Soil Mechanics and Foundations

• • SOILM AND F5(j}!ANICS DATIONS R A U C SE A FO SOIL MECHANICS AND FOUNDATIONS MUNI BUDHU JOHN WilEY & SONS, I...

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• •

SOILM AND F5(j}!ANICS DATIONS

R A U C SE A

FO

SOIL MECHANICS AND FOUNDATIONS MUNI BUDHU

JOHN WilEY & SONS, INC. "lew York / Chichestcr IlVeillheim I Brisballe I 5111gopore I TorOIlfO

Editor Wayne Anderson Marketing Manager Katherine Hepburn Semor Production Manager Lucille Buonocore Production Editor Leslie Surovick Cover Designer Lynn Rogan Illustration Editor Sigmund Malinowski Illustration Studio Radianl Illustration II:- Design Cover Photo CORBISlRogu Wood Th is book was set in 10112 Times Ten by UG I GGS Informati pnnted and bound by RR DonnelJey/Wiliard . The cover was p Corporation. This book is printed on acid-free paper.

e

The paper in this book was manufactu red by a include sustained yield harvesting of tts timbedan principles ensure that the numbers of of new growth.

tf'J''t';'
in a remeval system or Ira.nsmillcd • scanning United States permission o( tile Publisher. or , per-copy fee to the Copyright MA 01923. (SOB) 750-8400. fu for permission should be addressed 10 the &. Sam. Inc_. 605 Third Avenue, New York. NY fax (2!2) 850-6008, e-mail: PERMREQ®WILEY.COM . I l·8O(j-CALL· WILEV (225·5945). LiI',• ., .,t,~" g)"'C","I,>g".g ill Publicalio" Dala: s.;:1 n"'~!>!;:'" 'nd foundations I by Munt Budhu.

p. ISBN 0-47!·25131·X (aU::. paper) 1. Soil mechanu;s. 2. Foundations. I. Tilte.

TA710.B765 1999 624.1'513622] Printed in the United States of Amenca

10 9 8 1 6 5 4 3 2

99-050184

F

PREFACE

• Professor Hilary 1. Inyang, University of Massachusetts-Lowell • Professor Derek Morris, Texas A&M University • Professor Cyrus Aryani, California State University

111181

v

PREFACE

• Professor Shobha K. Bhatia, Syracuse University • Major Richard L. Shelton, United States Military Academy • Professor Colby C. Swan, University of Iowa • Professor Panos I(jousis, University of Arizona • Professor Carlos Santamarina, Georgia Institute of Tech • Dr. William Isenhower, Ensoft, Inc. Mr. Wayne Anderson and his staff, and Leslie S Sons were particularly helpful in getting this book done. to my wife and children who have contributed 1!mifican this book.

FO

vi

NOTES for Instructors I would like to present some guidance to assist you i graduate geotechnical engineering courses.

FO

DESCRIPTION OF CHAPTERS

tlement is interpreted and the parameter required to determine time rate of settlement. The oedometer test is described and procedures to determine the various parameters for settlement calculations are presented. Chapter 5 deals with the shear strength of soils and the tests (laboratory and field) required for its determination. The Mohr-Coulomb failure criterion is discussed using the student's background in strength of materials (Mohr 's circle) and in statics (dry friction). Soils are treated as a dilatant-frictional material

vii

viii

NOTES FOR INSTRUCTORS

rather than the conventional cohesive-fric tional materifll. Typical stress- strai n responses of sand and clay are presented and discussed. The implications of drained and undrained conditions on the shear strength of soil s are discussed. Laboratory and field tests to determine the shear strength of soils are described. Chap ter 6 deviates from traditional undergraduate textboqk-topics by dealing with soil consolidation and strengt h as separate issues. I~'i.his chap\er, deformation and strength are integrated within the framewor k 0 b ritical stilte soil mechanics using a simplifi ed version of the modified c m ay d!!!#fhe emphasis is on understanding the mechanical behavior 0 ~ls ra her than present· ing the mathematical form ul ation of critical state soil mec a~cs M the modified cam-clay model. The amoun t of mathematics is k e~o the ~!p jmum needed for un derstand ing and clarification of im portant co ceptS!--P rojection geomet ry is used to illustrate the different responses 0 soils wh the "'loading changes under drained and undrained loading. Althoug: thisap e eals with a simplification and an idealization of real soils, the real neftt i a simple framework, which allows the stlldent to think abou possJbte i sponses if can itions change from those originally conceive sis usuau n e gineeri ng ~tice It also allows them to bette r interpret soil test re ults. ,I Chapter 7 deals wi bearing cae.aci1Y and seJ.t'ement of footings. Here bearing capacity and seuleme.pl are tteated as a i n~e tO~ iC. n the design of foundations. the geotechnical ngineer must be sa ' sfied at th earing capaci ty is sufficient and ~ttlem nl at working load is lera . In deed. for most shallow foo tings. it is settlf lT].ent that gover -Iii 1 n. ot bearing capaci ty. Limit equilibrium analysis is int roduced to illu trate th e method that has been used to fi n the pop~tieari n g cap}' Ity eq u ~ ion.s..and LO make use of the SIlldent's ba' tgr nd in statics (equ ili rium) to mtrod uce a simple but powerful analytical t~o . Three sets of bearin capacity equations (T erzaghi as modified by Vesic, M~yerh of. and Sk pton) , lh inft uence of groundwater level , and ntri c load'ton bearing apacity are discussed. T hese equations are simpli fi ed b breaking them down 1m catego ries-one relating to dra ined conditions. the o r er to undra ipe~ condi ·ons~laS{ic. one-dimensional consolidation. and Sk: ~pton and Bj erQ!...nLs m thod'of determin ing settlement are presented . T he elastic methodA~ findingSe-tt ement is based on work done by Gazettas (1985), who descri1)ed p blems a ociated with the Janbu, Bjerrum , and Kj aernali (1956) m 1\od that conventionally quoted in textbooks. P ile fo dalio are described and discussed in C hapter 8. Methods for finding beari ng ~pa city and sett le ment of single and group piles are presented. Chapter 9 is about [wo-dimensional steady sta te flow th rough soils. Solutions to two-dimensional flow using flow nets and the finite difference technique' are discussed . Emphases are placed on seepage. pore wate r pressure, and instability. This chapter normally comes early in most current textbooks. The reason for placing this chapter here is because two-di mensional flow influences the stability of earth structures (retaining walls and slopes), discussion of wh ich follows in Chapters 10 and 11. A student would then be able to make the practica l connection of two-di mensional flow and stability of geotechn ical systems readi ly. Lateral earth pressures and their use in the analysis of earth retaining systems and excavat ions are presen ted in Chapter 10. Gravit)' and flexible re laining walls. in addition \0 reinforced soil walls, arc discussed. Guidance is provided as to what strength parameters to use in d rained and undrained cond itions.

~

a

~

twi>

NOTES FOR INSTRUCTORS

ix

Chapter 11 is about slope stability. Here stability conditions are described based on drained or undrained conditions. An appendix (Appendix A) allows easy access to frequently used typical soil parameters and correlations.

F

CHAPTER LAYOUT

dvent 0 ersonal computers, learning has become more visual. Some , studies hav epo[ ed that visual images have improved learning by as much as 400% . This tex-t ok is accompanied by a CD ROM that contains text, interactive animation, images, a glossary, notation, quizzes, notepads, and interactive problem solving. It should appeal, particularly, to visual learners. A quiz is included in appropriate chapters on the CD ROM to elicit performance and provide feedback on key concepts. Interactive problem solving is used to help students solve problems similar to the problem-solving exercises. When an interactive problem is repeated, new values are automatically generated. Sounds are used to a limited extent. The CD ROM contains a virtual soils laboratory for the students to conduct geotechnical tests. These virtual tests are not intended to replace the necessary hands-on experience in a soil laboratory. Rather , they complement the hands-on experience, prepare the students for the real experience, test relevant prior knowledge of basic concepts for the interpre-

X

NOTES FOR INSTRUCTORS

tation of the test results, guide them through the evaluation and interpretation of the results, allow them to conduct tests that cannot otherwise be done during laboratory sessions, and allow them to use the results of their tests in practical applications.

ABET REQUIREMENTS The United States Accreditation Board for Engi (ABET) has introduced new criteria for accreditation p this book has the author's judgment on how it satisfies ence (ES) and engineering design (ED) crit mended percentages allocated to ES and

FO

COURSE MATERIAL

NOTES for Students and Instructors

FO

PURPOSES OF THIS BOOK echanics and its appli-

technical engineering. The goals of this te as follows:

o characterize soil properties, cs to analyze and design simple geo-

ng this textbook you should be able to: • Descr be soi and determine their physical characteristics such as, grain size, wate ~ ontent, and void ratio • Classify soils • Determine compaction of soils • Understand the importance of soil investigations and be able to plan a soil investigation • Understand the concept of effective stress • Determine total and effective stresses and pore water pressures • Determine soil permeability • Determine how surface stresses are distributed within a soil mass • Specify, conduct, and interpret soil tests to characterize soils

xi

xii

NOTES FOR STUDENTS AND INSTRUCTORS

• Determine soil strength and deformation parameters from soil tests, for example, Young's modulus, friction angle and undrained shear strength • Discriminate between "drained" and "undrained" conditions • Understand the effects of seepage on the stability of strucrure Estimate the bearing capacity and settlement of structures unded on soils Analyze and design simple foundations • Determine the stability of earth structures, for examFi slopes

et '

in~

Distribution of Main Topi

Foundation and earth structures Description

6

7

8

9

10

11

0

0

0

0

0

0

FO

0

• •

Stresses in soils

D

Drained and undrained conditions

U

Settlement and deformation

C

Shear strength

T

Bearing capacity and settlement of foundations

0

Stability of earth structures

N

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.

0

0

•

Seepage

0

0

• •

0

0

0

•

0

0

0 0

• •

NOTE S FOR STUDENTS AND INSTRUCTORS

xiii

ASSESSMENT You will be assessed aD how well you absorb and use the fundamentals of soil mechanics. Three areas of assessment are incorporated in the Exercise sections of this textbook. The first area called "Theory" is intended fog y-ou to demon· strate your knowledge of the theory and extend it to uncovec1!Cwt elat ionships. The questions under "Theory" will help you later in your cf.~eec to address unconven tional issues using fundamental principles. The selo n a cea ca~ d " Problem Solving" requires you to apply the fundamenta prin i les and--concepts to a wide variety of problems. These problems willtes o ur un erstandi ng and use of the fundamental principles and concep ts. The thirtl ea caUl d " Practical" is intended to create practical scenarios for you to use "a t 'o.n~y the subject matter in the specific chapter but prior mate rial/ i"h ya.u "havl' encountered. These problems try to mimic some aspects of r~tuatl \fls and give you a feel for how the materials you have studied so far t"~ be pplied 'o practice. Communications ese "Pract ica l" are. at leas!. as important as the technic detail~ 1n many 0 prob lems you are placed in a sit atlOn I co iv' ce stakeholders f you r technical competence. A q uiz (multip) choice) on each chapter' • d ud~d in the CD \0 tcst you r general knowledge of t subj ct matter in that chapter: h,e questions on the quiz are re lated to the sec Io n Question.!J6 Guide Your eading,"' included in each chapter. ~ ~

SUGGESTIONS F R PRO.B EM

SOL\lING~

Enginee ring is, ~ ,a bout propl em 5"' I~ n ~ r most engineering problems, there i~~ nique method or procedure for fi nding solutions. Often, there is no unique .s~ u Ion to an engineering ~;~bl~~A suggested problem-solving pro~ cedure is ~t1ined be low.

,

1. Rea/ the

prObleJ.1l...c a ref~ UY; ate or wrile down what is given and what you

~ required to find.

2. Draw clear diagrams or e tches wherever possible. 3. Devise trategy tOtn the solution. De termine what principles, concepts, and.rq ua tl ns a e needed to solve the problem. 4. p.lrfocm calcuJations making sure that you are using the correct units. 5. Chcc 'hethl r your results are reasonable. The units of measure ment used in this textbook follow the SI system. Engineering calculations are approximations and do not result in exact numbe rs. A ll cal<:u lations in th is book are rounded. at the most. to two decima l places except in some exceptional cases, for example. void ratio.

SUGGESTIONS FOR USING TEXTBOOK AND CD-ROM This textbook is accompanied by and integrated with a C D-ROM. Not all sections of the textbook are covered in the CD-ROM. The textbook provides significantly more details on the subject matter than the CD-ROM. The CD-ROM

xiv

NOTES FOR STUDENTS AND INSTRUCTORS

provides animations, interactive problem solving, quizzes, virtual laboratories, special modules (for example, a computer program to find stresses within a soil) , spreadsheets, videos, a notepad, a glossary, a list of notations, and a calculator. CD icons in the textbook have Inset numbers that are intend1...d to alert you to sp~cial features present on the CD-ROM. The numbers hv ~ following meamng: 1. Interactive animation 2. Virtual lab

~

3. Interactive problem solving 4. Spreadsheet 5. Video

6. Computer program utility

PRESENTATION

F

normally given any projects, the

SHEAR STRENGTH OF SOILS • Resistance to shear forces • Coulomb's law Now, let us start our study of soil mechanics and foundations.

CONTENTS

CHAPTER 1

INTRODUCTION TO SOIL MECHANICS A

1.0 1.1 1.2

FO

CHAPTER 2

Introduction 1 Marvels of Civil Engineering-The Hi :d Geotechnical Lessons from Failu es 4

PHYSICAL CHARACTE INVESTIGATIONS 6

2.0 2.1 2.2 2.3

Introduction

22

oarse-Grained Soils 22 t

2.6 2.7

2.8 2.9 2.10

ation of Soils Based on Particle Size 25 29 31

Fall Cone Method to Determine Liquid and Plastic Limits 32 2.7.4 Shrinkage Limit 33 Soil Classification Schemes 36 Engineering Use Chart 39 One-Dimensional Flow of Water Through Soils 42 2.10.1 Groundwater 42 2.10.2 Head 42 2.10.3 Darcy's Law 44 2.10.4 Empirical Relationships for k 45 2.10.5 Flow Parallel to Soil Layers 49 2.10.6 Flow Normal to Soil Layers 50

xv

xvi

CONTENTS

2.11

2.12

51

Dry Unit Weight-Water Content Relationship 2.12.1 Basic Concept 58 2.12.2 Proctor Compaction Test 58 2.12.3 Zero Air Voids Curve 59 2.12.4 Importance of Compaction 60 2.12.5 Field Compaction 60 2.12.6 Compaction Quality Control 2.12.6.1 Sand Cone 61 2.12.6.2 Balloon Test

2.13 2.12.2 2.13.3 2.13.4 2.13.5 2.13.6 2.13.7 2.13.8

6

2.14 Exercises

FO

0

3.

3. 3.3

81

~

3.4

3.5

3.6

3.7

3.3.3 hear Stresses and Shear Strains 83 Idealized Stress-Strain Response and Yielding 84 3.4.1 Material Responses to Normal Loading and Unloading 84 3.4.2 Material Response to Shear Forces 86 3.4.3 Yield Surface 87 Hooke's Law 88 3.5.1 General State of Stress 88 3.5.2 Principal Stresses 89 3.5.3 Displacements from Strains and Forces from Stresses 89 Plane Strain and Axial Symmetric Conditions 90 3.6.1 Plane Strain 90 3.6.2 Axisymmetric Condition 91 Anisotropic Elastic States 94

FO

CONTENTS

3.8

3.9

3.10 3.11

Stress and Strain States 96 3.8.1 Mohr's Circle for Stress States 96 3.8.2 Mohr's Circle for Strain States 98 Total and Effective Stresses 100 3.9.1 The Principle of Effective Stress 100 3.9.2 Effective Stresses Due to Geostatic Stress Fie ' s 3.9.3 Effects of Capillarity 103 3.9.4 Effects of Seepage 104 Lateral Earth Pressure at Rest 109 Stresses in Soil from Surface Loads 110 3.11.1 Point Load 111 3.11.2 Line Load 112 3.11.3 3.11.4 3.11.5 3.11.6 3.11.7 3.11.8

xvii

02

3.12

3.13

133

4.0 4.1 4.2 4.3

',on 141 Definitions of Key Terms 142 Questions to Guide Your Reading 143 Basic Concepts 144 4.3.1 Instantaneous Load 145 4.3.2 Consolidation Under a Constant Load-Primary Consolidation 145

4.3.3 4.3.4 4.3.5 4.3.6 4.3.7

Secondary Compression 146 Drainage Path 146 Rate of Consolidation 147 Effective Stress Changes 147 Void Ratio and Settlement Changes Under a Constant Load 148

xviii

CONTENTS

4.4

4.3.8 Effects of Vertical Stresses on Primary Consolidation 148 4.3.9 Primary Consolidation Parameters 149 4.3 .10 Effects of Loading History 150 4.3.11 Overconsolidation Ratio 151 4.3.12 Possible and Impossible Consolidation Soil States Calculation of Primary Consolidation Settlement 152 4.4.1 Effects of Unloading/Reloading of a Soil Samp e Taken am the Field 152 4.4.2 Primary Consolidation Settlement of Fine-Grained Soils 153 4.4.3 Primary Consolidation Settlement of 0 e Grained Soils 153 4.4.4 4.4.5

4.5

4.5.3 4.6 4.7 173

FO

~ O

~

~

4.8 4.9

De rrnination of the Secondary Compression Index 79

R 1"a . nship Between Laboratory and Field Consolidation 182 Typica-:l al~J of Consolidation Settlement Parameters and Empirical

Relatio~ps

183

4.10 Sand Drains 184 4.11 Lateral Earth Pressure at Rest Due to Overconsolidation 4.12 Summary 188 Practical Examples 189 Exercises 195

CHAPTER 5

5.0 5.1

SHEAR STRENGTH OF SalLS 199 Introduction 199 Definitions of Key Terms 200

188

FO

CONTENTS

5.2 5.3

5.4 5.5 5.6 5.7 5.8

5.9 5.10

xix

Questions to Guide Your Reading 200 Typical Response of Soils to Shearing Forces 201 5.3.1 Effects of Increasing the Normal Effective Stress 203 5.3.2 Effects of Overconsolidation Ratio 205 5.3.3 Cemented Soils 206 Simple Model for the Shear Strength of Soils Using 206 Interpretation of the Shear Strength of Soils Mohr-Coulomb Failure Criterion 213 Undrained and Drained Shear Strength 2 Laboratory Tests to Determine Shear Streng 5.8.1 Shear Box or Direct Shear £' 21 5.8.2 5.8.3 5.8.4 229 5.8.5 5.8.6 Pore Water Pres sur 240

5.11

249 252

MODEL TO INTERPRET SOIL BEHAVIOR

6.0 6.1 6.2 6.3

Definitions of Key Terms 263 Questions to Guide Your Reading 264 Basic Concepts 264 6.3.1 Parameter Mapping 264 6.3.2 Failure Surface 265 6.3.3 Soil Yielding 266 6.3.4 Prediction of the Behavior of Normally Consolidated and Lightly OverconsoJidated Soils Under Drained Conditions 267 6.3.5 Prediction of the Behavior of Normally Consolidated and Lightly Overconsolidated Soils Under Undrained Condition 269

XX

CONTENTS

6.3.6

6.4

6.5 6.6 6.7

6.8 6.9 6.10

Prediction of the Behavior of HeaVIly Overconsolldated Soils 270 6.3.7 Critical State Boundary 271 6.3.8 Volume Changes and Excess Pore Water Pressures 271 6.3.9 Effects of Effective Stress Paths 272 Elements of the Critical State Model 273 6.4.1 Yield Surface 273 6.4.2 Critical State Parameters 273 6.4.2.1 Failure Line in (q, p') Spa 27'3 6.4.2.2 Failure Line in (q, e) Space. 275 Failure Stresses from the Critical Sta le Model 6.5.1 Drained Triaxial Test 277 6.5.2 Undrained Triaxial Test 1,9 Soi l Stiffness 287 Strains from the Cntical Stat 6.7. 1 Volumetric Strains 6.7.2 Shear Strains 91 Calculated Stress-51 81 ResponfF 296 6.8.1 Drained Co mpr~ '[tits 296 6.8.2 Undrain Compr n Tests 2f>7 K..-Consoli ared i\ 'nse 304 R elatio ~i Between Simple Soil Tests, ~ itic S e Parameters, and Soil STrength 307 6.10. 1 ndrained Shear Strengt h6f'(~l'ays~ar-f e Liquid and Plastic bimi ~ 307 .' - ' "<'~,., Vertical Effective resses a he ;':lquid and Plastic Limits ~7

6.1.3 6.1.4

ti.lO.5

7.0 7.1 7.2 7.3

7.4 7.5

7.6

Undrained Shear 51 th- ertica l Effective Stress Relatio 'mit 308 Compr ibili J~dicies (A and Cd and Plasticity Index 308 Undtai ned ear S ength, Liquidity Index, and Sensitivity 0 09

In troduction 318 Definitions of Key T erms 319 Questions to Guide Your Reading 320 Basic Concepts 321 7.3.1 Colla pse and Failure Loads 321 7.3.2 Fail ure Surface 322 ollapse Load from Limit Equilibrium 323 Bearing Capacity Equalions 325 7.5. 1 Terzaghi's Bearing Capacity Equations 325 7.5.2 Skempton's Bearing Capacity Equalion 327 7.5.3 Meyerhors Bearing Capacity Equation 327 Choice or Bearing Capacity Equations and Friction Angles 328

CONTENTS

7.7

7.8 7.9 7.10 7.11 7.12

7.13

Allowable Bearing Capacity and Factor of Safety 329 7.7.1 Calculation of Allowable Bearing Capacity 329 7.7.2 Building Codes Bearing Capacity Values 330 Effects of Groundwater 330 Eccentric Loads 336 Bearing Capacity of Layered Soils 339 Settlement 340 Settlement Calculations 342 7.12.1 Immediate Settlement 343 7.12.2 Primary Consolidation Settlement Determination of Bearing Capacity and Sett e oarse-Grained Soils from Field Tests 350 7.13 .1

7.13.2 7.13.3 7.14 Horizontal Elastic Dis 7.15 Summary 358 Practical Examples 358 Exercises 365 CHAPTERS

8.0 8.1 8.2 8.3

1

FO

374

8.7 8.8 8.9 Consolidation Settlement Under a Pile Group 400 8. 10 Procedure to Estimate Settlement of Single and Group Piles 401 8.11 Piles Subjected to Negative Skin Friction 404 8.12 Pile-Driving Formula and Wave Equation 406 8.13 Summary 408 Practical Example 408 Exercises 411

CHAPTER 9

9.0 9.1

xxi

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS 415 Introduction 415 Definitions of Key Terms 416

xxii

CONTENTS

Questions to Guide Your Reading 416 Two-Dimensional Flow of Water Through Porous Media 417 Flow Net Sketching 419 9.4.1 Criteria for Sketching Flow Nets 419 9.4.2 Flow Net for Isotropic Soils 420 9.4.3 Anisotropic Soil 421 Interpretation of Flow Net 422 9.S 9.5.1 Flow Rate 422 9.5.2 Hydraulic Gradient 423 9.5.3 Static Liquefaction, Heaving, Boiling, 423 9.5.4 Critical Hydraulic Gradient 4 3 9.5.5 Pore Water Pressure Distribu(t ----..·,9.5.6 Uplift Forces 424 9.6 Finite Difference Solution for 'D 429 9.7 Flow Through Earth Dams 43 9.8 Summary 440 Practical Example 440 Exercises 442

9.2 9.3 9.4

CHAPTER 10

10.0 10.1 10.2 10.3 10.4 10.5

Definiti

FO

fir Total Stress Analysis 462 art Pressures to Retaining Walls 465 nd Modes of Failure 467

472

10.10

482 484

10.11 Braced Excavation 496 10.12 Mechanical Stabilized Earth Walls 500 10.12.1 Basic Concepts 500 10.12.2 Stability of Mechanjcal Stabilized Earth Walls 502

CONTENTS

10.13 Summary 510 Practical Examples 510 Exercises 516

SLOPE STABILITY 522

CHAPTER 11

11.0 11.1 1l.2 11.3 11.4

Introduction 522 Definitions of Key Terms 523 Questions to Guide Your Reading 523 Some Types of Slope Failure 524 Some Causes of Slope Failure 525 1l.4.1 Erosion 525 11.4.2 Rainfall 525 1l.4.3 Earthquakes 525 11.4.4 Geological Features 1l.4.5 External Loadin 11.4.6

11.4.7 11.5 11 .6 11 .7 11.8

550

FO

551

~

P.

~APPENDIXB APPENDIXC

DISTRIBUTION OF SURFACE STRESSES WITHIN FINITE SOIL LAYERS 562 LATERAL EARTH PRESSURE COEFFICIENTS (KERISEL AND ABSI, 1970) 566 ANSWERS TO SELECTED PROBLEMS 571 REFERENCES 573 INDEX 578

xxiii

Notation Note: A prime (') after a notation for stress denotes A

effecti~

stress.

Area

B

Width

Co

Cohesion Compression index Recompression index Horizontal coefficient of consolidation

Hydra · cen coefficient a

ctivityor ~eability

ctive lateral ear
Critic

D

Diam

o

o

Df

F

Dr

Ending bearing stress Skin friction

N

Standard penetration number

N c , N q, Ny

Bearing capacity factors

n

Porosity

NCL

Normal consolidation line

OCR

Overconsolidation ratio

p

Mean stress

q

Deviatoric stress or shear stress

qa q,

Allowable bearing capacity

Factor of safety

G

xxiv

Shear modulus

Surface stress Ultimate bearing capacity

Mobilization factor for ct> Mobilization factor for

Modulus of volume compressibility

5"

qu

Flow rate Flow, quantity of flow

Specific gravity

End bearing or point resistance

Pressure head

Skin or shaft friction

Elevation head

Point bearing resistance

Head

Ultimate load capacity

Height

Ultimate group load capacity

FO

NOTATION

Temperature correction factor

XXV

Deflection or settlement I:

Normal strain

Resultant lateral force

Volumetric strain

Overconsolidation ratio with respect to stress invariants

Deviatoric stil'ain

Undrained shear strength

5

Degree of saturation

51 5PT T

Sensitivity Standard penetration test Sliding force or resistance Time factor

u

Pore water pressure

u

Average degree of consolidation

UC

Uniformity coefficient

URL

Unloading/reloading

1

Velocity

v Wall friction coefficient Poisson's ratio Elastic settlement Primary consolidation Psc

Secondary consolidation settlement

cr

Normal stress

T

Shear stress Critical state shear strength Shear strength at failure

w..

Peak shear strength

z

Residual shear strength

ex

Dilation angle

Velocity potential

Slope angle Adhesion factor

Rotation of principal plane to the horizontal

Skin friction coefficient for drained condition

Stream potential

Plastification angle for piles

R A U C SE A D O E N M LY I C

FO

CHAPTER

1

INTRODUCTION TO SOIL MECHANICS AND FOUNDATIONS 1.0 INTRODUCTION

~ateria1. o~anceslOrs

Soil is the oldest and most cOIllp,le ' ngine used soils as construct ion material to%uild burial s· es, fl ood pro lt ctio ,and shellers. Western civilization credits tHe omans {or recogn izing the-imp~tance of soils in the stability of structures. Roma enjinee rs. espec· alJy VhruviUs 0 served du ri ng the reign of Empt'fOr-A ugust In the first ce ury B.C pai great attention to soil types (s nd, gr v';' e t;")and to the esign and co truction of solid founda tions. The as no tieore ical basis for design; ex rie nce from trial and error was relied upon. ,,____ Coulomb tt 773) is credited as the iirst ftC rson to use mechanics to solve soil problems. H, as a/fllember of ~nch' ~al Engineers, who were inte rested rOlecting old fortresses that fen asHy from cannon fi re. To protect the (ortre es'fr~rtillery attack, loping maf scs of soil were placed in front of them (Fig..l). The enemy had to tu LbeMw the soi l mass and the fortress to attack. O f c rse, the ene then became an easy target. The mass of soil applies a ateral force to the fa s ;"at'!:cUld cause it to topple ove r or could cause it tCi:llifle--away from ~sod ass. ulomb attempted to determine the lateral ree so that he coifld evaluate th sta bi lity of the fortress. He postulated that a wedge of soil AB Fig. 1 would fa il along a slip plane Be and [his wedge wou ld push tlie II out or overtopple it as it moves down the slip plane. Mo t'meot 0 the edge along the slip plane would occu r only if the soil resistan e ~ng the edge were overcome. Coulomb assumed that the soil resista nce is pro ided by friction between the particles and the problem became one of a wedge s wing on a rough (frictional) plane. which you may have analyzed in your physics or mechanics course. Cou lomb has tacitly defined a fail ure crilerio n for soils. Today, Coulomb's failure criterion and method of analysis still preva il. From the ea rly 20th century. the rapid growth of cities. industry, and commerce required a myriad of building systems: for example, skyscrapers, large publ ic build ings. dams fo r electric power generation and reservoirs for water supply and irrigation. lUnnels. roads and railroads, pon and harbor facilities, bridges, airpo rts and runways, mining activities, hospita ls, sanitation systems, drainage systems, and towers for communication systems. These bui lding systems requ ire stable and economic fo undations and new questions about soils were asked. For example, what is the state of st ress in a soi l mass, how can one design safe and economic foundations, how much wou ld a bu ilding settle, and what is

1

2

CHAPTER 1 INTRODUCTION TO SOIL MECHANICS AND FOUNDATIONS

A

Soil mass for protecti on of the fortress

Unprotected fortre ss that was felled easily by cannon fire

FIGURE 1.1

Slip plane

Unprotected and protected fortress .

FO

ontinue to ask these v confronted us. Some of

1.1 MARVELS OF CIVIL ENGINEERINGTHE HIDDEN TRUTH The work that geotechnical engineers do is often invisible once construction is completed . For example, four marvelous structures-the Sears Tower (Fig. 1.2), the Empire State Building (Fig. 1.3), the Taj Mahal (Fig. 1.4), and the Hoover

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R

1.1 MARVELS OF CIVIL ENGINEERING-THE HIDDEN TRUTH

FIGURE 1.4 Taj Mahal. (© Will & Deni Mclntyre/Photo Researchers .)

3

4

CHAPTER 1 INTRODUCTION TO SOil MECHANICS AND FOUNDATIONS

r.e-alQ>-\?fJ;~eG la ~n ~ron,

F~ OM

1.2

U.S. Department of

FAILURES

&t uctures that , e foun eel earth rely on our ability to design safe and r CQflomic found a-Gons. Becml e f the natural vagaries of soils, failures do occur. . orne failures hav bee ca strophic and caused severe damage to lives and properties· thers hav be n insidious. Failures occur because of inadequate site and soil inves . ations unforeseen soil and water conditions; natural hazards; poor engineering n:} sis, design, construction, and quality control; postconstruction activities; and usage outside the design conditions. When failures are investigated thoroughly, we obtain lessons and information that will guide us to prevent similar types of failure in the future . Some types of failure caused by natural hazards (earthquakes, hurricanes, etc.) are difficult to prevent and our efforts must be directed toward solutions that mitigate damages to lives and properties. One of the earliest failures that was investigated and contributed to our knowledge of soil behavior is the failure of the Transcona Grain Elevator in 1913 (Fig. 1.6). Within 24 hours after loading the grain elevator at a rate of about 1 m of grain height per day, the bin house began to tilt and settle. Fortunately, the structural damage was minimal and the bin house was later restored . No borings were done to identify the soils and to obtain information on their strength.

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IS;.

5

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1.2 GEOTECHNICAL LESSONS FROM FAILURES

more to the development of this science than is generally recognized ." We have come a long way in understanding soil behavior since its fatherhood by Terzaghi in 1925. We continue to learn more daily though research on and experience from failures and your contribution to understanding soil behavior is needed. Join me on a journey of learning the fundamentals of soil mechanics and its applications to practical problems so that we can avoid failures or, at least, reduce the probability of their occurrence.

CHAPTER

2 -------------------------,

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

) ES ED 90 10

2.0

INTRODUCTION

The purpose of this chapter is to inLt u ~ou 115. You will learn some basic descriptions of sOLis and some fundamen,lal p.hysical soi l oper(ies that you should retain for fu ture use ;n this [ l(t a~dj,in geotechnical engmee g ractice. Soils. derived from the weathering o f y, are \'ery cCf1 plex materia and va ry widely. There is no certain . ha a so I will have th arne pro rl ies within a few centimeters of its curre L at.i,9n. , One of the primary tasKS of a geotec hnical eng' eer a collect, classify, and investig'l le ttie PhYSiCa~1 pr Denies of sgil this apter, we will deal wit h descript ions of soil'S tests t dete rmine the J2.hYS~ ·cal propert ies of soils, soil classification, one-dime io fl ow of w er thrOugh o· s, and methods of soi l invest igation ~all:t soils lnvestigati ns are c ndu ted only all a fraction of n proposed sit ea it would be prohl itively $.Xpensive to conduct an extensive Investigation f a whole sile. e then ha~ make est imat es and judgmcnfs sed o n infor alion from lim 'ted set o f observations and field and labora tory les \:lata. n-.e ngineering, ~ isasse bl omplex systems into parts and then study each p art and its re ationship to e wh ole. We will do the same for soils. Soils will,!)e dismantled in to th ree st ituents and we will exa mine how the proporions of eachl onstl ent s;.hllJ cterize soils. When you complete this chapter, you "1'..." should be ' Ie to: ......... • • • • • •

Describe and classify soils Determme particle size distribution in a soil mass Determine the proportions of the main constituents in a soil Determine index properties of soils Detennine the rate of flow of water through soils Determine maximum dry unit weight and optimum water conlent Plan a soi l investiga tion

Sample Practical Situation A highway is pl"Oposed to link the city of Noscu t to the village of Windsor Fores!. The highway rou te will pass through a terrai n Ihat is relallve!y flat and is expected to be flooded by a 100 year SlOrm even t.

6

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2.1 DEFINITIONS OF KEY TERMS

7

Degree of sa r (ion (5) is the ratio of the volume of water to the volume of void. Bulk unit weight ('Y) is the weight density, that is, the weight of a soil per unit volume. Saturated unit weight ('Ysat) is the weight of a saturated soil per unit volume. Dry unit weight ('Yd) is the weight of a dry soil per unit volume. Effective unit weight ('Y') is the weight of soil solids in a submerged soil per unit volume. Relative density (Dr) is an index that quantifies the degree of packing between the loosest and densest state of coarse-grained soils. Effective particle size (D 10 ) is the average particle diameter of the soil at 10 percentile; that is, 10% of the particles are smaller than this size (diameter).

8

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

A verage particle diameter (Dso) is the average particle diameter of the soil. Liquid limit (wLd is the water content at which a soil changes from a plastic state to a liquid state. Plastic limit (WPL) is the water content at which a soil changes fro , to a plastic state . Shrinkage limit (wsd is the water content at which a soil c a to a semisolid state without further change in volun . Groundwater is water under gravity in excess of th pores. Head (H) is the mechanical energy per unit weig 1. Coefficient of permeability (k) is a propo flow velocity of water through soils.

2.2

soil tests required to characterize soils? t ?

FO

l4'? 1' b';/ Ul,V'"

U. What is a s 1 in vestigation?

13. How do you plan a soil investigation?

14. What are the effects of water on the unit weight of soils? 15. What factors affect the compaction of soils?

2.3

COMPOSITION OF SOILS 2.3.1 Soil Formation Soils are formed from the physical and chemical weathering of rocks . Physical weathering involves reduction of size without any change in the original composition of the parent rock. The main agents responsible for this process are

2.3 COMPOSITION OF SOILS

9

exfoliation, unloading, erosion , freezing , and thawing. Chemical weathering causes both reductions in size and chemical alteration of the original parent rock. The main agents responsible for chemical weathering are hydration, carbonation, and oxidation. Often, chemical and physical weathering take place in concert. Soils that remain at the site of weathering are called resid · R soils. These soils retain many of the elements that comprise the parent r ck. Alluvial soils, also called fluvial soils, are soils that were transported by river and strea s. The composition of these soils depends on the environmen del' hich t ey were transported and is often different from the parent r k. rofil of alluvial soils usually consists of layers of different soils. Muc our ities has been and is occurring in and on alluvial soils. 01 'al ils are soils that were transported and deposited by glaciers. M oils deposited in a marine environment.

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2.3.2 Soil Types

• Glacial till is a soil that consists mainly of coarse particles. • Glacial clays are soils that were deposited in ancient lakes and subsequently frozen. The thawing of these lakes reveals a soil profile of neatly stratified silt and clay, sometimes called varved clay. The silt layer is light in color and was deposited during summer periods while the thinner, dark clay layer was deposited during winter periods. Gypsum is calcium sulphate formed under heat and pressure from sediments in ocean brine. • Lateritic soils are residual soils that are cemented with iron oxides and are found in tropical regions.

F

10

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOilS AND SOil INVESTIGATIONS

• Loam is a mixture of sand, silt, and clay that may contain organic material. • Loess is a wind blown, uniform fine-grained soil. • Mud is clay and silt mixed with water into a viscous fluid.

2.3.3 Clay Minerals soil.

• and (a) Single

Aluminum

(c) Single octahedrons

0

= Silicon

(b) A tetrahedral

o

and

= Oxygen or Hydroxyl 0 = Aluminum (d) Octahedral sheet

(a) Silica tetrahedrons, (b) silica sheets, (c) single aluminum octahedrons, and (d) aluminum sheets.

FIGURE 2.2

2.3 COMPOSITION OF SOILS

'---'

Silica sheet

L-l""'_~:':"-. Hydrogen bonds

(a) Kaolinite

FIGURE 2.3

-...

O

F

~

'+-

11

Sili ca sheet

Alumina sheet ~-"""""", Silica sheet ~-=:a::::::E!I~ Potassium ions

(b) Illite

(c)

Structure of kaolinite, illite, and montmoril

per gram while ar of 45 gram 0 i.s their large sur ac s, surface forces significantly influence the behavior of finegrained so· l~ compa d to coarse-grained soils. Th s fa ce charges on fine-grained soils are negative (anions). These negative surface rge attract cations and the positively charged side of water molecules from surrounding water. Consequently, a thin film or layer of water, called adsorbed water, is bonded to the mineral surfaces. The thin film or layer of water is known as the diffuse double layer (Fig. 2.4). The largest concentration of cations occurs at the mineral surface and decreases exponentially with distance away from the surface (Fig. 2.4) . Drying of most soils, with the exception of gypsum, using an oven for which the standard temperature is 105 ± 5°C, cannot remove the adsorbed water. The adsorbed water influences the way a soil behaves. For example, plasticity, which we will deal with in Section 2.6, in soils is attributed to the adsorbed water. Toxic chemicals that seep into the ground contaminate soil and groundwater. The surface chemistry of fine-grained soils is important in understanding the migration, sequestration, re-release, and ultimate removal of toxic compounds from soils.

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Mineral surface

c

.2

~ c

'"uc o

(.)

Diffuse double layer

FIGURE 2.4

Diffuse double layer.

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12

(a) Flocculated structure--saltwater environment

(b) Flocculated structure-freshwater environment

(c) Dispersed structure

FIGURE 2.5

Soil fabric.

2.3 COMPOSITION OF SOILS

13

parallel to each other. A flocculated structure, formed under a freshwater environment, results when many particles tend to orient perpendicular to each other. A dispersed structure is the result when a majority of the particles orient parallel to each other.

00 a · -bearing capacities and good drainage qual. an their stren an ' olum change characteristics are not significantly flected by chang 111 oistu l' c nditions. They are practically incompressible w en dense, b '/? Sl . C n ol
i

.

FO )

CHAPTER 2

1. 2. 3. 4.

5.

6.

7.

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14

PHVSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

The essential points are: Soils are derivedfrom the weathering of rocks and are commonly described by textural terms such as gravels, sands, silts, and clays. Particle size is used to distinguish various soil textures. Clays are composed of three main types of mineral-kaolinite, illite, and montmorillonite. The clay minerals consist of silica and alumina sheets that are combined to form layers. The bonds between layers play a very important role in the mechanical behavior of clays. The bond between the layers in montmorillonite is very weak compared with kaolinite and illite. Water can easily enter between the layeJ'S in montmorillonite, causing swelling. A thin layer of water is bonded to the mineral surfaces of soils and significantly influences the phy sical Jlnd mechanical characteristics of fine-grained soils. Fine-grained soils have much la rger surface areas than coarse-grained soils and are responsible for the major physical and mechallical differences between coarse-grained andfine-grained soils. The engineering p roperties offine-grained soils depend mliinly on mineralogical factors.

Soil is composed of solids, liquids, and gases (Fig. 2.6a). The solid phase may be mineral , organic matter, or both. As mentioned before, we will not deal with organic matter in this textbook. The spaces between the solids (soil particles) are called voids. Water is often the predominant liquid and air is the predominant gas. We will use the terms water and air instead of liquids and gases. The soil water is commonly called pore water and it plays a very important role in the behavior of soils under load. If all the voids are filled by water, the soil is saturated. Otherwise, the soil is unsaturated. If all the voids are filled with air, the soil is said to be dry. We can idealize the three phases of soil as shown in Fig. 2.6b. The physical properties of soils are influenced by the relative proportions of each of these

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15

2.4 PHASE RELATIONSHIPS

Idealization

• (a) Soil

FIGURE 2 .6

(b) Idealized soil

Soil phases.

(2.1)

where

(2.2)

w" x

=

W,

100%

(2.3)

The water con nt o~r1 is found by weighing a sample of the soil and then placing 'o( in an ov aNifO ::t: SOC until the weight of the sample remains constant; that is, all e abso bed water is driven out. For most soils, a constant weight is achieved in u 4 hours. The soil is removed from the oven, cooled, and then weighed. Example 2.2 illustrates the measurements and calculations required to determine the water content.

2. Void ratio (e) is the ratio of the volume of void space to the volume of solids. Void ratio is usually expressed as a decimal quantity. (2.4)

3. Specific volume (V') is the volume of soil per unit volume of solids:

I V'

=

f

=

1 + e

I

(2.5)

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

This equation is useful in relating volumes as shown in Example 2.3 and in the calculation of volumetric strains (Chapter 3).

solids to the

(2.8)

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16

Let m 1 be the mass of the container; m2 be the mass of the container and dry soil; m 3 be the mass of the container, soil, and water; and m 4 be the mass of the container and water. The mass of dry soil is ms = m2 - mi , the mass of water displaced by the soil particles is m s = m4 - m 3 + ms> and G s = m ) m s.

6. Degree of saturation (5) is the ratio, often expressed as a percentage, of the volume of water to the volume of voids: or

Se

=

wG.

I

(2.9)

If 5 = 1 or 100%, the soil is saturated. If 5 = 0, the soil is bone dry. It is practically impossible to obtain a soil with 5 = O.

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2.4 PHASE RELATIONSHIPS

17

TABLE 2 .1 Typical Values of Unit Weight for Soils Soil type Gravel Sand Silt Clay

'Y.at (kN/m 3 )

'Yd (kN/m 3 )

20-22 18-20 18-20 16-22

15-17 13-16 14-18 14-21

7. Unit weight is the weight of a soil per unit vol bulk unit weight, 'Y, to denote unit weight: (2.10)

Special Cases (a) (2.11 )

(b) Dr; (2.12)

of a saturated soil, sur-

(2.13)

8. Rela! e de sity (Dr) is an index that quantifies the degree of packing between the loosest and densest possible state of coarse-grained soils as determined by experiments: (2.14)

where emax is the maximum void ratio (loosest condition), emin is the minimum void ratio (densest condition), and e is the current void ratio. The maximum void ratio is found by pouring dry sand, for example, into a mold of volume (V) 2830 cm 3 using a funnel. The sand that fills the mold is weighed. If the weight of the sand is W, then by combining Eqs. (2.10) and (2.12) we get emax = Gs'Yw(VIW) - 1. The minimum void ratio is determined by vibrating the sand with a weight imposing a vertical stress of 13.8 kPa on top of the sand. Vibration occurs for 8 minutes at a frequency of 3600 Hz and amplitude of 0.064

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOilS AND SOil INVESTIGATIONS

TABLE 2.2 Description Based on Relative Density Dr (%)

Description

0-15 15-35 35-65 65-85 85-100

Very loose Loose Medium dense Dense Very dense

Prove the (a)

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18

Strategy he proo s of these equations are algebraic manipulations. Start with the basic ni ,. n and then manipulate the basic equation algebraically to get the desired form. Solution 2.1 (a) For this relationship, we proceed as follows:

Step 1:

Write down the basic equation,

Step 2:

Manipulate the basic equation to get the desired equation. You want to get e in the denominator and you have V v . You know that Vv = eVs and Vw is the weight of water divided by the unit

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2.4 PHASE RELATIONSHIPS

19

weight of water. From the definition of water content, the weight of water is wWs' Here is the algebra:

(b) For this relationship, we proceed as follows :

Step 1:

Write down the basic equation,

Step 2:

ap ipulate the b Sle equation-to get the new form of the equation.

+ Vs +

Ws

"I =

Ww

+ wWs Vs + V v

Ws

Vv

W,( l + SetG s ) VsU + e)

G s'Yw(1 + Se/G,) 1+ e

Gs'Yw(1 + w) 1+ e

or

_(0.,1 ++ ese)

"I -

"I,.

•

EXAMPLE 2.2 A sample of saturated clay was placed in a container and weighed. The weight was 6 N. The clay in its container was placed in an oven for 24 hours at lOSOC. The weight reduced to a constant value of 5 N. The weight of the container is 1 N. If G s = 2.7, determine the (a) water content, (b) void ratio, (c) bulk unit weight, (d) dry unit weight, and (e) effective unit weight.

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Strategy Write down what is given and then use the appropriate equations to find the unknowns. You are given the weight of the natural soil , sometimes called the wet weight, and the dry weight of the soil. The difference between these will give the weight of water and you can find the water content by using Eq. (2.3). You are also given a saturated soil, which means that S

Solution 2.2 Step 1:

Write down what is given. Weight of sample + container = 6 N. Weight of dry sample

+ container

=

Step 2:

t.; f dry soil:

5=14N~

, =6

i

_ 5-1~ ~ Step 3:

=!4 x

00

Step 4:

10A s%

Determi

(see Example 2.1)

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20

Gs ) (1 + e

_

'/w -

2.7 1 + 0.675

x

_ 3 9.8 - 15.8 kN/m

or =

'I d

Step 7:

'I

(1

+ w)

=

19.7 = 15.8 kN/m 3 1 + 0.25

Determine the effective unit weight. 'I' =

(~s: e1 )'/w =

2 (1 ;

~6~5)

x 9.8 = 9.9 kN /m

or 'I' =

'/sat -

'/W = 19.7 - 9.8 = 9.9 kN/m 3

3

•

2.4 PHASE RELATIONSHIPS

21

EXAMPLE 2.3

C

An embankment for a highway is to be constructed from a soil compacted to a dry unit weight of 18 kN/m3 . The clay has to be trucked to the site from a borrow pit. The bulk unit weight of the soil in the borrow pit is 17 kN/m 3 and it natural water content is 5% . Calculate the volume of clay from the borrow p.' re uired for 1 cubic meter of embankment. Assume Gs = 2.7.

Solution 2.3

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Step 1:

pit and embankment. Let oid ratio, respectively, of borrow pit clay

17 _ 3 1 + 0.05 - 16.2 kN/m

But

and therefore eJ

= Gs

'Yw 'Yd

-

-

9.8 ) = 2.7 ( -6-

1

1 .2

- 1

=

0.633

Similarly, ez = Gs

~: -

1 =

2.7(~.:)

-

1 = 0.47

F

22

CHAPTER 2

Step 3:

PHVSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Determine the volume of borrow pit material.

V;

1+

e1

Vz

1

+

e2

Therefore V' 1

=

V' 1. 2 1

+ e1

+

e2

=

1(1 + 0.633) 1 + 0.47

•

EXAMPLE 2.4

Solution 2.4 Step 1:

Step 2:

• is a is ribution of particles with various sizes. ces t response of soils to loads and to flow d n the laboratory to find particle sizes in

2.5.1 Particle Size of Coarse-Grained Soils The distribution of particle sizes or average grain diameter of coarse-grained soils-gravels and sands-is obtained by screening a known weight of the soil through a stack of sieves of progressively finer mesh size. A typical stack of sieves is shown in Fig. 2.7. Each sieve is identified by a number that corresponds to the number of square holes per linear inch of mesh. The particle diameter in the screening process, often called sieve an alysis, is the maximum particle dimension to pass through the square hole of a particular mesh. A known weight of dry soil is placed on the largest sieve (the top sieve) and the nest of sieves is then placed on a vibrator, called a sieve shaker, and shaken. The nest of sieves is dismantled, one sieve at a time. The soil retained on each sieve is weigh ed and the percentage of soil retained on each sieve is calculated. The results are plotted on a graph of

23

2.5 DETERMINATION OF PARTICLE SIZE Of SOILS

FIGURE 2.7 Stack of sieves.

~

~E~~~~,:~~~~.~e;,~:e"t Fig. re aUled) as the 2.8. The re,,~~;t;:::;.:I :;, the gradation i(

{(c~:U'""~~~t

sca le for particle

the smallest in asoil

ilh sieve (rom the top of the

,;:~':..'-",-~~:,.:~~lt-~,,:!p~e~,~c."e~n~t weight retained is (2.15)

100

;

2: (% ,.,

Retained on ith sieve)

(2.16)

weight. The unit of mass is grams or kilograms. 100 90

80

, 0

•"

t

+

,I

70 60

'. 50

3.

2. 10

•

100 PartIcle SIze tmm) - qal"llllmlc sea%!

FIGURE 2.8 Particle size distribution curves.

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOilS AND SOil INVESTIGATIONS

2.5.2 Particle Size of Fine-Grained Soils

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24

w+""--t-

FIGURE 2.9

Hydromeler

Hydrometer in soil-water suspension .

25

2.5 DETERMINATION OF PARTICLE SIZE OF SOILS

sophisticated methods are available (e.g., light scattering methods). The dashed line in Fig. 2.S shows a typical particle size distribution for fine-grained soils.

2.5.3 Characterization of Soils Based on Particle Size

(2.18)

which 60% of the particles are

FO

(2.19)

TABLE 2.3 to uses Soil type

f the soil particles for which 30% of the particles are

Soil y pe, Descriptions, and Average Grain Sizes According

Description

Average grain size

Gravel

Rounded and/or angular bulky hard rock

Coarse: 75 mm to 19 mm Fine: 19 mm to 4 mm

Sand

Rounded and/or angular bulky hard rock

Coa rse: 4 mm to 1.7 mm Medium: 1.7 mm to 0.380 mm Fine: 0.380 mm to 0.075 mm

Silt

Particles smaller than 0.075 mm. ex hibit little or no strength when dried

0.075 mm to 0.002 mm

Clay

Particles smaller than 0.002 mm. exhibit significant strength when dried; water reduces strength

< 0.002 mm

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

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26

The essential points are: 1. A sieve analysis is used to determine the grain size distribution of coarse-grained soils. 2. For fine-grained soils, a hydrometer analysis is used to find the particle size distribution.

2.5 DETERMINATION OF PARTICLE SIZE OF SOILS

27

3. Particle size distribution is represented on a semilogarithmic plot of % finer (ordinate, arithmetic scale) versus particle size (abscissa, logarithm scale). 4. The particle size distribution plot is used to delineate the different soil textures (percentages of gravel, sand, silt, and clay) in a soil. 5. The effective size, D 10, ;s the diameter of the particle$ of which 10% of the soil is finer. D 10 is an important value in tegulatitrg flow through soils and can significantly influence the mechanical behavior of soils. 6. D50 is the average grain size diameter pi the soil. 7. Two coefficients-the uniformity coefficient and the coefficient of curvature-are used to characterize the particle size distribution. Uniform soils have uniformity coefficients <4 and steep gradation curves. Well-graded soils have uniformity coefficients >4, coefficients of curvature between 1 and 3, lind flat gradtuion curves. Gap-graded soils have coefficients of curvature 3, and one or more humps on the gradation curves.

FO

R

90.1 181.9 108.8

6.1

par cle size distribution curve. Determin (1 ) the effective size, (2) the average particle size, (3) the uniformity coefficient, and (4) the coefficient of curvature. (c) Determine the textural composition of the soil (i.e., the amount of gravel, sand, etc.). (d) Describe the gradation curve.

Strategy The best way to solve this type of problem is to mak e a table to carry out the calculations and then plot a gradation curve. Total mass of dry sample (M) used is 500 grams but on summing the masses of the retained soil in column 2 we obtain 499.7 grams. The reduction in mass is due to losses mainly from a small quantity of soil that gets stuck in the meshes of the sieves. You should use the "after sieving" total mass of 499.7 grams in the calculations.

28

CHAPTER 2 PHVSICAL CHARACTERlsnCS OF SOilS AND SOIL INVESTIGATIONS

Solution 2.5 Step 1: Tabulate data Sieve no.

4 10 20 40 100 200

10

Mass retained (grams) M,

obtain % fi ner. % Retained IM,IM ) x 100

I.( % Retained}

0

0

0

3.0~

14.8 98.0

19.6

90.1

18.0

181 .9

36.4

10 8.8

21.8

6. 1 TOl al mass M '" 499.7

1.2 100

p,"

Slep 2:

~

Frner

l ob\: 0 - 100 3.0

22.6

- 2 2.6

100 00

77.4

~.6 =

T77 =

59.4

23.0

- 98.8 - 1.2

PIOl the gradalion curve. See Fig. E2.5 for a plo

Slep 3:

Slep 4:

Gravel 3% ) Sand = 9 -.&% SilL - n~ (= 1.2%

Step 5: Ca l u1ate'iiC and Cc. c

~ C

._,.;,.

Dw = 0.45 = 4.5 '~ 01 2 \f~JIJ)2 = 0.18 = 0.72 DloD6iJ 0.1 x 0.45 Sand~

Gravels

80

10

~

•

60 50

40 30 20 10 0 0.001

0.01 Partlc.~

s,ze (mm) - logarithmIC !.Cll1e

FIGURE E2 .S Pa rticle size distribut ion cu rve.

•

2.6 PHYSICAL STATES AND INDEX PROPERTIES OF FINE-GRAINED SOILS

29

2.6 PHYSICAL STATES AND INDEX PROPERTIES OF FINE-GRAINED SOILS

FO

(2.20)

A

E

D

::J

I

I I I

I

Semisolid:

I I I

I I I

g l----"r Solid

Plastic

Liquid

WPL

Water content

FIGURE 2 _10 Changes in soil states as a function of soil volume and water content.

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Table 2 .4 Description of Soil Strength Based on Liquidity Index Description of soil strength

Values of IL

IL < 0

Semisolid state-high strength, brittle (sudden) fracture is expected

0
Plastic state-intermediate strength, soil deforms like a plastic material

h> 1

Liquid state-low strength, soil deforms like a v' cou fluid

G (2.21)

FO

30

A

lp

=--------'.---

Clay fraction ( % )

You should recall that the clay fraction is the amount of particles less than 2 f.lm.

TABLE 2.5 Typical Atterberg Limits for Soils Soil type

WLL(%)

Sand Silt Clay

WPL (%)

Ip

(%)

Nonplastic

30-40 40-150

20-25 25-50

10-15 15-100

2.7 DETERMINATION OF THE LIQUID, PLASTIC, AND SHRINKAGE LIMITS

2.7 DETERMINATION OF THE LIQUID, PLASTIC, AND SHRINKAGE LIMITS 2.7.1 Casagrande Cup Method

'0

small differences in apparatus.

FO

5

31

Cam

Hard rubber base

\SlJ H

2 mm

FIGURE 2.11

of Geotest.)

Cup apparatus for the determination of liquid limit. (photo courtesy

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

60

\

55

~ C 50

2c

w LL =

0

'-'

2

-

45

\

I

I

c- Best-fit straight line

I called the liquid state lin e l

- -'- '

.

\

46.2%

-

:

,

\

'" ~

\.

40

\

35 10

20

25 30

! 40

50 60 708090 100

Number of blows - logarithmic scale

FIGURE 2.12

Typical liquid limit results fro

2.7.2 Plastic Limit Test

FO

32

FIGURE 2.13

Fall cone apparatus.

e cup method.

2.7 DETERMINATION OF THE LIQUID. PLASTIC, AND SHRINKAGE LIMITS

60

33

I

I

80 gram co ~e 55

~ c ~

Best-fit straignt line 50

c

45

2 '" 3:

40

:3

V

W LL

= 40%

)

35 30 10

/

)'

11 / /

~

~

V 24

gram cone

V

30 40 50 60 70 20 Penetration of con e (mm) - loga ri thmic scale

FIGURE 2 .14 Typical fall cone test resul

.

water co 1: liquid sta 80 gram c (2.22)

FO

R

~

The shrinkage limit is determined as follows. A mass of wet soil, m l , is placed in a porcelain dish 44.5 mm in diameter and 12.5 mm high and then oven-dried . The volume of oven-dried soil is determined by using mercury to occupy the vacant spaces caused by shrinkage. The mass of the mercury is determined and the volume decrease caused by shrinkage can be calculated from the known density of mercury. The shrinkage limit is calculated from (2 .23)

where m l is the mass of the wet soil, m 2 is the mass of the oven-dried soil, VI is the volume of wet soil, V 2 is the volume of the oven-dried soil, and g is the acceleration due to gravity (9.8 m/s2).

CHAPTER 2

1. 2. 3. 4.

5. 6.

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

The essential points are: Fine-grained soils can exist in one offour states: solid, semisolid, plastic, alld liquid Water is the agent that is responsible for changing the stat~ of soils. A soil gets weaker if its water content increases. Three limits are defined based on the water content that causes a change of state. These are the liquid limit-the water content that caused the soil to change from a liquid to a plastic state; the plastic limit-the water content that caused the soil to change f rom a plastic to a semisolid; and the shrinkage limit-the water content that caused the soil to change from a semisolid to a solid state. A ll these limiting water contents are found from laboratory tests. The plasticity index defines the range of water content for which the soil behaves like a plastic maltrial. The liquidity index gives a measure of strength.

EXAMPLE 2.6, A liquid limit test c results:

FO

34

iquid limit, you must make a semi-logarithm plot of water content versus n of blows. Use the data to make your plot, then extract the liquid limit (water content on the liquid state line corresponding to 25 blows). Two determinations of the plastic limit were made and the differences in the results are small. So, use the average value of water content as the plastic limit.

Solution 2.6 Step 1: Step 2:

Plot the data . See Fig. E2.6. Extract the liquid limit. The water content on the liquid state line corresponding to a terminal blow of 25 gives the liquid limit. WLL

=

38%

FO

35

2.7 DETERMINATION OF THE LIQUID, PLASTIC, AND SHRINKAGE LIMITS

6C ~------~----~--.--r-r-.-'"

55 .

~ 501 ---~ c

~ 45 ~-----~~----r-~r-~-r-r~~

8iii 40 I~-~~~.:j::~=-IIIIIIIt1

~ 35

30~-------~~'--~~r-~-r-r~~

25L-______~~__L-~~~_L_L~~ 10 20 25 30 40 50 60 708090100 Number of blows - logarithmic scale

FIGURE E2.6

cup method. Step 3:

Calculate plastic limit. The plastic limit is 2Q. PL =

Step 4:

L

+ 0.8 2

=

i

20.6"1~

Calculate lp.

Step 5:

Step 6:

• Parameter

Penetration (mm) Water content (%)

80 gram cone

5.5 39.0

7.8 44.8

14.8 52.5

240 gram cone

22

32

60 .3

67

8.5 36.0

15

21

45.1

49.8

35 58.1

Determine (a) the liquid limit, (b) the plastic limit, (c) the plasticity index, and (d) the liquidity index if the natural water content is 36%.

Strategy Adopt the same strategy as in Example 2.6. Make a semilogarithm plot of water content versus penetration. Use the data to make your plot, then extract the liquid limit (water content on the liquid state line corresponding to 20 mm). Find the water content difference between the two liquid state Jines at any fixed penetration. Use this value to determine the plastic limit.

36

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

70 65 . ~ 60 C 2c 55 0

u

(;j

ro

s:

50 45 40 35

10

1

20

100

Solution 2.7 Step 1:

Plot the data. See Fig. E2. 7.

Step 2:

Step 3: 10 = 17%

36 - 17 = 0.45 42

•

FO

~.:#,--W.....:P--"L

1

2.8

SOIL CLASSIFICATION SCHEMES A classification scheme provides a method of identifying soils in a particular group that would likely exhibit similar characteristics. Soil classification is used to specify a certain soil type that is best suitable for a given application. There are several classification schemes available . Each was devised for a specific use. For example, the American Association of State Highway and Transportation

37

2 .• SOIL CLAS$JFI(AnON SCHEMES

Officials (AAS HTO) developed one scheme that classifies soi ls according to thei r use fulness in roads and highways while the Un ified SOl i Classifica llon System (USeS) was originally developed for use in airfield construcllon but was later modified fo r genera l use . We will study on ly the uses because it is neither too ela rate nor 100 simplistic. The uses uses symbols for the particle size grou s. These symbols and their represe ntations are : G- gravel, S- sand. M-sih, lay. Tr ese are combincd with other symbols expressing gradation charlcten ics- for well graded and P for poorly graded- and plasticity characteri ics-H': or high and L for low, and a sy mbol. 0, indicating the presence 0 ganic- aterial . A typical classification of CL means a clay soi! wi th low plasticit)': hile S.P means a poorly graded sand. The flowcharts shown in Figs. - ,0 pro ide ystematlc means of classifying a SOl i accord ing 10 the USCS. A Expe ri menta l results from soi ls lesttcl fn m di rent parts of the world were plotted o n a grap h o[ plasticity index 0 in te) r s liq uId ill}lit (abscissa) . It was fou nd that clays. silts, and o r i 501 ' lie ' distinct regions of the graph. A line defi ned by the eq uation ~;;-:;:;;(2.24)

Soll 11 111le "allied

PToc;eed 10 1l0000hart 01 Fli. 2.15b

Between 5% and 12%?

'"

IS clay "ilCtlon > SIlt tractIon? Yes

"1"$1 Ifll.f "G. UC;t4 .nd 1 !i:CC!i:3

11

PIII$llc tl.yey"non s.cond Jettei' C

Non -pl~1C $Illy t,non Second lett... M

1I1,,~lt!tt ... ~G.U-C:il:4

"'.......

CIess,loellion

."d lSec s3

second l.n_ os W. 0Ihe!w!,. YCond lett ...

~

P

II IusllltiefosS,VC}'6 end lSec...:3 §l!(:ond lett_

1$

W. 01 ........." second letter

~

P

GW...(IC

No

1",rsll"Ie, IS G. UC~4."dlsCCs3

CI_,Ioc:'-:1OII OS: GP GC

CluSI'it.lton IS: GW-GM Oth_,H. Classll't.Ioon Is: GP...(IM

"'''$Ilttttf .. S.

11',rsI (.tt... " S,

"'........

Clnsrt'ClI,on

IS:

UC~6 .ndl~ct$3 Ct-.'WIOll ~: SW-SC

CIa5,loeallOll

~:

SP -sc

UC2'6 ,"d I SCC!i:3

"'--.

15

SW-SM

CI,ultIClt,O" 05. SP-5M

FIGURE 2.15. Unified Soil Classification flOWChart for coarse-grained soils.

38

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOilS AND SOil INVESTIGATIONS

Are 50% of particles < 0.075 mm? ----'---No

Is

"'LL IVLL

(oven·dried)

< 0.75%

(not dried)

daries between clays (above the line) and silts and org ic soils (below e line) n in Fig. 2.16. A second line, the -Hne expre~ed as Ip = 0 (WL - 8) , defines the upper limit of the correlation be een plasticity index .frt ligilfc Ii it. If the results of your soil tests fall above

0'

FO

50 -

, e, you sho»

of your cesults and ,epea! younes!s,

,

~ ><

50

I--t--t-~~

40

I- -t--+--t---+--+/ C-.-+--+

/ - 1 - - + - -1

OJ

-0

.!: .~

30 f--- - f - - + - - - \ - --bo'' - - - l - --V''-:-

u

~ '" 20 I- - t - - + --Y-'---+--:>"'f-- + - -r-'--II--t--i n::

OL-~~~L~_~_L-~_~_-L_-L_~~

o

10

20

30

40

50

60

Liquid limi t (%)

FIGURE 2.16

Plasticity chart.

70

80

90

100

2.9 ENGINEERING USE CHART

ENGINEERING USE CHART You may ask: "How do I use a soil classification to select a soil for a particular type of construction, for example, a dam?" Geotechnical engineers have prepared charts based on experience to assist you in selecting a soi 0 particular construction purpose . One such chart is shown in Table 2.6. Th numerical values used to rovide 1 to 9 are ratings with No.1 the best. The chart should 0 y guidance and to make a preliminary assessment of the tabi1' of a oil for a particular use. You should not rely on such descriR strength or "negligible" compressibility to make fin decisions. We will deal later (Chapters 4 and 5 with determine strength and compressibility prope·i-t"· .. "'-EXAMPLE 2.8

Particle size analyses were carried 0 particle size distribution curves e s the two soils are:

l

Soil A

B

il Classification Scheme . ......: . ~~"'.e two soils is a better material for

, se t e partlc e

ntageo g a?

FO

2.9

39

Cla~

100 90

l

Silt

I

.-..•.. -

I

;:

'I

111

/

'"c:

60 50 - - .

if!.

40

if

10

o 0.001

V

Soil B

, I

/'

I I L... I I .t" I 0 .01

0.1

Particle size (mm) - logarithmic scale

FIGURE E2.8

I

/

So il A

u::

20

Gravel

/

80 70

30

Sand

10

40 TABLE 2 .6

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Engineering Use Chart (After Wagner, 1957)

Group symbols

Permeability when compacted

Workability as a construction material

Well-graded gravels, gravel-sand mixtures, little or no fines

GW

Pervious

Excellent

Poorly graded gravels, gravel sand mixtures, little or no fines

GP

Very pervious

Silty gravels, poorly graded gravel-sand-silt mixtures

GM

Clayey gravels, poorly graded gravel-sand-clay mixtures

GC

Well-graded sands, gravelly sands, little or no fines

SW

Poorly graded sands, gravelly sands, little or no fines

SP

Silty sands, poorly graded sandsilt mixtures

SM

Typical names of soil groups

Clayey sands, poorly graded sand-clay mixtures Inorganic silts and very fine sands, rock flour, silty or clayey fine sands with slight pi as ieit\!

, Organic clays of medium plasticity

Good

Negligible

Good Good Excellent

Excellent

Fair Low

Fair

Low

Good

Medium

Fair

Medium

Good to fair

Poo r

Medium

Fair

Fair to poor

High

Poor

Poor

High

Poor

Poor

High

Poor

~

FO

Inorganic clays of low to medium plastici ty , gravelly c sandy clays, silky clays, lean clays

Negligible

ML

Impervious

. ! lat and other highly organic soils

the different percentages of each soil type and then follow the flowchart. To determine whether your soil is organic or inorganic, plot your Atterberg limits on the plasticity chart and check whether the limits fall within an inorganic or organic soil region.

Solution 2.8 Step 1:

Determine the percentages of each soil type from the particle size distribution curve.

2.9 ENGINEERING USE CHART

Con$tituent Percent o f particle greater than 0.075 mm Gravel fraction (%) Sand fraction (".) Sill fraction 1%1 Clay fraction 1%)

Soil A

Soil B

12 0 12

80 16 64

59 29

20 0

41

42

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Step 2:

Use the flowchart. Following the flowchart, Soil A is ML and Soil B is SM.

Step 3:

Plot the Atterberg limits on the plasticity chart.

Step 4:

Use Table 2.6 to make a preliminary assess Soil B with a rating of 5 is better than Soil B dam core.

from two-dimensional flo

F

1 4

2.10.2 Head Darcy 's law governs the flow of water through soils. But before we delve into Darcy's law, we will discuss an important principle in fluid mechanics- Bernoulli 's principle-which is essential in understanding flow through soils. If you cap one end of a tube, fill the tube with water, and then rest it on your table (Fig. 2.17), the height of water with reference to your table is called the pressure head (h p ). Head refers to the mechanical energy per unit weight. If you raise the tube above the table, the mechanical energy or total head increases.

FO

2.10 ONE·OIMENSIONAL FLOW OF WATER THROUGH SOILS

43

hp = ufy",

Pressure head

Pressure head Datum-top of table

FIGURE 2 . 17

You now have two components 0: elevation head (hJ. If water ere to a under steady state condition, tH given as u2 !2g. The total head (so to Bernoulli's principle i (2.25)

(2.26)

ere u = hp is pore w er pressure. Consider a lind~;'('~taining a soil mass with water flowing through it at a consta t'G-ate as piMi~ Fig. 2.18. If we connect two tubes, A and B, called piezomete , t a dis ance I apart, the water will rise to different heights in each of the tubes. 19ht of water in tube B near the exit is lower than A. Why? As the water flows through the soil, energy is dissipated through friction with the soil particles, resulting in a loss of head. The head loss between A and B, assumw,

Tube A

I-

FIGURE 2 . 18

Tube B

I ---+j

Head loss due to flow of water through soil.

44

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

ing decrease in

h e~d

is posi tive and our _d
o f the cy linder, is !J.H = l(h,,)B - (hp)" I.

i 2 . 10.3 Darcy' s Law Darcy (1856) proposed that average flow velocity through soil to the grad ien t of the total head. The flow in any direclion.;1. is dH

UJ

= k' -d

x,

where u is the average flow velocity, k is a coefficient

(2.27) 0

(228)

where j = !J. Hli is the hydra ul ic gract.!ent. jls if the flow is laminar (Reynolds umber < 1) The average veloci ty, cal laled from Eq. 0.;.'2.8) l~ fo r the cross-sectional area normal 10 Ih;;~tion \' fto . Flow IhrotlgJ] soils, how ver. occurs only through the in.tcrc:onnected. vOi9fi ..The ve l oc~tr through..the Old spac~s is called seepage velOCity ~.) and ij olhamed by dJV1~g the-a erage velocllY by the ~ ~ porosity of the soil ·

t® u~

The vol ~me rale of fI aver ge velocity and the

"

- I.

(2.29)

~qll' or. simpl y. flow rate is the product of the i.P al area: (2.30)

e un it o f meas re le II is ml/s or cml/s. The conservation of flow (law f con tinuity) stlP ales ~hat e volume rate of inflow (qv),n in to a soil element must e ~ual tne v01u: \ rate of outflow. (q... )o.... or, simply. inflow must eq ual ou tRow. (qll)ou\, The coefti n f permeability depends o n the soil type , the particle size d istribution. the structural arrange ment of the grains or void ratio, and the wholeness (homogeneity, layering, fissuring , e tc.) of the soi l mass. Typical value ranges of k . for variolls soil types are shown in Table 2.7. Homogeneous clays are practically impervious. Two popular uses of " im-

(tJ,n

TABLE 2 .7 Coefficie nt of P e rmeabilitv fo r Common Soil Type. Soil type Clean gravel Clean sands, clean sand and gravel mixtures Find sands, sil ts, mixlures comprising sands. sillS, and clays Homogeneous clays

>1.0 1.0 to 10- 3 10- 3 to 10- 1 <1 0 _7

45

2.10 ONE-DIMENSIONAL FLOW OF WATER THROUGH SOILS

pervious" clays are in dam construction to curtail the flow of water through the dam and as barriers in landfills to prevent migration of effluent to the surrounding area. Clean sands and gravels are pervious and can be used as drainage materials or soil filters.

2.10.4 Empirical Relationships for k For a homogeneous soil, the coefficient of permeabilit on its void ratio. You should recall that the void ra i is

(2.31)

1.

2. 3.

FO

R

4.

The essential points are: The flow 0/ water through soils is governed by Darcy's law, which states that the average flow velocity is proportional to the hydraulic gradient. The proportionality coefficient in Darcy's law is called the coefficient of permeability or hydraulic conductivity, k. The value of k is influenced by the void ratio, particle size distribution, and the wholeness of the soil mass. Homogeneous clays are practically impervious while sands and gravels are pervious.

n diameter is placed in a tube 1 m long. A constant supply flow into one end of the soil at A and the outflow at B is of water is a 1 e collected by a bell er (Fig. E2.9). The average amount of water collected is 1 cm 3 for every 10 seconds. The tube is inclined as shown in Fig. E2.9. Determine the

Table

FIGURE E2.9

CHAPTER 2

PHVSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

(a) hydraulic gradient, (b) flow rate, (c) average velocity, (d) seepage velocity, if e = 0.6, and (e) coefficient of permeability.

Solution 2.9 Step 1:

Define the datum position. Select the to

Step 2:

Find the total heads at A (inflow) a d

Step 3:

Step 4: collected, Q = 1 cm 3 , t = 10 seconds

FO

46

Step 6:

u =

A =

Au 'IT

X (diam)2

4

=

X 102

4

= 78.5 cm

2

u = qu = ~ = 0.0013 cmJs A 78.5

Determine seepage velocity. u n

uS

=-

n

= --

e

1

+e

0.6

= - - - = 0.38

0.0013

Us

Step 7:

'IT

+ 0.6

1

= 038 =

0.0034 cmJs

Determine the coefficient of permeability. From Darcy's law u = ki. k

v

0.0013

i

1.2

= - = -- =

10.8

X

. 10- 4 cm/s

•

2. 10 ONE·DIMENSION AL fl OW OF WATE R TH ROUG H SOILS

47

EXAMPLE 2.10 A drainage pipe (Fig. E2.lOa) became comple te ly blocked during a srorm by a plug of sand , 1.5 In long, followed by anothe r plug of a mixture of clays, silts. and sands, 0.5 m long. When the storm was over. the wa ter level above ground was l m. The coefficient ofpenneability of the sa nd IS 2 t lmet t ~ the mlxtUie of days. SlitS, and sands (a) Plot the variation of pressure , elevlltlon, and rot s

ver the l ength of

a

rhe pipe. e r 0 t ~ sand plug and (2) the center of the mixture of clays, siits, and an~s. ,

(b) Calculate the pore water pressure al (J) the e

(c)

Ste p L:

t£e

Select exit j t datum ..........,

drainage pipe as the

e rmine heads at A aqd B.

Step 2:

(h,)7= 0 m , th)A - 0 h')B - 0

~:~\.. = 0 m,

2

+

:: 33m, HA

=

0

+ 3.3

== 3.3 m

Hn - 0 m

Determine t~~~.ll 6sS In each plug. wee ~ A al).d B - IHe - H ... I = 33 m (decrease 10 head H ead 10 take n a positive.}. be I1HI! LJ> 1<.1' and ql be the head loss, length, coejficieht of pe"'~eabil ity, and

r----~~"-~----'a Stick wall

1m

G,,,te

Clay $Oil

20

m

C

Dram pope

o3m:i=I~~A~~-~~~-~~~-~~~~B~-2o:.:":m~~l-s.",

FIGURE

J.5m

05m '

tA .. tu'l! of SiltS. and ~nds

clays

E2.10a Illustration of blocked drainage pipe.

48

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

flow in the sand; let I:l.H2 , L 2 , k 2 , and q2 be the head loss, length, coefficient of permeability, and flow in the mixture of clays, silts, and sa nds. Now, D.H\

D.H\

ql = Akl - - = A X 2k2 - L\ Ll

D.H2 q2 = Ak2 - - = A L2

From the continuity equation, ql

X

M-I2 L2

k2 - -

= q2·

Solving, we get

(1)

(2)

FO

Step 4:

tiD

~

= 2.31 x

'Yw

= 2.31 x

9.8 = 22.6 kPa

3.5 3 2.5

E

2

"ro

1.5

CI)

I

0.5

Distance (m)

FIGURE E2.10b

Variation of elevation, pressure, and total heads along pipe .

2.10 ONE·DIMENSIONAL FLOW OF WATER THROUGH SOILS

49

Let E be the center of the mixture of clays, silts, and sands.

UE =

Step 7:

0.66 x 9.8

=

6.5 kPa

Find the average hydraulic gradients.

FO

•

(2.33)

~ Horizontal !low kl

~

k2

~

k3

FIGURE 2 . 19

l ? ?

Vertical flow

Flow through stratified layers.

50

CHAPTER 2

PHVSICAL CHA RACTERISnCS OF SOILS AND SOIL INVESTIGATIONS

2.10.6 Flow Normal to Soil layers For flow norma1to the soil layers, the head loss in the soil mass is the sum of the head losses in each layer: (2.34)

whe re I1H is the total head loss. and I1hl to I1h" are (he head sses in et ch o f the n layers. The velocity in each layer is the same. The anilogx to electr city is flow of curren t through resistors in series. From Darcy's aw we o~ (2.35)

wherc

k t(c q)

is the eq uivalent permeability i

k :" are the vertical permeabilities of the fir

and (2.35) leads to (2.36)

Values of less.

k t(cq\

arc gener

EXAMPLE 2 .1

.!

PI1Y.:$how~n

A canal is cur into so il a st rati gr iFig. E2.11 . Assuming flow takes plac laterally and vertically th,i0ugh n,} id970f the ca nal and vertically below the ~. ~ermine the eq uivalent pe eability in the horizontal and vert ical direC{lons. Calculate the ratio N he;.Auivalent horizontal permeability 10 the equiva'p r vertical7 e-ability [or flow through the sides of the canal .

,-I. Strate'"

Use til depth o f the

~d t~ eq uivalent horizontal permeabili ty over -?then use Eq . (2.36) to find the equivalent anal. To make the calculations easier, convert

Canal

k=0_3"10~~

k=0.8)( tolc~

FIGURE E2 .11

T 1 3.0m

51

2 .11 DETERM INATION O F THE CO EFA CIENT OF PERMEABI UTY

Solution 2.11 Step 1:

Find

k.« c q)

and k* "l) for flow through the sides of the canal. H ..

=

3m

From Eq. (2.33), k.(.q)

= -

I

Ho

(z l k .. 1 + z2k n ... ... +

- ~( l x 0.23 x

LO

G -j

z . k , ~)

j.5 xS.2

= 3 x 10- 6 cmls

Fro m Eq . (2.36).

1

3

1 ()- 6 ( 0.23

0 .61 x

1.5

IO -~

emts

0.5)

+ ""2 + 2

~ _. 49 ) SI Cp 3: 5.7

• What's next . . .In m'eal5i rity k. We

or~t ih

ca1

ate fl ow, we need to know the coefficient of perdiscuss how this coefficient is determined in the laboratory and (f

in t he fie ld .

2

2 . 11 DETERMINATION OF THE COEFFICIENT OF PERMEABILITY 2 . 11 .1 Constant-Head Test The consta nt -head lest is used to de termine the coefficient of permeability of

coarse.grained soils. A typical constant-head apparatus is shown in Fig. 2.20. Water is allowed to flow through a cylindrical sam ple of soil under a constant head (II) . The outflow (Q ) is collected in a graduated cylinder at a convenient duration (t ).

CHAPTER 2

T

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Mariotte bottle

h

Coarse-grai ned soil

FIGURE 2 _20

A constant-head apparatus _

With reference to Fig. 2.20,

uantity of water

(2.37)

(2.38)

FO

52

T

ter, T is the temperature in DC at which the mea= fL7"c/fL2fPC is the temperature correction factor

! RT =

2.42 - 0.475 [neT)

I

(2.39)

2.11.2 Falling-Head Test The falling-head test is used for fine-grained soils because the flow of water through these soils is too slow to get reasonable measurements from the constanthead test. A compacted soil sample or a sample extracted from the field is placed in a metal or acrylic cylinder (Fig. 2.21). Porous stones are positioned at the top and bottom faces of the sample to prevent its disintegration and to allow water to percolate through it. Water flows through the sample from a standpipe attached to the top of the cylinder. The head of water (h) changes with time as flow occurs through the soil. At different times, the head of water is recorded.

53

2.11 DETERMINATION OF THE COEFFICIENT OF PERMEABILITY

~

T

dh

T

I+- Standpipe

h

/TL

Fine-grained soil

...... 1. -:::

FIGURE 2_21

:~ To

beaker

A falling-head apparatus.

Let dh be the drop in head over a tim loss in the tube is

FO

R

the cross-section al area, s length of the soil sample, and h is t. The continuity condition requires that (qu)in = dh h = Akdt L

-Q -

and the solution for k in the vertical direction is (2.40)

The essential points are: 1. The constant-head test is used to detennine the coefficient ofpenneability of coarse-grained soils. 2. The falling-head test is used to detennine the coefficient of permeability offine-grained soils.

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

EXAMPLE 2.12

A sample of sand, 5 cm in diameter and 15 cm long, was prepared at a porosity of 60% in a constant-head apparatus. The total head was kept constant at 30 cm and the amount of water collected in 5 seconds was 40 cm 3 . The tesr,.,temperature was 20°C. Calculate the coefficient of permeability and the seep' ~e locity.

Strategy From the data given, you can readily apply D t cy

Solution 2.12 Step 1:

Step 2:

Calcula cm/s

ki n

0.2 x 2 0.6

- = - - - = 0.67

FO

54

Initial head Final head

= =

cmls

•

90 em 84 cm

Duration of test = 15 minutes Diameter of tube = 6 mm Temperature = 22°C Determine k.

Strategy Since this is a falling-head test, you should use Eq. (2.40). Make sure you are using consistent units.

2.11 DETERMINATION OF THE COEFFICIENT OF PERMEABILITY

Solution 2 . 13 Step 1:

A

=

=

12 -

Step 2:

G

Calculate the parameters required in Eq. (2.40). a

'IT

X

~6110? =

80 cm2 tl

= 15

0.28 cm

2

(given) X

55

60 = 900 seconds

Calculate k. k =

aL A(lz - t l )

In(hl) = 0.28 X 10 In h2

80

X

900

6:\.=.

~lb.. .

27

•

FO

What's next . .. In the constant-head ·

ell penetrates through the water-bearing stratum and is perthe section that is below the groundwater level. 2. The soil mass is homQgeneous, isotropic, and of infinite size. 3. Darcy's law is valid. 4. Flow is radial toward the well. 5. The hydraulic gradient at any point in the water-bearing stratum is constant and is equal to the slope of groundwater surface (Dupuit's assumptions) .

Let dz be the drop in total head over a distance dr. Then according to Dupuit's assumption the hydraulic gradient is .

dz dr

[=-

56

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Observation wells

Pumping well

I

------~----e-----~-- --- ------------I

l--- r

!- '2 '

1------. ,

...---_ _- , I , - - _ - ,

The area of flow

where

O

FO

an

"

z is

need to rearran o 'f, and hl an Ii '

ation and integrate it between the limits rj

Completing th k = qo In(r2Ir[) TI(h~ - hi)

(2.41)

With measurements of rl> r2 , h1' h2' and qv (flow rate of the pump), k can be calculated from Eq. (2.41). This test is only practical for coarse-grained soils. Pumping tests lower the groundwater, which then causes stress changes in the soil. Since the groundwater is not lowered uniformly as shown by the drawdown curve in Fig. 2.22, the stress changes in the soil will not be even. Consequently, pumping tests near existing structures can cause them to settle unevenly. You should consider the possibility of differential settlement on existing structures when you plan a pumping test. Also, it is sometimes necessary to temporarily lower the groundwater level for construction. The process of lowering the groundwater is called dewatering.

2.11 DETERMINATION OF THE COEFFICIENT OF PERMEABILITY

I

Observation wells

57

Pumping well

---- - -~----e--- --~- ----------------!

!-15m~

~30m

+

15 m

Impervious

FIGURE E2. 14

soil bed of thickness ate of pumping was 10.6 cated at 15 m and 30 m from the

EXAMPLE 15 m and the center

.4

FO

ketc

Step 1: Step 2:

e measurements to directly apply Eq. (2.41) to f the pump test to identify the values to be

tch of the pump test with the appropriate dimensions-see Substitute given values in Eq. (2.41) to find k. 1'2 =

hI

k

30 m ,

1'1

= 15 m,

= 15 - (1.9 + 1.6) =

qu In(rzlr1) 1T( h~ - hD

=

=

15 - (1.9 + 1.4)

h2 =

=

11.7 m,

11.5 m 3

10.6 X 10- In(30/15) = 50 X 10-2 I 1T(1l.72 - 11.5 2) 10 4 ' em s

•

What's next . .. Water, although regarded as the "foe" in geotechnical engineering, can be used to improve soil strength, reduce soil deformations under loads, and reduce the permeability. Next, we will study how water can assist in the improvement of soils.

58

CHAPTER 2

PHVSlCAl CHARACTERISTICS Of SOILS ANO SOlllNVESTlGATIONS

2.12 DRY UNIT WEIGHT- WATER CONTENT RELATIONSHIP 2 . 12.1 Basic Concept Let 'S examine Eq . (2.12) for dry unit weight, that IS , '1" =

(~)'Y. = C+ ~.'G~S)'Yw

(2.42)

The extreme right· hand side term was obtained by rc la ci n~ b~ e = wGjS. How can we increase the dry unit weight? Examination iE . (2~) reveals [hat we have to reduce the void ratio; tha i is, wlS mllsLbe r du .' T he theoret iea! maximum dry unit weight is ob tained 1 tha is

W 7.:.

e~

(2. 43 )

2 . 12.2 Proctor Compact) n Te A laboratory test . ca lled the Proci was developed to delive landa rd amou nt of mechanical ener (comp~c i e effort) t i:lelermine th maximum d ry unit weight of a soi l. In the fi and,V'd Proctor te • a r soi l s~im en is mixed with water and com acted a c:tlmdrical mold vOlu 044 x 10-· m) (sta ndard Procto mold) by r Beated blows fro m th ass a hammer, 2.5 kg, falling freely (fa a height jpf 05 mm (Fig.....2. ). . IS compacted in th ree layers, each of w his su iected to 2~s. A nfodified Pr test was de eloped or c paction of airfields to support hea aire f I ads. In the modifi ed Pro tor tesl, a hamme r with a mass 4.54 kg falls eely om a height of 4 mm. e SOil is compacted in five layers with 25 blo per layer in l)Atandard Proctor mold. Fou r or ore tests a r~o ~ duct ed on the soil using different water con tents. T h last test is idenl lf\e w e additional water causes the bulk uni t weight of e soil fO decrease fThe resu l ared,lo!led as dry unit weight (ordinate) versus wa r'content (ab ssa ). T ieal unit weighl- water content plots are shown in f ig. 2.24. Clays usually yjeld l=shaped curves. Sands do nOl often Yield a clear bell ;

dry

Yo) Moltl

(b) Hamme.

FIGURE 2.23 Compaction apparatus . (photo courtesy of Geotesr.)

2.12 DRY UNIT WEIGHT-WATER CONTENT RELATIONSHIP

59

12 10L---------~L---------~----~~~~~~~----~

4

6

FO

R

FIGURE 2 .24

un aturated at the maximum dry unit weight, that is, can det mine the degree of saturation at the maximum dry unit weight using q. ( -A2). We. know 'Yd = ('Yd)max and W = WopI from our Proctor test results. If Gs is know w. an solve Eq. (2.42) for 5. If G s is not known, you can substitute a value of 2.7 with little resulting error in most cases. Equation (2.42) can be used to plot a series of theoretical curves of dry unit weight versus water content for different degrees of saturation (Jines of constant degree of saturation) as shown in Fig. 2.24 for 5 = 100% and 5 = 80 % . You plot these curves as follows:

1. Assume a fixed value of 5, say, 5 = 1 (100 % saturation). 2. Substitute arbitrarily chosen values of w, approximately within the range of water content on your graph. 3. With the fixed value of 5 and either an estimated value of Gs (= 2.7) or a known value, find 'Yd for each value of W using Eq . (2.42) and plot the results of 'Yd versus w. 4. Repeat for a different value of 5.

60

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOilS AND SOil INVESTIGATIONS

OJ

"

Increasing compaction

1 Water content

Effect of increasing compaction efforts on water content relationship . FIGURE 2.25

FO

The curve corresponding to S = 1 is k voids line. This line represents the mini ... _-.: ...._ ._ water content [Eq. (2.43)]. The achievement of zero Proctor test, using higher leve.l mum dry unit weight at a lower 0 (Fig. 2.25). The degree 0 aturation i than the standard compactIDlnii"'t,~I"""".i·

a uration line or zero air ble at a given

,

2.12.5 Field Compaction A variety of mechanical equipment is used to compact soils in the field. You may have seen various types of rollers being used in road construction . Each type of roller has special mechanical systems to effectively compact a particular soil type. For example, a sheepsfoot roller (Fig. 2.26a) is generally used to compact finegrained soils while a drum type roller (Fig. 2.26b) is generally used to compact coarse-grained soils.

2.12.6 Compaction Quality Control A geotechnical engineer needs to check that field compaction meets specifications. Various types of equipment are available to check the amount of compac-

61

FO

R

2.12 DRY UNIT WEIGHT-WATER CONTENT RELATIONSHIP

Two types of machinery for field compaction. (Photos courtesy of Vibromax America, Inc.)

FIGURE 2.26

tion achieved in the field . Three popular apparatuses are (1) the sand cone, (2) the balloon, and (3) nuclear density meters.

2.12.6.1 Sand Cone A sand cone apparatus is shown in Fig. 2.27. It consists of a glass or plastic jar with a funnel attached to the neck of the jar. The procedure for a sand cone test is as follows: 1. Fill the jar with a standard sand-a sand with known density-and determine the weight of the sand cone apparatus with the jar filled with sand

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Jar

Ottawa sand

FIGURE 2 .27

STM) recom2. 3.

FO

62

D ry umt . welg . h t:

'fd

W" = 11

2 . 12.6.2 Balloon Test The balloon test apparatus (Fig. 2.28) consists of a graduated cylinder with a centrally placed balloon. The cylinder is filled with water. The procedure for the balloon test is as follows : 1. Fill the cylinder with water and record its volume, VI' 2. Excavate a small hole in the soil and determine the weight of the excavated soil (W). 3. Determine the water content of the excavated soil (w).

2.12 DRY UNIT WEIGHT-WATER CONTENT RELATIONSHIP

63

Air re lease va lve

Ba lloon Pump

FIGURE 2.28

Balloon test device.

4.

5. Record the volume of wat

6. Calculate the unit weigb;

FO

nt of a soil at a particular site.

FIGURE 2 .29

Nuclear density meter. (Photo courtesy of Seaman Nuclear Corp.)

CHAPTER 2

1. 2.

3. 4. 5.

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

The essential points are: Compaction is the densijication of a soil by the expulsion of air and the rearrangement of soil particles. The Proctor test is used to determine the maximum dry unit weight and the optimum water content and serves as the reference for field specijications of compaction. Higher compactive effort increases the maximum dry unit weight and reduces the optimum water content. Compaction increases strength, lowers compressibility'; and reduces the permeability of soils. A variety offield equipment is used to check the dry unit weights achieved in the field Popular fiel!1 equiwnent includes the sand cone apparatus, the balloon apparatus.., an the nu lear density eter.

EXAMPLE 2.15 The results of a standard

5 % standard compaction?

n plot the results of 'Yd versus

FO

64

w

(%) . Then

Zero air voids

Water content (%)

Bulk unit weight (kN/m3)

Dry unit weight (kN/m 3)

Y 'Yd=1+w

Dry unit weight (kN/m3) Water content

Yd = (, +

~Gs/5)YW;

(%)

5=1

22.8

6.2

16.9

15.9

6

8.1

18.7

17.3

8

21 .8

9.8

19.5

17.8

10

20.8

11.5

20.5

18.4

12

20.0

12.3

20.4

18.2

14

19.2

13.2

20.1

17.8

2.13 SOil INVESTIGATION

--1~ 23

.I

22

~ J

19

,

~ 18

.

Zero .Ir I'OIds

~ \ _ 18.4 kNim

M .... Mum unit...,

95% =pachon

On

.~

I

I' 15 6

65

I

,

7

8

9

10

11115 12 ~ 3

i

r-.

Wale< conlen! (%)

FIGURE E2.1S

Step 2: Step 3:

Compaction test results.

Extract the desired

Step 4:

.7

What~
.. .W e have

(18.419.8) _ 0 7 ~ % 18.419.8 . 1 71

•

discu ss~

s,p.ll s w it h the tacit assum ption that we have obof sOils from th e field on which we conducted tests s h ave b e ~ escribed. So' Is are o bserved and recove red during a soil investigation of a pro oseCl site. A s9-i invest igati n'is an essential part of the design and construct ion of a p ro posed structu al sy te (bu ildings, dams, roads and highways, etc.) . You i!!...B.P
~

2.13

SOILiNVESTIGATION 2.13.1 Purposes of a Soil Investigation A soil investigat ion program is necessary to provide information for design and construction and for environmental assessment. The purposes of a soil investi· gation are:

1. To eva luate the general sui tabil ity of the site for the p roposed project. 2. To enable an adequate and economical design to be made. 3. To disclose and make provision for difficulties that may arise during con· struction due to ground and other loca l conditions.

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOilS AND SOil INVESTIGATIONS

2.13.2 Phases of a Soil Investigation The scope of a soil investigation depends on the type, size, and importance of the structure, the client, the engineer's familiarity with the soils at the site, and local building codes. Structures that are sensitive to settlement s ~ machine foundations and high-use buildings usually require a thorough oil investigation compared to a foundation for a house. A client may wish 1J ta e a great r risk than normal to save money and set limits on the type At e t of e site investigation. If the geotechnical engineer is familiar " a si te he/s-he may undertake a very simple soil investigation to confirm his/h per . Some local building codes have provisions that set out the extent 0 a . mandatory that a visit be made to the propoSP.,(l-''':' IJP:",,"~ In the early stages of a project, the avai :able in rmatJ!)n is often inadequate to allow a detailed plan to be made. A sti a ' 0; must be developed in phases.

importance of the structure, ports, topographic clippings.

disturbed samples for laboratory tests.

FO

66

Phase

rite a eport. The report must contain a clear description of the site, ethods of exploration, soil profile, test methods and results, and t location of the groundwater. You should include information and/or explanations of any unusual soil, water-bearing stratum, and soil and groundwater condition that may be troublesome during construction.

2.13.3 Soil Exploration Program A soil exploration program usually involves test pits and/or soil borings (boreholes) . During the site visit (Phase II) , you should work out most of the soil exploration program. A detailed soil exploration consists of: 1. Preliminary location of each borehole and/or test pits. 2. Numbering of the boreholes or test pits.

FO

2.13 SOil INVESTIGATION

67

3. Planned depth of each borehole or test pit.

4. Methods and procedures for advancing the boreholes. 5. Sampling instructions for at least the first borehole. The sampling instructions must include the number of samples and possible locait,·ons. Changes in the sampling instructions often occur after the first bo 'eho e.

6. Requirements for groundwater observations.

2.13.4 Soil Exploration Methods Access to the soil may be obtained by the fo!lowjng • Trial pits or test pits • Hand or powered augers • Wash boring • Rotary rigs The advantages and disa 2.8.

the foundatio oa s is less than 10% , whichever is greater. Borings should penetrate at least 1 m into rock. In very stiff clays, borings should penetrate 5 m to 7 m to prove that the thickness of the strata is adequate.

2.13.7 Soil Sampling The objective of soil sampling is to obtain soils of satisfactory size with minimum disturbance for observations and laboratory tests. Soil samples are usually obtained by attaching an open-ended thin-walled tube-called a Shelby tube or, simply, a sampling tube-to drill rods and forcing it down into the soil. The tube is carefully withdrawn, hopefully , with the soil inside it. Soil disturbances occur from several sources during sampling, such as friction between

68

CHAPTER 2

TABLE 2.8

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Advantages and Disadvantages of Soil Exploration Methods

Method Test pits A pit is dug either by hand or by a backhoe.

Advantages

Disadvantages

• Cost effective • Provide detailed information of stratigraphy • Large quantities of disturbed soils are available for testing • Large blocks of undisturbed samples can be carved out from the ~~"'",..J~

Hand augers The auger is rotated by turning and pushing down on the handlebar.

FO

Power augers Truck mounted and equipped continuous flight augers that bar a hole 100 to 250 mm i ai a eter. Augers can have a solid stem.

pies can be cl ay deposit Cannot be used n rock, stiff nd, or caliche

~...._ _ _....:

ept limited to about 15 m . At greater depth drilling becomes difficult and expensive • Site must be accessible to motorized vehicle

• Depth limited to about 30 m • Slow drilling through stiff clays and gravels • Difficulty in obtaining accurate location of groundwater level • Undisturbed soil samples cannot be ob tained uick • Can drill through any type of soil or rock • Can drill to depths of 7500 m • Undisturbed samples can easily be recovered

• Expensive equipment • Terrain must be accessible to motorized vehicle • Difficulty in obtaining location of groundwater level • Additional time required for setup and cleanup

2.13 SOIL INVESTIGATION

(a)

FO

R

FIGURE 2 .30

The essential points are: 1. A site investigation is necessary to determine the nature of the soils at a proposed site for design and construction. 2. A soil investigation needs careful planning and is usually done in phases. 3. A number of tools are available for soil exploration. You need to use judgment as to the type appropriatefor a given project.

69

70

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

So il profile

Atterberg limits water conte nt, %

Effective stresses (kPa)

400

o -5

-10

FO

Silty clay (marine de posit)

2.14

" " . • • _ •. "M'"

.ple of a boring log. (Redrawn from Blanchet et aI., 1980.)

SUMMARY We have dealt with a large body of basic information on the physical characteristics of soils and methods of soil exploration. A brief summary of what we covered follows . Soils are derived from the weathering of rocks and consist of gravels, sands, silts, and clays in decreasing order of particle size. Soils are conveniently idealized as three-phase materials: solids, water, and air. The physical properties of a soil depend on the relative proportion of these constituents in a given mass. Soils are classified into groups through their particle sizes and Atterberg limits. Soils within the same group are likely to have similar mechanical behavior and construction use. Flow of water through soils is governed by Darcy's law. A soil mass can be made denser by removing the air constituents through

2.14 SUMMARY

71

mechanical effort (compaction). The main physical parameters for soils are the particle sizes, void ratio, liquid limit, plastic limit, shrinkage limit, plasticity and liquidity indices, and the coefficient of permeability. Water can significantly change the characteristics of soils.

FO

Practical Examples

EXAMPLE 2 . 16

(a) The weight of sandy soil from

to

embankment. (b) The number of 10.0 m3 truck construction. (c) The weight of wate (d) Strategy The

Solution 2.1 Step 1:

x V

C"Yd),equi,ed

~ x 45

X 103 15.7 51.6 x 103 m3

=

C"Yd)borrow pit

=

e number of trucks required. Number of trucks

Step 4:

=

51.6

x

10

=

5160

Determine the weight of water required. Weight of dry soil in one truckload : Wd Weight of water:

Step 5:

10 3

WWd =

0.15

X

=

10 x 15.7

=

157 kN

157 = 23.6 kN

Determine the degree of saturation. S

=

wG = 0.15 x 2.7 e 0.69

=

0.59 = 59%

•

EXAMPLE 2.17

An earth dam requires 1 million cubic meters of soil compacted to a void ratio of 0.8. In the vicinity of the proposed dam, three borrow pits were identified as

72

CHAPTER 2

PHVSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

having suitable materials. The cost of purchasing the soil and the cost of excavation are the same for each borrow pit. The only cost difference is transportation cost. The table below provides the void ratio and the transportation cost for each borrow pit. Which borrow pit would be the most economical? Borrow pit

Void ratio

Transportation cost IS/m3)

1.8 0.9 1.5

$0.60 $1.00 $0.75

#1 #2 #3

r . ~

Strategy The specific volume is very usef volume of borrow pit material. Solution 2.17 Step 1:

and

Step 2: VI 2

FO

V3

=

1,555,555 m3

= 1,055,555 m 3 =

1,388,888 m 3

ransportation cost = Volume Pit 1 = $933,333 Pit 2

=

$1 ,055 ,555

Pit 3

=

$1,041 ,667

X

Borrow pit 1 is therefore the most economical.

EXERCISES Theory 2.1

Prove the following relations: _ Gs"Y,,, () a "Yd - 1 + e wGs(l - n) (b) S = -"-'-------'-n

$/m3

•

EXERCISES

2.2

Show that

2.3

Tests on a soil gave the following results: G s = 2.7 and e of saturation versus water content for this soil.

2.4

A spherical soil particle of radius r is settling in water, which is a' velocity, u. If the drag force for a sphere is 6'lTr"T]u, show that

=

1.96. Mak

r=

2.S

FO

R

fl> n for k is

n FIGURE P2.5

s

"~

-$----

I

Pumping well

, ------------------

Impervious

TH ...L Impervious

73

74

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOilS AND SOil INVESTIGATIONS

Problem Solving Assume Gs = 2.7, where necessary, for solving the following problems. 2.6

The mass of a sample of saturated soil is 520 grams. The dry mass, aft en drying, is 405 grams. Determine the (a) water content, (b) void ratio, (c) salurat a unit weight, and (d) effective unit weight.

J

2.7

A soil sample has a bulk unit weight of 19.6 kN/m 3 at a water 0 n nt of f)~o ~ etermine the void ratio, percentage air in the voids (air voids) , and t degre f saturation of this sample.

2.8

2.9

2.10

2.11

2.U

0.25

Pan

0.074 6.8

2.1

FO

2.6

200

~

Mass retained (grams) Sieve no.

4 10 20 40 100 200

Pan

Opening (mm)

Soil A

Soil B

Soil C

4.75 2.00 0.85 0.425 0.15 0.075

0 20.2 25.7 40.4 18.1 27.2 68.2

0 48.2 19.6 60.3 37.2 22.1

0 15 98 90 182 109 6

75

EXERCISES

2.14

The following results were obtained from a liquid limit test on a clay using the Casagrande cup device.

Number of blows

6

Water content (%)

52.5

12

20

28

32

47.1

43.2

38.6

37.0

(a) Determine the liquid limit of this clay. (b) If the natural water content is 38% and the plastic limit index.

isl2~

Yo. cal

(c) Do you expect a brittle type of failure for this soil? W

2.15

A fall cone test was carried out on a soil to det eflrurl~ cones of masses 80 grams and 240 grams. The

Penetration (mm)

8

Water content (%)

2.16

FO

2.17

10 kPa

t 0.4 m

ttt s

t

1m

Ll.-===:.J-_-.l-~

FIGURE P2.16

30 55.1

76

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

1 l

10m

k, = 2.3

X

10-2 em/s

2m-tf~____________~k~l_=~5~.7~X~1~O~~C~nU~s~__________~. 10 m

1 __

k,

=9.2 X 10-7 cm/s

FIGURE P2.20

2.20

Calculate the equivalent coefficient of permeabil"t . profile shown in Fig. P2.20.

2.21

FO

2.22

2.24

f'

Pract" A Proctor co actio test in the laboratory on a borrow pit soil gives a maximum dry unit weight of 19 k 1m3 and optimum water content of 11.5%. The bulk unit weight and water content of the soil in the borrow pit are 17.2 kN/m 3 and 8.2%, respectively. A highway fill is to be constructed using this soil. The specifications require the fill to be compacted to 95% Proctor compaction. (a) How many cubic meters of borrow pit material is needed for 1 cubic meter of highway fill? (b) The water content at 95% compaction is 11 % on the dry side and 12% on the wet

side of the compaction curve. What is the minimum additional weight of water per unit volume to achieve 95% Proctor compaction? (c) If the soil were expansive, what is the additional weight of water per unit volume to achieve 95% Proctor compaction? (d) How many truckloads of soil will be required for a 100,000 m3 highway embankment? Each truck has a capacity of 10 m3

EXERCISES

77

(e) Determine the cost for 100,000 m 3 of compacted soil based on the following: purchase and load borrow pit material at site, haul 2 km round trip, and spread with 200 HP dozer = $2S/m J ; extra mileage charge for each km = $3.10/m J ; round trip distance = 10 km; compaction = $1.02/m 3 . 2.25

Operation

Compaction

FO

R

Miscellaneous

100 90

.-.t.

80 70 Qj c

u:

60 -

'u.s.~a~oo c 258

~

I'

~

t

!I II

l-

4

10

Pi! 2

II

/

I' 1

l/pij 1

!

--

i

I

l

i

--

1j!

If

40

/

i

30

-rI

20

i

10

FIGURE P2.25b

20

' ''j'

~

o

i'l'

j

50

0.001

40

100 I

I

if

I!

I

I I

I

Y

1/ 1

. }I

I

I

I

0 .01 0.1 Particle size (mm) - logarithmic scale

10

78

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

~_ _R_iv_e_r ----....

&------[Qt-Excavation

8m-+hF--2~~------ ~ i Not to scate~

Cross section a-a. FIGURE P2.26

FO

2.26

An excavation is pj,l6piillXl P2.26. It is expect~~~

CHAPTER

3

STRESSES, STRAINS, A ND ELA STIC DEFORMATIONS OF SOILS

3 .0

) ES

ED

90

10

INTRODUCTION You would have studied in mechanics he s resses imp ~sed on homog eous, elastic, rigid bodies by externa fo rces, So are not ho ogeneous" elast ic, rigid bodies, so the determination 6{ s e ~ ~n d strains) lSO l is a pa icularly diffi cu lt task. You may s~ 1{ so s are not elastic mate 'als, t hy do I have 10 study e lastic mcttiods of anal~ 's?" Here are some...reaso ns wny a knowledge of elastic analysis is ~dvantage'o s. isotropic e I in ~ ves only two constantsAn elastic ana l~is-of us and Poisson's ratio-and thu S~we assume that soi ls are isoYoung's m tropic elastic rna na hen we ha ve a werfu l, ut simple, analytical tool to pred ict a soil's espo se under loading. We wjll ve to determine only the two elastk constants om our labo tory or fie ld tests. geot echnical enginee us~ sure that a geotechnica l st ructu re must not Iia ode)" any anli~!l:ted a-cTi ng c odition and that settlement under workad) must be within tolerable limits. We ing I!;ad (a fraction o~ lli~ olla w ula prefer the settl ~ eT working loads 10 be elastic so that no perma~enl seulemen ~ uld oceu a calculate the elastic settlement , we have to r e mpie, in designing foundations on coarse-grained use an elasticp nalysis. soils, we nonna)' assum that the settlement is elastic and we then use elastic analysis to calcula the ettlement. An importa nt task of a geotechn ical engineer is to determine the st resses and strains that are imposed on a soil mass by externa l loads. It is customary to assume that the st rains in the soils are small and this assumption allows us to apply our knowledge of mechanics of elastic bodies to soils. Small strains mean infi nitesimal strains. For a realistic description o f soils, elastic analysis is not satisfactory. We need soil models that can duplicate the complexity of soil behavior. However, even for complex soil models, an elastic analysis is a fi~t step. In this chapter, we will review some fundamental principles of mechanics and st rength of materials and apply these principles to soils as elastic porous materials. This chapler contains a catalog of a large number of equations for soil stresses and strains. You may become weary of these eq uations but they arc necessary for the analyses of the mechanical behavior of soils. You do not have . to memorize these equations except the fundamental ones.

1

79

80

CHAPTER J

STRESSES, STRAINS, ANO ELASTIC OEFORMATIONS OF SOILS

Wh en you complete this chapter, you should be able to: • Calculate stresses and strains in soils (assuming elastic behavior) from extemalloads • Determine stress states • Determine effective stresses • Determine stress paths • Calculate elastic settlement You will lise the (ollowing principles lc_ ,.,;:","o;;;:,,: mate ri als: Stresses and strains Mohr's circle E lasticity-Hooke's law

'cs and strength of

)'

Sample Practical Situation l' 0 stprage tanks . " ''0, be our c11~ nl and the layer of sti ff saturated cI aeed an es imate of designi ng the pipe they are . B) :use o f land

a deep who is tanks when desires that

FIGURE 3.1 The "kissing" silos IBozozuk. 1976. permission from National

Research Council of Canada).

3.2 QUESTIONS TO GUIDE VOUR READING

81

the tanks be as close as possible to each other. If two separate foundations are placed too close to each other. the stresses in the soi l induced by each foundation overlap and cause in tolerable tilting of the structures and their foundations. An example o f tilting of structures caused by stress overlap is shown in Fig. 3.1. These silos tilted toward each other at the top because stresse the soi l overlap at and near the internal edges of their fou ndations. The fo ndations are too close to each other.

3.1

DEFINITIONS OF KEY Stress or inte nsity of loadi ng is the load per unit of stress is the ratio of the fo rce AP ....... v. ~ AS when AS tends to zero; A denotes a Effectil'e l'tress

definit ion the area of the plane

«(T; is the stress c"Ticd~'Y_lhe ~iI p" !lklle s..

Total stress (u) is the stress in the voids.

,.(

and the I quid\ and gases

Strain or intensity of d~,~~;,;{~;;:~~;~~ the original dimension I

ch'(PI!e in a dimension al length .

10

Sires:,' (strain) ""teaf'aix,int all pla nes passing IIij,o",gh sen t st ress (strain) Mean stress. p is the ' ",i'li!(e stresses in lhr~o ns. ~el';alOriC slre~q, IS

the shea

are aler:pres!J'urt; u,~ p

sure ) the water held in the soil pores .

.\"I~/h is a graphi 81 re resent1! 'on of the locus o f stresses on a body.

Isdtr/Pic means It('{.. arne ~a l propert ies in aUd irections and also the same ~ di n g

in all d 'rection,

AI/isotropic me~ s the ~aterial propenies are different in different directions and also the loadm a different in d iffere nt directions. Elastic material:,' are materials that return to their original configu ration on unloading and obey Hooke 's law.

3.2 QUESTIONS TO GUIDE YOUR READING 1. 2. 3. 4.

What are normal and shear stresses? What is stress slate and how is it determined? Is soil an elastic material? What are the limitations in analyzing soils based on the assump tion that they (soils) are elastic materials?

82

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

5. What are shear strains, vertical strains, volumetric strains, and deviatoric

strains? 6. How do I use elastic analysis to estimate the elastic settlement of soils and

what are the limitations? What are mean and deviatoric stresses? What are the differences between plane conditions? How do I determine the stresses and strains/di soil mass by external loads? What is effective stress?

7. 8. 9. 10.

3.3

STRESSES AND STRAII\J 3.3.1 Normal Stres

(3.1)

cu e compressed by /)'x, /)'y, and these directions, assuming they

in the are small (in

/),Z

FO

~ 0 33 ,

0

o

x

.11). £=y

Y

(3.2)

2VO (3.3)

FIGURE 3 .2

Stresses and displacements due to applied loads.

83

3.3 STRESSE S AND STRA INS

3 .3 .3 Shear Stresses and Shoar Strains Let us consider, for simplici ty, the XZ plane and apply a force Fthat causes lhe square to distort into a parallelogram as shown in Fig. 3.3. The force F is a shearing fo rce and the shear stress is

I,·FI xy

34)

distortio~.f~'J

Simple shear strain is a measure of the angu lar s nng forces. If the horizontal dIsplacemen t is L\.x. the shearst raln or sim Ie shear strai n, "'lUI is

For small strains, Ian "'Iv. =

"y p

and Iherefp (3.5)

If the shea r stress on a plane is zero, he nogna l stress on that plane c, Ued a pn nclpal stress. We will discuss princi I resses late ! In geotechnical engi· neering. compressive stresses in ils a assumed t.9 P,.OSitive. Soils cannOI sustain any appreciable"tensile y:~sse and we norrMUy a me al the tensile strength of soils is nlgJigible. St~ns can be compressiv or I I e,

J,

2. J. 4.

5. 6.

The ~ential points DTt!: A norm" 1 SIIUS is Ihe load per unit area on a plane normal to the direction oj the load. A shear sQ"e.JS is Ihe load p er unit area on a plane paraUdto the direction of Ihe shear fo rce. .1Vorm .. # Jtressu compress or e/ong,a te a malerial; shear strrssu distort a material. ,A normal strain is the change in length divided by the original length in the direction oj the original length. Princigal ..strt!:SSu an normal strtSSu on planes oj zero shear Siress. Soils can only sustain compressive stresses.

FIGURE 3 ,3

Shear stresses and shear strains.

84

CHAPTER J

STRESSES. STRAINS, ANO ELASnc DEFORMATIONS OF SOILS

What's next , . .What happens when we apply stresses to a deformable mate rial ? From the last section, you may answer that the material deforms, and you are absolutely correct. Different materials respond d ifferently to applied loads. Next, we will examine some typical responses of deformable materials to apR led loads to serve as a base to characterize the load ing responses of soils. ,

3.4 IDEALIZED STRESS-STRAIN RESPONS AND YIELDING 3.4.1 Material Responses to Normal Unloading If we apply an incremental vertical load, ll P, a d~ f9nnablc cylinder (Fig. 3.4) of cross-sectional area A, the cyl inde will compT9S by. say, 6. nd the radius will increase by 11,. The loading co (litia n we apply here is ca llt ui aXiall Oading. The change in vertical stress i , J

(3.6)

The vertical aod ra (3.7)

(3.8)

where Ho is t original lengJlfa-nd '0is the original radius. The ratio of the radial (0 ateral) sttain to the rtical train is ca ll ed Poisson's ratio, v. defined as =

--

- 6e,

".

(3.9)

Typical values o~jsson's ratio for soil are listed in Table 3.t. We can plotalrap~t = L l1u t versus Et =! I1E,. lffor equal increments of I1P, we g the same value of 11<;, then we will get a straight line in the graph of U t versus £l shown by OA in Fig. 3.5. If al some stres.<; point, say. at A (Fig. 3.5), we un load t e cylinder and it returns to ils original configura tion, the ma~

.,

T

T

1 1+T -,--_O,j glnal coofjguralion

1

Delorml!d

"'.uv -

configuration

r~f--

FIGURE 3.4

Forces and displacements on a cylinder .

3.4 IOEALIZED STRESS-STRAIN RES PON SE A ND YIELDING

85

TABLE 3 . 1 Typical Values of Poisson's Retio Soil type

."

Descripti on Soh Medium Stiff loose

Clay

Sand

Medium

Dense

0.35-0.40 0.30-0.35 0.20-0.30 0.15-0.25 0.25-0.30 0.25-0.35

"These values are effeclivs va lues,

~'

(see

d,scussions later in [his chapter. SSClions 3.12 eod 3.13).

{erial comprising (he cylinder is called a line {-x. elastic ma terial uppose fo r equal increments of 6. P we get different ue ('6.z, but 0 unloading the en a plot 0 stress-strain cylinder it relOms to its original 0 figuratio. relationsh.ip will be a curve as ilIusu* ed b OB in Fig. 3.5 . In this !foe, the material comprising the cylihder is calleCi a nonlinear! elastic material . If we apply a load P t t hai ca uses a d placement 6.z 1 on an elasti a teria and a second load P z that causes a ~l@cem nt 6. z 2 , lhen the tota1 displa me t"fs 6..<.: = 6.z 1 + 6.'<':2' Elastic materials obey the pc' ciple of sUPJ.(pQsition he rdec in which the load is ap plied is no important; we could apply ~ rst an t hen P I bu t Ihe fi nal displacement would the s~ me. Some ~a terial s, so IS one of them, do not et Ii' to their original configurations after unl ing. They exhibit a Iress-str in relationship similar 10 Ihat depicted in Fig!'3.6, where OA is the I'oad ing r sponse, AB the unloading response, and B the reload in l! ponse. T he strai ns that occu r during loading, ( A, nsist of wo pa rts- e las' or recoye rable part. BD. and a plastic or unre OVer8 Ie pa ri , D B. uc aleria behavior is ca lled eJaslopiaslic. Pan of ~~"",the o adlng response i e astic, th otper plastic. engineers, e are p icularly interested in the plastic strains since these ar ttie result of pef,{'lanent de orma lions of the material. But to calculate the

plirmanent d" tmatm)

Linearly

e l a~tlC

A

w

ust know the elastic deformation. Here , elastic

S lope IS E, . tile langent elastic modulus

8

S lope '5 E, . the socanl elastoc modu lus

o FIGURE 3 . 5

Li near and nonl inear stress-strain cu rves of an elastic material.

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

Slope is E, the initial tangent elastic modulus C

Elastic response during unloading

o

B D r-Plastic+Elastic--j

FIGURE 3 .6

Strain

Idealized stress-strain curves 0

F

86

Tzx

Initial tangent shear modulus, G j

Secant shear modulus, G

FIGURE 3.7 Shear stress-shear strain response of an elastoplastic material.

3.4 IDEALIZED STRESS- STRAIN RESPONSE AND YIELDING

a,

87

Y,eld surface

Elastic

reg;oo

B

"I FIGURE 3.8

Ibl

Elastic, yield, and elaSloplastic stress stales.

(0) is the slope of a line from the desired shear SI e origin of the,.u versus"'lu plot (Fig. 3.7).

ain point to the

3.4.3 Yield Surface Let us consider a more complex sit ation t~~ th uniaxial loading 0 a cylinder (Fig. 3.8a). In this case, we are gt;'i to appJ~ increments of--veftlc and ra dial messes. Since we are not applying any hear stresses, the axial stresses 81ld radial stresses are pri ncipal stresses; - 0"1 = I ti.O"t and (1, = L flu ., respectively. Let us, for example, se t (13 t ~O .~d increase %.The ateria will yield al some value of 0"1, w iell...we w caJl (1 1)y, and pl6ts~ POlflo.! n Fig. 3.Sb. If, alternatively, we se't(11 = 0 an jncrease 0"3. tlIe maten al wi1y yield at (0" 3),. and is represen ted by pbint B in F~g.J3.8b. We can the subjecHl\e cylinde r to various combi nations of (11 d.. CJ) axes.

AB,

T he essen.h" poiot$..are: An elastic rna trial recO.J'ers its original configuration on unloading; an elas/oplastic material undergoes both elastic (reeotJerable) and plastic (p ennanentj defonnation during loading. 2. Soils are dasloplaslie materials. J. A t small sIra/ns, soils behatJe like an elastic material alld thereafter like an eiastoplastie material. 4. The locus oj the stresses at which a soil yields is called a yield l·uiface. Strel'ses below the yield stress cause Ihe soil 10 respond elmtically; stresses beyond the yield l·tress cause Ihe soitlo respond elal·toplastically.

What's next . , .I n the next two sections, we will write the genera l expression for Hooke's law, wh ich is t he fu ndamenta l law for linear elastic materials, and then conside r two loading cases appropriate to soils.

88

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

HOOKE'S LAW 3.5.1 General State of Stress Stresses and strains for a linear, isotropic, elastic soil are related t law. For a general state of stress (Fig. 3.9), Hooke's law is 1 -v -v

ex

-v

ey ez

1 -v

-v -v

0

0

0

0

1

0

0

0

o 2(1

+ v)

'{yz

0

0

0

0

'{zx

0

0

0

'{xy

1 E

(3.10)

(3.11 )

where

FO

3.5

CHAPTER 3

(3.12)

shown in Table 3.2.

z y

~--------L-----+X

FIGURE 3 .9

General state of stress.

3.5 HOOKE'S LAW

TABLE 3 .2

Typical Values of E and G Description

E-,MPal

G- (MPal

Clay

Soft M edium Stiff

Sand

loose Medium

1-15 15-30 30- 100 10-20 20- 40 40- 80

0.5-5 5-15 15-40 5- 10 10-15 15-35

Soil type

89

Dense

"These ale .velage seCln! elu!;c mod>! li for !he d .. ined condition (~ee discussions letel in Ihis c~ep!e., Sections 3.12 and 3.131.

3.5.2 Principal Stresses If the stresses applied to a soil are

iifclOa l stre e , then Hooke': (w reduces

to

(3.13)

The matrix on the Thc invcrse of Eq.

-calledoillhe- compl iance

matrix.

(3.14)

matrix on the righ led the stiffness matrix. If you know the s'8nu the mate I 1 parameter E and v, you can usc Eq. (3.13) to calcula te U:!'e st ains; or if Y9 know:ne s~ il1 s. E, and v. you can use Eq, (3 .14) to calcu late t e stresses. 'V'" tre

...._"

3.5.3 Displacements from Strains and Forces from Stresses The displacements and forces are obtained by in tegration. For example, the vertical displacement. 6.z. is (3.15)

and the axial force is (3.16)

where dl. is the heigh t or thickness of the element and dA is the elemental area.

90

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

The essential points are: 1. Hooke's law applies to a linearly elastic material. 2. As a first approximation, you can use Hooke's law to calctJlO:le stresses, strains, and elastic settlement of soils. 3. For nonlinear materials, Hooke's law is used with an.ap oximate elastic modulus (tangent modulus or secant modul s an he cal~ula tions are done for incremental increases in stres

What's next , ..The stresses and strains in three when applied to real problems. For practical puq~ tion, we will discuss two conditions that sim

3.6 PLANE STRAIN AND A"".~... SYMMETRIC CONDITIONS

FO

(3.17) (3.18)

Reta ining wall

Z (1 )

r

;

(2)

~X(3) FIGURE 3 .10 Plane strain condition in a soil element behind a retaining wall.

3.6 PLANE $TRAIN ANO AXIAL SYMMETRIC CONOITIONS

and 10'1 .., t.{0'1 + a,}

I

91

(3.19)

In matnx form, Eqs. (3.17) and (3.18) become

{,,} . ~ [' -, E

tl

- ],/

-,

1 -],/

]{"'} O'J

(3.20)

The inverse of Eq. (3.20) gives ~~----~--~~

{::J -

(I +

,)~

2,) [' : '

(3.2l)

3.6.2 Axisymmetric Condition T he other co ndition that occurs in practica l ble is axial symmetry or the axisymmerric condi tion where two Slres a re u61. Let us consi er a water tank or an oil tank founded on a mass ;11 trated in Fi 3.U . The rad ial stresses (0",) and clccumfer ntial stresses (cr,) on a lindrical element of soil directly und the ce n t~ qt-the ta nk are equal because of ax ial symmetry . The oil tank will pply 8 unifo rm vertic· J ( xial) stress at the soil surface and the soi l element w~1I be subjected to an merea in aJial stress, 6 0, "" 60"10 and an incre€e" in radial stress, 60 , "" 6 0, "" crl' ill a soil element under the edge of j\e tank bj under an axis): e. ndition? The answer is no, since the stres ' s at th., edge of the ta nk c...e all differen t; the re is no "" symmetry. Hooke' la 'fo r the axisymmetric condition is (3.22) (3.23)

or, in p

tri x form , (3.2 4)

z

, Tank

,I

1 FIGURE 3.11 tank.

I ""

Axisymmetric condition on

a soil element u nder the center o f a

CHAPTER 3

STRESSES. STRAINS. AND ELASTIC DEFORMATIONS OF SOILS

The inverse of Eg. (3.24) gives

t:}

= (1 +

v~1 _ 2v) [1 ~ v~v]{::}

The essential points are:

(3.25) i

1. A plane strain condition is one in which the strain in one or more ai-

rections is zero or small enough to be neglected. 2. An axisymmetric condition is one in which two stre$Ses are qual. EXAMPLE 3.1

A retaining wall moves outward causing a strain of 0.05% on a soil element loc e(:l

FO

. e stliess condition and write the appropriate

Step 2: ~(Jl

~(J3

= 9615.41(0.7 x 0.0005) + [0.3 x (-0.001)]1 = 0.5 kPa = 9615.41(0.3 x 0.0005) + [0.7 x (-0.001)]! = -5.3 kPa; the negative sign means reduction

Step 3:

Calculate the lateral force per unit length. ~cr3

=

~Px =

~(Jx

f ~(Jx

dA = - (

5.3 (dx

x 1) =

-[5.3x]8 = -31.8 kN/m

•

EXAMPLE 3.2

An oil tank is founded on a layer of medium sand 5 m thick underlain by a deep deposit of dense sand . The geotechnical engineer assumed, based on experience,

93

3.6 PlAlIIE STRAIN AND AXIAL SYMMETRIC CONDITIONS

Tan k

.ok"" 2OkPa _

T

2.5m

.1

--

5m

Med,um Sind

FIGURE E3. 2

/'.

that the sertlement oC the lank would occurArom ~ ttlem t in the mcdium sand. The vertical and lateral stresses al the middle f em dlum sand directly under nd a, re ~. Ivcly. The valu s of E and the center of the tank are 50 k Pa ~ 1/ are 20 M"Pa and 0.3, respectively ssuming a li.pear, isotropi elas c material vertical behavior, calculate the strains im ed on t e medium sana and I settlement. Strategy You have to decide the stress condition on the il element directly under the c~ter of th tan . Once you mak e yo deCision, use the appropriate equatioils to find th~ trains and then integrate e vertical strains to ca lculate the set emen!. D a.Jr a diagram i6us~ng the roblem.

Solution 3.

,

Y

Ora a diagram of the proble ~g . E3.2. Decid ~.Jm a stress cop dition. The element is dir t U r he center of the tank , so Ihe .... ' _'axi ymmclric conditio

~~~~~.s the2~:pe { E)

== 20

llE) = 20

~

t()l [1

= 20

~

101

.6.£3

WI~06l(50}

~ to' [-~.3

0.7

20

e gel

Using alg bra

Step 4:

u.lion, ,nd

x SO - 0.6 x 201 = 1.9 x 10 - 3

[- 0.3 x SO + 0.7 )( 20] - -5 )( lO-s

Calculate vertical displacemcnt. 6£) = 6(, 6z

=

f

6£., dz - [1.9

X

10- 3

z15 = 9.5

)( 10- 3

m=

9.5

mm

•

What's next . ..We have used the elastic equations to ca lcul ate stresses, strains, and displacements in soils assuming that soils are linear, isotropic, elastic materials.

94

CHAPTER 3

STRESSES. STRAINS. AND ELASTIC DEFORMATIONS OF SOilS

Soils, in general, are not linear, isotropic, elastic materials . We will briefly discuss anisotropic, elastic materials in the next section.

F

3.7

ANISOTROPIC ELASTIC STATES Anisotropic materials have different elastic parameters' Anisotropy in soils results from essentially two causes

1.

2. The difference in stresses in induced anisotropy.

s Poisson's ratio determined from the ratio of the strain (X direction) to the strain in the vertical direction (Z direction) with ad applied in the vertical direction (Z direction). In the laboratory, the direction of loading of soil samples taken from the field is invariably vertical. Consequently, we cannot determine the five desired elastic parameters from conventional laboratory tests. Graham and Houlsby (1983) suggested a method to overcome the lack of knowledge of the five desired elastic parameters in solving problems on transverse anisotropy. However, their method is beyond the scope of this book. For axisymmetric conditions, the transverse anisotropic, elastic equations are irecti

(3.26)

95

3.1 ANISOTROPIC ELASTIC STATES

where the subscript z de notes vertical and , denotes rad ial. By su perposit ion, \1,/ "'" = EJ E•.

The essential poinls are: J. Two forms of allisotropy an praenl ill soils. One is struc/ural anlsolropy, which is nlatuJ to the history of 10000l11g and enfJlronmental conditiolls during chposilion, 4IId Ihe other is slrtSS-induced onisolropy, which raults from differmcrs in strrsses in different directions. 2. The pNwulenl form of structural anisQtropy In soils is /ralU ven 'e anisolropy; 'he soil properties and the soil rtSponse ;n tire la/eral directions are the slime but lire different/rom those in the vertical direction. 3. You need 10 Ji.d Ihe elastic parameters In different directions of a soil mass to dele,..",/ne elaJlic strrsses, strains, and displacements.

EXAMPLE 3 .3 Redo .Example 3.2 but now the soil un ~e 0 11 ta nk is an anisotropIc maten al Wit h E~ = 20 MPa, E~ MPa v'r == 0.15, v" "" 0.\

Strategy The (326)

SOI~f this problen~ IS a stralA~wara app"

aSllc

tiOn of Eq.

,

Solution 3 .3 Step 1:

Determine

( by superpositio

1-2 0.,,]

Find the ~!~~

use' . ~3.,

{""J At,

...

to-

,[

x

20 -0.12 (l

25

20

The solu tion is E~ 0.03%.

Step 3:

=

~50.3) {~}

2.26 X 10- 3 = 0.23% and €, = 0.26 x 10-' -

De termine vert ical displacement. 4z =

fE,

dz -

(2.26 x lO-lzl&

=

11.3 X 10 -) m

=

11 .3 mm

The ve rt ical d isplacemenl in the anisotropic case is about 19% more than in the isotropic case (Example 3.2). Also. the radial strain is te nsile fo r the isotropic case but compressive in the anisotropic case for this problem. •

96

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

What's next . . .We now know how to calculate stresses and strains in soils if we assume soils are elastic, homogeneous materials. One of the important tasks for engineering works is to determine strength or failure of materials. We can draw an analogy of the strength of materials with the strength of a chain. The cain is only as strong as its weakest link. For soils, failure may be initiated at a poi w ithin a soil mass and then propagate through it; this is known as progressive fai reo The tress state at a point in a soil mass due to applied boundary forces y equal t o the strength of the soil, thereby initiating failure. Therefore, as e g1 ed to know the stress state at a point due to applied loads. We will using your knowledge in strength of materials.

3.8

STRESS AND STRAIN STATES

FO

1

rY,

T

~

0

M

'~ 1-0,

Plane on wh ich the

Txz

(J

II'

(rY,. (a)

FIGURE 3 .12

-T,) (b)

Stresses on a two-dimensional element and Mohr's circle.

l .8 STRESS AND STRAIN STATES

97

and 3. The stresses at these points are the major principal stress, (TI! and the minor principal stress, (Tl' The principal stresses are related to the stress componen ts (1. . (T... 'T u by cr, +

2

CT~ =

a, + )(CT, -2 aA) ~ + r.

,~

+

0-, a~ -, --

J(" -2 '<)' + , T ,~

The angle between the major pri ncipal stress plan is tan'" -

(3.'8)

rita I plane (1/1)

(3.'9)

(3.30)

(3.31)

The stresses o n a pia

orient d at an all (3.32)

0:

(3.33)

above eq uations

IS

~ maX1l11U~ h ar...st

_ 'T.... ~ -

0'1 -

2

a}

(3.34)

For the stresses sb,.bwn 19 FIg. 3.9, we would get three cIrcles but we have simplified the problemby-p10tt1l1g one cIrcle fo r stresses on all planes perpendIcula r to one pri ncipal direction. The stress CT, acts on the horizontal plane and the stress CT.., acts on the vertica l plane for our case. If we d ra w these planes in Mo hr's circle , they intersect at a point, P. Point P is ca lled lhe pole of the stress circle. It is a specia l poi nt because any line passing through the pole will intcr.)ect Mohr's circle at a point that represents the stresses on a plane parallel to the line. Let us see how this works. Suppose we wan t to fi nd the stresses on a plane incli ned at an angle a to the horizon ta l plane as depicted by MN in Fig. 3.12a. Once we locate the pole , P, we can draw a line parallel to MN through P as shown by M 'N' in Fig. 3.l2b. The line At' N' intersects the circle at N' and the coordinates of N' , (<1&, T.). represent the normal and shear stresses on MN.

100

CHAPTER J

Step 5:

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

D etermine '!T. Draw a line from P to IT. and measure the angle between the horizontal plane and this line, qr = 22,50

Alternatively, the angle AOC "" 21J1 = 4SO :. 'If .. 22,50

Slep 6:

Determ ine the stresses on a plane inclined at 3 principal stress plane. Draw a line M' N' through P wIth an inclination the major principal stress plane, angle CPN'. TIle ~ at at point N' is (470, 120).

Alternatively:

St,p 1,

Use Eqs. (3.27) to (3.29)

nd-1' .,.

2 258.6 "" 141.4 kPa

Ild eYe and " •.

,

• IWQ"t"s next . ..The stresses w e have calcu lated are for soils as solid elastic materials,

We

have not accounted for the pressure within the soil pore spaces. In the ned section, we will discuss the principle of effective stresses that acco unts for the pres· sures within the soi l pores. This principle is the most important principle in soil mechanics.

3.9

TOTAL AND EFFECTIVE STRESSES 3 .9 . 1 The Principle of Effective Stress The deforma tions o f soils are similar to the deformations of structural framework such as a truss. The truss deforms from changes in loads carried by each member. If the truss is loaded in air o r submerged in watt!l, the deformations under a given

3.9 TOTAL AND EFFECTIVE STRESSES

a

101

Externa I force or tot aI stress

• .-/' Internal resistance • .....--from water or pore water pressure Contact area Internal resistance from sol ids or effective stress

a

FIGURE 3.14

Plane on which effective stress is calculated

Effective stress.

equation is (3.38)

so that (3.39)

(3.40)

where U a is the pore air pressure, U", is the pore water pressure, and X is a factor depending on the degree of saturation. For dry soil, X = 0; for saturated soil, X = 1. Values of X for a silt are shown in Fig. 3.15. To determine the effective stress in a soil mass, the pore water pressure must be known. The pore water pressure at a particular point in a soil mass is

102

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

0'

---

/'

0.' 0'

./

0.6 X 0.5

./

O'

0' 0.2 0.1

o

0

20

"

60

Degree of saturation

80 (%)

/': ~~

FIGURE 3.15 Va lues of X for a silt at dlffe r8~ e9,eeS of saturation.

[he depth of water above that poi t.mu!t1. ' p ied by the unit eig 0 water. Pore water pressu res are measured by p .. wa r pressure transducers (Fi~0"3 .16) or

by piezometers (Fig. 3.17~ln a pore tee pressure transducer, water passes through a porous material and p u stt~gainst a metal diaphragm which a strain gauge is anached. 'he...s tra~ gaug~is usuaUy wire 'nto a W~atstone bridge. The pore water p essure tran ducer is calibrateclb a p~lyi ng known pressures and measuring t electric; ! oJtage aurpul f.(,2m--r:he Wheatstone bridge. Piezometers are por tuby that allow Jh~p ss a ~ of water. In a simple piczometer, you C8Jl measure the height of wft er in t¥ tube from a fixed elevation and then calcula e ~ water prcssur by multi ~ly in g the height of waler by the unit weight Qf water (Chapt er 2). A bOI hole cased to a cert ain depth acts lik e a pje20meter~ odern piezo etcrs are equipped with pore water pressurc lransdue rs for elc Ironic read' an dala acquisition.

9'e...

The effect"

stress in a soil mass not subjected to external loads is found from t of the soil and the depth of grou ndwater. Consider a soil element

Poroos elemellt

FIGURE 3 .16

Schema tic of a pore

water pressure transducer.

FIGURE 3 .17

Piezometers .

3.9 TOTAL AND EffECnVE STRESSES

GrQund

~

103

Ground !ewe!

'I (;WL. Ground .... a!ef leve!

,

._-L

.,----'-

,.,

'bi

FIGURE 3 .18 Soil element at a depth

z with

groundwater level at

( a )~und

level

T

and (b) below ground level.

at a depth l below the ground surface an g4!.le gr u ndw,rer level (G WL) is at ground surface (Fig. 3.18a) . The total verticaf ess is (3.41)

Thc pore water pressure is (3.42)

(3.43)

=

(j'

=

~ ==

= ";J'z ...

+

"Y ...(z -

,(<: - <: ... ) - "Yw(z ~

+ h ••, """"'j ... )(z - lM )

= ""it ...

+

z .. )

z...)

"Y (z -

tM)

.3 Effec(s of C8l~ i a rity In si lts and 6ne ds, th soil above the groundwater ca n be saturated by capill ary action. You wouldl ave e ncountered capillary action in your physics course when you studied menisci. W e can get an understanding of capillarity in soils by idealizing the conti nuous void spaces as capillary tubes. Consider a single idealized tube as shown in Fig. 3.19. T he height at which water will rise in the tube can be fou nd from statics. Summing forces vertically (u pward forces are negative). we get I.F, = weight of water - the tension forces (rom capillary actioD

tha t is,

.d'

""""4"

[ ."Y .. -

'II"

dT cos

0. =

0

(3.44)

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOilS

~ Pore water pressure d istri bution

Solving for

Zc>

Idealization

we get (3.45)

O. Since T, a, and 'Yw are

(3.46)

FO

104

3 .9.4 Effects of Seepage In Section 2.9, we discussed one-dimensional flow of water through soils. As water flows through soil it exerts a frictional drag on the soil particles resulting in head losses. The frictional drag is called seepage force in soil mechanics. It is often convenient to define seepage as the seepage force per unit volume (it has

U TOTAl AND EffECTIVE STRESSES

(
105

VJ) UlIWilrd K1I~

Oown_d Sftp.1g1!

FIGURE 3.20 Seepage in soils. units similar to unit weight). which we will denot~f the ~"ead loss over a flow distance. L. is tlh , the seepage force is (3.47)

th n the scepa~st(e ses are in stresses. From st atic t:quilibrium )

stre f.l:!lHl~a

(3.48)

s arc in the opIC equilibrium

the

(3.49)

ces pay a very impor in destabil izi ng geotechnical structures. For e ample, a canti e er recai mng all , shown in Fig. 3.21 , depends o I dept h of e mbed m en~ - r ils lability. The retai ned soil (left side of wa ll) plies..an a tward lateroJ. pr e to the wall, which is resisted by an inward lateral r~ i ance [rom leSoil on he Ight side o f the wall. [f a steady quan tity o wa r Is available the-lett side j> the wail , for example, from a busted water p i ~e , then water wi flow from t e left side to the right side of the wall. The path followed by a p,article f wllte depicted by AB in Fig. 3.21 and as water flows from A to B. bad loss urs. The seepage stresses on the left side o f the wall

B

FIGURE 3.21

Effects of seepage on the effective stresses nea r a retai ning wall.

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

are in the direction of the gravitational stresses. The effective stress increases and, consequently, an additional outward lateral force is applied on the left side of the wall. On the right side of the wall, the seepage stresses are upward and the effective stress decreases. The lateral resistance provided by the embedment is reduced. Seepage stresses in this problem playa double ro increase the lateral disturbing force and reduce the lateral resistance) in red ing the st bility of a geotechnical structure. In Chapters 9 through 11, you ill t dy the ffects of seepage on the stability of several types of geotechni

FO

106

1.

2. 3. 4. 5. 6.

The essential points are: The effective stress represents the average-SIre carried" y the soil solids and is the difference between the t tal stre s antJ.the pore water pressure. The effective stress principle applies 0 11. to no mal stresses and not to shear stresses. Deformations of soils are e.. to effective not total str s. Soils, especially silts and fine ds, dan be affected by capillary action. Capillary action results iwnega five pore wate.r pressures a d increases the effective slresses. Downward seepage in(:r6(lSeS the resulta sl ess; upward seepage decreases the resultant effective st

Ights from the given data and you should water level is not saturated.

Above grounilwater level

_ (G1 ++ ese) _G1(1 ++ew) 'YIV s

s

'Y -

'( W -

Se = wG" 'Y =

:. e =

2.7(1 + 0.3) 1 + 1.35

X

0.3

X

0.6

9.8

=

2.7

= 1.35

14.6 kN/m

3

Below groundwater level Soil is saturated, S = 1 e = wGs = 0.4

X

2.7 = 1.08

_(G~ + e) _ (2.7 + 1.08) _ 1 + 1.08 9.8 - 17.8 kN/m s

'{sat -

'Yw -

3

J-9 TOTAL ANO EFFECTIVE STRESSES

107

C'QUrd II'VeI

1

20m

S .0.6 .. ", 30%

----'-'-'-~,'---~

5m ... .. 40%

.~

FIGURE E3.5

Step 2:

Calculate the effecti ve stress. Total stress:

Alternalively, = 2 X 14. + 3(17.8 - 9. ) = 53.2 kPa

•

EXAMPLE ig. £ 3.6a. Plot the distri-

v~l, ne ~t sa

30

5,'

w'th s,lt

",,,, ~,S " 0%

2.0

<""""-Fine sa

sat urated

by

'ol;;;"'"' ... " 12%

Soft blo.>e clay .... " 28%

20.6

FIGURE E3.6a

caprllary act,on

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

Solution 3.6 Step 1:

Calculate the unit weights.

0-2m S = 40% = 0.4;

w

= 0.05

s = 0.05 x 2.7 = 034 e = wG S 0.4 .

"Y

=

G s (1 + w) 1 + e "Yw

=

2.7(1 + 0.05) 1 + 0.34 9.8

2-5.4 m

5.4-20.6 m

Step 2: cr' =

Depth

u (kPa)

(m)

o

(I -

(kPa)

0

-1

o

x 9.8 = -9.8

2.4 x 9.8 = 23.5 23.5 + 15.2 x 9.8 = 172.5 or 17.6 x 9.8 = 172.5

51.6 ~.8

94.1 238.5

FO

108

25L-~-L~L-~-L~--~-L~~

FIGURE E3.6b

•

U

3.10 LATERAL EARTH PRESSURE AT REST

,"

109

,

FIGURE E3. 7

EXAMPLE 3.7

Water is seeping downward through a soil I~y er a in Fig. E3.7. Two piezometers (A and B) located 2 m apart> (ve IC~ showed a head loss of 0.2 m. Calcu late the resulta nt vertical eft ctiv€:;. stre~ for a soil element at a depth

of 6 m as shown in Fig. E3.7.

Strategy You ha ve to

"

."

caJctJ1!t~epy.ge stress. But

oob ain this you must

know the hydraulic gradient, hieb yo can find from {he data ~n.

Solution 3 .7 Slop 1,

F;"d the

~.

hYdca~:?::ent.

u

Step 2:

the average for the soil mass; then

• What' next . _ .We

h~

onl Y.£2Psidered vertical stresses. But an element of soil in he 9 ound is als subject d to"fatera l stresses. Next, we will in trod uce an eq uat ion that re lates the ve ieal an late ral effective stresses.

3.10

LATERAL EARTH PRESSURE AT REST The ratio of the horizontal principal effective stress to the vertica l principal effective stress is called the lateral earth pressure coefficient at rest (K that is, Q ),

K

o

= O'}

cr,,

(3.50)

The at-rest condi tion implies that no deformation occurs. We will revisit the atrest coefficient in Chapters 5, 6, and 10. You must remember that Ko applies only to effective not total stresses. To fi nd the latera l total stress, you must add the pore water pressu re. Remember that lhe pore water pressure is hydrostatic and, at any given depth, the pore water pressures in all direCli(ms are equal.

110

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

EXAMPLE 3.8

Calculate the lateral effective stress and the lateral total stress for the soil element at 5 m in Example 3.5 if Ko = 0.5.

F O

Strategy The stresses on the horizontal and vertical planes on the soil element are principal stresses (no shear stress occurs on these planes). You need to apply Ko to the effective stress and then add the pore water pressure to get the lateral total stress. Solution 3.8 Step 1: Ko =

Step 2:

, (J3

cr 1'

= (J~ I

(J z

. '

c~ •

6

esulting~eforIjlations .

The distribution of surface stresses within a soil is deter-

nlli1'e~bY a~fning thdt the soil is a semi-infinite, homogeneous, linear, isotropic, elastic tn\atlr'ial. A semi-infinite mass is bounded on one side and extends infinitely all other directions; this is also called an "elastic half-space." For soils, th~ofizontal surface is the bounding side. Equations and charts forseveral types of surface loads based on the above assumptions are presented. A computer program to compute stresses from surface loads is available in the CD ROM. Most soils exist in layers with finite thicknesses. The solution based on a semi-infinite soil mass will not be accurate for these layered soils. In Appendix B, you will find selected graphs and tables for vertical stress increases in onelayer and two-layer soils. A comprehensive set of equations for a variety of loading situations is available in Poulos and Davis (1974) .

iN

111

3.11 STRESSES IN SOIL fROM SURfAce LOADS

3.11.1 Point Load Boussinesq (1885) presented a solution (or the distribution of st resses for a point load applied on the soil surface. An example of a point load is the vertical load ........ transferred to Ihe soil from an electric power line pole. The increases in stresses on a soil element located al &nf'A (Fig. 3.22a) due to a point load , Q, are ,-------::l (3.51)

(3.52)

(3.53)

(3.54)

where v is Poi, n~ ta ti practice. Equ lion (3.51)

r

. ~ stress is needed in (3.55)

uence factor, an .----~.;_----::l

)~

1 1+

The

(,/z)~

(3.56)

distribution~he in

in vertical stress from Eq . 3.55 reveal that the d Teases with depth (Fig. 3.22b) and radial distance

Derease in v.ert~~1

(Fig.3.22c).

P."' ~")

Q

Q

\---"""--'_ t

Constant depth

M,

---------1----------;:0. Constant depth

-----.... '~'I------'::;='" -----------~,

CoR$\&nt depth

A M. (II)

FIGURE 3.22 distance.

(bl

lei

Point load and vertical stress distribution with d epth and radial

112

CHAPTER 3 STRESSl:S. STRAINS. AND ELASTIC DEfORMATIONS Of SOilS

LillO': I~ .

Q (/acM'rt)

l Ine IQ.)Q ,(

(Ioteelm )

1--"

... 'H·i .,:,::",-~p.

'. I I}

.L

""" -~~ao. ~I

(b )

FIGURE 3 .23 (a) lin e load and (b) line load nsar a rJ

3.11.2 Line Load With reference to Fig, 3.23a, the increases in..&uesses due to a hne I length), are

..-f2Qa .

M , :

1T1l2

L\(1~'"

"

t. Z~2

2 Q~2 l 1T(X2 + Z2)2

2g"L,

.o.Tu '"' 1T(.

lLL .

A practit;lI eXamt.~.a line load is tbe l'

"'d.

from~al.g bri ck wall.

Q~~ . (3.57)

(3.58)

(3.59)

1.3 Line Load Noar • BU,n ed

Eanh R8~ining Struc~u r8 / . The i~rease in lale~a sl(.ess ! ta _buned earth relaining structure (Fig. 3.23b) due to; 8' line load of ntensi

.J

Q (forcellength) is

The

4Qa2 b

6 ,

incr~ in l a terai force is

(1,

..

-rr Ho(a2

+

b1)1

r------::2Q :o---, !J.P~ =

;(Ql + 1)

(3.60)

(3 61)

3.11 ..JStrip Load A str(;'oad is Ihe load transmitted by a structure of finit e wid th and infinite length o n a soil surface. Two types of strip loads are common in geotcchnical engineering. One is a load th at imposes a uniform stress o n the soil, for example , the middle section of a long emban kmcnl (Fig. 3.24a) . The other is a load tha t induces a triangular stress distri bulion over an area of widlh B (Fig. 3.24b) , An example o f a st rip load with a triangular stress distribut ion is the stress unde r the side of an embankment.

311 STRESSES IN SOIL FROM SURFACE LOADS

113

, 8------1

.

q,uo!cef~rea)

" '0'

(>,

8--1

8

'~

I

,T

,,'

tWl

~ posmg (a) a u niform ~ess (1d~ a linear varying td)

FIGURE 3.24 triP load slress. (e) Sir ~ load near from a st rip load.

w~nd-{d )

etaini ng

lateral Jorce on a retaining wall

e stress q. (force/area) are as follows: s (Fig. 3.24a) ~""·"·o

+

Sin 0

cos (0

+ 213)J

(3.62)

(3.63)

6.T" ""

1! [sin ~

0 sin(o:

+ 213)]

(3.64)

where q. is the applied surface stress. (b) Area lransmitling a triangular sfress (Fig. 3.24b)

qt

(3.65)

'Rl! + Sin 2fl )

(3.66)

80 -

qt

B

80"~ = -

'IT

8T

'2)13

6.0", "":;

...

-

B

%SIn

a - - In "2 R~

"..!h... (1 + cos21l2-.r

2!a)

B

(3.67)

114

CHAPTER 3

STRESSES. STRAINS. AND ELASTIC DEFORMATIONS OF SOILS

(c) Area transmitting a uniform stress near a retaining wall (Fig. 3.24c,d)

~()".x

=

2qs (13 - sin 13 cos 2et)

(3.68)

7T

The lateral force and its location were derived by Jarquio (1981) and are

~P.x

z=

~6 [Ho(82

-

(3.69)

8 1)]

F O

_

=

lf2;,(8 z - 8]) - (R l - R 2 ) -+:;

..

where

ass is a circular

(3.71)

)3/2J

1 1

qs [

()"e

="2

+ (rolz)Z

2(1 + v) 1 (1 + 2v) - [1 + (r o/z)2]112 + [1 + (rolz?p/2

J

(3.72)

tic settlement at the surface due to a circular flexible loaded area

Below center of loaded area:

Below edge: I ~z =

I

~ qsD(l 7T

qsD(lE- vZ) I

~z =

2

- v

E

where D = 2r0 is the diameter of the loaded area.

(3.73)

)

(3.74)

l .,1 STRESSES IN SOIL FROM SUR FACE LOADS

115

3 _11 .6 Uniformly loaded Rectangular Area Many st ructural foundations are rectangular or approximately rectangular in The increases in stresses below t he comer of a rectangular area of width 8 and length L are

sh up~ .

.---~----~~~' 60' _!l!.. [Ian- I LB + LBl. (-'- + 2...

t

<.R)

R~

R)

(3.75)

(3.76)

(3.77)

(3.78)

where RI = (L 2 + '-'- -•.l{, These equations (3.79)

, (3.80) (3.81)

/

' " - q,1,

(382)

~~factor. The influence factor for the vertical stress

(3.83)

=

where m I and tJ = L It.. You can program your calculator or usc a spreadsheet program to find It. You must be careful in the last te rm (Ian - I) in progra mming. If ml + tJ l + 1 < m 2n 2 , then you have to add 'IT to the bracketed qua ntity in the last term. A chart for J, produced by a spreadsheet program is shown in Fig. 3.25. You would have to calculate m = BIz and n = LIz and read I~ from the chart; m and n are interchangeable. In general, the vertical stress increase is Jess than 10% of the surface stress when z > 3B. The vertical elastic settlement at the ground surface under a rectangular flexible su rface load is •

_ q,8(1 - ",) I

,

E

'

(3 .84)

116

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

0.30 0.251 . 0 .201 I, 0. 15

I I I I I II

t,,+iV

1.

F mz--/

z

1.5

~ I~

0.10 I

0.2

I

I

0.000.1

f7ftT! II II1

I v" ;1 I I I II " II "'-""""'" 10

I

I

I

m

FIGURE 3 .25 Influence factor for calculating t ttl the corner of a rectangle.

e: j Is == 0.31

~B---l\

L

----

1

7

L+z

]\1/

B + z----

FIGURE 3.26

a rectangle.

In(~) + 0.56

F O

0.05

Dispersion of load for approximate increase in vertical stress under

3.11 STRESSES IN SOIL fROM SURfACE LOADS

117

called the 2: I method). The su rfnce lond on an area, B x L, is dispersed at a depth z over an area (B + z) x (L + z) as illustrated in Fig. 3.26. The vertical stress increase under the center of the load is .0.0'

q,Bl.

=

,

(3.85)

(8 + z)(L + z)

The approximate method is reasonably accu rate (compart, B. '-' 3.11.8 Vertical Stress Below Arbitraril

Shap d Areas

Newmark (1942) developed a chart to deter me il rease in vertical stress due to a uniformly loa ded area of a~~pe. e chart consists of concentric circles divided by rad ial lines (Fig. J~ . The area..o each segment represents an equal proportion of the applied surface st ess~ a depth z ~ow the surface. If there are 10 concentric eirel (Only 9 are Ilown beca u e th,~ 10th extends to infinity) and ZO radial line . stress ea h circle is q nJ on each segment is qj(10 x ZO). The rad ius to th r tio of the fi rst (inner) c' 10 is found by setting au, = O.l qs in Eq. (3.71), 0.1 ,

~ q,[ L~~r] I -

from which r = O.Z7. i
Depth stille

I' N° 000' I

FIGURE 3.27 Newmark's chart for increase in vertical stress.

118

CHAPTER 3

STRESSES. STRAINS. AND ELASTIC DEFORMATIONS OF SOILS

The procedure for using Newmark's chart is as follows: 1. Set the scale, shown on the chart, equal to the depth at which the increase 2.

3.

5.

F O

4.

in vertical stress is required. We will call this the depth scale. Identify the point on the loaded area below which the stress is required. Let us say this point is A. Plot the loaded area using the depth scale with point A at the center of the chart. Count the number of segments (Ns ) covered by the scaled loaded area. If certain segments are not fully covered, you can~6Stimate what fraction is covered. Calculate the increase in vertical stress i

The essential points are: 1. The increases in stresses below"a surjp ce load are found by assuriUng the soil is an elastic, semi-infinite 2. Various equations are available~r thl Increases in -stresses friJJh sorface loading. 3. The stress increase qt any dtifJth depends on the shOP; a nd distribution of the surjace IQad. 4. A stress applied ~t fhe...surjpce of a soil "rass by a loaded area decreases w~th depth and lateral distance away fropt. the center of the loaded ,area.,

5. Th! vertk al stn!s~ increases are--generq.lly le~ than 10% of the surface stress when_"the depth to widJh ratio.is ereater than 3 .

. Determine the vertical stress increase

De~rmine

the load type.

~ssume the load from the pole can be approximated by a point load. Use the equation for a point load. Use Eq. (3.55): Z

r

= 5 m, Q = 200 kN; Under load , r = 0, ... - = 0 z

From Eq. (3.56):

Q

Clcr, = ~ I =

200

52

r -

=

X

0.48 = 3.8 kPa

z

0,

I

=

0.48

3.11 STRESSES IN SOIL FROM SURFACE LOADS

Step 3:

119

Determine the venical Stress at the radial distance.

,

r -

2 m.

0.0', '"

200

""S'1

l -

2

5"

= 0.4.

I - 0.33

x 0.33 - 2.6 kPa

•

~,

EXAMPLE 3 . 10

A rectangular concrete slab. 3 m x 4.5 m. rests on Ii ~~ a soil mass. The load on the slab is 2025 kN. Determine the vertica sl ss increase at a depth of 3 m (3) under the ceoter of the slab. point i'I (Fig. 3.1 . (b) under point B ( Fig. EJ.10a). aod (c) al a distance of 1.5 m fro~ C.Q neT, poiot C (Fig. E3.10a).

~ e~ulions

Strategy The slab is rectangular aa(j for a un iformly loaded rectangular area are for the c~of 'the a~e . You shoultf ivide the area so that the point of inlerest is ...-:£orner- a r$C angle(s). YO'u m ~ have to extend the loaded area if the pOion interest )s outside it ( Io/cted--ar is fictitious so you hay to subtracr tb clitious inc~ ase in stress area. \..

.,

~

I

3

1 3m

m-----J

.

"",

'. Sethon

FIGURE E3.10a

,,'

.

'c

The extension e extended

120

Cl-iA PTER J

STRESSES . STRAINS. AND eLASTIC DE FORM ATI ON S OF SOltS

3m

m

~

I_

,.,

3m

I,

~

~C

1. 5m ~

FIGURE E3.1 Ob, e

A.

Step 2: Divide the rectangle so that ~e ~~~ CQrner. In .this problem, all four recta~g!es after thii subd~vision are ~pal (Fig. E3. l0b), so you onlvfue;~ fin~stress Inc~Oi rectangle of size B = 1.510, ~ ~25 m and mul~iply lhe results"'liv 4. L 2.25 m· n : - = ~ O.75

~ 15

,

" ~ :::''''''~ = O.5;

c4 t,...•

0.11

Fro m the ~!)g. 3.26, I z = Step 3: Find the stre~crease at the centeJ of the\J~b (point A , F ig.

3. 0b~';'

~ A.

~

4~

2025

q.

=

3

0 ." a

'~' r-.Jj, ~: x 1

x 0 105- 6JkP,

~~: The appr ~m~{Eq. (3.85)] gIves a =

•

q;8L B "\ f)(L + z)

,.

2025 = 45 kPa (3 + 3)(4.5 + 3)

which is..ab~t 3~'SS than the e lastic so lution. Step 4: jfnd ~e stresv'ifcrease for point B. Po~t,B is at the corner of two rectangles each of widtb 3 m and length2.25 m. You need to find the stress increase for one rectangle ~rtd. m~fiply the result by 2.

(

3 m = - = t· n 3 ·

2.25 = 0.75 3

=- -

From the chart in Fig. 3.25, I . = 0.158.

Au, = 2 x 150 x 0. 158 = 47.4 kPa

You shou ld nO te that the ve rt ical stress increase at B is lower than at A , as expected.

R

O

122

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

Solution 3.11 Step 1:

Find the centroid. 1.0 x 10.0 x 5.0 + 1.5 x 2.0 x 1.0 + ~ x 8.0 x 1.5 x (2.0 + ~ x 8.0) 81 x = 1 = - = 4.26 (1.0 x 10.0) + (1.5 x 2.0) + 2 X 8.0 x 1.5 19

_

Y=

1.0 x 10.0 x 0.5 + 1.5 x 2.0 x (1.0 + 1.5/2) + ~ x 8.0 x 1.5 x (1.0 ~ 1.5/3) = 16.25 = 0.86 (1.0 x 10.0) + (1.5 x 2.0) + 2 X 8.0 x 1.5 19

ill

F O

Step 2:

ill

Scale and plot the foundation on a New~rK c art. The scale on the chart is set equal to the; depth. y entroid is located at the center of the chart ~ID!..he . un dati ' n is scaled u the depth scale (Fig. E3.11b) .

Step 3:

Step 4:

•

3.12

Ip

=

O' j

+

~2

+

0'3 =

O'x

+

i

+

0',

I

(3.86)

On a graph with orthogonal principal stress axes, at> a 2, a 3, the mean stress is the space diagonal, that is, a line oriented at equal angles to all three axes (Fig. 3.28). Mean stress causes volume changes.

3.12 STRESS AND STRAIN INVARIANTS

FIGURE 3.28

123

Mean and deviatoric stresses.

3.12.2 Deviatoric or Shear Stress (3.87)

3.12.3 Volumetri (3.88)

3.12. (3 .89)

FO

R

e ling with two conditions-axisymmetric and con , . ions, the stress and strain invariants are as

p' =

(I' I

+

2(I' 3

3

Iq =

Therefore, q

(II -

=

(I 3;

q' =

(I i -

(I 3 =

and p =

«(II -

(II

+

2(I3

(3.90)

--'--_----=.

3

6.u) -

«(I3 -

6.u) =

(II -

(I3 1

(3.91)

q' ; shear is unaffected by pore water pressures.

12p

=

21

I 2q = ~(2J

+ -

223 1

(3 .92)

I

(3.93)

23)

124

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOilS

3.12.6 Plane Strain, t2 = 0 p'

=

u'1 + u'2 + u'3

and

p =

+

Ul

3

U3

=P

- u

I

(3.94)

(3.95)

[(u; - (2)2 + (U2 - (3) 2 + (U3 -

or

u;fl 1l2 I

(3.96)

F O

~

+

3

IP' q' = q =

U2

3.12.7 Hooke's L Stress and StrBin are related as follows: (3.99)

n·lOO)

I

IG

=

£~ 3~ =

G' = 2(1

[K'0 {p'} q =

q

(3.101)

I

~

v')

0] {£~}

3G

£~

I

(3.102)

(3 .103)

Equation (3 .103) reveals that for a linear, isotropic, elastic material, shear stresses do not cause volume changes and mean effective stresses do not cause shear deformation.

3.12 STRESS AND STRAIIIIINVARIANTS

125

We can generate a generalized Poisson's ratio by eliminati ng £' from Eqs. (3.100) and (3.102) as follows: Equation (3.100): £' Equation (3.lOZ): E'

= 3K'(1 - Zv') = ZG(l

+ v')

3K·(1 - 2v') :. 2G(1 + v') = 1

and

,

I ~

3K ' - 2G 2G + 6K '

(3.104)

The essenlial poinls are: I. Stress and slrain invariants are Indepnrdent 0/ the chosen axis system. 2. Stress and strain invariants are convenient measures to tktennine the effects 0/ a general state 0/ stresses and strains on soils. 3. Mean Slress represents th~ averagt stNfss on a soil while deviatoric stress represents the average shear or distortional stress. EXAMPLE3 .~

A cylindrical sample of soi,U O mm in diameter anCl 100 m long is subjected to an ax.ial effecti e principa lltress of 400 k'P' and a tadia1 effective principal st ress of 100 kPa. Th axial ~ d radial dis iacemen,u are 0.5 mm and - 0.04 mm, respective-I Assurrun --the soil is a isotroPic, elastic material, calculate (a) the mean an ~ . tork stresses, (b) he volumetric and shear (disto rtiollal) strains, and (c) t e's hear, bulk, and elastic-moduli.

st~~forward

Strategy This is a problem. You only need to apply the equations gi ven in the pre...(olls sectIon. T he negative sign for the radial displacement indi a es an expap l tQ. Solution ~1

0"; -

O".i -

",-:::;:::::r,y,~ q = q' =

Step 2:

=

(7; -

400 kPa,

0"3 -

0";

-= 100 kPa

400 + 2 x 100

O"~

= 200 kPa 3 = 400 - 100 = 300 kPa

Calculate the volumetric and shear strains. 6. z =

(b) €~ = 10 1 =

0.5 mm, fj, r =- - 0_04 mm, r - 5012 - 25 mm,

Il z

L

=

t:J.r

E,

=

£) = - ; =

E;

=

Ez + 2E,

e;

= § ( E~

=

- E,)

0.5 100

=

- 0.04 ----zs

0.005 =

- 0.0016

x 0.0016 = 0.0018 = 0.18% + 0.0016) = 0.0044 = 0.44%

0.005 - 2 = ~(O.005

L - lOOmm

126

CHAPTER 3

Calculate [he moduli.

Step 3:

~ p'

K'

(e)

STRESSES. STRAINS. AN D ELA STIC DEFORMATIONS OF SOilS

e;

200 0.0018 = 111,111 kPa

G~'L 300 3e~ - 3 x 0.0044

=

22,727 kPa

but £' and v' = 3K' - 2G = 3 x 111 ,11 1 - 2 x 22,727 = 04 2(1 + v') 2G + 6K' y--2~7 + 6 x ] 11,111 . :. E' = 2G(1 + v') = 2 x 22,727(1 + o.~ '"' 63,636. kPa • G ,..

EXAMPLE 3.13

(

- \,

A sample of soil was subjected to simple shear;W~ch consists Of-4 fo~gU cuboidal sample into a parallelepi~(l:tg. E3.lJ). The strains, in.,!.hy.fanaY directions are zero. The sample si,* i$ lO~ ?( 100 mm X~nuplhi~~. The maximum lateral displacement applied a t e top is O.2.t'mm an d ~~~ in a vertical displacement of O~m and a ~b ar force of z'1cli. Calculate the principal, volumeteic, and dejiatOri1 ains.

Strategy You ar? ::give iQhe disp l acemen ~-q~a n use--tJiis information to calculate strai~. The samp'le iffeformed unde r the pla~e strain condition. Therc-

ENe, the apgrop
$tep 1:

Calculate

. -.,'l £,

El

and

EJ.

'

0.04, h~ 20 mm, tlx ""0.2 mm _ M _~ , , _ _ _ _ tlx _ 0.2 _ h - 20 £~ - £r - O , 'h, - h - 20 '

-De

<.

EquatiQ~125) ~l

r p,.,,_ J_

=

0,0022 + 0+ )("0022 - 0)' + (OOt)' 2

£ = 0.002 + 0 - )(0.002 '2 2

= 02mm

r V. . ;._.

20[

f - 100 mm------+l FIGURE E3.13

-.l M = o.04 mm T "".m.op""""

V L

om

(nol to

scal e~

X

0)' + (0.~1

= 0.006

)' = - 0.004

t

3.13 STRESS PATHS

127

Ca(culalc the volumetric and deviatoric strains.

Ste.p 2:

tp -

tl

+ tJ

0.006 - 0.004 = 0.002

-

Also, tp = t~

Eq =

+

htf

t, + t . = 0 + 0 + 0.002 - 0.002 + €~ - €)€l)112 "" ~[ O.OO6~ + ( - 0.004)2 - (0.006

e

What's next , . ,We have examined how applied surface st~ sses'li distributed in soils as if soi ls were linear, isotrop ic, elastic materia ls Differ~ nt ~ t ures will im pose different stresses and cause t he soil to respond diffe enl ly Fo r exa mp le, an element of soi l under the cent er of an o il tank w ill expe.!ience a ntinuous increase/decrease in vertical stress while the tank is being fill d7em~d . oweve r, th e soil nea r a retaining earth structu re will suffer a reduction rNateral stress if t he wall moves out. If the soil were the same for both struc ure • t he-tfifYJfent loa ding condit ions would cause the soil to respond differently. erefore, w~n e ed to trac the istory of st ress increases/decreases in soi ls to ev alua e ossible soi l responses and to conduct tests that replicate the loading histo V. of the L sit soi l. In the next sectiony l m ethod of keeplOg track of the loading his or o f a sqil is described I

~

3.13 STRESS P THS

r

3 . 13 1 Basic Conc ept

conSid~Wales representi

-

,

par~Jcles

two of a coarse-grained soil. Let us rble 10 a hemispherica ole anH stack the other on top of it (Fig. 3.29a). e constructin one-dimens1Onal system in which relative displace~e nt of the two marbl will occur at the contact. Let us incrementally apply a \lertic~, concent~ ic orce Ft , on the top marble. We wilt calt this loading "A". he forces al Ih ntact re edt'al to the applied loads and the marbles are ced together erticall o'Ate't~tive displacement between the marbles occurs. or the syst ~ to becom unstable or fail, the applied forces mus t crush the marbl es ~e ca make a lot of our loading by arbitrarily choosing an axis system . r...er.trschoose Carlesian system with the X axis representing the horizontal force and t e Z a. s representing the vertical fo rces. We can represent loading " A" by a Jin as shown in Fig. 3.29c. The line OA is called a load path or a force path. fU one

z

(.(

(0)

FIGURE 3.29 Effects of force paths on a one-dimensional system of marbles.

128

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

F O

Let us now apply the same force at an angle e to the X axis in the ZX plane (Fig. 3.29b) and call this loading "B". There are now two components offorce. One component is Fx = F cos e and the other is Fz = F sin e. If the frictional resistance at the contacts of the two marbles is less than the horizontal force, the top marble will slide relative to the bottom. You should recall from your mechanics or physics course that the frictional resistance is fLFz (Coulomb's law), where fL is the coefficient of friction at the contact between the two marbles. Our one-dimensional system now has two modes of instability or failure-one due to relative sliding and the other due to crushing of the marbles. The force path for loading "B" is represented by OB in Fig. 3.29c. The ssential point or principle . is that the response, stability, and failure of the sy~ Qend on the force path. Soils, of course, are not marbles but the u derlyin ~ciple is the same. The soil fabric can be thought of as a spa~m with th~ soil particl~ senting the members of the frame and e particl 0 tacts represe~tng" the joints. The response, stability, and failnr~f<m:e:s:ail abric or th~pac~frame depend on the stress path. Stress paths are presented i~ PlO~t)} Q1Wirl'{i:he relationS:hlP-:'l;)~~@ess parameters and provide a conv. enie~.aN ay allow a ge9 technical e12~eer to study the changes in stresses in a soil c u by loadin~i nditions. - e can, for example, plot a two-di . S10 ' ah,graph of (J" 1 versus 3 0 . 2, which will give us a relationship between these parameters. r, he stress invariants, being independen~f tH~'l.xis sysi!m, are more_convem~ ' use .

f

~Ss

rl~

.161 =.163 = .16 6

.163 =.16

~~

I

o

• P

A

FIGURE 3.30

Stress paths.

l .U STRESS PATHS

129

The loading condition we are applying is called isot ropic compression ; that is, the stresses in all directions are equal (6 CT. = I1U2 = 60-). We wiU call this load ing condition, loading "1." It is oft en convenient to work with increments of stresses in determi ning stress paths. Conseq uently, we are going to usc the incremental form of the stress invaria nts. The stress invariants a ~ otropic com· pression are

and are shown as coor~ reA in . g. 3 30 The . fo, ;sotmp" ,om p.,ess;o~e- Io'pe of OA " •

~~: 0

J

ding "2" b keep~l 0') nstant, that IS, ll<1J '" 0, but (1 •• thai is, .1.0". > 0 O~ se figure labeled "2" in Fig. 3.30). Increaseyn thEVSlress invanants fa loadm "2" are

Let u£"-now apply cO lltin ~

Jncr~e

6(11 + 2

. . ~~, :>:

x0

6U1 -]0 _

AU~

<1U1

-3-

n~ Yend of loading "2" are p z = P1 + 6pz ::::: 6u. ~

+

q z = q.

<1q~ '"

3llU1 = •3"60"1

0 + 6(1'1 = 60-.

Point B in Fig. 3.30 represents (Q2, P2) and the line AB is the stress path for loading "2." The slope of A B is

Let us make another change to the loading cond itions. We will now keep > 0) as illustrated by the inset fig ure labeled "3" in Fig. 3.30. The increases in stress invariants are

a1 consta nt (lla l = 0) and then increase aJ (lla )

6.

Pl -

0 + 2611)

3

=

260"]

3

130

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

The stress invariants at the end of loading "3" are P3

= P2 + q2 +

q3 =

D.P3

= 5D. Ul + ~D.U3

D.q 3 =

~Ul -

~U3

The stress path for loading "3 " is shown as BC in Fig. 3.30. The slope of BC is ilq3

-~U3

3

~P3

~~U3

2

F O /

"'4¥10<"..!., \,£'an

is,

q

I I

I I I /

']3 I

.' 1

I

I

I

,, \

,TSP

3~\ 2 , ESP

\

r--- LI u

, 'c

C' 3

-------+j

TSP

I

I I I

I I

0'

A

FIGURE 3.31

' p ', p

Total and effective stress paths.

total stress; that

J .13STRESSPATHS

131

water can not drain [rom the soil. The implication is that the volume of our soil sample remains constant. In Chapter 5, we will discuss drained and undrained load ing conditions in more detail. For loading " 2," the total stress path is AB. In this book, we will represent total stress paths by dashed lines. If OU f soil were an isotropic, elastic material, then accordi to Eg. (3.99), written in incremental form 6£< ,

6 ' _....E.... K'

= 0

(3.105)

The solution of Eg. (3. 105) leads to either 6 p' = 0 Oi' ' = tO~ ere is no reason why K' should be 00. T he act of preventing the drainl ge 0 th~cess pore water can not change the (effective) bulk modul s...~l the ~il . ids. Remember the truss analogy we used for effective stresseb..Thl'same.~ alogy is applicable here. The only tenable solution is 6p' = 0 ~e ca n al~ite Eq. (3.105) in terms of ~ , total Hresses; that is, (3.106)

where K = £ j3(1 - 211~ ) and t su cript u de~tes undraine ondition. In this case, fip cannot be o..since t . is the cha, #in mean lotal s tress from the applied loading. \herefo~~he{>Jlly tenable solution K = t '!. = 00, which leads to li" = 0.5. T eslmplicattons or Egs. (3.105) an (3.1 o r a linear, isolropic, elastic soil un er undrain ·cond itions aro:::-_..;:! 1,

~Change · n m2neffec.tiv tre~ s ath is vertical. sS\.zey- d, consequently, the effective tl

2, Ttl

ndriined bulk modulu

The 8 eviatOflc str ss is lmaffected by pore wate r pressure changes. We can

~q . (3.102»ter~ta~stress pa rameters as G = G "

..

EM

2( 1 + v.. )

Since G E. 2(1 + v,,)

E'

2(1 + v')

and, by substituting v,, = 0.5, we obtain c----~

E..

1.5E' + v')

= (I

(3.107)

Fo r many soils, v';;;;:! and, as a result, E" =: 1.1£'; that is, the undrained elastic modulus is about 10% greater than the effective elastic modulus. The effective stress path for loading "2," assuming our soil sample be haves like all isotropic, elastic materi al, is represented by A B' (Fig. 3.31); the coordinates of 8 ' are

pi

=

pi + top; == Pl + 0 - 6.01 + uQ2 = 0 + /lUI == 6.0"1

q2 = q )

132

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

q do-I >

~ l:J

°

+- 60-3 = 0, £3=0

I

FIGURE 3 .32

F O

0 ""'-=

p'

One-dimensional compression stres

The difference in mean stress betwel1. of q is the change in excess pore water horizontal line between the TSP a¥!lE pressure. The maximum change i ing "2" is

>tnpression. ffect~€" stress for an increment of vertical stress

given by Eq. (3.50) as Ll0'3 = Llcr;

3

= Ko LlO'{.

,(1 +

+ 2Ko Llcr' = Llcr1

-'-------=----

LlO'~

2Ko) 3

Llq = Llq' = Llcri - Ll0"3 = Llcri - Ko LlCTi = LlO"i (l - Ko)

e\TSP is equal to the slope of the ESP; that is Llq

Llq

Llp

!1p'

3(1 - Ko) 1 + 2Ko

e-dimensional compression stress path is shown in Fig. 3.32.

The essential points are: 1. A stress path is a graphical representation of stresses in stress space. For convenience, stress paths are plotted as deviatoric stress (q) on the ordinate versus mean effective stress (p') and/or mean total stress (p) on the abscissa.

3.13 STJlESS PATHS

133

2. Th e efftctive stress path lor a linear, elastic soil undu the undrained condition is vertical; that is, 4p ' = O. 3. rhe mean srral differrnce belween the 10 Iai siress palh and the effective slress path is the excess pore water prtssurt!. 4. Th e response, stability, and f ailure of soils depend on stress paths.

3 .13.3 Procedure for Plotting Stress P A summary of the procedure for plotting SIr ss path is as;oltows: 1. Determine the loading conditionyirained ~d rained o r both .

2. Calcul ate the initial loading va.I~. of p:;' p ,,nd qQ' 3. Set up a graph of p' (an~ if..you e &ping to also plA't the total stress path) as the abscissa and q as fhe:Ordmafe. Pl ot the i itial values of (p~ , q~)

and (Po, qo)' 4. Dete rmine the increase in I>e negative. 5. Ca lculate the incre;]. in ress invarian inva riant ta~ negative . 6.

calculat~ :;ln~,ess inv"i. n

e

stresses can Ap

s ~I1'

Ap, a d tJ.q. These stress 6p ', P = Po + 6p, and

q ::: qo q. current value of p cannot be negative but q ca n be negative. 7. PI t ll}.e rent stress im an nls (p', and (p, q). 8. Co ect t e points id lltifyin effective stresses and do the same for total

y

. ext loading cond it ion. O. The pore ""f e pressure attn desired level of dcviatoric Slress IS the mea n stress difference betwee the total stress path and the effective stress path Remem dramed

or ~lned

r that loadmg condition , ESP = TSP, and for an un · ndibon, he ~S P for a Imear, elastiC soil IS vertical.

Two cylindrical specimens. A and B. of a soil were loaded as follows. Both spec· imens were isolropically loaded by a stress of 200 kPa under drained conditio ns. Subsequently, the radial stress applied on speci men A was held constant and the axial stress was incrementally increased to 440 kPa under undrained conditions. The adal stress on specimen B was held constant and the radial stress incrementally reduced to 50 kPa under drained conditions. Plot the total and effective stress paths fo r each specimen assu nung the soH is a linea r, isotropic, clastic materia l. Calculate the maximum excess pore water pressure in specimen A.

Strategy The loading condit ions on both specimens are axisymmet ric. The easiest approach is to write the mean stress and deviatoric st ress equations in terms of increments and make the necessary substitutions.

134

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

Solution 3 . 14 Slep 1: Determine loading cond ition. Loading is axisymmetric and both drained and undrained conditions are specified. Step 2: Calculate initial stress invari ants for isotropic loading path. For axisymmetric, isolropic loading under drained condilions, 6u = 0,

.p, -

+

AO"~

260";

.0.(1";

240"j _ 4(J j _ 200 kPa

-t

3

J

Slep 3:

Slep 4:

'eased to 440 kPa .

Ip'e(!ime~

Drai~d loa~ing (6/1 = Q.k.the~re, TSP = ESP.

Axial s'{fSs held cons.J.(nt,A(J"~; 50 kPal that is •

It) - 4~ -

.I Slep S:

taJcu lale

Ihe-~ cre~"),s iy

200 - - ISO kPa

260") = 240 + 2 X 0 ... 80 kPa

3 .0.(11 -

0; radial stress decreases to

tress in\'ari.mts.

·0'1 .,.

oz1q '"

=

3 .o.O'J =

240 - 0 - 240 kPa

Aq 240 Slope of total stress path - Ap "" 80

B·t - SO kl!a ......... . B

: ESP . 1

~1:l,o

I 50

ISO p', p

FIGURE E3.14

.r

l i~ 100

r~p.)

:

:

TSP

'260

2SO

300

-

3

135

314SUMMARY

Specimen 8 .. '

/1p : up '"

/1(1;

+ 26 0') - 0 + 2 X (- ISO) =- 100 kP a

3 6 q - /1(11 - .0.0) '" 0 - (- ISO) - 150 kPa llq 150 Slope of ES P (or TSP) '" - - - '" - I S Ap' - 100

Step 6:

3

J

Calc ulate the current stress invariants.

~

SI.ccimen A

+ 6 p - 200 + 80 ... 280 kPa, p' .. Po + Ap' - 200 + 0 '" 200 kPa p - Po

q - q' - \ (e~ ic..SOi1)

+

.

240 '" 240 kPa

•

Specimen B p "" p ' = Po

q "" q.. + 6q -

Step 7:

Ap =~QO -

+'IsO

.. 100 kPa, 150 kPa

Plot the curren t sire

Step 8: e total

cS&-path and A ll' shows the

s he ES P and TSP.

Step=t':

Determi

e excess ~re water pressure . Specimeo..A ~ BB hows the maximum excess pore water pressure. The mean Stress differe ce is.2&) - 200 = 80 kPa.

3.14

•

SUMMARY Elastic theory provides a simple, fi rst approximation to calculate the deformation of soils at small strains. You arc cautioned that the elastic theory cannot adequately describe the behavior of most SOils and mo re involved theories arc re~ quired. T he most important principle in soil mechanics is the principle of effective stress. Soil deformat ion is due to effective not total stresses. Applied surface Slresses are distribu ted such that their magnitudes decrease with depth and distance away from their points o f app1ication. Stress paths provide a usefu l means through which the Illstory of loading of a soil can he followed. The mean effective stress changes for a linear, isotropic, ehtstic soil are ze ro under undrai ned loa ding and the effective stress path is a vector parallel to the de viatoric stress axis with the q ordinate eq ual to t he cor· respondi ng state on the tota l stress pat h. The difference in mean stress between

13 6

CHAPTER 3

STRESSES, STRAIN S, AND ELASTIC DEFORMATIONS OF SOILS

th e total stress path and th e effective stress path gives the excess pore waler pressure at a desired value of deviatoric stress.

Practical Example EXAMPLE 3 .15

A building foundation o f wid th 10 m and length 40 m transmi ts a load of 80 MN to a deep deposit of stiff satu rated clay (Fig. E3.1 5a). The elastic modulus o f the clay varies with de pth (Fig. E3.15b) and v' = 0.32. Estimate the elastic settlement of the clay under the center of the foundation assumi ng (1) d rai ned con dition and (2) undrained condition. I

......

Strategy The major decision in this prob~s W:~ deptJr to use to dettr:in. an appropriate elastic mod ulus. O ne opti'n': is t~se average elastic T odulus over a depth of 3B. Beyond a depth oflB.~ 1 ress i ncrease-.is l~ than 10%.

J

Solution 3.15 Step J:

Find Ihe applieo--ve"rt;al surface stress.

x

80

q,

Step 2:

A

=

1(]l

10 x 40 ~ kPa

D e t e r~ine.the'l ~c modulus. ~ Assume,an effective depth of 3 = 3 ~ 10 1 30 m. ' h ~erage value of £ ' ~ 'a. y-

FrO~(3.107) wiU~'" '"' O.

sreo3:

'" 1.1 E ' '" 1.1 x 34.5 = 38 MPa

Calculaf6 the ve nical se~I11ent.

" 30

o ,"

32

E' (M pa) 3. 16 i

38

.0

I

10 11Nd 11111 I l " , IIII ~ II I I g ~

",I I 1111I ttl I

,al Found.tlon pl.n . 1Id sectIOn

FIGURE E3.15

(~)

V.lltoon at elasllc modulus Wit h depth

EXERCISES

137

(1) Drained condition.

~~

_ 200 x ~.; ~11~ 0.32

1

1.98 = 5l.S x 10- 6 m = 51.5 X 10- ] mm

)

(2) Undrained condition. 200 6 ;: =

x 5 x (1 - 0.52 ) 38 x 1(J6

~

Theory 3.1

•

l.98 - 39.1

An elastic soil is confined laterally and is axi-t!y. co'!ypressed unde rained conditions. In soil mechanics, the loading imF.~ d t e called Ko oomp ession or consolidation. Show that under the K~ondition,

"1$

...,;O:r

~

". 1-

v'

•

saturaled~~ giv~

3.2

The increase in P.9 e-water p essur in a b !J.u = Ao:] + A(6<.J"1 6<.J"). Show that (the soil is a i ar, isotropic eJas tic..material, ~ = l for tbe axisymmetric condition.

3.3

il dy creased and the radial stress is }ow.lIlat the stress path has a slope q/p

3..

The initial ean effective Me on a sal IS 0 and Ihe deviatoric stress q = O. If the soil i~ a linear, ik)tropic, elas~ mateaal, plot the lolal and effective stress paths for the following axisymmetric undra· e 100000ing condi tion: (a) 60:3 - jAO"l and (b) 6a) = - !.o.O"l.

A long embankme is li>ca d on a soil profile consisting of 2 m of medium clay followed by 8 m sl fu edium to dense sand on top of bedrock. A vertical settlement of 5 mm al the center of th~mbank~ent was measured during construction. Assuming all the settlement is elastic and to the medium clay, detennine the average stresses imposed on the medium clay under the cenler of the embankment using the elastic equations. Th.e elastic parameters are E. - 15 MPa and v. = 0.45. {Hint: Assume the lateral strain is zero.] 3.6

An element of soil (sand) behind a retaining waU is subjected to an increase in \<ertical stress of 60 kPa and an mcrease In lateral stress of 25 kPa. Determine the increase in vertical and lateral strains, assuming the soil is a linearly elastic material with E' = 20 M Pa and II' = 0.3.

3.7

A cylindrical specimen of soil is compressed by an 811.ial principal stress of 500 kPa and a radial principal stress of 200 kPa. Plot Mohr's circle of stress and determine (a) th.e maximum shear stress and (b) the normal and shear stresses on a plane mclined at 30" clockwise to the horizontal.

3.8

A soil specimen (100 mm x 100 mm X 100 nun) is SUbjected to the forces shown in Fig. P3.8. Determine (a) the magnitude of the principal stresses, (b) the orientation of the

138

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

8 kN

4 kN

FIGURE P3.S

F O

principal stress plane to the horizontal, (c) the maximum shear stress, and (d) the normal and shear stresses on a plane inclined at 20° counterclo~s~o the horizontal. 3.9

3.10

Plot the distribution of tot

o 3.11

3.12

3.B . The effeHive size of a fine sand is 0.15 mm. Estimate the distance in the sand, above the gr~ndJiter level, that would be saturated by capillary action. Determine the vertical effedive stress in the sand at depths of 1 m above and 3 m below the groundwater level for the following cases: (a) groundwater level 3 m below ground surface and (b) groundwater level 1.5 m below ground surface. The void ratio of the sand is 0.6. Assume that the degree of saturation is 90% for soil layers that are unsaturated. 3.141

A soil profile consists of a clay layer underlain by a sand layer as shown in Fig. P3.14. If a tube is inserted into the bottom sand layer (Fig. P3.14) and the water level rises to 1 m

EXERCIS ES

139

FIGURE P3. 14

and pore water prestal lateral stresses at A. ~IS

A rectangular foundation 4 m x 6 m (Fig. ~15) tr~ its a stressp,f 150 kPa on the surface of a soil deposi t. Plot the di~tribution r U:;cre'tscs of vt:rtlcartitresses wlIh depth

m

under poin ts A , D, and C up to a epth 01 m. ~hat dept h ' the mcrease in vertical stress below A less than 10% 0{ surface re ?

•

r

I~~'~m;==J,.1'2m--l GURE P3. 15

elerm ine IheAner foundation show in Fi

In 'Y

Y

rti al slfess at a dep th of 5 m below the centroid of the

6.

1--2.5m~

I

1 1 10m

1-3.75 m-J

t

' •• '00

2.5m

L I FIGURE P3.16

150--

.e,

140

CHAPTER 3

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

1-3 m---+-2.5 m-+-l.I....., - - 6 m - j

TAB

1

5m -- _

- - ......

q, = 120kPA

q., =90 kPA C

fl

'--_ _ _ _ _---'

1 5m

1

T

3m

".'00k'A

I~

r-- 4 m----+j FIGURE P3_17

tress in-

3.17 3.18

FO

3.19

otechril l e e r for a proposed office building in a densely clustered city. The of,ti . building ill oe constructed adjacent to an existing office complex. The soil at the site Is deposi, of very dense sand with E' = 45 MPa and v' = 0.3. The sand rests on a deep de . of dense gravel. The existing high rise complex is founded on a concrete slab, 100 m X 120 m located at 2 m below ground surface, that transmits a load of 2400 MN to the soil. Your office foundation is 50 m X 80 m and transmits a load of 1000 MN. You also intend to locate your foundation at 2 m below ground level. The front of your building is aligned with the existing office complex and the side distance is 0.5 m. The lesser dimension of each building is the frontal dimension. The owners of the existing building are concerned about possible settlement of their building due to your building. You are invited to a meeting with your client, the owners of the existing building, and their technical staff. You are expected to determine what effects your office building would have on the existing building. You only have one hour to make the preliminary calculations and you are expected to present the increase in stresses and the amount of settlement of the existing office complex due to the construction of your office building. Prepare your analysis and presentation.

CHAPTER

4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS 4 .0 In this chapter. we will conside r on4 ensional e!Ol:dation 'Settleme"nt of finegrained soi ls. We will restrict seulemen #tera tion to/ he vertica l tion. Under load, all soi ls will set! ausing settlement of jU"uctures fo unded on or within them. If the sett lement n t ~pr to a tolera,hl'e li~ the d sired use of the structure ma y be j 'pal ed an the design life of tfi~ ucture pay be reduced. Structures may settl uniformlr nonuniforml 1'he..Jat ~s called differential settlement and is of en the erueia1 design conslde tion. T he total sett le nt us ally consis { r e ~ -immed i ate or elastic compression, 'mary consolidation, an secondar cQm pression. We have considered elastic etile in Chapter 3 and we will conSider some modificati ons to the elastic an ysis or practica~plicai'iQ[l.!.i.!!..S3hapter 7. Here we will develop I h~a sic conccp f consolid ion-'and sho;'ll'j)w these concepts can be applied ..i~cu late the consol idatio . ttl en fro m applied loads. After that, we will ~ fo m4~ the! theory of one-dim nsiona consolidation and use it to predict the , . ." ate of settlemen t <})fer you hav st

eh

6r

• H ave a basic underslantiing of the consolidation of soils under vertical loads • Be able 10 calcurnte one-dimensional consolidation seulement and lime rale of settlement You will make use of the fo llowing concepts learned from the previous chapters and your courses in mechaojcs and mathematics. Stresses in soils-e ffective Stresses and vertical stress increases from surface loads (Chapter 3) Strains in soi ls-vertica l and volumetric strains (Chapter 3) Elasticity (Chapte r 3) • Flow of water through soi ls (Chapter 2. D arcy's law) Solutions of partial differential equations

141

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

FO

142.

4.1

DEFINITIONS OF KEY TERMS Consolidation is the time-dependent settlement of soils resulting from the expulsion of water from the soil pores. Primary consolidation is the change in volume of a fin e-grained soil caused by the expulsion of water from the voids and the transfer of load from the excess pore water pressure to the soil particles.

4-2 QUESTIONS TO GUIDE YOUR READING

143

Secondary compression is the change in volume of a fine-g rained soil caused by the adjustment of the soil fabric (internal structure) after primary consolidation has been completed. F.J:cess pore water pnssur~ 6.u, is the pore water pressure in excess of the current equilibrium pore water pressure. For example, if the pore water cess re in a soil is U o and a load is applied to the soil so that the existing pore ater 2P es ure increases to "l> then the excess pore water pressure is Au = ~- . . Drainage path, 1I., is the longest vertical path that a wa e'r pai: ide reach the drainage surface.

11 ta e to

a;u

Preconsolida(;on stren or past maximum effect ive stress. lSJ.he maximum vertical effecti ve stress that a soil was subjected t610 It. a~t. Normally consolidated soil is one that has eve xpe r ~ n ced vertical effective stresses greater than its current vertical effec jv tr (u lO = t:1;c) A Overconsolidated soil is one that bas ex~eri ~ce(i vertical effect l e stresses greater than its existing vertical e~tive str (u"Yo < O"~c). Overconsolidatiott ratio, OCR, is-t1~-at'io b which the current vertical e ffe ctive CR = o-;c1p (0)' stress in the soil was exceed in the past

,

Compre!iSiott index, Cet is the s £PC of t he normal ohd tion !ige in a plol of the loganthm of verf cal ~ffecthctress versus void ratl~ ~ Ultloadiuglrefoadi iudex a} ,Tcompreuio,tf';;;;;., "i~he average slope of the unloadinglreloa1:!.in c u ~s in a plo e loganthm of vertical effective id ratIo. stress vcrSUS~ Modulus ofv st ress pomts in

4.2

a

Ie mpressibilify, m is the S pe of the curve between p'lot of vertical e.(fective sir ersus vertical stram

IWO

ES1)ONS TO~I DO P R READING I. 2.

o"'t-ioil consolidation? ce between consolidation and compaction?

3. Wh at is the go e.rni ng equation in one-dimensional consolidation theory? 4. How is the excess pore water pressure distributed within the soil when a load is applied and aftcr various elapsed times? 5. What factors determine the consolidation settlement of soils? 6. What are the ave rage degree of consolidation, time fa clor, modulus of volume compressibility, and compression and recompression indices? 7. What is the difference between primary consolidation and secondary compression? 8. What is the drainage path for single drainage and double drainage? 9. Why do we need to carry ou t consolida tion tests, how are they conducted, and what parameters are deduced from the test results?

144

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

10. How is time rate of settlement and consolidation settlement calculated? 11. Are there significant differences between the calculated settlements and

field settlements?

4.3

BASIC CONCEPTS In our development of the various ideas on con 'olia assume: • A homogenous, saturated soil • The soil particles and the water • Vertical flow of water

FO

o ,

Pore water pressure read out unlls E::::J1==='

Burette

pressure transducer Porous stone

FIGURE 4 .2a Experimental setup for illustrating basic concepts on consolidation ,

4.3 BASIC CONCEPTS

145

from the applied loads. Since the side wall of the con taine r is rigid, no radial displacement can occur. The lateral and circumfercntial Slrains are then equal to zero (€, = E~ = 0), and the volumetric strain (Ep = €( + Ee + €,) is equal to the vertical strain , E. = tJ.l./f1 o' where I.\z is the change in height or thickness and H o is the initia l height or thickness of the soil.

4.3. 1 Instantaneous Load

4 .3.2 Consolidation Un e"f 7 Primary Consoli

- ";'

Let us now open the valve an ~ a lo w the inili ~~ !?Of wa er to drain. The total volume of soil % time I I s ecreases by the.amount of excess pore water that drains from i --as indica d ~ he change on volu 0 water in the burette (Fig. 4.2c). At the t ~ ~ttom of the soil ample, t e e cess pore water pressure is zero because ~ese¥c the drainage bo daries. The decrease of initial excess pore waler pressure at the middl f the s i tion C) is the slowesl because ~ particle.rg st travel f:S; 0 th~ iddle of the soi l 10 ei ther the top or bottom a n ry to exll the system. a u may have nO li d tha e se ifement of the soil (.6.z) with time I (Fig. 4~c at linea r. Most 0 set c meol occurs shortly after the valve was o~ne . The rate a ettlement, zl, is also much fa ster soon after the valve was op'€ned compa!{d Wit l at~es. Before the valve was opened. an initial hy· draulic hcad •.,Au,('Y... , wa created by the applied vertical stress. When the valve was opened, the ' itial ex ess pore water was forced out of the soil by this initial

~

r \.

llou

( kP~ )

FIGURE 4.2b Instantaneous or initia l ellcess pore water pressure when a vertical load is applied .

FO

146

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

Excess pore water pressure at time 11 O.----==->i~:----.._____-----,

125

l

FIGURE 4 .2c

l -

hange in height (m m)

= chan ge in excess pore water pressure (kP a)

Excess pore

consolidation.

when L1u o = O. The later time settlement response is called secon ary com ession or creep. Secondary compression is the change in volu e of a fin -gr {fled soil caused by the adjustment of the soil fabric (internal struct after rimary consolidation has been completed . The term consolidation is re or the process in which settlement of a soil occurs from changes in effective stresses resulting from decreases in excess pore water pressure. The rate of settlement from secondary compression is very slow compared with primary consolidation. We have separated primary consolidation and secondary compression . In reality, the distinction is not clear because secondary compression occurs as part of the primary consolidation phase especially in soft clays. The mechanics of consolidation is still not fully understood and to make estimates of settlement, it is convenient to separate primary consolidation and secondary compression.

4.3.4 Drainage Path The distance of the longest vertical path taken by a particle to exit the soil is called the length of the drainage path. Because we allowed the soil to drain on

4.3 BASIC CONCEPTS

147

the top and bottom faces (double drainage) , the length of the drainage path , H dn is

FO

R

4.3.5 Rate of Consolidation

The essenb'a poin s are: 1. When a load is applied to a saturated soil, all of the applied stress is supported initially by the pore water (initial excess pore water pressure),. that is, at t = 0, ..:1uo = ..:1O'z or ..:1uo = ..:1p. The change in effective stress is zero (..:10': = 0). 2. If drainage of pore water is permitted, the initial excess pore water pressure decreases and soil settlement (..:1z) increases with time; that is, ..:1u(t) < ..:1u o and ..:1z > O. The change in effective stress is ..:10'; = ..:1O'z - ..:1u(t). 3. When t - co, the change in volume and the change in excess pore water pressure of the soil approach zero; that is, ..:1 V - 0 and ..:1u o - O. The change in vertical effective stress is ..:10':' = ..:1u".

148

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

4.

5. 6.

4.3.7 Void Ratio and Settlement Changes Under a Constant Load

(4.2)

a relationship

(4.3)

(4.4)

01 + eo

consolidation settlement rather than ~z,

so (4.5) .

FO

the consolidation under load P is

,

e

=

e - 6.e o

=

6. z e - - (1 0

rIo

+e) 0

(4.6)

4.3.8 Effects of Vertical Stresses on Primary Consolidation We can apply additional loads to the soil and for each load increment we can calculate the final void ratio from Eq. (4.6) and plot the results as shown by segment AB in Fig. 4.3. Three types of graph are shown in Fig. 4.3 to illustrate three different ways of plotting the data from our test. Figure 4.3a is an arithmetic plot of the void ratio versus vertical effective stress. Figure 4.3b is a similar plot except the vertical effective stress is plotted on a logarithmic scale. Figure 4.3c is an arithmetic plot of the vertical strain (sz) versus vertical effective stress. The segment AB in Figs. 4.3a,c is not linear because the settlement that occurs for each increment of loading brings the soil to a denser state from its initial state

4.3 BASIC CONCEPTS

(b )

A

(c)

A

A

Impossi ble state

u

<5

>

E

E

,,', Vertical effe ct ive stress

FIGURE 4.3

a'zc

Verti ca l eff ective stress (log scale )

Three plots of settlement data from

se with typical material responses to loads

F

(a)

149

4.3.9 Primary Consolidation Parameters The primary consolidation settlement of the soil (settlement that occurs along path AB in Fig. 4.3) can be expressed through the slopes of the curves in Fig. 4.3. We are going to define two slopes for primary consolidation. One is called

FO

150

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

the coefficient of compression or compression index, C e , and is obtained from the plot of e versus log a~ (Fig. 4.3b) as C c

=-

e2 - el I (CT~)2 og-( ') U l Z

It>.el

= I

(CT~h

(no units)

(4.7)

og -(crzl ')

n the NCL.

(4.8)

(4.9)

m vn

and is ( 4.10)

bitrarily selected points on the URL. ung's modulus of elasticity is (4.11)

( 4 .12)

The slopes co Cn m v , and m ur are taken as positive values to satisfy our sign convention of compression or recompression as positive.

4.3.10 Effects of Loading History In our experiment, we found that during reloading the soil follows the normal consolidation line once the past maximum vertical effective stress is exceeded. The history of loading of a soil is locked in its fabric and the soil maintains a memory of the past maximum effective stress. To understand how the soil will respond to loads, we have to unlock its memory. If a soil were to be consolidated to stresses below its past maximum vertical effective stress, then settlement would be small because the soil fabric was permanently changed by a higher stress in

4.3 BASIC CONCEPTS

1 51

FO

the past. However, if the soil were to be consolidated beyond its past maximum effective stress, settlement would be large for stresses beyond its past maximum effective stress because the soil fabric would now undergo further change from a current loading that is higher than its past maximum effective stress. The practical significance of this soil behavior is that if the loadin imposed on the soil by a structure is such that the vertical effective stress i the soil does not exceed its past maximum vertical effective stress, the settle me t of the s uc· ture would be small, otherwise significant permanent settle en ould 0 cur. The preconsolidation stress defines the limit of elastic be 0 s that are lower than the preconsolidation stress, the soil will URL and we can reasonably assume that the soil will behave like a stresses greater than the preconsolidation stress soil elastoplastic material.

ratio, OCR,

(4.13)

The essential points are: 1. Path AB (Fig. 4.3), called the normal consolidation line (NCL), describes the response of a normally consolidated soi/-a soil that has never experienced a vertical effective stress greater than its current vertical effective stress. The NCL is approximately a straight line in a plot of e versus log and is defined by a slope, Cc , called the compression index. 2. A normally consolidated soil would behave like an elastoplastic material. That is, part of the settlement under the load is recoverable while the other part is permanent.

u:

1 52

CHAPTER"

ONE-OIMENS10NAL CONSOUOATION SETTLEMENT Of FINE-GRAINEO SOILS

3. An overconsolidated soil has experienced vertical eJJective stresses greatu than its CUTTent vertical eJJutive stress. 4. An overconsolidaled soil willjollow paths such as CDE (Fig. 4.3). For stresses below the preconsolidillion stress, an ovucpnsolidilted soli would approximately behave lib an elastic malerlal and selllement would be small. However,jor stresses greater Ilian the preeonsolidalion SITtss, an overconsofidaled soil will b,thav(, lik e an eluslOplastlc material, similar 10 U normall), cons lidaled soil.

What's next . ..Next. w e w ill co ns ider how to us,the asie-qm cepts t o ca lcu late o n e~ dimensional sett lement.

3

4.4 CALCULATION OF PIII~AR CONSOLIDATION SETTLEMe NT 4 .4 . 1 Effects of a Soil Samp e TalteiFt?ilJlfi

and the urrcnl void ra li can be undJrom 'Y••, using Eq . (2.11). On a plot of g 0';, the cue en ve rtical effeC tLve stress ca n be represented as A as

e verSllS

depicted

In

Fig. 4.4b

c.

F

OCR.! OCR:> I

FIGURE 4.4 (al Soit sample at a depth z below ground surface. (b) Expected onedimensional response.

4.4 CALCULATION OF PRIMARY CONSOLIDATION SETTLEMENT

re e in vertical stress due to pie, is Au z ' (Recall that you 3.11.) The final vertical stress

FO

R

to construct a the building a can find Au z USI

153

OCR

=

1

(4.14)

4.4.3 Primary Consolidation Settlement of Overconsolidated Fine ~ Grained Soils If the soil is overconsolidated, we have to consider two cases depending on the magnitude of AO'z . We will approximate the curve in the log u~ versus e space as two straight lines, as shown in Fig. 4.5. In Case 1, the increase in Au z is such that O' ~ in = u~ o + AO'z is less than u ~c (Fig. 4.5a) . In this case, consolidation occurs along the URL and (4.15)

154

CHAPTER 4

ONE·DIMENSIONAL CONSOlIDATION SETTLEMENT OF FINE·GRAINED SOilS

,

,

'"

)

'0 C,

C,

URe

'"

0';.

O" bn

Ioi a;

0';,

"'";&

(4.16)

o- ; , ~

> 0-;<

(4.17)

4 .4 .4 .,procedure ~Calc "
he-pcoeedurC

alc

la'feP7ary eonsohdalion selliemeni ;s as follows:

l. Ca\Cu latell:fe.c:.~~1n vertical effective st ress (cr;o) and the curren t void ratio

(eo}al ~e cen ~f the soi l layer for which settlemen t is required. ;""€alculate he applied vertical stress increase (60'. ) at (he center of the soil la ero. usin.,e th e appropriate method in Section 3.11. 3. Calc~the final venical effective stress O'r;n = 0';0 + 6O'~. 4. Calculate the primary consolidation settlement. 3. If the soil is normally consolidated (OCR = 1), the primary consolidation settlement is

b. 1f the soil is Qverconsol idated and consolidatio D settlement is

(Jr,n<

O';c. the primary

4.4 CALCULATION OF PRIMARY CONSOLIDATION SETIlEMENT

c. If the soil is overconsolidated and

(J ria >

(J ~c>

155

the primary

consolidation settlement is Ppc

Ho = -

1 + eo

( C log ( 0 CR) + Ce log -O'lin) , O'ze

where Ho is the thickness of the soil layer. You can also calculate the primary consolidation settleITJ4lt -Il!gm u ' ever, unlike C e , which is constant, m u varies with stress le~ ~S\..You ul pute an average value of m v over the stress range (J ~ t.o (Jrin he primary consolidation settlement, using m u , is (4.18)

FO

R

( 4.19)

soil Jayer is subjected to hIgher vertical stress increases.

EXAMPLE 4.1

The soil profile at a site for a proposed office building consists of a layer of fine sand lOA m thick above a layer of soft normally consolidated clay 2 m thick. Below the soft clay is a deposit of coarse sand. The groundwater table was observed at 3 m below ground level. The void ratio of the sand is 0.76 and the water content of the clay is 43%. The building will impose a vertical stress increase of 140 kPa at the middle of the clay layer. Estimate the primary consolidation settlement of the clay. Assume the soil above the water table to be saturated, C c = 0.3 and G s = 2.7 ..

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

".----,,--------. 3m

1

7.4 m

Fin e sand

~ 1 2m

Clay • • • • • • •

~

Coarse sand

FIGURE E4_1

,~ 2. 7

..L

i

fclay~

wn'olidation E (Fig. .3). The appropriate

o -

FO

156

s -

wG, = 2.7 x 0.43 = 1.16;

1)

_ (2.7 - 1 ) _ 3 1 + 1.16 9.8 - 7.7 kN/m

~ -V w -

tical effective stress at the mid-depth of the clay layer is CY~o =

(19.3

x

3) + (9.5

x

7.4) + (7.7

x

1)

=

135.9 kPa

Step 2:

Calculate the increase of stress at the mid-depth of the clay layer. You do not need to calculate l1a z for this problem. It is given as l1a z = 140 kPa.

Step 3:

Calculate a;inCY;in

Step 4:

=

CY~o

+

aCYL

= 135.9 + 140 = 275.9 kPa

Calculate the primary consolidation settlement. Ppe = 1

Ho CY;in + eo Ce log CY~o

=

2 275.9 1 + 1.16 X 0.3 log 135.9

= 0.085

m

= 85

mm

•

157

4.4 CALCULATION OF PRIMARY CONSOLIDATION SETTLEMENT

EXAMPLE 4.2

Assume the same soil stratigraphy as in Example 4.1. But now the clay is overconsolidated with an OCR = 2.5, W = 38%, and C,. = 0.05. All other soil values given in Example 4.1 remain unchanged. Determine the primary con olidation settlement of the clay. Strategy Since the soil is overconsolidated, you will have 00 c the preconsolidation stress is less than or greater than the vertical effective stress and the applied vertical stress at This check will determine the appropriate equation to use unit weight of the sand is unchanged but the clay h s chan

Solution 4.2 Step 1: settlement obtained in Ex

lre

that the

Step 2:

FO

Step 3:

•

EXAMPLE 4 .3 Assume the same soil stratigraphy and soil parameters as in Example 4.2 except that the clay has an overconsolidation ratio of 1.5. Determine the primary consolidation settlement of the clay. Strategy Since the soil is overconsolidated, you will have to check whether the preconsolidation stress is less than or greater than the sum of the current vertical effective stress and the applied vertical stress at the center of the clay. This check will determine the appropriate equation to use.

FO

158

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

Solution 4.3

Step 2:

Calculate (J"~o and eo. From Example 4.2, (J" ~o = 136.4 kPa. Calculate the preconsolidation stress.

Step 3:

Calculate

Step 1:

(J" fin .

Step 4:

Step 5:

1

1

+ 1.03

•

Strategy"

vertical ss mcreas in the clay layer from the building load. Since the clay layer ·s finit , we i: ave to use the vertical stress influence values in Appendix B. J; e assume a rough base, we can use the influence values specified by Milovic and To nier ( 971) or if we assume a smooth base we can use the values specified by So! } -c (1961) . The clay layer is 10 m thick, so it is best to subdivide the clay layer into sublayers s::2 m thick.

I

Foundation: width B

= 10 m,

length L

= 20 m I

200 kPa

HHHHH .--'T+'--;- ,1m=--

r---L---, 1--10 m-----I

10 m Clay

FIGURE E4_4

!

4.4 CALCULATION OF PRIMARY CONSOLIDATION SETTLEMENT

159

Solution 4.4 Step 1:

Find the vertical stress increase at the center of the clay layer below the foundatioD . Divide the clay layer into five sublayers, each of thickness 2. Athat is, Ho = 2 m. Find the vertical stress increase at the middl ~f~~'ch sublayer under the center of the rectangular foundation. A:ssume a rough base and use Table B1 (Appendix B). B = 10 m, L

=

L 20 m, 13

=

2,

qs

=

z

-

8

Layer

z(m)

2

3

0.3

3

5

0.5

4

7

0.7

5

9

Izp

0.1

Step 2:

FO

d results from the bias toward the top

,

•

EXAMPLE 4.5

A laboratory test on a saturated clay taken at a depth of 10 m below the ground surface gave the following results: C c = 0.3, C = 0.08, OCR = 5, w = 23% , and G s = 2.7. The groundwater level is at the surface. Determine and plot the variation of water content and overconsolidation ratio with depth up to 50 m.

Strategy The overconsolidation state lies on the unloadinglreloading line (Fig. 4.3), so you need to find an equation for this line using the data given. Identify what given data is relevant to finding the equation for the unloading! reloading line . Here you are given the slope, C" so you need to use the other data to find the complete question. You can find the coordinate of one point on the unloading!reloading line from the water content and the depth as shown in Step 1.

160

CHAPTER. ONE-DIMENSIONAL CONSOLIDATION SEnLEMENT OF FINE -GRAINED SOILS

Solution 4 .5 Step J:

Determine eo and

(1;0'

, (G, - ') ( ~ "I~,

"I '"

Step 2:

-

\

2.7 - 1 )

eo = G,II!

=

2.7 x 0.23 ". 0.621

0;0

=

10.3 x 10 = 103 kP

=

..,'z

.

+ 0.621 9.8 "" I

Dele rmine (he preconsolidation Siress

Step 3:

,,

c

'"r~C'~ . B

•

-

a:.

,\og-; cr"

en = 0.621 - 0.08 log(5) .. 0.565 Henc~ '; on 0' 'he unlo,d ;ng/<eload;ng I;n. ;, e - 0.565 .... 0.08 Jog(OCR)

(1)

~h~tUting

,;;;S"

wGs (G s = 2.7) and OCR = 515ly'z (y' = 10.3 IeNI at;p );n Eq, (I) g;ves IV ""

0.209

+ 0.03 log (5 0) z

You can now substitute va lues of z from 1 to 50 and find \'\I and substi tute e = wG, in Eq . (1) to find th e OCR. The table below shows the calculated values and the resu lts, which are ploued in Fig. E4.5b . Depth

w (0/0)

OCR

5 10 20 30 40 50

26.0 23.9 23.0 22.1 21.6 21 .2 20.9

51.4 10

,

2.' 1.7 1.2 1.0

... ('!I:.) ~nd

flO

tOO

to

§

20

-=

30

~

I OCR

40 50

60

FIGURE E4.5b

200

OCR

300

I

,

Wilter toI1 tent

" ' >----

400

soO

[

+

600

4.5 ONE-DIMENSIONAL CONSOLIDATION THEORY

161

You should note that the soil becomes normally consolidated as the depth increases. This is a characteristic of real soils. •

What's next . . .So far, we have only considered how to determine the fin imary consolidation settlement. This settlement might take months or years occur, depending essentially on the permeability of the soil, the soil thickness d ainage c nditions, and the magnitude of the applied stress . Geotechnical e ine s hav to know the magnitude of the final primary consolidation settlemeflit an also t e rate of settlement so that the settlement at any given time can be ev ated. The next section deals with a theory to determine the settl Several assumptions are made in developing this the How that many of the observations we made in Section theory.

4.5 ONE-DIMENSIONAL CONSOLIDATION THEORY

F

ns made in Section 4.3:

2. At any depth , the change in vertical effective stress is equal to the change in excess pore water pressure at that depth. That is, a(J"~ = au.

q, +

T dz

1.····

L

oq, dz -az

~---------

0(d

T y

q, d.x

FIGURE 4.6

One-dimensional flow through a two-dimensional soil element.

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

For our soil element in Fig. 4.6, the inflow of water is qu dA and the outflow over the elemental thickness dz is qu + (3qj3z) dz dA. The change in flow is then (aq)3z) dz dA. The rate of change in volume of water expelled, which is equal to the rate of change of volume of the soil, must equal e change in flow. That is,

av = at

aqu dz dA

az

( 4.20)

Recall [Eq. (4.2)] that the volumetric strain £p =

Substituting Eq. (4.21) into Eq. (4.20 a d

SI

(4.22)

The one-dimensional flow (4.23)

(4.24)

The

a2h az aqu

az

F

162

=

2

1

a2 u

'I~

az 2

a2u 2 'Iw az k:

(4.26)

( 4.27)

.22) and Eq. (4.27) , we obtain

au

~

a2 u

at

mu'lw

ilz 2

(4.28)

We can replace

,-------------------------------------------,

~ mv'lw

by a coefficient Cu called the coefficient of consolidation

The units for Cv are length 2 /time, for example, cm 2 /min. Rewriting Eq. (4.28) by substituting Cu , we get the general equation for one-dimensional consolidation as (4.29)

4.5 ONE-DIMENSIONAL CONSOLIDATION THEORY

163

This equation describes the spatial variation of excess pore water pressure (Llu) with time (t) and depth (z). It is a common equation in many branches of engineering. For example, the heat diffusion equation commonly used in mechanical engineering is similar to Eq. (4.29) except that temperature, T, replaces u and heat factor, K, replaces Cv ' Equation (4.29) is sometimes called the Terzag i onedimensional consolidation equation because Terzaghi (1925) develo ed it. In the derivation of Eq. (4.29), we tacitly assumed that k z an mv are co th Qid spa es stants. This is usually not the case because as the soil consolida are reduced and k z decreases. Also, mv is not linearly rel~-e d 6'~ ( ig . c). The consequence of k z and m v not being constants is thaf , is n a onstant. In practice, Cu is assumed to be a constant and this assumpti n ' eas9'nable only if the stress changes are small enough such tha and not change significan tl y.

The essential point is: 1. The one-dimensional consolida'on equation a{(ows us to prediat the changes in excess pore water m sure at ari us depths within the soil with time. We need to know the excess pore water pressure at a desired time because we have to determine the )lerticai effectIve stress to cal'Culate the primary consolidatIOn ettl menlo

fro

F

in programmin

ensional consolidation the top and bottom

eme. The latter is simpler and in sp e dsheet app ~ c tions for-a ny boundary condition.

1 erential e uation requires a knowledge of the boundary co aitions. By.J ecificatl 0 t e initial distribution of excess pore water pressures at the bOu'n ries, we can obtain solutions for the spatial variation of excess pore water pressu wit time and depth. Various distributions of pore water pressures within a soil ayer are possible. Two of these are shown in Fig. 4.7. One of these is a uniform distribution of initial excess pore water pressure with depth (Fig. 4.7a) . This may occur in a thin layer of fine-grained soils. The other (Fig. 4.7b) is a triangular distribution. This may occur in a thick layer of fine- grained soils . The boundary conditions for a uniform distribution of initial excess pore water pressure in which double drainage occurs are

When t = 0, Llu = Llu o = LlCT z . At the top boundary, z = 0, Llu

=

O.

At the bottom boundary, z = 2Hdn Llu = 0, where H dr is the length of the drainage path.

FO

164

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

Soft thin clay layer

(a) Uniform distribution

uniform distribution with depth in a depth in a thick layer.

(4.30)

(4.31)

where

is known a the ·me factor ; it is a dimensionless term. o gl e the variation of excess pore water pressure with . Le s examine Eq. (4.30) for an arbitrarily selected o fme factor Tu as shown in Fig. 4.8. At time t = 0 (Tu = 0), the · itia1 excess are water pressure, flu o, is equal to the applied vertical stress hrou oil layer. As soon as drainage occurs, the initial excess par · ater pre ure will immediately fall to zero at the permeable boundaries. The m imum xcess pore water pressure occurs at the center of the soil layer because tn ainage path there is the longest, as obtained earlier in our experiment in Section 4.3. At time t> 0, the total applied vertical stress increment flu , at a depth z is equal to the sum of the vertical effective stress increment flu ~ and the excess pore water pressure flu , . After considerable time (t ~ co) , the excess pore water pressure decreases to zero and the vertical effective stress increment becomes equal to the vertical total stress increment. We now defin e a parameter, U z , called the degree of consolidation or consolidation ratio, which gives us the amount of consolidation completed at a particular time and depth. This parameter can be expressed mathematically as u

Uz

=

1 - -6.u z -_ 1 6.u a

~ -2 L.J

m ~O

M

(Sin - -MZ) - exp ( ' ..-2T ) H -

dr

I VI

U

(4.32)

165

4.5 ONE-DIMENSIONAL CONSOLIDATION THEORY

If.-' - - - - duo = .10', - - - - + 1 .I

I

I~~__

.1a---.J;

FIGURE 4 _8 An isochrone illustrating the theoretical excess1>or distribution with depth.

one-

(4.33)

F

(4.34)

Triangular initial excess pore water pressure

U",lmm 'ml'"

water pressure I

0.2

0.4

' '~,

I -~

P

- -t

"""

-I

Arrows show drr nage direction

0.6

0.8

1

U(%)

T,

0 10 20 30 40 50 60 70 80 90

0 0.008 0.031 0.071 0.126 0.197 0.287 0.403 0.567 0.848

1.2

T,.

FIGURE 4 _9 Relationship between time factor and average degree of consolidation for a uniform distribution and a triangular distribution of initial excess pore water pressure.

1.4

F

166

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

and

I Tv =

1.781 - 0.933 log(lOO - U)

for U ;::: 60 %

I

( 4.35 )

The time factor cor,responding to every 10% of averag.-hdegree of consolidation for double drainage conditions is shown in the insr l5t€ in Fig. 4.9. The time factors corresponding to 50% and 90% consolidati ~ are often used in interpreting consolidation test results. You should re e m r that To = 0.848 for 90% consolidation , and Tv = 0.197 for 50% consoliCi ion.

4.5.3 Finite Difference Solution of th Governing Consolidation Equafl~t:-....

(4 .36)

and (4.37)

(4.38)

Equation (4.38) is valid for nodes that are not boundary nodes. There are special conditions that apply to boundary nodes. For example, at an impermeable boundary, no flow across it can occur and, consequently, aulaz = 0 for which the finite difference equation is

au 1 az = 0 = 2~z

(Ui- l .i -

Ui + l )

=0

( 4.39)

and the governing consolidation equation becomes U i.i + !

=

U i.i

+

Cv

~/

~ (2U i-l)

-

2Ui.J

( 4.40)

4.5 ONE -DIMENSIONAL CONSOLIDATION THEORY

F

1.

T

Ll z

~ ~--+-~~-+--~--+

1 + - - - - - 1 = nLlI - - - ----'"i

FIGURE 4 . 10 /

,/

Division of a soil layer into a depth (row)-time (column) grid.

167

168

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

4. Calculate the excess pore water pressure at interior nodes using Eg. (4.38) and at impermeable boundary nodes using Eg. (4.40) . If the boundary is

permeable, then the excess pore water pressure is zero at all nodes on this boundary.

The essential points are: 1. An ideal soil (isotropic, homogeneous, satur.bted) is- assumed in developing the governing one-dimensional con~olida tiQn equation. 2. Strains are assumed to be small. 3. Excess pore water pressure dissipation depen(ls on the time, soil thickness, drainage conditions, and per.mea.fJflity of tire soil. 4. The decrease in initial excess pore ater rcssure causes an equivalent increase in vertical effective strtsS and ~ettlement increases. 5. The average degree of e6ns Iidatio . conventionally used to find the time rate of settlemeftt.

EXAMPLE 4.6 rs of sand. The initial

FO

n the increment in applied stress and the degree of at he initial change in excess pore water pressure is lie ertical stress. From the data given decide on the n this case is Eg. (4.32) .

te the initial excess pore water pressure.

~

l1u o = 110-, = 100 kPa

Step 2:

Calculate the current excess pore water pressure at 60% consolidation. From Eg. (4.32), l1u, = 11110(1 - UJ = 100(1 - 0.6) = 40 kPa

Step 3:

Calculate the vertical total stress and total excess pore water pressure. Vertical total stress:

110-,

= 200 + 100 = 300 kPa

Total pore water pressure: 100 + 40 = 140 kPa

Step 4:

Calculate the current vertical effective stress. o-~

= o-z - l1u , = 300 - 140 = 160 kPa

4.5 ONE ·DIMENSIONAL CONSOLIDATION THEORY

169

Alternatively: Step 1:

Ca lculate the initial vertical effective Stress. Initial vertical effective stress = 200 - 100 :: 100 kPa

Stt!p 2:

Same as Step 2 above.

Step 3:

Calculate the increase in vertical effective stress at ~ consolidatiQ 60-;

=

100 - 40

=

60 kPa

.

"'

Step 4:

• 4

EXAMPLE 4 .7 A layer of soft clay, 5 m thick, is draine t t top Swce only. The initial excess pore water pressure from an applied oad at i e i'-:. 0 is distrib ted according to duo = 80 - 2z 2 , where z is the p,th m~ sured from the top undary. Determine the distribution a excess po~; ler pressu re with depth fter 6 months using the finite differencc e thod.1f"he coeffi cient o f conSO , ·dation, CIt, is 8 X 10- 4 cm 2 /s. Strategy Divide t e clay la~r into, say, five .€quat \aye 1 m thickness and find the val ue of the !.(Jaial exc 5S pore watef (!ressure at each node at time t = o using duo = 80 - 2zl'. nd a time tep tJ.) at w· lead to Q
Divide the soil

e'

0

a grid .

Divide' ' he d7 ,'O five lay"" ti..z = - =lm 5 C~ = 8 x 10-· cm 2/s = 8 X 10- 8 m2/s == 2.52 m2 /y r

Assume 6.t

=

0.1 yr. = C~ 61 _ 0:

Step 2:

6 z2

2.52 x 0.1 _ 025 12

.

05

<.

Identify boundary conditions. The bottom bou ndary is impermeable, therefore Eg. (4.40) apphes to the nodes along this boundary. The top boundary is pervious, therefore the excess pore water pressure is zero at all times greater than ze ro .

170

CHAPTER 4

Step 3:

Step 4:

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

Determine the distribution of initial excess pore water pressure. You are given the distribution of initial excess pore water pressure as I:!u o = 80 - 2Z 2. At time { = 0 (column 1), insert the nodal values of initial excess for water pressure. For example, at r w 2, column 1 (node 2), I:!u o = 80 - 2 X 12 = 78 kPa. The inii ex ess pore water pressures are listed in column 1; see the table low. ode 0 the grid. vR.i..u>-,undary, is

Let us calculate the excess par node located at row 2, colu 2. U 2. I+l

= 7

.~_..~·~ ng

a spreadsheet program are ults are plotted in Fig. E4.7 .

FO

5

30.0

61 47 39

0.20

0.30

0.50

0.0 46.3 65.0 60.0 48.5 43.0

0.0 39.4 59 .1 58 .4 50 .0 45.8

0.0 34.5 54.0 56.5 51 .0 47 .9

Excess pore water pressure (kPa )

00

FIGURE E4.7

20

40

60

•

What's next . . .We have only described primary consolidation settlement. The other part of the total consolidation settlement is secondary compression, which will be discussed next.

4.6 SECONDARY COMPRESSION SETTLEMENT

SECONDARY COMPRESSION SETTLEMENT You will recall from our experiment in Section 4.3 that consolidation settlement consisted of two parts. The first part is primary consolidation, which occurs at early times. The second part is secondary compression, or creep, w ., takes place under a constant vertical effective stress. The physical reason for secondary compression in soils are not fully understood. One plausible planatiom is the expulsion of water from micropores; another is viscous fon of he soil structure. We can make a plot of void ratio versus the logan experimental data in Section 4.3, as shown in Fig. 4.11. Pri assumed to end at the intersection of the projectio of the the curve. The secondary compression index is (4.41)

(4.42)

F

4.6

171

Primary consolidat ion

Secondary Gompression

/

ep

e,

I

A

Slope = C" : I I

' - - - -- -- - - - - - L --

--L----+-Iog I

Ip

FIGURE 4 . 11

Secondary compression.

172

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

4.7 ONE-DIMENSIONAL CONSOLIDATION LABORATORY TEST 4.7.1 Oedometer Test The one-dimensional consolidation test, called the oedome

FO

2

CHAPTER 4

Displacement gauge

Porous sto ne (b ) Fixed ring ce ll

Displacement gauge

Ring

Soi l sample

Porou s stone (c) Fl oating ring ce ll

(a) A typical consolidation apparatus (Photo courtesy of Geotest.) (b) a fixed ring cell and (c) a floating ring cell.

FIGURE 4.12

4.7 ONE·DIMENSIONAL CONSOLIDATION LABORATORY TEST

173

consolidation load completely, a negative excess pore water pressure that equals the final consolidation pressure would develop. This negative excess pore water pressure can cause water to flow into the soil and increase the soil's water content. Consequently, the tinal void ratio calculated from the final water content would be erroneous. The data obtained from the one-dimensional consolidation est are as follows: 1. Initial height of the soil, Ho, which is fixed by the hei fl. t 0 2. Current height of the soil at various time intervals un settlement data) . 3. Water content at the beginning and at the en of the soil at the end of the test. You now have to use these data to deter We will start with finding Cu'

FO

4.7.2 Determination of the Coefficient of Consoli

-':,i----->-YT:

(ff) A1 1'-------'--

1.15( ff)A

FIGURE 4.13

Correction of laboratory early time response to determine Co.

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDA liON SETTLEMENT OF FINE-GRAINED SOILS

for which Tv = 0.848 (Fig. 4.9). If point C were to lie on a straight line, the theoretical relationship between U and Tv would be U = 0.98~; that is, if you substitute Tv = 0.848, you get U = 90%. At early times, the theoretical relationship between U and Tv is given by Eq . (4.34); that is,

The laboratory early time response is represente 4. 13. You should note that 0 is below t e initi

tion is achieved

1. Plot 2.

at

i 'tial part of the curve intersecting and the abscissa (Vtime) at A.

it is

-VC:;.

FO

174

L -_ _~~_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _~

~

115V'i;;

FIGURE4.14

Roottime method to determine Cu '

4.7 ONE-DIMENSIONAL CONSOLIDATION LABORATORY TEST

175

6. The intersection of the line OB with the curve, point C, gives the displacement gauge reading and the time for 90% consolidation (t90)' You should note that the value read off the abscissa is ~. Now when U = 90%, Tu = 0.848 (Fig. 4.9) and from Eq. (4.31) we obtain

c u

=

a.848m, (90

where H dr is the length of the drainage path.

4.7.2.2 Log Time Method In the log time method, th lace ent gauge readings are plotted against the logarithm of time . typic I c e obtained is shown in Fig. 4.15. The theoretical early time s ttle ent ponse in a plot of logarithm of times versus displacement gau ding's a parabola (Section it a p arabola and a correc4.3) . The experimental early time curve is R'O tion is often required. The procedure, with reference dary

2.

FO

R

3.

~

ell el 2

~ ~

'"

""~

el 50

CD C

E

~

a~

d 100 /+---------I--I------"I----1t-------=:....--

L - - -_ _ _ _-L~~L-_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __+logt

I)

41) 150

FIGURE 4 . 15 Log time method to determine Cu'

176

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

5. You will recall (Fig. 4.9) that the time factor for 50% consolidation is 0.197 and from Eq. (4.31) we obtain

c

=

O.197H~r

u

( 4.44)

150

or each loading

1

·tial void ratio, eo = efin

+

l1e, where l1e is found from

FO

R

the

efin

e e

=

+ (Llz)fin H

0

1 _ (~Z)rin

He

4. Calculate e for each loading step using Eq. (4.6).

4.7.2.4 Determination of the Preconsolidation Stress Now that we have calculated e for each loading step, we can plot a graph of the void ratio versus the logarithm of vertical effective stress as shown in Fig. 4.16. We will call Fig. 4.16 the e versus log a~ curve. You wiJl now determine the preconsolidation stress using a method proposed by Casagrande (1936).

4.7 ONE-DIMENSIONAL CONSOLIDATION LABORATORY TEST

177

Horizontal line

--------------

- -Bisector

c,

FIGURE 4 . 16

Determination of the preco

method.

2. 3. 4.

S.

F

ed in practice is to project the straight mpres 'o curve to intersect the backward projec. , e at F as shown in Fig. 4.17. The abscissa of

'-------'-------+Ioga; O"~c

FIGURE 4 . 17 A simplified method of determining the preconsolidation stress.

CHAPTER 4

ONE·DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

Both of these methods are based on individual judgment. The actual value of (T~e for real soils is more difficult to ascertain than described above. Degradation of the soil from its intact condition caused by sampling, transportation, handling, and sample preparation usually does not produce the ideal curve shown in Fig. 4.16.

4.7.2.5 Determination of Compression and Recompression Indices The slope of the normal consolidation line, EA, gl To determine the recompression index, C, draw a way between the unloading and reloadi b /:l.Tv.·...... 'l r line is the recompression index ( C). Field observations indicate t total settlement and the rate of sett

pression index, Ce • approximately mid'1-.16). The slope of this

The

FO

178

C, without correction

0.42 eo I-------r~

' - - - - - - - ( 3 - ' - ,- - - - - - - > - log (3;

zo

FIGURE 4 . 18 Schmertmann's method to correct Cu for soil disturbances.

4.7 ONE-DIMENSIONAL CONSOLIDATION LABORATORY TEST

179

4.7.2.7 Determination of the Secondary Compression Index The secondary compression index, Ca , can be found by making a plot similar to Fig. 4.11. You should note that Fig. 4.11 is for a single load. The value of C", usually varies with the magnitude of the applied loads and other fac · such as the UR.

What's next . ..Three examples and their solutions are present how to find various consolidation soil parameters as discusse examples are intended to illustrate the determination of the com how to use them to make predictions. The third examg . lustr using the root time method. EXAMPLE 4.8

At a vertical stress of 200 kPa, the vo' an oedometer is 1.52 and lies on t e vertical stress of 150 kPa compresse

(3)

(4) an e versus log (J~ curve. Use

o n in Fig. E4.8.

F

-(1.43 - 1.52) log(350/200)

1. 52 1-------''1.' 1.4 5 t:=====t:::s~ 1.43

~-2..L. O O-3-"50 --50'-O-- log

FIGURE E4_8

0"; (kPa )

=

0.37

CHAPTER 4

Step 2:

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

Determine C. C,. is the slope of BC in Fig. E4.8. C,

Step 3:

- (1.43 - 1.45) log(3501200)

=

=

0.08

Determine the overconsolidation ratio. Preconsolidation effective stress:

Step 4: The void ratio at 500 kPa . consolidation line (Fig. E

• EXAMPLE 4.9

ertical strain. You know the in-

FO

180

6..£

6.. z

= -

z

Ho

0 .5 18

=-

= 0.Q28

te the modulus of volume recompressibility. m . v'

Step 3:

6..£, t;rr;

= -

0.028 =- = 1.9 x 10- 4 m2 IkN 150

Calculate the constrained elastic modulus. E~ = -

1

m vr

=

1 1.9 x 10-

4

= 5263 kPa

•

EXAMPLE 4.10

The following readings were tak en for an increment of vertical stress of 20 kPa in an oedometer test on a saturated clay sample, 75 mm in diameter and 20 mm thick. Drainage was permitted from the top and bottom boundaries.

4.7 ONE-DIMENSIONAL CONSOLIDATION LABORATORY TEST

Time (min)

0.25

t).H

0.12

0.23

2.25

4

9

0.33

0.43

0.59

16

068

25 0.74

36 0.76

181

24 hours

0.89

(mm)

Determine the coefficient of consolidation using the root time met

Strategy Plot the data in a graph of displacement rea follow the procedures in Section 4.7.2.1.

Solution 4.10 Step 1: Step 2:

FO

Step 3:

Make a plot of settlement (decreas shown in Fig. E4.10. Follow the procedures out From Fig. E4.10,

Calculate C

o .----,~~----~--~~--_.--__.

o

0.1

1-'~-+--;"~--I----+---+--1

0.2 f - -""''I.- - - ' - - - - - - I - - - - - + - - - + - - j

E .5

0.3 f ----I--¥<,__i-----I----+----+--- j

~ 0.4 I-----I---~~----I---+----+---j E

<1>

~

0.5

I----+---t--"I),;,,,--+----+---~---l

0.6

f----I---+---'~_r_""-r_----+-----_I

OJ

\----t-----i---+f'.c--'"

Vl

0.8 ' -__- L_ _ _'--_-'--+--~~~---'----' 6 o 2 5

FIGURE E4.10

•

182

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

What's next . . .We have described the consolidation test of a small sample of soil and the soil consolidation parameters that can be obtained. What is the relationship between this small test sample and the soil in the field? Can you readily calculate the settlement of the soil in the field based on the results of your consolidation test? The next section provides the relationship between the small test sam e and the soil in the field.

4.8 RELATIONSHIP BETWEEN LABORAT_n......... AND FIELD CONSOLIDATION

( 4.45)

and , by simplification, ( 4.46)

o u how to find the expected field a i,9U lar degree of consolidation.

FO

settlement fr

Strategy You are given all the data to directly use Eq. (4.46). For part (1) there is double drainage in the field and the lab, so the drainage path is one-half the soil thickness. For part (2), there is single drainage in the field, so the drainage path is equal to the soil thickness. Solution 4.11 (1) We proceed as follows:

Step 1:

Calculate the drainage path. (Hdr)lab

=

20

2

=

10 mm = 0.Q1 m;

" .9 TYPICAL VALUES OF CONSOLIDATION SETILEMENT PARAMETEA S

183

Slep 2: Ca lcula te the field time uSing Eq . (4.46). _ II'b( ~') h
y~ a rs

(2) We proceed as foll ows:

Step I:

)

Calculat e the d ra in age path . ( H~') I'b '"

20

2" -

10 mrn .. 0.01 m;

(H

Sle11 2: Ca lculate fi eld time us ing Eq . (4.46). lr.dd ""

l .b( Hl.)r..", '" 15 X

10: A'5:.x l

om (

(~')I'b

,

on one bou~ary rather a en for a giten ~ cent co n.-

ex,1 ts on

than bo th bounda ries of the clay 19 solid D. tion in the field is four timett

"""'0.... er.

,

~tlo

Wha", next , _,Several empirical eq rameters to simple , less tj~e .cJn~umi

I nkl~g

s ,,}L lable conso idal;on pa,son tests suc h as t eAt e cbe rg lim its and

water content. In the next settion. some.,f these rela tionsh ips re present ed .

4.9

TYPICAL~~~~O~F'~C~O~N~S~O~L=D:A~T~IO[N~_ _ _ _ __

'~j~~~;~~a~~~~,

I

may

nOI

c, C,' '"' C..IC. =

'1 ests and conso lidation seUlement param. You sho d be ut ious in using these relationships be-

betweeq,.;(mpl

b aQ cable

ur

soil type.

to C.J~O Ol.l~.07

Em pirical Relationships (Skempton, 1944) (Azzouz e l aI., 1976) C, "'" O.Ol (1V - 5) (A zzOUl el al.. 1976) C, '" 0.37(1"" + 0.003wu. - 0.34) (Azzouz et at, 1976) C, - O.OO234wLL G, (Naga raj and Murth y, 1986) C, - 0. 15(e" - 0.007) ( A ZZOU1. ct al. 1976) C, - 0.003(14' + 7) (Azzouz ct a!. . 1976) C," 0.126(1"" t 0.00310''-1 - 0.(6) (Azzouz el aI., 1976) C, - 0000463wl.LG, (Naga raj and Murt hy. 1985) It-' is th e natural water Content (%), "'LL is t h~ liquid limit (%0). e" is the inlllaJ void ral io

C, - O.OO9(wll. - 10) C~

- 0.40(1"" - 0.25)

184

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

What's next . ..Sometimes, we may have to build structures on a site for which the calculated settlement of the soil is intolerable. One popular method to reduce the consolidation settlement to tolerable limits is to preload the soil and use sand drains to speed up the drainage of the excess pore water pressure. Next, we will discuss sand drains.

FO

4.10

SAND DRAINS

Lateral

(a)

~ 2R

t--O~6·

J'

' --

9 - -,

,

--0-0--0-'r-s-j R

~ I~

= 0.56s (b)

(c)

FIGURE 4.19 (a) Vertical section of a half-closed sand drain . (b) Plan of a square grid sand drain. (c) Plan of a triangular grid sand drain.

4.10 SAND DRAINS

185

sional consolidation two-way drainage reduces the time fo r a given degree o f consolidation by four limes compa red with single drainage. The d iameter of sa nd drai ns ranges from about 200 to 600 mm. The diameter required must only be large enough to d rain the pore wate r and prevent prematu re clogging from fi nes in the soil to be drained. Filter fab rics are now commonly used at the interface of the natural soil and the backjin to prevent clogging. The spacing of sa nd dra ins depends on the permeabili({ 9 f the soil and the desired time to achieve the required degree of consolidatio n. he spacing of ·ng:ro nges the drains must be less than the thickness of the soil layer. yp·ca l s from 2 ill to about 5 m in eit her a squa re or triangular gri Fig . . 9,). 'J'he excess pore water from an applied IO0< 0.56 s f~a...s~ ~ gHQ' ·nd R .... 0.53 s for a triangular grid; s is spacing of drain . The cons lid ·on 0 Ihe soil has two components-one component is due 10 vert" al a ra in a~ and the o ther compo, nen t is due to radial drainage. The governing equa tion for axisymmetri adia rainage is (4.47)

and tbe initi I ndition IS I = O. U .: U ~ where is tne radius of the drains. , is tim e. and R I~e a IUS of tbe cyli ndrical inHuence zone (Fig. 4.19). Richart (1959) repOrle solutions 10 Eq . 4.47) fo t' oca5es-free strai n and eq ual strai n. F. strain occ when Ihe~ft.! e load is uniformly distributed (flexible foun ati ) and the resulting s ac ttl menl is uneven , as shown in Fig. 4.19. ~ l!al~t:Qj occurs w ~ the {ace IIlemen t is forced to be uniform (rigid fou ~ ~on) and the su lti n surf load is not uniformly dist ribu ted. Richart (1?59) showed lha~~ fferenc in the two cases are small and the solution equal stra in is 0 ,~!ypractice and shown in Fig. 4.20 .

y

100

90 80

70

100 50

--1

60

~ 50

20

.1

~.

'" 30

20 iO

0

0.001

0.01

01 T.

FIGURE 4.20 Time factor for radial conso lidation.

10

FO

186

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

The time factor for consolidation in the vertical direction is given by Eg _ (4.31) while the time factor for consolidation in the radial direction (T,) is (4.48)

and Ch to account for the smear z09' . The average degree of cons@flJ ti pore wa ter pressure (V ur) is (4.49)

where V is the avera

1.

erms of n by solving Tr = Ch tl4R 2 radius of the drain and n = Rlrd _

=

Ch tl4(nr d) 2, where rd is the

8. You now have to find a value of T, such that the degree of consolidation matches Vr calculated in Step 6. You do this by iteration since you do not know either T, or n. Select a value of n from the values shown in Fig. 4.20 and find Tr. Determine Vr from the curve corresponding to the selected value of n . Compare this value of V, with Vr calculated in Step 6. If they are not approximately equal «10% difference) then reiterate until you get a satisfactory solution. You may interpolate from Fig. 4.20 for intermediate values of n from 5 to 100. You can also plot a graph of n versus V r and find the n value corresponding to the desired V , (Step 6) . 9. Calculate the spacing, s, depending on the grid you desire. For a square grid s = 1.8R = 1.8nrd and for a triangular grid s = 1.9R = 1.9nrd.

4.10 SANO DRAINS

187

EXAMPLE 4.12 A foundat ion for a st ructu re is to be constructed on a soft deposit of clay. Below the soft clay is a stiff overconsolidated clay 20 m thick . The calculated settlemen t cannot be tolerated and It was decided that the soi l should be preeo solida ted 1 re: by an emba nkment equivalent to the building load. The data avaj[ ~ C" '"" 6 m2/yr. Cit = 10 m2/yr, m u = 0.2 m21MN

J

The founda tion sizt: is 10 m X 10 m and CIs = 400 kPa. The da n....d ia e~OO mm. Determine the spacing of a square grid of the sand dfains ac teve W % consolidation in 4 months.

to

Strategy Follow the procedure outlined in Sec 1

Solution 4 .12 Step 1:

The given va lues are: 2

G'", = 10 ~~ r , mu - 02 mZ

Cu = 6 m /yr

m'J I:: 400 kPa

Ho =

Step 2:

Calcu late the Pri mar'~SO lida 'o n settJemeu etermine the average 1Ilcrease in a ~ ~d v. ical stress a idthe center ~oun atlon 10

B

"- - 2' m

Step 4:

!:1J12) ,

5

L

"2 =

111 ,

5 10

~ --=-

10 -

""

5

-

Dcterml Til' The draina-ge system is half·dosed (stiff clay layer at bottom is assumed imper vious). Therefore. H dr ::: N o = 20 m. 4

Tu = Step 5: Step 6:

n

Determine U. From Fig. 4.9, U - 8% for

Cu i

m," T~ =

6 x 12 (20f

- 0'()05

0.005.

Find U,. From Eq. (4.49), U "" 1 _ I - U~, =- 1 _

,

l

V

I - 0.9

1 0.08

=

0.89

188

CHAPTER 4

ONE·DIMENSIONAL CONSOLIDATION SETILEMENT Of FINE·GRAINED SOILS

Step 7:

Determine T,.

Step 8:

Determine n.

From Fig. 4.20 with T,

'=

1.48, U,'" 1

>

Step 9:

• coeffi~nt

't:-

The lateralfa'fth ; lure 3' LO~ r normally co nsoltdateff soil, eq ua on u ted by Jaky (1944 as

=

0': 0-;. was presented

Ko = o~

IS

In Section reasonabl y predicted by an

['§o~~~'~in~~~~J

(450)

whe re ~ IS a fundJli enta o i constant th a t will be discussed In Chapter 5 ..._", The value ot I cons t. During unload ing or reloadi ng, the soil stresses m ust adjust I e in eqllli ium with the applied stress. This means that stress 01 oJ vertically but also horizon tally. From Chapter 3. you changes take a know tha , or a give ~urface stress, the changes in horizontal stresses and vertic~tresses a diffeTenL Therefore, Ko for overconsolidated soils, denoted by K o , ""a uld not e a constant. Various eq uations have been suggested linking K': to ~O n equation that is popular and fo und to match test data reasonably well is an equation proposed by Meyerhoff (1976) as

K:

~ K': 4 . 12

=

K:C(OCR)1I1 - (I - sin

¢l~)(OCR)'I2 !

(4.51)

SUMMARY Consolidation se ll iement of a soil is a lime-depende m process thai depends on the soil 's permeabi lity, thickness and the drainage conditions. When an increment of ve rtica l stress is applied to a soil, the instantaneous (i nitial) excess po re water pressure is equal to the vert ical stress increment. Wit h time, the ini tial excess pore water pressure decreases, the vertical eHective stress increases by the amount o f decrease of the in itial excess pore water pressure , and settlement

189

4.12 SUMMARY

increases. The consolidation settlement is made up of two pa rIs-th e ea rly time response called primary consolidation and a later time response called secondary comptession. Soils retain a memory of the past maximum effective stress, which may be erased by loading to a higher stress level. If the current vert ical effect jve st ress on a soil wa s never exceeded in the paSI (a normally consolidated 11), '"I would behave elastoplastically when stressed. If th e current vertica l eff live stresf on a soil I'o'as exceeded in Ihe past (an overconsolidaled soil) , it "",oul behave la s· tically (approximalely) for stresses less thau its past maxim m trect,,,,~" , PractiClll Examples

Vertical stress [kPa ) San lemenl (m rn)

.15

,

Time-tmin)

o

0"

16 0.79

36 0.86

"

0.91

100 0.93

he IV K. when full " ,im vertical stresses of 90 kPa and 7S kPa al the top (I b'OttOIll of th e.gralne~~'lla yer. respectIvely. You may assume that Ihe ~.'''''I~,verl ical Slre~nearl~IS .t) led 10 thIS layer. a

(a) Determine the pr ,ary consolidation settlement of the fine.gralned soil layer wh en tbe-tank is fu ll. (b) Calculate and plot the selt lement-time curve.

2.

- --- -----"" r.

18 1>.Nfm1

-"LO'WoL'-_ _ _ _ _ _ r,,\ = t 9.4 kNlm' lm F.ne·i!ai!1e4 SO\I 3m

[clay

~nd

... .. 62%

FIGURE E4. 13a

sill m..!ur..s)

F

190

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

Strategy To calculate the primary consolidation settlement you need to know Cc and C, or m v , and (J~o , ~a-, and (J~C" Use the data given to find the values of these parameters. To find time for a given degree of consolidation, you need to find Cu from the data.

Solution 4.13 Step 1:

Find Cu using the root time method. Use the data from the 240 kPa load step t Vtime curve as depicted in Fig. E4.13b. ~ in section 4.7 to find Cu' From th

_ 0.2

~ 0.3 :;::; 0.4 c

E 0.5

-

.'!' 0.6

'iii

c/)

07 . 0.8 0.9 t+r+-t\t\~~H-t+lfl"'!"t"-;-:...~~~++-l....l.-j

=

59.3

X

10-6 m 2 /min

void ratio at the end of each load step. eo = wG. = 0.62 x 2.7 = 1.67

'tial v d ratio: Equation (4.6):

e

=

~z

e - - (1 - e) o Ho 0

=

~z

1.67 - - (1 20

= 1.67 - 13.35 x 10- 2

+ 1.67)

~z

The void ratio for each load step is shown in the table below.

a; (kPa) Void ratio

15 1.66

A plot of e - log

30 1.66

a- ~

60

120

1.64

1.52

240 1.38

480 1.25

versus e is shown in Fig. E4.13c.

4.12 SUMMARY

1.75

'"§

1.55

II

1.45

I

-0

g

' ,

Ii i II

1.65 o

I

U ,~

i\

I o-;c

Co = 0.45

",

1.35 1.25 10

191

1000

100

0-; (kPa)

FIGURE E4. 13c

Step 3:

Determine cr~e and Ce . Follow the procedures in Sectio

Step 4:

FO

R

Step 5:

Divide the clay layer into three sublayers of 1.0 m thick and compute the settlement for each sublayer. The primary consolidation settlement is the sum of the settlement of each sublayer. The vertical stress increase in the fine-grained soil layer is 90 - (

90 - 75) 3 Z

=

90 - 5z

where z is the depth below the top of the layer. Calculate the vertical stress increase at the center of each sublayer and then the settlement

192

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

from the above equation. The table below summarizes the computation.

Layer

z

u ~o at center of

(m)

sublayer (kPa)

au z 87.5

0.5

48.7

2

1.5

54.9

82.5

3

2.5

61.1

77.5

(kPa)

Step 6:

t=

FO

26.3Tu (days)

~

80 90

0.2 0.8 1.9 3.3 5.2 7.6 10.6 14.9 22.3

20.8 61.3 81.6 102.1 122.5 142.9 163.3 183.7

0.567 0.848

Time (days)

5

E

5

c

Q)

E

50 100

Q)

:;:::

OJ 150

(/)

~

'"

200

FIGURE E4_13d

10

15

20

25

I

I

I

~ I

---

~

1

-

•

EXAMPLE 4.14 A geotechnical engineer made a preliminary settlement analysis for a foundation of an office building, which is to be constructed at a location where the soil strata contain a compressible clay layer. She calculated 50 mm of primary cons lidation settlement. The building will impose an average vertical stress of 15 a in the clay layer. As often happens in design practice, design changes are equired. n this case, the actual thickness of the clay is 30% more than the 0 ·gin soil pro Ie indicated and, during construction, the groundwater table ha be I by 2 m. Estimate the new primary consolidation settlement.

Strategy From Section 4.3, the primary consolidation se tl tional to the thickness of the soil layer and also t tl1 ·n [see Eq. (4.18)]. Use proportionality to find tile ne settlement.

Solution 4.14 Step 1:

Estimate the new primary con increase in thickness.

Step 2: ue to lowering of water table

FO

= 2 x 9.S = 19.6 kPa

•

IE XAMPLE 4.15 The foundations supporting two columns of a building are shown in Fig. E4.15. An extensive soil investigation was not carried out and it was assumed in the design of the footing that the clay layer has a uniform thickness of 1.2 m. Two years after construction, the building settled with a differential settlement of 10 mm. Walls of the building began to crack. The doors have not jammed but by measuring the out-of-vertical distance of the doors, it is estimated that they would become jammed if the differential settlement exceeded 24 mm. A subsequent soil investigation showed that the thickness of the clay layer was not uniform but

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

780 kN

FIGURE E4.1S

varies as shown in Fig. E4.1S. The a expected total differential setJ;l:e~~·t.. become jammed.

FO

194

780 (LlO"z)B = (1.5 + 2.8)(1.5 + 2.8) = 42.2 kPa

Step 2:

Note: For a more accurate value of ~(J"z you should use the vertical stress increase due to surface loads on multilayered soils (Poulos and Davis, 1974). Calculate the primary consolidation settlement. Use Eq. (4.18), Ppc = Homu ~(J", to calculate the primary consolidation settlement. (PPC)A = 1.2 x 0.7 x 10- 3 X 30 = 25.2 X 10- 3 m = 25.2 mm (PpC)B = 2.8 x 0.7 x 10- 3 x 42.2 = 82.7 x 10- 3 m = 82.7 mm

EXERCISES

Slep 3:

195

Calculat e the differential settlement. Differential settlement: 0 - 82.7 - 25.2 """ 57.:5 mm

Slep 4:

Calculate the time for 24 mm differential settlement to occur.

Current differential settlement: 0, = 10 mm &~

10

8

57.5

Degree of consolidation: U - - = From Eg. (4.34).

4

4

•

•

=

0.17

T" '"' - Ui = - X 0.17< ., 0.037

. H_._ ' From Eg. (4.31): , - _T_

C.

" 7;::';:;~=

Therefore, in the next 10.25 ears, the total differential sell l e~ t would be 24 mm.

•

EXERCISES For all prob le

Theory cia soil of thie ness H is a e ~ rain on Ihe lOp boundary through a thin sand I er A vet . al stress of A waS'1fplied ~ the clay. The excess pore water pressure ~"'lIIIid~ ,i st~b uttOn was linear i fne'lo il laycl--~ a value of II, a{ the top boundary and u~ (Uh u,) ft'"tbe bottom bf u Q:dfl ry~ex.s:tss pore water pressure at the top boundary was n t zero because t ~na layer wa part ially blocked. Derive an equation for Ihe excess p re water pre ~ re disti but" n i h soil thickness and time. 4:1

A soil layer o~kness H o as only single drainage through the top boundary. The excess

pore water pressur istri tion when a vertical stress. CT, is applied va ries parabolically with a va lue of zero at the top boundary and uh al Ihe bOllom boundary. Show thai

, 4.3

Show that. for a linear elaslic soil. m~

4.4

=

(1

+ v')(l - 2v') v')

£'(J

Show thaI, If an overconsolidflted soil behaves like a linear elastic matcnal,

K':

=

(OCR)K~C

- - ' -' -, (OCR - 1) )

-

,

196

CHAPTER 4

4.5

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

The excess pore water pressure distribution in a 10 m thick clay varies linearly from 100 kPa at the top to 10 kPa at the bottom of the layer when a vertical stress was applied. Assuming drainage only at the top of the clay layer, determine the excess pore water pressure in 1 year's time using the finite difference method if Cu = 1.5 m 2 /yr.

4.6

4.7 me that the present

Problem Solving 4.8

4.9 4.10

FO

4.11

4.12

test on a soli sample for an

to a sample

4

9

16

36

64

100

0.49

0.61

0.73

0.90

0.95

0.97

urs the settlement was negligible and the void ratio was 1.20, corresponding of 18.2 mm. Determine C using the root time and the log time methods.

A sample of saturated clay of height 20 mm and water content of 30% was tested in an oedometer. Loading and unloading of the sample were carried out. The thickness H f of the sample at the end of each stress increment/decrement is shown in the table below.

cr~ (kPa)

100

200

H,(mm)

20

19.31

400 18.62

200

100

18.68

18.75

(a) Plot the results as void ratio versus log (J"~. (b) Determine Cc and C. (c) Determine mv between

(J"~

= 200 kPa and

(J"~

= 300 kPa.

EXERCISES

4.13

197

A sample of saturated clay, taken from a depth of 5 m, was tested in a conventional oedometer. The table below gives the vertical stress and the corresponding thickness recorded during the test. cr~ (kPa)

h(mm)

100 19.2

200 19.0

400 17.0

800 14.8

1600 12.6

800

400

13.1

14.3

100. A

15~

The water content at the end of the test was 40% and the initial h (a) Plot the graph of void ratio versus log cr~. (b) Determine Cc and C. (c) Determine mu between cr~ = 400 kPa and cr~ = 500 (d) Determine the relationship between e (void rat~nd (e) Determine cr~c using Casagrande's method. 4.14

The following observations were recorded in an oe in diameter and 30 mm high. Load

(N)

Displacement gauge reading (mm )

o 1.4

(a) (b) (c) Calculate t

FO

(d)

An oil tank is to be Jted 0 a soft alluvial deposit of clay. Below the soft clay is a thick layer of stiff clay. It w decided that a circular embankment with sand drains inserted into the clay would be constructed to preconsolidate the soil. The height of the embankment is 6 m and the saturated unit weight of the soil comprising the embankment is 18 kN/m 3 . The following data are available: thickness of clay = 7 m, mu = 0.2 m 2 /MN, Cu = 3.5 m 2 /yr, Ch = 6.2 m 2 /yr, diameter of drain = 300 mm. The desired degree of consolidation is 90% in 6 months. Determine the spacing of a square grid of the sand drains such that when the tank is constructed the maximum primary consolidation should not exceed 20 mm.

Practical 4.18

Fig. P4.18 shows the soil profile at a site for a proposed office building. It is expected that the vertical stress at the top of the clay will increase by 150 kPa and at the bottom by 90

198

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

4.19

'/so, =

17_5 kN/m 3

FO

lidated clay,

Vertical stress (kPa) Void ratio

50 0_945

100 0.895

200 0_815

400 0.750

800 0.705

Calculate the primary consolidation settlement. Assuming that the primary consolidation took 5 years to achieve in the field, calculate the secondary compression for a period of 10 years beyond primary consolidation. The secondary compression index is Cj6. [Hint: Determine ep for your Urin from a plot of e versus log u~.l

CHAPTER

5

SHEAR STRENGTH OF SOILS ED 15

INTRODUCTION

FO

5.0

d determine the shear strength d be able to: • Determine the shear strength Qf soils • Understand the differences between drained and undrained shear strength • Determine the typ o f shear test that best simulates field conditions • Interpret laboratory and field test results to obtain shear strength parameters principles learned from previous chapters and other

of stresses and strains, and stress paths • Friction (statics and/or physics) Sample Practical Situation You are the geotechnical engineer in charge of a soil exploration program for a dam and housing project. You are expected to specify laboratory and field tests to determine the shear strength of the soil and to recommend soil strength parameters for the design of the dam. In Fig. 5.1 a house is shown in a precarious position because the shear strength of the soil within the slope near the house was exceeded. Would you like this to be your house? The content of this chapter will help you to understand the shear behavior of soils so that you can prevent catastrophes like that shown in Fig. 5.1.

199

200

CHAPTER 5

SHEAR STRENGTH OF SOilS

1

5.1 Shears"e~~~~!oll

lstance to applied shearing

forces .

FO

shear strength of soils.

5.2

the change in volume of a soil when it is distorted by

QUESTIONS TO GUIDE YOUR READING 1. What is meant by the shear strength of soils? 2. What factors affect the shear strength?

3. How is shear strength determined? 4. What are the assumptions in the Mohr-Coloumb failure criterion? 5. Do soils fail on a plane? 6. What are the differences between peak, critical, and residual effective friction angles?

5.3 TYPICAL RESPONSE OF SOILS TO SHEARING FORCES

201

7. What are peak shear strength, critical shear strength, and residual shear

8.

9. 10. 11. 12.

strength? Are there differences between the shear strengths of dense and loose sands or normally and overconsolidated clays? What are the differences between drained and undrained Under what conditions should the drained shear strengtl or the u drained shear strength parameters be used? What laboratory and field tests are used to dete mi What are the differences between the results of tests?

5.3 TYPICAL RESPONSE OF S...--__. TO SHEARING FORCES

FO

s 0 soil by applying simple shear ample, which we call Type I, rep-

z ~===~ Expansion . ................ I LIz

to (a)

Original soil sample

"-------.L-----x (b) Simple shear defor mat ion of

Type I soils

FIGURE 5.2

(c)

Simple shear deformat ion of Type II soils

Simple shear deformation of Type I and Type II soils.

202

CHAPTER 5

SHEAR STRENGTH OF SOILS

on soil behavior. Since in simple shear Ex"" £7 "" 0, the volumetric strain is eq ual to the ve rtical strain , E. = 6 zlH o, where az is the vertical displaccment (posi tive for compression) and H o is the initial sample height. The shea r st rain is the small 's t e horizontal angular distortion expressed as 'Y:.t = tu l Ho, where 6x ~ displacement . We are going to summarize the important features of the esponses of these two groups of soils when subjected to a constant vertica l (n rm 1 effective slress and increasing shea r strain. We will consider the shear,...s ess e rsu ~ e shear strain, the volumetric strain versus the shear strain, a the v id ralio versus the shear strain responses, as illustrated in Fig. 5.3. When a shear band(s) develops in som~s 0 ov TC9nsolidated clays, the particles become o rien ted parallel to the dlr~ 1 0 the..shear band, causi ng the fina l shear stress of these clays to decrease belo the-critical state shear stress. We will call this type of soil, Type II·A . and he fin I~ h ear stress attained the residual shear stress, 'T,. Type I soils at ve mal effective slress can also exhibi t a peak shear stress duriny)'ie.aring.

Type I

Type II SOII$

1<,

1>,

FIGURE 5.3 Response of soils to shearing.

5.3 TYPICAL RESPONSE OF SOilS TO SHEARING FORCES

Type I soils-loose sands, normally con clays (OCR :s 2)-are observed to:

203

overconsolidated

until a

Type II soils-d observed to:

FO

• Sho

The critical state shear stress is reached for all soils when no further volume change occurs under continued shearing. We will use the term critical state to define the stress state reached by a soil when no further change in shear stress and volume occurs under continuous shearing.

5.3.1 Effects of Increasing the Normal Effective Stress So far, we have only used a single normal effective stress in our presentation of the responses of Type I and Type II soils. What is the effect of increasing the

204

CHAPTER 5 SHEAR STRENGTH Of'SOILS

normal effective stress? For Type I soils. the amount of compression a nd t he magnitude of th e critical state shear stress will increase (Figs. S.5a.b). For Type II soils. the pea k shear stress tends to disappear, the crit ical shear stress increases and the change in volume expansion decreases (Figs. S.Sa.b). If we were to pial the peak shear stress and the critical s ~h ear stress for each constant normal effective stress for Type I and II soil we would get: 1. An approxima te straight line (OA. Fig. S.5c) that shea r stress values of Type ' and Type 11 soils. tween OA a nd the a;, axis. the critical sta te fric ti will be cal1ed the failu re envelope because any sh a critica l sta te shear stress.

all the edt.J.,1 state e wi call th~ angle bea ngle, • ..,.The line OA al" tressti'hatlies on it is

ij

2. A cu rve (OBCA . Fig. SSc) that lin ~ pea hear-stress values for Type I I soils. We will call OBC (the cu~ ,p a)1..Qf ?B}?A). the peak shear stress envelope because any shea r stress thaYUes'o it is a peak s ~ear st ress, fo r 'Ii pe n soils is ap ~rs}S a poin t (point 9) located on OA (fig. 5. • r Type 11- soils, the r~ua l shear st resses would he on a lin OI>::beloVOA , We wil II the an~le ' betwee n 00 and the a~ axis. the...residuaJ iriCl n angle, ~; s th norm ' eUective slress increases. the cri 'fal void ratio decreases (Fig. 5,5dl.T hus,j)le critical void ra tio

At large normal effective resses, pel'l.k shea r sIr sup pre~d and only a criti cal sta te..; hear stress is observed an

u t of the norll'

is dependent on he rna,

eii~"r
Type II SOIls

• Type I 5O,ls

o

a. 'd

•

~

~

~ ~~;::~~~T:"'~';~" ~

I

lllCre.1$I"i nOflNI

effet'~ $InISS

l'"

T,-pe l150ils

I

notm~1

IIlClUSl"l effectIVe s[.esJ;

' - - - - - - - - :-

to,

'd)

'.

FIGURE $.5 Effects of i ncreasi ng norma l effective stresses on the response of

soils.

S.l TYPICAL RESPONSE OF SOILS TO SHEARING FORCES

205

5.3.2 Effects of Overconsolidation Ratio The init ial state of the soi l dictates the response of the soil to sheari ng forces. For exa mple, two overconsolida ted soi ls wi th different ove rconsolidation ra tios but the sa me mineralogica l composition would exhibi t different peak shear stresses and volume ex pansion as shown in Fig. 5.6. The highe ve onsolidated soil gives a higher pea k shea r strength and grea ter volume e ansion .

I.

2.

J. 4.

5.

6.

7.

The essential points are: Type I soils-Ioost sands and normally consolidlUed and lightly over· consolidated clays- strain harden to a critical Skltt slt",r stress and comp'YSS toward a cril/cal void ratio. Type II soils-dem'e sands and o.~rconsolidated days-naclt a peak shear stnss, strain soften to a critical stille sll«u stnss and txpmuJ 10ward a critical void ralio afler an i,dtial compression al low shear strains. The peak shear slress o/ Type II soils ;s suppressed and 'he volume txplJnsion dec«ases when Ihr normal effective stress is large. All soils 'YOM a criticaluate. irresptclive oj ,lte;r inilial Slate, at which continuous shearing occurs wUhoUl changes In sMar sinsS and J.·olum£. AI large strains, Ihe p articles 0/ some overconsolidaled days become orlentul parallel to the direction oj shear bands and the jinal shear stress attained is lower Ihan the cridcld st4fe sllear Slress. The critical slOIe shear stnss and Ihe critical void ralio tkptnd On the normal effective stress.. Higher nonnal effective strnsa mW/1 in Itigher critical -stale shear stlY!SJes and lower cril/call'oid ,.tios. Higller overconsolidation ratios nsull;n higher peak shear slrrsses and grealer l'oillme up4nsion.

FIGURE 5.8

Effects of OCR on peak strength and volume exp ansion.

206

CHAPTER 5

SHEAR STRENGTH OF SOILS

T Co

~L--_ _ _ _ _- +

FIGURE 5.7

Peak shear stress envelope for cemented s

5.3.3 Cemented Soils

FO

What's next . . .You shaul

(5.1)

H= f-lW

t of static friction between the block and the table and

(5.2)

w

,--T

/Sli P plane

H ---+

N

;\ R (a)

FIGURE 5.8

(b)

(a) Slip of a wooden block. (b) A slip plane in a soil mass.

FO

5.4 SIMPLE MODEL FOR THE SHEAR STRENGTH OF SOILS USING COULOMB'S LAW

207

where T f (= TIA, where T is the shear force at impending slip and A is the area of the plane parallel to T) is the shear stress when slip is initiated, (u:')f is the normal effective stress on the plane on which slip is initiated . The subscript f denotes failure, which according to Coulomb's law occurs when rigid body movement of one body relative to another is initiated. Failure doe ot necessarily mean collapse but the initiation of movement of one body relative to another.

that is, ' ansi on. We ar goin to use our knowledge of statics to investigate impending sliding of particles up or down a plane to assist us in interpreting the shearing behavior of soiJs using Coulomb's frictional law. The shearing of the loose array can be idealized by analogy with the sliding of our wooden block on the horizontal plane. At failure (impending motion),

(5.3)

ROW 20 _~_0

Row 1

(0)

FIGURE 5.9

loose

~ Row2 ~R owl (b)

Dense

Packing of disks representing loose and dense sand.

208

CHAPTER 5

SHEAR STRENGTH OF SOILS

w

z

,·,LL ,..

(,, ) S'reues on Iillhlle pl,ne

X

e/iol Simulated sheanng 01 8 dense . .

FIGURE 11.10 Simulation of failure in dense sand

Consider two particles A and B in tj1e d ~n se a e bly and draw the freebody diagram of Ihe stresses at the slidi st o tact 1ktween A and as depicted in Fig. S.lO. We now appeal to our wooden ock r an analogy describe the shearing behavior o f the dense array. Fa the d; nse array, II e woe<Jen block is placed on a plane oriented at a "gle a t~ he horizontal ( Fig .~.lOb). Our goal is to find the horizontal force to inii1ate movemenl of lhe block up e incli ne. You may have solved this f fob lem in .sties. Anywa we are gain to solve it where N is the normal t ree. Using the again. At impending mario, T. force eq ui librium eq \jons the I. and Z directib s, w e t 'LF. ... . H - Nsin ~ . 0 I.F "'0: Ncos ~. a-W - O

Solving Co

and w,-we obtain

e

(5.5)

, cos a)

(5.6)

W - N(cos a-- f.I. sin Q)

(5.7)

ding Eq. (S.6) by Eq.. .-(S ) n~ sf H W

(5.4)

+ an 1

).I.

plifying, we obtain =- tao 4l' + tan Q J Ian 4>' Ian 0.

Q

tan 0

l~emblY' we can replace H by TJ and 130$'+ 13001 (O'~)J 1 A..'

tan,+, tan

0.

::

, (O'~)J

,

tan(¢, + 0.)

W by

«(f~)J' (5.8)

Let us investigate the impl ications of Eq . (5.8). If a = 0, Eq . (S.8) reduces to Coulomb's frictional eq uation (S.2). If Cl increases, the shear strcngth , "), gets larger. For instance, assume 4l' = 30" and (a~)J is constant; th en for (II ". 0 we gel TJ = O.S8(a~ )f' but i f (II - 10" we get Tf = O.84(O'~)f' that is, an increase of 4S% in shear st rength for a 10% increase in Q. If the normal effective stress increases on our dense disk asse mbly, the amou nt of " riding up" of the disks wiU decrease. In fact. we can impose a sufficien tly high normal effective stress to suppress Ihe "riding up" tendencies of the de nse disk assembly. Therefore, the abilily of the dense disk assembly to e xpand depends on the magnitude of the normal effective stress. The lower the normal effective stress, Ihe grea ter the value of (II. The ne t effect o f a due 10 normal effective st ress increases is th at the fai lure envelope becomes curved as illustra ted

5. 4 SIMPLE MODEL FOR THE SHEAR STRENGTH OF SOILS USING COULOMB'S LAW

209

Curved CoulQrnl) lall ure envelope caused by dllal lon A

o

G·

FIGURE 5 . 11

~

Effects of dilation on Cou"lomb's failu reaflve lope

others occu r (5.9)

where the positive sign refe to soils in wfiic -rhe net ovemen t of the particles is initialed up \ plane and the " n lti e--sign e rs to net particle movement lane. down I W dl caJUhe angle 0: the I ation ang e. It is a measure of Ihe change in volumetri trair7"wilh respecl lo th change In shear strain. Soils Ihal have pos-i.tive value f a expand during shearing whi le soils with nega tive values of ~ ~ contract uring shearfng. I hr's circle of st rain (Fig. 5.12). the dilation ngleJ (5.10)

fiGURE

5.12 Mohr's circle of strain and angle of dilation.

210

CHAPTER 5 SHEAR STRENGTH OF SOilS

where L'l denotes change. The negative sign is used because we want Ct to be positive when the soil is expanding. You should recall that compression is taken as positive in soil mechanics. The angle Ct is also the tangent to the curve in a plot of volumetric strain versus shear strain as illustrated for simple shear in Fig. 5.3b. ~ If a soil mass is constrained in the lateral directions, th~ilation angle is represented as (5 .11)

Dilation is not a peculiarity of soils but ocetu:S in example, rice and wheat. The ancient trade s 0 er phenomenon of volume expansion of grai it was Osborne Reynolds (1885) who described the phenomena brought it to the attention of the scientific community. For cemented soils, Coulo (5.12)

where

FO

o

Co

is called cohesio

The essential points are' 1. Shear failu fe of soils can be modeled using Coulomb'sfrictionallaw, Tf = (u~)r tlm (cf> , ± It) where Tr is the shear stress when slip is initiated;, (u~)r is the normal effective stress on thlJ slip plane, cf>' is the friction angle and it is the dilation angle. 2. The effect of dilation is to increll$e the shear strength of the soil and cause the Coulomb'sfoilure envelope to be curved. 3. Large normal effecti e strt s$es tend to suppress dilation. 4. At the critical slatt:, the dilati(m angle is zero. S. For cemented soils, CouloJ1(b'sjrictionallaw is Tr = Co + (u~), tan(cf>' + a) where Co is called cohesion.

hat's next . . .In the next-section, we will define and describe various parameters a interpret the shear strength of soils. It is an important section, which you should read carefully, because it is an important juncture in our understanding of shear strength of soils for soil stability analyses and design considerations.

1

5.5 INTERPRETATION OF THE SHEAR STRENGTH OF SOILS The shear strength of a soil is its resistance to shearing stresses. In this book, we will interpret the shear strength of soils based on their capacity to dilate. Dense sands and overconsolidated clays (OCR > 2) tend to show peak shear stresses

5.S INTERPRETATION OF THE SHEAR STRENGTH OF SOilS

Z 11

and expand (positive dilat ion angle), while loose sands and normally consolid ated and lightly overconso lidated clays do not show peak shea r stresses except at very low normal effective stresses and tend to compress (negative dila tion angle). In our interpretation of shea r st rength, we will describe soils as dilating soils when they exhibit peak shear stresses at 0: > 0 and nondilating soi ~ n they exhibit no peak shear stress and attain a maximum shear stress O. However. a nondilating soil does not mean that it does not change volume (expand or contract) during shearing. The terms di lating and nondilating 0 Iy refer t particular stress stales (peak and critical) during soil deformatIon, We will refer to key soil shear strength pa eters. using the follow ing notation. The peak shear strength, Tp. is the peak he strf$Ss attained by a dilating soil (Fig. 5.3). The dilation angle a eak,..sh r strc s will be denoted as O:p. The shea r stress attained by all soils lar~ he r slla'ins (Yu > 10%), when the dilation angle is zero, is the criti I Slate sh r strength denoted by""" The void ratio corresponding to the critil\! ~ai'e sh . r strength' the critical void ratio denoted by eC$' The effccl;l'e...frictlon angle correspon 'ng to the critical state shear strength and critictvora ratio is $;,. The peak effeclive fric ('0 angle for { dila ting soi I S,_ .....

ara. =:

(5.13)

Tesl results ~' 19

J

$; - ...

.

(5,14)

W e will continuu J se Eq. (5.1.3 fb~:1 " rae icc you can make (Eq. ( -4)):ugg, sted by Bolto ~986) .

the adj ustment

T1 ical 'talues o f $~, ~; and if:l.' for

ils are shown in T able 5.1. effecti e~~scribi ng friction angle and accept it by default s c:h tha t effej {lVe critical state friction angle becomes critical state friction angle, $;'. and "Cf cit e pe'ak friction angle becomes peak fric tion angle, We will drop the ter

for the shear strength of soils are: (5, IS) (5.16)

TABLE 5 . 1

Ranges of Friction Angles for Soils (degrees)

Soil Type Gravel Mixtures of gravel and sand with fine-grained soils Sand Sitl

or sitty

Clays

sand

~~.

.;

30-35 28-33 27-37' 24-32 15- 30

35-50 30-40 32-50 27- 35 20-30

5-15

"Higher val...esI32"-37"1 in the range ~re for sands with signlflcant amount of feldspa r (Bolton, t966t . lower values 127'"-32"1 in tile range are 10' quam sands.

CHAPTER 5

SHEAR STRENGTH OF SOILS

The Coulomb equation for soils that exhibit residual shear strength is

FO

212

(5.17)

1

(5.18)

state condition in all soils:

I = T; I FS

(5.19)

The essential points are: 1. Thefriction angle at the critical state, q,;., is a fundamental soil parameter. 2. The friction angle at peak shear stress for dilating soils, q,~, is not a fundamental soil parameter but depends on the capacity of the soil to dilate. 3. Coulomb equation only gives information of the soil shear strength when slip is initiated. It does not give any information on the strains at which soil failure occurs.

5.6 MOHR·COULOMB FAILURE CRITERION

213

What's next .. .Cou lomb's fric t iona l law fo r fi nding the shea r strength of soi ls requi res that we know t he frict ion angle and t he norma l effective st ress on the slip plane. Both of t hese are not readily known for soils because soils are usua lly subjected to a variety of st resses. You shou ld recall from Chapter 3 t hat Mohr's circle can be u sed to determine the stress w ith in a soil mass. By combinill Moh r' s ci rcle for finding st ress states wit h Coulomb's frict ional law we can dev op a generalized fai lure criterion.

5.6

MOHR-COULOMB FAILURE

CRITERIO~

Let us draw a Cou lomb fric tional fai lure line I st[8 ted tsy AB in Fig. 5.13 and subject a cylindrical sam ple of soil to p n ci pa ~t resses so that M ohr's circle touches the Coulomb fail ure line. lrrr.... . V

Of cou rse, several circles ca n share A"Bas e common tangent but we will show only one for simplicity. n:~ intt; ta gency is at B''rfv.' (0-;,),] and t he cen ter of the circle is ill O. are gOIn$; to iscuss the to halfl of the circle; the bOltom half is a reflection 0 the op ha The major and minor-p n~a l effective stresses at failure arc (aJ )1 and ri it ur obj ective is to find a r ationship be· tween the principal e ff Cl ~ s tresses and <1/ al (a Jure. We will discliss the ap· propriate ~' lat~ n thi rr~io n . From th l.eo etry ~ Ohr' s circle, C

~"r"'-; -;)-_ .

~O~ OA

Sin Ijlo ..

("0'))[

'-J

aD'

+ (0-;) 2

(O'il, - (0';), (0"1), + (0'»[

FIGURE 5.13 The Mohr-Coulom b failure envelope.

(5.20)

CHAPTER 5

SHEAR STRENGTH OF SOILS

Rearranging Eq. (5.20) gives

,----------------(rr;)! = 1 + sin <1>' = K (rr3)! 1 - sin <1>' P

(5.21)

or (rrD! = 1 - sin <1>' = K (rrD! 1 + sin <1>' a

where Kp and Ka are called the passive and active eartH Chapter 10, we will discuss Kp and Ka and use them analysis of earth retaining walls. The angle BOD ~e re res of the failure plane or slip plane to the plan on ie. t e ajor principal effective stress acts in Mohr's circle. Let us . d relat" nshtp between e and cP'. From the geometry of Mohr's circle (Fig. . 3)

(5.23)

FO

214

(rr;)} _ 1 •

A-. , _

SIn't'

-

(rr))! (

')

rrJ ! (rr])!

--- + 1 The implication of this equation is that the Mohr-Coulomb failure criterion defines failure when the maximl!m principal effective stress ratio , called maximum effective stress obliquity,

((rr~»f,

is achieved and not when the maximum shear (J3 f is achieved . The failure shear stress is then less than the

stress, [( (Ji - (J~)/2lmax' maximum shear stress. You should use the appropriate value of cP' in the Mohr- Coulomb equation [Eq. (5.20)]. For nondilating soils, cP' = cP~s; while for dilating soils, cP' = cP;·

5,6 MOHR,COULOMB FAILURE CRITERION

Z

15

The essential poinlS are: 1. Coupling Mohr's circle with Coulomb'sfr;ct;onallaw allows Ul' tO define shear failure based on the Sll'ess state of the soil, 2. The Mohr- Coulomb failure criterion is l 'ln

4J '

(CTa, - (ui ),

=';'-'~-'--7-C,," (CT[), + (uJ),

(CT[),

1

+ £in ifJ '

-( ,,-;-), = I - "'£In.p ' 0' (ui ),

(Ut),

1 - sin ~' = I + sin til'

3. Failure occur.;, according to the Mohr-Coulomb failure criterion, when the soil reaches the maximum effective stress obliquity, that is, (un, (un, 4. Thefailure plane or slip plane ;l' inclined ql an angle (J = 'IT/4 + 4> '/2 10 the plane on whit h the major principlll effect/tie stress acts. 5. The moximum shear suess, 'Tm .... = [(~; - ( 5)12/"'
E .1

;..

A cylind cal soil sam p1'e w§s su bjected to axial principa l stresses (0-;) and radial prin~pal st resses (uD,\ The sob could not suppo rt addi liona l st resses when u; - 300 kPa a~loo..kP Det.,m; ne the fr;ct;on angle and the ;ncHnahon

7 f the sl ip pl an

to tne:::horiV'mal. Assume no significan t dilational effec ts.

Strates y nus e (5.23 . ince dilation

IS

e is a straight forward applicat ion o( Eqs. (5.20) and neglected $' 4>;,.

=

Solution 5 . 1 Step 1:

Find 4>;" From Eq. (5 .20) ,

!

sin <1>' = ('1;), - (O'i), _ 300 - 100 = ~ = eo (O'il, + (O'~)! 300 + 100 4 2 ~~

Step Z: Find

- 30'

e.

From Eq. (5.23),

8 - 45.....

~.

~=45 ..

2

30"

+--6O' 2

•

CHAPTER 5

SHEAR STRENGTH OF SOILS

EXAMPLE 5 .2

Figure E5.2 shows the soil profile at a site for a proposed building. Determine the increase in vertical effective stress at which a soil element at a depth of 3 m, under the center of the building, will fail if the increase in lateral effe tive stress is 40% of the increase in vertical effective stress. The coefficient lateral earth pressure at rest, Ko, is 0.5. Strategy You are given a uniform deposit of sand an data given to find the initial stresses and then use the to solve the problem. Since the soil element is under th c axisymmetric conditions prevail. Also, you are that fore, all you need to do is to find LlO'~.

er 0'3

b equation he building, O.4LlO'~. There-

0

Solution 5.2 Step 1:

Th(

The subscript

Step 2:

0

de ote

~I;; (a~)

pres.",e is =

Find LlO'l' t failure:

vertical effective stress to bring the soil to

FO

216

34.4 + Llai 17.2 + O.4Lla;

-----'-- = 3

Ground surface

'I'

--

r

FIGURE ES.2

•

5.1 UNDRAINED AND DRAINED SHEAR STRENGTH

217

What's next . . .In the next section, we will consider two rather extreme condition sdrained and undrained conditions-under which soil is loaded and the effects these load ing conditions have on the shear strength . Drained and undrained conditions are the bounds to evaluate soi l sta bi lity.

5.7 UNDRAINED AND DRAINED SHEAR STRENGTH stress paths in C hapter 3. Drained cond itio curs e excess pore water pre ssu re developed du ring loading of ~il slpat I.e., nil = O. Undrained condition occurs when th e excess pqre wa ter p e~ u re cannot dra in, at least quick ly, from the soil; that is, nu *- 0. e t ,,'Thte"n of either cW1dilion- drained or undrained-depends on the soil tYR:' ~olog1cal fo rmation ~ ssures, sa nd layers in clays, etc.), and the ratrof load{ng. The ratc of load ing unoel' the un~ rained condition'is o n uch fa ster than s re wate r pressure and the l ume change the rate of d issipation of the ex ~ e result a is suppressio is a change in tendency of the soil is upP'resseCl . excess pore water press reXfutio shearing. .xs0~1 ith a te deney 10 compress during drained I j;ling ~i e~ibil an incre~~~n e ess pore waler press ure (positive excess pore wa~# pressu re, Fig.-S.14J u~der . ndrained condit ion resulling in a d rease in ~ff ctive stress. Aso~hat expan ds during drained loading will exhibit a c ease n excess p~te r pressure (nega tive excess pore wa ter press ~ , Eig. 5.14) under undraint d conc!.i Ion ulting in an increase in effective stress~ ~~§-e.. h ges in excess P\'~e water ressure occur because the void ratio does not C'h ange during undrained ~ hat is, the volume of the soil rema ins ~ constant. Dunng t he life a ge ~c qnical structure, ca lled the long-term condition, the ex<:ess pore ~ ter r sure de veloped by a loading dissipates and drained coRail ion appli r Oays us ay take many years to dissipate the excess pore aler press u ~es . D urt g onst ruction and shortly afler, called the short -term condition , soil~ ·th low per eabili ' y (finc-grain ed soils) do not have sufficient time

""•

.~

,[

•

0

u

• a •

;;

w

, (a) Dra ined COM ttion

FIGURE 5.14 Effects of drained and undrained condi tions on volume changes.

CHAPTER 5

SHEAR STRENGTH OF SOILS

for the excess pore water pressure to dissipate and undrained condition applies. The permeability of coarse-grained soils is sufficiently large that under static loading conditions the excess pore water pressure dissipates quickly. Consequently, undrained condition does not apply to clean coarse-grained soils under static loading but only to fine-grained soils and to mixtures of c ~e and fine grained soils. Dynamic loading, such as during an earthquak , is imposed so quickly that even coarse-grained soils do not have sufficient i to dissip te the excess pore water pressure and undrained condition appv . The shear strength of a fine-grained soil under un the undrained shear strength, SUo The undrained shea of the Mohr total stress circle; that is, (5.24)

FO

218

T

__________ Jl_~ ___ ~~~:' Su

1 (a) Undrained shear strength

"'

I

,

,

I

-"', ,

, \

"

\

, \I

\

I,

(b) Increase in undrained shear strength

from increase in confining pressure

FIGURE 5.15

Mohr's circles for undrained conditions.

5.8 LABORATORY TESTS TO DETERMINE SHEAR STRENGTH PARAMETERS

1.

2. 3.

The essential points are: Volume changes that occur under drained condition are suppressed under undrained condition. The result of this suppression is that a soil with a compression tendency under drained condition wi ll respond with positive excess pore water pressures during undrained condition, and a soil with an expansion tendency during drained condition will respond with negative excess pore water pressures during undrqined condition. For an effective stress analysis, the shear streflgth parameters are cfJ:. and cfJ~. For a total stress analysis, which applies to fine-grained soils, the shear strength parameter is the undrained shear strength, suo The undrained shear strength depends Oil the confining pressure. It is not a fundamental soil shear strength parameter.

FO

4.

21 9

p-

I'

~

5J

Possible failure zone

~.1U.J.4=-/. or fail ure plane 0.

An;

T

;]..!:-

FIGURE 5. 16 Shear box.

220

QlAPTER 5

SHEAR STRENGTH

OF SallS

(.ral or by weights through a pulley system for load conlroL Most shear box tests are conducted llsing displacement control because we can get both the peak shear force and the critical shear force. In load control tests. you cannot gel data beyond the maximum or peak shear force.

The horizontal displacement, 6.x , the vertical displacement, ~he vertical loads, P~ . and the horizontal loads, p.. , are measured. Usually, I b~e e or more tests are carried Qui on a soil sample using three different ccnstan ertical f~rces. Failure is determined when the soil cannot resist any furt~ increment- IT horizontal force. The stresses and strains in the shear box Ie ( are di cult to calculate from the forces and displacements measured. The stres In the ntln (dimension unknown) const rained failure zone (Fig. 5.16) are...not un for ly dist ributed and strains cannot be de termined. Th e shear box apparatus cannot prev~drai ge, but one can get an estimate of the undrai ned shear strength oMrays l run 'ng the shear box test at a fast rate of loading so that the test is completed )VIick ly. Howev.t:r, the shear det~n ation of the un rained shear box test shou ld not be used for an a · ur strength of soi ls. T In summary. drained tests are s...enerally/onducted in a shearbo test. Three or more tests are pe rform ~ on a so T)!e soil samp in each test sheared under a constant vertical force hich is dIfferent in ~ test. Thi data recorded for each test are the horizon ~I dlsP!atements, th ~orizo tal forces, the vertical displacements, an &Ie nSla I vertical force under ich e{est is conducted. From th e recQrde data, yo q.rf find the fo~ ar- mete rs: Tp. To<. $;. (K" a (and SU ' if fine-g r ' ned soils re tested quickly). hese parameters are generally determi ned..from pi e data, as i lustra t d in')!-B. 5.17 for sand .

,

/

0."

A"",

J2,

~

'0'--"2--~c--"cf\C~'"ILO--~12 t(mm)

!

~

1000 500

o

V i ......

o

1000

'.

~"" en! " II!

~~

2000

JOOCl

4000

Ver1iclIl force IN)

1<1 0.3

E §

!

02 0

1

I....

~

0

~

-0. 1

,

~ -0.2

0

100

--

H« Fan',,1diSillace

f'-

e nl (mm}

1

~

~""

12

.-

~ -0.3

-
FIGURE 5.17 Resu lts from a shear box test on a dense and a loose sand.

50:).)

5.8 LABORATORY TESTS TO DETERMINE SHEAR STRENGTH PARAMETERS

221

Only the results of one test at a constant value of P z are shown in Figs. 5.17a,b. The results of (Px)p and (PX>cs plotted against P z for all tests are shown in Fig. 5.17c. If the soil is dilatant, it would exhibit a peak shear force (Fig. 5.17a, dense sand) and expand (Fig. 5.17b, dense sand), and the failure envelope would be curved (Fig. 5.17c, dense sand) . The peak shear stress is the ak shear force divided by the cross-sectional area (A) of the test sample; th (5.25)

The critical shear stress is (5.25)

FO

R

(5.27)

A- ' _ 'i'p -

t

an

-1

(Px)p P

(5.28)

z

of horizontaf

'ne the peak dilation angle directly for each test from a plot isp' cement versus vertical displacement, as illustrated in Fig. dilation angle is 0: p

= tan- 1

(-6.z) 6.x --

(5.29)

We can find cx.p from (5.30)

EXAMPLE 5 .3

The shear box test results of two samples of the same soil but with different initial unit weights are shown in the table below. Sample A did not show any peak values but Sample B did.

CHAPTER 5

Soil

A

B

SHEAR STRENGTH OF SOILS

Test number

Vertical force (N)

Horizontal force (N)

Test 1 Test 2 Test 3 Test 1 Test 2 Test 3 Test 4

250 500 750 100 200 300 400

150 269 433 98 175 210 248

Determine the following: (a) ~s (b) (c) Strategy To obtain the des· force versus horizontal force.

Solution 5.3 ntal forces for

Step 1: Step 2:

n a straight line through the

Step 3:

, .. . . . . . (

FO

222

_tan _1(175) 200

p 200 N -

(p') 400

N =

ap (a)2oo N

(a)4oo

N =

0

0 tan _1(248) 400 = 31.8

= ~

=

= 41.2

-

~,

41.2 - 30

= 11.20

31.8 - 30

=

Note that as the normal force increases

(Y.p

1.80

decreases.

~ 500r-----,-----.---~~----,

11!1~ :r:

0

200

400

VerticaLforee (N)

FIGURE E5.3

600

800

•

5.8 LABORATORY TESTS TO DETERMINE SHEAR STRENGTH PARAMETERS

223

EXAMPLE 5.4 The critical state friction angle of a soil is 28°. Determine the critical state shear stress if the normal effective stress is 200 kPa.

Strategy This is a straightforward application

's frictional

equation.

Step 1:

• EXAMPLE 5.5 The data recorded during a shear 3 cm, at a constant vertical fo negative sign denotes vertical

(b) (c)

Horizontal force (N)

FO

R

6.10 6.22 6.48

178 2.03 2.41 2.67 3.30 3.68 4.06 4.45 4.97 5.25 5.58 5.72 5.84 5.97

0.00 0.00 0.D1 0.00 571.29 631.62 663.54 759.29 807.17 844.47 884.41 928.35 939.34 950.32 977.72 982.91 988.29

0.00 -0.01 -0.03 -0.05 -0.09 -0.12 -0.16 -0.21 -0.28 -0.31 -0.34 -0.37 -0.37 -0.40

6.60 6.86 7.11 7.37 7.75 7.87 8.13 8.26 8.51 8.64 8.89 9.14 9.40 9.65 9.91 10.16 10.41 10.67

988.29 988.29 993.68 998.86 991.52 999.76 1005.26 1002.51 994.27 944.83 878.91 807.50 791.02 774.54 766.30 760.81 760.81 758.06 758.06 758.06 755.32

Vertical displacement (mm)

- 0.40 -0.41 -0.45 -0.46 - 0.49 - 0.51 -0.53 -0.57 -0.57 -0.58 -0.58 -0.58 -0.59 -0.59 -0.60 - 0.59 -0 .59 -0.60 -0.59 -0.59 -0.59

224

CHAPTER 5

SHEAR STRENGTH OF SOILS

After you plot the graphs, you can get an idea as to wh~t h er you have a loose or dense sand. A dense sand may show a peak horizontal force in the plot of horizontal force versus horizonta l displacement and would ex pand.

Strategy

Solution 5.5 Step I:

Plot gra phs. See Fig. E5.5. Step 2: Determine whether the sand is dense o r loose, The sand appea rs to be dense-i t sho a. a dilated. Step 3: Extract the req uired val ues. Cross-sectional area of sam I · A (el) Tp

10

= (P.), = ~o-, = OO~. kPa A 1O-1 ~'

(<2) ' . =

(<3) . , =

(P~). = 1715'il8"N~~:j

~n-~1

I~r---~--~~----'-~-r~~

: Ol$placement

000

~

~ -0.30 li ~ -0.40

:>

-0.50 -060

,

:--..,

...0,\0

l...o20

i

0

I.,

•

I

6

,

, ,,

8

10

12

iO

12

(m ml

8

=- F ,-,' ......0. -1'\

'

r-

--

- 0.70

FIGURE ES.S

I

I

'"

10 -

mlal force and

; oo d- 1O-~ m1

5 .8 LABORATORY TESTS TO OETERMINE SHEAR STRENGTH PARAMET"ERS

Normal stress:

«'4)

x

10- 3 = 120 kPa

¢I' = ,.

lan-l ("-) - tan- l(loo·t =
$' =

tan - I( ' ") _ tan- I(75"J1

co

225

39.9"

= 3230 120 '

a~

Also. 0.,. -

a; -

•

0"';' = 39.9

5.8 .2 Conventional Triaxial Apparatus A widely used appa ratus to determ in e shear strength parameters and the stress-strain behavior of soils is the t naxia i"~pa ~us. The name is a misnome r 1)illce two, not three. stresses ca n be conlrbUed. n the triaxial lest , a cylindrical sample of soil. usually with a lengtltto di el ralio of 2, is subjected to ei ther coni rolled increases in axiaJ... tresses 0 axl I displacemertts and radial st resses. The sa mple is laterally confine'ct by a m mbrane and r{oial strtsses are applied by pressu ring waler i a chambe ,<&g. 5. 18). stresses Ie applied by loading a plunger. If t e xial streJS"is greater . stress, Ihe soil is co mpressed \lerl~cally and the I is called If the rad ial stress is grea ty'"than the xials ress, the soil is I and the test is called triaxIa l extens~fn The ap, st res; s (axial and . I stresses and the loading

+tt--

Water

- f t - - Cell pressure, ",

Acryhc cyhnder

FIGURE 6 .18 Schematic of a triaxial cell.

CHAPTER 5

SHEAR STRENGTH OF SOILS

condition is axisymmetric. For compression tests, we will denote the radial stresses (J,. as (J3 and the axial stresses (J , as (Jr. For extension tests, we will denote the radial stresses (J ,. as (Jl and the axial stresses (J, as (J3 . The average stresses and strains on a soil sample in the triaxial apparatus for compression tests are as follows:

G

A,i,' to'" ",""

! 0.

~ ~ + :' I "'"'

[5.31)

D"i"oci, ",,'"

10. - 0, ~ :,f~\..~

(5.32)

(5.33)

.----'"t-1----L_ _

~

( .34)

~

(5.35)

.L

5

(5.36)

(5.37)

FO

226

2) is t initial cross-sectional area and H is the current height of the sample. di- tion angle for a triaxial test is given by Eq. (5.10) . The triaxial apparatus is versatile because we can (1) independently control the applied axial and radial stresses, (2) conduct tests under drained and undrained conditions, and (3) control the applied displacements or stresses. A variety of stress paths can be applied to soil samples in the triaxial apparatus. However, only a few stress paths are used in practice to mimic typical geotechnical problems. We will discuss the tests most often used, why they are used, and typical results obtained.

5.8.3 Unconfined Compression (UC) Test The purpose of this test is to determine the undrained shear strength of saturated clays quickly. In the UC test , no radial stress is applied to the sample ((J3 = 0). The plunger load, Pz , is increased rapidly until the soil sample fails , that is, cannot

227

5.8 LA80RATORY TESTS TO DETE RMINE SHEAR STRENGTH PARAMETERS

q

.0

(b) TOI') I slJ~~ ~Ih

Tf.-___ __

~'''" otal

siress ellcle

.

, ,, 1'-------'--:!----''-7---: tel Mohr',

(tIes

FIGURE 5.19 Stresses. st res p aths and MOh ~ rc~ for UC test.

a~nalloth~ading· Oli ~iC kl

, uppo" any • so that the poce watec cannot drain f m the sj il; the sample is sti rea at co sia n! volume. The stress 'S'appjied on the s \l saQ1 ple d the lota l stress path followed are sho in...Figs. m a,b. 111e eff clive strli.ss p th is unkn own si nce pore water pressure h g re not normally easured Mohr's circle using total stresses is depicted ~ Fig. S.19c. If Ihe excess re.,\ ~ I e r pressures were to be measured , ,hey woul~ be negative ~he theo retical reason for nega tive excess pore water ~ressures is as fo llowS#" mee )..= 0, then from the principle of effective stresses, 0:' - 6u = 0 u = Au. T effective radia l stress. a; . cannot be negative because soi ls cannot sustain n Ion. Therefore, the excess pore water press ure jIlUSI be negatiVe at ~$ i positive. Mohr's ci rcle of effective stresses would be to the righ f the tota stress circle as shown in Fig. 5. 19c. Ttl undra in d s ar strength is

r-----.

(5.38)

where. from Eq. (5.37), A = A J (1 - £ \) (no vol ume change, i.e" ep undrained clastic modulus. Eo., is determined from a plot o f el versus The results from UC tests are used to:

=

0) . The

0"1.

Estimate the sho rt-term bearing capacity of fine-gra ined soils for found atio ns Estima te the short-term stability of slopes Compare the shear strengths of soils from a sile to establish soil strength variability quickly and cost·effectively (the UC lest is cheaper to perform than other triaxial tests) Determi ne the stress-strain characteristics under fast (undrained) load ing condit ions

CHAPTER 5

SHEAR STRENGTH OF SOILS

EXAMPLE 5.6 An unconfined compression test was carried out on a saturated clay sample. The maximum load the clay sustained was 127 N and the vertical displacement was 0.8 mm. The size of the sample was 38 mm diameter X 76 mm Ion Determine the undrained shear strength . Draw Mohr's circle of stress for th o test and locate SUo

Strategy Since the test is a UC test, 0'3 = 0 and (0'1) You can find Su by calculating one-half the failure axia

Step 2:

Step 3:

FO

228

401--~---+--4---+-~~-+--~

,

2 0 ~~---+--4---+-~~-+--~

ro

~ ..,

I

o~~---+--~--+---~~--, 1 0 : llO l ' 0

-20 I-':,. --f-----ir--+-+_

I

a (kPa)

'\ '\

... .....

- ---'

-60 ~-+--~--+_--~~--_+--1

-80 '----'---'--"---- -'---'----'----'

FIGURE E5.6

•

229

5.8 LABORATORY TESTS TO DETERMINE SHEAR STRENGTH PARAMETERS

5.8.4 Consolidated Drained (CD) Compression Test The purpose of a CD test is to determine the drained shear strength parameters, ~ and ~ , to analyze long-term loading of a soil mass. The effective elastic moduli for drained condition £' and £~ are also obtained from th~ test. A consolidated drained compression test is performed in two stages. ~ rst stage is consolidating the soil to a desired effective stress level by pres urizing the water in the cell and allowing the soil sample to drain until t e pressure dissipates. In the second stage, the pressure in · cel ell pl' ssure or confining pressure) is kept constant and additional ~lloa or isplacements are added very slowly until the soil sample fails. The dtsRlaceme ate (or strain rate) used must be slow enough to allow the e cess p e t pressure to dissipate. Because the permeability of fine-grain 'd Olls-i · uc lower than coarsegrained soils, the displacement rate for t ti g fin -grai ed soils is much lower than for coarse-grained soils. Drainage the xcess . e water is permitted and the amount of water expelled is measurea t s cus mary to perf~ a minimum of three tests at different cell pre ur . :rli r ses on the SOIL sample for the two stages of a CD test are as lows: 0

0

re for a sa tura ted At the end of the sure diSSipates; that is, 6.u = O. 0'3'

6.0-3 = 0

6.0-3 = 6.o-J = 0;

FO

R

O'

6.q 6.p'

,

and

6.q 6.q = - = 0 6.p' 6.p

-

6.u = 0

6.q 6.p

-=-=3

3 '

-

ses an stress path applied are illustrated in Fig. 5.20. The change in volume of soi is measured by continuously recording the volume of water expelled. The volumetric strain is (5.39)

where 6. V is the change in volume and Vo is the original volume of the soil. Also, the axial displacements are recorded and the axial strain is calculated as Cl = 6. zl Ho oThe radial strains are calculated by rearranging Eq. (5.39) to yield (5.40)

The maximum shear strain is

I

()'zx)max

=

(El -

E3)max

I

230

CHAPTER 5 SHEAR STRENGTH OF SOI LS

00' ) '"

..l.,

, dll . 0

tal

Stag~

(/) Stage 2

L Consohdallon ph,)$e

q Stage 1

Stage 2

ESP " TSP

3

p ·. P (e)

Stre-.s

th

y

FIGURE 5.20 Stre ses and str

paths during a CD te t.

.4,,395. "Xl

/

-

/ " .33'

....qen$e W llo

300

"

200

25

100

"• ~

0

h

V

" -100 - ' 00

-0.0 1 p

c

-o02 i-----l--'' '-+---t--+---j -0.03 f---j--"l.--+---+ -
I

,,\

FI'

02 fl 3

,\

'\. ~

!O ' r.!~I· "'0810'

'/

1- /

-300 -400 L lOOSO'! SIIIld

'
FIGURE 5.21 Results from CD tests on dense and loose sand. (From 'The Measurement of Soil Propenies i n the Triaxial Test: by Bishop and Henkel, Edward Arn old 1962.)

I flO

S 8 LABORATORY TESTS TO OETERMINE SHEAR STRENGTH PARAM ETERS

which, by substitution of Eq. (5.39), gives

I

('u)_ =

\(3£, - ',)

I

23'

(5.41)

Since the CD test is a drained lest, a single test can take several days if the permeability of the soil is low (e.g.• fine-grained soils). Typical res Its of consolidated drained tests on a sand are shown in Fig. 5.21. f The elastic moduli for drained conditions, E' and~ ' are"obtained from the CD lest from the plot o f devialoric stress, (0'; - 0')), a r inale nd ~s abscissa as shown in Fig. 5.21a. The results of CD tests are :-;cd I e termlOe the long· term stability o r slopes, foundations, retaini ng w I excav (fons and other earthworks.

r

EXAMPLE 5 .7

The cesullS of ,heee CD ,ests on a SOl! ;' {~e " clas follows: T&s1

-

numbe,

1

100 180 300

2 3

The detailed esults , expansion.

~

"' 7

250 (pea 162 (peak)

i ~~peak observed)

l lst

tare aF-~ 0!1o:i"""

sign indica tes

, Plunger load A ~l mmJ

.152

~

9.76 1.52

6V(cmf)

0.00 0.02 0.03

- 0.'\'1....t _ 1. S sO, F~

(~~:o 94.3

12 2 1.5 257.5 292.9 298.9

2.28

2.66 3.04 3.' 4.56 5.32 6.08 684 7.6 8.36

Pk IN)

- 2.59 - 2.67 - 2.62 - 2.64 - 2.66 - 2.63

298.0 279.2 268.4 252.5 238.0 229.5

223.2 224.3

The in it ial size of the sample is 38 mm in d iameter and 76 mm in length. (a) Determine the friction angle for each test. (b) Determine Tp. Te:s, E', and E; al peak shear mess for Test 1.

FO

232

CHAPTER 5

SHEAR STRENGTH OF SOILS

(c) Determine

¢~s .

(d) Determine cx.p for Test 1.

Strategy From a plot of deviatoric stress versus axial strain fo Test 1, you will get Tp ' T es , £ ' and £;. The friction angles can be calculat or ound from Mohr's circle.

Solution 5.7 Step 1:

Determine the friction angles. Use a table to d

Test No.

kPa

kPa

kPa

Test 1 Test 2 Test 3

100 180 300

250 362 564

350

a; -

(13

O"~

(1;

360

- 360 '----'---'

'-~--'---'---'--'

FIGURE E5.7a

Step 2:

Determine Tp and Tcs from a plot of deviatoric stress versus axial strain response for Test 1. TID 2 0 The initial area is: A o = __ 4

v

= o

TID oHo 4

2 TI X 38 - - - = 1134 mm 2 4 2

TI

X

38 X 76 = 86193 mm 3 4

233

5.8 LABORATORV TESTS TO DETERMINE SHEAR STRENGTH PARAMETERS

See lhe table for calculations and Fig. ES.7b for a plOl of the results.

Aa = 1134 mm 2

'.

ro, !! z/ Az(mml

H.

0.00 0.15

0.00 0.20 0.30 0.50 1.00 2.00 3.00 3.50 4.00 5.00 6.00 7.00 8.00 9.00

0.23 0.38 0 .76 1.52 2.28 2.66 3.04

3.80 4.56 5.32 6." 6.84 7.60 8.36

1O ~ 11

;a:

A

q = P./A

~V(cmll

(6.V/Vg l

(mm 2 1

IkPal

0.00 0.02 0.Q3 -0.09 - 0.50 - 1.29 - 1.98 - 2.24 - 2.41 -2.55 -2.59 - 2.67

0.00 0.02 0.03 - 0.10 - 0.58

1134 1136 t 137 1140

)

a 53.8

229~3

246.7 250

- .62

1245 125

-21""

~ 127~

- 2 .~

-3.06

191 .2 182.; > 176.4

'{?

300

""

~ 200

•

~"

~150

O ~OC---1r e.~~r---~~--'8----'IOO----71·' """'-

€l (%)

~. 7b

Extract 'rp and "l" p

Step 3:

= (0"; -

2

Te •.

O"~)p = 250 = 125 kPa 2

. ~

=:

(0"; - (13)"" _ 175.8 = 879 kP

2

2

.

a

Determ ine £ ' and £ 3. The initial slope of Fig. ES.7b gives E' and the slope of the line fro m {he origin to 2Tp gives £;. •

54

£ = 0.002 - 27.000 kPa

E; =

250 0.035 .. 7143 kPa

FO

234

CHAPTER 5

Step 4:

SHEAR STRENGTH OF SOILS

Determine ~s' The deviatoric stress and the volumetric change appear to be constant from about £1 = 10%. We can use the result at £1 = 11 % to determine ~ . (a3)cs = 100 kPa , (aDcs = 175.8 + 100 = 5.8 kPa. ,j.. '

,+, cs

=

.

SIn

- I

(
Step 5:

Determine

+

SI'O-1 _ _ __

C:i. p

• 5.8.5 Consolidated Undrained Co

Therefore,

~
- !1u

3 ~
--~u

3

3~u

1- -

~
While the to al stress path is determinate, the effective stress path can be determined only if we measure the changes in excess pore water pressures. The stresses on the soil samples and the stress paths in a CU test are illustrated in Fig. 5.22. The effective stress path is nonlinear because when the soil yields, the excess pore water pressures increase nonlinearly causing the ESP to bend. In a CU test, the volume of the soil is constant during the shear phase, Therefore,

which leads to (5.42)

5.8 LABORATORY TESTS TO DETERMINE SHEAR STRENGTH PARAMETERS

235

+ - - Lla3= Llai Llu = 0

(a) Stage 1: Consolidation phase

q

Stage 2

Stage 1

(b) Stage 2:

Stage 2

_--+___ \

Stage 1

FIGURE 5.22

FO

R

(5.43)

pressure is positive. Each Mohr's circle of total stress is associated with a particular value of SI< because each test has a different initial water content resulting from the different confining pressure, or applied consolidating stresses. The value of Su is obtained by drawing a horizontal line from the top of the desired Mohr's circle of total stress to intersect the vertical shear axis. The intercept is Suo The value of SI< at a cell pressure of about 830 kPa is 234 kPa, as shown in Fig. 5.23. Alternatively, you can calculate SI< from

CHAPTER 5

SHEAR STRENGTH OF SOILS

140 120 ~ 100 C

t5' t

t)

80 60

~

ormally onso li ated c l (}3

/'

I

40 20

/

- LOTI(

..,---

3 ' . 5 kPa Heav y overc nsolia. eo clay' ,(}3

./"'" /

/

o/ o

~ 5

10

15

20

25

30

Axial strai n (%) (a ) 150 ,--.--~--,--.---.--,--"

r (kPa)

s" = 234 kPa

RE 5 .23

Triaxi a

rtle, In th~'

FO

236

ear str ngth parameters (~ and <1>;) are found as described in the previous se ion d aling with the CD test. You would normally determine s" at the maximum anttcipated stress level in the field. The CU test is the most popular triaxial test because you can obtain not only Su but ~ and <1>;, and most tests can be completed within a few minutes after consolidation compared with more than a day for a CD test. Fine-grained soils with low k values must be sheared slowly to allow the excess pore water pressure to equilibrate throughout the test sample. The results from CU tests are used to analyze the stability of slopes, foundations , retaining walls, excavations, and other earthworks.

EXAMPLE 5.8 A CU test was conducted on a saturated clay soil by isotropically consolidating the soil using a cell pressure of 150 kPa and then incrementally applying loads on the plunger while keeping the cell pressure constant. Failure was observed

5.8 LABORATORY TESTS TO DETERMINE SHEAR STRENGTH PARAMETERS

237

when the stress exerted by the plunger was 160 kPa and the pore water pressure recorded was 54 kPa. Determine (a) S u and (b) ¢~s. Illustrate your answer by plotting Mohr's circle for total and effective stresses. Strategy You can calculate the effective strength paramp~y using the Mohr- Coulomb failure criterion [Eg. (5.20)] or you can de;rr;in; thi m from plotting Mohr's circle. Remember that the stress impose b e plunger is not the major principal stress al but (al - a 3) = (a~ - a ') .

Solution 5 .8 Step 1:

Calculate the stresses at failure.

Step 2:

FO

R

Step 3:

~ =

27°

s" = 80 kPa 160 120 80 40

ro ~ ....

0 -40 --80 -120 -160

L _--'-----"L..-....l...----'_-'---'-_...L..----'._-L_---'

FIGURE E5.8

•

238

CHAPTER 5

SHEAR STRENGTH OF SOILS

5 .8.6 Unconsolidated Undrained (UU) Test The purpose of a UU test is to determine the undrained shear strength of a saturated soil. The UlJ test consists of applying a cell pressure [Q the soil sample without drainage of pore water followed by increments of axial stress. The cell pressure is kept constant and the test is completed very quickly be{ause in neither of the twO stages- consolidation and shearing- is the excess pore water pressure allowed to drain. T he stresses applied are:

Stage 1: Isotropic compression (not consolidation)

8~q ... ~

80"1 = 80"3,

8p =

80"]0

~a 'ie

;t'-

Stage 2: Shearing phase

a~ 0

0,

)

8q = 3

"p

Two or more s~ mples of he s me soil and the s e initial void ratio are normally tested a!,1different cell p essures. Each Mohr's eire e is the same size but th e circles a~t translated ~ rizontally by the.difference.>tn the magnitude of the cell pressure,.g, Mohr's icircies, stresses, ano:a.s fress p(lths for a UU test are shown in Fig. 5.1 O nly4 he total s gpath""~own since the pore water pressures not measured to e nabJ the ca l lacis:>n of the effecti ve stresses.

p

(a) Sheanng stresses

(bl Stress path

,

,I

,,

, ,, 1 <13 \

,,

Test 1

,,

... _....

, ,, ,,, Test "2 , ,,"', "" ,,

,,

,

,,0, " , ..._ ....

(c) Moh r's circles

FIGURE 5.24

Stresses, stress path, and Mohr's circles for UU tests.

5.8 LABORATORY TESTS TO DETERMINE SHEAR STRENGTH PARAMETERS

239

The undrained shear strength, su, and the undrained elastic moduli, Eu and (E u ) " are obtained from a UU test. The UU tests, like the UC tests, are quick

and inexpensive compared with CD and CU tests. The advantage that the UU test has over the UC test is that the soil sample is stressed in the lateral direction to simulate the field condition. Both the UU and UC tests are us ful in preliminary analyses for the design of slopes, foundations, retaining ails, excavations, and other earthworks. EXAMPLE 5.9

A UU test was conducted on saturated clay. The cel failure occurred under a deviatoric stress of 20.kPa. shear strength.

Solution 5.9 Step 1:

5 0

FO

~-+.-+---+.-+\-'.... ~~+-~-+.'--+·6-fI<
Step 2:

Dete mine the undrained shear strength. Draw a horizontal line to the top of Mohr's circle. The intersection of the horizontal line with the ordinate is the undrained shear strength . Su

= 110 kPa

By calculation:

Su

-_

((Tj)t - ((T)! __ 220 -_ 110 kPa 2 2

•

What's next . .. In the UU test and sometimes in the CU test, the excess pore water pressures are not measured. However, we need to know the magnitude ofthe excess

240

CHAPTER 5

SHEAR STRENGTH OF SOilS

pore water pressures to calculate effective stresses. Next, we will present a method to predict the excess pore water pressure from axisymmetric tests.

5.9 PORE WATER PRESSURE UNDER AXISYMMETRIC UNDRAINED LOADING Pore water pressure changes in soils are due to' the cha deviatoric stresses. Skempton (1954) proposed the fall · mine the pore water pressure under axisymmetric co

aJ and 1

(S.44) 0"3

is the deviatoric aturated soils

(S.4S)

ing hether a soil is saturated in 4) by dividing both sides by ~0"3,

FO

(S.46)

(S.47 )

.,.

TABLE 5 .2

A, Values

Type of clay Normally consolidated Compacted sandy clay Lightly overcon solidated clays Compacted clay-gravel Heavily overconsolidated clays

0.5 to 1 0.25 to 0.75 o to 0.5 -0.25 to 0.25 -0.5 to 0

SOURCE: From 'The Measurement of Soil Properties in the Triaxial Test: by Bishop and Henkel, Edward Arnold 1962.

-0.5

FIGURE 5 .25

Overconsolidation ratio

Variation of OCR with A,.

5.10 OTHER LABORATORY DEVICES TO MEASURE SHEAR STRENGTH

241

pore water pressure change must be equal to the increase of confining pressure. Equation (5.47) then provides a basis to evaluate the level of saturation of a soil sample in an axisymmetric test. The coefficients A and B are referred to as Skempton's pore water pressure coefficients.

The essential points are: 1. Under an axisymmetric loading condition, the pore water press'tlre can be predicted using Skempton's pore water pressure coefficients, A and

B. 2. For a saturated soil, B = 1.

What's next . . .In the next section, we will apparatuses to determine the shear strengt complex and were developed to more el'''' ,rR r,r.d;" laboratory soil sample than the tria ,j and .

FO

5.10 OTHER LABORA~~ab7 MEASURE SHEAR TR~

shear test is to determine shear strength parameters and o s il under loading conditions that closely simulate Q the .ncipal axes of stresses and strains to rotate. ccur in the direct shear test but are indeterminate. T he stress s i es i.n..soilsi any geotechnical structures are akin to simple shear. There aT: two~ s of commercially available simple shear devices. One deform an initia cub0idal sample under plane strain conditions into a parallelepiped ~ . 5.26 . The sample is contained in a box made by stacking square hollow pia et een two platens. The top platen can be maintained at a fixed height for constant-volume tests or allowed to move vertically to permit volume

t:!1 0l

0,_

r;:

Soil

(a)

:::~~~n Top platen

Hollow plates +- stacked vertically

r+Wire reinforced rubber membrane Bottom platen

Bottom platen (b)

(c)

FIGURE 5.26 Cuboidal simple shear apparatus: (a) simple shear box, (b) stresses

imposed on samples, and (c) direct simple shear.

CHAPTER 5

FO

242

SHEAR STRENGTH OF SOilS

(5.48) (5.49)

and (5.50)

shear loa (a) Plot Mohr's cIrcles for total and effective stresses. (b) Determine the friction angle and the undrained shear strength , assuming the soil is nondilating. (c) Determine the failure stresses. (d) Find the magnitudes of the principal effective stresses and the inclination of the major principal axis of stress to the horizontal. (e) Determine the shear and normal stresses on a plane oriented at 20° clockwise to the horizontal.

Strategy Draw a diagram of the forces on the soil sample, calculate the stresses, and plot Mohr's circle. You can find all the required values from Mohr's circle or you can calculate them. You must use effective stresses to calculate the friction angle.

5.10 OTHER LABORATORY DEVICES TO MEASURE SHEAR STRENGTH

Solution 5.10 Step 1:

Determine the total and effective stresses. P, (J"z

=

A = Px

(J" x

Step 2: Step 3:

= A =

500 X 10(0.05)2

3

00 kP

__

2 3

375 X 10(0.05? = 150 kPa

Draw Mohr's circle of total an See Fig. ES .I0. Determine ~ and SUo Draw a tangent to M the axes.

FO

R

from

240 2 0 3 I 0 I

(90 - 60)

d (] (k ~a)

(1 50, - 0)

-90 f---+--~--+-~---r--+-~---+--1--~

-120 - 1 50 ~~--~--~-~--~--~~--~--~~

FIGURE E5. 1 0

a

243

CHAPTER 5

Step 4:

SHEAR STRENGTH OF SOILS

Determine the failure stresses. At the point of tangency of the failure envelope and Mohr's circle of effective stress, we get Tf

Step 5:

= 54

(~)f

kPa,

= 79

kPa

Determine (j;, (j;, and tV. From Mohr's circle of effective stresses, we get (J"; = 180 kPa

and

(J") = 50 kPa;

2l~

= 66.

Step 6:

Equation (3.27):

FO

244

60 cos 40° -

140 - 90 . 2 Sin 40°

=

29.9 kPa

•

EXAMPLE 5.

A cuboidal soil sample, with 50 mm sides, was tested under drained conditions and the maximum shear stress occurred when the shear displacement was 1 mm and the vertical movement was -0.05 mm. (a) Plot Mohr's circle of strain. (b) Determine the principal strains.

(c) Determine the maximum shear strain. (d) Determine

Ct .

Strategy Calculate the vertical and shear strains and then plot Mohr's circle of strain. You can determine all the required values from Mohr's circle or you can calculate them.

5.10 OTHER LABORATORY DEVICES TO MEASURE SHEAR STRENGTH

245

Solution 5.11 Step 1:

Determine the strains. E

z

= !1z = _ 0.05 = -0.001 Ho 50 (negative sign because the sample expands; com

Ex = Ey =

"Y zx

=

!1x

Iio

=

a 1 50 = 0.02

Step 2:

See Mohr's circle of strain, Fig.

By calculation,

FO

R

Equation (3.35):

FIGURE E5. 11

•

5.10.2 True Triaxial Apparatus The purpose of a true triaxial test is to determine soil behavior and properties by loading the soil in three dimensions. In a true triaxial test, a cuboidal sample is subjected to independent displacements or stresses on three Cartesian axes. Displacements are applied through a system of rigid metal plates moving perpendicularly and tangentially to each face, as shown by the arrows in Fig. 5.27a. Pressure transducers are fixed to the inside of the faces to measure the three principal stresses. Like the conventional triaxial apparatus, the directions of principal stresses are prescribed and can only be changed instantaneously through

CHAPTER 5

SHEAR STRENGTH OF SOILS

-L

-r

Sample

l~ Movab le plate

T (a) Pl an view

FIGURE 5.27

soil.

emeasu:%i1 ;triaxial

an angle of 90°. The stresses and test are shown in Fig. 5.27b.

5.10.3 Hollow CylinQItJeA.op

FO

246

--

(0)

FIGURE 5 .28

(b)

(a) Hollow cylinder cell and (b) stresses on an element of soil.

5.11 FIELD TESTS

247

If the internal and external radial pressures are equal, then

The shearing stress applied is (5.51)

where M is the applied torque and r1 and r3 are the in er can obtain ¢;, ¢~s, su, and G from the hollow cylinde st.

FO

5.11

adii. We

FIELD TESTS

(5.52)

i1e h is the height, and d is the diameter of the

Sheath

Section Y - Y Vane

Vane probe in protective sheath

FIGURE 5.29

Vane extended and ready for testing

Shear vane tester.

FO

248

CHAPTER 5

SHEAR STRENGTH OF SOILS

-== ~~:ii~~;

lifterconnector interface C~=~H:riaiiTmiiTm;eerr-Connectors to cathead Guide rod r:::::::ll::::==-- Cushion hammer rod interface Ground surface

J

Determ i nes net input energy to dnll rod

Segments of drill rod Joint in drill rod Drilling mud Borehole with /without casing

Determines transfer mechan ism to sampler

Split spoon sampler Soi I-sampler interface

, \ "<~

Displacement due to net useful energy at tip

FIGURE 5.30 Standard penetration test.

5.11.2 The Standard P

e st-andard penetration test. Sevorld to release the hammer. Also,

are given in ou should be cautious in using the correlation in Tables 5.3 TABLE 5.3 Correlation of N. N so • 'Y. Dr. and
N

0-5 5-10 10-30 30-50 >50

y

Dr

Nso

Description

(kN/m 3 )

(0/0)

(degrees)

0-3 3-9 9-25 25-45 > 45

Very loose

11-13 14-16 17-19 20-21 > 21

0-15 16-35 36-65 66-85 >86

26-28 29-34 35-40· 38-45" >45"

Loose Medium Dense Very dense

"These values correspond to

q,~.

5.11 FIELD TESTS

TABLE 5.4 Correlation of and Su for Saturated FineGrained Soils Nso

Su

N60

(kPa)

< 10 10-25 25-50 50-100 100-200 > 200

Very soft Soft Medium Stiff Very stiff Extremely stiff

FO

R

0-2 3-5 6-9 10-15 15-30 > 30

Description

249

(5.52)

where N k is a cone factor that depends on the geometry of the cone and the rate of penetration. Average values of N k as a function of plasticity index can be estimated from

I .

N = 19 k

I - 10 - p- _ .

5'

Ip > 10

(5.54)

250

CHAPTER 5

SHEAR STRENGTH Of SOILS

Cooe

'e~'5t ance

q, (MPa)

"",,'

$o.t de5l;.iption

0

2

0

Pore reS i$tMce u (MPal

310 0 25 0.50 0.75 100

Fill

-'-COnnecting

,,'"

5

10

Reda.med

~nd

uPP'" mallne ,'~ (S, . 42)

15

Cone

pte5sure measurement

i"a

20

<

~ (II)

Dutch cone

(I» PltlOCone

,

•\

••

m~r"'e

""

(S, = 3.9)

0,0

c) Core

FIGURE 5.31 (a ) Dotch cone and (b) p

-- >

f~ultS

ocone le) Cone results. IF rom Chang,

1988.)

(5.55)

5.11.4 Pressure Meters The Mena rd pressure meier (Fig. 5.32a) is a probe thai is placed at the desired depth in an unlined borehole and pressure is appl ied to a measuring cell of the probe. The stresses near the probe are shown in Fig. 5.32b. The pressure applied is analogous to the expansion of a cylind rical cavity. The pressure is raised in stages at constant lime intervals and volume changes are recorded at each stage. A pressure-volume change curve is lhen drawn from which the elastic modulus, shea r modulus. and the undrained shea r strength may be estima ted. One of the disadvantages of the Menard pressure meter is that it has to be inserted in to a pred rilled hol e and consequently the soil is disturbed. The Cam-

5 .11 FIELD TESTS

/"\

/ '\

Water pressure to expand membrane

j

_-- --1- --- 1-_

",/

I

I

\

///

\ '\

:

\

I

I

\

>< / \

A

i

dO',

I

v'; ,'" \

\

__

.... \,...\...............

\

\

I I

I

>

/ '\ \ /

_

II

I' I

I

\

4S' +~'12

",(

_ /

\

\

\

~

/-

II' II'

V \

Pressure cell

I

\ d O'o//-

II

Gu ard cell

'-',

t

as Ie zone

~ Ex~::tn~

-

~_____

/

EI

_----

Gas pressure to / ' , /"\ In f late guard ce ll s Dlrect lonsof prin cipal stresses;' \

,

251

;' " \

I I' \

'''''''-l I

I' \

-

I

I

v,

I ,

Guard ce ll

I

I "

-.../

I

(a) Vertical secti on

FO

R

FIGURE 5.32

Cutaway view

-

FIGURE 5.33

Wroth,1972.)

~-'---

Auger

Schematic of Cambridge Camkometer. (Redrawn from Hughes and

252

CHAPTER S SHEAR STRENGTH OF SOILS

An excellent source on the inlerpretation of the pressure meter test is Wroth (1984). The essential points art

I . Variousfield leslS are used 10 deltrmine soil sIrength parameters. 2, The most simplefil!ld Itst, and Ihe mOSI popular, j~' lh e standard penetration ust (SPT).

What's next . ..Severa l em p irical relat io nships hav b eelL!" opqse d to obtain soil strength parameters from laboratory tests, for example, h e- er6erg limits, or from statistical analyses of field and laboratory test esults. 5 .Q:.1 ~ of t hese re lationships are presented in t he next secti on. T 6.12 EMPIRICAL RELATIONS,,{,S FOR SHEAR _ ... , STRENGTH PARAMETERS ) some-suggested empirical rei X r the in Table 5.5. Thes. elationr'ips stlould only nary design cal ations. , . TABLE

~.5

Emp ~Soil

Strengt

Shea~[ength ~f so~ are shown ed as

gu'Ke and in prelimi-

Relations hips Referenee

Normally consolidated fays Overconsotidated clays

~ "

: 0. 11 + 0.0037 1"

•

Is

~I ...

.. {OCRlo,

·...u,

(sJa..:

Skemplon 11957) Ladd 61 al. (1977) Jamioll::owski et 81. (1985)

A ll clays

~ O.22 Wo. + 3D,(10 - In pil - 3, where pi is the mean effective stress at fa ilure !in kPa) and D, is rela tive density, This equation should only be used if 12 > (cI>~ - 4>~.) -:> O.

<j)~ .. <j)~.

Mesri 119751 Bolton (1986)

The strength of soi ls is in terpreted usi ng Cou lomb's frictio nal law. All soi ls regardless of their in itial state of stress will reach a critical state characterized by conti nuous shearing al constant shear stress and constant volume. The ini tial void ratio of a soil and the norma l effective stresses determine whet her the soil will dila te or not. Dilating soi ls often exhibit (1) a peak shear stress and then stra in soften to a constant shear stress, and (2) ini tia l contraction followed by expansion

5.13 SUMMARY

253

towa rd a critical void ratio. Nondilating soils (1) show a gradual increase of shea r stress, ultimately reaching a constant shea r stress, a nd (2) contract toward a critica l void ratio. The shear st rength parameters are the fri ction angles (
Pract;cal Examples EXAMPLE 5.12 fal load of 5 MN 10 a deep, = 4 an d

.. : +

Z2"'f",

= 71. 1

+ 4 x 9.8

d!

=

110.3 kPa

_",Jc::==~I""=1=~'F:oo:..;:al'on. 4mx 5m {, . 1 m

l-o"~--~+i--'CI .~

Samp!e

FIGURE E5. t2

•

z~ '" 4 m

-~~~--,,"-'--

)1,., -

254

CHAPTER §

Step 2:

SH EAR STRENGTH OF SOILS

Determine the vertical su ess increases al z = 5 m under the center of the rectangular foundation. Use Eq. (3.79) or rhe computer program on the CD. l!.u , = 71.l kPa.

Step 3:

l!.0', - 5.1 kPa

Neglect the effect of shear stresses (a Tu)' Determine imposed shear stress for short-te rm 10 din Current vertical total stress: «(Jllr - 0', + l!.0'. Current horizontal total stress. (o ,)r = uJ + A Cu rrent shear su ess: T~

(o-Ilr - (0-,)

= ""!.!..'2--"7'iIIIII~iji~-"=

The soil will nOI reach the fai l

rC;

Step 4:

o ng-term loading. ume lhat th-e excess pore

For long-term loading. we pressure dissipated.

~ a te r

-<

.. ().o-t = ~8 + 71~ 121.9 kPa

~"'''~-''+ ~ = . . nctwn

d~' 0 b·'· Ilze:

-

I

q~i,

0',

\

-

76.2 kPa

. 1(121.9 - 76.2) sm12 l.9 + 76.2

- 1JJ. .'

2Jo, which is grea{er than 4>'. .!

ill ) stale angle of friction is refore, fai lurer ld not eGur .

J

FS = tan 24' = 1.9

felY :

tan 13.3'

•

o /MPLE5.13

m.. ~oP~~cUor

An earth dam a site consisting of a homogeneous stiff clay. as shown in E5 .13a YoV,the geotechn ical engineer, are required to specify soil strength ie~"'to dete Imi ne the stabi lity of the dam. One of the critica l situations is the possible failm of the dam by a rotational slip plane in the stiff clay. Wh at laborato ry strenglh tests would you specify and why?

Strategy The key is to determine the stress states for s ho rt-term and longte rm cond it ions along the failure surface .

FIGURE E5.1 38

S.13 SUMMARY

255

./'---SI,P $\lIf,ce leI AlIiS\'mmelTlc exten$ion

(8) Simple 51-. or direcl shear

FIGURE 1:5. 1 3b

~

Solution 5 .13

Determine stresses along the ~tl sur ceo Let us select three pomts~B:;nd ~ n the possi!;Jle failure surface The rotational slip..su?iac wilJ.introduee consRresslon on element A , shear on emen JJ an extension 01 elemen t C FIg. E5.13b. The slresse elemen~ A are analogQ a r:iaxial compresSIon tes!. Elemen;;,!: B deform In a manner c0.e!Patibl e with simple shear, \\'hile eleme wi]! s uffc~ am an upwanP"thrust that can be simulated by t-rj lal extension t Step 2: Makep'&mme~ations. '""The fo llowing s;;.ing th tests ar%.eromm~ded . (a) T ria i~ CU co\n pression tests wltll, pore wa r pressure meas reme ~ Paramete~i ed re
ih

te}l~Uld

dratn ed be carned out at the maxImum antiCIpated stress on thCsoll a u c~ n determJl1e the stress Increases from the dam USll,t lh.e rri'e hods (fescribed in Chapter 3 •

You ha~ontrac d a aboratory to conduct soil t~sts for a si te . which consists of a laye sand. m thick. wilh "f ..., = 18 k N/m ~. Below the sa nd is a deep. lIoi fh ""Y'~I = 20 kN/ml (Fig. E5.14). The site is in a remOle area. soft. bluish Groundwnter level was located at 2.5 m below the surfa ce. You specified a co nsolida tion lest and a triaxia l consolidated undrained test for samples of the soil

2.smI

..L.

• FIGURE E5. 14

256

CHAPTER 5

SHEAR STRENGTH OF SOILS

taken at 10 m below ground surface. The consolidation test shows that the clay is lightly overconsolidated with an OCR = 1.3. The undrained shear strength at a cell pressure approximately equal to the initial vertical stress is 72 kPa . Do you think the undrained shear strength value is reasonable, assuming t . OCR of the soil is accurate? Show calculations to support your thinking. Strategy Because the site is in a remote area, it is likel existing soil results from neighboring constructions. I empirical relationships as guides but you are warned having variable strengths.

Solution 5.14 Step 1:

Determine the initial effectiv s the groundwater level is satur:a e

Step 2:

Step 3:

FO

0.27(OCR)08 to 0.19(OCR)0.8 =

0.27(1.3)° 8 to 0.19(1.3)°8

=

0.33 to 0.23 < 0.63

Su

--;- =

0.22 < 0.48

0' zc

nces between the reported results and the empirical relationships are substantial. The undrained shear strength is therefore suspicious. You should request a repeat of the test.

•

EXERCISES Theory 5.1

A CD triaxial test was conducted on a loose sand. The axial stress was held constant and the radial stress was increased until failure occurred. At failure the radial stress was greater than the axial stress. Show, using Mohr's circle and geometry, that the slip plane is inclined at 71'/4 - ¢~/2 to the plane on which the principal stress acts.

5.2

The initial stresses on a soil are 0'; = O'~ and (J'~ = Ko(J'~ , where Ko is the latenl earth pressure coefficient at rest (see Chapter 3). The soil was then brought to failure by re-

EXERCISES

257

ducing u) while keeping ui constant. At failure. 0") .. Ku;, where K is a lateral earth pressu re coefficienl. Show that Te.

-

T..

where stress.

5.3

T,. IS

I + K . , = ---5111 $", 1 - Ko

the maximum shear stress under the imtlal stresses and

"L l h

failure shear

Sand is placed on a clay slope as shown in Fig. P5.3. (a) Sho ~~at sa Q will b unSlable (i.e., fail by Sliding) if e > $'. (b) Does the thic kness of the c'ay la r inti. mpending failure? (c) If $' - 25 0 and 9 = 23°, determine the factor O'i'safety.

FIGURE PS.3

Problem Sol .in9 5.4

A structure will Impose a norrlla effective s t r~5,gl ' lcPa an a shear stress of 30 kPa on a plane inclin~d al 58" to the ho rizonlal...,Ihe fnion ) angle of the soil is 25°. Will the SOIl fail? r not, wh~actor of sai {Y'!

5.5

Figure P5. sho tILe stress-strain befavior of 8cJoi . Determi ne the peak and crit ical shear sires . ESltmale Ihe peak and c(tical slat' friction angle and the dilation angle. The normal effect ive streSS/~kPa.

2

,

6

Sl>ear

~tr:iHn

8

10

1'%)

"

FIGURE PS.S

5.6

The follo wi ng results we re obtained from three shear box tests on a sampk of sand), clay. The cross section of the shear box is 6 cm X 6 em. Normal force IN, Shearing force (N)

1250

1000

500

250

506

405

'"

2"

(iii) Determine the critical state friction angle. (bl If the soil IS dilalant. de termine til; at a nOTlllal force of 250 N.

258

CHAPTER 5

5.7

SHEAR STRENGTH OF SOilS

The results of a direct shear test on a dense sand are shown in the table below. Determine ¢~, ¢~s and Ci p • The vertical force is 200 N and the sample area is 100 mm X 100 mm.

Horizontal displacement (mm) Vertical displacement (mm) Horizontal force (N)

0.00

0.25

0.38

0.64

0.76

1.02

3.18

3.68

4.06

4.57

5.08

5.59

7.11

7.62

8.13

8.64

9.14

9.65

0.00

0.00

-0.02

-0.03

-0.04

0.09

0.14

0.15

0.19

0.22

0.28

0.28

0.28

0.28

0.28

0.0

17.7

19.3

33.4

137.4

148.8

155.9

161.7

172.9

163.5

161.2

160.8

2.92

2.67

6.86 0.06

.04

0.28

40.3

127.2

133.8

177.7

176.4

5.8

5.9

A CD test was conducted

0

Axial (N)

Change in volume (em 3 )

load

,·

604.3

-2.7

5.2

593.0

6.4 7.1 9.7

576.0 564.8 525.2

-3.2 -3.4 -3.5 -3.6

12.9

497.0

- 3.7

15.5

480.0

-3 .7

G:

FO

0 ~

5.10

1. 1.6 1.9

------~~~-----&--------~----------------------------

CU tests were carrie out on two samples of a clay. Each sample was isotropically consolidated before the axial stress was increased. The following results were obtained.

!l.uf

(kPa)

Sample No.

I

420

320

205

II

690

365

350

(a) Draw Mohr's circles (total and effective stresses) for each test on the same graph.

(b) Why are the total stress circles not the same size?

(c) Determine the friction angle for each test. (d) Determine the undrained shear strength at a cell pressure, ((f3)j of, 690 kPa. (e) Determine the shear stress on the failure plane for each sample.

EX ERCISES

S.H

259

eu tests were conducted on a compacted clay. Each sample was brought to a saturated state before shearing. The results. when no further change in excess pore water pressure or devialonc stress occurred, are shown in the table below. Determine (a) $';'" and (b) $~ at a cell pressure of 420 kPa. 10, - 0, ]

5.12

O", lkPa )

(kPa)

140 280 420

100.

~u

636

(kPa)

71 -51.2 -19.4

1323

t::-&f.Ve1co~i~ns. !pW.

Three samples of a loose sand were tested u r:;...U excess pore water pressures for each sampt; are

T he failure stresses and

Sample lu,), (kPa)

No.

2

3

2 10

;];1'

360 685

""'62 323

(.) Plot Mohr's

d e of

ef~ctive stress from thes

angle for eae test. Determine the the an erel o be subjected t a vertical ffee!ive stress of 300 k Pa, what magnitude ~onzon!al effective stress ill cause failure? (d ) De term e the orientati~f (1) Ihe (al \lte plane and (2) the plane of maximum shear stress to Ihe ho r~laL e) l~ failure stre~~ximum sht:ar stress? Give reasons. (b)

( c) If

eu tri axial iesl ~rried~~Si1IY day thaI was isotroplcally consolidated with II pressure

11

jrp

Stress

Stress imposed by plunger (kPal

0 5.5 11.0 24.5 28.5

foll owing data were obtained:

impose d by plunger IkPa)

Axial strain, C, (%1

(kPa)

0

35.0

0.56

34.8

'.0 '.6 19.1 29.3

50.5 85.0 105.0 120.8

1.08 2. 43 4.02 9. 15

41.0 49.7 55.8 59.0

••

\kPal

0 0.05 0.12 0.29 0.38

••

(a) Plot the deviatoric stress against axial strain and excess pore water pressure against axial stram. (b) Determine the undrained shear strength and the fric tion angle. (c) Determme E' and £;. (d) Determine Skempton's A"

260

CHAPTER 5

FO

5.14

~

Three CU tests with pore water pressure measurements were made on a clay soil. The results at failure are shown below.

(1, -

UU CU

5.18

SHEAR STRENGTH OF SOilS

(13

(kPa)

210

105 246

57 133 210

74.2 137.9 203

Au (kPa)

To be determined To be determined 11.9 49.0 86.1

The failure stresses in a simple shear constant-volume test are shown in the table below.

Total normal stress on the horizontal plane

300 kPa

Total normal stress on the vertical plane

200 kPa

Total sh ear stress on the horizontal and vertical planes

100 kPa

Pore water pressure

50 kPa

EXEROSES

2 61

(8) Draw Mohr's circles of lotlll and dfectlve Stresses and determine the magnitude of the principal effective stresses lind dlreClion of the major principal effective stresses. (b) DetermlOe the fnctlon lingle IIssuming the soil is nondilalional. (c) Detenmne the undrained shear stre ngth.

Practical 5. 19

You are in charge of desigmng a retaining wall. for the backfill soil? GIve reasons.

s.zo

The following results were obtained from CU tests on a material for an embankment:

U"J

IkPal

11, -

U"J

(kPa )

300 400

331

600

487

IS the foundation

.:111 ]kPal

111

42.

Recommend the shear str h p'aramele to be used yscs. The maximum confinly e t (cell pressure,

s hort. len~

and ong·term anal· inte rest IS 300 kPa.

de ~

CHAPTER

6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHA VIOR ED 10

6.0

INTRODUCTION So far, we have painted individual pi ctu e~\oi lt.be avior. We looked at the physical characteristics of soil in Chapter 2, effective stresses all stress paths in Chapter 3, one-dimensional co r;fidat~ in apter 4, an shC;Jlr strength in Chapter 5. You know that if y consoliClate soil to a hjg r 'S tress state than its current one, the shear streng h of the.soil will Increase. But ht-am oun t of increase depends on the ~I type, th,\ l,pfd ing cond.!}i6ns (drained ¥Undrained condition), and the stress paths. Th.!;,refore, the in5;: Vi ual pict ~es should a1\ be linked together: Bu how? ' , In this chaf er, we are &,oing to take [he indl idua lcturcs and build a mosaic that will " rovide a bate for us to int~R.ret a~licipate soil behavior. Our mosaic is m lillY int oded to unite conso"h ation and shear strength. Real soils. of tourse, require a complex ;;nosaic t a t because soils are natu ral , complex mar~s.but also because the loads d oading paths cannot be ant icipated aecu a'feiyY" We are §9ing to build a"'mosaic to fa ae a simple framework to describe, in erp~et. an~ antid~at~. it res~nses to \'.a ~ious loadin~s . The fr~mewo.r~ is esse ntla ll a theoretL~r m dcd based on cnllca] state SOIl mechanlcs-cntleal state mOdel (SchofielCl nd \-V RL,hA"%8). Laboratory and field dala, especially r I s from sof ~rma\ly con,a(jdated clays. lend support to th e underlying co ceprs embo d In the de eloprnent of the critical siale model. The emphasis in this chagfer will...,Qe 0 us ng the critical sla te model to provide a gene ralized understap ing of so~behavior rather than on the mathematical formulation . The cntical stat~ model (CSM) we are going to study is a simplification and an idealization olson'" bebavior. However. the CSM captures rhe behavior of soils that are of greatest importance to geotechnical engineers. The central idea in the CSM is that all soils will fail on a unique failure surface in (q, p', e) space. Thus, the CSM incorporates volume changes in its failure criterion unlike the MohrCoulomb failure criterion, whi ch defines fa ilure only as the attainment of the maximum stress Obliquity. According to the CSM. the failure stress state is in sufficient to guaran tee fa ilure: the soil structure must also be loose enough. The CSM is a tool to make est imates of soil responses when you cannot conduct sufficlenl soil tests to completely characterize a soil at a site or when you have to predict the soil's response from changes in loading duri ng and after con struction. Although there is a debate on the application of the C5M to rea l soils, the ideas behind the CSM are simple. Tt is a very powerful tool to gel insights into soil behavior, especially in the case of the " what -if " situation. There is also a plethora of soi l models in the literature that have critical sta le as their core. By

1

6.1 DEFINInONS OF KEY TERMS

263

stud ying the CSM, albeit a si mplifi ed version in this chapter, you will be able to better unders tand these other soil models. When )'ou have studied this chapte r, you should be able to: • Estimate failure stresses for soil • Estimate strains at failure • Predict stress-strain characteristics of soils from a few parameters obtained from simple soil tests • Evaluate possible soil stress states and failure if the loading on a geotechnical system were to change You will make use of all the mater."~s. you fudi pa rticularly:

in Chapters 2 to 5 but

Sampla Prac ca Sit lion An 0;1 tank ;5 l cted on a 50ft al luvial cia)'. II as decide"- t t the clay ~6lIJ be r loa ed with a circular embankmen t im sing a strAss equal to, at leaSt, Ihe 10lal applied st ress of the tank when /ilJed, ~an fai are .to ~ to cce eTale the consolida tion process. The fou da ' o for the tan k IS a CI e ular sla~ of toncrete and the purpose of the preloadi . to e duce the total se tlement the founda tion. You are required to <'I dvise e owners on how".the tank~urd be filled du ring preloading to p reem prem; re failure. fter.,preloading. the owners decided to increase the eight of the lank.. Yo re'qU e~d 10 de termine whether the soil has enough sheaLs re ngth to uP-p'ort [l add 'tlOnal increase in lank heigh t, and if so the amou nt of se ttle ent tha caI\)ie expected. The owners do not wanl 10 finan ce any fun he, : snBand 5011 test;ng.

tx:

9f

6 . 1 DEFINITIO S 0

KEY TERMS

Overconsolidarion ratio (R,) is the ratio by which the current mean effective st ress in the soil was exceeded in the past (Ro = p;lp~ where p ~ is the past maximum mean effective stress and p~ is the cu rren t mean e ffective stress). Comprrssion index ()..) is the slope of the normal consol idation line in a plot of the natural logarithm of void ra tio versus mea n effective stress.

Unloadi"gl reloadi"g index or rrcomprusion ;nde.\' (K) is the average slope of the unloading/reloading curves in a plot of the natural logari thm of void ratio versus mean effe ctive stress. Criticlli Slllll! nne (CSL) is a line that represents the fa ilure state of soils. I n

(q, p') space the critica l state line has a slope M . which is related 10 the friction angle of the soil at th e critical slate. In (e. In p') space, Ihe critical state line has

264

CHAPTER 6

A CRrTlCAlSTATE MODEL TO INTERPRET S Oil BEHAVIOR

a slope A, which is parallel to the normal consolidation li ne. In th ree-dimensional (q, p' , e) space, t he critical sta te line becomes a cri tical state surface.

J . What is soil yieldi ng?

What is the difference between yielding a n~~ llu in What parameters affect the yield ing and fail of so Does th e fail ure stress depend on the consoli a What are the crit ical state parametlrs and ho from soil tests? 6. Are stra ins important in soil f 7. of soils due to differen l stress pa ths?

2. 3. 4. 5.

,

6.3

BASIC CONCEPT

I n our deve qpmenl 0 IH asic cOllce ~ critic~sta te , we are going to map certa in plots lwe have s died in C1~ ters 4 ~ 5 using stress and strain invariants and 'fncentra te on sa lUrated .if~l un ax* ymmetric loading. However, the conce ts nd..method ho ld for a ~y loading ~n B itio n . Rather than plott ing T ver· sus a~ ~r l' w.F will plot the dat as q verjUs p' (Fig. 6. l a). This means that you must kn w the principa ~esses cting:;6'n the element. For axisymmet ric (triaxial) cop Ilion, you 0e,ly n~d to know two princi pal stresses. The Mohr-Co b ail u ~ line in (T, a ;) space of slope q,~. "" tan -I [Tal a ;;rt is now din fJ p ' space as a line of slope M = q1ipi, where the s.ubscripl f denp tes failure. I stead of a plot of e versus a~, we will plot the da ta Jl'a s e versus (Fig_6.1bj an instead o f e versus log a ;. we will plot e versus In p ' (Fig. 6.1c). will {i.e~te the slope of the norma l consol idati on line in Ih e plot of e versus In p s ~Yn d the unloading/reloading line as 1(. There are now rela· t ionst ip~bet~ee~
m JP

(6.1)

c,

I(

C,

=- ~ '" 2.J =-

C

0.434 ,

(6.2)

Both}.. and

K are posi tive for compression. For many soils. KIA has values within the range 10 ~. We will formu late the relationship be tween
to

R o

p~

p~

(6.3)

6.3 BASIC CONCEPTS

265

q Failure line or critical state line: M = . e: ¢e,, --t a n - l ~ Failure lin (" ;),

~ P,

==+ P' (a )

e

F O

,,;

Y C

Ro =

1 + 2K~c OCR 1 -L ')fl"~-

ill be required to prove this equation in Exercise 6.1.)

6.3.2 Failure Surface The fundamental concept in CSM is that a unique failure surface exists in (q, p', e) space, which defines failure of a soil irrespective of the history of loading or the stress paths followed. Failure and critical state are synonymous. We will refer to the failure line as the critical state line (CSL) in this chapter. You should

266

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

, CS l~

, esc 101

,

, NCL

~.

CSC

,.

~ )

recaJll hat c,,'tical state is a c ~ta nt streSSSht e characterized by continuous shea r d~ ormation at co n sta n ~o lum,In stress space (q , p') the CS L is a straight line of slo-pe...,.M = Me' for com ression, nd M = Me. for extension (Fig. 6.2a). Ex tension 'does not m an ensio bu refers to the case where the lateral stress IS gr (er than the v\l".ti cal"'"mess~ There is a correspond ing CSL in (p', e) space ( ,g. 6.2b) oe (. p") ? ig. 6.2c) Ihat is parallel to the noemal co(\sol;dat;on line. W represe, ti e CSL in a single three-dimensional plot with axes q, p', e (see b cov~, ' but we will usc the projections of the fai lu re su rface in the (q. p') space and the (e, p ' ) space for simplicity.

6 .3 .3 Soil Yielding You should recall from Cha pter 3 (Fig. 3.8) that there is a yield surface in stress space lha l separates su ess states that produce elastic responses from stress states that produce plasti c responses. We are going to use the yield surface in (q, p') space (Fig. 6.3) rath er than (0"1. Cl" l) space so that our interpretation of soil responses is independent of the axis system. The yield surface is assumed to be an ell ipse and its initial size or major axis is determined by the preconsolidation stress. p;. Experimenta l evidence (Wong and Milchell, 1975) indicates that an ell iptical yield surface is a reasonable approximation fo r soils. The highe r the preconsolidation Slress, the larger the

6.3 BASIC CONCEPTS

267

q A- elastic stress state

B- initial yielding C- elastoplastic

_C / /~ B "-.

Expanded yi el d su rface

. In

.

compression

p

Expansion of the yield surface.

F O

p~

FIGURE 6 .3

. ..

Initial Yield surface

~

Y C

~on to illustrate the ideas presented so far. y to""P.redicrbow a sample of soil of initial void ratio eo will de~ ained condition in a triaxial apparatus, that is, a respond whenA'eSte~Kch~ uld'\eca that the soil sample in a CD test is isotropically firen a 'lalloads or displacements are applied, keeping the cell preSS\.l:f~lSon.~ anL"~re going to consolidate our soil sample up to a maximum me~. effe~ stress p~, and then unload it to a mean effective stress p~ such that ~~ p"';l~~ < 2. We can sketch a curve of e versus p' (AB, Fig. 6.4b) during n~ n ~$ 1! on phase. You should recall from Fig. 6.1 that the line AB is the norn~'l c0)1s01idation line of slope A. Because we are applying isotropic loading, the li\~ AB (Fig. 6.4c) is called the isotropic consolidation line. The line Be is ~IOading/reloading line of slope K. The preconsolidated mean effective stress, p~, determines the size of the initial yield surface. A semi-ellipse is sketched in Fig. 6.4a to illustrate the initial yield surface for compression. We can draw a line, as, from the origin to represent the critical state line in (q, p') space as shown in Fig. 6.4a and a similar line in (e, p ' ) space as shown in Fig. 6.4b. Of course, we do not know, as yet, the slope M , or the equation to draw the initial yield surface. We have simply selected

268

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

q

q

s

E)

:; :: :: :: ::

FO

O

~

10 1

arbitrary valu s ater, we a e going to develop equations to define the slope M , the shape f the yl d s e, and the critical state line in (e ,p ' ) space or (e, In p') space. she r the soil sample at its current mean effective stress, p~ , by Let us increasing the a;loa stress, keeping the cell pressure, 0'3 , constant and allowing the sample to drain. You should recall from Chapter 5 that the effective stress path for a CD test has a slope q/p' = 3. The effective stress path is shown by CF in Fig. 6.4a. The effective stress path intersects the initial yield surface at D . All stress states from C to D lie within the initial yield surface and, therefore, from C to D on the ESP the soil behaves elastically. Assuming linear elastic response of the soil, we can draw a line CD in (q , £1) space (Fig. 6.4c) to represent the elastic stress-strain response. At this stage, we do not know the slope of CD but later you will learn how to get this slope. Since the line BC in (e, p ' ) space represents the unloading/reloading line (URL), the elastic response must lie along this line. The change in void ratio is t.e = e c - e D (Fig. 6.4b) and we can plot the e versus el response as shown by CD in Fig. 6.4d. Further loading from D along the stress path CF causes the soil to yield.

269

6,3 BASIC CONCEPTS

F O

The initial yield surface expands (Fig, 6.4a) and the stress-strain response is a curved path (Fig, 6.4c) because the soil behaves elastoplastically (Chapter 3), At some arbitrarily chosen loading point, E, along the ESP, the size (major axis) of the yield surface is Pc corresponding to point G in (e, p') space, The total change in void ratio as you load the sample from D to E is DE (Fig, 6.4b), Since E lies on the yield surface corresponding to a mean effective stress p ;', then E must be on the unloading line, EC', as illustrated in Fig. 6.4b. If you unload the soil sample from E back to C, the soil will follow an unloading path, EC' , parallel to BC as shown in Fig. 6.4b. We can continue to add increments of loading along the ESP until the soi fails, For each load increment, we can sketch th~ t'r~ss-strain curve and the pa followed in (e, p') space. Failure occurs when teES ~fntersects the critical state line as ind,icated ~y F i,n ~ig, 6.4,a. The failure tre~ses ~ f and qf., '~~,~ and the failure VOId ratio IS ef (FIg. 6.4~~~ ea . mcrement of loadn g, we.~ determine !:J.e and plot 8 1 versus 2:!:J.e [PI f.fJ. .~_.(~LS ~~ 1 + eo)] as;;h
6.3.5 Prediction Consolidated Soils Under Un

Y C

270

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

q

FO

q

o

c:

p', p

(a)

FIGURE 6 .5

So far we hav e nSloered a lightly overconsolidated soil (Ro < 2). What is the verconsolidated soils, that is, Ro > 2? We can model situation regarding eavt a heavily consolI atea soil by unloading it so that p~ /p~ > 2 as shown by point C in FIg . 6.6a . Heavily overconsolidated soils have initial stress states that lie to the Ie t the critical state line in the e versus p' plot. The ESP for a CD test has a slope of 3 and intersects the initial yield surface at D. Therefore, from C to D the soil behaves elastically as shown by CD in Figs. 6.6b,c, The intersection of the ESP with the critical state line is at F (Fig. 6.6a), so that the yield surface must contract as the soil is loaded to failure. The initial yield shear stress is analogous to the peak shear stress for dilating soils. From D, the soil expands (Figs. 6.6b,d) and strain softens (Fig. 6.6c) to failure at F. The CSM simulates the mechanical behavior of heavily overconsolidated soils as elastic materials up to the peak shear stress and thereafter elastoplastically as the imposed loading causes the soil to strain soften toward the critical state line. In reality, heavily overconsolidated soils may behave elastoplastically before the peak shear stress is achieved but this behavior is not captured by the simple CSM described here.

6.3 BASIC CONCEPTS

,

ESP

"

271

s CSL D

q, F

"

--~--.

,

o . c:

c

"

~I

•

••

,

"

FIGURE,

I",

ll~ r8t1Ve predicted res'UI'ts fro~test IRo > 2) using the CSM.

ca;~f" ,,1t~

( , Ihe cu he4 .rconsOlidat.d so ils. Ih. path 10 failuc. in (e, p') space is CF as 5~ow.,\in Fig. 6.7b. Initial yielding is atlained al D and failu r~aJ, F. The excess pOTe wate~pressures at initial yield. flu>" and al failure, 4,l/f' ar.eshown in lhi in~\ of Fig)V,a. The excess pore waler pressure at failu re is negalh. (PI f'7J,"' -~

8 .3 .7 CritiCal S

6 :,1.8 Vo lume Changes and Excess 'o re Water Pressures If you com pare the responses of soils in dr
by the CS M. you will notice that compression in drained tests translates as positive excess pore water pressures in undrained tests. and expansion in drained tests translates as negative excess pore water pressures in undra ined tests. The

272

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

q

TSP

q

CSL

o

FO

avil ' overconsolidated soils strain soften to fail,e CSM then qualitatively match observed

,

~

ive Stress Paths f a soil epends on the ESP, Effective stress paths with slopes less g, 6.8) will not produce shear failure in the soil because the ESP will never intersect the critical state line. You can load a normally consolidated or a lightly overconsolidated soil with an ESP that causes it to respond like q ESP that causes a lightly overconsolidated soil to respond like a heavi Iy overconsol idated soi I

o FIGURE 6_8

p'

Effects of effective stress paths on soil response,

6.4 ELEMENTS Of THE CRITICAL STATE MODEL

273

an oveTconsolidated soi l as shown by OB in Fig. 6.8. Effective stress paths si milar to OB are possib le in soi l excavation. Remember tha t a soil must yield before it fails.

1. 2.

3. 4. 5. 6.

The essential points are: There is a unique critical slate line in (q, p') space and a corresponding critical siale. lin~ in (e, p') spac~ for soils. There;s an initial yield surface for l·oils. The size oJthe initial yield surface depends on the preconsoliciation mean effective litress. The yield surface expands Jor Ro ~ 2 and contracrs for R" ~ 2 when the applied effective stresses exceed the initial y ield 5IUSS. The soil will behave elastically Jor stresses that are within the yield surface and elaslOplru·tically for SlresSes outside the yield surface. EI·ery stress !Uale must lie on an expanded or contracted yidd surface and on a cornsponding URL. The critical slate model qualitali.·d)' captu~· the essenliallealures of soil response.f under drained and undrained loading.

What's next . . .You we re sential ingredients of ample, yo u d id not yield surface. In

Remember tures of con s£.j\da~

usi ng ? rOje: \n goo meuy of the os· . T h e r~ were maQY unkn owns. For exth e criticaf'St'tjte. . '~ne and the equation of th e will de~ l op equation s to fi nd th ese unknowns. bui ld a slnJple ~!;cOUPHng the essential fea' l;UJlear strength . .A. <_

_

•

6 .4 ~

-tI

6 .4.1 Yield ,urf~4t& Th e cquati9([Jr fh(Yi~ rface is an ellipsze giveln by (p')2 _ p'p~

+~

=

0

(6.4)

tiieoretical basis for the yield su rface is presented by Schofield and

rO\~_l 96J} and Roscoe and Bu rland (1%8). You can draw the ini tia l yie ld s~ac~ m

the initial stresses on the soil if you know the va lue o f M .

6),.2 Critical State Parameters "'6.4.2 . f Failu,eline in (q, p ' ) Space The Mohr- Coulomb fai lure criterion for soils as descri bed in Chapter 5 can be wri tten in terms of stress invariants as

Iq, ~ " pi I

(6.5)

where q, is the deviatoric stress al failure (si milar to '"t). M is a friction constant (sim ila r to tan 4t~), and PI is the mea n effective stress at failu re (simi lar to O"~).

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

For compression, M = Me and for extension M = Me. The critical state line intersects the yield surface at p~/2. Let us find a relationship between M and ~s for axisymmetric compression and axisymmetric extension. Axisymmetric Compression

M = qf

pj

e

We know from Chapter 5 that

Therefore, (6.6)

or (6.7)

f

=

ea; ; a~}

qf = (a; - a~ )f

FO

274

M

( 2a~a3 + 1)

f

= qf <

pj

6sin:"

(6.8)

or . SIO

, 3M. ¢ cs = 6 - Me

(6.9)

An important point to note is that while the friction angle, ~s , is the same for compression and extension, the slope of the critical state line in (q , pi) space is not the same. Therefore, the failure deviatoric stresses in compression and extension are different. Since Me < Me, the failure deviatoric stress of a soil in extension is lower than that for the same soil in compression .

Y

C

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

The essential critical state parameters are: A-Compression index, which is obtained from an isotropic or a onedimensional consolidation test. K-Unloading/reloading index or recompression index, which is obtained from an isotropic or a one-dimensional consolidation test. M-Critical state frictional constant, which is a function of cf>:S and is obtained from shear tests (direct shear, triaxial imp!e shear., tc.).

To use the critical state model, you must also kno example, p ~ and p ~ , and the initial void ratio o'

EXAMPLE 6 . 1

_ O"t - 0"3
FO

276

M =

c M = e

Step 4:

6 sin c\>~ 3 - sin c\>~

=

=

140 260 + 120

6 x 0.37 3 - 0.37

=

=

0.37

0 84 .

6 sin c\>~s = 6 X 0.37 = 0.66 3 + sin c\>~s 3 + 0.37

Find qf for extension. qj

0.66 0.84

=-

x 140 = 110 kPa

•

EXAMPLE 6 .2 A saturated soil sample was isotropically consolidated in a triaxial apparatus and a selected set of data is shown in the table. Determine A, K, and er.

6.S FAILURE STRESSES FROM THE CRITICAL STATE MODEL

Cell pressure IkP,]

Condition Loading Unloading

277

Final void ratio

200

1.72

1000

1.20

500

1.25

Strategy Make a sketch of the results in (e . Inp ') space to provide a visual aid for solving this problem.

Solution 6.2 Step 1: Make a plot of In p ' ve rsus e. See Fig. E6.2.

1m uta 1.1 4

5Pl

5

~

'

.

FIGURE E6.2

Step 2:

Step 3:

c~ate K. From Fig. E6.2.

_ I(

. ,.01

In(~;Jp~)

_ 11.20 - 1.251 _ -

In(~)

- o.ro

Slip 4: tr

eo>

c

-I(

In

p~

]000

- 1.25 + (0.32 - O.ro) In - 2-

+ 0.07 In 500 '"" 3.24

• n.~w

:0

'e know the key parameters to use in the CSM . Next, we will predict the shear strength of soils.

6,5 Fl!\;,OJRE STRESSES FROM THE CRITICAL STATE MODEL 6.5.1 Drained Triaxial Test Let us consider a CD test in which we iSOlfopically consolida te a soil to a mean e ffective stress p; and unload it isotropically to a mean effecti ve stress of p ~ (Figs.

278

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

q

q,

r--+----------.r

FO

(a)

~

q! = 3(p j - p ~ )

(6.14)

The equation f ritical state line , using a generic M, which for compression is Me and for extensIon is Me, is q! = Mp j

(6.15)

The intersection of these two lines is found by equating Eqs. (6.14) and (6.15), which leads to , p!

3p~ =

3 - M

(6.16)

and q!

qf ~

=

, MP!

3Mp ~

= 3-M

(6.17)

Let us examine Eqs. (6.16) and (6.17). If M = Me = 3, then pi ~ 00 and Therefore, Me cannot have a value of 3 because soils cannot have infinite

00.

6.5 FAILURE STRESSES FROM THE CRITICAL STATE MODEL

279

F

strength. If Me > 3, then PI is negative and qj is negative. Of course, PI cannot be negative because soil cannot sustain tension. Therefore, we cannot have a value of Me greater than 3. Therefore, the region bounded by a slope q/p = 3 originating from the origin and the deviatoric stress axis represents impossible soil states (Fig. 6.10a). For extension tests, the bounding slope is q/p = -3. Also, you should recall from Chapter 4 that soil states to the right of the normal consolidation line are impossible (Fig. 6.1Gb). We have now delineated regions in stress space (q , p') and in void ratio space versus mean effective stress-that is, (e, p') space, that are possible for soils. Soil states cannot exist outside these regions.

6.5.2 'Undrained Triaxial Test

8)

(6.19)

( 6.20)

p'

F ' A' ~SL CSL

p'

FIGURE 6 . 11

Failure in CU tests.

In p'

280

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOil BEHAVIOR

For a CU test, the TSP has a slope of 3 (Fig. 6.11). For the elastic range of stress, the ESP is vertical (6.p' = 0) up to the yield stress and bends toward the critical state line as the pore water pressure increases considerably after yield. The undrained shear strength, denoted by Sr; , is defined as one-half the deviatoric stress at failure . That is,

'. ~ ¥e,p(¥)

r .

'-.)

(621 )

For a given soil, M, A., and er are constants and the on in Eq. (6.21) is the initial void ratio. Therefore, the undrained shear a particular or in ia ter content. You saturated soil depends only on the initial void r should recall that we discussed this in Chap' er 01 nO' show any matherna tical proof. We can use Eq. (6.21) to compa samples of the same soil tested at different Id rati or to predic he undrained shear strength of one sample if w kn the other. Consider two samples, a d B, of drained shear strength is

as

FO

(6.22)

~

(S,,)A (S,JB

=

1.20

That is, a 1 % increase in water content causes a reduction in undrained shear strength of 20% for this soil. The implication on soil testing is that you should preserve the water content of soil samples, especially samples taken from the field , because the undrained shear strength can be significantly altered by even small changes in water content. For highly overconsolidated clays (R o > 2) or dense sands, the peak shear stress (qp) is equal to the initial yield stress (Fig. 6.7) . Recall that the CSM predicts that soils with Ro > 2 will behave elastically up to the peak shear stress (initial yield stress). By substituting p' = p~ and q = qp in the equation for the yield surface [Eq. (6.4)], we obtain 2

" + M2 qp - 0 (Po')2 - PoPe -

6.5 FAILURE STRESSES FROM THE CRITICAL STATE MODEL

281

which simplifies to ----------------------------, I

qp =

MP~~ = Mp~~;

(6 .23)

Ro > 2

and

Is" =

~P~~;

Ro >

21

(6.24)

From the TSP,

F O

The excess pore water pressure at failure is found from the difference between the mean total stress and the corresponding mean effective stress at failury, that is,

Therefore, (6.25)

The essentiill. poiots ~r". 1. The intersection oj the ESP and the critical state line gives the failure st"esses~

2. Th e undrained shear strength depends on.ly on the initial void ratio. 3. Small changes in water content can significantly alter the undrained shear strength.

of a clay were each isotropically consolidated under a r a a¥ then unloaded isotropically to a mean effective stress teSVto be conducted on specimen A and a CU test is to be '111~~en B. Estimate, .for each speci~e~, (a), the yield ,stresse~, (3)y; and (b) the faIlure stresses Pf' qf, (aJ)f' and (a3)f ' EstJpie B the excess pore water pressure at yield and at failure. The soil p~ra~<\!ers )?e A = 0.3 , K = 0.05 , eo = 1.10, and
St7itegy Both specimens have the same consolidation history but are tested er different drainage conditions. The yield stresses can be found from the Ihtersection of the ESP and the initial yield surface. The initial yield surface is known since P~ = 300 kPa, and M can be found from
282

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOil BEHAVIOR

Solution 6.3 Step 1:

Calculate MeMe =

6 sin 30°

3 -

.

SIn

00 = 1.2 3

Step 2:

Calculate er. From Eg. (6.13),

Step 3:

Make a sketch or draw a scaled plot ~.•~~.~~ versus p' graphs. See Figs. E6.3a,b. Find the yield stresses.

Step 4:

(1 )

From the ESP,

FO

(2)

Initial yield .. surface I

~

400

500

400

500

(o )

2.5 2 e 1.5

B

I 100

200

300

p' (kPa) (b)

FIGURE E6.3a,b

F

6.5 FAILURE STRESSES FROM THE CRITICAL STATE MODEL

283

is not possible because we are conducting a compression test. The yield stresses are then p~ = 246.1 kPa, qy = 138.2 kPa. Now, qy

=

(o-i)y -

(o-~)y =

(o-~)! =

138.2 kPa;

200 kPa

Solving for (aD f gives (0-1)! = 138.2 + 200 = 338.2 kPa

F O

Undrained Test The ESP for the undrained test is vertical for the region of stress paths below the yield stress, that is, I1p' = O. From the yield surfac [Eg. (6.4)] for p' = p; = p ~, we get

o~

and

From the TSP,

Y (o-D y

\

=

143.4 kPa

> .(0-3)y = (o-3) y + !::J.uy = 143.4 + 56.6

C

. EquatIOn (6.16): Equation (6.5):

PI

=

3

x

200

=

= 200 kPa

333.3 kPa

q! = 1.2 X 333.3 = 400 kPa

Now, qF = (o-D! - (o-~)f = 400 kPa

and

(o-~)! =

Solving for (o-;)f' we get (o-i)!

=

400 + 200

=

600 kPa

200 kPa

F

284

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

Undrained Test Equation (6.19):

Now,

pi

=

CUi)

1 f

+ 2(u' ) 3

3

!

= 1586 kPa .

q j = CuD! - (U3)f = 190.4 kPa

Solving for (aD! and (aD!, we find (uD!

=

285 .5 kPa

or

normal consol-

e

pi

= 1.08

+ 0.05 In ~ pi

The equation for the critical state line in (e , pi) space is e

=

2.62 - 0.3 In pi

Now you can plot the normal consolidation line, the unloading/reloading line, and the critical state line as shown in Fig. E6.3b. Plot Initial Yield Surface

The yield surface is

(p')2 - 300p' +

(l:~? = :.q

For pi

=

°

=

° Up'

lOOpi -

1

to 300, plot the initial yield surface as shown in Fig. E6.3a.

Y

C

286

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

EXAMPLE 6.4

Determine the undrained shear strength in (a) a CU compression test and (b) a CU extension test for a soil with Ro = 5, p~ = 70 kPa , and

Strategy Since you are given
Calculate Me and Me.

Step 2:

Calculate SUo Use Eq. (6.24) .

~ a

FO

= 51.8 kPa

•

t - t is problem is a straightforward application of Eq.

Step 1: Determine the difference in Use Eq. (6.22) .

Su

values.

(Su)l ab = ex (2.7(0.48 - 0.44»)

(S,,)field

p

=

2.3

0.13

The laboratory undrained shear strength would probabl y show an increase over the in situ undrained shear strength by a factor greater than 2. •

What's next . . .We have discussed methods to calculate the failure stresses. But failure stresses are only one of the technical criteria in the analysis of soil behavior. We also need to know the deformations or strains. But before we can get the strains from

6.6 SOIL STIFFNESS

287

the stresses we need to know the elastic, shear, and bulk moduli. In the next section, we will use the CSM to determine these moduli .

6.6

SOIL STIFFNESS

F O

The elastic modulus, E', or the shear modulus, G, and the bulk modulus, K', characterize soil stiffness. In practice, E' or G, and K' are commonly obtained from triaxial or simple shear tests. We can obtain an estimate of E' or G and K' using the critical state model and results from axisymmetric, isotropic consol' dation tests. The void ratio during unloadinglre).eadjng is described by

(6.28)

(6.29)

E' = 3K'(1 - 2v')

In p'

FIGURE 6 . 12 Loading and unloading/reloading (elastic) response of soils in (e-p' In) space.

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

Therefore, £' = 3p'(1

+ eo )(l - 2v')

(6.30)

K

G

Also, from Eq. (3.102), E'

G = 2(1 + v')

Therefore,

(6.31)

~::§:(:so:t!::l@!l.le s based on the level

usually < 0.001 %), the d the soil behaves like a tra ns between 0.001 % and 1 %, il behavior is elastoplastic (nont{ffness decreases slowly to an ape strains ('y > 00), the s t e soil approaches critical state. At the critical or

£d

FO

288

Small strains

Intermediate strains

Large strains

0.001

FIGURE 6.13

strain levels.

Schematic variation of shear, bulk, and Young's elastic moduli with

6.6 SOIL STIFFNESS

289

F O

soil stiffness varies within the loaded region of the soil. Consequently, large settlements and failures are usually initiated in the loaded soil region where the soil stiffness is the lowest. In conventional laboratory tests, it is not practical to determine the soil stiffness at shear strains less than 0.001 % because of inaccuracies in the measurement of the soil displacements due to displacements of the apparatuses themselves and to resolution and inaccuracies of measuring instruments. The soil stiffness at small strains is best determined in the field using wave propagation techniques. In one such technique, vibrations are created at the soil surface or at a prescribed depth in the soil, and the shear wave velocity (V sh ) is measured The shear modulus at small strains is calculated

(6.33)

Idriss (1970) for sands

G

0.4 0.5 0.6 0.7 0.8 0.9

k,

Dr (%)

k,

484 415 353 304 270 235

30 40 45 60 75 90

235 277 298 360 408 484

=

kjW MPa

290

CHAPTER 6 A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

What's next . . .Now that we know how to ca lcula te the shear and bulk modu li, we can move on to determine strains, which we will consider next.

6.7 STRAINS FROM THE CRITICAL STATE MODEL 6.7.1 Volumetric Strains The total change in volumetric strains consists of two par . the t overable parr (elastic) and the unrecoverable part (plastic). W e can wnte n expression for the total change in volumetric strain as 6Ep = I.\E~

6£~

~

(6.35)

where the superscripts e and p denote ela ic'a~d ~Iastic, respecrively. Let us consider a soil sample that is isotr~ a consolid fe'd to a mean e~eclive stress p~ and unloaded to a mean effective stre¥ p~ .a s represe nt by~BC in Figs. 614a,b In a CD test, the soil will eld at q. Let us now consl(je r-~mal1 increment of stress, DE, Whic~ auses th leiK su rface to expand as sho n in FIg. 6 14a.

"'.

.

~

p ·.t,

, ,

". ~

~l V ; : ~I: - - --~~"--

~D

~I:

. :- ' : c" . ,'

Pi;,

' B:

- - - - - - - -- - - - - - - - - - - -

------------- •• •

-.---'": p"

-·; t

'

,

•

"!"

,

FIGURE 6 .14 Determ ination of plastic strains.

,E

Pc

,

6.7 STRAINS FROM THE CRITICAL STATE MODEL

The change in void ratio for this stress increment is ~e = 6.14b) and the corresponding total change in volumetric strain is 6£ p

=~ 1 + eo

=

(Ie c - eDI ) = -A-In PE 1 + eo

1 + eo

Po

leE - e 01

291

(Fig.

(6.36)

We get a positive value of llf~ becaus pansion) from E to D' , we compute. t The volumetric elastic straiH is,

F O

The volumetric elastic strain component is represented by ED'. That is, if you were to unload the soil from E back to its previous stress state at D, the rebound will occur along an unloadinglreloading line associated with the maximum mean effective stress for the yield surface on which unloading starts. The elastic change in volumetric strain from E to D is

(6.39)

bange is zero. Consequently, (6.40)

(6.41)

eviatoric strains, we will assume that the resultant plastic stW n ~l'e\ent, ~E P, for an increment of stress is normal to the yield surface ~~i~~.14a~~ ormally, the plastic strain increment should be normal to a plastic po.!en~'aJ. f~ction but ~e are assuming here that the plasti~ potenti.al funct!on he~d surface (YIeld functIOn, F) are the same. A plastIc potential functIOn is a\ 'hlar quantity that defines a vector in terms of its location in space. Classical plast~city demands that the surfaces defined by the yield and plastic potential Dfn'cide. If they do not, then basic work restrictions are violated . However, modern soil mechanics theories often use different surfaces for yield and potential functions to obtain more realistic stress-strain relationships. The resultant plastic strain increment has two components- a deviatoric or shear component, ~f~, and a volumetric component, M:~, as shown in Fig. 6.14. We already found ~f~ in the previous section. Since we know the equation for the yield surface [Eq. (6.41)] , we can find

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

the normal to it by partial differentiation of the yield function with respect to p i and q. The tangent or slope of the yield surface is dF

~

2p' dp' - p; dp ' + 2q :;:,

dq

=

(P~/2 -

q/M2

0

r.

(642)

~\.)

Rearranging Eq. (6.42), we obtain the slope as

dp '

~

pI)

(6.43)

From Fig. 6.14a, the normal, in (6.44)

which leads to (6.45)

(6.46)

FO

292

A sample of clay was isotropically consolidated to a mean effective stress of 225 kPa and was then unloaded to a mean effective stress of 150 kPa at which stress eo = 1.4. A CD test is to be conducted. Calculate (a) the elastic strains at initial yield and (b) the total volumetric and deviatoric strains for an increase of deviato ric stress of 12 kPa after initial yield. For this clay, A = 0.16, K = 0.05, ~s = 2SS, and v' = 0.3. Strategy It is best to sketch diagrams similar to Fig. 6.4 to help you visualize the solution to this problem. Remember that the strains within the yield surface are elastic.

293

6.7 STRAINS FROM THE CRITICAL STATE MODEL

Solution 6.6 Step 1:

Calculate initial stresses and Me p~

= 225

kPa, p~

= 150

kPa

225 Ro = 150 = 1.5

6 sin $';' Me = 3 - sin $';'

Determine the initial yield stresses. The yield stresses are the stresses at the~intersection of the initial yield surface and the effective stress dth~

F

Step 2:

6 sin 25S 3 - sin 25S

Y .

Pr = 150 +

q;

3

= 150 +

90

3

= 180 kPa

·, tic strains at initial yield.

Step 3:

. •

. .

C

Ytic volumetnc strams:

nE~

K

P;

0.05

180

= - - In - = - - In = 38 1 + eo P~ 1 + 1.4 150

X lO-

rternatively, you can use Eg. (6.38). Take the average value of p ' from p~ to p; to calculate K'. p ~ + p;. ' = -P av 2 -

=

150 + 180 = 165 kPa 2

K' = 3p'(1 + eo) = 165(1 + 1.4) = 7920 kPa K 0.05

n pe = C

np' = 180 - 150 = 38 x 10-4 K' 7920

4

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

Elastic shear strains

G = 3p ' (1 + eo )(l - 2v' ) = 3 2K(1

M"

=

q

Step 4:

I::.q

3G

=

165(1 + 1.4)(1 - 2 X 0.3) = 3655 kPa 2 X 0.05(1 + 0.3)

X

+ v ')

90 = 82 X 10- 4 3 X 3655

Determine expanded yield surface. After initial yield: I::.q

... I::.p ' = - 3

12 3

=

-

4 kPa

=

The stresses at E (Fig. 6.4) are p £: and

Step 5: "\-+----"--

+ 1.4

FO

294

Step 6:

184 In 180

=

15

X

10- 4

ains constant, we can calculate the elastic shear

12

I::.ce - I::.q _ 0

q -

3G - 3

X

- 11 X 10-4

3655 -

Calculate total strains. Total volumetric strains : £p = 1::.£; + Total shear strains:

£q = I::.£~

+

M~

I::.£~

= (38 + 10)10- 4 = 48

X

= [(82 + 11) + 16]10- = 109 4

EXAMPLE 6.7

Show that the yield surface in an undrained test increases such that

10- 4 X

10- 4

•

6.7 STRAINS FROM THE CRITICAL STATE MODEL

295

eB

eD ~--:--------:---------: -

.

t

t

p

P~rev

t t (p;J pcev p;

F O

FlfREE6.7

.A

tion given.

Solution 6.7 Step 1: Step 2:

diagram.

~

'~!"~~:<')

(1)

(2)

(3)

ull'.from the normal consolidation line BD, we get leD - eB

I = 'A In

Lp~~ceJ

(4)

Substituting Eq. (4) into Eq. (3) and simplifying gives

p~

=

(p') c prey

(P~rev) <1(1,-<) p'

•

What's next . . .We have calculated the yield stresses, the failure stresses, and strains for a given stress increment. In the next section, a procedure is outlined to calculate

296

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

the stress-strain, volume change, and excess pore water pressure responses of a soil using the critical state model.

6.8

CALCULATED STRESS-STRAIN RESPONSE

FO

4

You can predict the stress-strain response, volume changes, water pressures from the initial stress state to the failm t ess methods described in the previous sections. The re~ ired p~, eo, p~ or OCR, A, K, ~s> and v'. The procedures for a 'ven S' follows.

6.8.1 Drained Compression Te

(6.47) (6.48)

2.

e in'tial yield point and the failure point into a em ts. Small increment sizes «5 % of the stress y) tend to give a more accurate solution than

reconsolidation stress, p~, for each increment; that is, you are calculating the major axis of the ellipse using Eg. (6.4), which gives

6. Calculate

, Pc

7. 8. 9. 10.

11.

, =

P

q2

+ M2p'

(6.49)

where p' is the current mean effective stress. Calculate the total volumetric strain increment using Eg. (6.36). Calculate the plastic volumetric strain using Eg. (6.39). Calculate the plastic deviatoric strain increment using Eg. (6.45). Calculate the elastic deviatoric strain increment using Eg. (6.46). Add the plastic and elastic deviatoric strain increments to give the total deviatoric strain increment.

297

6.8 CALCULATED STRESS-STRAIN RESPONSE

12. Sum the total volumetric strain increments (£p). 13. Sum the total deviatoric shear strain increments (£,J 14. Calculate

+

3£q

£p

£p

3

£1 =

= Eq

+ '3

(6.50)

15. If desired, you can calculate = 2

q

+

3

p'

f

and

The last value of mean effective stress instability in the solution.

0'3

=

P

I

_

s~
I

~

3

F O

0';

about O.99pt to prevent

"-" -

6.8.2 Undrained Compressio

""

1.

(6.51)

Y n effective stress, calculate the following: reconsolidation stress after each increment of mean effec-

C

p~ =

(P') c prt!v

(P-~,cv) K/(I' -K) p'

here the subscript " prev" denotes the previous increment,p~ is the current preconsolidation stress or the current size of the major axis of the yield surface, and p' is the current mean effective stress. 7. Calculate q at the end of each increment from q

=

Mp'

' F

~ ~

p'

1

8. Calculate the volumetric elastic strain increment from Eq. (6.37).

298

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

9. Calculate the volumetric plastic strain increment. Since the total volumetric strain is zero, the volumetric plastic strain increment is equal to the negative of the volumetric e lastic strain increment; that is, M:~ = - Lle~ . 10. Calculate the deviatoric plastic strain increment from Eq . (6. 45) . 11. Calculate the deviatoric elastic strain increment from Eq. .46). 12. Add the deviatoric elastic and plastic strain increment total deviatoric strain increment. 13. Sum the total deviatoric strain increments. E eJ

=

ed'

14. Calculate the current mean total stress froJR the TS . the current value of q from Step 7. For f0 t ,

mber you know = rp~

+ q/3.

15. Calculate the change in excess pore rent mean effective stress from th

EXAMPLE 6.8

to conditions

6 sin 24° = 0.94 3 - Sin 24°

FO

4

+I 0.2S - O.OS) In

2S0 2 + O.OS

In 200 = 2.38

Each step corresponds to the procedures listed in Section 6.8.

Sample A, Drained Test Step 1: p~ =

(M2p~)

+

18p ~)

+

Y(M 2p ~

+ 18p~)2 - 36(M2 + 2(M 2 + 9)

9)(p~) 2

(0.94 2 X 2S0 + 18 x 200) + y'"7:(0:-:.9::-:4"2-X-2=-=S=0-+-----:-18O:--x-2=-=0:-::-0'"")2------::-36-:-(=0--;::.9~42,---+------:9,,--)(=2-;:-00::7"'
qy =

224 kPa 3(p~

-

p~)

= 3(224 - 200) = 72 kPa

299

6.8 CALCULATED STRESS-STRAIN RESPONSE

Step 2: , _ 3Mp~ . Pf-3-M'

~ ~.~~

PI = :

= 291.3 kPa, qf = MPI = 0.94 x 291.3 = 273.9 kPa

Step 3:

F O

200 + 224 = 212 kPa

p~, =

2

x (1 - 2 x 0.3) 0.3)

=

C5

Step 4:

Step 5:

4207 kPa

Let 6.p' =

Y £p p

C 6

(J

-

£q -

6

p

£p

= ~ In P' = (0.25 - 0.05) In 228 = 1.6 1 + eo 1 + 1.15 224

P;

- 16 X 10- 3

q

M2(p' - p~/2) -

M~

=

6q 3G

.

=

3

X

X

10-3

83.9 - 16 0.94 2 (228 - 262.9/2) - .

12 4207 = 1.0

X

10-

3

Step 11: 6£" = 6£~ + 6£~ = (1.0 + 1.6) X 10- 3 = 2.6 X 10- 3

X

10- 3

CHAPTER 6

A CRITICAL STATE MODEl TO INTERPRET SOil BEHAVIOR

Step U: (2.6 + 2.1) x 10- 3 = 4.7

X

10- 3

Step 13: £q

(t.£~)initi"1 + Mq = (5.7 + 2.6)

x

10- 3

Step 14:

The spreadsheet program and the stress-strai plo and Figs. E6.8a,b. There are some slight d', er ues shown above and the spreadsheet b Drained Case Given data

M

),.

0.25

K

0.05

12 kPa

q,~s

24

4207.0 kPa

eo

1.15

~.~---...::...

p~ p~

v'

0.3

(x

FO

300

<1-: 10- ' )

G (kPa)

0.0 0.0 1.6

240.0

48.0

119.9

307.6

2.0

10.7

1.6

4207.0

de: (x

10-')

(x

<1<. 10-')

Eq:::;

(x

~a£q

10-')

"'

(x 10 '~

0.0

0.0

0.0

0.0

5.7

5.7

5.7

6.6

1.6

4484.4

0.9

2.5

8.2

9.8

1.9

4563.8

0.9

2.7

10.9

13.2

2. 1

4643.1

0.9

3.0

14.0

16.9

2.5

4722.5

0.8

3.3

17.3

20.8 25.1

244.0

60.0

131.9

324.4

1.9

12.6

1.5

2.8

4801.9

0.8

3.6

20.9

248.0

72.0

143.9

342.2

1.9

14.5

1.5

3.2

4881.3

0.8

4.0

24.9

29.7

252.0

84.0

155.9

360.8

1.9

16.3

1.5

3.7

4960.7

0.8

4.5

29.4

34.8

256. 0

96.0

167.9

380.3

1.8

18.2

1.5

4.2

5040.1

0.8

5.0

34.4

40.5

260.0

108.0

179.9

400.5

1.8

20.0

1.4

4.9

5119.4

0.8

5.7

40.1

46.7 53.9

264.0

120.0

191 .9

421.4

1.8

21.7

1.4

5.8

5198.8

0.8

6.6

46.6

268.0

132.0

203.9

443.1

1.7

23.5

1.4

6.9

5278.2

0.8

7.7

54.3

62.2

272.0

144.0

215.9

465.4

1.7

25.2

1.4

8.6

5357.6

0.7

9.3

63.6

72.0

276.0

156.0

227.9

488.4

1.7

26.9

1.4

11.0

5437.0

0.7

11.7

75.4

84.4

280.0

168.0

239.9

512.0

1.7

28.6

1.3

15.1

5516.4

0.7

15.9

91.3

100.8

284.0

180.0

251 .9

536.2

1.6

30.2

1.3

23.7

5595.8

0.7

24.4

115.7

125.7

288.0

192.0

263.9

561.0

1.6

31.9

1.3

52.0

5675.1

0.7

52.7

168.3

179.0

291.0

201.0

272 .9

579.9

1.2

33.1

1.0

290.2

5745.0

0.7

290.9

459.3

470.3

6.8 CALCULATEO STRESS-STRAIN RESPONSE

301

300 250 _200

••

~

•

~-+--t-

150

100 SO 0

0

100

200

300

SOO

' 00

tL )< 10"'"

1,1

200

100 5

,

~

,

~

10

15

.:. 20 25

30

35 FIGURE E8. 8a,b

ample D,

ndrQincd Test

Step 1:

~~ ~ . R'

0.94 x 200 / 250 - 1 \ 200

~

94 kP,

Seep 2:

~ "p(

P

IP)

+ eo)(1 - 2v') 2K(1 + v')

=

0.~4

=exp 2.38 -251.15) _ L37 kPa ( A q, -_ Mp,' = x 137 '" 128.8 kPu

er -eo

)

3 X 200(1 + 1.15) x (1 - 2 x 0.3) 2 x 0.05(1 + 0.3) "" 3969.2 kPII

Slep 4:

(ll~}"""" Step 5:

Le1 tlp '

=

!1q _

== 3G - 3

94

x 3939.2

3 kPa .

First stress increment after the initial yield follows.

= 7.9 X IO- l

C

303

6.8 CALCULATED STRESS-STRAIN RESPONSE

Step 11: t:..q 97 - 94.1 t:..£e = = = 024 X 10-3 q

Step U:

b.£q

=

£~

Step 13: £q = £1 Step 14: P = p~ Step 15: b.u = p

3G

3 X 3969.2

b.£~

.

= (0.24 + 0.54) X 10- = 0.78 X 10- 3 = (b.£~)initial + b.£q = (7.9 + 0.78) X 10- 3 = 8.7 + q/3 = 200 + ¥ = 232.3 kPa - p' = 232.3 - 197 = 35.3 kPa +

3

X 10- 3

F

The spreadsheet program and the stress-strain plots are shown in the table be and Figs. E6.8c,d. Undrained Triaxial Test Given data

A

0.25

M

K

0.05

~s

24

Ro ecs

eo

1.15 200 kPa

p~ p~

250 kPa

v

0.3

eq

Y 188.0

253.9

.;~ '

,1£.

Ix 10-')

'( ~ ')9

0.6

=

L d£q Ix 10-')

E,

Ix 10-')

P IkPa)

dU IkPa)

0.0

0.0

0.0

0.0

3969.2

7.9

7.9

7.9

7.9

231.4

31.4

3939.5

0.2

0.8

8.7

8.7

232.3

35.3

3879.9

0.2

0.8

9.5

9.5

233.2

39.2

3820.4

0.2

0.9

10.4

10.4

234.1

43.1

C

0.7

3760.8

0.2

0.9

11.3

11 .3

234.9

46.9

0.8

3701.3

0.2

1.0

12.3

12.3

235.7

50.7

0.9

3641.8

0.2

1.1

13.4

13.4

236.4

54.4

0.4

1.0

3582.2

0.2

1.2

14.5

14.5

237.1

58.1

0.4

1.1

3522.7

0.2

1.3

15.8

15.8

237.7

61.7

-0.4

0.4

1.2

3463.2

0.2

1.4

17.1

17.1

238.3

65.3

-0 .4

0.4

1.3

3403.6

0.2

1.5

18.7

18.7

238.9

68.9

-0.4

0.4

1.5

3344 .1

0.2

1.7

20.3

20.3

239.4

72.4 75.9

-0.4

0.4

1.7

3284.5

0.2

1.9

22.2

22.2

239.9

121.1

-0.4

0.4

2.0

3225.0

0.1

2.2

24.4

24.4

240.4

79.4

122.5

-0.4

0.4

2.4

3165.5

0.1

2.5

26.9

26.9

240.8

82.8

266.4

123.7

-0 .4

0.4

2.9

3105.9

0.1

3.0

29.9

29.9

241.2

86.2

267.8

124.8

-0.5

0 .5

3.5

3046.4

0.1

3.7

33.6

33.6

241.6

89.6

149.0

269.1

125.9

-0.5

0.5

4.6

2986.8

0.1

4 .7

38.2

38.2

242.0

93.0

146.0

270.5

126.9

- 0.5

0.5

6.3

2927.3

0.1

6.4

44.7

44.7

242.3

96.3

143.0

271 .9

127.8

-0.5

0.5

9.9

2867.8

0.1

10.0

54.6

54.6

242.6

99.6

140.0

273.3

128.6

-0.5

0.5

21 .4

2808.2

0.1

21.5

76.1

76.1

242.9

102.9

137 .4

274.6

129.2

-0.4

0.4

636.6

2752.7

0.1

636.7

712.9

712.9

243.1

105.7

•

304

CHAPTER 6

A CRITICAL STATE M ODEL TO INTERPRET SOIL BEHAVIOR

What's next . ..W e have concenlrated on isotropic consolidat io n of soi ls and axisymmetr ic co nditions du ring shearing. The concepts and m et hodology deve loped are equa lly applicable to plane strain or other load ing cond itions. In nat ure, most so ils are one-dimensiona lly co nsolidated, ca lled Ko·consolidatio n. Next. we w ill cons ider Ko·consolidatio n using the critical state model.

6 .9

Ko·CONSOLlDATED SOIL RESPONSE When a soil is one-d imensionally consolidated. anisotrop' is conJcrred on the soil structure. T he soil properties are no longer the same · a,JJ d irections. We can use our simple cri tical state model to provide j~h inJO Ko-conso lidated soils although the model, as described, cann r-handle-amsotropy. We will assume that the yield surface is unaltered, that is remains an ellipse, for Ko·consolidated soils. The normal consolidation line for a ~nsordated soi l is hitted to the left of the normal consolidation l.i9 0 an i t (O lcally consoliC! ed soil (Fig. 6.15b) because p' for a Ko·consolfdated soil is

3:

ically cons9 lrdated'soil. ca ll that Ko is at rest ~

"

!ation path

" , \-~i---+--:- K.-eoosohdatlon pa th ~--:--"'I$()f.ople consoli(!anon path

, (bJ

FIGURE 6 .15 Comparison between a K,,-consolidated so il and an isotropicaUy conso lidated soil.

6.9 Ko-CONSOLIDATED SOIL RESPONSE

305

Let us compare the probable response of two samples, sample A and sample B, of a soil. Sample A is Ko-consolidated while sample B is isotropically consolidated. Both samples are normally consolidated to a void ratio e. The Ko-consolidated sample requires a lower mean effective stress to achieve the same void ratio as an isotropically consolidated sample (Fig. 6.15). The ESP from the isotropically consolidated sample is OB and for the Ko-consolidated sample it is OA (Fig. 6.15a). You should recall from Chapter 3 that the stress path for isotropic consolidation is qlp' = 0 and for Ko-consolidation is

UI

= (u;)o +

ui

=

U3 =

u)

=

(u;)o

F

!l.. _ 3(1 - Ko) pi - 1 + 2Ko

(6.52)

flu)

+ flu I

-

flu

Ko( uDo + fl<J3 Ko(uDo + flu) - flu

(6.53) (6.54) (6.55)

306

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

For a saturated soil, Skempton's coefficient B

= 1, and

from Eq. (5.44) (6.56)

Substituting Eq. (6.56) into Eq. (6.55) gives

Solving for

~(J' J -

~(J'3 ,

we obtain

G ~

(6.57)

At failure,

(6.58)

(6.59)

At failure ,

FO

(6.60)

The essential points are: 1. A Ko-consolidated sample of a soil is likely to have a different undrained shear strength than an isotropically consolidated sample of the same soil even if the initial confining pressures before shearing are the same and the slopes of the stress paths are also the same. 2. Failure stresses in soils are dependent on the stress history of the soil. 3. Stress history does not influence the elastic response of soils.

What's next . ..We have established the main ideas behind the critical state model and used the model to estimate the response of soils to loading. The CSM can also be used with results from simple soil tests (e.g., Atterberg limits) to make estimates of the soil strengths. In the next section, we will employ the CSM to build some

6.10 R.ELATIO NSHIPS BETWEEN SIMPLE SOIL TES TS, CR ITICAL STATE PARAMETERS. AND SOIL

307

rel atio ns h ips am o ng res ults from s imple soil tests, cri tica l state para m eters, a nd soi l strength s.

6 . 10 RELATIONSHIPS BETWEEN SIMPLE SOIL TESTS, CRITICAL STATE PARAMETERS, AND SOIL STRENGTHS Wood and Wroth (1978) and Wood (1990) used the critical Siale model to cor· relate results from Atte rberg limi t tests with various engi neering prope rties 2' fi ne·grained soi ls. We are goi ng to present somc::.!>f these correlations. T h'tS1 correlations are very uSeful when lim ited test datR"r<;. avai lable during the pre· liminary design of geotechnical systems or whgn you need 10 evaluate~ualily~ of test results. The correla tions Ulj l iZej.-~er~~ tent,lwhjCh at ~fis acc~le to 0.1 %. Most often wa ter content ~su llS are ned to ther nearest whole number and consequent ly sjgnifica~nt ~n~s an occur be.\teen~he asuial test results and the correlatjo!!t"e~ciallXr-~e involvin ex ?entlal? :srnce we are using CSM and inde~ proper~ies. ine rela li onships-onl pertain to fe· molded o r d Istu rbed soils. ~ ~

(1965) a nd Dumbleton

~ =R

~'

(6.61)

whe re R depends on actjyfty (~h apter 2) and varies between 30 and tOO, and the ~r.ipts PL and LL de note p~~tic limit and liquid lim it, respectively. Wood , pd Wroth (1 978~f~fo~end ~alue of R = 100 as reasonable for most soils. l~ recommen<~d v" ue oGtr;h. I,..' culled fro m the published data, is 2 kPa (the tg t data ShO~ vwt~~etwee n 0.9 and 8 kPa) and that for (S,,)PL is 200 kPa. Since most s its ar~ witBin the plast ic ra nge these recommended val ues place lower (2 k·Pa) QS up,p"fr (200 kPa) limits o n the undrained shear strength of dislur6ed br r~d.ed clays.

(

·e rtical Effective Stresses .,.,he, tquid and Plastic Limits WOOd (1990) used resu lts from Skempton (1970) and recommended that (O"; )LL '" 8 kPa

(6.62)

The test results showed that (<1;)I,..L va ries from 6 to 58 kPa. Laboratory and field data also sho wed that the undrained shea r strength is proportiona l to the vertical effective stress. T herefore I-

~ ( .~;~ ),-,-.--:R~(.~;)-,,-~-: 800 ::-k~P~·-'I

(6.6))

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

6.10.3 Undrained Shear Strength-Vertical Effective Stress Relationship Normalizing the undrained shear strength with respect to the vertical effective stress we get a ratio of

~= ~ (J"~

200

8 or 800

= 0.25

(6.64)

Mesri (1975) reported, based on soil test results, that s , cr;c good agreement with Eg. (6.64) for normally consolid soils

6.10.4 Compressibility Indices (A and Plasticity Index

and (6.65)

(6.66)

FO

308

t

the compression index increases with plasticity

In

FIGURE 6.16 Illustrative

<J;

graph of e versus

In~~ .

6.11 SUMMARY

309

6.10.5 Undrained Shear Strength, Liquidity Index. and Sensitivity Let us build a relationship between liquidity index and undrained shear strength. The undrained shear strength of a soil at a water content w, with reference to its undrained shear strength at the plastic limit, is obtained from Eq. (6.22) as (Su)w ( (Su)PL = exp G s (WPL"A.-

W»)

we get

F

Putting G s = 2.7, A. = O.6Ip in the above equation and recalling that

"IiII )

are significantly greater shs. The term sensitivity, ar strength to the re-

(6.68)

(6.69)

IlL=

1.2 'oglo 51 I

(6.70)

6.1 is chapter, a simple critical state model (CSM) was used to provide some .inSl'ght into soil behavior. The model replicates the essential features of soil beavior but the quantitative predictions of the model may not match real soil values. The key feature of the critical state model is that every soil fails on a unique surface in (q , p' , e) space. According to the CSM, the failure stress state is insufficient to guarantee failure; the soil must also be loose enough (reaches the critical void ratio). Every sample of the same soil will fail on a stress state

310

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

that lies on the critical state line regardless of any differences in the initial stress state, stress history, and stress path among samples. The model makes use of an elliptical yield surface that expands to simulate hardening or contracts to simulate softening during loading. Expansion and conine of the traction of the yield surface are related to the normal consolidat" soil. Imposed stress states that lie within the initial yield surta will cause the soil to behave elastically. Imposed stress states th at lie outsid he initia yield surface will cause the soil to yield and to behave elastopla I all ach i ' posed ac a C\ on an stress state that causes the soil to yield must lie on aiel unloadinglreloading line corresponding to the precons . atio stress associated with the current yield surface. The CSM is not intended to replicate all soils but to serve as a simple framework fr derstand the important features of soil b .

Practical Examples EXAMPLE 6.9

~

FO

0 capacity. Wha evels of water will cause the soil to yield t the end of the nsolida ' e owners propose to increase the sedon on top of the existing tank. However, the n 0 idation or soil tests. What is the maximum ou recommend so that the soil does not fail a mm? The dead load per meter height of the ro osed additi~ sectiem i !4 kN. The unit weight of the oil is 8.5 kN/m 3 .

StrategyR he so~ on dimensionally consolidated before the tank is placed on it. The 'Gao( from t e tank will force the soil to consolidate along a path that depends on the Ii CI stress increments. A soil element under the center of the tank will be subjected to axisymmetric loading conditions. If the tank is loaded quickly, then undrained conditions apply and the task is to predict the failure stresses and then use them to calculate the surface stresses that would cause failure. After consolidation, the undrained shear strength will increase and you would have to find the new failure stresses.

Solution 6.9 Step 1:

Calculate initial values. eo

= wGs = 0.55

2.7 = 1.49

X

K ~c =

1 - sin

K ~c

K~C (OCRy J 2 =

=

1

~

sin 26° = 0.56 0.56 X (1.2)1 /2 = 0.61

311

6.'1 SUMMARY

"1' =

G - 1

27-1

'Yw = '

_ s -

1 + eo

1 + 1.49

X

9.8

6.69 kN/m3

=

u~ o = -y'z = 6.69 x 3 = 20. 1 kPa

u;o = u ~c

K ~cu ~ o

= 0.6] x 20.1 = 12.3 kPa u~o

= OCR x

,

Po =

1 +

= 1.2 x 20.1 = 24.1 kPa 1 + 2 x 0.61 3 x 20.1

2K ~c,

3

qo = (1 -

u zo = K ~C )u ~o =

(1 - 0.61) x 20.1

=

=

14.9 kPa = 15 kPa

7.8 kPa

(p~ )o =

1 + 2K"c 1 + 2 x 0 56 3 0 u ;c = 3 '

(qcL

(1 -

=

M = c

K ~C )u~c =

x

(1 - 0.56)

F

The stresses on the initial yield surface are:

~

x 24.1 =

o~

24.1

6 sin
Step 2:

= 49 kPa

Y

9\ 1 =

="2

qs (

="2 =

C ~P

=

~q =

43.7 - 11 .8

(1

=

=

+ 2v)

2

J J 0.78qs =

2(1 + v) 1 ) [1 + (rlzfl1l2 + [1 + (rlzrp l2

1)

0.21qs

0.21 x 56

=

11.8 kPa,

224 kPa .

31.9 kPa

. . Slope of TSP = ESP dunng consolidatIon:

Step 3:

2

2(1 + 0.5) (1 + 2 x 0.5) - [1 + (5fl l /2 + [1 + (5)2P/2

~ur =

43.7 + 2 x 11 .8 3

r r

~C -~rlzf

lJ~r1 - C+ 1(%)2

qs (

a;r

350 = ~tpa 50.~ 7 =1tkPa

~q

A up

31.9 22.4

= -

=

1.42

Calculate the initial yield stresses and excess pore water pressure at yield.

FO

312

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

q

o

FIGURE E6.9

+e (p ' ) '+ ( t\)2o oPe

(qc)~ M2

=

0

+ (lO.6?/(l.03)2 = 0 and solving for P~ we get

Pe

tresses (point C, Fig. E6.9) are found from

qy

=

p; =

j0

Mp ~ Pc - 1

=

p~

p~ =

15 kPa,

32 15

~

1.03 x 15 _ . - 1 !:1qy

=

qy - qo

=

=

11.4 kPa

11.4 - 7.6

=

The excess pore water pressure at yield is !:1u =!:1p Y

!:1qy 1.42

= y

3.89 1.42

= -

=

2.7 kPa

The vertical effective stress and vertical total stresses are (!:1CJ;)y

= !:1p' + ~!:1q =

(!:1CJ z)y

=

(!:1CJ ~ )y

+ !:1u y

0 + ~ x 3.8 =

2.5 + 2.7

=

2.5 kPa

= 5.2

kPa

3.8 kPa

FO

Y

IC

CHAPTER 6

Step 8:

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

Calculate the equivalent surface stress and load.

. EqUivalent surface stress:

flqs

(tlCfz)F = - 0.78

=

53.4 -0 = 68.5 kPa .78

+

Surface load applied during consolidation:

350

Possible additional surface load: 68.5

= 68.5

Hy •.A = 350 =

Total surface load:

Step 9:

2813.2 + 3443.6

=

X

A

X

50.27

+5

X

2813.2 k

= 3443.

6256.8 kN

Find the additional height to bring the soil t consolidation. Let !J.h be the additional heigH

and !J.h

=

8.1 m.

Step 10:

75 =

6000 1 + 1.20

.. . p' = 40.8 kPa

flp'

FO

314

=

p' - pc' = 40-:8--"37.4

=

3.4 kR ,

flqs =

!lh

=

fl(J z 13.4 -08 = -8 = 17.2 kPa .7 0.7 17.2 8.5 = 2.0 m

Since the tank was preloaded with water and water is heavier than the oil, it is possible to get a further increase in height by (9.8/8.5 1)5 = 0.76 m. To be conservative, because the analysis only gives an • estimate, you should recommend an additional height of 2.0 m.

EXAMPLE 6.10

You requested a laboratory to carry out soil tests on samples of soils extracted at different depths from a borehole. The laboratory results are shown in Table E6.l0a. The tests at depth 5.2 m were repeated and the differences in results were about 10%. The average results are reported for this depth. Are any of the results suspect? If so, which are?

6.11 SUMMARY

315

TABLE E6.1 Oa W

W pl

Wll

Su

Depth

(%)

(%)

(%)

(kPa)

2.1

12

32

102

0.14

15

31

10

0 .12

15 17

29

10

35

13

22

35 47

0 .09 0.1

6.4

22 24 29 24 17

0 .07

8.1

23

12

27

85

0.1

3 4.2 5.2

h

F O ~

Solution 6.10 Step 1:

Step 2:

Y Calculated results

C

12 15

Su

(kPa)

h

Ip

32

102

31

10

0.14 0.12

20 16

"

h

(kPa)

0.50

0.12

0.56

0.096

20.1 15.0

29

15

29

10

0.09

14

1.00

0.084

24

17

35 22 27

35

18

0.39

0.108

47

0.1 0.07 0.1

9 15

0.44

85

0.054 0.09

17

13

8.1

23

Average

23.2

12 14.0

3.5

1.8

STD a

Su

WLl

(%)

'STD is standard deviation.

29.3 4.1

0.73

2.0 33.4 25.9 6.9

316

CHAPTER 6

A CRITICAL STATE MODEL TO INTERPRET SOIL BEHAVIOR

the borehole, the Su values predicted (=1 kPa) would be much lower than the laboratory results. You should repeat the tests for the sample taken at 6.4 m. The SIt value at 8.1 m is suspect because all the other values seem reasonable at these depths. •

EXERCISES Assume G s

=

2.7, where necessary.

Theory 6.1

Prove that

6.2

Prove that

6.3

FO

6.4

6.8

= 32°.

A fill of height Sm with "Isat = 18 kN/m 3 is constructed to preconsolidate a site consisting of a soft normaJly consolidated soil. Test at a depth of 2 m in the soil gave the following results: w = 4S% ,

q2

+ M2 = 0

(b) The fill is rapidly placed in lifts of 1 m. The excess pore water pressure is allowed to dissipate before the next lift is placed. Show how the soil will behave in (q , p' ) space and in (e, p') space.

EXERCISES

317

Problem Solving 6.9

The following data were obtained from a consolida tion test and K.

p'

• 06. 10

6.1l

(kPa)

25 1.65

50 1.64

200 1.62

400

800

1.57

1.51

1600 1.44

011

a clay soil. Determine A

8~

,, 1 .~~

400 1.46 ]

200 1.47

T he water con lent of a sample of salUraied soil at a ate
C

~d

(b) a CD les for the condiTions

6, 12

A CU triaxialte$t was conducted n ~o~ a!ly consolidated sa mple 0 turaled clay. The waler con ten t of tl1~--day was"5Qy :md Its undramed shear stren th was 22 kPa. Estimate the undnuned s~~ r tre ngth of a sample ofth da y if - 15. 1<1 : 30%. and the initial s tresse~~re the me the sample tha was ted. e parameters fo r the normally consol~ clay e A - 028. K = 0.06, an(f'
6 .13

Two samples 0 a soft cl~ re to be tes te I a conlle n o nal triax lnl apparatus Both samples were i~opjcal 11 COnSOJjda~ e er a ceqpressure of 250 kPa and then allowed to swe ~ back to amean effective str of 175' a. Sample A is to be tested under drained conditi0tls--w ile-sa)pple B is to be tesl d under u dra1ned conditions. Estimate the streSlistrain. v~ etns>slrain (sample A). 3 a excess re water pressure (sample B) responses = 0.15, K = 0.04. $~ = 26.r. eo =- 1.08. for the !w samples. The so Ra ramete r~ and II' = 0

.14

Dete ~in e and plol the stT strain.Jq lIersus rd and volume change ([" lIersus (1) responses for,a~ ove~~lidat~.~ru nd~r~ C D test. The soil p~ame ters are" = 0.11, 0.04, <\I.. ~ 2St... v':'=--()3.. eT O.92. p~ ~ 280 k Pa, and OCR ~ 8.

6. 6.16

Exe~14

Repeat for results ifh the Unj ain

an~.mdrained triaxial compression (CU) test and compare the riaxial extension test.

iso~OPicallY

consolidated to a mean effect ive stress of 300 kPa
Practical 6.11

A tank of diameter 5 m is to be located on a deep deposit of normally consolidated homogeneous clay, 25 rll thick. T he vertical Slress imposed by lhe ta nk at the surface is 75 kPa. Calculate the excess pore water pressure al depths of 2, 5. 10. and 20 m if the \ertical stress were TO be applied inslantaneously. T he soil pa rameters are A - 0.26, K = 0.06. and 41;' = 24°. The allerage water content is 42% and groundwa ter lellel IS at I m below the ground surface.

CHAPTER

'1--------------------------,

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS 7.0

ES

ED

75

25

INTRODUCTION

FO

able loadin 2. Settlement

• Estimate the settlement of shallow foundations • Estimate the size of shalJow foundations to satisfy bearing capacity and settlement criteria You will use the following concepts learned from previous chapters and from your courses in mechanics. • Statics • Stresses and strains-Chapter 3 • Shear strength-Chapter 5

Sample Practical Situation The loads from a building are to be transferred to the soil by shallow foundations. You are required to recommend the sizes of shallow foundations so that there is a margin of safety against soil failure and 318

1.1 DEFINITIONS OF KEY TERMS

FIGURE 7.1 Construction of a shallow courtesy of John Cernica .)

319

. (Photo

settlements of the buildihg arC-within tOlerabl ~ imi t The construction of a shallow foundation for a tanki( shQ.wn in Fig. 7. 1.

7.1

DEFINITION$\,O F Kj V TERM<S_..... '~-:-_ Found

·0

is a struct ure that Ir

smits

_ _ _ __

:~he underlying soils.

Foo,;ng a fou ndation consisting a sma. I slab for transmitting the struct ural load to the underl ying Footi ngs can be individual slabs supporting single olumns (Fig. 7.23) or oombiped to support two or more columns (Fig. 7.2b), or a I ng strip of co ncrete srab gfg. 7.2c, width B to lengt h L ralio is small , i.e., it approaches zeJ'6) supporti~ ,I~ad bearing wall, or a mat (Fig. 7.2d).

pfl.

l.al/ow JowftJa on is one In which the ratio of the embedmenl depth 10 the minimu, plan imension, which is usually the width, is Di E ::5 2.5.

fDJ ) is the depth below the ground surface where the base of

Embermnll depth the founda! n rests.

Ultimate bearing capacity is the maximum pressure that the soil can support. Ultimate net bearing capacity (q.. lj IS the maximum pressure that the soil can support above its current overburden pressure. A llowable bear;"g capacit)' or safe bearing capacity (qJ is the working pressure that would ensure a margin of sa fety agai nst collapse of the structure from shear failure. The allowabl e bearing capacity is usually a fraction of the ultimate net bearing capacity. Factor of safelY or safelY factor (FS) is the ratio of the ultimate net bearing capacity to the allowable bearing capacity or to the applied maximum ve rtica l stress. In geotech nical engineering, a factor of safety between 1.5 and 5 is used to calculate the allowable bearing capacity.

320

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Load bearing wall

~I'-------- L --------~

(a) Individual

(b) Combined

FIGURE 7.

FO

iting shear stress that should not be exceeded 10 ding during the design life of a foundation

.2

es a limiting deformation or settlement of a founill impair the function of the structure that it

QUESTIONS TO GUIDE YOUR READING 1. What are the ultimate net bearing capacity and the allowable bearing capacity of shallow footings? 2. What are the differences between the various methods for calculating a soil's ultimate net bearing capacity? 3. How do I determine the allowable bearing capacity for shallow footings? 4. What are the assumptions made in bearing capacity analyses? 5. What soil parameters are needed to calculate its bearing capacity? 6. What effects do groundwater and eccentric loads have on bearing capacity? 7. How do I determine the size of a footing to satisfy ultimate and serviceability limit states?

321

7.3 BASIC CONCEPTS

BASIC CONCEPTS 7.3.1 Collapse and Failure Loads In developing the basic concepts we will use a generic friction angle, ' , and later discuss whether to use ; or ~s' Failure in the context of bear' · capacity means the ultimate net bearing capacity. For dilating soils, failure corresponds to the peak shear stress, while for nondilating soils failure cor " sp ds to tlJl critical state shear stress. To distinguish these two states, we wffl fer t~e f Ilure load in dilating soils as the collapse load and reserve th erm j ure load for nondilating soils. Thus collapse load is the load at peak sn a nile failure load is the load at critical state. Collapse means sudde capacity of a soil. Let us consider two separate soil sand; the other is a deep deposit of t now take two similar square blocks of on the surfaces of the two soil ,.....J ~. ,.~~ ..... the same magnitude of ver their settlements using dl'S;Dl'lCe~l:!nt "'~.~ ....~~.

Collapse

Load

FO

7.3

c '" E

~

Q) if)

Failure (bl Loc al shear failure

Q"fl

V C

'"

E

~

Q) if)

(c) Punching

FIGURE 7.3

Failure mechanisms.

Failure

Load

FO

322

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDAIIQNS

Let us represent the concrete block on the dense sand layer as DS and that on the loose sand as LS. As we load the blocks, which block will settle more? Intuitively, and from your experience playing with sand, you may have guessed correctly that the loose sand would. Here is a list of expected responses as we load the blocks

3.

The pressure from the load on ea or failure is the soil's ultimate be . foundation so that the applied

7.4 COLLAPSE LOAD FROM LIMIT EOUILIBRIUM

323

within the soil layer below the base of the footing and extend laterally. The collapse in the dense sand is termed general shear failure , as the shear planes are fully developed . The failure in the loose sand is termed local shear failure, as the shear planes are not fully developed. Another type of failure is possible. For very loose soil, the failure surfaces may be confined to th urfaces of the rigid wedge. This type of failure is termed punching shear.

1.

2. 3. 4. 5.

The essential points are: Dense soils fail suddenly along well-defined slip plan'eS"r.esulting in a general shear failure. Loose soils do not fail suddenly and the slip planes are not well defined, resulting in a local shear failure. Very loose soils can fail by punching shear. More settlement is expected in loose 'Soils than in dense soils. The expected failure surface for gtneral shear failure consists of a rigid wedge of soil trapped beneath the footing bordering radial shear zones under Rankine passive zoneS.

FO

consider these bearing footing resting 0. a clay equilibrium meth o .

FIGURE 7.4

Circular failure mechanism.

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

ity, we will neglect the weight of the soil. Step 1 of the limit equilibrium method requires that we either know or speculate on the failure mechanism. Since we do not know what the failure mechanism is because we have not done any testing, we will speculate that the footing will fail by rotating about the inner edge A (Fig. 7.4), so that the failure surface is a semicircle with radius B Step 2 is to determine the forces on the failure surface. ong the circumference of the semicircle, there would be shear stresses (T] an ormal s resses (j~). We do not know whether these stresses are uniforrp:l is ' uted ver the circumference, but we will assume that this is so; othe ~ e, ha 0 perform experiments or guess plausible distributions. Since fa ' re oce n: d, the maximum shear strength of the soil is mobilized and therefo e ~ sjlear stresses are equal to the shear strength of the soil. Now w · ead 0 n6 ve to Step 3. The moment due to the normal force acting o~ se iClrci about A is zero since its line of action passes through A. :he mome ~e ~i ~fium equation is then P"

-.

'IT

x/=

0

.L

i

~

and the collapse load is

(7.1)

(7.2)

(7.3)

(7.4)

s on two variables, Rand 8, and as such there is a

FO

324

ap" = 4s"R('IT - 28)(R cos 8 - B) = 0 aR (2R cos 8 - B)2

P"

t OR

"e;~ , " t (In

FIGURE 7 .5

Circular arc failure mechanism.

(7.5)

7.5 BEARING CAPACITY EQUATIONS

325

2R cos 9 + TIR sin 9 - 2R9 sin 9) (2R cos 9 - B? = 0

(7.6)

and aP 4s R2(B ae = u

u

The solutions of Eqs. (7.5) and (7.6) are e = 23.2 and R = B e , that is, point o is directly above A . Substituting these values in Eq. (7.4) we obtain the collapse load as 0

Pu

=

5.52Bs"

(7.7)

This is a better solution because the collapse load i an Eq. (7.2) but we need to investigate other possible mechan i·sms wH ch yield yet a smaller value of P u . The exact solution to our prob em, sing r complex analysis than the limit equilibrium method, gives (7 .8)

FO

7.5

7.5.1 Terzaghi's Bearing Capacity Equations Terzaghi (1943) derived bearing capacity equations for a footing at a depth D f below the ground level of a homogeneous soil. For most shallow footings, the depth, D f , called the embedment depth, accounts for frost action, freezing, thawing, and so on. Building codes provide guidance as to the minimum depth of embedment for footings. Mat foundations can be embedded at a depth D f such that the pressure of the soil removed is equal to all or part of the applied stress.

FO

326

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

For such a case, the mat foundation is caJled a compensated mat or a raft foundation. Terzaghi assumed the following:

1. The embedment depth is not greater than the width of the B). 2. General shear failure occurs. 3. The angle

e in the wedge (Fig. 7.6) is cj>'.

4.

(7.9) (7.10)

(7.11) (7.12)

Sq

= 1 +

B, tan ,

L

s-y

=

1 - 0.4

B

L

(7.13)

ring capacity is the ultimate pressure that the soil can tings, the width B in Eq. (7.10) is replaced by the diameter D . For square and circular footings, BIL = 1; for strip footings, BIL = O. We will call Eqs. (7.9) and (7.10) the Terzaghi bearing capacity equations. These equations are not the original equations proposed by Terzaghi; others have modified Ground surface

•••• Dr

:\--.l

Terzaghi neglected shear along this . surface but Meyerhof considered it.

FIGURE 7.6

Failure surface assumed by Terzaghi.

7.5 BEARING CAPACITY eQUATIONS

it, especially Vesic (1973). The factors s-y and 1970).

Sq

327

were proposed by deBeer (1967,

7.5.2 Skempton's Bearing Capacity Equation

a on a TSA, for

Skempton (1951) proposed a bearing capacity equation, bas rectangular and square footings resting on a clay. Skempt n's

(7.14)

(7.15) (7.16)

=

(7.17)

5. 14su dele

FO

R

ult

(7 .18)

I N.., =

(Nq

-

1) tan(l.4<\>')

I

(7.19)

The shape (s), depth (d), and load inclination (i) factors are

I Se = 1 + 0.2 Z, I de +

1 + 0.2

!?;,

Sq

= S.., = 1 +

O.lKp

ZI

d q = d'Y = 1 + 0.1 VK;,

!?;

(7.20)

I

(7.21)

where K = tan2(450 + <\>') = _l_+_si_n--,<\>_' p 2 1 - sin <\>'

(7 .22)

328

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

For loads inclined at an angle e to the vertical in the direction of the footing width, the inclination factors are ic = iq =

r

(1 - 9~O i~ = ( 1- :'

r

For loads inclined at an angle e to the vertical in the directi length for a surface footing CDf = 0) , the inclination factor Me moto, 1987) are ic

where

0. 0

,

=

cos

l

e[ 1- (1 - ~ 2) sin e 'IT

the adhesion factor, is usually ~ t

(7.23)

ooting Kou-

(7.24) 2

or

7.6 CHOICE OF B EQUATIONS ~D F The major G H . en es among the va 'ous beari g capacity equations are the values of N~ an the geometric ctors (s pth and load inclination factors). Fo values 0 < 35°, the iff . rences between N~ among the bearing capacity eq ations are not practi fic nt. However, for <1>' > 35°, the differences ntial. Fort ely, st so shave <1>' < 35° so that for many practical e s. p ~ms there wo d not be differences between the predictions of the var. ·o us method . h cwera h ice is the value of <1>'. Should it be ~ or ~? The beann apa 'ty e uations have been derived based on the existence of slip plan s. Slip pes, ow ever, have been only observed in dense sands and heavily ove solida ed clays. Most foundations are constructed on compacted soils Cexception inc nde lightly loaded foundations, for example a foundation for a single story house, on normally consolidated clays or medium sands). It is expected that for these compacted soils slip planes would develop and the appropriate value of <1>' to calculate the collapse load should be <1>;. However, the attainment of <1>; depends on the ability of the soil to dilate, which can be suppressed by large normal effective stresses. Since neither the loads nor the stresses induced by the loads on the soil mass are certain, the use of <1>; is then uncertain. Slip planes, if they develop, do not occur suddenly. They are likely to develop progressively at different levels of stresses. If you reexamine the slip plane for our shallow footing on dense sand (Fig. 7.3), you would realize that the normal effective stress at C is much larger than at E. Indeed, at E, the normal effective stress is zero. The implication is that while the peak shear stress is being mobilized near E, it has already been surpassed at C; that is, the critical state shear stress I

FO

O

~

~

7.7 ALLOWABLE BEARING CAPACITY AND FACTOR OF SAFETY

329

FO

ext . . . So far, W have st tlied how to calculate the ultimate net bearing . If a footin IS desi e sed on the calculated ultimate net bearing capacity inent nger of failing. And, of course, we do not want that to happen. We then have o.b in a design bearing capacity value called the allowable , earing capacity, qa' The a lowable bearing capacity allows a margin of safety against collapse or failure. We will examine how to calculate the allowable bearing capacity next.

7.7 ALLOWABLE BEARING CAPACITY AND FACTOR OF SAFETY 7.7.1 Calculation of Allowable Bearing Capacity The allowable bearing capacity is calculated by dividing the ultimate bearing factor capacity by a factor, called the factor of safety, FS. The factor of safety or

330

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

TABLE 7.1

Allowable Bearing Capacity (UBC, 1991)

Soil type

Sandy gravel/gravel (GW and GP)

96

Sand, silty sand, clayey sand, silty gravel (SW, SP, SM, SC, GM, GC) Clay, sandy clay, silty clay, and clayey silt (CL, ML, MH, and CH)

safety factor is intended to compensate for assumptio bearing capacity equations, soil variability, inaccurate s of loads. The allowable bearing capacity is (7.25)

(7.26)

FO

Capacity Values

~

7.8

EFFECTS OF GROUNDWATER For all the bearing capacity equations, you will have to make some adjustment for the groundwater condition. The term "VDf in the bearing capacity equations refers to the vertical stress of the soil above the base of the foundation. The last term "VB refers to the vertical stress of a soil mass of thickness B, below the base of the footing. You need to check which one of three groundwater situations is applicable to your project. Situation 1: Groundwater level at a depth B below the base of the footing. If the groundwater level is at a depth B below the base of the foundation, no modification of the bearing capacity equations is required.

7.8 EFFECTS OF GROUNDWATER

(a)

Groundwater within a depth B below base

FIGURE 7.7

Groundwater effects.

FO

R

EXAMPLE 7.

Step 1:

Calculate the bearing capacity factors and geometric factors for Terzaghi's method.

FIGURE E7.1

331

FO

332

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETILEMENT OF SHALLOW FOUNDATIONS

Use a table for ease of calculation and checking. Parameter

Nq

=

2

(

45° +

~')

33,3

Nq

-

1

32,3

N,

=

2(Nq + 1) tanq,'

48,0

Sq =

s,

Step 2:

e~tanq,' tan

1+

B

1,70

L tan q,'

= 1 - 0.4

18.4

0,6

B

L

Calculate the ultimate net be

I

0.5 x 18 x 2 x 22.4 x 0.6

o able bearing capacity of the soil is about S = 3 and using
itimate et bearing capacity for Example 7.1 using
Solution 7.2 Step 1:

Calculate bearing capacity numbers and, shape and depth factors. N q = e'IT tan <1>' tan 2 ( 45°

Nq

-

1 = 33.3 - 1

Ny

=

(Nq

-

=

+ 4>'/2) =

e'IT tan 35- tan2 ( 45"

+ 35°/2) = 33.3

32.3

1) tan(l.44>')

=

32,3 tan(1.4

X

35°)

=

37,2

7.8 EFFECTS OF GROUNDWATER

Kp = tan 2 (4SO Sq

+ <1>'/2)

= s., =1 + O.lKp

d q = d., = 1

=

tan 2 (45°

B

L=

+ 0.1 vK;:

D

Bf

+ 3S0/2)

333

3.7

=

2

1 + 0.1 X 3.7 X

2:

= 1 + 0.1 VD

2:

= 1.37

1

= 1.09

FO

R

Step 2:

N q - l)sq dq + 0.5"Y'BN.,s., d., = 8 X 1 X 32.3 X 1.37 X 1.09 + 0.5 = 1323.7 kPa

X

8.2 x 2 x 37.2 x 1.37 x l.09

Cd) Groundwater level at 1 m below the base. In this case, the groundwater level is within a depth B below the base and will affect the last term in the bearing capacity, where you should use 'I' B = "YsatZ

+ "y'(B

- z) = 18

x

1

+ 8.2 x

(2 - 1) = 26.2 kN/m 3

Thus, quIt

= "YDf(Nq - l)sq dq + 0.5('1' B)N.,s., d.,

=

18 x 1 x 32.3 x 1.37 x 1.09 + 0.5 x (26.2) x 37.2 x 1.37 x 1.09

= 1596.9 kPa

CHAPTER 7

Step 3:

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Compare results. We will compare the results by dividing (normalizing) each ultimate net bearing capacity by the ultimate net bearing capacity of case (a).

Groundwater level at (b) Ground surface

~X100 (qultla

851 - - x 100 = 46% 1868

(c) Base (d) 1 m below base

• EXAMPLE 7.3

FO

334

• EXAMPLE 7.4 Determine the size of a rectangular footing to support a load of 1800 kN. The soil properties are c!>; = 38°, c!>~s = 32°, and "'!sat = 18 kN/m 3 • The footing is to be located at 1 m below the ground surface. Groundwater level is 6 m below the ground surface. Strategy You have to select a method. A good choice is Meyerhof's method. Neither the footing width nor the length is given. Both of these are required to find qa. You can fix a length to width ratio and then assume a width (B). Solve

7.8 EFFECTS OF GROUNDWATER

335

for qa and if it is not satisfactory [qa ~ (aa)ma,J then reiterate using a different B value. We will use¢' = ¢~s and a factor of safety of 1.5.

Solution 7.4 Step 1:

Calculate bearing capacity numbers and shape and epth factors . Assume, LIB = 1.5; that is, BIL = 0.67, and B = .5 m.

FO

R

Step 2:

dq

D

= d'l = 1 + O.I~!:::.J.B

=

,~1

1 + 0.1 v3 .25 - = 1.13 1.4

The new allowable bearing capacity is _ -yD[(l\lq - l)sq dq + 0.5-yBN..,s'l dq FS

qa -

18

+ 'YD[

x 1 x 22.2 x 1.22 x 1.13 + 0.5 x 18 x 1.4 x 22 x 1.22 x 1.13 1.5

+ 18 x 1 (
=

640 kPa

Applied load 1800 Area = 1.4 x (1.5 x 1.4) = 612 kPa < qa'

The footing size is then 1.4 m

X

2.1 m.

•

336

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

EXAMPLE 7.5

Using the footing geometry of Example 7.1, determine qa with the load inclined at 20° to the vertical along the footing width, and <1>' = ~s ' Assume the groundwater level is at ground surface and FS = 1.5.

Strategy You need to use Meyerhof's equation for incl'ne ing capacity numbers and shape factors remained unch a- ed. culate the depth factors and inclination factors.

A~~ -"=~:>",

bearto cal-

Solution 7.5 Step 1:

Calculate the inclination factors

Step 2:

FO

•

'7.9

Theoretical solutions for eccentric (off-centered) loads are very complicated. Meyerhof (1963) proposed an approximate method to account for loads that are located off-centered. He proposed that, for a rectangular footing of width Band length L , the base area should be modified with the following dimensions: B'

=

B - 2eB

and

L'

=

L - 2eL

(7.27)

where B' and L' are the modified width and length, and es and eL are the eccentricities in the directions of the width and length, respectively (Fig. 7.8) . From your course in mechanics, you should recall that (7.28)

FO

7.9 ECCENTRIC LOADS

337

p

y

~P

T -.-.~.~-r 1 : , 8

L-

··X

~ ~->1.1

FIGURE 7.8

Footing subjected to a vertical load and

where P is the vertical load and My and axes, respectively, as shown in Fig. 7. eccentricitye (moment = Pe) are (f a (7.29)

where I is the second to the outer edge, A is t a rectangular sectfen,

aXIs. tresses along the X axis are
=

~- ~=

:L (1 - 6;8)

(7.30)

Let us exam~e 0: m' If eB = B/6 or eL = L/6, then (]min = O. If, however, eB > B/6 or e L > L7 , then (] min < 0, and tension develops. Since the tensile strength of soil is approximately zero, part of the footing will not transmit loads to the soil. You should try to avoid this situation by designing the footing such that e B < B/6 or eL < U6. The bearing capacity equations are modified for off-centered loads by replacing B with B'. The ultimate load is (7.32)

EXAMPLE 7.6 Redo Example 7.2 using Meyerhof's method. The footing is subjected to a vertical load of 500 kN and a moment about the Y axis of 125 kN . m. The groundwater level is 5 m below the ground surface. Calculate the factor of safety.

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Strategy Since we are only given the moment about the Y axis, we only need to find the eccentricity, eB' The bearing capacity factors are the same as those in Example 7.2.

Check if tension develops.

Step 2:

B 6

Step 3:

Step 4:

Step 5:

FO

338

y

1m

f+:-B=2m~ + .

.

.

'1'

4m

.~ FIGURE E7.6

I

1 .10 BEARING CAPACITY Of LAYERED SOILS

Step 7,

339

Calculate the factor o f safety . FS =

q~h (0". ).... , -

"" "'/ 0/

1567 = 7.8 219 - 18 x 1

•

7 . 10 BEARING CAPACITY OF LAYERED SOILS (

EXAMPLE 7 .7 The soil profile at a site is shown in Fig. E7.7. A square footing 2.5 m wide is located at 1.5 m below ground level in the stiff clay. Determine the safety factor ror sho rt-term loading for an applied load of 1000 kN.

Strategy Skempton"s equation is an appropriate choice for days, especia lly stiff clays. Check the factor of safety of the stiff clay assuming that the soft clay layer does not exist. We can then use an artificial footing on top of the sort clay and calculate the faclOr of safelY.

340

CHAPTER 7 BEARING CAPA CITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

P .lOOO ~N S~I 'f

I

day

I•• 120 ~Pa . Y.... 20 kNlmJ

". Soft clay s• • 2O kPa.

r... " 18 kNfml

Dense sand

FIGURE E7.7

Solution 7 .7 Step L:

Step 2:

17 8 kPa

• What's next . ..The size of many shallow founda ti ons is governed by settlement rather than bearing capacity considerat ions. That is, serviceability limit slate governs the design ra t her th an u ltimate limit state. Next, we will consider how to determine settlement for shallow foundations.

7.11

SETTLEMENT It is practically impossible lO prevent settlement of shallow foundation s. At least,

elastic settlement will occur. Your task as a geotechnical engineer is 10 prevent

341

7.11 SETTLEMENT

TABLE 7 .2

Serviceability Limit States Examples

Architectural damage (damage to appearance)

Tilting of structures (chimneys, retaining walls) and cracking in walls

Loss of serviceability

Cracked floors, misalignment of machinery, dis ocation of pipe joints, jammed doors and windows

Structural damage (collapse)

Severe differential settlement of footings columns and overstressing beams

FO

Serviceability limit state

r-1-j

'----1

f+H

'----1

U-rrutl (a) Uniform settlement

FIGURE 7 .9

(b) Tilt or distortion

Types of settlement.

(c) Nonuniform settlement

TABLE 7 .3 Limits

Distortion

Type of damage

e

o.Istortlon, . 0

Architectural

300 Structural

150

342

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

TABLE 7.4 RValues and Maximum Allowable SettlementS Foundation type Soil type

Clay

Values

R Pmax (mm)

Sand

R Pma, (mm)

Isolated footing

Rafts or mats

22,500 75 15,000 50

30,000 100 18,000 60

"The original R and Pm.. values were for use in English units. The author converted these values for SI and adjusted them for ease of use in practice.

o rrelate distortion to the 956) propos that the max·

(7.33)

FO

and soil types. Pm ax suggested

The essential points are: 1. Distortion caused by differential settlement is crucial in design because it is responsible /or cracking and damage to structures. 2. The distortion limits wen observed for old structures and may not be applicable to modern structures. 3. To estimate the totaf maximum allowable settlement, multiply the appropriate R values in Table 7.4 by the distortion value.

section, we are going to discuss methods to calculate

settlement of foundations.

7.12

SETTLEMENT CALCULATIONS The settlement of shallow foundations is divided into three segments-immediate or elastic settlement, primary consolidation settlement, and secondary consoJjdation settlement (creep). We have already considered elastic settlement (Chapter 3) and consolidation settlement (Chapter 4). However, we have to make some modifications to the methods described in those chapters for calculating settlement of shallow foundations. These modifications are made to the method of calculating elastic and primary consolidation settlements.

7.12 SETTLEMENT CALCULAT10NS

343

7.12.1 Immediate Settlement We can use the theory of elasticity to determine the immediate or elastic settlement of shallow founda tions. In the case of a uniform rectangular nexible load, we can use Eq. (3.84). However, the elastic equations do not account for the shape of the footing (not just LIB ratio) and the depth of embedment, which significantly intluence settlemen t. An embedded foundati n has the following effects in comparison with a surface footing: """- 1 1.. Soil stiffness generally increases with dePtl ~ ~ ootm oads will be t ransmitted to II stiffer soil tha n the surface Th.is :l!!l result in smaller seu lemcnts. .' 2. Normal stresses from the soil abov~ ot! g leve l have been shown (Eden, 1974; Gazetas and Stoke '991) to~duc the seulement by providing increased confinement on e de f1 alfspace. This is called the trench effect or embedment effec .

Jil

3. Part of the load on the ~ay~ be transmitted through the side walls depending on til a mou~t'f'f sJfear res i sla~ bilized at the soilodatlo n of part of the loaao }· resistance wall interface. The aceo reduces the vcr~a l settleme IThis has beeb called the sme wall -contact , • effect.

spc

Gazeta~(1985

considered an arbitrari!Y. shar .d rigid footing embedded in a dee p homogeneous soil (Fig. 7. 0 an pwposed the following equation for the elastic

ttle7t:·r-_---,~~~ c-'''',,-- , (7.34)

•

Plan

~-2 8-1

~ect ion

I

0,

1

\-}

p

1

SIde wllll--soil conlilct height ~

...-'

s.,,,•••'"

Elastic lIomopneous 1lalI.o$pIICl!

7.10 Geometry to calculate elastic settlement of shallow footings. (On behalf of the Institution of Civil Engineers.)

FIGURE

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

where ,P is total vertical load, Eu is the undrained elastic modulus of the soil, L is one-half the length of a circumscribed rectangle, Vu is Poisson's ratio for the undrained condition, and /l-s, /l-e mb, and /l-wall are shape, embedment (trench) , and side wall factors given as I fJ-s = 0.45 ( ~ )

fJ-emb

= 1 - 0.04

D[ 1 B f

-0.38 I

+

3'4

(A4L

G

(7.35)

(7.36)

(7.37)

(7.38)

(7.39)

FO

344

TABLE 7.5 Values of Ab/4L2 for Common Footing Shapes

Footing shape

Square Rectangle Circle Strip

Ab 4L'

,

BIL 0.785 0

7.12 SETTLEMENT CALCULATIONS

345

deviatoric stress equal to one-half the maximum shear strength. However, for immediate settlement it is better to determine Eu over the range of deviatoric stress pertaining to the problem. In addition, the elastic modulus is strongly dependent on depth while Eqs. (7.34) and (7.38) are cast in terms of a single value of Eu' One possible solution is to divide the soil into sublayers and use a weighted harmonic mean value of Eu (Chapter 4). The full wall resistance will only be mobilized if suffici nt settle ent occurs. It is difficult to ascertain the quality of the soil-wall ad sion. Conse ently, you should be cautious in relying on the reduction of ettle ent f, I ng from the h n f-lwall = 1 and wall factor. If wall friction and embedment are f-lemb = l. Equations (7.34) and (7.38) strictly a term loading. For long-term loading in Hn.esoils, you should use £' and v' instea of u an VU' EXAMPLE 7.8

Strategy You have (7.38). The side all ef really is no w

Ab

=4

wall

x 6 =

-

FO

R

(ProOf:

FIGURE E7.8

~2 =

~

= 3 m, B

=

~

=

= 1, since there

2m

2: : z:- = Z= ~ = 0.67)

CHAPTER 7

Step 3:

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Calculate the immediate settlement. _

2)

P

P. - E L (1 -

_

VII i-Lsi-LembfLwall -

It

x

4000 2 15 OOOX 6 (1 - 0.45 ) ,

0.52

x

0.94

x

•

1 = 0.018 m = 18 mm

EXAMPLE 7.9

Determine the immediate settlement of the foundatio undrained elastic modulus varies with depth as shown in nd w1dth (2B) of a circumscribed rectangle. The undrained elastic v ri ith depth, so you need to consider the average value of Eu for e c of t e layers an4 hen find the harmonic mean. You also need to nd so aramete~

Solution 7.9

. .

~

Step 1:

ctangle.

FO

346

Layer 1

8000 kPa

-------------+----t-------"<.....-...,.

Layer 2

10,000 kPa

8m

_ _ _ _ _ _ _--'----_ _ _ _---L._ _---L.._ _ _ _ _ _ _- - - '

Eu

FIGURE E7.9

30,000 kPa

347

1,12 SETTlEMENTCALCULATIQNS

Slep 2:

Determine Ewo Layer I £~

4

al base level - 8" x 8000 = 4000 kPa;

(£..)••~ -

4000 + 8000 2

£~

at bottom of layer = 8000 kPa

"" 6000 kPa

Layer 2

E" al lOp of layer = 10,000 kPa:

E.. 31 ~ tom of ayer - 30,000 kPa

C) = 10,000 + 30,000 = 0 ( I;. .. " l 2 2,

Step 3: Find the weighted harmonic

"_

lI.e·a

0l8n Ew' ,000) = 10,667

ke~

Step 4: Ab

'1

4L l - 4x 62

05 -

Stcp 5:

176

A:~ 72 - 2'> ,11-011 Calculate

o

p,

I I1.e1mm~ iat settlement.

-~E~I -

v!

\~~(1

rm,,~_·.I'

-

0.45') x 0.52 X 0.47 X 0.74 - O.oII m - II mm •

7 .12.2 Primary Consolidation Settlement The method described in Chapter 4 ca n be used to calculate the primary consolidation settleme nt of clays be low the footin g. However. these equations were obtained for one-dimensional consolida tion where the lateral strain is zero. In practice, lateral strains are significant except for very thin layers of clays or for situ ations when t he ratio of the layer thickness to the lateral dimension of the loaded area IS small (a pproaches zero). We also assumed that the initial excess pore water pressure is eq ual to the change in applied stress at the insta nt the load is applied. Theoretically, this is possible if the la teral stresses are equal to the vertical stresses. If the lateral strai ns are zero, then under undrained condition (a t the insta nt the load is applied), the verl ical setllemen t is zero .

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Skempton and Bjerrum (1957) proposed a method to modify the onedimensional consolidation equation to account for lateral stresses but not lateral strains. They proposed the following equation: {HO (7.40) (Ppc)SB =)0 mu flu dz where l1u is the excess pore water pressure and Ho is the thi

(7.41)

By substituting Eq . (7.41) into Eq.

(7.42)

FO

348

C

E

:;::;

'"

(/)

- - Circular footing - - - - Strip foot ing Overconsol idated

Very sensit ive clays

I

--+----4<- Normally consol idated

O.2L-----~L-----~------~------~----~~------~

o

0.2

0.4

0.6

0.8

1.0

1.2

Pore-pressure coefficient. A

FIGURE 7.11 1963.)

Values of

fJ.SB

for circular and strip footings. (Redrawn from Scott,

7.2 SETTLEMENT CALCULATIONS

349

V Ahr, where A is area of the rectangle or the square. Equation (7.42) must be used appropriately. It is obtained from triaxial conditions and only applies to situations where axial symmetry occurs, such as under the center of a circular footing. EXAMPLE 7.10 Determine the primary consolidation settlement under in Fig. E7 .1O using the Skempton-Bjerrum method.

Strategy First you have to calculate the one-di tion settlement and then determine f.LSB from the c finite, you should calculate the vertical Appendix B.

Solution 7.10 Step 1:

Step .

FO

R

D =

A

= 0.2;

from Fig. 7.11 ,

j§; fJ.SB

= 1.13 m

= 0.41

ry consolidation settlement. x 72 x 4 x 0.41 = 0.012 m = 12 mm

FIGURE E7.10

•

350

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

What's next . . . Often, the recovery of soils, especially coarse-grained soils, for laboratory testing is difficult and one has to use results from field tests to determine the bearing capacity and settlement of shallow foundations. Some of the field methods used for coarse-grained soils are presented in the next section.

7.13 DETERMINATION OF BEARING CAPACIT AND SETTLEMENT OF COARSE-GRAINED SOJl-: FROM FIELD TESTS, ~ e nt f . ...... .... ..Jte

7.13.1 Bearing Capacity and Settle Standard Penetration Test (SPT) It is difficult to obtain undisturbed sam

. These are (7.43)

(7.44)

FO

.

(7.45)

the groundwater table, D f is the footing depth, and B is e depth of the groundwater level is beyond B from the

IN

cor = CNcW N

I

(7.46)

The allowable bearing capacity for a shallow footing (qa) is I

qa

= 0.41Ncor Pa kPa

I

(7.47)

where Pa is the allowable settlement in mm. In practice, each value of N in a soil layer up to a depth B below the footing base is corrected and an average value of N cor is used in Eq. (7.47). Meyerhof (1965) proposed that no correction should be applied to N values for the effects of groundwater as these are already incorporated in the measurement. Furthermore, he suggested that qa calculated from Eq. (7.47) using N cor = cNN be increased by 50%.

7.13 DETERMINATION OF BEARING CAPACITY AND SETTLEMENT OF COARSE ·GRAINEO SOilS

351

Burla nd and B urbridge (1985) did a statistical ana lysis of settlement records from 200 foo tings located in sa nd and in gravel. Th ey proposed the following equa tion (or the se ttlement of a footing in a normally consolidated sand at the end of construction : (7.48)

wh", P is the sett lement (mm).

f.

1.25L1~ .

(

LIB + :'i~

= Shape factor =

II

= (Holz1)(z - Ho/z)) is a correction factor if the t i.!:knes§ ( Ho) of the sand stratum below the footi ng base is less than N n8uen e de th Zit Z is depth (m) from the ground surface, a~ is the ve r tic~ re ~ pp 1 by the footing o r allowable bearing capacity (kPa), B and L are"fhe idtn and length of the foo ting (m), respectively, ~~.. .~ ..

Ie = CDmprej

;l

ty I

X

I.7J

(

= N l4

7.49

)

and N is the uncorrected N val The mfl ue nce depth is the dC,plb be low the footing that wi ll influence the sen mf nt and bea\!.ng capacity. If""N increases with depth or N is approxl tcl.,v nstant, theJ rtfi ence dep"th is taken as 4, == Bo.163 . If N tenc;rt;b ecreas Wl~ depth, the il'ifl,uenc eptllAs I, = 2B . If the sa n is overco~~idated.

4

(7.50)

(7.51)

'fhe procedure fo the

Bu rland~ B ur b ri d ge

method is as follows:

[)elermine !pJ\nftue~Pfh 41' Fi nd Ihe.;"'r:age,.-,,. r-vall!.e wit hin the depth 41 below the footing.

ca lculft}.~rOtn E gJ(7.49).

4. D rmine p~r~e a ppropriate equation [Eq. (7.49) or (7.50) or (7 .51)] or, If is sper ed, you can determine a". EXAMPLE

., (

T he SPT results at various de pths in a soil are shown in Ta ble E7. lla . TABLEE7.'1. Depth (m )

0.6

0.9

1.2

I.'

2.1

2.7

3

N (blow s/h I

2S

28

33

29

28

29

31

33 3S

4.2 41

Determine the allowable beari ng capacity for a footing loca ted at 0.6 m be low the surface. The tolerable sett lement is 25 mm. T he groundwater level is deep and ils effects can be neglected.

Strategy The question that arises is what value of N to use. If the size of the footing were known , then you cou ld estimate the thickness of the soil below the

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Solution 7.11 Step 1:

TABLE E7. 11b Depth (m)

~

Vertical effective stress (kPa)

-~

0.6 0.9

9 .6

1.2 1.5

2.1 2.7 3 3.3

47 49

•

FO

352

t e Burland-Burbridge method for a footing 3 m X

Strategy You ave to determine whether the sand is normally consolidated or overconsolidated. No direct evidence is provided to allow you to make a decision as to the consolidation state of the sand . One way around this problem is to use Table 5.3 to make an estimate of the consolidation state.

Solution 7.12 Step 1:

Determine the consolidation state and find Z! . Within a depth equal to B (3 m), the average N value is 29. From Table 5.3, the sand can be classified as medium in the range 1030). A reasonable estimate of the consolidation state is normally consolidated.

eN

z\ = BO. 763 = 3°·763 = 2.3 m

7.13 DETERMINATION OF BEARING CAPACITY AND SETTLEMENT OF COARSE-GRAINED SOILS

353

Find an average N for a depth 2.3 m below the base. Average N value over a depth of 2.3 m below the base is 29. (Note: 2.3 m below the base is equivalent to a depth of 2.9 m, so use the N values up to 3 m.) Step 3: Calculate IC' Step 2:

Step 4:

Calculate qQ' L

4

B = 3'

= 1.33;

p

(J'a

=

•

ff 1 BO.? [ c j'

7.13.2 Settlement fro

strains; that is

FO

R

(7.52)

e

c{

=

hfactor 1- 0.5 =

Creep factor

= 1.0 +

(J"

_

2=

1.

qnet

0.5

I~ I, 0.1

A loglo

!3 is cone factori = 2.5 for square footing (axisymmetric condition), !3 = 3.5 for strip footing (plane strain condition)], qne t is the net footing pressure in kPa (applied stress minus soil pressure above the base of footing) , (J'~ is the vertical effective stress in kPa , t is time in years (t 2: 0.1), A is an empirical factor taken as 0.2, LlZI is the thickness of the ith layer, (lco); is the influence factor of the ith layer given as: Axisyounetric leo

l eo =

=

0.1 +

C.;5B) (~)

0.5 - (3~)(~)

z :5 -B B 2

for -

for 2B

(7.53)

z > -B

2: -

B

2

(7.54)

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETILEMENT OF SHALLOW FOUNDATIONS

Plane strain

1

Ico

I

=

= co

0.2

+ (3.3 3B)

0.5 - ( - 1 ) -Z 6B B

(~)

for i.

B

(7.55)

for 4B ;:::: -Z > B B

(7.56)

B

0:;

(qJ; is the cone tip resistance for the ith layer, and n is The units of B is meter. The procedure to determine the settlement from 1.

2.

EXAMPLE 7.1

constructio

influence depth is 2B. You need to divide Eq. (7.52).

FO

354

Thickness of each of 4 sublayers:

o

5

10

q, (MPa) 15

20

25

30

O ~~~~~r---~----~--~----I

2~-~·-=---~--~--+---~--4 4 ~~~~~~~----~--+---~---1

E

6 f--~"""_

.c:

8 f--_,.....-

~ 10 r--~--~-+----~----+---r--~ o 12 f---~~~-+--~-~~~~r---4 14 ~-~-~~~~~----+-----~-~ 16 ~-~--~--~--+---~--4 18 ~-~----L- - - -- - ~--·L---L--~

FIGURE E7.13

6 Llz = -4 = 15m .

1.1 3 DETEflMINATION Of BEAfl lNG CAPACITY AND SETTlEMENT OF COARSE-GRAINEO SOILS

Step 2:

355

For each su blayer, find the average value of qc. l eo. and p. Use a spreadsheel program.

Liver

,

2 3

,

..

-• B

I~

q.

0.75 2.25 3.75

0.25 0.75 1.25

0.3 0.42

5.8 7 8.5

::'.£b

1.15

0 .09

·" 1 center of flyer. qnel ""

.

0.25

,.,

-I~ q.

..

0.08 0.09 0.'"

I

200 kPa

c, - J Step 3:

~~:=:·"".-ziQ. x 0.22

p

=

20.3 = 20

mm •

Tests on full sized foo t~ · n gs e desira Ie.. expensive. T he al terna tive is to carry .12) to sim ulate the load-se ttlement behavior of a real ut plate load tests (Fi footin;. T he plates ~rc de"fto steel wi th sizes varying from 150 to 760 mm. C!wo common pia eCS1z.c s arc usecYln practice. One is a square plate of widlh 300 rnm and the o t h ~ is a circular plate of diameter 300 mill. The test is carried o ut in a pit of d ~ h at least 1. ffi . Loads are applied in increments of 10% 10 20 % of the eSYmate a llo,,:,!9J' bearing capacity. Each load increment is held until settlemeRt. cease~e~al settleme nt at the end of each loading increment is recorded. adi~gJl -contln ued until the soil fai ls o r settlements are in excess of 10% of the plat tHameter. The max imum load should be at least 1.5 times the estimated allowable bearing capacity.

r

I- -.-1 FIGURE7. 12

Plat e load test.

Stress or IOI1d

356

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

If the sand were to behave like an elastic material, then the settlement can be calculated from (7.57)

where Pp is the plate settlement, CT ap is the applied stress, B Po diameter of the plate, V I is Poisson's ratio, £1 is the elastic mo Ius, and J:, is an influence factor (0.82 for a rigid plate). The settlement of footin · (p) of width B is related to the plate settlement by p =

ppC + ~p/B

r

(7.58)

In the limit BplB --7 00 , p/pp --7 4. Equation (7.57) is only valid if the st inS ar are several problems associated with th late

FO

1. The test is reliable only if the

water pressure.

What's next . . .We have considered only vertical settlement under vertical and inclined loads. Horizontal displacements and rotations are importantfor structures subjected to significant horizontal loads. Next we will consider horizontal displacements and rotations.

7.14 HORIZONTAL ELASTIC DISPLACEMENT AND ROTATION Structures such as radar towers and communication transmission towers are subjected to significant horizontal loads from wind , which can lead to intolerable

7.14 HORIZONTAL ELASTIC DISPLACEMEN T AND ROTATION

357

lateral displacemen t and rotation of their fo undations. Gazelas and Hatzikonstantinou (1988) proposed equations based on an isotopic linearly elastic soil to determi ne th e elastic horizontal displacemen t and rotation of an arbitrarily loaded foundation. The equations were obta ined by curve fitting lheoretical e lastic solutions for

B;

:s; 2. A summa ry of these

eqUatio~~resented

Table 7.6 You must be cau tious (see Section 7.12.1) in usi g the eqJ ations Table 7.6.

in III

TABLE 7 .6 Equations for Estimating the Horizont,t] Oisp laca men t and Rotation of Arbitrarily Shaped FoundationslSource: Gazets :and Hatzik.ftnstantinou, 19881. LOldlllg

r-----

L

2B

1

H,

-

80

dj'Kl;OIl.

M,

.i.. :_ .... " '_' \

I,

--

~'V

1.12

, "')]..... -

~

_~1 - C.14

(tfD)"

e_ ~ t,

iI.".. -

0 43 .....0.

Longltudinll dir.ctio...

Nons: Subsl ilul e the appropriate villues lor HJlind M in the above equations. For eKllmple, if vou are co"sidering the X direction. H, - H. end M = M. Also use the appropriate values for .... end e, . For e ~ample. in the X direction. "'. - "' •• and e, _ ."" For shen ter m loading use the undrained vBlues 01 E and ~ while for long term loading use the effective ... Iues. The terms I, aJld '. are ,he '$«lot! moment or areas about the X and Y allel respe<:'iV{l IV ftnd e is the rotation Clused bv the moment.

358

7.15

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

SUMMARY In this chapter, we described ways in which you can calculate the bearing capacity and settlement of shallow foundations. The critical criterion for design of most shallow foundations is the serviceability limit state. Because of piing difficulties, the bearing capacity and settlement of coarse-grained J~ iIS are often determined from field tests. We showed how to use the result~f ~e SPT, t ~ cone test, and the plate test to estimate the bearing capacity and~ttle nt 0 hallow foundations on coarse-grained soils.

Practical Examples

FO

EXAMPLE 7 .14

Assu ~

a wi
T

2.5 m -",,0

-.-L

~=_

t :,::,: T

Sand

'~o 32-, yo 16 kNlm', y~ 0 17 kNlm', E' 0 40 MP', .. 00.35

s" = 40 kPa, 4>; = 20°, E" = 8 MPa, E' = 6.5 MPa, OCR = l.3, C, = 0.45, C, = 0.09, w = 55%, G, = 2.7

Dense gravel

FIGURE E7. 14

359

7 .15 SUMMARY

Step 2:

Calculate the elastic settlement. Sand

Neglect side wall effects; that is, I

f-lcm b

Is

Df ( 1 + :34 LB) B

=

1 - 0.08

=

0.62 InCUE) + 1.12 E'

Pe

= 1

1.5 ( 1 + :3 4 1 - 0.08 3

=

1.12

=

= qsB[l - (V i?] I '

fLwall

=

s f-lemb

P[l - (V')2] E' L }.

b

2

- 500(1 - 0.35 ) 2 0 91 - 40 X 103 X 3 x 1.1 x .

Clay

Find the equivalent footin the depth from the ba e of Equivalent width anlI f-lemb

=

1 - 0.0

= 3

+ 2.5

= 5.5 m

2.5

x 0.92 = 8.8 x 10- 3

8.8 mm

=

R

Step 3:

~"""~~"Ie ll'

FO

vertical effective stress (overburden pressure) at lay layer.

'0

=

3 x 16 + 1(17 - 9.8) + 1(16.5 - 9.8) = 61.9 kPa

the stress increase at the center of the clay layer (z =

m

(J' ~o

+

=

n

=

(!)

3~5 = 0.43;

f1(J'z =

4 X

500 Y

f1(J'z =

61.9

+ 15.1

(J' ~c =

Ppc

=

OCR x

=

0.068

X 0.068 = 15.1 kPa =

(J' ~o =

77 kPa 1.3 x 61.9 = 80.5 kPa >

~ C 10 (J'~o 1 + r g eo

2 1 + 1.49

from stress chart (Fig. 3.25) , l z

X

+

(J'~o

+

f1(J' z

f1(J' z

I

cr zo

77 0.09 log -69 = 6.9 1.

X

10- 3 m

=

6.9 mm

CHAPTER 7

Step 4:

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Find the total settlement. Total settlement:

P

= (Pe)sa nd + (Pc)clay + Ppe 3.7 + 8.8 + 6.9 = 19.4 mm

=

Step 5:

Check the bearing capacity. The groundwater table is less than B = 3 m bel groundwater effects must be taken into acc 1. Calculate the bearing capacity using M

ase so

ESA (sand)

Kp =

.2

00

32 = 55.6 kPa 692

-F:r-~~~-~~-5-5.-6---7.-2-X-1.-5

FO

360

Equivalent footing width : B (~(T),". x

=

+

15.4 > 1.5 (okay)

'1

= 3 + 2.5 = 5.5 m

500

= 5.5 2 = 16.5 kPa

TSA (clay) Use Skempton's equation (7.14). quit

= 5S u ( 1 + 0.2

i)

= 5 X 40 X (1

(1 +

0.2~)

+ 0.2 ~) (1 + 0.2

304 FS = - 6 = 18.4 > 3 (okay) 1 .5

~)

= 304 kPa

7.15 SUMMARY

3E

ESA (clay)

Use Meyerhof's method . K = 1 + sin 20° = 2' p 1 - sin 20° '

N

q

Ny

=

e"'tan20>

=

(N q

-

tan 2 45 + -20) 64' N q 2=" (

1) tan(1.4

X

20)

=

2.9

Settlement governs EXAMPLE 7 . 15

a building. Your

satisfy ultimat the ground sur

FO

R

ickness of the clay. The Skempton-Bjerrum settlement is suitable for this problem.

Inside edge ~~~r-~~------ 20m --------------~~--~

Plan

I

Footing A

Footing B

Vertical load = 500 kN Moment = 0

Vertical load = 250 kN

Moment = 100 kNm

Clay: Ysa' = 17 kN/m 3 , m .
=

FIGURE E7.15

=

= 0.00045 m2/kN,

E"

= 25

MPa, v"

= 0.45,

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Solution 7.15 Footing A Step 1: Assume a width. Assume a square footing of width B = 3 m. Step 2: Calculate the elastic settlement. From Eqs. (7.35) to (7.37):

From Eq. (7 .34):

FO

362

Step 3:

(3/2)

=n=--

z

1

0.175 0.045 0.018

0.33 0.2

Equivalent diameter of footing:

D

.i0' z = 4qs/z (kPa)

I.

=

~=

j!

38.9 10.0 4.0 L52.9

=

1.69 m

He = He = _9_ = 3 B D 1.69 5.

H JB = 5.3 is outside the plotted limits in Fig. 7.11. We will use H olB = 4, which will result in an overestimation of the primary consolidation settlement. From Fig. 7.11, f.LSB = 0.58. Ppc

= "Lmv

~(J'z

H ofLsB = 0.00045 x 52.9 x 3 x 0.58

= 41.4 x 10- 3 m = 41.4 mm Total settlement = 4.5 + 41.4 = 45.9 mm < 50 mm

Total settlement of footing B is okay.

7.15 SUMMARY

363

Footing B Step 4:

Calculate the eccentricity. M

100

e = p = 250 = 0.4 m

Step 5:

Assume a width. For no tension e/B < 6; that is, B > 6e. B min = 6

Step 6:

x 0.4 = 2.4

m;

Calculate the elastic settlement fLemb = 1 - 0.04

1 ( 1 + 3(1 4 x 1.25

Pe=-----

1

= 5.3

X

10- 3 m

= 5.3

mm

Step 7:

/ ical stress

L 1.25 n=-=-

R

z

FO

0.83 0.56 0.33

z

0.42 0.28 0.17

~= B

0.10 0.055 0.025

N

Ho = Ho D

I.

=

.1<1 z

= 2q.l.

15.7 8.6 3.9 2:28.2

= 1.41 m

~

=

6.4

1.41

HjB = 6.4 is outside the plotted limits in Fig. 7.11. We willtise HJ B = 4, which will result in an overestimation of the primary consolidation settlement. From Fig. 7.11, f.LSB = 0.58. Ppc

Step 8:

~(J' , H ofLsB

= 0.00045 x 28.2 x 3 x 0.58

=

"Zmu

=

22.1 x 10- 3 m = 22.1 mm

Calculate the total settlement. P = Pe

+

Ppc =

5.3

+

22.1 = 27.4 mm < 50 mm

Total settlement of footing B is okay.

364

CHAPTER 7

Step 9:

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

Calculate the distortion.

o = 45.9 - 27.4 = 18.5 mm o 18.5 1 -= = -1- < _ . 3

e

FO

Step 10:

(20 - 1.25) x 10

1014

350 '

Check the bearing capacity. Use Meyerhof 's method because of the eccentr" B.

ting

Footing A Factors

Equation (7.20):

a,

Equati

d,

+ 0.2G) , 41.07;

~ ~ ~i.~ d.,

1

1.05

80 x 1.2 x 1.07 = 528 kPa

.5 x 17 x 3 x 5.7 x 1.24 x l.05) = 380 kPa 380 6 = 9.8 > 1.5 (okay) 55. - 17 x 1

Footing B B' = B - 2e = 2.5 - 2 x 0.4 = 1.7 m

The only factors that would change are the depth fac tors . de = 1 +

0.2(~) 1.7

= 1.12; d q = d.., = 1 + 0.1

V2.37(~) 1.7

= 1.09

TSA quI<

Equation (7.26):

= 5.14 x 80 x 1.2 x 1.12 = 553 kPa

553 FS = 78.4 _ 17 x 1 = 9 > 3 (okay)

EXERCISES

365

ESA q uit

= -yDf(Nq_j)sq dq =

+ (0.5 FS

Step 11:

=

+ 0.5-yB 'N-ys-y d-y

(17 x 1 x 8.6 x 1.24 X

310 78.4 _ 17

17

X

1.7

X

1

=

X

X

5.7

1.09) X

1.24

X

1.09)

=

5 > 1.5 (okay)

Recommend footing sizes. Footing A : Footing B:

•

EXERCISES Theory 7.1

FO

~

: ) FIGURE

7.2

A strip footing, wide, is founded on the surface of a deep deposit of clay. The undrained shear strength of the clay increases linearly from 3 kPa at the surface to 10 kPa at a depth of 5 m. Estimate the vertical ultimate load assuming that the load is applied at an eccentricity of 0.75 m from the center of the footing 's width. [Hint: Try a circular failure surface, determine the equation for the distribution of shear strength with depth, and integrate the shear strength over the radius to nnd the shear force .]

7.3

The centroid of a square foundation of side 5 m is located 10 m away from the edge of a vertical cut of depth 4 m. The soil is a stiff clay whose undrained strength is 20 kPa and whose unit weight is 16 kN/m 3 . Calculate the vertical ultimate load.

Problem Solving 7.4

Calculate the ultimate net bearing capacity of (a) a strip footing 1 m wide, (b) a square footing 3 m X 3 m, and (c) a circular footing 3 m in diameter using Terzaghi 's and Mey-

366

CHAPTER 7

BEARING CAPACITY OF SOILS AND SETTLEMENT OF SHALLOW FOUNDATIONS

erhof's methods. All footings are located on the ground surface and the groundwater level is at the ground surface. The soil is coarse-grained with '{sat = 17 kN/m 3 ,
A strip footing, founded on dense sand (
7.6

7.7

7.8 7.9

FO

7.10

5m

9m

FIGURE P7.10

~1 r-3m-j

Dense sand eo = 0.52

E'=55MPa v' = 0.35 4>~ = ¢~s

36°

= 32°

R A U C SE A O N L

FO

CHAPTER

PILE FOUNDATIONS 8.0

INTRODUCTION

and moments. Pile foundations are used w apacity to sup-

onuniform structural loads

shallow foundation on a firm soil layer are dif-

FO

y the bearing capacity (load capacity) and es under axial loads. When you complete this

• Determine the allowable axial load capacity of single piles and pile groups • Determine the settlement of single piles and pile groups

You would need to recall the following: • Effective stresses-Chapter 3 • Consolidation-Chapter 4 • Statics Sample Practical Situation A structure is to be constructed on a deposit of soft soil. Shallow foundations were ruled out because the estimated settlement exceeded the tolerable settlement. The structure is to be supported on piles. You are required to determine the type of piles to be used, the configuration of the

368

8.1 DEFINITIONS OF KEY TERMS

369

FO

8.

stres or point resistance stress (fb) is the stress at the base or tip of Ultimate load capacity (Q uit) is the maximum load that a pile can sustain before soil failure occurs. Ultimate group load capacity [(Qult) g] is the maximum load that a group of piles can sustain before soil failure occurs. Skin friction or shaftfriction or side sltear (Qf) is the frictional force generated on the shaft of a pile. End bearing or point resistance (Qb) is the resistance generated at the base or tip of a pile. End bearing or pOint bearing pile is one that transfers almost all the structural load to the soil at the bottom end of the pile.

370

CHAPTER 8

PILE FOUNDATIONS

Friction pile is one that transfers almost all the structural load to the soil by skin friction along a substantial length of the pile. Floating pile is a friction pile in which the end bearing resistance is neglected.

8.2

QUESTIONS TO GUIDE YOUR READING 1. What are the differences between the different ty

2. How is a pile installed in the ground? 3. How do I estimate the allowable load ca groups? 4. How do I estimate the settlement 0 5.

8.3

30- 135 em

FO

(5 Fini shed grade

Butt

Soil plug

+ H-seetion steel pile

(c) Steel plate

Tip or base (d) Timber pile

FIGURE 8.2

Pile types.

8.3 TYPES OF PILES AND INSTALLATION

371

magnitude of the loading, the soil type, and the environment in which the pile will be installed, for example, a corrosive environment or a marine environment.

FO

R

8.3.1 Concrete Piles

8.3. Timber pile e been used since ancient times. The lengths of timber piles depend on the types of trees used to harvest the piles, but common lengths are about 12 m. Longer lengths can be obtained by splicing several piles. Timber piles are susceptible to termites, marine organisms, and rot within zones exposed to seasonal changes. A comparative summary of the different pile types is given in Table 8.1.

8.3.4 Pile Installation Piles can either be driven into the ground (driven piles) or be installed in a predrilled hole (bored piles or drilled shafts). A variety of driving equipment is used in pile installations. The key components are the leads and the hammer. The leads are used to align the hammer to strike the pile squarely (Fig. 8.3).

FO

w

~

TABLE 8.1

..... N

Comparisons of Differe

(")

:l: l>

Section Pile type Cast-in-place concrete

Precast concrete

Advantages

(m)

0.15-0.45

(,an sustain hard driving, resistant to marine organisms, easily inspected, length can be changed easily, easy to h a r:I dre::ai'td. s hip

(diameter)

Steel pipe

0.6-2.3 0.2-1

0.2-1

Cost (expensive)

<35

Needs treatment for corrosive environment

Similar to steel pile

900

(diameter) Steel:

62-83 Steel H-pile

Webs: 1-3 Flange:

< 60

900

59-83

12-35

250

5.5-8.5

0.2-0.35 Timber

C

(diameter)

Y

0.1250.45

"

00

...,

;= m

o"

c: 2

< 60

< 35

::!l m

o

(diameter) (diameter)

Concrete-filled pipe

Concrete can arch during placement, can be damaged if adjacent piles are driven before concrete sets Cutting and lengthening of piles can be expensive, handling is a problem, shipping long piles is expensive, may crack during driving

0.15-0.25 (width)

Raymond cylinder

Disadvantages

~

5;::: f/)

B.3 TYPES OF PILES AND INSTALLATION

373

Hammer

FIGURE 8.3

Key components 0

d 15 kN or

FO

R

The essential points are: 1. The selection of a pile type dep ends oh the structural loads, availability, and the environrhent at the site. 2. Piles can be installed using simple drop hammers but, most often, they are installed using Steam or pneumatic hammers.

r;:::======::~~

Upper val ve

Valve

I

Ram

r---'-----'----(

Ram

Cushion Drive ca p (a) Single act ing

FIGURE 8.4 Two types of hammer.

Anvil Drive cap (b ) Double acti ng

374

CHAPTER 8

PILE FOUNDATIONS

What's next . . .We have described the various types of piles that are in common use. In the next section, methods of estimating the load capacity are presented. The load capacity of piles has been studied extensively. But no single satisfactory method of determining the load capacity has evolved . Pile load capacity is mostly based on empirical equations, experience, and judgment.

LOAD CAPACITY OF SINGLE PILES

Method of parameter determination

Simple in situ or laboratory tests with correlations Routine, relevant in situ tests-may require some correlations

FO

8.4

3

S OURCE:

s for 2A, but theory is nonlinear (deformation) or elastoplastic (stability). 3A

Based on theory using sitespecific analysis, uses soil mechanics principles. Theory is linear elastic (deformation) or rigid plastic (stability).

3B

As for 3A, but nonlinearity is allowed for in a relatively simple manner.

3C

As for 3A. but nonlinearity is allowed for by way of proper constitutive models of soil behavior.

On behalf of the Institution of Civil Engineers.

Careful laboratory and/or in situ tests that follow the appropriate stress paths

8.4 LOAD CAPACITY OF SINGLE PILES

375

QUi'

1

Pile shaft

1 1 1

L

1t 1

Pile base or tip

Qb

FIGURE 8.5

Pile shaft and end bea(

~} ~y

'------_ .

(8.1)

FO

R

(8.2)

Shaft fric ~l) for coarseg!

Embedded length, L

Plastic zone

FIGURE 8.6

Failure surtace due to end bearing

Load transfer characteristics.

376

CHAPTER 8

PILE FOUNDATIONS

FO

of the pile. For coarse-grained soils, the load transfer is approximately linear with depth (higher loads at the top and lower loads at the bottom). In order to mobilize skin friction and end bearing, some movement of the pile is necessary. Field tests have revealed that to mobilize the full skin friction a vertical displacement of 5 to 10 mm is required . The actual vertica a· lacement depends on the strength of the soil and is independent of the pil length and pile diameter. The full end bearing resistance is mobilized in drive piles whl n the vertical displacement is about 10% of the pile tip diame l' F bored 'lies or drilled shafts, a vertical displacement of about 30% 0 the Qt e ·p:z!ia eter is required. The full end bearing resistance is mobilized n sir 0. ailure zones similar to shallow foundations are formed (Fig. 8.6). T d be~ring resistance datio s. important bearcan then be calculated by analogy with shallow ing capacity factor is N q • The full skin friction and full end

The essential points are: 1. Pile load capacity depends on the soil type, method of installation, and construction practice. 2. The critical state friction angle should be used in estimating the longterm load capacity of piles, except you should use the residual friction angle for overconsolidated clays with a predominance offlat particles.

What's next . . .A variety of methods are available to determine Of and Ob. We will deal with four methods: • Pile load test • Statics-(\'- and f3-methods

8.5 PILE LOAD TEST

• Pile-driving formulas • Wave analysis We begin with the pile load test.

PILE LOAD TEST The purposes of a pile load test are: • To determine the axial load capacity of a sing

FO

8.5

Reaction pile

FIGURE 8 .7

Test pile

A pile load test setup.

377

378

CHAPTER 8

PILE FOUNDATIONS

Load

Q uit

A -::.::a-----= ' - W - e-II--d-ef-in-ed- u - Itimate load

- - - - - -

0 "---- - - - - - - - - - - - -....

Settlement

FO

FIGURE 8 .8

~

Load-settlement curves.

The essential points are: 1. A pile load test provides the 10(l,d capacity and settlement of a pile at the working load 4t a particular location in a job site. 2. Various criteria and techniques are used to determine the allowable load capacity from pile load tests.

The results of a loa test on a 0.45 m diameter pile are shown in the table below. Determine (a) the ultimate pile load capacity, (b) the allowable load for a factor of safety of 2, and (c) the allowable load capacity at 10% pile displacement. Displacement 0.0 Imm)

1.3

Load IkN)

200 350

0

2.5

5.1

7.6

670 870

10.2

12.7

15.2

17.8

20.3

22.9

25.4

27.9

30.5

33.0

1070 1250

1400

1500

1600

1700

1750 1780

1810

1830

Displacement Imm)

35.6

38.1

40.6

43.2

45.7

47.0

Load (kN)

1860

1870

1890

1890

1900

1905

B.6 METHODS USING

STAncs

379

2500

2000

/

Q. Z 1500

V

~

•

.3 1000

-~I'

500

o

0.0

-

100

200

I ' 00

300

-.oo~

FIGURE ES. 1

Strategy Plot a graph of displacement l\rsus ccdure in Section 8.4. .,/ ...

d and t hen fol low the pro·

Solution 8.1 Step 1: Step 2:

Plot a displacement- load gra h. See Fig. E8.1. • . De termine the lJ.ima~;.-pile load caPleity. The fa1lure load l S ill defined. Locate til inter

'ne the allowable

1780 = 890 kN k!i 2 Determ~e pi ~.('capacity at 10% pile diameter.

Q = Q...., FS

Step 4:

>=

DIsplacement = 450 x 0.01 "" 4 5 mm From Fig. EB.l: Q. - 510 kN

•

What's next .. . Pile

oWests are expensive and are not often conducted in t he pre· liminary stages of a design . To get an est im ate of the pi le load capacity, at least in the preliminary design stages, recou rse is made to statics and to correlations us ing soil test results. In the next section, we w i ll examine how statics is applied to obtain an estimate of pile load capacity and some correlations between field soil test results and measured p ile load capacity.

8.6

METHODS USING STATICS 8.6.1 OJ·Method The a-method is based on a total st ress analysis (TSA) and is normally used to eSlimate Ihe short· term load capacity of piles embedded in fine-grained soils. In

380

CHAPTER 8

PILE FOUNDATIONS

the (X-method, a coefficient, (xu, is used to relate the undrained shear strength, su, to the adhesive stress Us) along the pile shaft. The skin friction , Qf' over the embedded length of the pile is the product of the adhesive stress Us = cxus,,) and the surface area of the shaft (perimeter X embedded length). Thus, }

Qf

=

2:

(cxuMsJ; x (Perimeter);

X

(Length);

(8.3)

i= l

a cylindrical pile of uniform cross section and diame er, D, mogeneous soil, Qf is given by ,..---- - - - - - ,

(8.4)

TABLE 8.3 Soil type

Clay

Semple and Rigden (1984)

FO

Fleming et al. (1985)

Burland (1973) ~

=

~

= 0.44

~

= (KIKa) Ka tan('cj>j/')

.15-0.35 0.10-0.24

(compression) (tension)

for cj>' = 28° 0.75 for cj>' = 35° 1.2 for cj>' = 37°

cj>i/' depends on installation method (range 0.5-1.0) KIKa depends on installation method (range 0.52.0) Ka = coefficient of earth pressure at rest and is a function of OCR Uncemented calcareous sand SOURCE:

~

= 0.05-0.1

On behalf of the Institution of Civil Engineers.

McClelland (1974) Meyerhof (1976)

Stas and Kulhawy (1984)

Poulos (1988)

8.6 METHODS USING STATICS

381

model piles installed in a uniform deposit of soil. The major problems with these laboratory tests are: 1. It is difficult to scale up the laboratory model test results to real piles. 2. The soils in the field are mostly nonuniform compared

pared uniform soils in the laboratory. 3. Pile installation in the field strongly influences rately duplicated in the laboratory.

<X

c

ests are expensive are performed. The n linked to <Xu but these coefficient of correlation.

(8.5)

Reference

Skempton (1959) Cl. u

FO Uncemented calcareous sand SOURCE:

Fleming et al. (1985)

displacement pile Cl. u

= 0 for z s, 1.5 m and z> L - 0 Cl u = 0.55 for all other points

Reese and O'Neill (1988)

K is lesser of Ko or 0.5{1 + Ko) KlKo = ~to 1; Ko is a function of

Fleming et al. (1985)

OCR;
fs =

Sand

= 0.7 times value for driven

fs =

f3 =

f3
13
60 kPa :s:

Is:S:

100 kPa

On behalf of the Institution of Civil Engineers.

for ' = 35° 0.35 for
Stas and Kulhawy (1984)

Meyerhof (1976)

f3 = FtanW - 5°) where F = 0.7 (compression) = 0.5 (tension)

Kraft and Lyons (1974)

[3 = 0.5-0.8

Poulos (1988)

382

CHAPTER 8

TABLE 8 .5

PILE FOUNDATIONS

End Bearing Resistance Factors, Nc

Soil type

Equation

End bearing factors

Nc = 9 for LID ~ 3 Nc = 6[1 + O.2(LIDb )l;

Clay

Reference

Skempton (1959 )

is the diameter at the base of the pile, Nc ,;;; 9, and fb';;; 3.8 MPa ,..,------.,----,--,N = (tan <1>' + V 1 + tan 2 <1>')2 Db

6)

q

exp(2ljJp tan <1>')

Sand

Uncemented calcareous sand

Poulos (1988) Dutt and Ingram (1984)

FO

o the pile. Ranges of values for Nc are shown

j

Qf

=

L

((r~); tan <jJ; X (Perimeter); x (Length);

(8.6)

i= l

The lateral effective stress is proportional to the vertical effective stress a coefficient K (Chapters 3 and 4). Therefore, we can write Eq. (8.6) as

(a~)

by

}

Qf

=

L i= l

K(rr z ) ; tan <jJ ;

X

(Perimeter); x (Length);

(8.7)

FO

8.6 METHODS USING STATICS

We can replace the two coefficients K and tan

~;

by a single factor

383

!3 to yield

j

Qf = ~ \3;(0";); x (Perimeter); x (Length);

(8.8)

i= ]

and for overconsolidated soils

A general expression for !3 is 13

=

K tan <1>:

(8.9)

=

(8 .10)

is Nq

=

(tan <1> ' +

VI +

1 tan 2 <1>')2 exp(2tjJp tan <1> ')

(8.11)

The value of \jJp varies from \jJp :s 'IT/3 for soft, fine-grained soils to \jJp :s O.S8'IT for dense, coarse-grained soils and overconsolidated fine-grained soils. For soft, compressible soils, the pile tip may easily penetrate the soil without causing significantly large plastic zones. For these types of soils, \jJp should not exceed 'IT/3 . In dense, coarse-grained soils, the plastic zones could be substantial but it is recommended (Janbu, 1976) that for these soils \jJp should not exceed 'IT/2. Berezantzev et al. (1961) proposed a relationship between N q and ~' as shown in

CHAPTER 8

PILE FOUNDATIONS

190 rr==========;---r~ 180 170

a,

160

0.8

150 140 130 120 110

100 N' q

90

-;--- I ! I I a, is a correction factor for N q

~-+-- I

N q = a, N~ 80 r-'--r--r-~-+--~H--+--1 701--1--+--~4--+--~~4--+--' 601--~--+--r-~-+~~~~~1

50 r-;--+--r-~-+~~~'

40 f---+--+----4--+--.H 30 f---;--~--~A--+-4~

20 f---1--+~~~-+--~~~~1 10 f---~-+--~4_-+--~~_+--1 O'-----'------'----1~--'-------'---+--F___----'.-----'

22 FIGURES.

FO

384

. (1961).

(8.12)

(8.13)

and (8.14)

where G is the shear modulus, fp is the volumetric strain, and «(j~)b is the vertical effective stress at the base. The original equation for N q proposed by Vesic (1975) contained s" in the denominator of Eq. (8.14). The author removed s" from the original equation since we are considering an effective stress analysis. A list giving values for N q proposed by various investigators is shown in Table 8.S. Coarse-grained soils are difficult to sample. A number of correlations be-

8.6 METHODS USING STATICS

TABLES.6 Pile type

Correlations B Between Skin Frictional Stress, f A

Soil type

Driven displacement

Cast in place

B

s'

and SPT Values

Remarks

Coa rse-gra i ned

0

2.0

fs =

Coarse-grained and fine-grained

10

3.3

Pile type not 502: N 2: 3

Fine-grained

0

10

Coarse-grained

30

385

average val ue over shaft N = average SPT along shaft Halve f s for small displacement pi l

Reference Meyerhof (1956)

Shioi and Fuku i (1982) Yamashita et al. (1987)

0 Fine-grained

0

Findlay (1984) Shioi and Fukui (1982)

FO

Bored

Wright and Reese (1979) Shioi and Fukui (1982) Decourt (1982) pentonite 50 ::s N::s 3 f s ::s170kPa

30 > N > 15

fs

Fletcher and M izon (1984)

::s 250 kPa

ues in blows/ft or blows/O.31 m.

SOURCE:

On behalf of the Institution

tween field soil tests and pile load capacities have been proposed for coarsegrained soils. Some of these correlations for the SPT are shown in Tables 8.6 and 8.7 . Other correlations exist for some penetrometers and pressure meters.

The essential points are: 1. The a-method is based on a TSA and is used to estimate the shortterm pile load capacity in fine-grained soils. 2. The p-method is based on an ESA and ;s used to estimate the shortterm and long-term pile load capacities in all soil types. 3. The actual va.lues of au, /3, and N q are uncertain.

386

CHAPTER 8

PILE FOUNDATIONS

Correlations B Between End Bearing Resistance,

TABLE 8.7 Pile type

Soil type

Driven displacement

c

Sand

0.45

Sand

0.40

Sand

I b, and

Remarks

SPT Values Reference

N = average SPT value in loca l failure zone

0040. . 0

Silt. sandy silts

0.35

Glacial coarse to fine silt deposits

0.25

Thorburn and MacVicar (1971 )

Residual sandy silts Residual clayey silts Clay Clay Shioi and Fukui (1982)

All soil s

Shioi and Fukui (1982)

Cast in place

Yamashita et al. (1987)

10 =

0.09 (1 + 0.16L,) L.. = tip depth

where

Yamashita et al. (1987)

(m)

FO

Shioi and Fukui (1982) Shioi and Fukui (1982) Hobbs (1977)

EXAMPLE 8.2

A cylindrical pile of diameter 400 mm is driven to a depth of 10 m into a soft, normally consolidated clay. The soil parameters are Su = 20 kPa, ~s = 28°, and "Isa, = 18 kN/m 3 . Groundwater level is at the surface. Estimate the allowable load capacity for a factor of safety of 3. Is the pile a friction pile?

Strategy The solution is a straightforward application of the pile load capacity

equations.

8.6 METHODS USING STATICS

Solution 8.2 Slep J:

Select

0:"

and

~.

Table 8.3: s. "" 20 kPa < 25 kPa. s. "" 20 kPa < 35 kPa . the refore

(I: ..

1

(Se np

i3 ;; (1 -

Sin

q,;')(OCR)0.5 tan ¢I' •

For OCR "" 1 (normalJy consol idated

SOil~d 41'

13 "" (1 - si n 28·)~ Ian 2l- ""

Slep 2:

Calcu late Q. using a TSA.

Q, ..

~~(1TD ) L

0 x (1.26)

10;; 252 kN

Qt .. 9s~". : 9- 0 x 0.126 = 22-:1 .. Q, ~52 + 22.7 =><~ 7 kr-.

yu,

252 -Q, .. .......... 27} .7

0.92 > 0.8. -'Ti;W.-roo~~ There have a friction pile.

274.7 Ts -,-

Step 3:

alcul ate Q. using a n

/

ES~ ....~~

Use Ja nbu's equation for N q .

For 41;' "" 28" and assuming WI' "" -rrf3 (soft clay), we find N• .. ( Ian

ZS- + V I + lan2(28°) P exp(Z; tan 28") .. 8.4

Qb '" 10.3 X 8.4 .. 86.S kN Q~1o

- Q, + Qt "" 144.6

. , ,

Q

c:::

+ 86.5 .. 23l.J kN

Q..... = 231.1 .. 77 kN

387

CHAPTER 8

PILE FOUNDATIONS

The load capacity from an ESA is less than that from a TSA. Therefore , use the allowable pile load capacity from the ESA, that is, Q a = 77 kN. •

EXAMPLE 8.3 A square concrete pile 0.3 m X 0.3 m is required to supp with a factor of safety of 3. The soil stratification consi$ normally consolidated clay (su = 25 kPa,
conditions

Solution 8.3

=

Step 2:

1 (soft clay)

= 0.5 (stiff clay)

FO

388

~ Y = 17.8 kN/mJ

Ysa • = 18 kN/m3

Stiff clay 3 = 18.5 kN/m

rsa.

FIGURE E8.3

450 kN

389

8.6 METHODS USING STATICS

= <\>; = <\>::., we find the following:

For <\>'

Soft clay:

~ =

Stiff clay:

~ = (1 -

=

Perimeter

=

(1 - sin

0.27

sin 24°)(4)°5 tan 24° = 0

+ B) = 2(0.3 + 0.3)

Step 3:

Length

In

(Qj )SOft

cl ay

soft clay

=

5m 2

= CL u 5" X Perimeter X

X 1.2 X 5 = 150 kN

Stiff clay

Ultimate load capaci y

FO

R

That is,

3(18 - 9.8)

f),01

(Qf}stiff

u~ J

- 9.8) = 60.6

0.27 X 40.1 X 1.2 X 5

=

0.53 X (60.6 + 4.35L I ) X 1.2 X (5 + L 1 )

=

2.8(L J)2

+ 52.4L j + 192.7 <\>~ s

+ VI + tan 2 (24°)j2

= 24° and

expe;

ljip = TI/2 gives

tan 240)

=

9.6

+ 3(18 - 9.8) + L1(18.5 - 9.8) = 60.6 + 8.7L1 Nq«(J~)vAb = 9.6 X (60.6 + 8.7L J) X 0.09 = 52.4 + 7.5LJ

(U~)b = 2 X 18

Qp

=

+ 4.35LJ

65 kN

=

Using Janbu 's equation with <\>' = N q = [tan 24°

LJ "2 (18.5

Perimeter X Length

clay = clay

+

Ultimate load capacity QUi'

=

(Qj ) SOft clay

+

(Qj ) st lff clay

+ Qp

That is, 450 = 65

+ 2.8(L J)2 + 52.4L + 192.7 + (52.4 + 7.5L 1 ) j

CHAPTER 8

PILE FOUNDATIONS

Rearranging and simplifying, we get

+ 59.9LI + 310.1 + 59.9Ll - 139.9 = 0

450 = 2.8Lr 2.8Li

Step 5:

Solution is Ll = 2.1 m. Decide on the length required. TSA: ESA:

Use the larger of the two values, LI

=

4.9 m

Total length

EXAMPLE 8.4

FO

390

Y"'t = 17 kN /m

L

r--

----j

(S,.lb = 0 .22 y·L

FIGURE ES.4

3

=

5 +

•

8.6 METHODS USINGSTAnCS

391

Conside r an element of thickness dz , The surface area is 2'm' dz an d t he ski n friction is

Q, Step 3:

=

2".ro~

r (O.22"Y~)

dz

=

2.24[

~IS = 252 kN

Calculate the end bearing capacity.

A~ =

". X

4

D2 '"

11'

2 X 0.45 = 4

Q/> = 9(sull> A I> = 9 x 23.8 x O.

Stel.4:

)

o~ =

34,

.l

Calculate the allowable load capacjl'~~...

86.1 kN

• EXAMPLE 8.5 A 450 mm diameter pile-is driven " sand praql¢" 10 depth of 1 m. The SPT results are shown in the a ble bela . Estimate !OWabl? ad capacity (or a factor of safety of

ti"&.

5 25

3

18

Strat89Y\ lhe N

6 ,~. 20

value~.a re blo w

ft no

11 39

"

"

lows/m . Use one of the correla tions

own in the table a bove"!'"

)

Step 2:

Determine the load capacity. Skin fril-liun

Perimeter = 'TID - 11' x 0.45 = 1.41 Tabte 8.6: Q, = (A + BN) x Perimeter x Length

USlflg Meyerho f 's method . A Q j:'

=

O. 8 = 2.

(0 .,. 2 x 25 ) x 1.41 x 10 = 705 kN

E nd bearing Z

A ~ - -".D 4- =

'TI

1

X 0.45 ;;;; 0 159 m' 4 .

392

CHAPTER 8

PILE FOUNDATIONS

Using Table 8.7 and Meyerhof's method, L D

Qb

=

0.04N

Qb

=

10 0.04 X 36 X -0- X 0.159

Check that

Ab :::;

OANA b , where N

A5

Qb :::;

O.4NA b :

=

O.4NA b

.

IS

the SPT value at the base.

5.09 MN

= 0.4

X

36

N<

Qb '

• What's next . . .Piles are rarely used sin Next, we will discuss how to determin

PILE GROUPS

FO

8.7

9 0

000

o 0

(c) 7 piles-{)ctagonal

arrangement

_____ ...J.,......

.,.I;;:;;;;_;;:- Pi Ie cap

-

I+-

Diameter or width D

(d) Pile cap resting on ground surface

FIGURE 8.10

Pile groups.

1iIIP.,.-I11111"'--

Pile cap

Diameter or width D

(e) Pile cap elevated above ground surface

8.7 PILE GROUPS

393

The load capacity for a pile group is not necessarily the load capacity of a single pile multiplied by the number of piles. In fine-grained soils, the outer piles tend to carry more loads than the piles in the center of the group. In coarsegrained soils, the piles in the center take more loads than the outer piles. The ratio of the load capacity of a pile group, (Q ult)g () e total load capacity of the piles acting as individual piles (nQult) is called e efficiency factor, '11. ; that is, (8 .15)

where n is the number of piles in the

group lCn-l.:~~1i

ESA (8.16)

FO

R

(8 .17)

(8.18)

(Q,,/, )gb

- -- - --

-7- - - - - -- -:..: - :,.:. - .-. .. - - -

Bl oc k fai lure Single pile failure

Num ber of piles

FIGURE 8.11

Block failure mode.

CHAPTER 8

PILE FOUNDATIONS

The group efficiency in fine-grain ed soils is defined as 'TJe = (QuIJ9b [(Q"'I)~b

+ (nQullfr l/2

(8.19)

The values of I/Ip to use in determining N q in lanbu's equatio the sID ratio and the friction angle. lanbu (1976) showed that

15 =

1 + 2 sin ljip(tan <1> ' + V I + tan 2 <1>' ) exp(IjiR tan

The value of I/Ip is not significantly affected by sID

(8 .20)

:S:

The essential points are: 1. The ultimate load capacity of a pile grQUp is not necessarily the ultimate load capacity of a single pile multiplied by the number of piles in the group. 2. A pile group can either fail by the group failing as a single unit, called block failure mode, or as individualpiles, called single pile failure mode. EXAMPLE 8.6

FO

394

D = 0.4 m ; Perimeter = 'ITD = 'IT X 0.4 = 1.26 m ; 'ITD2 0.4 2 Ah = = 'IT X = 0.126 m 2 4 4 Perimeter = (2s + D) = 4[2(1.2) + 0.4] = 11.2 m;

Single pile:

Group:

Base area = (A b)g = (15 + D)2 = 2.8 2 = 7.84 m 2

Step 2:

Calculate the ultimate load capacity using TSA.

TSA-B1ock Failure Mode Soft clay-skin friction Table 8.3: (Qj)soft

Ci" c la y

=

1

for

Su =

20 kPa

= CtuSu X (Perimeter)g X Length =

1 X 20 X 11.2 X 8 = 1792 kN

8.7 PILE GROUPS

~tt

395

m Soft clay

m Stiff clay

(] 0-0 0-0-

J:1.2 ] 1.2

m m

FO

R

FIGURE E8.6

r:i.uS u

X Perimeter X Length

(Qj)S1iff

day

= 0.5

=1

X 20 X 1.26 X 8

X 90 X 1.26 X 2

=

113.4 kN

StitT clay-end bearing CQp)"iff clay

= 9s u Ab = 9

X 90 X 0.126

= 102.1 kN

Single pile load capacity Qull = (Qj) soft Qull

= 201.1 +

clay

+

+ (Qp)stiff clay = 416.6 kN

(Qj) stiff clay

113.4 + 102.1

Group load capacity Number of piles:

n = 9

(Qul,) g = nQul1 = 9

X

416.6

=

3749 kN

= 201.1 kN

FO

396

CHAPTER 8

Step 3:

PILE FOUNDATIONS

Calculate the ultimate load capacity using an ESA.

ESA-B1ock FaiJure Mode Assume OWL will rise to the surface and <1>; =

<1>~.

Table 8.3: Soft clay: Stiff clay: Soft clay: Stiff clay: Skin friction

Soft clay: Stiff clay:

1048.6 + 9565.6 = 11466.5 kN

n since the only difference in the calculation is the

Soft clay: Qf Stiff clay:

=

1.26 852.3 x 11.2

=

95.9 kN

1.26 Qf = 1048.6 x 11.2 = 118 kN

End bearing-stiff clay Qp

=

0.126 9565.6 x -8- = 153.7 kN 7. 4

Group load capacity Quit

=

95.9 + 118 + 153.7

=

367.6 kN

(Qu/t)g = nQul1 = 9 x 376.6 = 3308.4 kN

B.B ELASTIC SETTLEMENT OF PILES

Step 4:

397

Decide which failure mode and conditions govern. Load capacity (kN) Analysis

Block mode

TSA

9150.4

ESA

11466.5

Single pile mode 3749 3308.4

The lowest ultimate load capacity is 3308.

:€SA.

•

FO

8.8

(8.21)

where G is the shear modulus and is is the skin frictional stress. The shear strains can be integrated over the pile length to give the elastic settlement (Pes) resulting from skin friction ; that is, Pes =

1 (L G(z) Jo T( Z) dz

(8.22)

where (z) means that the parameter in front of it varies with depth. To solve Eq. (8.22), we need to know how G and T vary with depth but this we do not know.

CHAPTER 8

PILE FOUNDATIONS

FIGURE 8.12

bearing.

(8.23)

and (i) is the shear stress at an ordinate i. We , . ment to solve Eqs. (8.22) and (8.23). Stress

FO

398

Qa I

Pes = E L

(8.24)

so

where 1 is an influence factor and Qa is the design load. An approximate equation for 1 is 1 = 0.5

+

IOg(~)

(8.25)

Soft soils tend to have elastic moduli that vary linearly with depth; that is, (8.26)

where m is the rate of increase of Eso with depth (units: kPa/m). For soft soils, the elastic settlement is (8.27)

8.8 ELASTlC SETTLEMENT Of PILES

399

where

(8.28)

(8.29) n in flu ence factor

.13.

(8.30)

(8.31 ) ngth and Ap is the area of the cross section of the pile.

03

I Q.

'.

Values 01 X~,

T 1 --iL1--

0.'

10'

L

0 1

10'

Q. P" =E D/~

•

0

1

10' 10

100

I; FIGURE 8.13 Influence fact or for vertical settlement of a single fl oating pile . (After Poulos, 1989.)

400

CHAPTER 8

PILE FOUNDATIONS

Piles in a group tend to interact with each other depending on the spacing between them. The smaller the spacing, the greater the interaction and the larger the settlement. Pile group settlement is influenced by spacing to diameter (or width) ratio (sID), the number of piles (n) in the group, and the length to diameter ratio (LID) . For convenience, pile group settlement is re to a single pile settlement through a group settlement factor Rs defined as Settlement of group

Rs = Settlement of single pile at same

(8.32)

An empirical relationship between Rs and n (Fleming (8.33)

8.9 CONSOLIDATION SETTLEJVrENl\ UNDER A PILE GROUP

vertical stress at

F

(8.34)

l.L 3

Load transferred to this level

"

FIGURE 8 .14 Assumed distribution of load for calculating settlement of a pile

group.

401

8.10 PROCEDURE TO ESTIMATE SETTLEMENT OF SINGLE AND GROUP PILES

where Qag is the design group load, Bg is the width of the group, Lg is the length of the group, and z is the depth from the load transfer point on the pile group to the location in the clay layer at which the increase in vertical stress is desired. The primary consolidation settlement is then calculated using the procedure described in Section 4.4.

FO

8.10 PROCEDURE TO ESTIMATE SETTLEME OF SINGLE AND GROUP PILES The procedure to estimate the settlement of single 1. Obtain the required parameters Qa 2.

0

5. Calculate R, usin 6. 7.

8.

The esseJttill1 points are: 1. Settlement of piles is determ ined -using numerical analyses assuming the S(){[ is an elastic material. The equations givenJ()r p ile settlement will only give an estimated settlement.

Strategy This is a straightforward application of the procedure in Section 8.10.

Solution 8.7 Step 1:

Determine the influence factor.

~

=

..!2. = 25 0.4

I

=

0.5 +

D

Equation (8.25):

IOg(~) 20 X 10 5000

0.5 + log(25)

= 6

=

4000

From Fig. 8.13, Ip = 0.09 for Kps = 4000 and LID = 25.

=

1.9

CHAPTER 8

Step 2:

PILE FOUNDATIONS

Calculate the elastic settlement. Equation (8.24):

Qa 70 Pes = ELI = 5000 X 10

X

so

= 2.7 X 10- 3 m = 2.7

1.9

IIUll_ - .

Equation (8.29):

Difference between the two solutions is 0.5 m of 3 mm. Step 3:

Calculate the elastic shortening of t e Equation (8.30):

Step 4: = 3.0 mm

is approximately 3 mm.

•

FO

402

Dense sand

2m

y = 17 kN/m 3 , r.. , = 17.5 kN/m 3 39°,
t/J~ =

E' = 30 MN/ m2

10m

E;o = 30,000 MN/m2

4:L=

3m

Imr-----~~~~~~~~~~~~~

m, = 3 .5 x 10-4 m2 /kPa Gravel

FIGURE E8;8

8.10 PROCEDURE TO ESTIMATE SETTLEMENT OF SINGLE AND GROUP PILES

403

Strategy Since the sand is dense, driving of the piles will likely loosen it around the piles. You should use

~s

in calculating the pile load capacity.

Solution 8.8 Step 1:

Determine the geometric parameters and D = 0.4 m, Kps

=

Ep E; o

L

D=

~

and N .

10 0.4 = 25; n = 9 piles, s ,-( 1

30 X 106 30,000

= 1000

FO

R

Step 2:

Step 3:

capacity: Quit

=

Qf + Qb

(Q ult)S = nQult

=

9 x 675.1 = 6076 kN

=

410.1 + 265 = 675.1 kN

DeteTmine the block failure mode load capacity by proportion. Skin friction : (Qf)g = 410.1 End bearing:

(Qb)g = 265

X

X

9.6 - 6 = 3124.6 kN 1.2

5.76 -0 6 .12

=

12114.3 kN

Ultimate load capacity: (Q",J g =

Step 4:

(Qf)g + (Qb)g= 3124.6 + 12114.3

Calculate the factor of safety. The single pile failure mode governs. ... FS

=

6076 3000

='

2

=

15238.9 kN

402

CHAPTER.

Ste p 2:

PILE fOUNDATIONS

Calculate the elastic settlement.

70 X 19 Equation (8.24): P.. -- ~/ E,,,,L - 5000 x 10 . =

Equation (8.29):

2.7 X 10- 3

In =

Q.

-

70

P.. = E 0 I~ "" 5000 =

3.2

X

2,7 mm

x

1O-J m

Difference between the two solutions is 0.5 of

Step 3:

average value

3 mm.

Calculate the elast;e shorten ;ng ~ Equation (S,30). Pp "" ~

0-.) mm

nd can be Step 4:

\

stiJij.1ted tota l elastic

l1i - ~ x 100 "" 0.75% a load of 3 (b) Calcula

r = 17~Nlm', 1'... . 17 .5 1
_;_39" ,,;'_31'

E;.. .. 30 MNlm 2 E, = 30,000 MNlml

FIGURE EB.&

•

8.10 PROCfOU RE TO ESTIMATE SETI LEMENT OF SIN GLE AN D GROUP PILES

403

Strategy Since the sand is dense, driving of the piles will likely loosen it around the piles. You should use 4>~ in calculati ng the pile load ca pacity. Solution 8.S

f3 and

Step 1: De termine the geometric parameters and D = OA m '

L

O•

10

'"

. ",, 25; n = 9piles,s O4

)

£p 30 x lW KP1 .... £ ;0 '" 30,000 = 1000

Table 8.3:

~ b~ line~ inttr~olalion .Of Meyer~Of (197

"

ure 8.9

(!1 1

N. =

Step 2:

p

D e ermine

a,N~

40.

•

Cr, -

.55 x 40 =

th~ single

0.55

rOr D

values]

25,

2~

pile fai lure

~ load capacity.

3

7'lJ - 9.8 • 7.7 1m Ce ter of sand layer ithin emb$dmenl length of lhe pile: 'I

= 2 x 17 ~ x 7.7 ~a At base: j.0'~)" = 2 X 17 + 8 x 7.7 = 95.6 kPa

q.

Skin

fric~; ~O'~ x

Perimeter x Length 57. 1. x 10 = 410.1 kN E ~be.ari q ': N9 (q;)~ A ~ = 22 x 95.6 x 0.126 = 265 kN Ultimate loa capacity: Qw, :: Qf + Qo = 410.1 + 265 = 675.1 kN

"" ts

(Q,.,,)& = nQ,." "" 9

Step 3:

x 675.1

= 6076 kN

Determ ine the block failure mode load capacity by proportion. Skin friction :

:.~~

= 3124.6 kN

g~7266

- 12114.3 kN

(Q, ), - 410.1 X

End bearing: (Q,,), - 265 X

Ultimate load capacity: (Q.,J,), - (Q, ), + (Qb),= 3124.6 + 12n4.3 :: 15238.9 kN

Step 4:

Calculate the factor of safety. The single pile fail ure mode governs. FS :: 6076

3000

-= 2

CHAPTER 8

Step 5:

PILE fOUNDATIONS

Calculate the elastic settlement of the pile. Assume the full design load is carried by skin friction and the load is equally shared by each pile. 0

Q o = 3°9°

= 333.3 kN

Equation (8.25):

L For 15

Equation (8.24):

p

= 25,

I

= 0.5 + log(25) =

Qo 333.3 - I = X EsoL 30,000 X 10

= -

es

Neglect elastic shortening of the pile s Assume = 0.5 .

Step 6:

•

FO

404

Piles located in sett mg soil layers (e.g., soft clays or fills) are subjected to negative skin friction called downdrag (Fig. 8.15). The settlement of the soil layer causes the friction forces to act in the same direction as the loading on the pile. Rather than providing resistance, the negative friction imposes additional loads on the pile. The net effect is that the pile load capacity is reduced and pile settlement increases. The allowable load capacity is given, with reference to Fig. 8.15, as

I

Qa

=

Q b + Qj

FS

- Qnr

I

(8.35)

For soft, normally consolidated soils, the negative skin friction is usually calculated over one-half its thickness. Negative skin friction should be computed for long-term condition; that is, you should use an ESA.

8.11 PILES SUBJECTED TO NEGATIVE SKIN FRICTION

-- - ------ - ---

---- - ---_ .. _-- --- ---------=-=::---=--=.:-- -- ---

FIGURE B. 15

Negative skin friction.

EXAMPLE 8.9

FO

R

Determine the allowable load E8.9. The fill is recent

Recent fill r = 18.5 kN/m 3 Soft normally consolidated clay 5"

=30 kPa , ¢;", =26°

rsat = 17.5 kN/m 3

·' 1

15 m

-.

FIGURE EB.9

~ 3 rsat = 18 kN/m ¢~

= 36°,

¢;",

= 30°

405

406

CHAPTER 8

PILE FOUNDATIONS

Solution 8.9 Step 1:

Determine

13, N q , and other relevant parameters.

15L =

25 004 = 62.5; Perimeter = TID = OATI = 1.26 TID2

Ab = -

4

=

TI X 004 2

4

= 0.126 m2

Clay

13

=

(1 - sin ~,) tan cj>~s (OCR)0 5 = (1 - sin

Sand

Step 2:

8.7 x 4 + 7.7 x 6 + 8,2 x 7.5 = 142.5 kPa

FO

15

=

1400,5 kN

Step 4: Qa

= (Qj)sand + (Qb),and _ Q = 1400.5 + 329 _ 82 = 58 kN FS nf 3 11. 4.3

•

8.12 PILE-DRIVING FORMULAS AND WAVE EQUATION A number of empirical equations have been proposed to relate the energy delivered by a hammer during pile driving and the pile load capacity. One of the earliest equations is the ENR (Engineering N ews Record) equation, given as WRh

Q " I51! = + c-l

(8.36)

8.12 Pl LE·DRIVING FORMU LAS ANO WAVE EQUATION

TABLE B.8

401

Hammer

Efficiency Hammer type

'h

Orop hammer

0.75-1.0 0.75-0.85 0.85 0.85-1.0

Single-acting hammer Double·acting hammer Oiesel hammer

(8.37)

B.8. down the pil the pi le ave

~

rss wave. which moves (8.3B)

whe re Et = du/az is the vertical str-a ~p is Young's modulus of the pile, A p is the cross·~ctional area-6f t')tPile, and u is the pile displacement From Newton 's second law.

r

(IF

'f

rr-u

-= A - al P gor 2

(8.39)

where g is the ccel£rw n due 10 gravity and f is time. Setting Eq. (8.39) equal

tOE~ei"'l"'"

ifu "f flu EA - = - A P 'oZ2 g Pat"

(8.40)

which, by simplification, leads IO r:C-_ _-,--,

a u '" ~ rfu a(l ~ azl 2

(8.41)

where (8.42)

is the vertical wave propagation veloci ty in the pile. The solution of Eq. (8.41) is found using appropriate boundary conditions and can be modified to accouot fo r soil resistance. Computer programs (e.g.,

408

CHAPTER 8

PILE FOUNDATIONS

WEAP) have been developed for routine wave analysis of pile-driving operations. These programs are beyond the scope of this book. Driving records can provide useful information on the consistency of a soil at a site. For example, if the number of blows to drive a pile at a certain depth is A blows and B blows at another location at the same site, then tIl oil stratification is different. You should re-examine your design and the soil borin records and make the necessary adjustments, for example, incr ase r decrea e the pile length .

The essential points are: 1. A number of empirical equations are avajlable to estimate the pile load capacity from driving records. 2. Driving records can provide some useful inform a,twn regarding the character of the soil at a site. 3. Careful judgment and significant e:xperi't!flce are required to rely on pile load capacity from pile-driving I) erqlions.

SUMMARY

FO

8.13

A fish port fac· t is be constructed near a waterfront area as shown in Figs. E8.10a ,b. A soil Investigation shows two predominant deposits. One is very soft, normally consolidated clay and the other is a stiff, overconsolidated clay. Soil data on the two deposits are shown in Fig. E8.10c. Determine the pile configuration (single or group piles) , the pile length , and the expected settlement to support a design column load (working load) of 500 kN at A (Fig. E8.l0) . Timber piles of average diameter 0.38 m and average length 18 m are readily available. The elastic modulus of the pile is Ep = 20,000 MPa. The settlement should not exceed 0.5% of the pile diameter. The pile shafts within the tidal zone will be treated to prevent rot. From experience on this site, it is difficult to drive piles beyond 8 m in the stiff, overconsolidated day. Driving tends to damage the pile head. You should allow for a 0.25 m cut out from the pile head.

Strategy The very soft silty clay is likely to cause downdrag on the pile. You should determine the single pile capacity assuming the full available length of

8.13 SUMMARY

409

Wharf

16om t;

Fender piles @ 3 m center

D

A 600 kN Plant

~~ "~;~-------------------------,~ '::'I,'f~r------- 100 - - - - -"="'1.

m

(a) Plan

-3.5m-----'-------::I

-5.9 m - - - - - - - - - - ,

-10.9 m - - - - - ,

R

consolidated clay w

= 32%,

W LL

= 68%,

W PL

= 25%

s" = 120 kPa, ¢:o, = 25°, Ysa ! = 18 kN/m 3

FO

(E, )" = 90 MPa, OCR = 8

(c) Borehole

#1

the pile will be used. If a single pile is not capable of carrying the load, then you should design a pile group. Use an ESA when you are considering downdrag.

Solution 8.9 Step 1:

Determine the geometric parameters, D Ab

=

0.38 m; TID2

Perimeter 2 TI X 0.38

4

4

= --

=

=

=

TID

= TI

0: , ~,

and N q .

x 0.38

=

1.19 m;

0.113 m 2

From borehole #1, the length of piJe in soft clay: Length of pile in stiff clay:

L1

=

7.4 m

L2 = 18 - 7.4 - 3.5 - 0.25 = 6.85 m.

CHAPTER 8

PILE FOUNDATIONS

The value 0.25 m is the cut-off length from the pile head and 3.5 m is the distance from the finished elevation to the surface of the soft soil. Soft clay, Table 8.3:

!3 = (1

- sin 24°) tan 24°(1) 112

= 0.26

Stiff clay, Table 8.3:

Assume tVP = TI/2. Then Nq

Step 2:

=

(tan 25° + VI + tan 2 (2SOW ex p ( 2

~

Determine the negative skin friction for a sin ~ A conservative estimate is to assume ne~ative whole length of the pile in the very s0f t ca .

Step 3:

Step 4:

FO

410

. . 600 Number of plies requIred = 180.6

=

3.3

Try 4 piles in a 2 x 2 matrix at a spacing of 1 m: s

15 =

1 0.38 = 2.6

+ D = 1 + 0.38 = 1.38 m ; Perimeter = (1.38 + 1.38) X 2 = 5.52 m; Ab = 1.38 2 = 1.9 m 2 Soft clay: Qnf = 0.27 X 13.3 X 5.52 X 7.4 = 146.7 kN

Step 6:

Bg = Lg =

S

Stiff clay:

(Qf)g

=

0.76 X 81.4 X 5.52 X 6.85

Stiff clay:

(Qb)g

=

10.7 X 109.5 X 1.9

=

=

2339 .2 kN

2226.1 kN

Calculate the allowable load capacity for block failure mode. (Qa)g = (Qj) g : (Qb)g _ Qnf

=

2339.2 ; 2226.1 - 146.7 = 1375 kN

EXERCISES

Step 7:

411

Calculate the allowable load capacity for single pile failure mode. n

=

4 piles

(Qa)g = nQa = 4 x 180.6 = 722.4 kN

Therefore, a 2 X 2 pile group is adequate. Single pile failure mode governs the design. Step 8:

Calculate the settlement. Assume the full design load of Qa = 600 k friction (floating pile) within the stiff clay' . 60 Load per pile (Qap) = 4 -

FO

•

red to support the load shown in Fig. P8.1. The diameter A"'f.act()r of safety of 3 is required .

Soft clav = 30 kPa , y = 17 kN/m 3

S"

¢~

L

FIGURE PB. 1

=30°, OCR = 1.2

Stiff clay = 60 kPa, J'sat

Su

¢;, = 24°,

OCR

= 18 kN/ m3 =4

412

CHAPTER 8

8.2

PILE FOUNDATIONS

Determine the allowable load for a pile, 0.4 m in diameter, driven 20 m into the soil profile shown in Fig. P8.2. Groundwater is at 2 m below the surface but you can assume it will rise to the surface. A factor of safety of 3 is required.

2

Soft normally consolidated clay = 15 kPa, Y= 18 kN/m 3

$"

10 m

1 FIGURE PS.2

8.3-

18.5 kN/m 3 ,

¢J'cr =

25°, OCR = 2

F

, rsa, =

s" = 100 kPa,

r"" = 18.5 kN/m 3 , ¢J'cs = 25°, OCR = 4

FIGURE PS.3

8.4

Estimate the allowable load capacity of a 0.5 m diameter pile embedded 17 m in the soil profile shown in Fig. P8.4. The factor of safety required is 3. The N values are blows/ft (blows/0.31 m).

EXERCISES

Depth (m)

413

N va lues

0

2 4

11 5

20 Sand

10

19

20 Rock

FIGURE PS.4

8.5

FO

R

of 3 is required.

8.6

A soil profile consists of 3 m of a loose fill (K, = ISO, "'Isat = kN/m 3 ) over a stratified overconsolidated clay. The shear strength of the overconsoJidated clay increases linearly with depth. At the top of the clay, s" = 90 kPa and at a depth of 20 m (17 m into the clay), s" = 180 kPa . The critical state friction angle is ~s = 28°, OCR = 8 and "'Isat = 18 kN/m 3 Determine the ultimate load capacity of a drilled shaft (bored pile) 1 m in diameter embedded 18 m into the clay. Groundwater is at 8 m below the ground surface. You may assume that the soil above the groundwater level is saturated.

8.7

Estimate the allowable load capacity of a 3 x 3 pile group. Each pile has a diameter of 0.4 m and is driven 15 m into a soft clay whose undrained shear strength varies linearly with depth according to s" = 0.25 u~. The critical state friction angle is 30° and "Isat = 17.5 kN/m 3 The spacing of the piles is 1.5 ffi. Groundwater level is at ground surface. A factor of safety of 3 is required.

414

CHAPTER 8

PILE FOUNDATIONS

Practical

FO

8.8

The soil profile and soil properties at a site are shown in the table below. A group of 12 concrete piles in a 3 x 4 matrix and of length 12 m is used to support a load. The pile diameter is 0.45 m. Determine the allowable load capacity for a factor of safety of 3. Calculate the total settlement (elastic and consolidation) under the all able load. Assume Ep= 20 X 106 kPa.

Depth (m)

0-3

Type of deposit

Sand

3-groundwater level

3-6

Sand

6-15

Clay

15-17

>17

CHAPTER

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

FO

9.0 . INTRODUCTION

• Calculate flow under and within earth structures • Calculate seepage stre sses, pore water pressure distribution, uplift forces, hydraulic gradients, and the critical hydraulic gradient

• Statics • Hydraulic gradient, flow of water through soils (Chapter 2) • Effective stresses and seepage (Chapter 3)

Sample Practical Situation A deep excavation is required for the construction of a building. The soil is a silty sand with groundwater level just below ground level. The excavation cannot be made unless the sides are supported. You, a geotechnical engineer, are required to design the retaining structure for the excavation and to recommend a scheme to keep the inside of the excavation dry. Figure 9.1 shows the collapse of a sewer and a supported excavation by seepage forces. You should try to prevent such a collapse. 415

416

CHAPTER 9

lWO-DIMENSIONAl FLOW OF WATER THROUGH SOilS

9.1

FO

'Iar t frictional stress in pipes) imposed on a soil vi or of a soil as a viscous fluid when seepage re-

QUESTIONS

UIDE YOUR READING

1. What is the governing equation for two-dimensional flow and what are the methods adopted for its solution for practical problems? 2. What are flow lines and equipotential lines? 3. What is a flow net? 4. How do I draw a flow net? 5. What are the practical uses of a flow net? 6. What is the critical hydraulic gradient? 7. How do I calculate the pore water pressure distribution near a retaining structure and under adam? 8. What are uplift pressures and how can I calculate them?

9,3 TWO-DIMENSIONAL FLOW OF WATER THROUGH POROUS MEDIA

417

9. What do the terms static liquefaction, heaving, quicksand , boiling, and piping mean? 10. What are the forces that lead to instability due to two-dimensional flow ? 11. How does seepage affect the stability of an earth retaining tructure?

FO

9.3 TWO·DIMENSIONAL FLOW OF WATER THROUGH POROUS MEDIA The flow of water through soils is described by Lapla ' through soils is analogous to steady state h flew a mogeneous conductors. The popular for . if L sional flow of water through soils is (9.1)

the X and Z directio s. changes of hydraulic gr other directions. The assu-..··.." ,....

aterial then k x = k z and Laplace's equation ,PH

-

~

dX 2

,PH

+-

dZ 2

= 0

(9.2)

lution any differential equation requires knowledge of the boundary conditio . h boundary conditions for most "real" structures are complex, so we cannot 0 ain an analytical solution or closed form solution for these structures. We have to resort to approximate solutions, which we can obtain using numerical methods such as finite difference, finite element, and boundary element. We can also use physical models to attempt to replicate the flow through the real structure. In this chapter, we are going to consider two solution techniques for Laplace's equation. One is an approximate method called flow net sketching; the other is the finite difference technique, which you have encountered in Chapter 4. The flow net sketching technique is simple and flexible and conveys a picture of the flow regime. It is the method of choice among geotechnical engineers. But before we delve into these solution techniques, we will establish some key conditions that are needed to understand two-dimensional flow. The solution of Eq. (9.1) depends only on the values of the total head within

CHAPTER 9

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

the flow field in the XZ plane. Let us introduce a velocity potential describes the variation of total head in a soil mass as I;

=

kH

= k x. aH = al;

Vx

ax

=

Vz

k aH Z

az

(~),

which (9.3)

where k is a generic coefficient of permeability. The velocities of and Z directions are

ax

0

in the X

(9.4)

al;

(9.5)

az

(9.6)

(9.7)

FO

418

Streamline or flow line

t.h = head loss or -¥-potential drop

.--..=-____ h +t.h

h + 2Llh

FIGURE 9 .2

Illustration of flow terms.

Constant total head or equipotential line or a line of constant piezometric head

9.4 flOW NET SKETCHI NG

419

Since flow lines are normal to equipotential Jines, there can be no flow across flow lines. T he rate of flow between any two flow Jines is constanl. The area between two flow lines is called a flow channel (Fig. 9.2). Therefore. the rale of flow is constant in a flow channel.

1. 2. 3. 4. 5. 6. 7.

The essential points are: Streamlines or flow lines repN!SentfloM' paths oj particles nfwat ,.r. The area between two flow lines ;s allied a flo w chann el. The rare of flow in a flow channel is constallC. Flow cannot occur acroSl-jlow lines. Th e velocity ojflow is normal to the equipo tential fin e. Flow li/JeJ.· and equipotential finIS are orthogona l (perpendicular) to each other. Th e difference in head between tWO equip otentiallinel' is called the p otential drop or head [OS-f.

What's next . .•The flow co ditio ~ e e ablished i e p rev.ious ection allow us to use a graph ica l method'to lnd s o uti o ns to two-di mens.i..0 na o w p roblems. In the next section, we wil l describe flow n sketching ~ a provid&g uid ance in interpreti ng a flow net to determ ine low thro ug h soi ls, the distrib ution of' pore water pressures, and the hydra ulic. g radie n t~."'"

9.4 Flow Nots flow net is a grephical rep,c ntalion of a flow field that satisfies Laplace's eAuation and comR!.ise.s a fa mily of flow lines and equipotcmiallines. A flow net must7 'thc following criteria: 1. The 3.

4. 5. 6. 7.

st intersect equipotentia l lines at righl angles. The area between flow lines and equipotential lines must be cu rvilin ear squares. A curvihnear square has the property that an inscri bed circle can be drawn to touch each side of the square and cont inuous bisection results, in the limit, in a point. The quan tity of flow through each flow channel is constant. T he head loss between each consecutive equipotential line is constant. A !low line ca nnot in tersect another ftow line. An equipotential line can not intersect another equipotentia l line.

An infinite number of flow lines and equipotent ial li nes can be drawn to satisfy Laplace 's equation. However, only a few are required to obtain an accurat e solution. The procedure for constructing a flow net is described next.

42«)

CHAPTER 9

TWO·DIMENS IONAL FLOW OF WATER TH ROUGH SOILS

Scale

10 m

FIGURE

,

9.3 Flow net for

s heet pile.

>

9.4 .2 Flow Net for Iso t. Draw [he SI ~ re

icC:! soil mass to a suita~ scale.

2. I den ~fY--im pe n ne8 Ie and perr.:eable una)fles. The soil- impermeable bouna r i er aces are flow 1i ~ because water can ftow along Ihese in· terfac . Th soil-permeable bO'l.ndarr. nterfaces are equipotential lines because , e 10lal hea~.)co n slan l 110ng these interfaces. Sketch a series of fJ~Jines (fqur or five ) and then sketch an appropriate ".._~u"" ber of equi tentia Jfnes such that the area between a pair of flow lines and a pair of e uipotential Bpis' (cell) is approximately a curvil inear square. P ~~,~ S l (he flow lines and eq uipotential lines 10 make ou would h a curvi linear s u ares~ should check t hat the average wid th and the av-

:tg;~ii ~

,, ,, , ,,

,, ,, ;

,,

,

,, ,, , ,, ,, , ,

,, ,,

,,

,

Flow net under a dam wil h a culoff curtain (sheet pile) on t he upstream end .

FIGURE 9.4

9.4 FLOW NET SKETCHING

421

Drainage blanket

Drainage pipe

FIGURE 9.5

Flow net in the backfill

0

FO

R

blanket.

9.4.3 Anisotropic Soil Equation (9.1) is Laplace's two-dimensional equation for anisotropic soils (the permeabilities in the X and Z directions are different). Let us manipulate this equation to transform it into a form for which we can use a procedure similar to the one for isotropic soils to draw and interpret flow nets. Put C = V kzlk x and Xl = Cx. Then aXl ax

=

aH ax

=

c aH aXl ax! ax

=

c aH ax!

CHAPTER 9

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

and

Therefore, we can write Eq. (9.1) as

which simplifies to (9.8)

cs e 11 use the procedure for oils b scaling the _ distance by

Equation (9.8) indicates that for anisot , flow net sketching described for v'kzlk.c That is, you must draw th horizontal distances by v'k/kx-

ow dom: " y ~ing th'

(9.9)

FO

422

l

= (b

!1h b I:lH b = k !1h - = k - L L Nd L

x 1)k -

where band L fined as shown in Fig. 9.3. By construction, bl L therefore the total flow is

(9.10) =

1, and

(9.11)

where N f is the number of flow channels (number of flow lines minus one); in Fig. 9.3, N f = 9. The ratio NflNd is called the shape factor. Finer discretization of the flow net by drawing more flow lines and equipotential lines does not significantly change the shape factor. Both N f and Nd can be fractional. In the case of anisotropic soils, the quantity of flow is q

Nf

= I:lH -

Nd

Ykxkz

(9.12)

9.5 INTERPRETATION OF FLOW NET

423

9.5.2 Hydraulic Gradient You can find the hydraulic gradient over each square by dividing the head loss by the length, L, of the cell; that is,

Ii =

~I

(9.13)

You should notice from Fig. 9.3 that L is not con ant. draulic gradient is not constant. The maximum hyd uh L is a minimum; that is, (9.14)

where L m in is the minimum length of occurs at exit points or around c at these points that we can get t Lmin

9.5.3 Static Liquefac and Piping ownstream end of element is upward.

FO

R

(9.15)

9 .5 .4 Critical Hydraulic Gradient We can determine the hydraulic gradient that brings a soil mass (essentially, coarse-grained soils) to static liquefaction. Solving for i in Eq. (9.15) when cr~ = 0, we get . I

.

'Y '

= l cr = -:;: =

(G1 h1) -:;: 1 Gh1 s -

'Y w

s -

=

(9.16)

where ier is called the critical hydraulic gradient, G s is specific gravity, and e is the void ratio. Since G s is constant, the critical hydraulic gradient is solely a

424

CHAPTER 9

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOilS

function of the void ratio of the soil. In designing structures that are subjected to steady state seepage, it is absolutely essential to ensure that the critical hydraulic gradient cannot develop.

9.5.5 Pore Water Pressure Distribution The pore water pressure at any point j is calculated as foil 1. Select a datum. Let us choose the downstream

(Fig. 9.3). 2. Determine the total head at j: H j = 6. ' number of equipotential drops at po', '; ( ample, at B, He = 6.H - 16.5 6.h. 3.

W C)

For point B, (h p )8 =

4. The pocc

(9.17)

LH ~--

(9.18)

r es

0 ·

FO

n

~

Pw

=

L

(9.19)

Uj tlX j

j= t

where P w is t orce per unit length, uj is the average pore water pressure J' and n is the number of intervals. It is convenient to use over an interva Simpson's rule to calculate Pw: Pw

=

3tlx

(

Ul

+

Un

n +2~ odd

U;

tI)

+ 4 ;~

U;

(9.20)

even

EXAMPLE 9.1 An excavation is proposed for a site consisting of a homogeneous, isotropic layer of silty clay, 12.24 m thick , above a deep deposit of sand. The groundwater is 2 m below ground level (see Fig. E9.1). The void ratio of the silty clay is 0.62 and its specific gravity is 2.7. What is the limiting depth of the excavation to avoid heaving?

9.5 INTERPRETATION OF FLOW NET

425

'I

12.24

1 FIGURE E9. 1

Solution 9.1 Step 1:

Calculate icr.

- h

.

=

12.24 - h

!1H

[=-

FO

R

h

1 as = 12.24 - h .

h

•

A dam, sho fl 19. E9.2a, retains 10 m of water. A sheet pile wall (cutoff curtain) on the upstream side, which is used to reduce seepage under the dam , penetrates 7 m into a 20.3 m thick silty sand stratum. Below the silty sand is a thick deposit of clay. The average coefficient of permeability of the silty sand is 2.0 X 10- 4 cm/s. Assume that the silty sand is homogeneous and isotropic. (a) Draw the flow net under the dam. (b) Calculate the flow , q.

(c) Calculate and draw the pore water pressure distribution at the base of the dam. (d) Determine the uplift force. (e) Determine and draw the pore water pressure distribution on the upstream

and downstream faces of the sheet pile wall.

CHAPTER 9

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

k =2

X

10-4 cm/s

G

FIGURE E9.2a

(f)

water. (g) (h)

the sheet pile

(i)

wall?

Strategy and calculate the

FO

426

Step 4:

l'

ree flow lines and then draw a suitable number of equip lines. Remember that flow lines are orthogonal to equipotential lines and the area between two consecutive flow lines and two consecutive equipotential lines is approximately a square. Use a circle template to assist you in estimating the square. Adjust/ add/subtract flow lines and equipotential lines until you are satisfied that the flow net consists essentially of curvilinear squares. See sketch of flow net in Fig. E9.2b. Calculate the flow. Select the downstream end, EF, as the datum. !J.H

=

10 m

Nd

=

14, Nt

=

Equation (9.11):

4

Nt q = k !J.H - = 2 Nd

X

10- 4

X

(10

X

102)

X -

4

14

= 0.057 cm 3 /s

427

9.5 INTERPRETATION OF FLOW NET

FIGURE E9.2b

Step 5:

L

Determine t

re

to, con,",'; nce

:te pressure at each nodal point. Use a table etteF et, use a spreadsheet. t:..H

10

M = = - = 0.714 m Nd 14

FO

R

~ _

tion in the table below was done using a spreadsheet

Flow under a dam:

t:..h = 0.714 m

Under base of dam

Parameters x(m)

Nd(m) Ndah(m) h,(m)

0

3.06

6.12

9.18

12.24

15.3

18.36

21.42

24.48

27.54

30.E

5.60

5.80

6.20

6.90

7.40

8.00

8.80

9.40

10.30

11.10

12.5C

4.00

4.14

4.43

4.93

5.28

5.71

6.28

6.71

7.35

7.93

8.93

-2.40

-2.40

-2.40

-2.40

-2.40

-2.40

-2.40

- 2.40

-2.40

-2.40

- 2.4C

hp (m) = aH - Ndah - h,

8.40

8.26

7.97

7.47

7.12

6.69

6.12

5.69

5.05

4.47

3.4E

u (kPa) = hp'lw

82.3

80.9

78.1

73.2

69.7

65.5

59.9

55.7

49.4

43.9

34.1

Plot the pore water pressure distribution. See Fig. E9.2c.

428

CHAPTER 9

lWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

x (m)

5

10

15

20

25

30

35

--

10 20 30

:,..

ro a. 40

~

--

" 50 60 70 80 90

-~

---

.

....... ~

~

./'" -1 -

FIGU HE E9.2c

~

:~~

Step 6:

FO

Step 7:

~

Parameters

z(m)

Back of wall

Front of wall

0

1.17

2.33

3.50

4.67

5.83

7.00

7.00

0.70

1.00

1.30

1.60

1.90

2.40

3.00

5.00

5.60

0.71 -3.57

0.93 -4.73

1.14

1.36

1.71

- 5.90

-7.07

-8.23

2.14 -9.40

3.57

hz(m)

0.50 -2.40

-9.40

4.00 -2.40

hp (m) = !J.H - Nd!J.h - h z

11.90

12.85

13.81

14.76

15.71

16.52

17.26

15.83

8.40

u (kPa) = hp'lw

116.6

126.0

135.3

144.6

154.0

161.9

169.1

155.1

82.3

Front

Back

Difference

1011.7

830.9

180.8

Nd(m) Nd!J.h(m)

Pw (kN/m)

0.00

429

9.6 FINITE DIFFERENCE SOLUTION FOR TWO,DIMENSIONAL FLOW

..

600rc~

u (kPa)

1~ Pal

____~1'~O~____""200

r1~ 'O~__~1~OOe,'--"~r-__~OO

J

1 -

2 -

§ •

3

l4

)

,' f - 7 '--_ _ _---l~---'

(41

FIGURE E9.2 Pore water pressure d istri butio n (d) a

of wall.

Step 8:

Determine the

maXimum..hY~l!

The sma llest val ue of Lm ,,, = 2 m.

occur

grad'ent. .-( By measurement,

t the xi I.

0.714

Step 9:

'f PiPlt

E,

inee i m u <

l e..

•

0 l!ld occur.

"' " (9 16)

i" ~

piping will not

J

Faclor of safet agai

effeQ~':ft2 'n

I piping

~:

=

2.5

State the the depth of penet ration of the shee t pile wall. "If the de h is red uced, le va lue of Ill! increases and i m u is likely to

•

<-9.6

FINITE IFI:ERE~CE SOLUTION TWO-DIMENSION~LOW

FOR

In Chapler 4, we used the finite difference tech nique to solve the governing one, di mensional panial differentia l equa tion to determine the spatial va n at ion of excess pore water pressure. We will do the same 10 solve Laplace's equa tion to determine two-dimensional confi ned How through soils. Let us conside r a grid of a How domain as shown in Fig. 9.6, where (i ,j) is a nodal point. Using Taylor's theorem, we have

(9.21)

430

CHAPTER 9

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

,---------------------+x Co lumn s

--+-

-- i- I Row,

i + 1

+ j - 1

------t--------t--------t-- - - - - Row L h i - 1. i - I

hi. i-I

h i + I. i-I

z j +! ------~------~------~----~ hi _ 1. j + 1 hi, j + 1

~

'~r.: -.' ~n ..

into a square

(9.22)

FO

(9.23)

~

(9. 24)

The finite difference form of Eq. (9.24) is ah ax

Therefore, h i. i + 1 =

hi,i-l

1

= 2~x

(h i. j +l

-

h i. j-I)

=0

(9.25 )

and, by substitution in Eq. (9.22) , we get

Ih

i. j

= hhi+ l .j

+ hi-I ,j + 2hi.j- 1)

I

(9.26)

Various types of geometry of impermeable boundaries are encountered in practice, three of which are shown in Fig. 9.7. For Figs. 9.7a,b the finit e difference equation is (9.27)

431

9.6 FINITE DIFFERENCE SOLUTION FOR TWO-DIMENSIONAL FLOW

i, j - 1

i, J - 1

o

[J

I

i - I,j

i, j - 1

Impermeable

i, j

l i,j

i + l,j

~J--D -- O

i + 1, j Impermeable

Impermeable (a)

FIGURE 9 _7

and, for Fig. 9.7c, r---------------------~~~~r_~~

(9.28)

(9.29)

FO

R

(9.30)

(9.31)

o a a by considering a vertical plane across the flow dop row and K be the bottom row of a vertical plane defined Rig. 9. . Then the expression for q is kx q = 4

hi+ I .L

-

hi-l. L

+

K-l

2

.2:

} = L+I

(h i+ 1• i - hi-l.)

+ hi+ 1 ,K

) -

h i - I .K

(9.32)

The procedure to determine the distribution of potential head, flow , and pore water pressure using the finite difference method is as follows:

1. Divide the flow domain into a square grid . Remember from Chapter 4 that finer grids give more accurate solutions than coarse grids, but are more tedious to construct and require more computational time. If the problem is symmetrical, you only need to consider one-half of the flow domain. For example , the sheet pile wall shown in Fig. 9.8 is symmetrical about the wall

432

CHAPTER 9

TWO·DIMENSIONAL FLOW OF WATER THROUGH SOILS

Equipotential bounda ry

:--2D (minimum) I

I

J.. I

Flow line ~ :

I I

~ ~-----T

I

I I

D

I I

HI

I I

C~ I ====F= low==lin=e====~F=======-====-oFIGURE 9 .8

A sheet pile wall:

ample , if D is the thickn~ width of the left half of the

3.

FO

o

6.

ntt tie new value at a node differs from the old value by tolerance, for example, 0.001 m.

7.

el t a sequential set of nodes along a column of nodes and calculate the flow, q, using Eq. (9.32). It is best to calculate q' = q for a unit permeability value to avoid too many decimal points in the calculations.

8. Repeat items 1 to 6, to find the flow distribution by replacing heads by flow q '. For example, the flow rate calculated in item 7 is applied to all nodes along AC and CF (Fig. 9.8) . The flow rate at nodes along BE is zero. 9. Calculate the pore water pressure distribution by using Eq. (9.29) . A spreadsheet program can be prepared to automatically carry out the above procedure. However, you should carry out "hand" calculations at selected nodes to verify that the spreadsheet values are correct.

9.6 FINITE DIFFERENCE SOLUTION FOR TWO-DIMENSIONAL FLOW

4

433

EXAMPLE 9.3

Determine the flow under the sheet pile wall (Fig. E9.3a) and the pore water pressure distribution using the finite difference method.

T

t ~--,-----,-.-.,---,..~~=,--....-l

4 m

Datum

m

6m

t

6m

l _........r Impermeable

FIGURE E9.3a

Strategy

FO

R

.3 and in Fig. 9.3c.

roo m is symmetrical, perform calculations for only se the left half of the flow domain of width

I

A

f - - - - l - t - - - - - t -- - - - - t - - - - + - - B-

12 m

l

eT

f--t-------+--t+-+------+----t--t-+-----i 6 m

~~~~~~~~~~~~ Dl E

~--~--------24m--------------~

2 m x 2 m grid

FIGURE E9 _3b

TABLE E9.3 A

-

-

B

E

0

C

G

F

K

L

M

N

~ ~

20

22

24

3.00

3.00

3.00

2.7~ 2.68

2.64

H

I

1

Example 9.3

2

Flow under a sheet pile wall using finite difference method

3

H,

4m

4

H2

1m

5

f'..H

3m

6

H3

6m

7

0

J

12 m

Cell size 2m 8 t--9 k 0.0002 cm/s 10

Equ ipotentials

11

zlx

0

2

4

6

8

10

12

3.w(

0

3.00

3 .00

3.00

3.00

3.00

3.00

2

2.96

2.96

2.96

2.95

2.94

2.92 1~0

14

4

2.93

2.92

2.92

2.90

2.88

2~f' ~ 1\2.~ 12.69

15

6

2.90

2.89

2.88

2.86

2.83

16

8

2.87

2.87

2.85

2.83

2.80

17

10

2.86

2 .85

2.84

2.81

2~7;. ,-"'2.72

18

12

2.85

2.85

2.83

2)3:.'!

2.7\

~tr

19

-

20

q'

21

~

1.63 m 3/s ~e r unit valu t

~~

,

~5

2.55 2.53 ~

FO ~~~

1'32

-

33

24

39 .2

39.2

55.7

55.3

77 .4 I ~)y~ 1~1.b \"76.6

76 .1

75.4

74.5

73.4

71.9

70.6

~

j~30 t--31

22

-"

39.2

58.3

116.~

20

') 11

56.3

58.4 1SS.4 77.7 77 .6

116.3

1.50

56.8

2~ ....&,8.4

1 ~1 ~

2,93

39 .2

24

~ h1¢

1.50

1.78

39.2

39.2 1, 39.2

ff7

{ 80

57 .3

39.2

\'28

2. q5

57.6

39.2 { 39 .2

96.8

1.50

2 .~ 'IJ .26 ~O ~"24

~

57 .9

39 .2

~g'6 .9

1.50

~ ~~

0

)17.7

2.21

18

23

A

2.34

2.4~ ~~ i\.1 ·99

17

0

_ 6~ ~~O

~

2.60 A 2 .49 2.12

,

),

2.80

2.32

2.48

~,

zlx

26_

2ft4

~ .6f 2.57

~ni ~ .63

0''1

Pore water ~r~'ss,!! re ~fa) 6 -" f"' 8

2,~

2.74

~~

2.75'- ~..?8 1 2.59

22

25

-

3.09~ 3.9 0' ~ .OO

12 13

10

~ ~8.1

JJ,~ ~96.4 ~6.9; ' 95.4 94.7 93.8 92 .5 90.7 88.1 83.3 1~'5r9- I~' l ji'l 114.5 113.6 112.5 111.0 109.0 106.3 102.9

13 5.8 135.8 13~6' 1>J.35.4 135.0' 134.4 133.7 132 .8 131.5 129.9 127.9

125.4 122.5

155.4 155..3 155.2 1'S~19 IWl(S 153.9 153 .2 152.2 151.0 149.3 147 .3 144.8 142.1 Flow lin: s

'V

q' (upstream) = 1.63 m 3/s per value unit of k q ' (downstream) = 0

34 35

zlx

0

2

4

6

8

10

12

14

16

18

20

22

24

36

0

1.63

1.58

1.54

1.49

1.43

1.35

1.26

1.14

1.00

0.81

0.58

0.31

0.00

37

2

1.63

1.59

1.54

1.49

1.43

1.36

1.27

1.16

1.02

0.83

0.61

0.32

0.00

38

4

1.63

1.59

1.55

1.51

1.45

1.39

1.31

1.21

1.07

0.90

0.68

0.38

0.00

39

6

1.63

1.60

1.56

1.53

1.48

1.43

1.37

1.28

1.17

1.03

0.83

0.53

0.00

40

8

1.63

1.61

1.58

1.56

1.53

1.49

1.44

1.38

1.30

1.20

1.07

0.91

0.77

41

.10

1.63

1.62

1.60

1.59

1.57

1.56

1.53

1.50

1.46

1.41

1.35

1.28

1.24

42

12

1.63

1.63

1.63

1.63

1.63

1.63

1.63

1.63

1.63

1.63

1.63

1.63

1.63

9.6 FINITE DIFFERENCE SOLUTION FOR TWO-DIMENSIONAL FLOW

435

Pore water pressure (kPa)

lOrO__-,8,O__-,6,O__-.40~__,20__~Oo

2E ~ (l

3~

(l)

1!

4;; .<::

5~ 0 6

FIGURE E9.3c

Step 2:

OW

Step 3:

lines.

Dete

FO

R

Step 4:

Step 5:

Step 6:

pply a priate equations. the im ermeable boundaries-B13 to B18 (corresponding to AE), Cl to Q (corresponding to ED), and N13 to N14 (corresponding to BC)~pply Eq. (9.26). You should note that some nodes (e.g., B18) are common. Apply Eq. (9.22) to all other cells except cells with known heads. Carry out the iterations. In Excel, go to Tools ~ Options ~ Calculation. Select the following: (i) Automatic. (ii) Iteration, insert 100 in Maximum Iterations and 0.001 in Maximum

change. (iii) Under Workbook Options, select Update remote reference, save

external link values, accept labels in formulas. You can then click on Calculate Now (F9) or Calc.sheet.

CHAPTER 9

Step 7:

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

Calculate q. Use Eq. (9.32) to calculate q' for a unit value of permeability. In the spreadsheet for this example, q' is calculated in cell C20 as

{(8l3 - 815) + N13 - N15 - 2 * SUM(C15:M15)}/4

+

2)

*

(SUM ~M:; )

The actual value of q is q

=

kq '

= 1.62 x 2 x 10- 4

=

3.2 X 10- 6 m/s

X

(10- 2

10- 2 is used

to

to m/s)

0 Yert

Step 8:

Step 9:

•

FO

436

H x .;:;

~--· D----------------------X~M~--~N ~-~~--~ . ~

1+1------------- b - - - - - - -- 11+-/ -+j~ / -+ FIGURE 9.90 Phreatic surface within an earth dam .

Cl

437

9.7 FLOW THROUGH EARTH DAMS

the upstream slope at the water surface. From the basic property of a parabola, we get (9.33)

Solving for z, we obtain (9.34)

or (9.35)

R

(9.36)

FO

. = (a sin 13 X l)k dz = (a sin l3)k tan 13 dx

ndition at sections KM and GN, qKM dz

zk dx

=

.

(a

Sin

a

Sin

=

(9.37)

qCN; that is,

l3)k tan 13

(9.38)

13 tan 13

(9.39)

which simplifies to dz zdx

a sin

=

.

We now integrate Eq. (9.39) within the limits Xl H.

r. and Zz =

I

H

a

:. HZ -

. Sin

13

z dz

=

a sin

13 tan 13

= a

cos

lb a

r. and Xz =

dx CQs!3

aZ sin Z 13 = a sin 13 tan l3(b - a cos 13)

b,

Zj =

CHAPTER 9

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOilS

0.4

N

0.3

'1:'

I ~~

0.2

~

i

0.1

60

30

90

f3 FIGURE 9.10

120

~

150

180

(degrees)

Correction factor for downstream 'a

Simplification leads to b

a=--cos

(9.40)

~

(9.41)

FO

438

+Z

1 + - - - -FIGURE 9 .11

L - --

A horizontal drainage blanket at the toe of an earth dam.

--+I

9 .7 FLOW THROUGH EARTH DAMS

439

and does not intersect the downstream face of the dam. Therefore, no correction to the basic parabola is required on the downstream end of the dam . The flow through the dam is q

=

. dz Akl = Ak-

dx

where dzldx is the slope of the basic parabola and the 9.11). From the geometry of the basic parabola, FJ = basic parabola at J is, from Eg . (9.34),

Therefore, the flow through a dam with (9.42)

The procedure to dra erence to Fig. 9.9, is as foHows: 1.

rom the bottom of

FO

R

2.

of the discharge face, a, using b cos

a = -- -

13

cos

I3Yb 2

-

H2 coe 13; 13:S 30°.

For (3 > 30°, use Fig. 9.10 and (a) measure the distance TF, where T is the intersection of the basic parabola with the downstream face ; (b) for the known angle ~ , read the corresponding factor AalL from the chart; and (c) find the distance a = TF(l - Aal L) . 10. Measure the distance a from the toe of the dam along the downstream face to point G. 11. Sketch in a transition section, GK. 12. Calculate the flow using q = ak sin ~ tan ~, where k is the coefficient of permeability. If the downstream slope has a horizontal drainage blanket as shown in Fig. 9.11, the flow is calculated using q = 2fk.

440

CHAPTER 9

9.8

SUMMARY

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

The flow of water through soils is a very important consideration in the analysis, design, and construction of many civil engineering systems. The governing equation for flow of water through soils is Laplace's equation. In thi chapter, we examined two types of solution for Laplace's equation for two-d· ensional flow . One is a graphical technique, called flow net sketching, which onsists 0 a network of flow and equipotential lines. The network (flow t) linear squares in which flow lines and equipotential line a other. From the flow net, we can calculate the flow rate e di ·bution of heads, pore water pressures, seepage forces , and the maximu n raul'i gradient. The other type of solution is based on the finite di erence et and requires, in most cases, the use of a spreadsheet or a co ut r-Q _Instability of structures embedded in soils could occur if the ydr ulic gradient exceeds the critical hydraulic gradient. Practical Example

EXAMPLE 9.4

FO

ort the sides of

~ 3m (b)

FIGURE E9.4

441

9 .8 SUMMARY

There was a long delay before construction began and a 100 mm laycr of silty clay with k "" 1 X 10- 6 cm was depositcd at lhe site. What effect would this si lty clay layer have on the factor of safe ty aga inst piping?

Strategy The key is to draw a How ne Eand determine whether lhe maximum e presence of the hydraulic gradient is less than Ihe critical hydraulic gradie nt. si lty clay would resu lt in significant head loss withi n it, ana conseqqently the factor of safety against piping is likely to increase. Sin the cOfferdjfn is symmet rical about a ve rtical plane. you only need to d2'-w tn How tor one-half of the co fferdam. ,

Solution 9 .4 Ste p 1: Step 2:

Draw the cofferdam See Fig. E9.4b. Determi ne the flow.

10

ketch 1 J How net.

scale£

)

Ill! :: 6 m:

S tep 3:

aximum

(th s is an at t he exit

m'" .. OJ m

I

Calculote

=-- - --!1h ",,"

.... ~ /

tl

NdL onuo

~ra~aJue of the flow length 0

he sheet pile) 6

10 x 0.3 - 2

th ~~L~~I~' ~':d~';t_ t_

107

,

or q .

S te p 5:

h Ydfaulic~aQient .

l+e

l + O.59

piping is likely to occur; the factor of safety is

Del fmi e the effects of the silty clay laye r. ConslCe r the one-dimensional flow in the flow domain, as shown in Fig. E9.4c. The head loss through 9 m of sand (6 m ou tside and 3 m inside of excavation) is 6 m in the absence of the sil t layer. Let us find the new head loss in the sand due to the presence of the silt layer. From Darcy's law and the continuity condition: kOnd •

6h ,""nd _

or x,on
-

n nd

T herefore ,

k

Ilh .."

.." L

" II

' ..n
k,," i ....

442

CHAPTER 9

TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

Water .:- .

-1

Silty clay

-

T

0.1

3m

ml

T Sand

FIGURE E9_4c

But I1h sand + I1hsill 3.16 m. Therefore,

=

6 m; that is

• EXERCISES Theory

FO

9.1

8m

t

6m

Sand

Impervious layer

FIGURE P9_2

(a) Draw the flow net. (b) Determine the flow rate if k = 0.0019 cm/s. (c) Determine the pore water pressure distributions on the upstream and downstream faces of the wall. (d) Would piping occur if e = O.55?

EXERCISES

443

9.3

Repeat Exercise 9.2 using the finite difference method.

9.4

The sheet pile waJl supporting 6 m of water has a clay (almost impervious) blanket of 3 m on the downstream side, as shown in Fig. P9.4.

3 m

--+j

Clay

8m

Sand

Impervious layer

FIGURE P9.4

(a) (b) Determine (c) (d)

FO

R

9.5

Impervious clay

FIGURE P9.5

(a) Draw the flow net under the dam. (b) Determine the pore water pressure distribution at the base of the dam.

(c) Calculate the resultant uplift force and its location from the upstream face of the dam .

444

CHAPTER 9 TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS

9.6

Draw the phreatic surface for the earth dam shown in Fig. P9.S. Determine the flow rate within the dam, if k = kx = 1 X 10- 6 em/s.

9.7

Gravel drainage blanket

FIGURE P9.7

Practical

FO

9.8

~--------------14m~--~~------~

8m

I

! Sand , silt and clay

I I I I I

Sand.and grave' FIGUREP9.8

·

R U S

FO

CHAPTER

STABILITY OF EARTH

RETAINING STRUCTURES 10.0

INTRODUCTION

• Understand and determme lateral earth pressures

FO

• Understand the forces that lead to instability of earth retaining structures • Determine the stability of simple earth retaining structures

• • • •

Static equilibrium Effective stresses and seepage (Chapter 3) Mohr's circle (Chapter 3) Shear strength (Chapter 5)

• Two-dimensional flow of water through soils (Chapter 9) Sample Practical Situation A retaining \vall is required around a lake. You, a geotechnical engineer, are required to design the retaining wall. An example of a retaining structure in a waterfront area is shown in Fig. 10.1

446

10.2 QUESTIONS TO GUIDE YOUR READING

447

10.1

FO

structure moves away (by a

10.2

M.echanical stabilized earth is a gravity type retaining wall in which the soil is reinforced by thin reinforcing elements (steel, fabric, fibers, etc.).

QUESTIONS TO GUIDE YOUR READING 1. What is meant by the stability of earth retaining structures? 2. What are the factors that lead to instability? 3. What are the main assumptions in the theory of lateral earth pressures?

4. When shall I use either Rankine's theory or Coulomb's theory?

10.3 BASIC CONCEPTS ON LATERAL EARTH PRESSURES

449

Mohr-Coulomb failure line

\ Pole for active state

FO

R

(T'

Back of wall

11"v'Y](-i'<'!-- Rankine active zone

~~

Wa ll after rotation about base Base

(a)

(b)

FIGURE 10.4 Slip planes within a soil mass near a retaining wall.

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

K (lateral earth pressure coefficient) Passive

Passive

FIGURE 10.5

Rotation required to

What happens to th 10.2) when the wall is rota ment but the lateral effectiv

FO

450

' -==-'"-=' 1 sin <1>' 1 + sin <1>'

= tan 2(450 _

<1>')

2

=

K

(10.1 ) a

ctive lateral earth pressure coefficient. Similarly, for circle

(O";)j = (O"~)j = 1 + sin <1>' = tan2(4SO + <1>') = K (O":;)j (O"~)j 1 - sin <1> ' 2 p

(10.2)

where Kp is the passive earth pressure coefficient. Therefore,

If, for example, <1> ' = 300, then Ka = ~ and Kp = 3. The stress states of soil elements A and B are called the Rankine active state and the Rankine passive state, respectively (named after the original developer of this theory, Rankine, 1857). Each of these Rankine states is associated

451

10.3 BASIC CONCEPTS ON LATERAL EARTH PRESSURES

with a family of slip planes. For the Rankine active state, the slip planes are oriented at

I ea

=

45° +

~'

(10.3)

I

to the horizontal, as illustrated in Fig. 10.4b and proved in C er 5. For the Rankine passive state, the slip planes are oriented at

Ie

p

=

45" -

~I

(lOA)

to the horizontal, as illustrated in Fig. 10.4a. The lateral earth pressure for the Rankl (10.5)

(10.6)

(10.7)

FO

R

(10.8)

q,

HHHHt

Pp =

t Kpy"H~ t

~ 3

J.(a )

Passive

I (b)

Active

(c )

Hydrostati c pressure

(d)

Surface stress

FIGURE 10.6 Variation of active and passive lateral earth pressures, hydrostatic pressure, and a uniform surface stress with depth.

FO

452

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

If groundwater is present, you need to add the hydrostatic pressure (pore water pressure) to the lateral earth pressure. For example, if the groundwater level is at distance hw from the base of the wall (Fig. lO.6c), the hydrostatic pressure is

and the hydrostatic force is

..L

i

~

and

(10.11)

(10.12)

The essential points are: 1. The lateral earth pressures on retaining walls are related directly to the vertical effective stress through two coefficients. One is the active earth pressure coefficient, K.

= 11 +- sin sin

t/J' t/J'

= tan

2

( 45° - -t/J') 2

and the other is the paisive earth pressure coefficient,

= 1 + sin

K p

t/J'

1 - sin t/J'

= tan2(450 + t/J') = ~ 2

Ka

2. Substantially more movement is required to mobilize the full passive earth pressure than the full active earth pressure. 3. A family of slip planes occurs in the Rankine active and passive states. In the active state, the slip planes are oriented at 45° + 4>'/2 to the horizontal, while for the passive case they are oriented at 45° - 4>'/2 to the horizontal. 4. The lateral earth pressure coefficients, developed so far, are only valid for a smooth, vertical wall supporting a homogeneous soil mass with a horizontal surface. 5. The lateral earth pressure coefficients must be applied only to effective stresses.

10.3 BASIC CONCEPTS ON LATERAL EARTH PRESSURES

453

EXAMPLE 10. 1

Determine the active lateral earth pressure on the frictionless wall shown in Fig. EIO.la. Calculate the resultant force and its location from the base of the wall. Neglect seepage effects.

Strategy The lateral earth pressure coefficients can only e applied to the effective stresses. You need to calculate the vertical effee ive s ;.ress, ag ly Ka, and then add the pore water pressure. Solution 10.1 Step 1:

Calculate Ka.

Step 2:

Step 3:

FO

R

Step 4:

"" = ~ ;)aHo + ~uHo = (~ x 17 x 5) +

T 5m

rsat = 20 kN/m2 1/1' = 30 0

1

r-------4

r----------+i

(b) Lateral pressure

(c) Hydrostatic pressure

17 kPa

(a) Wall

from soil

FIGURE E10. 1

49 kPa

Gx

49

x

5) = 165 kN

FO

454

CHAPTER 10

Step 6:

STABILITY OF EARTH RETAINING STRUCTURES

Determine the location of the resultant.

EXAMPLE 10.2

For the frictionless wall shown Fig. El0.2a, determin (a) (b) (c) The magnitudes and locations of

(d) (e)

1

1

3'

q, = 20 kPa

Lateral passive pressures (kPa)

ttt tt ttHtt HI = 2 m

8.2

Kp = Kn = 3

Lateral active pressures (kPa)

2

Ysal = 19 kN/m 1jJ' = 2~"___ _

-JilIl- _r-t-----'!....-.:=

H2 = 4 m Ysat = 20 kN/m2 1jJ' = 30" Base

(j) Pore water

FIGURE E10.2

(e) Soil

(a) Wall

(b) Surface stresses

(c) Soil

(d) Pore water

455

10.3 BASIC CONCEPTS ON LATERAL EARTH PRESSURES

Step 2:

Calculate the active and passive lateral earth pressures. Use a table as shown below to do the calculations or use a spreadsheet. Depth"

Active Surcharge Soil

u (kPa)

(m)

CT z

(kPa)

0

0

2-6

0

--~""' ''' 20 = 6.7

0

0

o

2-

0

2+

0

6

"YwH2 = 9.8 x 4 = 39.2

20

20 = 8.2

~

x 38

=

~ x 78.8 =

12.7 26.3

Depth Passive Soil

u (kPa)

(m)

o

0 4

2 m.

FO

R

calculations. Active moments are assumed to

Moment arm from base (m)

Moment (kN·m)

4 + 1 = 5 = 4.42

x x 15.6 x 12.7 x x 13.6 x 6.7

3 4 5 6 (water)

~

2-6 2-6

~

2 = 15.6 4 = 50.8 4 = 27.2

2-6

7

8 (water)

Depth

Force (kN)

(m)

2- 6 2-6

~

+ 4 = 4.67 ~=2 4

:3 4

78.4

L Active lateral forces = 215.2

Passive Area

~=2

4 = 26.8

~

~

-101.6 - 36.3 -104.5

L Active moments = -450 .9 Moment arm from base (m)

Moment (kN·m) 4

x 122.4 x 4 = 244 .8

:3

78.4

:3

L Passive forces = 323.2

-82.0 - 53.6 -72 .9

4

326.4 104.5

L Passive moments = 430.9

456

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

Location of resultant active lateral earth force:

Za=

2: Moments = 450.9 = 2.09 m 2: Active lateral forces 215.2

Step 5:

R, = Step 6:

Pp - Po

= 323.2

- 2152

=

Calculate the ratio of moments. . Ratio of moments

=

2: Passive" '<' L.J

Act~

Since the active moment is gre will rotate.

10.4

FO

6

onsider vertical, frictionless wall of height Ho, supporting a soil r-izontal surface (Fig. 10.7a). We are going to assume a dry, homass with a mogeneous soil nd postulate that slip occurs on a plane inclined an angle

Failed soil wedge

P,

(0) Relaining wall

FIGURE 10.7

(b) Free-body diagram of failed soil wedge

Coulomb failure wedge.

10.4 COULOMB'S EARTH PRESSURE THEORY

457

e to the horizontal. Since the soil is dry, -y' = -y. We can draw the free-body diagram as shown in Fig. 10.7b and solve for Pa using statics as follows: 'iFx = Pa + T cos e - N sin 'iFz = W - T sin e - N cos

e= e=

0

0

The weight of the sliding mass of soil is W

=

hm cot e

At limit equilibrium,

Solving for Pa we get (10.13)

To find the maximum thrust a to differentiate Eg. (10.13) It

0

aP ae-= n a

1

which leads to

(10.14)

Substitu ng R

FO

R

(10.15)

his is the same sult obtam d arlier from considering Mohr's circle. The solu io fro im: t equilibrium method is analogous to an upper bound soluti n because it ives a solution that is usually greater than the true solution. he rea n is is that a more efficient failure mechanism may be the 0 e we postulated. For example, rather than a planar slip surpossibl face we co I a postulated a circular slip surface or some other geometric form, and we could have obtained a maximum horizontal force lower than for the planar slip surface. For the Rankine active and passive states, we considered the stress states and obtained the distribution of lateral stresses on the wall. At no point in the soil mass did the stress state exceed the failure stress state and static equilibrium is satisfied, The solution for the lateral forces obtained using the Rankine active and passive states is analogous to a lower bound solution-the solution obtained is usually lower than the true solution because a more efficient distribution of stress could exist. If the lower bound solution and the upper bound solution are in agreement, we have a true solution, as is the case here. Poncelet (1840), using Coulomb's limit equilibrium approach, obtained expressions for Ka and K" for cases where wall friction (8) is present, the wall face I

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

(10.16)

(10.17)

. You should note that Kpc -:/= 11Kac. Recall e c nts are applied to the effective not total gr undwater, the total and effective vertical

FO

458

(sin ' cos 0)112 ] [ cos '(sin( ' + 0)}1I2 -+ tan '

(10.1S)

where the positive sign refers to the active state (6 a ) and the negative sign refers to the passive state (6 p ). Wall friction causes the slip planes in both the active and passive states to be curved. The curvature in the active case is small in comparison to the passive case. The implication of the curved slip surface is that the values of Kac and Kpc from Eqs. (10.16) and (10.17) are not accurate. In particular, the passive earth pressures are overestimated using Eq. (10.17). For the active state, the error is small and can be neglected. The error for the passive state is small if 1) < <\> '/3. In practice, 1) is generally greater than <\>'/3. Several investigators have attempted to find Ka and Kp using nonplanar slip surfaces. For example, Caquot and Kerisel (1948) used logarithm spiral slip surfaces while Packshaw (1969) used circular failure surfaces. The Caquot and Ker-

10.4 COULOMB'S EARTH PRESSURE THEORY

459

TABLE 10. 1 Correction Factors to be Applied to KpC to Approximate a Logarithm Spiral Slip Surface for a Backfill with a Horizontal Surface

of<\>' <\>'

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

0.91 0.86

0.88

0.85 0.79

0.83

15

0.96

0.93

20

0.94

25

0.91

0.90 0.86

30 35

0.88 0.84

40

0.78

0.82

0.81

0.76

0.71

0.69

0.63

0.75

0.75 0.67

0.60

0.36

0.68

0.59

0.51

0.26

0.81

FO

nse of the inclination of passive it arge vertical loads tend to move ion the frictional force on both sides

Wall

Active

I

Movement

Pa

Movement

FIGURE 10.9 present.

Directions of active and passive forces when wall friction is

FO

460

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

tion equipment is generally used. The wall pressures from using heavy compaction equipment can be in excess of the active earth pressures. You should account for the lateral stresses in designing retaining walls. You may refer to Ingold (1979) , who used elastic theory to estimate the lateral str sses imposed by construction equipment. Some practicing geotechnical engin rs P' efer to account for the additional compaction stresses by assuming that t e resultan active lateral earth force acts at OAHo or O.SHo rather than ~Ho the base of the wall. Alternatively, you can multiply the active b-y: factor (= 1.20) to account for compaction stresses.

The essential points are: 1. Coulomb's analysis of the lateral fo ces on a tetaining structure is based on limit equilibrium. 2. Wall friction causes the slip p lanes to curve which leads to an overestimation of the passive earth pressure using Coulomb}s alllllysis. 3. For calculation of the latel'lI.I earth pressure coefficients you can use Eqs. (10.16) and (10.17), and torrf!ct Kpc using the factors lI)ted in Table 10.1. 4. The active and passive.fJ rces jlre inclined Ii an angle j) f rom the normal to the 'fan face.

<::::;;;J Rankine (1857) established the principle of stress states or stress field in solving stability problems in soil mechanics. We have used Rankine's method in developing the lateral earth pressures for a vertical, frictionless wall supporting a dry, homogeneous soil with a horizontal surface. Rankine (1857) derived expressions for Ka and Kp for a soil mass with a sloping surface that was later extended to include a sloping wall face by Chu (1991). You can refer to Rankine 's paper and Chu 's paper for the mathematical details. With reference to Fig. 10.10, the lateral earth pressure coefficients according to Rankine 's analysis are

K

cos(f3 - TJ)Yl + sin 2 <1>' - 2 sin <1>' cos cos TJ(cos 13 + Y sin <1> ' - sin 13)

eo

-----~~~---r~~==~~-2 2 2

aR -

(10.19)

10.5 RANKINE'S LATERAL EARTH PRESSURE FOR A SLOPING BACKFILL AND A SLOPING WALL FACE

461

- Normal to sloping face

T HQ 3"

~

FIGURE 10. 10 Retaining wall with sloping soil surfa

interface, and sloping back for use with Ranki e's met

and (10.20)

(10.21)

(10.22)

FO

orizontal. The sign conventions for

t;"

=

~

= tan- 1

p

tan- 1 (

sin ' sin eo ) 1 - sin ' cos eo

sin ' sin ep ( 1 + sin ' cos

(10.23)

)

(10.24)

e

p

to the normal of the wall face. The angles ~a and ~p (reminder: the subscripts a and p denote active and passive) are not interface friction values. In the case of a wall with a vertical face, T] = 0, Eqs. (10.19) and (10.20) reduce to KaR =

_1_ = (cos f3 - Vcos ~ - cos ') K pR cos f3 + V cos 2 f3 - cos 2 ' t

2

(10.25)

and the active and passive lateral earth forces act in a direction parallel to the soil surface, that is, they are inclined at an angle [3 to the horizontal.

462

CH APTER 10

STABILITY OF EARTH RETAINING STRUCTURES

Th~ essential points ON!.: I. Rankine used slUSS slala of a soil mass 10 dt/~rmin~ Ih~ laleral earlh pressures on a frictionless wall. l. The active and passilte lateral earthforees au incl/nul at f .. and €p, "speclively, from Ihe normal to the wall face. If the wall face is vern· col, the active and passive lateral earth forces are para lie/ to the soil surface.

What's next . .. W e conside red the lateral earth press~a rv $oil mass, w hich is analogous to an effective stress analysis. Next, we w ill co side ,...\ otal stress ana lysis.

10.6 LATERAL EARTH PRESS RES FOR A TOTAL STRESS ANAL 'IS Figure 10.11 shows a smooth, vertica w~ support in homogeneo soil mass under undrai ned conditions: ing th ~ li mit eq uilibJ;i method· we will assume, fo r the active state, that a 511 pia!! IS formed ~ an a Ie e t the horizontaL The forces o n th~ wedge are sfiown III Fig 10.

USing s ta UC ~qUllibri r v e obtain th '-'"

hhe 0 ces along the slip plane:

u

p~ cos z:. ~sr~ 0

s ..
tJt!

p ~ ! .~ ~y-

0

hH~ co e. gquation (10 26) then yields s" H..

-

(1026)

I

H2

Is. Ho

sin 6 cos 6 - 1"Y ,,- sin 20

T o find the maximum actfve la ~l earth force. we differenti ate P with respect to e ana.-1et the resul t-eq~al to zero givi ng Q

p.

= 4s"H" COt 29 esc 29 -

0

T he soluti { is 9 = e "" " 5°.

r ,·t H

H"

•

3

H

1«-",-FIGURE 10.11

Fo rces on a re taining wall for a to t al stress ana lysis.

10.6 LATERAL EARTH PRESSURES FOR A TOTAL STRESS ANALYSIS

463

By substituting e = 45 0 into the above equation for Pa, we get the maximum

active lateral earth force as

I

Pa =

hH~ -

2s u H o I

(10.27)

If we assume a uniform distribution of stresses on th active lateral stress is (10.28) .

Let us examine Eq. (10.28). If (ax)a = 0, for exa vation, then solving for z in Eq. (10.28) gi es (10.29)

(10.30)

pressure, "{

sli , lane, as shown in Fig. 10.12; no shearength of the slip plane above the depth of

FO

tion, the active lateral force is also zero. From

(10.31)

/ 'Tension crack. This crack can be filled by water. When this .. happens, the critical depth then extends to /

1

FIGURE 10. 12 Tension cracks behind a retaining wall.

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

If the excavation is filled with water then

IH'

cr

= 4su 'V'

I

(10.32)

(10.31). The correct solution lies somewhere between these design practice, a value of

Ct

is used for unsupported excavation in fine-grain~ with water, (10 .34)

(10.35)

(10.36)

We can w 't earth press

active and passive lateral (10.37) (10.38)

'Kau and K p'

dr 'ned active and passive lateral earth pressure case, or a sooth wall supporting a soil mass with a horizontal Iiace, Kou = Kpu 2. Wall tare e bedCled in fine-grained soils may be subjected to an adhesive stress s at th wall face. The adhesive stress is analogous to a wall-soil interface friction an effective stress analysis. The undrained lateral earth pressure coefficients are modified to account for adhesive stress as

FO

464

FE

Kau = K pu = 2 1 + -

w

(10.39)

Su

The essential points are: 1. Lateral earth pressuresfor a total stress analysis arefound using apparent lateral earth pressure coefficients Kau and Kpll" These coefficients are applied to the undrained shear strength. For smooth, vertical walls, Kau = Kpu = 2.

10.7 APPLICATION OF LATERAL EARTH PRESSURES TO RETAINING WALLS

465

2. Tension cracks of theoretical depth 2su/y, or 2su/Y' if water fills the tension cracks, are usually formed in fine-grained soils and they modify the slip plane. If water fills the cracks, it softens the soil and a hydrostatic stress is imposed on the wall over the depth of the tension crack. You must pay particular attention to the possibility af theformation of tension cracks, and especially so if these cracks can be filled with water, because they can initiatefai/ure of a retaining struc,ure. 3. The theoretical maximum depth of an unsupported vertical cut in finegrained soils is Her = 4su/y or, if the cut is fille.d wllh water, H:r = 4sJy'.

What's next . ..You were introduced to C u active and passive lateral earth pressure Co curacy of, in particular, Kpc because w fncto c from a plane surface. Several inveS' ~ curved failure surfaces. The question t to address this question. You ar geotechnical engineer u y ha with a particular metho .

is/her experience

FO

10.7 APPLiAIl''r1 PRESSURES Tn~'Y' Field and aboratory te s ha e not confirmed the Coulomb and Rankine theoies. In particular, fie'fd n labo atory test results showed that both theories v-er:e timate the ,9-as 've la e al e rth pressures. Values of Kp obtained by Caquot d Kerisel (19MO low til assive lateral earth pressures but they are still J igher than xpe cimenla esults. Other theories, for example, plasticity theory (Rosenfa b an he 1 :n), have been proposed, but these theories are beyond the sea of this ok nd they too do not significantly change the Coulomb and Rankine p ive la eral earth pressures for practical ranges of friction angle and wall geome Hi. Rankine's theory was developed based on a semi-infinite "loose granular" soil mass for which the soil movement is uniform. Retaining walls do not support a semi-infinite mass but a soil mass of fixed depth. Strains, in general, are not uniform in the soil mass unless the wall rotates about its base to induce a state of plastic equilibrium. The strains required to achieve the passive state are much larger than for the active case (Fig. 10.5) . For sands, a decrease in lateral earth pressure of 40% of the at-rest lateral earth pressure can be sufficient to reach an active state, but an increase of several hundred percent in lateral earth pressure over the at-rest lateral earth pressure is required to bring the soil to a passive state. Because of the large strains that occur to achieve the passive state, it is customary to apply a factor of safety to the passive lateral earth pressure.

F

466

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

We have assumed a generic friction angle for the soil mass. Backfills are usually coarse-grained soils compacted to greater than 95% Proctor dry unit weight. If samples of the backfill were to be tested in shear tests in the laboratory at the desired degree of compaction, the samples may show peak shear stresses resulting in <1>;. If you use <1>; to estimate the passive ral earth pressure using either the Coulomb or Rankine method, you are lkely to overestimate it because the shear strains required to develop the assive I teral earth pressure are much greater than those required to m lize (1 The se of <1>; in the Rankine or Coulomb equations is one fo t e disagreement between the predicted passive lateral earth p tal results.

Sw = Sw =

0.5s" or 0.5s" or

Sw ~

50 kPa

(10.40)

Sw ~

25 kPa

(10.41)

The interface nction between the wall face and the soil depends on the type of backfill used and construction methods. If the surface texture of the wall is rougher than D so of the backfill, the strength characteristics of backfill would control the interface friction . In such a case, the interface friction angle can be taken as equivalent to ~s' If the wall surface is smooth compared with Dso of the backfill, the interface friction value can be assumed , in the absence of field measurements, to be between ~~ and ~~s . A layer of coarse-grained soil is often used in construction to rest the base of gravity retaining walls (see Section 10.8) founded on clays. The interface friction angle for sliding would then be the lesser of the interface friction between the layer of coarse-grained soil and the wall base, and the interface friction between the layer of coarse-grained soil and the clay.

10.8 TYPES OF RETAINING WALLS AND MODES OF FAILURE

467

What's next . .. In the next three sections, we will analyze retaining walls to determine their stability. We will consider an ESA (effective stress analysis) and a TSA (total stress analysis). We begin by considering the possible failure modes.

10.8 TYPES OF RETAINING WALLS AND MODES OF FAILURE There are two general classes of retaining walls. 0 of concrete walls relying on gravity for stability ( 1 flexible and consists of long slender members of eit

FO

tructures and as temerally present and must

(b) Cantilever rigid retaining wall

(c) Counterfort wall

(d) Buttress wall

FIGURE 10.13 Types of rigid retaining walls.

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

Soil

1----_--f.~.~"iI Anchor T ie rod

Wall

Soil So il Base

Base

(a) Cantilever

Prop

Soil

FIGURE 10.14 Ty

FO

468

(b) Rotation and beari ng

capac ity failure

I

I I

I

,

I

I

,, \..

"

,

.... .. -------- ... ,~', ~

(c) Deep-seated failure

I

, 'Slip plane

(d) Structural failure

Failure modes for rigid retaining walls (the dotted lines show the original position of the wall).

FIGURE 10.15

10.9 STABILITY OF RIGID RETAINING WALLS

469

(a) Deep·seated failure

FIGURE 10. 16

STAB ssive concrete walls. Their stability

FO

10.9

Longitudinal drain (a) Simple vertical drain

FIGURE 10. 17

(b) Inclined drain (after Sibley, 1967)

Two types of drainage system behind rigid retaining walls.

470

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

of safety to prevent excessive translation , rotation, bearing capacity failure, deepseated failure, and seepage-induced instability.

10.9.1 Translation A rigid retaining wall must have adequate resistance against trans the sliding resistance of the base of the wall must be greater tha lateral force pushing against the wall. The factor of safety gai s transl ion, (FS)r, is

(FS)T =

T

p~<;

(FS)T

~

(10.42)

1.5

(10.43)

FO

+{3

:;~

~

:1-

Xn

""1":.

'; x",

Ho

Heel

w,

l!

Zn

J_

Toe

(a) Gravity

Forces on rigid retaining walls .

y

Pa.<

1 I

FIGURE 10. 18

Xs

vi: .....

+t!b

B (b) Cantilever

Heel

10.9 STABILITY OF RIGID RETAINING WALLS

471

where W,v is the weight of the wall, W , is the weight of the soil wedge, Paz and Pax are the vertical and horizontal components of the active lateral force , and 8" is the inclination of the base to the horizontal (8 b is positive if the inclination is counterclockwise, as shown in Fig. 10.18). If Elb = 0 (base is horizontal), then (FSh = (Ww + Ws + Paz ) tan <1>; Pax

(10.44)

_ swBlcos 8b FS) T-------"---"-----'=-""r--~ Pax cos 8b + (Woo. + Ws +

(10.45)

For a TSA, I (

If Elb = 0, then (10.46)

FO

R

(10.47)

a lve lateral earth force from the toe. The wall is

WwXw + Wsxs + Pozx a - PaxZa W w + Ws + Paz

(10.48)

10.9.3 Bearing Capacity A rigid retaining wall must have a sufficient margin of safety against soil bearing capacity failure. The maximum pressure imposed on the soil at the base of the wall must not exceed the allowable soil bearing capacity; that is, (10.49)

where (Jrnax is the maximum vertical stress imposed and qa is the allowable soil bearing capacity.

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

10.9.4 Deep-Seated Failure A rigid retaining wall must not fail by deep-seated failure, whereby a slip surface encompasses the wall and the soil adjacent to it. In Chapter 11, we will discuss deep-seated failure.

10.9.5 Seepage A rigid retaining wall must have adequate protection from Qun The pore water pressures and the maximum hydraulic adie t seepage must not cause any of the four stability criten tated lated and static liquefaction must not occur, tha

10.9.6 Procedures t"-~'»~iib~e

1.

2. 3. 4.

FO

472

5. 6.

7.

8.

ails

The essential steps in determining thi!- sIabmry of rigId retaining walls are as follo ws: Calculate the active lateral earth force and l is components. If the wall is smooth" use Rankine's equations because they are simpler than Coulomh's equations to calculate the active lateral earth pressure coefficient Determine the weight of tht wall and soil above the base. Use Eq. (10.43) or Eq. (1 0.44) to find (FShlise Eq. (10.47) or Eq. (10.48) to determine the location of the resultant vertiCllI force, R z , from the toe of the wall. Check that the eccelltl'icity is less than B/6. If it is, then the wall is unlikely to fail by rotation. Determine the maximum soil pressure from U max = (Rz/A) (1 + 6e/B). Calculate the ultimate bearing capacity, qui" using one of the methods described in Chapter 7. Since, in most cases, Rz would be eccentric, Meyerhof's bearing capacity equation is preferable. Calculate the factor of safety against bearing capacity failure: (FS)B = qul'/Umar ,

EXAMPLE 10.3 A gravity retaining wall, shown in Fig. E10.3a, is required to retain 5 m of soil. The backfill is a coarse-grained soil with "{sal = 18 kN/m 3 , ~s = 30°. The existing soil (below the base) has the foJlowing properties: "{sa l = 20 kN/m3 , ~s = 28°.

473

10.9 STABILITY OF RIGID RETAINING WALLS

The wall is embedded 1 m into the existing soil and a drainage system is provided as shown. The groundwater level is 4.5 m below the base of the wall. Determine the stability of the wall for the following conditions: (a) Wall friction is zero. (b) Wall friction is 20°. (c)

The unit weight of concrete is 'Yc

=

24 kN/m 3 .

Strategy For zero wall friction , you ca e friction, you should use Coulomb's met tP. neglected in rigid retaining walls. Sine onl acti Ka from the Rankine and Coulomllf th a s sli groundwater is below the base,

a 've resistance is normally

Solution 10.3 Step 1:

20) 3~O) ~~-.:...~~

FO

R

y be Pa

= ~ Kac"Ysa tHo = ~ x 0.3 x 18 x 52 = 67.5 kN

Ie 0

=

20° to the horizontal (see Fig. EIO.3b) .

Backfi ll

FIGURE E10.3a

m-

-

a , tl from Eq. (10.16)

use the ground surface is horizontal.

Drainage blanket

~-- 4.2

3

->J'\

Concrete, Yc = 24 kNl m

3

474

CHAPTER 10 STABILlTV OF EARTH RETAINING STRUCTU RES

P.c: - 67 5 kN (Coulomb)

,--r.:2O" w,

} .. t

O~ b, _36m

1b _06mI'" l S _4_2m---I1

FIGURE E10.3b

HOrizontal component of p. e : ( P,..) Vertical compone nt or P.c: (P

~"J'I~~ . /

Step 3:

~

"" 1f!' 63.' 67.5 sin 20" = .<23.1

8 ,)

Determjne wall st ilil Consider a..u ·t length of wall. W, -

k o'l< . ,

W2

i b ,NQ"'I, '"

5x

u.•.:-"

x 24 - 2 16 kN

= W, + W - 72 J

"'

H .2

IV - i( B + ~) Ho"'< :::

calculat~ the loca~ . n Ihv RanJc me.

0

esultant

~ - p. ... z.

W,X,

kN

6

kN

0.6) x 5 x 24 - 288 kN

~rom 0 (Fig. ElO.3b).

1-

- 72(3.6 + 0.3) + 216 x (J x 3.6) - 75 x

i-

674.2 kN.m

R. - W~ N .....M :;:£o =

69.4.2 = 2.34 m

Coulomb! 10 = IV,x\ + WJ-fl + (P.,k x B - (p.. >C x~. = 72(3.6"t- 0.3) + 21 6 x x 3.6) + 23.1 x 4.2 - 63.4 x ~ - 790.6 kN

(J

R, = IV

.t -

Mo

+ (P",)c

Ii; ,..

= 288

+ 23.1

K

311.1 leN

790.6 31 Ll = 2.54 m

Base resistance: T = R, tan

cjl~,

where R, is the resultant venical force

Assume $'; = ~ 4>~ = ~ x 28 = 18,7°. Rankme: T = 288 x Ian 18.7" = 97.5 kN Coulomb; T "" (288 + 23, 1) x tAn 18.r.: \05.3 leN

T 97.5 Rankine: (FS)r - -p - = 1.3 < 1.5: _It 75

,

I ~ erefo re unsAusfacrory

10.9 STABILITY OF RIGID RETAINING WAllS

Coulomb:

(FS)T

=

T (Paxk

=

105.3 63.4

=

475

1.7 > 1.5; therefore satisfactory

With wall function, the factor of safety against translation is sa tisfactory.

B 6

-

4.2 6

= -

=

0.7 > e

The resultant vertical forces1.6 lie within the middle one-thlrCt is unlikely to occur

FO

Determining Factor of resultant vertical force is locate develop in the soil.

where B '

quit

(FS)B = ( Coulomb: (F S) B = (

C

365.2 0 f . .. . = --6 == 3. ; there ore, beanng capacIty IS satIsfactory 120.

R

qUI!

)

(Jm . .

Step 4:

=

409.2 = -9 - = 4.4 > 3; therefore satisfactory 2.1 ~ X 20 X (4.2 - 2 X 0.44) x 11 = 365.2 kPa

)

quit

B - 2e

= ~ X 20 x (4.2 - 2 x 0.24) x 11 = 409.2 kPa

quI! (Jmax

=

Determine the effects of water from the rainstorm. Using Rankine's method (zero wall friction)

PaR

=

!Ka'y'Ho + h wHo =

!X

~ X (18 - 9.8) X 52

+

~ X 9.8 X 52

= 34.2 + 122.5 = 156.7 kN

476

CHAPTER 10

ST ABILITY OF EARTH RETAINING STRUCTURES

I.ocation of resultant from 0 Mo -

w 1x , +

W~Xl - P~R"l.a = 72(3.6

+ 0.3) + 216 x - 156.7 X

538 187 .r = 28S = . m

dx

i

3.6) = 538 kN.m

v

Trllnslluion

97.S (FSh .. 156.7 = 0.62

The wall will fai l by translation. Rotation

0.23)

x 11 - 168.7 kPa

r'l1l"illlJcittHanal A. (FS)o =

~l68.

{max

~9 < 3

1

it! not fail by beanns. capaSity failure but the factor of satetv s inadequate. ......,

EXAM PLE 10.4

•

\.

etermine the slab' ~f the an !fever, gravi ty retaining wall shown in Fig. cia and the backfill is a coarse-grained soil. The 1 3. The ex IS 'n soil i ase of the wall w' 1 rest on a) O mm thick. compacted laye r of the backfill. T he . . ."" lIlterface fri tion be ~e n~ base and the compacted layer of backfill is 25°. Strategy Yo should use Coulomb's method to determine the lateral earth pressure because o( the presence of wall friction . The he ight of the wa ll for calcula ting the lateral eart h pressure is the vertical heigh t from the base of the wall to the soil surface. You should neglect the passive resistance of the 1.0 m of soil behind the wall.

Solution 10.4 Step I:

Determine lhc active lateral force and its location. See Fig. E lO.4b. We are given TJ = 0, j3. = go, 0 = 15°, and~;" = 2SO. T herefore, from Eq. (10. 16),

10.9 STABILITY OF RIGID RETAINING WALLS

0.4 m

H 11

Batter

1:20

rsat = 18 kN/m 3 3
f--1.8 m - 1 / +---- 3 m

R

FIGURE E10.4a

FO

0.41 X 20 X 7.42 X sin 15° = 15.7 kN

rsat = 18 kN/m 3
2

p'

"'I

6.1 m Pac

-L.--., P ---r •

I

ax

2.7

0.9 m

f--1.8 m-+t+- -- 3 m - - - - . . j Heel

FIGURE E10.4b

m

477

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

Resultant force components

Pax = Fax + F, = 196.3 + 58.8 = 255 .1 kN Paz = Faz + Fz = 52. 6 + 15.7 = 68.3 kN

Step 2:

Determine the resultant vertical force per unit length an ocation. A table is useful to keep the calculation tidy and easy check.

2

3 4 5

Moment arm from toe (m)

Force (kN/m)

Part

0.5 x 0.42 x 3 x 18 = 11.3 3 X 6.1 x 18 = 329.4 0.4 x 6.1 X 23 .5 = 57.3

3.80

0.5 x 0.36 x 6.1 x 23.5 = 25.8 0.9 x 4.8 x 23.5 = 101.5

6

7

196.3

X

7.42 - 3- + 58.8

7.42

X

2

-Pr-- - - - - - - - - =

196.3 + 58.8

2.75 m

nt vertical component of force from the toe

FO

478

x=

e

Step 4:

'ZMo = 1334.6 = 2.03 m Rz 656

B 4.8 - x= 2 03 2 2- ·

= -

=

0 .37 m

Determine the stability. Rotation

B

"6

=

64.8

= 0.8 m > e (= 0.37 m);

therefore rotation is satisfactory

Translation T = R z tan

T

(FS) T = Pax

~ =

656

305.9 =

255.1

=

X

tan 25°

=

305.9 kN/m

1.2 < 1.5; therefore, translation is not satisfactory

479

10.10 STABILITY OF FLEXIBLE RETAINING WALLS

In design , you can consider placing a key at the base to increase the factor of safety against translation. Bearing capacity 656 ( = ---1 1 + 4.8 x

Urn,.

6 x 0.37) 4.

8

= 199.9

Use Meyerhof's equation. For

¢ ~s

= 30°, N-y = 15.7;

Groundwater level is within B from the ba bearing capacity equation. quit

=

(FS)B =

h'B' N-y quit

Urn. ,

=

=

~

~;

x (19 - 98)

293.2 = 1

199.9

• What's next . .. In the next section, we w ' flexible retaining walls.

tabilityof

FO

10.10 STABILI RETAINING WAL

10.10.1 Analysis of Sheet Pile Walls in Uniform Soils In analyzing sheet pile walls, we are attempting to determine the depth of embedment, d, for stability. The analysis is not exact and various simplifications are made. The key static equilibrium condition is moment equilibrium. Once we determine d, the next step is to determine the size of the walL This is done by calculating the maximum bending moment and then determining the section modulus by dividing the maximum bending moment by the allowable bending stress of the material constituting the sheet pile, for example , steel, concrete, or wood. An effective stress analysis is generally used to analyze sheet pile walls and as such we must evaluate the pore water pressure distribution and seepage pressures. We can use flow net sketching or numerical methods to determine the pore water pressure distribution and seepage pressures. However, approximate meth-

480

CH APTER 10

STABIUTYOF EART H RETAINING STRUC TURES

Backfl ll

Resultant h)'drostallc force IS zero fa) Water level on both

~es

equal

c ~~~~~~~

I~n

a":'t'

(bl Water front of wa different

Sackf,n

pt,~;;;:'N~;'; ,pressure distribut ions behind

~pH

"'!'...,

~ctice. ~~roundwater

are often used level on both sides of a sheet al l is the sa,m~ t-he1 resu ltant pore water pressures and seepage presures a~e_zero (FI~ 1?:9a). au ca~ then neglect the effects of groundwater. in dete rmining the st a ~i ht 0 sheet pile walls. Howeve r, you must use effective stresses irfy T calcul lions of the lat eral ea rth forces. The appT i~ distribution of pore wate r pressures in front of and behind sheet pile walls fo r-conditions in which the water tables are different is obtained by assu mi ng a steady state seepage condition and uniform distribution of the total head. Approximate resultan t pore water pressure distributions for some comm on cond iti ons (Padfield and MaiT, 1984) are shown in Fig. 10.19. The maximum pore water pressu res (lis), maximum pore water forces (P... ) and their locations (loy), and the seepage force per unit volu me (j~) are as follows: Case (a)-Fig. 10.1911 Resu](ant pore wa ter pressure is zero and the seepage force is ze ro. Cllse (b)- Fig. IO. l 9b (10.50)

10.10 STABILITY OF FLEXIBLE RETAINING WALLS

ad(a + d) Pw =

a

+ 2d

(10.51)

'Yw

IZW=TI ~'Yw I

I js

481

(10.52)

=

Case (c)-Fig. lO.19c (10.54 )

(10.55)

(10.56)

(10.57)

FO

R

i d the wall) and is decreased scussed in Chapter 3.

H

--1.+-_ _ _ _ _'-_________ __ -_ - - - - - - - - - --

Sand

Clay

FIGURE 10.20 Sheet pile wall penetrating different soils.

482

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

a total stress analysis should be used for the fine-grained soil. For long-term conditions, an effective stress analysis should be carried out for both soil types.

10.10.3 Consideration of Tension Cracks in Fine-Grained Soils If a sheet pile wall supports fine-grained soils, you should co..,psid

ation

of tension cracks. The theoretical depth of a tension crack"iis>-' Zcr =

2s u

-

or

'Y

FO

rm e the stability of sheet pile Iffer in the wa the lat Is esses are distributed on the wall and th w ' factor of safet is applie in solving for the embedment ~ hOdS in j.s b k. In the first method, called the depth. We WI discuss three m , you would determine an embedment depth ·ae ored mom , t method (j~ to tisfy moment equilib;dtfrn b app ing a factor of safety (FS)p on the passive ret'l c he factor f>.safe , (FS)w' s usually between 1.5 and 2,0, In the second ethod, ca 1~ ne factored strength method (FSM), reducion 'actors are a : Ii the r strength parameters, These reduction factors re called mobIll2i tion facto s because they are intended to limit the shear _~, strength p:.;: meters va p es that are expected to be mobilized by the design loads. A mo li.etion actor, F¢, is applied to the friction angle,
~s F q,

and

where 1 Fq, = 1.2 to 1.5

and

Fu = 1.5 to 2 1

The results from the FSM are sensitive to F¢ and Fu'

10.10 STABILITY OF FLEXIBLE RETAINING WALLS

483

Excavati on level

Net passive pressure

FO

FIGURE 10.21

Net active pressure

Net pressures for the NPPM.

(a) "Stiff" flexible wall

(b) Less "stiff" flexible wall

Lateral pressure distributions expected on a "stiffer" flexible wall and less "stiff" flexible wall. (After Padfield and Mair, 1984.)

FIGURE 10.22

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

Rowe (1957) developed a method, based on laboratory tests, to reduce the calculated maximum bending moment to account for the effects of wall flexibility on the bending moment. Rowe's moment reduction is applicable when a factor of safety has been applied on the passive resistance as in the FMM. here is some debate on the applicability of Rowe's method. Some engineers pr er 0 calculate the maximum bending moment at limit equilibrium ((FS)p = F = (FS)r = FLl = in this Hook. 1) and use it as the design moment. This is the preferred e th To account for soil-wall interface friction, you ne a us 1<:6 However, the active and passive coefficients derived QY Ca om or a flexible (1948) are regarded as more accurate than those of retaining wall, only the horizontal components of the late al t~S are important. In Appendix C, the horizontal components of ac e nd assive coefficients of Caquot and Kerisel (1948) as tabulated rise and ~ bsi (1990) are plotted for some typical backfill slopes and so~ all t e1 ac riction angles. We will use the values of the lateral earth pressur efficie nts in Appen ix C in some of the example problems in this en pter

fn

a distance do from the excavation level. ac ive an passive earth pressures using the FMM or FSM or

FO

484

Excavation level

1

o Base

(a) Cantilever wall

Active (b) Pressure distribution

(c) Approximation of pressure

distribution

Approximation of pressure distributions in the analysis of cantilever flexible retaining walls (padfield and lVlair, 1984). FIGURE 10.23

10.10 STABILITY OF FLEXIBLE RETAINING WALLS

485

3. Calculate the net pore water pressure (u) distribution and the seepage force

per unit volume Us). The effective unit weight is increased by is in the active zone and is decreased by is in the passive zone. 4. Determine the unknown depth do by summing moments about O.

s. 6. 7.

8. Check that R is greater than (Pp)nel' If no , exten t and recalculate R. 9.

R

10.10.6 Anchored

FO

using the FMM or FSM or

~---~

Anchor Limit equilibrium->Working condition

Idealized earth +--- pressure distribution

(a) Deflected position. point of rotation, and idealized earth pressure distribution

IFIGURE 10.24

Mair, 1984).

(b)

Bending moment

Free earth conditions for anchored retaining walls (Padfield and

CHAPTER 10

FO

486

STABILITY OF EARTH RETAINING STRUCTURES

~------La ------~

T

d,

.....- ....- -.....~.. ll

FIGURE 10.25

Location of anchor plates.

5.

6.

7.

2Ta

x (FS).

-y'(Kp - Ka)

(10.58)

A passi develops in front of the anchor plate and an active wedge develops behi nd the retaining wall. The anchor plate must be located outside the active slip plane. The minimum anchor length (La) of the anchor rod, with reference to Fig. 10.25, is La

=

(Ho + d) tan(4SO - <1>'/2) + d z tan(45° + <1>' /2)

(10.59)

8. Calculate the spacing of the anchors. Let s be the longitudinal spacing of the anchors and ha be the height of the anchor plate. If ha 2: d z/2, the passive

resistance of the anchor plate is assumed to be developed over the full depth d z • From static equilibrium of forces in the horizontal direction, we obtain (10.60)

10.10 STABILITY OF FLEXIBLE RETAINING WALLS

487

9. Calculate the maximum bending moment (Mmax) using the embedment depth at limit equilibrium (unfactored passive resistance, unfactored strength values, or (FS)r = 1).

10. Determine the section modulus, Sx

=

Mmax1fa, where f " is the allowable

bending stress of the wall material.

FO

4

EXAMPLE 10.5 Determine the depth of embedment required for st pile wall shown in Fig. E10.5a. Compare the results 0 FSM, and NPPM-using (FS)p = 2.0, Fq, = 1.25, a S)r maximum bending moment for each of these metHo dS! the base of the wall.

~

.5. Calculate the

Strategy You should use the Ca Q a Appendix C) and either the Caguot an

px

for factored

Solution

ax

and

Kpx

(Appendix C).

(des;gn = 24°, 0 = ~deSign, I3/des;gn = 0) (design = 24°, 0 = ~desigm I3/desisn = 0) Kax = 0.28 (~ = 30°, 0 = ~~, 13/~ = 0) (~s = 30°, 0 = !des;sn, 13/~ = 0)

r

3m

FIGURE E10.5a

Coarse-grained soil 'Ysat

= 18 kN/m 3

tp' = 30°

oc: ttp~s

488

CHAPTER 10 STA81L1TY OF EARTH REl AINING STRUCTURES

...---'--'_R FIGURE E10.5b

Step l:

Determine the lateral earth pressurc,distri bu Ions. The la teral pressure dimib fiO'ns for the .,FMM and FSM have the sa me shape but diJfereo rmagnillldes geca use of the diffe enllaleral earth pressure coefficien (Fig. 0.5b). T he lalerafl'res ure distribution for the NPPM i sho n in Fig. E1,O.5c. Since grOl,ll1dwater is not within the d. ~~lt h of the etaining wa l! • .y' "" 'YIO' "" 'Y. With ref to 'Fi • b: ~

r

Active ca e P

,

"" ! K....

_

l.=

-yUto+ do)1 = ! XKM

H/ t to 3

_ _;PlIssive case

FIGURE E10.5c

- 9Ko..(3

= ~

(3 + ~

1;; ~o)l

3 + d

oJ:(!:!") ~ 3K~(3 +

d.)'

+ do)1

10.10 STABILITY OF FLEXIBLE RETAINING WALLS

Step 3:

489

Find do. All forces are calculated per meter length of wall. FMM

The passive pressure is factored by (FS)p. K a", = 0.28 and Kpx = 4.6 (MoL = 3 X 0.28(3 + do? = 0.84( ( Mo )p

_ 3 X 4 .6d ~ _ 3 X 4.6

(FS)p

-

-

:. 0.84(d~ + 9d; + tdo

+

which simplifies to By trial and error or by

USl

do = 2.95 m.

FSM

w

FO

R

polynomial function on a calculator,

do "3

(Mo)p

=

PP

(FS)

=

(Mo)p (M o )"

r

=

3

13d o

For (FS)r = 1.5, 1.5(7.6d~

+ 22.7d o + 22.7)

= 13d~

Rearranging, we get 7.67d~

- 7.6d ~ - 22.7d o

-

22.7 = 0

By trial and error or by using the polynomial function on a calculator, do = 2.41 m.

F

490

CHAPTER 10

Step 4:

STABILITY OF EARTH RETAINING STRUCTURES

Calculate the design depth. FMM: NPPM:

Step 5:

R = Ppx - Pax do = 2.95 m; R

FSM: NPPM:

d

= 1.2do = 1.2 x 2.41 = 2.89

Determine R.

FMM:

=

9 x 4.5

X

2.95 2

do = 2.75 m; R = 9 x 3.3 X 2.75 do

= 2.41

m; R

=9

m 0

d = 1.2do = 1.2 x 2.95 = 3.54 m d = 1.2do = 1.2 x 2.75 = 3.3 m

FSM:

2

9 x 0.28 x (

-

.95)

:71 .1 kN/m

9 x 0.36 x (3 ~

x (4.6 - 0.28) x 2.

FMM

Average

= 16.4 kPa

herefore, depth of penetration is satisfactory FSM

Net lateral pressure Net force = 338.3

=

x

357.9 - 19.6

=

338.3 kPa

0.2do = 186.1 kN > R (= 109.4 kN) ;

therefore, depth of penetration is satisfactory

NPPM

+ 1.1do) = 4.6 x 18 x (3 + 1.1 x 2.41) = 467.9 kPa

Average passive lateral pressure = Kpx'l(Ho Average active lateral pressure = Kax'l

X

1.1do = 0.28 x 18 x 1.1 x 2.41 =

Net lateral pressure

=

13.4 kPa

467.9 - 13.4 = 454.5 kPa

Net force = 454.5 x 0.2do = 219.1 kN > R (= 166.7 kN) ; therefore, depth of penetratio n is satisfactory

10.10 STABILITY OF FLEXIBLE RETAINING WAllS

Step 6:

491

Determine the maximum bendi ng moment. Maximum bending moment fo r (FS)p = F& = 1. Let z be the location of the point of maximum bending moment (point of zero shear) such that z > H o • M,

"l K~1Z' x ~ - l K" '"

~x

=

O.84zl - 13.8(z -

0.28 x 18 x

,(z -

0

~ - ! )( ~.6

W

l-¥,ol

11 1' x 1

(Z

-;JJ

~

To find z at which the bending moment is 'iraximum e need to d ifferentiate the above equation with respec to a d set the result equal to zero. dM dX' - 0 - 2.52<:2 - 41. ( - 3 - 38y - 248.4z + 372.6 = 0 Solving for z, we get z ..5ii2:.6& m r 2.5 m. The corrd t answer is 3.68 msince zero shear ca hot6cc'hr a ve the exca ation~~evel in this problem (positive s ~ in the clive zone ani et etl uced below the excava tion level).

~.84 x 3 ~

-

13.8(3.~W

= 36 kN ' m

FOl'-this pro (em, is easy to u ~1.~S to termine the maxi~ndins moment For most p (lblem ' you will have to find the s~ear force di~lribut ion wi ~ide lif (lr calculate the point of zero shea r, nd then calculate t

Determifl, the mbedmen t depth and the shown in Fig. EIO.6a u si ~the FS .

,

~imum

bendi ng moment

•

nchor force of the tied-back wall

You sholJ,l~se Caquot and Kerise l (1948) passive pressures (see dix C) and either Caquot and Kerisel (1948) or Coulomb active presSdt~~~~~undwa er level on -it sides of the wall is the same, so seepage will

I I,,""'-trategy

-ws

T

im

I--...J._.....-

8m

1'" 18 kNlm) .~.30"

S ..

t··..

r..,. 20 ~Nlnr' ,~" 30"

/) .1..' ... J F." 12

FIGURE E10.6a

Tie rod at 0.5 m

centers

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

Solution 10.6 Step 1:

Determine Kax and KpX"

~eslgn = ~: = ~.~ = 25

0 ;

From Appendix C, Kax = 0.42 K px = 3.4

Ka.<= 0.31 K px = 4.6

Step 2:

Moment (kN·m) + )

Part

-(2 .1d 2 + 29.4d+ 100.81

~He - h =

2 ~

x 8 - 1

=

-1047.4

4.33

d Ho -h+-= 2

- (30.3d 2

+ 423.5d)

11.5d3

+ 121 .1£12

d 2

7+-

FO

492

He - h + ~d = 7 + ~d

~

He - h + ~d = 7 + ~d

Kpx"Y'd 2 = ~ X 3.4 X 10.2 X d = 17.3d 2 X

2

5

"LM = -(10.1d 3

Step 3:

+ 74d 2 +

452.9d + 1148.21

Determine d. Equate the sum of moments to zero (simplify equation by dividing by the coefficient of d 3 ). d3

+ 7.3d 2

-

44.8d - 113.7 = 0

By trial and error or by using the polynomial function on a calculator, = 5.38 m.

d

10.10 STABILI TY OF FLEX IBLE RETAINING WALLS

,-

493

--.III--

= 1m

H~=6 m

+~--- S"rChafge

2 3

FIGURE E10.6b

Ste p 4: Determine d for the ul!fLetored . ~ en t va lues. To calculate the new deirtfi 0 ~ne alion for unfactored strength values, use proportlO ality, for example. Active moment = Active orne fo r factored strength valu Unfac lOred K •• F;'ictored K~~ 0 3J x 0:42

<.

.}

'52~+ 11.~ ,

urn 0 moments: l fl o)p + 14.5d ' + 129.Jdl S lving, we getr = 3.38 m. Determine the n 0 fd ce for d = 3.38 m.

~ve fo r

(Z.1tf

pa~rces -

1

-

334.3d - 847.5

+ 64 .7d + 275.5) x 0°·31 - 381.9 kN .42

J7 .3d x

~:: =

265.8 kN

T. - 381.9 - 265.8 = 116.1 kN (T~)
4

EXAMPLE 10.7

- 2 x 116, L = 232.2 kN

•

Determine the embedment depth and the design anchor force required for stabil ity of the sheet pile wall shown in Fig. ElO.7a using the NPPM.

Strategy In the NPPM. you must use the unfactored strength values to calcu late K"x and Kpx and then determine the net active and net passive latera l pressures. To calculate d . you have to do iterations. A simple approach to solve fo r d IS to sel up the forces and moments in terms of the unknow n d and t hen assume values of d until you find a d value that gives the required factor o f safety «FS)r iI! 1.5). A spreadshee t program or a programmable ca lculator is very helpful in solv ing this type of problem. JII Excel. for example, you can' use the Goal Seek function to find d. II is quite easy to make e rrors in calculations. so you

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

r

~1.5m ...-------'--~ Anchor

6 m

Ysat

tfi'

= 18 kN/m = 30°

3

"c: -ttfi'cs

(FS), = 1.5

{j=~'" 2 'Pcs

FIGURE E10.7a

Solution 10.7 Step 1:

Step 2:

FO

494

is

=

C: 2d) ~w (5 :

I\----+-+ Tn

Lateral earth pressure from soil

FIGURE E10.7b

3

= ~sa( - ~w + i,. = 18 - 9.8 + is = 8.2 + is kN/m 3 ~' = ~s.( - ~w - is = 18 - 9.8 - is = 8.2 - i,. kN/m 3

~'

Passive zone:

2d)9.8 kN/m

= anchor force

Pore water pressure

10.10 STABILITY OF FLEXIBLE RETAINING WALLS

Depth {ml

Passive pressure (kPa)

Active pressure (kPal

o

o

o

6

K."Ih w + K. b '

Ka"l h w = 0.28 =

6 + d

495

x 18 x 1

o o

= 5

+ js)(Ho - hw) 5.0 + 0 .28 x (8.2 + jsl x 5

=

16.5 + 1.4js

16.8 + 1.4js

Water UB

2(ad) 2d "I w

=a +

98 .1d 2d

=5 +

Step 3:

Moment (kN . ml

FO

R

d=7m

+ hw =

h- h

d= 5.75 m

-2.1

-2.2

-50.0

- 50.0

-109.2

-113.4

'2 = 4.5 + '2

-1126.2

-876.0

Ho + d - h - Zw 5 + 2d 8.5 + d =4.5+ d - - - = - 3 3

-1119.3

-871 .9

I Md

-2402.6

-1909.2

=

5179.8

2912.8

2.M, (FS) , = I Md

2.16

1.53

3

W

0.5 + 0.33 = 0.83

'!. 2

(h - hw)

2a

3" -

3

= 2.5

- 0.5 = 2

(h - hwl =

10

3" -

0 .5

= 2.83 4

(16.5 + 1.4isl x d = (16.5 + 1.4isl d Water:

ad(a + dhw a + 2d

49d(5 + d) 5 + 2d

Ho - h -

d

d

Rx = 56.3 + 3.8is 49d(5 + d) + (16.5 + 1.4is) d + 5 + 2d 5

0.5 x (35.4 - 4.88i,1 d 2 = (17.7 - 2.4is) d 2

Ho - h

2d

2d

+ 3" = 4.5 + 3" I M,

496

CHAPTER 10

Step 4:

STABILITY OF EARTH RETAINING STRUCTURES

Calculate the anchor forces for (FS)r = l. For (FS)r = 1, d = 4.62 m. Substituting d = 4.62 m, we get Active zone

is =

(5 + 2

5 X

4.62)9.8

=

3.44 kN/m

3

Rx = 56.3 + (3.8 x 3.27) + (16.5 + 1.4 x 3.44)4.62 =

320.1 kN

Passive Passive lateral force = (17.7 - 2.4

1'a

= Active lateral force - Pa

•

10.11

FO

ation consist of sheet pd driv . into the soil to form the sides of an excavatio ig. 10.26a) . cb as in the construction of bridge piers and abutts. As ex avation pro ds . hin the area enclosed by the sheet piles, struts dde to keep the hee i es in ace. Tne top struts are 'nsta owed by others at lower depths. The wall cements be a h 0 st~ · ts are installed are usually very small but get ar er as the exca ation gets eeper. The largest wall displacement occurs at the base of th~ iXcavatI n (~-AO.26a) . Wall displacements are inconsistent with all the earth 'PT- sure th ries. The criti des' n elements in a braced excavation are the loads on the struts, which are usually different because of different lateral loads at different depths, the time between excavations, and the installation procedure. Failure of a single strut can be catastrophic because it can lead to the collapse of the whole system. The analysis for the forces and deflection in braced excavation should ideally consider the construction sequence, and numerical methods such as the finite element method are preferred. Semi-empirical methods are often used for shallow braced excavations and in the preliminary design of deep braced excavations. The finite element method is beyond the scope of this book. We will only discuss a semi-empirical method. Lateral stress distributions for use in the semi-empirical method are approximations from field measurements of strut loads in different types of soil. The lateral stress distributions used for coarse-grained and fine-grained soils are shown in Figs. 10.26b-d. These lateral stress distributions are not real but average

10.11 BRACED EXCAVATION

r---B--1a

497

b

Wale

Sheeting

~~~~~d+-------' -}, C \

Heaving

D Soft, normally \ consolidated soil ' ,...... , (

" _ ..../

(a) Braced excavation

FIGURE

FO

R

approximat tress distrib 10 s. The laferal stress distribution for coarse-grained ils (Fig. O.26b) was ~ t d from strut loads measured for dense sand .ace t to the exca-.vati . he a propriate value of friction angle is
(FS )h e a ve

=

N

s"

C

'Y

H + 0

qs

(10.61 )

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

where Nc is a bearing capacity coefficient given by Skempton (1951). The coefficient Nc can be approximated, for practical purposes, by

(

Nc = 6 1 + 0.2

Ho) B

for

Ho,:; 2.5 B

(10.62)

and

ic equilibrium equations on

5.

uts f

the braced excavation in soft, normally 0.8a.

FO

498

Soft, normally consolidated soil = 24 kPa, Ho = 6 m

Su

r

= 20 kN/m 3

4.5 m

-1104.6 kPar(a)

FIGURE E1 O_Ba.b

(b)

10.11 BRACED EXCAVATION

499

Solution 10.8 Check for stability against bottom heave.

Step 1:

~o = ~ = 1.5, qs

=

~o)

= 6(1 + 2 x 0.15) = 7.6,

0

24

S"

(FS)heave = Nc H "/

Nc = 6( 1 + 0.2

0

+ qs

= 7.6 - 2 06' = 1.52 > 1.5; X

Step 2:

Step 3: Step 4:

~5 + 1.5)

FO

R

~

+

B

~ ~

x 104.6 x 1.5 + 1.5 x 104.6

7.3 = 98.1 kN/m

Leve B2 = C1 =

'f

1.5 m

-t

2m

1.5 m

1

+ 1.5 x 104.6 x 0.75 ]

BI l_ 104.6 kPa

FIGURE E10.8c

1

104.6 2

x

2

= 104.6 kN/m

82

1 1C

j

l-

-I

104.6 kPa

FIGURE E10.8d

= 235.4 kN/m

500

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

r

1m

L _I'

,I

104.6 kPa

FIGURE E1 O.8e

Level 3 (Fig. EIO.8e)

Step 5:

Calculate .

FO

•

MECH EARTH WALLS Mechanical stabilized earth (MSE) walls (Fig. 10.27a) are used for a variety of retaining structures. Metal strips (Fig. 10.27b), geotextiles (Fig. 10.27c), or geogrids (Fig. 10.27d) reinforce the soil mass. A geotextile is a planar, textile, polymeric product. A geogrid is a polymeric product formed by joining intersecting ribs. MSE walls are generally more economical than gravity walls. The basic mechanics of MSE walls is described in the next section.

10.12.1 Basic Concepts You should recall from Chapter 5 that if a load is applied to a soil mass under axisymmetric undrained conditions, the lateral strain (£3) is one-half the axial strain (£j) as expressed by Eg. (5.42). If the undrained restriction is lifted, then

10.12 MECHANICAL STABILIZED EARTH WALLS

501

+ - - - Backfill

Facing panels Geogrids

(a)

(b)

Metal strip

'f61~metci1...st (il)J (C)

FIGURE 10.27

geotextile, and (d)

FO

R

geogrid.

I point in MSE walls is that the reinforcement serves as an nfinement that allows the soil to mobilize more shearing

Stress state following an increase in effective lateral stress with the vertical effective stress remaining constant <1'

FIGURE 10.28

Effects of increasing the lateral soil resistance by reinforcement.

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

10.12.2 Stability of Mechanical Stabilized Earth Walls

(10.63)

(10.64)

FO

502

y

__----+--' Reinforcing element

1

z

l "~/'~=~~-:+--N-a-tiV:~Oil

lB~ ---

--------- •. ' G

f+-- L ~ FIGURE 10.29

MSE wall using Rankine's method.

10.12 MECHANICAL STABILIZED EARTH WALLS

503

(10.65)

where (FS){ ranges from 1.3 to 1.5. The total length of reinforcement is

IL =

Le + LR

I

where LR is the length of reinforcement within the Because LR is zero at the base of the wall, forcement at the base is often the shortest. This caleN for internal stability, is often inadequate fo ternal stability). You should check whe t er ment at the base is adequate for trans 1. Calculate the maximum I .

FO

R

2.

reg, ired for translation. For shortnee is T = LiJSw, where L b is the and Sw is the adhesion stress. By s ability against translation T = P = (FSh, where ety a ainst translation; usual range 1.5-3.0. The ree gainst translation for short-term loading in clays

(10.66) n

2: Wi tan
soil layer i, n is the number of layers, and <1>;' is the effective interfacial friction angle between the reinforcement and the soil at the base. Assuming a uniform soil unit weight throughout the height of the wall, then

The length of reinforcement at the base required to prevent translation under long-term loading is

(10.67)

504

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

where (Kac)x is the horizontal component of Coulomb's active lateral earth pressure coefficient. You should use the larger value of L b obtained from Eqs. (10.66) and (10.67) . The procedure for analysis of a reinforced soil wall using rials is as follows:

1. Calculate the allowable tensile strength per unit polymeric material. (10.68)

meaning:

2. (10.69)

FO

0

3.

~

eng of reinforcement required at the base for external . (10.66) and (10.67).

4.

tal length of reinforcement at different levels.

TABLE 10.2 Typical Ranges of Factor of Safety Factor of safety (FS)ID (FSlcR (FSlcD (FS)BD

Range

1.lt02.0 2.0 to 4.0 1.0 to 1.5 1.0 to 1.3

FO

10.12 MECHANICAL STABILIZED EARTH WALLS

505

where (10.70)

5. Detennine the external stability (translation and bea . g apacity). Remember that translation is already satisfied from item 3 above. Overturning is not crucial in MSE walls because these walls re flexible a d cannot satisfied if

(10.71)

(10.72)

e zone (Fig. 1O.30a) is for

1-

I lt HO

L. 04H

I

0.6Ho

z > O.6Ro

,

:- , -

6m

: failure suface -------8

-----

L

~ne~r apprOXimatIOn

(a) Failure surface for the coherent gravity method

FIGURE 10.30

for

:5

:____Pn

1"-:tfr.==1==::::o:=:.-, Logarlth m IC SPI ral

O. 2Ho

~J

z

z (b) Variation of active lateral stress coefficient with depth

Coherent gravity method.

(10.73)

(10.74)

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

The procedure for analyzing a MSE wall using low extensible materials is as follows: 1. Select the spacing of the reinforcement in the Z and Y direcfons and the width of the reinforcement. Use a manufacturer's catalog to -r o ,ide information on standard sizes. 2. Calculate the required maximum thickness of the reinfDrc

3.

4. Determine the totallengtn

0

is ~fermined from e 11eral earth pres5. EXAMPL

FO

506

pr cedure for mechanical stabilized earth walls using a

Solution 10.9

Step 1:

Calculate the allowable tensile strength of the geotextile. From Table 10.2, use (FS)ID = 1.5, (FS)CR = 2, (FS)CD = 1.3, and (FShD = 1.3. Tall =

Step 2:

1.5

X

2

58.5 X 1.3

X

1.3

=

11.5 N/mm

=

11.5 kN/m

Calculate the vertical spacing. KaR

=

tan 2 ( 45 - ¢~/2)

=

tan 2 ( 45 - 30/2)

Lateral stress due to surcharge:

=

~

Kaqs = ~ X 15 = 5 kPa

+ KaRqs = h' z + 5 = ~ X 18 + 5 = 6 X 4 + 5 = 29 kPa

(J'x

=

KaR(J'~

((J'x)max

=

6 X Ho

X z

+5

=

6z

+5

507

10.12 MECHANICAL STABILIZED EARTH WALLS

From Eg. (10.69) with (FS)sp = 1.3, we get 11.5 29 x 1.3

=

0.305 m

=

305 mm

Check spacing requirement at midheight (z = 2): O'x

= 6 x 2 + 5 = 17 kPa

S

=

z

11 .5 17 X 1.3

=

0.520 m

=

You should try to minimize the number spacing to easily measurable values. Use half of the wall and 500 mm for · ha Step 3:

Determine the length of reint rc me translation . From Eg. (10. Ka c

=

os 0 = 0.3 x cos 20 = 0

0.3

x

FO

R

0.28

=

Le(m) LR (m) = (Ho

-

z)

tan(45° - ~s/2)

1.00 1.50 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00

0.50 0.50 0.50 0.50 0.25

2.02 1.73 1.44 1.15 1.01

0.25 0.25 0.25 0.25 0.25 0.25

0.87 0.72 0.58 0.43 0.29 0.14

0.25

0.00

KsRS.(FS), 2 tan

1\>/

L(m)

0.3 0.3 0.3 0.3 0.15 0.15

2.3 2.0 1.7 1.5

0.15 0.15 0.15

0.9 0.7 0.6 0.4 0.3 0.2

0.15 0.15 0.15

1.2 1.0

15

x

4

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

Since Lb is greater than the total length required for internal stability at each level, use L = L b = 3.5 m. The total length can be reduced toward the top of the wall but for construction purposes, it is best to use, in most cases, a single length of polymeric product. Step 5:

Check external stability. Stability against translation is already satisfied in Step 13. Check the bearing capacity.

TSA

Use Skempton's method [Eq. (7.14/

~

ESA

0].

•

FO

508

Strategy You ave to guess the spacing of the ties. You can obtain standard widths and properties of ties from manufacturers' catalogs. Follow the procedure for the coherent gravity method. Solution 10.10 Step 1:

Assume spacing and width of ties. Assume 5,

Step 2:

= 0.5

m, Sy

=1

m, and w

Calculate required thickness of reinforcement. Ko

=1-

sin ~

KaR = tan 2 ( 45 -

=1-

~/2)

sin 32°

= 0.47

= tan 2 ( 45 - 32/2) = 0 .31

= 75

mm

FO

509

10.12 MECHANICAL STABILIZED EARTH WALLS

0.31(16.5 Wfy = {corros io n (des ign

283

X

10- 5 m

=

X 6 + 15) 0.075 X 2.5

X X

0.5 105

X

1

X

3

2.8 mm

= Annual corrosion rate X design life = 0.025 X 50 = 1.25 = Calculated thickness + Corrosion thickness = 2.8 + 1 5 = 4.05 mm

Step 3:

Lb =

From Eg. (10.67):

Step 4:

z(m)

LIm)

Recommended L(m)

0.50

7.2 7.0

7.5 7.5

6.7 6.5 6.2

6.5 6.5

6.0 5.8 5.4 5.0

6.5 6.5 6.5 5.0

4.6

5.0 5.0 5.0

1.00 1.55 1.47 1.38 1.30 1.22 1.00 0.75 0.50 0.25 0.00

1.50 2.00 4.86 4.55 4.39 4.23 5.00 5.50

4.08 3.92 3.76

6.00

Step 5:

4.2 3.8

7.5

Check for external stability. Translation is satisfied because L used at the base is greater than Lb' Check bearing capacity. (uz)max = "tHo = 16.5

X

6 = 99 kPa

Using Terzaghl's method [Eq. (7.10) with D f = 0]

For

<j> ~ =

N~ =

16.7

qui' = !B"tN~ = ~ X 8.4 X 18 X 16.7 = 1262.5 kPa

ESA:

(FS)B

28°,

=

~ = 1262.5 = 12.8 > 3;

(uz)max 99 therefore, bearing capacity is satisfactory

•

510

CHAPTER 10

10.13

SUMMARY

STABILITY OF EARTH RETAINING STRUCTURES

PRACTICAL EXAMPLES

FO

4

I I

I

Q.5 m+

o

I HD = 6.8 m I

i

Ore pile y = 22 kN/m 3 4> ~ ::; 35°

1 m 3.3 m

FIGURE E10. 11a

5.7 m

:-

Tie rods

511

10.13 SUMMARY

for the geometry shown in Fig. EIO.lla and (b) the force in the tie rods, assuming they resist all the horizontal loads. If the tie rods were not present, would the walls be safe? Do you expect any alignment problems with the gantry crane?

Strategy The maximum slope angle would be the frictio gle. Since the storage is symmetrical, each wall will carry identical loads. Y; u will have to make an assumption regarding wall friction. You can assume J> J -~s or 8 ~~s and use Coulomb 's method to determine the lateral force ~

Solution 10.11 Step 1:

Determine the lateral forces . The maximum admissible slop

Step 2:

Moment (kN·m)

Part

Wall

= 2.4

+252 .2

1.7+~XO.4=1 . 97

6/2 = 3.00

+69.0 +324.0

1.7 + 1 + (~ X 3.3) = 4.9 1.7 + 1 + 3.3/2 = 4.35 6

+411.1 +2147.6 +1350 .6

2.4

+57.6

8.05

-36.2

FO

R

1.7 + 0.4 + 0.6/2

Pz Lateral forces

Gantry

4 .5

Px

0.75 + 6.8 + 0.5

=

Soil

PBX

Step 3:

714 Rx = L Lateral forces = 718 .5

9.87/3 = 3.29 -2349.1 M = L Moments = 2226 .8

Determine stability. Rotation M Rz

2226.8 1074.8

:x = - = - - = 2.07

m

512

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

-1 r- b = 0.6 m

ct

124 kN ±4 5 kN

J:..

0.5 m

T

:

~

Ho = 6.8 m

J

Ore pile

: f3 = 3 5,0

8.05 m

5 '1 Hb = 0.75

: :

mt'4 .tIlil• •;;;;;;;;;::;::;;;::;~~=~~:-= 1. 7 m +-...-+1--+ 1 - ---t-- -""'""I 1 m 3.3 m 5.7 m

FIGURE E10.11b

e=

2"B - x = 26 -

2.07 = 0.93

B 6 = - = 1 m· 66'

-

Sliding

FO

T=

~

Because the resultant force is eccentric, use Meyerhof 's method. B' = B - 2e = 6 - 2 x 0.93 = 4.14 m

tan 2 (45° + 35"/2) = 33.3

Equation (7.11) : N q =

e"ITlanJ5'

Equation (7.19):

(33.3 - 1) tan (1.4 x 35°)

Ny

=

=

37.2

quI! = 0.5-yB' Ny quit = 0.5 x 18 x 4.14 x 37.2 = 1386.1 kPa = (J max

2:.R,

B X1

(1 +

6e) B

1386.1 (FS)B = 345.7 = 4 > 3;

=

1074.8 ( 6 XI

I

+

6

x60.93)

=

57 kP

34 .

a

therefore , bearing capacity is satisfactory

10.13 SUMMARY

513

Summary of Results 1. The wall is unlikely to rotate. 2. Without the tie rods, the wall will translate. 3. The design tie rod force is =687 kN/m.

4. The soil bearing capacity is adequate . 5. Assuming that the base slab is rigid and the loading is symmet ical, there

should be no alignment problem. 4

•

EXAMPLE 10.12

it pressure coefficients. ~~s)

::. = 30°,

1)

=

(<1>~, = 27°,

1)

= ~~);

1) =

~~)

=

27°,

FO

R

(<1>::'

S

= - a= ( 3 . 5 ) 9.8 = 34.3 kN /m3 a + 2d "Yw 3.5 + 2d 3.5 + 2d

1.0 m Normal water~ 0.5 m level flu ctuation 1.0 m

. r>at .

d

1 FIGURE E10.12a

Sj,

= 18.8

kN/m 3

.

= 55 KPa; ~~s =2r

• . ........ ....

3

. Ysa\=.19:4kNlm ..

:': '.'

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

Active state Sand:

(

Clay:

(

9 +

34.3 ) = 65.S + lSd 3.5 + 2d 3.5 + 2d

9.6 +

34.3 ) = 67.9 + 19 3.5 + 2d 3.5 +

Passive state Clay: -y jp UB

19 .2d - O. =

=

Step 3:

FO

514

FIGURE E10.12b

Step 4:

Calculate the anchor force. Tn = 317.1 - 20S.7

=

10S.4 kN /m

Assume a factor of safety of 2. Design anchor force

=

TaCFS) x Anchor spacing

=

10S.4 x 2

X

3 = 650.4 kN

•

R

FO Pressure (kP.)

Part

PUl

D = 5.57 m

Moment arm from 0

~-h = ~-1=2

= KIJAs = 0.3

x 10

Moment

= Force x

Moment arm

Moment (kN'm/m)

D= 4.39 m Moment (kN'm/m)

Force (kN/m)

(- )33.6

-33.6

- 33.6

16.8

(-) 11.0

-11.0

-11.0

16.5

(- )149.7

-149.7

-149.7

46.1

463 + 126.8d 1-)----

- 74.6

-77 .5

20.2

-1173.1

-869.2

120.8

-673.0

-5 12.4

96.7

I = -1653.4

2 = 317.1

+1653.9

20B.7

2

= 3

P:J2 = K:JK"Yhw

= 0.3 x lB.B x 2.5 = 14.1 P03

= P .lI'2 -

14.1

PI14 = K u,/Yjt;la

E

3.5 + 2d

65.8 + 18d) 0.3 ( - -- - 3.5 3.5 + 2d e 9.1 + 16.9d

3.5 + 2d P;s

!!!!!

PI)' + Pa2

=

3

+ PD3 + P04

+ 14.1 + 14.1 + 69.1 + 18.9d 3.5 + 2d

178.3 + 81.3d

3.5 + 2d Water

u.

6B.6d = 3.5 + 2d

Pp6 = KPJ«YiP d - Kw/'fJfI d =

19.2d - 0.7) (67.9 + 19d) - 0.34 - -- [ 3.8 ( 3.5 + 2d 3.5 + 2d

= (66.4d - 25 .7) d

3.5

U'I ~

U1

+ 2d

p w

=

(1 20 + 34.3d1 d ---:3--: .5:-+---::2-d'---

l (66.4d - 25.7) d ., p".d - 2 3.5 + 2d 1

_

=

(33.2d - 12.9)d 2 3.5 + 2d

~ d + (H o

516

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

EXERCISES Theory 10.]

Show, using Mohr's circle, that the depth of a tension crack is Zcr = 2sj-Ysal fer a saturated clay.

10.2

Show that a tension crack will not appear in a saturated clay if a su a is present.

Problem Solving Plot the variation of active and passive lateral pre · shown in Fig. PIO.3. Ysat = 17 kN /m3 ifJ;, = 20 0

Ysat = 18 kN /m

¢;, = 30

3

0

5m

1 arts so· of saturated unit weight 18 kN/m 3 , ~ = 30°. arge 0 15 kPa . Calculate the active force on the wall _ '<._ ~_ _ _ is rough (8 = 20°). Groundwater is below the

FO

10.3

•. 2

3

4 q,

q,

t

'"

.' ..'

tH :. Basement

=. 5

FIGURE P10.6

6

7

8

EXERCISES

51 7

10.5

A retaining wall 5 m high was designed to stabilize a slope of 15°. The back of the wall is inclined 10° to the vertical and may be assumed to be rough with 8 = 20°. The soil parameters are ~s = 30° and 'isat = 17.5 kN/m 3 . After a flood , the groundwater level, which is usually below the base of the wall , rose to the surface. Calculate the lateral force on the wall. Neglect seepage effects.

10.6

Figure PIO.6 shows rigid walls of height Ho with different geomet es. Sketch the distribution of lateral earth pressures on each wall; indicate the location and direation of the resultant lateral force . Show on your diagram what other f~ es a on the all, for example, the weight of the wall (W .. ) and the weight of the 6i.r . two cases of soil-wall friction: (a) 0 > 0 and (b) 8 = O.

10.7

Which of the two calculations.)

T Ho

1

y

= 18 kN /m3

t/>~s = 30° 8 = 15°

FIGURE P10.7

FO

R

to.8

y = 18.5 kN/m 3 cp~s=30°

8 = 20°

FIGURE P10.9

518

10.10

CHAPTER 10

STABILITY OF EARTH RETAINING STRUCTURES

The drainage system of a cantilever wall shown in Fig. PIO.I0 became blocked after a heavy rainstorm and the groundwater level, which was originally below the base, rose to 1.5 m below the surface. Determine the stability of the wall before and after the rainfall. Neglect seepage effects. 5 kPa

Y..,

= 17.9 kN/m 3

",;.=29° . /j

~Filter

= 14°

. Weep hole

'-'-_ _ _ _ _ _- , ~

1m

~L__ _ _ _ _ _ _ _ _ _ _ _~

If+---- 4.5 m -

- --

FIGURE P10.10

r the cantilever ) withFq, = 1.25.

FO

10.11.

I

1.5 m

+- ~~~ 1.0 m

~

----- - ---- -

- - - --------- - ---------- -------------- - - ------- ---- - - - - ._ - - - - - - - - - - - - - - - - - ..-

FIGURE P1 O. 12

EXERCISES

10.13

An anchored sheet pile wall is shown in Fig. PlO.13 . Determine the embedment depth, the maximum bending moment, and the force on the anchor per unit length of wall. Use either FSM (PcP = 1.25 , P" = 2) or the NPPM with (FS)r = 1.5 . Assume the soil above the groundwater to be saturated. q, = 10 kPa

I

~~ HHHHHH

A" h.bloc k

1m

t

4m

FIGURE P1 0 . 13

Determine the depth of emlbeldment~ ~ld retaining wall shown in water to be saturated.

ll.JI'l..0"""

FO

R

10.14

519

10.15

for the braced excavation shown in Fig. PIO.IS .

£]---+----{::J

Coarse-grained soil

r sa , = 17 .5 kN/m 3 ¢~s=3 1 °

1.5m

t

!

0.8 m

FIGURE P10.15

520 10. 16

CHAPTER 10

S TABIUTY OF EARTH RETAINING STRUCTURES

A braced excavation is requIred in a soft clay as shown in Fig. PIO.16. Determine Ihe load on Ihe struts and Ihe factor of safcly against bottom heave.

18m

1.8m

j

I

1.8 m

I m FIGURE P10. 16

10.1 7

A 6 m high geote)(lile wall is req ui~ 10 uppar! a coa rse-grame backfill with 17.5 kN/m' and~;' = 'lfJ". The fou datipn (base) soil ~ a clay with "y••,# t8 kN /mJ , ~~ - 22",5" = 72 kPa, and cf" 6". Th IIimate st ren.&dl.o f t he ge textile is 45 kN/m and the soil-geolext ile inlerfa ~c n-angle is 20". }:h~ pe 'll ~en l s charge is IS kPa.

"Y.., -

Delerm ine the spat ' g alld len~ of geotextile required:: or Sia illt

10,18

Redo Exercise 10 7 using ga\v~ed sleel tie -wide wit a yield strength 'l.S X lOs kPa, lie-soil int rface fric tion of2a', and rate ol'\qrrosion 0 0.025 mm/yr. The design life is 50 years. "'"

A section of a pproach are to be: sup Orled m m..a ~e readi ly ability. he i st r~h is 2.5 x

to. ~" ;" ~'

is shown 10 Fig. PIO.l9, The sides of Ihe app roach I m X 1 m facmg panels. Steel ties of width Icnglh and thickness of the lies required for ralc of corrosion is 0.Q25 mmlyr. The yie ld

T

'm

lifa~1 .i~~~:;~~~1 Compacted y . 17kNl~,.,,"J2. Normally coo50lidaled

'~~3tl~i~:'1 :~

-'-;III:i1l1i1iiilliilliitiliis~"",,.,he:avi".,iilililii' overCOl1soiodal e:c Clay

FIGURE P10. 1,.

""=

J

r.., \7,5C,,,,O.05 kNlm C<",O,4, OCR = 1,5, .. ,, 40%

EXERCISES

521

20 kPa

FIGURE P10.20

FO

10.20

A cantilever sheet pile wall is required temll access road as shown in Fig. PIO.20. Determ' the maximum bending moment. SeJ compare the results.

, and NPPM, and

CHAPTER

SLOPE STABILITY

FO

11.0 INTRODUCTION

• Estimate the stability of slopes with simple geometry and geological features • Understand the forces and activities that provoke slope failures • Understand the effects of geology, seepage, and pore water pressures on the stability of slopes

• Shear strength of soils (Chapter 5) • Effective stresses and seepage (Chapter 3) • Flow through dams (Chapter 9)

Sample Practical Situation A reservoir is required to store water for domestic use. Several sites were investigated and the top choice is a site consisting of clay soils (clay is preferred because of its low permeability-it is practically impervious). The soils would be excavated, forming sloping sides. You are required to determine the maximum safe slope for the reservoir. Slope failures occur frequently, one of which is shown in Fig. 11.1. Your job is to prevent such failure.

522

'1.2 QUESTIONS TO GUIOE YOUR REAOING

11.1

DEFINITIONS OF

523

TERMS

Slip or /ai/u", one is a Q.1 in. zone of soil tb reaches e crit ical state or residual state and result in m.21'ment of t~p"per s0'l.m~ss. Slip pi I or.../..ait:::'p lane or sli sur/ac sliding.

r J{iilm: !mrface is the surface of

Sliding m 's is the mass ~oil withm t e sl ip plane and the ground surface. lope al/gle (0:.) is th6n~f 'ncl ination of a slope to the horizontal. The s10 angle is elint re erred to as a ratio, fo r example, 2: 1 (~iz.onlal: verli I).

Pore water p " u~r,,) is the ratio of pore water force on a slip surface to the tota l eight

11 .2

f-th~~land any external loading.

QUESTIONS,O GUIDE YOUR READING 1. 2. 3. 4.

What types of slope failure are common in soi ls? What factors provoke slope failures? What is an infin ite slope fai lure? What methods of analysis are used to estimate the factor of safety of a slope? S. Whal are the assumptions of {he various metbods of analysis? 6. How does seepage affect the stability of slopes? 7. What is the effect of rapid drawdown on slope stabili ty?

524

SLOPE STABILITY

SOME TYPES OF SLOPE FAILURE Slope failures depend on the soil type, soil stratification, groundwater, seepage, and the slope geometry. We wiU introduce a few types of slope failure that are common in soils. Failure of a slope along a weak zone of soil is ed a translational slide (Fig. 11.2a). The sliding mass can travel long dista ces before coming to rest. Translational slides are common in coarse-grained ils. A common type of failure in homogeneous fine-grai d so' l is a ro ational slide that has its point of rotation on an imaginary ai is p aile to t e slope. Three types of rotational failure often occur. One type, ~ ed a slide, occurs by an arc engulfing the whole slope. A soft soil layer 0 a stiff layer of soil is prone to base failure (Fig. 11.2b). The se 1.' a tational failure is the toe slide, whereby the failure surface pa th ug~ toe of the slope (Fig. 11 .2c). The third type of rotational failu IS t , slop s -' de, whereby the failure surface passes through the slope (Fig. . (}). A flow slide occurs when internet ana do

FO

11.3

CHAPTER 11

(c) Toe slide

(d) Slope slide

(e) Fl ow slide

(j) Block slide

FIGURE 11.2

Some common types of slope failure.

11.4 SOME CAUSES OF SLOPE FAILURE

525

Block or wedge slides occur when a soil mass is shattered along joints, seams, fissures, and weak zones by forces emanating from adjacent soils. The shattered mass moves as blocks and wedges down the slope (Fig. n.2f).

What's next . .. What causes the slope failures that we briefly desc ;j;6 e above? The causes are many and varied. In the next section, we will descri e some ammon causes of slope failure.

SOME CAUSES OF SLOPE FAILURE

11.4.1 Erosion

11.4.2 Rai e e soils. Water enters into exI I yers, leading to failure, for ex-

FO

11.4

11.4.4 Geological Features Many failures commonly result from unidentified geological features. A thin seam of silt (a few millimeters thick) under a thick deposit of clay can easily be overlooked in drilling operations or one may be careless in assessing borehole logs only to find later that the presence of the silt caused a catastrophic failure. Sloping, stratified soils are prone to translational slide along a weak layer(s) (Fig. 11.3e). You must pay particular attention to geological features in assessing slope stability.

CHAPTER 11

SLOPE STABILITY

C

Crest

(c) Rainfall fills crack and introduces

Y

seepage forces in the thin, weak soil layer

Movement

q

Critical state line for compression

FO

526

p', p (h) Reservoir stresses

During rapid drawdown the restraining water force is removed High water level

(i) Rapid drawdown

FIGURE 11 .3

(j) Groundwater seepage

Some causes of slope failure.

11.4 SOME CAUSES OF SLOPE FAILURE

527

11.4.5 External Loading Loads placed on the crest of a slope (the top of the slope) add to the gravitational load and may cause slope failure (Fig. l1.3f). A load placed at the toe, called a berm, will increase the stability of the slope. Berms are often used to remediate problem slopes.

11.4.6 Construction Activities Construction activities near the toe of an existing sl e can use faIlure because lateral resistance is removed (Fig. l1.3g). We can co ienfl-X ,lvide slope failures due to construction activities into two cases. The rs ~~ IS excavated slope and the second case is fill slope. Y 11.4.6.1 Excavated Slopes Wh n exc ationr curs, the total stresses are g era ted in thy soil. With time reduced and negative pore water pressu~ ar~ oi sip e , using a decfease in effective the negative pore water pres~ stresses and consequently 10~\l~g the ear rength of th soil. f slope failures were to occur, they would take lace aft r construction IS comp eted. We can use our nowledge f st ess paths (C . apter 3) to p~ide insight on the possible effects cavatio on slope s ~ tijty. Let us consider a con'tY 'nvolvi ~c-av ion of a n~ally~onsorated fine-grained struction activ~ soil to construc ~erv~r(Fig. l1.3h). Let ~s~ nsid a element of soil, X, at a depth z b ow the sur~ . The initi I vertical effective stress is '0 = 'Y ' z and the lateral effective stresse are u ' Kou~o' ThC ·ti Ror water pressure is U o = 'Ywz. The n..ts are I

I

P =

I

)

(1 + 2K~ 0 = 3

0

0)

+ u o, and qo =

~~0(1 - Ko)

stress space !P' (p), 4 trej 'l iai total stresses are represented by point A and 'ni ial effective~sses~ repJ!esented by point A The excavation will cause ~duction in u .J;(i.e.,' D.u x ut very little change in u z (i.e., D.u z == 0) and ¥ (i.e., D.u y ¥ ). 11a a III mean total stress is then D.p = -D.u)3 and the change in de'v1a~ric stress D.q = D.ux . The total stress path (TSP) is depicted as AB in Fig. 11.3 1\ ough B is near the failure line, the soil is not about to fail betau~ failure s dictated by effective not total stresses. In some cases, the slope fails b~.Y s the undrained shear strength is exceeded. We will discuss slope failure unoer undrained condition later in this chapter. If the soil were a linear, elastic material, the ESP would be A BIf (recall that for elastic material D.p' = 0 under undrained condition). For an elastoplastic soil, the TSP will be nonlinear. Assuming our soil is elastoplastic, then A' B' would represent our ESP. The ESP moves away from the failure line and the excess pore water pressure is negative. Therefore, failure is unlikely to occur during the excavation stage unless the undrained shear strength of the soil is exceeded. After the excavation, the excess pore water pressure would dissipate with time. Since no further change in q occurs, the ESP must move from B' to B, that is, toward the failure line. The implication is that slope instability would occur under drained condition (after the excavation). The illustration of the excavation process using stress paths further demonstrates the power of stress paths to provide an understanding of construction events in geotechnical engineering.

R

FO

3~z

I.

fs

I

528

CHAPTER 11

SLOPE STABILITY

11.4.6.2 Fill Slopes Fill slopes are common in embankment construction. Fill (soil) is placed at the site and compacted to specifications, usually greater than 95% Proctor maximum dry unit weight. The soil is invariably unsaturated and negative pore water pressures develop. The soil on which the II is placed , which we will call the foundation soiJ, mayor may not be saturt e . f the foundation soil is saturated, then positive pore water pressures will b generated from the weight of the fill and the compaction process. The effe iv tresses decease and consequently the shear strength decreases. With tim the ROst lVe- . e water pressures dissipate, the effective stresses increase and doe he shear strength of the soil. Thus, slope failures in fill slopes are likely diately after construction.

11.4.7 Rapid Drawdown

water would occur and the 1l.3j).

The essential points are: 1. Geological f eatures and environmental conditions (e.g., external loads and natural forces) are responsible for most d ope failures. 2. The common modes of slope fa ilure in soils are by translation, rotation, flow, and block mQvements.

FO

J

seuss the methods of analyses used to

1.5 TWO-DI yS¢TABILlTV ANAL Slope stability can be analyzed using one or more of the following: the limit equilibrium method, limit analysis, finite difference method, and finite element method. Limit equilibrium is often the method of choice but the finite element method is more flexible and general. You should recall (Chapters 7 and 10) that in the limit equilibrium method, a failure mechanism must be postulated and then the equilibrium equations are used to solve for the collapse load. Several failure mechanisms must be investigated and the minimum load required for collapse is taken as the collapse load. The limit equilibrium method gives an upper bound solution (answer higher than the " true" collapse load) because a more efficient mechanism of collapse is possible than those postulated. The limit analysis makes use of the stress-strain characteristics and a failure criterion for the soil. The solution from a limit analysis is a lower bound (answer lower than

1'.SINFIN1TE SLopes

529

Ihe ",rue" collapse load). The finite element method requires the discretization of the soil domain, and makes use of the stress- strain characteristics of the soil and a failure criterion to identify soil regions that have reached the failure stress state. The finite element method does not requ ire specula tion on a possible (ailure surface. We will concentrate on the lim it equil ibrium meth . this book because of its simplicity. We will develop The analyses using generic friCTion angle,
I

What's next ,. ,In the next section. we w ill study how to extent. We will make use of the limit equilibrium Jrl etho long-term and short-term conditions. ,-

ana ~vz e..··'Y' l ope

o f infinite f anal ysis and consider

11.6 INFINITE SLOPES Infini te slopes have dimensions tha :
FIGURE 11.4

Forces on a slice of soil in an infinile stope.

FO

530

CHAPTER 11

SLOPE STABILITY

ratio of the available shear strength of the soil, Tf' to the minimum shear strength required to maintain stability, 'l"m ' that is, (11.1)

The shear strength of soils is governed by the Mohr-Coulom failure cj terion (Chapter 5): that is, 'l"J = (J~ tan <\>' for an effective stress a. aly I and'l"f Su for a total stress analysis. The factor of safety is then (11.2)

(11 .3)

tan

(11.4)

tan as

(11.5)

Since seepage is parallel to the slope, i

=

sin

01. 5 '

From statics, (11.6)

and Tj = Wi sin a s + i s = "I' bjzj sin as + 'YwbjZj sin as = (-y' + 'Y w)bjzj sin a s = (-Ysa,)bjzj sin a s

From the definition of factor of safety [Eg . (11.2)] , we get N' tan
FS = --=J _Tj

'Y'b/l j cos as tan
as

l

ta n
"'Is. , tan

Cis

(11.7)

11.6 INFINITE SLOPES

At limit equilibrium , FS

=

531

1. Therefore, tan a , =

i

>••

tan c/l'

(11.8)

For most soils, Y',-yu., .... !. Thus, seepage parallel to the slope educes the bmiting. slope of a clean , coarse-grai ned soi l by abou t one-half. The shear stress o n the slip plane for a TSA , W Ich applicable to the short· term slope stability in fine-grained soi ls, is T, T, -

I, -

(11.9)

The fac tor of sa fety [Eq . (11.3) 1 is (lUO)

(11.11)

~

of l occurs whe a = 4S that

(IU 2) IS.

Th e ~sential points art: I. The maximum slab/~ slope in a coarse-grained soil, In Ih, absnrce of leepag~J Is equal 10 I.e friclion angle. 2. The muimum stable slope ;n coorse-grainul soils, In Ihe pmmce of seepage parallel 10 Ihe slope. is approximalely one-JuJIj Ihe friclion angle. 3, The critical slope angle infine-groined soils is 45°for an infinite slope failure mechaniJm. The ailielll depl' is '''~ deplh of lensioll cracks, Ihal is, 28./ "Y.

I

CHAPTER 11

SLOPE STABILITY

EXAMPLE 11.1

Dry sand is to be dumped from a truck on the side of a roadway. The properties of the sand are <1>' = 30°, 'Y = 17 kN/m 3 , and 'Ysat = 17.5 kN/m 3 • Determine the maximum slope angle of the sand in (a) the dry state, (b) the sat! ated state, without seepage and (c) the saturated state if groundwater is pres nt and seepage were to occur parallel to the slope. What is the safe slope in t e dry stat for a factor of safety of 1.2S?

Strategy The solution to this problem is a straightf (ll.S) and (1l.8).

Solution 11.1 Step 1:

Sketch a diagram. See Fig. E1l.l.

FIGURE E11.1

Step 2:

Determ

~m:~pe l n~d~e;::;=~~

ilie m

==: Ct

FO

532

s

tan

tan <\>' = tan 30° = 24.80 FS

l

Ct s

=

Ct s

= -

Ct s

= 14°

1.25

tan <\>'

"'Isa' "'I' = "'Isa, - "'Iw = 17.5 - 9.8 = 7.7 kN/m 3

tan

"'I'

"'Isa'

7.7

tan <\>' = -

17.5

tan 30° = 0.25

•

EXAMPLE 11.2

A trench was cut in a clay slope (Fig. Ell.2) to carry TV and telephone cables. When the trench reached a depth of 2 m, the top portion of the clay suddenly failed, engulfing the trench and injuring several workers. On investigating the failure, you observed a slip plane approximately parallel to the original slope. Determine the undrained shear strength of the clay.

533

11 .7 ROTATIONAl SLOPE f AILURES

2m TrenCh

1

J

FIQURE E1 1.2

Strategy The fa ilure observed ca n be ana lyzed as~ n nfi ni ~ slope failure in a fine-gra ined soil unde r undrained cond i tio~ :yOU sh r! conside r a TSA.

Solution 11 .2 Step 1:

Determine s~. 5" ::z

l z(sm 20.) FS 2

• rota-

SLq~~ILURES ~l1'continue t II the ~it e! ilibrium method but instead of a plana r slip

11.7 ,..RqTATIOJlSAL

I

~

sMace of infinite tent.-we wll assume circula r (Fig. 11.5a) and noncircular (Fig. l r.5b) slip surCiices of-1i.mte extent. We will assume the presence of a phreatic surface within th slidi~ ass. A ee-body ~iagram of the postulated circu lar failure mechanism would show the w ght of he soi l within the sliding mass acting at the center of mass. Tf seepage is pc I. then the seepage forces, J H which may vary along the fl ow pa lh. are presen!. The forces resisting outward movemen t o f slope arc the shearing forces mobilized by the soil along the slip surface. We must now use statics to determi ne whether the disturbing forces and moments created by Wand J. exceed the resisting forces and momentS due 10 the shearing forces mobilized by the soil. Ho wever. we have several problems in dete rmining Ihe forces and moments. Here is a Jist. • It is cumbersome , if not difficull. to determine rite location of the center of

mass especia ll y when we have layered soi ls and groundwater. • The problem is stalically indeterm inate. We do not know how Ihe mobilized shear strength , 'J m , of lhl:: soil va ries along lhe sli p surface.

534

CHAPTER"

SLOPE STABILITY

o R

FIGURE 11.5

soil.

FO

o

One ap ach t solve our problem is to divide the sliding mass into an arbitrary numo r slices and then sum the forces and moments of each slice. Of course, the larger the number of slices, the better the accuracy of our answer. Dividing the area inside the sliding mass into slices presents new problems. We now have to account for the internal forces or interfacial forces between the slices. Let us consider an arbitrary slice, ABeD (Fig. 11.5a), and draw a free -body diagram of the forces acting on the slices as illustrated in Fig. 11.5c. The forces have the following meaning: • Wi is the total weight of a slice including any external load. • E j is the interstice lateral effective force .

• (Is)j is the seepage force on the slice. • Nj is the normal effective force along the slip surface.

11 .7 ROTATIONAL SLOPE FAILURES

535

T, is the mobilized shear for(:e along the slip surface.

X, is the IOterslice shear force. V, is the fo rce from the pore water pressu re. l,

is the location of the interslice lateral effective force .

l .. IS (II

the location of the pore water force .

is the location of the normal effective force along t

b l is the width of the slice. I, IS the length of slip surface along the slice. OJ is the inclination of the slip surface within the

The side Be is assumed to be a st raightjjne. We now have to obtain the val ueS of 3 pa amete rs. We can find Wi> V" (1$)" hI' l ... and 0, from the geome{TY he slic the un it w~ghts of the soils, and the location of the ph reatic urface e h ve six unknotfns for each slice and only three equilibrium e~lIa tlons ur le m is. the sta i~ally indete rminate. To solve ou r problem. ie- have to a e assumptions. . idmg three of the unknown pa rameters. Several so tion me thods have evolved de tiding on the assumptions made abo t the unk n n parameter and which equi libnum condi tion (force, moment. ~)' satisfied. -~ble .1 pro~ides a summary of methods that ha en r:6'"~CI. '"""' We will escribe th :tlPethods deve lopc!:d by ishop ( 1955) and by Janbu (1973) beca u thCY t r popul ar rnethbds;"3nd req . e only a calculator or a mputer pr ar commercially available for all the spreadsheet pc ram. rne th, lis ted in fu e 11 .1. Be! e you th programs, yo u should under-

'I'

TABLE 11 . 1

°

Slope Analy.e. Method. Sued on Limit Equilibrium I

\-

Equilibrium Flilu.e surlace

Bishop', melhod (Brshop.

~quation

5ltisfit!d

Circula.

Moment

Calculalor

Circular

Moment

Csicullno.

Circular

Momen t

Any shape

""

Ca lculato r' comp ut er Compu".

Any shspl!

All

Compl.lte.

Any shape

""

Compu tlt.

Noncin:ul"

Horilonta l kI.ees

Calculal0'

1955)

Morgen,te.n .nu Price 119(5)

S~ncer

(1967)

Bell'. method (Bell. 1968)

J.nbu (1973) Satm. (1976)'

R.latum.h, p between f snd X of the lo.m)( " ~/IJ(IE./("" is. luncllon .. 1, ~ is a seale feelo,. J. ~ 0 Inte •• liee fo.c" are pa.lllel: J ... 0 Assum&
'Se.ma·, method '""luOn selsmo(; fOfte'S.

Solution by

Any 51101pe

.,

Computer

536

CHAPTER 11

SLOPE STABILITY

stand the principles employed and the assumptions made in their development. The methods of Bishop (1955) and lanbu (1973) were developed by assuming that soil is a cohesive-frictional material. We will modify the derivation of the governing equations of these methods by considering soil as a dilatant-frictional material. We will also develop separate governing equations ~ II effective stress analysis and a total stress analysis. However, we will re ain the names of original developers. For long-term or drained condition, w h e to con~uct an effective stress analysis. For short-term or undrained CQ . io I finel rained soils, we have to conduct a total stress analysis.

METHOD OF SLICES 11 .8 .1 Bishop's Method Bishop (1955) assumed a circular sli the equilibrium equations to the that Ej and Ej + 1 , Vj and V j + 1 a e that is, at 1/2, and (Is)j = O. Summing forces vert (11.14)

(11.15)

(11.16)

FO

11.8

(11.17)

pore water pressure ratio . Substituting Eq . (11.17) into (11 ,18)

Bishop (1955) considered only moment equilibrium such that, from Fig. 11.4a, (11 .19)

where Xj is the horizontal distance from the center of the slice to the center of the arc of radius R , and Tj is the mobilized shear force . Solving for Tj from Eq. (11.19) and noting that X j = R sin 8j , we get 'ZTJ

W jXj

.

= 'Z - R = 'ZWI Sill eJ

(11.20)

n .8 METHOD Of S LICES

537

Recal l rrom Eq. (11.1) that the factor of safety is defined as

'T,

(T')i

T...

'0

FS=- = -

(11.21)

where Tj is the soil shear rorce at failure . In developing the ~rning equation for FS, we will. firs tly, consider effective mess and later tOta:rstre~. For an ESA, (11.22)

By rearranging Eq. (11.22). we get

T, ~ ~ . )PS )

(11.23)

•

Substitut ing Eq . (l L23) into (11. 18) Yle s /

N; cos 0, ~h - r~) Solving for

Ni~tan(~'), sin '1.:.,...,._, FS +-

(11.24)

N;, we get (11.25)

cos 9, + ·e-.C8,n

tan Ib' , sin 6, FS

(11.26)

. N J' write ~" ')i ,. [W,( I - fy)

+ (X, - X,. 1)lml

(11 .27)

Sub!ltiluting Eq. ~ 1. 2) IOta ( 11 .20) gives !

N' 130($') ·

)

FS

, =!W .

e

,5m I

( 11.28)

Combining Eqs. (11.27) and (11.28) yields

=-

~~ )~+~(~ X~'~-~X~I.~'~)I~"=-"~($~'~)r ~, FS - r[~w~,(~I_-~ IW, sin 9,

(1 1.29)

Equation (11.29) is Bishop's equation for a n ESA . Aishop (1955) s howed Ihal neglecting (X, - X, II ) result ed in about 1% of error. Therefore. neglecting (X) - XJ ~ I). we get (11.30)

CHAPTER 11

SLOPE STABILITY

Equation (11.30) is Bishop's simplified equation for an ESA. If groundwater is below the slip surface, ru = 0 and (11.31)

Let us now consider a TSA. The mobilized shear force

0

is

(11.33)

Since bj = lj cos 8j , Eq. (11.33) b r~~---'\-1-----,

(1134)

(Pi . 11.6a). The forces acting ered equilibrium of horizontal

lanbu (197

FO

538

FIGURE 11.6

(a) Noncircular slip surface

Failure surface proposed by Janbu and forces on a slice of soil.

11,8 METHOD OF SLICES

539

SA

f. Ll Aswmed

~hp

surfxt

0. 1

FIGURE 11 .7 Correction fac tor for Janbu'

(1135)

(11.36)

(11.37)

(11 .38)

For a

FS ~ =c-c-oc~:::(",.",I,,,b,--cc---,,. I[Wj + (X, - Xj.dl tan a,

(11.39)

Re placing (X, - X , .. l) by a correction factor f 0 (Fig. 11.7), we get FS

~

f

I (s~)lbi

o

nv, tan 9/

(IIAO )

The essential points are: I. Bil'hop (1955) assumed a circular slip plane and considel'td 0"1)' moment equilibrium. He neglected seepage fOKes and a5Numed chat the lateral normal forces an collinear. In BINhop's simplified method, the resullant Inler/ace shear is assumed 10 be zero.

540

CHAPTER 11

SLOPE STABILITY

2. lanbu (1973) assumed a noncircular failure suiface and considered equilibrium of horizontal forces. He made similar assumptions to Bishop (1955), except that a correction factor is applied to replace the inteiface shear. 3. For slopes in fine-grained soils, you should conduct both an ESA and a TSA for short-term loading and an ESAfor long-term loading. For slopes in coarse-grained soils, only an ESA is necessa ry f or short-term and long-term loading provided the loading is static. In most problems, you wouldfind that an effective stress analysis would yield the minimum factor of safety.

FO

11.9

z, 2s, lcr =y

Consider only this zone

when tension crack is presen t

FIGURE 11.8 Effect of tension crack on the slip surface.

541

1110 PROCEDURE FOR T HE METHOO OF SLICES

Comp lete drawUO'oO"tl

FIGURE 11 .9

Drnw dow n in a resorvoi r.

I

tension crack to the cen ter of rotation. The factor plified method becomes (11AI)

(1 1.42)

Third . th,.[ ion rack provides a chann fo r at to reach underlying soil layers. The water can in.traduce seepage...fo.tc~and eaken these layers. T he locations of the tensi0l) ctack and the Crit~al slip pthne are not sensitive to the loca tion of the hreat"t surface. ~ ms and cut supporli ng eservolt\l npc subjected 10 rapid d rawdown. Consider tlJe rlh dam shown in ig. l1.9. hen the reservoir is fu ll. the groundwater Ie ti withi n the da ~ iIl eqll ·librate- jth thc reservoir water level. If wate r is wi lhd r~n rapidly, I e Water level in the reservoir will drop bu t very li n le change i,," thc grou n ater )e.ye\ in the dam wi ll occur. In flOe-gra ined soils, a few eeks of d rawdown be' ppid beca use of the low permeability o f these soils. Because t e r€straini ~~Heral force of the water in the reservoi r is no lo nger present d Ilu!...PQr water pressure in the dam is high. the FS will be reduced. Trie orst case scenario is rapid , comp lete drawdown. If a partia l drawdown oct:urs an is m rifta ined, then the phrea tic surface will keep changing and see pa e o rces (resulting from pore water pressure gradients) are presen t in addition to h P.:Qre water pressu res.

rn

11.10 PROCEOURE FOR THE METHOD OF SLICES T he procedure to determi ne the fac tor of safety of slopes using the method of slices, with reference to Fig. 11.10, is as follows: 1. D raw the slope to scale and note the positions and magni tudes of any external loads. 2. Draw a tria l slip surface and identify its poin t of rotation. J . Draw the phreatic surface, if necessary (Chapter 9).

CHAPTER 11

FO

542

SLOPE STABILITY

Grid

Negative slope

5. Divide the

501: mass above the slip surface into a convenient number of slices. More than five slices are needed for most problems. 6. For each slice: (a) Measure the width, b i . (b) Determine Wi-the total weight of a slice including any external load. For example, for the two-layer soil profile shown in Fig. 11 .10, the weight of slice (1) (j = 2) is W2 = b 2 (qs + ZaCY)s2 + Zb(-vsat)s2 + Zc(-YsaJsl) , where Sl and S2 denote soil layers 1 and 2, qs is the surface load per unit area, and Za, Z", Zc are the mean heights as shown in Fig. 11.10. (c) Measure the angle 8i for each slice or you can calculate it if you measure the length, Ii [6 i = cos-1(b/l)] . The angle 8i can be negative. Angles left of the center of rotation are negative. For example, the value of 6 for slice (J) is negative but for slice (1), e is positive.

11.10 PROCEDURE FOR THE METHOO Of SLICES

, •, "',

543

~

, +, liIn-; .-r

t

•

, -,,.

0

~,.

,. "

FlGUREl • . "

"'or.' 20"

The m, 'O be

o.hod .

Alternatively. negative alues of the slope of Ih

rface in

a slice

(.)

7.

( "" O. O I) . If a tension crack is present, set the ro but keep the term W sin 9, for the slices you are co nsideri ng an ESA; for a TSA , set SIt abo... lh te nsion crack but keep the term W, sin er

~

10

= 0 for th{ SIi

8. Repe [ theR oce~ur from item 2 10 item 7 until Ihe smallest faclor of saf 's fo un Th"'ere are several techniques that are used to reduce the numb f trial slip surfa ces. One Si mple technique is to draw a grid and se lective y s&l he nodal points as centers of rotation. Commcn;;ially available programs use different methods to optimize the search for Ihe slip plane with the least factor of safety. Th ~

essentilll poialS an.'

of 41 ' to lise is .p:." aaptfor fissured ol'~rcon· where you should w e 41' = tb;· 2. Tension crocks in Jin~.graillftl soils rrdua t'e factor of SIIfely of a )'/op~. Tension cracks may also provld~ clrannels for waltr to introduce seepag~ forus and weaken underlying soil layers. J . For slopes adjacent to bodln of water, you should consider t"~ effects of operating and enJ'irolUflDltal ronmtions on t"~ir stability. I.

T'~ QPpropriat~ l'Qlu~

solidat~d Jin~·groinw soils,

544

SLOPE STABILITY

EXAMPLE 11.3

Use Bishop's simplified method for the factors of safety of the slope shown in Fig. E11.3a. Assume the soil above the phreatic surface to be saturated. Consider th,ee case" Case 1-no tension c,"ck; Case 2-tension crack; C )_the tension crack in Case 2 is filled with water.

R

FO

4

CHAPTER 11

Step 2:

Step 3:

Step 4: Step 5:

2s u

Zcr

= -:; =

2

x

30

~

= 3.33 m

Divide the sliding mass into slices. Find a height from the crest to the failure surface that equals Zcr and sketch in the tension crack. Use this location of the tension crack as a side of a slice. In Fig. E11.3b, the sliding mass is divided into six slices. Set up a spreadsheet. See Table E11.3. Extract the required values. Follow the procedure in Section 11.10. The weight W = 'Ysatbz. When the tension crack is considered, the shear resistance of slice ® is

11.10 PROCEDURE FOR THE METHOD OF SLICES

545

em

,

)

FIQUREE11 .3b

al~~A. assu

For beco

Bi~ou:'.

TABLE E'1 .3

s equal to

Simplified Method

7

'-

.' ,.

)

"Y •••

'" 51ie.

, ,

,

•5

, 6

•,

a value of ~ and thell ch<\llge this value until it

e ca1culatedva e.

•' ,m e"

," , , , , " "

•

'm'

,

.." 36

"

" ,., " OS

No tension craek

W- l' k IkNI

'"

182.0 1656

,." 2340 2484 24<: .8

238.5

'"

'.

•

ES'

'"

, ,..

Id.;1

m,

Wsln 8

- >J

38.3

1~9.1

,.. " ,.,

->0

1.4 7 1,14

-34~

0,54

-281

54 .6

76.l

>0,

00

." 62. 1

0<,

"

36

5 55

'-

, ,

'" " ". ,,',. , '30 5,' e.5

,",6 49.5

0.03

"

'"08. ,'85

315

.

1204

'90

181 4

1.02

S,m

68.4

1551

Wll -

r~lI lln

72.2 60.1

",'I "',

,.,blcO$ ,

... ,, ... 62.1

41.8

816 97.5 211.6

71 .8 115.5 1136

536.8

510.9

194 8

FS

,..

,,
546

CHAPTER 11

TABLE E11 .3

SLOPE STABILITY

(continued) Tension crack

FS

= 1 assu med TSA

b Slice

z 1m)

W=-ybz IkN)

2.5

3.6

2

1m)

ru

162.0

3.6

4.6

165.6

4.6

0 .54

4.9 3

f)

zw 1m)

88.2

Ideg)

mj

0.54

-23

1.50

- 34.5

0.54

-10

1.15

-28.1

76.2

a

1.00

0.0

60.0

Wsin6 159.7

4

2

5.6

201.6

5

0.49

9

0.92

5

2

6.5

234.0

5.5

0.46

17

0.87

6

2

6.9

248.4

5 .3

0.42

29

7

2

6.8

244.8

4.5

0.36

39.5

77.8

8

2.5

5.3

238.5

2.9

0.30

49.5

115.5

9

1.6

1.6

46.1

0.1

0.03

65

60.7 62 .7 68.6

0.0 681.2 1.27

R = 14.3 m TCM/R = 23 .7 kN FS = 0.95 assumed TSA

b Slice

1m)

2

2.5

z 1m)

sub/cos 6

4.9

39.9 55.3

76.2

4.6

49.0

60.0

61 .5

60.7

70.9

62 .7

FO

3

159.7

3.6

Step 6:

77.8

68.6

0.83

155.7

84.3

77.8

0 .86

181.4

93.0

115.5

0.96

41.8

0.0

0.0

Sum

536.6

531.7

681.2

FS

0.95

1.22

Compare the factors of safety. FS

Condition

ESA

TSA

Without tension crack With tension crack

1.06 1.00

1.27

Tension crack filled with water

0.95

1.22

1.48

The smallest factor of safety occurs using an ESA with the tension crack filled with water. The slope, of course, fails because FS < 1.

•

11.10 PROCEOURE FOR THE METHOD OF SLICES

547

EXAMPLE 11.4

Determine the fac tor of safety of the slope shown in Fig. El1.4a. Assume no tension crack.

Strategy Follow the same strategy as described

Example

In

)

Solution 11 .4 Step 1: Redraw the figure to scale. See Fig. E l1.4b . Step 2: Divide the sliding mass into slices as snown

I

,

--- ® --,

,, , '(1) ,, , \® ,,, ,, \ r ...

'm

Sca le

I-- 4 FIGURE E11 .4b

m--l

548

CHAPTER 11

TABLE E11 .4

Sv

<1> ' -y~

"YSCII

FS

Slice

Soil 1 30 33° 9.8 18 1.01

SLOPE STABILITY

Three Soil Layers Soil2 42 29' 17.5 assumed

Soil 3 58 kPa 25° kN/m3 kN/ m3 17

b

z,

Z.

Z3

W= -ybz

z_

(m)

(m)

(ml

(m)

(kN)

(m)

TSA (}

'v

(deg)

mj

1.49

svb/cos

4.9

1

0

0

88.2

1

0.54

-23

2

2.5

2.3

1.3

0

160.4

3.6

0.55

-10

76.2

3

2

2.4

2.2

0

163.4

4.6

0.55

0

60.0

4

2

2

3.6

0

198.0

5

0.49

9

5

2

0.9

4.1

1.5

226.9

5.5

0.48 0.43

159.7

60.7 62.7 68.6

6

2

0.8

4.1

2

240.3

5.3

7

2

0

3.7

3.1

234.9

4.5

8

2.5

0

1.5

3.8

227.1

2.9

161.7

9

1.6

0

0

1.6

43.5

0.1

219.6

FO

a

108.9

978.1 1.91

Step 3:

Set up See Tab

Step 4:

assumed, as using an ESA.

as laced on saturated clay. A noncircular slip surface was Fig. Ell.5a. Determine the factor of safety of the slope T~roundwater level is below the assumed slip surface.

Strategy Since a noncircular slip surface is assumed, you should use Janbu's method. Groundwater is below the slip surface; that is, ru = O. Solution 11.5 Step 1: Redraw the figure to scale. See Fig. Ell.5b. Step 2: Divide the sliding mass into a number of slices. In this case, four slices are sufficient. Step 3: Extract the required parameters. Step 4: Carry out the calculations. Use a spreadsheet as shown in Table Ell.5.

Assumed failure surface

I

3 2

5m

FIGURE E11.Sa

I

5m

FO

R

•

ESA b

z,

ZZ

W= ",bz

I)

Sliee

(m)

(m)

(m)

(kN)

(deg)

mj

1 2 3 4

2 3.5

1 2 1 0

0.7 2.5 4.3 2.5

59.8 274.8 182.2 123.3

-45

3.03 1.00 0.92 0.95

-59.8 0.0 182.2 212 .6

71 .0 152.3 65.9 38.9

Sum

335.0

328.0

2 2.9

0 45 59.9

Wtan

FS

I)

Wtan <jI'(eas I))mj

1.04

549

550

CHAPTER 11

SLOPE STABILITY

What's next . ..Charts can be prepared to allow you to quickly estimate the stability of slopes with simple geometry in homogeneous soils. In the next section, we present some of the popular charts .

11.11 STABILITY OF SLOPES WITH SIMPLE GEOMETRY 11.11.1 Taylor's Method Let us reconsider the stability of a slope using a • u-'...._. a"We can rewrite Eg. (11.34) as

FO

(11.43)

70

60

50

40

30

Slope angle,

Cl,

20

10

(degrees)

T

Rock

FIGURE 11 . 12 Taylor's curves for determining the stability of simple slopes.

11.11 STABILITY OF SLOPES WITH SIMPLE GEOMETRY

551

The procedure to use Taylor's chart to determine the safe slope in a homogeneous deposit of soil using a TSA, with reference to Fig. 11.12, is as follows:

1. Calculate n, ~ DJHo, where Do is the depth fcom the ct~op of the stiff layer and Ho is the height of the slope. 2. Calculate No = FS(-yHolsu). 3. Read the value of

O's

at the intersection of nd and

o'

If you wish to check the factor of safety of a slope, the procedure is as follows:

L Calculate

nd =

DoIHo·

2. Read the value of No at the inJ;

FO

R

3.

e assumed slope angle, determine the values of m and

4. 5.

5

5

4

4

3 m

n

2

0 2; 1

2

3 :1

4:1 Slope

F IGURE 11 . 13

5: 1

3:1

4:1

5:1

Slope

Values for m and n for the Bishop-Morgenstern method.

CHAPTER 11

SLOPE STABILITY

EXAMPLE 11.6

Determine the factor of safety of the slope shown in Fig. E11.6.

Strategy

Step 2:

Step 3:

•

FO

552

Strategy Use the Bishop and Morgenstern (1960) charts and equations. Follow the procedures in Section 11.11. Since you are given Yu , you only need to do Steps 4 and 5.

FIGURE E11. 7

EXERCISES

553

Solution 11 .7 Step 1:

Determine m and II . From Fig. 11.13, m = 1.73 and

1/

= 1.92 for a slope of 3:1 and ~~. =

3~" .

Step 2: Calculate FS. FS c: m - nr,, "'=

•

11 . 12 SUMMARY I n this chapter, we examined the stability of sim pl stQp'es. lope fa ilures are lion ana alhs. Slopes usually often catastrophic and may incur extensive 4 thq es) and by construction fail from natural causes (erosion. seepag~d activities (excavation, change of land) rface, etc. pte analyses we considered were based on limit equilibrium , whiCh qr res Simplifying assumpt ions. Careful judgment and experience are needed ~ el alua~s l ope stabilj The geology of a site is of particular i mport ~e e ml~'n'i slope sta ~ility. au should consider both an effective stress analysis an a total stress affa ly 0 slopes in fine grained soils and an effective s a alysis for slopes in coars~ ined so ils. The main sources of er,ors-in slope ability anal IS are the, shca'r strength parameters. especially s". and de terminat ion o(tn re water pressures.

ress

EXERCISES Theo 1t.1

n or the factor of safety of the slope in f ig. Pll .2 usmg the mechanism

FIGURE P'1 ,2

554

CHAPTER 11

11.3

SLOPE STABILITY

Figure Pl1.3 shows the profile of a beach on a lake. It is proposed to drawdown the lake by 2 m. Determine the slope angle of the beach below the high water level after the drawdown. You may assume an infinite slope failure mechanism.

High water level

y = 16.5 kN/m 3

28°

2m

1

33°

FIGURE P11 .3

Problem Solving A cut for a highway is shown in Fig. using an ESA and a TSA. Assume through the toe of the slope.

6m

slope shown in Fig. Pll.S using an ESA and a TSA. y 0 and the line representing the top of the stiff soil

FO

11.4

1-5 m-l

O.

T

3m

-r--------------------- ---------~ -

-_.--

FIGURE P11 .5

m

-

Ysa' = 18 kN/m3 s" = 25 kPa - - ~~s = 28°

-

EXERCISES

11.6

555

A compacled earth fill is constructed on a soft, saturated clay (Fig. PI 1.6). The fill was compacted to an average dry unit of 19 kNlm 3 and water content of 15%. The shearing strength of the fill was determined by CU tests on samples compacted to representative field conditions. The shear strength parameters are Sk = 35 kPa, q,; = 34°, and q,~, = 28~. The variation of undrained shear strength of the soft clay with d ~lh as determined by simple shear tests is shown in Fig. PU.6 and the friction 3n!¥rit the critical state is "'; , = 30". The average water content of the soft clay is 55% . C ompute the factor of safety using Bishop's simplified method. Assume that a tensi :~ ill de velop in the fill.

T 6m

s. (kP~) 15 30

o

I;;[]j FIGUREP'"~ 1 1.7

)

cl i~ of a canal is shown in Fig. P 1.7. Determine the factor of safety for (a) the eXisting condition and ( b) a ra 1\:1 drawdo wn of the water level in the canal. Use Bishop~elhod. T he d nter of rota tion of the sliding mass is at coordinates x = 113 m aud y = 133 m. The-r&:V-rfi!C\is tangent to the slip plane. The properties of the soil a ~'II s follows: ~

A crq, s

Soil

A

B

.I

Descrip tion

t € IBY 1,IBY

'Y... (kN / m ',

s"lkPa)

17.8 18.0

34

30 28

21

1-- - - - - 35 m ------:::1'~ll35::, . 115) 10

~ ,

____4.5m i-____ _

....... ;;t~

ZL:o:'.'.OOi'iliiil.llillIlI••I!!lIillllIi•••1i 4.5m

~~~~-~----------- -r- --- -SOil B Rock

FIGUREP11 .7

6m

556

CHAPTER 11

lUI

SLOPE STA81UTY

Use Janbu's method 10 delermine (he faclor of safelY of the slope shown in FIg. PII .S .

• ,~. 20" '12. 18.5 kN/l'n l

11.9

safety of the slope shown

i~ .

PI 1.9.

r ----------

-;.;.Io":.4F'!P"='9

'm

1-..............

FIGUREP11 . 10

11 .11

Delermlne the fac lor of safety of a 2: 1 slope wIth Morgenstern method

r~ -

0.25,
EXERCISES

557

Practical 1l.U

The sO!'1 at a site' one possible slip plane is shown . ~:to~n Fig. Pl1.12 A ermme10the factor of'saf:t~~ was made as shown and

40

Distance (m)

~~=:-,~~-,~6~0~r-~~~~

FO

R

~

80

100

120

140

APPENDIX

A COLLECTION OF FREQUENTLY USED SOIL PARAMETERS AND CORRELATIONS TABLE A . 1 Typical Values of Unit Weight for Soils Soil type

"Is.t (kN/m 3 )

Gravel Sand Silt

FO

Clay

:t pes, Description, and Average Grain Size According to

uses

Soil type Gravel

Average grain size

Description Rounded and/or angular bulky hard rock

Coarse: Fine:

Sand

Rounded and/or angular bulky hard rock

75 mm to 19 mm

19 mm to 4 mm

Coarse:

4 mm to 1.7 mm

Medium : 1.7 mm to 0.380 mm Fine: 0.380 mm to 0.075 mm

558

Silt

Particles smaller than 0.075 mm exhibit little or no strength when dried

0.075 mm to 0.002 mm

Clay

Particles smaller than 0.002 mm exhibit significant strength when dried; water reduces strength

< 0.002 mm

APPENDIX A

A COlleCTION OF FREQUENTLY USED SOIL PARAMETERS AND CORRELATIONS

TABLE A .4 Description of Soil Strength Based on liquidity Index De5(:riplion 01 soil strength

Values oilL h 1

G

Semisolid IIlale-high slrengtn, brittle (sudden) fracture e>;pected Plastic stata-intermediate strength, soil deform like a plastic malerial liquid stale-low slrength, soil deforms ite viscous fluid

•

TABLE A.5 Atterbe rg limits for Tvpical Soils Soil type Sand Sill Clay

10~15

15~100

) Coefficient of Perm •• blli~ for Common Soil

5011 tvpe

r. V (

Clean grave l Clean sands, cl an sand e ~d gr vel miduros Fme sand Its, IXluresjOmp rls ing sands, silts, and claVs Homogeneo u c lays~ (

.

TABLE A .7 Typica l Values of Poisson's Ratio Soil type Clay

Sand

Descript ion Soh Medium Stiff Loose Medium Dense

.'

0.35-0.4 0.3-0.35 0.2- 0.3 0.15- 0.25 0.25-0.3 0.25--0.35

:> 1.0

1.0 to

10 ~3

10-' 10 10 -'

< 10- 7

559

A COLLECTION OF FREQUENTLY USED SOIL PARAMETERS AND CORRELATIONS

Typical Values of E and G

TABLEA.8 Soil type

Description

Clay

Soft Medium Stiff

Sand

Loose Medium Dense

P(MPa)

G(MPa)

1-15 15-30 30-100 10-20 20-40 40-80

0.4-5 5-11 11-38 4-8 8-16 16-32

C

APPENDIX A

"These are secant values at peak deviatoric stress for dense and stiff soils, and when the maximum deviatoric stress is attained for loose, medium, and soft soils.

TABLEA.9 for Soils Soil type Gravel Mixture of gravel and sand fine-grained soils

WI

h

Sand Silt or silty sand Clays

FO

560

-~ to ~

-1 to 0

TABLE A.11 Correlation of IV, 1V6o , "I, D" and Coarse-Grained Soils

N

0-5 5-10 10-30 30-50 > 50

N60

Description

0-3 3-9 9-25 25-45 >45

Very loose Loose Medium Dense Very dense

"These values correspond to

~.

cP'

for

Dr

'Y (kN/m 3 )

(%)

11-13 14-16 17-19 20-21 >21

0-15 16-35 36-65 66-85 >86

(degrees)

26-28 29-34 35-40" 38-45" > 45"

R A U SE O

FO

APPENDIX

FO

DISTRIBUTION OF SURFACE STRESSES WITHIN FINITE SOIL LAYERS

(b) Under corner

B1.2 Rectangular Area with Rough Rigid Base (Milovic and Tournier, 1971)

q, 1¢.~~~Yf t

Hj

IB I I

v = 0 .3

C

Rough rigid base

LIB

2

5

Center

zlB

0.000 0.100 0.200

!!I.uJqs

1.000

1.000 0.974

0.992

0.943

0.977

00400

0.842

0.600

0.690

0.800

0.570

0.250 0. 2-50

'-250

Y

H,IB

1.000

0.233

0.000 0.100 0.200 2

0.247

0.400 0.800

0.230 0.340

0.183

0.208 0.188

0.269

0.160

0.168

0.250

0.250

0.250

0.250 0.249

O. 0

R

~ o. oo

FO

~

5

0.250 0.249

0.200 0.884

00400

0.650

0.241 0.203

0.250 0.246 0.222

0.246 0.225

0.492

0.158

0.191

0.197

0.395

0.163

0.174

0.333

0.122 0.098

0.141

0.156

0.171

0.281

0.078

0.120

0.138

0.137

0.123

0.238

0.064

0.102

0.000

1.000

1.000

1.000

0.250

0.250

0.250

0.100

0.970

0.98 1

0.990

0.250

0.250

0.200

0.930 0.798

0.961

0.969

0.250

0.870

0.881

0.249 0.241

0.250 0.249

0.800 1.200

00450

0.641

0.200

0.219

0.258

0.594 0.394

00475

0.153

1.600 2.000

0.271

0.368

0.114

0.184 0.151

0.191

0.162 0.111 0.075

0.195

0.296

0.125

0.143

0.235 0.193

0.123 0.108

0.165

0.050 0.040

0.100 0.082

3.500

0.056 0.044

0.139 0.105 0.085

0.087 0.064

0.069

0.097

4.000

0.037

0.071

0.144

0.034

0.060

0.089

4.500

0.032

0.062

0.128

0.030

0.082

5.000

0.027

0.0 53

0.113

0.026

0.053 0.047

00400

2.500 3.000

0.245

0.245 0.221 0.164

0.075

564

APPENDIX B

DISTRIBUTION OF SURFACE STRESSES WITHIN FINITE SOil LAVERS

B1 .3 Rectangular Area with Smooth Rigid Base (Sovine, 1961)

0.2

0.4

0.6

0.8

1.0

C

o

0.5

l.0

" 1

z

L

2.0

FO

2. 5 UL.I----'-'------'----'------"----"----'a-L~~~'------'------'

~

Y

l.5

APPENDIX B

DISTRIBUTION OF SURFACE STRESSES WITHIN FINITE SOil LAYERS

565

82 VERTICAL STRESSES IN A TWO-LAYER SOIL UNDER THE CENTER OF.. A UNIFORMLY LOADED CIRCULAR AREA (Fox, 1948)

E,/E2 10

100

1000

Rough interface

z

a/H,

0 H, 2H, 3H, 4H,

FO

1/2

0 H, 2H, 3H, 4H,

Ilu z /q.

0.284 0.087

0.101

0.0403 0.023

0.0105 0.0083

0.0148 0.646 0.2 0.046 0.036

0.136 0.089 0.062

0.0428

1.025 0.869 0.596 0.396

0.677 0.576 0.421 0.302

0.271

0.22

0.0051 0.0045 0.0038 0.0033 0.0029

0.082 0.068 0.0525 0.0409 0.0326

0.019 0.0172 0.0151 0 .0133 0.0117

0.249 0.225 0.186 0.15 0.122

0.067 0.063 0.057 0.051 0.0454

APPENDIX

FO

LATERAL EARTH PRESSURE COEFFICIENTS (KERISEL AND ABSI, 1970)

566

APPENDIX C

567

LATERAL EARTH PRESSURE COEFFICIENTS (KERISEL AND ABSI, 1970)

10

---

K""

I

y~

~

I

I

20

25 ~.

Kax

0.40 0.20 0.00 -0.20 ·

-...;

l.00

-0.80

a--~~' - 3

a=-f

I

30

35

40

45

35

40

45

(degrees)

(3W

a=~~' 3

0.80 0 .60 0.40 0.00 -0.40 L....----'..L..._--'--_--'-_-'--_---'-_---'-_--' -0.80 20 25 30 35 40 ¢. (degrees)

FO

15

O. 00 0.40

~

----

0 .1 10

0.80 O. 80 O. 40

0.1

L-_~_~_~I_ _L-_~_~_~

10

15

25 30 I/!' (degrees)

35

40

FIGURE C.1 Horizontal component of the active lateral pressure coefficient. (Plotted from data published by Kerisel and Absi, 1990.)

568

APPENDIX C

LATERAL EARTH PRESSURE COEFFICIENTS (KERISEL AND ABSI, 1970)

10 ~

2 ~' 3

//

//

7?

....-:::::::: ~

0.80 0.60 0.40 0.00

~

fIf-f--

0.40 0.80 ~~~~~~~=F~~~~~

0.60 0.40

~~

L_

0.00 -0.40

_l~~~~

15

FO

0 .1 10

20

25

30

0 .80

35

40

45-D·80

45

¢!' (degrees)

FIGURE C.2 Vertical component from data published by Kerisel a

0

APPENDIX C

ff

100

r-

1.00

{3(+ve)

/ '0.60

:;(::~

/V

I/L'

./

~v

............ ~~

~ ~~ :..--!----

~ l.----

DAD

Kpx

0.00 1- 0 .20

0.00

----:------ -0.20

-DAD

-oAO

-0.60

-

~~

F?==

10

fJl4>' 1.00

(j- 0

- -

~ ~; I-

1

569

LATERAL EARTH PRESSURE COEFFICIENTS (KERISEl AND ABS!, 1970)

0.60 15

20

25

4>'

30

35

40

40

45

-0.80 45

(degrees)

§.!L

1.00 100 -

- t--:7"'i-r-74r-----:l 0.00

ll~~~~~~~I~~~~'~-0.20

-0.40

~;::.<"v~.,...,~-t-=-t=-t-"""""==j -0.60

~~~~==:.t=:~==t==t=~ -0.80 20

25

30

35

40

45

4>' (degrees)

ff (j=-4>'

F

1.00 0.60 OAO

0.00 -0.20 -OAO

-0.60 -0.80 1L-~L-~~~--~

10

15

20

25

__~__~__~

30

35

40

45

4>' (degrees)

Horizontal component of the passive lateral pressure coefficient. (Plotted from data published by Kerisel and Absi, 1990.)

FIGURE C.3

570

APPENDIX C

LATERAL EARTH PRESSURE COEFFICIENTS (KERISEL AND ABSI, 1970)

1.00 -j-- - , - - - , L-/--:;/ 0.60

10

0.40

FO

0.0 1 '---'---"'----'---'-------'----'-----' 10 15 20 25 30 35 40 45

40

45

¢' (degrees)

(3 /¢ '

1.00 - 0.60 0.40 000 -0.20 . -0.40 -0.60 -0.80

20

25

30

35

40

¢' (degrees)

nt the passive lateral pressure coefficient. Kerisel and Absi, 1990.)

45

ANSWERS to selected problems Answers to Even-Numbered Problem Solvin Questions in Chapter 2 to Chapter 6. Chapter 2 2.6

(a) 28%

2.8

(a) 38%

2.10

Dr = 39%, S

(b) 0.76

=1

2.U SP 2.14 2.16

2.18 2.20

2.22

- 8.9

10- 4

X

FO

(b) 23° u (kPa) 0 0 49 3.12

(c) 283 kPa

,

O'z

(kPa) 0 85 .5 133 .0

19.6 kPa

3.14 Position

0" z

(kPa) 17.3 28.1 42.4

A B C 3.16

57 kPa

3.18

TSP:-

~q ~p

3 2

= --

u (kPa) 39.2 53.9 73.6

ESP:

Cd) 626 kPa,

:;1 ~ 00

,

O'x

O'x

(kPa) 8.7 14.1 21.2

(kPa) 47.9 68.0 94.8

~u

= 133.3 kPa 571

ANSWERS TO SELECTED PROBLEMS

Chapter 4

4.8 4.10

4.U 4.14 4.16

Chapter 5

FO

572

5.4

Soil will not fail. FS = 1.55

5.6 5.8 5.10

~s =

5.U 5.14

5.16 5.18

31.1°

22°

<1>; = 45.6°

REFERENCES Adamson, A W. (1982). Physical Chemistry of Surfaces. Wiley, New York. API (1984). API Recommended Practice for Planning, Designing and Constructing Fixed Offshore Platforms. American Petroleum Institute, Washington, D.C. Atterberg, A (1911). "Uber die Physikalishe Bodennuntersuchung und uber die Plastizitat der Tone." Int. MiU. Boden, 1, pp. 10-43. Azzoz, A S., Krizek, R. J., and Corotis, R. B (1976). "Regression analysis of soil compr sibility." Soils Found. , 16(2), 19-29.

32). ' esearch on the Atter" Public Roads, 13(8), 121-

FO

-'-0""';'.3;',.

Bjerrum, L. (1954). "Geotechnical properties of Norwegian marine clays." Geotechnique, 4(2), 49-69. Bjerrum, L., and Eggestad, A. (eds.) (1963). "Interpretation of loading tests on sand. " Proceedings of the European Conference on Soil Mechanics and Foundation Engineering, Wiesbaden, Vol. 1, pp. 199-203. Bolton, M. D. (1986). "The strength and dilatancy of sands." Geotechnique, 36(1), 65-78. Boussinesq, J. (1885) . Application des Potentiels a l'Etude de Ie l'Equilibre et du Mouvement des Solides Elastiques. Gauthier-Villars, Paris.

Casag nde, ed.) (1936). "The determination of the reconsolidation load and its practical signifi nee." 1st International Conference on Sili echanics and Foundation Engineering, Cambridge, Massachusetts, Vol. 3, pp. 60-64. Casagrande, A (1937). "Seepage through dams." 1. N. Engl. Water Works Assoc., L1(2) , 131-172. Casagrande, A, and Fadum, R. E. (1940) . "Notes on soil testing for engineering purposes." Soil Mechanics Series, Graduate School of Engineering, Harvard University, Cambridge , MA, Vol. 8(268), p. 37. Chang, M. F. (ed.) (1988). "Some experience with the dilatometer test in Singapore." Penetration Testing, Orlando, 1,489-496. Chu , S. (1991). "Rankine analysis of active and passive pressures in dry sands." Soils Found. , 31(4),115-120. Coulomb, C. A "Essai sur une application des regles de maximia et minimis a quelques problemes de statique relatifs a l'architecture." Memoires de [a Mathematique et de Physique, presentes a {'Academic Royale des Sciences, par divers savants, et Ius dans ces Assemblees. L'Imprimerie Royale, Paris, pp. 3-8. Darcy, H. (1856). Les Fontaines Publiques de la Ville de Dijon. Dalmont, Paris.

573

574

REFERENCES

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Drnevich , V. P. (1967). Effect of Strain History on the Dynamic Properties of Sand. PhD. Thesis, University of Michigan. Dumble ton, M. J., and West, G. (1970). "The suc tion and strength of remoulded soils as affected by composition." LR306, Road Research Laboratory, Crowthorne. Dutt, R. N., and Ingra, W. B. (ads.) (1 8 . ~ils. " 0" Jackup rig sitting in calcare ceedings of the 16th Annual OTC, Houst , Texas , pp. 541-548.

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Roscoe, K. R , and Burland, J . B. (1968). "On the generalized stress-strain behavior of 'wet' clay." Engineering Plasticity, J. Heyman and F. Leckie (eds.) . Cambridge University Press, Cambridge , pp. 535-609. Rosenfard, J . L. , and Chen, W analysis solutions of earth p Fritz Engineering Lab0'r at Lehigh University, them Rowe, P. W. (1957) . ' Proc. Inst. Civ. En

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Reese, L. c., and O'Neill , M. W. (1988). Drilled Shafts: Construction Procedures and Design Methods. Federal Highway Administration , Washington, DC. Reynold, O . (1885). " On the dilatancy of media composed of rigid particles in contact; with experimental illustrations." Philos. Mag. , 20 (5th Series) , 469-481. Richart, F . E . (1959) . "Review of the theories for sand drains." Trans. ASCE, 124, 709-736. Robertson, P. K., and Campanella, R . E . (1983). "Interpretation of cone penetration tests. Part I: Sand ." Can. Geotech. J, 22(4) , 718-733.

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FO

Skempton, A. W., and MacDonald, D. H. ~ 6). "The allowable settlement of buildings." PI' Inst. Civ. Eng. Part III, 5, 727-768.

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INDEX

of piles, 374 of soils, 318 bearing capacity equations accuracy of, 329 differences among, 328-329 in general use, 323

FO

A adhesive stress, 464 A-line equation, 37, 38 allowable bearing capacity, 322 calculation of, 329-330 equation for, 330 and factor of safety, 329-330 allowable soil bearing capacity, 471 alluvial soils, 9 cr-method based on total stress analysis, 379-380 use of, 379-380, 385 alumina sheets, 10-11 American Association of State

axial symmetric condition, 91-92, 274 axial total stress, 226 axisymmetric condition , 91-92

C B

backfills, 466 balloon test, 62-63 base failure, 524 base slide, 524 bearing capacity see also allowable bearing capacity of layered soils, 339

578

calcareous soil, 9 calcium carbonate, 201 caliche, 9, 201 Cambridge Camkometer, 250-251 cantilever retaining walls, 469, 471 cantilever sheet pile walls analysis of, 484 determining stability, 485 capillary action, 104

coarse-gra ined soils comparison with fine-grained soils, 13 determining bearing capacity, 350 determining settlement, 350 particle size of, 22-23 sampling of, 385 volume changes in , 13 coefficient of concavity, 25 coefficient of consolidation, 173-176 coefficient of curvature, 25, 27 coefficient of permeability for common soil types , 559 in Darcy's law, 44-45 determining, 8, 51-53, 55 coherent gravity method procedure for, 505-506 use as low extensible reinforcing materials, 504-505 cohesion, 200,206,210 collapse load in dilating soils, 321 versus failure load, 321 from limit equilibrium, 323- 325 compaction of backfill soil, 459 definition of, 64 equipment for, 60 importance of, 60

INDEX

FO

quality control, 60-64 stresses, 460 compensated mat foundation, 326 compliance matrix, 89 compression axisymmetric, 274 downward movement, 207 compression index, 276 and correlation with plastic index, 308 definition of, 143,263 determination of, 178 compression tests, 226- 239 concrete piles cast-in-place, 371 precast, 371 types of, 371 cone data, 354 cone penetration test, 353 confined flow , 430 confining pressure, 218, 329 consolidated drained (CD) compression test duration, 231 failure, 278 purpose of, 229 stages of, 229 stresses, stress paths and strai s during, 229-231 using the results, 231 consolidated undraine compression test failure , 279 and Mohr's circle, 235 purpose of, 234 les, 305

consolidation settlement basic concept of, 144-147 estimation of, 400-401 parts of, 188-189 under a pile group, 400-401 consolidation settlement, onedimensional estimation of, 348 of fine-grained soils, 141 consolidation settlement parameters, 183 consolidation soil, 151

consolidation theory, onedimensional, 161-168 . constant-head test, 51, 52 constant interface friction value, 501 constitutive relationships, 214 construction activities, 527 Coulomb, Charles (1736-1806), 1 Coulomb's earth pressure theory , 456-460 Coulomb's equations, 207 [or shear strength of soils, 211 as slip is initiated , 212 Coulomb's failure criterion, 1 Coulomb's failure envelope, 0 Coulomb's failure wedge 456-457

r

579

crystalline materials, 10 curved slip surface, 458

D

dispersed struc 2-13 displacement gal,lge to meastLre settlement, 321 r testi settlement rate , 229 dis rli settlement, 341 _____~~d,ow lJ! rag, 404 :o- - _.nr'3 inage blankets, 438 drainage path definition of, 143 length of, 146-147 drained compression test, 296-297 drained condition, 130, 217 drained shear strength , 217-219 drained triaxial test, 277 drum type roller, 60 dry unit weight, 8, 17, 58- 60 dry unit weight-water content, 58-59 Dupuit's assumption, 437 "Dutch " cone penetrometer, 249-250 dynamic liquefaction, 525

as an estimation tool , 262 failure stresses from, 277 and failure stress state, 309 intention of, 310 to interpret soil behavior, 262 key feature of, 309 predicted responses from , 272 strains from, 290-295 use of, 262 critical state parameters, 273-276 the essen tial , 276 and test correlations, 307-309 critical state shear stress, 203, 204

E early time response, 174 earth dams, flow through, 436-439 earth pressure, lateral, 109, 188 earth pressure coefficient, active, 447, 450,452 earth pressure coefficient, lateral, 460-462 earth pressure coefficien t, passive, 447, 450, 452 earthquakes, 525 earth retaining structures methods of analysis, 446 stability of, 446

580

INDEX

engineering use chart, 39, 40-41 ENR equation, 406-407 equilibrium equations, 529-530 equipotential line, 416 erosion, 525 excavated slopes, 527 excess pore wa ter press ure definition of, 143 dissipation of, 217-218 volume changes in, 217 excess pore water pressure, negative, 466 expansion, 207 expansive soils, 9 external loading, 527 external stability, 502 extrema, 324

Empire State Building, 2 empirical soil strength relationships, 252, 561 end bearing, 374 end bearing capacity, 383 end bearing pile, 369, 397 end bearing resistance, 375, 382 end bearing stress, 369 engineering design economy of, 2 stability of, 2 engineering soils, 13

field tests, 247-252 fill slopes, 527, 528 filter fabrics , 185

F

13

FO

eccentric loads, 336-337 effective friction angle, 200 effective stress for calculations of lateral earth forces, 480 definition of, 81 due to geostatic stress fields , 102-103 horizontal principle, 109 principle of, 100-103 vertical principle, 109 effective stress analysis (ESA) for bearing capacity, 325 for drained conditions, 218 for sheet pile walls, 479 for slopes, 553 effective stress changes, 147 effective stress path, 130-131 effects of, 272-273 and excess pore water pressures, 234 intersecting the critical state line, 269 and soil behavior, 267 elastic analysis, 79 elastic equations for soils, 88, 344-345 elastic horizontal displacement and rotation, 356-357 elastic materials definition of, 81 and stress-strain response 84-86

circular and noncircular, 534 concept in CSM, 265 definition of, 523 failure versus coUapse, 207 failure zone, 523 fall cone apparatus, 32 fall cone method determining liquid limits, 32-33 determining plastic limits, 32-33 falling-head test, 52-53 field compaction, 60

calculations for, 440 definition of, 416 examples of, 420--421 interpretation of, 422-424 for isotropic soils, 420--421 procedure for constructing, 420 flow net sketching, 417, 419-422, 479 flow path, 418 flow rate, 52, 422 flow slide, 524 fluvial soils, 9 footings definition of, 319 shallow types of, 320 forces and moments problems determining, 533-534 provoking failures, 415, 446 foundations circular and uniformly loaded, 114 definition of, 319 designing, 4 intolerable displacement and rotation, 356-357 rectangular and uniformly loaded, 115-116

INDEX

stability conditions, 318 stable and economic, 1 as a structure, 318 foundation settlement, 341 foundation soil, 528 foundation stress, 330 free earth conditions, 485 free earth method, 485 frictional failure line, 213 frictional forces , 459 friction angle at critical state, 212 as a fundamental soil parameter, 212 at peak shear stress, 212 range of, 211 selection of, 328- 329 for soil mass, 466 friction pile, 370 full hydrostatic pressure, 482 full skin friction, 376

procedures to analyze, 472 type of, 469 groundwater definition of, 8 effects of, 330-331 and gravitational flow, 42 lowering of, 56 groundwater seepage, 472 groundwater situations, 330-331

FO

isotropic compression, 129 isotropic consolidation, 130, 305 isotropic consolidation line, 267 isotropic consolidation phase, 229 ]

H hammer types, 407 hand augers, 68 head, 8, 42-43 head loss, 43, 50,418, 422

G general shear failure, 323 geogrids, 500, 501 geological features , 525 geomechanics, 2 geotechnical engineer assessing slope stability, 522. concerns of, 165 considerations whe 218 descriptions of, 6 design of geotechnica 212

geotechnics, 2 geotextiles, 500, 501 glacial soils, 9 glacial till , 9 governing consolidation equation finite difference solution of, 166-168 procedure of applying the finite difference form, 167- 168 using the finite difference method, 166 using the Fourier series, 163-166 gravity retaining wall definition of, 447

581

initial yield shear stress, 270 initial yield surface and effective stress path, 268-269 expanding, 269 intersecting, 268 for soils, 273 instability causes, 415 instantaneous load, 145 interface friction angle, 466, 470 interface friction value, 466, 501 internal stability, 502 isotropic, 81 isotropically consolidated soil , 304-305

for two-dimensional flow, 417-418 lateral earth pressure and applications, 510 assumptions of, 448 basic concepts of, 448 coefficient, 109 passive, 322-323 at rest, 109 on retaining walls, 465 for total stress analysis, 462-465 lateral earth pressure, passive to estimate, 466 lowered by Caquot and Kerisel values, 465 lateral earth pressure coefficients, 460-462, 566- 570 lateral stresses, 460 lateritic soils, 9 law of continuity, 44 layered soils defining soil properties for, 339 determining soil properties for, 339 leaning tower of Pisa, 142 length of reinforcement, 503 limit analysis, 528 Limit equilibrium , 456

582

INDEX

FO

limit equilibrium method, 528 for collapse load, 324-325 essentia I steps, 456 for infinite slopes, 529 purpose of, 318 solution from, 457 steps for, 323-324 using, 462 line load, ll2 liquidity index, 30, 34, 309 liquid limit definition of, 8, 34 determination of, 31 versus plasticity index, 37 liquid state line, 31, 33 load, instantaneous, 145 load capacity allowable, 378 interpretation of, 377 of piles, 374 load cell, 249 loading history, 150-151 loading methods, 377 loading rate, 217 load-settlement behavior, 355 load transfer, 376 loam, 10 local shear failure, 323 loess, 10 log time method determining consolidatio coefficient, 173 plotting readings of, 175 procedure for, 175-176 versus roo tI e method, 17 lower bou solu i n, 457

metal strips, 500, 501 method of slices, 536-538, 542 application of, 540-541 determining the factor of safety, 543 procedure for, 541-543 methods of analyses applying factor of safety, 482--483 to determine stability, 482-483 types of, 482-483 Meyerhof's bearing capacity equations, 327-328 Meyerhof's bearing capacity factors, 327-328

one-dimensional consolida tion test, 172- 179, 173 one-dimensional flow, 42-44 overconsolidated soil, 143

f coarse-grained S J s, 22-23 i tribution urve, 23 o . n -g. ~a_i ~ soils, 24 rallO A-_ _ _ _'.:.e~~dila

.ron angle, 221

",£,--peax.~ective

pas ive pressure method (NPPM), 483 _"'---.::J_wark 's chart, 118 n imum active lateral earth nondilating soils, 253 force , 462--463 normal consolidation line, 149 maximum tolerable settlement, normal effective stress, 204 342 normally consolidated soil, 143 mean stress, 81 normal strain, 82 mechanical stabilized earth, 447 normal stress, 82 mechanical stabilized earth nuclear density meter, 63 (MSE) walls numerical methods, 479 basic concept of, 500-501 essential point in, 501 o stability criteria for, 502 oedometer test, 172 stability of, 502-506 oedometer test apparatus, 172 types of, 500- 501 one-dimensional consolidation use of, 500-506 equation, 162-163,165,348 using Rankine's method , 502 one-dimensional consolidation settlement, 141, 144-147 Menard pressure meter, 250

friction angle , 211 peak shear force, 221 p. ak shear stress, 221 for dilating soils, 270 dilation angle at, 211 for Type II soils, 204 for Type I soils, 204 peak shear stress envelope, 206 phase relationships, 14-18 phreatic surface, 436, 541 presence of, 533 procedure to draw, 439 piezocone,249,250 pile cap, 392 pile-driving formula , 406-407 pile-driving operation, 373 pile-driving records , 408 pile foundations definition of, 369 indica tions for use, 368 types of, 368-369 pile groups connection of, 392 load bearing distribution, 393 load capacity for, 393 use of, 392-393 pile installation equipment used in, 371, 373 implication of, 376 methods for , 371 pile load capacity analyses of, 374 determining allowable, 378

INDEX

FO

factors of, 377 reduced effect, 404 pile load test, 377 piles comparison of types, 372 definition of, 368 load capacity of, 374 subjected to negative skin friction, 404-405 types of, 370-371 pile se ttlement components of, 397 increased effect, 404 pile shaft, 375 piping, 423 plane strain condition , 90-92 plastic behavior, 29 plasticity index, 29-30, 37 plastic limit definition of, 8, 32-33 determination of, 26, 33 plastic strains, 290-291 plate load test problems associated with, 356 settlement from, 355 point bearing pile, 369 point load, 111 Poisson's ratio , 84, 559 pore water, 14 pore water pressure, 101-102, 103

triangular distribution, 16 , variation of initial excess, 167 pore water pressure ratio, 536 pore water pressure transducer, 102 porosity, 7,16 power augers, 68 Prandtl's theory , 322, 325 preconsolidation effective stress, 176-178 preconsolidation pressure , 176-178 preconsolidation stress, 143, 151, 176- 177 pressure head, 42-43 pressure meters, 250- 251

583

primary consolidation under a constant load, 145-146 definition of, 142 effects of vertical stress on, 148-149 slopes for, 149-150 primary consolidation parameters, 149-150 primary consolidation settlement, 148 calculation of, 152-155 method for calculating, 347-348 of normal and

Rankine's principle of stress sta tes, 460-462 Rankine states, 457 active and passive, 450-451 family of slip planes in, 452 and lateral forces, 451 passive zones, 323 rapid drawdown , 541 rate of consolidation, 147 Raymond cylindrical prestressed pile, 371 recompression index, 263, 276 relative density, 18 residual friction angle, 540

sa e bearing capacity, 319 safety factor, 319 sampling tube, 67, 69 sand cone test, 61-62 sand drains analysis of, 186 construction of, 184-185 diameter of, 185 half-closed drain, 184 procedure for spacing, 186 purpose of, 184 and radius of influence, 185 spacing of, 185 triangular grid , 184 two-way drainage, 184-185 and use of filter fabrics, 185 saturated unit weight, 17 saturation, degree of, 7, 16 Schmertmann 's method , 178, 353 Sears Tower, 2 secondary compression, 146 secondary compression index, 171 , 179 secondary compression settlement, 171 secondary consolidation settlemen t, 171 seepage effects of, 104-106 and retaining wall instability, 472 in soils, 105

584

INDEX

sheepsfoot roller, 60 sheet piles, 496 sheet pile walls analysis of, 479-482 definition of, 447 mixed analysis of, 481-482 in mixed soils, 481-482 types of, 479-481 in uniform soils, 479-480 Shelby tube , 67, 69 shrinkage limit, 8, 29, 33 side shear, 369 side-wall contact effect, 343 sieve, 22-23 sieve analysis , 22-23, 26 silicate minerals, 10-11 silicate sheets, 10-11 silica tetrahedrons, 10

; ~ r strength of soils ' based on dilation capacity, 210-211 importance of, 199 interpretation of, 210-213 simple model using Coulomb's law, 206-210 shear strength parameters, 540 calculating for weakest layer, 339 empirical relationships for, 252 types of tests to determine, 219-241 shear stress, 83 shear vane device, 247 shear wave velocity, 289

methods of analyses, 528-529 using simple geometry, 550-551 slope stability analysis, 553 slope stability calculations, 540 smear zone, 186 ~ soil(s) calcuating weight of, 15 calculating lume of, 1 class' Clatio 38

r. .:::

~~~:::::::i.t~r~s~s-strain characteristics, 199

FO

seepage force, 480-481 seepage forces, 104- 105, 467, 529, 530 seepage-related failures, 472 seepage stress, 416 seepage stresses, 105-106 semi-empirical methods, 496 sensitivity, 309 serviceability limit state, 320, 341 settlement amount of, 13 rate of, 13, 148 of shallow foundations , 340-341, 342-349 settlement calculations, 342-349 settlement changes, 148 settlement of piles, 397 settlement of single and group piles, 401 shaft friction, 369 shaft friction stress, 369 shallow footings, 343 shallow foundation, 319 shallow foundations, 318, 319 shallow foundation settlement, 342-345,347-349 shape factors, 326 shear band(s), 202, 203 , 205 shear box apparatus , 219-220 shear box test, 219-220 shear failure, 200, 323 shearing forces, 201-206 shearing phase, 229, 234, 238 shear modulus, 289 _ 2, 235, 291

uniform distribution of stresses on, 463 slip surface, 523 slope analyses, 535 slope angle, 523 slope failure , 553 causes of, 525- 528 from natural causes, 553 types of, 524-525 slopes, 522, 530 slope stability, 522 determining, 553 effects of excavation, 527

types of, 9-10 typical response to shearing forces, 201-206 as a viscous fluid, 423 soil bearing capacity failure , 471 soil behavior at the critical state, 288 interpreting and anticipating, 262, 267-269 soil deformation, 135 soil description, 558 soil fabric structural arrangement of, 209 types of, 12-13 soil failure, 393 soil tilter, 26 soil formation, 8-9 soil grains, 209 soil groups, 40 soil investigation, 5, 6, 65-69 and exploration methods, 66-68 and identification, 67 phases of, 66 purposes of, 65 soil layers, 155 soil parameters , 329, 558- 561 soil phases, 15 soil properties, 339 soil-reinforcement interface, 502 soil responses, 262 soils, heavily overconsolidated , 270

INDEX

FO

soil sampler, 69 soil sampling, 67, 69 soil states, 29 soil stiffness, 287-289 effect of shear strains, 288 regions of, 288 using wave propagation techniques, 289 soil strength based on liquidity index, 559 soil strengths, 307-309 soils under drained conditions, 267-269 soils under undrained conditions, 269 soil tests, 5 soil types, description and grain size, 558 soil-wall interface friction , 484 soil water, 14 soil yielding, 266-267 specific gravity, 16 specific volume, 15, 16 stability, 446 stability of slopes, 550-551 standard penetration number, 248 standard penetration test corrections applied to, 248 purpose of, 249 steps in performing, 248 standard penetration t sKS PT) and correction facto 350 determining bearing c from , 350

strain invariants, 123 strains, shear and volumetric in evaluation of deformation of soils, 212 in evaluation of shear strength, 212 required to achieve passive state, 465 strain states, 98 stream function, 418 stress, 81, 82-84, 89, 100-103 stress distributions, 496-497 stresses in soil, 110-111, 113-114 stress history, 305-306

stress invariants, 122, 123 stress path basic concept of, 127-133 definition of, 81 plotting of, 128-132 procedure for plotting, 133 stress states definitioo of, 81 Mohr's circle for, 96-97 Rankine's use of, 460-461 , 462 of soil elements, 450-451 stress-strain relationships, 214, 291 stress-strain response, 84-87 /l stress wave, 407 strip load, 112-114 structural loads, 368 stru t loads, 498 struts, 496

585

U

U-line' 38 ~

ultimat bea ng capacity in laj:lse d failure, 322 definition"'of 319 ,322-323

ultimy e load capacity '~-""-I:-:--_-:,,-_-:,,-_-_~_de1ln ition of, 369 parts of, 374 of piles, 374 ultimate net bearing capacity, 319, 326 un cemented soils, 201 unconfined compression (UC) test and Mohr's circle, 227 purpose of, 226-228 using the results, 227 unconsolidated undrained (UU) test advantages of, 239 purpose of, 238 stages of, 238- 239 undrained compression test, 297-298 undrained shear strength, 217-219 definition of, 200, 218 effects on volume changes, 217 and initial void ratio, 281 and relationship with liquidity index, 309 undrained triaxial test , 279 Unified Soil Classification System (USeS) categories of, 25 flowchart for coarse-grained total stress path, 130,234 soil, 37 Transcona Grain Elevator, 4-5, 329 flowchart for fine-grained soil, transition curve, 437 38 translation, 470, 471 uniformity coefficient, 25-26, 27 translational slide, 524 uniform settlement, 341 trench effect, 343 unit weight, 17

586

INDEX

unit weight, typical values, 558 unloading/reloading index, 143, 263 unloadinglreloading line (URL), 268 unloading/reloading of soil, 152-153 unstable slopes, 522 uplift forces , 424 upper bound solution , 457

FO

V velocity of flow , 418 velocity potential , 418 vertical component of the active lateral pressure coefficient, 568 of the passive lateral pressure coefficient, 570 vertical consolidation, 186 vertical dra inage, 186 vertical effective stress, 147 vertical effective stresses, 307 vertical elastic settlement, 114-115 vertical stress below arbitrarily shaped loads, 117

effects of, on primary consolidation, 148- 149 Newark's chart for increase in, 117 vertical stresses in a finite soil layer on a circular area, 562 on a rectangular area, 563, 564 vertical stresses in a two-layer soil,565 vertical total stress, 147 vertical wave propagation velocity, 407 Vesic's equation, 384 void ratio, 45 change in, 269 under a constant loap definition of, 7, 15

W wall adhesion , 466 wall displacements, 496 wall friction , 458, 460,466

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SOIL MECHANICS AND FOUNDATIONS CD-ROM The accompanying CD-ROM complements this textbook and contains multimedia interactive animation of the essential concepts of soil mechanics and foundations. virtual laboratories. a glossary. notation. quizzes. notepads . interactive problem solving. spreadsheet links and computer program utilities. Not all sections of the textbook are covered in the CD-ROM. The author has selected only those sections of the textbook that can be enhanced by the application of multimedia. You should refer to the CD-ROM section of "NOTES FOR fNSTRUCTORS" on page ix and the section on SUGGESTIONS FOR USfNG TEXTBOOK AND CD-ROM in "NOTES FOR STUDENTS AND INSTRUCTORS" on page xiii for further details. Installing the Soil Mechanics and Foundations CD To install the CD I. Start Windows if it is not already running on your computer. 2. Insert the CD into your CD-ROM drive. 3. Locate the file. setup.exe. on your CD. 4. Double click on setup.exe Alternatively From Windows. Click Start .-Run and enter D:/setup.exe where 0 is the drive letter of you ' drive has a different letter. for example E, then you should enter E:/setup.exe. 5. Follow the instructions presented by the installation program. 6. An Application shortcut named SoilMechanics will be created in the start menu.

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