*X1(I) Xll(I+1)=ALFA*Y1(I+l) Y11(I+1)=ALFA*X1(I+1)

A89


IF (I .GT. NNP-NMP) THEN X11(I)=FX1(I)+X11(I) X11(I+1)=FX1(I+1)+X11(I+1) ENDIF IF (I .EQ. NNP) THEN WRITE(9,5160) I,Xll(I),Yll(I) GO TO 1117 ENDIF WRITE(9,5160) I,Xll(I),Yll(I),I + l,Xll(I + l), Yll(I + l)

1919 CONTINUE C 1117 IF (ZEL .EQ. 1) GO TO 859

WRITE (*,*) 'DISPLACEMENTS OF BASE NODAL POINTS^' WRITE (9,5133) NUP DO 123 J=1,NNP+1-NMP,2*NMP

22 WRITE (*,*) 'DELTA(X) & DELTA(Y) OF NODE', J,'= READ (*,*) X1(I),Y1(I) IF (J .EQ. NNP+1-NMP) GO TO 860 WRITE (*,*) 'DELTA(X) & DELTA(Y) OF NODE', J+NMP,'= READ (*,*) Xll(I),Yll(I) WRITE (*,*) 'FOR CHANGING DATA TYPE 1 WRITE (*,*)'FOR CONTINUE TYPE 2 READ ( * , *) P IF (P.EQ.l) GO TO 22 WRITE (*,11)

860 IF (J .EQ. NNP+1-NMP) THEN WRITE(9,5160) J,XI(I),Yl(I) GO TO 859 ENDIF WRITE (9,5160) J,X1(I),Y1(I),J+NMP,X11(I) ,Y11(I)

123 CONTINUE C C VARIATION OF SEEPAGE FORCE C 859 WRITE (*,11)

WRITE ( * *) - j * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 1

WRITE (9,5010) 1 WRITE (*,*) 'NUMBER OF PHREATIC SURFACE SEGMENT END POINTS=?' READ (*,*) NWAT WRITE (9,5010) NWAT DO 124 J=1,NWAT WRITE (*,*) 'X-CORDINATE OF NODE', J,'= ' READ (*,*) XI(J) WRITE (*,*) 'PRESENT LEVEL (Y-COORDINATE) OF THE PHREATIC SURFACE & AT NODE', J,'= READ (*,*) Xll(J) WRITE (*,*) 'NEW LEVEL (Y-COORDINATE) OF THE PHREATIC SURFACE AT &NODE', J,'= READ (*,*) Yll(J) WRITE (*,11) WRITE(9,5165) XI(J),Xll(J),Yll(J)

124 CONTINUE 11 FORMAT {1111111111111111111111) 19 FORMAT (/) 5000 FORMAT(20A4) 5010 FORMAT(15I5) 5015 FORMAT(2F10.4) 5020 FORMAT(16I5) 5040 FORMAT(6I3,2X,1A3 0) 5060 FORMAT(40I2) 5080 FORMAT(II0,7D10.5) 5085 FORMAT(I10,3D10.5,I10) 5090 FORMAT(6D10.5) 5100 FORMAT(8D10.5) 5120 FORMAT(110,6D10.4) 5130 FORMAT(515,5D10.4) 5131 FORMAT(4D10.3) 5132 FORMAT(2D10.3) 5133 FORMAT(2I5) 514 0 FORMAT(4I5,2F10.5,2D10.5,F10.5) 5160 FORMAT(2(I10,2D10.2))

A90


5165 FORMAT(3D10.2) 522 0 FORMAT(3X,I4,lP2D14.5,4X,I4,2D14.5) 1050 RETURN

END C Q ****************************************************************** C * C * MAIN PROGRAM * C * * Q ******************************************************************

c COMMON /TWO/ GIS(900,5),B(2000),DF(2000) ,X(999),Y(999),PD(999) ,

& BM(900),ET(900),PP(999),DIX(999),DIY(999),IL(900,5),NA(2 000), Sc IC(200) ,NP(60) ,LE(30,2) ,KC(15,3) , NUT (15) , NUS (15) COMMON /THREE/ E(40),AO(40),FR(40),GAM(40),XB(40),BF(40),PI(40), 6c XP(40) ,CE(40) ,EN(40) ,TN(40) ,AL(40) , HC (4 0 ) ,UL(40) ,IDR(40) ,GUE(40) COMMON /FOUR/ BR(100,9),STS(199),STN(199),CJ(40),FJ(40),PJ(40), Sc TJ(40) ,SC,CSA,SNA,CM,DC,IB(100,3) ,ITP,IDT(40) ,STI(40) ,STF(40) , Sc SFK40) ,SFF(40) , INO, JDN(40) ,REJ(40) COMMON /FIVE/ SE(10,10),ST(3,10),HED(20),D(3,3),P(10),Q(4),STCR(3) Sc ,R,DEl,DE2,VOL,GAMW,PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1,SIG3,TEP,NN, NTP, Sc SLT,MOD(40,15) COMMON /SEVEN/ XW(30),FL(30),PL(30),SNL(2,4) COMMON /EIGHT/ 1D(8),IDS(900),N,MQ,NC,NQ,NSN,NX,NY,INT,ITD,NL,NMOD k ,NXY,NRT,IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP,IHZ,1ST,NUP,NBR,NMT, k NNP2,NSL,NOP COMMON /NINE/ NBEAM,NBTYP COMMON /TEN/ N1T,N2T,N3T,N4T DIMENSION VS(IOOOOO),HEDCS(15) OPEN (5,FILE='dam.in',STATUS='OLD') OPEN (6,FILE='dam2.out',STATUS='OLD') READ(5,1000) (HED(I),1=1,20)

10 00 FORMAT(20A4) READ(5,1010) NNP,NEL,NJT,NBR,NBEAM,NBTYP,NC,NMOD,INT,IHZ, ITD

1010 F0RMAT(15I5) READ(5,1015) GAMW,PATM

1015 FORMAT(2F10.4) READ(5,102 0) NMT,NSL,NAP,NCT,N1T,N2T,N3T,N4T,ITP,INO

1020 FORMAT(16I5) WRITE(6,20 00) (HED(I),1=1,20),NNP,NEL,NJT,NMT

2000 FORMAT(//1H1,4X,20A4,//10X, 'NUMBER OF NODAL POINTS = ' ,I10/10X, 'NU MBER OF ELEMENTS = ',I14/10X,'NUMBER OF INTERFACE ELEMENTS = ',14) WRITE(6,2010) NC

2010 FORMAT(1OX,'NUMBER OF LOADING STEPS = ',16) DO 2 1=1,NNP

2 IDS(I)=2 NDOF=0 DO 3 1=1,NNP

3 NDOF=NDOF+IDS(I) IDS(NNP)=NDOF+l-lDS(NNP) N1=NNP-1 DO 4 1 = 1, Nl J=N1+1-I

4 IDS(J)=IDS(J+1)-IDS(J) NNP2=NDOF DO 5 1=1,NMT MOD(I,1)=0

5 CONTINUE GO TO 140 DO 10 J=1,NC READ(5,1060) (MOD(I,J),1=1,NMT)

1060 FORMAT(40I2) 10 CONTINUE 140 WRITE(6,2080) 2080 FORMAT(//3X,'MATERIAL',7X,'GAMMA',5X,'COHESION',10X,"PI",4X, & "TEN. STRGTH',7X,'K0'/) DO 2 0 N=1,NSL READ(5,1080) IDR(N),GAM(N),CE(N),PI(N),TN(N),AO(N),XP(N),HC(N) 1080 FORMAT(110,7D10.5) READ(5,1090) UL(N),FR(N),EN(N),XB(N),BF(N),E(N),GUE(N),AL(N)

A91


1090 FORMAT(8D10.5) WRITE(6,2100) N,GAM(N),CE(N),PI(N),TN(N),AO(N)

2100 FORMAT(4X,I4,4X,4F13.2,F12.3) IDT(N)=0

2 0 CONTINUE READ(5,112 0) N,X(N),Y(N),PP(N),PD(N)

200 READ(5,1120) N,X(N),Y(N),PP(N) , PD (N) 1120 FORMAT(II0,6D10.4)

L=L+1 LM1=L-1 DUM=DFLOAT(N-LMl) DX=(X(N)-X(LM1))/DUM DY=(Y(N)-Y(LM1))/DUM DELP=(PP(N)-PP(LM1))/DUM DELT=(PD(N)-PD(LMl))/DUM LM1=L-1 X(L)=X(LM1)+DX Y(L)=Y(LM1)+DY PP(L)=PP(LM1)+DELP PD(L)=PD(LM1)+DELT L=L + 1 WRITE(6,2200)

2200 FORMAT(//1H1,4X, ' ** ERROR **: NODAL POINT DATA INPUT INCORRECTLY') STOP

300 WRITE(6,2220) 2220 FORMAT(///5X, 'COORDINATES OF NODAL POINTS'

k //11X, 'NODAL POINT',6X, 'X-COORDINATE',6X, 'Y-COORDINATE'/) DO 33 0 M=1,NNP WRITE(6,2240) M,X(M),Y(M)

2240 FORMAT(11X,15,2(9X,F7.3)) 33 0 CONTINUE 340 READ(5,1020) NY,NX,NXY,NRT

IM=NY+1 IN=NY+NX IO=IN+l IP=IN+NXY IOO=IP+NXY+l IPP=IOO+NRT-l IF(NX .EQ. 0) GO TO 460 READ(5,1020) (IC(N),N=IM,IN) WRITE(6,2300) (IC(N),N=IM,IN)

2300 FORMAT (//5X, 'NO X-MOVEMENT M0I5/18X, 10I5/18X, 10I5/18X, 1015 k /18X,10I5/18X,10I5)

420 DO 360 N=IM,IN IC(N)=IDS(IC(N))

3 60 CONTINUE 460 IF(NXY .EQ. 0) GO TO 510

READ(5,1020) (IC(N),N=IO,IP) WRITE(6,2320) (IC(N),N=IO,IP)

2320 FORMAT(//5X, "NO X OR Y MOVEMENT',10I5/18X, 10I5/18X,10I5/18X, 1015 k /18X,10I5/18X,10I5)

480 1=0 DO 490 N=IO,IP 1=1 + 1 IC(N)=IDS(IC(N)) IC(IP+I)=IC(N)+1

490 CONTINUE 510 IF(NRT.EQ.O) GO TO 520

READ{5,1020) (IC(N),N=IOO,IPP) WRITE(6,2130) (IC(N),N=IOO,IPP)

2130 FORMAT(5X,'NO Z-ROTATION',20I5/23X,20I5/23X, 2015) 490 DO 500 N=IOO,IPP

IC(N)=IDS(IC(N))+2 500 CONTINUE 520 IF(NEL.EQ.O) GO TO 660

N=0 540 READ(5,1020) M,(IL(M,I),1=1,5) 560 N=N+1

IF(NEL-M) 600,640,540 600 WRITE(6,2340) 2340 FORMAT(/1H1,4X,'*** ERROR *** : INCORRECT ELEMENT DATA INPUT')

A92


STOP 64 0 WRITE(6, 236 0)

2360 FORMAT(//5X,'ELEMENT DATA' & /5X, 'ELEMENT',8X, 'I',5X, 'J',5X, 'K',5X, 'L',4X, 'MATERIAL'/) DO 65 0 M=1,NEL

^nn WRITE(6,2380) M, IL (M, 1) , IL (M, 2 ) , IL (M, 3 ) , IL (M, 4 ) , IL (M, 5) 2380 FORMAT(8X,I4,3X,4I6,8X,I4,6X,I4) 650 CONTINUE 660 IF(NBR .EQ. 0) GO TO 700

DO 680 N=1,NBR READ(5,1140) M, (IB(M,I),1=1,3), (BR(M,I),1 = 1,5)

114 0 FORMAT(4I5,2F10.5,2D10.5,F10.5) 680 CONTINUE

WRITE(6,2400) 24 00 FORMAT(/5X,'REINFORCEMENT DATA'/'REINFORCEMENT',5X,'I',5X,'J",4X,

k "TYPE",5X, "PRESTRESS' ,5X, 'DISP TO'/4X, 'NUMBER',38X, 'ACTIVATE'/) WRITE(6,2420) (N,(IB(N,I),1=1,3),BR(N,3),BR(N,5),N=l,NBR)

2420 FORMAT(7X,I4,10X,I5,3X,I5,4X,1PD10.3) 700 CALL NDF

DO 710 N=1,NNP DIX(N)=0 DIY(N)=0


CALL EBTEDA(VS) DO 1001 MQ=1,NC WRITE(6,2440) MQ

2440 FORMAT(//5X, ****************************** */5x, 'STAGE NUMBER',13) DO 760 1=1,NMT

760 CONTINUE KSB=1 DO 780 I=1,NNP2 DF(I)=0

780 CONTINUE IF(KC(MQ,1).NE.3 .AND. KC(MQ,2).NE.3 .AND.KC(MQ,3).NE.3) GO TO 880 CALL SEEP

880 IF(KC(MQ,1).NE.5 .AND. KC(MQ,2).NE.5 .AND.KC(MQ,3).NE.5) GO TO 960 READ(5,1020) NCARDS WRITE(6,2560) NCARDS

2560 FORMAT(///5X,'THE FOLLOWING',13,' REINFORCEMENTS ARE ADDED' klI'REINFORCEMENT NUMBER',5X,'I',5X,'J',5X,'DISP. TO ACTIVATE'/) DO 900 N=l,NCARDS READ(5,114 0) M, (IB(M,I) ,1 = 1,3) , (BR(M,I),1 = 1,5) WRITE(6,242 0) M,(IB(M,I),1=1,2),BR(M,5) BR(M,6)=DIX(IB(M,1)) BR(M,7)=DIX(IB(M,2)) BR(M,8)=DIY(IB(M,1)) BR(M,9)=DIY(IB(M,2))

900 CONTINUE NBR=NBR+NCARDS CALL NDF

960 IF(KC(MQ,1).NE.8 .AND. KC(MQ,2).NE.8 .AND. KC(MQ,3).NE.8 .AND. k KC(MQ,1).NE.9 .AND. KC(MQ,2).NE.9 .AND. KC(MQ,3).NE.9) GO TO 999 WRITE(6,2600)

2600 FORMAT(///5X,'FORCE AND/OR DISPLACEMENT LOADING IS SPECIFIED FOR T &HIS INCREMENT'//3X,'NODE',8X,'X-LOAD',8X,'Y-LOAD',4X,'NODE',8X,"X-&LOAD',8X,'Y-&LOAD'/)

965 READ(5,1020) NUMNDE NCARDS=(NUMNDE-1)12 + 1 DO 980 1=1,NCARDS READ(5,1160) M,X1,Y1,N,X2,Y2 WRITE(6/2620) M,X1,Y1,N,X2,Y2

2 62 0 FORMAT(3X,I4,lP2D14.5,4X,I4,2D14.5) NY1=IDS(M)+1 NX1=NY1-1 DF(NX1)=DF(NX1)+X1 DF(NY1)=DF(NY1)+Y1 CONTINUE

999 NSP=NUS(MQ) NUP=NUT(MQ) NQ=1

A93


970 CALL TSSM(VS) CALL SSMILV(VS) CALL TANESH IF(NQ .GE. NUP) GO TO 1100 NQ=NQ+1 GO TO 970

1001 CONTINUE STOP END

0********************************************************************* SUBROUTINE EBTEDA(VS)

Q* ******************************************* * *********************** * COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999) ,PD(999) ,

k BM(900),ET(900),PP(999),DIX(999),DIY(999),IL(900,5),NA(2000), & IC(200),NP(60),LE(30,2),KC(15,3),NUT(15),NUS(15) COMMON /THREE/ E(40),AO(40),FR(40),GAM(40),XB(40),BF(40),PI(40), Sc XP(40) ,CE(40) ,EN(40) ,TN(40) ,AL(40) ,HC(40) ,UL(40) ,IDR(40) ,GUE(40) COMMON /FOUR/ BR(100,9),STS(199),STN(199),CJ(40),FJ(40) , PJ(40), k TJ(40) ,SC,CSA,SNA,CM,DC,IB(10 0,3),ITP,IDT(4 0) ,STI(4 0) ,STF(40) , k SFK40) ,SFF(4 0) , INO, JDN(40) ,REJ(40) COMMON /FIVE/ SE (10,10),ST (3,10),HED(2 0),D(3,3),P(10),Q(4) ,STCR(3) Sc ,R,DEl,DE2,VOL,GAMW,PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1,SIG3,TEP, NN, NTP, Sc SLT,MOD(4 0,15) COMMON /EIGHT/ ID(8) ,IDS(900) ,N,MQ,NC,NQ,NSN,NX,NY,INT,ITD, NL,NMOD Sc ,NXY,NRT, IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP, IHZ , 1ST, NUP, NBR, NMT, Sc NNP2,NSL,NOP COMMON /NINE/ NBEAM,NBTYP COMMON /TEN/ NIT,N2T,N3T,N4T DIMENSION VS(1) IF(NEL.EQ.O) RETURN MQ=1 NQ=1 NTP=0 NUP=1 NSP=1 KSB=1 DO 40 N=1,NEL DO 2 0 M=l,5 GIS(N,M)=0

2 0 CONTINUE MTP=IL(N,5) IF (MTP -GT. NSL) GO TO 160 GNU=AO(MTP)/(l+AO(MTP)) IF (GNU.GT.0.49 .AND. GNU.LE.0.5) THEN GNU=0.49 ELSE IF (GNU.GT.0.5 .AND. GNU.LT.0.51) THEN GNU=0.51 END IF IF (MTP.EQ.NAP .OR. MTP.EQ.NIT .OR. MTP.EQ.N2T .OR. MTP.EQ.N3T k .OR. MTP.EQ.N4T) THEN ET(N)=1 ELSE ET(N)=1.D5 END IF BM(N)=ET(N)/(3*(1-2*GNU)) GO TO 180

160 STS(N)=1.D8 STN(N)=1.D8

4 0 CONTINUE DO 60 N=1,NNP2 DF(N)=0

60 CONTINUE DO 80 N=1,NNP DIX(N)=0 DIY(N)=0 80 CONTINUE CALL TSSM(VS) CALL SSMILV(VS) CALL TANESH 240 NN=0

A94


INU=0 IF(NEL.EQ.O) GO TO 57 0 DO 12 0 N=1,NEL MTP=IL(N,5) IF(MTP .GT. NSL) GO TO 560 PPAVG=(PP(IL(N,1))+PP(IL(N,2))+PP(IL(N,3))+PP(IL(N,4)))*0.25*GAMW IF(IL(N,1) .EQ. IL(N,4)) PPAVG=(PP(IL(N,1))+PP(IL(N,2))

Sc +PP(IL(N,3) ) )*GAMW/3 IF(INT .EQ. 1) GO TO 280 DO 100 1=1,4 Q(I)=GIS(N,I)

100 CONTINUE CALL PSTMS GO TO 460

280 Q(4)=0 IFdDR(MTP) .EQ. 1) GO TO 300 IF(IHZ .EQ. 0) GO TO 3 80 GO TO 340

300 GIS(N,2)=GIS(N,2)-PPAVG IF(IHZ .EQ. 1) GO TO 320 GIS(N,l)=AO(MTP)*GIS(N,2) GO TO 4 00

320 GIS(N,1)=GIS(N,1)-PPAVG 340 DO 360 1=1,3

Q(I)=GIS(N,I) 3 60 CONTINUE

CALL PSTMS GO TO 460

380 GIS(N,l)=AO(MTP)*(GIS(N,2)-PPAVG)+PPAVG 400 GIS(N,3)=0

DO 410 1=1,3 410 Q(I)=GIS(N,I)

CALL PSTMS 4 60 CALL VSE

GIS(N,4)=Q(4) GIS(N,5)=DMAX1(SIG1,GIS(N,5))

120 CONTINUE 570 IF(NJT .EQ. 0) GO TO 640

INU=0 DO 62 0 N=1,NJT XC=(X(IL(N,1))+X(IL(N,2)))12 YC=(Y(IL(N,1))+Y(IL(N,2)))12 PPAVG=(PP(IL(N,1))+PP(IL(N,2)))*0.5*GAMW IF(INU .GT. 0) GO TO 580 WRITE(6,2040) (HED(I),1=1,20)

2040 FORMAT(//1H1,4X,20A4//5X,"INITIAL INTERFACE STRESSES' k //"ELEM NO',3X,"X",3X,"Y",2X,"NORM. STRESS',2X,'SHEAR STRESS",2X, Sc'NORM. STIFF',2X, 'SHEAR STIFF'/) INU=60 GO TO 600

580 INU=INU-1 600 WRITE(6,2060) N,XC,YC,(GIS(N,I),1=1,2),STN(N),STS(N)

2060 F0RMAT(I4,1X,2F7.2,1P5D12.3) 62 0 CONTINUE 64 0 IF(NBR .EQ. 0) GO TO 72 0

WRITE(6,2080) (HED(I),1=1,20) 2080 FORMAT(//1H1,4X,2 0A4 //5X,'INITIAL REINFORCEMENT STRESSES'

& //5X,'REINFORCMENT',5X,'I',5X,'J',4X,'TYPE',3X,'COMPR FORCE',3X, Sc • COMPRESSION' , 5X, ' STIFFNESS ' / ) DO 680 N=1,NBR MTP=IB(N,3) CALL SBE IF(INT .EQ. 0) GO TO 660 DC=0 CM=BR(N,3)

660 WRITE(6,2100) N,(IB(N,I),1=1,3),CM,DC,SC 2100 FORMAT(9X,I4,2(2X,I4),4X,14,1P3D14.6,0P2F10.5,6X,14) 680 CONTINUE 720 IF(INT.EQ.O) GO TO 760 755 INT=0 760 NTP=0

A95


RETURN END

p* ****************************************** * ************************ *

SUBROUTINE SSMILV(VS) p* ******************************************************************* *

COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999) ,Y(999) ,PD(999) , Sc BM(900) ,ET(900) ,PP(999) ,DIX(999) ,DIY(999) ,IL(900,5) ,NA(2000) , Sc IC(200),NP(60),LE(30,2),KC(15,3) , NUT (15) , NUS (15) COMMON /THREE/ E(40), AO (40),FR(40),GAM(40),XB(40),BF(40) , PI(40) , Sc XP(40) ,CE(40) ,EN(40) ,TN(40) ,AL(40) ,HC(40) ,UL(40) ,IDR(40) ,GUE(40) COMMON /FOUR/ BR(100,9),STS(199),STN(199),CJ(40),FJ(40),PJ(40), & TJ(40),SC,CSA,SNA,CM,DC,IB(100,3),ITP,IDT(40),STI(40),STF(40), Sc SFK40) ,SFF(40) ,INO,JDN(40) ,REJ(40) COMMON /FIVE/ SE(10,10),ST(3,10),HED(20),D(3,3),P(10),Q(4) , STCR(3) Sc ,R,DE1,DE2,V0L,GAMW,PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1,SIG3,TEP,NN,NTP, Sc SLT,MOD(40,15) COMMON /SEVEN/ XW(30),FL(30),PL(30),SNL(2 , 4 ) COMMON /EIGHT/ ID(8) ,IDS (900),N,MQ,NC,NQ,NSN,NX,NY,INT,ITD,NL,NMOD

Sc ,NXY,NRT,IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP,IHZ,IST,NUP,NBR,NMT, Sc NNP2,NSL,NOP COMMON /NINE/ NBEAM,NBTYP COMMON /TEN/ N1T,N2T,N3T,N4T DIMENSION VS(1) NEQ=NNP2 NEQQ=NEQ-1 ILL=1 NAJP=NA(1) DO 140 J=2,NEQ NAJ=NA(J) JK=NAJ-J IF=1-JK+NAJP IF(IF .GE. J) GO TO 120 IF1=IF+1 KF=JK+IF KL=NAJ-1 AA=0 DO 100 K=KF,KL NAI=NA(IF) CC=VS(K)/VS(NAI) AA=AA+VS(K)*CC VS(K)=CC IF=IF+1

100 CONTINUE VS(NAJ)=VS(NAJ)-AA

120 ILL=ILL+1 NAJP=NAJ

14 0 CONTINUE DO 160 N=1,NEQQ N1=N+1 I1=N1+1 KL=N NAIP=NA(N) DO 240 I=N1,NEQ NAI=NA(I) II=I1-NAI+NAIP 11=11+1 KL=KL+1 NAIP=NAI

240 CONTINUE DO 260 I=N,NEQ NAI=NA(I) B(I)=B(I)/VS(NAI!

2 60 CONTINUE J=NEQ NAJ=NA(NEQ) DO 32 0 I=1,NEQQ NAJP=NA(J-D JKA=NAJP+1 II=J-NAJ+JKA IF(II .GE. J) GO TO 300

A96


KL=J-1 BB=B(J) DO 280 K=II,KL B(K)=B(K)-VS(JKA)*BB JKA=JKA+1

280 CONTINUE 300 J=J-1

NAJ=NAJP 32 0 CONTINUE

RETURN END

0********************************************************************* SUBROUTINE TSSM(VS) C*********************************************************************

COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999),PD(999), Sc BM(900) ,ET(900) ,PP(999) ,DIX(999) ,DIY(999) ,IL(900,5),NA(2000), Sc IC(200) ,NP(60) ,LE(30,2) ,KC(15,3) , NUT (15) , NUS (15) COMMON /THREE/ E(40),AO(40),FR(40),GAM(40),XB(40),BF(40),PI(40), Sc XP(40) ,CE(40) ,EN(40) ,TN(40) ,AL(40) ,HC(40) ,UL(40) ,IDR(40) ,GUE(40) COMMON /FOUR/ BR(100,9),STS(199),STN(199),CJ(40),FJ(40),PJ(40), Sc TJ(40) ,SC,CSA,SNA,CM,DC, IB (100, 3) , ITP,IDT(40) , STI (40) ,STF(40) , Sc SFI (40) ,SFF(40) , INO, JDN(40) , RE J ( 4 0 ) COMMON /FIVE/ SE(10,10),ST(3,10),HED(2 0),D(3,3),P(10),Q(4),STCR(3) Sc ,R,DEl,DE2,VOL,GAMW,PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1,SIG3 , TEP,NN,NTP,

k SLT,MOD(40,15) COMMON /SEVEN/ XW(30),FL(30),PL(30),SNL(2,4) COMMON /EIGHT/ ID(8),IDS(900) ,N,MQ,NC,NQ,NSN, NX,NY, INT, ITD,NL,NMOD Sc ,NXY,NRT, IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP, IHZ , 1ST, NUP, NBR, NMT, Sc NNP2,NSL,NOP COMMON /NINE/ NBEAM,NBTYP COMMON /TEN/ NIT,N2T,N3T,N4T DIMENSION VS(1) DO 2 0 1=1,NSN VS(I)=0

2 0 CONTINUE IFL=0 IST=0 IF(NBR .EQ. 0) GO TO 160 DO 14 0 N=1,NBR IF(DABS(BR(N,4)-0) .LE. l.D-6) GO TO 140 ID(1)=IDS(IB(N,1)) ID(3)=IDS(IB(N,2)) ID(2)=ID(1)+1 ID(4)=ID(3)+1 MTP=IB(N,3) CALL SBE DO 120 1=1,4 IROW=ID(I) DO 12 0 J=l,4 ICOL=ID(J) IFdCOL .LT. IROW) GO TO 12 0 IADR=NA(ICOL)-(ICOL-IROW) VS(IADR)=VS(IADR)+SE(I,J)

120 CONTINUE 14 0 CONTINUE 160 IM=NY+NX+NXY+NXY+NRT

IF(INT .EQ. 1) GO TO 17 0 IF(KC(MQ,1).EQ.8 .OR. KC(MQ,2) .EQ.8 .OR. KC(MQ, 3) .EQ.8) GO TO 260

170 DO 240 M=1,IM J=IC(M) DF(J)=0 IF(J .EQ. 1) GO TO 200 ISTRT=NA(J-1)+1 IEND=NA(J)-1 IFdSTRT .GT. IEND) GO TO 2 00 DO 180 IADR=ISTRT,IEND VS(IADR)=0

18 0 CONTINUE 200 VS(NA(J))=1

IF(J .EQ. NNP2) GO TO 240

A97


JSTRT=J+1 KTR=0 DO 22 0 ICOL=JSTRT,NNP2 KTR=KTR+1 IADR=NA(ICOL)-KTR

24 0 CONTINUE 260 DO 280 I=1,NNP2

B(I)=DF(I) 28 0 CONTINUE

DO 400 M=1,IM J=IC(M) FDJ=DF(J) JSTRT=J+1 KTR=0 DO 3 60 ICOL=JSTRT,NNP2 KTR=KTR+1 IADR=NA(ICOL)-KTR CONTINUE

380 VS(NA(J))=1 4 0 0 CONTINUE

DO 420 M=1,IM J=IC(M) B(J)=DF(J)

42 0 CONTINUE RETURN END

£********************************************************************* SUBROUTINE NDF

£********************************************************************* COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999) ,PD(999) ,

Sc BM(900),ET(900),PP(999) ,DIX(999) ,DIY(999) ,IL(900,5),NA(2000), & IC(200),NP(60),LE(30,2),KC(15,3),NUT(15),NUS(15) COMMON /FOUR/ BR(100,9),STS(199),STN(199),CJ(40),FJ(40),PJ(40), S: TJ(40) ,SC,CSA,SNA,CM,DC, IB (100, 3) ,ITP,IDT(40) , STI (40) ,STF(40) , k SFK40) ,SFF(40) , INO, JDN(40) ,REJ(40) COMMON /EIGHT/ ID(8),IDS(900),N,MQ,NC,NQ,NSN,NX,NY,INT,ITD,NL,NMOD

Sc ,NXY,NRT,IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP,IHZ,IST,NUP,NBR,NMT, Sc NNP2,NSL,NOP COMMON /NINE/ NBEAM,NBTYP DO 10 J=1,NNP2 NA(J)=J

10 CONTINUE IF(NBR .EQ. 0) GO TO 180 DO 40 N=1,NBR ID(1)=IDS(IB(N,1)) ID(3)=IDS(IB(N,2)) ID(2)=ID(1)+1 ID(4)=ID(3)+1 IDMIN=ID(1) DO 2 0 1=2,4 IDMIN=MIN0(IDMIN, ID(I) )

2 0 CONTINUE DO 30 1=1,4 NA(ID(I) )=MIN0 (IDMIN,NA(IDd) ) )

3 0 CONTINUE 4 0 CONTINUE 18 0 IDIADR=1

DO 60 J=2,NNP2 IDIADR=IDIADR+J-(NA(J)-l) NA(J)=IDIADR

60 CONTINUE NA(1)=1 NSN=NA(NNP2) WRITE(6,202 0) NSN RETURN

2020 FORMAT(/////5X,'SIZE OF STIFNESS MATRIX = ',17) END

Q********************************************************************* SUBROUTINE ESM

n******************************************************************** * COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999) ,PD(999) ,

A98

RSDAM PROGRAM LISTING APPENDLX G

Sc BM(900) ,ET(900) ,PP(999) ,DIX(999) ,DIY(999) , IL (900 , 5) , NA (2000 ) , Sc IC(2 00) ,NP(60) ,LE(30,2) ,KC(15,3) , NUT (15) , NUS (15) COMMON /THREE/ E(40),AO(40),FR(40),GAM(40),XB(40),BF(40),PI(40), Sc XP(40) ,CE(40) ,EN(40) ,TN(40) ,AL(40) ,HC(40) ,UL(40) , IDR(40) ,GUE(40) COMMON /FIVE/ SE(10,10),ST(3,10),HED(2 0),D(3,3),P(10),Q(4),STCR(3) Sc ,R,DEl,DE2,VOL,GAMW,PATM COMMON /EIGHT/ ID(8),IDS(900),N,MQ,NC,NQ,NSN,NX,NY,INT,ITD,NL,NMOD Sc ,NXY,NRT,IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP,IHZ,IST,NUP,NBR,NMT, Sc NNP2,NSL,NOP COMMON /TEN/ NIT,N2T,N3T,N4T DIMENSION SS(4),TT(4) DATA SS/-1,1,1,-1/,TT/-1,-1,1,1/ MTP=IL(N,5) DFAC=ET(N)/(l-GUE(N)*GUE(N)) D(1,D=DFAC D(1,2)=GUE(N)*D(1,1) D(2,D=D(1,2) D(2,2)=D(1,1) D(3,3)=DFAC*(1-GUE(N))/2

15 DO 20 J=l,10 P(J)=0 DO 20 1=1,10 SE(I,J)=0

2 0 CONTINUE I=IL(N,1) J=IL(N,2) K=IL(N,3) L=IL(N,4) VOL=X13*Y24-X24*Y13 IF(VOL .LE. 0) RETURN IF(MTP .NE. NCT) GO TO 3 0 DTAVG=(PD(I)+PD(J)+PD(K)+PD(L))/4 IF(I .EQ. L) DTAVG=(PD(I)+PD(J)+PD(K))/3 DE1=(D(1,1)+D(1,2))*(1+GUE(N))*DTAVG*AL(MTP) DE2=(D(2,1)+D(2,2))*(1+GUE(N))*DTAVG*AL(MTP)

30 IF(1ST .EQ. D GO TO 120 DO 100 11=1,4 S=SS(II)*0.577 T=TT(II)*0.577 XJ=VOL+S*(X34*Y12 - X12*Y34)+T*(X23*Y14 - X14*Y23) XJAC=XJ/8 SM=1-S SP=1+S TM=1-T TP=1+T FAC=XJAC XS=0.25*(-TM*X(I)+TM*X(J)+TP*X(K)-TP*X(L)) YS=0.25*(-TM*Y(I)+TM*Y(J)+TP*Y(K)-TP*Y(L)) XT=0.25*(-SM*X(I)-SP*X(J)+SP*X(K)+SM*X(D) YT=0.25*(-SM*Y(I)-SP*Y(J)+SP*Y(K)+SM*Y(L)) XC=-2*(T*SM*SP*XS-S*TM*TP*XT)/XJAC YC= 2*(T*SM*SP*YS-S*TM*TP*YT)/XJAC DO 40 IM=1,3 D1=D(IM,1)*FAC D2=D(IM,2)*FAC D3=D(IM,3)*FAC

4 0 CONTINUE DO 60 IM=1,10 D1=ST(1,IM) D2=ST(2,IM) D3=ST(3,IM)

60 CONTINUE IF(INT .EQ. 0) GO TO 100 DUM=-GAM(MTP)*FAC P(2)=P(2)+0.25*DUM*SM*TM P(4)=P(4)+0.25*DUM*SP*TM P(6)=P(6)+0.2 5*DUM*SP*TP P(8)=P(8)+0.25*DUM*SM*TP P(10)=P(10)+DUM*SM*SP*TM*TP

100 CONTINUE IF(IST .EQ. 0) GO TO 160

A99


120 DO 140 IM=1,3 STCR(IM)=0 D1=D(IM,1) D2=D(IM,2) D3=D(IM,3) Tl=( Dl*Y24-D3*X24)/VOL T2=(-D1*Y13+D3*X13)/VOL T3=(-D2*X24+D3*Y24)/VOL T4=( D2*X13-D3*Y13)/VOL

14 0 CONTINUE RETURN

160 DO 180 NM=1,2 LM=10-NM MM=LM+1 SEMM=SE(MM,MM) DO 180 IM=1,LM DUM=SE(IM,MM)/SEMM P(IM)=P(IM)-DUM*P(MM) DO 180 JM=1,LM SE(IM,JM)=SE(IM,JM)-DUM*SE(MM,JM)

180 CONTINUE DO 200 IM=1,4 KM=IDS(IL(N,IM))+l JM=KM-1 MM=2*IM LM=MM-1 DF(JM)=DF(JM)+P(LM) DF(KM)=DF(KM)+P(MM)

2 00 CONTINUE RETURN END

p * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

SUBROUTINE SBE Q*********************************************************************

COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999),PD(999), Sc BM(900) ,ET(900) ,PP(999) ,DIX(999) ,DIY(999) ,IL(900,5),NA(2000), Sc IC(200),NP(60),LE(30,2),KC(15,3) , NUT (15) , NUS (15) COMMON /FOUR/ BR(100,9),STS(199),STN(199),CJ(40),FJ(40) , PJ(40) , & TJ(40),SC,CSA,SNA,CM,DC,IB(100,3) ,ITP,IDT(40),STI(40),STF(40) , Sc SFK40) ,SFF(4 0) , INO, JDN(4 0) ,REJ(40) COMMON /FIVE/ SE(10,10),ST(3,10),HED(20),D(3,3),P(10),Q(4),STCR(3) Sc ,R,DEl,DE2,VOL,GAMW,PATM COMMON /EIGHT/ ID(8),IDS(900) ,N,MQ,NC,NQ,NSN,NX,NY,INT,ITD,NL, NMOD Sc ,NXY,NRT,IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP,IHZ,IST,NUP,NBR,NMT, Sc NNP2,NSL,NOP I=IB(N,D I1=IDS(I)+1 12=11-1 J=IB(N,2) J1=IDS(J)+1 J2=J1-1 SNA=BR(N,2) CSA=BR(N,1) SC=BR(N,4) IF(MTP .EQ. 1) GO TO 120 DISPXI=DIX(I)-BR(N,6) DISPXJ=DIX(J)-BR(N,7) DISPYI=DIY(I)-BR(N,8) DISPYJ=DIY(J)-BR(N,9) IF(NQ .NE. NUP .OR. 1ST .NE. 1) GO TO 80 DISPXI=DISPXI-B(I2) DISPXJ=DISPXJ-B(J2) DISPYI=DISPYI-B(ID DISPYJ=DISPYJ-B(Jl)

80 IF (I.EQ.J) THEN COMPR=DISPXI*CSA+DISPYI*SNA ELSE COMPR=(DISPXI-DISPXJ)*CSA+(DISPYI-DISPYJ)*SNA END IF IF(MTP .EQ. 3) GO TO 100 IF(COMPR .LT. BR(N,5)-1.0D-6) SC=0

A100


GO TO 12 0 100 IF(-COMPR .LT. BR(N,5)-1.OD-6) SC=0 14 0 IF (I.EQ.J) THEN

DC=B(I2)*CSA+B(I1)*SNA ELSE DC=(B(I2)-B(J2))*CSA+(B(I1)-B(J1))*SNA END IF IF(DABS(BR(N,4)-0) .LT. l.D-6) DC=0 CM=BR(N,3)+DC*SC IF(NQ .EQ. NUP) BR(N,3)=CM RETURN END

Q********************************************************************* SUBROUTINE SIE

Q* ******************************************************************* * COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999),PD(999),

Sc BM(900) ,ET(900) ,PP(999) ,DIX(999) ,DIY(999) ,IL(900,5),NA(2000), Sc IC(200),NP(60),LE(30,2),KC(15,3) , NUT (15) , NUS (15) COMMON /FIVE/ SE(10,10),ST(3,10) ,HED(2 0) ,D(3,3),P(10),Q(4) ,STCR(3) Sc ,R,DEl,DE2,VOL,GAMW,PATM COMMON /EIGHT/ ID(8),IDS(900),N,MQ,NC,NQ,NSN,NX,NY,INT,ITD,NL,NMOD Sc ,NXY,NRT, IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP, IHZ , 1ST, NUP, NBR, NMT, Sc NNP2,NSL,NOP DIMENSION BC(4,4),COSIN(2,2) DO 2 0 J=l,8 DO 20 1=1,8 SE(I,J)=0

20 CONTINUE I=IL(N,1) J=IL(N,2) DELY=Y(J)-Y(I) DELX=X(J)-X(I) VOL=DSQRT(DELY* * 2 +DELX* *2 ) IF(VOL .LE. 0) RETURN CKS=STS(N)*VOL/6 CKN=STN(N)*VOL/6 DO 40 11=1,4 IN=2*II IS=IN-1 DO 4 0 JJ=1,4 JN=2*JJ JS=JN-1 SE(IS,JS)=CKS*BC(II,JJ) SE(IN,JN)=CKN*BC(II,JJ)

4 0 CONTINUE IF(DELY -EQ. 0) RETURN COSIN(1,1)=DELX/VOL COSIN(l,2)=DELY/VOL COSIN(2,l)=-COSIN(l, 2) COSIN(2,2)=COSIN(l,D DO 100 M=l,4 MT2=2*M DO 60 1=1,8 J=MT2-1 TEMP=SE(I,J) DO 60 K=l,2 SE(I,J)=TEMP*COSIN(1,K)+SE(I,MT2)*COSIN(2,K) J=J+1

60 CONTINUE DO 80 1=1,8 J=MT2-1 TEMP=SE(J,I) DO 80 K=l,2 SE(J,I)=TEMP*COSIN(1,K)+SE(MT2,I)*COSIN(2,K) J=J+1


RETURN END

Q******: SUBROUTINE VSE

:***************************************************************

A101


o********************************************************************* COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999),PD(999),

Sc BM(900) ,ET(900) ,PP(999) ,DIX(999) ,DIY(999) ,IL(900,5),NA(2000), Sc IC(200),NP(60),LE(30,2),KC(15,3) , NUT (15) , NUS (15) COMMON /THREE/ E(40),AO(40),FR(40),GAM(40) ,XB(40),BF(40) ,PI(40), Sc XP(40) ,CE(40) ,EN(40) ,TN(40) ,AL(40) ,HC(40) ,UL(40) ,IDR(40) ,GUE(40) COMMON /FIVE/ SE(10,10),ST(3,10),HED(2 0),D(3,3),P(10),Q(4),STCR(3) Sc ,R,DE1,DE2,V0L,GAMW,PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1,SIG3,TEP,NN,NTP,

Sc SLT,MOD(40,15) COMMON /EIGHT/ ID(8),IDS(900),N,MQ,NC,NQ,NSN,NX,NY,INT,ITD,NL,NMOD Sc ,NXY,NRT, IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP, IHZ , 1ST, NUP, NBR, NMT, Sc NNP2,NSL,NOP COMMON /TEN/ NIT,N2T,N3T, N4T IF(MTP.NE.NCT.AND.MTP.NE.NAP.AND.MTP.NE.NIT.AND.

ScMTP.NE.N2T.AND. DABS (XP (MTP)-0 ) . GT . 1 . D-5 . AND . ScMTP.NE.N3T.AND.MTP.NE.N4T) GO TO 20 ET(N)=E(MTP) BM(N)=ET(N)/(3*(1-2*GUE(MTP))) SL=0 Q(4)=0 GIS(N,5)=0 SLT=0

5 RETURN 20 IF (NQ .EQ. NUP) THEN

SG3=SIG3 ELSE SG3=(PSG3+SIG3)/2 RD=(((PSG1+SIG1)/2)-SG3)/2 END IF PH=PI(MTP)*1.745329251994329D-2 RDF=(2*CE(MTP)*DCOS(PH)+2*SG3*DSIN(PH))/(1-DSIN(PH)) IF(SG3 .GE. 0) GO TO 50 IF(DABS(SG3) .GE. TN(MTP)) GO TO 40 PHIT = DATAN2 (CE(MTP),TN(MTP)) RDFT = ((2*DSIN(PHIT))*(TN(MTP)+SG3))/(1-DSIN(PHIT)) RDF = DMINKRDFT, RDF) GO TO 50

40 RDF=2*RD 50 IF (RDF.GT.l.D-5) THEN

SL=(2*RD)/RDF ELSE SL=1 END IF MQM=MQ+NN IF (SG3.GT.0) THEN BM(N)=BF(MTP)*PATM*(SG3/PATM)**XB(MTP) IF (PI(MTP).GE.0.1) THEN STRSS=(SG3+CE(MTP)/DTAN(PH))/PATM ELSE STRSS=(SG3+CE(MTP)*5.73D2)/PATM END IF STST=SL*STRSS**0.2 5 SLT=Q(4)/STRSS**0.25 IF (SL.GE.l .AND. MOD(MTP,MQM).EQ.0) THEN ET(N)=E(MTP) ELSE IF (SL.GE.SLT .OR. MOD(MTP,MQM).EQ.1) THEN EI=HC(MTP)*PATM*(SG3/PATM)**XP(MTP) EI=DMAX1(EI,EN(MTP)) FRSL=FR(MTP)*SL IF (FRSL.LT.l) THEN ET(N)=((1-FRSL)**2)*EI ELSE ET(N)=E(MTP) ENDIF ELSE IF (SL.LE.0.75*SLT .OR. MOD(MTP,MQM).EQ.2) THEN EI=UL(MTP)*PATM*(SG3/PATM)**XP(MTP) ET(N)=DMAX1(EI,EN(MTP)) ELSE EI=HC(MTP)*PATM*(SG3/PATM)**XP(MTP) EI=DMAX1(EI,EN(MTP))

A102


FRSL=FR(MTP)*SLT IF (FRSL.LT.l) THEN ETL=((1-FRSL)**2)*EI ELSE ETL=E(MTP) ENDIF ETL=DMAX1(ETL,E(MTP)) EI=UL(MTP)*PATM*(SG3/PATM)**XP(MTP) ETH=DMAX1(EI,EN(MTP)) ET(N)=4*((SL-0.75*SLT)*ETL+(SLT-SL)*ETH)/SLT END IF ELSE IF (NQ.LT.NUP .AND. PSG3.GE.0) THEN BM(N)=BM(N)/4 ELSE BM(N)=0 END IF IF (DABS(SG3).GE.TN(MTP)) THEN IF (NQ.LT.NUP .AND. PSG3.GE.0) THEN ET(N)=ET(N)/4 ELSE ET(N)=E(MTP) END IF STST=Q(4) SLT=1 ELSE STST=SL*STRSS**0.25 SLT=Q(4)/STRSS* * 0.2 5 IF (FRSL.LT.l) THEN ETL=((1-FRSL)**2)*EN(MTP) ELSE ETL=E(MTP) ENDIF ETL=DMAX1(ETL,E(MTP)) ETH=EN(MTP) ET(N)=((SL-0.75*SLT)*ETL+(SLT-SL)*ETH)/(0.2 5*SLT) END IF END IF END IF Q(4)=DMAX1(Q(4),STST) SLT=DMAX1(SLT,SL) ET(N)=DMAX1(ET(N),E(MTP)) BM(N)=DMIN1(BM(N),1.7D1*ET(N)) SIG5=GIS(N,5) IF (SIG1.GT.SIG5 .AND. PI(MTP).GT.2.2) THEN BM(N)=DMAX1(BM(N),(2-DSIN(PH))*ET(N)/(3*DSIN(PH))) ELSE IF (SIG1.LE.SIG5 .AND. PI(MTP).GE.1.6) THEN TEMPV=(1-DSIN(PH))*(5-5**DSIN(PH)) BM(N)=DMAX1(BM(N),(4+TEMPV)*ET(N)/(3*(4-TEMPV))) ELSE BM(N)=1.7D1*ET(N) END IF RETURN END

0* ******************************************************************* * SUBROUTINE STIE

0* ******************************************************************* * COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999),PD(999),

Sc BM(900) ,ET(900) ,PP(999) ,DIX(999) , DIY (999 ) , IL (9 00 , 5) , NA (2 000 ) , Sc IC(2 00) ,NP(60) ,LE(30,2) ,KC(15,3) , NUT (15) , NUS (15) COMMON /FOUR/ BR(100,9),STS(199),STN(199),CJ(40) , FJ(40),PJ(40),

Sc TJ(40) ,SC,CSA,SNA,CM,DC,IB(10 0,3) ,ITP,IDT(40) ,STI(40) ,STF(40) , Sc SFI(40) ,SFF(40) , INO, JDN(40) ,REJ(40) COMMON /FIVE/ SE(10,10),ST(3,10),HED(20),D(3,3),P(10),Q(4) ,STCR(3)

& ,R,DEI,DE2,VOL,GAMW,PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1,SIG3,TEP,NN,NTP,

Sc SLT, MOD (4 0,15) I=IL(N,1) J=IL(N,2) MTP=IL(N,5) SLN=DSQRT(DELX**2+DELY**2)

A103


CSN=DELX/SLN SNE=DELY/SLN DO 2 0 K=l,4 K1=K*2 K2=K1-1 L=IL(N,K) L1=IDS(L)+1 L2=L1-1 P(KD=-B(L2) *SNE+B(L1) *CSN P(K2)= B(L2)*CSN+B(L1)*SNE

2 0 CONTINUE RDN=0.5*(P(8)-P(2)+P(6)-P(4)) RDS=0.5*(P(7)-P(1)+P(5)-P(3)) DO 30 11=1,3

30 Q(II)=GIS(N,II) 60 AA = 0.5

IF(INT .EQ. 1 .OR. NQ .EQ. NUP) AA=1 Q(l) = GIS(N,1) - AA*STN(N)*RDN IF (INT.EQ.l .AND. JDN(MTP).EQ.1) Q(1)=Q(1)-PPAVG IF (INT.EQ.l .AND. IHZ.EQ.O) THEN Q(2)=0 ELSE Q(2) = GIS(N,2) - AA*STS(N)*RDS END IF

100 IF (MTP.EQ.INO .OR. MTP.EQ.ITP) THEN SL=0 SLT=0 Q(3)=0 GOTO 150 END IF IF (Q(l).GT.-TJ(MTP)) THEN STN(N)=SFI(MTP) PH=PJ(MTP)*1.74532925D-2 IF (Q{1).GT.0) THEN SHRST=CJ(MTP)+Q(1)*DTAN(PH) IF (PJ(MTP).GE.0.1) THEN STRSS=(Q(1)+CJ(MTP)/DTAN(PH))/PATM ELSE STRSS=(Q(1)+CJ(MTP)*5.73D2)/PATM END IF ELSE SHRST=(TJ(MTP)+Q(D)*CJ(MTP)/TJ(MTP) END IF SL=DABS(Q(2))/SHRST STST=SL * STRSS * * 0.2 5 Q(3)= DMAX1(Q(3),STST) SLT=Q(3)/STRSS**0.25 MQM=MQ+NN IF (NQ.LT.NUP .AND. GIS(N,1).GT.-TJ(MTP)) THEN STN(N)=DMAX1(SFF(MTP),STN(N)*REJ(MTP) ) STS(N)=DMAX1(STF(MTP),STS(N)*REJ(MTP)) ELSE STN(N)=SFF(MTP) STS(N)=STF(MTP) END IF END IF

150 IF (NQ -NE. NUP) GO TO 210 DO 200 11=1,3

200 GIS(N,II)=Q(ID 210 IF (INT.EQ.l) RETURN 230 IFdTD .LE. 0 .AND. NQ .NE. NUP) RETURN

WRITE(6,2000) N,XC,YC,Q(1),Q(2),STN(N),STS(N) RETURN

2000 F0RMAT(I4,2F8.2,1P4D13.3) END c********************************************************************* SUBROUTINE SEEP c********************************************************************** COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999),Y(999),PD(999), & BM(900),ET(900),PP(999),DIX(999),DIY(999),IL(900,5),NA(2000), Sc IC(200),NP(60),LE(30,2),KC(15,3),NUT(15),NUS(15)

A104


COMMON /THREE/ E(40),AO(40),FR(40),GAM(40),XB(40) , BF(40) , PI(40 ) , Sc XP (4 0 ) , CE (4 0 ) , EN (4 0 ) , TN ( 4 0 ) , AL (4 0 ) , HC (4 0) , UL (4 0 ) , IDR (4 0 ) , GUE (40) COMMON /FIVE/ SE(10,10),ST(3,10),HED(20),D(3,3),P(10),Q(4) ,STCR(3)

Sc , R, DEI, DE2, VOL, GAMW, PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1,SIG3,TEP,NN,NTP,

Sc SLT, MOD (40,15) COMMON /SEVEN/ XW(3 0),FL(3 0),PL(30),SNL(2 , 4) COMMON /EIGHT/ ID(8),IDS(900),N,MQ,NC,NQ,NSN,NX,NY,INT,ITD,NL,NMOD

k ,NXY,NRT, IFL,KSB,MTP,NAP,NCT,NSP,NEL,NJT,NNP,IHZ,1ST,NUP,NBR,NMT, Sc NNP2,NSL,NOP COMMON /TEN/ NIT,N2T,N3T,N4T WRITE(6,2000)

2000 FORMAT(///5X,'SEEPAGE LOADING IS SPECIFIED FOR THIS INCREMENT') READ(5,1000) NCODE

1000 FORMAT(16I5) IF(NCODE .NE. 0) GO TO 60 DO 40 N=1,NNP PTEM=PP(N)+PD(N) IF(PTEM .GE. 0) GO TO 2 0 PD(N)=-PP(N) PP(N)=0 GO TO 40

2 0 PP(N)=PTEM 4 0 CONTINUE

GO TO 2 00 60 READ(5,1000) NWAT

WRITE(6,1020) NWAT 1020 FORMAT(6D10.2)

READ(5,1020) (XW(I),PL(I),FL(I),1=1,NWAT) WRITE(6,2040) (XW(I),PL(I),FL(I),1=1,NWAT)

2 04 0 FORMAT(5X,F10.2,7X,F10.2,3X,F10.2) DO 180 N=1,NNP DO 8 0 1=2,NWAT IF(DABS(X(N)-XW(I)) .LT. l.D-5) GOTO 100 IF(X(N) .LT. XW(I)) GO TO 120

80 CONTINUE GO TO 140

100 TFL=FL(I) TPREL=PL(I) GO TO 140

120 IM1=I-1 DELX=DABS(XW(I)-XW(IM1) ) DELF=FL(I)-FL(IM1) DELP=PL(I)-PL(IM1) DX=DABS(X(N)-XW(IM1)) TFL=(DX/DELX)*DELF+FL(IM1) TPREL=(DX/DELX)*DELP+PL(IM1)

14 0 PD(N)=TFL-TPREL IF(TFL .LT. TPREL) GO TO 160 IF(Y(N) .GT. TFL) PD(N)=0 IF(Y(N) -GE. TPREL .AND. Y(N) .LE. TFL) PD(N)=TFL-Y(N) GO TO 180

160 IF(Y(N) .GE. TPREL) PD(N)=0 IF(Y(N) -GE. TFL .AND. Y(N) .LT. TPREL) PD(N)=Y(N)-TPREL

180 PP(N)=PP(N)+PD(N) 200 DO 300 N=1,NEL

MTP=IL(N,5) ,mrs _ IF(MTP GT. NSL .OR. MTP .EQ. NCT.OR.MTP.EQ.NAP .OR. MTP .EQ. Sc N1T.OR.MTP.EQ.N2T.OR.MTP.EQ.N3T.OR.MTP.EQ.N4T) GO TO 300 DO 210 1=1,2 DO 210 J=l,4 SNL(I,J)=0

210 CONTINUE II = IL(N,D JJ=IL(N,4) DO 220 1=1,4 n ^r 24 0 H=(SNL(1,1)+ SNL(1,2)+SNL(1,3)+ SNL(1, 4))* 0 . 2 5 V=(SNL(2,D+SNL(2,2)+SNL(2,3)+SNL(2,4))*0.25 IF(II -NE. JJ) GO TO 260 H=(SNL(1,1)+SNL(1,2)+SNL(1,3)+SNL(1, 4)) /3 V=(SNL(2,l)+SNL(2,2)+SNL(2,3)+SNL(2,4))/3

A105


260 DO 280 J=l,4 IF(II .EQ. JJ .AND. J .EQ. 4) GO TO 300 J2=IDS(IL(N,J))+l J1=J2-1 DF(J1)=DF(J1)+H DF(J2)=DF(J2)+V

280 CONTINUE 3 00 CONTINUE

WRITE(6,2060) 2060 FORMAT(//5X,'THE CUMULATIVE EQUIVILENT NODAL FORCES GENERATED AT T

ScHE SPECIFIED DEGREES'/5X, ' OF FREEDOM TO SIMULATE THE SPECIFIED PHR ScEATIC LEVEL CHANGES FOLLOW' //5X, ' NODE ', 8X, ' X-FORCE ', 8X, ' Y-FORCE '// ) DO 32 0 1=1,NNP IY=IDS(I)+1 IX=IY-1 IF(DABS(DF(IX)).LT. l.D-4 .AND. DABS(DF(IY)).LT. l.D-4) GO TO 320 WRITE(6,2080) I,DF(IX),DF(IY)

2080 F0RMAT(5X,I4,1P2D15.6) 32 0 CONTINUE

RETURN END

p* ******************************************************************* *

SUBROUTINE PSTMS Q*********************************************************************

COMMON /FIVE/ SE(10,10),ST(3,10),HED(20),D(3,3),P(10),Q(4),STCR(3) Sc , R, DEI, DE2, VOL, GAMW, PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1,SIG3,TEP,NN,NTP, Sc SLT,MOD(40,15) CNTR=(Q(l)+Q(2))/2 HL=(Q(l)-Q(2))/2 RD=DSQRT(HL**2+Q(3)**2) SIG1=CNTR+RD SIG3=CNTR-RD RETURN END

Q**************************************** ******************** ********* SUBROUTINE TANESH

Q* ************************************** ****************************** COMMON /TWO/ GIS(900,5),B(2000),DF(2000),X(999) ,Y(999),PD(999),

Sc BM(900) ,ET (900) , PP (999) , DIX (999) , DIY (999) , IL (900,5 ) ,NA (2000) , & IC(200) ,NP(60) , LEO 0,2) , KC (15 , 3 ) , NUT (15) , NUS (15 ) COMMON /THREE/ E(40),AO(40),FR(40),GAM(40),XB(40),BF(40),PI(40), Sc XP(40) ,CE(40),EN(40) ,TN (40 ) , AL (40 ) , HC ( 40 ) , UL (40 ) ,IDR(40) ,GUE(40) COMMON /FOUR/ BR(100,9),STS(199),STN(199),CJ(40),FJ(40),PJ(40), & TJ(40),SC,CSA,SNA,CM,DC,IB(100,3),ITP,IDT(40),STI(40),STF(40), & SFK40) ,SFF(40) ,INO,JDN(40) ,REJ(40) COMMON /FIVE/ SE(10,10),ST(3,10),HED(20),D(3,3) , P(10),Q(4),STCR(3) Sc , R, DEI, DE2, VOL, GAMW, PATM COMMON /SIX/ RD,SL,RDF,HL,PH,SG3,PSG1,PSG3,SIG1, SIG3 , TEP,NN,NTP, Sc SLT, MOD (40, 15) COMMON /NINE/ NBEAM,NBTYP COMMON /TEN/ N1T,N2T,N3T,N4T IF(INT .EQ. 1 -OR. NQ .LT. NUP) GO TO 120 INU=0 J=l DO 100 N=1,NNP IX=IDS(N) IY=IX+1

20 DIX(N)=DIX(N)+B(IX) DIY(N)=DIY(N)+B(IY)

4 0 IFdNU .GT. 0) GO TO 6 0 WRITE(6,2000) MQ,NQ,NUP

2000 FORMAT(//5X,'DISPLACEMENT RESULTS FOR STAGE',13,4X,'ITERATI ScON' 12 ' OF',I2//5X, 'NODAL',5X, 'X',7X,'Y',9X, 'TOTAL',9X, 'TOTAL',9X Sc, 'PORE' /5X, 'POINT' ,25X, ' UX' ,12X, ' UY ' , 10X, 'PRESS' ) INU=1000 GO TO 80

60 INU=INU-1 80 CONTINUE

WRITE(6,2020) N, X (N) , Y(N) , DIX(N),DIY(N),PP(N) 2020 FORMAT(5X,I5,2F8.2,1P2D14.5,OPF11.2)

A106


100 CONTINUE 120 IST=1

NN=1 IF(NQ .NE. NUP .OR. MQ .EQ. NC .OR. KSB .LT. NSP) NN=0 INU=0 IF(NEL.EQ.O) GO TO 610 DO 600 N=1,NEL MTP=IL(N,5)

6 00 CONTINUE 610 IF(NJT .EQ. 0) GO TO 680

INU=0 DO 660 N=1,NJT IF(INT .EQ. 1) GO TO 64 0 IF(ITD .LE. 0 .AND. NQ .NE. NUP) GO TO 640 IF(INU .GT. 0) GO TO 62 0 WRITE(6,2080) MQ,NQ,NUP

2080 FORMAT(//5X, 'INTERFACE ELEMENT RESULTS FOR STAGE',13 , 4X, Sc' ITERATION' ,12, 'OF' ,12//'ELEM NO' ,2X, 'X' ,5X, 'Y' ,2X, 'NORMAL STRESS' Sc,2X, "SHEAR STRESS" ,2X, 'NORMAL STIFF', 2X, ' SHEAR STIFF'/) INU=200 GO TO 64 0

620 INU=INU-1 640 CALL STIE 6 60 CONTINUE 680 IF(NBR .EQ. 0 .OR. INT .EQ. 1) GO TO 740

IFdTD .LE. 0 .AND. NQ .NE. NUP) GO TO 700 WRITE(6,2100) MQ,NQ,NUP

2100 FORMAT(//5X, 'REINFORCEMENT RESULTS FOR STAGE',13 , 4X, Sc' ITERATION' ,12, ' OF' ,I2//5X, 'REIN. NUM. ' ,4X, 'I' ,4X, ' J' ,4X, 'TYPE' , Sc3X, "COMPR FORCE' , 4X, ' INCR COMPR', 5X, ' STIFFNESS'/)

700 DO 720 N=1,NBR MTP=IB(N,3) CALL SBE IF(ITD .LE. 0 .AND. NQ .NE. NUP) GO TO 720 WRITE(6,212 0) N,(IB(N,I),1=1,3),CM,DC,SC

2120 FORMAT(9X,14,2(2X,I4),4X, 14,1P3D14.6,0P2F10.5,6X, 14 ) 720 CONTINUE 740 RETURN

END

A107

aspects of design and analysis of reinforced soil dams(thesis)

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fundamentals of design

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