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Geomechanics

IN SOIL, ROCK, AND ENVIRONMENTAL ENGINEERING

JOHN C. SMALLThe University of Sydney, New South Wales, Australia

CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

© 2016 by Taylor & Francis Group, LLCCRC Press is an imprint of Taylor & Francis Group, an Informa business

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v

Contents

Preface xixAcknowledgments xxi

1 Basic concepts 1

1.1 Introduction 11.2 Basic definitions 1

1.2.1 Submerged unit weight 31.3 Soil tests 4

1.3.1 Triaxial tests 41.3.1.1 Unconfined compression test 41.3.1.2 Unconsolidated undrained test 51.3.1.3 Consolidated undrained test with

pore pressure measurement 61.3.1.4 Consolidated drained test 61.3.1.5 Alternative failure plots 7

1.4 Direct shear tests 81.5 Consolidation tests 91.6 Permeability 12

2 Finite layer methods 15

2.1 Introduction 152.1.1 General concepts 15

2.2 Approximation of Fourier coefficients 192.3 Formulation 212.4 Solution procedure 222.5 Three-dimensional problems 232.6 Consolidation problems 252.7 Fourier transforms 262.8 Examples 28

3 Finite element methods 31

3.1 Introduction 313.2 Types of elements 31

vi Contents

3.2.1 Finding shape functions 323.2.2 Isoparametric elements 343.2.3 Infinite elements 353.2.4 Finite element meshes 36

3.3 Steady state seepage 373.3.1 Governing equation 373.3.2 Finite element formulation 383.3.3 Approximation of total head h 383.3.4 Finite element equations 393.3.5 Calculation of flows 393.3.6 Flow lines 403.3.7 Calculation of flow using the stream function 413.3.8 Determining the stream function 413.3.9 Pumping or extracting fluid 42

3.4 Stress analysis 423.5 Consolidation analysis 45

3.5.1 Effective stress analysis 453.5.2 Volume balance 47

3.6 Numerical integration 503.7 Elastic–perfectly plastic models 52

3.7.1 Formulation 523.7.2 Examples for a specific failure surface: Mohr–Coulomb 54

3.8 Work hardening models 553.9 Effective stress analysis using Cam Clay type models 56

3.9.1 Normally consolidated clay 583.9.2 Overconsolidated clay 59

3.10 Cam Clay type models 603.10.1 Cam Clay yield surface 60

3.11 Undrained analysis 633.12 Finite element analysis 65

3.12.1 Examples 67Appendix 3A: Shape and mapping functions for various element types 68Appendix 3B: Global matrix assembly and boundary conditions 73Appendix 3C: Boundary conditions 74

4 Site investigation and in situ testing 77

4.1 Introduction 774.2 Exploration methods 774.3 Site investigation 784.4 Object of site investigation 784.5 Category of investigation 784.6 Planning an investigation 794.7 Preparing cost estimates for the work 804.8 Detailed exploration 804.9 Presentation of information (logs) 80

Contents vii

4.10 Excavation or drilling methods 814.10.1 Test pits 814.10.2 Excavations 814.10.3 Drilling 81

4.10.3.1 Hand augers 814.10.3.2 Wash borings 834.10.3.3 Rotary drilling 834.10.3.4 Auger boring 83

4.11 Sampling methods 864.11.1 Thin-walled sampler (or Shelby tube) 874.11.2 Split spoon sampler (SPT sampler) 884.11.3 Piston sampler 884.11.4 Air injection sampler 884.11.5 Swedish foil sampler 88

4.12 Rock coring 894.13 Field tests 904.14 Vane shear test 914.15 Standard penetration test 91

4.15.1 Equipment 924.15.2 Sampler 924.15.3 Drive hammer 934.15.4 Rods 934.15.5 Test procedure 944.15.6 Properties of sands 954.15.7 Properties of clays 964.15.8 Liquefaction 98

4.16 Pressuremeters 1014.16.1 Types of pressuremeters 1014.16.2 Interpreting test results 102

4.17 Dilatometers 1034.17.1 Type of soil 1054.17.2 Shear strength of clays 1054.17.3 Other quantities 105

4.18 Cone penetrometers 1064.18.1 Equipment 1064.18.2 CPT equipment 1074.18.3 CPTu equipment 1074.18.4 Pushing equipment 1094.18.5 Calibration 109

4.19 Interpretation of cone data 1114.19.1 Soil classification 1114.19.2 Relative density of sands 1134.19.3 Friction angle of sands 1144.19.4 Constrained modulus of sands 1164.19.5 Young’s modulus of sands 1164.19.6 Undrained shear strength of clays 116

viii Contents

4.19.7 Undrained modulus of clays 1174.19.8 Permeability 118

4.20 Liquefaction potential 1194.21 Geophysical methods 121

4.21.1 Seismic surveys 1224.21.2 Reflection surveys 1234.21.3 Seismic refraction 1244.21.4 Rippability of rock 125

4.22 Resistivity 1254.22.1 Electrical sounding method 1284.22.2 Push-in resistivity instruments 129

4.23 Magnetic surveying 1304.24 Ground probing radar 1304.25 Seismic borehole techniques 130

4.25.1 Down-hole seismic testing 1304.26 Cross-hole techniques 1314.27 Other seismic devices 132

5 Shallow foundations 135

5.1 Introduction 1355.2 Types of shallow foundations 135

5.2.1 Strip footings 1355.2.2 Pad footings 1355.2.3 Combined footings 1365.2.4 Raft or mat foundations 136

5.3 Bearing capacity 1365.3.1 Uniform soils 138

5.3.1.1 General formulae 1435.3.1.2 Soil layers of finite depth 147

5.3.2 Non-uniform soils 1485.3.2.1 Strength increasing with depth 1485.3.2.2 Fissured clays 1515.3.2.3 Footings on slopes 1515.3.2.4 Layered soils 1545.3.2.5 Working platforms 157

5.4 Numerical analysis 1615.5 Settlement 162

5.5.1 Limits of settlement 1635.5.2 Settlement computation 1645.5.3 Theory of elasticity 1645.5.4 Rate of settlement 1665.5.5 Settlement of footings on sand 1695.5.6 Methods based on settlement and bearing criteria 172

5.6 Numerical approaches 1745.6.1 Layered soil: Finite layer approaches 174

Contents ix

5.6.2 Finite element methods 1755.6.3 Estimation of soil parameters 178

5.7 Raft foundations 1795.7.1 Strip rafts 1805.7.2 Circular rafts 1805.7.3 Rectangular rafts 1815.7.4 Raft foundations of general shape 183

5.8 Reactive soils 1845.8.1 Pad or strip footings 1875.8.2 Rafts on reactive soils 189

5.9 Cold climates 189Appendix 5A 190

6 Deep foundations 191

6.1 Introduction 1916.2 Types of piles 191

6.2.1 Driven piles 1926.2.2 Driven and precast piles 1926.2.3 Jacked piles 1926.2.4 Bored piles 1936.2.5 Composite piles 1936.2.6 Grout injected piles 195

6.3 Installation 1956.3.1 Types of displacement piles 1956.3.2 Small displacement piles 195

6.4 Pile driving equipment 1956.4.1 Piling rigs 1966.4.2 Piling winches 1966.4.3 Piling hammers 1966.4.4 Helmet, driving cap, dolly, and packing 199

6.5 Problems with driven piles 1996.5.1 Problems from soil displacement 199

6.6 Non-displacement piles 2006.6.1 Precautions in construction and inspection of bored piles 2006.6.2 Continuous flight auger piles (or grout injected piles) 201

6.7 Design considerations 2016.8 Selection of pile type 2026.9 Designs of piles 2026.10 Single piles 202

6.10.1 Piles in clay 2036.10.2 Piles in sand 2046.10.3 Lambda method 206

6.11 Methods based on field tests 2076.11.1 Correlations with standard penetration test (SPT) data 207

x Contents

6.11.2 Correlations with cone data 2086.11.3 Seismic data 209

6.12 Pile groups 2106.12.1 Piles in clay 2116.12.2 Piles in sand 212

6.13 Piles in rock 2136.14 Settlement of single piles 213

6.14.1 Closed form solutions 2146.14.2 Settlement of single piles 2146.14.3 Soil modulus increasing with depth 215

6.15 Interaction of piles 2166.15.1 Use of interaction factors for pile groups 2206.15.2 Simplified method for pile groups 222

6.16 Assessment of parameters 2236.17 Lateral resistance of piles 225

6.17.1 Single piles 2266.17.2 Piles in clay 2266.17.3 Piles in sand 226

6.18 Laterally loaded pile groups 2276.19 Displacement of laterally loaded piles 230

6.19.1 Linear elastic solutions (single piles) 2306.19.2 Constant soil modulus with depth 2306.19.3 Soil modulus linearly increasing with depth 2336.19.4 Non-linearity 234

6.20 Deflection of pile groups 2366.20.1 Interaction methods 237

6.21 Estimation of soil properties 2396.22 Load testing of piles 2406.23 Pile load tests 242

6.23.1 Static load tests 2426.23.2 Types of static load tests 2446.23.3 O-cell tests 2456.23.4 Lateral load testing 2456.23.5 Measurement of deflection 246

6.24 Dynamic pile testing 2476.24.1 Dynamic pile test 2476.24.2 Statnamic testing 248

6.25 Pile integrity tests 2496.25.1 Cross-hole sonic logging 2496.25.2 Sonic integrity test 2496.25.3 Gamma logging 251

6.26 Capabilities of pile test procedures 2516.27 Number of piles tested 2526.28 Test interpretation 253

6.28.1 Ultimate load capacity 2546.28.2 Pile stiffness 254

Contents xi

6.28.3 Acceptance criteria 2546.28.4 Other quantities 255

6.29 Monitoring of piled foundations 2566.30 Measurement techniques 256

6.30.1 Deflection 2566.30.2 Pressure cells 2576.30.3 Strain gauges 2586.30.4 Piezometers 2586.30.5 Extensometers and inclinometers 2586.30.6 Frequency of measurements 258

6.31 Comparison with predicted performance 2596.31.1 Emirates twin towers, Dubai 259

6.32 Interpretation and portrayal of measurements 2646.33 Piled rafts 2646.34 Uses of piled rafts 2656.35 Design considerations 265

6.35.1 Design process 2666.36 Bearing capacity of piled rafts 2666.37 Analysis of piled raft foundations 268

6.37.1 Numerical modelling 2686.37.2 Finite layer techniques 2696.37.3 Non-linear behaviour 271

6.38 Example of the finite layer method 2726.39 Applications 273

6.39.1 Westend Strasse 1 tower 2746.40 Structural stiffness 277

7 Slope stability 281

7.1 Introduction 2817.2 Slip circle analysis 282

7.2.1 The method of slices 2837.2.2 The Swedish, Fellenius, USBR, or Common Method 2857.2.3 Bishop’s method and simplified method 2857.2.4 Spencer’s method 2857.2.5 Finding the critical circle 2867.2.6 Water pressures 2867.2.7 Surface loads 2887.2.8 Computer programs 2887.2.9 Three-dimensional failure surfaces 288

7.3 Non-circular failure surfaces 2897.3.1 Morgenstern–Price method 2897.3.2 Janbu’s method 2897.3.3 Sarma’s method 290

7.4 Wedge analysis 2907.5 Plasticity theory 292

xii Contents

7.6 Upper- and lower-bound solutions 2927.7 Finite element and finite difference solutions 2937.8 Seismic effects 2947.9 Factors of safety 2957.10 Slope stabilisation techniques 296

7.10.1 Control of surface water 2967.10.2 Horizontal drains 2977.10.3 Stabilising piles 2977.10.4 Toe fill 2977.10.5 Retaining structures 297

7.11 Stability charts 298

8 Excavation 303

8.1 Excavation 3038.2 Types of excavation support 303

8.2.1 Steel ‘H’ piles and lagging 3038.2.2 Sheet piles 3038.2.3 Bored pile walls 3058.2.4 Diaphragm walls 306

8.3 Stability of excavations 3068.4 Base heave for cuts in clay 309

8.4.1 Shallow excavations (H/B < 1) 3098.4.2 Deep excavations (H/B > 1) 3108.4.3 Excavations of rectangular shape in plans 3108.4.4 Base failure in sands 311

8.5 Ground settlement caused by excavation 3128.5.1 Effect of shape of excavation 313

8.6 Forces on braced excavations 3148.7 Stability of slurry-filled trenches 316

8.7.1 Wedge analysis 3178.7.2 Purely cohesive soil 3178.7.3 Cohesionless soil 320

8.8 Numerical analysis 3218.8.1 Finite element analysis 321

8.8.1.1 Non-linear analysis 3238.8.2 Finite difference approach 323

8.9 Excavation including groundwater 3248.9.1 Example excavation problem (no drawdown) 3278.9.2 Excavation involving drawdown of the water surface 328

8.10 Soil models 329

9 Retaining structures 333

9.1 Introduction 3339.2 Earth pressure calculation 333

9.2.1 Rankine’s theory 333

Contents xiii

9.2.1.1 Inclined backfill 3379.2.2 Coulomb’s theory 337

9.2.2.1 Active case 3379.2.2.2 Passive case 3389.2.2.3 Surface loads 3399.2.2.4 Uniform materials 3399.2.2.5 Active earthquake forces 340

9.2.3 Log spirals 3419.2.3.1 Passive earthquake forces 341

9.2.4 Upper- and lower-bound solutions 3439.3 Effect of water 3449.4 Surface loads 346

9.4.1 Compaction stresses 3469.5 Sheet pile walls 3489.6 Anchored walls 350

9.6.1 Anchors 3509.7 Reinforced earth 352

9.7.1 Sliding 3559.7.2 Bearing failure 3559.7.3 Rupture of the reinforcement 3569.7.4 Pull-out of the reinforcement 3579.7.5 Overall slip failure 3589.7.6 Excessive deformation 358

9.8 Computer methods 3589.8.1 Limit equilibrium methods 3599.8.2 Finite element methods 361

10 Soil improvement 363

10.1 Introduction 36310.2 Soft soils 36310.3 Surcharging and wick drains 364

10.3.1 Surcharging 36410.3.2 Field observations 36810.3.3 Sand or prefabricated vertical (wick) drains 37010.3.4 Vacuum consolidation 375

10.4 Vibroflotation 37510.5 Vibro-replacement 376

10.5.1 Bearing capacity analysis 37710.5.1.1 Bearing capacity of single columns 37710.5.1.2 Bearing capacity of column groups 380

10.5.2 Settlement analysis of column groups 38310.5.2.1 Flexible foundation 38310.5.2.2 Rigid foundation 384

10.6 Column-supported embankments 38410.6.1 Collin beam method 38810.6.2 BS 8006 method 390

xiv Contents

10.7 Controlled modulus columns 39110.8 Dynamic compaction 392

10.8.1 Impact rollers 39410.9 Deep soil mixing 39610.10 Jet grouting 39710.11 Grouting 39810.12 Other methods 400

10.12.1 Ground freezing 40010.12.2 Electro-osmotic or electro-kinematic stabilisation 401

10.13 Numerical analysis 40110.13.1 Three-dimensional analysis 40210.13.2 Equivalent two-dimensional analysis 402

11 Environmental geomechanics 405

11.1 Introduction 40511.2 Landfills 405

11.2.1 Liners 40511.2.2 Covers for landfills 407

11.3 Compacted clay liners 40911.3.1 Compaction of clay 40911.3.2 Compaction method 41111.3.3 Compaction control 41211.3.4 Permeability of clay 41311.3.5 Measuring permeability of CCLs 414

11.3.5.1 Ring infiltrometer 41411.3.5.2 Borehole test 41511.3.5.3 Lysimeters 41611.3.5.4 Porous probes 416

11.4 Flexible membrane liners 41711.4.1 Types of geomembranes 418

11.4.1.1 High-density polyethylene 41811.4.1.2 Very low density polyethylene 41811.4.1.3 Polyvinyl chloride 41911.4.1.4 Chlorosulfonated polyethylene 419

11.4.2 Placing geomembranes 41911.4.3 Seaming 419

11.5 Geosynthetic clay liners 41911.5.1 Types of GCLs 42011.5.2 Manufacturing 42011.5.3 Placement 42011.5.4 Examples of use 421

11.6 Stability of liners 42211.6.1 Tension in the membrane 42211.6.2 Factor of safety 424

11.7 Processes controlling pollutant transfer 42511.7.1 Advective transport 425

Contents xv

11.7.2 Diffusive transport 42611.7.3 Dispersive transport 42611.7.4 Sorption 42711.7.5 One-dimensional transport 427

11.7.5.1 Ogata–Banks solution 42811.7.5.2 Booker–Rowe solution 429

11.8 Finite layer solutions 43211.8.1 Three-dimensional solutions 43211.8.2 Boundary conditions 434

11.8.2.1 Boundary condition at the base 43511.8.2.2 Boundary condition at the surface 435

11.8.3 Assembly of finite layer matrices 43711.8.4 Inversion of transforms 43711.8.5 Solutions for a three-dimensional problem 437

11.9 Remediation 43811.9.1 In situ leaching and washing/flushing 43911.9.2 In situ chemical treatment 440

11.9.2.1 Oxidation 44011.9.2.2 Chemical reduction 44011.9.2.3 Polymerisation 440

11.9.3 In situ biological treatment 44011.9.3.1 Microbial treatment 440

11.9.4 Soil venting 44111.9.5 Thermal desorption 44211.9.6 In situ stabilisation/solidification 44211.9.7 Electro-remediation 44311.9.8 In situ vitrification 444

11.10 Mining waste 44411.10.1 Properties of tailings 44511.10.2 Tailings dam construction 445

11.10.2.1 Upstream method 44511.10.2.2 Spigotting 44511.10.2.3 Cycloning 447

11.10.3 Centreline method of construction 44811.10.4 Downstream method 44911.10.5 Embankments built entirely of borrowed materials 45011.10.6 Tailings storages 45011.10.7 Control of water 451

11.10.7.1 Inflows 45211.10.7.2 Outflows 452

11.10.8 Stability of embankments 45311.10.9 Piping 453

11.10.9.1 Filters 45411.10.10 Cut-offs and barriers 455

11.10.10.1 Controlled placement of tailings 45611.10.10.2 Clay liners 456

xvi Contents

11.10.10.3 Cut-off trenches 45711.10.10.4 Slurry cut-off walls 45711.10.10.5 Grout diaphragm walls 45911.10.10.6 Grouting 460

11.10.11 Synthetic liners 46111.10.12 Seepage return systems 462

11.10.12.1 Collector ditches 46211.10.12.2 Collector wells 46211.10.12.3 Collection and dilution dams 463

11.10.13 Acid rock drainage 46311.10.13.1 Factors affecting acid generation 46411.10.13.2 Control of acid generation 46511.10.13.3 Control of acid migration 465

11.10.14 Remediation 46611.10.14.1 Hydrological stability 46611.10.14.2 Long-term erosion stability 46811.10.14.3 Vegetation 46811.10.14.4 Riprap 46811.10.14.5 Prevention of environmental contamination 468

12 Basic rock mechanics 471

12.1 Introduction 47112.2 Engineering properties of rocks 471

12.2.1 Point load strength index 47112.2.2 Unconfined compression test 47312.2.3 Modulus of rock from unconfined compression test 47312.2.4 Confined compressive strength 47312.2.5 Sonic velocity test 474

12.3 Failure criterion for rock 47512.3.1 Hoek–Brown criterion for intact rock 47612.3.2 Hoek–Brown criterion for rock mass 476

12.4 Classification of rocks and rock masses 47812.4.1 Classification on strength 47812.4.2 Classification by jointing 47812.4.3 Rock quality designation 47912.4.4 Classification of individual parameters used

in the NGI tunnelling quality index 48012.4.5 Rock mass rating method 480

12.5 Planes of weakness 48112.5.1 Stereographic projections 48112.5.2 Roughness of joints 483

12.6 Underground excavation 48412.6.1 Support systems 48412.6.2 Design process 48512.6.3 In situ stresses 485

Contents xvii

12.6.4 Stresses around underground openings 48712.6.4.1 Circular tunnel 48712.6.4.2 Elliptical tunnel 488

12.6.5 Support design 48812.6.5.1 Q index method 49012.6.5.2 RMR method 491

12.6.6 Support types 49112.6.7 Rock bolts and shotcrete 493

12.7 Rock slopes 49312.7.1 Planar sliding 49412.7.2 Wedge failure 496

12.8 Foundations on rock 49612.8.1 Surface foundations 49712.8.2 Shafts in rock 497

12.8.2.1 Base resistance 49812.8.2.2 Shaft resistance 49812.8.2.3 Lateral capacity 50012.8.2.4 Uplift capacity 50112.8.2.5 Piles on sandstones and shales of the Sydney region 502

12.8.3 Deformation of foundations on rock 50212.8.3.1 Vertical deformation 50212.8.3.2 Lateral deformation 503

12.9 Vibration through rock 50312.10 Numerical methods 504Appendix 12A 506Appendix 12B 510Appendix 12C 512Appendix 12D 515

References 517Index 537

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xix

Preface

Since the arrival of the twenty-first century, analysis and design of geotechnical facilities such as foundations, excavations, tunnels, and slopes has advanced rapidly, mainly due to the availability of computer-based techniques and the ease by which computer software can be purchased over the Internet.

The use of software makes complex analysis very simple and designs can be improved and refined through its application. However, there is still a need to look carefully at the results of any analysis and decide if the results are correct. Careful checking of input parameters and the output is essential in all design work, and the introduction of advanced graphical input and output into computer codes has made this process easier and has had the effect of reducing errors.

However, there is still a strong need for engineering judgment and this still has to be applied even though the tools for analysis are becoming more and more powerful. Therefore, in this book, although there are many references to advanced software, and for Internet locations for obtaining the software, there are many hand-based calculation methods pro-vided. These enable simple and swift checking of more advanced analysis. If the problem can be simplified and checked by hand, then this is a useful way of deciding whether to accept the results of numerical computations.

As many numerical techniques are commonplace today, an introduction is given to finite element and finite layer techniques in the initial chapters. Then, the chapters that follow deal with the application of geomechanics to soil and rock mechanics; the use of these numeri-cal techniques is mentioned where they are applicable. For example, finite layer techniques can be applied to settlement of foundations, consolidation problems, piled raft analysis, and pollution migration problems. Finite element techniques are applicable to most geotechni-cal problems; stress analysis, consolidation, seepage, soil–structure interaction, and rock mechanics are some applications.

New techniques that involve the use of optimisation are also mentioned where applicable. These techniques involve finding solutions to stress or velocity fields at discrete points that minimise a quantity such as the applied load or the power dissipation. These methods can be used in a whole range of geotechnical problems, and for problems involving collapse, provide upper and lower bounds to the collapse load. The real solution must lie between these bounds.

Many classical problems in soil mechanics have only partially been solved, and in recent years new solutions have become available that either show that the past solutions were of reasonable accuracy and may still be applicable, or show that these methods should no longer be used. An example is the Coulomb wedge method for obtaining passive earth pres-sures. These solutions have been shown to be grossly in error especially for large friction angles, and so there is no longer any need to use them, as more accurate modern values are available.

xx Preface

Today, the environment is becoming increasingly important and so one chapter is devoted to environmental considerations. Municipal and industrial waste is an increasing problem for large cities, and sites for landfills are rapidly being used up. The problem of isolating the waste so that it does not produce further pollution will become a major issue in the future and is one that needs to be addressed today.

Mining waste will also become a future problem as it has to be disposed of safely. The practice of dumping waste into rivers and the sea with no treatment or containment is no longer acceptable today, and so this aspect of geomechanics is addressed in Chapter 11.

Rock mechanics is a logical extension to soil mechanics, and often it is overlooked in texts on geomechanics. Here, an introduction is provided to rock mechanics that may assist students and professionals who wish to obtain some basic understanding of the design of foundations and excavations in rock. More detailed texts are available, and where possible the reader is encouraged to obtain the papers and textbooks referred to in the chapter on rock mechanics.

The material contained in this book is information that the author has found useful in his time as a university professor and as a consulting engineer, and it is hoped that it will serve as a source of reference for the reader as well.

John C. Small, BSc(Eng) Hons I, PhD, FIEAust, MASCE

xxi

Acknowledgments

I wish to express my thanks to my teachers from whom I have learned so much: Professor Bob Gibson of King’s College London, Professor John Booker, Professor Ted Davis, Professor Harry Poulos, and Dr. Peter Brown at the University of Sydney.

I also extend my thanks to all of my research students who over the years have gen-erated much new material, some of which is presented in this book. Finally, I gratefully acknowledge the support given to me by my parents throughout my career.

1

Chapter 1

Basic concepts

1.1 INTRODUCTION

In later chapters of this book, various terminologies are used and standard soil mechanics tests are referred to, therefore, this chapter provides some background on these basic tests and concepts. More details on testing of soil and rock are provided in the National and International Standards that are referred to in this and later chapters.

1.2 BASIC DEFINITIONS

Soil is considered to consist of the soil grains and the voids that exist between them. The voids can contain air and water, and so may be considered fully saturated if the voids are full of water or partially saturated if air and water fill the voids. Sedimentary rocks also consist of soil grains, but cementing that is present makes the rock stronger and less deformable than a soil.

Generally, soils subjected to moderate stress levels deform through changes in the void space in the soil, although calcareous soils will undergo volume change through particle crushing. At high pressures, soil grains will eventually crush, but in most soil mechanics applications, the particles are considered non-deforming.

The concept of a soil for engineering purposes is therefore as shown in Figure 1.1.The void ratio e of a soil is defined as the volume of the voids to the volume of the solid

soil particles and is defined as

e

VV

V VV

v

s

a w

s

= = +

(1.1)

where Va is the volume of air in the voids, Vw is the volume of water in the voids, Vs is the vol-ume of solids, and Vv is the volume of voids in a given total volume VTot of soil. Sometimes it is more convenient to work with the porosity n of a soil where the porosity is defined as

n

VV

V VV V V

v a w

a w s

= = ++ +Tot

(1.2)

A very useful property of a soil is its water or moisture content that is usually given the symbol w or m and is usually expressed as a percentage. If the weight of water is ww and the weight of solids is ws, then we can write

m

ww

w

s

= × 100%

(1.3)

2 Geomechanics in soil, rock, and environmental engineering

The degree of saturation S is a measure of how much of the voids of the soil are filled with water

S

VV

w

v

= × 100%

(1.4)

Another important measure of a soil’s properties is its unit weight. The saturated unit weight γsat is calculated for a soil with voids fully saturated with water, that is, S = 100%.

γ sat

Tot

= +w wVs w

(1.5)

The dry unit weight γdry of a soil is calculated for no water in the voids

γ dry

Tot

= wV

s

(1.6)

The soil may be in a state of saturation between the dry and totally saturated case (0 ≤ S ≤ 100%), where the voids contain some air and some water. In this case, the bulk unit weight γbulk is calculated from

γ bulk

Tot

= +w wVs w

(1.7)

Unit weights are expressed in kN/m3 or lbf/ft3 as they are expressed in terms of the weight as a force. A typical unit weight for a soil may be 19 kN/m3 or 120 lbf/ft3. The unit weight of water is 9.81 kN/m3 as the acceleration due to gravity is 9.81 m/sec2.

The density of a soil ρ is sometimes used in calculation and the density is defined as the mass of a soil per unit volume. For example, the bulk density may be calculated from

ρbulk

Tot

= +m mVs w

(1.8)

VA

VW

wA (= 0)

wW (=VW γW)

wS (=VS G γW)

Voids AirWater

Vs

Figure 1.1 Three-phase soil model.

Basic concepts 3

The density of a soil is expressed as the mass of soil per unit volume and has units of kg/m3 or lb/ft3. A typical value might be 2000 kg/m3 or 125 lb/ft3. The density of fresh water is approxi-mately 1000 kg/m3 or 62.4 lb/ft3 (as it depends on temperature).

With sands and gravels, often the term ‘relative density’ is used rather than the unit weight. The relative density Id gives a measure of where the void ratio of the soil is relative to the minimum emin (densest state) and maximum emax (loosest state) void ratios, that is,

I

e ee e

d = −−

max

max min (1.9)

It may be seen from the definition, that Id must range between 0 (loose) and 1 (dense).The specific gravity G of the soil grains is defined as the ratio of the density of the soil

grains ρs to the density of water ρw, that is,

G s

w

s

w

= =ρρ

γγ

(1.10)

1.2.1 Submerged unit weight

The concept of submerged unit weight comes from the calculation of stresses in the ground. According to the effective stress law that was originally proposed by Terzaghi (1923), the total stress in the ground is equal to the sum of the stress in the soil grains or ‘skeleton’ (called the effective stress) plus the water pressure. This can be expressed as

σ σ= ′ + u (1.11)

where σ is the total stress in the soil, σ′ is the effective stress, and u is the water pressure.This equation is one of the most important in soil mechanics as it explains how soils

behave in both drained and undrained conditions. It governs the strength and deformation of soils under all drainage conditions as it is the effective stresses in the soil that govern soil behaviour, and as can be seen by Equation 1.11, the effective stresses depend on the water pressures that are acting.

If we have a totally saturated layer of uniform soil with a static water table at the surface, the total vertical stress acting at any depth z is given by

σ γv z= sat (1.12)

The water pressure will be given by γwz at depth z and so we can calculate the vertical effective stress as

′ = − = ′σ γ γ γv wz z zsat (1.13)

where γ ′ or γsub is equal to the difference in the saturated unit weight and the unit weight of water, that is, γ ′ = γsat − γw. We can therefore calculate the effective stress directly by using the submerged unit weight, and this is why it arises. If there is any doubt in calculating an effective stress, it is recommended that a return to basics is used, and the water pressure subtracted from the total stress.


1.3 SOIL TESTS

In the following chapters, reference is made to soil properties that are obtained from lab-oratory tests, and so in this section some of the more common laboratory soil tests are described. Often the soil properties measured are also found from field testing by using cor-relations, as this is a faster and cheaper way to obtain the soil property. Because the values are obtained from correlations, they are generally not as reliable as laboratory values. For example, the undrained shear strength of a soil may be obtained from triaxial tests, or from shear vane tests, or static cone tests in the field (see Chapter 4).

1.3.1 Triaxial tests

One of the most valuable and commonly performed laboratory tests is the triaxial test. A cylindrical shaped sample of the soil is placed into a cell and covered with a latex mem-brane that is attached to end caps with rubber rings. The cell is pressurised by water and therefore applies an all-round pressure to the soil specimen called the cell pressure σ3. The triaxial cell is shown in Figure 1.2.

A loading ram through the top of the cell allows a load to be applied to the top of the soil specimen through the top cap. This load can be divided by the area of the specimen to calcu-late the stress being applied in the axial direction σ1–σ3 called the deviator stress. (Because there is a cell pressure of σ3 acting down on the top cap as well as the deviator stress from the ram, the total stress acting on the top cap is therefore σ1.)

Provision is made through porous stones placed at the top and bottom ends of the speci-men for drainage and pore pressure measurement. By knowing the pore water pressure in the sample, the effective stresses in the sample can be calculated.

There are several types of tests that can be performed with the triaxial cell, and some of the more common are examined in the following.

1.3.1.1 Unconfined compression test

As the name suggests, in the unconfined compression test, there is no cell pressure applied and there is no need to use the outer cell or membrane. The test is only applicable to clays as a clay sample is self-supporting. The unconfined compressive strength su is given by

s

PA

u =2

(1.14)

Drainage and pore pressuremeasurement

Rubber membrane

Porous disc

Porous discRubbersealing ring

Rubbersealing ring

Cell water Soil sample

Cell pressure

Pressure gauge

Loading ramBleed cock

Top cap

Axial load

Figure 1.2 Triaxial cell.

Basic concepts 5

where P is the applied axial load at failure and A is the cross-sectional area of the sample (corrected for increase in diameter after loading; A = A0/(1 − ε) where A0 is the original cross-sectional area and ε is the axial strain in the specimen). In such a test, the initial stress state of the sample is not controlled and so the strength obtained is for an unknown effec-tive stress state. However, the test is quick and easily performed and is therefore commonly used (ASTM D2166 2013).

1.3.1.2 Unconsolidated undrained test

If three similar specimens of the same clay material are tested under three different cell pres-sures in an undrained state (i.e. they are loaded in the triaxial cell until failure), then the Mohr’s circles for each of the samples can be plotted in terms of total stress at failure. The minor total principal stress is the applied cell pressure σ3 and the undrained strength su at failure is given by the deviator stress

s

PA

u = = −2 2

1 3σ σ

(1.15)

If the soil sample is not saturated, then consolidation can occur when the cell pressure is applied even though the sample is undrained. In this case, the undrained strengths will be different for samples tested at different cell pressures and the envelope to the Mohr’s circles at failure is often curved. If the failure surface is approximately linear, it can be represented in terms of total stresses by

τ σ φ= +cu n utan (1.16)

where cu is the intercept on the vertical axis of the plot and ϕu is the undrained angle of shearing resistance as shown in Figure 1.3. In applying this failure criterion in design, it should be noted that total stresses should be used.

If the soil samples tested are totally saturated, then any increase in cell pressure on the sample causes an equal rise in pore pressure inside the sample and, for each of the tests, the undrained strength su will be the same (because the effective stress in the sample is the same) and so the undrained angle of shearing resistance will be ϕu = 0. The test is described in ASTM D2850-15 (2015).

Undrained failure surfaceϕu

ϕu = 0 if soil saturated

su

τ

cu

σ3 σ1 σ

Figure 1.3 Undrained strength envelope from an unconsolidated undrained test.


1.3.1.3 Consolidated undrained test with pore pressure measurement

In this test, soil samples are allowed to consolidate under each cell pressure applied, and then tested undrained.

If the pore pressures are measured during each stage of the consolidated undrained test, it is possible to calculate the effective stresses at each stage of the test. Sometimes, three stages are performed on the same sample, but ideally three separate samples are used. The test is described in AS 1289.6.4.2 (1998), ISO/TS 17892-9 (2004), and ASTM D4767-11 (2011).

To obtain the effective stresses by subtraction of the pore pressure, the sample needs to be saturated, and so a test of saturation is performed by shutting the drainage valves and increasing the cell pressure. If the soil is perfectly saturated, then the increase in pore pres-sure should be equal to the increase in cell pressure. However, this is rarely achieved because of air existing in the pore water.

To overcome this problem, de-aired water is used to saturate the specimen and the pore water in the sample is pressurised under a back pressure to force air into the solution. For instance, if the back pressure is 200 kPa and the cell pressure is 300 kPa, then the net consolidation pressure on the sample is 100 kPa. A back pressure of up to 900 kPa may be required for this to be effective in some cases.

If the increase in the cell pressure is Δσ3 and the increase in pore pressure measured is Δu, the pore pressure parameter B can be calculated from

B

u= ∆∆σ3

(1.17)

The value of B should be greater than 0.95 before testing; otherwise, the back pressure should be increased (AS 1289.6.4.2 1998).

As the effective stress is known in this test, both the drained and undrained strength envelopes can be plotted. The drained strength envelope as plotted in Figure 1.4 is given by

τ σ φ= + ′ ′′c tan (1.18)

The deviator stress at failure for each stage is the same for the total stresses and the effec-tive stresses since

( ) ( ) ( )

PA

u u= − = + − ′ + = ′ − ′′σ σ σ σ σ σ1 3 1 3 1 3

(1.19)

1.3.1.4 Consolidated drained test

For this test, the sample is consolidated under the action of a cell pressure and water is allowed to drain from the sample. Back pressure may be used to keep the soil sample satu-rated as was explained in Section 1.3.1.3. The test is performed at a slow rate of shearing such that no excess pore pressure is generated in the soil sample and this requires drainage of the sample at all times during the test.

The test can be performed on sands and clays, although it can be carried out more quickly on sands because they have relatively high permeability. Back pressure may be required for dense sands that tend to dilate and cause a drop in the pore pressure. Tests on clays can last

Basic concepts 7

for several days, and so the test is not often used for fine-grained soils. Instead, the consoli-dated undrained test described in Section 1.3.1.3 is used.

The test is performed for three (or more) different cell pressures, and for each, the deviator stress at failure is found from

σ σ σ σ1 3 1 3− = = ′ − ′

PA

(1.20)

The effective stress ′σ3 at failure is the difference between the cell pressure and the back pressure ub and may be found from ′ = −σ σ3 3 ub and so the Mohr’s circles for each stage can be plotted in terms of the effective stresses. The result is the same as shown in Figure 1.4 for the failure surface found in terms of effective stresses from the consolidated undrained test with pore pressure measurement. The test is described more fully in ASTM D7181-11 (2011).

1.3.1.5 Alternative failure plots

An alternative form of plot from a triaxial test is a p − q plot. This form of plot is useful as the stress path plot goes vertically upward when the soil is elastic in an undrained test, and when the soil yields, the plot deviates towards the failure line (see Figure 3.28). The slope of the failure line in this type of plot is given by M where

M =

−6

3sinsin

φφ′

′ (1.21)

and so the drained angle of shearing resistance can be found from the plot.

sin ′ =

+φ 3

6M

M (1.22)

More is provided in Section 3.8 on this type of plot, and further information is given in the books by Atkinson (2007) and Atkinson and Bransby (1978).

Mohr–Coulomb failure surface ϕ′

c′

τ

σ′σ′1σ′3

Figure 1.4 Mohr–Coulomb failure envelope showing the Mohr’s circle at failure for each of the three stages.


1.4 DIRECT SHEAR TESTS

The shear strength available on a surface of shearing can be found directly in a shear box test. The box consists of two halves (Figure 1.5), and the bottom half is held fixed while the upper half is pushed, thus shearing the soil. The shear force is applied through a curved yoke so that the force is not applied eccentrically to the upper box. The shear force on the yoke is measured with a proving ring or load transducer (see Figure 1.6). The outer box into which the shear box is placed can be filled with water to keep the sample saturated during slow shearing.

For sands and gravels, the shearing rate is not of great importance as they are normally tested in a dry state. For clays, the test can be performed in the drained or undrained state.

For drained shearing of a clay, the speed of the machine is set to a very slow shearing rate so that the excess pore pressures that are generated have time to dissipate. When the normal load is placed onto the lid of the shear box, the settlement of the lid is monitored to see when the lid stops settling thus indicating that the pore pressures have dissipated (see ASTM D3080 or AS 1289.6.2.2 1998).

The residual strength of a clay can be found by reversing the shear box. The soil is firstly sheared forwards and then backwards until the shear force does not change with further shearing. This type of test may be required for soils that are liable to undergo large shear displacements where the shear strength drops to the residual value.

To perform an undrained test on a clay, the normal loads are placed on the hanger and the soil immediately sheared at a fast rate.

The test has the drawback that the pore pressure in the box cannot be controlled as it can in a triaxial test and so it is not possible to know what pore pressures are being generated in clay soils.

Liftinghandles

Lifting screw

Shear forceapplied to yoke

Clamping screw

Upper half of box

Lower half of box

Clearance holes

Gap

Figure 1.5 Two halves of a shear box.

Water (if necessary)

Lid for vertical loadWorm drivepushing outer box

Upper and lowershear box halves

Spacer block (onlyif box is reversed)

Loading ring dialgauge

Figure 1.6 A shear box in a loading frame.

Basic concepts 9

For each different normal stress applied to the lid of the box, the shear stress at failure can be measured and a plot is made as shown in Figure 1.7 (for a drained test). The same Mohr–Coulomb failure line is obtained as found in the triaxial test, but from the measure-ment of the shear stress and normal stress on the plane of shearing.

1.5 CONSOLIDATION TESTS

One-dimensional compression tests on clays are performed in an oedometer. The oedom-eter has a ring that is used to cut a sample of soil so that it is a tight fit inside the ring (Figure 1.8). Because the soil cannot expand sideways, the compression only occurs verti-cally (one-dimensional).

Porous disks are placed at the top and bottom of the soil specimen so that it can drain, and loads are placed on the sample so as to compress it. The time–settlement behaviour is recorded so as to obtain the rate of consolidation of the soil sample. As well, the amount of compression of the sample under the applied load is measured.

The vertical load applied to the sample is doubled at each load increment and a plot is made of the void ratio of the sample versus the effective stress applied (to a logarithmic scale) as shown in Figure 1.9. This can be used to compute the settlement of soil layers as outlined in Section 10.3.1 in Chapter 10.

The rate at which consolidation occurs is recorded at each loading stage, and a plot is made of settlement against either the square root of time, or the logarithm of time. A log time plot is shown in Figure 1.10.

Various parameters can be found from the tests that are used in the calculation of the magnitude of settlement or rate of settlement.

In the log time versus settlement plot, it may be seen that the final part of the plot is a straight line. This is due to creep of the sample after the primary consolidation has finished

Mohr–Coulomb failure surfaceϕ′

σ′n1 σ′σ′n2 σ′n3

τ

τ3

τ2

τ1

c′

Figure 1.7 A plot of shear strength versus a normal load for a drained shear box test.

Vertical load

Porous stonesCutting ring

Soil

Figure 1.8 A section through an oedometer.


and excess pore pressures are almost zero. Two straight lines are drawn as shown in Figure 1.10 (one through the steepest part of the curve and one through the final straight line part), and where they intersect is deemed to be 100% primary consolidation. Two points are then chosen on the initial curved part of the plot: one at a time t and the other at 4t. Because the first part of the curve is a straight line on a plot against the square root of time, the settle-ment from 0 to t is the same as from t to 4t. Using this fact, the location of zero settlement can be found as shown in Figure 1.10. The 50% consolidation point is then midway between the 0% and 100% consolidation points.

Theoretically, the 50% degree of consolidation takes place at a time factor Tv equal to 0.197, where

T

c tH

vv= 2

(1.23)

t2 10t2

50%

100%

0%

16 min8421 min30 s15 s7.5100 1000 10,000

24 ht1 4t1

2.1

2.0

1.9

1.8

1.7

1.6 ΔHα

1.5Vert

ical

def

orm

atio

n di

al g

auge

read

ing

(mm

)

Elapsed time t

10

yy

t50

Figure 1.10 A settlement versus log time plot for an oedometer test.

Pressure (kN/m2)1 10 100 1000 10,0001.10

1.30

1.50

1.70Vo

id ra

tio

1.90

2.10

2.30

2.50

Figure 1.9 Void ratio versus logarithm of effective pressure plot e − log σ′v for an organic silt.

Basic concepts 11

and t is time, H is the half-depth of the specimen (because of two-way drainage), and cv is the coefficient of consolidation.

We can therefore solve for the coefficient of consolidation cv that has units of m2/yr or ft2/yr as shown in Equation 1.24.

c

T Ht

Ht

vv= =

2

50

2

50

0 197.

(1.24)

In the above equation, t50 is the time for 50% consolidation and H is the average half-depth of the specimen during consolidation.

The rate of creep can also be calculated from the coefficient of secondary compression Cα defined as

C

HH

α = ∆0

(1.25)

where ΔH is the change in thickness of the specimen over one log cycle of time (for the last linear part of the curve), and H0 is the initial height of the specimen.

From the void ratio versus log of the applied effective pressure plot (Figure 1.9), it can be seen that there is a distinct kink in the loading part of the curve. This is the point at which the stress in the specimen reaches the pre-consolidation pressure (or maximum past pressure). The slope of the curve before this point is called the recompression index Cr and is defined as

C

e ep p

epp

ri

c

c

log loglog= −

′ − ′= ′

′

0

1 1

∆

(1.26)

The soil becomes more compressible after the pre-consolidation pressure is reached, and for this part of the plot the compression index Cc is calculated as

C

e ep p

epp

ci

c clog loglog= −

′ − ′= ′

′

2

2

2∆

(1.27)

Figure 10.3 shows the definitions of effective pressures p′ and void ratios e used in the above definitions.

Sometimes, the coefficient of volume compressibility mv is calculated from the oedometer test where

mv

v

v

=′

∆∆

εσ

(1.28)

and Δεv is the vertical strain that occurs due to a change in vertical stress ∆ ′σv on the sample.Hence, mv is like an inverse modulus, and if the soil gets stiffer, the value of mv reduces.

This means that mv has to be calculated for the stress range that it is to be used for as it is not constant, but varies with stress level. We can also write

m

Ce

vc

v

=+ ′

0 4351 0

.( )∆σ

(1.29)


In terms of a conventional elastic modulus of the soil, we can write

m

Ev = + ′ − ′

− ′ ′( )( )

( )1 1 2

1ν ν

ν (1.30)

where ν′ is Poisson’s ratio of the soil and E′ is its elastic modulus. As well, we can write cv in terms of elastic constants,

c

km

k Ev

w v w

= = − ′ ′+ ′ − ′γ

νγ ν ν

( )( )( )

11 1 2

(1.31)

This expression (1.31) is often used to find the permeability k of a clay as a conventional permeability test cannot be used for very low permeability soils.

The use of the parameters found from the oedometer test is explained more fully in Section 10.3.1. Details of the test are provided in relevant standards such as AS 1289.6.6.1 (1998), ISO/TS 17892-5 (2004), and ASTM D2435 (2011).

1.6 PERMEABILITY

For sands, the permeability can be found by direct means (for clays use Equation 1.31) by allowing water to flow through a sample of the soil at the appropriate void ratio. For soils with permeabilities of between about 10−2 and 10−5 cm/sec, the falling head permeameter can be used. The apparatus is shown in Figure 1.11, where water in a tube is allowed to drop as it flows through the soil. The permeability k is then given by

k

aAt

hh

=

ln 1

2 (1.32)

Soil sample

dt

ℓ

h1

Ds

h2

Figure 1.11 A falling head permeameter.

Basic concepts 13

where a is the cross-sectional area of the piezometer tube ( ),a dt= π 2 4/ A is the cross- sectional area of the sand sample ( ),A Ds= π 2 4/ ℓ is the length of the soil sample, and t is the time for the total head difference across the sample to drop from h1 to h2.

For more permeable soils, a constant head permeameter can be used (ASTM D2434-68 2006). For this type of permeameter, the head difference across the soil sample is kept constant through a supply of water as shown in Figure 1.12. The permeability k is given by

k

QthA

=

(1.33)

where an amount of water Q is collected from the outlet in a time t, ℓ is the sample length, A is the cross-sectional area of the sample, and h is the head difference across the sample length. The permeability is dependent on temperature and can be corrected back to a standard value at 20°C (k20°C = kTμT/μ20°C), and μT is the viscosity of the soil at temperature T.

These tests can also be conducted using a flexible wall permeameter (which contains the soil specimen in a rubber membrane such as a triaxial test sample – see ASTM D5084 2010).

Typical permeabilities of soils are shown in Table 1.1.

Water

Overflowh

Filter discsSoilsample

Q

ℓ

Figure 1.12 A constant head permeameter.

Table 1.1 Typical permeabilities of soils

Soil type Permeability range (cm/sec)

Gravels >1Sands 10−3 to 1Silts 10−6 to 10−3

Clays <10−6

15

Chapter 2

Finite layer methods

2.1 INTRODUCTION

There are many problems in the field of geomechanics, where the soil profile may be ide-alised as consisting of horizontal layers for which the geometry and material properties do not vary in one or two spatial directions.

Examples include (1) pavements that are constructed by placing horizontal layers of dif-ferent pavement materials (see Figure 2.1) and (2) sedimentary soils that are horizontally layered because sediments are deposited under a vertical gravity field (Figure 2.2). In the latter case, the layering may be in one direction only if the sediments are laid in an ancient stream bed or valley (Figure 2.3).

2.1.1 General concepts

In order to convey the general concepts of the method, suppose we consider a problem that is common in geomechanics, that of a strip loading applied to the surface of a layered soil as shown in Figure 2.4.

A function such as the loading function shown in Figure 2.4, may be represented by a Fourier series, where the addition of various periodic sine or cosine functions are added together to approximate the original function. Here, the function is a step function that is equal to the applied uniform load q between −a and +a if the load is half-width of a as shown in Figure 2.5 (Cheung 1976).

Because the loading function chosen is an even function of x, we only need use the cosine part of the Fourier series and to represent the loading function as a sum of cosine terms.

q x Q x

n L

QL

q x x dx n

Q

n

nn

n

n

L

n

( ) cos( )

( )cos( )

( )

( )

=

=

= >

=

∞

∑

∫

0

0

2

20

α

α π

α

/

(( ) ( )n

L

Lq x dx n= =∫1

00

(2.1)

We are therefore stating that the load function is periodic with a period L (Figure 2.6).


Clay

Clay

Bedrock

SiltPeat

Sand

Structure

Figure 2.2 Soil that has been deposited horizontally by sedimentation.

Surface loading

Void

Material interface

Figure 2.3 Soil layering that is horizontal in one direction only.

z

x

q

Figure 2.4 Strip loading applied to a layered soil.

Tyre load

Basecourse

Sub-base

Figure 2.1 Wheel loading applied to a pavement.

Finite layer methods 17

Provided that we make L large enough and that we use enough terms n in the Fourier series, we will be able to synthesise an isolated strip loading. Figure 2.7 shows the effect of adding terms to the series used to approximate the strip loading. By the time 50 terms are used in the series, the sum of the cosine terms is beginning to take the shape of the strip load-ing. The plot in Figure 2.7 is for a spacing of the loads (the period) of L/a = 20. If the spacing is made larger, it takes more terms in the Fourier series to get a good approximation of the step function, so in applying the method in practice, a judicious choice of spacing needs to be made so that fewer terms can be used in the solution, and the solution time is faster.

We can now make use of the theory of superposition (provided the problem involves linear elasticity). Suppose that instead of applying the strip loading, we apply the individual cosine loadings as shown in Figure 2.8.

For each of the applied cosine loads, we will get a periodic response, that is, the displace-ments in the soil, and the stresses will be distributed with the same period as the load.

If we add up the applied cosine loads (plus the one-dimensional or constant n = 0 term), we get the applied strip load. If we add up the corresponding displacements, stresses, or strains, we get the displacements, stresses, and strains for the strip load. We may therefore write for the vertical uz and horizontal ux displacements

u U x

u W x

xn

n

n

zn

n

n

=

=

=

∞

=

∞

∑

∑

( )

( )

sin( )

cos( )

α

α

0

0

(2.2)

Note that the horizontal displacements ux are represented by a sine series, as they will be in the positive direction (if x is positive) or in the negative direction (if x is negative).

z

q(x) = q

x–a +a

Figure 2.5 Strip-loading function.

L L

Figure 2.6 Periodically spaced loadings.


No. of termsin series

q(N

)q

1050

5

2

1

La = 20

aq

L

0

–0.50.5 1.0 1.50

0.5

1.0

1.5

xa

∞

Figure 2.7 Approximation of uniform strip loading by Fourier series.

Periodic loading

uz (same period asloading)

x

z

Figure 2.8 Response of soil to periodic load.


The stresses also may be represented in this form:

σ α

σ α

τ α

xxn

n

n

zzn

n

n

xyn

n

H x

N x

T x

=

=

=

=

∞

=

∞

∑

∑

( )

( )

( )

cos( )

cos( )

sin( )

0

0

nn =

∞

∑0

(2.3)

The direct stresses in the horizontal x and vertical z directions, σxx and σzz, are represented by a cosine series as they will have the same sign for all x values, but the shear stress τxz is represented by a sine series as the shear will be of an opposite sign if it is on the negative or positive side of the x-axis.

2.2 APPROXIMATION OF FOURIER COEFFICIENTS

For a single component in the Fourier series for the displacements, we can write

u

u

uU x

W xx

z

nn

nn

=

=

( )

( )

sin( )

cos( )

αα

(2.4)

The strains can then be found since εxx = ∂ux/∂x, εzz = ∂uz/∂z, γxz = (∂ux/∂z) + (∂uz/∂x)

εεεγ

αα( )

( ) ( )

( )

(

cos( )

cos( )n

xx

zz

xz

nxn

n

zn

n

xz

E x

E x

E

=

=nn

nx) sin( )α

(2.5)

where

E

E

E

Ez

z

Un

x

z

xz

n n

n

( )

( )

=

= ∂∂

∂∂

−

α

α

0

0(( )

( )

n

nW

(2.6)

The stresses may be related to the strains since

σστ

νν ν

νν

νν

xx

zz

xz

n

E

= −+ −

−

−

( )

( )( )( )

( )

( )1

1 1 2

11

0

11 0

0 0(( )

( )

( )

1 22 1

−−

νν

εεγ

xx

zz

xz

n

(2.7)


or in a more compact form

σ ε( ) ( )n n= D (2.8)

The Fourier coefficients for the displacements U(n), W(n) will vary with depth z (as the displacements must vary with depth). In order to obtain a numerical solution, suppose that we approximate these coefficients by assuming they vary linearly between nodal values (see Figure 2.9).

We can then write

W z

W z z W z zz z

i i i i

i i

( )( )

( )( )= − + −

−+ +

+

1 1

1 (2.9)

A similar linear interpolation may be carried out for the Fourier coefficient U

U z

U z z U z zz z

i i i i

i i

( )( )

( )( )= + −

−−+ +

+

1 1

1 (2.10)

For the nth Fourier term, we therefore have

U

W

a b

a b

U

W

U

W

n

n

i

i

i

i

n

( )

( )

( )

=

+

+

0 0

0 0 1

1

(2.11)

a

z zz

bz z

zi i= − = −+( ) ( )1

∆ ∆

x

z

ElementsNodes

q

(i + 1)

W(z)z

Wi+1 zi+1

zi

Wi

Δz

(i )

Figure 2.9 Use of linear interpolation functions.


or in matrix form

U C n( ) ( )n = δ

We can now use Equation 2.6 to obtain

E

a b

z

za z

zb

U

W

U

W

n

n n

n n

i

i

i

( ) =−

−

− −

+

α α

α α

0 0

0

1

1 0

1

1

1

∆∆

∆∆

ii

n

+

1

( )

(2.12)

Or, in matrix form

E B( ) ( )n n= δ

2.3 FORMULATION

For a layered system, we may write the virtual work equation as

V

T

S

Td dV d S∫ ∫=ε σ u T d

(2.13)

This equation states that if the body is in equilibrium, the work done by the external loads will be equal to the energy stored in the body under small virtual displacements.

If we take a single Fourier component (the nth) and substitute it into Equation 2.13, we have

d dV I d dz

d dS I d

n T n

V

nn T n T n

h

n T

S

nn

ε σ δ δ

δ

( ) ( ) ( ) ( ) ( )

( ) ( )

∫∫

∫=

=

B DB

u T

0

TT nT ( )

(2.14)

where

I x xn n n

LL

= = ∫∫ cos ( ) ( )2 2

00

α αsin

(2.15)

Substituting the expressions in 2.14 into 2.13, we have (since dδ(n) are arbitrary variations and not necessarily zero)

z

z

n T n n n

i

i

dz+

∫ ( ) =1

B DB T( ) ( ) ( )δ

(2.16)


or writing the above equation in more compact form

K Ten n

en( ) ( ) ( )δ = (2.17)

where

K B DBen

z

z

n T n n

i

i

dz( ) ( ) ( )=+

∫ ( )1

δ

(2.18)

As B(n) contains terms that are a function of z, we need to carry out the integration of Equation 2.18. This may be done numerically or analytically.

For all the linear elements, we must sum the effects giving

K K

T T

( ) ( )

( ) ( )

nen

e

ne

n

e

=

=

∑∑

(2.19)

As the only node that is loaded in this example problem for a strip footing is the top node, we will only have one entry in the T (n) vector, corresponding to the top node, for example,

T ( ) ( )( , , , , , )n n TQ= 0 0 0 0… (2.20)

where Q(n) is the nth Fourier coefficient in the series for the applied load.

2.4 SOLUTION PROCEDURE

The solution procedure involves solving the set of equations

K T( ) ( ) ( )n n nδ = (2.21)

for each of the Fourier components. This involves setting up the T (n) vector for each of the Fourier coefficients in the series for the applied load (see Equation 2.20) and the correspond-ing K(n) matrix. This matrix is a function of αn and must be set up for each Fourier term. Usually, this is a very fast process as we are dealing with only a few linear elements.

Equations 2.21 are then solved to give the Fourier coefficients of the displacements at the nodes δ(n)

Once these coefficients are known, the actual displacements at the nodes may be found since (for example, the displacement ux)

u U xxn

n

n

==

∞

∑ ( ) sin( )α0

(2.22)

In practice, only a few terms in the series are required to obtain a good approximation of the displacements. The number of terms depends on the spacing (or period L) of the


applied loads, however 20–30 terms would generally be enough. For larger L, more terms are needed.

The solution procedure is shown schematically in Figure 2.10.

2.5 THREE-DIMENSIONAL PROBLEMS

For three-dimensional problems, we may make use of the double Fourier series. For exam-ple, we could represent the uniform loading shown in Figure 2.11 by a double cosine series (Equation 2.23).

q x y Q x y n L m Mm n

mn n m n m( , ) cos( )cos( )= = ==

∞

=

∞

∑∑0 0

2 2α β α π β π/ /

(2.23)

This implies that the loadings are periodic, and hence in the case of general shaped loading, we would have a series of loads spaced at the centre of L and M as shown in Figure 2.12.

Form stiffness matrixand load vector forparticular harmonic

Approximate loadsusing Fourier series(vertical load only)

Solve stiffnessequations

Do forn = 0, …., N

Calculate fieldquantities at anydesired position x

q = ∑ Q(n) cos αnxn=0

N

ux = ∑ U (n) sin αnxn=0

N

uz = ∑ W (n) cos αnxn=0

N

K (n) δ(n) = T (n)

δ(n) = K (n)–1 T (n)

Tnδ(n) = (U1, W1, U2, W2, ....)

Q = Q(n)

α = αn

Figure 2.10 The process for solving a problem for each term in the Fourier series.


We may now express the displacements and stresses as double Fourier series

u V x y

u W x

y

m n

mnn m

z

m n

mnn

=

=

=

∞

=

∞

=

∞

=

∞

∑∑

∑∑0 0

0 0

cos( )sin( )

cos( )co

α β

α ss( )

cos( )cos( )

β

σ α β

m

yy

m n

yymn

n m

y

S x y==

∞

=

∞

∑∑0 0

(2.24)

If we again approximate the Fourier coefficients’ variation with depth by using linear interpolation between the nodal values, exactly the same set of finite layer equations arises as is shown in Equation 2.21 for the two-dimensional case with the stiffness matrix set up using ρ α β= +( ) /

n m2 2 1 2 in place of αn:

K T( )ρ δmn mn= (2.25)

This means that we may solve the three-dimensional problem by use of a simple one-dimensional discretisation. It is necessary to set up the stiffness matrix for each value of αn,

y

z

x

Figure 2.11 Three-dimensional loading.

Periodically spaced loadings

2B = M

2A = L

z

x

y

Figure 2.12 Periodic three-dimensional loadings.


βm and to solve for all the Fourier coefficients. Again, it is not necessary to carry out the full summation, but to use only a finite number of terms in the series. If there are N terms in the αn summation and M terms in the βm summation, we would have to solve the simple set of equations (M × N) times to obtain all of the Fourier coefficients. Once these have been found, summations of the type shown in Equation 2.24 may be used to obtain the actual field values at any point x, y.

2.6 CONSOLIDATION PROBLEMS

The same finite layer techniques may be used for problems involving consolidation of horizontally layered soil deposits. For example, if we take the strip-footing problem again and represent the excess pore pressures p that are induced into the soil as a Fourier series

p P xn

nn=

=

∞

∑0

( ) cos( )α

(2.26)

we obtain a set of equations

K L

L q

h( ) ( )

( ) ( )

( )

( )

( )

(

( )n T n

n n

n

n

n

t

t

t

−− −

=

θδ

∆ Φ∆∆

∆∆ Φ nn n t) ( )( )q

(2.27)

where q(n) is the vector of nodal pore pressure coefficients for Fourier term n, δ(n) is the vector of displacement coefficients for Fourier term n, K(n) is the stiffness matrix (for term n), L(n) is a coupling matrix, Φ(n) is the flow matrix, and h(n) is the force matrix that is completely analogous to the force matrix in Equation 2.21.

All of the matrices in Equation 2.27 are fully explained in the paper by Small and Booker (1979).

To obtain a complete solution for pore pressures and displacements at any time, a ‘march-ing type’ process is used. A time step Δt is taken and the changes Δδq(n), Δq(n) are found. These rely on the solution for the excess pore pressure coefficients at the start of the time step q(n)(t) (as may be seen in Equation 2.27 as these form part of the right-hand side of the equations.

Solutions at a later time may then be calculated from

δ δ δ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

n n n

n n n

t t t

t t t

+ = +

+ = +

∆ ∆

∆ ∆q q q

(2.28)

The whole solution may be ‘marched’ forward in this fashion for each Fourier component of the load vector Δh(n). To obtain the solution at any time t, we must carry out a summation using the appropriate Fourier coefficients (that are now functions of time).

Equation 2.26 may more correctly be written

p t P t x

n

nn( ) cos( )( )=

=

∞

∑0

α

(2.29)


2.7 FOURIER TRANSFORMS

If we apply Fourier transforms to the field variables (i.e. the displacements, pore pressures, and stresses) instead of using Fourier series to represent them, we may again reduce two- and three-dimensional problems to ones involving only a single spatial dimension.

The Fourier integral in effect replaces the Fourier summation and we are left with a very similar set of equations to solve.

If we apply a Fourier transform to the loading function shown in Figure 2.5, we would have

Q q x x dx

q x x dx

qa

a

a

( ) ( )cos( )

( )cos( )

sin( )

απ

α

πα

απ

=

=

=

−∞

+∞

−

+

∫

∫

12

12

aa

(2.30)

In Equation 2.30, the first infinite integral is the Fourier transform of the step function q(x), and as the strip function is an even function of x, a cosine function is used in the trans-form. The transforms may also be applied to the displacements and stresses, for example, if we take a strip-loading problem,

U u x x dx

W u x x dx

x

z

( ) sin( )

( ) cos( )

( )

( )

απ

α

απ

α

=

=

−∞

+∞

−∞

+∞

∫

∫

12

12

(2.31)

For an elastic problem, we would finally have a set of equations

K f( ) ( )α αδ α= ( ) (2.32)

The force vector f(α)would contain the transform of the applied load, for example,

f 0

q aa

T

( ) ,sin( )

0α απ

= 0 0

, , , ,

(2.33)

and the vector δ(α) would contain the transforms of the displacements,

δ = ( , , , , )U W U W T1 1 2 2

As we can write the inverse transforms as say

u U x dx =

−∞

+∞

∫ ( )sin( )α α α

(2.34)


u W x dz =−∞

+∞

∫ ( )cos( )α α α

(2.35)

once the transformed quantities are found from Equation 2.32, we can carry out the inte-gration to obtain the final results.

This is usually difficult to do, and so numerical integration is used. Gaussian integration of a function of α may be carried out as follows:

−∞

∞

=∫ ∑=f d f

i

G

i i( ) ( )α α ω α1

(2.36)

where the αi are the values of α at the selected Gauss points, and the ωi are the Gaussian weights. G is the number of Gauss points. These weights and Gaussian coordinates may be found from tabulated values in maths books.

Hence, to numerically invert a function of α, we need to know its value at the Gauss points αi. This means that we need to solve Equation 2.32 at these selected values, that is,

K f( ) ( ) ( )α δ α αi i i= (2.37)

Now we may use numerical integration to find all the required values, for example,

u U xx

i

G

i i i==

∑1

ω α α( )sin( )

(2.38)

The Gaussian integration may be carried out in several blocks as shown in Figure 2.13.

−∞

+∞

−

+

∫ ∫≈f d f dA

A

( ) ( )α α α α

(2.39)

We do not need to integrate all the way from −∞ to +∞, but we can truncate the integral at some large value A. As long as A is large enough, a good approximation to the integral may be obtained.

Block 1

f (α) f (α)

α

Block 2 Block 3

Figure 2.13 Numerical integration scheme.


2.8 EXAMPLES

As an example of finite layer analysis, the program FLEA (Finite Layer Elastic Analysis – Small and Booker 2014) is used in the following. A simple problem involving a uniform circular load-ing q = 100 kPa of radius a = 1.0 m applied to a finite layer of uniform soil of thickness h = 2.0 m is analysed. The elastic modulus is taken as E = 10,000 kPa and Poisson’s ratio as ν = 0.3.

The vertical displacements along the axis of the footing are shown in Figure 2.14, where it can be seen that the settlement at the surface is about 12 mm.

A more complex problem is one of a square footing 2 m × 2 m in plan. The soil consists of two layers of soil: the upper layer being 2 m thick and the lower layer 4 m thick. The soil is anisotropic, and the layers have the properties shown in Table 2.1.

The square footing has unit loads of 1 kPa applied laterally in the x-direction and 1 kPa applied vertically on it, and the results are calculated throughout the depth of the layer beneath the point x = 0.5 m, y = 0.5 m.

The stress in the x-direction σxx is shown in Figure 2.15, where it may be seen that the stress has a discontinuity in it between the layers. A plot of the vertical displacement (Figure 2.16) shows that the displacement is continuous although there is a kink at the interface of the two layers (z = 2.0 m).

Displacement (mm)

Dep

th (m

)

4 8 1202.0

1.6

1.2

0.8

0.4

0

Figure 2.14 Vertical displacement beneath the centre of circular loading.

Table 2.1 Properties of layered soil

LayerThickness

(m)Modulus Ex (kPa)

Modulus Ez (kPa)

Shear modulus Gz (kPa)

Poisson’s ratio νxx

Poisson’s ratio νzx

1 2 1500 1000 450 0.25 0.22 4 12,000 4000 2000 0.1 0.3


Stress σxx (kPa)0 0.4 0.8 1.2

6

5

4

3

2

1

0

Dep

th (m

)

Figure 2.15 Stress in x-direction σxx beneath the square surface loading at x = 0.5 m, y = 0.5 m.

0

1

2

3

Dep

th (m

)

4

5

6

Displacement (mm)1.20.80.40

Figure 2.16 Vertical displacement beneath the square surface loading at x = 0.5 m, y = 0.5 m.

31

Chapter 3

Finite element methods

3.1 INTRODUCTION

Because soils are in general complex materials that consist of the soil skeleton, the pore air, and pore water and exhibit non-linear stress–strain behaviour and perhaps time-dependent behaviour, it is not always simple to obtain analytic solutions to problems. Analytic solu-tions are those that contain closed form solutions that may be a simple algebraic formula that can be evaluated. Some solutions are semi-analytic, for example, the solution may be written as an integral that is not able to be inverted through analytic means and has to be evaluated using numerical integration. Finally, there are the numerical solutions that rely on numerical approximation rather than algebraic manipulation for their solution.

Although there are many very useful analytic solutions available for use in geomechan-ics, more complex problems often arise that need the power of numerical means to obtain a solution. One of these numerical methods, the finite element method, is presented in this chapter.

3.2 TYPES OF ELEMENTS

The basis of the finite element method is that the volume of interest is divided up into dis-crete elements. The elements may be of different shapes and some of the shapes are shown in Figure 3.1. As shown in this figure, the elements have ‘nodes’ (shown as black dots), and different element types may have different numbers of nodes. For example, in Figure 3.1, the triangular shaped element is shown with three or fifteen nodes.

It is assumed that the field quantities are known at the nodes, but can be found within the element by mathematical interpolation functions. In geomechanics, the field quantities may be displacements or water pressure head for instance.

The mathematical functions that are used within the element are called ‘shape functions’ as they depict the shape of the field quantity within the element if plotted. The number of shape functions used is equal to the number of nodes in the element as the shape function can be thought of as giving the shape of the field quantity within the element for a unit value of the field quantity at one of the nodes.

For instance if we give one node a unit displacement, the shape function gives the dis-placement within the element. Figures 3.2 and 3.3 show how a six-node triangle will deform for a displacement at node 1 or at node 2, respectively. If all of the nodes displaced a unit amount, then we superimpose the displacements given by the shape functions for each node. If the displacement is not unity, we can scale the displacement by multiplying the shape functions by the actual displacements at each of the nodes and then adding the result.


3.2.1 Finding shape functions

Suppose, as an example, we have an element with six nodes and the values of some quantity of interest (w), such as displacement, head, and temperature, are known at each of the nodes. It is assumed that within the element the variation of w at position (x, y) can be approxi-mated by a polynomial expression:

w x y a a x a y a x a xy a y( , ) = + + + + +1 2 3 42

5 62

(3.1)

3-node triangle 8-node isoparametric

20-node isoparametric 3D element

15-node triangle

Figure 3.1 Different types of finite elements.

1

2

3

4

5

6

Figure 3.2 Shape function for the first node of a six-node triangle.

Finite element methods 33

where ak are polynomial coefficients. Because there are six nodes, we have six unknown coefficients. The functions of x and y are chosen to give a quadratic variation of the function in both the x and y directions. If the element were three-dimensional, functions of x, y, and z would be needed. If the element has m nodes, we would need m terms in the polynomial.

We can therefore write

w x y x y x yT T( , ) ( , ) ,( )= = ⋅a f f a (3.2)

where a = a1, a2, …, ak, …, a6T and f(x, y) = 1, x, y, x2, xy, y2T.Suppose that the element nodes are located at the points (x1, y1), (x2, y2), …, (x6, y6). At

the ‘kth’ node, the value of the quantity w is

w a a x a y a x a x y a yk k k k k k k= + + + + +1 2 3 42

5 62

(3.3)

Equation 3.3 holds at each of the six nodes. These equations may be written in matrix form as follows:

w Ca= (3.4)

where

w C= =( , , , )w w w

x y x x y y

x y x x y y

x y xT1 2 6

1 1 12

1 1 12

2 2 22

2 2 22

3 3 32

1

1

1… and

xx y y

x y x x y y

x y x x y y

x y x x y y

3 3 32

4 4 42

4 4 42

5 5 52

5 5 52

6 6 62

6 6 62

1

1

1

(3.5)

The solution of Equation 3.4 is

a C w= −1 (3.6)

1

2

3

4

5

6

Figure 3.3 Shape function for the second node of a six-node triangle.


When a from Equation 3.6 is substituted into Equation 3.2, it is found that the quantity of interest can be expressed in the form of

w x y x yT( , ) ,( )= ⋅ = ⋅−f C w N w1

(3.7)

The vector N must contain the shape functions as it relates the values at the nodes of the element w to the value at any point within the element w(x, y).

Hence,

N f C= = ⋅ −( , , , ) ,( )N N N x yT T1 2 6

1… (3.8)

and we can find the shape functions for any element in this manner simply from a knowledge of the coordinates of the nodes of the element (as they are used to establish the C matrix). The inverse of the C matrix can be found numerically or if it is possible to find the algebraic values of the inverse, the shape functions can be written as algebraic expressions. Some shape func-tions for different types of elements are given in Appendix 3A, but others may be found in spe-cialized books on finite elements such as Zienkiewicz (1977) or Potts and Zdravkovic (1999).

3.2.2 Isoparametric elements

An isoparametric element is one where the geometry of the element is also determined by a shape function. The term ‘isoparametric’ comes from the fact that for this particular type of element, the shape functions used for determining the shape of the element are the same as those used to interpolate the field quantities within the element. They do not have to be the same, and could be different, but using the same functions simplifies the analysis.

An eight-node isoparametric element with curved sides is shown in Figure 3.4. By allow-ing the element to have curved sides, shapes such as circular tunnels and curved geometries can be more easily discretised.

The shape functions are now written in terms of local parameters ξ, η because the coor-dinates x, y cannot be used. The local parameters range between −1 and +1, for example, at node 1, the local coordinates are ξ = −1 and η = −1. The shape functions are given in Appendix 3A but, for example, the shape function for the first node is given by

N1

14

1 1 1= − − − − −( )( )( )ξ η ξ η

(3.9)

3

4

5

(a) (b)

67

21

8

1

2

3

87

6

5

4

ξ

η

Figure 3.4 (a) Local coordinates. (b) Curved shape of an eight-node isoparametric element.


The coordinates within the element are given by

x N x N x N x

y N y N y N y

= + ++

= + + +

1 1 2 2 8 8

1 1 2 2 8 8

(3.10)

The problem now is to differentiate the shape functions with respect to x, y because the actual coordinates are also functions of the local coordinates. Often the shape functions need to be differentiated to obtain the strains from the displacements or the gradient of the total head for example. This is done by using the chain rule for differentiation

∂∂

= ∂∂

∂∂

+ ∂∂

∂∂

∂∂

= ∂∂

∂∂

+ ∂∂

∂∂

N Nx

x Ny

y

N Nx

x Ny

y

1 1 1

1 1 1

ξ ξ ξ

η η η

(3.11)

or in matrix form

∂∂

∂∂

=

∂∂

∂∂

∂∂

∂∂

∂N

N

x y

x y

N1

1

1

ξ

η

ξ ξ

η η

∂∂

∂∂

=

∂∂

∂∂

x

Ny

Nx

Ny

1

1

1

[ ]J

(3.12)

The matrix in the above equation [ J] (Equation 3.12) is called the Jacobian, and by inverting the Jacobian we can obtain the differentials of the shape functions with respect to the actual coordinates. Hence,

∂∂

∂∂

=

∂∂

∂∂

−

Nx

Ny

N

N

1

1

1

1

1

[ ]Jξ

η

(3.13)

This can be done for any isoparametric element; if it is three-dimensional then the Jacobian is a 3 × 3 matrix as there are three axis directions.

3.2.3 Infinite elements

As can be seen with an isoparametric element, the shape of the element can be deter-mined by using shape functions. If we choose the shape functions (now called mapping functions) so that the coordinates at one side of the element become infinite, we can cre-ate an infinite element as shown in Figure 3.5. Infinite elements are useful when the side boundaries of the problem at hand need to be a long way from the region of interest so that they do not influence the solution. They can also reduce the size of the mesh required and therefore save a good deal of computation time, especially for three-dimensional problems.


For example, a five-node infinite element that can be used with the eight-node isoparamet-ric element shown in Figure 3.4 can have mapping functions that decay with ξ−1

x M x M x M x

M

M

M

= + + +

= −−

= + −−

=

1 1 2 2 5 5

1

2

3

11

0 5 1 11

( )( )

. ( )( )( )

ξη η

ξξ η

ξ00 5 1 1

1

11

2 11

4

5

2

. ( )( )( )

( )( )

( )( )

+ +−

= − +−

= − −−

ξ ηξ

ξη ηξ

ξ ηξ

M

M (3.14)

It may be seen from the mapping functions that the coordinate in the direction of ξ goes to infinity as ξ approaches +1. The shape functions relating displacement (say) to nodal dis-placement are the same as for the eight-node element. To differentiate a shape function with respect to x and y, the same procedure can be used as for ordinary isoparametric elements, but with a Jacobian formed by differentiating the mapping functions. It may be noted that if the infinite element is used at the side of a finite element mesh to extend the mesh laterally to infinity, then the y coordinate does not go to infinity, only the x coordinate. In this case, the normal shape function is used for the y coordinate mapping.

For mapping functions that decay as ξ−2 and for three-dimensional infinite elements, see Appendix 3A.

3.2.4 Finite element meshes

Individual elements are joined together to form meshes covering the field of interest for the problem. A mesh of elements is shown in Figure 3.6 for a three-dimensional problem involv-ing a piled raft on a layered soil (1/4 of the mesh is used because of symmetry). The mesh of Figure 3.6 is made up of different types of elements: (i) solid 20-node elements for the soil, (ii) 8-node two-dimensional elements for the raft, and (iii) infinite 12-node elements to extend the soil boundaries to large distances. The elements can be used to model different behaviours. For example, a raft element will behave differently to a solid soil element. To

ξ

ξ = +1 at ∞

12

3

4

5

η

Figure 3.5 A five-node infinite element.


model the behaviour of different materials, we need to formulate the problem and incorpo-rate laws that best model the behaviour of a material.

Formulation of finite element equations for different problems is discussed in the following sections.

3.3 STEADY STATE SEEPAGE

Finite element analysis of steady state seepage involves the analysis of flow that is continuous and for which the total head or permeabilities within the field of flow do not change. If the location of the water table changes (e.g. when the stored water in a dam drops) or the soil is being wet by an advancing front, then the seepage is not steady state. Here, steady state seepage will be examined first, and then non-steady flow problems will be addressed.

3.3.1 Governing equation

The governing equation for steady state flow is established by assuming that for any small volume in the soil, the volume of water that flows out must be equal to the volume that

Material types

Piled raft (5 × 5 pile group) – Solid elements for piles

Infinite elements

Materials12345

xy

z

Figure 3.6 Example of a finite element mesh made up from different types of elements.


flows in. This implies that water is not being pumped into or out of the soil at the point, and that there is continuity of flow.

This leads to the equation

∂∂

∂∂

+ ∂∂

∂∂

+ ∂∂

∂∂

=x

khx y

khy x

khz

xx yy zz 0

(3.15)

where kxx, kyy, and kzz are the permeabilities of the soil in the three axis directions x, y, and z. In Equation 3.15, h is the total water pressure head which is made up of the elevation head hE and the water pressure head hw, that is, h = hE + hw. It is necessary to work in terms of the total head as flow depends on the elevation of a point in the soil as well as the water pressure head that exists there.

3.3.2 Finite element formulation

If we approximate the total head h over the volume of interest, then the governing equation will not be zero everywhere but will contain some error, and this is called the ‘residual’.

To get the best approximation to the real function, we try to make the residual as small as possible everywhere within the region. This can be done by integrating the residual times a small change in the approximating function over the region occupying the volume V and setting the result to zero.

δhx

khx y

khy x

khz

xx yy zz∂

∂∂∂

+ ∂∂

∂∂

+ ∂∂

∂∂

=∫ dV

V

0

(3.16)

This leads to the equation

δi k iT

V

V[ ] d =∫ 0

(3.17)

where i = (∂h/∂x, ∂h/∂y, ∂h/∂z)T = ∇h is the hydraulic gradient of the pore fluid and

[ ]k =

k

k

k

xx

yy

zz

0 0

0 0

0 0

is the matrix of permeabilities for the soil.

3.3.3 Approximation of total head h

We can use the usual finite element interpolation functions to approximate the total head h within an element. Suppose the element has n nodes and Ni are the shape functions.

h N h N h N h N h N hn n= + + + ++1 1 2 2 3 3 4 4 (3.18)


Then the value of the hydraulic gradient for the element ie would be

ie

n

n

n

Nx

Nx

Nx

Ny

Ny

Ny

Nz

Nz

Nz

=

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

1 2

1 2

1 2

h

h

hn

1

2

(3.19)

or in a more concise form

i Ehe = (3.20)

3.3.4 Finite element equations

Equation 3.17 can now be written for a single element e

δh E k EheT T

e e

V

dV

e

[ ] =∫ 0

(3.21)

As δhe is an arbitrary variation in head and not necessarily zero, then we must have

E k EhTe e

Velem

dV

e

[ ] =∫∑ 0

(3.22)

or

Φh = 0 (3.23)

where

Φ = ∫∑ E k ETe

Velem

dV

e

[ ]

(3.24)

The matrix is called the global flow matrix for the problem and is assembled for all of the elements in the finite element mesh. The vector h contains the total head at each node in the finite element mesh. By solving the set of Equations 3.23 subjected to some boundary condi-tions, the total head at each node can be found. Assembly of the element stiffness matrices and the application of boundary conditions is examined in Appendix 3B. Because in flow problems, there is only one variable per node (i.e. the total head), the set of equations to be solved is generally not large and the bandwidth of the equations is small.

3.3.5 Calculation of flows

Returning to Equation 3.17 and introducing Darcy’s law (i.e. v = [k]i), we can write


δ δ

δ

δ

δ

i k i h h

h v

h v

h

T

V

T

T

V

T T

V

T

S

dV

dV

dV

v ndS

[ ]

( )

∫∫∫∫

=

= ∇

=

= ⋅

Φ

∇

by the divvergence theorem

(3.25)

Hence, we have

Φh = ⋅∫ v ndSS

(3.26)

and so by multiplying the flow matrix by the solution for the heads, we get the flows at the nodes of the finite element mesh. This is done for each element in the mesh and the flows at the nodes are summed.

Internal nodes will have no flow, but the boundary nodes will exhibit flows if they are on a permeable boundary. There should be a flow balance in the finite element mesh with flow in = flow out as shown in Figure 3.7.

3.3.6 Flow lines

In time dt, the distance travelled by a particle (with flow velocity components vx and vy as shown in Figure 3.8) of fluid is

In the x-direction = vx ⋅ dtIn the y-direction = vy ⋅ dt

Therefore,

dxdy

v dtv dt

vv

v dx v dy

x

y

x

y

y x

= =

⋅ − ⋅ = 0

(3.27)

Flow in Flow out

Figure 3.7 Confined seepage showing the flow balance.


If we define a function Ψ(x, y) that is constant along the flow line we would have on differentiating

∂∂

⋅ + ∂∂

=⋅Ψ Ψx

dxy

dy 0

(3.28)

Equations 3.27 and 3.28 suggest that

vy

vx

x

y

= − ∂∂

= ∂∂

Ψ

Ψ

(3.29)

3.3.7 Calculation of flow using the stream function

From Figure 3.8, it can be seen that the flow across the face dQ is given by

dQ v dy v dx

ydy

xdx

x y= ⋅ − ⋅

= − ∂∂

⋅ − ∂∂

⋅Ψ Ψ

(3.30)

and so the total flow across a boundary between points A and B can be found by integration.

Q dQ A B

A

B

= = −∫ Ψ Ψ

(3.31)

This means that the values of the stream function can be found relative to a point on the boundary (say A) by summing the flows at the nodes around the edge of the finite element mesh.

3.3.8 Determining the stream function

From Equation 3.29, we can show that

v khx y

v khy x

x x

y y

= − ∂∂

= − ∂∂

= − ∂∂

= ∂∂

Ψ

Ψ

(3.32)

vx

dy

vy

dx

dQ

v

Figure 3.8 Section of a flow line.


and so the hydraulic gradients are

ihx k y

ihy k x

xx

yy

= ∂∂

= ∂∂

= ∂∂

= − ∂∂

1

1

Ψ

Ψ

(3.33)

By substituting back into the original equation (Equation 3.17), used to formulate the problem, we can obtain a set of equations for the stream function. These equations are the same as the finite element equations for the flow, but with

kk

kk

h

xy

yx

→

→

→

1

1

Ψ

(3.34)

Hence, we can solve the same set of finite element equations as for the total head (Equation 3.23), but using the reciprocal of the permeabilities. The solution will be the stream func-tion, and these values can be contoured to plot the flow lines (e.g. see Figure 3.7).

To solve the finite element equations, we need some boundary conditions, and these are the values of the stream function around the edges of the finite element mesh. As mentioned before (Equation 3.31), the value of the stream function at the nodes around the outside of the mesh can be computed by summing the flows, and these flows are computed in the first part of the analysis by solving for the total heads and applying Equation 3.26.

3.3.9 Pumping or extracting fluid

If fluid is being pumped from or injected into the ground, Equation 3.15 does not hold, and the right-hand side of the finite element equations will not be zero. Equation 3.23 will become Φh = f where f = (0, 0,…,qn,…,0)T and qn is the quantity of fluid being injected or extracted (depending on sign) at node n.

3.4 STRESS ANALYSIS

The finite element equations for problems involving a single-phase elastic material can be established by using the principle of virtual work. This states that if a body is in equilibrium under a set of applied forces, then for any virtual displacements applied to the body the energy stored in the body will be equal to the work done by the external forces and any body forces (e.g. self-weight of the material).

For soil and rock mechanics, the body considered is the soil mass and applied forces would be due to loads from structures or embankments for example.

The internal energy WI stored per unit volume due to a virtual displacement to the body is given by

dW dI = σ ε (3.35)


where σ is the stress at a point in the body and dε is the increment in strain caused by the virtual displacements. For general three-dimensional problems, the stress vector consists of the six components of stress at a point, namely the three normal stresses and three shear stresses. The strain vector also contains six strain components. The energy stored in the whole body involves the integral of the work per unit volume over the volume of the body, that is,

W d dVI

V

= ∫σ ε

(3.36)

The work done by the surface tractions (such as applied uniform or point loads) is given by

W d dS d dV dF

VS

= + +∫∫ q u u P uγ

(3.37)

where q is a traction (or load) applied over an area of the surface dS and so must be inte-grated over the area where the loading (which may not be uniform) is applied. P is a point or concentrated load applied at a point on or in the body. The increment in displacement at points in the body is du. For general three-dimensional loading, the displacement vector has three components corresponding to the three displacements in the three axis directions, as does the point load vector P. The loading traction q also has three components as one normal and two shear stresses can act on a surface. The body force vector γ has three com-ponents, as body forces can act in all three-axis directions (these can be due to gravity or centrifugal forces).

Equating internal and external work, we can write

σ εd dV d dS d dV dVSV

= + +∫∫∫ q u u P uγ

(3.38)

For linear elastic materials, the stress–strain relationship can be found simply from Hooke’s law which may be written as

σ ε= DE (3.39)

where DE is a matrix of elastic constants. For an isotropic elastic material and general three-dimensional conditions, the DE matrix is a 6 × 6 matrix as shown in Equation 3.40.

DE

G

G

=

++

λ λλ λ

λλ

λ λ

2

2

0

0

0 0

0 0

0 0

2 0

0

0 0

0 0

0 0

0 0

0 0

0 0

0

0

λ + G

G

G

GG

(3.40)


where E is the elastic modulus of the material and ν is its Poisson’s ratio. G is the shear modulus relating shear stress to shear strain and λ is Lamé’s parameter defined as

λ νν ν

νσ σ σ τ τ τ

ε

=− −

=+

=

=

E

GE

xx yy zz xy yz zxT

xx

( )( )

( )

( , , , , , )

(

1 1 2

2 1

σ

ε ,, , , , , )ε ε γ γ γyy zz xy yz zxT

(3.41)

In the above equations, σ are the normal stresses to a plane and τ are the shear stresses. Likewise, ε are the normal strains and γ are the shear strains.

We can now make the finite element approximations of the field quantities (displacements u = (ux, uy, uz)T) through use of the element shape functions N.

u N= δ (3.42)

where the displacements at the nodes of the element are contained in the vector δ.By differentiating the displacements to obtain the strains (see Section 2.2), the strain–dis-

placement relationship can be obtained.

ε δ= B (3.43)

By substituting the finite element approximations for the strains and the stresses, we have

d dV d dS d dV dT TE e

T T T Te

T T

VelemSVelem ee

δ δ δ δ γ δB D B N q N N P= + +∫∑∫∫∑

(3.44)

Since the virtual displacements are arbitrary and not necessarily zero, we must have

B D B N q N N PTE

T T T

VSV

dV dS dVδ γ= + +∫∑∫∫∑

(3.45)

or in more compact form

K f f fδ = + +q Pγ (3.46)

whereK B D B= ∫Σ V

TE ee dV is the stiffness matrix for the body

fq = ∫SNTqdS is the force vector for uniform loadsf Nγ = ∫Σ V

Tee dVγ is the force vector for body forces

fP = P is the force vector for point loads

The point load force vector reduces to simply the values of the point forces as the shape functions are equal to unity at the nodes of an element, and point loads can only be applied at the nodes.


3.5 CONSOLIDATION ANALYSIS

Consolidation analysis is a combination of stress analysis and flow analysis, as the flow of water out of the pores of the soil allows the soil to undergo a volume change. Excess pore pressures caused in the soil due to external loads dissipate during consolidation and the applied stress is transferred from the pore water pressures to the soil skeleton.

3.5.1 Effective stress analysis

We can set up the finite element equations for the stress analysis part of the equations by again using Equation 3.36. This time however, use can be made of the effective stress prin-cipal that states that the total stresses σ are made up of the effective stresses σ′ (the stress in the soil skeleton) and the pore water pressures p. The pore pressures only add onto the direct stresses not the shear stresses and so we can write

σ σ= ′ + pj (3.47)

where j = (1, 1, 1, 0, 0, 0)T.Substituting the total stress into Equation 3.38 gives

( )′ + = + +∫∫∫ σ ε γp d dS d dV dVSV

j d dV q u u P u

(3.48)

If the finite element approximations of the field quantities are now made, we obtain

d dV d pdV d dS d dV dT TE e

T Te

T T T Te

T T

VSV ee

δ δ δ δ δ γ δB D B B j N q N N P+ = + +∫∑∫∫∑∑∫∑Ve

(3.49)

where BTj = d a vector that is the sum of the first three rows of the B matrix. It therefore contains the interpolation functions for the volume strain, as the volume strain is the sum of the three linear strains, that is, θ = εxx + εyy + εzz. As before, ∑ denotes a summation over all of the elements in the mesh, and Ve is the volume of an element.

The pore water pressures can be interpolated within the element by using the shape fac-tors (these are the same shape factors as used for the displacements) that will be contained in a vector a to distinguish these factors from the displacement shape factors. Some inves-tigators like to use different interpolation functions for the pore water pressures and the displacements, but in the author’s experience this is not necessary. We can therefore write

p =

=

a q

d

T

Tθ δ (3.50)

If as before we note that the variations in the displacements are arbitrary and not neces-sarily zero, then we must have

K L q f f fTδ + = + +q Pγ

(3.51)


where the force vectors have the same definitions as in Equation 3.46, and the matrix L has the definition

L daT Te

V

dV

e

= ∫∑

(3.52)

In incremental form, the above equation may be written as

K L q f f fT∆δ ∆ ∆ ∆ ∆+ = + +q Pγ

(3.53)

The pore water pressure may be written in terms of the pressure head as the total head h is equal to the sum of the elevation head hE and the water pressure head hw, that is,

h h hE w= + (3.54)

The water pressure p may therefore be substituted by

p h h hw w w E= = −γ γ ( ) (3.55)

The same vector of interpolation functions can be used for the total head and the water pressure and so

p Tw

TE= = −a q a h hγ ( ) (3.56)

Any changes in the pore pressure can now be calculated, and since changes in the eleva-tion head are zero (if we are dealing with small strain analysis), we have

∆ γ ∆q h= w (3.57)

This enables us to write the incremental equations in terms of total head as

K L h f f f∆δ ∆ ∆ ∆ ∆+ = + +γ γw

Tq P

(3.58)

For consolidation problems, we need to take care of the signs of the terms if we want compression to be positive, so that compressive effective stresses and compressive pore water pressures are positive.

If a compressive load is downward as shown in Figure 3.9, then it is opposite to the posi-tive y-axis and is therefore negative. A downward movement will therefore need to yield a positive stress and so we must write

σ = −Dε (3.59)

Hence, Equation 3.58 will need to have a change in the signs of the force vector and the stiffness matrix giving finally

K L h f f f∆δ ∆ ∆ ∆ ∆− = + +γ γw

Tq P

(3.60)


3.5.2 Volume balance

If it is assumed that the soil is saturated, then any change in volume of the pores of the soil due to water flowing out of the pores is compensated for by a change in the volume of the soil that is equal to the volume of water lost from the pores. This is making the assumption that the pore water is not compressible and that there is no air in the pores of the soil.

In examining the flow of water through the soil, we saw previously that the quantity of water flowing from an element of soil (Equation 3.15) was balanced by the water flowing into the element for steady state seepage. The amount of water flowing out of the element of soil is not zero when consolidation is occurring, and is equal to the rate of the volume strain. We can therefore write

[ ]k ∇2h

t= ∂

∂θ

(3.61)

where, as before, [k] is the matrix of permeabilities (see Equation 3.17), ∇ is the gradient vector ∇ = (∂/∂x, ∂/∂y, ∂/∂z)T, and h is the total head. The operator

∇ = ∂

∂+ ∂

∂+ ∂

∂

= ∇ ∇22

2

2

2

2

2x y zT ( )

(3.62)

As before (Equation 3.16), we can integrate the flows over the volume to minimise the error in the flows:

δ δ θh h dV h

tdVT

VV

[ ] ( )k ∇ ∇ − ∂∂

=∫∫ 0

(3.63)

δ δ θi k iT

VV

dV ht

dV[ ] − ∂∂

=∫∫ 0

(3.64)

If we now substitute the finite element approximations for the volume strain and the gra-dient of the total head, we have

δ δh E k Eh h a dT Te

T Te

VV

dVt

dV

ee

[ ] − ∂∂

=∫∑∫∑ δ0

(3.65)

x

yq

Figure 3.9 Sign convention for compression positive.


We can write this equation in more compact form (noting that the increment in total head δh is not necessarily zero and so the rest of the equation must be zero) as

Φ δ

h L− ∂∂

=t

0

(3.66)

and the definitions of the matrices are

Φ =

=

∫∑

∫∑

E k E

L a d

Te

V

Te

V

dV

dV

e

e

[ ]

(3.67)

The flow matrix Φ is the same as before for steady state seepage, however, we now have an extra term in the finite element equations because the flow of water at a point in the soil is not zero as it was for steady state seepage. Once again, ∑ indicates a summation for all the elements in the mesh.

Equation 3.66 can now be integrated numerically by noting that if we integrate a function f(t) with respect to time t we are finding the area under the curve of f(t) versus t. The integral can be approximated by the following formula

f t t f t t f t tt

t t

( ) ( ) ( ) ( )= ⋅ + − ⋅ ++

∫ α α∆ ∆ ∆∆

1

(3.68)

This is illustrated in Figure 3.10, where it can be seen that this is an approximation to the area under the curve (the areas of the two shaded rectangles), and α is a quantity that can lie between 0 and 1.

If we apply this integration scheme to Equation 3.66, then we have

α∆ Φ ∆ Φ ∆δα ∆ Φ∆ ∆ Φ ∆δ

∆t h h L

h h Lt t t

t

t

t t

+ − − =− − − − =

+( )

( )

1 0

1 0

α

(3.69)

f (t)

f (t)f (t + Δt)

t t + ΔtαΔt (1 – α) Δt

Figure 3.10 Numerical integration scheme.


and so combining the volume balance equation (Equation 3.69) and the effective stress equation (Equation 3.60), we obtain the final set of equations involving consolidation in an incremental form.

K L

L h

f f f

h−

− − −

=

+ +

γγ α ∆ γ Φ

∆δ∆

∆ ∆ ∆∆ γ Φ

wT

w w

q P

w tt( )1 tγ

(3.70)

The solution can therefore be ‘marched’ forward, with the solution started off by a knowl-edge of the total head at time zero. The increments in the displacements and heads are found, and then the new displacements and heads at time t + Δt are found, that is,

δ δ ∆δ∆

∆

∆

t t t

t t t

+

+

= += +h h h

(3.71)

The new values of total head are used in the right-hand side of the equations to allow a new solution to be found and so on. Errors can accumulate with solutions of this kind, and so it is of interest to know under what circumstances the ‘marching’ scheme is stable. This has been examined by Booker and Small (1975) where it was found that the scheme is unconditionally stable if α ≤ 0.5. The scheme can be made stable with values of α greater than 0.5, but this depends on the eigenvalues of the equations, and so this is not practical.

An example of a solution for consolidation of a soil layer under a uniform circular load-ing q is shown in Figure 3.11 where the vertical permeability kv is different to the horizontal permeability kh. The properties used to obtain the solutions are shown in Table 3.1.

By comparing the solution of the numerical equations of consolidation with an analytic solution from the program CONTAL (Small 2012), it may be seen that the closest agreement

1

kv= 10.kh

kv= 0.1.kh

α=0.0α=0.5

α=0.5α=0.0

F.E.solutions

–0.080

–0.075

–0.070

–0.065

–0.060

–0.055

–0.050

–0.0450.001 0.01 0.1 1 10 100 1000 10,000 100,000

Settl

emen

t (m

)

Time (days)

CONTAL

kv = 10.kh

kv = 0.1.kh

α = 0.0α = 0.0α = 0.5

α = 0.5α = 0.0

F.E. solutions

Figure 3.11 Effect of the value of α on the finite element ‘marching’ solution. Comparison with an analytic CONTAL solution.


between the solutions is obtained when α = 0, and so this is recommended for numerical analysis.

The accuracy of the numerical solution also depends on the size of the time step Δt chosen as for many other forms of time-dependent numerical solutions. Consolidation problems tend to need small time steps at the beginning of consolidation when pore pressure gradients are high and consolidation is rapid. As time goes on, the consolidation process slows down as pore pressures tend to even out. Therefore, it is necessary to increase the size of the time step to allow a solution to be obtained with a reasonable amount of computational effort.

As can be seen from Equation 3.70, if the time step is changed, the set of equations has to be set up again and the consolidation matrix has to be refactored (for a Crout–Cholesky factorisation) or a new Gaussian elimination performed if these kinds of solution methods are being used. Once this is done a simple back substitution is performed at each time step until the time step is changed.

3.6 NUMERICAL INTEGRATION

In order to obtain the matrices such as the ones in Equation 3.70, there is a volume integral to perform. For two-dimensional elements, this reduces to an area integral. In some cases, this integration can be carried out algebraically, but in most cases, this is complicated and the integration can be carried out numerically.

As the shape functions used are generally powers of the coordinates x and y, the mul-tiples are powers of the coordinates as well. Integration of such functions can be performed exactly using Gaussian numerical integration, and so this is a popular means of obtaining the integrals when forming the finite element matrices.

Gaussian integration is based on fitting polynomials to the curve of the function being integrated and finding the area under that curve. If the function being integrated is the same order of polynomial, the area will be exact, but if it is a different polynomial, say one of lesser order, then the integration is approximate. Lower order Gaussian integration is some-times used to obtain a better performance of elements, for instance eight-node isoparametric elements can sometimes ‘lock’ when used for plasticity problems.

The area A under a function f(x) (or its integral) is given by

A f gi i

i

n

==

∑ω ( )1

(3.72)

where ωi are the Gaussian weights, gi are the Gaussian coordinates, and n is the number of Gauss points. For higher order schemes, there are more Gauss points used. These values

Table 3.1 Properties used in finite element analysis

Quantity Value

Drained modulus of elasticity 10,000 kPaDrained Poisson’s ratio 0.35Radius of load a 8 mDepth of layer h 16 mHorizontal permeability kh 0.0001 m/dayUniform load q 80 kPa


are given in tables that are provided in various books (see Zienkiewicz 1977) for different integration schemes. The coordinates are generally provided so that they range between −1 and +1; for instance for a three-point scheme, the values in Table 3.2 may be used. To integrate between a and b as shown in Figure 3.12, we can make a transformation xi = (a + b)/2 + (b − a)gi/2.

If we wish to integrate over an area as in a two-dimensional problem we would need to integrate with respect to x and y, but we can relate the real coordinates to the local coordi-nates and perform the integration as follows:

I f d d=−

+

−

+

∫∫ ( , )ξ η ξ η| |J1

1

1

1

(3.73)

where |J| is the determinate of the Jacobian matrix (Equation 3.12). For three-dimensional problems, we have to integrate over three local coordinates and the Jacobian is a 3 × 3 matrix.

This is shown in Figure 3.12 for a three-point Gauss scheme. It may be seen that the Gauss points lie between the integration limits of a and b.

For an eight-node isoparametric element, the integration needs to be performed in both the x and y directions, and so the Gauss points run in both directions. This is shown in Figure 3.13 where a 3 × 3 Gaussian integration scheme is used.

For triangular elements, the Gauss points are located at prescribed locations within the triangle. The Gauss points are normally given in terms of the ‘area coordinates’ of the ele-ment. There are three area coordinates that define a point within the element, and they each range from 0 to 1, and the sum of the coordinates is 1. At a corner of the triangle, one area coordinate will be 1 and the others 0. At the centroid of an element, they are all 1/3.

Table 3.2 Gaussian weights and coordinates for a three-point rule

Weight ωi Gauss coordinate gi

−0.77459 66692 0.55555 555550.0 0.88888 88888+0.77459 66692 0.55555 55555

f (x)

x

x1 x2a bx3

w1 w2 w3

Figure 3.12 A three-point Gaussian integration scheme.


3.7 ELASTIC–PERFECTLY PLASTIC MODELS

When the stresses in a soil or rock reach a combination that will cause failure, the stresses are said to lie on the failure surface for that material, and the soil/rock will then undergo plastic behaviour. Within the failure surface, the material is assumed to behave as if elastic, but once the stress state reaches the failure surface, the stress increments have to stay on the surface (and run along it) as the stress state cannot go outside the surface.

If unloading occurs, the stress state can move inside the failure surface and the material will behave like an elastic material once again.

3.7.1 Formulation

When the stress state in the soil is within the failure surface (i.e. f(σ) < 0 where f is the func-tion for the failure surface), then the material is assumed to behave in an elastic manner and the stress–strain relationship may be written as

σ ε= DE (3.74)

where DE is the stress–strain matrix for an elastic material.Once the stress state is such that the stresses reach the failure surface, the material becomes

plastic and the strain rate can be expressed as

ε ε ε= +E p (3.75)

The total strain rate ε is made up of the elastic εE and plastic εp strain rates where the elastic strain rate is given by

ε σE = −D 1 (3.76)

During plastic flow, the stresses must remain on the failure surface f = 0 and so any change in the stress state must be such that

df = 0 (3.77)

This means that we can write

df

fd= ∂

∂=

σσ 0

(3.78)

x

x x x

x xx

x x

Gauss points

Nodes

Figure 3.13 An eight-node isoparametric element with a 3 × 3 Gauss point scheme.


or

sT σ = 0 (3.79)

where

sT

n

f f f= ∂∂

∂∂

∂∂

σ σ σ1 2

, ,…

(3.80)

In Equation 3.79, the dot product of the two vectors is zero, so this indicates that the vector sT is perpendicular to the increment in stress, σ, and as the increment of stress is along the failure surface, the vector sT must be perpendicular to the yield surface.

The plastic material is assumed to have a flow rule such that the plastic strain rate vector is given by

εp = λr

(3.81)

In the above equation, r is a vector giving the direction of the plastic strain increment, and λ is an unknown scalar multiplier.

Equation 3.75 then becomes

ε σ= +−D rE1 λ (3.82)

Rearranging gives

σ ε= −D D rE Eλ (3.83)

We can now multiply by the vector sT and noting the result of Equation 3.79, we have

0 = −s D s D rTE

TEε λ (3.84)

and solving for the unknown λ, we have

λ = s D

s D r

TE

E

εT

(3.85)

Substituting the value of λ into the stress–strain law of Equation 3.83 results in

σ = DIε (3.86)

where the incremental plasticity matrix DI is given by

D D

sI E T= − αβα

T

(3.87)


and where

αβ

==

D r

D sE

E

3.7.2 Examples for a specific failure surface: Mohr–Coulomb

In the general derivation of the incremental stress–strain relationship in the previous sec-tion, the equation for the failure surface f could be any function of the stresses. One failure criterion that is of interest in soil and rock mechanics is the Mohr–Coulomb failure surface. For a two-dimensional problem involving plane strain, we can write the failure criterion (for the failure line as shown in Figure 3.14) simply as

f cx y xy x y= − + − + +( ) sin ( cot )σ σ τ φ σ σ φ2 2 2 24 2

(3.88)

We can therefore differentiate this equation for the failure surface with respect to the stresses to obtain the s vector.

s =

∂∂∂

∂∂

∂

=− − −

f

f

f

x

y

xy

x y xσ

σ

τ

σ σ φ σ2 2 2( ) sin ( σσ φσ σ φ σ σ φ

τ

y

y x x y

xy

c

c

+− − − +

2

2 2 2

8

2

cot )

( ) sin ( cot )

(3.89)

The vector r is the direction of the incremental plastic strain. This vector was traditionally made perpendicular to the yield surface in the plasticity of metals so that no plastic work was done. This is called an associated flow rule, as in this case the r vector and the s vectors

τ

σ

φ

τxy

σy

σxc

Mohr–Coulomb failureenvelope

Figure 3.14 Mohr–Coulomb failure criterion for a 2-D stress state.


are the same. However, for soils, the flow rule can be such that the r vector is not perpen-dicular to the failure surface, and in this case, we have

r =

∂∂∂

∂∂

∂

=− − −

g

g

g

x

y

xy

x y xσ

σ

τ

σ σ ψ2 2 2( ) sin (σ σσ ψσ σ ψ σ σ ψ

τ

y

y x x y

xy

c

c

+− − − +

2

2 2 2

8

2

cot )

( ) sin ( cot )

(3.90)

where g is called the plastic potential of the soil. This is a surface like the failure surface, and the vector r will be perpendicular to it. The angle ψ is the dilation angle of the material and can be seen to be equivalent to ϕ in Equation 3.89. Figure 3.15 shows the velocity vectors calculated for the case of ϕ = 0, ψ = 0.

3.8 WORK HARDENING MODELS

With simple elastic-perfectly plastic models like those discussed in Section 3.7, the soil is elastic and then suddenly becomes plastic once the stress state at a point reaches the failure

0.0 50.0 100.0 150.0 200.0 250.0 300.0 350.0–50.0

0.0

50.0

100.0

150.0

200.0

250.0

300.0

Velocity vectors

Rigid strip footing

Phi = 0Psi = 0Uniform soil

Scale

1.00E–01

Figure 3.15 Velocity vectors computed for strip footing: ϕ = 0, ψ = 0.


surface. For most soils and rocks, this is not what is observed, as there is a non-linear behav-iour before final failure.

To account for this, models have been proposed where the soil can yield before it fails, and these models are generally classed as work hardening (or softening) models.

Unlike Equation 3.77, for a work hardening material there can be a change in the location of the yield surface as work hardening occurs, that is,

df

fd

fh

dh= ∂∂

+ ∂∂

=σ

σ 0

(3.91)

where h is the hardening parameter. It could be a change in some parameter such as the plastic volume strain. In this case,

h HVp= =ε εd p

(3.92)

where H = (1, 1, 1, 0, 0, 0)T for a general three-dimensional case.If c = ∂f/∂h, Equation 3.91 can be written as

s ch

s D s D r ch

T

TE

TE

σ

ε λ

+ =

− + =

0

0 (3.93)

and so solving for λ gives

λ ε=

−s D

s D H

TE

TE

r rc

(3.94)

Hence,

σ ε ε ε= −−

⋅ =Ds D

s D r HrD r DE

TE

TE

E Ic

(3.95)

and now the incremental plasticity matrix DI is given by

D D

s HrI E

T

T c= −

−

αβα

(3.96)

The Modified Cam Clay Model is a hardening model that uses the plastic volume strain as the hardening parameter. This is discussed in the following, Section 3.9.

3.9 EFFECTIVE STRESS ANALYSIS USING CAM CLAY TYPE MODELS

It is very important in using constitutive models with finite element analysis to under-stand how they work as this may influence the results. One such model that is worthwhile


examining is the Cam Clay Model which is often used to simulate soft clay behaviour. Similar models are often implemented in finite element codes.

Soils consist of solids (the soil particles or skeleton), with liquid and air in the voids. If the soil has a low permeability (i.e. a clay or a silty clay), the fluid cannot flow out of the soil rapidly when the soil is loaded, and the soil behaves as an undrained material. If the soil is saturated, it will deform at constant volume, as the water in the pores is incompressible, and so only shear deformations can occur.

If a soil is loaded slowly, the fluid in the pores has time to flow from regions of high excess pore pressure to regions of low excess pore pressure, and the soil behaves as a drained material. At intermediate rates of loading, the behaviour is somewhere between drained and undrained. The behaviour of a soil therefore is dependent on the loading rate as well as the permeability.

Figure 3.16 shows the results of triaxial tests on a soil consolidated isotropically, that is, ′ = ′ = ′σ σ σ1 2 3.The stress path is plotted using the mean effective stress ( )′ = ′ + ′ + ′p σ σ σ1 2 3 3/ and

the deviator stress ′ = ′ − ′q σ σ1 3. During drained loading, the stress path will follow the straight broken line (which will be at a slope of 3 vert:1 horiz) until it hits the failure line. If the soil is loaded undrained, it will follow a curved path as shown in Figure 3.16 until it hits the failure line. This is because pore pressure u is generated during loading, and the difference between the drained path and the undrained path is u. This is because

( ) .p u u u p u= ′ + + ′ + + ′ + = ′ +σ σ σ1 2 3 3/The failure line in a plot such as the one in Figure 3.16 is the usual Mohr–Coulomb failure

line, but has a slope given by M, where

M =

−6

3sinsin

φφ′

′ (3.97)

and ϕ′ is the angle of shearing resistance of the soil.

σ ′1

σ ′3

σ ′2

p′

Drained

Undrained

Partiallydrained

q′

2su

Isotropicconsolidation

Excess porepressure = u

Failure line

Figure 3.16 Triaxial test stress paths.


If the soil is partially drained, the stress path will follow the curved broken line shown in Figure 3.16.

For undrained loading, the point at which the stress path intersects the failure line gives twice the undrained shear strength of the soil, as the undrained shear strength su is given by

s

qu = − = − =σ σ σ σ1 3 1 3

2 2 2( )′ ′ ′

(3.98)

3.9.1 Normally consolidated clay

The effective stress within a soil will increase with depth due to the overburden pressure. The lateral stress is related to the vertical stress by the earth pressure coefficient ′K0. This effective stress variation with depth is shown in Figure 3.17, where the vertical effective stress and the lateral effective stress are seen to increase with depth. At any point in the soil, we therefore have

pK K

q K K

v v v

v v v

′′ ′ ′ ′ ′

′ ′ ′ ′ ′ ′

= + = +

= − = −

σ σ σ

σ σ σ

23

1 23

1

0 0

0 0

( )

( )

(3.99)

If we plot the initial stress points for soil at two depths A and B, then these plot as points A′ and B′ as shown in Figure 3.18a. The points lie on what is called the K0 line as it is the line that the initial stress states lie on under K0 conditions (or one-dimensional consolidation conditions).

The void ratio e at point B would be less than at point A because the soil is under greater stress (i.e. it is deeper), and so it would plot as point B′ as shown in Figure 3.18b which is a plot of the natural logarithm of p′ versus the specific volume υ = 1 + e. This is how the void ratio versus mean effective stress is conventionally plotted in critical state soil mechanics, but is equivalent to the conventional e v– log10 ′σ plots made for oedometer tests.

In Figure 3.18b, the isotropic consolidation line (which is the line the stresses in the soil would follow if the soil is consolidated under vertical and lateral stresses that are all equal) is shown along with the critical state line (CSL). The CSL is the line showing the void ratio–mean effective stress relationship at failure.

σ′v

σ′h = K ′0σv

Effective stress

Dep

th z

Point B

Point A

Figure 3.17 Effective stress variation with depth in a normally consolidated soil.


Upon undrained loading, the stress path at the point A in the soil would be from A′ to A, and at point B the stress path would be from B′ to B. The void ratio would remain constant as the soil is undrained (Figure 3.18b) and so the path in this figure is horizontal (because there is no change in void ratio) until it hits the critical state line at failure.

We can see therefore that the undrained strength depends on the initial stress at a point in the ground and the path taken to failure. At point B in the ground, the stresses are higher and so the stress path leads to a higher undrained shear strength at failure at B (suB) than for point A that follows a path to A (suA). This is what is observed in reality, that the undrained shear strength in a soil deposit will normally increase with depth.

3.9.2 Overconsolidated clay

If a clay is overconsolidated, which means that it has been loaded in the past to a greater pressure than that it exists at a point in the ground at present, then the stress path taken by the soil in the ground will be as shown in Figure 3.19a.

During undrained loading, the volumetric behaviour is as shown in Figure 3.19b, where once again the void ratio remains constant during loading and the path followed is such that point A finishes on the critical state line. The path is now such that the CSL is approached from the left instead of from the right.

This time the undrained shear strength is higher at the same point in ground A, because of the different stress path taken on loading. The undrained shear strength can be seen to be different because the stress path hits the failure surface at a greater value of q′.

Hence, the undrained shear strength of a soil depends not only on the initial stress state but also on whether it is overconsolidated or not (and the degree of overconsolidation).

p′

K0 lineB

q′

2suA

Failure line

2suB

AB′

A′

ln(p′)

B

1 + eA

B′

A′

Isotropicconsolidation line

Critical state line

(a)

(b)

Figure 3.18 (a) Undrained stress paths to failure at different depths in a soil. (b) Void ratio during undrained loading.


3.10 CAM CLAY TYPE MODELS

In order to try to simulate the behaviour of real soils described in the previous sections, models have been developed that can be put into finite element codes, and one such model is the Modified Cam Clay Model (Roscoe and Burland 1968) so called because it was a modi-fication of the original Cam Clay Model proposed by Roscoe and Schofield (1963).

The features of this model are that it can simulate the yield of a soil and the increased compressibility of the soil once it has yielded. The model is based on effective stress analysis, and therefore the excess pore pressures generated are important as they affect the stress path as seen in the previous section.

3.10.1 Cam Clay yield surface

With the Modified Cam Clay Model for soft soils, there is both a yield surface and a fail-ure surface. The yield surface has been noted from triaxial tests on soft clays for which the model was developed, and can be seen to be an ellipse in Figure 3.20.

The model is a work hardening model such that the ellipse expands when the stress path reaches the yield ellipse (see the dotted line in Figure 3.20). If the stress path goes back inside the ellipse, the elliptical surface stays at its maximum size (i.e. the maximum ′p0 ). In this way, the pre-consolidation pressure that is exhibited by soils can be modelled. The stress path has to reach the new location of the yield surface before it becomes plastic again.

The equation for the yield surface ellipse is

M p p p q2 20

2 0( )′ − ′ ′ + = (3.100)

p′

K0 line

q′

2suA

Failure line

A

A′

ln(p′)

1 + eAA′

Isotropic consolidation line

Critical state line

(a)

(b)

Figure 3.19 (a) Undrained stress path for an overconsolidated soil. (b) Volumetric behaviour of an overconsolidated soil during undrained loading.


Eventually, the stresses will reach the failure surface, which is given by the equation

q M p= ⋅ ′ (3.101)

and the stress state must then stay on this failure surface.For a work hardening plastic material, we can write the incremental stress–strain matrix

as in Equation 3.93, and for the Modified Cam Clay Model, the hardening parameter h is the plastic volume strain. We can therefore write for a two-dimensional problem

c

fh

fp

ph

M pe

p= ∂∂

= ∂∂ ′

∂ ′∂

= ′+−

′

0

0 2 00

1λ κ

(3.102)

The parameters λ and κ are the slopes of the normal and recompression lines if we plot the specific volume υ = (1 + e) versus p′ as shown in Figure 3.21 (e is void ratio).

In order to work out the s vector, the function for the yield surface has to be differenti-ated with respect to the stresses. For a two-dimensional plane strain problem, there are four stress components σx, σy, σz, τxy as we have to take into account the stress in the zero strain direction σz.

M = 6 sinϕ′/(3 – sinϕ′)

q = (σ′1 – σ′3)

p′ = (σ′1 + 2σ′3)/3

p0′

p′0/2

New location of yield surface

q = M . p′

Figure 3.20 Modified Cam Clay Model showing a yield ellipse and failure line q = M · p′.

ln p′

Slope = λ

Slope = κ

υ

Critical state line

Normal compression line

Figure 3.21 Volumetric behaviour. Specific volume υ = 1 + e versus ln p′ plot.


s = ∂∂

= ∂∂

∂∂

+ ∂∂

∂∂

= ′ − ′ ⋅

f fp

p fq

q

M p p

σ σ σ′′

202

1313130

( )

+ ⋅′ −′ −′ −

23

2

2

qq

p

p

p

x

y

z

xy

σσσ

τ

′′′

(3.103)

In the Modified Cam Clay Model, the plastic potential g is the same as the yield surface, that is, it has an associated flow rule.

While the stresses are inside the yield surface, they deform along the recompression line that has a slope of κ in the plot of specific volume υ = 1 + e versus ln p′. Once the yield sur-face has been reached (the ellipse in Figure 3.22a), the soil yields. If we are moving along the x-axis or along the hydrostatic compression line, once yield occurs, we would move along a line that has a slope of λ as shown in Figure 3.22b. This is what is observed for real soils where after yielding, the soil is more compressible. If we were to perform a one-dimensional compression test such as an oedometer test, the slope of the compression line would also

Yield surface at failure

q = M . p′

p′0/2

q = (σ′1 – σ′3)

A

A′

X

υ = 1 + e Slope = λ

Slope = κ

Critical state line

Normalcompression line

A′

XA

ln p′

p ′0

(a)

(b)

p′ = (σ′1 + 2σ′3)/3

Figure 3.22 (a) Stress path taken during drained loading showing an expanding yield surface. (b) Volumetric behaviour during the drained loading path of (a).


be λ after the pre-consolidation pressure is reached. The relationship between the more familiar parameters Cc (the compression index) and CR (the recompression index) obtained from an oedometer test is given by

λ

κ

=

=

C

C

c

R

2 3

2 3

.

.

(3.104)

Suppose we perform a drained triaxial test on a slightly overconsolidated soil from an initial stress state at point A in Figure 3.22a. The stress state would hit the yield surface at the point X and then the yield surface would move with the stress path and expand as load-ing occurs.

If unloading occurs, the soil returns to being ‘elastic’ inside the yield surface. This allows overconsolidation to be modelled, as the soil will not now yield on reloading until it once again hits the (expanded) yield surface. The stress state will eventually hit the failure surface at point A′.

In terms of the volumetric behaviour, the soil will follow the path shown in Figure 3.22b where it can be seen to yield at point X before continuing on to failure at point A′ on the critical state line.

If an undrained analysis is performed on the same clay starting from the same point on the p′ – q′ plot (i.e. point A), the stress path will move vertically upward hitting the yield surface at point X. It will then move to the left intersecting the failure surface at point A′ as shown in Figure 3.23a. The volumetric behaviour is shown in Figure 3.23b where it can be seen that there is no volume change as the loading takes place and so the path from point A to A′ is horizontal.

Hence, the success of an effective stress analysis considering pore pressure will depend on

1. Having the correct initial stress state (which depends on ′K0) 2. Choosing the correct permeabilities kv and kh and the correct boundary conditions as

the excess pore pressures control the effective stress path 3. Having the correct loading rate 4. Having the correct overconsolidation (OCR) ratios 5. Obviously, the soil strength ϕ′ parameter (which is used to compute the slope of the

failure line M) and the soil deformation parameters λ, κ play an important role as well

The undrained behaviour of a clay in a numerical analysis will be such that the soil behaves according to the model, and for a Cam Clay Model, the yield surface is an ellipse with the top of the ellipse on the critical state line. Other shaped ellipses are possible to get different soil behaviours.

3.11 UNDRAINED ANALYSIS

In conventional slip circle analysis or in an undrained analysis using a finite element pro-gram (say), the soil strength is treated as having undrained strength parameters su and ϕu = 0. As we have seen, the undrained strength su is a function of the initial stress state and the overconsolidation ratio.


According to Cam Clay theory (Wroth 1984), the undrained shear strength of a clay under plane strain conditions (with plain strain angle of friction ϕPS) is given by

suPS

v

PS

σφ

′

Λ

= +

sinc

c( )2

12

2

(3.105)

where Λ = (1 − Cr/Cc), Cc is the virgin compression index and Cr is the recompression index of the clay, and c = 1/(1 − 2 sin(ϕPS)). If we take typical soil parameters such as ϕ′ ≈ 26–27° and Λ in the range 0.7–0.8 for clays of low to medium sensitivity, then this gives

sOCRu

vσ ′

= 0 27 0 8. .

(3.106)

Data based on back analysis and field observation have led to empirical equations such as

s

ss

OCR

sOCR

uN C v

uO C

uN C

uO C

v

/

/

/

/

≈

=

′

=

0 22

0 22

0 8

0 8

.

( )

.

.

.

σ′

σ

(3.107)

p′ = (σ′1 + 2σ′3 )/3

q = (σ′1 – σ′3 )

p′0

p′0/2

Yield surface atfailure

q = M . p′

A

A′

X

XA

υ = 1 + e

Slope = λ

Critical state line

Normalcompression line

A′

ln p′

(a)

(b)

Figure 3.23 (a) Stress path taken during undrained loading from the K0 line (slightly O/C soil). (b) Volumetric behaviour during undrained loading from the K0 position.


where the subscript N/C denotes the normally consolidated value of undrained shear strength and O/C denotes the overconsolidated value. These formulae are similar to the theoretical equation (Equation 3.106), from the Cam Clay Model that gives a slightly higher value of su.

The su values change depending on soil type, stress paths, and initial stress states, but the empirical equations (Equation 3.107) apply in many practical situations. A plot of the undrained shear strength ratios versus overconsolidation ratio OCR is shown in Figure 3.24 (Ladd et al. 1977).

3.12 FINITE ELEMENT ANALYSIS

If we use a finite element analysis (Britto and Gunn 1987) for an undrained problem of loading of a soft clay, then we would start off with a stress state at point X in Figure 3.25 if the soil is normally consolidated and the stress state would lie on the yield surface. Upon loading, the stress would follow a path to point A′ on the critical state line. The undrained shear strength is then half of the q′ value at point A′.

If we take the angle of shearing resistance of the clay as ϕ′ = 27° and assume ′ = − ′ =K0 1 0 546sin . ,φ then we have an initial stress state (at point X in Figure 3.25) of

q K

pK

v v

v v

′ ′ ′ ′

′ σ ′′

= − =

= − =

σ σ

σ

( ) .

( ).

1 0 454

1 23

0 697

0

0

(3.108)

M =

−=6

31 07

sinsin

.φ

φ′

′ (3.109)

OCR1 2 5 10

2

1

3

4

5

6

Varvedclay

Clays

(s u/σ′ v0

) OC

(s u/σ′ v0

) NC

Figure 3.24 Ratio of undrained shear strength for an overconsolidated clay to undrained shear strength for normally consolidated clay versus OCR. (From Ladd C.C. et al. 1977. Proceedings of the 9th International Conference on Soil Mechanics and Foundation Engineering, Tokyo, Vol. 2, pp. 421–494.)


The actual stress path is shown in the finite element results of Figure 3.26. In this case, su ≈ 13.6 kPa when ′ =σv 50 kPa or su = 0.27 ′σv as predicted by the Cam Clay formula of Equation 3.106. This is slightly higher than the values usually measured for normally con-solidated clays that are observed to be about 0.22 ′σv as we saw in the previous section (ratio = 0.27/0.22 = 1.23).

This may be critical in the analysis of embankments being constructed on soft clays, as the predicted height to failure may be higher than it should be. In cases such as this, it is best to use a slip circle analysis with the variation of undrained strength with depth as a check of the collapse height. This of course can only be done if the distribution of the undrained strength has been measured. If the embankment loading is such that it is kept constant for a period of time to allow consolidation, then the effective stresses and the undrained shear strengths will increase and the initial distribution of undrained shear strengths cannot be used in a slip circle analysis.

Yield surface atfailure

X

su = q′/20.454σ′v

q′ = (σ′1 – σ′3)

p′ = (σ′1 + 2σ′3)/3

q′ = M . p′

q′ = 0.454σ′v

p′ = 0.697σ′v

A′

p′0

Figure 3.25 Stress path taken by normally consolidated clay under undrained loading in a Cam Clay Model.

10 20 30 40 500Mean stress p (kPa)

Dev

iato

r str

ess q

(kPa

)

10

20

30

40

50

Stress path

Figure 3.26 Stress path taken under plane strain conditions.


3.12.1 Examples

Some examples of the finite element implementation of the Cam Clay Model are shown in Figures 3.27 through 3.29. These show the behaviour of a clay specimen tested in triaxial compression in either the drained or the undrained condition.

Figure 3.28 shows the behaviour of a lightly overconsolidated soil in an undrained tri-axial test. The stress path (Figure 3.28b) goes vertically upward until it hits the yield locus and then it turns towards the critical state line, where failure occurs. The load–deflection curve (Figure 3.28a) shows the failure at a deviator stress of about 35 kPa.

Figure 3.29 shows the results of an undrained triaxial test on a heavily overconsolidated clay. The stress path once again goes vertically up until it hits the yield surface (passing above the critical state line). It then turns towards the CSL and ends on it at failure. Figure 3.29a shows the load–deflection curve for this case, where it may be seen that strain soften-ing occurs after yielding. The final deviator stress at failure is about 31 kPa.

More on Critical State Soil Mechanics can be found in texts by Atkinson and Bransby (1978) and Schofield and Wroth (1968).

Critical stateline

Mean stress p (kPa)

Dev

iato

r str

ess q

(kPa

)

20 6040 80 1000

0

20

40

60

80

100

Unload–reload

Unload–reload

App

lied

pres

sure

(kPa

)

80

60

Deflection (m)

40

20

00 0.4 0.8 1.2 1.6

log (p)50 55 70 80

Void

ratio

1.60

1.55

1.65

1.70

1.75

1.80

1.85

60 65 75

(a) (b)

(c)

Figure 3.27 A drained triaxial compression test: (a) stress path in p – q space (finite element result); (b) load–deflection; (c) e – log p′. Cam Clay Model showing load–unload behaviour.


APPENDIX 3A: SHAPE AND MAPPING FUNCTIONS FOR VARIOUS ELEMENT TYPES

Shape functions for the six-node triangle

t = 1.0

s = 0

t = 0

s = 0.5

t = 0.5

s = 1.0

1 2 3

4

5

6

App

lied

pres

sure

(kPa

)

Deflection (m)0.120.080.040 0.16

10

0

20

30

40(a)

Mean stress p (kPa)

Dev

iato

r str

ess q

(kPa

)

200 40 6010 30 50 70

10

20

30

40

50

60

70(b)

Stress path

Figure 3.28 An undrained triaxial test on lightly overconsolidated soil: (a) load–deflection; (b) p – q plot.

Dev

iato

r str

ess q

(kPa

)

Mean stress p (kPa)0 10 20 30 40 50 60 70

10

20

30

40

50

60

70

Stress path

0

(a)

Deflection (m)

App

lied

pres

sure

(kPa

)

0 0.05 0.15 0.2 0.250.10

5

10

20

30

35

25

15

(b)

Figure 3.29 An undrained triaxial test on heavily overconsolidated soil: (a) load–deflection; (b) p – q plot.


Area coordinates are s, t, u = (1 − s − t)

N s t u

N u s

N s s

N st

N t t

N u t

1

2

3

4

5

6

1 2 2

4

2 1

4

2 1

4

= − −== −== −=

( )( )

( )

( )

( )

( )

Shape functions for the eight-node isoparametric element

1 2 3

8

7 6 5

4ξ

η

N

N

N

1

22

3

0 25 1 1 1

0 5 1 1

0 25 1 1

= − − − − −

= − −

= + −

. ( )( )( )

. ( )( )

. ( )(

ξ η η ξ

ξ η

ξ η))( )

. ( )( )

. ( )( )( )

. (

ξ η

η ξ

ξ η η ξ

− −

= − +

= + + + −

=

1

0 5 1 1

0 25 1 1 1

0 5 1

42

5

6

N

N

N −− +

= − + − + −

= − −

ξ η

ξ η ξ η

η ξ

2

7

82

1

0 25 1 1 1

0 5 1 1

)( )

. ( )( )( )

. ( )( )

N

N

Mapping functions for five-node infinite element with ξ−2 decay

ξ

η

ξ = +1 at ∞

1 2

3

4

5


M

M

M

1 2

2

2

2

3

2

2 11

0 5 1 11

0 5 1 1

= −−

= + −−

= + +

ξη ηξ

ξ ηξ

ξ η

( )( )

. ( ) ( )( )

. ( ) ( )(11

2 11

4 11

2

4 2

5

2

2

−

= − +−

= − −−

ξξη η

ξξ η

ξ

)

( )( )

( )( )

M

M

Shape functions for 15-node triangle

1

6

2 3 4 5

7

8

9

10

13 14

1511

12

Area coordinates are s, t, u = (1 − s − t)

c

c c

1

2 1

1023

4

=

=

c

c

t s

t s

t s

t t

t t

t t

3

4

1

2

3

4

5

6

64

128

0 25

0 5

0 75

0 25

0 5

=

=

= −

= −

= −

= −

= −

=

.

.

.

.

.

−−

= −

= −

= −

0 75

0 25

0 5

0 75

7

8

9

.

.

.

.

t u

t u

t u


N c s t t t

N c s t t t

N c s t t t

c s t t t

N

N

1 1 1 2 3

2 2 1 2

3 3 1 4

4 2 4 5

5

= ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

== ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

c t t t t

N c t u t t

N c t u t t

N c t u t t

N

1 4 5 6

6 2 4 5

7 3 4 7

8 2 7 8

9 == ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

c u t t t

N c s u t t

N c s u t t

N c s u t t

1 7 8 9

10 2 7 8

11 3 1 7

12 2 1 2

NN c s t u t

N c s t u t

N c s t u t

13 4 1

14 4 4

15 4 7

= ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅

Shape and mapping functions for a 12-node infinite element

12

3

45

67

89

1011 12

∞

∞

∞

∞

t = −1

t = +1u = +1

u = −1

s = −1

Mapping functions:

Ms u t t u

s

Mt u s

s

1

2

0 5 1 1 11

0 25 1 1 11

= − − − − − −−

= − − +−

. ( )( )( )( )

. ( )( )( )( )

MMt u s

s

Ms u t t u

s

3

4

0 25 1 1 11

0 5 1 1 11

= + − +−

= − − + − + −−

. ( )( )( )( )

. ( )( )( )( ))


Ms t u

s

Ms u t

s

Ms u t

5

2

6

2

7

2

1 11

1 11

1 1

= − − −−

= − − −−

= − − +

( )( )( )

( )( )( )

( )( )(( )

. ( )( )( )( )

. ( )( )(

10 5 1 1 1

10 25 1 1 1

8

9

−

= − − − − − +−

= − + +

s

Ms u t t u

s

Mt u s))( )

. ( )( )( )( )

10 25 1 1 1

110

−

= + + +−

s

Mt u s

s

Ms u t t u

s

Ms t u

s

11

12

2

0 5 1 1 11

1 11

= − + + − + +−

= − − +−

. ( )( )( )( )

( )( )( )

The shape functions are

(i) Mid-side nodes

N s t u

N s t u

N t

22

32

52

0 25 1 1 1

0 25 1 1 1

0 25 1

= − − −

= − + −

= −

. ( )( )( )

. ( )( )( )

. ( ))( )( )

. ( )( )( )

. ( )( )( )

1 1

0 25 1 1 1

0 25 1 1 1

62

72

− −

= − − −

= − − −

s u

N u s t

N u s t

NN s t u

N s t u

N

92

102

12

0 25 1 1 1

0 25 1 1 1

0 25 1

= − − +

= − + +

=

. ( )( )( )

. ( )( )( )

. ( −− − +t s u2 1 1)( )( )

(ii) Corner nodes

N s t u N N N

N s t1 2 5 6

4

0 125 1 1 1 0 5

0 125 1 1 1

= − − − − + += − + −

. ( )( )( ) . ( )

. ( )( )( uu N N N

N s t u N N N

N

) . ( )

. ( )( )( ) . ( )

− + += − − + − + +

0 5

0 125 1 1 1 0 53 5 7

8 6 9 12

11 == − + − − + +0 125 1 1 1 0 5 7 10 12. ( )( )( ) . ( )s t u N N N

The derivatives of the shape functions are then computed in the usual way (see Equation 3.13) but where derivatives are found for x, y, and z.

∂∂

= ∂∂

∂∂

+ ∂∂

∂∂

+ ∂∂

∂∂

Nx

Ns

sx

Nt

tx

Nu

ux

i i i i


and since

x M M M

x

x

x

=

[ , , , ]1 2 12

1

2

12

…

we can find ∂x/∂s, ∂x/∂t, ∂x/∂u, and obtain the inverse derivatives by forming the Jacobian matrix and inverting it as explained in Section 3.2.3.

APPENDIX 3B: GLOBAL MATRIX ASSEMBLY AND BOUNDARY CONDITIONS

Once the element stiffness matrix is formed by multiplying element matrices together and numerically integrating the result, the element matrices need to be assembled into a global matrix. This is done by assigning a number to each node in the mesh. If the node has only one variable associated with it, for instance in flow and seepage problems, then the vari-able can be assigned the same number as the node. For instance, node 22 has unknown h22 (total head) at the node in a seepage problem. If it has three variables per node as in two-dimensional consolidation problems (the horizontal u and vertical w displacement and the total head h or pore pressure), then the node has three unknowns. Again if we take node 22, the variables in the global matrix are 3 × 22 − 2 for u, 3 × 22 − 1 for w, and 3 × 22 for h.

If, for example, we had an element with nodes 6, 13, 22, then for a problem with three variables per node we would have the following correspondence between the positions in the element matrix and the global matrix.

Element position Global position

1

2

3

4

5

6

7

8

9

16

17

18

37

38

39

64

65

66

→

So, an entry in position 9,9 in the element matrix becomes entry 66,66 in the global matrix, and entry 3,9 in the element matrix becomes entry 18,66 in the global matrix.

It can be seen that the further apart the node numbers are for the element, the further apart the locations in the global matrix are. The bandwidth of the global matrix depends on how far apart the element numbers are.

From a 12-node infinite element, we can see that the element matrices will add into the global matrix along the diagonal, and provided the numbering of the mesh is done


judiciously, the numbers all fall within a banded region down the diagonal of the global matrix, and the width of this band is called the bandwidth. The half width of the banded region is called the half-bandwidth of the global matrix.

Often problems are such that the global matrix is symmetric, and so we only need to store the upper half of the matrix. In addition, we can store only the upper half of the band because outside of the band all values are zero. This is often done in finite element work to increase the efficiency of storage. A more efficient method called ‘skyline’ storage is also used where each row in the upper half of the band is stored but only to its maximum length. If zeros exist at the end of a row, they are not stored.

If the problem is not symmetric, the whole bandwidth has to be stored. This occurs for seepage or consolidation problems where the permeability is anisotropic, or in plasticity problems where the soil has a non-associated flow rule (see Section 3.7.2).

APPENDIX 3C: BOUNDARY CONDITIONS

To solve the global set of finite element equations for the unknowns at each node, we usually have to apply some boundary conditions at the nodes of the finite element mesh. A common condition is that a displacement is zero at a boundary or a total head is prescribed in a flow problem.

If we know what a prescribed value for a variable is, then it is not an unknown for the problem and we can multiply it by a column of the global matrix and subtract it from the right-hand side of the equations.

For example, if we have a set of equations Aδ = f and we know that the 24th variable of the solution is Q, then we can multiply the 24th column of the matrix by Q and subtract it from the right-hand vector f as shown in Figure 3B.1. This can be done repeatedly if there

Zeros

Zeros

Element matrices

Global matrix

Half-bandwidth

Figure 3B.1 Placing element matrices into a global stiffness matrix.


are more than one prescribed value. If the prescribed value is zero, then this obviously does not need to be done, as zero will be subtracted. All subtractions must be done for each pre-scribed variable before proceeding to the next step as described in the next paragraph, as the matrix must not be modified while multiplying the columns by the known values.

However, it should be noticed that this makes the matrix non-symmetric, therefore one way to make the matrix symmetric again is to set the row and column (in this example row 24 and column 24) equal to zero, and place a 1 on the diagonal for each prescribed bound-ary condition. The prescribed value is entered at row 24 on the right-hand side so that there is now an equation that states 1 × δ24 = Q. An alternate method is to delete all the rows and columns for the prescribed variables, and reduce the size of the global matrix, but this is not addressed here (Figure 3B.3).

24

A

Zeros

24 1 δ24 = Q

Figure 3B.3 Replacing row and column with zeros and inserting a unit value on the diagonal.

δ24 = f – × Q A

24

Figure 3B.2 Applying a prescribed boundary condition to a set of equations.

77

Chapter 4

Site investigation and in situ testing

4.1 INTRODUCTION

Most of the analytical and numerical methods that are presented in the following chapters depend on obtaining the appropriate material properties for use in the analysis. The prop-erties may be obtained from laboratory tests, but are more often found from field tests as these can be performed in situ and are often a more cost-effective way of obtaining soil properties.

The correlation of the field test result with the soil property often depends upon the type of analysis being performed. For instance, some computer programs may need an initial modulus, while for others a tangent modulus may be more suitable. Some may require a small-strain modulus if this is appropriate.

In this chapter therefore, the most common field tests are described, and where applicable, methods of obtaining the appropriate material property from the field test are presented. As field tests are performed as part of a site investigation programme, common site investiga-tion techniques are also described in this chapter.

The most common field tests are the standard penetration test (SPT), the dynamic cone penetration test (DCP), the static cone penetration test (CPT), the pressuremeter test (PMT), and the dilatometer test (DMT). Various standards outlining test methods have been devel-oped in different countries and these should be referred to when performing the test as there are often slight differences in test procedures among the different standards. The various field test methods are described in the following sections.

4.2 EXPLORATION METHODS

For some in situ test methods, a hole has to be drilled or formed using wash boring, and the test is conducted at the base of the borehole. Such tests are the standard penetration test, the Ménard pressuremeter test, or seismic cross-hole tests.

For SPT tests, the base of the borehole needs to be cleaned (e.g. using a blank bit) so that loose material that has fallen back into the hole or has been loosened by drilling does not interfere with the test result.

For other in situ tests, no borehole is needed. For tests such as cone penetration tests, or dilatometer tests, the device is pushed into the ground using hydraulic pressure. Seismic refraction tests can be performed by simply striking the ground with a sledge hammer and recording the seismic waves returning to the ground surface.


4.3 SITE INVESTIGATION

In order to carry out a geotechnical design, it is necessary to acquire information about the soil or rock properties pertinent to the type of design to be carried out and to obtain an idea of the distribution of the various soil and rock types that exist at a particular site. A brief overview is provided in the following of the most common techniques used to obtain site and subsoil information.

4.4 OBJECT OF SITE INVESTIGATION

There may be a number of different reasons for performing a site investigation, and it is nec-essary to establish what the object of the investigation is so that the appropriate information can be collected when performing the work. Some reasons for performing an investigation are listed as follows:

1. To determine if the site is suitable for the proposed works. 2. To enable the design of structures, earthworks and excavations. 3. To allow prediction of any difficulties arising during construction. a. It may be necessary to check the bearing capacity of proposed foundations or to

determine settlements. In this case, it would be necessary to obtain soil samples which could be tested or to carry out field tests as part of the investigation, so that soil strengths and compressibilities are known. It would also be necessary to deter-mine the soil types and their distributions.

b. If cuts are being made, the stability of the slopes would need to be considered. c. If excavations are to be made, information would be required to enable the support

systems to be designed. d. Dewatering may be required at the site, and if so, information on materials, their

permeabilities, and their distributions would be required. 4. To locate or determine the availability of construction materials, for example, rockfill,

clay, and filter materials for a dam. 5. To assess the effect of changes brought about by construction, for example, excavation

may affect adjacent sites; construction may alter runoff patterns. 6. To investigate soil/groundwater aggressiveness on buried structures. 7. Environmental and health issues may require an investigation. If the site is known to

be contaminated, the extent of the contamination may need to be determined. Today, it is common for combined geotechnical/environmental investigations to be carried out, as use of the same boreholes for both types of investigations leads to obvious cost savings.

4.5 CATEGORY OF INVESTIGATION

Investigations may need to be performed on ‘green field’ sites (ones that have not been sub-ject to previous construction or filling) or on ‘brown field’ sites (ones that have had some form of previous development carried out). Investigations can be categorised as follows:

1. New Work Examples of new work include a. Buildings, bridges, roads, railways, airfields, port facilities, industrial plants

Site investigation and in situ testing 79

b. Dams c. Mines (underground and open cut) and mine infrastructure 2. Existing Facilities Existing facilities include a. Pavement which may need repair or upgrade b. Structures which need extensions or modifications, for example, adding more sto-

ries to an existing building 3. Failures and Remedial Work Repair or remedial work may need to be carried out for a. Slope instability or failures b. Retaining wall collapses c. Dams (which may suffer collapse or excessive leakage) d. Structural damage due to swelling or shrinking soils, or subsidence due to mining

or withdrawal of groundwater

4.6 PLANNING AN INVESTIGATION

A good deal of time and money can be saved by making use of pre-existing information about the site and planning the investigation before going into the field.

Some of the following activities may be useful when planning and executing an investigation:

1. Define the aim of the investigation or of the problem to be solved. 2. Initial Office Study a. For large areas, use can be made of topographical and geological maps or aerial

photographs. Internet-based satellite photographs are also commonly used. Stereo pairs of photographs allow terrain types to be identified, for example, alluvial flats, shallow slopes, and steep scarps. Large slips can also be identified. Old aerial photographs of industrial areas can be used to determine previous positions of buildings, waste storage, or dumping (fill) areas.

b. For smaller sites, information can be obtained from geological maps or from previ-ous site investigations in the area. Most consulting firms which have been in exis-tence for any length of time, have comprehensive records of past site investigations, and investigations could have been previously carried out on adjacent sites or on the same site. Geographical information systems (GIS) have been set up for some areas and these contain data such as borehole, cone, and test data for all investiga-tions in the area covered.

c. Maps of services (water, gas, electricity) need to be obtained from councils or other government agencies.

Excavating or drilling through such services will cause costly delays and possible disruption to surrounding buildings.

3. Site Inspection a. A preliminary reconnaissance is necessary to confirm features seen on maps and

aerial photographs. b. At this stage, a more detailed investigation can be planned. The location of drill

holes or test pits can be marked on a map of the site. These are usually along the line of proposed foundations (dig test pits to one side so as not to create settlement problems later).


It may be more convenient to use other forms of site investigation than drilling/excavating. At this stage, the decision may be made to use cone penetrometers or to use geophysical methods (i.e. seismic refraction or resistivity methods).

c. It may be necessary to locate electricity supplies or water supplies for some equip-ment which is to be used.

d. Buildings similar to those planned for the site may exist in the area. It is a good idea to inspect these buildings for signs of damage, for example, cracking due to swelling soils. Cuts in the area may be at slopes similar to those proposed and their performance can be assessed.

4.7 PREPARING COST ESTIMATES FOR THE WORK

Some of the main costs in carrying out the site investigation include the following:

1. Cost of establishing plant, for example, may need to drive the drilling rig to country areas and back

2. Cost of drilling: Drilling with augers and diamond drilling is charged by the metre; excavators are charged by the hour

3. Cost of supervision by the engineer 4. Cost of laboratory testing 5. Cost of preparing the report

4.8 DETAILED EXPLORATION

1. The detailed exploration depends on the type of structure and the type of material revealed during the exploration, for example, if soil is uniform, fewer holes need to be bored.

2. The investigation needs to be taken to a depth where shearing of the soil due to the applied structural loads is small, for example, to depths of at least 50% greater than the width of the structure. However, if soft clay exists below this, then the investiga-tion should be taken to a greater depth. For buildings, usually preliminary borings are taken to rock (if it is at depth less than the required depth), and if no soft clay layers are found, holes are taken to 6–10 m. For light industrial buildings, backhoe holes to depths of 3–4 m are usually sufficient, while for tall buildings with deep pile founda-tions, some boreholes may need to be taken to depths of 100 m.

Spacing of boreholes may vary from 10 m in erratic strata to 60 m in uniform material.

4.9 PRESENTATION OF INFORMATION (LOGS)

The log of a borehole, test pit, or excavation contains information about subsurface condi-tions with depth. Log layouts vary from company to company, but contain similar infor-mation about soil types, groundwater conditions, field test, and samples taken. Typical information included on the logs is

1. The soil type, classification symbol (usually the Unified Classification System is used), soil colour, plasticity (clays), density (sands), particle size (fine, coarse), and secondary components


2. The position at which samples are taken 3. The depths at which in situ tests are performed 4. The moisture condition of the soil and water levels

An example of a borehole log for a hole drilled in soil is shown in Figure 4.1.

4.10 EXCAVATION OR DRILLING METHODS

Physical methods of investigation of foundations or subsoil may involve either drilling bore-holes or excavating a hole in the ground so as to expose the soil profile for visual examination.

4.10.1 Test pits

Test pits are an effective method for investigating conditions for shallow foundations or proposed road alignments. Pits are dug with a backhoe or excavator (see Figure 4.2) and allow visual inspection of the soil profile. Samples of soil can be cut from the sides of the pit or undisturbed samples can be taken with a thin-walled sampling tube. Water levels can be recorded if the pit is left open for a period of time.

The sides of test pits can collapse suddenly, especially for pits dug in sandy soil that is below the water table. For this reason, various government bodies restrict entry into test pits unless they are properly shored to prevent collapse.

The soil can be visually classified and a log made of the excavation. In addition, a sketch of the side of the pit can be made showing the extent of all materials exposed. Test pits are economical as they can be excavated quickly; however, they are limited in depth to about 3.5 m with a backhoe. Larger excavators can dig deeper holes if they have a long reach bucket mounted on them.

The test pit should be dug to one side of the foundation alignment so as not to disturb the soil and cause foundation problems later (i.e. excess settlement). The pit is generally filled in at the completion of the investigation with the front mounted bucket of the backhoe so as not to be of danger to humans or animals that may fall into the pit.

4.10.2 Excavations

Excavations can be made on large projects, for example on dam sites, where a large trench is excavated with a bulldozer. The exposed walls of the trench allow examination of any faults, fissures, or intrusions in the foundation and their extent. Detailed logs of the founda-tion cross section can be made by looking at the sides of the trench.

4.10.3 Drilling

Various drilling techniques are available to enable deeper investigation of the subsoil. Unlike excavation methods, where large cross sections of the foundation are exposed, drilling only gives an idea of strata encountered in the borehole and interpolation must be made between boreholes to obtain a geological cross section.

4.10.3.1 Hand augers

Hand augers or post-hole diggers (see Figure 4.3) can be used on small jobs, for example, investigation of pavement failures or foundations of houses. A borehole log can be made by


JayCee GeotechnicsGeotechnics/Environment/Groundwater

CLIENT: Piling Constructions Pty LtdPROJECT: Shopping CentreLOCATION: Sydney, Australia

CLAY – Yellow brown clay with ironstone gravel

CLAY – Light grey to greyclay with ironstone bands

SANDSTONE – Extremely low-strength, fine-grained sandstone with ironstone bands

SANDSTONE – Medium then high strength, moderately weathered and fresh stained, slightly fractured and unbroken, brown then light grey, fine- grained sandstone

BOREHOLE No: 1APROJECT No: JC123AADATE: 12/3/2015SHEET: 1 of 1

SURFACE LEVEL: 3.7m AHDEASTING: 333720NORTHING: 6252303DIP/AZIMUTH: 90°/--

BOREHOLE LOG

RIG: DT 100 DRILLER: JCS LOGGED : JCS CASING: HW to 4.0 mTYPE OF BORING: Solid flight auger to 4.0 m; Rotary to 5.5 m; NMLC coring to 8.65 mWATER OBSERVATIONS : No free groundwater observed whilst augeringREMARKS: Used sucker truck in soil section

LAMINITE – High then very high strength, fresh, slightly fractured, light grey to grey laminate with approximately 50% fine-grained sandstone laminations and bands.8.45–8.65: Very high strength

Descriptionof

strata

Depth(m)

Degree ofweathering

Rock strength

B = Bedding J = JointS = Shear F = Fault

Sampling and in situ testingTest results

and comments

Note: Unless otherwise stated, rock is fractured along rough planar bedding dipping at 10°

PL(A) = 0.8

PL(A) = 2.2

PL(A) = 1.1

PL(A) = 2.1

PL(A) = 4.4

C

C 100

100 100

91

2.0

4.0

5.5

7.4

5.5–5.58 m: clay band

5.68 m: B0°, fe

JC

7.81 m: 80°, clay, vn

7.4 m: J20°, pl, ro, fe

8.65

1

2

3

4

5

6

7

8

9

6.68 m: B5°, fe, clay, vn6.82 & 6.88 m: B5°,–10° fe, clay, vn

RL

Gra

phic

log

Wat

er

Fracturespacing

(m)

Type

Core

Rec %

RQD

%

Discontinuities

0.05

0.10

0.50

0.01

1.00

HW

EW MW

SW FRFs

SAMPLING and IN SITU TESTING LEGENDA Auger sample G Gas sample PID Photo ionisation detector (ppm)B Bulk sample P Piston sample PL(A) Point load axial test IS (50) (MPa)BLK Block sample Ux Tube sample (xmm dia) PL(D) Point load diametral test IS (50) (MPa)C Core drilling Water seepage pp Pocket penetrometer (kPa)D Disturbed sample Water level V Shear vane (kPa)

Ex L

owV.

Low

Low

Med

ium

Hig

hV.

Hig

hEx

Hig

h

32

10

–1–2

–3–4

–5–6

Figure 4.1 Example of a borehole log.


withdrawing the post-hole digger or auger and noting the soil type adhering to the device and the depth from which it was recovered.

4.10.3.2 Wash borings

This method is less commonly used than auger boring (see Section 4.10.3.3). Water is forced through inner rods and it exits from the chopping bit, which is moved up and down and rotated (see Figure 4.4). The water washes the loosened soil to the surface and enables the hole to be advanced. A casing is advanced in the wash hole as the chopping bit is advanced. A change in colour of the wash water at the surface indicates a change of soil type. If a soil sample is required, the bit can be withdrawn and samples can be taken with thin-walled samplers.

4.10.3.3 Rotary drilling

With rotary drilling, the drill rod and cutting bit are rotated mechanically and advanced by hydraulic pressure. Water is forced through the drill rods and it emerges at the cut-ting bit which may be of the tricone roller type shown in Figure 4.5. The soil cuttings are flushed to the surface with the wash water or the drilling mud, and changes in soil type may be noted by observing changes in colour of the wash water. Biodegradable drilling fluids such as ‘Revert’ can be used for keeping the drilled hole open as well as drilling mud or casings.

4.10.3.4 Auger boring

Augers are turned mechanically, usually from a truck-mounted rig. They are advanced hydraulically into the soil so that hard to stiff clays or dense sands can be penetrated.

3–4 m deep

Front shovel to fill hole

Figure 4.2 Excavation of a test pit using a backhoe.


Lifting bail

Water swivel

Drill rods

Wash tee

Wash tubCathead

Water hoseWater pump

Chopping bit

Casing drive shoe

Casing

Figure 4.4 Wash boring.

Figure 4.3 Hand augers used for shallow investigations.


Continuous flight augers consist of separate flights which are generally 100 mm (4 inches) in diameter and 1.5 m (5 ft) or 1.8 m (6 ft) long. Sections are added to the drill string as drilling proceeds (Figure 4.6).

Depths of drilling are usually (depending on the power of the motor) 35 m in sand, or 25 m in clay after which the drag on the augers will cause refusal of the drilling motor.

Figure 4.5 A tricone roller.

Figure 4.6 Auger used for drilling in soil showing a ‘V’ bit.


Types of bits used are

1. Blank bits which are suited to soft soils and cleaning out the base of the drill hole 2. Steel ‘V’ bit which can be used in hard clay or soft rock 3. Tungsten carbide bit which can be used for soft rock 4. Fishtail bit (sands) 5. Drag bit (clay)

To calculate the depth drilled

1. Multiply the number of flights in the ground by the length of each flight. (A good check is to count the number of flights on the rig before drilling. The number in the ground can be calculated by subtracting the number left on the rig from the total number.)

2. Add 0.1 m for ‘V’ or tungsten carbide bit. 3. Subtract the distance from the top of the flight (which remains above ground) to the

ground level.

Sampling material using the auger (disturbed samples) involves

1. Using the blank bit, drill in low gear and adjust the speed of penetration so that the auger ‘winds’ itself into the ground (i.e. spirals do not appear to be going up or down).

2. Stop the flight after about 1 m penetration (stiff clay 0.5 m) and withdraw the auger without rotation.

3. Cut the contaminated material away and log the material extracted.

As drilling proceeds, the ease of drilling can be noted as a guide to the strength of the soil (this can range from easy drilling to refusal).

When pulling up the drill string (especially in sands), keep the hole full of water and pull the drill string up slowly. If the water level drops, the bottom of the hole can blow up and SPT readings will be erroneous.

4.11 SAMPLING METHODS

As well as performing in situ tests and obtaining soil properties, soil samples may be recov-ered and taken back to a laboratory for testing.

The site investigation code (AS 1726 1993) recommends that at least one undisturbed sample should be taken from every different stratum encountered. If the stratum is uniform, one sample can be taken at every 1.5 m, which is the length of the auger flights.

Samples may be disturbed or undisturbed depending on the type of test to be performed.Disturbed samples are those for which the soil fabric is disturbed during sampling. Such

samples are normally placed in a plastic bag and sealed so that no moisture is lost. A tag or label is placed on or inside the bag identifying the soil sample (i.e. date, depth, borehole number).

Disturbed samples can be used for

1. Compaction tests: 20–30 kg 2. Laboratory or visual classifications, for example, sieve analysis or index tests: 1–5 kg 3. Chemical tests: 1–5 kg 4. Moisture content: 1 kg


Undisturbed samples are those for which the soil fabric has been disturbed as little as pos-sible, although some disturbance will inevitably occur.

Undisturbed samples are required for

1. Oedometer tests 2. Triaxial tests 3. Swell tests 4. Unconfined compression tests

4.11.1 Thin-walled sampler (or Shelby tube)

The tube (see Figure 4.7) is usually 50 mm (2 inches) in diameter (so the specimen is called a U50 or undisturbed 50 mm diameter specimen). Larger and smaller tube sizes are available with the larger sizes used if sample defects need to be studied. Tubes are about 0.5 m long.

Use of thin-walled sampling tubes:

1. Ensure that the cutting edge has not been bent (e.g. from previous use in gravelly or hard soil).

2. Ensure that the tube is not rusty and that there is no old soil or wax in the tube from previous use. Lubricate the tube with oil or water.

3. Attach the ball valve adaptor. 4. Lower the tube to the required depth. The hole may need to be cleaned out first with

a blank bit, as it may be difficult to distinguish soil which has fallen into the borehole

Figure 4.7 Thin-walled sampling tube.


from the sample back in the laboratory. The tube is pushed at a constant rate into the soil by the hydraulic ram of the drilling rig.

5. Turn the drill rods once to break off the sample and then withdraw the sampler. 6. Record the depth and length of samples on an identification tag and record the sample

on the borehole log, for example, 1.8–2.1 m; 0.3 m recovery. 7. Tubes are sealed at each end with wax, plastic caps, or end seals to prevent moisture

loss before testing. 8. Ensure that the samples are not left in the sun or broken up during transport.

4.11.2 Split spoon sampler (SPT sampler)

Samples are taken as part of the standard penetration test described in Section 4.15.Because the sampling tube has a thick wall, samples are disturbed more than samples

taken with a thin-walled tube. Samples recovered in this way are suitable for index tests and identifying soil types.

4.11.3 Piston sampler

The piston sampler has a piston inside the tube which prevents the sample from slipping out of the tube. The piston also prevents unwanted material entering the tube as it is lowered (see Figure 4.8b).

4.11.4 Air injection sampler

An air line is provided so that air can be introduced to the underside of the sample facilitat-ing its retention in the sampling tube (see Figure 4.8c).

4.11.5 Swedish foil sampler

When a tube is pushed into the soil, friction between the walls of the tube and the soil cause the length of the sample which can be removed to be limited. This also causes disturbance to the soil samples.

Drill rods(a) (b) (c)

Sampler headBall valveFixing screws

in-walled tube

Piston

Air line

Figure 4.8 Sampling tubes. (a) Simple push sampler, (b) piston sampler, and (c) air injection sampler.


A specialised piece of equipment, the Swedish foil sampler, removes the side friction by means of foil ribbons which line the sides of the sampler. Lengths of sample up to 18 m can be obtained with this device.

4.12 ROCK CORING

To obtain samples of rock, it is necessary to use diamond drilling equipment. Rock core can be inspected for fissuring, weathering, and defects or can be used to obtain rock strength by testing the recovered core.

The core is obtained by using a triple or double core barrel which has a diamond studded bit attached at one end. A bit being removed is shown in Figure 4.9. Water is passed down the drill string and it emerges at the bit. This has the dual effect of cooling the bit and of flushing away the cuttings. However, the wash water will also wash away any weathered or soft rock or soil and it will be lost.

Core barrels can be of the triple or double type. The triple core barrels have a thin split inner sheath into which the core slides, and surrounding this two outer core barrels are found. This arrangement gives better core recovery. In order to retain the core in the core barrel, a core catcher device is used which can retain the core stopping it from slipping out when an attempt is made to lift the core barrel. Different sizes of core may be recovered, but the common sizes are N size multi-leaf coring (triple core barrel) NMLC (52 mm) and H size HMLC (63 mm).

The core, when recovered, is placed in boxes (usually galvanised steel sheet boxes) in a standardised way. Each section of the box is approximately 1.05 m long and represents l m of core (50 mm extra is allowed for expansion). The core is placed in the first box and run from left to right (the top of the core being to the left) and continued onto the next row in the box (Figure 4.10). Once the first box is full, the core is placed into the second box and so on. All boxes are labelled to indicate the depths of the core contained in them.

The core is boxed in the following way:

1. Suppose a length of core is retrieved by drilling from 1.4 m depth to 2.6 m depth. The recovered core is placed in the core box and the depth of the bit at the end of the run is marked with a piece of wood or polystyrene foam which shows the depth, for example, 2.6 m.

Figure 4.9 Diamond-drilling bit being removed.


2. The core is removed from the barrel and placed in the box after measuring, say 1.0 m long.

3. This means that the core loss is 0.2 m (i.e. 2.6–1.4 = 1.2 m should have been obtained) and this is marked with a red piece of polystyrene (Figure 4.11). Generally, core loss is shown at the top of the run, but if its true position is known this may be indicated by a red marker in the correct position. Core loss can be due to cracks in the rock, voids, or soft material being washed away.

4.13 FIELD TESTS

Various tests may be carried out in the field to obtain soil data rather than taking samples back to a laboratory for testing. In some cases, field tests can produce better results than laboratory tests as the soil has been tested in situ and not disturbed. This is especially true for sands where sampling will change the void ratio of the sand, and unless the material can be reconstituted to the correct void ratio, results may not be accurate.

A

B

C

Figure 4.10 Method of boxing the cores.

Figure 4.11 Fractured core in a box with polystyrene markers showing core loss.


4.14 VANE SHEAR TEST

Shear vanes are used in soft clay deposits where sampling of the soil would be difficult or cause disturbance. The vane consists of two blades at right angles to each other (see Figure 4.12) and is pushed into the soil at the base of a borehole. Vanes come in various sizes and usually have a height to diameter ratio of 2.

A torque is applied to the vane and increased until the vane shears off a cylindrical volume of soil (which is contained within the blades). The undrained shear strength of the soil can then be calculated from the torque at failure T from

s

Td h d

u =+π 2 2 6( )/ /

(4.1)

where h is the height of the vane, d is the diameter, and su is the undrained shear strength of the clay.

Corrections can be applied to the field values of shear strength to give values which can be used for design purposes. The field values are multiplied by a factor λ to give the corrected design value. Values of λ have been given by Bjerrum (1972) and by Aas et al. (1986) and are shown in Figure 4.13. The vertical effective stress σ′v, is the present vertical effective pressure acting at any point in the soil.

4.15 STANDARD PENETRATION TEST

The standard penetration test first originated with the Raymond Pile Company in the United States in about 1927 and today is widely used in site investigation work. The test originally

Sheared cylindrical surface

Four-bladed vane

Rod

d

h

Figure 4.12 Shear vane.


involved hammering a thick-walled sampler into the soil so that a soil sample could be taken, but it was soon realised that the number of blows that were taken to drive the sampler into the ground were indicative of the density of sands or the stiffness of clays. The weight used to drive the sampler and the distance over which the sampler was driven 300 mm (1 foot) were eventually standardised.

The original test involved manually lifting a weight with a rope, and dropping the weight onto an anvil connected to the drill rods. More modern equipment uses an automatic trip hammer that does not have a rope attached and so results from the two types of equipment vary. International Standards have tried to make the test more repeatable by specifying the equipment that should be used and the method of performing the test.

4.15.1 Equipment

The equipment consists of a split sampler, rods, and falling weight that is used to drive the sampler. The equipment specifications are outlined in Australian Standard AS 1289.6.3.1 (2004), American Society for Testing of Materials ASTM D1586-11 (2011), and British Standards Institution BS EN ISO 22476-3 (2011) documents.

The test is performed at the bottom of a borehole and so auger drilling equipment (usually a truck-mounted auger) or a wash boring rig is required to drill the hole.

4.15.2 Sampler

The SPT sampler consists of a split cylindrical barrel of outside diameter 50 mm (2 inches) and inside diameter 35 mm, making the wall thickness about 7.5 mm. Because of the wall thickness, samples taken will be disturbed, and so are not to be used for tests requiring undisturbed samples (e.g. triaxial tests). At one end of the sampler, a cutting shoe is attached and at the other end is a ball valve adaptor. The ball valve consists of a ball bearing that will allow air and water to be forced out of the sampler as soil moves into the barrel, but will prevent a return of air or water if the sampler is lifted. This creates a suction above the

1.2

1.0

0.8

0.6

0.40 20 40 60 80 100

Plasticity index PI

Corr

ectio

n fa

ctor

(λ)

Corr

ectio

n fa

ctor

(λ)

0.6

0.2

1.0

1.4

(b)(a)

0.2 0.4su(field)/σv

0.6 0.8 1.00

Normallyconsolidated clay

Overconsolidated clay

′

Figure 4.13 Correction factors for vane shear strength. (a: After Bjerrum, L. 1972. Proceedings of the ASCE Conference on Performance of Earth-Supported Structures, Purdue University, Vol. 2, pp. 1–54.) (b: After Aas, G. et al. 1986. Proceedings, In Situ 86, American Society of Civil Engineers, pp. 1–30.)


sample, and helps break off and lift materials like soft clays that would otherwise slip out of the sampler. The sampler is shown in Figures 4.14 and 4.15.

4.15.3 Drive hammer

The drive hammer consists of a 63.5 kg (140 lb) weight that can free fall onto an anvil such that the drop is 760 mm (30 inches). The hammer must be able to fall freely onto the anvil and not jam on the guide or be restricted by ropes. The hammer is generally dropped by a self-tripping mechanism so that it does fall freely and delivers most of the energy from the drop to the sampler (called the ‘efficiency’). A trip hammer commonly used in the United Kingdom is shown in Figure 4.16.

One way to account for the efficiency of the hammer (and the effect of other factors) is to standardise the blow count to 60% efficiency. This is discussed in Section 4.15.5.

4.15.4 Rods

The rods that connect the anvil to the sampler should be stiff enough so that there is no loss of energy through bending and compression of the rods. The Australian Standard requires

50 mm

Sampler headDriving shoe

Split barrel

Four vents

Ball valve

35 mm

Figure 4.14 Split spoon sampler used in a standard penetration test.

Figure 4.15 Split spoon sampler showing thick-walled split tube and cutting shoe.


AW rod, with holes deeper than 15 m having steadies at intervals of 6 m or stiffer rod. The rod diameter should not exceed 66.7 mm.

4.15.5 Test procedure

The hole is first cleaned out making sure that in sands the water table is not allowed to fall. If the water table did fall, water could flow upwards into the borehole and loosen the soil (especially sand) where the test is to be conducted. The test should also be performed below any casing that is being used.

Trip mechanism

63.5 kg weight (140 lb)

Anvil

Lifting assembly

Figure 4.16 Drop hammer showing trip mechanism.


The sampler and rods are then lowered and three intervals of 150 mm (6 inches) are marked off along the drill rods. The sampler is driven 150 mm to seat the sampler, and then two further drives of 150 mm (making 450 mm in all) are performed. The sum of the blow counts is then made for the last two drives of 150 mm to get the blow count N.

For example, if the blow counts were 10, 13, and 14, then we add the last two to get N = 27 as the blow count.

If the full 450 mm is not penetrated, the blow count and the penetration achieved over the last two drives are recorded, for example, 50 blows over 200 mm. If not more than the first 150 mm is penetrated, the blows and distance are recorded, for example, 60 blows for initial 100 mm of penetration.

The test is usually stopped once 50 blows are recorded, however when soft rock or very stiff or dense materials are encountered, blow counts higher than 50 are recorded by some operators.

In order to account for energy losses due to

• The use of a cathead and winch plus rope• Loss of energy between the hammer and anvil• Differences in drop height, especially when a cathead is being used by an operator

the SPT blow count can be corrected to values for a system where 60% of the theoretical free-fall energy is imparted to the rods. This correlates to the safety hammer with cathead and winch that is commonly used in the United States and on which many correlations are based.

Correction of the raw SPT blow count Nfield to 60% efficiency N60 can be done by use of the following formula (Equation 4.2).

N ER Nm60 60= ( )/ field (4.2)

where ERm is the efficiency of the hammer in (%). This procedure has been examined by Kulhawy and Mayne (1990). When using charts and correlations with material properties, it is necessary to know if the correlation is performed with N60 or some other value.

To correlate the SPT blow count with some soil properties, it is necessary to correct the blow count for the effects of overburden pressure as well, for instance when assessing lique-faction potential of a soil (see Section 4.15.8). A curve giving the correction factor with depth (or effective stress) needs to be provided and the corrected blow count is then called (N1)60.

4.15.6 Properties of sands

The SPT test is able to indicate the relative density and the drained angle of shearing resis-tance of sands. The variation of the angle of shearing resistance ϕ′ with the corrected blow count (N1)60 (see Section 4.15.8) is shown in the plot of Figure 4.17.

The relationship in the figure is given by

′ = +φ [ . ( ) ] .15 4 201 600 5N

(4.3)

Tests carried out by different investigators have shown that the relative density of sands depends on the overburden pressure, and so plots can be made of relative density for sands in terms of the effective vertical stress and the SPT blow count N. This is shown in Figure 4.18 where results obtained by Gibbs and Holtz (1957) and Bazaraa (1967) are shown. The curves of Gibbs and Holtz tend to underestimate the relative density and so the plots of Bazaraa are more generally accepted.


4.15.7 Properties of clays

Because the SPT is a dynamic test, correlations of the blow count with soil properties for clays are approximate only, since the driving process can set up pore pressures in a clay and these will affect the blow counts.

Correlations have been made of the blow count with the undrained shear strength su of clays, but the relationship depends upon the plasticity of the clay. Clays with a lower

55

50

45

40

35

30

25

20

Fric

tion

angl

e ϕ ′

(deg

rees

)

SPT blow count (N1)60

ϕ′ = [15.4(N1)60]0.5+ 20°

Sand (SP and SP–SM)

Sand fill (SP to SM)SM (Piedmont)H&T (1996)

20 30 40 50 60100

Figure 4.17 Peak friction angle of sands from SPT resistance. Note: The normalised resistance is (N1)60 = N60/(σ′vo/pa)0.5, where pa = 100 kPa = 1 bar = 1 tsf. (After Hatanaka, M. and Uchida, A. 1996. Soils and Foundations, Vol. 36, No. 4, pp. 1–10.)

SPT blow count (N1)60

Gibbs and Holtz Bazaraa

Vert

ical

pre

ssur

e (kP

a)

0 10 20 30 40 50 60 70

50

100

150

200

250

0100

100

Index

Density

in %

60

80

40

4060

80

Figure 4.18 Relative density of sands as a function of effective vertical stress.


Plasticity index PI has a higher undrained shear strength for a given blow count than a high plasticity clay. Charts like shown in Figure 4.19 (which is based on data from several inves-tigators) show the large spread in results for clays of different plasticities and so need to be used with caution. Figure 4.20 also shows the variation of undrained shear strength with the plasticity of a clay.

Sowers

(high

PI) lo

wer bo

und

Sanglerat (sandy clay); T

erzaghi and Peck

Sowers (medium PI) u

pper bound

Chicago clay (Schmertmann)

Golder

Sanglerat (silty clay)

Sowers (low PI) upper bound

Sanglerat (clay); Illinois (Loess);Golder

Houston (U

SBR)

clay

India

Yugoslavia

20

15

10

5

25

00 0.5 1.0

Su/pa

1.5 2.0

SPT

N va

lue

Figure 4.19 Correlation of undrained shear strength of clays su to SPT blow count N (pa = 100 kPa = 14.5 psi = 1 tsf).

×

×

×××

Plasticity index (%)

s u/N

(kPa

)

20 301000

2

4

6

8

10

40 50 60 70

Boulder clayLaminated claySunnybrook tillLondon clay

Oxford clayBracklesham beds Finz

Kimmeridge clayWoolwich and Reading clayUpper-Lias clayKeuper marl

Figure 4.20 Ratio of undrained shear strength to SPT blow count versus plasticity index (PI). (After Stroud, M.A. 1974. Proceedings European Symposium on Penetration Testing, ESOPT, Stockholm 1974. National Swedish Building Research, Vol. 2.2, pp. 367–375.)


Correlations based on local knowledge can give a reasonable correlation once a database of information is built up.

4.15.8 Liquefaction

If cohesionless soils that are saturated are vibrated very rapidly (e.g. by an earthquake), excess pore pressures can be generated and this leads to the phenomenon of liquefaction. This is an important consideration when designing structures in areas prone to earthquake, as foundation liquefaction has led to widespread damage. The Niigata earthquake of 1964 is an example.

Soils most susceptible to liquefaction are loose soils of uniform grain size. Fluvial depos-its, and colluvial and aeolian deposits or man-made hydraulic fills when saturated are most prone to pore pressure build up that will lead to liquefaction.

SPT tests have commonly been used in the United States and many other countries to determine whether a soil deposit is liable to liquefy. Seed et al. (1985) have presented a plot of the corrected blow count versus cyclic shear stress ratio for clean sand (Figure 4.21). In the figure, the cyclic shear stress ratio CSR at which the soil will liquefy under an earth-quake of magnitude M = 7.5 is indicated by a solid black symbol, and where it does not, by an open symbol. The data are based on field data from Japan, China, and Pan America and it can be seen that above the line on the plot, the soil generally liquefied and below the line, liquefaction did not generally occur.

Because the data were collected from many different countries using different techniques for performing the SPT, it was necessary to correct the blow counts to an energy ratio of 60% or N60 for use with the charts (see Table 4.1). The corrected blow count must also be determined for a standard effective overburden pressure of 1 ton/ft2 (approx. 100 kPa) and so is called (N1)60. The correction factor CN for determining the overburden correction is shown in Figure 4.22. The corrected blow count can then be found from the measured blow count N as follows

N C NN1 = (4.4)

To use the chart, therefore, a knowledge of the corrected SPT blow count (N1)60 and the cyclic shear stress ratio CSR is required. The cyclic shear stress ratio may be computed at any depth required from the equation

τσ

σσ

av max

′= ⋅

′⋅

v

v

vd

ag

r0

0

00 65.

(4.5)

whereamax is the maximum acceleration at the ground surfaceg is the acceleration due to gravityσv0 is the total overburden pressure at the depth considered

′σv0 is the effective overburden pressure at the depth consideredrd is a stress reduction factor that varies as shown in Figure 4.23

For sands containing more than 5% fines, liquefaction is less likely to occur. For this case, Seed et al. (1985) have presented an alternative chart for determining liquefaction potential as shown in Figure 4.24.


(N1)60

0 10 20 30 40 500

0.1

0.2

0.3

0.4

0.5

0.6

CSR M

=7.5

Fines content ≤ 5%Chinese building code (clay content = 0)

Pan-American dataJapanese dataChinese data

MarginalLiquefaction

NoLiquefactionLiquefaction

Figure 4.21 Relationship between cyclic stress ratios causing liquefaction and (N1)60 values for clean sands in M = 7.5 earthquakes. (After Seed, H.B. et al. 1985. Journal of Geotechnical Engineering, ASCE, Vol. 111, No. 12, pp. 1425–1445.)

Table 4.1 Energy ratios for different test conditions

Country (1)Hammer type (2) Hammer release (3)

Estimated rod energy (%) (4)

Correction factor for 60% rod energy (5)

Japana Donut Free fall 78 78/60 = 1.30Donutb Rope and pulley with special throw release 67 67/60 = 1.12

United Statesb Safetyb Rope and pulley 60 60/60 = 1.00Donut Rope and pulley 45 45/60 = 0.75

Argentina Donutb Rope and pulley 45 45/60 = 0.75China Donutb Free fallc 60 60/60 = 1.00

Donut Rope and pulley 50 50/60 = 0.83

Source: Adapted from Seed, H.B. et al. 1985. Journal of Geotechnical Engineering, ASCE, Vol. 111, No. 12, pp. 1425–1445.a Japanese SPT results have additional corrections for borehole diameter and frequency effects.b Prevalent method in the United States today.c Pilcon type hammers develop an energy ratio of about 60%.


CN

DR = 60%–80%

DR = 40%–60%

0.40 0.8 1.2 1.6

400

300

200

100

0

Effec

tive o

verb

urde

n pr

essu

re (k

Pa)

Figure 4.22 Chart for correction values CN.

1.0

Dep

th (m

)

Average values

Range for different soil profiles

Dep

th (f

t)

Stress reduction factor (rd)

5

0

20

0

10

15

20

25 80

60

40

100300.2 0.4 0.6 0.80

Figure 4.23 Reduction factor rd to estimate the variation of cyclic shear stress with depth below level or gently sloping ground surfaces. (After Seed, H.B. and Idriss, I.M. 1971. Journal of the Soil Mechanics and Foundations Division, ASCE, Vol. 97, No. SM9, pp. 1249–1273.)


4.16 PRESSUREMETERS

Pressuremeters are basically devices which consist of a cylindrical rubber membrane which can be lowered into a borehole and inflated against the sides of the borehole. Measurements can be taken of the radial expansion of the membrane as it is inflated by gas pressure (water or oil is used in some types).

4.16.1 Types of pressuremeters

The types of pressuremeters commonly in use are

1. The Ménard type which is lowered into a pre-formed borehole. The test method is described in ASTM D4719-07 (2007) or British Standard BS EN ISO 22476-4 (2012).

2. The self-boring pressuremeter (or Camkometer) which has a cutting head to enable the instrument to be inserted into the ground at the bottom of a borehole caus-ing very little disturbance. Such a device is shown in Figure 4.25. The name of the device comes from its original name the Cambridge K0 meter as it was developed at Cambridge University and could be used to measure the coefficient of earth pressure at rest K0.

Modified Chinese code proposal (clay content = 5%)

Pan-American dataJapanese dataChinese data

MarginalLiquefaction

NoLiquefactionLiquefaction

0 10 20 30 40 500

0.1

0.2

0.3

0.4

0.5

0.6

(N1)60

CSR M

=7.5

Percent fines = 35 15 ≤5

Fines content ≥ 5%

Figure 4.24 Relationship between cyclic stress ratios causing liquefaction and (N1)60 values for silty sands in M = 7.5 earthquakes.


3. Push in pressuremeters which are pushed into the ground at the bottom of a borehole. 4. The main advantage of the device is that it is able to test soil in a relatively undisturbed

state (especially the Camkometer), however the disadvantage is that the membrane is being expanded laterally and so the response of the soil is being obtained laterally.

4.16.2 Interpreting test results

A plot is made of the radial strain of the cavity against the pressure inside the membrane. Such a plot is shown in Figure 4.26 where it can be seen that there is initially very little strain

A

Total pressure cell

Pore pressure cell

Electric/gas cable

Feeler

CutterTapered cutting head

Flushing waterReturn flow

Water flush

Rubber membrane

Ax

Figure 4.25 The Camkometer.

8

6

4

4 6 8

2

200

1410 12 16 18 20Cavity strain (%)

Pres

sure

p (M

Pa)

Figure 4.26 Plot of cavity strain versus pressure for pressuremeter showing unload – reload loops.


as the pressure increases back to the geostatic lateral stress existing in the ground σh = K0σv at which point the cavity begins to expand.

Usually, some load and reload loops are carried out so that the reload modulus of the soil can be obtained as this is generally accepted to be the most appropriate value for design.

a. Shear modulus The shear modulus G of the soil may be computed from the formula

G

dpd

dpdc c

=

≈

12

120

ρρ ε ε

(4.6)

where p is the internal membrane pressure εc is the cavity strain (Δρ/ρ0) and the term ρ/ρ0 is the ratio of the expanded radius of the cavity to the initial radius and may be approximated as unity.

The shear modulus may therefore be found from the plot of cavity strain versus pres-sure since it is merely 1/2 the slope of the curve as may be seen from Equation 4.6.

b. For clays, the undrained shear strength may be found from Gibson and Anderson’s (1961) method of analysis. A plot is made of the natural logarithm of the volume strain ΔV/V of the cavity versus the pressure.

Since

p p s

VV

L u= +

ln∆

(4.7)

the slope of the plot will give the undrained shear strength of the clay. This is shown in Figure 4.27 where it can be seen that the theoretical limit pressure pL occurs where Δ V/V = 1.

4.17 DILATOMETERS

This instrument (which is shown in Figure 4.28) was developed by Marchetti in Italy (Marchetti 1980). It consists of a flat stainless steel blade 14 mm thick and 95 mm wide

0.1

PL = 7.5 MPa

10.010.001

1

2

ΔV/V

Pres

sure

p (M

Pa)

3

4

5

6

7

8

Figure 4.27 Plot of log ΔV/V versus membrane pressure for pressuremeter test in very dense silty sand.


which is pushed into the ground. A 60 mm diameter membrane which is on the side of the instrument is inflated by gas pressure (see Figure 4.29) after the dilatometer has been pushed to the required depth. Tests have to be performed at discrete intervals of depth and so tests are usually conducted at distances of 0.15–1.3 m apart.

The pressure required to inflate the membrane, p0 and the pressure required to lift the membrane 1 mm p1 are recorded (and corrections for the instrument are applied).

Flexible membrane

95 mm

60 mm

Pneumatic tubing

Wire

Flexible membrane

1.1 m

P0 P1

14 mm

Figure 4.28 Marchetti dilatometer.

Figure 4.29 Marchetti dilatometer with pressure gauges.


More details for performing the test may be found in British Standard BS EN ISO 22476-5 (2012).

The following quantities are then calculated:

i. The material or deposit index ID

I

p pp u

D = −−

1 0

0 0 (4.8)

where u0 is the static water pressure.

ii. The horizontal (or lateral) stress index KD

K

p uD

v

= −′

0 0

0σ (4.9)

where σ′v0 is the in situ vertical effective stress.

iii. Dilatometer modulus

E p pD = −34 7 1 0. ( ) (4.10)

4.17.1 Type of soil

The type of soil can be identified by using the chart developed by Marchetti and Crapps (1981). The chart makes use of the dilatometer modulus ED and the material index ID to identify the soil type (see Figure 4.30).

4.17.2 Shear strength of clays

The undrained shear strength of a clay su may be calculated from (Marchetti 1980)

s Ku v D= ′0 22 0 501 25. ( . ) .σ (4.11)

If ID ≤ 1.2; if ID > 1.2 the soil is cohesionless.

4.17.3 Other quantities

Other soil properties such as K0 (the coefficient of earth pressure at rest), ϕ′ (the angle of shearing resistance) can be found from the test (Marchetti 1980).

The formula for computing K0 in clay is given by Equation 4.12.

K

KD0

0 47

1 50 6=

−.

..

(4.12)

The angle of shearing resistance of sand can be found from Marchetti’s equation (Marchetti 1997).

′ = ° + ° − °φDMT D DK K28 14 6 2 1 2. .log log (4.13)

More relationships of soil properties with the dilatometer test results can be found in the book by Schnaid (2009).


4.18 CONE PENETROMETERS

Cone penetrometers can be of two types: mechanical or electrical. For cone penetration test-ing, it is not necessary to drill a hole as the cone is pushed into the ground using a hydraulic ram. Cones can be mounted on small trailers that are anchored to the ground by screw-in anchors, or can be truck mounted using the weight of the truck as a reaction against which to push the cone into the soil.

4.18.1 Equipment

Most of the modern cone penetrometers consist of a cone which has a 60o apex angle and a diameter of 35.7 mm (giving a cross-sectional area of 1000 mm2) as well as a friction sleeve of 10,000 or 15,000 mm2 in surface area. Cones also come in other sizes, but these dimen-sions are the more commonly used.

Cone specifications and the procedures that should be used during testing are outlined in Australian Standard AS1289.6.5.1-1999 for both the mechanical cones and the elec-tric cones, in ASTM D 5578-95 for electric friction cones and piezo-cones, or in British Standard BS EN ISO 22476-1 (2012) for piezo-cones and electric cones.

The older mechanical cones are more rugged than the electric cones, but the advantages of the electric cone are

• Better accuracy and repeatability of results, particularly in weak soils• Better delineation of thin strata• Faster speed of operation

1 MPa 10

20

50

100

200

500

1000

12

0.15

0.2 0.5 1 2 5 10

0.3 0.6 0.9 1.2 1.8 3.3

Material index ID

Dila

tom

eter

mod

ulus

ED

/pa

Very softClay/peat

CLAY

Hard

(2.05)

(1.90)

Medium

(1.80)

Soft(1.80)

(1.70)

Silty(2.10)

(1.95)

(1.80)

(1.70)

(1.60)

(1.80)

(2.00)

Loose

(1.70)

Medium

(1.90)

SILTClayey sandy

SAND

Very dense (2.10)Silty

Notes:a – Number in parenthesis

is normalised unit weight (γ /γw)

b – If PI > 50, (γ /γw) isoverestimated by about 0.1

Figure 4.30 Chart for identification of soil type using a dilatometer.


• It is possible to measure pore pressure on some devices• Continuous electronic recording of data

Two types of cones are available, those with pore pressure measurement devices used for CPTu (cone penetration test with pore pressure measurement) tests and those without pore pressure measurement used for CPT tests (cone penetration testing).

4.18.2 CPT equipment

The older mechanical cones are pushed into the soil using hydraulic pressure and readings of the pressure versus the depth are recorded. These devices can have a conical tip that is advanced by an inner rod while the outer rods are held stationary. Cones can also have a sleeve (as shown in Figure 4.31). By pushing an inner rod, the cone is firstly advanced, and then the sleeve is advanced. The sleeve friction is obtained by subtracting the cone pressure from the combined pressure measured for both the cone and the sleeve.

Electric cones are constructed as shown in Figure 4.32 where it may be seen that the cone tip is connected to a load cell with strain gauges being used to measure the strain. The output from the strain gauges can be calibrated so as to give a direct readout of the cone value qc that is commonly expressed in MPa. A cone with the sleeve and tip removed is shown in Figure 4.33.

The friction sleeve is also connected to a load cell so that the friction on the sleeve of the penetrometer can be automatically recorded.

4.18.3 CPTu equipment

As mentioned above, some electric cones have the ability to measure pore pressures, and this provides extra useful information. Such cones are called piezo-cones, and can have the pore pressure sensors mounted in different locations, at the tip, on the face of the cone, behind the cone tip, or behind the sleeve of the cone. This is shown in Figure 4.34a,b. The response to pore pressure is very different depending on the location of the filter and trans-ducer. The highest pressures are measured when the piezo-element is mounted at the face of the cone. The pore pressure depends on the type of soil, with highly overconsolidated soils producing higher pressures at the face.

Figure 4.31 Mechanical cone showing friction sleeve.


A small pressure sensor is mounted behind the porous element to measure pressure changes. It should be such that a minimal volume change in fluid passing the filter is required to operate the transducer to give a fast response time.

In addition (for a fast response time in the pore pressure measuring system), the fluid used should have a low viscosity and compressibility, and a high permeability filter should be used. The filter itself should not be compressible, as this leads to filter compression effects, that is, the compressing filter would cause an increase in pore pressure.

The filter can be made of porous plastic, ceramic, or sintered stainless steel. Ceramic filters are generally damaged when pushed into dense sands, but polypropylene can survive being pushed into dense sands and gravelly soils.

Pore pressures measured in the CPTu test rely very strongly on the filters being satu-rated, and not containing any air. General practice is to saturate the filter elements in the

Cone

‘O’ ring seal

Cone strain gauge load

Friction sleeve

Friction sleeve strain gauge load cell

‘O’ ring seal

Signal cable

Figure 4.32 Typical electric friction cone penetrometer.


laboratory by placing them under a high vacuum. Some cone users submerge filters in warm glycerin in an ultrasonic bath under a vacuum. The voids in the cone itself are de-aired by flushing with a suitable fluid (e.g. either glycerin or water) and the cone can be kept de-aired by placing a latex sheath over it while it is placed on the push rods.

Results of a piezo-cone test are shown in Figure 4.35 where the cone resistance (pressure on the end of the cone), the friction ratio (see Equation 4.14), and the pore pressure are plot-ted with depth. The type of soil can be identified from these results as in this plot, the low friction ratios correspond to sand and silty sand, whereas the high friction ratios (at about 3 m and 7.7 m) correspond to clay layers. At the location of the clay layers, the pore pres-sures measured also show high values.

4.18.4 Pushing equipment

Truck-mounted rigs are generally specifically built for the equipment, but sometimes anchored trailer mounted rigs are used. The thrust capacity required is usually between 10 and 20 tonnes, although a capacity of 10 tonnes is enough to carry out most penetration testing to 30 m depth. Trucks ballasted to about 15 tonnes are suitable for this type of testing.

Hydraulic power for pushing the cones comes from the truck engine and the cone must be pushed at the standard rate of 20 mm/s (AS 1289.6.5.1-1999 allows 10–20 mm/s). Sections of the rods are added as the cone is pushed deeper, and rigs of this sort can produce up to 250 m of testing in one day.

Friction reducers that consist of an expanded coupling can be used behind the cone to reduce friction on the rods, or natural or polymer drilling fluid can be pumped down the drill rods and allowed to flow up the outside of the rods about 1.5 m behind the cone to reduce friction and therefore pushing force.

4.18.5 Calibration

The cones are normally calibrated by using a load cell before each sounding is made. The calibration should be done with O-rings and seals in place as would be the case when the cone is in use.

Figure 4.33 Electric cone with friction sleeve and the cone tip removed.


Temperature can have an effect on the calibration and this can be overcome by pushing the cone into the ground about 1 m and leaving it for about half an hour before calibration.

Pore pressure calibration should be done with a pressure chamber that completely encloses the cone and is sealed at a point above the friction sleeve. Measurement of the tip stress and friction sleeve stress at applied pore pressures will allow direct determination of unequal

(a)

(b)

Piezo-element behind friction sleeve

Friction sleeve

Piezo-element behind tip

Piezo-element on face

Electronics housing

Cone tip

Figure 4.34 (a) Location of piezo-elements on piezo-cone. (b) Piezo-cone showing possible locations of piezo-elements.


end effects (i.e. loads caused by pore pressures that may affect the cone readings such as water pressure at the back of the cone tip).

4.19 INTERPRETATION OF CONE DATA

Cone data can be used for estimating many different soil properties and the type of soil through which the cone is being pushed. The cone values are affected by the lateral in situ stress and this should be considered where possible. However, this information is not often available, and cannot be applied in the interpretation.

4.19.1 Soil classification

Charts such as the ones developed by Douglas and Olsen (1981) or Robertson et al. (1986) (for a standard CPT) can be used to classify the soil by the use of the friction ratio Rf and the

qt (MPa) Rf (%)

Cone resistance

u (kPa)

Friction ratio Pore pressureD

epth

(m)

Dep

th (m

)

Dep

th (m

)

0 2 4 6 8 10 0 50 100 150 2000 4 8 12 16

12

13

11

10

9

8

7

6

5

4

3

2

1

0

In situ water level

1 1

0 0

2 2

3 3

13 13

12 12

11 11

10 10

9 9

8 8

7 7

6 6

5 5

4 4

Figure 4.35 Pore pressure distributions for a piezo-cone – different soil types.


cone resistance qc. The friction ratio is the ratio of the sleeve friction fs to the cone resistance expressed as a percentage

R f qf s c= ( )/ %

(4.14)

From such a chart, the soil type can be identified as can be seen from Figure 4.36.However, the cone resistance depends on the depth of the test and at shallow depths and

depths over 30 m the values measured are affected by overburden stress and soils may not classify correctly using the chart of Figure 4.36. Therefore, Robertson (1990) proposed a normalised chart that could be used with both CPT and CPTu tests where a normalised cone resistance Qt is used.

Q

qt

t v

v= − σ

σ0

0′ (4.15)

Zone qc/N Soil behaviour type(1) 2 Sensitive fine grained(2) 1 Organic material(3) 1 Clay(4) 1.5 Silty clay to clay(5) 2 Clayey silt to silty clay(6) 2.5 Sandy silt to clayey silt(7) 3 Silty sand to sandy silt(8) 4 Sand to silty sand(9) 5 Sand(10) 6 Gravelly sand to sand(11) 1 Very stiff fine grained*(12) 2 Sand to clayey sand*

* Overconsolidated or cemented

Friction ratio (%) Rf

Cone

bea

ring

pres

sure

qc i

n ba

rs

Undrained

Partially drained

3

2

1

45

6

78

9

10

11

12

1000

100

10

110 2 3 4 5 6 7 8

Drained

Figure 4.36 Simplified soil classification chart for a standard electric friction cone. (After Robertson, P.K. et al. 1986. Proceedings of In Situ 86, ASCE Specialty Conference, Blacksburg, Virginia.)


In the formula, the cone value is corrected for the effect of pore pressure that acts down-wards on the cone tip if there is a piezo-element behind the tip (Campanella and Robertson 1988). The correction is given by

q q a ut c= + −( )1 (4.16)

where a is the area ratio of the cone and u is the water pressure acting at the piezo-element. a = d2/D2 where the diameters are shown in Figure 4.37.

Soil type can also be found based on pore pressure measured in the CPTu test by charts such as the chart presented by Robertson 1990 (see Figure 4.38). In the chart, the pore pres-sure parameter ratio Bq is defined as

B

uq

qt v

=−∆

σ 0 (4.17)

whereΔu is excess pore pressure measured behind the tip u − u0

qt is cone resistance corrected for pore pressure effectsσv0 is the total overburden stress

The pore pressure data are not always reliable for classification and so it is recommended that the pore pressure ratio chart be used in conjunction with the friction ratio chart in identifying soil type.

4.19.2 Relative density of sands

Relative density of sands is a function of the effective stress state of the sand, and can be determined from the cone resistance qc. Relationships have been developed from calibra-tion chamber work for predominantly quartz sands (Jamiolkowski et al. 1985) as shown in Figure 4.39.

The compressibility of the sand has an effect on the result as can be seen from the figure, and it should be noted that the results are only applicable to normally consolidated, uncemented, unaged predominantly quartz sands. The cone resistance and vertical effective stress used with the chart should be expressed in tonnes/m2, as the ratio qc/(σ′)0.5 is not non-dimensional.

u

d

D

Piezo-element

Friction sleeve

Figure 4.37 Area acted upon by water pressure at the cone tip.


For young uncemented silica sands, Kulhawy and Mayne (1990) suggest that the relative density can be found from the more simple formula

D

Qr

tn

v

2

0350=

′σ (4.18)

and Qtn = (qt/pa)/(σ′v0/pa)0.5 is the normalised cone value and pa = 100 kPa is the atmospheric pressure used in the non-dimensionalisation.

The value of 350 is closer to 300 for fine sands and 400 for coarse sands.

4.19.3 Friction angle of sands

The friction angle for sands changes depending on the compressibility of the sand and so no unique relationship between cone resistance and angle of shearing resistance exists. However, for sands that lie in the zones 7,8,9 of the classification chart of Figure 4.36, the angle of friction can be estimated by using the chart of Figure 4.40. For overconsolidated sands, this figure may overestimate the angle of friction by about 2o.

Alternatively, the angle of friction ϕ′ for clean rounded uncemented quartz sands can be found from the following simple expression (Kulhawy and Mayne 1990).

′ = ° +φ 17 6 11. ( )log Qtn (4.19)

where Qtn is defined in Equation 4.18.

2

1

34

5

6

78

9

3

4

OCR

St

5

6

7

1

10

100

1000 1000

100

10 0.4 0.8 1.21 100.1

Normalised friction ratio × 100%

Nor

mal

ised

cone

resis

tanc

e

1

=

Pore pressure ratio Bq

10

u0 u

u − u0

σv0 qt

q t – σ

v0

qt – σv0

σv0′

fs

φ′ Normally consolidated

Increasi

ng

OCR, age,

cementat

ion

Increasi

ng

OCR and ag

e

Increasing

sensitivity

qt – σv0

Bq

1. Sensitive, fine-grained soils 6. Sands – Clean sand to silty sand2. Organic soils – Peats 7. Gravelly sand to sand3. Clays – Clay to silty clay 8. Very stiff sand to clayey sand*4. Silt mixtures – Clayey silt to silty clay 9. Very stiff, fine grained*5. Sand mixtures – Silty sand to sandy silt

* Heavily overconsolidated or cemented

Figure 4.38 Proposed soil behaviour type classification system from CPTu data. (After Robertson, P.K. 1990. Canadian Geotechnical Journal, Vol. 27, No. 1, pp. 151–158.)


Rela

tive d

ensit

y D

r (%)

Moderate compressibility sands

= −98 + 66 log10 [ ′ ]0.5

Edgar sandHilton mine sandHokksund sandOttawa 90 sandTicino sand

Probable lower limit (low compressibility sands)

Probable upper limit (high compressibility sands)

2SD

2SD

SD = standard deviation

expressed in tonnes/m2

10 20 40 60 80 100 200 400 600 800 1000

25

15

35

45

55

65

75

85

95

100

Drqc

qc

σv0

[ ′ ]0.5qc

σv0

′σv0

Figure 4.39 Relationship between relative density and cone resistance of uncemented, normally con-solidated quartz sands. (After Jamiolkowski, M. et al. 1985. Theme Lecture, 11th International Conference on Soil Mechanics and Foundation Engineering, San Francisco, Vol. 1, pp. 57–153.)

Cone pressure qc bars

Vert

ical

effec

tive s

tres

s v0′

bars

1 bar = 100 kPa

φ ′= 48°

46°

44°

42°

40°38°36°34°32°

30°

100 200 300 400 5000

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

σ

Figure 4.40 Proposed correlation between cone resistance and peak friction angle for uncemented quartz sands. (After Robertson, P.K. and Campanella, R.G. 1983. Canadian Geotechnical Journal, Vol. 20, No. 4, pp. 718–733.)


4.19.4 Constrained modulus of sands

The constrained modulus Mt of a sand is the modulus under one-dimensional conditions and is equal to 1/mv (mv is the coefficient of volume change as found from an oedometer test – see Section 1.5).

Charts have been presented by Baldi et al. (1981) that enable the constrained modulus to be found from the cone resistance (noting that 1 bar = 100 kPa) as shown in Figure 4.41. This chart only applies to uncemented, normally consolidated quartz sands.

4.19.5 Young’s modulus of sands

A relationship similar to that developed for the constrained modulus has been developed for the secant elastic modulus of a sand (Baldi et al. 1981). The secant modulus depends on the load level and so the secant modulus is given for load levels of 25% and 50% of the ultimate load in Figure 4.42.

It may be seen from Figure 4.42 that the assumption that E = 2qc (i.e. the straight line on the plot) is not a bad average of the data for normally consolidated sands. This is the assump-tion made by Schmertmann (1970) in his original approach for calculating settlement of surface footings. He modified this to E = 2.5 or 3.5qc in his later work (Schmertmann et al. 1978) to allow for the shape of footings (i.e. whether square or strip). This is discussed in Section 5.5.5.

4.19.6 Undrained shear strength of clays

A cone being pushed through a clay is like a deep footing loaded to cause failure of the soil. We should therefore expect that the cone resistance can be expressed by a formula like the bearing capacity formula

q N s N qu c u q s= +

(4.20)

Cone bearing pressure qc in bars

Medium dense, Dr = 46%Dense, Dr = 70%Very dense, Dr = 90%

Baldi et al. (1981)Normally consolidated Ticino sand

Cons

trai

ned

tang

ent m

odul

us M

t in

bars

1000 200 300 400 500

0.5 bar

єvє

1 bar

2 bars

4 bars

v0′

′

= 8 bars

Mt = qc

Mt = 3qc Mt

2000

1500

1000

500

0

σv′σv

Figure 4.41 Relationship between cone resistance and constrained modulus for normally consolidated, unce-mented quartz sand. (After Baldi, G. et al. 1981. Proceedings of the 10th International Conference on Soil Mechanics and Foundation Engineering, Stockholm, Vol. 2, pp. 427–432.)


whereNc and Nq are the bearing capacity factors (Nq = 1 for a clay)su is the undrained shear strength of the clayqs is the surcharge

Here, the surcharge is the total vertical stress σv0 at cone level and so we can write

s

qN

uc v

k

= − σ 0

(4.21)

Nk is a bearing factor like Nc and is generally obtained from empirical correlations.For normally consolidated marine clays, Nk ranges between 11 and 19 with an average

of 15.For non-fissured overconsolidated clays, Nk is about 17; for stiff fissured marine clays, it

ranges from 24 to 30; and for glacial clays, it ranges from 14 to 22.Values for Nk should be developed for individual areas based on local experience and cor-

relations with measured su values. In Sydney, values of Nk in the range 12–15 are used for normally consolidated clays and values of 15–30 for overconsolidated clays.

4.19.7 Undrained modulus of clays

As for sands, the undrained modulus of a clay Eu depends on the stress level and so it may be estimated at say 25% of the failure load.

One way to obtain the undrained modulus of a clay is to relate the modulus to the und-rained shear strength su that can be determined from the cone as discussed in Section 4.19.6. The ratio of Eu/su depends on the overconsolidation ratio and the plasticity index of the clay and so a knowledge of su is not enough to find the undrained modulus.

Cone bearing pressure qc bars

Dra

ined

seca

nt Y

oung

’s m

odul

us at

50

% fa

ilure

stre

ss le

vel E

50 (b

ars)

0 100 200 300 400 500

Dra

ined

seca

nt Y

oung

’s m

odul

us at

25%

failu

re st

ress

leve

l E25

(bar

s)

E25 = 2qc

v0 = 4 bars

2 bars

1 bar

0.5 bar

Medium dense, Dr = 46%Dense, Dr = 70%Very dense, Dr = 90%

Baldi et al. (1981)Normally consolidated Ticino sand

0.5σDmax0.25σDmax

σD

σD

σDmax100

200

300

400

500

600

0

150

300

600

900

750

450

0єa

є50є25

єa

′σ

Figure 4.42 Secant Young’s modulus values for uncemented, normally consolidated quartz sands. (After Baldi, G. et al. 1981. Proceedings of the 10th International Conference on Soil Mechanics and Foundation Engineering, Stockholm, Vol. 2, pp. 427–432.)


The recommended procedure is therefore

1. Obtain the value of su for the clay from the cone resistance as described in Section 4.19.6. 2. Estimate the stress history by computing (su/σ′v0)NC for a normally consolidated soil

from Skempton’s (1957) chart (Figure 4.43). 3. Calculate the actual su/σ′v0 for the overconsolidated soil. 4. Use the relationship of Figure 4.44 to estimate the OCR of the deposit. 5. Use the relationship of Duncan and Buchignani 1976 (Figure 4.45) to obtain the ratio

of Eu/su and hence Eu (that is really a value of Eu at 25% maximum load), because these values are for working loads that are at about 1/4 of the failure load.

4.19.8 Permeability

The permeability of a soil can be found by performing a dissipation test with a piezo-cone. The cone is advanced into the soil and this will generate a pore pressure. The cone is then

= 0.11 + 0.0037 Iw

Plasticity index Iw

Skempton (1957)Ladd and Foott (1974)

0 20 40

0.2

0.4

0.6

60 80 100 1200

s u/σ v0′

su /σv0′

Figure 4.43 Statistical relation between su/σ′v0 and plasticity index for normally consolidated clays.

max. pastpresentOCR = Overconsolidation ratio

Range of data for 7 NC and OC clays, with recommended average

1 1.5 2 3 4 5 6 7 8 9 101

2

3

4

5

6

σvm′ σv0′

s u/σ v0′

(s u/σ v0

) NC

′

Figure 4.44 Ratio of su/σv0 for overconsolidated soil to that for normally consolidated soil versus overconsolidation ratio.


held stationary and the reduction of the excess pore pressure with time is recorded. A plot can be made of the percentage dissipation of excess pore pressure with time and the time at which 50% dissipation occurred is noted (called t50). As the permeability is dependent on soil stiffness, a chart can be plotted which gives the permeability versus t50 for various nor-malised cone resistances as shown in Figure 4.46 (Robertson 2010). The chart is for cones where the piezo-element is located behind the cone tip as shown in the inset to Figure 4.46 (qt is defined in Equation 4.16 and Qt in Equation 4.15).

4.20 LIQUEFACTION POTENTIAL

In a similar fashion to the methods used for evaluating liquefaction potential using SPT data, cone data may be used. The approach presented by Stark and Olsen (1995) is to com-pute the corrected cone value qc1 to correspond to the cone value at 100 kPa (1 ton/ft2) by the equation

q C qc q c1 = ⋅

(4.22)

The overburden correction factor Cq can be calculated from

Cq

v

=+ ′ ′

1 80 8 0

.. ( )σ σ/ ref

(4.23)

500

1000

1500

01 2 3 4 5 6 7 8 9 10

E u/s u

Overconsolidation ratio

Ip > 50

30 < Ip< 50

Ip < 30

Figure 4.45 Ratio of undrained Young’s modulus to shear strength against overconsolidation ratio. (After Duncan, J.M. and Buchignani, A.L. 1976. An Engineering Manual for Settlement Studies. Department of Civil Engineering, University of California, Berkeley.)


whereσ′ref is a reference stress equal to one atmosphere (100 kPa)σ′v0 is the vertical effective stress at the depth of interest

The corrected cone value can then be used with the chart of Figure 4.47 (for clean sands) to determine whether the sand deposit is liable to liquefy. To do this, the seismic shear stress ratio called the SSR needs to be computed. This can be found from

SSR av max=

′= ⋅

′⋅τ

σσσ0

0

0

0 65.a

grv

vd

(4.24)

whereamax is the maximum acceleration at the ground surfaceg is the acceleration due to gravityσv0 is the total overburden pressure at the depth considered

′σv0 is the effective overburden pressure at the depth consideredrd is a stress reduction factor that may be computed fromrd = 1 − (0.012z) and z is the depth in metres

Soils with SSR values above the line will have the potential to liquefy (in an M = 7.5) earthquake, whereas those below will not.

For soils that contain some fines, Stark and Olsen (1995) have presented the chart shown in Figure 4.48. It may be seen from the chart that sands with >5% fines have a greater resis-tance to liquefaction because it is more difficult for collapse of the sand structure to occur.

qtQtn ≥ 14

Qtn = 2

Qtn =qt − σv0

u2

356 mm (10 cm2)

v0′

′

σv0′

= 100 kPa

t50 (min)

Perm

eabi

lity k

(m/s

)

0.1

5

10

1 10 100 1000 10,000

10–11

10–12

10–10

10–9

10–8

10–7

10–6

10–5

σ

v0 = 50 kPaσ

Figure 4.46 Permeability as a function of time for 50% consolidation t50.


4.21 GEOPHYSICAL METHODS

Geophysical methods offer a rapid means of determining subsoil information by measuring some property such as the speed of seismic waves or the electrical resistivity of the soil. As such, the methods are largely non-destructive, that is, holes do not have to be dug or drilled.

Seism

ic sh

ear s

tres

s rat

io

0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6

M = 7.5

Liquefaction No liquefaction

Proposedrelationship

0.25 < D50 (mm) < 20F.C. (%) ≤ 5

Field performanceLiquefactionNo liquefaction

Corrected CPT tip resistance qc1 (MPa)

Figure 4.47 Relationship between seismic shear stress ratio (SSR) triggering liquefaction and qc1 values for clean sand and M = 7.5 earthquake.

M = 7.5

Sandy siltD50 (mm) ≤ 0.10F.C. (%) ≥ 35

Corrected CPT tip resistance qc1 (MPa)

Seism

ic sh

ear s

tres

s rat

io

0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6Silty sand0.10 ≤ D50 (mm) ≤ 0.255 < F.C. (%) ≤ 35

Clean sand 0.25< D50 (mm) < 2.0F.C. (%) ≤ 5

Figure 4.48 Relationship between seismic shear stress ratio (SSR) triggering liquefaction and qc1 values for sands containing fines and an M = 7.5 earthquake.


Down-hole or cross-hole techniques are the exception, where boreholes are used to deter-mine what lies between the holes.

Engineering geophysical methods can be used to determine thickness of strata, to map contamination plumes, find depth to bedrock, to find channels and cavities, and find buried waste among many other things.

Geophysical surveys can also be used to plan conventional investigations that involve boreholes or to interpolate data between boreholes.

The type of geophysical method used will depend on the type of problem that is to be inves-tigated. Table 4.2 shows some of the various geophysical techniques, and their applications.

Offshore geophysical methods are in some ways different to those used on land, and include echo sounding, the use of pingers and sparkers, and side scan radar.

Waves generated can be compression (P) waves or (S) waves. For a P wave, the motion of particles is in the direction of propagation of the wave, whereas for a shear wave, the motion of particles is in a direction perpendicular to the direction of travel of the wave.

4.21.1 Seismic surveys

Seismic surveys involve creating a shock wave in the soil or rock through means such as a sledge hammer or explosives, and recording the arrival of P waves generated by geophones.

Waves generated by the seismic source, travel through the soil and can be either reflected or refracted by different soil interfaces.

Table 4.2 Summary of geophysical methods (on land)

Method Principal characteristics Applicability and limitations

Seismic refraction

Refraction of seismic waves at interfaces of different materials

Preliminary investigation for major stratigraphic units and depth to bedrock. Limited by hard layer over soft layer

High resolution reflection

Arrival times of seismic waves reflected from interfaces of adjoining strata

Deep bedrock identification. Useful for locating groundwater

Vibration Travel time of transverse shear waves Stratigraphy in terms of soil type and thickness and dynamic properties

Up-hole, down-hole, and cross-hole

Measurement of travel times. Geophones on surface and energy source down-hole or vice versa. Energy in central hole and geophones in surrounding holes

Dynamic ground properties at small strain, rock quality, cavity detection. Unreliable for irregular strata or soft strata with high gravel content

Electrical resistivity

Difference in electrical conductivity or resistivity of various strata measured from surface

Horizontal extent and variation of strata and depths to 30 m. Granular material searches. Fresh/salt water boundaries, clay over bedrock, corrosivity of soils. Polluted regions

Drop in potential

Ratio of potential drops between electrodes as a function of current imposed

Similar to resistivity but gives clear definition on vertical or inclined boundaries

E-logs Difference in resistivity and conductivity measured in boreholes

Correlation of units between boreholes

Magnetic Highly sensitive proton magnetometer measures Earth’s magnetic field

Delineation of faults, bedrock, buried utilities, and steel drums in ground

Gravity Differences in density of subsurface materials as indicated by changes in gravitational field

Voids or cavities in bedrock, tracing steeply inclined and irregular features such as faults, intrusions, and domes


4.21.2 Reflection surveys

Figure 4.49 shows the principle of a reflection survey. The waves from the source travel down to an interface and are reflected back up to the surface and recorded by the geophones.

The time for arrival at the geophone at a distance x from the source is given by

T

v vh xx = = +2 2

41 1

12 2SR

/( )

(4.25)

or

T T x vx2

02 2

12= + / (4.26)

where

T h v0 1 12= / (4.27)

A plot of T 2 versus x2 will yield a straight line as shown in Figure 4.50 where the slope is 1 1

2/v and the intercept is given by 4 12

12h v/ .

0S G G′

R′RR0

Tx′

T

Tx

T0

x′x

Figure 4.49 Seismic reflection.

Slope = 1/v12

Intercept= 4h1

2/v12

T 2

x2

Figure 4.50 Graphical method of obtaining wave velocity and depth to firm stratum.


The slope therefore yields the velocity of the wave and this can be used to obtain the thick-ness of the layer h1 from the intercept.

4.21.3 Seismic refraction

In the case of seismic refraction, a seismic wave travels down until it hits a stratum with a different seismic velocity. It will then be refracted at the interface between the two layers, and then back up to the surface. If the velocity of travel in layer 2 is faster than in layer 1, the refracted wave will arrive back at the surface first, that is, before the direct surface wave.

This is shown in Figure 4.51 where the time of arrival at the geophones T is plotted against the distance from the source x. This will also occur for a third layer if the velocity in that layer is higher, that is, v3 > v2 > v1.

Velocities of travel in different materials are given in Table 4.3.For a two-layer system, the thickness of the upper layer can be determined from

h

x v vv v

11 2 1

2 12= −

+ (4.28)

where x1 is the distance to the first crossover point. The velocities are found from the slope of the plot (see Figure 4.51).

1

h1

h2

Ti3

Ti2 v1

v3 > v2 > v1

1 v2

1 v3

v1

v2

v3

Figure 4.51 Results of seismic refraction survey.

Table 4.3 Seismic velocities in various soil and rock types

Material Velocity (m/s)

Loose sand (above water table) 250–600Hard clay 600–1200Soft shale 1200–2100Soft sandstone 1500–2100Basalt 2400–4000Granite 3000–6000


The method can be generalised for multiple layers as long as the velocity in each of the layers is increasing with depth.

Th v v

v vi n

j

nj n j

j n( ) =

−

=

−

∑21

1 2 2

(4.29)

The times Ti(n) are the intercepts on the time axis as shown in Figure 4.51. Equation 4.29 can then be used to progressively calculate the layer thicknesses from the velocities (which are obtained from the slope of the segments of the plot).

Cross shooting can be used to obtain two sets of data for the soil profile. The seismic source (i.e. the sledge hammer) is used at either end of the array of geophones. A plot like the one shown in Figure 4.52 is obtained. This data can be used to identify dipping strata as the forward and reverse profiles will not be the same in this case (see Sharma 1997).

4.21.4 Rippability of rock

Seismic velocity data can be used for determining when rock can be ripped with a bulldozer or when drilling and blasting will be necessary. Higher velocities indicate harder rock that cannot be ripped. Figure 4.53 shows a chart developed by the Caterpillar Tractor Company that is applicable to a D9 bulldozer.

Rippability of rock also depends on the jointing in the rock, as highly jointed rock con-sists of small blocks of rock that can be ripped out individually. This needs to be taken into account as well as seismic velocity.

4.22 RESISTIVITY

The resistivity of different materials is shown in Figure 4.54. Dense rocks have high resistivi-ties while sands or gravels that contain water will have low resistivities. Salt water has a very low resistivity, and so this method can be used to locate salt water intrusions.

400

200

500 1000 150000

Tiu

T (ms) T (ms)

x (m)Tid

1780

m/s 2250 m/s

2870 m/s3200 m

/s

Figure 4.52 Reversed profile for strata that is slightly dipping.


Rocks with fractures and fissures that contain water will also exhibit higher conductivi-ties. Therefore, low conductivity can be an indication of faulting, shearing, weathering, or hydrothermal alteration of rocks.

Resistivity of soil and rock can be measured by passing an electrical current through the ground, and measuring the voltage between two electrodes. Common layouts of the elec-trode arrays are the Wenner and Schlumberger layouts (see Figure 4.55).

TOPSOILCLAYGLACIAL TILLIGNEOUS ROCKS

SEDIMENTARY ROCKS

METAMORPHIC ROCKS

MINERALS and ORES

GraniteBasaltTrap rock

ShaleSandstoneSiltstone

Limestone

Claystone

Caliche

ConglomerateBreccia

SchistSlate

CoalIron ore

Rippable Marginal Non-rippable

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Velocity in feet per second × 1000

Figure 4.53 Seismic rippability chart for a D9 Caterpillar bulldozer with a No. 9 ripper.

1 10 1000.10.01

100100010,000100,000 10

Graphite

Gravel and sandClays

1000 10,000 100,000

1 0.1 0.01

TillsConglomerateSandstoneShales

Fresh water Permafrost

Conductivity (ms/m)

Salt water

Glacialsediments

Sedimentaryrocks

Water,aquifers

Sea ice

Metamorphic rocks

Igneous rocks: mafic

Massivesulphides

Unweatheredrocks

DuricrustWeatheredlayer

Resistivity (Ohm-m)

Dolomite, limestoneLignite, coal

Igneous andmetamorphic rocks

Felsic

Figure 4.54 Typical resistivities of soil and rock.


i. For the Wenner configuration, the resistivity (ohm m or ohm ft) is given by

ρ π=

2 aVI

(4.30)

where I is the current and V is the voltage measured between the electrodes.

ii. For the Schlumberger configuration

ρ π= L

IV2

2 (4.31)

The resistivity that is calculated is an apparent resistivity as it is a weighted average of the material within the zone where the current is flowing. The depth of material affected is roughly the same as the spacing of the electrodes.

With the electrical profiling method, the spacing of the electrodes is kept constant, and the whole configuration is moved (see Figure 4.56). At each new position, the resistivity is found and may be plotted as shown in Figure 4.57.

(a)

(b)

Voltage

Current (Amps)Battery

2L

Current (Amps)

Voltage

a a a

Battery

Figure 4.55 Wenner (a) and Schlumberger (b) electrode layouts.

a

a

a

1 2 3 4Location of lines

Ground surfaceElectrodes

Figure 4.56 Electrical profiling method.


An idea of the change in material type can be obtained from this method and the outline of each material can be marked on a map.

4.22.1 Electrical sounding method

For this approach, the spacings of the electrodes are increased about the same central posi-tion. As the spacing is increased, so is the depth of the area through which the current flows. This is shown in Figure 4.58.

If on spreading the electrodes, the resistivity becomes higher, there must be a layer of material with high resistivity below a soil with a low resistivity. If there is a material with low resistivity below one having a high resistivity, then the opposite will occur, that is, the resistivity will rise as the electrode array is spread. This is shown in Figure 4.59 for a number of different cases.

The electrical profiling method is therefore capable of detecting softer layers beneath harder layers, something that the seismic refraction method cannot achieve.

App

aren

t res

istiv

ity ρ

(ohm

-feet

)

Position of centre of electrode array (feet)

100

200

300

a = 20 feet (6.1 m)400

0100 200

a = 50 feet (15.2 m)

App

aren

t res

istiv

ity ρ

(ohm

-m)

500

300 400 500 600 700 800 900

Position of array (m)100 200

100

50

Figure 4.57 Resistivity versus position of array (profiling method).

e position of the centreof the array is fixed

a1 a1

a2

a3 a3 a3

a2 a2

a1

Figure 4.58 The electrical sounding method.


4.22.2 Push-in resistivity instruments

Resistivity can be measured between the anode and cathode mounted on push-in instruments such as the piezo-cone and the dilatometer. One such device called an RCPTu (resistivity cone penetration test with pore pressure measurement) is shown in Figure 4.60 (Campanella 2008). The module can measure the electrical resistance to current flow in the ground on a continuous basis, and is useful in environmental work as it can be used to detect regions in the ground where pollution may be present.

Ground surface

Layer 1

Layer 2Lower

Higher ρ

a

Ground surface

Layer 1

Layer 2

Lower

Higher

ρ

a

ρ

a

Ground surface

Layer 1

Layer 2

High

Low

Layer 3Medium

Ground surface

Layer 1

Layer 2

Low

Medium

a

ρ

Layer 3High

Figure 4.59 Resistivity curves for different subsurface conditions.

ConePushed at2 cm/s

Insulator

Electrodes

150 mm350 mm

650 mm

Accelerometer

10.5 sq cm‘isolated’ resistivity

15 mm

U3

U2

U1

Figure 4.60 Resistivity module attached behind a piezo-cone.


4.23 MAGNETIC SURVEYING

Magnetometers are used for mapping the intensity of the earth’s magnetic field at various points. A contour map can be made showing the intensity of the magnetic field and areas of magnetic intensity can be used to identify certain types of rocks, that is, those high in iron content. Laboratory measurement of magnetic susceptibility of rock types in the area can be used to interpret results.

Engineering applications include investigations over landfills where buried barrels, pipes, and domestic rubbish can be identified by more intense magnetic zones.

4.24 GROUND PROBING RADAR

Ground probing radar may be used for imaging at shallow depths. It makes use of electro-magnetic waves in the frequency band of 10–1000 MHz. The electromagnetic waves are detected after they are reflected from the different strata in the soil, for example, dry soil, wet soil, or bedrock.

The method is most effective with low-attenuation media such as ice, sand, crude oil, bedrock, or fresh water, but is less effective for materials such as wet clay or silt, or salt water.

The radar pulses are reflected from different materials in much the same way as seismic reflection signals, and therefore can be processed using similar methods (see Section 4.21.1).

4.25 SEISMIC BOREHOLE TECHNIQUES

Seismic techniques can be used without having to drill a borehole such as those discussed in Section 4.21.1. However, there are methods that make use of boreholes into which a seismic detector is lowered and seismic waves generated either at the ground surface or in another borehole.

The seismic shear wave velocity can be used to compute the small-strain shear modulus of the soil since the amount of strain generated by the waves is very small. Small-strain mod-uli can be used in foundation vibration analysis, earthquake analysis, and in the analysis of retaining structures. The shear modulus can be reduced in magnitude and used for large strain analysis such as pile settlement. This is discussed further in Section 6.11.3 in Chapter 6.

4.25.1 Down-hole seismic testing

Down-hole seismic testing is performed by drilling a borehole down in which a probe can be lowered (see Figure 4.61). The vertical seismic shear wave test is performed using pre-drilled holes (50 mm minimum diameter) which may be cased in PVC and grouted around the out-side, or uncased. Seismic shear (S) waves are generated at the surface near the top of the hole using impacts on a wooden plank generally weighted by a vehicle and about 1 m from the hole. Seismic compressional (P) waves are generated separately by impacting a metal plate. Both types of waves are detected using an in-hole, geophone probe which is air-packed or wedged against the hole wall at a pre-selected depth. Individual P and S wave travel times to the detector are used to compute average P and S wave seismic velocities between the surface sources and in-hole detector (see ASTM D7400-08 2008).

The seismic velocity of either the shear S wave or the P wave (at two locations down the hole) is simply calculated by dividing the distance between the two readings in the hole


Δd by the difference in arrival times Δt of the waves at the two locations, for example, vs = Δd/Δt. The S wave velocity can be used to compute the small-strain shear modulus as shown in Equation 4.33.

4.26 CROSS-HOLE TECHNIQUES

For the cross-hole test, two boreholes are drilled typically 3 m apart. By creating a shear wave in one borehole (by using an impulse rod at the bottom of the hole), and recording the arrival time in a neighbouring borehole, the shear modulus of the soil can be determined (see Figure 4.62). Once again, this modulus is a small-strain modulus as the shear waves cause only small shear strains in the soil.

The velocity of the shear waves can be found from

v

Lt

s =

(4.32)

and from the shear wave velocity, the shear modulus can be found.

G

vgs=2γ

(4.33)

Seismograph

BoreholeShear wavepropagates down

Wooden plankweighted downby vehicle

Hammer with impactswitch

Geophone held againstside of borehole

Direction of particlevibration

Figure 4.61 Down-hole seismic test.


wherevs is the shear wave velocityG is the shear modulus of the soilγ is the unit weight of the soilg is the acceleration due to gravityL is the spacing of the boreholest is the time for the wave to travel the distance L

Poisson’s ratio ν can be found if the P wave velocity vp is known

ν =

−−

( ))(

v vv vp s

p s

//

2

2

22 2

(4.34)

The standard for performing this test is ASTM D4428/D4428M-07 (2007) and more details on the test can also be found in Stokoe and Woods (1972).

4.27 OTHER SEISMIC DEVICES

Seismic sources may also be generated by push-in devices such as the dilatometer or the cone penetrometer. One type of seismic cone test is very similar to the down-hole seismic test where the surface of the ground is excited by a sledge hammer, but where the geophone receiver is attached to the cone (usually a piezo-cone).

Borehole Borehole

Down-holehammer

Geophone

Sidewall clampDirection of particle motion

Direction of wavepropagation

Seismograph

Cable tomovingweight inhammer

Figure 4.62 Cross-hole seismic survey method.


A seismic dilatometer SDMT is the combination of the standard flat dilatometer (DMT) with a seismic module. The module is a probe outfitted with two sensors, spaced 0.5 m apart, for measuring the shear wave velocity vs. Again, the source of seismic waves is a surface-based source such as a sledge hammer. The seismic velocity is calculated from the distance between the two sensors and the time difference that it takes for the shear wave to reach each of the sensors.

135

Chapter 5

Shallow foundations

5.1 INTRODUCTION

A foundation is usually termed ‘shallow’ if the base of the foundation is at a depth that is less than the breadth of the foundation. This type of footing is suitable where structural loads are not high, or the soil is strong enough to support the applied loads with a foundation of moderate depth.

The purpose of a shallow foundation is to apply the structural loads to the foundation by spreading the load from walls or columns over a larger area. The size of the footing can be selected so that the contact pressure between the footing and the soil is small enough so as to reduce settlement to acceptable levels and provide an adequate factor of safety against the possibility of a bearing failure.

Shallow foundations may be of several different types, and the selection of the appropriate type will depend on the structural loads and the foundation conditions. Some commonly used shallow foundation types are discussed in Section 5.2.

5.2 TYPES OF SHALLOW FOUNDATIONS

5.2.1 Strip footings

A strip footing is one that is long compared to its width. Such foundations (see Figure 5.1a) are typically used to support continuous masonry walls.

Strip footings are constructed by excavating a trench with a backhoe (or by hand), placing a reinforcing cage and pouring concrete into the trench. If the trench is able to stand open without collapsing (e.g. clayey soils) then formwork is not needed or may only be needed to form that part of the foundation that is above ground level. For sandy soils, the trench has to be excavated and fully formed before concrete can be poured as the trench will not stay open.

5.2.2 Pad footings

For support of isolated loads such as column loads or support of raised slabs or flooring, pad footings may be used (Figure 5.1c). The process of construction is similar to that for strip footings as the hole dug for the footing can be filled with concrete without the need for formwork (if the hole will stay open without collapsing). The hole should be inspected before placement of concrete to ensure that it does not contain any water or soil that has fallen in from the sides.


It is also generally recommended that in sandy areas or on sites that may be subjected to wind or water erosion, footings should be constructed at least 300 mm below the surface to prevent any undermining.

5.2.3 Combined footings

A combined footing may be necessary when a column load is close to a property boundary, and the foundation cannot be made large enough to support the load without encroaching onto the neighbouring property (Figure 5.1b). In this case, the footing is combined or tied to a nearby footing that provides additional support.

5.2.4 Raft or mat foundations

For special conditions (i.e. in areas of swelling soils) or where pad footings would need to be large to support loads, a raft or mat foundation may be used. Rafts can be reinforced concrete slabs of uniform thickness, which for tall buildings can be several metres thick.

For smaller scale construction, rafts can be stiffened under walls and columns by making the raft thicker at the point of loading. In areas where reactive soils exist, it is common to construct a ‘waffle’ slab, so called because it consists of stiffened ribs running perpendicular to each other under the slab (Figure 5.1d). The upper flat portion of waffle slabs can be cast on polystyrene blocks so that swelling of the soil does not affect the slab or raft but merely compresses the polystyrene.

5.3 BEARING CAPACITY

Theoretical solutions for the bearing capacity of shallow foundations are generally based on the theory of plasticity where the soil is assumed to behave as a rigid plastic material that fails according to the Mohr–Coulomb failure criterion. Shallow foundations may be defined

Property line

Slab fabric

Void or polystyrene

Combined footingStrip footing

Waffle slabPad footing

Plan

(a) (b)

(c)

(d)

Figure 5.1 Types of shallow foundations.

Shallow foundations 137

as those which are founded at a depth Df which is less than the full width B of the founda-tion (B may be taken as the width of a strip, the diameter of a circle or the smaller dimension of a rectangular footing). Solutions are obtained by establishing the stress characteristic fields beneath a foundation. The stress characteristics are lines that show the orientation of the planes on which the stresses in the soil have reached the critical combination τf and σf as shown in Figure 5.2. The method is described in Davis and Booker (1971) and in the book by Hill (1983) and in Wu (1976).

Many solutions to bearing capacity problems have been obtained using classical plasticity theory; however, for more complex problems (i.e. layered soils, eccentric loadings), approxi-mate methods may need to be used to obtain solutions. Such methods include those based on the bound theories of plasticity whereby upper and lower bounds to the collapse load may be found. The true collapse load therefore lies between the bounds that have been established. An example of the use of this technique is given by Sloan and Yu (1996) who use linear programming techniques to establish upper and lower bounds to bearing capacity problems.

Other methods include finite element techniques (see Zienkiewicz 1977), finite differ-ence methods as used in the commercially available computer program FLAC3D (Ver. 5.0) (ITASCA Consulting Group, Inc. 2014) or optimisation codes (OPTUMG2 2014). Obviously, a great deal more effort is involved in setting up finite element meshes or finite difference grids and computing ultimate loads than in using the results of a theoretical solution, and so such numerical techniques are limited to problems involving complex geometry, loading, or material properties. A further advantage of numerical techniques is that advanced con-stitutive laws governing deformation and plastic behaviour of the soil can be incorporated. Analyses can therefore give predictions of the load-deformation behaviour of a foundation, rather than just the collapse load.

Solutions have been found to particular types of bearing capacity problems using differ-ent solution methods. In the following, solutions are presented to various bearing capacity problems for surface footings, and where applicable, the solution technique is mentioned.

ϕ

45° + ϕ/2

45° + ϕ/2 45° – ϕ/2

45° – ϕ/2

Log spiral

Smooth

B

qs

| τf | = σf tan ϕ along these lines

A

ϕ

Pole

τf

−τf−τf

σf , τf

Poleqs

Direction of failure planes

At point A At point B

Qf

ττ

σf σσ

Figure 5.2 Slip lines for smooth strip footing on cohesionless, weightless soil.


5.3.1 Uniform soils

For soil layers that can be considered to be of infinite depth and having uniform strength with depth (i.e. the cohesion c and angle of shearing resistance ϕ are constant) the well-known Terzaghi equation may be used (Equation 5.1 – Terzaghi 1943). This equation applies to strip footings of width B subjected to vertical loading as shown in Figure 5.3a.

q cN BN q Nu c s q= + +1

2γ γ

(5.1)

wherequ is the ultimate pressure that can be carried by the foundationqs is the surcharge acting on the surface of the soilB is the full width of the stripγ is the unit weight of the soil below the foundation levelc is the cohesive strength of the soilNc, Nγ, and Nq are bearing capacity factors

The bearing capacity factors depend on the angle of shearing resistance of the soil ϕ. Values have been presented by Terzaghi (1943) but were approximate. More modern values of the bearing capacity factors have been computed using slip line theory (Martin 2005) and are shown in Figure 5.4 and are presented in Table 5.1.

The formula of Equation 5.1 can be seen to consist of three terms, the first arising from the cohesive strength of the material, the second from the self-weight of the soil beneath the foundation, and the third from the surcharge considered to be applied at foundation level (see Figure 5.3a). For foundations that are not on the surface of the soil, the surcharge is considered to be equal to the pressure exerted by the soil above foundation level. Plastic failure is therefore not considered in the soil above foundation level and so some error exists because of this assumption, however, for most practical purposes the error is small. Figure 5.3b shows the slip lines for the case where the soil above footing level is considered. Shearing would have to take place through the surcharge in this case, as can be seen from the figure. Terzaghi used superposition to add together the effects of the three terms as shown in Equation 5.1. This is an approximation, and improved estimation of the bearing capac-ity can be obtained by using the theory of plasticity and slip lines, and performing a single calculation considering all components at the same time. Computer code for performing the calculation has been provided at the Web site http://www.eng.ox.ac.uk/civil/people/cmm by C. Martin. An example of the slip lines for a smooth strip footing is shown in Figure 5.5, where the parameters used are c = 5 kPa, ϕ = 30°, γ = 18 kN/m3, qs = 20 kPa, and B = 2 m.

The calculated bearing capacity for this case is 745.9 kPa for a smooth footing and 936.8 kPa for a rough footing. Note that for this calculation, a greater number of slip lines was used than shown in Figure 5.5, to improve the accuracy of the computed bearing capacity.

B

Da

bc

d

qs = γDqu

(a)

D

(b)

Figure 5.3 Failure mode for footing at depth D: (a) Terzaghi’s assumption; (b) actual failure mode.


The ultimate bearing capacity of a foundation depends upon whether it is loaded rapidly so that the soil does not drain or whether it is loaded slowly so that all pore pressures have dissipated.

Undrained Case

For the undrained case, the undrained cohesion su and angle of shearing resistance ϕu are used with Equation 5.1 (i.e. c = su, ϕ = ϕu). If ϕu = 0, then Nc = 5.14 = (2 + π), Nγ = 0, and Nq = 1 and so the bearing capacity equation becomes

q s Du u= +5 14. γ bulk (5.2)

where D is the depth of the foundation and γbulk is the bulk unit weight of the soil above foundation level.

Drained Case

For the drained loading case, the drained strength parameters are used with Equation 5.1 (i.e. c = c′, ϕ = ϕ′), however the drained case is a little more complex than the undrained case

0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60

ϕ (in

deg

rees

)

Nq or Nc

Nc

Nq

0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60 70 80 90

ϕ (in

deg

rees

)

SmoothRough

Nγ

(a)

(b)

Figure 5.4 Bearing capacity factors for use in Equation 5.1 (i.e. using superposition). (After Martin, C.M. 2004. Program ABC [Analysis of Bearing Capacity] V1.0. http://www.eng.ox.ac.uk/civil/people/cmm/software.)


Table 5.1 Bearing capacity factors obtained from plasticity theory

ϕ Nc Nq Nγ

(o) δ/ϕ = any δ/ϕ = any δ/ϕ = 0 δ/ϕ = 1/3 δ/ϕ = 1/2 δ/ϕ = 2/3 δ/ϕ = 1

0 5.14159 1.00000 0.00000 0.00000 0.00000 0.00000 0.00001 5.37926 1.09390 0.0106339 0.0110586 0.0112596 0.0114539 0.01182402 5.63160 1.19666 0.0242179 0.0257878 0.0265319 0.0272506 0.02860453 5.89977 1.30919 0.0408212 0.0443525 0.0460261 0.0476371 0.05062954 6.18504 1.43250 0.0607622 0.0672331 0.0702954 0.0732293 0.07859165 6.48882 1.56770 0.0844649 0.0950574 0.100057 0.104821 0.1133716 6.81264 1.71604 0.112443 0.128586 0.13618 0.143368 0.1560207 7.15820 1.87892 0.145304 0.168723 0.179693 0.190002 0.2077708 7.52736 2.05790 0.183757 0.216532 0.231807 0.24605 0.2700549 7.92217 2.25475 0.228629 0.273262 0.293945 0.313066 0.34454010 8.34493 2.47144 0.280879 0.340379 0.367775 0.392867 0.43316411 8.79814 2.71019 0.341627 0.419603 0.455253 0.487578 0.53817512 9.28461 2.97351 0.412173 0.512957 0.558674 0.599689 0.66219113 9.80746 3.26423 0.494036 0.622817 0.680737 0.732111 0.80825914 10.3701 3.58556 0.588986 0.751982 0.824616 0.888264 0.97993915 10.9765 3.94115 0.699096 0.903758 0.994049 1.07216 1.1813916 11.6309 4.33511 0.826793 1.08205 1.19345 1.28852 1.4174817 12.3381 4.77215 0.974928 1.29147 1.42803 1.54291 1.6939318 13.1037 5.25764 1.14685 1.53753 1.70397 1.8419 2.0174619 13.9336 5.79771 1.34653 1.82673 2.02861 2.19325 2.3960020 14.8347 6.39939 1.57862 2.16686 2.41065 2.60618 2.8389421 15.8149 7.07076 1.84869 2.56721 2.8605 3.09162 3.3573722 16.8829 7.82112 2.16332 3.03892 3.39057 3.6626 3.9644923 18.0486 8.66119 2.53035 3.59535 4.01573 4.33468 4.6760424 19.3235 9.60339 2.95919 4.25262 4.75384 5.12649 5.5108025 20.7205 10.6621 3.46108 5.03017 5.62641 6.06038 6.4913126 22.2544 11.8542 4.04956 5.95158 6.65942 7.16328 7.6446727 23.9422 13.1991 4.74097 7.04550 7.88433 8.46773 9.0035828 25.8033 14.7199 5.55510 8.34686 9.33941 10.0132 10.607629 27.8605 16.4433 6.51599 9.89841 11.0713 11.8475 12.505030 30.1396 18.4011 7.65300 11.7527 13.1371 14.0294 14.754331 32.6711 20.6308 9.00208 13.9744 15.6069 16.6306 17.427532 35.4903 23.1768 10.6074 16.6437 18.5673 19.7393 20.613133 38.6383 26.092 12.5237 19.8603 22.1254 23.4649 24.420334 42.1637 29.4398 14.8188 23.7485 26.4145 27.9427 28.984935 46.1236 33.2961 17.5771 28.4643 31.6012 33.3421 34.476136 50.5855 37.7525 20.9049 34.2044 37.8947 39.8748 41.105937 55.6296 42.9199 24.9357 41.2180 45.5591 47.8083 49.141638 61.3518 48.9333 29.8388 49.8224 54.9296 57.4811 58.921939 67.8668 55.9575 35.8302 60.4242 66.4339 69.3250 70.878740 75.3131 64.1952 43.1866 73.5471 80.6214 83.8937 85.565641 83.8583 73.8969 52.2656 89.8703 98.2022 101.902 103.69742 93.7064 85.3736 63.5316 110.280 120.100 124.280 126.203

(Continued)


if the water table is in the vicinity of the foundation. Figure 5.6 shows several different cases for the level of the water and for each case, the method of calculating the total ultimate pres-sure that can be applied at foundation level is given in Equations 5.3.

Case :

Case :bulk bulk

sub

1 0 5

2 0 5

q c N BN DN

q c N Bu c q

u c

= ′ + += ′ +

.

.

γ γγ

γ

NN DN

q c N BN D d d Nq

u c q

γ

γ

γγ γ γ

+= ′ + + − +[ ]

bulk

sub bulk subCase :3 0 5. ( ) ++= ′ + + += ′ +

γγ γ γγ

w

u c q w

u c

d

q c N BN DN D

q c N

Case :

Case :sub sub4 0 5

5 0

.

.. ( )5γ γ γγsub subBN DN D hq w+ + +

(5.3)

Table 5.1 (Continued) Bearing capacity factors obtained from plasticity theory

ϕ Nc Nq Nγ

(o) δ/ϕ = any δ/ϕ = any δ/ϕ = 0 δ/ϕ = 1/3 δ/ϕ = 1/2 δ/ϕ = 2/3 δ/ϕ = 1

43 105.107 99.0143 77.5929 135.943 147.525 152.243 154.3044 118.369 115.308 95.2519 168.401 182.075 187.396 189.59245 133.874 134.874 117.576 209.715 225.876 231.874 234.21346 152.098 158.502 145.996 262.657 281.784 288.541 291.02647 173.640 187.206 182.449 330.993 353.662 361.270 363.90748 199.259 222.300 229.584 419.882 446.795 455.359 458.15049 229.924 265.497 291.056 536.469 568.482 578.119 581.06750 266.882 319.057 371.967 690.752 728.912 739.755 742.86351 311.752 385.982 479.523 896.883 942.480 954.679 957.94752 366.660 470.304 624.024 1175.14 1229.77 1243.49 1246.9253 434.421 577.496 820.392 1554.95 1620.59 1636.04 1639.6354 518.805 715.074 1090.56 2079.65 2158.80 2176.18 2179.9355 624.924 893.484 1467.23 2814.00 2909.77 2929.35 2933.2556 759.793 1127.44 2000.05 3856.36 3972.74 3994.8 3998.8557 933.170 1437.96 2765.60 5358.81 5500.86 5525.73 5529.9358 1158.83 1855.52 3884.45 7560.90 7735.20 7763.26 7767.5859 1456.54 2425.08 5550.24 10847.9 11063.0 11094.7 11099.160 1855.10 3214.14 8081.21 15853.6 16120.6 16156.5 16161.0

Source: After Martin, C.M. 2004. Program ABC (Analysis of Bearing Capacity) V1.0. http://www.eng.ox.ac.uk/civil/people/cmm/software.

Note: δ is the angle of friction between the footing and the soil ϕ is the angle of friction of the soil.

Figure 5.5 Slip lines computed for a smooth strip footing. (After Martin, C.M. 2004. Program ABC [Analysis of Bearing Capacity] V1.0. http://www.eng.ox.ac.uk/civil/people/cmm/software.)


whereγsub = the submerged unit weight of the soilγw = the unit weight of waterγbulk = the bulk unit weight of the soilD, d, h = the distances shown in Figure 5.6

It may be noted that qu is the total stress that can be applied at foundation level and therefore in Equations 5.3, the effective ultimate bearing pressure has been calculated and the water pressure added if the water is above foundation level.

If the water level is at a depth of less than one foundation width B beneath the footing, but below the footing base (i.e. intermediate of Cases 1 and 2), simple empirical corrections are sometimes made to allow for the changed unit weight of the soil beneath the footing. This assumes that the water level only begins to affect the unit weight of the soil being pushed aside by the foundation when it reaches a distance of B beneath it, and that as the water table rises further, the overall unit weight of the soil can be computed from a simple linear interpolation, for example,

γ γ γ γ= + −sub bulk sub

zBm ( )

(5.4)

where γbulk is taken as the unit weight of soil corresponding to the minimum water con-tent above the water table and zm is the distance of the water level beneath the footing (see Figure 5.6).

The unit weight calculated from Equation 5.4 is used in the self-weight term (Nγ term) of the bearing capacity equation.

Effect of Footing Shape

The Terzaghi bearing capacity factors were computed for strip foundations (i.e. the length of the foundation is large compared to its width) and therefore do not apply to footings of other shapes. For footings that are square or circular in plan, Terzaghi and Peck (1967) proposed the following formulae

q cN RN q N

q cN BN q Nu c s q

u c s q

= + += + +

1 2 0 6

1 2 0 4

. .

. .

γγ

γ

γ

circular

square (5.5)

where R is the radius of a circular footing and B the full width of a square footing.The effect of footing shape is discussed more fully in Section 5.3.1.2.

Ground level5

4

2

1

3

Water level

h

D

zm

d

Intermediate case0 ≤ zm ≤ B

Well below foundationzm ≥ B

B

qu

Figure 5.6 Location of a water table beneath a footing.


Net Bearing Capacity

The value of bearing capacity qu as computed from equations such as Equations 5.1 or 5.3 give the ultimate total pressure that can be applied at foundation level. If a hole is exca-vated for a foundation, the stress at the foundation level is reduced by an amount equal to the overburden pressure. Therefore, we should be able to apply a stress at foundation level equal to the overburden pressure to return to initial conditions. It is only stresses above the overburden pressure that will contribute to a bearing failure.

This observation leads to the concept of net bearing pressure that is useful in comput-ing the allowable loads that can be applied to a foundation. The net bearing capacity is defined as

q q qu unet ob= − (5.6)

whereq unet = the ultimate net loadqu = the bearing capacityqob = the total overburden stress at foundation level

The allowable load qall that can be applied (at foundation level) can be computed from

q

qF

quall

netob= +

(5.7)

where F is the factor of safety.Equation 5.7 shows that if the factor of safety were infinitely large, then the allowable

load is equal to the overburden pressure. This implies that if a load equal to the excavated soil were applied, there is no possibility of failure (i.e. F is infinite) since the initial conditions have been reinstated.

5.3.1.1 General formulae

Loads on foundations may not always be vertical or applied at the central point of the foun-dation or the foundation may not be on level ground. In order to take into account many other factors that may affect the bearing capacity, the following general bearing capacity equation (Equation 5.8) has been proposed by Vesic (1973, 1975).

q cN BN q Nu c c s q q= + +ζ γ ζ ζγ γ

12

(5.8)

where the correction multipliers ζ are obtained from Table 5.2 by multiplying together indi-vidual correction factors if they are applicable, for example,

ζ ζ ζ ζ ζ ζ ζζ ζ ζ ζ ζ ζ ζζ ζ

β δ

γ γ γ γ γβ γδ γ

c cs ci cd c c cr

s i d r

q qs

=== ζζ ζ ζ ζ ζβ δqi qd q q qr

(5.9)


and the additional subscripts s, i, d, β, δ, and r indicate that the correction factors apply for foundation shape, inclination, depth, slope of soil surface, slope of foundation base, and soil rigidity, respectively, (see Figure 5.7) and are given in Table 5.2.

The bearing capacity factors Nc, Nq, and Nγ are the factors for a strip footing given in Table 5.1. The method is approximate, but is accurate enough for practical application where often large factors of safety (e.g. three or more) are applied to the ultimate bearing capacities that are computed. For eccentric loads, it is assumed that the load acts at the cen-tre of a foundation of reduced size. For example, if the eccentric load V shown in Figure 5.7, were acting at the point X at eccentricities eB, eL, then the footing would be treated as one having a reduced area of B′ by L′ where the reduced dimensions are given by

′ = −′ = −

B B e

L L eB

L

2

2 (5.10)

Table 5.2 Vesic Correction Factors

Factor Cohesion (c) Self-weight (γ) Surcharge (q)

Foundation shape (s) ζcs

q

c

BL N

= + ′′

1N

ζγsBL

= − ′′

1 0.4 ζ φqsBL

= + ′′

1 tan

Inclined loading (i) ζ φci

c

mHB L cN

= −′ ′

=1 0( )

( )tan

( )= −−

>ζζ

φφqi

qi

cN1

0

ζγ φi

mH

V B L c= −

+ ′ ′

+

1cot

1

ζφqi

mH

V B L c= −

+ ′ ′

1

cot

Foundation depth (d)

ζ ξ φcd = + =1 00.4 ( )

= −−

>ζζ

φφqd

qd

cN( )

tan( )

10

ζγd = 1.0 ζqd = 1 + 2 tan ϕ(1 − sin ϕ)2ξ

Surface slope (β) β π< /4

β is in radians

ζ βπ

φβc = −+

=12

20( )

= −−

>ζζ

φφβ

βq

q

cN( )

tan( )

10

ζγβ = (1 − tan β)2

Note: Nγ β φ= − =2 0sin ( )aζγβ = (1 − tan β)2

Base tilt (δ) δ π< /4

δ is in radians

ζ δπ

φδc = −+

=12

20( )

= −−

>ζζ

φφδ

δq

q

cN( )

tan( )

10

ζγδ = (1 − δtan ϕ)2 ζqδ = (1 − δ tan ϕ)2

Rigidity (r) (see Appendix 5A for Ir)

ζ

φ

cr

r

BL

I

= −

+ =

0.32 0.12

0.60 log ( )10 0

ζ ζζ

φφcr qr

qr

cN= −

−>

tan( )

10

ζ ζγr qr= ζqr A B= exp[ + ]

ABL

= − +

4.4 0.6 tan φ

BIr=

+3.07sin log

sinφ

φ( )102

1

Note: V = vertical load; H = horizontal load; B = foundation width; L = foundation length (L > B); eB = eccentricity parallel to B; eL = eccentricity parallel to L; ′ = −B B eB2 ; ′ = −L L eL2 ; m = (2 + x)/(1 + x); x = B/L if H parallel to B; x = L/B if H parallel to L; ξ = D/B if D/B ≤ 1; ξ = >−tan ( )1 D B D B/ if / 1.

a q cN BN q Nu c c s q½ ( )= + + =ζ γ β φβ γ cos 0ο if only applying surface slope correction.


If the footing is subjected to a moment ML in the L direction and a moment MB in the B direction and has an applied vertical load V an equivalent loading system can be obtained by placing the vertical load at eccentricities eL and eB (see Figure 5.7) where

eMV

eMV

LL

BB

=

=

(5.11)

The correction multipliers in the general bearing capacity equation (Equation 5.8) can also be computed from values given by Meyerhof (1953, 1963) or by Hansen (1970). Hansen’s correction factors include the effects of the slope of the soil surface and the slope of the base of the foundation. The correction multipliers may be computed by multiplying the correc-tion factors together as for the Vesic case (Equation 5.9).

EXAMPLE 5.1

A square footing, 1.5 m by 1.5 m in plan, is founded at a depth of 1 m in a deep layer of clay. The footing is loaded by a vertical load and moment loadings such that the load can be considered to act eccentrically at a point 0.3 m away from each of the centrelines of the footing.

The water table is at foundation level and the saturated unit weight of the clay may be taken as γsat = 17 kN/m3 (as can the bulk unit weight of the clay above water level). If the magnitude of the inclined load is 100 kN and it is inclined at 60° to the vertical, compute the bearing capacity of the footing

1. When it is loaded rapidly so that the soil is in an undrained condition 2. When it is loaded very slowly

L

MBML

V

H

BL

H

B

eB

= MBV eL = ML

V

XL′

L

B

B′

X

qs = γD cosβ

B/2B/2

δH

V

β

D

eL

eB

Figure 5.7 Footing subjected to vertical and horizontal loads plus moments.


The strength properties of the clay are

s

cu u= =′ = ′ =

60 0

0

kPa φφ ο28

Solution

Because the load is eccentric, we must calculate the reduced length and breadth of the footing

′ = − = − ×′ = − =

L L e

B B eL

B

2

2

1.5 0.3 0.9

0.9 m

2 m=

The components of the load are H = 100 cos60° = 50 kN; and V = 100 sin60° = 86.6 kN

1. Undrained Loading

For ϕu = 0, the bearing capacity factors from Figure 5.4 are Nc = 5.14, Nq = 1, Nγ = 0. Using the Vesic Correction Factors of Table 5.2, we have

ζ ζ ζ ζ ξc cs cd ciq

c c

B NL N

mHB L cN

= = + ′′

+ −′ ′

= +

1 1 1( )

(

0.4

1 00.194 0.267 0.3

1.058

)( )( )1 + −=

1

ζq = 1 0.

and so the ultimate bearing capacity qu is given by

q s N DNu u c c q q= += × × + × × ×=

ζ ζγ60 5.14 1.058 17 1.01 1

343 kPa

2. Drained Loading

For ϕ′ = 28°, the bearing capacity factors from Figure 5.4 are Nq = 15, Nγ = 11. In this case, we have to compute the factor m that depends upon whether the hori-

zontal load is parallel to side B or L. Because the loading is at an angle to both sides for this problem, it is necessary to compute m = mB sin2θ + mL cos2θ (as suggested by Vesic) where θ is the angle of the inclined load to the long side L of the foundation.

As the footing is square, mB = mL = (2 + 1)/(1 + 1) = 1.5 and the angle is 45° so that m = 1.5.

We now may compute the correction factors from Table 5.2

ζ ζ ζ ζζ ζ ζ ζγ γ γ γ

q qs qd qi

s d i

= = + + == =

( . )( . )( . ) .

(

1 0 53 1 0 199 0 274 0 503

1 −− =0 4 1 0 0 116 0 0696. )( . )( . ) .

and so the bearing capacity for drained conditions is

q BN DNu q q= +

= × − × × × + × ×

12

γ γζ ζγ γsub

(17 ) 11 17 1 10.5 9.81 1.5 0.0696 55

132 kPa

×= +=

0.503

4.1 128.3

Therefore, for this particular problem, the drained bearing capacity is the lowest.


5.3.1.2 Soil layers of finite depth

The solutions for the bearing capacities of foundations mentioned so far have been for soils that are of infinite depth. It is of interest to determine the effect of the bedrock underlying a soil layer if the bedrock (or a very stiff layer) is at a finite depth.

Solutions to this problem have been found by Mandel and Salençon (1969) who consid-ered a strip footing on a uniform layer of soil (i.e. strength parameters are constant) having a rough base so that the maximum friction has to be mobilised on the base before slip can occur.

The conditions assumed for the interface between the rigid base and the layer of soil is important because it influences the collapse load, and Mandel and Salençon have also con-sidered other base conditions. The rough based solution is more applicable to soils where it would be expected that there would be some shearing resistance between the soil and the underlying rock and so only results for the rough base are considered here.

The bearing capacity of strip footings having a full width B may be computed from Equation 5.12.

q cN F BN F q N Fu c c s q q= ⋅ + ⋅ + ⋅1

2γ γ γ

(5.12)

where the bearing capacity coefficients are the same as for the Terzaghi equation, but are modified by the factors Fc, Fγ, and Fq which depend upon the depth h of the soil layer.

Tables 5.3–5.5 give values of the factors Fγ, Fc, and Fq, respectively. The values in the tables may be seen to depend on the angle of shearing resistance and the ratio of the footing width to the layer depth B/h. It may be noted that the correction factors were computed for a smooth footing base and by considering the effects of cohesion, self-weight of the soil, and surcharge separately.

Table 5.3 Values of factor Fγ

ϕ ↓ B/h → 2 3 4 5 6 8 10 15 20 30 40

30° B/h ≤ 1.3 Fγ = 1

1.20 2.07 4.23 9.90 24.8 178 1450 3.81 × 105 1.3 × 108 1.95 × 1013

20° B/h ≤ 2.14 Fγ = 1

1.07 1.28 1.63 2.20 4.41 9.82 97 340 2.6 × 105 7 × 105

10° B/h ≤ 4.07 Fγ = 1

1.01 1.04 1.12 1.36 2.28 4.33 20 113

Fγ = 1 for all ϕ = 0°

Table 5.4 Values of factor Fc

ϕ ↓ B/h → 1 2 3 4 5 6 8 10 15 20 30

30° B/h ≤ 0.63 Fc = 1

1.13 2.50 6.36 17.4 50.2 150 1444 1.48 × 104 5.81 × 106

20° B/h ≤ 0.63 Fc = 1

1.01 1.39 2.12 3.29 5.17 8.29 22.0 61.5 905 1.50 × 104

10° B/h ≤ 1.12 Fc = 1 1.11 1.35 1.62 1.95 2.33 3.34 4.77 11.7 29.40° B/h ≤ 1.414 Fc = 1 1.02 1.11 1.21 1.30 1.40 1.59 1.78 2.27 2.75 3.72


5.3.2 Non-uniform soils

Inhomogeneity can be due to the soil strength increasing with depth, or perhaps being stron-ger at the surface due to overconsolidation. The soil beneath a foundation may also consist of different layers of material or the soil strength may simply be different at different places on a site. In the latter case, the footing should be designed for the soil strength at the loca-tion of the footing, or if this is not known, a conservative approach is to design footings for the lowest strength.

Solutions have been found for the case where the soil strength varies uniformly with depth or where the soil is layered, and these solutions are presented in the following sub-sections.

5.3.2.1 Strength increasing with depth

Strip Loading on Infinitely Deep Soil Layer

Solutions to the undrained problem (ϕ = ϕu = 0) where the cohesion of the soil varies linearly with depth have been produced by Davis and Booker (1973) for the case of a strip footing.

The ultimate load per unit length that can be carried by the foundation is given by

QB

F cBu = + +

( )24

0π ρ

(5.13)

wherec0 is the cohesive strength at the surface of the soilρ is the rate of increase of cohesive strength with depthB is the full width of the footingF is a factor that is presented in Figure 5.8a

In the figure, the factor F is presented for both rough FR and smooth FS foundations.Factors were also presented by Davis and Booker for a soil having a crust. The crust is

assumed to have a cohesion c0 to a depth where it intersects the line c = ρz (see inset to Figure 5.8b). The factor F is presented in Figure 5.8b for both rough and smooth footings, where the factor is again used with Equation 5.13.

Figure 5.9 shows the stress characteristics or slip lines for the case of a clay, whose strength increases linearly with depth for both a (a) smooth and (b) rough based strip footing computed using the program ABC by Martin (2004).

Soil Layers of Finite Depth

The case of an undrained clay (ϕ = ϕu = 0) with a cohesion that varies with depth has been examined by Matar and Salençon (1977) for the case of a strip footing on a soil layer of finite depth (overlying a rough, rigid base).

Table 5.5 Values of factor Fq

ϕ ↓ B/h → 1 2 3 4 5 6 8 10 15 20 30

30° B/h ≤ 0.63 Fq = 1

1.12 2.42 6.07 16.5 47.5 142 1370 1.40 × 104 5.50 × 106

20° B/h ≤ 0.86 Fq = 1

1.01 1.33 1.95 2.93 4.52 7.14 18.7 51.9 763 1.26 × 104

10° B/h ≤ 1.12 Fq = 1 1.07 1.21 1.37 1.56 1.79 2.39 3.25 7.37 17.9 92.3

Fq = 1 for all ϕ = 0°


The bearing capacity of the strip is given by

q q c N

Bc

u s c c= + ′ +

µ ρ0

04 (5.14)

whereqs is the surchargeN′c is the bearing capacity factor given by Figure 5.10μc is the correction factor given in Figure 5.10c0 is the cohesion at the surfaceρ is the rate of increase in cohesion with depth

0 4 8 12 16 201.0

1.2

1.4

1.6

1.8

2.0

2.2(a)

F n

0.05 0.04 0.03 0.02 0.01 0ρB/coco/ρB

B

qu co su

zRough FR

Smooth FS

FR/FS

co + ρz

qu = Fn [(2+π)co + ρB/4]

0.7

0.8

0.9

1.0

1.1

1.2

1.3(b)

F n 4 8 12 160.02

ρB/co

0.01co su

ρzz

200.05 0.04 0.03

FSC (Smooth strip)

qu = Fn [(2+π)co + ρB/4]Fn = FRC (rough)or FSC (smooth)

co/ρB

FRC (Rough strip)

Figure 5.8 Undrained bearing capacity factor Fn for a strip footing on an infinitely deep layer: (a) shear strength increasing linearly with depth; (b) with a crust. (After Davis, E.H. and Booker, J.R. 1973. Géotechnique, Vol. 23, No. 4, pp. 551–563.)


(a)

(b)

Figure 5.9 Slip line field for a (a) smooth and (b) rough strip footing on a clay whose strength increases with depth ρB/c0 = 4.

0.1 1 10 102 103 104

10

20

30

30 25 20 15 10

⇓

zh

Bqs

su = co + ρz

5π + 2 √2

B/h

Rough footing – Rough based layer

1.715 1.72 ρB/co

1.10

1.15

1.2

1.251.3

1.41.5

1.61.7

μc = 1.05

N′c

Figure 5.10 Curves of µc and ′Nc . Strip on finite layer of soil with undrained strength increasing with depth. (After Matar, M. and Salençon, J. 1977. Sols et Fondations, No. 352, Juillet-Août, pp. 95–107.)


5.3.2.2 Fissured clays

Some clay deposits, especially overconsolidated clays, can contain fissures which have a lower strength than the matrix due to shearing and weathering. Some solutions have been obtained to this problem (Davis 1980, Booker 1991) for the case of a strip footing.

Solutions to the problem have also been presented by Lav et al. (1995) for the case of a smooth strip footing resting on a weightless, purely cohesive (ϕu = 0) material that has a regular set of fissures. The problem considered is shown in Figure 5.11. The angle between the sets of fissures is ωi while the fissures lie at an angle ω to the vertical.

If the cohesive strength of the fissures is cf and the strength of the solid material between the fissures is cm, then the bearing capacity of the strip footing can be expressed in terms of cm. Results are shown in Figure 5.12a–d which show the effects of the fissures on the bear-ing capacity for joint sets with an included angle ωi of 90, 60, 45, and 30 degrees. Where there is one set of fissures, for example, Figure 5.12a, the inclination ω of that set is used. Where two sets of fissures exist (Figure 5.12b–d), the mean inclination of the two sets of fissures is used (ω1 + ω2)/2 to indicate the inclination of the fissure sets. Solutions have been evaluated for several values of the ratio of the fissure strength to the strength of the solid material cf/cm.

5.3.2.3 Footings on slopes

The bearing capacity of footings on slopes (rather than on level ground) is of interest to engineers as abutments to bridges are often founded on valley walls or are at the top of a slope. Shields et al. (1990) have produced design charts that are based on the centrifuge tests of Gemperline (1988) and other model tests.

Figure 5.13 shows the locations of footings with respect to the slope and definitions of the non-dimensional terms λ, η that give the location of the strip footing with respect to the slope. Design charts for two slopes (1V:2H) and (1V:1.5H) on cohesionless soil are shown in Figure 5.14a,b. The charts show contours of the percentage capacity P of a footing on level ground that the footing in a slope can carry.

ωiω

qu

B

Smoothinterface

Matrix shear strength, cmFissure shear strength, cf

ω ω

2 Orthogonalsets

Single set 2 Non-orthogonalsets

ωi

ω1ω2

ω

Figure 5.11 Strip footing on fissured clay.


The bearing capacity of a footing in a slope is therefore given by Equation 5.15 (for a cohesionless soil)

q BN q= 1

2γ γ

(5.15)

N qR

Bγ

φ φ= − −( )( )log10 10 100.1159 2.386 0.34 0.2

(5.16)

where Nγq may be found from Nγq = PNγqR (P is from Figure 5.14a,b).

0 10 20 30 40 50 60 70 80 900

1

2

3

4

5

6

q u/c m

cf /cm =0.1

0.2

0.3

0.4

0.50.60.70.80.9

0 10 20 30 40 50 60 70 80 900

1

2

3

4

5

6

q u/c m

cf /cm =0.1

0.2

0.3

0.4

0.5

0.6

0.70.80.9

0 10 20 30 40 50 60 70 80 900

1

2

3

4

5

6

q u/c m

cf /cm =0.1

0.2

0.3

0.4

0.50.60.70.80.9

0 10 20 30 40 50 60 70 80 90Fissure slope, ω (degrees)

0

1

2

3

4

5

6

q u/c m

cf /cm =0.10.20.30.40.50.60.70.80.9

Fissure slope, (ω1 + ω2)/2 (degrees)

Fissure slope, (ω1 + ω2)/2 (degrees) Fissure slope, (ω1 + ω2)/2 (degrees)

(a)ωi = 90° or single set

ωi = 30°(d)

ωi = 60°(b)

ωi = 45°(c)

Figure 5.12 Undrained bearing capacity of strip footing on fissured clay deposit: (a) one or two orthogonal fissure sets; (b) two fissure sets with included angle of 60°; (c) with included angle of 45°; (d) with included angle of 30°. (After Lav, M.A., Carter, J.P., and Booker, J.R. 1995. Proceedings of the 14th Australasian Conference on the Mechanics of Structures and Materials, Hobart, pp. 38–43.)


The value of ϕ is in degrees and the value of the footing width B is in inches in the above formula.

More recent solutions have been provided by Leshchinsky (2015) for c′–ϕ′ materials, where the footing is at the edge of the slope.

(i) (ii)

BB

b(–) b(+)

DO

Dβ

Legend:β = angle of slope with respect to horizontalD = depth of footing with respect to the level of the horizontal groundB = footing widthb = horizontal distance leading edge of footing is away from crest of slope

Note:λ = b/Bη = D/B

a

a

Figure 5.13 Definition of terms for a footing on a slope. (i) Footing on slope; (ii) footing at top of slope.

–6 –4 –2 0 2Note: Contours give percent capacity, P

3

2

1

0

η

21

5070

90110

130 150170 190

use 200%

–6 –4 –2 0 2

3

2

1

0

λ

Slope surface

–6 –4 –2 0 2Note: Contours give percent capacity, P

3

2

1

0

η

1.5

1 30 5070 90 11

013

0 150170 190

use 200%

–6 –4 –2 0 2

3

2

1

0

λ

Slope surface

(a)

(b)

Figure 5.14 (a) Suggested design factors for 2:1 (26.6°) slope. (After Shields, D., Chandler, N., and Garnier, J. 1990. Journal of Geotechnical Engineering, ASCE, Vol. 116, No. 3, pp. 528–537.) (b) Suggested design factors for 1.5:1 (33.7°) slope. (After Shields, D., Chandler, N., and Garnier, J. 1990. Journal of Geotechnical Engineering, ASCE, Vol. 116, No. 3, pp. 528–537.)


5.3.2.4 Layered soils

The bearing capacity of footings on layered soils presents a more difficult problem than that of a footing on a uniform soil, and so solutions are less plentiful and often approximate in nature. Problems may be divided into two categories, those involving sand and clay layers, and those involving clay layers of different strengths.

Sand and Clay Layers

If a loading is applied to a foundation that is constructed on a sand layer overlying a clay layer, then the footing can punch through the sand into the clay layer beneath, especially if the clay layer is soft. This type of failure has been examined by Meyerhof (1974), Meyerhof and Hanna (1978), Hanna and Meyerhof (1980), Michalowski and Shi (1995), Burd and Frydman (1997), and Shiau et al. (2002).

Solutions to this problem should be bounded by the solution for a footing on clay when the sand layer becomes thin and by the solution for a footing on sand when the sand layer becomes thick (as the failure would occur through the surface sand layer).

The solutions of Michalowski and Shi (1995) were obtained through the use of a kine-matic approach that will yield an upper bound to the bearing capacity of a footing, and so one would expect the collapse loads to be somewhat higher than the true solution. These solutions are presented in Figure 5.15a–d. The numerical results of Burd and Frydman (1997) were obtained from finite element- and finite difference-based computer codes, and so may still contain some error. These results (for an angle of friction of the sand of 40°) are presented in Figure 5.16.

In these figures, the full width of the foundation is B, the depth of the sand layer is D, the undrained shear strength of the clay is su, the angle of friction of the sand is ϕ′, and the unit weight of the sand is γ. The ultimate bearing pressure that can be carried by the footing is qu.

The accuracy of the results can roughly be assessed by looking at the limiting cases as mentioned above. For deeper sand layers, the value of qu/γB should approach 1/2Nγ because the failure would occur through the sand alone. The rigorous plasticity solutions of Martin (2004) (see Table 5.1) show that this value should be about 43 for a rough strip footing on a sand layer having an angle of friction of 40°. The charts of Michalowski and Shi show a value of 60 for this case while Burd and Frydman’s solution shows a value of about 50.

For thin sand layers, the results should be that for a clay layer alone so that a plot of pu/γB versus cu/γB should be a straight line with a gradient of 5.14 as in this case qu = Nc · su and Nc is equal to the classical value of (2 + π). This is certainly true for the Michalowski and Shi results; however, it is not so easy to see on the plot of Burd and Frydman because their plot is qu/γB versus su/γD.

Solutions have been found by Shiau et al. (2002) for the problem of a frictional material (sand or gravel) overlying a clay layer through use of the finite element based upper- and lower-bound theorems. In the plots, B is the full width of the footing, D is the thickness of the sand layer beneath the footing, q is the surcharge (at footing level), and γ is the unit weight of the sand. The plots of Figure 5.17 are based on the average between the upper and lower bounds.

Shiau et al. show that the results of Burd and Frydman, Michalowski and Shi and Hanna and Meyerhof are slightly overestimating bearing capacities, although if appropriate factors of safety are used, then the charts should be a reasonable guide for foundation design.

Clay Layers of Different Strengths

For clay layers of different stiffnesses, some solutions are available for the undrained case where the angle of shearing resistance can be taken as ϕu = 0 (Vesic 1975). For the case where


a soft clay layer (having a cohesion su1) overlies a deep stiffer clay layer (having cohesion su2), the bearing pressure can be found from

q s N qu u m s= +1 (5.17)

where the coefficient Nm can be found from Figure 5.18 for square or circular footings.For the case where a stiff clay layer (having undrained shear strength su1) overlies a deep

soft clay layer (having strength su2), then the bearing capacity can be estimated from

N N Nm c c c c= + ≤1

βκζ ζ( )

(5.18)

where κ = su2/su1, β = BL/[2(B + L)H], and ζc is the bearing capacity shape factor for the footing (see Equation 5.9). (Figure 5.18 shows the dimensions B, H. L is the length of the footing.)

0

5

10

15

20

25

30ϕ = 30° ϕ = 35°

ϕ = 40° ϕ = 45°

t/B = 0 0.5

2.0

t/B = 00.5

1.0

3.0 2.0

0 1 2 3 4 50

10

20

30

40

50

60

70

80

0 1 2 3 4 5

5.0 4.0

3.02.0

1.0

0.5

5.0

0.5t/B = 0

1.0

(a) (b)

(c) (d)

0

5

10

15

20

25

30

0

40

80

160

200

3.0

2.0

4.0

1.0

suγB

suγB

q u γB

120

q u γB

q u γB qu γB

t/B = 0

Figure 5.15 Dimensionless limit pressure for strip footing on a sand layer overlying clay. Case of no sur-charge. (After Michalowski, R.L. and Shi, L. 1995. Journal of the Geotechnical Engineering Division, ASCE, Vol. 121, No. GT5, pp. 421–428.)


210 3 4 5 6

0 0

00

5

5

10

10

10

20

20

40

40

30

50

15

1520

20

25

2530

30

35

35

4540

60

70

80

60

80

100

120

140

su/γ B210 3 4 5 6

su/γ B

210 3 4 5 6su/γ B

210 3 4 5 6su/γ B

ϕ′ = 50°ϕ′ = 40°ϕ′ = 30°ϕ′ = 50°ϕ′ = 40°

ϕ′ = 30°

D/B = 0·25Rough base

q/γ B = 1q/γ B = 0 ϕ′ = 50°

ϕ′ = 40°

ϕ′ = 30°ϕ′ = 50°

ϕ′ = 40°

ϕ′ = 30°

Rough baseD/B = 2.0

q/γ B = 1q/γ B = 0

ϕ′ = 50°

ϕ′ = 50°ϕ′ = 40°ϕ′ = 40°ϕ′ = 30°

ϕ′ = 30°

D/B = 2.0Rough base

q/γ B = 1q/γ B = 0

ϕ′ = 50°

ϕ′ = 40°ϕ′ = 50°ϕ′ = 30°ϕ′ = 40°

ϕ′ = 30°

Rough baseD/B = 0.5

q/γ B = 1q/γ B = 0

p γB p γBp γBp γB

Figure 5.17 Ultimate bearing capacity p for rough strip footings on a frictional material overlying a deep clay layer.

0 2 4 6 8 10 120

10

20

30

40

50

60

q u/γB

su/γD

B/t1.510.750.50.33

Figure 5.16 Bearing capacity for strip footing on a sand layer overlying a clay layer. ϕ″sand = 40°. (After Burd, H.J. and Frydman, S. 1997. Canadian Geotechnical Journal, Vol. 34, No. 2, pp. 241–253.)


Solutions have been found to the problem of a strip footing on a layer of clay overlying a deep layer of clay. The solutions were obtained by Merrifield et al. (1999) by using an optimisation technique that yields an upper and lower bound to the correct solution. Some solutions to the problem shown in Figure 5.19 for a footing of width B resting on a layer of clay with undrained shear strength su1 and thickness H which overlies another layer of clay having strength su2 and which is of great (infinite) depth.

The bearing capacity is given by

q N su c u= *1 (5.19)

The roughness of the footing only makes about 2% difference in the bearing capacity, and so the charts can be used for rough and smooth footings.

The solutions are shown in the plots of Figure 5.20a–d.

5.3.2.5 Working platforms

Often engineers are required to design platforms to support tracked machines. For exam-ple, if piling is being placed in soft soil that cannot support the weight of the piling rig, a platform of granular material is placed to provide sufficient bearing to support the rig. Geotextile reinforcement may also be placed at the base of the granular fill to increase the bearing capacity.

The British Research Establishment (BRE) document 2004 has been widely used for the design of working platforms. The approach is based on calculating the bearing capacity of

1 2 3 4 5 6 7 8 9 10 ∞6

7

8

9

10

11

12

13

Bear

ing

capa

city

fact

or N

m

Undrained strength ratio, κ = su2/su1

qs

H B

qu

su1su2

Square or circularfooting

≤4 6.17

qult = su1Nm + qs

B/H = ∞B/H = 40

20

16

12

8

Figure 5.18 Modified undrained bearing capacity factor Nm for square or circular footings on two layers of purely cohesive soil. (After Vesic, A.S. 1975. Foundation Engineering Handbook, New York.)


a rectangular shaped loading applied to a granular layer (the working platform) overlying either a cohesive layer (a clay) or a frictional layer (sand). The punching type bearing fail-ure of the loading through the upper-granular platform is assessed by using the theory of Meyerhof (1974).

Meyerhof and Hanna (1978) presented an improved method for calculating the bearing capacity of a granular material overlying a weaker material, and this could be used to cal-culate the bearing capacity rather than relying on the BRE tables.

The BRE document distinguishes between two load cases: Case 1 – where the crane or rig operator is unlikely to be able to aid recovery from an imminent platform failure; or Case 2 – where the rig or crane operator can control the load safely (e.g. by releasing the load or reducing power) if there is imminent platform failure. The amounts by which the loads are factored up when calculating the thickness depend on the loading case. Table 5.6 shows the factors recommended by the BRE.

The aim of designing a platform is to work out the thickness of a platform required to support the tracked vehicle. For a cohesive subgrade where the undrained shear strength of the clay is between 20 and 80 kPa, the thickness of the platform D is given by Equation 5.20, where

D W Fq s N s T W K su c c p p p= − − [ ( )] [ tan ]2 / / γ δ

0.5

(5.20)

Note that Equation 5.20 will give a negative value within the square brackets if the soil has high strength and it will not be possible to take the square root. This generally occurs if no platform is needed.

If the subgrade is a non-cohesive material such as sand then the platform thickness is given by Equation 5.21.

D W Fq WN s T W K ss s c p p p= − − [ ( )] [ tan ]

.½ / /γ γ δγ 2

0 5

(5.21)

B

su1 ≠ su2

∞

su1γ = 0

γ = 0su2

r

H

Assumed circular failuremechanism of Chen (1975)

θ

Figure 5.19 Strip footing on a two-layered clay deposit.


2

0

4

6

7

5

3

1

2

0

4

6

5

3

1

0 1.0 2.0 3.0 4.0 5.0

Upper bound Chen (1975)

Finite element upper boundFinite element lower boundSemi-empirical Meyerhof andHanna (1978), empiricalBrown and Meyerhof (1969)















H/B = 0.125

10

8

8

6

6

4

2

0

2

0

4

2

0

4

6

5

3

1

2

0

4

6

5

3

1

2

0

4

6

5

3

1

2

0

4

6

5

3

1

H/B = 0.375

H/B = 0.25

H/B = 0.5

H/B = 0.75

H/B = 1.5

H/B = 2

H/B = 1.0

N* c

N* c

N* c

N* c

N* c

N* c

N* c

N* c

su1/su2

0 1.0 2.0 3.0 4.0 5.0su1/su2

0 1.0 2.0 3.0 4.0 5.0su1/su2

0 1.0 2.0 3.0 4.0 5.0su1/su2

0 1.0 2.0 3.0 4.0 5.0su1/su2

0 1.0 2.0 3.0 4.0 5.0su1/su2

0 1.0 2.0 3.0 4.0 5.0su1/su2

0 1.0 2.0 3.0 4.0 5.0su1/su2

(a) (b)

(c) (d)

Figure 5.20 Variation of bearing capacity factor Nc*: (a) H/B = 0.125 and H/B = 0.25. (b) H/B = 0.375 and

H/B = 0.5. (c) H/B = 0.75 and H/B = 1. (d) H/B = 1.5 and H/B = 2.


In Equations 5.20 and 5.21,

q is the applied pressure from the tracked vehicleW is the width of one of the tracksF is the loading factor from Table 5.6 depending on the loading caseT is the ultimate tensile strength of the geogrid used divided by a factor of 2, for

example, T T= ult /2 (units are force per unit length, e.g. kN/m)γp is the unit weight of the platform materialsc is a shape factor for a rectangular track sc = (1 + 0.2W/L) and L is the track lengthsp is a shape factor for the geometry of the punching shear sp = (1 + W/L)Kp tan δ is the shearing resistance coefficient for the platform (see Figure 5.21)Nc is the bearing capacity factor for clay (generally Nc = 5.14)Nγs is the bearing capacity factor for a sand subgradeγs is the unit weight of a sand subgrade (use submerged unit weight γsub if sand is

submerged)

Several other checks should be performed when designing a platform

• The bearing capacity of the platform material alone should be checked to see if it is adequate.

• The bearing of the subgrade alone should be checked to asses if a platform is required at all.

2

35

4

0

6

40 45 50

K p tan

δ

8

10

12

18

16

14

20

ϕ′

Figure 5.21 Design values of Kptan δ. (BRE 2004. Working Platforms for Tracked Plant: Good Practice Guide to the Design, Installation, Maintenance and Repair of Ground-Supported Working Platforms, British Research Establishment, Garston, Watford.)

Table 5.6 Load factors F applied to track pressure from the vehicle

Loading conditionPlatform required

Yes No

Case 1 2.0 1.6Case 2 1.5 1.2


• The track loading should not come closer than half of the machine width from the edge of the platform.

• There may be soft spots in the subgrade, for example, backfilled trenches. These should be eliminated if possible.

• Slopes should ideally be less than 1 in 10 so that plant does not become unstable.

5.4 NUMERICAL ANALYSIS

For difficult problems involving complex material properties, complex geometry, or where structural elements such as retaining walls or anchors, finite element or finite difference methods are often used. Generally, care needs to be taken, as the collapse loads calculated are greater than the exact theoretical loads (see Sloan and Randolph 1982). However, for engineering purposes, the flexibility of numerical methods is of advantage and these meth-ods are often used.

Computer programs such as the finite element (FE) codes that are specifically written for geotechnical applications such as PLAXIS2D (2014) and PLAXIS3D Foundation or PLAXIS3D Tunnel, CRISP (2013), Phase2 (2013), or the more general FE code ABACUS (2014) may be used. The finite difference codes FLAC (Ver. 7.0) or FLAC3D (Ver. 5.0) may also be used (ITASCA Consulting Group 2014).

If a margin of safety against collapse of a geotechnical structure is required, a “c–ϕ” reduc-tion can be carried out. The process involves reducing the strength of the soil until failure occurs. The amount by which the strength parameters are reduced gives the margin of safety.

Alternatively, load can be applied and displacement at some point monitored until there is evidence of large displacement occurring which indicates collapse. An example is shown in Figure 5.22 for a strip footing with a total width of 2 m resting on a uniform weightless soil with strength parameters c′ = 10 kPa and ϕ′ = 20°. A non-associated flow rule was used with a Mohr–Coulomb failure surface where the angle of dilation for the soil was taken as ψ = 10°, the modulus of elasticity for the soil as E = 30,000 kPa and Poisson’s ratio as ν = 0.25. The load may be seen to level off at about 153 kPa, which is slightly more than the rigorous plasticity solution of 148.3 kPa (3% difference).

0

20

40

60

80

100

120

140

160

180

0.00 0.01 0.02 0.03 0.04 0.05

Ave

rage

pre

ssur

e (kP

a)

Deflection (m)

Figure 5.22 Load–deflection curve for strip footing on weightless soil with c′ = 10 kPa, ϕ′ = 20°, ψ = 10°. Rigorous plasticity solution for collapse load = 148.3 kPa.


For such a model, the numerical analysis has been noted to become unstable if there are large differences between the angle of friction ϕ and the dilation angle ψ (e.g. ϕ = 40° and ψ = 0°). This is often the cause of instability in calculation with commercial codes as well.

The plot of velocity vectors at failure shows a wedge of material moving down beneath the footing and material being pushed up to the side of the footing as shown in Figure 5.23. In the figure, the extent of the mesh that is composed of eight-node isoparametric elements may be seen. Increments of displacement are specified for the nodes beneath the footing and the loads at the nodes are back figured to obtain the corresponding load.

5.5 SETTLEMENT

If a structure settles substantially with very small differential settlements (settlements are almost uniform everywhere), then the structure is unlikely to sustain any damage, although if settlement is too large, the structure will be subject to serviceability problems (i.e. water, gas, and sewer pipes and other services may be damaged). It is the differential settlements that cause cracking and damage to structures (i.e. movement of one part of a structure rela-tive to another). Tilt may also be a problem for crane rails or machinery that has to remain level. It is therefore necessary to try to limit the differential settlements as well as the overall vertical movements of a structure.

The magnitude of allowable settlements depends on the type of structure or the purposes for which the foundation is to be used. Here, only static loadings are considered (i.e. not cyclic or vibratory loads).

0.0 5.0 10.0 15.0–2.0

0.0

2.0

4.0

6.0

8.0

10.0

12.0

Velocity vectors

Rigid strip footing

–0.00021A

Scale

5.00E–04

Figure 5.23 Velocity vectors at failure for strip footing on weightless soil with c′ = 10 kPa, ϕ′ = 20°, ψ = 10°. (Half of problem shown due to symmetry.)


5.5.1 Limits of settlement

Allowable deflections (or angular distortions) of different structures have been proposed by Skempton and MacDonald (1956), Bjerrum (1963), and Grant et al. (1974). More recently, Boone (1995) has proposed a method for determining damage to structures that relies on many factors such as the ground movement profile, the geometry of the structure, and the critical strains in the building materials.

Allowable angular distortions for different types of structures and construction materials have also been presented by Polshin and Tokar (1957) and these are shown in Table 5.7. It may be seen from this table that in general, stiffer structures such as those constructed from

Table 5.7 Allowable settlements and deflections

Item no. Description of standard value

Subsoil

Sand and clay in hard condition

Clay in plastic condition

1 Slope of crane way, as well as tracks for bridge crane truck

0.003 0.003

2 Difference in settlement of civil and industrial building column foundations:

(a) for steel and reinforced concrete frame structures

0.002ℓ 0.002ℓ

(b) for end rows of columns with brick cladding 0.007ℓ 0.001ℓ(c) for structures where auxiliary strain does not

arise during non-uniform settlement of foundations (ℓ = distance between foundation centres)

0.005ℓ 0.005ℓ

3 Relative deflection of plain brick walls:(a) for multi-storey dwellings and civil buildings

at /H< 3 0.0003 0.0004

at /H 5 (ℓ = length of deflected part of wall,H = height of wall from foundation footing)

0.0005 0.0007

(b) for one-storey mills 0.0010 0.00104 Pitch of solid or ring-shaped foundations of high rigid

struc tures (smoke stacks, water towers, silos, etc.) at the most unfavourable combination of loads

0.004 0.004

Item no. Kind of building and type of foundation Average settlement (cm)

1 Buildings with plain brick walls on continuous and separate foundations with the wall length ℓ to the wall height H (H counted from the foundation footing): /H >2.5 /H<1.5

8

102 Buildings with brick walls, reinforced with reinforced

concrete or reinforced brick belts (not depending on the ratio of ℓ/H)

15

3 Framed buildings 104 Solid reinforced concrete foundations of blast

furnaces, smoke stacks, silos, water towers, etc.30

Source: After Polshin, D.E. and Tokar, R.A. 1957. Proceedings of the 4th International Conference on Soil Mechanics and Foundation Engineering, Vol. 1, pp. 402–405.


masonry, are more likely to be damaged than flexible structures such as those having steel frames with sheet cladding.

5.5.2 Settlement computation

When a load is applied to a footing on a non-cohesive, free draining soil like a sand or a gravel, the soil will deform immediately and will reach its final settlement (provided there is no creep settlement). However, when a load is applied to a saturated clay, the load is partly supported by the soil skeleton and partly by the pore water. There is therefore an increase in the pore water pressure beneath the foundation and water will begin to flow from areas of high water pressure to areas of low water pressure, and as it does, the soil beneath the footing undergoes a volume reduction. The footing will therefore settle with time until all excess pore pressures have dissipated, at which time the footing will have reached its final settlement (again if creep is not considered).

When the load is first applied, the footing will undergo an initial settlement with the soil deforming at constant volume (since the water does not have time to flow through the soil). This can only occur if the load is not one-dimensional so that shear deformation can take place.

Generally, it is the final settlement of a foundation that is of interest as this is the largest settlement that will occur and therefore this leads to the largest differential settlements. However, the load–deflection behaviour may be of importance in some cases.

5.5.3 Theory of elasticity

Working loads applied to foundations are usually selected so that they are well below the ultimate bearing capacity of the building. It is therefore possible to treat the soil as an elastic material, provided that the elastic modulus used in any computations is appropriate for the stress range that occurs in the ground beneath the footing.

For obtaining the initial settlement Si of the footing, the undrained elastic modulus Eu and Poisson’s ratio νu of the soil are needed. As the soil is assumed to undergo zero volume change, the value of νu is taken to be 0.5 as this can be shown to be the requirement for zero volume change in an isotropic material. For computing the final settlement Sf, the drained elastic modulus E′ and Poisson’s ratio ν′ are used. For free draining sands and gravels, only the drained parameters will apply.

For one-dimensional loadings (i.e. where the load is large in extent compared to the thick-ness of the soil layer such as a surcharge), the elastic theory of Terzaghi (1923) may be used. One-dimensional theory is based on the assumption that the strain will be entirely vertical and that there will be no lateral strains. Vertical settlement S is then given by

S m hi

N

vi vi i= ′=

∑1

∆σ

(5.22)

wheremvi is the coefficient of compressibility of layer i of the soil∆ ′σvi is the increase in vertical effective stress in layer i of the soilhi is the thickness of the ith layer of soilN is the number of layers into which the soil is divided

The soil layer is divided into horizontal layers each of thickness hi and the stress increases are computed at the centre of each layer. The coefficient of compressibility mv is found from


an oedometer test and is calculated from the strain in the specimen divided by the stress change (Section 1.5). As the value of mv is dependent on the stress level, the appropriate value should be used for each layer. A value corresponding to the mean stress (i.e. intermediate of the initial stress and the stress after application of the load) is appropriate. For strictly one-dimensional problems, the stress increase ∆ ′σvi should be the same for all layers and equal to the applied uniform load, however, vertical stress increases are often calculated for rectan-gular or circular loading shapes using three-dimensional solutions for the stress increases. This is in effect carrying out a pseudo three-dimensional analysis as stresses are calculated from three-dimensional solutions and then the one-dimensional equation (Equation 5.22) is used to compute settlements.

Although Equation 5.22 allows the compressibility of the soil to be stress dependent, it does not allow for the compressibility to change within a stress increment within each layer. An alternative method that uses the compression Cc and recompression Cr indices can be used. It is observed from laboratory testing that the soil becomes more compressible after the previous pre-compression pressure ′σp is reached. The pre-compression pressure is the greatest pressure that the soil has been subjected to in the past and may be due to layers of soil being eroded away that were once present, or is often due to swelling and shrinking of the surface layer of the soil that is subjected to wetting and drying. The settlement S is given by Equation 5.23 when the soil is divided into N horizontal sub-layers in order to perform the calculation.

SC

eC

ei

Nr

i

pi

i

c

pi

fi

pi

=+

′′

++

′′

=∑

1 010

010

1 1log log

σσ

σσ

⋅ hi

(5.23)

In Equation 5.23

′σ0i is the initial vertical effective stress in the soil for layer i′σpi is the pre-consolidation pressure in layer i′σ fi is the final effective stress in layer i after loading

e0i is the initial void ratio in layer iepi is the void ratio in layer i at the pre-consolidation pressure

For problems that are three-dimensional in nature, elastic solutions for the settlement of foundations of various shapes and stiffnesses and subjected to various types of load-ings, are available and may be found in books such as the one by Poulos and Davis (1974). Provided that the appropriate solution can be found, the settlement of the foundation can be calculated.

The settlement of flexible footings that are rectangular in plan (of length L by breadth B) and founded at depth D in uniform soil layers of finite depth may be computed from the charts of Christian and Carrier (1978). These charts are reproduced in Figure 5.24 and apply for only the undrained case (νu = 0.5). Elasticity solutions are normally presented in non-dimensional form so that they apply to a wide range of problems. In addition, solutions for strip or rectangular shaped loadings are often presented for the edge (strip) or corner (rectangle) because the solutions may then be superimposed. For example, the deflection at the centre of a rectangle may be obtained by dividing the rectangular loaded area into four equal rectangular areas and summing the deflections under the corners of each.

Figure 5.25 shows solutions for a rigid circular foundation constructed on a layer of soil of finite thickness. This plot may be used to determine both initial and final settlements, as


the solutions have been evaluated for a range of Poisson’s ratios. The solutions apply only to foundations on the surface of the clay layer and may also be used for square footings since the settlement of a square footing may be approximated as the settlement of a circular foot-ing having the same area in plan with very little error.

5.5.4 Rate of settlement

The rate at which footings settle may be determined by carrying out a consolidation analy-sis. It is generally assumed that the soil is totally saturated, and that as water flows from the soil, the soil skeleton slowly deforms. The volume change in the soil skeleton is assumed to be equal to the volume of water that flows from the pores in the soil.

If the problem is one-dimensional, there can be no initial settlement (i.e. upon application of a load) as the pore water is incompressible, but for three-dimensional problems, shear deformation of the soil can take place therefore the soil can undergo an initial undrained deformation at constant volume.

Plots have been produced to allow estimates to be made of the rate of consolidation of footings, and an example of such a plot is shown in Figure 5.26 (Davis and Poulos 1972). This plot is for a circular footing of radius a resting on the surface of a clay layer of thick-ness h, which may drain across its upper surface and through its lower surface. It is assumed

0 5 10 15 200.8

0.9

1.0

D/B

μ 0

0.1 1 10 100 10000

0.5

1.0

1.5

2.0

μ 1

H/B

L/B = ∞L/B = 10

L/B = 5

L/B = 2

Square

Circle

ρ = Average settlement

ρ =μ0 μ1qBE

D

H B

qL = Length ν = 0·5

Figure 5.24 Chart for evaluating settlement of embedded rectangular footings. (After Christian, J.T. and Carrier, W.D. 1978. Canadian Geotechnical Journal, Vol. 15, pp. 123–128.)


h/a

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

I ρ

a/h

r

Pa

h

z

pav =P

πa2

Settlement

S =pava Iρ

E

ν = 0

0.20.4

0.5

00 0.2 0.4 0.6 0.8 1.0

1.0 00.20.40.60.8

Figure 5.25 Chart for evaluating settlement of rigid circular footing on a soil layer of finite thickness.

11.0

0.8

0.6

0.4

0.2

0

Up =

Us

10–4 10–3 10–2 10–1

c1th2Tv =

a

h

251020

50

0 (One-dimensional)

Values of h/a1

0.5

Figure 5.26 Degree of settlement versus time factor for circular footing on a soil layer drained at the upper and lower surfaces. (After Davis, E.H. and Poulos, H.G. 1972. Géotechnique, Vol. 22, No. 1, pp. 95–114.)


that the load is applied to the footing at time t = 0 and then held constant. In the plots, the degree of consolidation U is plotted versus the time factor Tv where

T

c th

vv= 1

2

(5.24)

and cv1 is the one-dimensional coefficient of consolidation that may be measured in an oedometer test.

If the extent of the load is large enough, it can be seen that the consolidation curve reaches the one-dimensional curve (h/a = 0).

The degree of consolidation U is defined as

U

S SS S

t i

f i

= −−

(5.25)

where St is the settlement at any time, Sf is the final settlement, and Si is the initial settle-ment. Initial and final settlements can be calculated using the appropriate elastic moduli and Poisson’s ratio as described above (Section 5.5.3).

Settlement at any time t can therefore be computed (from Equation 5.25) once the degree of consolidation is known and the initial and final settlements have been found.

EXAMPLE 5.2

A square footing having a side of 2 m is constructed on a layer of soft clay that is 4 m thick and has the following properties

′ = ′ = == =

E c

Ev

u u

8 1MPa m /yr

12 MPa

2νν

0.3 0.8

0.5

The footing is constructed on a granular base layer of permeable material, so that the surface of the clay layer may be considered drained. The base of the clay layer is underlain by sand, therefore drainage can take place at the base of the clay layer also.

If a vertical load of 400 kN is applied to the footing, what would be the expected settle-ment of the footing after 1 year?

Solution

• The pressure applied to the footing would be pav = 400 kN/(2 m × 2 m) = 100 kPa• The square footing may be treated as a circular footing of area equal to the square,

therefore πa2 = 2 × 2 giving a = 1.128 m• The thickness of the soil layer to the radius of the footing would be

h a a h/ /4 m/ m or= ==1.128 3.55 0.28

1. Drained Case

Figure 5.25 gives (for Poisson’s ratio ν′ = 0.3) an influence factor Iρ = 1.1

Therefore, the final settlement Sf is given by

S

p aIE

fav=

′= × × =ρ 100

81.128 1.1

15.5 mm


2. Undrained Case (or Initial Settlement)

Figure 5.25 gives an influence factor of Iρ = 0.84 (for νu = 0.5)

The initial settlement Si is given by

S

p aIE

iav

u

= = × × =ρ 10012

1.128 0.847.9 mm

3. At 1 year, the time factor can be computed as

T

c th

vv= = × =2

116

0.80.05

and so from Figure 5.26, the degree of consolidation U can be found to be 0.82. We can therefore write

U

S SS S

St i

f i

t= −−

= −−

=7.915.5 7.9

0.82

and solving for St, gives the settlement at 1 year St ≈ 14 mm.

5.5.5 Settlement of footings on sand

Often footing design for sand foundations is based on field-test results (e.g. standard pen-etration tests [SPT] or cone penetrometer results) because of the difficulty of sampling and testing sand samples in the laboratory (i.e. testing samples at the correct density). The design must again be based on both settlement criteria and bearing criteria.

Generally, a small or narrow footing subjected to an increasing pressure will have an unacceptably small margin of safety against a bearing capacity failure before it settles too much, whereas a larger or wider foundation will reach a given settlement criterion before the margin of safety against a bearing capacity failure is exceeded. Therefore, for smaller foundations, bearing capacity influences design, but for larger foundations, settlement is of greater importance.

Vibration can cause footings on loose sand to settle unduly as the sand will tend to densify (or even liquefy if it is saturated); however, this aspect is not considered in the following. Footings to be constructed on loose sands, which are often the result of hydraulically placed fills, should be considered in the light of potential vibration or earthquake hazard and den-sification of the sand deposit may be necessary by means such as vibroflotting.

Methods Based on SPT

D’Appolonia et al. (1970) presented a method based on obtaining the elastic modulus of the sand from the average SPT blow count taken over one footing width B beneath the foun-dation. The settlement of a footing can be found once the modulus is obtained, by use of the theory of elasticity (see Section 5.5.3). The relationship between elastic modulus E′ and the blow count is shown in Figure 5.27 (ν′ is Poisson’s ratio of the sand).

Burland et al. (1977) presented a compilation of observed data that can be used to deter-mine the settlement of footings on sands of different densities. The deflection of a foun-dation per unit applied pressure is presented for different footing widths B as shown in Figure 5.28. Suggested upper limits for the deflection are shown (as lines) on the plot for dense, medium dense, and loose sands. The density of the sand can be determined from the SPT blow count as shown in the inset to Figure 5.28.


Parry (1977) presented a method based on SPT results in which the likely settlement of strip or rectangular footings of width B may be computed.

The SPT blow count at a representative depth of 3/4B beneath the base of the footing is used in the design, and this is not corrected for depth or for the effects of the water table. If Nm is the SPT blow count at the representative depth, then the settlement of a footing may be estimated from

ρ = 300

qBNm

(5.26)

where q is in MPa, B is in metres, and ρ is the settlement of the footing in mm.

0 10 20 30 40 50 60 700

25

50

75

100

E′/(1

–ν′2 )

(MN

/m2 )

Average measured SPT resistance depth B below footing (blows/ft)

All data for footing foundations on cleansand or sand and gravel

Pre-loaded sand

Normally loaded sandor sand and gravel

Figure 5.27 Correlation between blow count and modulus of sand. Blow count is the average over a depth B below the footing. (After D’Appolonia, D.J., D’Appolonia, E. and Brissette, R.F. 1970. Journal of the Soil Mechanics and Foundations Division, ASCE, Vol. 96, No. SM2, pp. 754–762.)

Upper limit formedium dense

Upper limitfor dense

Key NLoose < 10Medium 10–30denseDense > 30

0.1 1.0 10.0 100.01.0

10

100

0.1

1.0

mmkN/m2( )

Breadth B (m)

L

mm

kg/c

m2

()

Settl

emen

t per

uni

t app

lied

pres

sure

Tentative – Loose sand

Figure 5.28 Observed settlement of footing on sand of various relative densities. (After Burland, J.B., Broms, B.B., and de Mello, V.F.B. 1977. State-of-the-Art Review, IX International Conference on Soil Mechanics and Foundation Engineering, Tokyo, Vol. 2, pp. 495–546.)


To be able to use any of these methods for deflection calculation, the load applied to the foot-ing must be well below the collapse load, therefore a bearing capacity check would also need to be carried out to verify that the loading level had an adequate factor of safety against collapse.

Method Based on Static Cone Penetrometer

The approach of Schmertmann (1970) and Schmertmann et al. (1978) is based on cone penetrometer results. The soil is divided up into N depth increments or layers and the expected settlement ρi is then computed from the formula

ρi

i

Ni

siziC C p

zE

I==

∑1 2

1

∆ ∆

(5.27)

whereC1 is a correction to allow for strain relief from embedmentC2 is a correction for time dependent increase in settlementΔp is the net applied footing pressure Δp = p − p0 (see Figure 5.29b)Δzi is the depth incrementIzi is the influence factor for soil layer i (see Figure 5.29a)Esi is the elastic modulus of soil layer i

6B

0 0.2 0.4 0.6 0.8 1.0

Rela

tive d

epth

bel

ow fo

otin

g le

vel z

/B

Rigid footing vertical strain influence factor = Iz

Plane strainL/B > 10

For plane strainuse Es = 3.5 qc

AxisymmetricL/B = 1

Izp = 0.5 + 0.1 √Δpσ′vp

........(*)For axisymmetricaluse Es = 2.5 qc

(a)

Modified strain influencefactor distributions

Explanation of pressure termsused in equation (*) above

B

p

σ′vp

p′o

B/2 (axisym)B (pl. str.)

Depth to Izp

4B

3B

2B

B

0

B/2

Δp = p – p′o

(b)

Figure 5.29 Strain influence factors for use in Schmertmann’s method. (After Schmertmann, J.H., Hartman, J.P., and Brown, P.R. 1978. Journal of the Geotechnical Engineering Division, ASCE, Vol. 104, No. GT8, pp. 1131–1135.)


Factor C1 allows for strain relief due to embedment of a footing and is computed from the ratio of the overburden pressure at foundation level p0 to the net pressure increase at founda-tion level Δp = p − p0 (p is the applied pressure).

C

pp

101= −

0.5∆

(5.28)

Note that C1 must be greater than or equal to 0.5.Factor C2 is used to allow for creep of the foundation and is given by

C2 101= +

0.20.1

logt

(5.29)

where t is the time in years.The soil is divided into a number of horizontal layers and the elastic modulus of the soil

is estimated from the cone penetrometer tip resistance qc. The elastic modulus may be found from the following correlations:

Long or strip footings /

Axisymmetric f

L B E qsi c≥ =10 3.5

oootings or /L B E qsi c= =1 2.5 (5.30)

L = length and B = breadth of a rectangular footing.

The strain distributions beneath footings are different depending on the footing shape and vary between the two extremes of the distributions for a strip or for a square footing. Therefore, different influence factor diagrams are used for square (or circular) and strip-footings (see Figure 5.29). For intermediate cases, that is, rectangular footings of length L and breadth B (L/B from 1 to 10), interpolation between the two diagrams may be used. Iz may be assumed to vary linearly between 0.1 and 0.2 on the Iz axis and z/B to vary from 2 to 4 on the z/B axis.

It may be noted that the peak value on the influence factor plots varies and is given by

I

pzp

vp

= +′

0.5 0.10.5

∆σ

(5.31)

The effective stress used in computing the peak value of the influence factor ′σvp is calcu-lated at the depth shown in the inset to Figure 5.29b.

To design footings using this approach, the field SPT results must be corrected to allow for the effect of depth. The correction factor CN is shown plotted against vertical effective stress in Figure 5.30, so that at any depth the corrected blow count Ncor can be obtained by multiplying the field value Nfield by the correction factor, that is,

N N CNcor field= × (5.32)

5.5.6 Methods based on settlement and bearing criteria

Some methods of footing design (for sands) are based on both settlement and bearing cri-teria. An example of this is the well-known approach of Peck et al. (1974). This method is


based on SPT results, and allows for a factor of safety of 2 against a bearing capacity failure or an allowable settlement of 25 mm (1 inch).

The average corrected blow count (using the correction factors of Figure 5.30) over a depth of one footing width beneath the foundation is then used to design the foundation using the charts shown in Figure 5.31. This is straightforward if the size of the footing is known and an allowable pressure is being obtained, but if the applied pressure is known and the footing size is to be determined, some trial and error is required as the average blow

450

00.4 0.8 1.2 1.6 2.0

50

100

150

200

250

300

350

400

Effec

tive v

ertic

al o

verb

urde

n pr

essu

re, k

Pa

Correction factor CN = NcorNfield

Figure 5.30 Chart for correction of N values in sand for influence of overburden pressure (reference value of effective overburden pressure = 1 ton/sq ft, 96 kPa). (After Peck, R.B., Hanson, W.E., and Thornburn, T.H. 1974. Foundation Engineering, 2nd ed. Wiley, New York.)

0

100

200

300

400

500

600

Soil

pres

sure

, kPa

0 0.4 0.8 1.2

(a) (b) (c)Df /B = 1 Df /B = 0.5 Df /B = 0.25

Width of footing, B, m

50

40

30

2015105

50

40

30

2015105

50

40

30

2015105

N = N = N =

1.60 0.4 0.8 1.20 0.4 0.8 1.2

Figure 5.31 Design chart for proportioning shallow footings on sand. (a) Df /B = 1; (b) Df /B = 0.5; (c) Df /B = 0.25. (After Peck, R.B., Hanson, W.E., and Thornburn, T.H. 1974. Foundation Engineering, 2nd ed. Copyright Wiley, New York.)


count over one footing depth may change with the size of the footing selected. Full details of the method are given in the book by Peck et al. (1974).

A correction is also made for the effect of the water table by multiplying the allowable pressure (from Figure 5.31) by the correction factor Cw, where

CD

D BD D B

C D D B

ww

w

w w

= ++

≤ +

= > +

0.5 0.5

1.0

(5.33)

In the above equation, B is the footing width, D is the footing depth, and Dw is the depth of water below the surface. It may be seen that the correction is 0.5 when the water table is at the surface and 1.0 when greater or equal to one footing width below the base of the foundation – anywhere between these two extremes, the formula (Equation 5.33) provides a linear interpo-lation. The reasoning is that the unit weight of the cohesionless material will be approximately halved when submerged and therefore the bearing capacity will be similarly reduced.

5.6 NUMERICAL APPROACHES

Finding the appropriate analytic solution based on the theory of elasticity may be difficult, especially if the soil is anisotropic or consists of layers that have different stiffnesses. In such cases, numerical solutions are often necessary and some of these methods are discussed in the following sections.

5.6.1 Layered soil: Finite layer approaches

Computer programs can be useful in the situation where the soil is layered (see Chapter 2). Finite layer programs (Small and Booker 1986, 1996) allow elastic anisotropic layered soils subjected to multiple surface loadings to be analysed. The surface loadings may be strip, circular, or rectangular in shape, and the programs may be used to find elastic deflections only, or the time–settlement behaviour of a loading. One advantage of the formulation is that solutions may be found for a Poisson’s ratio of 0.5 (i.e. the undrained solution). As for any solutions based on elasticity, the method can be used for prediction of settlement when the loads are not near failure but restricted to the initial (approximately linear) part of the load–deflection curve. Undrained soil parameters Eu and νu = 0.5 are used to compute the undrained settlements and the drained soil parameters E′ and ν′ are used to compute the drained settlements.

The approach can be used for estimating time–settlement behaviour as well. In this case, the soil is treated as a two-phase material (soil skeleton and pore water) and the settlements at required times can be calculated (Booker and Small 1987, Small 2012). The drained modulus and Poisson’s ratio and the soil permeability are needed for these calculations. The soil may have anisotropic permeability and the loads may be time dependent in this kind of analysis.

The finite layer method is particularly easy to use, as the only information required is the thickness of each layer and the material properties of each layer (Small and Booker 1984, 1986). The only restriction is that the material properties of each material layer do not vary laterally (but can vary from layer to layer).

A plot is shown in Figure 5.32 of the surface deflection computed beneath a single circular uniform loading applied to the surface of a soil layer consisting of two isotropic sub-layers where the modulus E1 of the uppermost layer is greater than that of the bottom layer E2. All the parameters used in the computation and the geometry of the problem are shown on this plot.


5.6.2 Finite element methods

For soft clays or soils loaded close to failure, the use of simple elastic models will not be applicable, and a more sophisticated means of analysis is needed to compute settlements. The most common methods used are finite element techniques or finite difference based methods. Many commercially available computer codes (see those listed in Section 5.4) can perform non-linear analyses for problems that are two- or three-dimensional in nature.

The analysis may be linear or non-linear. If the soil is treated as being elastic, the settle-ments can be computed for the case of rapid or undrained loading (using the undrained soil parameters Eu and νu) although generally a Poisson’s ratio of 0.5 cannot be used with finite element programs because they are based on a stiffness formulation. A value close to 0.5 (say, νu = 0.49) can be used instead. For slow loading or loading where there is no pore pres-sure build up as in sands or gravels, the drained parameters E′ and ν′ may be used.

One advantage of these numerical models is that complex geometries, loadings and mate-rial distributions can be included in the analysis. The soil can also be treated as a two-phase material (the pore water and the soil skeleton) and consolidation behaviour can be modelled. Time-dependent consolidation is of great importance for clays that have a low permeability so pore water pressures can build up when loads are applied to the clay layer. With time, the pore water pressures can dissipate as pore water flows from areas of high pore water pres-sure to areas of low pore water pressure, and as this takes place, the soil will consolidate and undergo settlement.

If loads are high enough to cause yield or failure of the soil, then non-linear models of soil behaviour can be used with numerical models. There are two types of analysis that can be performed:

• Single-phase elasto-plastic analysis where the soil is treated as consisting of a single elasto-plastic material and the appropriate soil model (or constitutive model) is used.

• For clay there are the undrained and drained cases. For the rapid loading or und-rained case undrained deformation Eu and νu = 0.49 and strength parameters su and ϕu = 0 are used. As the undrained strength of clays generally increases lin-early with depth, the strength increase in su is modelled. The drained case is less

0 0.5 1.0 1.5 2.01.00

0.75

0.50

0.25

0

Defl

ectio

n ×

E 1/p av

z/a

rpav

a/2

a

2a

E1, ν1

E2, ν2

z

E1E2

= 2ν1 = 0.3ν2 = 0.5

Figure 5.32 Vertical surface displacement beneath circular loading obtained from finite layer analysis.


common, but can be modelled using the effective strength parameters c′ and ϕ′ in the analysis and the drained deformation parameters E′ and ν′. A Mohr–Coulomb failure criterion can be used in this case.

• For permeable soils such as sands and gravels, the soil can be modelled using the drained soil parameters E′ and ν′ and c′ and ϕ′ with a Mohr–Coulomb soil model. More sophisticated models of soil deformation involving estimation of more soil parameters can be used.

• Two-phase elasto-plastic consolidation analysis may be used particularly when the loading is applied at different rates so that it is sometimes a fast undrained loading and sometimes a slow loading. An example may be an embankment being constructed on a soft clay where the embankment is built up to a point where pore pressures are becoming high, and then left for a period of time to allow the pore pressures to dis-sipate before loading begins again. Such analyses are mostly applied to clays as they have low permeabilities and allow pore pressures to build up.

The finite element equations that are solved when considering pore pressures using a ‘marching’ solution are given in Equation 5.34 (and Section 3.5).

K L

L q

f

q−

− − −

=

T

w t wt( )1 α ∆ Φ∆∆

∆∆ Φt / /γ

δγ

(5.34)

The stiffness matrix of the soil K is the same as for any finite element stress analysis and depends on the type of element used. The coupling matrices L arise from the coupling of the pore water pressures and the deformation behaviour of the soil. Φ is the flow matrix, and arises from the flow of water through the soil, and γw is the unit weight of water. The term α arises from integration of the pore pressure over a time step. These matrices are defined as

K B L E kE= = =∫ ∫ ∫V

T

V

T

V

dV dV dVDB ad Φ T

(5.35)

The matrix D in Equation 5.35 contains the constants relating stress to strain under con-ditions of elasticity or if the soil is yielding it is the incremental plasticity matrix, k is the matrix of permeability, B is the matrix relating strain to nodal displacement within an ele-ment, a is the vector of shape functions relating pore pressure within the element to its nodal values, d is the vector relating the volume strain within an element to its nodal displacements, and E is the matrix relating the gradient of pore pressure to the nodal values of pore pres-sure. The form of the matrices and vectors B, E, a, d depend on the form of element chosen.

In Equation 5.34, the changes in displacements, and excess pore pressures q (at each node of the finite element mesh) can be found from the excess pore pressure field at the previous time step qt. Initially, as the excess pore pressures are zero, the flow term in the right-hand side is zero, therefore the equations can be solved and new values of pore pressure found at time t = Δt and the process marched forward.

The values of displacement and excess pore pressure at each time are found by updating the previous solutions as shown in Equation 5.36.

δ δ δt t t

t t t

+

+

= += +

∆

∆

∆∆q q q

(5.36)


As with many ‘marching’ processes, errors in the previous solution can be magnified in the current solution and the process can become unstable. The stability criterion has been established by Booker and Small (1975) who showed that the process is unconditionally stable as long as the integration parameter lies in the range 0 ≤ α ≤ 0.5, but that it can be stable under other conditions that depend on the eigenvalues of the consolidation equations.

It may be necessary for problems where a constant load is applied and then held constant to increase the time step during calculation as the rate of consolidation slows down with time. However, when the time step in Equation 5.34 is increased, the set of equations has to be set up again and resolved.

An example of a finite element calculation using eight-node isoparametric elements to interpolate both the displacement field and the pore pressure field is given in Figure 5.33. The problem involves a circular uniform loading q applied to the region 0 ≤ r ≤ a on the surface of a soil layer of depth h. The entire upper surface of the soil is assumed permeable and the base is assumed to be underlain by a permeable layer of sand. The parameters used for the solution are presented in Table 5.8.

The settlement at the central point of the loading versus time is presented in Figure 5.33. In this figure, the finite element solution is shown plotted over seven log cycles of time. The time step has been increased by a factor of 5 nine times during the calculation as can be seen from the increased distance between the plotted circles where the time step is increased.

–0.075

–0.07

–0.065

–0.06

–0.055

–0.05

–0.0450.001 0.01 0.1 1 10 100 1000 10,000

Settl

emen

t (m

)

Time (days)

Figure 5.33 Settlement versus time for the central point of circular loading on a soil layer of finite depth – drained upper and lower surfaces of clay layer.

Table 5.8 Properties used in finite element analysis

Quantity Value

Drained modulus of elasticity 10,000 kPaDrained Poisson’s ratio 0.35Radius of load a 4 mDepth of layer h 16 mVertical and horizontal permeability kv = kh 0.0001 m/dayUniform load q 80 kPa


Non-Linear Solutions

It is necessary to use the correct model of soil behaviour in this case, as the model must be capable of giving the correct behaviour for all rates of loading. Models that contain a yield and a failure surface are best suited to this kind of analysis as they can correctly simulate the soil behaviour when undrained. For example, Figure 5.34 shows the effective stress path that may occur in a soil that was initially in an isotropic stress state. Where the stress path reaches the failure surface (at point ‘X’) the deviator stress is the undrained shear strength of the soil (i.e. su = qfailure). The stress path can be seen to deviate to the left after it intersects the oval-shaped yield surface.

Various soil models may be implemented, one of the most popular, being the Cam Clay Model that was developed for clays (see Britto and Gunn 1987). Commercial computer codes that include models that can be used when a clay is treated as a two-phase mate-rial (i.e. soil skeleton and pore water) are the Modified Cam Clay Model (CRISP, Phase2, ABACUS, FLAC) or the Soft Soil Model (PLAXIS2D). These models for clays were dis-cussed previously in Chapter 3.

5.6.3 Estimation of soil parameters

In order to use the analytical methods described previously, it is necessary to estimate the soil parameters. For many of the more complex non-linear soil models, this may require special laboratory testing, although for models like the Cam Clay Model, the parameters can all be determined from standard laboratory triaxial and oedometer tests.

Here, attention will be restricted to determination of elastic soil properties, as these form the basis of many of the hand-based methods of calculation mentioned previously. The soil parameters needed Eu, E′, and ν′ are generally determined either from laboratory tests or from in situ tests. Commonly used field tests are SPT tests, plate loading tests, pressuremeter

Mean stress p (kPa)

Dev

iato

r str

ess q

(kPa

)

0 10 20 30 40 50 600

10

20

30

40

50

60

70

Critical state line

X

su

Stress path

Figure 5.34 Stress path taken for undrained consolidation analysis of isotropically consolidated clay.


tests, dilatometer tests, static cone penetration tests, and screw plate tests. See Chapter 4 for more details on field testing and properties of soils.

These field tests often rely upon empirical correlations between the test result and a par-ticular parameter, for example, correlations between elastic modulus and the SPT or elastic modulus and the cone penetrometer resistance qc as was seen in the previous Section 5.5.5 on the settlement of footings on sand. As stated earlier, laboratory testing is rarely used for sands, as it is difficult to keep sand samples in an undisturbed state (at the same density) as in the field. Correlations between shear modulus of sands and field test results from pressuremeters, dilatometers, SPT tests, and cone penetration tests have been presented by Décourt (1994) for use in the computation of settlement of footings.

For clays, the elastic moduli can be found from triaxial testing or from field tests. Triaxial testing normally involves consolidating undisturbed soil samples back to the in situ stress state before loading them in an undrained condition (to obtain the undrained parameters) or in a drained condition (to obtain the drained parameters). Methods of testing have been described in Davis and Poulos (1968) and in Ladd and Foott (1974). Other methods have also been discussed in Chapters 1 and 4 of this book.

For clays also, a common procedure is to relate the undrained modulus Eu to the und-rained shear strength of the clay su. This depends upon the overconsolidation ratio of the clay, therefore one needs some knowledge of both the undrained shear strength and the overconsolidation ratio of the clay to predict undrained moduli. Often this relationship can be established through experience in certain areas, so it is only necessary to obtain the und-rained shear strengths to predict moduli. Correlation between the undrained shear strength of clays and the undrained modulus for various overconsolidation ratios has been presented by Duncan and Buchignani (1976) (Figure 4.45).

5.7 RAFT FOUNDATIONS

While settlements of pad footings may be calculated by treating the foundation as being perfectly flexible or perfectly rigid, it is generally necessary to take into account the actual stiffness of a raft or mat foundation. This is because differential settlements or moments and shears in the raft usually need to be calculated, therefore the structural rigidity of the foundation has to be considered.

Often solutions are found by treating the soil beneath the raft as consisting of a series of springs (a Winkler foundation) as this makes the analysis simpler; however, such methods are to be used with caution, as they cannot represent true soil behaviour. For example, a flexible raft carrying a uniform vertical loading (over the whole raft) would be predicted to settle uniformly with no differential settlement apparent if on a spring foundation, and this is obviously contrary to observed behaviour. A rigid raft would also make all the springs compress equally, and hence the load in each spring would be equal. Measurement of con-tact stress below rigid rafts shows that the contact stress is generally larger at the edges (for a uniform loading on the raft), so again, the load distribution in the springs would not be modelling the correct behaviour.

It is therefore desirable to treat the foundation as an elastic continuum, as the observed behaviour of rafts can be more closely simulated. Full finite element analyses may be neces-sary if the soil has highly non-linear properties, but generally, the theory of elasticity is suf-ficient, as rafts are not designed to carry loads that are going to cause yielding of the soil. A soil modulus representative of the stress levels in the soil can be used in such cases.

Solutions have been found to some problems involving rafts constructed on an elastic continuum, but the solutions are limited because it is difficult to present results for all raft


shapes and for different soil conditions such as layering or modulus increasing with depth. The book by Selvadurai (1979) contains analytic solutions to problems involving rafts on elastic continua; however, most solutions have been based on numerical procedures. Some of the available solutions are presented in the following sections.

5.7.1 Strip rafts

A strip raft may be classified as a raft that has a length L that is much greater than its width B. Solutions to such a problem have been presented by Brown (1975) for a strip raft, hav-ing a length to width ratio L/B of 10, resting on an infinitely deep elastic soil. The raft is considered to be loaded by a point load which is located at a distance sL from one end as is shown in Figure 5.35. The results are presented for a single point load as the solutions may be superimposed if more than one point load is acting on the raft.

Shown in Figure 5.36a are the solutions for the deflections in the strip raft while Figure 5.36b shows the moments in the raft. These particular solutions are for the case where the relative stiffness of the raft to the soil K has a value of 0.01. The relative stiffness is defined as

K

EE L

r s

s

= −16 1 2

4

I( )νπ

(5.37)

where Er is the elastic modulus of the raft, I is the second moment of area of the raft (I = Bt3/12 for a strip raft of thickness t), Es is the modulus of the soil, and νs is Poisson’s ratio of the soil.

Solutions for the moment or displacement in the raft are presented in non-dimensional form at various positions x along the strip raft, for a point load of magnitude P. If several point loads are applied along the raft, the appropriate values of P (corresponding to each load) should be used when computing the moments or deflections before superimposing (i.e. adding) them.

5.7.2 Circular rafts

Brown (1969) has also obtained solutions for circular rafts subjected to uniform vertical loading q over their entire upper surface. The raft is assumed to be of thickness t and radius a, and to rest on a layer of soil of depth h.

In this case, the relative stiffness of the raft K is defined as

K

E vE

ta

r s

s

= −

( )1 2 3

(5.38)

BL

sL P

x

Strip Er, νr

Soil Es, νs

Figure 5.35 Strip raft subject to point load.


Maximum moment in the raft occurs at its central point, and this moment is shown plot-ted in non-dimensional form against raft stiffness K in Figure 5.37. Solutions are presented for different depth to radius ratios h/a, and for two extremes of Poisson’s ratio ν = 0 and 0.5. In this case, the moments presented in the plot are actually moments per unit length (i.e. units are kNm/m run). Differential deflections in the raft are shown in Figure 5.38 for various soil layer depths and Poisson’s ratios. For a circular raft, the differential deflection is defined as the difference in displacement between the centre and the edge of the raft.

Booker and Small (1983) have extended the charts of Brown to include the effects of the walls as well as the base raft of a tank. The restraining effect of the walls at the outer cir-cumference of the base means that there is a moment generated there that is opposite in sign to the moment at the centre of the base and may be large if the walls are stiff enough to resist the rotation at the base–wall junction.

5.7.3 Rectangular rafts

Rafts of rectangular shape (in plan) that carry a uniformly distributed load q have been analysed by Fraser and Wardle (1976). The rafts are assumed to be constructed on a layered soil profile, and methods are given in the paper to enable an equivalent elastic modulus and

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0x/L

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0x/L

9876543210

–1–2

Disp

lace

men

t × L

E s/P(1

– ν

2 s)

0.01

0.1

0.2

0.30.4

s = 0.5

K = 0.01

–0.10

–0.05

0

0.05

0.10

Mom

ent/P

L

s = 0.01

s = 0.1

0.2

0.3 0.4 0.5

K = 0.01

(a)

(b)

Figure 5.36 Solutions for point load on strip foundation: (a) displacement distributions; (b) moment distri-butions. (After Brown, P.T. 1975. Geotechnical Engineering. Vol. 6, pp. 1–13.)


Poisson’s ratio to be determined for the layered system so that it may be treated as a uniform material.

The raft of length L and width B is shown as the inset to Figure 5.39. The relative stiffness K of the raft as defined by Fraser and Wardle is

K

E tE B

r s

s r

= −−

4 13 1

2 3

2 3

( )( )

νν

(5.39)

where the terms have the same meanings as for the strip raft (see Section 5.7.1) and νr is Poisson’s ratio of the raft. In Figure 5.39, points α, β, γ, and δ are identified on the raft. The deflections at these points and the differential deflections between them can be calculated by use of the influence factors shown on the figure. For instance, the curve labelled Iα would be used to compute the settlement at point α, whereas Iαβ would be used to compute the dif-ferential settlement between point α and point β.

To compute the deflection or differential deflection ρsi for an infinitely deep soil layer the following equation is used

ρ νsi s

s

qBE

I= −( )1 2

⋅

(5.40)

1 101

K

–0.04

–0.02

0

0.02

0.04

0.06

0.08

102 10310–2 10–1

h/a = 1

0.5

0.2

0.5

0.2νs = 0.5νs = 0

∞22

1

M qa2

(Mom

ent/u

nit l

engt

h)

Figure 5.37 Maximum moments in circular rafts of various stiffnesses. (After Brown, P.T. 1969. Géotechnique, Vol. 19, No. 2, pp. 301–306.)


where I is the appropriate influence factor as determined from Figure 5.39. To compute the settlement or differential settlement for a soil layer that is not semi-infinite, corrections need to be applied to ρsi. The actual displacement ρ is computed from the following expression where S is a correction factor that is found from charts like those shown in Figure 5.40 (for a raft with a length to breadth ratio L/B of 2 on a soil layer of depth d).

ρ ρ= S si

(5.41)

Similar charts have been provided by Fraser and Wardle to enable moments to be com-puted in the raft. This involves calculating moments for the case where the soil is infinitely deep, and then applying a correction for soil layers of finite depth, as was done for the displacements.

5.7.4 Raft foundations of general shape

When the raft is of general shape and is subjected to general loading (that may include point loads, moments, or uniform loads), it becomes difficult to provide solutions in the form of charts. For more complicated cases, it is necessary to use computer programs, and these may analyse raft behaviour by using finite difference methods or by use of finite element meth-ods. Analysis of the soil behaviour may be carried out by the use of elasticity theory or by

1 101 102

K

–0.2

0

0.2

0.4

0.6

0.8

Diff

eren

tial d

ispla

cem

ent ×

Es/(

1 –

ν2 s)qa

2

1

1

νs = 0.5νs = 0

0.5

0.5

0.2

0.2

10–2 10–1

2h/a = ∞

Figure 5.38 Differential deflections (centre to edge) in circular rafts of various stiffnesses. (After Brown, P.T. 1969. Géotechnique, Vol. 19, No. 2, pp. 301–306.)


use of finite element techniques. An early demonstration of the use of finite element methods for the raft and solutions based on the theory of elasticity for the soil were presented by Cheung and Zienkiewicz (1965).

For rafts on layered soils, the soil behaviour can be analysed using the finite layer method (see Chapter 2) and finite element methods can be used for determining raft behaviour. The raft is divided up into finite elements, for example, in the program FEAR (Finite Element Analysis of Rafts – Small 2013), eight-node isoparametric shell elements are used for the raft. This type of element allows the moments and shears per unit length to be computed in the raft as well as the deflections.

The process of solving the problem is given in Section 6.37.2 (that also includes piles), and involves obtaining the contact stress distribution between the raft and the soil. The use of finite elements for the raft allows point loads, moments and distributed loads to be applied to the raft and the raft can be of any shape and have different thicknesses in different parts of the raft.

The soil response can be calculated using finite layer techniques, which makes the analysis quick and simple to use, although the soil response could be calculated using finite element methods. Finite layer techniques allow layered soils to be used for the foundation although this means that the soil must be considered to have an elastic response.

An example of this is shown in Figure 5.41 where the computed displacement contours in the raft are shown. The centre of the raft can be seen to be deflecting more than the edges in Figure 5.41b.

5.8 REACTIVE SOILS

Soils such as plastic clays that contain clay minerals like montmorillonites or smectites tend to shrink when their moisture content decreases or to swell as their moisture content

10–40

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

I

10–2 100 102

Iα

Iβ

Iδ

Iγ

Iαγ

Iαδ

Iαβ

L/B = 2

βα

γδ

B

L

K

Figure 5.39 Settlement influence factors for rectangular raft on an infinitely deep layer (L/B = 2).


increases. Soil suctions exist in the uppermost partially saturated ‘active zone’ in the soil and it is changes in these suctions (due to moisture content variations) that cause shrink–swell behaviour.

Swelling can cause damage to structures especially if differential movements occur in the structure. For example, moisture changes may be small beneath the centre of a raft or mat foundation, thus little swelling or shrinking will occur. At the edges of the foundation, moisture changes are larger, therefore swell–shrink movements are large. This creates dif-ferential movements in the foundation, which are seasonal, and this can cause damage if the raft is not stiff enough to resist the movements.

0 0.2 0.4 0.6 0.8 1.0 0.8 0.6 0.4 0.2 00

0.2

0.4

0.6

0.8

1.0

S

Sαβ

Sαγ

Sαδ

Sα

0

0.2

0.4

0.6

0.8

1.0

S

Sαβ

Sαγ

Sαδ

Sα

0

0.2

0.4

0.6

0.8

1.0

Sβα

γδ

Sαβ

Sαγ

Sαδ

Sα

FlexibleK = 0.5Rigid

D/B B/D

(a)

(b)

(c)

νs = 0.5, L/B = 2

νs = 0, L/B = 2

νs = 0.3, L/B = 2

Figure 5.40 Settlement correction factor (L/B = 2).


Mx1

Mx2Mx3Mx4Mx5

Mx6

Mx7

Mx8

Mx9

Mx10

Mx11

Mx12

Mx13

Mx14

Mx15 Mx16 Mx17

Mx18

Mx19

Mx20

Mx21

Mx22

Mx23

Mx24

My1

My2My3My4My5

My6

My7

My8

My9

My10

My11

My12

My13

My14

My15 My16My17

My18

My19

My20

My21

My22

My23

My24

My25

My26

My27

My28

Point and distributed loads

Foundation for wind farm

Loads and pressures in (kN) and (kPa)

Raft – 800 mm/1200 mm thick

2 layers, 3 and 20 m

LC155 – full-size model

Uniform loads1.877E+032.110E+032.334E+032.512E+032.171E+033.184E+03

(a)

Contours of vertical displacement

Foundation for wind farm

Displacement in (m)

Raft – 800 mm/1200 mm thick

2 layers, 3 and 20 m

LC155 – full-size model

Contour legend0.0000.0050.0100.0150.0200.0250.0300.0350.0400.0450.050

(b)

Figure 5.41 (a) Finite element mesh and (b) displacement contours for raft on a layered soil. (From the Program FEAR [Finite Element Analysis of Rafts]).


5.8.1 Pad or strip footings

A fairly simple means of computing swell movements has been reported by O’Neill and Poormoayed (1980). The amount of surface heave y can be calculated by summing the swell occurring in a number of sub-layers in the active soil zone as shown in Equation 5.42.

(%)y s hi

N

i==

∑1

100/

(5.42)

where s (%) is the swell as a percentage (found from a swell test) in layer i. The thickness of each layer is hi and there are N layers in the active zone.

To obtain the swell s (%), undisturbed samples of soil taken from representative depths are placed in an oedometer under a pressure equal to the overburden pressure plus the anticipated surcharge. The sample is then inundated and sufficient extra load applied in small increments to prevent swelling. Finally, the sample is unloaded back to the initial pres-sure (overburden plus surcharge) in decrements and the swell measured (as a percentage of original sample height). O’Neill and Poormoayed (1980) have stated that samples should be taken during the construction period, so that the method will predict the expected swell from the time of construction if the soil is totally saturated (i.e. the worst case).

Other methods used for computing swell that are used in North America are reported in Settlement Analysis (1994) adapted from the U.S. Army Corps of Engineers, and are based on oedometer swell tests.

A method that is commonly used in Australia (Residential Slabs and Footings 2011, see also Mitchell and Avalle 1984) to compute the likely heave of foundations is based on determining the changes in soil suction beneath a foundation and relating these to soil movements. Soil suction occurring in partially saturated soils is commonly expressed in pF units where

pF suction in kPa)+1= log (10 (5.43)

Soil suctions can be measured by using commercially available psychrometers or by using filter paper techniques, although this is not necessary if the simplified methods outlined below are used.

An instability index Ipt is used to relate changes in vertical strain in the soil to changes in the soil suction. Soil suction changes are the result of wetting and drying of the soil in the zone above that part of the soil which is permanently saturated. The instability index is related to the swell–shrink index Iss of a soil and this index may be found from experiment by allowing the soil to swell when inundated with water and to shrink when dried.

The swell test is carried out in an oedometer (or consolidation cell) by loading a sample at natural moisture content with a pressure of 25kPa and allowing the sample to come to equilibrium under the load. The sample is then inundated with water and allowed to swell for a minimum of 24 h or until movements are only 5% of the total swelling up to the time of reading. The swelling strain ϵsw (in percent) is then computed from

εsw

R RH

= − ×( ) %1 2

0

100

(5.44)

whereR1 is the final dial gauge reading in mmR2 is the initial dial gauge reading (before inundating) in mmH0 is the initial specimen thickness in mm


The shrinkage test is performed on a cylindrical soil sample (at natural moisture content) of length 1.5–2 diameters. A drawing pin is pushed into each end of the sample, and the distance D0 between the rounded heads of the sample measured. The sample is then dried in an oven, and the final length is measured between the pins Dd. The shrinkage strain ϵsh (percent) is then calculated from

εsh

dD DH

= − ×( ) %0

0

100

(5.45)

where H0 is the average initial length of the sample (not including the drawing pins).The shrink–swell index Iss is then computed from

Iss

swsh

=

+ε ε2

1.8

(5.46)

The swell strain is divided by 2 to allow for the effect of lateral restraint in the swelling test. The factor of 1.8 is the change in pF value assumed to take place over the range of moisture contents experienced in the test, and can be replaced by an actual measured value if desired. The swell–shrink index is therefore a measure of the percentage strain per pF change in suction.

The instability index is related to the swell–shrink index by Ipt = α · Iss where α is intro-duced to take account of whether the soil is in a cracked (unrestrained) zone or in an uncracked zone.

1. In a cracked zone

α = 1.0

2. In an uncracked zone (where the soil is restrained laterally by surrounding soil and vertically by soil weight)

α = −2.0 z/5

where z is the depth (in metres) from the ground level to the point under consideration in the uncracked zone. As suction changes generally only occur over 4 m or less, α will have a positive value.

The design surface movement ys is then calculated from

y I udhs

H

pt= ∫1100

0

∆

(5.47)

whereΔu is the suction change at depth h (in pF units)Ipt is the instability indexH is the depth of the active zone


The integral of Equation 5.47 can be treated as a summation if so desired.The method therefore depends on being able to predict the soil suction changes with depth

for use in Equation 5.47, as well as a knowledge of the cracked zone. It is often assumed that soil suction Δu varies linearly with depth as shown in Figure 5.42 and the values of suction change at the surface and the depth of influence H can be established for certain areas (for example, in Adelaide Δu = 1.2 pF and H = 4 m unless the water table is higher in which case H = depth to water table). In Sydney, the values are Δu = 1.5 pF and H = 1.5 m. The depth of cracks can also be established on a regional basis (for example in Adelaide the depth can be taken as 0.75H).

5.8.2 Rafts on reactive soils

The swelling and shrinking of reactive soils can produce differential movements in raft foundations, and this is usually due to soil movements at the edges of the raft where mois-ture changes are most likely to occur. Rafts constructed in areas where soils are reactive are generally stiffened with ribs (i.e. the waffle slabs as shown in Figure 5.1d).

If swelling and shrinkage movements take place around the edges of a raft foundation, the raft will undergo uplift at the edges (swelling) or will be supported on a mound of soil if the soil shrinks away at the edges. Analysis of this type of behaviour has been examined by Sinha and Poulos (1996) using finite element analysis of the raft, and by Li et al. (1996) who have used finite element methods to analyse moisture flow and soil–structure interaction for a stiffened raft foundation.

5.9 COLD CLIMATES

In cold climates, when the ground begins to freeze, water can be drawn upward towards the surface by capillary action (from the water table below) and can form ice lenses. The lenses usually grow perpendicular to the direction of heat flow, which for most soils is parallel to the ground surface. The increase in volume of the ice will cause the surface of the soil to heave, and these upward movements can be very large (i.e. 300 mm or more) causing struc-tural damage.

Frost-susceptible soils are those that have pores small enough to produce ice lenses at the freezing front (i.e. the surface separating frozen and non-frozen soil). Silty materials, fine

HH

H

Consistent soil case Effect of bedrock Effect of groundwater

Water table

Soil surface level

Bedrock level

Δu Δu Δu

Figure 5.42 Change in soil suction Δu with depth.


silty sands with more than 15% of particles finer than 0.02 mm and lean clays having a plasticity index <12, varved clays and chalks are most susceptible to frost heave while grav-els and clean sands are not because of their high permeability. Clays are less susceptible to heave because their low permeability prevents the flow of water (and hence the formation of ice lenses).

For a soil to exhibit frost heave, there must be sufficiently low temperatures, water avail-able in the soil, and the soil must be of a type (as mentioned above) that is frost susceptible. Freezing tests may be carried out on soil samples to determine frost heave potential.

Freezing of a foundation may be due to sub-zero climatic conditions, or it may be due to man-made causes, for example freezer rooms or ice rinks (see Chapuis 1988). In the former case, resistance to frost heave may be engineered by draining the soil, founding footings below the depth of frost penetration or replacing frost-susceptible soils by non-susceptible materials such as gravels or sands. For man-made sources of low temperature, these meth-ods may also be used or good insulation provided between the soil and the structure.

Prediction of frost heave is not generally a simple matter, as any analysis involves com-putation of the heat flow in the soil, the thermal properties of the soil, the cooling rate and suction at the frost front. Such analyses have been reported by Konrad and Morgenstern (1982) and Nixon (1991). Field measurements and numerical predictions of the measured heaves have also been reported in Hayhoe and Balchin (1990).

APPENDIX 5A

For soils that are compressible, the bearing capacity solutions need to be modified to take account of punching failure. The classical bearing capacity equations are for rigid plastic materials, and therefore do not apply to softer or looser soils.

To deal with loose or compressible soils, Vesic (1975) introduced a rigidity index Ir defined as

I

Gc q

r =+ tanφ

(5A.1)

whereG is the elastic shear modulus of the soilc, ϕ are the soil strength parametersq is the vertical overburden pressure evaluated at a depth of B/2 below foundation level

(B is footing width)

The value of Ir is compared to a critical rigidity index Irc where

I

BL

rc = −

−

12

3 30 0 45 452

exp o. . cotφ

(5A.2)

When Ir > Irc the soil behaves like a rigid plastic material and there is no need to use the modifying factors (i.e. they can all be considered to be 1). If Ir < Irc the soil is considered compressible and punching shear may occur. The correction factors in Table 5.2 may then be used.

191

Chapter 6

Deep foundations

6.1 INTRODUCTION

Deep foundations are defined as those where the depth of the foundation exceeds the breadth. Such foundations are generally caissons, barrettes, or piles. The most common type of deep foundations is piles, but barrettes (which are deep foundations of rectangular cross section) or caissons (which are large hollow foundations) can be regarded as special types of piles.

Piles are generally used to transfer the loads of structures down to rock or to deeper stiffer layers. This improves the bearing capacity of a foundation or has the effect of reducing settlements. However, there are other reasons for the use of piles:

1. As anchors to resist uplift forces 2. To resist against lateral forces, for example, bridge piers, wharves 3. To provide a deep foundation in cases where scour may remove upper soil layers 4. As foundations where shrinking and swelling soils exist. The piles transfer loads down

below the active surface zone, where water content changes occur

6.2 TYPES OF PILES

Piles can be classified in different ways:

1. Classification by material from which the pile is made:• Steel: A steel pile may be an ‘H’ section or hollow circular section. Sheet piles have

sections that may be interlocked with the adjacent sheet pile• Concrete: Concrete piles are generally reinforced with steel or may be pre-stressed• Timber: Timber piles may be treated with chemicals to prolong their life• Composite: Composite piles may be combinations of steel and concrete such as

concrete-filled steel pipe piles 2. Classification by the effect of installation:

• Displacement piles (i.e. piles that displace the soil when driven such as a large closed end pipe pile)

• Low displacement piles (such as ‘H’ section piles)• Non-displacement piles (such as drilled piles)

3. Classification by method of installation:• Driven• Driven tube filled with concrete• Bored (drilled) piles• Composite piles


• Screwed (Atlas)• Pushed• Vibrated

6.2.1 Driven piles

Steel, concrete, or timber piles can be driven into the soil by a hammer. Examples are

• Timber piles: Treated with chemicals so as to resist insects and rotting. Such piles can be used in soils with pH ranging from 2 to 11.5. Piles can be spliced, and cut off to length. Treated piles can be expected to last in excess of 100 years. In fact, the first masonry London Bridge built in 1176 stood on untreated elm piles and lasted 600 years. Marine hardwood piles have to withstand attacks from borers and therefore are impregnated with chemicals. They are vacuum/pressure impregnated twice. The first treatment is with CCA (copper chrome arsenic) preservative and after re-drying of the wood, the second treatment is conducted with PEC (pigmented emulsified creosote).

• Precast concrete piles: Such piles can be solid or hollow and are reinforced with steel cages. Precast concrete piles are best where there are soft materials overlying rock or hard strata. The pile sections can be spliced if desired to extend their length. A precast pile is shown in Figure 6.1.

• Pre-stressed concrete piles: These can be solid or hollow sections and are pre-stressed in a production yard.

• Steel piles: Solid steel piles can be ‘H’ section, box, or tube (also interlocking sheet piling).

6.2.2 Driven and precast piles

A steel tube with a closed end is driven into the soil and the tube filled with concrete. The steel tube may be withdrawn or it may be left in place. There are variations where a steel shell is driven by a withdrawable mandrel and the shell is concreted.

6.2.3 Jacked piles

In this case, the pile is jacked into the ground against a large reaction. The rigs that are used to do this can be very heavy and may be difficult to get into areas of soft soil. They may also cause movement of existing foundations. Steel or concrete piles can be jacked.

Steel shoe Spiral steel cage

(a)

(b)

Figure 6.1 Precast reinforced concrete piles: (a) solid square; (b) hexagonal.

Deep foundations 193

6.2.4 Bored piles

There are various types of bored piles, but most are formed by drilling a hole, and then plac-ing the reinforcing cage before concreting. This is shown in Figures 6.2 and 6.3. Advantages of bored piles are

• The base of the hole can be under-reamed to provide an enlarged base.• A continuous flight auger (CFA) can be used (see Figure 6.4).• It is possible to drill into rock using a bucket with tungsten tipped teeth.• There is less vibration and noise than for a driven pile.• Soil is not displaced (such as for a large cross-sectioned driven pile), so there is less

chance of damaging nearby structures.• Piles can be of large diameter to support very heavy loads.

6.2.5 Composite piles

Piles can be combinations of materials, for example, steel piles placed into a concreted pile or an ‘H’ section steel pile encased in concrete.

Figure 6.2 Steel cages ready for placing inside a drilled shaft.

Figure 6.3 Drilled shaft pile being installed. Concrete is tremied into the shaft displacing water.


Figure 6.4 Continuous flight auger rig.


6.2.6 Grout injected piles

Jet grouted piles are formed by drilling to the required depth and then injecting high pres-sure grout that cuts away the soil and forms a mix of soil and cement slurry.

6.3 INSTALLATION

Piles can be classified according to installation methods, whether they are large displace-ment piles or piles that cause very little displacement.

If large diameter piles are driven into the soil, they will cause the soil to be pushed side-ways as the pile head advances, therefore the soil must undergo lateral compaction. Such piles are called ‘displacement piles’. If a hole is bored initially, then concreted, the soil under-goes minimal disturbance and the pile is termed a ‘non-displacement pile’.

6.3.1 Types of displacement piles

Types of displacement piles are

1. Timber piles that are often used for marine and smaller scale structures. 2. Steel tubes, that can be readily extended, but may corrode and are generally more

expensive. 3. Precast concrete piles: Common lengths are from 12 to 15 m and have the advantage

of lower cost. They are not suited to hard driving, and can be spliced, but not as easily as steel piles.

4. Proprietary types: Many use steel casings that are withdrawn as concrete is placed.

Common problems are

• Vibration from driving• Excess pore pressures generated during driving• Lateral soil movements occur• Access for driving rigs may be difficult

6.3.2 Small displacement piles

Types of small displacement piles are

1. ‘H’ sections and rolled steel sections: Such piles are useful for punching through thin hard layers. They can have a steel point to assist in driving. However, thin sections may bend about the weakest axis of the pile.

2. Steel tube piles: The plug of soil that is pushed up inside the pile can be removed after driving. Tube piles are often filled with concrete once cleaned inside.

3. Pre-drilled piles: Such piles are usually pre-bored over part of their depth, and then driven. They are useful if there are hard layers near the surface.

6.4 PILE DRIVING EQUIPMENT

If piles are to be driven into the ground, they are usually driven using a heavy drop hammer, although vibration can also be used. Piles can be driven vertically or at an angle (called a ‘raked pile’).


6.4.1 Piling rigs

A piling rig usually consists of a set of leaders mounted on a standard crane base. Leaders consist of a stiff box or tubular member which carries and guides the pile as it is driven into the ground. Leaders can be raked backwards or forwards to install raked piles. A piling rig is shown in Figures 6.5 and 6.7.

6.4.2 Piling winches

The piling winch is used for raising the pile and (if necessary) the hammer. It may be pow-ered hydraulically or by steam, diesel, or petrol engines.

6.4.3 Piling hammers

There are several different types of hammers that are used to drive a pile into the ground. Some hammers like drop hammers, simply rely on their weight to drive the pile, while others

Figure 6.5 Rig for driving piles.


like the diesel hammers provide the hammer with an extra force to accelerate the hammer onto the pile head. Types of hammers are

1. Drop hammers The drop hammer is simply a weight that is dropped onto the pile head. A typical ham-

mer mass is 1–5 tonnes. 2. Single acting steam or compressed air hammer Steam or compressed air is used to raise the hammer, and once at the required height,

the hammer is allowed to fall under the action of gravity. 3. Double acting hammers These hammers are used mainly for sheet piling. They can impart a rapid succession of

blows (more than 100 blows per minute) and rely on steam or air to raise and acceler-ate the hammer downward. They do not need a guide but can be attached to the head of the sheet pile by leg guides.

4. Diesel hammers The falling ram compresses air in a cylinder into which diesel fuel is injected. The die-

sel fuel ignites, forcing the hammer back up the cylinder and the reaction imparts an additional force to the pile head (Figure 6.6). Diesel hammers may not be effective in all soil types, for example, soft soils and may cause breakage of concrete piles if a hard layer is encountered.

5. Hydraulic hammers The hammer is raised by hydraulic fluid and then hydraulic pressure is used to accel-

erate the hammer onto the pile (Figure 6.8). Hydraulic hammers are less noisy than diesel hammers and can impart a wide range of driving energies.

Diesel fuelinjected ashammer falls

Hammer attop of stroke

Figure 6.6 Diesel hammer.


Figure 6.7 Precast concrete piles being driven.

Pile sleeve

Shock adsorber

Segmented ramweight

Actuator

Lift cylinder

Figure 6.8 Hydraulic hammer.


6. Vibratory hammers A vibrating head is attached to the top of the pile and vibration and downward force

causes the pile to be driven into the soil. This type of pile driver is best suited to sandy or gravelly soils. The technique is also useful for extracting piles such as sheet piles or the casing of bored piles.

6.4.4 Helmet, driving cap, dolly, and packing

Various components are placed on the pile head to stop it from suffering damage due to driving. These are

1. Helmet: This is placed over the top of a concrete pile. It holds the dolly and packing between the hammer and pile to prevent damage to the pile head.

2. Driving cap: The cap is used to protect the heads of steel piles, and is shaped for the particular type of pile being driven. The cap is fitted with a recess for a hardwood or plastic dolly.

3. Dolly: The dolly is placed in a recess in the top of the helmet. It can be made of timber or plastic. Under moderately hard driving conditions, plastic dollies can last for several hundred piles.

4. Packing: This is placed between the helmet and pile top to cushion the blow between the two. Materials used include hessian, paper sacking, thin timber sheets, coconut matting, and wallboards.

6.5 PROBLEMS WITH DRIVEN PILES

Driving piles into hard soil with a heavy hammer can cause damage to the pile and as the pile may not be visible above ground, this may not be detected. Types of damage can be

1. Steel piles: The top or shaft of the pile can be damaged by driving. Tubular piles may collapse or the pile toe may be damaged. Base plates may rise relative to the casing, or plugs and shoe may be lost in a cased pile.

2. Concrete piles: The head, shaft, or toe can be damaged. 3. Timber piles: The head or toe can be damaged.

6.5.1 Problems from soil displacement

Driving large displacement piles can cause problems because of the soil movements both vertically and laterally. Some of the problems caused by soil movements are as follows:

Vertical Soil Movements

1. Uplift causing squeezing, necking, or cracking 2. Uplift resulting in shaft lifting off base 3. Uplift resulting in loss of end bearing capacity 4. Ground heave lifting pile 5. Ground heave separating pile segments or units or extra tensile forces being placed on

joints


Lateral Soil Displacements

1. Squeezing or waisting of piles 2. Inclusions of soil forced into pile 3. Shearing of piles, or bends in joints 4. Collapse of casing prior to concreting 5. Movement and damage to adjacent structures

6.6 NON-DISPLACEMENT PILES

Non-displacement piles are commonly used where large loads are to be carried or vibration and noise from driving are not wanted. Common installation methods include drilling a dry hole, drilling the hole supported by a slurry or by drilling a hole supported by a steel casing.

Holes when drilled should be concreted preferably within the same day, as leaving the hole open can lead to softening of the soil along the shaft with resultant loss of bearing capacity. Slurries such as bentonite can form a cake along the sides of the drilled hole and this too can lead to loss of bearing capacity. Some of the advantages and disadvantages of drilled shaft foundations are listed below:

1. Advantages• Absence of ground heave• Absence of excessive noise and vibration• Piles can be installed where headroom is limited• Length and diameter can easily be varied• Base can be enlarged• Can obtain high capacities• Can inspect prior to concreting

2. Disadvantages• Loosening of sandy soils• Softening of clays• Waisting or necking of pile may occur in loose or soft soil if support from casing

or slurry not used• Inflow of water can damage concrete• Belled bases may be difficult to construct in sandy soils or clays containing silt or

sand lenses• Concrete should be placed as soon as possible after drilling

6.6.1 Precautions in construction and inspection of bored piles

Bored piles rely on construction methods being able to produce a defect free pile, and some of the precautions that need to be applied when constructing bored piles are summarised below:

1. The pile should be supported by casing through soft or loose soils to prevent shaft collapse.

2. A casing should be provided to seal off water-bearing soil layers. 3. Where drilling fluid is used, strict control of fluid density should be maintained and

cleaning and de-sanding carried out prior to re-use.


4. Compare soil or rock cuttings removed from pile with descriptions from site investigation.

5. Shear strength tests on soil from bottom of selected piles can be performed to check against design assumptions.

6. Plumb deep boreholes to bottom immediately before concreting. Compare plumbed depth to depth measured at end of drilling.

7. Proper measures should be taken for base cleaning, with video or visual inspection where possible.

8. Safety procedures must be followed strictly. 9. Time interval between completion of boring and placing of concrete should be as short

as possible, and no longer than 6 h.

6.6.2 Continuous flight auger piles (or grout injected piles)

1. The flight auger usually has a central hollow stem closed by a plug at the bottom (see Figure 6.4).

2. The borehole walls are supported at all times by soil rising within the flights. 3. On reaching the required depth, grout is injected down the hollow stem. This pushes

out the bottom plug, and the pile shaft is concreted by raising the auger (with or with-out rotation).

4. A fairly fluid grout with plasticiser to improve pumpability and an expanding agent to counter shrinkage during setting and hardening is used.

5. After removing the auger, a reinforcing cage is placed while the concrete is still fluid. Cage lengths up to 12 m long are usually achievable.

6. Strict control of construction is necessary, especially when end bearing resistance is required.

7. A check can be made of shaft soundness by recording the amount of concrete injected as the auger is withdrawn.

8. An indication of soil strength can be obtained by measuring the torque on the drill stem over the full depth of drilling.

9. Size limits: The diameter is restricted to less than 1.5 m but more commonly is less than 1 m. The length of the pile is governed by the length of the auger, and is com-monly less than 35 m.

6.7 DESIGN CONSIDERATIONS

The designs of piles is discussed in following sections, but may involve some of the following considerations:

1. Selection of pile type and method of installation. 2. Estimating the size and number of piles so as to obtain an adequate factor of safety. 3. Computing the settlement and differential settlement of piles or groups of piles to

check if they are within specified tolerances. 4. Effects of lateral loading including lateral load capacity and lateral deflection may

need to be calculated. 5. The effects of soil movements may need to be taken into account in design. Movements

may be due to nearby excavation, slope movement, or soil consolidation (causing downdrag).


6. Evaluation of pile performance through load testing is often carried out, and designs may need to be refined once this is done.

6.8 SELECTION OF PILE TYPE

As mentioned previously (Section 6.2), there are many different types of piles that the designer can choose from. Some of the factors which can influence the type of pile to use are listed below:

1. Location and type of structure 2. Loads being applied by the structure 3. Availability of piling equipment, and access of equipment to the site 4. Ground conditions 5. Durability requirements 6. Effects of installation on adjacent piles and structures 7. Relative costs

6.9 DESIGNS OF PILES

The ultimate load that a pile can carry can be estimated from a dynamic approach, that is, pile driving formulae or the wave equation. However, the load capacity can also be com-puted from knowledge of the soil strength. These methods are termed ‘static methods’, and there are several different ways of carrying out the computations ranging from empirical approaches to semi-analytical methods.

Soil properties can be found from laboratory tests or from field tests and correlations to field tests. The design methods also depend on the geotechnical conditions, as there are dif-ferent approaches taken for clays, sands, and piles driven to rock.

6.10 SINGLE PILES

The ultimate load Pu that can be carried by a pile is the combination of the base load and the shaft friction minus the weight of the pile itself, that is,

P P P Wu us ub= + − (6.1)

wherePus is the ultimate shaft loadPub is the ultimate base loadW is the weight of the pile

The shaft resistance can be obtained by integrating the available shear strength along the shaft, while the base resistance can be found from the bearing capacity for a deep foundation.

As for shallow foundations, there are two cases that can be considered, the drained and undrained cases. For clays, both undrained (short term) and drained (long term) analyses may be applicable, while for free draining materials like sands and gravels, only drained conditions apply.


The general formula for computing the bearing capacity of a single pile can therefore be written as

P C c K dz A cN N dN Wu

L

a v s a b c vb q= + + + + −∫0

0 5( tan ) .( )σ φ σ γ γ

(6.2)

whereC is the shaft circumferenceca is the adhesion between the shaft and soilϕa is the angle of friction between the shaft and soilσv is the vertical overburden stress at depth zσvb is the vertical overburden stress at pile base levelKs is the coefficient of lateral pressureAb is the area of the base of the piled is the diameter of the pileγ is the unit weight of the soilNc, Nq, and Nγ are the bearing capacity factors for a deep foundationW is the weight of the pile

6.10.1 Piles in clay

For piles in clay under undrained conditions, the angles of friction ϕa, ϕu can be taken as zero, therefore the general formula (Equation 6.2) becomes

P Cc dz A s N Wu

L

a b u c vb= + + −∫0

( )σ

(6.3)

Because the weight of the pile and the term for the overburden pressure are approximately equal, that is,

W LA A LAb vb b b= ≈ =γ σ γconc

these terms cancel.The final ultimate load for a pile in clay is therefore

P Cc dz A s Nu

L

a b u c= +∫0

(6.4)

For piles that are usually considered to have an L/d ratio high enough to be classified as a deep foundation, Nc = 9 and su is the undrained shear strength of the clay below base level.

The adhesion ca between the soil and the pile shaft, is related to the undrained shear strength of the clay su(z) at a depth z along the pile shaft. The relationship is given by

c sa u= α (6.5)

The method is sometimes called the ‘alpha’ method because of this. Relationships between α and the undrained shear strength su are given in charts such as the one shown in


Figure 6.9. It should be noted that there is quite a large scatter of the data that this chart is based upon.

For a pile with a constant diameter, the integral in Equation 6.4 is the circumference of the pile times the area under a plot of the adhesion versus depth.

In the long term, the effective stresses and drained soil parameters should be used to compute the shaft and base loads of a pile. In this case, the skin friction f is given by f v= ′βσ where β φ= ′Ks atan and is therefore often termed the ‘beta’ method. In the general equation (Equation 6.2) therefore, the shaft load may be computed by using ′ =ca 0 and Ks atan ′φ .

For normally consolidated clays, it has been suggested that

Ks R= − ′( sin )1 φ (6.6)

while for overconsolidated clays

K OCRs R= − ′( sin )1 φ (6.7)

where ′φR is the remoulded angle of friction of the soil and OCR is the overconsolidation ratio of the clay

6.10.2 Piles in sand

Piles constructed in sand can be designed by using the effective stresses (a drained analysis) with the general equation (Equation 6.2). The equation then becomes (if ′ =φ δa )

P C K dz A N dN Wu

L

v s b vb q= ′ + ′ + −∫0

0 5σ δ σ γ γtan ( . )

(6.8)

0 50 100 150 200 250 300Undrained shear strength, su (kN/m2)

0

0.2

0.4

0.6

0.8

1.0

1.2

Adh

esio

n fa

ctor

, α

Tomlinson (1957)(concrete piles)

α = 0.21 + 0.26 pa/su (≤1)

Shafts in upliftData group 1Data group 2Data group 3

Shafts in upliftData group 1Data group 2Data group 3

65U and 41Cload tests

Figure 6.9 Variation of α = ca/su against shear strength of clay su.


If it is assumed that the pile self-weight and the term in Nγ cancel (as they are of the same magnitude), then

P C K dz A Nu

L

v s b vb q= ′ + ′∫0

σ δ σtan

(6.9)

For a pile with a uniform diameter and a constant angle of friction δ between the pile and soil, the formula becomes

P CK dz A Nu s

L

v b vb q= ′ + ′∫tanδ σ σ0

(6.10)

The integral in this equation is just the area under a plot of the effective vertical stress dia-gram versus depth, and thus may be easily determined geometrically.

However, it has been noticed that the skin friction acting on the sides of piles in sand does not keep on increasing linearly with depth as Equation 6.10 may suggest. It has been noticed from pile test data that the skin friction increases up to a certain ‘critical depth’ and then does not increase in value a great deal beyond that depth (Vesic 1967). This is shown in Figure 6.10 where the skin friction variation depends on the density of the sand as well as the depth from the top of the sand layer.

0 1 2 3 4 5 6Average skin resistance (lb/in2)

140

120

100

80

60

40

20

0

Pile

leng

th (i

n)

Dense sand (G-3)

Field testsloose, moist,sand (G-4)

Medium densesand (G-2)

Loose sand (G-1)

0.1 0.2 0.3 0.4Average skin resistance (tons/ft2)

Figure 6.10 Variation of pile skin friction in sands of different densities.


Therefore to design piles in sand, the critical depth is determined (this depends on the density of the sand) and then it is assumed that the effective stress is constant below this depth. This pseudo-effective stress variation is then used to compute the skin friction. By doing so, the skin friction on the pile remains constant past the critical depth. This is shown in Figure 6.11 where it may be noted that the location of the water table is taken into account in computing the vertical effective stress variation.

We also need values of the coefficient of lateral earth pressure Ks to compute the skin fric-tion. Values of Ks tan δ vary for different sand densities (where δ φ= ′a is the angle of friction between the pile shaft and the sand), and therefore values have been tabulated for use with different sand densities as shown in Table 6.1.

The base bearing capacity of piles in sand also requires a knowledge of the bearing capac-ity factors, and some values of Nq have been suggested by Vesic (1967) as well. The value of

′σvb in this case is found from the truncated value of ′σvc in Figure 6.11.

6.10.3 Lambda method

A method mainly used for the design of offshore pipe piling is called the ‘lambda’ method reported by Vijayvergia and Focht (1972). The shaft friction for the pile is calculated from

P s Aus m um s= ′ +λ σ( )2 (6.11)

where′σm is the mean effective vertical stress along the pile shaft

sum is the mean undrained shear strength along the pile shaft

The value of lambda depends on the pile penetration and is shown plotted in Figure 6.12. The method applies to fairly uniform clay deposits and does not apply if sand layers are present.

W.T.

L

zcσ′vc

d

Figure 6.11 Effective stress variation used to compute skin friction for piles in sand.

Table 6.1 Parameters for calculating pile capacity in sand

Condition of soil zc/d

F = Ks tan δ Nq

DrivenBored or

cast in situ DrivenBored or

cast in situ

Loose (R.D. = 0.2–0.4) 6 0.8 0.3 60 25

Med. Dense (R.D. = 0.4–0.75) 8 1.0 0.5 100 60

Dense (R.D. = 0.75–0.9) 15 1.5 0.8 180 100


6.11 METHODS BASED ON FIELD TESTS

In many cases, the parameters such as shear strength and angle of shearing resistance between the pile shaft and the soil are not determined through laboratory tests, but field tests are used and base bearing and skin friction are found directly by correlation with test results.

Field tests of various types have been discussed in Chapter 4.

6.11.1 Correlations with standard penetration test (SPT) data

Empirical correlations with SPT data take the form:

f A B Ns n n= + (kPa) (6.12)

whereAn and Bn are empirical numbers, and depend on the units usedN is the SPT blow count

The base load (pressure) is computed from

f C Nb n b= (MPa) (6.13)

whereCn is an empirical factorNb is the average SPT below the pile base (typically 1–3 pile base diameters)

0 0.1 0.2 0.3 0.4 0.5

225

200

175

150

125

100

75

50

25

0

Pile

pen

etra

tion

(ft)

λ

Location Symbol Source

DetroitMorganzaClevelandDraytonNorth SeaLemooreStanmoreNew OrleansVeniceAllianceDonaldsonvilleMSC HoustonSan FranciscoBritish ColumbiaBurnside

HouselMansurPeckPeckFoxWoodwardTomlinsonBlesseyMcClellandMcClellandDarraghMcClellandSeedMcCammonPeck

λ = Psu

(σ′m + 2cm)As

Figure 6.12 Frictional coefficient λ versus pile penetration.


The most widely used correlations are those due to Meyerhof (1956) for piles driven in sand, where

An = 0Bn = 2 for displacement piles

= 1 for small displacement pilesCn = 0.3

Limiting values of fs of 100 kPa are recommended for displacement piles and 50 kPa for small displacement piles.

Décourt (1995) has developed more recent and extensive correlations between fs and N which take into account both the pile type and method of installation. For displacement piles An = 10 and Bn = 2.8, while for non-displacement piles An = 5 − 6 and Bn = 1.4 − 1.7.

For the base, values of Cn are given in Table 6.2.

6.11.2 Correlations with cone data

Empirical correlations with cone data often take the form:

f A qs q c=

(6.14)

whereAq is an empirical numberqc is the cone penetration resistance at a particular depth along the shaft

The base pressure is given by

f C qb q cb=

(6.15)

whereCq is an empirical numberqcb is the average cone penetration resistance below the pile tip

Figures 6.13 and 6.14 give values of shaft friction fs correlated to cone values for the dif-ferent soil types given in Table 6.3.

A useful adaptation of the method of Bustamante and Gianeselli (1982) has been sum-marised by Frank and Magnan (1995). The ultimate shaft friction and base capacity are given by

fqb

f

f k q

sc

sl

b c c

= ≤

=

(6.16)

Table 6.2 Factor Cn for base resistance

Soil type Displacement piles Non-displacement piles

Sand 0.325 0.165Sandy silt 0.205 0.115Clayey silt 0.165 0.100Clay 0.100 0.080


Tables 6.4 and 6.5 give values of the factors b and the limiting skin friction fsl as well as kc.

6.11.3 Seismic data

For design of piles, the soil modulus E can be found from the small-strain modulus Es-strain as found from cross-hole seismic tests. The suggested value of modulus is given by

E E= 0 3. s-strain (6.17)

0 5 10 15 20 25 300

50

100

150

200

250

300

Shaf

t res

istan

ce f s (

kN/m

2 )

Cone resistance qc (MN/m2)

54

32

1

3U

3U1L

3L

Curveno.

Application pile types(see Table 6.3)

1L1U23L3U45

Lower limit for IBUpper limit for IBIIBLower limit for IA and IIAUpper limit for IA and IIAIIIAIIIB

Note:Lower limit applies for unreliable constructioncontrol; Upper limit applies for very carefulconstruction control

Figure 6.13 Shaft resistance values for sand.

0 2 4 6 8 10 12 14 160

20

40

60

80

100

120

140

160

Shaf

t res

istan

ce f s (

kPa)

Cone resistance qc (MPa)

Curveno.

Application pile types(see Table 6.3) Note:

Lower limit applies for unreliable constructioncontrol; Upper limit applies for very carefulconstruction control

IIB, lower limit for IA, IB, and IIAUpper limit for IBIIIA, upper limit for IA, IIA, and IIIBIIIB

1234

432

1

Figure 6.14 Shaft resistance for piles in clay.


6.12 PILE GROUPS

Piles are usually used in groups rather than alone, and when loaded the piles will often act as a single block as the soil trapped between the piles will act in unison with the piles in the group. If the piles are closely spaced, this group action is more likely than if the piles are spaced more widely apart. Therefore, in calculating the bearing capacity of a group of piles,

Table 6.4 Ultimate shaft friction correlation factors for a CPT test

Clay and silt Sand and gravel Chalk

Pile type Soft Stiff Hard Loose Medium Dense Soft Weathered

Drilledb 75a – 200 200 200 125 80fsl, kPa 15 40 80a 40 80a – 120 40 120Drilled; removed casingb – 100 100b – 100b 250 250 300 125 100fsl, kPa 15 40 60b 40 80b – 40 120 40 80Steel; driven closed-ended

b – 120 150 300 300 300fsl, kPa 15 40 80 120 c

Driven; concreteb 75 150 150 150fsl, kPa 15 80 80 120 c

Source: Adapted from MELT 1993. Règles techniques de conception et de calcule des fondations de ouvrages de genie civil. CCTG, Fascicule No. 62, Titre V, Min. de L’Équipement du Lodgement et de Transport, Paris.

a Trimmed and grooved at the end of drilling.b Dry excavation, no rotation of casing.c In chalk, f can be very low for some types of piles; a specific study is needed.

Table 6.3 Classification of pile types

Pile category Description

1A Plain bored piles, mud bored piles, hollow auger bored piles, cast screwed pilesType I micropiles, piers, barrettes

1B Cased bored pilesDriven cast piles

IIA Driven precast pilesPre-stressed tubular pilesJacked concrete piles

IIB Driven steel pilesJacked steel piles

IIIA Driven grouted pilesDriven rammed piles

IIIB High pressure grouted piles (d > 0.25 m)Type II micropiles

Source: Adapted from Bustamante, M. and Gianeselli, L. 1982. Proceedings of ESOPT II, Amsterdam, Vol. 2, pp. 492–500.


it is usual to compute the capacity of the group acting as a block, as well as the sum of the individual capacities of the piles, and the lesser of the two taken as the group capacity.


For piles in clay, the group is firstly analysed as consisting of a number of individual piles and the bearing capacity estimated from

P nPgi u=

(6.18)

wheren is the number of piles in the groupPu is the ultimate load of a single pile

The group behaviour is then calculated for the piles acting as a single block using

P B W c dz BWN suB a

L

c ub= + +∫20

( )

(6.19)

whereB is the breadth of the group in planW is the length of the group in planL is the length of the pilesNc is the bearing capacity factor for a shallow foundation with an L/B ratio equal to

that of the groupsub is the undrained shear strength below the base of the group

Note that if the pile group has a non-rectangular shape in plan, that the plan perimeter of the group is used in Equation 6.19 instead of 2(B + W) and the base area instead of BW.

The adhesion ca acting on the side of the group can be taken as the undrained shear strength of the soil su if most of the shearing is soil to soil. A refined calculation can be made if the adhesion between the piles and soil around the perimeter of the group is included in the calculation for that part of the perimeter where piles exist, and the undrained shear strength used where soil exists.

Table 6.5 Base capacity factors for a CPT

kc

Soil type qc (MPa) Non-displacement pile Displacement pile

Clay silt A Soft <3B Stiff 3–6 0.40 0.55C Hard(clay) >6

Sand gravel A Loose <5B Medium 8–15 0.15 0.50C Dense >20

Chalk A Soft <5 0.20 0.30B Weathered >5 0.30 0.45

Source: Adapted from MELT 1993. Règles techniques de conception et de calcule des fondations de ouvrages de genie civil. CCTG, Fascicule No. 62, Titre V, Min. de L’Équipement du Lodgement et de Transport, Paris.


Note that the integral in Equation 6.19 is just the area under a plot of the adhesion versus depth. The integral can be evaluated simply by estimating the area of the plot and this can be done by using simple trapezoidal or rectangular areas to approximate the area under the plot.

Values of Nc need to be evaluated and a plot of the type shown in Figure 6.15 can be used for this purpose (where the group is W by B in plan and L deep).


Where pile groups are constructed in sand, the capacity of the group can be greater than the individual capacities of the piles. This can be due to the sand densifying as driving proceeds. Because of this, it is always desirable to drive the piles at the centre of the group first and work outwards, driving the perimeter piles last.

In general, the capacity of the group can be found from the sum of the individual pile capacities, as the group capacity may be 1.3–2 times the group capacity calculated from the individual pile capacities.

EXAMPLE 6.1

A storage tank, 25 m in diameter and having a total weight of 100 MN, is to be sup-ported on 400 equally spaced hollow steel pipe piles driven 30 m into a deep bed of clay with a saturated unit weight of 18 kN/m3. The piles are 0.4 m in diameter and weigh 0.77 kN/m run. Estimate the bearing capacity of the foundation and the factor of safety that it has against a bearing failure.

Solution

In this case, the weight of the piles is taken into account as the piles are hollow and the weight of the pile may not compensate for the surcharge term. The ultimate bearing capacity of a single pile is (from Equation 6.3)

P C c dz A s N N Wu a b u c vb q pile= + + −∫0

30

( )σ

(6.20)

0 1 2 3 4 5L/B

4

5

6

7

8

9

N* c

W/B = 1

2

3

∞

Figure 6.15 Bearing capacity factors for use with pile groups in clay.


For a pile in clay, the bearing capacity factors are calculated for ϕu = 0, so that Nc = 9 and Nq = 1.

The adhesion on the shaft of the pile is found from Figure 6.9, where for su = 30 kPa the alpha factor can be seen to be 1 and so ca = αsu = 30 kPa.

Hence, substituting these values gives

Pu = × × + × + × − ×= + × −

π π0 4 30 30 0 2 30 9 18 30 0 77 30

1130 9 0 125 810 2

2. . ( ) .

. . 33 1

1209 1

.

.= kN

For the group of piles, the bearing capacity is firstly calculated for the 400 individual piles acting independently, that is,

Pug = × =400 1209 1 483 640. , kN

The capacity is then calculated for the piles acting as a block that is 25 m in diameter and 30 m deep. For such a foundation, the L/B ratio (length to diameter) is 30/25 = 1.2 and so from Figure 6.15 the bearing capacity factor (for W/B = 1) is about 7.9. Generally, the skin friction acting around the perimeter of the group is taken as su since most of the shearing is between soil and soil rather than soil and pile. The weight of the block is approximately the weight of soil in the block if the weight of the piles is considered small in comparison, therefore

Publock = × × + × + × − × ×=

π π π25 30 30 12 5 30 7 9 18 30 12 5 30 18

70685

2 2. . .

.

( )

88 381408 9 265071 9

187 022 8

+ −=

. .

, . kN

Hence, the lower of the two values is the block value so the bearing capacity of the group is taken as about 187 MN, and hence the factor of safety is 187/100 = 1.87.

6.13 PILES IN ROCK

Piles that are socketed into rock can also be designed by calculating shaft friction and end bearing. For shaft friction, the roughness of the shaft affects the values used, and often the shaft is roughened up with a grooving tool to increase the values of the shaft friction. The design of rock-socketed piles and piers is discussed more fully Chapter 12, “Basic Rock Mechanics” rock (Section 12.8.2).

6.14 SETTLEMENT OF SINGLE PILES

Often it is the displacements or differential displacements that are of importance in the design of piled foundations rather than bearing capacity. Piles may be used singly or in groups, and if used in groups, they tend to interact with one another, so that if one pile is loaded, then it causes the surrounding piles to settle. Therefore, in order to compute the settlement of groups of piles, it is necessary to be able to estimate the ‘interaction’ between one pile and another.


This section deals with vertical loading only on piles and pile groups, and horizontal load-ing is left to a later section.

For single piles, there are several approaches that can be taken ranging from empirical methods to closed form solutions and charts. Generally, the solutions are based on the assumption that the pile deflection is linear which at working loads is a reasonable assump-tion. However, at higher loads the pile deflection will become non-linear, and methods such as hyperbolic functions, load transfer (t − z) methods, or finite element methods may be required to predict the non-linear behaviour.

6.14.1 Closed form solutions

Often solutions based on the theory of elasticity can be expressed as an analytic expression such as (Randolph and Wroth 1978)

s

PE d

L L L ds

sL

s

s

= + + −−

4 1 1 8 14 1

( ) ( ( ( ) )) (tanh( ) )( )( (

ν η πλ ν µ µν

ξ/ / // )))( ) ( )(tanh( ) )( )η πρ µ µξ ζ/ / / /+

4 L L L d

(6.21)

whereP is the load on the pileEsL is the modulus of the soil at pile base levelνs is Poisson’s ratio of the soilη is the ratio of the pile tip diameter to shaft diameter = db/dξ is given by EsL/Eb where Eb is the modulus of any hard layer below the tip of the pileρ is the ratio E/EsL and E is the average value of modulus along the shaft

and

ζ ρ ν ξ

λ ν

µζ

= + − −

= +

=

ln . [ . ( ) . ]

( )

0 25 2 5 1 0 252

2 1

22

s

sp

sL

Ld

EE

LLLd

(6.22)

6.14.2 Settlement of single piles

Charts have been developed by Poulos and Davis (1980) for single piles in a deep uniform elastic soil, or an elastic soil with a rigid underlying layer. The solutions are based on bound-ary element methods, and the problem considered is shown in Figure 6.16. In the figure, the following quantities are shown:

d is the shaft diameterL is the pile lengthh is the depth to the hard stratumP is the load carried by the piledb is the base diameterEp is the modulus of the pileEs is the modulus of the soil


Eb is the modulus of the firm stratumRA is the area ratio of the pile = area of pile cross-section/total area of pile

The settlement for a single pile is computed from

1. Floating pile

S

PE d

I R R Rs

K h= 1 ν

(6.23)

where I1 is an influence factor that is presented in Figure 6.17, and RK, Rh, and Rν are correction factors for the effects of pile stiffness, layer depth, and Poisson’s ratio and are shown in Figures 6.18, 6.19, and 6.21.

2. End bearing piles

S

PE d

I R R Rs

K b= 1 ν

(6.24)

For end bearing piles, the correction factor Rb is used in Equation 6.24 to correct for the effect of a stiffer underlying layer, and this factor is presented in Figure 6.20.

6.14.3 Soil modulus increasing with depth

Where the soil modulus is increasing with depth, the head deflection of a pile can be esti-mated by using the plot shown in Figure 6.22. Several sets of curves are shown, for different rates of increase of modulus with depth Nv. Settlements can be calculated from the influence factor Iρ from

s

PN dL

Iv

= ρ

(6.25)

d

L

h

Pile modulusEp

db

dL

Soil, Young’smodulus EsPoisson’s ratioνs

db

Stiffer stratum, Young’smodulus Eb

Rough rigid base

(a) (b)P P

Soil, Young’s modulus EsPoisson’s ratio νs

Pile stiffness factor K = RAEpEs

Figure 6.16 Definition of the problem: (a) floating or friction pile; (b) end bearing pile.


6.15 INTERACTION OF PILES

As mentioned in the chapter introduction, piles will interact with each other when used in groups, as if one pile is loaded, it will cause the surrounding piles to settle.

Therefore, it is useful to be able to calculate the interactions of piles in a group, and this can be carried out conveniently by looking up plots giving interaction factors for either float-ing piles or piles driven to a firm base. Some examples of interaction factor plots for two piles spaced at a distance s (see Figure 6.23) are given in Figures 6.24, 6.25, and 6.26 where the interaction factor αF is defined as

αF = Increase in settlement of pile 1 due to pile 2

Settlementt of pile 1 under its own load (6.26)

Interaction factors can be corrected using the factors in Figure 6.27 when the soil layer has a finite depth. Further plots of interaction factors can be found in the book by Poulos and Davis (1980).

0 10 20 30 40L/d

0.02

0.04

0.06

0.08

0.10

0.2

0.4

0.6

0.8

1.0

Values of db/d1

23

For L/d = 100I1 = 0.0254For 3 ≥ db/d ≥ 1

50

I 1

Figure 6.17 Pile settlement influence factor I1.


1 2h/L

0

0.2

0.4

0.6

0.8

1.0

R h

0.5 0L/h

Values of Ld

50

2510

5

2

1

Figure 6.19 Correction factor for the effect of layer depth.

10K

1

2

3

R K

102 103 104

Values of Ld

10050

25

10

5

2

1

Figure 6.18 Correction factor for the effect of relative pile stiffness.


1 10 100 10000

0.2

0.4

0.6

0.8

1.0

R b

Eb/Es

Values of K

100500

10005000

20,000

20,0005000

(e)

= 5Ld

1 10 100 1000Eb/Es

0

0.2

0.4

0.6

0.8

1.0R b

1 10 100 10000

0.2

0.4

0.6

0.8

1.0

R b

1 10 100 10000

0.2

0.4

0.6

0.8

1.0

R b

1 10 100 10000

0.2

0.4

0.6

0.8

1.0R b

Eb/Es

Eb/Es Eb/Es

5001000

5000

≥20,000Values of K

(a)

= 75Ld

100

500

1000

5000

≥20,000Values of K

(b)

= 50Ld

Values of K

100

500

1000≥10,000

(d)

= 10Ld

Values of K100

500

1000

≥2000

(c)

= 25Ld

100

Figure 6.20 Correction factors for the effect of a stiffer underlying layer. (a) L/d = 75; (b) L/d = 50; (c) L/d = 25; (d) L/d = 10; (e) L/d = 5.


5 7 10 15 20 30 40 50 70 100L/d

0

0.1

0.2

0.3

0.4

Values of Eb/Nv L

S = PNvdL

Iρ

I ρ

1

2

5

10

100

K = Ep/Nv d = 15,000K = Ep/Nv d = 4000

P

Ld

Es = Nvzz

νs = 0.3

Figure 6.22 Settlement of a single pile in soil where the modulus increases with depth.

0.1 0.2 0.3 0.4 0.5νs

0.75

0.80

0.85

0.90

0.95

1.00

R νK = 100

500

1000

2000

0

Figure 6.21 Correction factor for the effect of Poisson’s ratio of the soil.


6.15.1 Use of interaction factors for pile groups

The settlement of one pile i due to another loaded pile j is given by (see Figure 6.28)

s Pij j ij= ρ α1 (6.27)

where ρ1 is the settlement of a single pile under a unit load, and Pj is the load on pile j.Therefore, for a group of n piles where all the piles are the same, the settlement of a pile

k due to all the other loaded piles is given by

s Pk

j

n

j kj==

∑ρ α1

1

(6.28)

s

L

p p

dd

Pile 2Pile 1

P P

Figure 6.23 Interaction between two piles.

0 1 2 3 4 5s/d

0

0.2

0.4

0.6

0.8

1.0

α F

0.2 0.15 0.1 0.05 0d/s

Values of K

1000∞

500100

10

Ld

= 10

νs = 0.5

Figure 6.24 Interaction factors for floating piles L/d = 10.


For all of the piles in the group, the settlements can be written in matrix form where A is the matrix of the alpha factors:

[ ] s A P= ρ1 (6.29)

This is suitable for piles carrying different loads applied to the head of each pile. If the pile cap is rigid, the load carried by each pile is not known, but for symmetric vertical load-ing where the cap does not rotate, it is known that (1) the settlements of all piles are equal and (2) the sum of the loads on the piles is equal to the total load on the group PTot. The set

0 1 2 3 4 5s/d

0

0.2

0.4

0.6

0.8

1.0

α F

0.2 0.15 0.1 0.05 0d/s

Ld

= 100

νs = 0.5

Values of K

5000

∞

10100

1000500

∞5000

1000500

100 10

Figure 6.25 Interaction factors for floating piles L/d = 100.

0 1 2 3 4 5s/d

0

0.2

0.4

0.6

0.8

1.0

α F

0.2 0.15 0.1 0.05 0d/s

Ld

= 100

νs = 0.5Values of K1050

100200500

1000

5000

∞

Figure 6.26 Interaction factors for an end bearing pile L/d = 100.


of equations in Equation 6.30 can be set up and solved for all of the pile loads P and the unknown deflection of the group Δ.

ρ1

0

0[ ]A a

a

P

PT

−−

=

−

∆ Tot

(6.30)

The vector aT = ( , , , , , )1 1 1 1 1… . This of course is assuming that there is only a small reaction between the pile cap and the soil, and that the piles carry most of the load. Piled raft analysis is examined more fully in Section 6.37.

6.15.2 Simplified method for pile groups

For rapid estimation of pile group behaviour without recourse to a computer, the following simple formula due to Randolph (see Fleming et al. 1992) can be used:

S R S

R n

G S av

s

=

≈ ω

(6.31)

where SG is the settlement of the group, Sav is the settlement of a single pile at the average load of piles in the group, n is the number of piles in the group, and ω is an exponent depend-ing on the pile spacing, pile proportions, relative pile stiffness, and variation of soil modulus with depth.

Poulos has suggested approximate values of ω as 0.5 for piles in clay, and 0.33 for piles in sand.

1 5s/d

0

0.2

0.4

0.6

0.8

1.0

Nh

0.2 0d/s

Values of h/L5

∞

2.5

1.5

1.2

Figure 6.27 Correction factors Nh to interaction factors for the effect of the finite layer depth.


EXAMPLE 6.2

A reinforced concrete pile, 16 m long and 0.3 m in diameter is driven through a clay layer to a dense gravel stratum. Calculate the final settlement of the pile under a working load of 800 kN.

What would be the settlement if a 12 m long floating pile was used instead?The elastic modulus of the soil, pile, and gravel may be taken as

Soil Es = 7 MPaPile Ep = 15 × 103 MPa

Gravel Eg = 140 MPa

Solution

The length to diameter ratio of the pile is L/d = 16/0.3 = 53.3, and the ratio of the base modulus to the soil modulus is Eb/Es = 140/7 = 20. Because the pile is solid, its area ratio RA is 1.0, hence the pile stiffness ratio is given by

K

EE

Rpp

sA=

= × =15 0007

1 2143,

For an end bearing pile, settlement is given by Equation 6.24

S

PE d

I R R Rs

K b= 1 ν

where I1 = 0.043 (Figure 6.17), RK = 1.19 (Figure 6.18), Rb = 0.7 (Figure 6.20), and Rν = 1 (Figure 6.21) if it is assumed that the undrained Poisson’s ratio of the soil νu is 0.5.

Hence,

S =

×× × × × =800

0 3 70000 043 1 19 0 7 1 0 0 0136

.. . . . . m

or 13.6 mm.If the pile is a floating pile, then Equation 6.23 is applicable:

S

PE d

I R R Rs

K h= 1 ν

Figure 6.19 gives the value of Rh as 0.62 for a value of h/L = 16/12 = 1.33. The length to diameter ratio of the pile is now L/d = 12/0.3 = 40 and so I1 = 0.052 and Rk = 1.15. Hence, the settlement is given by

S =

×× × × × =800

0 3 70000 052 1 15 0 62 1 0 0 0141

.. . . . . m or 14.1 mm

As this solution is based on the assumption that the layer beneath the base is rigid, the settlement may be slightly more than 14.1 mm.

6.16 ASSESSMENT OF PARAMETERS

Soil modulus is the key parameter that needs to be estimated when predicting pile displace-ment. The soil modulus (or shear modulus) is a very variable quantity and depends on a number of factors including soil type, method of pile installation, stress level imposed by the pile or pile group, and whether long or short term analysis is required (i.e. drained or undrained analysis).


The best way to assess the soil modulus is to carry out a pile load test on a prototype pile and to back figure the modulus of the soil. As this is not possible in the preliminary stages of design, it is more common to estimate the soil properties through field tests such as the SPT or CPT (cone penetration test).

Four different values of modulus can be distinguished for pile analysis:

1. The value for the soil next to the shaft Es

2. The value immediately below the pile Eb

3. The initial tangent value of the soil between the piles Ei. This will affect the interaction between the piles

4. The value of Esl for the soil well below the tip; the influence of soft layers at depth will influence pile group behaviour and the group effects will reach deeper as the pile group gets bigger

Table 6.6 summarises some values of Es, Esi, and Esl that can be correlated to SPT data and to cone data. There is not a great deal of information for values of modulus that should be used below the base of a pile. Therefore, for clays the value of Esb can be taken as equal to Es and for sands, the value can be taken as 3–5 times Es.

It should also be noted that the values in Table 6.6 are secant values at typical load levels of one-third to one-half of the ultimate load.

Correlations can also be made to typical back-figured moduli from pile load tests such as the relationships shown in Figure 6.29.

Table 6.6 Summary of some correlations for drained Young’s modulus for pile settlement analysis

Near-shaft modulus, Es (MPa)Small-strain modulus,

Esi (MPa)Modulus well below pile tips, Es (MPa)

CLAYS (2.5 ± 0.5)N (Décourt et al. 1989) 14N (Hirayama 1991) (0.5 ± 0.2)N (Stroud 1974)Driven piles (15 ± 5)su (Poulos 1989) 49.4qc

0.695⋅e0−1.13 (7.5 ± 2.5)qc

(Mayne and Rix 1993)(500 ± 5)qc (Callanan and Kulhaway 1985)

1500su (Hirayama 1991)

Bored piles (150–400)su (Poulos and Davis 1980)

(150 ± 50)su

10qc (Christoulas and Frank 1991) (0.5–0.7)MSILICA SANDS

(2.5 ± 0.5)N (Décourt et al. 1989) 16.9N0.9 (Ohsaki and Iwasaki 1973)

7N0.5 (Denver 1982)

Driven piles (7.5 ± 2.5)qc (Poulos 1989) 53qc0.61 (Imai and Tonouchi

1982)(7 ± 4)qc (Jamiolkowski et al. 1988)

Bored piles (3 ± 0.5)qc (Poulos 1993)

Source: Adapted from Poulos, H.G. 2001. Geotechnical and Geoenvironmental Engineering Handbook, Chapter 10, Ed. Rowe, R.K., Pile Foundations.

Notes:

1. Values of Es and Esi for sands are for single isolated pile. In a group, the values may be increased, depending on pile spacing and initial density.

2. Below pile tip, Esb can be taken as equal to Es for clays and bored piles in sands; and 3–5 times Es for driven piles in sands.3. Above values of Es and Esb are for use in an elastic analysis. Higher values are appropriate for non-linear analyses (e.g. the

initial tangent values for a hyperbolic model should be 1.4–1.6 times the values in this table).4. N is the SPT value (blows per 300 mm), and should be corrected to a rod energy of 60%.5. qc = cone penetrometer resistance in MPa; su = undrained shear strength in MPa; eo = initial void ratio; M = constrained

modulus.


6.17 LATERAL RESISTANCE OF PILES

The lateral resistance of single piles or pile groups is of importance when piles are used as anchors or in offshore structures. The methods used for computing the ultimate resistance of piles under lateral load generally rely on computing the forces acting on the pile or pile group at the point of failure.

Pile i

Pile jsd( )

Settlement of pile i due to pile j

ij

= Sij = Pj . αij . S1

Figure 6.28 Interaction of piles in a group.

0 50 100 150 200 250 300su (kPa)

103

E′ (k

Pa)

Average fordriven piles

Average forbored piles

Driven pilesBored pilesBored piles in London clay

104

105

Figure 6.29 Back-figured soil modulus E′ for piles in clay.


6.17.1 Single piles

As for vertically loaded piles, the ultimate load that a single pile can carry is different for piles in clay, or piles in sand. However, in both cases, two types of failure need to be considered.

• Failure of the soil• Failure or yielding of the pile itself (i.e. structural failure)


For piles in clay, solutions by Broms (1964a) may be used. He assumed that the pile would rotate about a point and that the lateral soil pressure py in front of the pile and at the back of the pile would eventually reach a maximum value of

p sy u= 9

(6.32)

The pressure in front of the pile is not allowed to act over the full shaft length, but a zone of 1.5 pile diameters at the top of the pile is assumed to apply no pressure due to disturbance during installation (see Figure 6.30).

Results for the loading, obtained by Broms are shown in Figure 6.31, for the ‘short pile’ failure mode where the soil yields rather than the pile. Figure 6.32 shows the ultimate load for a ‘long pile’ where the pile will yield before the soil fails. Both modes of failure should be checked out, and the one giving the lowest lateral failure load should be selected.

In these figures, e is the eccentricity of the lateral load (i.e. its distance above the soil sur-face) and Myield is the yield moment of the pile.


For cohesionless soils, the ultimate pressure acting in front of the pile (where it is advancing into the soil) is assumed by Broms (1964b) to be equal to a function of the passive pressure of the soil (Figure 6.33). It is therefore a function of the depth below the ground surface and will increase with depth. Behind the pile (where it is moving away from the soil), the active pressure is assumed to be small and neglected.

For piles of rectangular or circular cross section therefore, Broms suggests that the lateral pressure be taken as

p py p= 3

(6.33)

where pp is the lateral passive earth pressure of the soil computed from

p Kp p v= ′σ

(6.34)

and

Kp = + ′

− ′11

sinsin

φφ

Again the pile can fail due to yield of the soil, or due to yield of the pile itself, thus there are two plots shown in Figures 6.34 and 6.35 for the ‘short pile’ and ‘long pile’ modes, respectively.


In Figures 6.34 and 6.35, γ is the unit weight of the soil, and D is the diameter of the pile. For loads of different eccentricities e, the ultimate lateral resistance P can be estimated for each of the failure modes, and the lowest one is selected.

6.18 LATERALLY LOADED PILE GROUPS

As for vertical loading on pile groups, under lateral loading, the whole group can behave as a single block if the piles are closely spaced, and the soil is trapped between the piles. Therefore, the group capacity under lateral loading is taken as the lesser of

1. The sum of the individual pile capacities 2. The capacity of a block containing the piles and the soil

For single pile analysis, it is still necessary to consider both ‘short pile’ and ‘long pile’ modes when calculating the group capacity.

g

f

1.5d

Myield

Myield Myield

f

1.5d

Mmax

Hu

Hu

HuMmax

1.5d

Myield

Mmax

Myield

9sud

9sud

9sud

Soil reaction Bending moment

L

d

Figure 6.30 Failure modes for laterally loaded piles in clay.


However, for block failure, it is only necessary to consider ‘short pile’ failure. For this mode, beware of direct use of Broms’ charts for piles in clay, as these charts assume that there is a ‘dead zone’ of 1.5 pile diameters (in this case the diameter of the block) and this is not realistic for a large block. It is suggested that a dead zone of 1.5 times single pile diam-eter be used in this case. Also, the Broms’ theory does not account for the shear on the two sides and base of a block moving forward, and it would only apply to a group of piles where the depth to width ratio is high.

For groups with low depth to length ratios that will move forward rather than rotate, the side and base shears should be estimated, and included in the calculation. The resistance

0 4 8 12 16 20Embedment length L/D

0

10

20

30

40

50

60

Ulti

mat

e lat

eral

resis

tanc

e P/s

u D2 e

P

D

L

e/D = 0

8

16

RestrainedFree headed

12

4

Figure 6.31 Ultimate lateral resistance of ‘short piles’ in cohesive soil (e.g. clay).

3 4 6 10 20 40 60 100 300 600Yield moment Myield/suD3

1

2

46

10

20

Ulti

mat

e lat

eral

resis

tanc

e P/s

uD

2

4060

100

12 4

16

e/D= 0


D

eP

L

8

Figure 6.32 Ultimate lateral resistance of ‘long piles’ in cohesive soil (e.g. clay).


Soil reaction

e

e

Mmax

Myield

Hu

Hu

3γdLKp

f

f

g

Bending moment

L

Figure 6.33 Free-head piles in cohesionless soil.

0 4 8 12 16 20Embedment length L/D

0

40

80

120

160

200

e/L = 0

e

L

P

D

0.2

0.4

0.8

1.5

3.0

Ulti

mat

e lat

eral

resis

tanc

e P/K

pD3 γ


Figure 6.34 Ultimate lateral capacity of ‘short piles’ in cohesionless soils (e.g. sands).


at the front of the block (for piles in clay) can be taken as 9su down the front of the group (allowing for a 1.5 pile diameter dead zone).

6.19 DISPLACEMENT OF LATERALLY LOADED PILES

There are several approaches to computing the lateral deflection of piles. A common analysis is the p–y analysis (Reese et al. 1974) in which the pile is represented as an elastic beam supported by non-linear springs for which the load–stiffness relationship is known. The p–y relationship can be different for each spring down the pile if desired.

The p–y relationships are generally found from pile load tests, and typical forms of these relationships have been presented by Reese et al. (1974) for sands, and Matlock (1970) for clays.

Other methods include solutions based on linear elasticity, and non-linear modifications applied to these elastic solutions.

6.19.1 Linear elastic solutions (single piles)

Solutions for the lateral deflections of single piles have been obtained by Poulos (1973) and are presented in the form of charts. The problem basically involves a pile as shown in Figure 6.36 of length L and carrying a moment M and lateral load H. A load at any eccentricity e can always be replaced with an equivalent moment and lateral force at the pile head, that is, M = He.

6.19.2 Constant soil modulus with depth

Where the soil modulus is reasonably constant with depth, the ground line deflection ρ of a free-head pile can be expressed as

ρ ρ ρ= ⋅ + ⋅H

E LI

ME L

Is

Hs

M2

(6.35)

where the factors IρH and IρM are presented in the charts shown in Figures 6.37 and 6.38.

1 10Yield moment Myield/D4 γKp

1

10

Ulti

mat

e lat

eral

resis

tanc

e P/K

pD3 γ

10–1 102 103 104

102

103

e

L

P

D

e/D = 0 1 2 4 8 16 32


Figure 6.35 Ultimate lateral capacity of ‘long piles’ in cohesionless soils (e.g. sands).


The rotation of the pile head θ is given by

θ θ θ= +H

E LI

ME L

Is

Hs

M2 3

(6.36)

In this case, the influence factors IθH and IθM are given in Figures 6.38 and 6.39.For a pile that has a fixed head (i.e. cast into a foundation or cap so that it does not rotate),

the ground line deflection is given by

ρ ρ= H

E LI

sF

(6.37)

where IρF is given in Figure 6.40.

H

M

L Pile diameteror width = d

Figure 6.36 Laterally loaded single pile.

1 10KR

1

2

5

10

20

50

I ρH

10–6 10–5 10–4 10–3 10–2 10–1

10025

10

νs = 0.5

50

Values of L/d

Figure 6.37 Values of IρH for a free-head pile in soil with a constant modulus.


1

10

I θM

102

103

104Values of L/d

10050

2510

1 10KR

10–6 10–5 10–4 10–3 10–2 10–1

νs = 0.5

Figure 6.39 Values of IθM for a free-head pile in soil with a constant modulus.

1 10KR

1

10

100

1000

I ρM an

d I θH

10–6 10–5 10–4 10–3 10–2 10–1

Values of L/d

100

50

2510

νs = 0.5

Figure 6.38 Values of IρM and IθH for a free-head pile in soil with a constant modulus.


To use these charts, it is also necessary to compute the dimensionless flexibility factor KR where

K

E IE L

Rp p

s

= 4

(6.38)

where Ip is the second moment of inertia of the pile cross section Ep is the modulus of the pile

6.19.3 Soil modulus linearly increasing with depth

Where the soil modulus increases linearly with depth, the ground line deflection and rota-tion are given by

ρ

θ

ρ ρ

θ θ

= ′ + ′

= ′ + ′

HN L

IM

N LI

HN L

IM

N LI

hH

hM

hH

hM

2 3

3 4

(6.39)

where Nh is the rate of increase of modulus with depth. In this case, the pile flexibility factor KN is given by

K

E IN L

Np p

h

= 5

(6.40)

The influence factors in this case can be found in Poulos and Davis (1980).

1

2

5

10

20

50

I ρF10050

2510

νs = 0.5

Values of L/d

1 10KR

10–6 10–5 10–4 10–3 10–2 10–1

Figure 6.40 Values of IρF for a fixed-head pile in soil with a constant modulus.


6.19.4 Non-linearity

For laterally loaded piles, non-linearity is more important than for vertical loading as the soil resistance can be small at the top of the pile, and the pile will tend to behave in a non-linear fashion. One way of dealing with this is to use p–y analysis as mentioned before.

Another approach is to compute the deflections from the theory of elasticity and to mod-ify these by dividing by a ‘yield factor’, that is,

ρ ρ

θ θρ

θ

=

=

el

el

F

F

(6.41)

where Fρ and Fθ are the yield factors for deflection and rotation, respectively, and the subscript ‘el’ denotes the deflection or rotation for an elasticity solution. The correction factors are shown in Figures 6.41 and 6.42 for free-head floating piles in a uniform soil, and Figures 6.43 and 6.44 for free-head floating piles in soil where modulus and strength increase linearly with depth. The correction depends on the load level H/Hu where Hu is the ultimate lateral resistance of a rigid pile.

For fixed-head piles, the pile head fixing moment MF is of interest because there is no rota-tion. The correction is then given by

M

MF

FFE

M

=

(6.42)

where MFE is the elastic moment at the pile head, and FM is the yield moment factor. Values of this factor and the displacement factor are shown in Figures 6.45 and 6.46.

Estimates of the moment at the head of the fixed-head pile can be found from Figure 6.47.

0 0.2 0.4 0.6 0.8 1.0H/Hu

0.01

0.1

1.0

F ρ

2.00.251.0

2.0

0.25

1.0

0.25

1.02.0

.250 Values of e/L

2.0

KR = 10–2

KR = 10–3

KR = 10–4

KR = 10–5

Figure 6.41 Yield correction factor Fρ for a free-head floating pile in uniform soil.


0 0.2 0.4 0.6 0.8 1.0H/Hu

0.01

0.1

1.0

F θ

2 1.2502

.25

021.25

.25

0

0 .251

2

12

1

0

Values of e/L 0 KR = 10–2

KR = 10–3

KR = 10–4

KR = 10–5

Figure 6.42 Yield correction factor Fθ for a free-head floating pile in uniform soil.

0 0.2 0.4 0.6 0.8 1.0H/Hu

0.01

0.1

1.0

F ρ′

KR ≥ 10–2

KR = 10–3

KR = 10–4

KR = 10–5

2

1.250

12

.250

2 1

.2502

0

Values of e/L

Figure 6.43 Yield displacement factor ′Fρ for a free-head floating pile in soil where the modulus and strength vary linearly with depth.


6.20 DEFLECTION OF PILE GROUPS

Deflection of pile groups under lateral loading is similar to that of pile groups under vertical loading in that the piles interact, and the deflections become bigger for a group than for an equivalent single pile under the average load of a pile in the group.

Simple methods for estimating the group behaviour exist, one being given by the formula:

ρ ρ ωG avn L= 1 (6.43)

where n is the number of piles in the group ωL is an exponent given in Figure 6.48 ρ1av is the deflection of a single pile under the average load of the group

The exponent can be seen to depend on the critical length of the piles and the pile spacing. For piles in uniform and Gibson (modulus increasing linearly with depth) soils, the critical lengths are given by

LL

K

LL

K

cR

cN

=

=

4 44

3 30

1 4

1 5

.

.

/

/

(uniform soil)

(Gibson soil)

(6.44)

0 0.2 0.4 0.6 0.8 1.0H/Hu

0.01

0.1

1.0

F θ′.25 1

20 .25 1

2

02

0.25

1

2

Values of e/L 0

KR = 10–5

KR = 10–4

KR = 10–3

KR = 10–2

Figure 6.44 Yield correction factor ′Fθ for a free-head floating pile in soil where the modulus and strength vary linearly with depth.


6.20.1 Interaction methods

Interaction factors can be used to compute the effect of one loaded pile on another. In the case of laterally loaded piles, this depends on where the unloaded pile lies relative to the loaded pile as shown in Figure 6.49.

The interaction factors have been computed for

αρH: Interaction factor for head deflection due to horizontal loadαρM: Interaction factor for head deflection due to momentαθH: Interaction factor for head rotation due to horizontal load = αρM

αθM: Interaction factor for head rotation due to moment

0

0.2

0.4

0.6

0.8

1.0

F M

(a)

0

0.2

0.4

0.6

0.8

1.0

F uValues of Lc/d

2cu

9cu

py

z

3.5d

Es = 500su 2010

52.5

(b)

0.4 0.6 0.8 1.0 2.0 4.0 6.0 8.0 10.0

0.4 0.6 0.8 1.0 2.0 4.0 6.0 8.0 10.0

2cu

9cu

py

z

3.5d

Es = 500su

Values of Lc/d

2010

52.5

Lc = π√2EpIpEs

HsudLc

HsudLc

Lc = π√2EpIpEs

( )

)( 1/4

1/4

Figure 6.45 Non-linear correction factors for a flexible fixed-headed pile in stiff clay: (a) deflection correc-tion factor Fu; (b) fixing moment correction factor FM.


An example of the interaction factors for horizontal loading is given in Figure 6.50. More plots can be found in the book by Poulos and Davis (1980).

To compute the deflection (say) of the head of pile j due to a horizontal load Hi on pile i, the following formula would be used:

ρ ρ α ρj i ij HH= 1 ( ) (6.45)

(b)

0.04 0.06 0.08 0.1 0.2 0.4 0.6 0.8 1.0

9cu

py

z

2010

52.5

(a)

0.004 0.006 0.008 0.1 0.2 0.4 0.6 0.8 1.0

20

10

52.5

9cu

py

z

su = nz

su = nz

Values of Lc/d

Values of Lc/d

Es = 500su

Es = 500su

HndL2

c

HndL2

c

F MF u

0

0.2

0.4

0.6

0.8

1.0

0

0.2

0.4

0.6

0.8

1.0

Lc = (4π4 EpIpEs

)1/5

Lc = (4π4 EpIpEs

)1/5

Figure 6.46 Non-linear correction factors for a flexible fixed-head pile in soft clay: (a) deflection correction factor; (b) fixing moment correction factor.


6.21 ESTIMATION OF SOIL PROPERTIES

Soil properties can be back figured from field load tests and fitting the observations to the theory, but in the absence of such data, the value of the quantities such as the soil modulus can be estimated from correlation with field tests.

Tables 6.7 and 6.8 show correlations between field tests and soil properties for use with laterally loaded piles. Correlations are shown for both clays and sands.

0 5 10 15 20 25Lc/d

0

0.1

0.2

0.3

0.4

0.5

0.6

ωL

Pile spacing s/d 2

4

6

Homogeneous soilPinned head pilesRr = nωL

Figure 6.48 Group factor exponent ωL for laterally loaded groups in uniform soil.

Fixing moment at headof fixed-head pile

Values of 50

210

10–6 10–5 10–4 10–3 10–2 10–1 1 10KR

–0.6

–0.5

–0.4

–0.3

–0.2

–0.1

0

Mf

HL

vs = 0.5

Ld

Figure 6.47 Values of the moment at the top of a fixed-head pile in a uniform soil.


6.22 LOAD TESTING OF PILES

Pile tests carried out in the field can be broadly grouped into two types of tests:

1. Tests performed to assess the integrity of a pile. Such tests are aimed at assessing if there are any defects in the pile such as voids or cavities in the concrete, or if the pile has been constructed to the designed dimensions with no ‘necking’ or thinner sections in the shaft for example.

H, M

H, M

β

Figure 6.49 Angle between a loaded and unloaded pile.

0

0.2

0.4

0.6

0.8

1.0

0

0.2

0.4

0.6

0.8

1.0

0

0.2

0.4

0.6

0.8

1.0

0

0.2

0.4

0.6

0.8

1.0

α ρ H

0 1 2 3 4s/d

5.2 .15 .1 .05

d/s0

0 1 2 3 4s/d

5.2 .15 .1 .05

d/s0

0 1 2 3 4s/d

5.2 .15 .1 .05

d/s0

0 1 2 3 4s/d

5.2 .15 .1 .05

d/s0

α ρ H

α ρ Hα ρ H

KR = 10–3

Values of L/d

10025

1002510

10

KR = 10

Values of L/d

10025

10

2510

100

KR = 10–5

Values of L/d

100

25

100

10

2510

νs = 0.5

β = 0β = 90

KR = 10–1

Values of L/d

100

10

2510 100

25

νs = 0.5

β = 0β = 90

νs = 0.5

β = 0β = 90

νs = 0.5

β = 0β = 90

Figure 6.50 Interaction factors αρH for laterally loaded free-head piles.


2. Load tests that are designed to assess the load that can be carried by a pile or to obtain the load–deflection behaviour of a pile. Generally, tests are performed on single piles, although there have been instances where small groups of piles have been tested.

For most large structures, pile testing is almost always carried out as there are many fac-tors including construction techniques and soil variability that may change the pile behav-iour from that predicted. In addition, pile integrity also needs to be confirmed and selected piles are tested to check if pile installation methods are resulting in piles that are free from defects.

Table 6.8 Empirical correlations for Young’s modulus in sands (laterally loaded piles)

Relationship Theory Reference Remarks

N Ihi D= 0 19 1 16. ,. MPa m−1

where ID = density index (%)

Non-linear subgrade reaction

Jamiolkowski and Garassino (1977)

Tangent value for driven piles in saturated sands

Condition Nhi, MPa m−1

Loose 5.4Medium 15.3Dense 34.0


Reese et al. (1974) Tangent value for driven piles in submerged sands

Nh = 8–19 MPa m−1 (av. 10.9)

Linear boundary element Banerjee (1978) Secant value Is

Esi = 1.6 N MPawhere N = SPT value


Kishida and Nakai (1977)

Tangent value

Table 6.7 Empirical correlations for Young’s modulus in clays (laterally loaded piles)

Relationship Theory Reference Remarks

Esi/su

300–600 Non-linear subgrade reaction

Jamiolkowski and Garassino (1977)

Initial tangent modulus for driven piles in soft clays

180–450 Non-linear boundary element

Poulos (1973) Tangent modulus from model tests on jacked piles

280–400 Non-linear subgrade reaction

Kishida and Nakai (1977)

Tangent modulus

100–180 Linear boundary element

Banerjee (1978) Secant value

Nhi MPa m−1

su, kPa Non-linear subgrade reaction

Sullivan et al. (1979)

Tangent values of rate of modulus increase Esi = Nhiz

0.8 12–252.7 25–508 50–10027 100–20080 200–400


6.23 PILE LOAD TESTS

Once piles have been designed, using some of the techniques mentioned in Sections 6.9 to 6.20, pile load tests are often carried out to assess if a pile is behaving as predicted. The pile load tests can be used to

1. Act as a proof load test, where the pile is taken above the working or serviceability load to ensure that the design load is adequate. The maximum load applied to the pile head is a multiple of the working load.

2. To allow load–deflection behaviour of the pile to be assessed, to see if it is similar to predicted load–deflection behaviour. If the measured response is not as predicted, the parameters used in design can be revised to give better fits to the observed load–deflection behaviour.

Pile load tests can be static or dynamic tests. In a static test, the load is applied slowly to the pile whereas in a dynamic test, the load is applied rapidly through a dropped weight or an explosive charge. The load may be applied to the pile:

1. Vertically (either in compression or in tension). This is the most common type of load test where the direction of loading is in the direction of the pile shaft

2. Laterally by applying load horizontally to the pile head 3. Using a torsional load, although this is not common

6.23.1 Static load tests

Static tests involve loading the pile against a reaction. The reaction can be supplied by

1. Dead weight (called ‘kentledge’), as shown in Figures 6.51 and 6.53 2. Reaction piles as shown in Figure 6.52 3. Ground anchors as shown in Figure 6.54 4. The pile shaft in an Osterberg cell test. In the Osterberg or ‘O-cell’ test, a hydraulic

jack is placed within the pile shaft, and the load is applied by the jack to the upper and lower halves of the pile shaft. In some cases, two jacks can be placed within the pile shaft (see Figure 6.55)

Kentledge

Universal beam

Universal beams

Support blocks

Test pileReference beam

Hydraulic jacksDial gauges

Figure 6.51 Reaction provided by kentledge.


Figure 6.54 Anchor cables being used as a reaction for a pile load test.

Reaction beam

Hydraulic jack

Test pile

Reference beam

Dial gauge ortransducer

Anchor piles

Tension cables

Figure 6.52 Reaction provided by anchor piles.

Figure 6.53 Kentledge being used as the reaction system.


Load is applied to the pile head by one or more hydraulic jacks and is applied according to the type of test (see Section 6.23.2).

The reaction system should not be too close to the pile that is being loaded or it will interact with the pile. Various codes specify minimum distances for the reaction system from the pile.

1. Anchorages: The Australian code AS 2159-2009 specifies that no part of the anchor shall be closer than 3 times the shaft diameter of the pile; the American code ASTM D1143-81 (1994) specifies a minimum distance of 5 times the pile diameter or 2 m.

2. Anchor piles: AS 2159-2009 specifies that the distance of the anchors from the loaded pile should be the greater of 5 times the pile diameter or 2.5 m; ASTM D1143-81 (1994) specifies a minimum distance of 5 times the pile diameter or 2 m.

3. Kentledge: AS 2159 specifies that no part of the kentledge support system shall be closer than 2.5 times the diameter of the pile head; ASTM D1143-81 specifies the sup-ports for the kentledge should be at least 1.5 m from the pile being loaded.

6.23.2 Types of static load tests

The main types of static load test are (1) the incremental sustained load test (ISL test) and (2) the constant rate of penetration test (CRP test).

Tell-tales

Casing

Strain gauges

LVWDT (transducers)

‘O’-cells (2 × 405 mm)

‘O’-cells (2 × 405 mm)

LVWDT (transducers)

Pile toe tell-tales

Break in pile

Break in pile

Figure 6.55 O-cell test using two levels of hydraulic cells.


1. Incremental Sustained Load Test In this test the load is applied in stages and maintained on the pile for a period of time

during the loading and unloading phases. The ASTM Standard Loading Procedure is an example of this type of test.

The load is applied to a single pile up to 200% of the anticipated working or service-ability load, applying the load in increments of 25% of the estimated working load. Each load increment is maintained until the rate of settlement is less than 0.25 mm/h but is held constant for a maximum of 2 h. At the maximum load, the load is held constant for 12 h and then removed if the settlement is less than 0.25 mm over a 1 h period. The maximum load is removed in any case if the load has been maintained for 24 h. The load is removed in decrements of 25% of the total test load with 1 h allowed between decrements.

In other loading procedures, the load is applied in increments up to a maximum, then the pile is unloaded in increments to zero load, and then reloaded before unloading again (i.e. two load–unload cycles). The load is maintained for a set period of time at each load step and readings of deflection taken during this time. With O-cell tests, the load is often applied in this manner with the load being applied over two load–unload cycles.

Codes or standards from different countries vary in how a test is conducted, there-fore, it is important to reference the regulations in the country where the test is being performed. This type of test is the most commonly performed test for piles supporting tall buildings.

2. Constant Rate of Penetration Tests As the name suggests, the pile is jacked into the ground at a constant rate and the

load–deflection behaviour of the pile monitored. Jacking rates specified by ASTM D1143-81 are between 0.25 and 1.25 mm/min for cohesive soil and between 0.75 and 2.5 mm/min for granular soils.

3. Other Types of Tests There are several other test methods where load is applied in increments and held for

a constant length of time with no rate of movement criterion, or cyclic loading tests. These types of test are less common than (1) and (2) type tests.

6.23.3 O-cell tests

The Osterberg cell test (Osterberg 1989) is commonly used for proof testing piles as it does not need kentledge for the reaction (which can be a safety concern as it has been known to topple over), and it is capable of providing additional information about the pile behaviour.

A typical O-cell test set up is shown in Figure 6.55 where two levels of hydraulic jack (O-cells) are used. In this case, the base of the lower O-cell assembly is 4 m from the toe of the pile and the upper assembly is 21 m from the toe.

Strain gauges are provided at 12 different levels within the pile to provide information on pile compression throughout its length. From this information, pile skin friction with depth can be back figured. Two tell-tales are installed to monitor pile toe movement and upper pile compression (end of tell-tale just above the upper O-cell).

6.23.4 Lateral load testing

In some instances, lateral load tests are performed if behaviour of the piles under lateral loads that can be applied by wind or earthquake are to be assessed. The lateral load test is described in ASTM D3966-90 (1995), and is generally performed by jacking one pile against


another so that both piles are loaded horizontally. Dial gauges are mounted so that the lat-eral deflection is measured.

For the standard type of loading test, the load is taken up to 200% of the working load and then reduced back to zero. The load is maintained for a set amount of time at each load increment (or decrement) and readings taken at the start and end of each time period.

6.23.5 Measurement of deflection

For piles loaded at the head, the deflection is measured by dial gauges, or electronic trans-ducers attached to a reference beam. Movements can also be measured with precise levels or laser beams placed at some distance from the pile head. A dial gauge used for measure-ment of deflection with reference to a support beam is shown in Figure 6.57. Generally, four gauges are placed at equal intervals around the pile head. They should be accurate enough to measure pile head deflection to about 0.25 mm and have 50 mm of travel.

Measurements are affected by temperature and this can be quite pronounced in regions where the early morning and midday temperatures vary widely. Care should be taken to shield the measuring equipment and the pile head from the sun in such circumstances.

When a pile is loaded, it causes the ground around it to deflect as well as the pile, and as a result, the location of the supports for the reference beam will be affected. For this rea-son, the supports for the reaction beam should be as far as possible from the loaded pile. Standards from different countries specify different distances that the reference beam sup-ports should be from the pile head. ASTM D1143-81 specifies that the supports should be at least 2.5 m from the pile.

If the soil properties or the pile head stiffness is to be back figured from the pile load test, the relative movement of the pile head and the reference beam supports can become critical. Erroneous values of soil modulus can be calculated if this is not taken into account.

Methods of correcting for interaction effects have been presented by Poulos and Davis (1980) for various pile reaction systems, and one such system is shown in Figure 6.56. The true settlement of the pile can be calculated from the measured settlement by multiplying by

L/d =1025

100

K = 1000νs = 0.5

F c

True settlement

Measured settlement= Fc ∙

r/L0.5 0.75 1.00 0.25

0.5

1.0

1.5

2.0

2.5

r

0

L

P

Figure 6.56 Correction factor for the effect of movement of reference beam supports.


a correction factor Fc. In Figure 6.56, the correction factor is plotted against the r/L factor where r is the distance of the support from the pile, and L is the pile length.

For O-cell tests, measurement of movement is performed using tell-tale rods that are placed inside casings within the pile. The end of the tell-tale rod can be placed at any level, and the movement recorded at that level. This allows movement of the base and top of the pile as well as at intermediate points, therefore more data can be collected on pile behaviour. The movements are still measured relative to a reference beam and so the supports of the beam are subjected to movement as they are in conventional tests.

O-cell tests also involve placing strain gauges at several locations within the pile shaft as mentioned previously (see Figure 6.55). Strains can therefore be measured over most of the pile shaft and load in the pile shaft back figured along with shaft friction on the pile.

6.24 DYNAMIC PILE TESTING

Dynamic tests are sometimes performed instead of static tests when it is not desirable to use large amounts of kentledge or to construct reaction systems.

6.24.1 Dynamic pile test

A hammer having sufficient energy to mobilise the pile resistance is necessary if a pile is to be tested dynamically. In this case, the energy of the blow applied to the pile should be large enough to mobilise the equivalent of at least 150% of the working pile load or in terms of limit state design 150% of the design action effect.

The pile head is instrumented with accelerometers and strain gauges as shown in Figure 6.58, and from the recorded values, a plot is made of force versus time and of velocity versus time.

The parameters for the soil and pile are adjusted using dynamic analysis software until a good fit to the measured force and velocity plots is obtained. There are several alterna-tive approaches available for doing this, but commonly used procedures are the ‘CAPWAP’ (Rausche et al. 1985) and the TNO procedures (Middendorp and van Weele 1986).

The rate of loading in a dynamic test is obviously much higher than the loads applied to piles beneath a tall structure but correction can be made for the rate of loading in the analy-sis to give estimates of the ultimate static pile load.

Figure 6.57 Dial gauge used to measure pile head deflection.


6.24.2 Statnamic testing

The Statnamic test was developed in Canada and the Netherlands (Middendorp et al. 1992, Birmingham et al. 1994). The principle of the test is illustrated in Figure 6.59.

Fast burning fuel is detonated in a chamber accelerating a mass upwards and the recoil applies a load to the head of the pile. Accelerations of about 20 g can be achieved, therefore

Force

102L/c

t (ms)Velocity

(b)

Counterweightreleasing deviceRam

Guide tube

Measurement equipmentAccelerometersand strain equipment

Cushion

(a)

Figure 6.58 Dynamic pile load test: (a) typical equipment; (b) typical force and velocity records.

Reaction mass

Pressurechamber

–Fstn

+Fstn

Foun

datio

n pi

le

(a)

Gravel container

Gravel

Reaction masses

Silencer

Cylinder

Platform

Laser sensor

Laser

Laser beam

Piston

Load cell

Pile to be tested

(b)

⇓Figure 6.59 Statnamic test setup: (a) principle of test; (b) test setup.


the reaction mass, which generally consists of rings of concrete or steel only needs to be about 5% of the load to be applied to the pile head.

During the test, a load cell and a laser sensor are used with a high speed laptop computer to take about 400 readings per second.

Some of the advantages of the test are

1. The test is quick and easily mobilised. 2. High loading capacity is available with small reaction loads. 3. It can be adapted to lateral loading. 4. The test is quasi-static (i.e. the load is applied more slowly than for a dynamic test).

During the test, the load on the pile head increases until it reaches a maximum, and then unloads back to zero. The reaction load is caught by gravel pouring in underneath the ram so that the weights do not fall back onto the pile head.

6.25 PILE INTEGRITY TESTS

There are a number of different tests that can be carried out on bored concrete piles in order to assess if the piles have been constructed correctly without defects. Common defects can be cavities in the pile shaft or inclusions caused by material falling from the sides of the drilled shaft. Pile integrity tests include

1. Drilling cores 2. Sonic tests 3. Radiometric logging 4. Vibration testing

6.25.1 Cross-hole sonic logging

Cross-hole sonic systems involve lowering two piezo-electric probes down parallel access tubes (steel tubes are preferred) inside the pile (see Figure 6.60). The tubes are filled with water prior to the test to ensure good acoustic coupling (ASTM D6760-02). One of the probes is an emitter, and the other probe is a receiver. The system is restricted to bored piles, and tests the integrity of the concrete between the tubes by measuring its effect on the propagation of the sonic wave between the emitter and receiver (Stain and Williams 1991).

Sound concrete shows good transmission characteristics, but the presence of voids, soil, or other foreign material affects the transmission of the signal. Generally, two tubes are installed in the pile, but three tube or four tube (placed at the corners of a square) layouts can be used.

6.25.2 Sonic integrity test

A simple and common test that does not require the pre-installation of equipment is the sonic integrity test. The test involves striking the head of the pile with a plastic mallet, and the measurement of the time interval for the reflected wave to return to a transducer con-nected to the pile head (Tchepak 1998). Details of the test are given in ASTM D5882-07.

If a pile is sound, the waves will travel to the base of the pile and will be reflected back to the pile head, with the travel time dependent on the wave velocity and the length of the pile. In Figure 6.61a, the wave can be seen being reflected at the pile toe which is at 14 m. If there


Signalgenerator

Signalprocessing

ermalprinter

Digitaloscilloscope

Sonic profileprintout

Receivedsignal

Electricalimpulse

Voltage proportional tothe depth of the test

Winchwithsensor

Concrete pile

ReceiverTransmitter

Zero

Figure 6.60 Elements of a cross-hole sonic logging system.

(b)

0

1 2 3 4 5 6 7 8 9

Stroke = 57%

Pile = 7.4 m Fil = 0.0 m Vel = 2200 m/s Exp = 10 ×

Pile = 14.0 m Fil = 0.0 m Vel = 3900 m/s Exp = 20 ×

Pile = 7.4 mVel = 2200 m/s

(a)

0

2 4 6 8 10 1214

16

Stroke = 72%Pile = 13.9 m

Vel = 3916 m/s

Figure 6.61 Typical results for sonic integrity tests on bored piles: (a) a sound pile; (b) an unsound pile.


is a defect in the pile, the reflected wave will return earlier as shown in Figure 6.61b where there is a change in impedance at about 2.7 m depth.

The test can identify:

1. Reductions in section 2. Increases in section 3. Shaft cracks 4. Zones of poor quality concrete 5. Large inclusions 6. Soil restraint 7. Toe level of pile

6.25.3 Gamma logging

Another technique that can be used for pile integrity testing is gamma logging, which uses a radioactive source and a detector that can be used to measure variations in the density of concrete in a drilled pile.

The radioactive source is generally Cesium 137 that emits gamma radiation. The detector is a Geiger–Mueller probe, and the source and detector are placed into different PVC pipes (50 mm diameter) that are cast into the pile in the same way as is done for cross-hole sonic tests. The PVC pipes must be free of any water, and so are sealed to prevent water ingress. The pipes should be inspected to make sure that they are free from water and other obstruc-tions before testing.

6.26 CAPABILITIES OF PILE TEST PROCEDURES

Based on the comments made above in relation to the various types of test, Tables 6.9 and 6.10 summarise the perceived capabilities of the various tests to satisfy the needs of the

Table 6.9 Summary of capabilities of various pile load tests with respect to the results obtained

Test procedure

Ult. axial geot.

capacity

Ult. lateral geot.

capacityLoad–

settlmentLateral

deflectionGroup effects

Struct. capacity

and integrity

Special loadings

Ground movements

Static uninstrumented

3 0 3 0 1 1 1 0

Static instrumented

3 0 3 0 2 2 2 2

Static lateral 0 3 0 3 1 2 2 0Dynamic (PDA) 3 0 2 0 0 3 1 0Osterberg cell 3 0 2 0 0 1 1 0Statnamic (uninstrumented)

3 2 2 2 2 1–2 1–2 0

Statnamic (instrumented)

2 2 2 2 2 2–3 2 1

3 = Very suitable

2 = May be suitable under some circumstances

1 = Possible but unlikely to be suitable

0 = Not suitable or not applicable


designer. It will be seen that no single test can satisfactorily supply all the information which the designer may require, and that the static load test, which is usually considered to be the ‘benchmark’ test, usually provides only single pile capacity and stiffness. In addition, no test can provide the ‘perfect’ load test without ‘side effects’, and as discussed in Section 6.23.5, the interpretation of the static test should allow for the interaction between the test pile and the reaction system.

The testing system chosen will depend on the information that is required from the test, the cost of the test and the availability of equipment to perform the test.

6.27 NUMBER OF PILES TESTED

For tall buildings that may be supported by large numbers of piles, the question arises as to the number of piles that should be tested so that the test results are representative of the whole pile group.

The Australian piling code AS 2159-2009 specifies the percentage of piles to be tested for serviceability conditions. This depends on the Risk Rating which is a number calculated from Risk Factors such as the variability of the geology and the extent of the site investiga-tion, experience with design in similar conditions, the extent of soil testing and the quality of construction supervision. From tables in the code, an average risk rating (ARR) can be calculated (higher risk has a higher ARR) and from this the number of piles to be tested estimated as shown in Table 6.11. Testing of piles is only specified if the strength reduction factor applied in design (which is an Ultimate Limit State Design) is greater than 0.4 (i.e. the soil strengths are factored down by a value >0.4).

AS 2159-2009 also gives guidance for integrity testing. The amount of testing depends on the pile type (e.g. precast or cast in place). Lower percentages of piles are specified for testing if the pile design load is governed by soil strength rather than pile structural capacity. For

Table 6.10 Summary of capabilities of various pile load tests with respect to the accuracy and relevance of the results

Test procedurePile loaded in same way?

Additional stress changes (side effects)

Accuracy of movement

measurement

Accuracy of load

measurement

Similar duration of loading to prototype?

Static uninstrumented 3 2 2 3 3Static instrumented 3 2 2 3 3Static lateral 3 2 2 3 3Dynamic (PDA) 3 2 1 1 1Osterberg cell 2 2 2 3 3Statnamic 3 3 3 3 2

3 = Good

2 = May be adequate

1 = Generally not good

Table 6.11 Pile testing requirements for serviceability (AS 2159-2009)

Average risk rating (ARR) 2.50–2.99 3.00–3.49 3.50–3.99 4.00–4.49 ≥4.5

Percentage of piles to be tested for serviceability 1 2 3 5 10


bored piles, the percentage of pile integrity testing depends on how carefully drilling fluid, base cleaning and concrete tremie pouring is monitored.

For example, for a bored pile constructed using a casing or drilling fluid with good con-struction monitoring and the design load governed by geotechnical capacity, 5%–15% of piles should be tested. If the design load is governed by pile shaft structural capacity and there is minimal construction monitoring, 15%–25% of piles need to be tested.

The Federation of Piling Specialists (United Kingdom) Handbook on Pile Load Testing (2006) gives guidelines for the number of piles tested, and these are shown in Table 6.12. The amount of testing depends on the amount of risk associated with the project.

The number of tests performed can also be estimated on a cost basis as described by Kay (1976). More tests mean that the pile sizes can be refined saving money, but too many tests will raise the cost due to the cost of testing. The formula of Equation 6.46 can be used to calculate cost C.

C

XFF mY

m=+0

(6.46)

where Fm is the factor of safety for m load tests, F0 is the original factor of safety for no pile load tests, X is the cost of the total number of piles, and Y is the cost of a single load test.

For tall buildings where the geological conditions are uniform and construction control is good, generally one or two vertical pile load tests are performed, with perhaps a lateral load test. Integrity testing may be performed on 10–15 piles and sonic integrity tests on 20–30 piles. A tension test may be required if some of the piles are subjected to uplift forces.

6.28 TEST INTERPRETATION

Information obtained from pile testing may be

1. The ultimate load capacity of a single pile 2. The load–settlement behaviour of a pile

Table 6.12 Pile testing requirements according to risk

Characteristics of the piling works Risk level Pile testing strategy

Complex or unknown ground conditions. High Both preliminary and working pile tests essential.No previous pile test data. 1 preliminary pile test per 250 piles.New piling technique or very limited relevant experience.

1 working pile test per 100 piles.

Consistent ground conditions. Medium Pile tests essential.No previous pile test data. Either preliminary and/or working pile tests can be

used.Limited experience of piling in similar ground.

1 preliminary pile test per 500 piles.1 working pile test per 100 piles.

Consistent ground conditions. Low Pile tests not essential.Previous pile test data is available. If using pile tests either preliminary and/or working

tests can be used.Extensive experience of piling in similar ground.

1 preliminary pile test per 500 piles.1 working pile test per 100 piles.

Source: Federation of Piling Specialists, Handbook on Pile Load Testing, 2006.


3. The acceptability of the performance of a pile, as-constructed, according to specified acceptance criteria

4. The structural integrity of a pile, as constructed

Such information may be used in a number of ways, including

1. Construction and quality control 2. As a means of verification of design assumptions 3. As a means of obtaining design data on pile performance which may allow for a more

effective and confident design of the piles

6.28.1 Ultimate load capacity

Once a loading test has been carried out, the load–deflection curve for the pile may be plotted. From this curve, it may be difficult to estimate where the pile reaches its ultimate load, as the deflection curve may continue to climb with the loading and not show any clear-cut failure. In this case, it is more usual to define the failure load as the load for a specific displacement. For example, for conventional compression load tests, Eurocode 7 (2004) defines the failure (or ‘limit’) load as that causing a gross settlement of 10% of the equivalent base diameter.

In the case where the load is well below the failure load and deflections are small, a fre-quently used approach is that of Chin (1970) which, in effect, assumes that the load–settle-ment curve is hyperbolic, and extrapolates the load–settlement data on this basis. It has been found commonly that Chin’s method tends to over-estimate the failure load, and it is occasionally modified so that the failure load is taken as a proportion (typically 90%) of the value derived from Chin’s construction. It is also possible to adopt a consistent approach and extrapolate the load–settlement curve via Chin’s approach, but to define the failure load as the value at a settlement of 10% of the diameter.

6.28.2 Pile stiffness

The deflection of a pile under load may be found from a pile load test and used to refine pre-dictions of pile group or piled raft behaviour. If the deflection at a working load is required, it may be adequate to back figure a secant pile stiffness, and to assume the pile has a linear load–deflection behaviour over the range of loads anticipated.

At higher loads, the load–deflection behaviour measured for the pile will be non-linear. Often, a hyperbolic relationship is used to model the pile stiffness in this case. Parameters for the hyperbolic relationship can be changed until a good fit to the measured load–deflec-tion behaviour of the pile is found.

In obtaining the pile stiffness whether linear or non-linear, it is necessary to allow for the interaction of the pile with the datum for the measuring system as discussed in Section 6.23.5. Misleading values of pile head stiffness may be obtained if this is not done.

Correction is also needed for interaction with the reaction system. Kitiyodom et al. (2004) have presented charts to allow correction of pile head stiffness found from pile load tests. The charts are for the case where anchor piles are used. Poulos and Davis (1980) also pres-ent charts to correct for the effects of the reaction system.

6.28.3 Acceptance criteria

In many cases, acceptance criteria are specified for quality control purposes, and are taken from a code, without necessarily being related directly to the design. Typical criteria as


specified in the Australian Piling Code (AS2159-2009), for example, are shown in Table 6.13. The design may be affected by the load testing in that, if the piles are deemed to be unacceptable, a decision then needs to be made on the future course of action by

• Re-design the pile foundation using more appropriate assumptions• Replacement of the piles which have shown inadequate performance• Addition of extra piles to compensate for the piles which have performed inadequately• Re-analysis of the proposed foundation with the inadequate piles carefully to assess

whether the performance of the foundation system as a whole will perform adequately

While there may be circumstances in which one of the first three options is inevitable, there may also be instances where the group action may allow re-distribution of some of the loads from the inadequate piles to the other piles, without causing unacceptable conse-quences to the group performance.

6.28.4 Other quantities

Other information can be obtained from pile load tests as well as the usual ultimate pile load and the load–deflection behaviour (or pile stiffness). With specially instrumented piles, the load in the pile as well as the skin friction along the pile shaft may be obtained.

Strain gauges can be attached to the reinforcing cage of a pile or concrete type gauges can be cast into the concrete. The strain gauges allow strains and therefore stresses to be calcu-lated at various depths within the pile shaft, and therefore the stress and the load in the pile shaft at that location. Tell-tales may also be used that consists of steel rods placed inside a casing. The end of the rod is cast into the pile at a chosen depth, and the movement of the head of the tell-tale monitored.

Extensometers may also be used to measure vertical movement within a pile shaft. Vibrating wire or DCDT displacement transducers are installed inside 51 mm steel or PVC sonic testing pipe. In one type of gauge, anchors at the top and bottom of the gauge can be expanded using compressed air to attach the gauge to the sides of the pipe. The relative movement between the top and bottom anchor of each gauge can be read to assess strains. The gauges can be retrieved after testing by releasing the air pressure in the anchors.

Table 6.13 Acceptance criteria for vertical pile load tests (AS2159-2009)

LoadMaximum settlement (mm)

Static load test

Serviceability load Ps PsL/AE + 0.01da

After removing serviceability load Max(0.01d,5)Factored up pile load Pg PgL/AE + 10 + 0.05dAfter removing the factored up pile load 10 + 0.05d

Notes:

1. The movement is to include no more than 2 mm creep over 3 h 45 min (after load has been in place for 15 min).

2. Ps is the pile serviceability load, Pg is the load comprised of the factored up load combinations, d is pile diameter, L is pile length, A is the pile cross-sectional area.

3. Loads are for no downdrag (or negative friction).


6.29 MONITORING OF PILED FOUNDATIONS

Monitoring of the performance of piled foundations can be carried out so as to confirm predictions of performance (i.e. settlements, pile loads), for reasons of safety or to provide a warning of impending problems. The data obtained from monitoring is invaluable as it may be used to refine soil models and analysis techniques for use in the design of similar structures in the same area or similar soil profiles.

For research purposes, very comprehensive instrumentation may be installed in the piles and beneath the raft of a foundation, and even in the surrounding soil. This may be per-formed when constructing tall buildings in an area for the first time, or when foundation conditions are very different across a site. For smaller structures or structures where there is experience with similar site conditions and the risk is low, monitoring may simply involve settlement measurements to confirm design predictions.

6.30 MEASUREMENT TECHNIQUES

There are many types of instruments that can be placed beneath or in piled founda-tion systems, and the extent of the instrumentation depends upon the purpose of the monitoring.

For example, the foundations of the Messe Turm in Frankfurt (which is supported by a piled raft) contained 13 contact pressure cells, 1 piezometer, 3 multi-point borehole exten-someters, and 12 instrumented piles. The piles contained strain gauges and load cells so that the pile loads in the shaft and at the base of the piles could be measured. Details are contained in Chapter 13 by Katzenbach et al. (2000) in the book by Hemsley (2000).

Katzenbach et al. (1995) describe the instrumentation placed beneath the Commerzbank Tower in Frankfurt (Figure 6.62) that included 300 strain gauges placed inside 30 piles, 15 piles had load cells at the pile toe, 5 also had load cells at the pile head, and 6 piles had small concrete load cells. Thirteen extensometers measured ground deformation down to a depth of 95 m below the raft level, and 13 contact pressure cells and 4 piezometers were installed beneath the raft.

These structures were the subject of extensive research as they were some of the first tall buildings constructed on piled rafts in the Frankfurt clay. Some of the types of instruments that may be used are discussed in the following sections.

6.30.1 Deflection

Probably, the most common and important measurements taken are of the movement of the foundation. As extra stories are added to the structure, the foundation will compress, and there may be immediate, consolidation or creep settlements of the soil, and elastic compres-sion of the pile that take place. In addition, the foundation settlement may not be uniform and the structure may rotate. Rotation can be serious for tall buildings since a small rotation at the foundation level may mean large lateral movements at the top of the structure. Often, corrections to the verticality of a tall structure are made as more stories are added if the building is moving away from the vertical.

Deflection measurements may be taken with accurate levels onto a measurement marker placed on the foundation. A benchmark that is not affected by the settlement of the struc-ture needs to be used as the datum for the measurements. Total station theodolites may be used to obtain both vertical and lateral movements of markers.


Other equipment such as lasers and electronic inclinometers can be used to record lateral movements and tilt of buildings. Dynamic behaviour of structures has been measured using GPS techniques that are capable of measuring distances to sub-centimetre accuracy and col-lecting data at 10 Hz (Luo et al. 2000).

6.30.2 Pressure cells

Pressure cells may be used to monitor the pressure beneath raft foundations or the load in piles. This may be of interest if the load sharing between the raft and the piles is to be mea-sured and compared with design estimates.

For measurement of pressure beneath a raft, Glötzl-type cells may be used as reported in Hemsley (2000). The Glötzl cell has a thin sealed chamber containing oil or fluid that causes a membrane to deflect when the fluid is pressurised. A fluid pressure is then applied to the membrane to return it to its null position, and that pressure is taken as the pressure applied to the cell.

Figure 6.62 Commerzbank Tower (Frankfurt, Germany).


Load cells for measuring pressure at the base of piles are available and consist of a fluid-filled cell between two plates. The pressure of the fluid is measured by pressure transducers.

In O-cell testing, the hydraulic pressure in the cell that is performing the loading within the pile shaft is used to back figure the load being applied.

6.30.3 Strain gauges

Strain gauges have been mentioned previously in the section on pile testing (see Section 6.24). The gauges allow loads in the pile shaft to be calculated. These types of gauges are generally used with bored piles where the gauge can be attached to the steel reinforcing cage, cast into the concrete, or extensometers can be placed in tubes within the pile shaft. For steel piles, strain gauges can be welded to the pile shaft.

6.30.4 Piezometers

Piezometers may be installed beneath rafts or piled rafts to monitor excess pore water pres-sures generated during loading. In the case where groundwater has been lowered to allow excavation of a basement, the total groundwater pressures will fall and rise again as the pumping is ceased. The water pressures need to be suppressed by pumping until the weight of the structure can counter the water uplift pressure, therefore water pressure monitoring is important.

Various types of piezometer may be used (see Dunicliffe 1993) including standpipe, hydraulic, and vibrating wire devices. The advantage of vibrating wire piezometers is that they are connected by electric wires to the readout location and have a short lag time (i.e. can register the pore pressure quickly).

6.30.5 Extensometers and inclinometers

Extensometers are sometimes placed beneath foundations to obtain the settlement of the foundation soils with depth. There are many different kinds of extensometers available commercially, but most involve a hollow tube that can telescope and move with the ground. Either magnets or steel rings are placed around the tube, and the posi-tion of these rings is detected with a probe lowered into the tube. The probe can accurately locate the position of the rings, so the soil movement at the locations of the rings can be found.

Inclinometers may be used to measure lateral soil movement. A plastic casing is placed into a borehole and grouted in place. The casing has grooves in the sides (generally two sets at right angles) in which the wheels of a probe can run. As the probe is lowered down the tube, an accelerometer takes readings of the inclination of the probe, and from these read-ings, the lateral movements of the casing may be found. Some inclinometers can have a dual role as an extensometer and an inclinometer.

Different types of extensometers and inclinometers are discussed in the book on instru-mentation by Dunnicliffe (1993).

6.30.6 Frequency of measurements

Measurements of displacements, pile loads, etc. need to be taken as the structure is increas-ing in height as the increased loads cause changes in the measurements of all instruments. Once the construction is complete, the structure may continue to settle due to consolidation


and creep of the foundation. Measurements may need to be taken for many years to assess if the rate of settlement is slowing down. During this period, there may also be changes in pile loads and raft moments.

It is therefore necessary to take several measurements for each storey that is constructed of a tall building. For the One Shell Plaza building constructed in Houston, Texas, for exam-ple (Focht et al. 1978), readings of instruments were taken at 2–4 month intervals during construction. After the structural frame was completed, observations were made every 4–6 months. Two years after completion of the structure, readings were taken at yearly intervals, up to 10 years post-construction.

6.31 COMPARISON WITH PREDICTED PERFORMANCE

Measurements from instruments installed on or in structural elements such as the piles and raft or in the soil beneath or beside the foundation of a tall building, can provide valuable information for use in predictive models.

The design and performance of the foundation can be carried out using finite element methods (PLAXIS 3D, ABACUS), combined finite element and boundary element methods like GARP (Small and Poulos 2007) or other simple hand techniques. Comparison of the predicted and measured behaviour of a foundation may be used to assess whether the foun-dation is behaving as predicted by the numerical models.

If the monitoring shows that the structure is not behaving as predicted (i.e. deflections are suddenly becoming larger than expected, or excessive tilting is occurring or pile loads are excessive) then remedial action may need to be taken. This may take the form of strength-ening the foundation by adding more piles or trying to re-analyse the foundation to assess whether the observed performance will result in an acceptable foundation for the particular structure.

Comparison of numerical predictions and measured performance allows design parame-ters to be refined and for less conservative designs to be carried out for similar soil conditions. Correlations used between results of field tests (such as SPT, CPT, seismic tests, and pres-suremeter tests) and pile design parameters (such as skin friction and end bearing pressures) can also be refined through comparisons with field performance data.

6.31.1 Emirates twin towers, Dubai

The Emirates Project is a twin tower development in Dubai (Figure 6.63), one of the United Arab Emirates. The towers are triangular in plan with a face dimension of approximately 50–54 m. The taller, the Office Tower, has 52 floors and rises 355 m above ground level, while the shorter, Hotel Tower, is 305 m tall (see Poulos 2009).

A comprehensive series of in situ tests was carried out. In addition to standard SPT tests and permeability tests, pressuremeter tests, vertical seismic shear wave testing, and site uni-formity borehole seismic tests were carried out.

Conventional laboratory testing was undertaken, consisting of conventional testing, including classification tests, chemical tests, unconfined compressive tests, point load index tests, drained direct shear tests, and oedometer consolidation tests. In addition, a considerable amount of more advanced laboratory testing was undertaken, including stress path triaxial tests for settlement analysis of the deeper layers, constant normal stiffness (CNS) direct shear tests for pile skin friction under both static and cyclic load-ing, resonant column testing for small-strain shear modulus and damping of the founda-tion materials, and undrained static and cyclic triaxial shear tests to assess the possible


influence of cyclic loading on strength, and to investigate the variation of soil stiffness and damping with axial strain.

The geotechnical model for foundation design under static loading conditions was based on the relevant available in situ and laboratory test data, and is shown in Figure 6.64. The ultimate skin friction values were based largely on the CNS data, while the ultimate end bearing values for the piles were assessed on the basis of correlations with unconfined com-pressive strength (UCS) data (Reese and O’Neill 1988) and also previous experience with similar cemented carbonate deposits (Poulos 1988).

Using the geotechnical data shown in Figure 6.64, predictions were made for pile load tests, and once the tests were complete, the predictions and measured pile responses were compared. Compression tests were performed using anchor cables as the reaction system. Other tests that were performed were tension tests, lateral load tests, and cyclic load tests.

Four main types of instrumentation were used in the test piles:

• Strain gauges (concrete embedment vibrating wire type) – To allow measurement of strains along the pile shafts, and hence estimation of the axial load distribution.

Figure 6.63 Emirates twin towers with construction almost complete (Dubai, United Arab Emirates).


• Rod extensometers – To provide additional information on axial load distribution with depth.

• Inclinometers – The piles for the lateral load tests had a pair of inclinometers, at 180°, to enable measurement of rotation with depth, and hence assessment of lateral dis-placement with depth.

• Displacement transducers – To measure vertical and lateral displacements.

Comparisons for one of the compression loading tests that were performed are shown in Figure 6.65 (load–deflection behaviour) and in Figure 6.66 (for pile axial load with depth). The predictions for this pile and for the other pile tests that were performed were considered to be reasonable.

Predictions were then made for the foundation of the towers. In the final design, the piles were primarily 1.2 m diameter, and extended 40 or 45 m below the base of the raft. In general, the piles were located directly below 4.5 m deep walls which spanned between the raft and the first level floor slab. These walls acted as ‘webs’ which forced the raft and the slab to act as the flanges of a deep box structure. This deep box structure created a relatively stiff base to the tower superstructure, although the raft itself was only 1.5 m thick.

SILTY SAND, somecalcarenite bands

As above

CALCAREOUSSANDSTONE

SILTY SAND

CALCISILTITE

As above

As above

100

40

125

700

125

500

90

700

30

500

100

400

80

600

0.2

0.2

0.1

0.2

0.2

0.3

0.3

18

73

200

150

450

200

450

0.15

1.5

2.3

1.9

2.7

2.0

2.7

0.1

1.5

2.3

1.9

2.7

2.0

2.7

1

2

3

4

5

6

7

80

70

60

50

40

30

20

10

0

Dep

th (m

)

EuMPa

E′MPa ν′ fs

kPafb

MPapu

MPa Unit

Figure 6.64 Geotechnical model adopted for design.


0 10 20 30 400

5

10

15

20

25

30

App

lied

load

(MN

)

Settlement (mm)

MeasuredPredicted

Figure 6.65 Predicted and measured load–settlement behaviour for Pile P3 (Hotel Tower).

0 5 10 15 20 25 30

–40–38–36–34–32–30–28–26–24–22–20–18–16–14–12–10

–8–6–4–2

02

Leve

l DM

D (m

)

Measured (15,000 kN)Measured (23,000 kN)Predicted

Load (MN)

Figure 6.66 Predicted and measured axial load distribution for Pile P3 (Hotel Tower).


Predictions of settlement were made using various computer programs, and these were compared with the measured vertical settlement of the towers. A comparison of the pre-dicted settlement and the measured settlement for the Hotel Tower is shown in Figure 6.67.

As can be seen from the plot in Figure 6.67, the predicted and measured settlements of the Hotel Tower were not in close agreement even though the predictions for the pile load tests were similar to those measured.

Two prime reasons for the larger prediction of settlement for the pile group were thought to be (1) the interaction of the piles and (2) the stiffness adopted for the ground below RL-53m. The calculation of interaction among the piles can have a large influence on group settlements when there are a large number of piles, so the estimates of the interaction factors were re-assessed. Pile groups stress the ground to greater depths than single piles, and if the soil modulus at depth is different than assumed, this can lead to inaccuracies. Lower strain levels in the ground at depth mean that a higher modulus should be used as the stiffness of the ground is strain dependent. These two reasons may explain why single pile predictions are reasonable, but the pile group predictions are not.

This experience was used when estimating the settlements of the Burj Khalifa where stiffer layers at depth were used and interaction factors were calculated assuming stiffer material existed between piles (and therefore the interaction was less). This gave reasonable predictions of the long-term settlements of about 74 mm. Measured settlements of the foun-dation before the tower was completed had reached about 42 mm, so the final settlement may be close to that predicted.

The monitoring programme was therefore very useful in indicating that the modelling pro-cedure for the piled raft had some shortcomings, and the experience gained from the Emirates project could be used to make better predictions for the foundations of the Burj Khalifa (Figure 6.81) which was founded in similar materials. This example demonstrates the value of monitoring programmes, especially when designing in new or unfamiliar ground conditions.

0 1 2 3 4 5 6 7 8 9 10 11 12

50

40

30

20

10

0

Settl

emen

t (m

m)

Predicted

T4

T15

1998Time (months)

Measured

Figure 6.67 Measured and predicted time–settlement behaviour for the Hotel Tower.


6.32 INTERPRETATION AND PORTRAYAL OF MEASUREMENTS

Measurements are often portrayed as a function of time. For instance, the settlement of a structure can be plotted against time thus showing how it increases with height of a struc-ture and how it continues to increase after construction is complete due to consolidation and creep.

Loads in piles may also be plotted against time to monitor increases with applied load increases. Loads in the pile shaft can be found from strain gauges, and this can be done for conventional top loaded piles or when performing O-cell tests. The load in the pile shaft can be used to make estimates of pile shaft skin friction and pile base load as is shown in Figure 6.68 for monitored piles of the Messe Turm building in Frankfurt.

Measurement deflections may be used to provide warning if the structure is not behaving as predicted. For example, the tilt of the building may be seen to exceed the allowable value or pile loads may exceed allowable values. In such cases, some remedial work may need to be carried out or decisions need to be made as to whether the measured values can be tolerated by the structure. For example, if some piles are overloaded, this may be acceptable if the deflections are not excessive, since the load can be shed to other piles in the group.

6.33 PILED RAFTS

If a surface foundation is not adequate to carry structural loads without excessive differen-tial deflections, piles may be needed. Both the raft and the piles then transfer load to the soil, and the interaction problem involves both the raft and the piles. In some cases, the piles are only placed beneath the raft to provide differential settlement control and are allowed to fail under load (Hansbo and Källström 1983).

It is important to realise that piles do not need to be uniformly placed over a foundation, but can be judiciously placed so as to carry the larger loads or to limit the differential deflec-tions. In this regard, it is useful to have a quick and simple computer program or simple design method that can be used in the design stage to determine the best layout of the piles

Outer pile-circleMiddle pile-circleInner pile-circle

OMI

M

M

M

O

O

O

I

I

I

Pile load (MN)

0

–26.9 m

–30.9 m

–34.9 m

Skin friction (kPa)50 100 150 2000151050

10

20

30

40

Dep

th (m

)

0

Figure 6.68 Distributions of the pile load and skin friction from monitoring of the piles beneath the Messe Turm building in Frankfurt.


beneath the foundation. For example, Horikoshi and Randolph (1997) have shown that the optimum design of a piled raft carrying a uniform load would involve piles placed under the central 16%–25% of the raft area.

A piled raft foundation therefore, combines a raft with piles so the overall performance of the raft can be improved.

• Piled rafts are useful where settlement or differential settlement of the raft is inadequate, even though the raft may have an adequate factor of safety against a bearing failure.

• Piles are used as settlement controllers or reducers.• There is a combined action of the raft and the piles – the piles do not carry all of the load.

6.34 USES OF PILED RAFTS

Rafts with piles placed so as to control settlements may seem an attractive alternative to piles with individual caps that carry column loads. However, piled rafts are not the best solution in every case, and the following sections discuss when piled rafts are favoured as foundations and when they are not.

1. Favourable circumstances• Where load capacity of a conventional piled foundation is adequate, but settlement

or differential settlement is not• There are relatively stiff clay profiles• Dense sand profiles exist• The foundation consists of layered profiles with no soft layers below pile tip level• Where soil movements due to external causes do not occur

2. Unfavourable circumstances• Where soft clays exist near the surface• Where loose sands exist near the surface• Where consolidation settlements may occur due to external causes• Where swelling movements may occur

3. Alternate strategies for piled raft design• Where piles operate at their normal load levels (a factor of safety of typically 2–3),

there is not a great potential for savings in design but if piles are used as settlement inhibitors and are allowed to yield (factor of safety reaches 1), there is potential for large savings

6.35 DESIGN CONSIDERATIONS

When designing a piled raft, the following design factors need to be taken into account:

• Bearing capacity of the piles and raft• Maximum settlement of the foundation• Differential settlement of the foundation• Raft moments and shears• Loads carried by the piles

There may also be circumstances where lateral loading due to wind loads or earthquake loads need to be taken into account.


6.35.1 Design process

The preliminary design process can involve estimating the number of piles required to sat-isfy overall bearing capacity and settlement requirements.

Then the detailed design can be used to determine

• The optimum pile locations (to limit differential settlement)• The differential settlements• The shears and moments in the raft

6.36 BEARING CAPACITY OF PILED RAFTS

Work by de Sanctis and Mandolini (2006) (based on 3D finite element analysis) suggests that the Factor of Safety for a piled raft can be calculated from the factor ξPR defined as

ξPR

G

QQ Q

FSFS FS

=+

=+

PR, ult

UR, ult G, ult

PR

UR (6.47)

whereQPR,ult is the ultimate load that the piled raft can carryQUR,ult is the ultimate load that the raft alone can carryQG,ult is the ultimate load that the pile group (with no raft) can carryFSPR is the factor of safety of the piled raft = QPR,ult/QFSUR is the factor of safety of the raft alone = QUR,ult/QFSG is the factor of safety of the pile group (with no raft) = QG,ult/QQ is the applied (working) load

Numerical analyses were carried out for different pile layouts (as shown in Figure 6.69) beneath piled rafts using a three-dimensional finite element program. The results of the analyses are shown in Table 6.14. In the table, H is the depth of the soil stratum, BR is the full width of the raft, s is the centre-to-centre pile spacing, L is the pile length, and d is the pile diameter.

It can be seen that the value of the ratio ξPR varies from a minimum of 0.82 up to a maxi-mum of 1.0. It was therefore recommended by de Sanctis and Mandolini to take the value of 0.80 for estimating the factor of safety of piled rafts in bearing, that is,

Q Q QPR ult UR ult G ult, , ,. ( )= +0 8

(6.48)

or dividing both sides of the equation by the applied load Q

FS FS FSGPR UR= +0 8. ( ) (6.49)

Hence, the suggested ultimate load that can be carried by a piled raft is 0.8 times the sum of the capacity of the raft alone and the capacity of the piles alone.

As much greater movement of the cap is needed to mobilise the bearing capacity of the cap than for the piles, it is suggested that the bearing pressure of the cap be taken as that of a strip footing with a width equal to the distance between the edge of the cap and the outer pile.


BR = 20 d

BR = 12 d

n = 9, 25

L/d = 20, 40

s/d = 4, 8

H/L = 2

n = 9

L/d = 20, 40

s/d = 4

H/L = 2

n = 9, 49

L/d = 20, 40

s/d = 4, 8

H/L = 2

BR = 28 d

AG

AG

AG

Figure 6.69 Pile groups examined by de Sanctis and Mandolini (2006).

Table 6.14 Ratio ξPR for piled rafts analysed by de Sanctis and Mandolini (2006)

Case L/d n s/d BR/d FSUR FSG FSPR ξPR

1 40 49 4 28 1.95 6.46 7.76 0.922 40 9 4 28 1.95 1.19 2.89 0.923 40 9 8 28 1.95 1.19 2.99 0.954 20 49 4 28 1.95 2.15 3.35 0.825 20 9 4 28 1.95 0.40 2.28 0.976 20 9 8 28 1.95 0.40 2.32 0.997 40 25 4 20 2.11 6.98 8.84 0.978 40 9 4 20 2.11 2.51 4.29 0.939 40 9 8 20 2.11 2.51 4.51 0.9810 20 25 4 20 2.11 2.33 3.64 0.8211 20 9 4 20 2.11 0.84 2.54 0.8612 20 9 8 20 2.11 1.84 2.70 0.9213 20 9 4 12 2.26 2.49 4.06 0.8614 40 9 4 12 2.26 7.47 9.71 1.00


6.37 ANALYSIS OF PILED RAFT FOUNDATIONS

In the past, piles were treated as groups that were rigidly joined at the head or carried equal loads, and the flexibility of the raft that joined the pile heads was ignored. The book by Poulos and Davis (1980) includes many of the methods for computing the settlement of piles or pile groups when the raft is assumed to be totally rigid or totally flexible (i.e. raft flex-ibility is one of two extremes). These solutions are based on treating the shear forces acting down the pile shaft as a series of uniform shear stresses acting over sections of the pile shaft. Mindilin’s (1936) equation for a sub-surface point load is integrated over the section of pile to obtain the solution for the effect of the uniform shear stress on deflections of the soil at other sections of pile for the pile itself or for other piles. Interaction between piles can there-fore be found using this technique often called a ‘boundary element’ technique.

Many different means of analysing piled raft foundations have been developed over the years (comprehensive reviews have been provided by Randolph 1994 and Hemsley 2000). Some of the methods that have been used to analyse piled rafts can be conveniently divided into the following groups:

1. Simple Plate on Spring Approaches These methods treat the piles as springs with the raft treated as a plate, and include the

methods of Clancy and Randolph (1993), Poulos (1994), and Viggiani (1998). 2. Boundary Element Methods These methods employ the technique described above and include solutions obtained

by Butterfield and Banerjee (1971), Brown and Weisner (1975), Hain and Lee (1978), Kuwabara (1989), and Chow (1986).

3. Finite Layer Techniques Ta and Small (1996) used finite layer techniques (see Chapter 2) to compute the behav-

iour of piled rafts, where the piles were driven into layered soils. Cheung et al. (1988) had previously used series to analyse the behaviour of pile groups in layered soils, and the method can be extended to piled rafts. Zhang and Small (2000) have extended these techniques to horizontal loading of a piled raft.

4. Simplified Finite Element Analyses Analyses can be carried out by approximating the piles as a two-dimensional or axi-

symmetric body and assigning ‘smeared’ material properties to the piles in order to approximate the actual three-dimensional behaviour. That is the solid continuous ‘pile’ in an axi-symmetric or 2D analysis is given a lower modulus to make it compress the same amount as the actual individual piles. Analyses of this sort include those of Desai et al. (1974) and Hooper (1973). Lin et al. (1999) have used a finite difference technique to compute the behaviour of the soil beneath a piled raft, and applied the theory to piled rafts in Bangkok clay using a two-dimensional finite difference grid.

5. Three-Dimensional Finite Element Analyses As computer storage has increased, full 3D analyses of piled rafts have been carried

out and early examples of this are given by Zhuang et al. (1991), Katzenbach and Reul (1997), Katzenbach et al. (1997), and Ottaviani (1975).

6.37.1 Numerical modelling

In the previous section, many different methods of piled raft analysis were listed, and the model chosen for a particular application would depend on the degree of sophistication required in the analysis. Spring models that treat the soil and piles as springs, with no interaction between the springs are inaccurate and it is recommended that they not be used


because the effects of interaction are large. Most of the simple techniques involve treating the soil as being a uniform elastic material so that there is interaction between the piles and the raft, and this leads to much improved solutions. In some cases, loads on the piles are limited by using a load ‘cut-off’ to limit pile loads to a maximum value, thus simulating yield of the piles.

Early analytic solutions such as those by Hain and Lee (1978) made use of the theory of elasticity to compute the settlement of a pile under load and its effect on other piles and on the ground surface. To compute the behaviour of a piled raft, they treated the contact stress beneath the raft as a series of blocks of uniform load, a technique originally used by Zhemochkin and Sinitsyn (1962). It is therefore also necessary to be able to compute the interaction of surface loads and the piles, and surface loads with each other. Hain and Lee did this using the theory of elasticity as well, and combined all of these interactions to compute the piled raft behaviour. For the analysis of the raft, they used the finite element technique. The loads were considered to be uniform over each of the rectangular elements in the raft, so an equal and opposite block of rectangular pressure was applied to the soil. The interactions required are shown schematically in Figure 6.70.

The solution involves calculating the deflections at each pile and block of load due to loadings on the piles or due to other surface blocks of load. This, along with the equations of equilibrium (i.e. the total upward contact force is equal to the total downward force of the applied loads and moment equilibrium) provides enough equations to solve for the mag-nitudes of all the contact stress blocks. Once these are known, the deflections and moments in the raft can be obtained. This approach can be extended to include both horizontal and vertical loading (Zhang and Small 2000).

Similar approaches have been used by Clancy and Randolph (1993), Ta and Small (1996), and Poulos (1994), with different means of calculating the interactions between the piles and the surface loadings from the raft being used to allow non-linearity or layered materials to be taken into account.

It is desirable to know the effects of assumptions made in the different types of analyses, therefore in the following sections, a limited examination is made of some aspects of the analyses listed.

6.37.2 Finite layer techniques

Finite layer techniques have been discussed in Chapter 2, and can be used to calculate the deflection of the surface of a horizontally layered soil due to a rectangular uniform load. This can be used to find the effect of the surface load on deflections of other surface patches of load or on locations down the pile shaft.

If the shear loads acting on the pile shaft can be treated as a series of ring loads, then the finite layer method can be used to compute the deflections at the locations of these ring loads and all other ring loads as well as at the location of the rectangular surface loads (generally the centre of the load is chosen).

This can also be carried out for horizontal ring loads down the pile shaft and for lateral loads over rectangular regions on the surface. This then enables both vertical and lateral loading to be applied to the piled raft as shown in Figure 6.71.

The method is similar to that originally used by Hain and Lee (1978) except that the finite layer method means the analysis of the piled raft can be performed for layered soils where each layer has different properties. Soil layers of finite depth can also be analysed.

The various ring and rectangular loads are shown in Figure 6.72 for general three-dimen-sional loading. Loads are also applied back to the raft and the piles (in the same locations as the loads on the soil) and the deflections of the structure are calculated.


By matching the deflection of the structure and the soil, plus using the equations of equi-librium for the piled raft, there are enough equations to solve for the magnitudes of all the unknown forces applied. These forces can be re-applied to the structure to calculate deflec-tions, moments, and shears in the raft and moment and deflections in the piles.

The raft can be analysed by the use of finite element techniques. Often it is treated as being a thin plate, so that thin plate theory (Timoshenko and Woinowski-Krieger 1959) can be used in the analysis. For rafts that are 2–3 m thick, this may lead to inaccuracy, although it is the width to thickness ratio of the raft that determines its flexibility.

Slip of the piles can be modelled by limiting the ring loads to a maximum value. In addi-tion, failure of the soil under the raft can be modelled by limiting the pressure to a maxi-mum value. Liftoff of the raft may be modelled by not allowing the contact stress to become negative. If it does become negative, it is set to zero.

L

d

S

aL

d

L

d

aS

S

P

q

P

S

(a)

(b)

(c)

(d) q

Figure 6.70 Interactions among piles and surface loads: (a) pile-to-pile interaction effect; (b) surface-to-pile interaction effect; (c) pile-to-surface interaction effect; and (d) surface-to-surface interaction effect.


6.37.3 Non-linear behaviour

If piles are designed to reach their maximum load, or are designed to carry a high propor-tion of their maximum load, then slip of the piles becomes important and non-linear behav-iour of the piles should be taken into account. This type of behaviour becomes important if piles are used to control differential deflections and are designed to yield or fail.

Full three-dimensional finite element analyses have been used to allow non-linear behav-iour of piled raft foundations (Katzenbach et al. 1997), but simpler techniques have been developed. Clancy and Randolph (1993) have presented a non-linear analysis for piles and this is incorporated into the computer program HyPR (Hybrid Piled Raft Analysis), and Bilotta et al. (1991) have presented two methods for computing the behaviour of piled rafts

MyMx Py

Pz

Px

Mz

Figure 6.71 Three-dimensional set of loads applied to a piled raft.

tr

Pz

Px

MxExternal forces

Pr

Pile

Soil

Interface forces transferred from piled raft to the soil

Interface forcesbetween thepiles and soil

x

z

Ring loadsacting on soilnodes

Circular loadsacting on pilebase

Interface forcesbetween the raftand soil

Figure 6.72 Loads applied to the soil and to the piles and raft.


where the pile may have a non-linear load–displacement relationship. The latter authors stress the point that if piles are designed to yield, then a non-linear analysis of the piles is essential.

Poulos (1994) has demonstrated the need to take pile non-linearity into account in analy-sis of centrifuge tests on piled rafts (Thaher and Jessberger 1991). Figure 6.73 shows the settlement and pile load predictions made for a piled raft with eight piles. If the pile is not allowed to fail, then the results of the analysis do not match the observed behaviour. However, if the skin friction on the piles is limited to 80 kPa so that the piles can yield, then the predicted values are much closer to the measured values for both settlement and percent-age load carried by the piles.

6.38 EXAMPLE OF THE FINITE LAYER METHOD

In order to test the accuracy of the Finite Layer Method, solutions were obtained from a three-dimensional finite element program, and from a finite layer program (APRAF – Zhang and Small 2000) for a piled raft with a horizontally applied loading.

The raft is shown in Figure 6.74 and consists of a 3 × 3 pile group with a raft in contact with the ground surface. The raft overhangs the piles by one pile diameter (around the perimeter). The finite element mesh used to model this raft is shown in Figure 6.75 where it may be seen that one quarter of the raft is modelled because of symmetry. The mesh extends further in the x-direction because loading is to be applied to the raft in that direction, and the boundary should not affect the results by being too close.

All of the properties of the piled raft are given in Table 6.15, and dimensions are shown on Figure 6.74.

Two horizontal point loads were applied to the heads of each pile (18 loads in all) making a total horizontal loading of 18 MN. For the purposes of comparison, no slip was allowed between the raft and the soil, or the piles and the soil. The deflection of the raft can be calcu-lated from the finite layer method, and a section (A–B in Figure 6.74) through the deformed raft is shown in Figure 6.76. In the figure it can be seen that the raft rotates under the hori-zontal loading and at its centre (x = 0) does not undergo vertical movement. The computed results from the finite layer and finite element methods can be seen to be in reasonably close agreement.

6

4

2

00 5 10 15 20

Settl

emen

t (m

m)

% Pi

le lo

ad

Pile diameter (mm) Pile diameter (mm)

100

50

0 0 5 10 15 20

Computed (piles elastic)Computed (piles skin friction = 80 kPa)Measured

Lp = 225 mm8-pile group

Figure 6.73 Centrifuge results compared with computed results.


The moments in the piles may also be computed, and are shown in Figure 6.77. In this figure, moments down the pile shaft are shown for pile 1 (the corner pile) and pile 5 (the centre pile), and it may be seen that there is very close agreement between the finite element and finite layer values.

6.39 APPLICATIONS

The application of various computer programs that have been developed for the analysis of piled raft foundations are examined in the following when applied to case histories.

1 4

52

3y

x B

Br

LrD

Overhang

Pxi

D

S

Soil

Pile

t

L

x

z

A

Figure 6.74 Piled raft analysed using either finite element or finite layer methods.

y

x

Figure 6.75 Finite element mesh used for piled raft analysis.


Full three-dimensional finite element programs can be used to model piled rafts, but there is considerably more effort involved than using programs based on continuum theory and soil–structure interaction as described in Section 6.38.

6.39.1 Westend Strasse 1 tower

A second example of the application of some of the analytic techniques mentioned in Section 6.37 is that of the Westend Strasse 1 tower in Frankfurt, Germany (see Figure 6.78). The building is 51 stories high (208 m) and has been described by Franke et al. (1994) and Franke (1991). The foundation for the building was a piled raft with 40 piles that were 30 m long as shown in Figure 6.79.

Table 6.15 Properties of a piled raft (3 × 3 group)

Quantity Value

Pile diameter d 0.5 m

Pile length L 10 m

Depth of soil 15 m

Pile spacing y; Raft width Lr s/d = 3; 9 m

Pile spacing x; Raft breadth Br s/d = 3; 9 m

Overhang of raft 0.5 m

Raft thickness 0.25 m

Soil modulus 10 MPa

Soil Poisson’s ratio 0.3

Raft modulus 30,000 MPa

Raft Poisson’s ratio 0.3

–60

–40

–20

0

20

40

60

–0.50 –0.40 –0.30 –0.20 –0.10 0.00 0.10 0.20 0.30 0.40 0.50

Defl

ectio

n (m

m)

Normalised distance x/Br

APRAFFiniteelement

Piled raft under pointhorizontal loadings s/d = 3

Figure 6.76 Deflection of the raft along centreline A–B.


The foundation was constructed in a deep deposit of the Frankfurt clay 120 m thick, and using pressuremeter tests reported by Franke et al. (1994), the modulus of the clay was assessed to be 62.4 MPa.

The ultimate load capacity of each pile was computed to be 16 MN and a total load of 968 MN was assumed to be applied to the foundation (this is greater than the ultimate capacity of the individual piles).

Six methods were used to predict the performance of the piled raft foundation:

1. The boundary element approach of Poulos and Davis (1980) 2. Randolph’s method (1983) 3. The strip on springs approach using the program GASP Poulos (1991) 4. The raft on springs approach using the program GARP Poulos (1994) 5. The finite element and finite layer method of Ta and Small (1996) 6. The finite element and boundary element method of Sinha (1997)

Measured values were available for the settlement of the foundation, the percentage of load carried by the piles, the maximum load carried by a pile in the group and the minimum load carried by a pile in the group. The results of the six different analysis methods are shown in the bar chart of Figure 6.80 compared with the measured values and the values reported by Franke et al. (1994).

0.0

0.2

0.4

0.6

0.8

1.0

–3 –2 –1 0 1 2

Nor

mal

ised

dept

h

Bending moment in pile (MN m)

Pile 1, finite elementPile 1, APRAFPile 5, finite elementPile 5, APRAF

Piled raft under pointhorizontal loadings s/d = 3

Figure 6.77 Moment variation with depth for piles beneath laterally loaded piled raft.


Maintower

Side building raft

Main towerraft40 piles

(b)

(a)

208 m

15 m

30 m

Sidebuilding

Figure 6.79 Layout of Westend Strasse 1 tower in Frankfurt, Germany. (a) Elevation; (b) plan. (Franke, E., Lutz, B., and El-Mossallamy, Y. 1994. Vertical and Horizontal Deformation of Foundations and Embankments, ASCE, Geotechnical Special Publication No. 40, Vol. 2, pp. 1325–1336.)

Figure 6.78 Westend Strasse 1 tower building in Frankfurt.


From Figure 6.80, it may be seen that

1. Most of the methods over-predicted the settlement of the foundation. However, this depends on the soil modulus chosen, and it can only be concluded that most of the methods gave a reasonable estimate of the settlement for the adopted soil stiffness.

2. Most of the methods over predicted the percentage of load carried by the piles, although the calculated values are acceptable from a design point of view.

3. All of the methods that are able to give a prediction of pile load suggest that the most heavily loaded pile is almost at its ultimate capacity, and this is in agreement with the measured value.

4. For the minimum pile load, there is a considerable variation in the calculated results, with three of the methods indicating a much larger value than was measured.

These results show that when some of the piles are carrying loads close to their capacity, there can be significant variability in the computed results, especially for simple methods and methods based on the theory of elasticity.

6.40 STRUCTURAL STIFFNESS

When analysing rafts, or piled rafts, inclusion of the stiffness of the superstructure will reduce the differential deflections in the raft, and this aspect may need to be addressed. The relative stiffness of the superstructure will determine the effect on deflections, but for very flexible structures, the raft alone can be analysed without great error. Neglecting the stiff-ness of the superstructure can be conservative, but it may be of importance where there are

Central settlement

Maximum pile load Minimum pile load

Proportion of pile load

0

50

100

150

200Se

ttlem

ent:

mm

Ta an

d Sm

all

GA

RP

GA

SP

Fran

ke et

al.

Poul

os an

d D

avis

Rand

olph

Sinh

a

Mea

sure

d

Ta an

d Sm

all

GA

RP

GA

SP

Fran

ke et

al.

Sinh

a

Mea

sure

d

Ta an

d Sm

all

GA

RP

GA

SP

Fran

ke et

al.

Sinh

a

Mea

sure

d

Ta an

d Sm

all

GA

RP

GA

SP

Fran

ke et

al.

Rand

olph

Sinh

a

Mea

sure

d

Method0

20

40

60

80

% Pi

le lo

ad

Method

Method Method0

5

10

15

20

Pile

load

: MN

0

5

10

15

20

Pile

load

: MN

Figure 6.80 Comparison of results from different analysis methods. Westend Strasse 1 tower in Frankfurt, Germany.


(a) (b)

Figure 6.82 (a) Raft Model 2. (b) Raft Model 3.

Figure 6.81 The Burj Khalifa, Dubai (United Arab Emirates).


stiff shear walls. These walls can be treated as stiff raft elements in some cases, thus model-ling the restraining effect of the stiff structural elements.

An example of this was an analysis carried out for the Burj Khalifa in Dubai (Russo et al. 2012). In order to investigate the effect on the computed settlement and differential settle-ment, and to try and obtain as accurate an estimate of the pattern of settlement, the stiffen-ing effect exerted by the superstructure on the raft was taken into account by increasing the bending stiffness of the raft in each wing (estimated by the structural designers to be equiva-lent to an increase of 25,200 kNm2 per wing). Three alternative methods of incorporating this increased bending stiffness were adopted:

1. Increasing the thickness of the whole raft (Model 1) 2. Increasing the raft thickness over the central part of the wings and on the core, as

shown in Figure 6.82a; this is denoted as Model 2 3. Increasing the raft thickness only below the shear walls (see Figure 6.82b), denoted as

Model 3

In addition, the actual pattern of loading via the columns and walls was applied. The pro-gram NAPRA was used to carry out the analysis (the details are given in Russo et al. 2012) and the deflections along one of the ‘wings’ (see the three wings in Figure 6.81) calculated.

As can be seen in Figure 6.83, the deflections computed were similar in magnitude to the values measured up until 2008. Further settlement is expected to occur in the future as was predicted in the original design profile (Figure 6.83).

0

10

20

30

40

50

60

70

80

0 20 40 60 80

Settl

emen

t (m

m)

Distance along wing (m) NAPRA Model 1

NAPRA Model 2

NAPRA Model 3

NAPRA – Usingaverage pile loadsNAPRA – Model 3modifiedOriginal designprofileMeasured (18 February2008)

Figure 6.83 Measured and computed settlements along Wing ‘C’ of the Burj Khalifa.

281

Chapter 7

Slope stability

7.1 INTRODUCTION

The stability of slopes is an important part of geotechnical engineering as some of the great-est damage to property and loss of life has occurred through landslips. Slips can be driven by gravity alone but the effects of earthquakes can also be a factor in causing a slip to occur.

Most importantly, it is the water that exists in a slope that can cause instability, and it is often after rainfall events that slope instability occurs. An example of a slope failure caused by the introduction of water is the slip that occurred in 1997 in the ski resort town of Thredbo in the Australian Alps (see Figure 7.1). The slip took place at 11.30 P.M. on the 30th July when skiers were asleep in their ski lodges. It demolished Carinya Lodge killing one person, before the earth flow pushed the wreckage of Carinya Lodge into Bimbadeen Lodge that was further downslope, killing another 17 people. There was only one survivor from those who were in the two lodges that night.

The ski lodges in that part of Thredbo village where the slip occurred were located on a fairly steep slope beneath a road embankment for the Alpine Way that had been constructed some 42 years earlier (in 1955) by pushing soil into place with a bulldozer. The road was only constructed as an access road for the Snowy Mountains hydro-electric scheme, and contained logs from trees that had been pushed over by the bulldozer. The fill was not prop-erly compacted, as it would be for a modern highway.

Carinya Lodge was constructed on a slope of about 30° that became much steeper (esti-mated to be about 45° similar to other slopes in the area) where the Alpine Way embank-ment sloped up towards the roadway. A cross section through the site is shown in Figure 7.2.

What caused the road to slip after so many years after its construction was the subject of an intense investigation, and the findings of the investigation were that the slope above the lodges had been moving slowly (creeping) for some time. The movement eventually became large enough to pull apart the coupling in a pipe that was buried in the downhill slope of the roadway. Once this occurred, water gushed into the slope making it totally unstable, and a rapid slope failure took place. The pipe coupling is shown in Figure 7.3.

This case illustrates two features of slope stability: (1) water has a profound effect on the stability of a slope and (2) slope movements can be very slow (creep) and can take place over several years before the stability of the slope or the cumulative movement of the slope becomes a problem.

The strength of the soil in the slope resists movement of the slope through shearing, and this is also of importance. Slopes that have been in place for a period of time behave in a drained fashion, and the effective strength of the soil and the pore water pressures are used in analysis. This is the most common type of analysis performed; however, there are slopes that have been cut rapidly for which an undrained analysis is more appropriate.


Various forms of analysis that may be applied to slope stability are discussed in the follow-ing sections.

7.2 SLIP CIRCLE ANALYSIS

It is often observed that when a slope fails it does so by undergoing a rotational failure with the soil shearing along a roughly circular failure surface. If there are planes of weakness in the soil lying on the slope, then the failure surface is likely to pass through one or more of the weak planes and the surface will not be circular but have some other shape. However for uniform soils, the circular failure surface may be considered, and this has been used in many simple methods of analysis including the method of slices. This is assuming that the slip is two-dimensional in nature, when in reality it will be a three-dimensional bowl shaped failure surface. Three-dimensional failure surfaces are addressed in Section 7.2.8.

1390 m1395 m1400 m1405 m1410 m1415 m1420 m1425 m1430 m1435 m1440 m

Alpine Way

Section ECarinya

19961946

0 5 10Metres

Figure 7.2 The Thredbo landslip. The slip occurred below Alpine Way and above the Carinya Lodge.

Figure 7.1 Destroyed ski lodges after the 1997 Thredbo landslip.

Slope stability 283

7.2.1 The method of slices

This method involves dividing the region above the supposed circular slip surface into a series of slices that can be of different thicknesses as shown in Figure 7.4. In some methods, the slices do not need to have vertical sides, but this is not addressed here.

The forces acting on each slice are then considered. As is known from testing of soils in shear (say in a direct shear box), the shear strength τ on a shear surface at failure is given by Equation 7.1 where c is the cohesive strength and σn is the normal stress acting on that surface.

τ σ φ= +c ntan( ) (7.1)

If the problem is one involving drained conditions, the angle of shearing resistance is ϕ′, and the cohesion is c′ while the effective stress σ′n is used for the normal stress. If the prob-lem involves undrained conditions then ϕ = ϕu is the undrained angle of shearing resistance, the cohesion c = cu and the total stress σn are used.

Figure 7.3 Fibre cement pipe at the top of the slope, thought to have leaked prior to the slide.

Circular slip surface

Slices do not need to beof equal thickness

Piezometric surfaceMay includetension crack

Surfaceloading

Figure 7.4 The method of slices used for a circular failure surface.


If the soil is not at the point of failure then the full soil strength is not mobilised, and the shear τf operating on a shear plane will be

τ σ φ

f ncF F

= + tan( )

(7.2)

This is assuming that the factor of safety F applies equally to the cohesive strength c com-ponent and the frictional strength tan(ϕ). This may not be the case, but it is a simplifying assumption. In addition, the same factor of safety may not apply all along a potential slip surface, and this too is an assumption.

Hence, the angle of mobilised friction θ is given by tan θ = tan(ϕ)/F. The resultant forces due to friction therefore act at an angle of θ to a face of the slice, as the full angle of friction is not operating if the slope has not slipped. This is shown in Figure 7.5, which shows forces acting on a single slice in the drained case. In the figure, Wn is the weight of the nth slice, U is the water force, N′n is the normal effective force on the base of the slice, and Tn is the shear force acting on the base of the slice. The forces acting on the sides of the slice are the normal En and the shear force Xn such that the resultant force is Zn acting at the angle θn to the normal. The locations at which the force at the base of the slice ℓn and at the sides of the slice yn act are not known, but various assumptions can be made to overcome this problem as will be seen later.

Various methods have been proposed by several researchers for finding the factor of safety of the overall assembly of slices above the slip circle, and some of these are listed below:

• The Swedish method for circular failure surfaces (Fellenius 1927)• Bishop’s simplified method; circular failure surfaces (Bishop 1955)• Spencer’s method; circular surfaces (Spencer 1967)

bnTn

En θn

XnZnyn

Tn

Nn′

Uℓn

αn

En+1

Xn+1 Zn+1

θn+1

Wn

WnN ′n

U Q

Zn+1Zn

θ1

Forcepolygon

Resultant

Nn = Nn′ + UTn = (Nn′ tan ϕ′+ c′bn sec α)/Ftan θ1 = tan ϕ′/F

Nn′ tan ϕ′

FF

c′bn sec α

Figure 7.5 Forces acting on a single slice.

Slope stability 285

7.2.2 The Swedish, Fellenius, USBR, or Common Method

This method (which is referred to by several names – most commonly the Swedish method) is the simplest method in which the assumption is made that the side forces on the slices cancel each other (i.e. are equal and opposite).

Then if moments are taken about the centre of the circle, and the overturning and resist-ing moments are assumed to be equal (for a static situation) then the factor of safety can be found (in terms of drained strength parameters) as shown in Equation 7.3.

Fc W U

W

i

n

i i i i i i

i

n

i i

=′ + ′−

=

=

∑∑

1

1

[ ( cos )tan ]

sin

∆ θ φ

θ

(7.3)

and Δℓi = bi/cos θi is the length of the base of the slice.

7.2.3 Bishop’s method and simplified method

Bishop (1955) developed a method that did not involve as many assumptions as the simple Swedish method. The simplified Bishop method makes the assumption that the vertical forces on the sides of the slices cancel, and the normal force at the base of the slice N ′ni can be found by resolving forces vertically. The resulting expression for the factor of safety (in terms of effective strength) is given in Equation 7.4.

Fc b W u b M

W

i

n

i i i i i i i

i

n

i i

=′ + ′−

=

=

∑∑

1

1

[ ( )tan ]

sin

( )φ θ

θ

/

(7.4)

M = +

F

i ii i( ) cos

tan tanθ θ θ φ1

It can be seen from the equation that the factor of safety is involved in both sides of the equa-tion as it is included in the Mi(θ) term. Therefore, an iterative approach needs to be taken, whereby an initial value of the factor of safety is chosen and used to calculate Mi(θ) and then a new value of F is found from Equation 7.4. The new value is then used to get a better estimate of Mi(θ), and the process is repeated until convergence is obtained.

In practice, the Swedish method is generally used to obtain the first estimate of F and then this is used as the starting value for the iterative process. If this is done, an accurate value of the factor of safety can be obtained in about three iterations.

Bishop’s full method requires values of the side forces Xn to be found so as to obtain force equilibrium of the slice; however, the error involved in omitting this in the simplified method is small.

7.2.4 Spencer’s method

Spencer (1967) made the assumption that all of the side forces were inclined at the same angle θi to the sides of the slice. He then found a factor of safety Ff that would give force equilibrium for various values of the angle θi and a factor of safety Fm that would give


overall moment equilibrium for different values of θi. The value of θi for which Ff = Fm is the required angle and the factor of safety for the slope is the corresponding value of F = Ff = Fm as shown in Figure 7.6.

7.2.5 Finding the critical circle

To find the slip circle with the lowest factor of safety, many different circles must be trialled, and the one with the lowest factor of safety found. This can be done fairly quickly with com-puter programs, and different search algorithms have been developed to aid in the search.

One method is to create a grid above the slope on which the centres of the circles lie. At each grid point, several circles of different radii are trialed before moving on to the next grid point (Figure 7.7).

Another method is to select points on the lines defining the top of the slope and allow circles to pass through these points as shown in Figure 7.8. There are other search options that have been developed to guide the search for the critical circle such as the algorithms by Nguyen (1985) and Arai and Tagyo (1985), among others.

7.2.6 Water pressures

The factor of safety may be seen to depend on the water pressure in drained analyses, as the water force can be seen in Equations 7.3 and 7.4 for example. The correct estimation of the water pressure is very critical, as the factor of safety is fairly sensitive to the vales used.

There are several ways that the water pressures can be specified and some are listed below:

1. Using a phreatic surface A free water surface may be specified to the computer program as a series of straight

lines, and then the distance from the free surface to the base of the slice (generally the

Fm

Ff

Fmo = 1.039

θi = 22.5°

Fi = 1.070

F

θ (degrees) 50 10 15 20 25

0.90

0.95

1.00

1.05

1.10

Figure 7.6 Variation of factors of safety for moment equilibrium and force equilibrium with θ.

Slope stability 287

central point) calculated. The water pressure can be computed as the distance to the base of the slice times the unit weight of water.

Some computer programs allow several different phreatic surfaces to be input, and water pressures in different materials to be computed from the free surface associated with that material. This allows for perched water table calculations.

2. Pore pressure ratio For problems involving the stability of embankments such as earthfill dam embank-

ments, it is convenient to use the pore pressure ratio ru which gives the ratio of the pore water pressure to the total vertical stress at a point in the soil as defined in Equation 7.5.

ruh

u =γ

(7.5)

The total unit weight of the soil is γ and the depth of the point beneath the surface of the soil is h.

Circles tangent to firmstrata should be tested

Different radii usedat each grid point

Centres of circleslie on grid points

Contours ofminimum factorof safety at eachcentre may beplotted

Figure 7.7 Grid of centres used for locating the critical slip circle.

Selected pointson slope

Circle centres lie alongthis perpendicular

Figure 7.8 Search method using points on the surface of a slope.


This allows for the fact that in earth fills, the water pressure generated is propor-tional to the depth of the soil above a point in the fill. The fill may not be saturated, therefore the value of ru will reflect this by being less than 1. Values of ru may be mea-sured in triaxial tests using the appropriate total stress increases applied to the sample that has been compacted at the appropriate moisture content.

3. Pore pressure grid For problems involving seepage (i.e. through embankments), the water pressures can

be found from a seepage analysis and then transferred to a grid of points. The grid of points and the water pressure at each point can then be used with a slip circle analysis by using interpolation between the grid point values. This is commonly done with finite element analysis of seepage.

Pore pressures from other analyses such as consolidation analysis can be transferred in the same way.

7.2.7 Surface loads

If there are loads applied to the surface of a slope as may happen if there are traffic loads, then the load can be added to the weight of the slice that it acts upon. The increased weight is then used in the formula for the factor of safety such as Equations 7.3 or 7.4.

In total stress (undrained) analysis of slopes that are submerged, the water can be treated as a force acting on the face of the slope and the water not considered as a separate material. Water can also be considered (in an undrained analysis) by introducing it as a material that has self-weight but no strength.

7.2.8 Computer programs

The summations shown in equations Equations 7.3 and 7.4, for example, can be carried out by hand, but today there are many commercially available computer codes that enable the calculations to be done rapidly with simple input of slope geometry and material properties. Different search algorithms are available for locating the critical slip circle (i.e. the one with the lowest factor of safety).

Generally, commercial programs allow different methods to be used such as the Bishop method, Spencer’s method, or the full method. They also allow non-circular failure surfaces to be used. Non-circular failure surfaces are discussed in Section 7.3.

Commonly used codes are SLOPE/W (2012), SVSLOPE (2014), and XSLOPE (2014). Several other commercial codes are available on the Internet.

7.2.9 Three-dimensional failure surfaces

As was mentioned previously, failure surfaces are in reality bowl shaped and not two-dimensional as assumed in the slip circle analyses mentioned in previous sections. The slip circle method can be extended to the three-dimensional case where the slices become three-dimensional columns. Early work on such analysis was reported by Chen and Chameau (1983) and Leshchinsky et al. (1985). These papers conclude that the factor of safety that is calculated using three-dimensional methods is generally higher than that from 2-D methods.

As three-dimensional analysis requires a great deal of computation, it is necessary to use a computer program to perform the analysis. Codes such as SVSLOPE/3D (2014) are able to perform the analysis.

Slope stability 289

7.3 NON-CIRCULAR FAILURE SURFACES

If there is some reason for the shape of the slip surface to be non-circular, for example, there is a weak plane along which failure can occur, then the non-circular analyses can be performed.

As for the circular surfaces, the region above the slip surface is divided up into strips. Trial surfaces are used to locate the one with the lowest factor of safety, and generally, this is done with input of the surface shape using graphical techniques (using a mouse) and a computer program.

There are several methods that are used for non-circular surfaces, and these are examined in the following sections.

7.3.1 Morgenstern–Price method

This method was described by Morgenstern and Price (1965). It satisfies all static equilibrium requirements, but the solution obtained must be checked for physical admissibility. The prob-lem is made determinate by assuming a relationship between the interslice shear force Xn and the interface normal force En. The form of the relationship is given in Equation 7.6.

X f x Ei i= λ ( ) (7.6)

The function f(x) determines the angle of the interslice forces, while λ determines their mag-nitudes. Various functions can be tried and there may be many functions that give admis-sible solutions.

The line of thrust (which is the line showing where the interslice forces act) and the nor-mal stress on the base of the slice are obtained as part of the solution. The line of thrust should lie within the slice and the stress at the base of the slice should be compressive. As well, the shear strength between the slices should not be exceeded.

7.3.2 Janbu’s method

In Janbu’s method, the line of thrust is assumed and then the equations of equilibrium are solved. Janbu (1973) states that the factor of safety so calculated is relatively insensitive to the location of the line of thrust as long as it is reasonable. The line of thrust (see Figure 7.9)

Line of thrust

Non-circular shear surface

External point loads

Distributed load

En+1

Eb

Ea

Xa

Xb

En

Xn+1

Xn

Slice

Water table Crack

P

Q

UTn

N′n

b

a

Figure 7.9 Non-circular slip surface showing the line of thrust.


should be at about one-third of the height of the slice from the base for cohesionless soils, and it should be below this height in the active zone and above it in the passive zone for cohesive soils.

Details of the method are given in Hirschfeld and Poulos (1972).

7.3.3 Sarma’s method

Sarma (1973) originally proposed a method of analysis whereby the stability of the slope was measured by the horizontal acceleration required to bring the soil mass into limiting equilibrium. If a conventional factor of safety was required, the solution could be found by successive reduction of the shear strength parameters until zero acceleration was required for equilibrium. The conventional factor of safety was the reduction of strength required.

The original method (1973) was based on vertical slices, and as the forces at the base of the slice were assumed to act at the centre of the slice, the result depended on the number of slices used. Sarma later presented a method (Sarma 1979) in which the factor of safety could be calculated for slices that do not have vertical sides, and the slices can be as large as possible as dictated by the geometry of the slip surface. Again, the method involves finding the horizontal acceleration on the slope to produce equilibrium.

7.4 WEDGE ANALYSIS

Hand analysis of slope failures may be undertaken if the failure mode is a wedge type of failure. This can occur where there is a weak plane due to a soft soil layer or a fracture or fissure. This approach may also be used when analysing the upstream slope of a sloping core rockfill dam (Sultan and Seed 1967, Seed and Sultan 1967).

A simple form of analysis is to assume that there are two wedges such as those shown in Figure 7.10. The upper wedge is the active wedge that is pushing down on the lower passive wedge that is sliding sideways.

The following assumptions are made (see Figure 7.10):

1. The critical failure surface bc passes through the slope. 2. On planes a−b, b−c the friction and cohesion are only partly mobilised, for example,

on b−c

tan

tanθ φ1

1= ′F

(7.7)

S

cF

11 1= ′

3. The shear strength which is mobilised between the two wedges may not be mobilised to the same degree as the shear strength on the base of the wedges. The angle of fric-tion δ could therefore lie between the limits of 0° and

δ φ=

′−tan 1 tanF

(7.8)

while the inter-wedge cohesion could lie between values of zero and c′/F.

Slope stability 291

One approach, therefore, is to assume that the degree of mobilisation is midway between the two extremes and to assume that both cohesive and frictional forces have this median value throughout the analysis.

If δ is assumed to be zero, the calculated factor of safety will be ≈25% on the conserva-tive side.

The procedure which may be used to find the factor of safety is as follows:

1. Assume two values of the factor of safety which are likely to span the true value, say F1, F2.

2. Determine for each block the mobilised angles of friction θ1, θ2 and the mobilised shear forces S1, S2 (see Equations 7.7) using the first factor of safety F1.

3. Draw the force polygon for each block, as shown in Figure 7.10. The magnitudes of forces P1, P2, P12, and P21 are unknown but their directions are known so that the polygons may be completed.

4. Values of P12, P21 may thus be found from the force diagram. 5. The correct solution is the one for which P12 = P21. If the two values are not equal,

another solution for P12 and P21 is then found using F2. Hence, the values of P may be plotted against the factor of safety (say F). Results are as shown in Figure 7.11 and linear interpolation may then be used to find the required value of F.

6. The positions of the lines a−b, b−c may be altered and corresponding factors of safety found. The lowest value is chosen.

This kind of analysis can be rapidly performed using non-circular slice methods, and therefore is not often performed by hand these days.

Wedge 2

Wedge 1

Trial slip surface

ad

bc

P1

U1

W1

S1

S2P2

U2

W2

W2

U2

tan θ1 = tan ϕ′1/FS1 = c ′1ℓ1/F

tan θ2 = tan ϕ′2/FS2 = c ′2ℓ2/F

ϕ′1, c ′1 applyon ab (= ℓ1)

ϕ′3, c ′3 applyon bd (= ℓ3)

ϕ′2, c ′2 applyon bc (= ℓ2)

θ1

θ2

θ1

P1W1

P2

S2

S12

U12

P12P21

U21S21

S1

U1

θ2

δδ

2

1

δ U12

P21

P12

S12

S21

U21

Force polygons

Figure 7.10 Two sliding wedges and force polygons for each wedge.


7.5 PLASTICITY THEORY

Solutions to the stability of slopes have been found using rigorous plasticity theory (Booker and Davis 1972). The lines along which slip is occurring may be seen in Figure 7.12 compared with a slip circle solution for the case where the soil is undrained (ϕu = 0) and the undrained strength varies with depth. In the plasticity solution, it may be seen that there is a wedge of material below OA with sides at 45° to the horizontal, and a curved section below the slope OB.

Booker and Davis showed that the slip circle factor of safety was less than 10% in error over a practical range of slope angles (angle from the horizontal >5°) for this particular problem.

7.6 UPPER- AND LOWER-BOUND SOLUTIONS

Solutions can be obtained to the slope stability problem by use of the upper- and lower-bound theorems (Sloan 2013). The stress field is interpolated from values at the nodes of triangu-lar elements using linear interpolation, and the failure criterion is represented as a series of

A

Critical circlefrom slip circleanalysis

O

B

Figure 7.12 Velocity field for a slope failure in undrained soil with strength increasing with depth.

Required value

P21

P12

P12

P21

F2F1 F

P

Figure 7.11 Factor of safety plot.

Slope stability 293

linear functions. Optimisation (linear programming) is then used to obtain the stress field that gives the lowest unit weight to cause failure, giving an upper bound.

The velocity field is then represented at the nodes of a triangular mesh (this can be the same mesh as for the stress field) and the velocities are optimised to give the lowest power dissipation. This gives the upper bound to the problem. The true solution then lies between the two bounds.

This type of solution does not depend on trial and error as do the limit equilibrium meth-ods, so the location of the failure surface is found directly. If there are weak layers, then the failure surface will be found to run along the weak layer as part of the analysis. Such a solution is shown in Figure 7.13, which was found by using the computer code OPTUMG2 (2014).

Solutions may also be found (using the upper- and lower-bound approach) where water pressures such as those caused by seepage in the slope are involved. Joints may also be included in the analysis.

7.7 FINITE ELEMENT AND FINITE DIFFERENCE SOLUTIONS

The stability of slopes can be found by using conventional finite element or finite difference (FLAC) solutions and a ‘strength reduction’ technique. This involves reducing the strength of an elasto-plastic material until collapse occurs, and then the reduction that is applied gives the factor of safety for the slope. The method has the advantage that different reduc-tion factors can be applied to the cohesive strength and the frictional strength if desired.

This is not a rigorous method of finding the factor of safety as the result depends on the mesh fineness, but it can give an indication of the failure mechanism as the failure surface is indicated by the region of high plastic shear strain.

An example is given by Griffiths and Lane (1999) for a slope in a two-layered undrained soil as shown in the upper diagram in Figure 7.14. The solutions show that there is a transi-tion in the failure mechanism as the ratio of the shear strengths in the two layers changes as can be seen in the lower plots of Figure 7.14.

Figure 7.13 Upper-bound solution for slope stability from program OPTUMG2 showing the location of the failure surface.


Finite element methods are easily extended to three-dimensions (Griffiths and Marquez 2007) and probabilistic methods can be applied to the soil properties. Different soil strengths can be applied to each element in the mesh according to statistical methods, and the cor-responding factors of safety found for the slope.

An early example of the use of finite difference methods for slope stability is that of Hammet, and this is reproduced in the paper by Cundall (1976).

Commercial programs such as PLAXIS (2014) and FLAC2D (ITASCA Consulting Group 2013) are capable of performing the strength reduction analyses. Stabilising features such as soil nails can be incorporated into such analyses.

7.8 SEISMIC EFFECTS

The simplest way to incorporate seismic effects into a slope stability analysis is to add an extra force vertically or horizontally to a slice to simulate the force due to ground acceleration.

2H

2H

21

2H

ϕu = 0

H

H

cu1

cu2

(a)

(b)

(c)

Figure 7.14 Deformed meshes for a slope with two layers of material with different strengths: (a) su2/su1 = 0.6; (b) su2/su1 = 1.5; (c) su2/su1 = 2.0.

Slope stability 295

The most common way to estimate the force is to make the force a multiple of the weight of the slice as shown in Equation 7.9.

F k W

F k Wh h i

v v i

==

(7.9)

The vertical or horizontal earthquake forces are multiples of the weight of the slice Wi where the horizontal and vertical seismic coefficients are kh and kv, respectively. This is suitable if the soil beneath or in the slope is not subject to liquefaction.

The horizontal force acting on a soil mass is shown in Figure 7.15 for a simple undrained analysis. By taking moments about the centre of the slip circle, the factor of safety against a rotational slide can be found if the undrained shear strength used is su/F.

Values of the seismic coefficient can be found for various regions from seismic maps. For example, the Australian standard AS1170.4 gives acceleration coefficients for different areas based on historical data. For Sydney, the coefficient is kh = 0.08. However, for seismically active areas such as Japan or the west coast of the United States, the accelerations are much higher. Seed (1979) recommended (for large slopes) kh = 0.1 for sites near faults capable of generating magnitude 6.5 earthquakes, and the acceptable pseudo-static factor of safety is 1.15 or greater, while kh = 0.15 is appropriate for sites near faults capable of generating mag-nitude 8.5 earthquakes (for an acceptable pseudo-static factor of safety of 1.15 or greater).

Full dynamic analysis can be performed by finite difference codes such as FLAC (ITASCA Consulting Group 2014), and these can be used for slope stability analyses of dam slopes. The onset of liquefaction in slopes of hydraulically filled dam walls such as tailings dams can also be analysed by these methods.

7.9 FACTORS OF SAFETY

If slope stability analysis shows that the slope does not have the required factor of safety against a slip, then remedial work may be necessary to improve the stability. Factors of safety that are chosen depend upon the consequences of failure and need to be chosen to reflect the risk. They also reflect how reliable the soil strength data is, and whether a mean value or a lower-bound value has been selected for design.

Hence, factors of safety are largely selected by engineers according to the perceived risk. However, some guidance is provided in the Geotechnical Manual for Slopes (Geotechnical Control Office 2011) as shown in Table 7.1. These values were developed for use in Hong Kong.

R

a

cb

kW

WArc lengthac = L

su

ℓ2

ℓ1

Figure 7.15 Static force representing earthquake force.


D’Andrea and Sangray (1982) quote a number of sources that give values of the conven-tional factor of safety lying between 1.25 and 1.5 depending on the particular conditions for the slope.

For tailings dams, the CANMET (1977) manual recommends the factors of safety given in Table 7.2. The factor chosen depends on the condition and should be checked for all of the different conditions that are likely to affect the dam.

7.10 SLOPE STABILISATION TECHNIQUES

Precautions can be taken to mitigate slope instability especially in the long term. It may be necessary also to modify a slope so as to increase the factor of safety that it has against failure. Generally, removing sources of water infiltration into a slope or lowering the water table are methods that can be applied to increase stability, but other methods can be used as presented in the following sections.

7.10.1 Control of surface water

Collection of water from above the slope can be carried out with a surface drain constructed at the top of the slope. Water collected by the drain must be channelled away from the slope and not allowed to infiltrate into the soil. Therefore, the drain should be lined, either with concrete or with other materials (e.g. plastic or steel half-pipe). Water caught in the drain can be drained off to either side of the slope or down to the foot of the slope in pipes.

Joints in the drain could open if the slope moves after installation and this could be a source of water influx into the slope.

Table 7.1 Recommended factors of safety for new slopes for a 10-year return period of rainfalla

Risk to life

Economic risk Negligible Low Highb

Negligible > 1.0 1.2 1.4Low 1.2 1.2 1.4High 1.4 1.4 1.4a The factors of safety in this table are recommended values. Higher or lower factors of

safety might be warranted in particular situations with respect to economic loss.b In addition to a factor of safety of 1.4 for a 10-year return period of rainfall, a slope in

the high risk to life category should have a factor of safety of 1.1 for the predicted worst groundwater conditions.

Table 7.2 Factors of safety for slopes of tailings dams

Slope Condition Required factor of safety

Upstream Rapid drawdown 1.2–1.4Upstream Rapid drawdown with earthquake 1.0–1.2Upstream End of construction 1.1–1.3Downstream End of construction 1.1–1.3Downstream Steady seepage 1.5–1.7Downstream Steady seepage plus earthquake 1.3–1.5

Slope stability 297

7.10.2 Horizontal drains

Horizontal drains are drilled into the slope at a slight angle to the horizontal (between −5° and +5°) so that water intercepted by the drain can flow out. The type of drain depends on the soil or rock type. In fractured rock, it may be adequate to drill open holes, but in soil slopes, drainage pipes are necessary.

Drains are normally 100–120 mm in diameter and can be lined with slotted PVC pipes wrapped in a filter geotextile to prevent ingress of soil. In some cases, the pipe is not slotted or is grouted near the exit to prevent blockage by tree roots and erosion near the exit.

Drains should be checked to determine if they are operative, and if not can be cleared with high pressure water.

7.10.3 Stabilising piles

Stabilising piles have been used to resist slope movements, and can be effective in supplying lateral force to retain the soil. Several methods have been developed for designing the piles, including those of Ito and Matsui (1975), Ito et al. (1982), Viggiani (1981), and Poulos (1995).

The Ito and Matsui results should be used with caution, as the loads tend to approach infinity when the pile spacing tends to zero (see Beer and Carpentier 1977).

7.10.4 Toe fill

In some circumstances, it is possible to remove the soil at the base of a slope, and to replace it with rockfill. The effect of the rockfill is to provide a free draining material with a high angle of shearing resistance that is much more stable than the original soil material on the slope.

This approach was used to stabilise the Hue Hue Road landslip (Fell et al. 1987) that occurred on the main highway leading north of Sydney. Features of the reconstruction were the use of filter fabric between the natural soil and the rockfill, catch drains at the top of the slope and a drainage trench at the base of the rockfill as shown in Figure 7.16.

7.10.5 Retaining structures

All types of retaining structures mentioned in Chapter 9 may be used to retain a slope. Gravity walls, reinforced earth walls, crib-block walls, soil nails, and anchored walls can be used.

1

Pavement Topsoil and vegetation

1 m ripped sandstoneCatch drain

Catch drain

Filter fabric

Rockfill

Drainage trench

Roadway

8 mCL

Gravel layer around filter fabric

Scale 13 m

1 in 21

1

100

LC

11.5

2 4 6 8

Figure 7.16 Stabilisation of a slope with rockfill at the toe.


If the slide is large in extent, walls cannot be used because of the large forces involved that would make the wall too large to be economical. In this case, water control methods are a more effective means of stabilising the slope.

7.11 STABILITY CHARTS

Stability charts for slopes have been developed by many authors, but one that is commonly used is that of Hoek and Bray (1981). Although, it is a simple matter to use slip circle computer programs for calculating Factors of Safety today, use of such charts is still quick and easy.

1

2

3

4

5

Chart numberGroundwater flow conditions

Fully drained slope

Surface water 8 × slopeheight behind toe of slope



Saturated slope subjectedto heavy surface recharge

Figure 7.17 Groundwater conditions: Cases 1–5.

Slope stability 299

2 1

3

4

4

tan ϕ γH tan ϕ

γHF

c

c

F

Figure 7.18 Method of using charts.

0.000.020.040.060.080.100.120.140.160.180.200.220.240.260.280.300.320.34

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

tanϕ

/F

4.0

2.01.5

1.0.90

.80.70

.60

.50.45

.40.35

.30

.25

.20.19.18.17.16.15

.14

.13

.12

.11

.10

.09

.08

.07

.06.05.0

4

∞

.03.02.010

80

90

7060

5040

3020

10

Slope angle (°)

Circular failure chart number 1

c/γH · tan ϕ

c/γHF

Figure 7.19 Stability charts for Cases 1–5. (Continued)


0.000.020.040.060.080.100.120.140.160.180.200.220.240.260.280.300.320.34

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

tanϕ

/F

4.02.0

1.5

1.0.90

.80.70

.60.50

.45.40

.35

.30

.25

.20.19.18.17.16.15

.14.13

.12

.11

.10

.09

.08

.07

.06.05.0

4

∞

.03.02.010

80

90

7060

5040

3020

10

Slope angle (°)


c/γH · tan ϕ

c/γHF

0.000.020.040.060.080.100.120.140.160.180.200.220.240.260.280.300.320.34

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

tanϕ

/F

4.02.0

1.5

1.0.90

.80.70

.60.50

.45.40

.35

.30

.25

.20.19.18.17.16.15

.14.13

.12

.11

.10

.09

.08

.07

.06.05.0

4

∞

.03.02.010

80

90

7060

5040

3020

Slope angle (°)


c/γH · tan ϕ

c/γHF

Figure 7.19 (Continued ) Stability charts for Cases 1–5.

Slope stability 301

0.000.020.040.060.080.100.120.140.160.180.200.220.240.260.280.300.320.34

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

tanϕ

/F

4.02.0

1.5

1.0.90.80

.70.60

.50.45

.40.35

.30

.25

.20.19.18.17.16.15

.14.13

.12

.11

.10

.09

.08

.07.06.05.0

4

∞

.03.02.010

80

90

7060

5040

Slope angle (°)


c/γH · tan ϕ

c/γHF

0.000.020.040.060.080.100.120.140.160.180.200.220.240.260.280.300.320.34

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

tanϕ

/F

4.02.01.5

1.0.90

.80.70

.60.50

.45.40

.35

.30

.25

.20.19.18.17.16.15

.14.13

.12

.11

.10

.09

.08

.07.06.05.04

∞

.03.02.010

c/γH · tan ϕ

80

30

7060

5040

Slope angle (°)

20

10

c/γHF


Figure 7.19 (Continued ) Stability charts for Cases 1–5.


A slope with different water conditions ranging from a dry slope (Case 1) to a fully satu-rated slope (Case 5) is shown in Figure 7.17. A choice is made of the condition closest to that existing in the field and then the appropriate circular failure chart selected.

The method of using the charts is shown in Figure 7.18. Firstly, a value of c/γH tanϕ is calculated, and then this value is found on the outside circular part of the chart. Then by moving down the radius of the circular chart, the slope angle of the slope is reached. The value of either tan ϕ/F or c/γHF can be read from the chart, and either can be used to cal-culate the factor of safety F of the slope. (The unit weight of material in the slope is γ, the height of the slope H and the cohesion and angle of shearing resistance of the slope material c and ϕ, respectively.)

The circular charts for each of the five groundwater cases are shown in Figure 7.19.Other specialist charts that may be useful are those of Morgenstern (1967) for rapid

drawdown (such as occurs if water in a dam drops); Bishop and Morgenstern (1960); and for three-dimensional surfaces, Cheng and Yip (2007).

303

Chapter 8

Excavation

8.1 EXCAVATION

Excavations may be unsupported or may require some kind of supporting structure that is designed to prevent the sides of the excavation from collapse. The base of an excavation may also fail through heave or piping and so must also be designed to resist failure.

If an excavation does not require support, the stability can be checked with slip circle techniques (see Chapter 7). If it does require support, the supporting structure can range from simple block work or brick retaining walls to concrete or steel pile walls held in place by anchors or props.

Support systems must be designed to reduce the load from groundwater, as this can be con-siderable. Walls are generally provided with weep holes (that need to be periodically checked to see that they are working) or drains that can channel any groundwater away from the back of the wall. If groundwater cannot be drained away from the wall (as may be the case for some basements), then the wall must be designed to withstand the water pressures.

8.2 TYPES OF EXCAVATION SUPPORT

When vertical cuts are made in soils, it is often necessary to brace or support the sides of the cut in some way so that collapse of the sides does not occur. Several techniques can be used for bracing the sides of cuts as described in the following sections.

8.2.1 Steel ‘H’ piles and lagging

For shallow cuts less than about 4 m in depth, vertical wooden planks can be used. The planks are braced by struts that run from one side of the cut to the other and bear onto wales that run along the face of the planks. Alternatively, prefabricated steel panels with steel bracing can be lowered into excavations or trenches that are not very wide (Figure 8.1).

For deeper cuts, sheet piles or ‘H’ piles with timber lagging that slots in between the piles may be used (see Figure 8.2). The ‘H’ section piles are driven first and the lagging is added as the excavation proceeds. Anchors may be needed or bracing may be used to increase stability if required. Anchors passing through horizontal cross-beams or wales are shown supporting a wall in Figure 8.3.

8.2.2 Sheet piles

Sheet pile walls are also a common means of support for retaining walls. The steel sections are driven or vibrated into the ground before excavation, each section interlocking with the


Figure 8.1 Prefabricated wall and bracing system.

Figure 8.2 Wall constructed in sand using ‘H’ piles with timber lagging.

Excavation 305

adjacent pile. The soil is then excavated leaving the sheet piles supporting the sides of the excavation. If additional support is required, anchors or bracing can be used as for ‘H’ pile walls.

8.2.3 Bored pile walls

Other construction techniques involve soldier pile, secant pile, or contiguous pile walls, where piles are constructed in a row. Piles are drilled and concreted before excavation and after excavation form the wall of the excavation. The different methods of construction are

1. Secant piles: In this case, the piles actually overlap each other. First, a series of ‘soft’ piles (made from a weak concrete) are drilled with a space between them and then a hole is drilled between two of the soft piles cutting into each of them. This pile is

Figure 8.3 Excavation supported by anchors.


concreted with a stronger mix and is therefore called a ‘hard’ pile. The result is a wall consisting of overlapping piles that is reasonably watertight (see Figure 8.4).

2. Soldier pile or king pile walls: For these walls, the piles are drilled and concreted at some distance apart, and as excavation proceeds the soil exposed between the piles can be left unfaced or can be shotcreted to prevent weathering.

3. Contiguous piles: Contiguous pile walls are constructed with the piles touching or nearly touching each other but not overlapping as for a secant pile wall. There are some small gaps left between piles, as it is difficult to drill precisely vertical shafts.

With all of these types of walls, it is possible to provide further support through anchors or bracing.

8.2.4 Diaphragm walls

Diaphragm or slurry walls are constructed by excavating a trench generally filled with benton-ite slurry. Bentonite is a clay that has a high liquid limit and forms a thick slurry when mixed with water. The slurry is able to support the sides of the trench and prevent collapse before concreting can take place. Two concrete guide walls are constructed at the top of the trench to guide the clamshell bucket that is used to excavate the soil. This helps keep the trench vertical (see Figure 8.5a,b). The bentonite slurry is circulated through centrifuges that can separate the heavier soil contaminants such as sand from the bentonite so that it can be used again.

The wall is constructed in sections with a vertical steel stop placed at the end of the new section so as to form up the end of the section. A plastic water stop (Figure 8.6b) can be placed between the previous section and the new section so that the joint is watertight. Concrete is tremied into the base of the wall section using a long pipe (Figure 8.6a), displac-ing the bentonite slurry upwards. Once the concrete has hardened, the excavation can take place exposing the concrete wall.

Anchors can be added as excavation proceeds, and concrete slabs can be keyed into the exposed concrete wall.

8.3 STABILITY OF EXCAVATIONS

There are a number of ways that failure of the support systems for an excavation can occur, and some of the failure modes are specific to the type of supporting structure. Some of these modes of failure are shown in Figure 8.7.

1. For gravity type walls made from concrete or brickwork, the wall can overturn due to excessive earth and water pressures at the back of the wall. Excessive bearing pressure may occur at the front of the wall (the toe) and bearing failure may take place. Sliding

Soft piles constructed first

Overlapping hard piles

Figure 8.4 A secant pile wall.

Excavation 307

(a) (b)

Figure 8.6 (a) Tremie being used to place concrete. (b) Plastic water stop.

(a) (b)

Figure 8.5 (a) Slurry in trench excavated between two guide walls. (b) Clamshell bucket used for excavating a trench.


forward of the wall can be another failure mode. Overall stability due to a slip failure beneath the wall can be checked using a slip circle analysis. All of these modes can be checked simply by hand calculation.

2. For sheet pile walls and diaphragm walls, the moments and shears in the wall at vari-ous depths need to be calculated and checked against allowable values. The capacity of the anchors and an outward failure of the toe of the wall due to a passive failure also need to be checked. Settlement behind the wall may be of importance and so may also need to be estimated.

Overstress

(a) Overturning

Check moment equilibrium

(b) Sliding

Toe

(d) General stability

Check force equilibrium Check bending and shear stresses

(e) Granular soil, associated with excessive upward seepage

Upward and inwardmovement of soil

(i) Bottom heave (j) Buckled struts (k) Overstress of foundations

Carry out overall (slope) stability analysis

(c) Overstress

(f ) Failure of anchor system (h) Settlement behind the wall(g) Bottom of piles move outward (passive resistance not sufficient)

Figure 8.7 Failure modes for support structures.

Excavation 309

3. Bottom heave can take place due to soil squeezing into the excavation under the action of the weight of soil outside. As well, if the water table is kept down within the excava-tion, the flow of water up through the base of the excavation can wash away the soil if the hydraulic gradient is high enough, and this can lead to failure.

Design of retaining walls can be carried out using computer programs. Although stability of a wall can be calculated by hand, soil layering and soil deflections may require the use of a program such as WALLAP (2013) (that uses finite element methods for calculating deflec-tions and limit equilibrium methods for calculating stability) or a finite element program such as PLAXIS or Phase2.

The modelling of excavation using finite element methods is discussed in Section 8.8, but it is necessary to decide whether the excavation is to be modelled treating the soil as a one-phase material (i.e. not treating the groundwater separately) or as a two-phase material (where pore water and the soil skeleton are considered in the model). If the soil is treated as a two-phase material, then the drawdown of the water table as excavation proceeds becomes an issue.

8.4 BASE HEAVE FOR CUTS IN CLAY

One way in which an excavation can fail is through heave of the base. This is most likely to occur in soft clays, but can occur in other types of soils.

Analysis of the base heave problem may be carried out for simple two-dimensional cases by using the method proposed by Terzaghi (1943).

8.4.1 Shallow excavations (H/B < 1)

In this case, we can assume that the failure surface reaches the ground surface. The assumed failure mechanism depends on whether the excavation has a firmer stratum below the base of the excavation or not. The two cases are shown in Figure 8.8; case (A) is where the exca-vation is quite narrow and there is no layer of stiff material close to the base of the excava-tion. Case (B) is where the excavation is wide and has a stiffer stratum close to the base.

Case AIn this case, the load per metre run due to the soil block (on a–b) is 0.7BHγ − Hsu and the bearing capacity per metre run will be suNc0.7B, therefore the factor of safety F is given by

F

s N BBH Hsu c

u

=−

0 70 7

.. γ

(8.1)

B(a) (b)

H0.7B

0.7B

H

D

D

45°

45°

su

a b

Figure 8.8 Failure modes for long shallow excavations in clay. (a) FH

s Ns Bu c

u

= ⋅−

10 7γ / .

narrow excavation with

no stiff stratum close to base. (b) FH

s Ns D

u c

u

= ⋅−

1γ /

wide excavation with stiff stratum close to base.


where B is the width of the excavation, H is the depth, Nc is the bearing capacity factor, and su is the undrained shear strength of the clay that is assumed to be uniform.

Case BFor the wide excavation, the load due to the soil block is DHγ − Hsu and the bearing capac-ity is suNcD, therefore the factor of safety F is given by

F

s NH s D

u c

u

=−( )γ /

(8.2)

8.4.2 Deep excavations (H/B > 1)

For a deep excavation, the failure surface will not reach the surface as shown in Figure 8.9. For a long excavation, the factor of safety F against a base heave failure is given by Equation 8.3 (Bjerrum and Eide 1956).

F

s NH

u c=γ

(8.3)

The bearing capacity factor Nc in the formula needs to be found for the appropriate value of H/B. For a long excavation, Figure 8.10 can be used where L/B = ∞.

8.4.3 Excavations of rectangular shape in plans

For cuts that are of width B and length L in plan, Bjerrum and Eide’s formula (Equation 8.3) may still be used, although the bearing capacity factor in the formula needs to be found for the appropriate value of L/B. The bearing capacity factors are given in Figure 8.10.

B

Hc, γ

c, γ

Figure 8.9 Failure mechanism for a deep excavation.

Excavation 311

8.4.4 Base failure in sands

With braced excavations in sands, the danger of base failure usually occurs when the water level inside the excavation is lowered so that an upward flow of water can take place. Piping of the sand can occur if the hydraulic gradient approaches a value of unity.

The hydraulic gradient at the base of the excavation can be estimated from a flow net such as the one shown in Figure 8.11. The exit hydraulic gradient iexit is given by

i

Ha

exit = ∆

(8.4)

9

8

7

6

5

40 1 2 3 4 5

23

L/B = 1

Nc

H/B

∞

Figure 8.10 Bearing capacity factors for rectangular excavations.

a

Impervious layer

Water level

Waterlevel

h

1

6345

7 28

Figure 8.11 Flow net used for estimating the factor of safety against piping failure.


where a is the smallest distance between the equipotential lines (in this case near the wall), and ΔH is the drop in total head between the equipotential lines. In Figure 8.11, the head drop between equipotential lines is given by

∆H

hN

h

d

= =8

(8.5)

where h is the difference in the water levels inside and outside the excavation, and Nd is the number of head drops (here there are 8). A factor of safety against a piping failure may be defined as shown in Equation 8.6.

F

ii i

pipingcr

exit exit

= ≈ 1

(8.6)

A factor of safety of 1.5 or more would generally be required to guard against a piping failure.

8.5 GROUND SETTLEMENT CAUSED BY EXCAVATION

Lateral movement of walls used to brace excavations results in vertical movement of the ground surface that is called ground loss. There is always some movement of retaining walls before the bracing can be applied and therefore some ground loss will inevitably occur.

Peck (1969) provided some information on expected ground movements, and these are shown in Figure 8.12. The amount of movement expected depends on the type of soil encountered, but is largest in very soft to soft clays as may be expected. The magnitude of the vertical deformation δv to depth of the excavation H depends on the distance from the edge of the excavation as can be seen from Figure 8.12. The settlements can be divided up into three regions called I, II, and III that give the maximum settlement envelopes for differ-ent soil types as are shown in Figure 8.12.

Tomlinson (1995) has presented some data collected for different excavations in soils of different types. He comments that the amount of horizontal movement that occurs is not dependent on the type of wall and bracing system, and that there is little difference in the movement of diaphragm walls and sheet pile walls. Plots of measured maximum horizontal movements divided by excavation depth are shown in Figure 8.13 for (a) soft-to-firm nor-mally consolidated clays; (b) stiff-to-hard overconsolidated clays; and (c) sands and gravels. Average ratios of maximum inward movement to depth are (a) 0.30% for soft-to-firm NC clays; (b) 0.16% for stiff-to-hard OC clays; and (c) 0.18% for sands and gravels.

Plots of vertical surface movements versus distance from the edge of the excavation have also been presented by Tomlinson and are shown in Figure 8.14. The curves are for differ-ent case studies in the same soil type, and show the range of movement that may occur. Also on the figure is Peck’s curve for very soft to soft clay, where it may be noted that unlike the other vertical settlement profiles, the curve reaches a maximum at the edge of the excavation. The other measured profiles reach a maximum away from the edge of the excavation. This is because of the upward movement of the soil as it is unloaded as soil is excavated, and is observed in both field cases and in finite element simulations of excavation. The older curves from Peck do not reflect this smaller vertical deflection near the edge of the excavation.

Clough and O’Rourke (1990) have also presented data for the vertical surface settle-ment δv with distance d from a braced excavation. Figure 8.15 shows the vertical surface

Excavation 313

settlement for excavations in sand. The data measured falls within a triangular region if the ratio of settlement to maximum settlement δv/δvm is plotted against the ratio of distance from the excavation to maximum excavation depth d/H as shown in the inset to Figure 8.15.

For soft to medium clays, the scatter of data for surface settlement versus distance from the excavation is shown in Figure 8.16. Also shown are Peck’s regions I, II, and III (see Figure 8.12). Again, if the settlement to maximum settlement is plotted, a plot can be made that encloses the range of data, although unlike the plot for sand this time the plot has a trapezoi-dal shape.

The settlement to maximum settlement δv/δvm has a limiting value of about 1.0 over a range of d/H of about 0.7 as can be seen from the inset to Figure 8.16.

Mana and Clough (1981) have also examined the relationship between the maximum horizontal movement δH(max) of the walls of a braced excavation and the maximum surface settlements δV(max). They have found that the vertical movement is about 0.5–1.0 times the horizontal movement as shown in Figure 8.17, that is,

δ δV H( (. .max) max) to = 0 5 1 0 (8.7)

8.5.1 Effect of shape of excavation

More recent data has shown that the shape of the excavation and the size of the excavation can have an influence on the deflections and forces that are observed (Tan and Wang 2013a,b).

For instance, if the excavation is cylindrical, the walls retaining the excavation can behave like an arch. However, when the diameter of the cylinder becomes large, the effect is less pronounced as may be expected (Tan and Wang 2013a).

I – Sand and soft clay and average workmanship

II – Very soft to soft clay. Limited in depth below base of excavation

III – Very soft to soft clay. Great depth below excavation

Distance from the braced wallH

Hδ v

III

II

I

1 2 3 400

1

2

3

(%)

Figure 8.12 Variation of vertical surface movement with distance from edge of excavation. (Adapted from Peck, R. 1969. Proceedings of the 7th International Conference on Soil Mechanics and Foundation Engineering, Mexico City, State-of-the-Art Volume, pp. 225–290.)


Large excavations may result in more extensive vertical ground settlements that extend further from the excavation than for smaller (in plan) excavations. This is shown in Figure 8.18 for the soft Shanghai clays (Tan and Wang 2013b) in a plot similar to that of Clough and O’Rourke (see Figure 8.16).

In Figure 8.18, it can be seen that the data for the Shanghai World Finance Centre (SWFC) Annex excavation (that is a large excavation of about 30,000 m2 in plan) extend further than excavations for the Metro excavations that are for narrow rail corridors. Building basement excavations that are smaller and the Metro data can be seen to extend to about a d/H value of 2 as predicted by Clough and O’Rourke, but the larger SWFC excavation the deflections extend to a d/H value of about 3.7.

8.6 FORCES ON BRACED EXCAVATIONS

Unlike ordinary retaining walls that can move away from the backfill and develop active pressures in the soil, braced walls cannot move easily and therefore different pressure distri-butions are developed on such walls.

Soft-firm NC clays

Max. inward movement/depth (%)

Max. inward movement/depth (%)

Sand and gravels

Stiff-hard OC clays

Average 0.18%

Average 0.30%Average0.16%

Exca

vatio

n de

pth

(m)

Exca

vatio

n de

pth

(m)

Diaphragm wall, anchoredDiaphragm wall, or secant pilewall, struttedSheet pile, soldier pile withconcrete infill, struttedSheet pile, soldier pile withconcrete infill, timber, anchored

# Excluded from average

#

0.2 0.4 0.6 0.8 0.2 0.40

10

10

0

20

20

30(a)

(c)

(b)30

20

10

000

Key

0.2 0.4 0.6 0.80

Figure 8.13 Observed maximum inward movement of braced excavations. (After Tomlinson, M.J. 1995. Foundation Design and Construction, 6th ed. Longman Scientific and Technical, Harlow, UK.)

Excavation 315

Peck et al. (1974) have presented some empirically developed pressure envelopes that are based on load measurements taken in struts. The apparent pressure envelope can then be used to compute the forces in a strut at any given level. This pressure envelope is not the true pressure distribution, but a device to allow computation of the forces in the bracing.

The forces in the bracing can be computed by finding the area under the pressure envelope for each strut. This is done by assuming that each strut carries the pressure applied to the

Distance from face of excavationDepth of excavation



Settl

emen

tEx

cava

tion

dept

h ×

100

Settl

emen

tEx

cava

tion

dept

h ×

100

Settl

emen

tEx

cava

tion

dept

h ×

100

Peck’s (1969) curve for2-level basements

Soft-to-firm normallyconsolidated clay

Stiff-to-hardoverconsolidated clay

Sands and gravels

0 1 2 3 4 5

0 1 2 3 4 5

0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0

0.2

0.4

0.6

0.8

1.0

0.10

0

0.20

7

8

1011

1614

2828

32

7 = Diaphragm wall, soft-to-firm N/C clay, strutted

9 = Sheet pile, soft-to-firm N/Cclay, strutted

11 = Diaphragm wall, soft-to-firm N/C clay, strutted

14 = Diaphragm wall, stiff-to-hard O/C clay, strutted

16 = Diaphragm wall, stiff-to-hard O/C clay, anchored

28 = Bored piles, sands andgravels, strutted

32 = Diaphragm wall, sandydecomposed rock, strutted

Figure 8.14 Maximum vertical movement of surface versus distance from edge of excavation. (After Tomlinson, M.J. 1995. Foundation Design and Construction, 6th ed. Longman Scientific and Technical, Harlow, UK.)


wall over the region going half way to the next strut as shown in Figure 8.19. If no strut is placed at the base of the excavation, it is assumed that part of the load is taken by the soil at the base of the excavation.

For cuts made in sand, the apparent pressure envelope may be considered to be of constant magnitude with depth and have a value of 0.65γHtan2(45° − ϕ). It should be noticed that this value only applies to dry or moist sands (see Figure 8.20).

For excavations made in clay, the pressure envelope depends on the parameter γH/su where H is the depth of the cut. If this value is less than 4, the envelope shown in Figure 8.17c should be used. In this case, the average magnitude of the apparent pressure envelope is about 0.3γH. If the ratio exceeds 4, the pressure envelope of Figure 8.20d should be used provided the envelope is greater than that in Figure 8.17c, otherwise the value of pressure from (c) is used. The value of the undrained shear strength su is taken as the average value over the depth of the cut.

8.7 STABILITY OF SLURRY-FILLED TRENCHES

As mentioned previously (Section 8.2.4), diaphragm walls can be constructed by excavating under a bentonite slurry that is thick (dense) enough to prevent the sides of the excavation from caving in. The density of the slurry needs to be high enough to prevent the sides of the trench from failing, but not so dense as to prevent excavation of soil by the clamshell grab.

Simple analytic techniques can be applied to estimate if a trench is likely to fail. The analysis is based on the assumption that the slurry supplies a hydrostatic force to the face of the excavation. This is a reasonable assumption as in most cases the slurry tends to cake against the sides of the excavation forming a water resistant seal and reducing water flow into the surrounding soil.

LegendHatfieldBershamra7th and G Sts.G St. Test Site8th and G St.OCC Bldg.Charter station

Settl

emen

t

Max

. exc

avat

ion

dept

h=

H(%

)Distance from excavationMax. excavation depth

= dH

Distance from excavationMax. excavation depth = d

H

Settl

emen

tM

ax. s

ettle

men

t=

0.5 1.0 1.5 2.0 2.5 3.000

0.1

0.2

0.3

0.5

1.0

1 20δv δvm

δv

0

Triangular boundson distribution ofsettlement

Figure 8.15 Measured settlements adjacent to excavations in sand. (After Clough, G.W. and O’Rourke, T.D. 1990. Design and Performance of Earth Retaining Walls, ASCE, Geotechnical Special Publication No. 25, pp. 439–470.)

Excavation 317

8.7.1 Wedge analysis

Shown in Figure 8.21a is the face of a slurry-filled trench, with a presumed wedge type failure occurring. If such a failure were to occur, then the force polygon can be drawn for the wedge assuming that it is in equilibrium, and the factor of safety (FoS) against a sliding failure can be computed for the wedge.

8.7.2 Purely cohesive soil

For a purely cohesive soil such as an undrained clay, the force diagram shown in Figure 8.21b is applicable. In this case, we can compute the magnitude of the mobilised cohesive force Cm since the vector b − c is equal to Pf and therefore vector a − b is given by (W − Pf).

C

W Pm

f=−( )

2 (8.8)

Distance from excavationMax. excavation depth

= dH

Settl

emen

tM

ax. s

ettle

men

t=

0.5 1.0 1.5 2.0 2.5 3.000

1

2

3

Zones I, II, III afterPeck (1969)

III

II

I

Settl

emen

tδ v

δ v

δ vmM

ax. e

xcav

atio

n de

pth

=H

(%)

LegendVaterland 1Vaterland 2Vaterland 3

Gronland 2Gronland 1

Olav Kyrres

Chicago TransitGrant ParkSocial SecurityHarris BankNorthern TrustWater TowerEmbarcadero III

Max. settlement

0.5

0.50

0 1.0 1.5 2.0 2.5 3.0

Transition zone

1.0

Figure 8.16 Measured settlements adjacent to excavations in soft-to-medium clay. (After Clough, G.W. and O’Rourke, T.D. 1990. Design and Performance of Earth Retaining Walls, ASCE, Geotechnical Special Publication No. 25, pp. 439–470.)


δv (max) = δH (max)

δv (max) = 0.5δH (max)

δH (max)

δ v (m

ax)

H(%

)

H(%)

1 2 300

1

2

3

ChicagoOsloSan Francisco

Figure 8.17 Relationship between maximum lateral and maximum vertical ground movement. (After Mana, A.I. and Clough, G.W. 1981. Journal of the Geotechnical Engineering Division, ASCE, Vol. 107, No. GT6, pp. 759–777.)

Clough and O’Rourke (1990)Max. settlement Transition zone

Max settlement Transition zoneTan and Wang (2013b)

Normalised distance behind retaining wall, d/He

00

0.2

0.4

0.6

0.8

1.0

δ v/δvm

1.0 2.0 3.0 4.0 5.00.5 1.5 2.5 3.5 4.5

SWFC-AnnexBuilding basement Metro excavations

Clough and O’Rourke (1990)Hashash et al. (2008)Hsieh and Ou (1998)Tan and Wang (2013b)

Figure 8.18 Vertical settlement adjacent to excavations of different sizes. (After Tan, Y. and Wang, D. 2013b. Journal of the Geotechnical and Geoenvironmental Engineering Division, ASCE, Vol. 139, No. 11, pp. 1894–1910.)

Excavation 319

The mobilised cohesion is therefore

c

CH

H H

Hm

m f= =−

2

2 2

2

2 2( )γ γ /

(8.9)

where γ is the unit weight of the soil, γf is the unit weight of the slurry, and H is the height of the trench. The factor of safety F against failure of the wedge may be defined as

F

sc

sH

u

m

u

f

= =−4

( )γ γ (8.10)

where su is the undrained shear strength of the soil. It may be seen from Equation 8.10, that the closer the unit weight of the slurry is to the unit weight of the soil, the higher the factor of safety.

Apparentpressureenvelope

Soil reaction

Strut

Strut

Strut

H

d1

d2

d3

d4d4/2

d4/2

d3/2

d3/2

d2/2

d2/2

Figure 8.19 Method of calculating strut loads from apparent pressure diagram. (Adapted from Peck, R.B., Hanson, W.E., and Thornburn, T.H. 1974. Foundation Engineering, 2nd ed. Wiley, New York.)

H

SandClay

0.75H

0.25H0.25H

0.25H

0.50H

≤ 4 > 4

0.65 γH tan2(45 − φ ⁄ 2)to

0.4γ

0.2γH

γH – 4su

γH su

γH su

(a) (b) (c) (d)

Figure 8.20 Apparent pressure diagram for calculating loads in struts of braced cuts. (a) Wall of height H. (b) Dry or moist sand. (c) Clays if γH/su ≤4. (d) Clays if γH/su 4 provided that γH/sub does not exceed about 4 (where sub is the undrained strength of the clay below excavation level). (Adapted from Peck, R.B., Hanson, W.E., and Thornburn, T.H. 1974. Foundation Engineering, 2nd ed. Wiley, New York.)


8.7.3 Cohesionless soil

Cohesionless soils such as sands can also be supported by slurry such as bentonite. With reference to the vector diagram of Figure 8.21a, we can see that the resultant force on the plane of slip is inclined at the mobilised angle of friction α. From the vector diagram of Figure 8.21c, we have

WPf

= − +

tanπ θ α2

(8.11)

The weight of the wedge is given by

W H= −

0 52

2. tanγ π θ

(8.12)

and the fluid force is given by

P Hf f= 0 5 2. γ

(8.13)

therefore Equation 8.11 becomes on substitution for these forces

γ tan /( )tan

π θγ

π θ α22

− = − +

f

(8.14)

It can be shown that the most critical plane occurs when θ = π/2 + α/2, so under these conditions Equation 8.14 becomes

γγ

π απ α

π θ αf

= ++

= − +

tan( )tan( )

tan/ // /4 24 2 2

2

(8.15)

Slurrylevel

Ground level

HC

CA

Pf

αR

B

H/3

(a) (b) (c)

RR

45°

45°

W

PfPf

90° – (θ−α )

C

θ

b

c

a

Figure 8.21 Assumed failure wedge (a) and force vector diagrams for (b) cohesive soil, and (c) cohesionless soil.

Excavation 321

Therefore,

tan( )αγ γ

γ γ=

−−

f

f2

(8.16)

The factor of safety F against collapse can be defined as

F

f

f

= ′ =′

−tantan

tanφα

γ γ φγ γ

2 .

(8.17)

If the free water level is near the ground surface, it may be difficult to maintain trench stability. It may be necessary to

1. Lower the water table in the vicinity of the excavation 2. Raise the level of slurry in the trench, or 3. Use a denser slurry

The factor of safety against a failure of the slurry-filled trench in this case is given by

F

f

f

=′ ′ ′′ − ′

2 γ γ φγ γ

. tan

(8.18)

where the submerged unit weights of soil and slurry γ ′, γ ′f are now used in the equation.


Many different means of analysing excavations supported by anchors or struts are available, and are today routinely used to design support systems. One such program WALLAP (2013) uses a combination of limit equilibrium methods to assess stability and finite element or subgrade reaction methods to estimate wall deflections (see Figure 8.22).

The program allows for different layers of soil (horizontally layered), surcharges at the back of the wall, struts or anchors, analysis of soldier or king pile walls, water table differ-ences on either side of the wall, and earthquake forces. Excavation can be applied in steps to allow calculation of wall deflections.

8.8.1 Finite element analysis

The method of computing the forces that are needed to simulate excavation using finite ele-ment methods was first published by Brown and Booker (1985). In an elastic medium, the forces due to excavation of some of the finite elements in the mesh may be calculated by noting that the finite element equations say that the stiffness matrix multiplied by the deflec-tions are equal to the body forces and the applied tractions. The definitions of terms in the following equations are explained in Chapter 3.

K f fδ = +b t (8.19)


Expanding the above equation leads to

V

T

V

T

S

TdV dV dS∫ ∫ ∫= +B DB N N tδ γ

(8.20)

or in terms of the stresses

V

T

V

T

S

TdV dV dS∫ ∫ ∫= +B N N tσ γ

(8.21)

Hence, for equilibrium of the finite element mesh, we must have

V

T

V

T

S

TdV dV dS∫ ∫ ∫− − =B N N t fσ γ nodes

(8.22)

Generally, the integrations of Equation 8.22 will yield zero forces fnodes as the mesh is in equilibrium, but if some of the elements are removed (as in excavation) the forces will not be in equilibrium at the nodes around the sides of an excavation, therefore these forces are the ones that need to be applied to the finite element mesh to simulate the excavation.

10

–2.00 –2.00

–0.90–0.90

Post Post

ConcreteConcrete

–0.50–0.40

0.00

5 0Water pressure (kN/m2)

Fill Fill

FillFill

Fill

1.70

5 10

Stage No. 1 Excav. to elev. 0.00 on passive side

Figure 8.22 Cross section of retaining structure analysed by computer program WALLAP.

Excavation 323

Forces will also be calculated at the location of fixed nodes (where boundary conditions are applied) and these forces must be reset to zero in the calculation.

If the material is a linear elastic material, then Brown and Booker (1985) show that the sequence of excavation does not change the final result, that is, if elements are removed from the whole excavation in one step or they are removed from the excavation in stages.

At the beginning of an excavation problem, it is necessary to set up the initial stresses in the ground due to the weight of overlying soil layers (and perhaps other causes). The lateral stresses σh can be set up through a knowledge of the stress regime in the ground. The lateral stresses may be due to the coefficient of earth pressure at rest K0 and sometimes locked in geostatic stresses σl due to earth movements and compression of the ground laterally. In general, the formula

σ σ σh v lK= +0 (8.23)

may be used, although more complicated lateral stress variations can be used if it is known that they exist. Different lateral stresses may exist in the two lateral axis directions for example, or surface loads may exist.

It is very important to get the stress regime correct as the forces that drive the excavation modelling are due to the initial stresses in the ground. If these are incorrect, then the whole excavation analysis will yield incorrect results.

8.8.1.1 Non-linear analysis

If the soil is modelled as an elasto-plastic material, then it can be shown (Brown and Booker 1985) that the same method of computing excavation forces can be applied as shown in Equation 8.22. The excavation process has to be modelled in incremental steps with layers of elements removed in sequence and equilibrium of the finite element mesh obtained before proceeding to the next step. Unlike excavation in an elastic material, the process depends on the excavation sequence.

The process involves removing some of the elements and calculating the forces to be applied to the sides and base of the excavation by integrating the stresses and self-weight over the remaining elements in the finite element mesh and applying any surface tractions. This ensures that the remaining elements in the mesh are in equilibrium as excavation proceeds.

If the mesh includes beam elements or other types of elements, then the stiffness of these elements must also be included as they contribute to the overall force balance in the mesh.

Shown in Figure 8.23, is a simple finite element mesh consisting of eight-node isopara-metric elements. Elements were removed from this mesh to simulate excavation of a tunnel (half of mesh is used because of symmetry). The material was treated as being elasto-plastic in this simple problem, and there are three layers of material.

Once excavation has taken place, the stress trajectories plot as shown in Figure 8.24 (a plot of the directions of the principal stresses), shows the major principal stresses parallel to the face of the excavation, and the minor principal stress almost zero perpendicular to the tunnel face as may be expected.

8.8.2 Finite difference approach

The finite difference code FLAC (ITASCA Consulting Group 2014) can also be used for excavation problems. Anchors and props may be added as excavation proceeds in the same way as may be done in finite element codes.


An example of an excavation problem involving a sheet pile wall with two sets of anchors is shown in Figure 8.25. The excavation and placing of the anchors was mod-elled, and after completion of the excavation, a c − ϕ reduction analysis was carried out. This involves reducing the magnitude of the strength parameters c (cohesion) and tanϕ (frictional shearing resistance) until failure occurs. The amount of reduction before fail-ure allows a factor of safety to be calculated (shown on the Figure as 1.52). In this way, the Factor of Safety against a collapse of the shoring system at the end of excavation can be estimated.

It may be noted that the solution shows the failure surfaces going behind the anchors that hold the wall back, with a wedge of soil moving down above the end of the anchors. In front of the wall, a passive wedge of soil is shown being pushed forward and upward.

8.9 EXCAVATION INCLUDING GROUNDWATER

Excavation of soil containing water in the pores may involve removing just the soil as would occur in an excavation under water, or removing both the soil and the water (included in the pores of the soil) as may occur when a saturated clay is excavated.

Where the soil is reasonably permeable, the groundwater around the excavation will flow into the excavation. This may occur through the sides and base of the excavation, but if the sides are retained by an impervious wall (say a diaphragm wall), flow may only occur through the base of the cut.

Figure 8.23 Finite element mesh for the tunnel excavation problem.

Excavation 325

Figure 8.24 Principal stress trajectories around a tunnel.

FLAC (Ver. 5.0)

Legend

14-Sep-11 12:35 step 53781 1.049E+01 <x< 4.443E+01–1.045E+01 <y< 2.350E+01

Factor of safety 1.52Max. shear strain-rate 0.00E+00 5.00E–07 1.00E–06 1.50E–06 2.00E–06 2.50E–06 3.00E–06 3.50E–06 4.00E–06

Contour interval = 5.00E–07Boundary plot

0 1E 1

Cable plotNet applied forces

–0.750

–0.250

0.250

0.750

1.250

1.750

2.250(*101)

1.250 1.750 2.250 2.750 3.250 3.750 4.250(*101)

Job title : .

Anchored sheet pile wallExcavation example

Figure 8.25 Failure mode for excavation using c − ϕ reduction (FLAC2D).


Finite element analysis of excavation in a two-phase soil (i.e. containing soil and water) can be performed through use of the equations of consolidation (see Equation 3.70). In Equation 8.24, we need to be able to compute the increment in force Δf due to removal of the soil (and pore water).

K L

L h

f

h−

− − −

=

γγ γ γ

wT

w w w tt( )1 α Φ∆δ∆

∆∆ Φ∆t

(8.24)

At any stage, the stress equilibrium is made up of the effective stresses and total ground-water pressure being in equilibrium with the body forces (due to self-weight) and any applied forces. We can therefore substitute for the total stress in Equation 8.22 as the total stress σ is equal to the effective stress σ′ plus the pore water pressure q. In flow problems, the total head hT is made up of the elevation hEL head plus the water pressure head hw, therefore we can write the water pressure as

h h hw T EL= − (8.25)

Hence, the water pressure may be obtained by multiplying the water pressure head by the unit weight of water. It may be noted that the water pressure can consist of any static water pressure plus excess pore pressures (say, due to loading).

q i i= = −γ γw w w T ELh h h( ) (8.26)

The vector i = (1,1,1,0,0,0)T is introduced as the water pressure is only added to the direct stress components and not the shear stress components.

The total stress is therefore given by

σ σ σ= ′ + = ′ + −q iγ w T ELh h( ) (8.27)

Substitution of the total stress into Equation 8.22 gives the forces to be removed from the sides of the excavation

V

Tw

V

TT EL

V

T

S

TdV dV dS∫ ∫ ∫ ∫′ − − − − = ∆B d a h h N N t fσ γγ ( ) nodes

(8.28)

Therefore, as excavation proceeds the increments in force to be applied to the sides of the excavation can be found from Equation 8.28 and used in Equation 8.24.

If the excavation is made underwater, and only soil is removed (as in say, a very perme-able sand), then only the effective stresses are used in computing the force increment, that is,

V

T

V

T

S

TdV dV dS∫ ∫ ∫′ − − = ∆B N N t fσ γ nodes

(8.29)

Once again, if there are structural elements in the finite element mesh, these need to be included in the calculation of the forces to maintain equilibrium of the finite element mesh.

Excavation 327

The process involves setting up the initial effective stresses in the mesh (generally at the Gauss points in the elements) and the static water pressures (generally at the nodes of the elements). Then any loads can be applied, and the excess pore water pressures that these cause is added to the static water pressures. Then as elements are excavated, the equilibrium equation can be used to compute the excavation forces to be applied to the mesh.

Because the equations of consolidation are written in incremental form in Equation 8.24, they can be used for materials that are non-linear. The soil may be an elasto-plastic material and the permeability of the soil may change with stress level for example.

Further information on simulation of excavation in a two-phase material is given in the paper by Hsi and Small (1993).

8.9.1 Example excavation problem (no drawdown)

Shown in Figure 8.26 is a tunnel that is excavated in a poro-elastic layered material. In this problem, the water table is assumed to be level with the ground surface (which is assumed to be permeable), and is held there and not allowed to drop. The sides of the tunnel are assumed to be permeable so that flow can occur into the tunnel, while the base of the finite element mesh is assumed to be impermeable. Because of symmetry, only half of the tunnel is analysed making the line of symmetry an impermeable boundary.

The solution shows the equipotential lines (lines of constant total head), and it may be seen that the equipotential lines are equally spaced with respect to height around the tunnel

–2.0 0 2.0 4.0 6.0 8.0 10.0–2.0

0

2.0

4.0

6.0

8.0

10.0

12.0

Total head

Isotropic soil – excavation in a poro-elastic material

Isotropic permeability

3 soil layers

Elastic materials

Contour legend3.000E+00

3.500E+00

4.000E+00

4.500E+00

5.000E+00

5.500E+00

6.000E+00

6.500E+00

7.000E+00

7.500E+00

8.000E+00

8.500E+00

9.000E+00

9.500E+00

Figure 8.26 Equipotential contours and flow vectors for tunnel excavation in a layered poro-elastic material.


boundary. This is because if the water pressure is zero (as specified), the total head is equal to the elevation head. If the total head increments are the same in the plot, then the elevation increments will be equal.

It may also be noticed from Figure 8.26 that the flow vectors (vectors showing the direc-tion and magnitude of the groundwater flow) are perpendicular to the equipotentials as they should be because the permeability was assumed to be isotropic in this example.

8.9.2 Excavation involving drawdown of the water surface

When an excavation takes place in a material that is reasonably permeable, such as a sand or a silt, the groundwater will flow into the excavation if the sides or base of the excavation are permeable. Even in low permeability soils, the groundwater will eventually flow into the excavation with time if the groundwater recharge is low enough so that the groundwater levels cannot be topped up, and the phreatic surface of the groundwater table will drop.

The drop of the phreatic (or free) surface involves the release of water from the elements that are intersected by the free surface as it drops from one location to another as shown in Figure 8.27.

Along the free surface, we have two boundary conditions: (1) the total head is equal to the elevation head and (2) the amount of water released from the pores of the soil as the water falls is equal to the flow across the free surface.

This may be mathematically written as shown in Equation 8.30.

h h f x y t

q Sht

d

T EL

w yT

= =

= ∂∂

( , , )

cosβ Γ

(8.30)

wherekn is the permeability normal to the free surfaceSy is the specific yield of the soil (has no units). It is the volume of water given up per

unit volume of soilβ is the angle of the free surface to the x-direction (Figure 8.28)Γ is the distance along the free surfaceqw is the flow rate of yielded water

This introduces an extra term G into the set of equations (see Equation 8.24).

K L

L G h

f

h−

− − − −

=

γγ γ γ

wT

w w w w tt( )1 α ∆ Φ∆δ∆

∆∆ γ Φt

(8.31)

Water released assurface drops

Figure 8.27 Drop of water table next to excavation.

Excavation 329

where the matrix G in Equation 8.31 is defined as

G aa= ∫Γ

ΓTyS dcosβ

(8.32)

The integration of Equation 8.32 can be performed numerically by evaluating the shape functions for the pore pressure in vector a along the free surface. The location of the free surface is found from the value of total head at the nodes of the element that is cut by the free surface. The values of total head can be used to compute the head within the element, and where the total head is equal to the elevation head is the location of the free surface. More details on this process are provided by Hsi and Small (1992).

In the region above the free surface, the permeability of the soil ks can be reduced gradu-ally by linking the permeability of the soil to the pore pressure in the soil. When the pore pressure becomes negative (as it will above the free surface), the permeability is reduced linearly (as shown down to 1/1000 of its value when saturated) at a limit pressure plimit. For higher suctions, the permeability remains at this limiting value (see Figure 8.29).

8.10 SOIL MODELS

When modelling excavation, the appropriate soil model needs to be used. The Mohr–Coulomb model can overestimate bottom heave and often can predict heave of soil behind

β

∆h

ΓFlow

Figure 8.28 Water released from a region above the free surface.

k

ks

Plimit P

ks : Permeability of saturated soil

klimit = ks/1000

– +

Figure 8.29 Pore pressure–permeability relationship.


the wall. Experience has shown that small-strain models give better prediction of the behaviour of retaining walls because the strain that occurs behind walls is generally fairly small.

This was pointed out by Simpson et al. (1979) for the stiff London clay during excavation for the New Palace Yard car park next to the Houses of Parliament in London.

The strains predicted from an elastic analysis are shown in Figure 8.30 where it can be seen that the strains are between about 10−4 and 10−3. Their finite element analysis using a small strain model as well as a plasticity model at larger strains gave good agreement with measured performance of the wall and of the surface settlement behind the wall.

Shown in Figure 8.31 is the strain that may be associated with various geotechnical works, and it can be seen that retaining walls undergo the smallest strains. The small-strain soil model is therefore best used with finite element analysis. In the commercial program PLAXIS, the HS-small model is recommended as it gives good bottom heave prediction independent of model depth and a more realistic settlement trough behind the wall (nar-rower and deeper).

The variation of shear modulus with shear strain in the PLAXIS HS-small model is based on the model proposed by Santos and Correia (2001) and is shown in Figure 8.32.

0.10%

0.07% 0.04% 0.01%

CL θ

Displacement vectors

Figure 8.30 Shear strain contours for a maximum wall displacement of 0.2% of wall height. (After Simpson, B., O’Riordan, N.J., and Croft, D.D. 1979. Géotechnique, Vol. 29, No. 2, pp. 149–175.)

Retaining walls

Foundations

Tunnels

Conventional soil testing Small strains

Larger strains Shear strains γs

Verysmallstrains

Local gauges

Dynamic methods

Shea

r mod

ulus

ratio

G/G

0

0

1

10–6 10–5 10–4 10–3 10–2 10–1

Figure 8.31 Range of shear strain associated with various geotechnical works.

Excavation 331

The equation for degredation of the shear modulus G is based upon the shear strain γ0.7 at which the shear modulus drops to 70% of its initial value G0 (at γ ≈ 10−6). The relationship is then given by

GG0 0 7

11 0 385

=+ . ( ).γ γ/

(8.33)

The constant 0.385 comes from fitting the degradation curve to experimental data. The small-strain shear modulus may be found from seismic field tests as discussed in Sections 4.25 and 4.26.

10–3 10–2

γ/γ 0.7

10–1 1 10 100 1000

G⁄G

0

0

0.2

0.4

0.6

0.8

1.0

Figure 8.32 HS-small model used in PLAXIS showing reduction in shear modulus with shear strain. (After Santos, J.A.D. and Correia, A.G. 2001. Proceedings of the 15th International Conference on Soil Mechanics and Geotechnical Engineering, Istanbul, Vol. 1, pp. 267–270.)

333

Chapter 9

Retaining structures

9.1 INTRODUCTION

The excavation of any kind of trenches or cuts such as basements for buildings or cuts for road works or railways may require some kind of support if the soil or rock is not of suf-ficient strength to stand safely on its own. The process of excavation is closely related to excavation support, therefore this chapter should be read in conjunction with Chapter 8.

There are a number of different systems that have been designed to support the sides of excavations and which type is used depends on the proximity of existing structures, the space available and the type of soil being retained.

9.2 EARTH PRESSURE CALCULATION

Pressures on the back of a retaining structure depend on the movement of the structure. If it is moving away from the backfill, then the pressures that act are called the active pressures. If the wall is being pushed into the soil, then the passive soil pressures apply. If the wall is rigid and not moving, then an intermediate case of at rest pressures are applicable.

Active pressures apply behind retaining structures that can move forward such that the soil begins to shear and the shear forces support some of the weight of the backfill. Passive forces occur in front of a wall moving forward and pushing into the soil. The forces are therefore being resisted by shear forces in the soil and are higher than the active forces.

For walls that are braced or are very rigid, and are not able to move forward, the at rest condition applies. The forces involved are intermediate between the active and passive forces.

9.2.1 Rankine’s theory

If a wall is smooth, then these forces can be calculated simply from Rankine’s theory (Rankine 1857). This is shown in Figure 9.1, where it is assumed that the soil fails according to the Mohr–Coulomb failure criterion. If the soil is at rest, then at any depth in the soil, the vertical stress σv0 is due to the overburden pressure and the horizontal stress σh0 is related to the vertical stress by the coefficient of earth pressure at rest K0, that is,

σ γσ σ

v

h v

z

K0

0 0 0

==

(9.1)


Equation 9.1 is written in terms of total stress, but if water is present, the effective stresses need to be used. The coefficient of earth pressure at depth may be found from the empirical relationship

K0 1= −( sin )φ (9.2)

If the soil is behind a wall and the wall moves forward, the lateral pressure σh0 will reduce until it reaches the value σha and the soil will fail since Mohr’s circle has reached the failure surface. If the lateral pressure is increased as it would as a wall pushes into a soil, then the lateral pressure will increase until it reaches the passive pressure σhp at which time the soil will fail as Mohr’s circle has reached the failure surface. It can be seen from Figure 9.1 that the passive pressure is much larger than the active pressure.

For a smooth wall, the direction of the failure planes can be found from Mohr’s circle at failure. For the passive case, the planes of failure (broken lines) pass through points C and D, while for the active case, the failure planes pass through A and B. The directions of the active planes of failure are therefore at 45° + ϕ/2 from the horizontal and the passive planes of failure are at 45° − ϕ/2 from the horizontal. The planes of shearing are shown in Figure 9.2.

From Mohr’s circle, we can find the magnitudes of the stresses σha and σhp using the fact that Mohr’s circle touches the failure surface. For a vertical wall with a backfill having a horizontal surface, we have

σ σha a v aK c K= − 2 (9.3)

σ σhp p v pK c K= + 2

(9.4)

where c is the intercept of the Mohr–Coulomb failure surface on the vertical axis (the cohe-sion) and

Mohr–Coulomb failure surface

B

A

C

D

σhpσv0σh0σha

τ

σ

Figure 9.1 Mohr’s circle at failure for active and passive cases.

Retaining structures 335

Ka = −

+11

sinsin

φφ

(9.5)

Kp = +

−11

sinsin

φφ

(9.6)

The angle of shearing resistance ϕ is the slope of the Mohr–Coulomb failure surface.Hence, at any depth down at the back of the wall, the vertical stress can be calculated, and

this can be converted to the horizontal stress through Equation 9.3. If there is a surcharge q over the surface of the backfill, this is added to the vertical stress, that is, σv = γz + q. In front of the wall, the vertical stress at any depth can be converted to the passive horizontal stress using Equation 9.4. Because the vertical stress increases linearly (and is equal to q at the surface), then the increase in lateral stress is linear with depth. Therefore, for a uniform soil, with unit weight γ and for a wall of height H, we have

P K H c K H qK Ha a a a= − +1

222γ

(9.7)

If the wall is not smooth, then there will be shear stresses acting at the back (or front) of the wall, and the stress state is not a point on the horizontal axis of Mohr’s circle diagram. The shear planes in the soil are then curved as near the back of the wall or at the front of the wall, the planes of failure will be at different angles than for the smooth wall case. The planes of failure are shown in Figure 9.3 for the active case.

Use of Rankine’s theory is shown in Figure 9.4 where a soil possessing friction and cohe-sion is saturated with water. First, the vertical effective stress is calculated and then the vertical effective stress is converted to the horizontal effective stress using Equation 9.3. This calculation results in negative horizontal stresses near the top of the wall (Figure 9.4c). If the soil cannot sustain tension, then the soil will form a crack and provide no tensile force on the wall (Figure 9.4d). If the water fills the cracks, then the water pressure must be added to the horizontal effective stress as shown in Figure 9.4f to find the final total stress on the wall.

45°+ φ/245°– φ/2

Activefailure

Passivefailure

Wallmovement

Failureplane at A

Failureplane at B

Failureplane at C

Failureplane at D

Front and backof smooth wall

See Figure 9.1 forpoints A, B, C, D

Figure 9.2 Directions of failure planes in active and passive cases (smooth wall).


B

A

C

F ′

F

SlipRough

Figure 9.3 Slip lines for active pressure on rough wall from plasticity theory.

2c′/ γsub Ka

H

Retaining wall

Verticaleffective stress

Horizontaleffective stress

Effective horizontaleffective stress for notension

Water pressure if noweep holes

Total stress =(d) + (e)

(a) (b) (c) (d)

(e)

φ′c′

σh′ = γH − γ wH σh′ = Ka σv′ − 2c′ σh′= γsubH

γwH σh(f)

√

Ka √

Figure 9.4 Calculation of active pressure on a vertical wall with horizontal backfill using Rankine’s theory.


9.2.1.1 Inclined backfill

If the backfill has an inclined surface (but the wall is vertical) then the solution of Mazindrani and Ganji (1997) may be used to find the active and passive pressures at a depth z below the top of the wall (Equation 9.8).

p p z cp a,coscos

cos sin [ ( )= + ± √ −βφ

γ β φ φ β β φ22 2 2 22 2 4cos cos cos cos γγ

φ γ β φ φ γ β

2 2

2 2 24 8

z

c c z z+ + −cos cos sin cos ] cos

(9.8)

where β is the angle of inclination of the backfill to the horizontal.

9.2.2 Coulomb’s theory

Coulomb’s theory (Coulomb 1776) is based on the assumption that the failure surface is planar, and while this is true for a smooth wall, it is not true for a rough wall where there is friction causing a shear force against the wall as was discussed in the previous section. If the backfill is sloping at an angle β to the horizontal, and the back of the wall is at an angle α to the horizontal, and the angle of friction between the wall and the soil is δ, then it is possible to draw the vector of forces for the soil wedge. In Figure 9.5, this is done for the active case and the wedge of soil ABC. For the passive case, the forces Pp and F are inclined at angles to the opposite side of the normal as the wedge is moving up and not down as for the active case (see Figure 9.8).

9.2.2.1 Active case

For the case where there is no water behind the wall, trial slip planes can be selected and the vector of forces drawn. This is shown in Figure 9.6, and the point at which the maximum active force on the wall is found corresponds to the critical failure plane. The maximum value of the active pressure is the value required.

If there is water behind the wall, then the water forces must be included in the force dia-gram as shown in Figure 9.7. It can be seen that the weight of water in the wedge is balanced

BA

C

Pa

Pa

R

W

(a)

WR

(b)

a

b

c

φθ

θ – φα

δ

β

ψ

ψ

Figure 9.5 Coulomb’s planar failure surface.


by the reactive forces pw and Rw, therefore it would be possible to carry out the analysis by omitting the lower part of the diagram aed and using the total weight of the soil W minus the weight of the water Ww. Trial wedges will give a maximum force ′Pa that is only the effec-tive force on the wall. The water pressure force must be added to get the total active force on the wall (see Figure 9.7).

9.2.2.2 Passive case

For the passive case, the force diagram can be drawn as shown in Figure 9.8. The vector dia-gram of forces can be drawn for trial wedges and this time, the minimum value is obtained. This gives the value of the passive force on the wall.

Criticalwedge

a

a

1 2 34

b

Pa′

Pa′

Pa′ maxb3

b2

b1

c1

c2

c3

c4b4

R ′

R ′ φ′δ

W

c

Figure 9.6 Coulomb wedge analysis (active case with no water).

b

c

a

e

d

Pa′

Pa′

Pa′

R ′

R ′

Pw Rw

Rw Ww

W

Pw

Pw

δ φ′

Figure 9.7 Coulomb analysis for case where there is water behind a wall (active case).


It may be noted that in this case cohesive forces have been included in the analysis. Cohesive forces acting on the back of the wall and along the trial slip plane can be included as extra forces when drawing the force diagram.

As for the active case, water pressures can be included if water is lying in front of a wall. Again, the effective passive force is found and added on to the water force on the wall to get the final passive force.

9.2.2.3 Surface loads

If there are surface loads such as distributed loads or line or point loads, these forces can be added into the weight of the soil when drawing the vector diagram. For some of the trial wedges of soil, part of a uniform load may not need to be included as the load is outside of the top of the wedge.

9.2.2.4 Uniform materials

For uniform soils that do not have cohesion and do not involve surface forces, the maximum (active) or minimum (passive) force can be calculated through the use of calculus. The soil has to be totally uniform, and so must be totally drained or totally saturated. By looking at force equilibrium on a wedge of soil, it is possible to calculate the critical wedge by writing the force on the wall in terms of the angle of the wedge θ to the horizontal, and then dif-ferentiating to find the minimum value of the force, that is, for the active case finding where ∂Pa/∂θ = 0. This leads to the result for the active case

P K H

K

a a

a

=

= +

− + + −

12

1

2

2

2

γ

α φ

α α δ φ δ φ βα

sin ( )

sin sin( )sin( )sin( )sin( −− +

δ α β)sin( )

2

(9.9)

Minimum

Pp′

Pp′Pp′

c2

c1

b1

d2

d1

a

Cw′

Cw′

C ′

R′

R′

cd

a

b

e

W

φ′

W

C ′

Figure 9.8 Coulomb analysis for passive case in cohesive soil.


and for the passive case

P K H

K

p p

p

=

= −

+ − + +

12

1

2

2

2

γ

α φ

α α δ φ δ φ βα

sin ( )

sin sin( )sin( )sin( )sin( ++ +

δ α β)sin( )

2

(9.10)

The angle of shearing resistance used in the Equations 9.9 and 9.10 is the drained angle of friction as the equations do not apply to cohesive soils, that is, ϕ = ϕ′.

The values given by the formulae of Equation 9.10 for the passive case can be largely in error, especially for high values of the angle of shearing resistance ϕ because of the assump-tion of a planar failure surface. For this reason, it is prudent to use the values given by the log spiral method or from the bound theorems given in Sections 9.2.3 and 9.2.4. The values of the active earth pressure coefficient are not so sensitive to error for higher values of ϕ and are sufficiently accurate for use in engineering applications.

9.2.2.5 Active earthquake forces

If there are earthquake forces acting on the fill behind the retaining wall, a pseudo-static analysis can be undertaken using the Coulomb method. The horizontal force Fh acting on any potential active failure wedge is assumed equal to the weight of the wedge times the seismic coefficient kh. Similarly, the vertical force Fv on the wedge is equal to the weight of the wedge times the vertical seismic coefficient kv.

F k W

F k Wh h

v v

==

(9.11)

The vector sum of the weight of the wedge and the two seismic forces gives a resultant force acting at an angle ψ to the vertical (see Figure 9.9) where

ψ =

−

−tan 1

1k

kh

v (9.12)

φ′

α

δ

β

Failureplane

Cohesionless soil

W

Wkv

Wkh

Wall

Figure 9.9 Seismic forces acting on Coulomb wedge.


The resultant force W ′ may be seen to be equal to

′ = −

W Wkv1

cosψ (9.13)

This leads to the Mononobe–Okabe solution (Mononobe and Matsuo 1929) and Okabe (1924) for the active earth force

P K Hk

K

a av

a

= −

= − +

− − +

12

1

1

2

2

2

γψ

α ψ φ

α α ψ δ

cos

( )

sin sin( )sin

sin

(( )sin( )sin( )sin( )

φ δ φ β ψα ψ δ α β

+ − −− − +

2

(9.14)

Solutions for active wall pressures have also been presented by Iskander et al. (2013) based on Rankine’s theory, and it is shown that these are not greatly different to the Mononobe–Okabe solutions. For the passive pressure case, see Section 9.2.3.1.

9.2.3 Log spirals

A log spiral shaped failure surface can be used to find values of the passive earth pressure coefficient, and this was employed by Caquot and Kerisel (1948) to calculate values for walls with a rough surface. The values of Kp are shown in Figure 9.11 for a range of values of the angle of shearing resistance ϕ and the angle of the soil surface to the horizontal β as shown in Figure 9.10. The plot of Figure 9.11 is for a vertical wall, that is, α = 90° and for a purely frictional material where the angle of shearing for the wall–soil interface is δ = ϕ. If the angle δ is less than ϕ, correction of the value of Kp can be made through use of the values of λ in Table 9.1 where Kpδ = λKp.

9.2.3.1 Passive earthquake forces

The effect of earthquakes on passive pressures can be somewhat in error if the Mononobe–Okabe method is used because of the assumption of planar failure surfaces. Solutions using a logarithmic spiral have been obtained by Subba Rao and Choudhury (2005) for this case.

φ′, γα

δ

β

Potential slip lineDrainedcohesionless

Figure 9.10 Passive pressure on retaining wall.


Table 9.1 Factor λ to be used with values from Figure 9.11

Values of δ/ϕ

ϕ (deg) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

10 0.978 0.962 0.946 0.929 0.912 0.898 0.880 0.86415 0.961 0.934 0.907 0.881 0.854 0.830 0.803 0.77520 0.939 0.901 0.862 0.824 0.787 0.752 0.716 0.67825 0.912 0.860 0.808 0.759 0.711 0.666 0.620 0.57430 0.878 0.811 0.746 0.686 0.627 0.574 0.520 0.46735 0.836 0.752 0.674 0.603 0.536 0.475 0.417 0.36240 0.783 0.682 0.592 0.512 0.439 0.375 0.316 0.26245 0.718 0.600 0.500 0.414 0.339 0.276 0.221 0.174

10 20 30 4001

2

3

45

–0.8

–0.6

–0.4

–0.2

0.2

0.4

0.6

0.81.0

H109876

40

30

20

80706050

0+β

δH/3 Log spiral

= −0.9

φ (in degrees)

σP = KpγH

K p

βφ

Figure 9.11 Passive pressure coefficient for vertical wall. (After Caquot, A.I. and Kerisel, J. 1948. Libraire du Bureau des Longitudes, de L’école Polytechnique, Paris Gauthier-Villars, Imprimeur-Editeur, p. 120.)


They compute the passive pressure coefficient for the self-weight of the soil, any surface surcharge and for cohesion separately and if all three are acting, then the effects are added together.

Values of the passive pressure coefficient Kpγd are shown in Figure 9.12 for a vertical wall with a horizontal backfill. More results are given in the paper by Subba Rao and Choudhury (2005). The force on the wall is given (for self-weight only) by

P H Kp p d.= 0 5 2γ γ

(9.15)

9.2.4 Upper- and lower-bound solutions

The upper- and lower-bound solutions of plasticity theory as discussed in Section 9.7.6 can be used to find bounds on the passive and active pressures. This has been done by Shiau et al. (2008) for the wall shown in Figure 9.10 (the passive case). Bounds for the passive earth pressure coefficients are shown in Figure 9.13 for the case where the wall is vertical and the surface of the soil is horizontal. Results are presented for different angles of friction δ acting on the front of the wall. To be conservative, the lower bound could be taken in any calculation without too much error. However, the upper bound gives an indication of what the error may be, as the solution must lie between the two values.

If the front of the wall is at an angle α to the horizontal (see Figure 9.10), then results can be found for the passive coefficient of earth pressure Kp from Figure 9.14 for a backfill with a horizontal surface. Variation of Kp with the slope of the backfill surface β (for one value of wall friction angle δ = 19.2°) is shown in Figure 9.15 for a wall with a vertical back.

Table 9.2 (Reddy et al. 2013) shows values of Kp from different sources and it may be seen that there are large differences especially when then angle of shearing resistance ϕ is large

φ = 40° φ = 40°

φ = 30°

φ = 20°

φ = 10°

φ = 30°

φ = 20°

φ = 10°α = 0° , β = 0° ,

δ/φ = 0.5α = 0° , β = 0° ,

δ/φ = 1.0

khkh

K pγd

K pγd

kv = 0.0khkv = 1.0kh

kv = 0.0kh

kv = 1.0kh

10

8

6

4

2

00 0.1 0.100.2 0.20.3 0.4 0.5 0.50.40.3

0

6

9

12

3

15(a) (b)

Figure 9.12 Passive earth pressure coefficient Kpγd for self-weight of fill for different horizontal earthquake acceleration factors kh. (a) δ/ϕ = 0.5; (b) δ/ϕ = 1.0.


along with the angle of friction δ on the wall. The values of Kp from the bound theorems are therefore the most desirable for use, as the error in the value is known.

9.3 EFFECT OF WATER

The formulae for calculating the pressures on retaining structures can apply to the drained or undrained case. If the retaining wall has a saturated clay backfill or the soil in front of a wall is a saturated clay, then any rapid loading will take placed under undrained conditions. In this case

φ φ==

u

uc c (9.16)

K p,h

β = 0°α = 90°, δ/φ′ = 1

α = 75°, δ/φ′ = 1

α = 60°, δ/φ′ = 1

α = 90°, δ/φ′ = 0

α = 75°, δ/φ′ = 0α = 60°, δ/φ′ = 0

Upper boundLower bound

20 25 30 35 400

16

14

12

10

8

6

4

2

φ′

Figure 9.14 Values of passive coefficient for different angles of wall friction δ and slope of back of wall α.

α = 90°β = 0° φ′ = 45°

φ′ = 40°

φ′ = 35°φ′ = 30°φ′ = 25°φ′ = 20°

δ/φ′


20

K p

30

10

0

40

50

0.0 0.2 0.4 0.6 0.8 1.0

Figure 9.13 Passive earth coefficients from upper- and lower-bound theorems.


Fang et al. (failure)

Fang et al. (S/H = 0.30)Fang et al. (S/H = 0.20)

Fang et al. (S/H = 0.10)

Fang et al. (S/H = 0.05)

K p,h α = 0°

δ = 19.2°φ′ = 30.9°


20

15

10

5

0–15 –10 –5 50 10 15 20

β°

Figure 9.15 Variation of Kp with slope of backfill β.

Table 9.2 Values of passive earth coefficient from different sources

Soil ϕWall angle of

friction δCoulomb (1776)

Caquot and Kerisel

(1948)

Kumar and Subba Rao

(1997)

Soubra and Macuh (2002)

Lancellotta (2002)

Reddy et al. (2013)

20 2.04 2.04 2.04 2.04 2.04 2.0425 2.46 2.46 2.46 2.46 2.46 2.4630 0 3.00 3.03 3.00 3.00 3.00 3.0035 3.69 3.69 3.69 3.69 3.69 3.6940 4.60 4.59 4.60 4.60 4.60 4.6020 2.41 2.35 2.38 2.39 2.35 2.4025 3.12 3.03 3.06 3.07 3.07 3.0830 1/3 ϕ 4.14 4.00 4.02 4.03 4.03 4.0535 5.68 5.28 5.42 5.44 5.44 5.4640 8.15 7.25 7.58 7.62 7.62 7.6220 2.64 2.60 2.50 2.57 2.48 2.6125 3.55 3.40 3.40 3.41 3.22 3.4630 1/2 ϕ 4.98 4.50 4.60 4.65 4.29 4.7335 7.36 6.00 6.60 6.59 5.88 6.7140 11.8 9.00 9.80 9.81 8.38 10.0020 2.89 2.65 2.73 2.75 2.58 2.8525 4.08 3.56 3.72 3.76 3.41 3.9130 2/3 ϕ 6.11 5.00 5.26 5.34 4.63 5.5735 9.96 7.10 7.78 7.95 6.51 8.3240 18.7 10.7 12.24 12.6 9.57 13.2720 3.53 3.01 3.07 3.13 2.7 3.425 5.60 4.29 4.42 4.54 3.63 4.9530 ϕ 10.1 6.42 6.68 6.93 5.03 7.5835 22.9 10.2 10.76 11.3 7.25 12.3340 92.6 17.5 18.86 20.1 11.03 21.64


and these values are used with the formulae for earth pressure along with the total stresses in the ground.

If the soil in front of or behind a retaining structure is very pervious such as a sand, or has been in place long enough to drain, then the drained strength parameters and the effec-tive stresses in the ground are used to compute the forces on the wall. If the soil is beneath the water table, these forces are the effective lateral stresses, so the water pressures must be added to them to obtain the total pressure on the wall. In this case

φ φ= ′= ′c c

(9.17)

The water pressures can be very large, therefore in the design of walls it is always essential to supply drains behind the wall to eliminate water pressures. The drains must be main-tained to ensure that blocking does not occur, as this would result in water pressures acting on the wall and possible failure.

9.4 SURFACE LOADS

The force exerted on retaining walls can be calculated using hand methods, and most of these are based on solutions that treat the soil as a uniform elastic body. For point and line loads, a distance x = mH from the wall, pressure distributions can be calculated from Figure 9.16. The magnitude Ph and location yp of the resultant force can be calculated for use in stability (overturning) calculations. The pressure distribution for a horizontal line load Qh is also given in the figure.

9.4.1 Compaction stresses

When soil is compacted behind a wall, the compaction can induce higher stresses than apply for a wall that has not had the soil specifically compacted. According to Ingold (1979), the locked in stresses due to compaction of each layer result in a stress diagram like that shown in Figure 9.17c.

In the figure, the compaction stresses are developed to a depth dc where

d

KQ w

ca

=

1 2 /πγ

(9.18)

The roller load is Q which includes dead and dynamic loads from the roller, the roller width is w, γ is the unit weight of the backfill, and Ka is the active earth pressure coefficient.

From the surface to a depth zc, the stress increase is linear. The depth zc is given by

z K

Q wc a=

2 /πγ

(9.19)

Horizontal stresses ′σha in the range zc ≤ z ≤ dc are given by the formula

′ =σ γ

πhaQw

2

(9.20)


(b)(a)

Lateral pressure on walldue to vertical line load, Qℓ

(c)

PhPh

Ph

Line load, Qh

Qp

Point load, QpLine load, Qℓ

Qℓ

x = mH x = mH

x

σhσh σh

σh

σhθ

θ

x = mH

z = n

H

z = n

H

HH

Wal

l He

z

y py p

A

Section A–AFor m ≤ 0 .4 For m ≤ 0.4 H2

Qp

For m > 0.4For m > 0.4

+ tan−1 1

= =

0.64Qℓ

=0.28n2

1.77m2n2 1.28m2n (m2 + n2)3(m2 + n2)2

(1 + m2)2(1+ m2) m (1 – m2)

m

(0.16 + n2)3= 0.20n(0.16 + n2)2

φ′

σh

2 )= 2 1 −

Resultant Ph = Qh

= σh cos2 (1.1 θ)σhθ

σh

Ph

σh

= 0.69Qp Ph = 0.55 Qℓ

Ph = 0.48Qp Ph =

H

Qℓ

H

0

0.2

0.4

0.6

0.8

1.00 0.2 0.4 0.6 0.8 1.0

Valu

e of n

=

z H

(d)

Pressure distribution dueto vertical line load, Qℓ

(e)

Pressure distribution dueto vertical point load, Qp

0.50 1.0 2.01.5

0.4 0.55 0.610.5 0.51 0.560.6 0.47 0.520.7 0.43 0.49

0.4 0.69 0.580.5 0.65 0.520.6 0.59 0.47

Symbol m

He = x tan (45° +

He

HeQp

z((

(

(()

)

))

H2

Qpσhσh (( ))

)

Lateral pressure on walldue to vertical point load, Qp

Lateral pressure onwall due to horizontalline load, Qh

A

H2

QpValue of σh ( )H

QℓValue of σh ( )

PhQp H

ypSymbol m PhQℓ H

yp

Figure 9.16 Pressure distributions on retaining walls due to point and line loads. (After Geotechnical Engineering Office, Geoguide 1 2000. Guide to Retaining Wall Design, Civil Engineering Department, The Government of Hong Kong Special Administrative Region.)


For walls that are unyielding, the active pressure coefficient should be replaced by the at rest coefficient K0.

Other methods include those of Duncan and Seed (1986) that involves a simple hand calculation to find the stress distribution. Chen and Fang (2008) have measured compaction stresses in sands. At the top of the compacted zone, they found that the lateral pressures were close to the passive earth pressure, but below this, they approached the at rest (Jacky) K0 pressure.

9.5 SHEET PILE WALLS

Often, temporary walls are made from sheet piling driven into the soil and then the soil is excavated in front of the sheet piling. For unsupported sheet piles, the wall can flex and rotate forward as shown in Figure 9.18. The soil behind the wall will mobilise the active

(a)

Horizontal earth pressure distributionin uncompacted fill resulting fromcompaction of surface layer only

(b)

Horizontal earth pressure distributionresulting from successively compactedlayers of fill

Dep

th b

elow

surfa

ce, z

Dep

th b

elow

surfa

ce, zResultant pressure

distribution due tocompacting surfacelayer

Locus of Phmfor successivecompactedlayers

Phm

z c z c

Ph′

(c)

Design diagram for horizontal earthpressure induced by compaction

z ch c

Dep

th b

elow

surfa

ce, z

Ph′ = Kγz

Ph′

P ′hm = 2Q1γπ

zc = K 2Q1πγ

hc = 1K

2Q1πγ

′ Phm′

Phm′

′

Figure 9.17 Compaction stresses. (After Ingold, T.S. 1979. Géotechnique, Vol. 29, No. 3, pp. 265–283.)


earth pressure and the soil in front of the wall will mobilise part of the passive pressure. Below the point of rotation, the wall will move back into the soil mobilising a passive force on the back of the wall.

One simple method that can be used to analyse the wall is to assume that the passive pres-sure at the front of the wall is only partially mobilised by applying a factor of safety of F to the full passive force.

The following assumptions are then made:

1. Full active pressure acts on the back of the wall. 2. Only part of the full passive pressure is mobilised; F can be taken as 1.5–2.0 depend-

ing on the risk involved. 3. The passive pressure behind the wall can be considered to be a force P2 acting on

the base of the wall. By taking moments about the base of the wall, the depth of embedment of the wall can be found without having to know what the value of the force P2 is.

With reference to Figure 9.19, the active Pa and passive Pp forces acting on the wall are

P K H

Ha a= ( )γ

2 (9.21)

Active

Passive

Passive

Point of rotation

X

Figure 9.18 Sheet pile wall.

h

d

Pp/F Pa

XP2

H

Figure 9.19 Sheet pile wall showing assumed force P2 at base of wall X.


P K d

dp p= ( )γ

2 (9.22)

where d is the depth of embedment and H is the overall length of the sheet pile.Each of these forces acts at the centroid of the pressure diagram, therefore by taking

moments about the base of the wall, we have

K dF

d K H Hp aγ γ2 2

2 3 2 3

=

(9.23)

If a value of H = d + h is substituted (where h is the depth of retained soil), we have

d hd

KK F

p

a

+

=

3

(9.24)

This equation can be solved for the depth of embedment d required. However, because of the approximations made in the analysis, the sheet piling is usually driven an extra 20% further than the calculated value.

If there is likely to be a difference in water level from one side of the sheet pile wall to the other (say, because of tides), then the water forces should be included in the analysis, as the water forces can be considerable.

9.6 ANCHORED WALLS

Walls of various types can be held back by anchors whether they are sheet pile walls or diaphragm walls such as those discussed in Section 8.2.4 in Chapter 8. The anchor cable is attached at one end to the wall and may be attached at the other end to an anchor block or may be grouted into the soil or angled downward so that it reaches a firm stratum or rock where it may also be grouted.

Hand calculations can be carried out by taking moments about the point of application of the anchor (assuming one row of anchors is used). This is the ‘free earth’ method of design for the wall. An anchored sheet pile wall is shown in Figure 9.20 where it has been assumed that there is a difference in water levels between the front and back of the wall due to a time lag in the water draining from behind the wall (see Table 9.3).

By choosing values of the factor of safety F and the depth of the wall d, the negative and positive moments can be calculated and the required value of d is the one that produces equilibrium.

9.6.1 Anchors

The force that an anchor can supply can be found from the upper- and lower-bound theo-rems. For strip anchors with a vertical face, solutions have been found by Merifield and Sloan (2006) for sands. The pull-out force depends on how deep the anchor is buried as shown in Figure 9.21. This is assuming that the anchor has been placed far enough back from the wall so that it does not lie within the active wedge of soil that would arise during failure.


In the charts, H is the depth to the base of the strip anchor and B is the height of the anchor. The pull-out force per unit length Qu is given in terms of the break-out factor Nγ from the following equation:

Q HBNu = γ γ (9.25)

Values of the break-out factor Nγ are given in Figure 9.21a for the lower-bound solution and in Figure 9.21b for the upper-bound solution. The true solution lies between the two extremes; however, the lower-bound solution gives a conservative estimate of the pull-out force.

H

d Pp

Pa3

Pa1

Pa2 Pw2Pw1

dw2

dw1 da

h2h1

F

AA

γsubdγsubh1

γdw1γwh2 γwh1

Figure 9.20 Anchored sheet pile wall.

Table 9.3 Forces and lever arms for taking moments about the anchor (point A)

Force number Magnitude of force Lever arm about A Moment

1 P K dp p sub= 12

2γ l H d dp a= + −23

PF

lpp×

2 P K da a w1 121

2= γ l d da w a1 1

23

= − − Pa1 × la1

3 P K ha a sub2 121

2= γ l h d da w a2 1 1

23

= + −( ) − Pa2 × la2

4 Pa3 = Kaγdw1h1 l h d da w a3 1 112

= + −( ) − Pa3 × la3

5 P hw w1 121

2= γ l h d dw w a1 1 1

23

= + −( ) − Pw1 × lw1

6 P hw w2 221

2= γ l h d dw w a2 2 2

23

= + −( ) Pw2 × lw2

Note: h1 = H + d − dw1.


For line anchors in clay, Merifield et al. (2001) have provided the solutions given in Figure 9.22. The pull-out force per unit length Qu of the anchor is given by

Q B s N Hu u c a= +( )0 γ (9.26)

where the bearing coefficient Nc0 lies between the upper and lower bounds given in the fig-ure. The undrained shear strength su of the soil is assumed to be constant with depth in the plot of Figure 9.22. Ha is the depth to the centre of the anchor.

9.7 REINFORCED EARTH

Reinforced walls consist of facing panels connected to reinforcing strips or geofabrics that are incorporated into the backfill (Figure 9.23). The facing panels can be made of steel or

RoughBH

RoughBH

100

10

120 25 30 35 40

100

120 25 30

φ′35 40

Nγ

Nγ10

qu = γHNγ

qu = γHNγ

H/B = 1

H/B = 1

H/B = 10

H/B = 10

(a) Lower bound

(b) Upper boundφ′

Figure 9.21 Upper- and lower-bound solutions to the vertical anchor break-out factor Nγ for different angles of shearing resistance (sands).


concrete, and the reinforcing may be made from steel or polypropylene or geosynthetic meshes. The original method (Terre Armée) that was invented by Henri Vidal in the 1950s used steel strips, as these were less extensible than geosynthetics. Steel strips can be galvan-ised to increase their service life or can be ribbed to give higher pull-out resistance.

Design of reinforced earth walls requires considering the failure modes that may occur and these are shown in Figure 9.24. These modes are

a. Sliding of the block either at the base or part way up the wall b. Failing in bearing c. Rupture of the reinforcement d. Pull-out of the reinforcement e. Overall slip failure or failure through the reinforced block f. Excessive settlement or tilt

Upper bound (five-variable)

Rough B H

Rough B

H

Finite element upper bound

Finite element lower bound

Finite element lower bound

Finite element upper bound

Finite element (Rowe 1978)

Meyerhof (1973) Das et al. (1980) Ranjan and Arora (1980)

10

8

6

4

2

0

0 1 2 3 4 5 6 7 8 9 10

10

8

6

4

2

Nc0

Nc0

(a)

(b)

H/B

Figure 9.22 Bearing coefficients for strip anchor in clay: (a) comparison with numerical results; (b) compari-son with experimental results.


There are several design approaches available, but as the British (BS 8006-1:2010) and Australian (AS 4678, 2002) Standards and the Roads and Maritime Services of NSW Australia (RMS Specification R57 2012) use partial factor methods this approach will be used here. In this approach, the soil strengths are factored down by ‘material factors’ and the loads are factored up.

If the forces causing failure are still less than the resisting forces, then the design is acceptable. This approach is more complicated than the single factor of safety method, but has the advantage that different factors may be applied to different kinds of load and to frictional and cohesive forces. The various failure modes are addressed in the following sections.

Wall facing panels

Compacted soil

Reinforcing strips

Figure 9.23 Reinforced earth wall showing reinforcing strips.

Rupture

Settlement

Translation

Rotation

(c)(b)(a)

(d) (e) (f)∆

L

H PWv

Figure 9.24 Modes of failure for reinforced earth walls.


9.7.1 Sliding

The reinforced block may slide as shown in Figure 9.24a where sliding may be on soil-to-soil contact or on soil-to-reinforcement contact. The force causing the sliding are assumed to be the active earth pressure behind the wall, and the resisting forces are those due to shearing along the base of the sliding block.

There are two cases to consider: (1) long-term stability where the drained strength param-eters are used and (2) short-term stability where undrained strength parameters are used.

For long-term stability, where there is soil-to-soil contact, we have

f R R

fcf

Ls h v≤ ′ + ′tanφms ms

(9.27)

whereRv is the resultant of all factored vertical load componentsRh is the factored horizontal disturbing forcefms is the partial materials factor applied to the soil strengthfs is the partial factor against base slidingL is the effective base width for slidingc′ and tan ϕ′ are the drained strength parameters

For long-term stability where there is sliding on a reinforcement-to-soil contact, then Equation 9.28 is used.

f R R

fc

fLs h v

bc≤ ′ ′ + ′ ′α φ αtan

ms ms (9.28)

whereα′ is the interaction coefficient relating soil/reinforcement bond angle with tan ϕ′,α′bc is the adhesion coefficient relating soil cohesion to soil/reinforcement bond

For short-term stability where there is soil-to-soil contact at the base of the structure

f R

sf

Ls hu≤

ms (9.29)

and su is the undrained shear strength of the soil.For short-term stability where there is reinforcement-to-soil contact

f R

sf

Ls hbc u≤ ′αms

(9.30)

where there are reinforcing strips, the sliding will partly occur on the strips and partly on soil, so the modifying factor α′ needs to reflect this.

9.7.2 Bearing failure

The weight of the reinforced earth block plus the vertical force due to surface loads plus the effect of any horizontal loads and the active earth pressure behind the block will create a


non-uniform pressure distribution across the base of the block. The resultant force can be treated as a uniform pressure block centred at an eccentricity e from the centreline of the block. The length of the pressure block is then L − 2e as shown in Figure 9.25.

The pressure acting qr is then (BS 8006)

q

RL e

rv=

− 2 (9.31)

where Rv is the resultant of all factored vertical load components.The imposed bearing pressure should be less than the ultimate bearing pressure as shown

in Equation 9.32.

q

qf

Dr ≤ +ult

msmγ

(9.32)

whereqult is the net ultimate bearing pressure of the soil beneath the reinforced earth block

(see Section 5.3.1)Dm is the wall embedment depthγ is the unit weight of the foundation soil

9.7.3 Rupture of the reinforcement

The load carried by the reinforcement Tj is assumed to be due to the horizontal force acting on a section of the facing at the level hj of the reinforcing that has a vertical height of svj as shown in Figure 9.25. Generally, svj is the vertical spacing of the reinforcing.

i

3

2 1

Self-weight

Resultant loading acting on rear of wall

L

Li

Ti

hi

SL

FL w1w2

Hℓ

Figure 9.25 Loads acting on reinforced earth block.


The horizontal force per unit length on a strip j is therefore given by

T K

cf

K sj a vj a vj= ′ − ′

σ 2ms

(9.33)

where ′σvj is the factored vertical stress acting on the reinforcing at depth hj

′ =

−σvj

vj

j j

RL e2

(9.34)

Rvj is the factored vertical load acting on the jth layer of reinforcement. Forces due to other loads such as surface loads can be included in the force Tj, and the methods for calculating these are given in BS 8006.

The tensile strength of the reinforcing needs to be high enough to withstand the tensile force applied and so we can write

T

Tf

jD≤n

(9.35)

whereTD is the design strength of the reinforcementfn is the partial factor for the economic ramifications of failure

9.7.4 Pull-out of the reinforcement

In the active wedge method of design, the reinforcement is checked for pull-out failure for the length of reinforcing that is outside of the active failure wedge as shown in Figure 9.26. This length for a reinforcing strip j is Lej.

1

2

3

i H La Lei

L

Resistant zone

Active zone

45° + φ/2

Figure 9.26 Pull-out of reinforcing strips of length Le behind an active wedge.


If the perimeter of the reinforcing strip is Pj, then the bond strength around the outside of the strip consists of frictional and cohesive components, and the force required to pull the strip out Fj is given by

F

PLf f

f h f wc Lf

jj ej

j sbc ej= + ′ ′ + ′ ′

p n

fs fms

( ) tanγ α φα

(9.36)

whereffs is the partial factor applied to soil self-weightff is the partial factor applied to dead loads ws

fp is the partial factor for reinforcement pull-out resistancefn is the partial factor for the ramifications of failure

9.7.5 Overall slip failure

Slip failure through or beneath the reinforced soil block is a possible mode of failure, so limit equilibrium methods can be used to check the stability of the reinforced soil block. This is discussed in Section 9.8.1, where the effect of reinforcing is included in an analysis using the method of slices. Circular and non-circular failure surfaces can be trialled cutting through either the reinforced soil block or ground beneath it as shown in Figure 9.24e.

9.7.6 Excessive deformation

The settlement of the block of reinforced soil can be calculated by hand using the simple methods outlined in Section 5.5.2 or computed using finite element methods. Guidance is provided in BS 8006 for the acceptable deformations of retaining structures and these are reproduced in Table 9.4.

9.8 COMPUTER METHODS

Today, hand calculations based upon simple methods such as those listed in the previous sections are used as checks on values obtained from computer programs or for simple cases. For more complex cases where excavation is being carried out in stages and anchors are being placed as excavation proceeds, computer based methods are used.

One such commercial code WALLAP is especially designed for performing calculations of pressures on retaining structures and has the added advantage that it will carry out moment and shear force calculations for sheet pile walls.

Table 9.4 Usually accepted tolerances for faces of retaining walls and abutments (BS 8006)

Location of plane of structure Tolerance ± 50 mm

Verticality ± 5 mm per metre height(i.e. ±40 mm per 8 m)

Bulging (vertical) and bowing (horizontal)

± 20 mm in 4.5 m template

Steps at joints ± 10 mmAlignment along top (horizontal) ± 15 mm from reference alignment


Figure 9.27 shows a sheet pile wall that was used to hold back water in a river while con-struction was carried out on the dewatered side of the wall. A small berm of rockfill was provided on the dewatered side to assist with stability of the wall.

Figure 9.28 shows the bending moments and shear forces (per m run) in the sheet pile wall and the predicted deflection of the wall. The large deflection predicted at the top of the wall of 350 mm was confirmed by monitoring of the wall deflection.

9.8.1 Limit equilibrium methods

Both circular and non-circular slip surface methods can be used to calculate the Factor of Safety that retaining structures have against an overall failure. Such an analysis is shown in Figure 9.29 where it can be seen that soil consisting of several layers and under the action of a surface surcharge can be analysed. Anchors can be included in the analysis as shown in the figure.

200 100 0 100

–10.00

–3.00

Sand Sand

Monoman sandMonoman sand

Upper alluviumUpper alluvium

–4.00

1.904.907.75

10.00

Fill/alluviumFill/alluvium

13.20

Rockfill

RoughWater pressure (kN/m2)

Smooth

200

Figure 9.27 Schematic diagram of sheet pile wall in a river.

1000 0 –1000 –0.4000 0 0.400012

6

Elev.

0

–6

Stage No. 2 Excav. to elev. 10.00 on passive side

12

6

Elev.

0

–6

–200.0 0Shear force (kN/m run)

Bending moment (kN m/m run) Displacement (m)

200.0

Active GLPassive GL

Figure 9.28 Moments and shears in a sheet pile wall.


0138140142144146148150El

evat

ion

(m)

152154156158160162164166

Clayey silt (II)

Clayey silt (I)

Silty clay

Silty sand/sandy silt

168170172

Sand and silt (II)Sand and silt (I)Sandy silt/silty sand fillSilty clay fill

General surcharge : 12 kPaElv. 171.2

Elv. 161.0

10.2 m

Track surcharge :108 kPa

10 20 30 40 50 60 70

Figure 9.29 Non-circular slip surface analysis of an anchored wall (SLOPE/W).

Phase 4

Phase 5

A A

A(a)

(b)

A A A

A A

Figure 9.30 Design of anchored sheet pile walls using finite element methods: (a) first set of anchors placed; (b) second set of anchors placed.


The forces that are available in the reinforcing are applied as forces to the slice through which the anchor or reinforcement passes at its base. This involves calculating the length of reinforcement that lies outside the slip surface and calculating the pull-out force that is acting on this length.

Another way to apply anchor loads is to distribute the anchor forces to all of the slices that the anchor intersects. This gives a more stable solution and does not create large normal forces on the base of the slice through which the reinforcement passes.

Limit equilibrium methods can be applied to problems involving

1. Anchors 2. Soil nails 3. Steel reinforcing (reinforced earth) 4. Fabric reinforced soil

The commercial stability analysis codes SLOPE/W (2014) or STARES (2014) can be used to estimate the stability of walls held back or reinforced by structural elements such as those mentioned above.

9.8.2 Finite element methods

Retaining wall design can be performed using finite element methods and the analysis of excavation was examined in Section 8.8.1 in Chapter 8. The advantages of the finite element method is that the excavation sequence can be modelled with sheet pile walls being added and anchors being added as the excavation is deepened. Anchors can have a free length (shown as black in Figure 9.30) or be grouted (grey in Figure 9.30). Loads can be applied to the surface of the soil, and lowering of the water table can be simulated. The moment and shear force can be calculated in the wall and the axial force in the anchors computed.

363

Chapter 10

Soil improvement

10.1 INTRODUCTION

If the ground beneath a site is not suitable for the purpose for which the site is intended, the site may either have to be abandoned and a more suitable one found elsewhere, or the soil may be improved by some means. The cost of improvement and the benefits achieved will be critical in the decision to go ahead with soil improvement.

Generally, it is soft clays that pose most problems, but loose sands and dispersive soils are other examples of soils that can be modified. Sites that are contaminated or landfill sites are not considered in this chapter, but are examples of unsuitable sites that can be modified. Some examples of rehabilitation of contaminated sites are given in Section 11.9 of Chapter 11.

10.2 SOFT SOILS

Soft clays are generally those that have been recently deposited and are typically found on the flood plains of rivers, in estuaries, and in mangrove swamps.

If the soil has never been loaded to a greater stress than the current overburden stress (i.e. due to the self-weight of the soil), then the soil is said to be normally consolidated. If the soil has been subjected in the past to higher stresses than the current overburden stress, then the soil is said to be overconsolidated. The maximum past pressure that has been applied to the soil is called the pre-consolidation pressure ′pc. The London clay and Frankfurt clay are examples of overconsolidated soils, as they were once loaded by stresses from thick ice sheets that have since receded hence unloading the clay layer. As a result of being loaded to greater pressures in the past, overconsolidated clays are stiffer and have greater shear strength than normally consolidated clays.

Normally consolidated clays can have an overconsolidated crust and this is commonly found in soft clay deposits. The overconsolidation is caused by wetting and drying of the surface layer of the soil. The swelling and shrinking cause compression and then unload-ing of the soil, and this causes it to be overconsolidated. The strength profile for a soft clay deposit in Malaysia is shown in Figure 10.1. It may be seen from Figure 10.1, that the undrained shear strength su is higher at the surface, reduces to a minimum, and then increases linearly with depth, which is due to the crust being overconsolidated.

Soft normally consolidated clays are generally too soft for civil works and some kind of improvement needs to be applied. Some of these techniques are discussed in the following sections.


10.3 SURCHARGING AND WICK DRAINS

A common way to improve soft soil is to remove water from the soil by causing the soil to consolidate under the weight of a surcharge. During consolidation, the effective stress in the clay increases and therefore the shear strength and stiffness of the soil increases.

The water in the soil may be left to drain to the surface of the clay layer or to any pervious layer of underlying soil (such as a sand layer), or the consolidation process may be speeded up by providing wick or sand drains in the clay layer. This greatly reduces the distance that any water in the soil has to move to reach a drainage boundary, and this speeds up the con-solidation process. As the time for consolidation depends on a square power of the drainage distance, a reduction of 10 in the drainage path will speed up consolidation by 100 times.

10.3.1 Surcharging

Often an embankment on soft soil is constructed to a height greater than the final height so as to apply a surcharge to the soil and thereby speed up consolidation. The surcharge is often removed before complete consolidation of the soil has taken place and monitoring of the settlement or of pore pressures can be used as a guide as to when the surcharge can be removed. A method commonly used to estimate when primary consolidation is complete is Asaoka’s method (Asaoka 1978). In this approach, the settlement at a time t (St) is plotted against the settlement at a time t − Δt (St−1) as shown in Figure 10.2. The time increments Δt are taken to be equal and this may require some interpolation between recorded settlement points. Where the line through the field data intersects the 1:1 line gives the point at which primary consolidation is complete.

2

12

10

14

16

18

20

4

0

6

8 Lab test BH2Lab test BH1Field tests 1–9Legend

Vane shear strength su (kPa)

Dep

th (m

)

0 20 40 60 80

×

××

××

×

××

×

×

×

×

×

+

+

+

+

+

+

+

Figure 10.1 Undrained shear strength versus depth for a normally consolidated soil with an overconsoli-dated crust.

Soil improvement 365

For wide embankments, the settlement can be calculated using Terzaghi’s one-dimensional theory and this may be done by testing a sample of the soil in an oedometer. An oedometer contains a ring holding the soil, and the soil is loaded vertically between two porous plates. Because the sample is confined, it is compressed under one-dimensional conditions (see AS 1289.6.6.1 1998 or ASTM D2435-04 2004).

From the oedometer test, we can make a plot of the void ratio e versus the vertical effec-tive pressure applied p′ (usually plotted to a log scale). The pressure applied in the oedometer test once all pore pressures have dissipated is the effective pressure p′ and this is what causes the soil to deform. The change in height of the sample (or its settlement S) is given by

S

ee

z=+∆

1 0

δ

(10.1)

where δz is the sample height. This equation can be used to calculate the settlement of a layer of soil where the stress goes from A to B as shown in Figure 10.3.

The soil will compress at slope Cr on a plot of void ratio e versus log p′ (the slope of line A–P) until it reaches the pre-consolidation pressure ′pc after which the soil will compress more readily at a slope Cc (the slope of line P–B). This information can be used to predict the amount of settlement with embankment height assuming that all pore pressures are allowed to dissipate. The definitions of the compression index Cc and the recompression index Cr are (see Figure 10.3) given by Equation 10.2.

Ce ep p

epp

Ce ep

ri

c

c

ci

log loglog

log

= −′ − ′

= ′′

= −′ −

0

1 1

2

2

∆

lloglog

′= ′

′

p

eppc c

∆ 2

(10.2)

20

70

120

170

220

270

320

20 70 120 170 220 270 320St–1 (mm)

S t (m

m)

1:1 line

Field data

Figure 10.2 An Asaoka plot used to estimate completion of primary consolidation.


We can find the change in void ratio Δe from Equation 10.2 depending whether the changes in void ratio occur before or after the pre-consolidation pressure and calculate the settlement by substituting for the void ratio change in Equation 10.1. If the original overburden pressure in the ground at any depth is ′σv0 and the increase in stress due to an embankment is ∆ ′σv so that the final stress is ′σ f , the vertical settlement of a layer of soil δz thick is given by

S

Ce

zp C

ez

pr c

v

c

i

f

c

=+

′′

++

′′

1 10 0

δσ

δσ

log log

(10.3)

The above equation is based on the assumption that the pre-consolidation pressure lies between the initial stress and the final stress in the ground, that is, we are going from A to B in Figure 10.3.

If the pre-consolidation pressure is greater than the final stress (the soil remains overcon-solidated), we would only use the first part of the equation, that is,

S

Ce

zr f

v

=+

′′

1 0 0

δσσ

log

(10.4)

If the soil is normally consolidated so that there is no pre-consolidation, we would only use the second part of the equation, that is,

S

Ce

zc f

v

=+

′′

1 0 0

δσσ

log

(10.5)

The process in applying the one-dimensional consolidation theory to embankments is to divide the soil up into layers δz thick and calculate the initial and final effective stresses at the centre of each layer. Then the settlement of each layer can be found from either

Void

ratio

Pre-consolidationpressure

A

B

P

Unload–reload path

p′1

e2

ei

e0

p′2p′c log p′

Figure 10.3 Plot of void ratio versus effective stress.


Equation 10.1 or Equation 10.3, and the settlement of all layers added up to obtain the overall settlement.

The rebound on unloading (surcharge removal) can be calculated using Cr as the soil will unload along the unload path shown in Figure 10.3.

The rate of primary consolidation can be found for wide embankments by using one-dimensional theory. The degree of consolidation can be found by calculating the time factor Tv and using a plot such as that of Figure 5.26 to find the degree of consolidation Uv. The settlement at any time can be estimated from St = UvS∞ where the final settlement S∞ is cal-culated from either Equation 10.1 or Equation 10.3. More advanced numerical methods are discussed in Section 10.13.

Additional settlement can occur through creep of the soil, especially in soft clays. Allowance can be made for this by calculating the creep settlement St from (Mesri 1973)

S

Ce

ztt

t =+

α δ1 0 0

log

(10.6)

where Cα is the coefficient of secondary compression or creep, t is the time, and t0 is the starting time of the creep (often taken as 90% of primary consolidation). From the for-mula, the creep settlement is linear on a logarithm of time scale, as this is often observed for clays.

The coefficient of secondary compression is related to compression index Cc and so esti-mates of Cα can be made from the compression index as indicated by Mesri and Godlewski (1977) and shown in Table 10.1 (see also Figure 1.10 for direct measurement).

Once a surcharge has been removed, the clay beneath an embankment is partially overconsolidated, and so the creep rate reduces. The formula of Equation 10.7 may be used (Wong 2007, 2010) to estimate the value of the creep coefficient for the overconsoli-dated soil Cαε(OC) from the value for a normally consolidated soil Cαε(NC) where OCR is the overconsolidation ratio.

CC

me

mOC

NCOCR n

α

α

ε

ε

( )

( )( )= − +−1

1

(10.7)

and C C eNC NCα αε( ) ( )= +/ 01 .The quantities m and n need to be found from experiment, but typical values are m = 0.1

and n = 6. If the OCR is large, then the ratio of Equation 10.7 tends to m, therefore m is equal to the ratio Cr/Cc. A plot of experimental data is shown in Figure 10.4, along with values from the expression of Equation 10.7 where m = 0.05 and n = 6.

Table 10.1 Relationship of Cα and Cc

Material Cα/Cc

Granular soils including rockfill 0.02 ± 0.01Shale and mudstone 0.03 ± 0.01Inorganic clays and silts 0.04 ± 0.01Organic clays and silts 0.05 ± 0.01Peat and muskeg 0.06 ± 0.01


The time at which creep begins ts after removal of a surcharge to the time that the sur-charge is removed tR is found to be given by

log . .

tt

Rs

Rs

= ′( ) ±0 0208 100 0 156

(10.8)

and ′ = −R OCRs 1. The rebound due to unloading is shown in Figure 10.5.

10.3.2 Field observations

Generally, embankments on soft soils are instrumented when they are constructed so that the progress of consolidation can be monitored. Settlement plates are often placed at the cen-tre of the embankment, and piezometers placed along the centreline. Other instrumentation

OCRf

Ng (1998)Ladd (1989)Ballina Bypass (2007) Gateway upgrade (2007)

C α (O

C)/C

α (N

C)

∋∋

Exponential fit (m = 0.05 and n = 6)

1.0

0.8

0.6

0.4

0.2

0.01.0 1.2 1.4 1.6 1.8 2.0

Figure 10.4 Reduction of coefficient of secondary consolidation with OCR. (After Wong, P.K. 2010. Australian Geomechanics Society Seminar on Ground Improvement, Perth, WA.)

tP – End of primary consolidationtR – Remove surcharge

tS – Start of secondary consolidation

C1

1C ′

0.01 0.1 1 10 102 103 104

log (t/tp)

Ps = Σ(HiCα′ log t/ts)

∋

υ

Figure 10.5 Pre-loading to reduce post-construction settlement.


such as slope indicators is placed at the toe of the embankment to monitor lateral movement with depth.

For embankments with no drains, Tavenas and Leroueil (1980) have shown that the pore pressure response in a soft clay foundation is typically as shown in Figure 10.6. Initially, the clay is in an overconsolidated state, and the pore pressure increases along the line shown as B1. The pore pressure coefficient is defined as the increase in pore water pressure Δu divided by the increase in vertical principal stress Δσ1, that is,

B

u= ∆∆σ1

(10.9)

Once the soil reaches its pre-consolidation pressure, then it yields and the pore pressure rise is about equal to the increase in applied pressure from the embankment. This is shown as B2 1 0= . in Figure 10.6. At failure, the pore pressures begin to increase more rapidly than the increase in embankment pressure, and the value of Bf >1.

This illustrates the need to construct trial embankments to a height greater than the pre-consolidation pressure in the soil (usually >2 m high), or there will be no yield of the soil and the embankment response will not be the same as for the full height embankment. In addition, when modelling a clay foundation, it is necessary to use a numerical model that allows yield and failure such as the Cam Clay type models (see Section 3.10).

Tavenas and Leroueil (1980) give several formulae for observed behaviour of embank-ments on soft clay. The first is the threshold height of an embankment Hnc at which the clay beneath the embankment first begins to yield

γ

σ σH

I Bnc

p v= ′ − ′−

( )( )

0

11 (10.10)

Applied vertical stress

Exce

ss p

ore p

ress

ure Δ

u

F

P

Δue

Δσ′v threshold

B 1

Δσ′p − σ′v0

Δσv threshold Δσve = IγH

Δσv = IγH

−

B 2 = 1.0

−

B = =

1.0

− B f > 1

.0

−

ΔuΔσ v

Figure 10.6 Pore pressure response beneath an embankment.


In the above equation, I is an influence factor that gives the vertical stress at any depth from elasticity theory, γ is the unit weight of the embankment fill, ′σp is the pre-consolidation pressure of the clay and ′σv0, is the initial effective stress in the ground.

During the initial construction of an embankment, when the clay is still in an overcon-solidated state, Tavenas and Leroueil (1980) suggest the settlement can be calculated from

S

Dm

nc p v

p

=

′ − ′′

σ σσ

0

(10.11)

where Dnc is the depth of clay in a normally consolidated state and m is the modulus number of the soil (which is of the order of 25–80 in the intact Champlain clays that Tavenas and Leroueil investigated).

Beyond the threshold height, the change in settlement ΔS of an embankment is approximately

∆SH Hnc−

= ±0 07 0 03. .

(10.12)

For changes in lateral movements Δym of embankments with side slopes in the range of 1.5–2.5 horizontal to 1 vertical, the following formulae are suggested:

a.

b

Below the threshold height:

Above th

∆ ∆y sm = ±( . . )

.

0 18 0 09

ee threshold height: ∆ ∆y sm = ±( . . )0 91 0 02 (10.13)

10.3.3 Sand or prefabricated vertical (wick) drains

If consolidation beneath an embankment or other structure is likely to be too slow, then the consolidation can be speeded up by the use of sand or wick drains.

Wick drains (often called prefabricated vertical drains PVDs) consist of a plastic core (to allow drainage) covered by a geotextile filter fabric (see Figure 10.7). The wicks are pushed into the ground with a steel mandrel by a special crane, and are wound from a reel as they are pushed into the ground.

Sand drains are formed by drilling a hole in the clay and filling the hole with free draining sand. Water squeezed from the clay by any surcharge can flow into the sand drain and then

Geotextile filter

Corrugated plasticcore for drainage

L

t

Figure 10.7 Section through a wick drain (PVD).


to the surface. Stone columns, as discussed in Section 10.5 also act as drains and help speed up consolidation as well as strengthening the soil. A typical surcharge and subsurface drain layout is shown in Figure 10.8.

Where settlements are large (2 m or more is common in soft clays), the PVDs may become kinked or bent. Care must be taken that this is avoided as the flow of water from the soil will be cut off and the drains become ineffective.

Consolidation theory may be applied to the prediction of consolidation rates of soils drained by wick or sand drains. Each individual drain can be considered to collect water from a cylindrical volume of soil around itself, therefore the problem reduces to that of analysing a circular cylinder of soil with an impermeable outer surface but for which flow can occur into the central drain or to the top and bottom surfaces. Such a drain is shown in Figure 10.9 where a soil layer of depth 2H is drained by a well of radius rw. If the drain is of the shape shown in Figure 10.7, the equivalent diameter can be calculated as the average dimension, that is, dw = 2rw = (L + t)/2. Depending on whether the drains are laid out in a triangular pattern or in a square pattern, the cylindrical region which is drained by each well can be determined and its diameter is found. The diameter of the cylindrical region of soil is given by de, where

d

de

e

==

1 05

1 14

.

.

××

well spacing for a triangular pattern

well sspacing for a square pattern (10.14)

This is shown for an isolated single drain in Figure 10.9.The equation governing the radial and vertical flow may be written as

∂∂

= ∂∂

+ ∂∂

+ ∂∂

ut

cu

r rur

cu

zr v

2

2

2

2

1

(10.15)

where u is the excess pore pressure, r is the radial distance, z is the vertical distance, and cr and cr are the coefficients of consolidation in the radial and vertical directions, respectively.

Wick or sand drain

Permanent fill

Settlement plateTemporary surcharge fill

Clay soil

Waterdrainagepattern

Piezometer gauges

Stability berms(if required)

Firm soil or rock

Sand drainage blanket

Piezometers

Figure 10.8 Typical wick drain (PVD) layout beneath an embankment.


Equation 10.15 may be solved in two parts; firstly for the radial flow (with cv = 0), and secondly for the vertical flow (with cr = 0). The vertical flow solution is merely the Terzaghi one-dimensional consolidation solution.

If the degree of consolidation for the radial consolidation is Ur and for the purely verti-cal flow is Uv, then the degree of consolidation when both vertical and horizontal flow are operative Urv is given by Carillo (1942) as

( ) ( )( )1 1 1− = − −U U Urv r v (10.16)

Plots that can be used to compute the degrees of consolidation for both radial and vertical flow are shown in Figure 10.10. The broken curve is Terzaghi’s one-dimensional solution for vertical flow only and is used with time factor Tv. The solid curves are used for the degree of radial consolidation Ur and are used with time factor Tr. The time factors are defined as

Tc tH

Tc tr

vv

rr

e

=

=

2

24

(10.17)

For the vertical time factor, the full soil depth is taken as 2H (two-way drainage) or H (one-way drainage) and t is time. For the radial time factor, re = de/2 is the effective radius.

WellSpacing Definitions

For triangular patternde = 1.05 (well spacing)For square patternde = 1.14 (well spacing)de = Effective diameter

of sand drain

rw = 1.5 ′

de

kv

kh

de

ks

Drain Well

Smear zone

No flow acrossouter surface

Section a-a

Flowpaths

Drain wells intrianglar pattern

kh = Horizontal permeabilityks = Shear zone permeability

n =dedw

=rerw

=Effective radiusRadius of drain

s =rsrw

= Radius of shear zoneRadius of drain

Example: To determine equivalentradius of drain without smear whoseeffect is equal to the actual drainwith smear

aa

rs = 1.8 ′re = ~7.5 ′

Actual sand drain

Estimated s = 1.2Determine neq from Figure 10.11

n = 5 kh/ks = 7

neq = 15 rw = 7.515

Equivalent sand drain no smear

rw = 0.5 ′re = 7.5 ′

Dra

in w

ell

Dra

in w

ell

h

rw

rsre

2H

= 0.5 ′

Figure 10.9 Allowance for smear effect in a sand drain (or PVD) design.


Several curves are provided for the radial flow solution, and these have been obtained for different values of the parameter n where

n

dd

rr

e

w

e

w

= =

(10.18)

and rw is the radius of the well. By knowing the radial and vertical time factors, the degrees of radial and vertical consolidation may be found from Figure 10.10, and then the degree of consolidation for both radial and vertical flow are found from Equation 10.16.

The above theory applied to problems involving wick or sand drains has sometimes been found to give inaccurate predictions of consolidation rates. This is due to the fact that when a hole is drilled or a wick drain is pushed into the soil, the sides of the hole are remoulded or ‘smeared’. The smeared zone has a lower permeability than the surrounding soil and slows down the flow of water into the drain.

The procedure for designing a drain layout considering smear can be summarised as follows:

1. The time factor Tv for vertical drainage after a time t is calculated, and the degree of consolidation for vertical drainage Uv found from the broken line on Figure 10.10.

2. If the degree of consolidation Uv without drains is greater than that needed, then drains are not needed. However, if Uv is less than the required degree of consolidation Ureq, then drains will be needed as vertical drainage to the upper (and perhaps lower) drainage boundaries will not provide adequate pore pressure dissipation.

3. The required degree of consolidation from radial drainage Ur may be calculated from Equation 10.16 as

U

UU

rv

= −−−

111

( )( )

req

(10.19)

0.004 0.01 0.04 0.10 0.40 1Time factor Tv, Tr

0

10

20

30

40

50

60

70

80

90

100

Exce

ss p

ress

ure,

u v, u r (

%)

0

10

20

30

40

50

60

70

80

90

100

Cons

olid

atio

n, U

v, U

r (%)

––

(a) Vertical flow(b) Radial flow

Values of n100

4010

5

Figure 10.10 Rate of consolidation for vertical and radial drainage by a sand drain (or PVD).


4. A drain diameter dw (and hence rw) is chosen and a drain spacing is selected. From Equation 10.14, the radius of influence may be found from knowledge of the drain layout pattern (see also Figure 10.9). Hence, the ratio n = re/rw may be calculated.

5. From the estimated values of s = rs/rw and kh/ks, an equivalent value of n called neq can be found by using Figure 10.11.

6. For this value of neq and Tr, the value of Ur can be found from Figure 10.10. The effect of using an equivalent n value instead of the actual n value with Figure 10.10 is to slow down the consolidation rate.

7. The procedures in Steps 4–6 can be repeated for different drain spacings (and drain diameters if desired) and a plot made of Ur versus drain spacing can be made.

8. From this plot (see Figure 10.12), the spacing necessary to achieve the required degree of consolidation (as calculated in Step 3) can be obtained.

1 2 3 4 5 7 10 20 30 4050 70 100Equivalent neq = n for drain with no smear

1

2

3457Ac

tual

n

200 400 700

10

20

30405070

100

Legends = 1.2s = 1.5s = 2.0

Relation of actual and equivalent n values

Example

s = 1

2 2 4 2 4 8 4 12 8 8

412 8

(kh/ks)

2 2 4 2 4 8

(kh/ks)

Figure 10.11 Allowance for smear effects in a sand drain (or PVD) design.

eoretical curve forchosen drain diameterRequired

spacing

Drain spacing s

Value required toachieve overalldegree ofconsolidation of Usp

Radi

al d

egre

e of

cons

olid

atio

n af

ter t

ime t

0

100

Figure 10.12 Finding the required drain spacing.


The permeability ratio can be found by performing oedometer tests on undisturbed soil (kh) and on remoulded soil (ks), and calculating the permeabilities from the tests. Vertical cv and horizontal ch coefficients of consolidation can be found by performing oedometer tests with the clay sample oriented in the appropriate direction.

10.3.4 Vacuum consolidation

Consolidation using PVD systems can be accelerated by applying a vacuum to the drains so as to create a greater pressure difference between the water in the drains and the clay. A typical vacuum setup is shown in Figure 10.13.

This system was trialled near Ballina in NSW (Kelly et al. 2008) where a vacuum of 80 kPa was applied to the PVD system in soft clay. The top 0.5 m of clay had an undrained shear strength of 5 kPa and below 0.5 m it increased at 1 kPa/m depth. An impervious mem-brane was used to seal the system of drains as shown in Figure 10.13.

As a result of the trial, it was found that embankments constructed using a vacuum con-solidation system were able to be built more rapidly and to greater heights than embank-ments constructed using standard surcharge filling methods because the vacuum pressure enhances the stability of the embankment.

10.4 VIBROFLOTATION

Loose sands are very suitable for densification through vibration or shock, since the sand particles will rearrange and form a more compact structure if vibrational energy is applied. Sands that have a low relative density will liquefy on being vibrated by earthquake, and therefore loose sand deposits are not suitable for foundations or as fill for wharf structures or tailings dam embankments wherever there is any risk of an earthquake.

Vibroflotation is a process that involves improving granular soils through the application of vibration, and was first established in Europe in the 1930s. The method involves push-ing the vibroflot into the ground with the aid of water jets and vibration. Loose granular materials such as sands and gravels can be densified with the vibration, and so their bearing capacity can be improved and they become less compressible.

Vertical vacuumtransmissionpipes

Horizontaldrains

Draininglayer

Surchargefill

Atmospheric pressureImperviousmembrane

Vacuum airwater pump

WatertreatmentstationPeripheral trenches filled

with bentonite andpolyacrylate

Vacuum gasphase booster

Airflow

Figure 10.13 Schematic diagram of vacuum consolidation.


The vibro-compaction process is shown in Figure 10.14a, where the vibroflot is advanced into the soil using water jets that emerge from the end of the device. When the vibroflot is raised vibration and the injection of water help compact the granular material around the vibroflot probe. In addition, extra soil is added to the hole to compensate for the compacted material.

The process of vibroflotation has also been adapted to clayey or cohesive soils that can-not be compacted by vibration. It involves replacing the cohesive material with a column of granular material that can be compacted in place by vibration. This process is generally referred to as vibro-replacement.

10.5 VIBRO-REPLACEMENT

Replacement of soft clays or silts with stronger, free draining materials can be used where it is necessary to reduce settlements under structures or embankments. The process can also

1 2 3

Feedof soil

material

Compactedcolumn

(b)

Vibro-replacement method

1 2 3

Feedof

soil

Waterfeed

Waterfeed

Waterfeed

Compactedzone

(a)

Vibro-compaction method

Figure 10.14 The vibroflotation process: (a) vibro-compaction; (b) vibro-replacement.


be used where it is necessary to speed up consolidation as well as strengthen the soil as the granular columns of material act as drains. A further use is for reducing the liquefaction potential of soils subject to seismic vibration. In this case, the granular column of material also acts as a drain and dissipates the pore pressures in the soil that can lead to liquefaction.

This section is mainly concerned with vibro-replacement processes which can be used for constructing stone or sand columns in soft clays and silts.

Figure 10.14b shows the vibro-replacement technique. The probe of the vibroflot is again advanced into the soft soil by vibration and water jets, however as it is withdrawn, granular material is introduced into the hole. The granular material can be coarse gravel, or crushed rock or slag generally graded from 20 to 80 mm. The diameter of the column formed depends on the properties of the clay. Larger diameter columns are formed in softer clays and smaller diameter columns in stiffer clays. Typical column sizes are from 0.5 to 1.0 m.

Another alternative to the crane hung vibrator is the Bullivant method that is used in the United Kingdom. The probe of the vibrator is tapered and the vibration is applied vertically. As well, a pull-down force is applied to the probe from winches on the probe mast. This method can achieve better densities in the material around the probe and can avoid the need to use pre-boring in stiffer soils.

Design of stone or sand columns requires both settlement and bearing capacity calcula-tions to be carried out in a similar fashion to a pile or surface foundation. These two aspects are examined in the following sections.

10.5.1 Bearing capacity analysis

The bearing capacity of stone or sand columns can be carried out individually or for the group. It may be necessary to compute the individual carrying capacity if only a few col-umns are used under a foundation or the single column capacity is used in formulae for the group capacity.

10.5.1.1 Bearing capacity of single columns

The load carrying capacity of stone columns is different to that of conventional steel or con-crete piles. This is because they compress and bulge under applied loads much more than a conventional pile. The column may undergo relatively large lateral distortion to a depth of four diameters under the ultimate load, and therefore an important component of the load carrying capacity is the maximum resistance that the soil can provide in the radial direction.

Greenwood (1970) assumed that the lateral resistance from the soil would create a triaxial stress system within the column, and that the lateral pressure would be the passive pres-sure of the soil (at failure). A more sophisticated design method is to use cavity expansion theory to obtain the lateral pressures on the column. If the lateral expansion of the column is idealised as the expansion of a cylinder in an elastic–perfectly plastic material, then the limiting radial stress σRL can be obtained from the theory of Gibson and Anderson (1961) for clay soils.

σ σRL R u e

u

sGs

= + +

0 1 log

(10.20)

whereσR0 = total in situ lateral stressG = shear modulus of the clay =E′/2(1 + ν′) = Eu/2(1 + νu)


E′, Eu = drained, undrained modulus of the claysu = undrained shear strength of the clayν′, νu = drained, undrained Poisson’s ratio for the clay

The results of quick pressuremeter tests (undrained tests) show that Equation 10.20 can be approximated by

σ σRL R uu s= ′ + +0 4 (10.21)

where′σR0 = effective in situ lateral stress

u = pore water pressure

If the gravel in the bulged zone has yielded, then from the Mohr–Coulomb failure criterion

′ = + ′

− ′

′σ φφ

σv RL11

sinsin

(10.22)

where′σv = vertical effective stress in the column

ϕ′ = angle of internal friction of the column′σRL = lateral effective stress

The ultimate load that the column can carry is then given by

′ = + ′

− ′

+ −σ φφ

σv R us u11

40sinsin

( )

(10.23)

Equation 10.23 relies on the assumption that the load is applied to the column only, whereas it may be applied to both the column and the surface of the soil. There is also the possibility that the column may fail like a pile, and in this case, the usual pile bearing capacity formula can be used. For clay, the ultimate base pressure may be taken as 9su and the ultimate shaft load computed using the full undrained shear strength of the soil.

This may be used to design a column so that it will fail in bearing and bulging at about the same load. This is not so straightforward if the shear strength of the clay varies with depth and a choice for the value of su for use in Equation 10.23 must be made.

Dunbavan and Carter (1994) noted that the radial effective stress ultimately increases by approximately four times the original undrained shear strength of the clay only under fully drained conditions. Research on driven piles (Carter et al. 1979) has shown that in the short term, the limiting value of the radial effective stress after cavity expansion is increased by approximately 2.5–3 times the undrained shear strength of the soil. Francescon (1983) has validated these predictions for normally and lightly overconsolidated inorganic clay. As some consolidation will take place during loading, the value is probably somewhere between 2.5su and 4su, therefore for the undrained case, Equation 10.21 with a value of 2.5su (replac-ing the value 4su) should be used.

′ = + ′

− ′

′ +σ φφ

σv R us11

2 50sinsin

( . )

(10.24)


Measured data by Bergado and Lam (1987) is shown in Figure 10.15 along with predic-tions of other investigators. The measured values are reasonably close to the values predicted by Equation 10.24 if the lateral soil effective stress ′σR0 is close to zero (as it would be near the surface where bulging would occur).

Thornburn (1975) has presented an empirical design method initially recommended by Thornburn and MacVicar in which the total building loads are supported entirely by the stone columns. Thornburn considers such an approach ensures adequate factors of safety against a bearing capacity failure and provides the ground with considerable stiffness. In Figure 10.16, the relationship between the allowable working load recommended for preliminary design, and the undrained shear strength of the cohesive soil is reproduced. This relationship was obtained from the Rankine theory of passive earth pressure modified for radial deformation and from correlating field measurements of the average diameters of stone columns with the undrained shear strengths of the soils in which they were constructed. This correlation was established using performance data from stone columns formed by Cementation and Keller vibrators.

Allowable stress that can be placed on stone columns is shown in Figure 10.17 together with the values derived from the theory of Hughes and Withers (1974). The ultimate load values as computed from Hughes and Wither’s theory have been divided by 3 to compute the allowable stress levels shown in the plot.

30

25

20

15

10

5

035° 45°40°

Bell (1915)

Greenwood (1970)

Hughes et al. (1975)

Hughes and Withers (1974)

Vesic (1972) Gibson andAnderson (1961)

Internal friction angle of granular material (ϕ′)

Field data

su

σv

ϕ′

σ v/s u

Figure 10.15 Ultimate load carried by a single stone column. (From Bergado, D.T. and Lam, F.L.L. 1987. Soils and Foundations, Vol. 27, No. 1, pp. 86–93.)


The methods described above for determining the load carrying capacity of a single isolated stone column should be sufficiently accurate for a wide variety of sites (Hughes et al. 1975). However, the predictions made using these methods will become less reliable if the cohesive soil deposit is not uniform, or the surface loading is actually applied to the surface of the soil.

10.5.1.2 Bearing capacity of column groups

The bearing capacity of a group of stone columns will be different to that of a single column just as the bearing capacity of a group of piles is different to that of a single pile.

10 20 30 40 50 60Undrained shear strength su (kN/m)2

50

75

100

125

150

Allo

wab

le lo

ad o

n st

one c

olum

n (k

N)

Allowable loadEffective diameter

1.0

0.9

0.8

0.7

0.6

Effe

ctiv

e dia

met

er o

f sto

ne co

lum

n (m

)

Figure 10.16 Allowable load on a stone column. (After Thornburn, S. 1975. Géotechnique, Vol. 25, No. 1, pp. 83–94.)

0 25 50 75Undrained shear strength su (kN/m)2

0

100

200

300

400

500

600

Allo

wab

le st

ress

on

ston

e col

umn

(kN

/m)2

orburn (1968)Hughes and Withers(1974)

Stone column

qA =

Circular footing5.14 × 1.3su

3

Figure 10.17 Allowable stress on stone columns.


Four modes of failure are possible for a group of stone columns:

1. A general bearing capacity failure similar to that which occurs for a conventional foot-ing placed on the surface of a deep clay layer.

2. Squeezing of the clay between the stone columns, that is, between the surface of the soil and the base of the layer.

3. A sliding failure through the material applying the load (e.g. an embankment or stock-pile) and through the stabilised clay.

4. An end bearing failure with the columns pushing into the clay and behaving as stiff piles.

10.5.1.2.1 Failure mode (1)

Bearing capacity factors are needed for a loading on a soil layer of finite depth, and such factors have been produced by Mandel and Salençon (1969). Generally, the layer thickness has to be taken into account as the loading on the stone columns is of large extent compared to the thickness of the soil layer.

The bearing capacity factors are a function of B/H where

B = footing widthH = thickness of soil layer

The bearing capacity coefficients are presented in Tables 5.3 through 5.5 in Chapter 5 on ‘Shallow Foundations’.

10.5.1.2.1.1 DRAINED LOADING

For long-term or slow-loading cases, the drained friction angle of the clay needs to be used. As the proportion of the area taken up by the stone columns is small and because the angle of friction of the column material is usually close to that of the soil (typically friction angles for the clay are 30° and that of the column material is 40°), the bearing capacity can be calculated using the angle of friction of the clay, and treating the column-reinforced soil as a uniform material.

10.5.1.2.1.2 UNDRAINED LOADING

A simple method of calculating the bearing capacity in undrained clay is to use the formula

σ α σ α σult = − +( )1 s c s s (10.25)

whereαs = cross-sectional area of stone/gross cross-sectional area.σc = maximum vertical stress that the clay can support. This corresponds to the value

calculated using bearing capacity theory for a uniform layer of clay without stone columns.

σs = maximum stress that the stone column can support. This corresponds to a bulging failure of the column.

It is recommended that Equation 10.24 be used when calculating the value of σs.



Where the width of the load to the depth of the clay layer is large, the clay can squeeze out sideways past the stone columns. In this case, the ultimate load that can be applied can be estimated from the same formula as given in Equation 10.25 but where in this case, the value of σc can be computed from the solution for a layer of cohesive material being squeezed between two rough rigid plates for which

σ πc us

wh

≈ + +

4

(10.26)

wheresu = undrained shear strength of the clayw = the width of the loaded areah = the thickness of the clay layer

The maximum vertical stress that the columns can support is calculated using Equation 10.24 that uses a value of 2.5su for the maximum lateral restraint provided by the clay.


The stability of the stone columns can be assessed by carrying out a slip circle analysis; however, care needs to be taken in doing so. This is because the stone columns are stiffer than the surrounding clay and so they carry a greater normal stress. This makes the columns stronger in resisting shearing, as the shear strength is proportional to the normal stress in a frictional material.

An equivalent cohesion and angle of friction may be used in the slip circle analysis (Priebe 1995) where the cohesion and angle of friction of the composite material ceq and ϕeq are given by Equations 10.27 and 10.28.

c cs ceq = −( )1 α

(10.27)

tan tan ( )tanφ φ φeq s cm m= + −1

(10.28)

and cc is the cohesion of the clay, ϕs and ϕc are the angles of shearing resistance of the stone column and the clay, respectively, and

αs

s

s c

sAA A

AA

=+

=

(10.29)

where As is the cross-sectional area of the stone column, A is the total area of a unit cell, and αs is the area ratio (area of stone columns to total area).

The stress carried by the stone columns is higher than just the stress from an embankment as the stress is concentrated into the stiffer stone columns. Priebe (1995) has suggested using Equation 10.28 in any slip circle analysis as it takes account of the stress concentration. Values of m are provided in the plot of Figure 10.18. Priebe (1995) recommends using the full lines in this plot rather than the broken lines as they allow for bulging of the columns.


He also suggests using a reduced equivalent cohesion (Equation 10.30) to allow for damage to the soil structure where m is taken from the full lines in Figure 10.18.

c m cceq = −( )1

(10.30)


This failure mode considers the column to behave as a stiff pile. Failure will occur if the load on the column is equal to or greater than the ultimate load that the column can carry. The ultimate load that the column can carry can be computed as for a pile, that is, the sum of the side friction force on the column (computed using the full shear strength su) and the base resistance = 9subase Abase.

10.5.2 Settlement analysis of column groups

When the load is applied to the stabilised soil deposit, the clay will deform under undrained conditions, while the stone columns remain drained. The deformation of the column group depends upon whether the load can be considered to be rigid or flexible.

10.5.2.1 Flexible foundation

Finite element analyses of stone columns using representative properties for the clay and the stone column (Balaam and Poulos 1978) have shown that elastic solutions for the settlement of the column is sufficiently accurate if the applied load q satisfies

q

h≤ γ2

(10.31)

whereγ = the unit weight of the clayh = the depth of the clay layer

Area ratio A/Ac

Prop

ortio

nal l

oad

m

Dashed lines:m = (n–1+Ac/A)/n

Solid lines:m = (n–1)/n

ϕc = 45°

1 2 3 4 5 6 7 8 9 100.0

0.2

0.4

0.6

0.8

1.0

ϕc = 42.5°ϕc = 40°

ϕc = 37.5°ϕc = 35°

vs = 1/3

Figure 10.18 Proportional load carried by stone columns. (Priebe, H.J. 1995. Ground Engineering, Vol. 28, No. 10, pp. 31–37. Reprinted with permission from Ground Engineering.)


Results for elastic solutions are shown in Figure 10.19. In this plot, the settlement ratio is defined as the ratio of the maximum settlement occurring if a column is present to the settlement that would occur if there were no pile.

10.5.2.2 Rigid foundation

A column-clay unit such as that shown in Figure 10.20 can be used to compute the settle-ment of the stone column group. A solution can be obtained by assuming that the unit remains in an elastic state (Balaam and Booker 1981).

Shown in Figure 10.21 is the reduction in settlement with respect to the spacing of the stone columns a/b. The reduction in settlement is given by εz/qAmvs which is equal to the strain in the column-clay unit divided by the strain in the layer of clay without any column.

It can be seen that there is a rapid reduction in settlement up to an a/b ratio of about 0.4 for all ratios of the column to clay stiffness ratios E Ep s/ .′

If the stone column yields (with the clay remaining plastic), then the settlement of the col-umns can be computed by treating the stone column as an elastic–perfectly plastic material. Correction factors that can be applied to the elastic solutions are presented in Figures 10.22 and 10.23. These correction factors are only for certain cases where de/d ≤ 3 and the angle of dilation of the granular material is zero (de is the equivalent diameter; see Figure 10.9).

10.6 COLUMN-SUPPORTED EMBANKMENTS

Stiff columns can be used to support embankments that are to be constructed in areas of soft clay. The columns can be installed to a depth where there is soil of reasonable strength, and then a layer of geotextile reinforced granular material placed over the heads of the col-umns to allow them to support the embankment without punching through the fill. Several layers of geosynthetic reinforcement can be used and such a foundation system is shown in Figure 10.24.

0 0.2 0.4 0.6 0.8 1.0Settlement ratio

1

10

100

h/d

de/d = 2 2 5 5 10 10

de

d

L hEp/Es′ = 20

L/h = 1.0L/h = 0.5

Figure 10.19 Settlement ratio from finite element analysis.


There are various ways in which such a column-supported embankment can fail and may include:

a. A slope failure at the edge of the embankment b. Lateral sliding c. Failure of the supporting columns d. Excess strain in the geosynthetics e. Excess settlement

ab

Smooth rigid

Smooth rigid

Outer boundarysmooth rigid

Uniform pressure qA

r

Stone column ClayEpνp

Esνs

Figure 10.20 A column-clay unit.

0 0.2 0.4 0.6 0.8 1.0a/b

0

0.2

0.4

0.6

0.8

1.0

∋

q Amvs

z

νp′ = 0.3νp = 0.3

mvs =(1 + νs′)(1 – 2 νs′)

Es′ (1 – νs′)

Ep/Es′10

2030

40

Figure 10.21 Vertical strain of column-soil unit for various spacings.


The British code BS 8006 and the FHWA (Federal Highway Administration in the United States) require that the column support goes within a distance Lp of the toe of the embank-ment as shown in Figure 10.25. This is to ensure that the edge of the embankment is stable, and this distance can be calculated from Equation 10.32.

L H np p= −[ tan( )]θ

(10.32)

0 1 2 3 4 5qA

0

0.2

0.4

0.6

0.8

1.0

δ elas

Ep/Es′= 40

δ

γh

30

20

10

de

h

d

qA

Figure 10.22 Correction factors for elastic settlement (de/d) = 2, ϕ = 40°, ψ = 0, νs = 0.3.

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1.0

Ep/Es′ = 40

qA

δ elas δ

γh

30

20

10

de

h

d

qA

Figure 10.23 Correction factors for elastic settlement (de/d) = 3, ϕ = 30°, ψ = 0, νs = 0.3.


wheren is the side slope of the embankment (n horizontal to 1 vertical)

′φemb is the angle of shearing resistance of the embankment material

θ φp = − ′

452emb is the angle shown in Figure 10.25

Lateral sliding or spreading is another way that the embankment could fail as shown in Figure 10.26. Both the British Standard BS 8006 and the FHWA give simple calculation methods for this case. The required tensile force that is needed in the reinforcement to pre-vent lateral spreading Tds is given by

T K H

Hwds a s= +

γ2

(10.33)

Embankment

GeosyntheticsGeosynthetic load transfer platform

Column caps

Vertical columns

Firm soil or bedrock

Figure 10.24 Column-supported embankment with geosynthetic reinforcement.

EmbankmentLs

Lp

θp

1n

Fill: ϕ′cv

H

Pile cap

Pile

Figure 10.25 Edge failure of a column-supported embankment.


whereKa is the active earth pressure coefficient = − ′( )tan /emb

2 45 2° φws is any surcharge on top of the embankmentH is the height of the embankment

It should be pointed out, however, that the geotextiles do not necessarily need to be designed to take this force as it may lead to a conservative design.

The minimum length of reinforcement required to stop the embankment sliding sideways across the geotextile Le (see Figure 10.26) is given by Equation 10.34.

L

TH

eds=

0 5. tanγ µ φemb (10.34)

where μ is the coefficient of friction for sliding of the embankment material against the geotextile.

The design of the load transfer platform can be done several ways, including the Collin method (Collin 2004, Collin et al. 2005), the tension membrane theory (BS 8006-1:2010) and the enhanced arching method (Guido et al. 1987).

10.6.1 Collin beam method

In the Collin method, the assumption is made that the thickness h of the load transfer platform is greater than one-half of the clear span between the columns s − d where s is the centre-to-centre spacing of the columns and d is the column diameter, and the distance between the layers of reinforcement is a minimum of 150 mm (6 inches). The method also assumes that there are at least three layers of geotextile used, and that the angle of the region of arching is 45° (see Figure 10.27).

The purpose of the reinforcing fabric is to provide lateral confinement in the select fill layer at the base of the embankment and to facilitate arching. This means that the vertical load from the embankment will arch onto the columns. The secondary function of the geo-textile is to support the weight of fill below the arch.

The weight of fill that each layer of reinforcing is required to carry is the weight of fill between that reinforcing layer and the layer above. Hence, the ‘uniform pressure’ WTn on layer n is given by

Soft foundation

Outward shear stress

Embankment

Reinforcement

Surcharge ws

Pile caps

Piles

Tds

PfillLe

Lb

Lp

H

Fill: γ, ϕ′cv

Figure 10.26 Sliding of an embankment on the geotextile.


W A A h ATn n n n n= + +[ ] /1 2γ (10.35)

where the height of each layer is hn and the unit weight of the select fill is γ. An is the area of reinforcement layer n, and this will be different for different column layouts (square or triangular) as shown in Figure 10.27.

The tension in the reinforcement Trpn at layer n is calculated from the applied pressure WTn, and the design span for the membrane D.

T W Drpn Tn= Ω /2

(10.36)

The design span for a tensioned membrane D is the diagonal length of a square (=1.41*side length) or the mid-side to corner distance for a triangle (=0.866*side length). The term Ω is a factor that comes from membrane theory and is a function of the strain in the membrane as given in Table 10.2.

Because the strain in the membrane is not known, the tension can be found from Equation 10.36 and plotted against a tension–strain diagram from the manufacturer. Where the two curves cross gives the strain to use in the calculation.

Square column spacing Triangular column spacing

For n = 2, 3, 4, etc.

Geogrid 3

Geogrid 2

Geogrid 1

Embankment fillWell graded granular fill h4

h3

L1 = (s – d)Angle of arching45° 45°

Ln = (s − d) − 2 hi/tan 45°i=1

n=1∑

(s – d) = Length between pile caps, L1 (s – d) = Length between pile caps, L1

(s – d)(s – d) (s – d)

h2

h1L2

L3

L4 h

Figure 10.27 Collin beam method of design for membranes. (Adapted from Collin, J.G. 2004. Proceedings 52nd Annual Geotechnical Conference, University of Minnesota, Mineapolis, MN, February 27, 2004.)


10.6.2 BS 8006 method

The method suggested for calculating the tension in the reinforcing in BS 8006-1:2010 is based on the tension membrane theory. First, the pile cap stress ′pc to embankment stress ′σv is calculated from

′′

=

p C aH

c

v

c

σ

2

(10.37)

where a is the size of the cap, H is the embankment height and Cc is an arching coefficient that can be found from Table 10.3.

The distributed load WT carried by the reinforcement can be found from the following equations where s is the centre-to-centre spacing of the piles.

For H > 1.4(s − a)

W

sf s as a

s ap

Tfs c

v

=−

−− ′

′

1 42 2

2 2. ( )γσ

(10.38)

For 0.7(s − a) ≤ H ≤ 1.4(s − a)

W

s f H f ws a

s ap

Tfs q s c

v

=−

−− ′

′

( )γσ2 2

2 2

(10.39)

But if sa

pc

v

2

2 ≤ ′′σ

WT = 0 (10.40)

Table 10.2 Values of Ω

Ω Reinforcement strain ε%

2.07 11.47 21.23 31.08 40.97 5

Table 10.3 Values for the arching coefficient, Cc

Pile arrangement Arching coefficient

End bearing piles (unyielding). .C

Ha

c = −1 95 0 18

Friction and other piles (normal) . .C

Ha

c = −1 5 0 07


The partial load factors used in BS 8006, ffs and fq can both be taken as 1 for the service-ability limit state and 1.3 for the ultimate limit state.

The tensile force Trp generated in the reinforcement per metre run is then given by

T

W s aa

rpT= − +( )

21

16ε

(10.41)

In Equation 10.41, ε is the strain in the geotextile. Again, the strain is not known and a plot of the strain versus the tensile force can be compared with the stress–strain curve for the geotextile being used. A typical plot is shown in Figure 10.28.

10.7 CONTROLLED MODULUS COLUMNS

Stiffening of soft soil can be carried out through the use of controlled modulus columns (CMCs) in a similar manner to stone columns. The columns are made of a weak concrete so that they will reduce settlement and provide extra strength to the soil. The modulus of the column is designed so that the load from an embankment or structure is applied to the original soil as well as the columns, unlike stiff piles that carry most of the load.

The columns are created by using a tapered auger (Figure 10.29) that is pushed down into the soil using a high downward thrust. The auger is then withdrawn, and grout is pumped from the base of the auger into the cavity thus formed. This method has the advantage that there is very little spoil produced, is vibration free, and compacts the soil laterally.

An example of the use of CMCs was the construction of 875 CMCs with a 450 mm diameter to depths of between 7 and 11 m under an approach embankment located on the left bank of the Macleay River near Frederickton as part of the Kempsey Bypass project.

0Strain (%)

2 4 6 8 10 12 14

20

80

60

40

100

Perc

enta

ge o

f ulti

mat

e ten

sile s

tren

gth

(%)

0

Figure 10.28 Short-term load–strain curve for a geotextile.


The design included six rows of ‘stepped’ columns, stopped 2 m short of the embedment level to act as a transition zone between pile reinforced soil and unreinforced soil. The mix design included a majority of 10 MPa concrete columns with the two rows at the periphery of the treated area made of 40 MPa concrete.

10.8 DYNAMIC COMPACTION

Dynamic compaction (DC) involves dropping a large weight from a crane onto the ground causing it to densify. The process was originally developed for sands or granular materials, but it has since been used for clayey soils as well.

The weights that are dropped can be made from steel or concrete typically weighing 4.5–18 tonnes (5–20 tonnes) from heights of up to 30 m (100 ft). The soil can be compacted up to 15 m with heavy weights and high drop heights (Mayne et al. 1984). In the initial ‘high energy’ phase, the weight is dropped on a square pattern and the area then levelled pushing soil into the craters. Then a second pass is performed over the area, dropping the weight at

Figure 10.29 Auger used for installing controlled modulus columns.


the centre of the initial square pattern. Finally, an ‘ironing’ phase is carried out using lower energy drops to compact the soil missed in previous phases.

For silts and clays, the impact creates pore pressures in the ground that eventually dissi-pate producing a reduction of volume in the ground and strengthening of the soil.

The potential energy available from the weight when raised is WH where W is the weight of the block that is dropped and H is the drop height. Field data indicates that the depth of compaction dmax that can be achieved is proportional to the square root of the potential energy, that is,

d

WHn

max = 12

(10.42)

The quantity n in Equation 10.42 is a unit factor which is equal to 1 tonne/m or 672 lb/ft.The data that the above relationship is based upon can be seen in Figure 10.30. The depth

of influence of the compaction is plotted versus the energy of the drop to a log–log scale from which it may be seen that the average of the data can be well represented by the square root relationship of Equation 10.42 although there is a good deal of scatter. The data for this plot has come from sites that were underlain by sands (50% of sites) but also silt, silty, clay, or clay as well as rubble fills.

The effects of the compaction process and the depth to which the improvement reaches can be assessed in the field by carrying out field tests before and after compaction. Tests such as SPT (standard penetration tests – Section 4.15) and CPT (cone penetration tests – Section 4.18) as discussed in Chapter 4, can be used.

An example of DC is the Penrith Lakes site in Sydney where a 20-tonne pounder was dropped from a height of 23 m. The spacing between drop locations was 4.5 m and the pounder was dropped 16 times in each location. This was successful in compacting the

Energy per blow = WH (tonne m)

Dm

ax =

Max

imum

dep

th o

f inf

luen

ce (m

etre

s)

100 1000 10,0001010.10.1

0.2

0.5

1

2

50

20

10

5

Dmax = 0.3(WH)0.5

Dmax = 0.5(WH)0.5

Dmax = 0.8(WH)0.5

Maximum observed depth of influenceInfluence greater than depth tested

Figure 10.30 Maximum depth of influence versus energy per blow.


uncontrolled fill (silty sand [SM], sandy clay [CL], and sandy silt [ML]) that had been placed at the site to depths of between 10 and 12 m (Moyle 2013).

10.8.1 Impact rollers

Impact rollers apply a dynamic force to the ground like a falling weight but they do so in a slightly different way. The roller can have three, four, or five sides, and is towed behind a tractor. Because the roller is not circular, it rises up on the high point of the roller and then falls down impacting the soil below. A three-side roller is shown in Figure 10.31.

Field data reported by Berry et al. (2004) shows that the peak densification occurs at a depth of 0.67–1.0 B where B is the width of the roller (typically 0.9 m) and the depth of influence of the roller is 2–3 B. In impact trials in South Africa, compaction settlements of over 500 mm and compaction down to depths of more than 2 m were achieved with a three-side roller.

Avalle (2004) gives the example of compaction carried out with a four-side impact roller for a new building at Adelaide Airport. The impact roller used was a 1.3 m wide, 1.5 m high square steel concrete-filled module that had a mass of approximately 8t. It was drawn in its 6t frame by a 200 kW four-wheel drive towing unit at a speed of 10–12 km/h.

The ground conditions at the site comprised 1.3–1.5 m of existing fill overlying firm to stiff clay and loose to medium dense sand with the water table at 2–1.5 m. The fill also con-tained bricks, concrete, and rock fragments.

The settlement of the site was measured for different numbers of passes of the roller, and these are plotted in Figure 10.32. Settlements of up to 85 mm were achieved with 40 passes of the roller as may be seen from the figure.

Figure 10.31 Three-side impact roller.


The vibration caused by impact rolling may be of concern in some cases, especially if the compaction is being carried out near residential areas or historic structures. Data on the peak particle velocity (PPV) versus distance from the compactor has been presented by Bouazza and Avalle (2006) for an 8-tonne, four-side impact roller (Figure 10.33). The area had been filled with refuse, observed to be about 3–4 m thick, and capped with 2–3 m of quarry overburden. Further information on the criterion for damage to structures in terms of PPV is given in Section 12.9 in Chapter 12.

0No. of impact roller passes

5 10 15 20 25 30 35 40

Average settlement

0

20

40

60

80 Polynomial trendAv

erag

e set

tlem

ent (

mm

)

Figure 10.32 Settlement versus number of passes. (Adapted from Avalle, D.L. 2004. Proceedings of the 23rd Southern African Transport Conference [SATC 2004], pp. 44–54.)

Distance (m)

Peak

par

ticle

vel

ocity

(mm

/s)

A

BC

1 10 100

100

10

1

0.1

Closest house at21 m

Figure 10.33 Peak particle velocity versus distance for an impact roller.


10.9 DEEP SOIL MIXING

In the deep soil mixing (DSM) process, an additive is mixed with the soil in either a dry state or a wet state. A mixing tool such as the one shown in Figure 10.34 mixes the grout that is ejected from the base of the mixer, with the surrounding soil. The mixer may be moved up and down to achieve a more complete and uniform mix. Often there are three mixers used side by side so as to create a wall of grouted soil. The blades of the mixers overlap (see Figure 10.34) so that there is no unmixed material between each of the mixing blades.

The grout mixture used depends upon the soil type and trials can be done to establish the best mix. Cement or lime or combinations of both have been used as admixtures. DSM carried out in the Shanghai clay (Chen et al. 2013) used water/cement ratios in the range 1.2–1.8 and amounts of cement in the range 360–450 kg/m3. Penetration speeds varied from 0.25 to 0.4 m/min with withdrawal speeds in the range 0.4–0.6 m/min.

Madhyannapu et al. (2010) report using DSM columns to stabilise expansive soils hav-ing a PI of 30% (Site 1) and 50% (Site 2). For a lime (3%)–cement (9%) binder used at the rate of 200 kg/m3, the field mixed samples had an unconfined compressive strength qu of 1140–1176 kPa and a stiffness Gmax of 108–114 MPa. The untreated soil had strengths of 105–300 kPa and stiffnesses of between 35 and 67 MPa. Laboratory mixed specimens of the clay had higher stiffnesses and strengths indicating that field-mixing trials give a better indication of actual performance (see Table 10.4).

Dry mixing can be used in soils with a high enough moisture content to allow hardening of the admixture used that can be cement, lime–cement, or blast furnace slag. The admix-ture is forced from the end of the mixer by compressed air. An advantage of the dry method is that it is cheaper to implement.

Cutter soil mixing (CSM) is a more recent process that involves using the double-headed rotor shown in Figure 10.35. The cutters do not need a guide wall (as for the construction of

Figure 10.34 Deep soil mixing process showing slurry injection at base of mixers.


diaphragm walls with a clamshell). On insertion, water and compressed air can be used to aid in the penetration of the cutting head. On reaching the design depth, the cutter is raised and cement slurry is injected between the cutting heads. Finally, steel reinforcement can be added to the grout–soil mix if desired either by penetrating under its own weight or with the assistance of vibration.

10.10 JET GROUTING

Jet grouting involves a process where grout is sprayed under high pressure (that can be up to 60 MPa) from the sides of a shaft (called a monitor) that is drilled into the soil. The jet of grout issues horizontally at high velocity (100 m/sec) and cuts into the soil churning it up and mixing the soil with the grout (Figure 10.36). The shaft is raised and rotated so that

Table 10.4 Strength and stiffness ratios of field and laboratory treatments

Site Gmax,field/Gmax,lab qu,field/qu,lab

1 0.43–0.67 0.67–0.702 0.56–0.65 0.83–0.86

Figure 10.35 Cutters used for cutter soil mixing (CSM).


a cylinder of soil mixed with the grout is formed. There are various types of jet grouting systems, some having two and three jets instead of one, and some having air shrouded grout jets or air shrouded water jets. The process is different to conventional grouting where the grout has to be injected into the pores of the soil, because the grout jet cuts into the soil and mixes with it.

There are many applications for the jet grout process, and it has been used for retaining systems, to seal the base of excavations in pervious material prior to excavation, to form cut-off barriers, and in underpinning for foundations. Care is needed when using the system for sealing the base of excavations as lack of overlap of the jet grout plugs can lead to water ingress and flooding of the excavation.

The system was used to repair the new Sydney Airport runway. The runway was con-structed of sand pumped from Botany Bay and was supported around the perimeter by a reinforced earth wall. Sand was being eroded between the gaps in the facing panels of the wall, therefore over 5000 jet grout columns were constructed behind the wall with diam-eters of between 1 and 2.7 m and to depths of 9 m to seal the sand backfill.

10.11 GROUTING

Grouting of soils can be performed by pumping a grout into the soil under pressure and allowing it to flow into the pores of the soil. Grouting of rock is different to soil grouting, as the grouts used are generally cement–water mixtures that are forced into fractures and fissures in the rock. These grouts are generally unsuitable for soils as the cement particles are too large to penetrate the voids in between the soil particles that act like a filter to stop the cement particles penetrating. However, they can be used for gravels and coarse sands.

Grout

Grout jet

Groutbackflow

Figure 10.36 Jet grouting system.


Many different types of grouts can be used and selection may be based on cost, the degree of penetrability and permanence of the grout. Penetrability is a primary factor and for grouts which consist of solid particles (such as cement and clay) the relative sizes of the grout particle and the void size are important. For grouts such as acrylates and phenols which do not contain solid particles, the viscosity of the grout determines ease of penetration.

The types of grouts and the soil types that they can penetrate are shown in Figure 10.37, however, further information is available in Baker (1985).

At present, the most satisfactory method which can be used for grouting deposits of allu-vium is the French tube à manchettes TAM (sleeved tube) process. The grouting pipe used in this process is shown in Figure 10.38. The PVC grouting pipe is placed into a borehole and grouted in with a weak clay cement grout. At regular intervals along the PVC pipe are holes surrounded with a rubber sleeve much like a large rubber band. This sleeve acts like a valve, stopping grout from coming back into the tube but allowing the grout to flow out. The grouting pipe is lowered to the desired depth and then packers are inflated to seal off a section of the PVC tube. The grout is then pumped out from the inner grout pipe and expands the rubber sleeve, thus flowing out into the soil.

The groutability of a soil using a particulate grout depends on its grain size distribution. A guide can be obtained from Mitchell and Katti (1981) who define an N value as given in Equation 10.43.

N

DD

= 15

65

(soil)

(grout) (10.43)

Particulate grouting is considered feasible if N > 24 and not feasible if N < 11.Although the process is expensive, it has the advantage that the same pipe can be re-

centred as often as desired to perform additional grouting or to use different grouts at dif-ferent levels. Alluvium such as sands, gravels, and cobbles has been grouted to depths of over 100 m by this process (e.g. Aswan, Terzaghi dams).

Cement, soil

Clay

Silicate chemicals

Chrome and lignins

Polymers (resins, phenoplasts)

Gravel Sand Silt Clay

10 1.0 0.1 0.01 0.001Grain size (mm)

Figure 10.37 Soil types that can be penetrated by various grouts.


10.12 OTHER METHODS

The methods of ground improvement that have been mentioned so far are the most com-monly used methods in the author’s experience. There are a number of other methods that can be used and some of these are listed below.

10.12.1 Ground freezing

The ground may be artificially frozen by inserting pipes and circulating a coolant through the pipes until the water in the ground freezes, thus forming a temporary hardened region. Calcium chloride (brine) is the most common cooling agent and is cooled down to between −15°C and −25°C before circulation. In some smaller operations, liquid nitrogen has been used because it can rapidly solidify the soil and it can produce high strength in the frozen material.

Ground freezing can be used on just about all soil types unlike other soil improvement methods. It has been used to construct vertical shafts, as crown support for tunnels, and horizontal tunnels.

Ou et al. (2009) describe the freezing of the ground at the face of a shaft where a tunnel-boring machine was to enter. Finite element analysis was undertaken based on the governing equation

∂∂

∂∂

+ ∂∂

∂∂

+ = ∂∂x

kTx y

kTy

QTt

x y λ

(10.44)

This heat flow equation is similar to the equation for steady state flow of groundwater, and can be solved numerically in a similar fashion. In the equation, kx and ky are the thermal conductivities in the x and y directions, Q is an applied flux, λ is the heat storage capacity, T is temperature, and t is time. The term λ is defined as

Packers inside63.5 mm pipe

Grout enters soilthrough cracks

Packers inside63.5 mm pipe

Brittle clay-cement inannular space

63.5 mm pipe withcircumferential rows ofholes each spaced at300 to 1000 mm

Grout pipe Holesin pipe

Rubbersleeve33

0 m

m

330

mm

140 mm

Figure 10.38 The tube à manchettes grouting pipe.


λ = + ∂

∂C L

wT

u

(10.45)

and C is the volumetric heat capacity (kJ m−3 °C−1), L is the latent heat of water (=334 kJ kg−1), and wu is the unfrozen volumetric water content.

10.12.2 Electro-osmotic or electro-kinematic stabilisation

Electro-osmosis involves applying a direct current to the soil through the use of electrodes implanted into the ground. It is most applicable to clays and silts where dewatering is required. As the soil contains charged particles, the particles are attracted to the anode (+ve) or the cathode (−ve). Generally, there is a net flow towards the cathode and water can be removed thus draining the soil and causing strengthening.

The velocity of flow vx created in the pore water (in one dimension) is given by Wan and Mitchell (1976)

v

k ux

kVx

xx

we= − ∂

∂− ∂

∂γ (10.46)

where ke is the electro-osmotic permeability and V is voltage in the soil. The first term in the equation is Darcy’s law for flow under a pore water pressure gradient in a soil with per-meability kx. At equilibrium, it can be shown that the pore pressure u(x) generated by the gradient in voltage is

u x

kk

V xe

xw( ) ( )=

γ

(10.47)

One system uses anodes and cathodes that are placed into a slope. The application of a current removes water from the soil and finally the anodes and cathodes are used as soil nails to further reinforce the slope.

Electro-kinematics involves using electro-osmosis to introduce stabilising chemicals into the ground, although the term seems to be used for electro-osmosis in many cases. One form of electro-kinematic stabilisation introduces calcium chloride at the anode and sodium silicate at the cathode.

In recent years, conductive polymer materials have been developed that do not suffer cor-rosion like steel or copper anodes.


Analysis of improved soil systems can be performed through the use of numerical means, such as finite element and finite difference analyses. Stability of embankments on soft soils and improved soils may be carried out by these means. As well, slip circle or non-circular analyses (Sections 7.2 and 7.3 in Chapter 7) may be employed, and where stone columns or CMCs are involved, uniform soils with equivalent properties can be used (Section 10.5.1.2).

However, it is difficult to analyse problems involving surcharging and consolidation with slip circle analyses because as the soil consolidates, the effective stresses in the soil increase, and this means that the undrained strength also increases. It is necessary to monitor pore


pressures so that the effective stress can be estimated and then the undrained strength can be found from a relationship of the kind (see Section 3.10 in Chapter 3).

s K OCRu vm

( ( )OC) = ′σ (10.48)

The undrained shear strength su(OC) is related to the vertical effective stress ′σv and the over-consolidation ratio OCR (see Ladd 1991).

For soft clays modelled using numerical (finite element) analysis, it is therefore essential to use the proper soil model, and the Cam Clay type models are most suitable since they allow yield of the soil to take place and can model normally consolidated and overconsolidated behaviour. Therefore, an embankment that is built up to an initial height and then allowed to consolidate before being raised again (with this cycle repeated if required) can be anal-ysed because the pore pressure build up and dissipation can be modelled in the analysis. This is addressed in detail in Section 3.12. A suitable model that can be used in the commercial code PLAXIS is the Soft Soil Model.

10.13.1 Three-dimensional analysis

For modelling embankments on soft soils, two-dimensional analysis is sufficient. Modern finite element programs allow the embankment fill to be added through the addition of elements. For analyses involving PVD or wick drains, a slice through the drains may be modelled using three-dimensional elements (see Figure 10.39). This allows edge movements and the three-dimensional nature of the drains to be considered (Borges 2004). Vacuum consolidation can also be modelled by specifying the negative pressure at nodes of the mesh in the location of the drains.

10.13.2 Equivalent two-dimensional analysis

To make the numerical analysis simpler, wick drains or stone columns can be treated approximately by smearing the individual columns into continuous two-dimensional rows.

5.3 m3.0

2.0

5.0

x

z

y

Figure 10.39 Three-dimensional mesh including vertical drains. (After Borges, J.L. 2004. Computers and Geotechnics, Vol. 31, No. 8, pp. 665–676.)


This is shown in Figure 10.40 where B is the half-width of the two-dimensional unit cell and R is the radius of the axi-symmetric unit cell. To calculate R for triangular and square patterns of well layout, the formulae of Equation 10.14 may be used.

Hird et al. (1995) suggested the use of an equivalent permeability for the two-dimensional drain where there are three alternatives. Firstly, the width of the equivalent plane strain unit cell can be calculated by keeping the permeability the same as for the axi-symmetric case; secondly, the permeability of the two-dimensional drain can be calculated keeping the half-width of the unit cell the same as the radius of the axi-symmetric cell; and finally, the half-width of the two-dimensional cell can be selected and a permeability calculated for the 2D case. Keeping the unit cell width equal to the unit cell radius B = R is probably the most useful in finite element work, therefore the formula for calculating the equivalent perme-ability kpl is given in Equation 10. 49.

kk R r k k r rs s s w

pl

ax ax/ / / /=

+ − 2

3 3 4ln( ) ( )ln( ) ( ) (10.49)

In the above equation, kax is the lateral permeability of the soil, subscripts (pl) and (ax) stand for plane strain and axi-symmetric, B is the half-width of the two-dimensional unit cell, and R is the radius of the axi-symmetric cell. As before, rs is the radius of the smeared zone, ks is the permeability of the smeared zone, and rw is the radius of the well.

If other matching formulae are required, these may be found in Hird et al. (1995).When used in a two-dimensional finite element analysis, the soil can be given a vertical

permeability that is the actual permeability of the soil, and an equivalent horizontal perme-ability (from Equation 10.49). There is no need to try to model the actual drain, only to use the equivalent parameters. Half of the unit cell is subdivided into a mesh, and pore pressure boundary conditions are applied at the top (and if needed at the bottom) of the mesh. At the location of the drain, the excess pore pressure is set to zero. Lateral movement is restricted at the sides and the base of the mesh (if rough). Alternatively, the unit cells can be used under a full two-dimensional representation of an embankment.

Oblique view

Plan view: Arrows indicate direction of flow

(a) (b)

Figure 10.40 Conversion of (a) a cylindrical drain to (b) an equivalent two-dimensional drain.

405

Chapter 11

Environmental geomechanics

11.1 INTRODUCTION

Today, the environment is playing an ever increasing role in geotechnical engineering, as waste has to be disposed of in belowground or aboveground repositories, and there is the likelihood of pollutants from the waste seeping into the groundwater, being washed into rivers, or being blown into the air by wind.

Waste from previous eras when regulations were not as strict as they are today, may exist in the ground and have to be cleaned up. Today, environmental investigations are often performed at the same time as routine site investigations, with the same borehole used for sampling pollutants as well as soil.

Waste from domestic sources and industry is generally placed into landfills, while waste from mining operations and large-scale industrial operations such as power station waste is placed into large dams. Both of these types of waste and the geotechnical aspects of dealing with the waste are examined in the following sections.

11.2 LANDFILLS

In most developed countries today it is required that the waste placed into landfills be con-tained so that pollution from the waste does not contaminate the surrounding areas. This generally requires a liner be constructed across the base and sides of the waste repository. If the waste is placed in a very low permeability layer of clay, this may be unnecessary, but in many cases a liner is required.

Once the landfill reaches its capacity, the waste needs to be sealed with a cover to keep the waste confined for many years into the future.

Designs of liners and covers are examined in the following sections.

11.2.1 Liners

Liners may be constructed of geosynthetic materials or from compacted clay that has a low permeability. The main purpose of these materials is to reduce the possibility of leachate (pollutants) passing through the liner and into the surrounding soil and groundwater. The various types of liner materials used are ideally resistant to any form of chemical attack or degradation so that they remain as a barrier to the leachate for a long period of time.

Leachate from the leachate collection layers is collected in slotted pipes (Figure 11.1) and can be placed back into the repository or treated and released elsewhere. Some different liner designs are shown in Figure 11.2 where a double barrier is used with leachate collection between the two barriers.


Liners can be constructed of compacted clay in combination with geomembranes such as high-density polyethylene (HDPE) sheet or geomembrane clay liners (GCLs). Permeable layers of sand or geonets can be used to collect the leachate. Types of liners are examined in more detail in Sections 11.3 through 11.5.

A clay liner being compacted is shown in Figure 11.3 and leachate collection pipes being laid in a trench surrounded by gravel is shown in Figure 11.4. A geotextile membrane being laid above the compacted clay layer is shown in Figure 11.5.

Once in operation, waste is compacted into the landfill with a heavy compactor (Figure 11.6) and generally covered with soil at each lift to prevent rubbish being blown around by the wind and to discourage birds and other vermin from being attracted by the waste.

Gas vent Cover soil Drainage layer Geomembrane Compacted clay Gas collection layer Waste

Primary leachate collection layer Geomembrane Compacted clay layer Leak detection layer Geomembrane Compacted clay layer Original soil

Drainage pipe

Figure 11.1 Liner and cover for landfill.

Sand leachate collection layerSand leachate detection layer

ClayClay

Geonet leachatecollection layer

Geonet leachatedetection layer

Protective soil layer

Composite top liner

Composite bottom liner

Compositebottom liner

Geomembranetop liner

Figure 11.2 Double liner designs.

Environmental geomechanics 407

11.2.2 Covers for landfills

Once a landfill has been filled to capacity the waste needs to be covered to prevent ingress of water and to stop gases produced by the waste from venting into the atmosphere. The cover may need to be designed to intercept and collected gases such as methane that are produced by the large volume of decaying vegetable matter in the landfill. In many cases, the gas is collected and burnt off, but in very large landfills, the methane can be used for commercial purposes such as generating power.

Figure 11.3 Impervious liner being constructed over the base of landfill.

Figure 11.4 Drainage pipe being laid in a leachate collection ditch.


An example of a cover is shown in Figure 11.7 where different types of protective layers are shown. Not all of these layers need to be used in every situation but potentially could be used in a cover design. In areas where wind erosion is likely to be a problem, cobbles can be used for layer 1 as they can resist strong wind. Otherwise, layer 1 may be a grassed topsoil layer to prevent erosion by runoff from rainfall. Runoff can be channelled down lined gutters and the water allowed to sediment into a pond before discharge from the site (see Figure 11.8).

A vent for burning off methane gas from a landfill is shown in Figure 11.9.

Figure 11.5 Membrane being laid on the base of a landfill.

Figure 11.6 Waste being compacted into a landfill.


The sides of the landfill may also have to be covered by an impervious seal and this can be done by extending the clay liner or by using a geomembrane in conjunction with a clay liner on the sides of the landfill. The geomembrane needs to be anchored at the top of the slope and this is often done by digging a trench along the top of the slope and placing the membrane into it. Backfilling the trench with soil will anchor the membrane (Figure 11.10).

Because the angle of shearing resistance between the soil and the geomembrane on the slope can be low, there is potential for a sliding failure to take place on the slope. This is discussed in Section 11.6.

11.3 COMPACTED CLAY LINERS

Compacted clay liners (known as CCLs) have been commonly used for creating barriers for waste containment in the mining industry and for landfills. Because the permeability of clays is low (10−6 to 10−9 cm/sec) any fluid takes a long time to move through the clay, and hence is contained.

Where there are naturally occurring clay deposits, like the old brick pits around Sydney, then the natural clay acts as the barrier, and waste can be placed into the pit directly. Where there is no natural clay barrier, then clay needs to be imported and compacted to form a lining for the waste. As mentioned before, often clay liners are used in conjunction with geomembranes to form the barrier.

11.3.1 Compaction of clay

Naturally occurring clays that can be used as a CCL are those that can be classified under the unified soil classification system as CH (high plasticity clay), CL (low plasticity clay), or SC (sand with clay content).

(a) Surface layer of topsoil or cobbles

(b) Protection layer of locally available soil

(c) Drainage layer of sand or geosynthetic

(d) Hydraulic/gas barrier layer (compacted clay liner, geosynthetic clay liner, or geomembrane)

(e) Gas collection layer (geosynthetic or soil)

(f) Soil foundation

(g) Waste

Cover soil

Figure 11.7 Typical cover design.


Figure 11.8 Channel for draining water from a landfill cover with a water treatment pond.

Figure 11.9 Vent for burning off methane produced by a landfill.


Clay can be mixed with existing soils to lower their permeability. For example, bentonite can be mixed with natural soil to create a composite material more resistant to seepage losses.

11.3.2 Compaction method

The compaction of clay liners is carried out in similar fashion to the compaction of soils for other purposes, in that the maximum dry density to be achieved is specified. For example, the specification may state that the clay is to be compacted to 98% of the maximum dry density found in the standard compaction test.

The standard (or modified) compaction tests are performed in the laboratory, and a plot of the water content of the soil versus its dry density is obtained. The maximum dry density is achieved at the optimum water content as shown in Figure 11.11.

It is then up to the contractor to meet the specifications by

1. Selecting the moisture content. If the clay is too wet it may need to be scarified and left to dry; if it is too dry, water can be added by a water truck.

2. Selecting the thickness of each lift of clay to be compacted. 3. Selecting the type of compaction equipment. Heavier rollers and rollers with studded

drums may be used to improve compaction. The number of passes of the roller also has to be determined.

Once the clay is compacted, it must not be allowed to dry out and crack as this will lead to paths through which leachate can leave the landfill.

Geotextile Primary HDPE geomembrane

Geonet or geocomposite

HDPE = high-density polyethylene

Secondary HDPE liner

0.61 m (24″)

0.61 m(24″)

0.91 m(36″) min

0.61 m(24″)

0.61 m(24″)

0.61 m (24″)

0.61 m(24″)

0.91 m(36″) min

Secondary HDPEgeomembrane

Primary HDPE liner

Figure 11.10 Single and double anchor trenches for geomembranes at the top of a side slope.


11.3.3 Compaction control

Once the clay is compacted, testing needs to be performed to determine if the specifications have been met. This is done by using field density testing, and some of the techniques are listed below.

1. Nuclear density testing: A nuclear source of fast neutrons is used to pass neutrons through the soil. Water slows these down and so the number of fast neutrons detected at the gauge is a measure of the water content m.

A gamma ray source is then used to determine the density γ of the soil either by placing a probe containing the source down into the soil (direct transmission) or by emitting the rays from the surface (backscattering). The two measurements allow the dry unit weight γdry to be calculated.

γ γ

dry =+1 m%

(11.1)

2. Sand replacement: A hole is dug in the compacted fill and the soil carefully collected so that its weight can be determined. Then the volume of the hole is found by filling it with sand of known density.

The weight of the soil and the volume of the hole are used to compute the bulk unit weight of the soil and from water content samples taken from the excavation, the dry unit weight can be found.

3. Balloon densometer: A water-filled balloon is used to find the volume of the hole exca-vated in the fill. The balloon is inflated by pumping water into it from a piston.

4. Other methods: There are many other methods in use, such as cutters that are driven into the clay, and the soil is extracted inside the cutting ring. The weight and volume (the volume of the ring after the soil is trimmed) give the bulk unit weight of the soil. Most of the methods employ different ways of determining the volume of the hole left by the excavated clay.

Zero air voids curve

Maximum dryunit weight

Water content m

Dry

uni

t wei

ght

(γd)

Optimum watercontent

γd max

mopt

Figure 11.11 Dry unit weight versus moisture plot.


11.3.4 Permeability of clay

Although clay that is compacted at optimum moisture content will have the greatest dry unit weight and therefore the greatest shear strength, the permeability is not necessarily the lowest at that point.

It can be seen from Figure 11.12 that the permeability is less on the wet side of optimum. The curves on the plot are for low compactive effort A, medium compactive effort B, and high compactive effort C. The data shows that heavier compaction equipment and therefore greater compactive effort gives lower permeabilities.

Therefore, it is desirable to compact slightly wet of optimum to get a lower permeability, but this may lead to the clay shrinking more when it dries out, and hence cracks forming in the CCL. If the clay can be kept moist with a sand cover or a geotextile cover before the waste is placed, then this can be an effective solution.

In addition, any clods in the clay should be broken up by more passes of the roller or heavier compaction equipment, as these can lead to paths along which water can flow (i.e. next to the clods).

Moulding water content, w (%)

Dry

uni

t wei

ght, γ d

(kN

/m3 )

Silty clay

S = 100%

A

B

C18

10–10

17

16

15

10–9

11 13 15 17 19 21 23 25

(b)

(a)

10–7

10–8 C

B

A

Hyd

raul

ic co

nduc

tivity

, k (m

/s)

Figure 11.12 Permeability versus water content for a silty clay. (a) Permeability versus water content; (b) dry unit weight versus water content for a silty clay. (Adapted from Mitchell, J.K., Hooper, D.R. and Campanella, R.G. 1965. Journal of the Soil Mechanics and Foundations Division, ASCE, Vol. 91, No. SM4, pp. 41–65.)


11.3.5 Measuring permeability of CCLs

The permeability of the clay liner is best determined in the field as laboratory testing can return incorrect values due to the limited size of specimen tested and the fact that in the field the clay may have defects that are missing in the laboratory specimens. Field methods for measuring permeability are examined in the following sections.

11.3.5.1 Ring infiltrometer

The ring infiltrometer is a large ring 0.5–2 m in diameter that is filled with water. The ring is sealed by embedding it into the clay so that water cannot leak out but must infiltrate down-wards into the liner. The rings may be single or double rings and sealed or open as shown in Figure 11.13.

The test takes a long time to perform, as infiltration of water into the liner is slow. However, the permeability can be calculated from (Daniel 1989)

k

QAti

=

(11.2)

wherek is the permeability of the clay linerQ is the quantity of water lostA is the area of the infiltrometer ringi is the hydraulic gradient

However, the problem is to determine the hydraulic gradient i as the difference in head driving the flow is uncertain, as the wetting front of the water as it infiltrates is not easy to determine. The suction in the soil at the wetting front that is used in calculating the head difference is also not easily found.

Tensiometers can be used to locate where the wetting front is, and the suction is often assumed to be zero. Alternatively, post-test excavation can be used to locate the wetting front. Then the hydraulic gradient can be calculated from

i

H LL

L f

f

=+

(11.3)

Clay liner

Clay liner

Water

Single ring; sealed

Single ring; open Double ring; open

Double ring; sealed

HL Lf

H

Figure 11.13 Infiltrometer rings.


whereHL is the depth of water in the infiltrometerLf is the depth to the wetting front (see Figure 11.13)

11.3.5.2 Borehole test

Another way of determining the permeability is to note the drop in water level in a standpipe connected to a casing set into the clay liner (Boutwell and Tsai 1992).

The water is first allowed to drop in the casing that is flush with the bottom of the bore-hole. The vertical permeability is determined from this test. Then the Stage II test is per-formed allowing water to infiltrate laterally as well as vertically into the soil. From this, both the lateral and vertical permeabilities can be deduced.

The process is as follows:

1. Falling head tests are performed and the hydraulic conductivity k1 from Stage I is com-puted using Hvorslev’s formula

k

dD t t

HH

1

2

2 1

1

211=

−

π( )

ln

(11.4)

In the formula, the heads H1 and H2 are measured at two times t1, t2 as the water falls down the tube, and D and d are the diameters of the casing and standpipe, respec-tively, as shown in Figure 11.14.

The test is performed several times (which may take up to 2 weeks) until a steady value of k1 is obtained.

2. The falling head test is performed in a hole deepened beyond the casing. In Stage II of the test, k2 is calculated from

k

AB

HH

21

2

=

ln

(11.5)

Clay Clay

Stage IIStage I

CasingGrout

HDD

z

d

H

L

Figure 11.14 Two-stage borehole test.


where

A dLD

LD

= + +

2

2

1ln

(11.6)

B D

LD

t tLD

= − − −

8 1 0 562 1 572 1( ) . .exp

(11.7)

Again, Stage II is repeated until a steady state result is obtained. The final stage is then to use Figure 11.15 to obtain a value of m from the ratio k2/k1. Finally, the permeabilities in the horizontal kh and vertical kv directions can be found from

k mk

km

k

h

v

=

=

1

11

(11.8)

11.3.5.3 Lysimeters

Large-scale seepage collectors (called lysimeters) can be installed below clay liners to collect seepage through the liner. They are basically a sand drain under the liner that is lined on the bottom and sides with an HDPE membrane as shown in Figure 11.16. The sand is about 15 m × 15 m in plan and about 20–30 cm thick.

The water is collected with drainage tubes and the permeability calculated from Q = kiA where the area of the lysimeter is A and Q is the quantity of water collected in time t. The hydraulic gradient i across the liner is the change in total head divided by the thickness of clay above the sand.

11.3.5.4 Porous probes

Porous probes, which are really standpipes, can be sealed into the clay liner and the perme-abilities can be calculated according to the formulae in Figure 11.17. The clay is assumed to be isotropic and saturated in deriving these formulae, and the test needs to be run for long enough to saturate the soil around the tip of the probe.

1 3 5 7 9 11 131

2

4

31.0

L/D = 2.0

k 2 k 17m

1.5

Figure 11.15 Graph for obtaining m for different values of L/D.


Constant head:

Falling head:/4

l

kq

FH

kd

F t tHH

=

=−

π 2

2 1

1

2( )n

=+ +( )

=+ +

Case A:

Case B:

/ /

/

FL

L D L D

FL

L D L

2

1

2

1

2

π

π

ln ( ) ( )

ln ( ) ( //DD

).

22 8( ) −

11.4 FLEXIBLE MEMBRANE LINERS

Flexible membrane liners (FMLs) or geomembranes are used to line landfills and prevent the escape of leachate. The various geomembrane materials are made from parent resins with other additives.

The membranes need to be as watertight as possible, thus the joining of individual sheets and the problem of puncturing of the membranes on site (due to trafficking) is of interest to engineers.

Landfill waste

Discharge Q

Leachate level

Clay liner

≈1.5 m

Flow of leachate

Subsoil

HDPEmembraneLysimeter sand ≈20–30 cm

Figure 11.16 Lysimeter used to determine the permeability of a clay liner.

Seal

d

HSeal

H

LL

D

d

(a) (b)

Figure 11.17 Porous probe tests: (a) Case A – Probe with permeable base; (b) Case B – Probe with imper-meable base.


The various types of geomembranes most commonly used in landfill applications are listed in the following sections.

11.4.1 Types of geomembranes

Several types of polymers are used in the manufacture of geomembranes, and the ones more commonly used in landfill applications are listed in the following.

11.4.1.1 High-density polyethylene

High-density polyethylene liners are known by the acronym HDPE liners. They are made from polyethylene resin, carbon black, and other additives.

The carbon black is added as a stabiliser against ultra-violet light and is generally added as a pellet concentrate to the resin or can be added in powder form. It is usually added in the proportion of 2%–3% by weight so as to give sufficient long-term UV protection. Additives are included to prevent oxidisation, to increase durability, and to act as a lubricant. It is called ‘high density’ because the density of the final product is in the range 0.941–0.954 g/cc.

The components of the mixture (the resin, carbon black, and additives) are fed into a continuous screw feed, where they are heated and then forced through a die. The HDPE sheet that is produced is typically 0.75–3.0 mm thick and about 4.5 m wide. Generally, the edges of the sheet are trimmed during manufacture to give an exact width of sheet before it is wound onto a spool ready for shipment.

Wider sheets can be produced by heat welding sheets together at the factory in a similar way to the way in which sheets are connected in the field.

Textured sheet can be produced that has a rough surface. This is an advantage for liners placed on slopes as the rough surface gives better sliding resistance between the membrane and any waste materials or soil. The roughened surface can be produced by (1) spraying hot HDPE particles onto the HDPE sheet, (2) laminating with a hot HDPE foam, (3) passing the hot sheet through a patterned roller, or (4) use of a blowing agent that expands and creates a textured surface (see Figure 11.18 for two of the methods).

11.4.1.2 Very low density polyethylene

Very low-density polyethylene (VLDPE) is so called because the density is lower (0.89–0.912 g/cc) than HDPE. It is made from a polymer resin of ethylene and other alpha-olefins. Carbon black is added at about 2%–3% by weight as for HDPE to give ultra-violet light protection.

Hot polyethylenefoam

Roll of smoothsheet

Roll of texturedsheet

Spreader bar

Roll of textured sheetExtruder Smooth

sheet

(a)

(b)

Figure 11.18 Adding texture to HDPE sheet: (a) lamination with polyethylene foam; (b) patterned calendering.


The sheet is produced in a similar way to HDPE sheet, by passing it through an extrusion die. The finished sheet should be free from any pinholes, surface blemishes, or carbon black agglomerates.

Texture can be added to the sheet in the same way as for HDPE sheet.

11.4.1.3 Polyvinyl chloride

Polyvinyl chloride (PVC) is made by mixing a PVC resin as a dry powder with a plasticiser together with a filler and mixing in a blender. The mixture is then heated to 180°C, which melts the mixture. The viscous mixture is then fed into rollers and rolled into sheet (called calendering).

The thickness of sheet produced is determined by the separation of the rollers. The thick-ness of the final sheet should not be less than 95% of the nominal thickness. The sheets of PVC are generally 1–2 m wide and are transported in rolls weighing up to 700 kg.

PVC sheet can be factory seamed, with one sheet being heat welded or chemically sealed to the next sheet.

11.4.1.4 Chlorosulfonated polyethylene

Chlorosulfonated polyethylene (CSPE) is made by mixing the CSPE resin with carbon black, fillers, and lubricants. The polymer is heated and made into a viscous material that can be calendered into a sheet.

In some cases, a woven fabric, called a reinforcing scrim, is introduced between two sheets of the CSPE to make it stronger. It is then called a CPSE-R sheet (R = reinforced). The reinforcing is commonly a woven polyester yarn.

11.4.2 Placing geomembranes

The geomembranes are rolled off the reels and placed onto the subgrade, which should be compacted to specification, and should not contain any ruts due to construction traffic.

The presence of sharp stone fragments will puncture the membrane, so stones over 12 mm, especially if angular, should not be present in the subgrade.

The sheets are placed or ‘spotted’ in their correct location, and are tack welded to keep them in position. This is done with hot air guns generally. At this stage, the geomembrane may be vulnerable to wind damage as wind can get under the membrane and lift it.

11.4.3 Seaming

The geomembranes can be joined together by (1) extrusion welding, (2) thermal fusion, (3) chemical fusion, or (4) adhesive seaming.

In extrusion welding, a ribbon of molten polymer is placed along the edge of the sheets to be joined and melts and fuses with the sheets. However, with fusion seams, a hot air gun is used to melt the polymers and fuse them together.

11.5 GEOSYNTHETIC CLAY LINERS

Geosynthetic clay liners (GCLs) consist of a clay core supported on its upper and lower faces by geotextiles. They are sometimes called ‘clay blankets’ or ‘bentonite mats’ because the predominant clay type used is bentonite, which is a clay from the smectite group.


11.5.1 Types of GCLs

There are several variations on the way in which the GCLs are manufactured.

1. The geotextile is attached to the clay with an adhesive. 2. The geotextile can be stitched to the clay that is provided in powdered form. 3. Needle punching can be used to push fibres into the geotextile and bond it to the lower

geotextile. 4. A geotextile can be attached by adhesive to just one side of the clay blanket.

The sheets are generally about 5–7 mm thick and weigh about 3.2–6.0 kg/m2. The width of rolls is generally 2–5 m.

Figure 11.19 shows some of the different types of geosynthetic clay liners.

11.5.2 Manufacturing

The GCLs are manufactured in different ways depending on whether they are glued, stitch bonded, or needle punched, and some of the methods used are shown in Figure 11.20.

The clay is dried and pulverised, and fed on top of a geotextile along with an adhesive in the first type of manufacturing process. The adhesive and clay are added in a number of layers. It then has the upper goetextile added and goes through rollers to produce a sheet of the desired thickness.

If the product is needle punched, the dry powder is fed onto the lower geotextile, then the upper geotextile added and needles punch the fibres through into the clay and lower layer.

A line may be drawn on each side of the GCL sheet to indicate the overlap that is needed in the field as the sheets cannot be joined as geomembrane liners can (i.e. by heat welding).

The GCL is then wrapped in a plastic sheet to prevent it from picking up moisture as this will make the clay swell as it is supposed to do when wetted in the field.

11.5.3 Placement

When the GCL is taken to the site, it needs to be stored off the ground so that it will not be affected by moisture. The rolls should also be treated carefully so as not to damage them, and not stacked high as this can thin the sheet.

Geomembrane

Clay plus adhesive~ 4.5 mm

~ 4–6 mm Needle punched fibres

Lower geotextile

Upper geotextileClay core

~ 5 mm Stitch bonding

Lower geotextile

Upper geotextileClay core

(a)

(b)

(c)

Figure 11.19 Different means of manufacturing geosynthetic clay liners: (a) adhesive bound clay to geotex-tile; (b) stitch bonded geotextiles; (c) needle punched geotextiles.


The GCLs are then laid on the subgrade that should be free of ruts from construction traffic. This is done by rolling the sheet from the spool, and overlapping the sheet onto the neighbouring sheet. Sharp gravel pieces are to be avoided as well, as these can puncture the sheet.

Overlap is generally 150–300 mm, and in some cases, bentonite is placed over the join to seal it. Any punctures or tears in the GCL can be repaired in the same way, that is, by plac-ing a patch over the hole, and sealing it with dry bentonite or bentonite paste.

The GCL is then covered with a geomembrane or other soil liner. This should ideally be done before any rain wets the liner as this will cause the GCL to hydrate and swell, making covering more difficult.

When the liner is placed on side slopes, it is better to cover the liner with soil working from the bottom up as this puts less tensile stress in the liner.

11.5.4 Examples of use

Some examples of where GCL liners have been used in Australia are:

1. Swanbank (Ipswich) Because the clay underlying the site of a proposed landfill was too permeable

(k = 1 × 10−7 m/sec), a GCL was used to cover the compacted clay, and then a geomem-brane was laid on top of the GCL.

The GCL used was a needle punched Bentofix X1000 liner (dry bentonite = 4000 g/m2 and k = 3 × 10−11 m/s) and the covering geomembrane was a 1.5 mm thick HDPE sheet. A Bidim geotextile was then placed on top of the HDPE liner to protect it from the leachate collection stone and waste that would tend to puncture the membrane.

Lower geotextileor geomembrane

Upper geotextile orgeomembrane (if used)

Bentonite

Adhesive

Bentonite hoppers

Needlepunching or stitching

To winding reel

To windingreel



(a)

(b)

Figure 11.20 Manufacturing processes used for the production of GCLs.


2. Gordonvale (near Cairns) A Bentofix X2000 GCL was used to cap a landfill rather than using a 400 mm thick

compacted clay cover. The GCL sheets were overlapped by 300 mm (no joining was necessary), but transverse joints were covered with bentonite paste. In all, 23,890 m2 of GCL was used in the project.

The GCL was finally covered by 400 mm of a mixture of green waste (50%) and sandy soil (50%), and monitoring has indicated that leachate production due to water infiltration has been greatly reduced from that of the uncapped landfill.

11.6 STABILITY OF LINERS

The use of liners at the side of a landfill can lead to a slip of the fill on the liner, and so some calculation of the stability of the slope needs to be performed before the slope is designed. Figure 11.21 depicts is a slope that has a membrane liner and a soil liner on top at the side of a landfill.

In the figure, γc is the unit weight of the clay liner, ϕc is the angle of shearing resistance of the clay, and cc is the cohesion of the clay liner. The clay liner runs across the base and up the side of the landfill and has a thickness tc and the side slope is at an angle β to the horizontal.

11.6.1 Tension in the membrane

A geosynthetic or geotextile or combinations of the two lie under the clay liner, and the minimum angle of shearing resistance of the geomembranes is ϕi. This is because if there are several types of geotextiles used in combination, slip will occur on the weakest interface.

Two cases are shown in the figure: the first is where the fill is not at the top of the liner (Figure 11.21a) and the second where the fill is above the top of the liner (Figure 11.21b). We can draw the vector of forces for a likely mode of failure, which is a two block failure as shown in Figure 11.22. The vector of forces (Figure 11.23) is drawn assuming that the inter-block force P is acting parallel to the slope, and the resultant force on the base of block 2 F2 is acting at the angle ϕi to the normal. An additional force α, which is the tensile force in the membrane, is shown acting parallel to the slope.

H

D

C

(a)

(b)

B

A

C

βϕi

ϕi

β

γ c, ϕ c, c c = 0

γ c, ϕ c, c c = 0

t c

t c

ϕc

ϕ c

D

A′

A

B

H

Figure 11.21 Liner on a side slope of a landfill. (a) Fill below top of liner; (b) fill above top of liner.


We can write the expressions for the weight of each block as shown in Equations 11.9 and 11.10.

W

t tcc

c c1

2 212 2

*

sin cos sin= =

β βγ γ

β (11.9)

W

tH

t t Ht

c c c c c

c2

2

2 22

1*

sin cos sincos= −

= −

γβ β

γβ

β

(11.10)

The magnitudes of F1 and F2 are not known but their directions are. They can be eliminated by taking a projection perpendicular to their direction.

CB

AD

B′

PP

W2

W1

F1

2

1

φiφc

Figure 11.22 Two block failure mechanism.

W1

W2

F2

F1

α

β – φi

P

1

2

Figure 11.23 Vector of forces acting on sliding blocks (α is the tension in the membrane).


W Pc c1* sin cos( )φ β φ= + (11.11)

W Pi i2* sin( ) ( )cosβ φ α φ− = + (11.12)

Eliminating P from the previous two equations gives

α β φ

φφ

β φ= − −

+W Wi

i

c

c2 1* *sin( )

cossin

cos( ) (11.13)

Substituting for the values of W1* and W2

* gives the value for the tension in the liner α.

α γ

ββ β φ

φφ

β φ= −

− −+

c c

c

i

i

c

c

t Ht

2

22

1sin

cos sin( )cos

sincos( )

(11.14)

The analysis shows that there is no need for reinforcement (in the way of a geogrid) if

H H< max (11.15)

where

Htc

c i

c i

max = ++ −

12

1cos

sin coscos( )sin( )β

φ φβ φ β φ

(11.16)

The fill can be placed in layers in the landfill so that Hmax is not exceeded, and this may require several cycles of filling and raising of the side liner.

11.6.2 Factor of safety

We can draw the vector diagram for forces on each of the sliding blocks (Figure 11.24) if we assume that the angles of shearing resistance are only partially mobilised on the lower triangular block and on the membrane on the upper block but assume that the same factor of safety applies to each block, that is,

tan

tantan

tanφ φ φ φcm

cim

i

F F= =;

(11.17)

W1

P1

F1 W2

P2

F2

Figure 11.24 Vector diagrams for forces on each block.


This enables the inter-block forces P1 and P2 to be found. Then a plot can be made of P1 and P2 versus the factor of safety (Figure 11.25). The required factor of safety is found at the point where the two P forces are equal.

11.7 PROCESSES CONTROLLING POLLUTANT TRANSFER

Pollutants can be carried through the soil by several processes. The first is by advection, where the pollutants move along with the pore water as it flows through the soil. The pol-lutants are therefore moving at the speed of the liquid.

The second is by diffusion. Chemicals dissolved in the groundwater can move from regions of high concentration to regions of low concentration even if the water is not flowing at all. If the velocity of flow of the groundwater is low, then the movement of pollutants may be of the same order of magnitude, and the effects of both need to be considered. For example, if water was flowing into a landfill due to advective flow and pollutants were moving out due to diffusion, then there could be a net inward movement of pollutants.

The third is due to dispersion which can be caused by turbulent flow and mixing of the pollutants in the groundwater.

11.7.1 Advective transport

The flow of polluted water through soil (leachate) from a landfill or other waste repository is governed by Darcy’s law as we saw previously in Section 3.3 in Chapter 3. The equation of flow is given for the x direction by

v k i k

hx

dx x x x= = ∂∂

(11.18)

wherevdx is the Darcy velocity in the x-directionkx is the permeability or hydraulic conductivity of the soil in the x-directionix is the hydraulic gradient, that is, the change in total head with distance in the

x-direction

Similar expressions can be written for the other two axis directions.

P1

P2

F of S

P

Actual

Figure 11.25 Forces acting on the block interface.


The Darcy velocity is merely an average velocity as it is measured in the laboratory by collecting a volume of water Q passing through an area A in a given time t, that is k = Q/At.

The real groundwater velocity v depends on the actual area through which the water is passing, and this depends on the porosity n of the soil as the water is flowing through the pores in the soil only. Hence

v

vn

k in

xdx x x= =

(11.19)

The mass of pollutant transported by advection per unit area per unit time is called the mass flux F and can be computed from

F nc v v vx y z= + +( )

(11.20)

where c is the concentration of pollutants (having units of mass per unit volume).

11.7.2 Diffusive transport

The mass flux F of pollutants due to diffusion is given by Fick’s First Law (1855) which states that the mass flux is proportional to the gradient of the concentration and may be written as

F D

cx

cy

cz

e= − ∂∂

+ ∂∂

+ ∂∂

(11.21)

where De is the effective (molecular) diffusion coefficient, generally taken to be the same in the three axis directions.

Because diffusion occurs mainly in the pores of a soil, Fick’s Law needs to be written:

F nD

cx

cy

cz

e= − ∂∂

+ ∂∂

+ ∂∂

(11.22)

11.7.3 Dispersive transport

Dispersion can occur when there is non-homogeneity in the flow of the water that can cause mixing and spreading of pollutants. Dispersion can be caused at molecular scale through Brownian movement or at larger scale by flow turbulence.

Dispersion is often dealt with by lumping together the coefficients of dispersion and diffu-sion through the use of a ‘coefficient of hydrodynamic dispersion’, D where

D D De m= + (11.23)

In the above equation, De is the coefficient of diffusion for the chemical of interest. The mechanical dispersion Dm depends on the velocity of flow v, and in aquifers the coefficient of dispersion can be described as

D vm = α (11.24)

where α is the dispersivity. This is a scale-dependent property and not a material constant.In aquifers, the flow rate tends to be relatively high and dispersion dominates diffusion.

Studies have suggested that the coefficient of hydrodynamic dispersion can be estimated from

D D dve= + 1 75. (11.25)


whered is the mean grain diameter of the soil in metresv is velocity in metres per year giving D in m2/year.

For clay liners, the velocity of flow is low, therefore we can assume that D = De.

11.7.4 Sorption

Due to reactions between the chemicals in the pore water and the clay particles in the soil, some chemicals may be removed from solution and sorbed onto the clay particle.

There are several chemical mechanisms responsible for adsorption. These include

• Ion exchange: For example, if the waste has high concentrations of sodium ions Na+ these can exchange with calcium cations Ca2+ from the clay causing a reduction in Na+ concentration but an increase in Ca2+ concentration in the pore water.

• Precipitation: An example is precipitation of heavy metal ions due to a change in pH.• Removal of radio nuclei.

There are several relationships that have been proposed for the mass of contaminant removed from solution S (the mass of solute removed from solution per unit mass of solid). The first is a simple linear relationship where

S K cd= (11.26)

The sorption is therefore proportional to the concentration of pollutant c and the constant of proportionality Kd called the distribution coefficient (having units of L3 M−1).

Another relationship is given by the Langmuir isopleth

S

S bcbc

m=+1

(11.27)

It can be seen that as the concentration becomes large, the sorption tends to Sm and so this term represents the maximum amount of mass that can be sorbed. The parameter b represents the rate of sorption of a pollutant species.

11.7.5 One-dimensional transport

The above equations can be simplified if the problem is one-dimensional in nature such as at the base of a wide repository. The total combined mass flux is then given by

F nvc nD

cz

etot = − ∂∂

(11.28)

Considering the conservation of mass within a small volume, we can write the following equation (if radioactive decay is not included):

n

ct

Fz

St

∂∂

= − ∂∂

− ∂∂

tot ρ

(11.29)


This equation states that the rate of increase in the concentration of pollutants in a small volume is equal to the increase in mass due to advection, diffusion, and sorption onto the soil particles. Substituting for the flux from Equation 11.28 and for the sorption using Equation 11.26 into Equation 11.29 gives

n

ct

nDc

znv

cz

Kct

d∂∂

= ∂∂

− ∂∂

− ∂∂

2

2 ρ

(11.30)

or in more compact form

∂∂

= ∂∂

− ∂∂

ct

Dc

zv

cz

* *2

2

(11.31)

wherec is the concentration of pollutantst is the timez is the depth

and

D

DR

vvR

R K nd* *, ( )= = = +and /1 ρ

11.7.5.1 Ogata–Banks solution

The Ogata–Banks solution (1961) to Equation 11.30 is for an infinitely deep deposit of soil with a constant velocity of flow v and a constant surface concentration, c0. The boundary conditions that apply are

c z z

c t c t

c t t

( , )

( , )

( , )

0 0 0

0 0

0 00

= ≥= ≥

∞ = ≥

(11.32)

The solution for the above boundary conditions is

c

c z vtDt

ez t

Dtvz D= −

+ +

0

2 2 2erfc erfc( / )

(11.33)

where erfc is the complementary error function.The solution of Equation 11.33 is presented graphically in Figure 11.26 so there is no need

to evaluate the complimentary error function erfc. For pollutants that are retarded due to sorption, this expression may be used by scaling the time of interest T so that t = T/R where R is the retardation coefficient (R = 1 + ρK/n).


11.7.5.2 Booker–Rowe solution

If the concentration of the contaminant is not constant, which it would not be if the con-taminant is leaving a landfill and not being replaced, then the Booker–Rowe solution is applicable (Booker and Rowe 1987).

If the finite mass of leachate in a landfill is represented by a height of leachate Hf, then the solution to the concentration of contaminants with depth z and time t is given by

c z t c e

bf b t df d tb d

ab b t( , )( , ) ( , )

( )( )= −

−

−0

2

(11.34)

where

f b t ea

tb t

f d t ea

td t

ab b t

ad d t

( , )

( , )

( )

( )

= +

= +

+

+

2

2

2

2

erfc

erfc

= +

=+

= +

a zn K

nD

b vn

D n K

dnDH

n KnD

d

d

f

d

ρ

ρ

ρ

4 ( )

− b

(11.35)

0.1 0.5 1.0 5 100.001

0.0050.01

0.050.10

0.30

0.50c 0c

z

c = c0

= 100νz

νtz

0.700.800.900.950.98

0.999

0.995

502010

5210.5

0.2

0.1

0.05

D

Figure 11.26 The Ogata–Banks solution to the dispersion–advection equation (constant surface concentration).


In order to evaluate the function f(p,t) where p can be either a or d, it may be noted that

f p t e xa t( , ) ( )( )/= − 2 4 φ (11.36)

and where

x

at

p t= +

2

(11.37)

The function ϕ(x) can be found from a plot once x has been calculated, therefore the solu-tion can be found by using a calculator. The plot of ϕ(x) versus x is shown in Figure 11.27.

EXAMPLE 11.1

Use the Booker–Rowe solution to find the concentration of contaminants in a layer of clay given the following data:

Downward Darcy velocity va = 0.002 m/annumDiffusion coefficient D = 0.01 m2/annumPorosity n = 0.4Sorption ρKd = 1.2Initial source concentration co = 2000 mg/LTime of interest t = 150 yearsDepth of interest z = 2 mHeight of leachate in landfill Hf = 1 m

v m a= =0 002

0 40 005

..

. /

0 1 2 3x

4 50

0.2

0.4

0.6

0.8

1.0

ϕ(x)

ϕ (x) ≅

ϕ (–x) = 2ex2 − ϕ(x)

5 ϕ (5) = 0.5535

2

For x > 5

Note:

f (p, t) = e–a2/4t ϕ(x)

where x = p

x x

√√

t +t

α

Figure 11.27 Function ϕ(x) used in the Booker–Rowe solution.


a z

n KnD

d= +

= +×

=ρ ½ ½. .. .

20 4 1 20 4 0 01

40

b v

nD n Kd

=+( )

=

× +

=..

. ( . . ).

½ ½

40 005

0 44 0 01 0 4 1 2

0 012ρ

55

d

nDH

n KnD

bf

d= +

− = × +×

−ρ ½ ½. . . .

. ..

0 4 0 011

0 4 1 20 4 0 01

0 01125 0 0675= .

f b tat

bta

t( , )

.

. ( . )

= −

+

=

exp2

40 5

0 0695 1 786

φ

φ

½½

From the graphical solution of Figure 11.27, ϕ(1.786) ≈ 0.3

∴ = × =f b t( , ) . . .0 0183 0 3 0 0208

f d tat

dta

t( , )

.

. ( . )

= −

+

=

exp2

40 5

0 0695 2 459

φ

φ

½½

From the graphical solution of Figure 11.27, ϕ(2.459) ≈ 0.21

∴ = × =f d t( , ) . . .0 0183 0 21 0 0146

c c ab b t bf b t df d t b do= − −[ ] −= ×=

exp

mg/

( ) ( , ) ( , ) / ( )

.

.

2

2000 0 0212

42 4 ll

Using the Ogata–Banks solution (with the same data), we have

R

Kn

d= + = + =1 11 20 4

4ρ .

.

Dvz

=×

=0 010 005 2

1.

.

In this case, we must use the transformed time because there is sorption, and real time T = 150 years.

t

TR

= = =1504

37 5.


vtz

= × =0 005 37 52

0 0938. .

.

From graph in Figure 11.26, c/co ≈ 0.05,Hence, the concentration for a constant surface concentration is

c = × =0 05 2000 100. mg/L

This is more than double the concentration than for the case where the concentration of leachate reduces as leachate flows into the soil (i.e. the Booker–Rowe solution).

11.8 FINITE LAYER SOLUTIONS

The finite layer method that was discussed in Chapter 2 can also be applied to the spread of contaminants. The process involves applying integral transforms to the governing equa-tions along with a Laplace transform. Once the equations are solved in transform space, then numerical inversion is employed to obtain the solutions in real time. The method is explained in detail in the book by Rowe et al. (1997).

11.8.1 Three-dimensional solutions

If we have a source of contamination as shown in Figure 11.28, pollutants can potentially flow downwards through the soil and into any more permeable layer lying beneath.

The equation governing the three-dimensional diffusion–advection of contaminants in soil is given by the following equation:

D

cx

Dc

yD

cz

vcx

vcy

vcz

Kn

xx yy zz x y zd∂

∂+ ∂

∂+ ∂

∂− ∂

∂− ∂

∂− ∂

∂= +2

2

2

2

2

2 1ρ

∂∂ct

(11.38)

1 23

j

zPermeable layer

Landfill

x

y

Hf

1

k

……

Figure 11.28 Three-dimensional contamination problem for layered soil.


The finite layer method can now be used to solve these governing equations by firstly applying a Laplace transform to simplify the time element of the equations and then a double Fourier transform to simplify the three-dimensional aspect of the equations. For example, the transform of the concentration c is given by

C ce d di x y=−∞

∞

−∞

∞− +∫ ∫ ( )η γ η γ

(11.39)

C Ce dtst=−∞

∞−∫

(11.40)

If the flow of groundwater is only in the z-direction (i.e. downward), then we have on transforming Equation 11.38

− +( ) + ∂

∂− ∂

∂= +

η γ ρ2 22

2 1D D C DCz

vCz

Kn

sCxx yy zz zd

(11.41)

It may be noted that this is an equation similar to the two-dimensional case if we make the substitution ξ2 = (η2 + γ2(Dyy/Dxx)).

D

Cz

vCz

D CKn

sCzz z xxd∂

∂− ∂

∂= + +

2

22 1ξ ρ

(11.42)

This equation has the simple solution

C Ae Bez z= +α β (11.43)

If we make a substitution for the right-hand side of Equation 11.42 of X, that is,

X D C

Kn

sCxxd= + +

ξ ρ2 1

(11.44)

then the solutions for the quantities α and β become

α = + +

vD

vD

XD

z

zz

z

zz zz

2

24

(11.45)

β = − +

vD

vD

XD

z

zz

z

zz zz

2

24

(11.46)

In a layered soil system, we want to ensure continuity of the flux Fz between one layer of soil and the next at the interface of these layers. The flux is given by

F nv c nD

cz

z z zz= − ∂∂

(11.47)


therefore the transformed flux equation becomes

F nv C nD

Cz

z z zz= − ∂∂

(11.48)

If the fluxes are matched at the top and bottom of the layers we can solve for the constants A and B and can write the concentration anywhere within the layer in terms of the concen-trations at the faces j and k of the layer, Cj and Ck

C C

e ee e

Ce

j

z z z z

z z z z k

z zk k

j k j k

j

= −−

+ −− −

− −

−α β

α β

α( ) ( )

( ) ( )

( ) eee e

z z

z z z z

j

k j k j

β

α β

( )

( ) ( )

−

− −−

(11.49)

and so we can write the relationship between the transformed flux at the top and the bottom of the layer

F

F

Q R

S TC

Czj

zk

k k

k k

j

k−

=

(11.50)

In the above matrix,

QnD e e

e e

RnDe e

SnD

kzz

kzz

kzz

= −−

= − −−

= − −

( )

( )

( )

β α

β α

β α

µβ µα

µβ µα

µβ µα

eee e

TnD e e

e ek

zz

µ β α

µβ µα

µβ µα

µβ µαβ α

( )

( )

+

−

= −−

(11.51)

where

µ = − −z zk k 1

This relationship is analogous to the finite layer matrices for stress analysis where the transformed displacements are related to the transformed forces at the top and bottom of each layer. Hence, we can assemble the layer matrices in the same way, here noting that the fluxes will cancel at the layer interfaces because of continuity requirements.

11.8.2 Boundary conditions

We need to apply some boundary conditions to the transformed equations and these can be of two kinds. First, at the top of the layered soil profile, we have the landfill and the concen-tration of contaminants can be specified there. These can be diminishing with time if there is only a finite amount of contaminant available in the landfill.

At the base of the layered soil profile, there can be an aquifer that carries the pollutants away through horizontal flow.


11.8.2.1 Boundary condition at the base

The condition for the concentration at the base is

cf

n hvn

cx

Dcx

Dcy

db

t

b

b

b

b

bxxb

byyb

b= − ∂∂

+ ∂∂

+ ∂∂

∫

0

2

2

2

2 τ

(11.52)

The subscript b denotes that the quantities are for the permeable base layer, and the thick-ness of the base layer is h. It is assumed that the concentration is constant across the depth of this base layer.

Applying the double Fourier transform and the Laplace transform to Equation 11.52 gives

C

sF

n hvn

i C D C D Cbb

b

b

bb xxb b yyb b= − − −1 2 2η η γ

(11.53)

Hence, we can find a relationship between the flux and concentration at the base

F Y Cb b b= (11.54)

where

Y n h s

i vn

DDD

b bb

bxxb

yyb

xxb

= + + +

η η γ2 2

11.8.2.2 Boundary condition at the surface

At the surface, we have the condition that the concentration in the landfill cLF is the initial concentration c0 minus a reduction in concentration due to flow of contaminant into the landfill. This can be expressed as (Rowe and Booker 1986)

c cA H

f x y dxdy dt

A

LFLF LF

LF

LF

= −

∫ ∫0

0

1( , , )τ τ

(11.55)

where ALF is the area in plan of the landfill, HLF is the height of the contaminant source, c0 is the initial concentration, and fLF is the flux of the contaminant into the surface of the soil.

Taking the Laplace transform of Equation 11.55 gives

ccs sA H

f x y dxdyA

LFLF LF

LF

LF

= − ∫0 1( , )

(11.56)

Suppose we can write the surface flux as a function of the surface concentration

F CLF LF= ψ (11.57)


where ψ is the transformed flux at the surface for a unit value of the transformed concentra-tion. The value of ψ can therefore be found by solving the Equations 11.66 for a unit value of the transformed concentration at the surface, and calculating the flux at the surface from Equation 11.50.

If the transform of the concentration can be written

C TcLFLF = (11.58)

we can calculate the transform parameter T, by taking the double Fourier transform of the concentration. For a rectangular landfill as shown in Figure 11.28 that has a total width B and total length L this can be written

C c e dxdyB

B

L

L

i x yLF LF=

−

+

−

++( )∫ ∫

/

/

/

/

2

2

2

2

η γ

(11.59)

so for this case, the value of T is

T

L B= 42 2sin( )sin( )η γηγ

/ /

(11.60)

The flux fLF can be represented by an inverse transform of FLF, hence

f F e d di x yLF LF=

−∞

+∞

−∞

+∞− +∫ ∫1

4 2πη γη γ( )

(11.61)

therefore if we substitute the value of fLF from the equation above into Equation 11.56, we have

ccs s A H

Tc e d dA

i x yLF

LF LFLF

LF

= −

∫ ∫ ∫

−∞

+∞

−∞

+∞− +0

2

14π

ψ η γη γ( )

dxdy

(11.62)

which becomes after performing the integral over the area of the landfill

ccs s A H

T T c d dLFLF LF

LF= −

−∞

+∞

−∞

+∞

∫ ∫02

14π

ψ η γ.

(11.63)

Solving for the transformed concentration at the surface gives

c

cLF =

+0

1 Λ (11.64)


where

Λ =

−∞

+∞

−∞

+∞

∫ ∫14 2s A H

T T d dfπ

ψ η γLF

.

(11.65)

Hence, we can find the solution of the finite layer equations for a unit value of the trans-formed concentration at the surface. The final solution of the finite layer equations is then multiplied by the value of cLF from Equation 11.64 to get the actual solution.

11.8.3 Assembly of finite layer matrices

The global matrix for all of the layers of soil may now be assembled by adding the individual layer matrices together (Equation 11.50) and noting that the flows will cancel at the inter-faces of the layers. The resulting matrix is a diagonal matrix with a full bandwidth of three, so that solution of the set of equations is very rapid.

Q R

S T Q R

S T Q R

S T Q R

S T Q R

S T

n n n n

n n n n

n n

1 1

1 1 2 2

2 2 3 3

2 2 1 1

1 1

++

++

− − − −

− −

++

−

Y

C

C

C

C

Cb

n

n

1

2

3

1

=

FT

0

0

0

0

0

(11.66)

11.8.4 Inversion of transforms

Once the finite layer equations are solved, the transformed variables need to be inverted to obtain the concentrations and fluxes in real time and at Cartesian coordinates. The inverse integrals are not easily evaluated algebraically, so numerical integration is used.

For the inversion of the double Fourier transform, Gaussian integration is used (see Section 3.7 in Chapter 3) and for the inverse Laplace transform the numerical algorithm due to Talbot (1979) is used.

11.8.5 Solutions for a three-dimensional problem

The solution was programmed by the author for a three-dimensional landfill that is rectan-gular in plan. The soil beneath the landfill was assumed to consist of a single layer of clay underlain by a more permeable sand layer. The depth of the upper clay layer was assumed to be 2 m thick and the underlying aquifer 1 m thick. The height of the leachate in the fill initially was assumed to be Hf = 1 m and it is assumed that there is a finite mass of contami-nant in the landfill so that the concentration diminishes with time. The properties shown in Table 11.1 have been assigned to the soil layers.

The results of the analysis are shown in Figure 11.29. The full line shows the concentra-tion to initial concentration ratio across the centreline of the fill (at y = 0) for a strip landfill


(i.e. very long in the y-direction). It can be seen from the plot (which is made at time = 300 years) that the concentration is largest just near the edge of the landfill (x = 100 m).

If we now consider a landfill that has all of the same soil properties and geometry, except that the landfill is 200 × 200 m in plan, then the concentration is given by the broken line. It may be seen that if the landfill is rectangular, the concentration of pollutants is less than for the strip case.

11.9 REMEDIATION

In situ treatment methods depend on the soil conditions, the extent of the contamination, and the type of pollution present. The treatment may be physical, chemical, biological,

0.04

0.08

0.12

c b c 0

0.16

0.20

0.24 Hf =1 m

–100 100 150 200 250 300 350 400–50 0 50

Square 200 m × 200 m

t = 300 years

Strip 200 m wide

C L

cb1

cb1

ρKd = 0

Lateral distance (m)

200 m

Clayzvb Sand1 m

2 m x

Figure 11.29 Concentration in aquifer across base of landfill at 300 years for two- and three-dimensional analyses.

Table 11.1 Properties of soil used in analysis

Layer Quantity Symbol Units Value

Clay Vertical Darcy velocity va m/a 0.0Porosity n 0.4Sorption potential ρKd 0.0Coefficient of hydrodynamic dispersion D m2/a 0.01

Sand Horizontal Darcy velocity vb m/a 1.0Porosity nb 0.3Sorption potential ρKd 0.0Coefficient of hydrodynamic dispersion

Horizontal DH m2/a 1.0Vertical Dv m2/a 0.2


thermal, or combinations of these methods (Vidic and Pohland 1995). The key feature of an in situ method is that the contaminated soil is treated where it is found, and is not moved elsewhere for treatment (CIRIA 1995).

The major difficulty in treating soils in place is ensuring effective contact between the contaminant and the treatment agents. This is affected by the permeability of the soil, and contaminants that can clog the pores. Methods such as hydraulic fracturing, pneumatic fracturing, electrokinetics, and ultrasonic methods plus a variety of other techniques can be used to try to improve the penetration of treatment agents.

11.9.1 In situ leaching and washing/flushing

The term ‘leaching’ applies to processes where the contaminants are dissolved in the leach-ing agent and are removed from the soil in this manner. Washing or flushing refers to a process where a fluid is used to flush the contaminants out (e.g. oil flushed with water).

Figure 11.30 shows some of the techniques that can be used for flushing contaminants. Some techniques make use of a hydraulic barrier, where water is pumped into the ground so as to raise the water table, and then water is extracted so as flow of water is always through the contaminated region and does not go outside (Figure 11.30b). Other tech-niques make use of a barrier wall that is constructed to contain the contaminated material (Figure 11.30a).

Acidic solutions are used for metal recovery (e.g. cadmium) or for basic organic materials (amines, ethers, and anilines). Surfactants aid the desorption of oily materials and form an emulsion that can be flushed from the ground.

Once extracted, the water or extraction fluid can be decontaminated by using materi-als that adsorb the contaminants, chemically reacts with them, or achieves microbial degradation.

Monitoringwell

Extractionwell Water infiltration

Containmentwall

Contaminatedzone

Water level

(a)

Infiltrationwell

Temporarysheet piling

Extractionwell

Ponded area

Monitoringwell

Contam-inated zone

(b)

Figure 11.30 Soil flushing systems.


11.9.2 In situ chemical treatment

Chemical treatment may involve oxidation, reduction, polymerisation, or precipitation. Whatever process is used, the treatment is aimed at destroying or detoxifying the pollutant in the soil.

The chemicals used can be sprayed or ponded on the surface so that they seep into the soil, or can be injected in wells.

11.9.2.1 Oxidation

An oxidising agent may be added to the soil such as ozone, hydrogen peroxide, or hypochlo-rates. One drawback is that oxidising agents will react with vegetable matter in the soil (if present) and be consumed by this non-target material.

Chemicals such as alcohols and glycols, cyanides, metals and metal compounds, phe-nols and cresols, chlorinated organics, and sulphides can be converted to more mobile products through the use of oxidants. For example, a formaldehyde spill was treated by using hydrogen peroxide and this reduced the concentration from 30,000–50,000 mg/kg to 500–1000 mg/kg.

11.9.2.2 Chemical reduction

Chemical reduction is the process in which a reducing agent (that is an electron donor) is added to the soil. An example is iron powder, which can be used to treat organic compounds such as chlorobenzene and cyclohexanol.

Another example is the use of ferrous sulphate to reduce hexavalent chromium to trivalent chromium for a site that had been used for a chromate smelter.

11.9.2.3 Polymerisation

Smaller molecules can be linked to form larger molecular chains by polymerisation. The larger molecules are less mobile and generally less toxic. Chemical polymerisation can be used for oxygenated monomers such as styrene, vinyl chloride, and isoprene. Agents such as sulphates can be used to initiate the polymerisation.

11.9.3 In situ biological treatment

Biological treatments mainly involve the use of microbial processes, but may include the use of enzymes or the use of plants that will uptake pollutants through their root systems.

11.9.3.1 Microbial treatment

One process is to use water conditioned with nutrients such as nitrogen and phosphates, oxygen, and (in some cases) biological agents such as bacteria. To do this, the soil must be permeable enough to allow the nutrients to infiltrate and the permeability should be >10−2 m/s. A simple system for achieving this is shown in Figure 11.31.

Sources of oxygen can be air, pure oxygen, hydrogen peroxide, or ozone. Of these, hydro-gen peroxide is the most favoured, but it is toxic to micro-organisms at high concentrations, so this must be controlled.

An example of bioremediation is the treatment of groundwater contaminated with methylene chloride, n-butyl alcohol, and dimethylene. Water extracted from the ground


can be treated with air, nutrients, and micro-organisms, and then pumped to a sec-ond tank where the micro-organisms settle out and are re-used. Water from the settling tank is then infiltrated back into the soil. This can reduce the contaminant levels by around 90%.

Biological treatments can be used on contaminants such as petrol, asphalt, jet fuel, and oil.

11.9.4 Soil venting

Soil venting or soil vapour extraction (SVE) is a process whereby injected air vapourises a contaminant that is then removed through extraction wells. The extracted vapour is either incinerated, condensed by cooling, adsorbed onto activated carbon, or allowed to vent into the atmosphere.

A PVC pipe is installed down to the zone of volatile pollutants and then (see Figure 11.32) is extracted by a vacuum pump and pumped to the treatment plant. The pipes need to have a screen (a slotted or perforated section) at the base that is installed in permeable material (i.e. sand) and then sealed in with bentonite along the upper part of the tube. This stops air leakage along the side of the pipe.

This method is applicable to volatiles such as petrol, diesel, kerosene, and gas oil, but there are a number of other liquids such as vinyl chloride, carbon tetrachloride, and toluene that can be vaporised by this method.

Discharge

Water extraction

Biological agentsand nutrients Monitoring wells

Gas Water Infiltrationtrench

ContaminantsContaminants

Figure 11.31 Simple in situ microbial treatment system.

Vapour–liquidseparator

Vapourtreatment

Groundwaterlevel

Vapourextraction

Groundwaterwell – totreatment

Pressuregauge

Monitoringinstallation

Contaminatedzone

Vacuumpump/blower

Figure 11.32 Vapour extraction system.


11.9.5 Thermal desorption

The process in general involves the injection of steam or hot air into the ground to strip vola-tiles from the soil and vaporise them. This can be done in two ways: by injecting the steam or hot air through a cutting/mixing head or injecting it straight into the ground.

Shown in Figure 11.33 is a typical arrangement whereby the steam is injected into the ground through the cutter bits and comes back to the surface where it is collected in the collector shroud. Hot air is injected down the outer pipe to assist in the recovery and the combination of hot air and steam raise the temperature of the soil to 80°C.

This system is applicable to organic compounds with high volatility such as petroleum wastes and halogenated solvents.

11.9.6 In situ stabilisation/solidification

Stabilisation involves mixing stabilising materials such as cement (with other additives), or lime or fly ash with the soil so that the permeability of the soil is reduced and the contami-nants are effectively bound up in the soil.

One such process is the deep mixing process where the grout is injected into the ground down the auger and is mixed into the soil by the mixing blades. This is shown schematically in Figure 11.34.

The group of augers (that may be three abreast) and mixing arms are attached to the boom of a crane. The augers are used to bore into the soil, and as they are lifted up, the stabilising agent is added and mixed with the soil by the mixing blades.

Another process is jet grouting. In this process, the grout is sprayed under high pressure from the central grout tube, and cuts into the soil and is mixed with it. Cement or cement/bentonite mixes can be mixed into the soil in this manner. Permeabilities can be lowered to between 10−6 or 10−9 m/s reducing advective transport of contaminants.

The process is suitable for containing polychlorinated biphenyls (PCBs), asbestos, inor-ganic cyanides, radioactive materials, and volatile and non-volatile metals.

Water Air

Aircompressor

Condensedvolatiles

Processtrain

Mixing blade Cutter blade

Cutter head

Shroud

Steamgenerator

Coolingtower

Figure 11.33 Steam injection system.


11.9.7 Electro-remediation

Low level DC currents are passed through the soil to remove contaminants through electro-kinetic and electro-chemical means. Electro-osmosis causes a movement of the groundwater towards the cathode, and this carries the contaminants with it. Positively charged species such as Cd, Pd, Cu, and As, Zn, will also move toward the cathode whereas negatively charged species will move toward the anode (e.g. CN−, CrO4

2−, and AsO43−) as shown in Figure 11.35.

Material migrating to the cathode and anode needs to be flushed to remove the contami-nants and permit adjustment of the pH there.

Water supply Generator

Cement

Cement storage Groutmixer

Groutpump

Agitator Mixerblade

Power supply

Figure 11.34 Slurry mixing unit for in situ stabilisation.

OH−SO4

2−

CN−

NO−3

F−

PO43−

Cl−

Cu2+

H3O+

Pb2+

Zn2+

H2OCd2+

Purification

Conditioning

Generator ormains supply

+ –

Pumps

Conditioning

Purification

Contaminatedsoil

Figure 11.35 Electro-remediation process.


The process works best in saturated soils, and can be used in fine-grained soils. As the groundwater migrates, it can carry pollutants with it and can be effective in removing phe-nol, benzene, and toluene.

11.9.8 In situ vitrification

In situ vitrification involves inserting electrodes into the ground and passing a current between them so as to melt the contaminated solids at temperatures of between 1600°C and 2000°C. Any organic compounds are evaporated or destroyed and inorganic pollutants are immobilised in the liquid mass when it solidifies.

A starter mixture is placed on the surface of the soil between the electrodes as shown in Figure 11.36. The current initially passes along the starter path and melts the soil and once this occurs, the molten soil will act as the conductor. The electrodes can then be low-ered melting the soil and contaminants. Once complete, the vitrified soil is left to cool and harden.

A hood is placed over the electrodes and molten soil and gas is extracted from within the hood and taken to a treatment plant, but in most cases the hot gases oxidise into less harm-ful compounds. The primary requirement is for a large amount of power (say, 13.8 kV at 4.25 MW) to pass through the soil. Silica or alumina need to be present as these are the best for creating a melt at the temperatures achievable.

The method has been used for immobilising mercury, pesticides, dioxins, and PCBs.

11.10 MINING WASTE

In many mining operations, the hard rock ore is mined and crushed to a fine sand consis-tency and the minerals are removed by floatation or chemical processes. The remaining fine material left over at the tail end of the process is called the tailings. This material, which often contains toxic or environmentally undesirable processing chemicals, has to be disposed of economically and safely. This is done by pumping the tailings away from the processing plant and storing them in a tailings impoundment.

The embankments which retain the tailings must be constructed as cheaply as possible, and as a result they are often constructed from the tailings themselves. Only the coarser (and therefore drier) fraction of the tailings is used to construct the embankments; the fines and water are confined in the lagoon.

In Australia, this type of construction, although used in some cases is not common, and tailings embankments are mostly constructed to higher standards from borrow materials or

Startermaterial

Electrodeslowered

BackfillElectrodes

Solidified soil andcontaminants

Moltensoil

Contaminatedzone

Figure 11.36 In situ vitrification.


from mine overburden rockfill much like conventional water storage dams. These dams are however staged (i.e. increased in height during the mining process) so as to spread the cost of construction throughout the life of the mining project and the design must be such as to allow the height of the embankment to be increased.

Embankments constructed from the tailings themselves are, by necessity, constructed in stages, with a small embankment or starter dike being constructed at the outset of a proj-ect. As more tailings are produced, the embankment is increased in height. This leads to economy of operation as not all of the cost of the tailings dam has to be outlaid at the start of the mining process.

With huge volumes of tailings being produced worldwide, embankment heights and impoundment areas have increased. The largest dams are the Syncrude Dam (Mildred Lake Settling Basin) in Alberta, Canada, holding 540 million m3 of tailings (in 2014) and are up to 88 m high and the New Cornelia tailings dam in Arizona, holding 209 million m3.

11.10.1 Properties of tailings

The tailings produced from different mining operations have quite different compositions and characteristics. Some examples are given below:

1. Gold: The ores are often treated with sodium cyanide and so the tailings are toxic. 2. Aluminium: ‘Red mud’ tailings are typically 35% sand-sized particles. 3. Coal: Washing of coal results in waste consisting of coarse shale particles, fine coal

particles, and clay. Water draining from tailings can contain high concentrations of sulphates.

4. Uranium: Waste may have a very low pH (i.e. are acidic) and high metal content. Seepage must be monitored and collected.

Tailings produced from different ores and different mills have quite a wide variation in the grain size distribution. Some typical grading curves are shown in Figure 11.37.

11.10.2 Tailings dam construction

Many different methods of construction are employed although they may be broadly grouped as upstream, centreline, or downstream methods.

11.10.2.1 Upstream method

With this method, the dam is built up with successive retaining dikes being constructed on top of previously deposited tailings. This is shown in Figure 11.38.

The dikes are formed from the coarser fraction of the tailings. Two methods may be used in construction.

11.10.2.2 Spigotting

The tailings are pumped onto a tailings ‘beach’ by a series of pipes running from a main feed pipe. As the tailings run down the ‘beach’, the heavier and coarser sand particles settle out first and remain on the downstream face of the dam, while the finer fraction (the slimes) run into the lagoon. To provide the required freeboard on the crest of the dam, drag lines, or bulldozers may be used to reclaim the material close to the crest.


Spigots discharging tailings into the tailings impoundment are shown in Figure 11.39.

Advantages 1. The advantage of the upstream method over other methods is simplicity of

construction

Disadvantages 1. Since the dike is built over the slimes as the embankment is raised, this leads to

reduced stability since a. Water in these layers will tend to seep towards the face of the embankment b. Slimes have less strength than the coarser fractions

Gravel sizes

Grain size (mm)100 10 1 0.1

00.01 0.001

10

Perc

enta

ge fi

ner

20

30

40

50

60

70

80

90

100U.S. sieve sizes6" 3" 1½" ¼"¾" 10 40204 100 20060

Sand sizesCoarse Med Fine Silt sizes Clay sizes

Typical range ofmaterial sizes

Figure 11.37 Gradation of typical mine tailings.

Free drainingstarter dike

Downstream face

Sand dikes

Spigot

Sand “beach”

Slimes

Figure 11.38 Upstream method of construction using spigotting.


c. There is little control over construction and therefore no real basis for engi-neering design

d. Earthquake or nearby blasting can shock and liquefy the sand in the dike

This method is rarely used in countries where earthquakes are likely and is no longer con-sidered satisfactory for major tailings dams in such areas.

11.10.2.3 Cycloning

A hydrocyclone is used to separate the coarse fraction (sands) from the fines (slimes). A sche-matic diagram of a hydrocyclone is shown in Figure 11.40. This shows the coarse material (called the underflow) and the fine material (called the overflow) being separated out from the slurry feed from the mill.

As the slurry spins inside the cyclone, the coarse material is centrifuged to the outside and exits from one end of the device, while the fines are forced to the centre and exit from the oppo-site end. The resulting sand fraction would typically contain from 10% to 20% of material finer than the 75 μm sieve. An embankment constructed in this manner is shown in Figure 11.41.

AdvantagesAn embankment constructed using cycloned coarse fraction has advantages over one

constructed using spigotting since 1. It is possible to form and control the width of the sand zone forming the down-

stream shell 2. The slimes do not come as close to the downstream face

Disadvantages 1. The downstream sand shell still has to be constructed over previously deposited slimes

Figure 11.39 Spigots discharging tailings.


This leads to reduced stability of the embankment compared to an embankment con-structed using the centreline or downstream methods.

11.10.3 Centreline method of construction

With this method, a cyclone is used to deposit sand above the centreline of a starter dike as shown in Figure 11.42. The cyclone is shifted vertically upward as the embankment is raised.

Advantages 1. Requires significantly less sand than the downstream method but the factor of

safety is only slightly less than for the downstream method 2. The embankment is much more stable than one built using the upstream method

as it is not built over the slimes

Hydrocyclone

Downstream face

SlimesStarter dike(free draining)

Figure 11.41 Upstream method of construction using a hydrocyclone.

Overflow (fine fraction)

Slurry feed

Underflow(coarse or sand fraction)

Figure 11.40 A hydrocyclone.


Disadvantages 1. Usually limited to where cyclones can separate sand and slimes 2. Large amounts of sands are needed in the tailings and the volume needed increases

with the height of the embankment. Generally, it is not practical to construct a dam by this method if the tailings contain more than 75% slimes

3. As the downstream slope is continually changing, it is difficult to apply slope pro-tection to prevent erosion

11.10.4 Downstream method

For this method, a hydrocyclone is used to produce sands which are placed on the downstream side of the starter embankment or dike (see Figure 11.43). The cyclones are moved downstream as the embankment rises and material is added to the downstream face of the embankment.

Advantages 1. The embankment is not built over the weaker slimes

Downstream face

Starter dike

SlimesSand

Hydrocyclone

Figure 11.42 Centreline method of construction.

Hydrocyclone

SlimesSand

Starter dike

Ultimate downstream face

Free draining rock toe

Figure 11.43 Downstream method of construction.


2. As the embankment is built entirely of competent material, it can be designed and constructed to the required factor of safety. Allowance can be made for the effects of seismic shock

3. Seals on the upstream face of the embankment or drainage layers beneath it can be added at various stages of construction

Disadvantages 1. As for centreline method of construction

11.10.5 Embankments built entirely of borrowed materials

Embankments can be constructed of waste rockfill or imported materials. Where rockfill is used, an impervious clay core with filter zones needs to be used as for water storage dams. Impervious upstream seals (e.g. bitumen, HDPE) may be used on homogeneous dams made from borrow materials (see Figure 11.44a,b).

11.10.6 Tailings storages

Storage lagoons may be constructed on flat ground, on hillsides, or in valleys.

1. Flat ground storages: Such storages are not subject to runoff inflows, only from direct precipitation. However, they require containing embankments which surround the entire impoundment area (see Figure 11.45a).

2. Hillside storages: Tailings storages constructed on hillsides are often tiered as shown in Figure 11.45b. Runoff from the slope needs to be intercepted and channelled around the tailings impoundment.

3. Valley storages: Storages built in valleys can consist of single embankments or multiple embankments however such storages are subject to large inflows of water from the sur-rounding countryside (see Figure 11.45c) and this can cause problems. It is best that

Compacted borrowmaterial

Drainage layer with filter

Tailings

Upstream sealand filter

(a)

(b) Sloping clay core Filters

Rock fillTailings

Figure 11.44 Tailings dams constructed from borrow or waste material: (a) homogeneous dam; (b) waste rock dam with clay core.


the impoundment is constructed at the head of the valley to minimise inflows; how-ever, diversion ditches can be placed on the valley sides to channel water away. Water from any stream in the valley can be channelled beneath the tailings in a conduit.

11.10.7 Control of water

Unlike a water storage dam, the tailings embankment is usually not designed to store water. Water in the tailings occupies valuable space which could be occupied by solid waste mate-rial and in many cases schemes (such as underdrains) are used to remove water from the solid waste. However, in some cases it is desirable to retain water in the tailings lagoon. For example, in the uranium mining industry, the tailings may be kept beneath a surface layer of water in order to prevent dangerous radon gas from escaping into the atmosphere (sub-aqueous deposition). This however can lead to the tailings having very low densities which makes rehabilitation of the waste more difficult. Sub-aerial deposition is most favoured for achieving high densities.

In other cases, water may be contaminated with toxic or undesirable chemicals and it needs to be prevented from overtopping the embankments, entering the groundwater system and streams or overflowing through a spillway system.

(a)

Stage I

(b)

Stage I

Stage II

Slimes

(c)

Retaining embankment

Tailings

Figure 11.45 Types of tailings storages: (a) flat ground; (b) hillside; (c) valley.


It is therefore important to be able to control the water entering or leaving the tailings lagoon. There are several ways in which water can enter and leave; these are shown sche-matically in Figure 11.46a,b.

11.10.7.1 Inflows

Inflow of water may come from several different sources:

1. Precipitation: It is not possible to control direct water influx by precipitation, however this is not usually a problem as in theory, 50 mm of rainfall will only cause a 50 mm rise in water level.

2. Water entering with tailings: The amount of water entering with the tailings can only be regulated by a small amount, as certain minimum levels are required for processing or pumping the slurry in pipes.

3. Runoff: The amount of runoff entering a lagoon depends on where the dam is con-structed. Large amounts of runoff would be expected if the dam is built in a valley. Three possibilities exist for control

a. Retain water in tailings lagoon and remove it through a decant system. b. Pass water around tailings lagoon by constructing a diversion system. c. Allow water to flow through an emergency system, that is, a spillway.

11.10.7.2 Outflows

Water can leave the tailings impoundment by several different mechanisms:

1. Evaporation: The rate of evaporation is governed by the pond area. Evaporative losses can be minimised by using a small area for the free pond water (i.e. deep pond with a small surface area is better than a shallow pond with a large surface area, if losses are to be minimised). Surface chemicals may also be used.

2. Decanting: Water may be removed from the surface by syphoning water or by using a floating pump. Alternatively, a buried decant line may be used. Water enters a pipe which is buried at the toe of the tailings embankment.

(3) With tailings(2) Direct precipitation

(1) Runoff

(1) Evaporation(a)

(b)

(4) Emergency spillway

(3) Decanting

(2) Seepage

Figure 11.46 (a) Water outflows from tailings pond; (b) water inflows to tailings pond.


3. Spillway: A spillway may be provided to deal with the maximum expected flood. For some types of tailings, which are potential pollutants, overtopping is not allowed. Sufficient freeboard must be allowed to accommodate the expected influx of water.

4. Seepage: Depending on the type of operation, it may be necessary to stop all seepage losses (i.e. toxic chemicals in tailings). In other cases, it is only necessary to con-trol seepage which can cause failure of the embankment due to piping. Water in the embankment lowers the factor of safety against collapse. Drains such as those described in Section 11.10.5 may be used to control seepage (i.e. toe drain or blanket drains).

Remedial action may be taken if seepage is observed discharging from the face of the tail-ings, by dumping filter material over the zone of emerging water (see Figure 11.47).

11.10.8 Stability of embankments

Stability of embankments can be assessed using any of the methods discussed in Chapter 7 on slope stability. Slip circle methods may be coupled with seepage analysis so that pore pressures in the embankment are more accurately modelled.

11.10.9 Piping

If a tailings embankment is constructed from borrow materials, then the soil should be tested to see if it is dispersive. Dispersive soils are those that easily go into suspension and therefore will wash away with any water seeping through the embankment. The soil is eroded through a small cylindrical hole or ‘pipe’ initially but eventually the hole will widen until collapse of the embankment occurs (see Figure 11.48).

There are many tests that can be performed on soils to discover if they are likely to be dispersive. A very simple test is the Emerson crumb test that involves dropping a crumb of soil into a beaker of water. If the soil is highly dispersive, it will begin to go into suspension and make the water cloudy. Depending on the reaction of the soil in water, the soil is placed into one of eight Emerson class numbers; Class 1 is the most dispersive and Class 8 is the least dispersive.

Details of how to perform the test are given in Australian Standard AS 1289.3.8.1 (2006) and ASTM D6572-13e1 (2013).

Another more direct test of a soil’s resistance to dispersion and erosion is the pinhole test. In this test, developed by Sherard et al. (1976), water is passed through a small hole in a sample of the soil. The rate of erosion determines the classification (see ASTM D4647).

Coarse filterFine filter

Seepage from toe ofembankment and fromfoundation

Figure 11.47 Remedial measure to prevent piping.


11.10.9.1 Filters

Filters can be used to resist erosion of the soil in a tailings embankment, even if the soil is dispersive although it is not good practice to use dispersive soil. A filter consists of a soil which has a grain size distribution that is fine enough to stop the particles of the protected soil from washing through it. Filters may be constructed in stages with a sand filter to pro-tect a clay or silt, and then a gravel filter to protect the sand filter.

Figure 11.44b shows an embankment dam constructed from waste rock with a sloping clay core. Downstream of the core is the filter zone that protects the core from erosion. A sloping clay core such as the one shown has the advantage that it is less prone to cracking as any settlement or compression in the core will result in the core and rockfill above moving down without cracks opening across the width of the core.

Filters can be designed by using the design approach of Sherrard and Dunnigan (1989). Filter requirements are given for the protection of four different soil groups.

Group 1: Fine silts and clays with more than 85% passing the No. 200 (75 μm) sieve.

For these soils, the filter requirement is

D Df b15 85 but not smaller than 0.2mm≤ ×9 ;

(11.67)

D15f is the diameter of soil particles for which 15% of the sample is finer, and the subscript f denotes the filter. This diameter can be found from a grading curve for the filter material. Similarly D85b is the grain size for which 85% of the protected (or base) soil is finer.

Group 2: Silty and clayey sands and sandy silts and clays with 40%–85% passing the No. 200 (75 μm) sieve.

D f15 0 7= . mm

(11.68)

Figure 11.48 Piping occurring in an earth dam embankment.


Group 3: Silty and clayey sands and gravelly sands with 15% or less passing the No. 200 (75 μm) sieve.

D Df b15 854≤ ×

(11.69)

Group 4: Soils intermediate between Groups 2 and 3.

For these soils, interpolation is used between Groups 2 and 3 requirements giving

D

ADf b15 85

4040 15

4 0 7≤ −−

× − +( )( )

( . mm) 0.7 mm

(11.70)

where A is the percentage of the base soil passing the No. 200 (75 μm) seive after regrading of the soil so that 100% passes the No. 4 (4.75 mm) seive.

Generally, for Group 1 and 2 soils, the condition that the filter contains less than 5% fines is stipulated so that the filter is permeable enough to drain water away from the filter, that is, <5% of sample is finer than the No. 200 (75 μm) seive.

Although the rules of Sherard and Dunnigan (1989) do not specify additional criteria other than those listed above, often the criterion

DD

f

b

15

15

4 5≥ or

(11.71)

is used to ensure that the filters are more permeable than the soils they protect and to pre-vent build-up of seepage forces and hydrostatic pressures. By applying this rule to the sand and gravel filters as well, the grading range for the filters can be established.

The U.S. Army Corps of Engineers (2004) basically use the above rules for design of filters with some small modifications. For Group 3 soils, they specify

D Df b15 854 5≤ ×( )to

(11.72)

while their permeability criterion allows a wider range than Equation 11.71, that is,

DD

f

b

15

15

3 5≥ to

(11.73)

An example of a filter design is shown in Figure 11.49. Filter ‘A’ is the sand filter to protect the clay core, and filter B is the gravel filter to protect the sand filter. The sand filter is placed downstream of the clay core, then the gravel filter placed downstream of the sand filter. Rockfill can then be placed downstream of the gravel filter.

11.10.10 Cut-offs and barriers

Pollutants from tailings storages can enter groundwater and rivers and may therefore harm the environment. Various schemes have been used to reduce seepage losses and these may be different to those used for municipal waste repositories in some cases. Some of the methods are listed in the following.


11.10.10.1 Controlled placement of tailings

This approach involves using the fine fraction of the tailings (the ‘slimes’) for lining the base of the storage thus providing resistance to seepage. If this is done, the whole of the base of the impoundment should be lined with slimes as water can flow around any partially con-structed liner.

For a slimes liner to be effective, it would be necessary to have at least 40% passing the No. 200 (75 μm) sieve. If spigotting is used, the area next to the spigot discharge will con-tain the sand fraction and so it is necessary to move the spigot around or to put an imperme-able membrane liner under the area which contains the sand fraction.

Slimes liners have the advantage that they are cheap to construct and are not susceptible to cracking or rupture as clay or synthetic liners may be. However, they are not able to reduce pollutant levels due to sorption as natural clay liners are able to do for some con-taminant species.

11.10.10.2 Clay liners

Clay liners for tailings impoundments are similar to those constructed for municipal waste repositories, and such liners have been discussed in Section 11.3. The liner may be exposed to the sun for a period of time before it is covered with tailings, and this can cause cracking and increased permeability. Covering the liner with 150 mm of clean or silty sand can pre-vent drying of the clay. Thicknesses of clay liners can vary from 150 to 900 mm thick, but only need to be 150–600 mm thick if properly constructed.

Underdrains can be used in conjunction with clay liners to reduce the required liner thick-ness as shown in Figure 11.50. The total head of water acting on the clay liner is greatly reduced from the height h (without drains) down to the height hmax (with drains).

681218 4 2 1½ 1

100 10 1 0.1

¾ ½U.S. standard sieve sizes in inches

3 4 6 10 14 20 30 40 50 70 100 200U.S. standard sieve numbers

Grain size in millimetres < 5% 0.01

d f a b

f ≥ 3 to 5a

d ≤ 4 to 5e

0.001

e d85

c

b ≥ 3 to 5c

3Hydrometer

10

20

30

Perc

enta

ge fi

ner

40

50

60

70

0

80

90

100

b ≤ 9 d85

Filter B (gravel filter)

Filter A (sand filter)

Impervious clay core

Figure 11.49 Design of filters to protect a clay soil. (Adapted from U.S. Army Corps of Engineers. 2004. General Design and Construction for Earth and Rockfill Dams, EM1110-2-2300, July 30.)


The pipes used for the underdrainage must be protected from clogging by filters. Suitable fil-ters can be designed by using the theory given in Section 11.10.9.1. Geotextiles are also used as filters, however they should not be exposed directly to the tailings or they may clog. This can be overcome by placing a sand layer between the tailings and the geotextile (Scheurenberg 1982).

11.10.10.3 Cut-off trenches

When the depth of permeable soil is not great, a barrier to seepage may be constructed by excavating out the permeable material and replacing it with material of low permeability (Figure 11.51). Such barriers are usually taken completely through the pervious foundation down to bedrock, but can be terminated at impervious material above bedrock. Common depths of cut-off are from 6 to 10 m as beyond this depth difficulty may arise in dewatering the trench and there may be problems with the stability of the sides of the trench. At greater depths, slurry trenches and other types of cut-off are likely to be more economic.

In order to prevent any loss of material from the rolled earth cut-off, a filter zone may be constructed on the downstream side.

11.10.10.4 Slurry cut-off walls

An excavation from 1.5 to 3 m wide is made with a backhoe, excavator, clamshell bucket, or dragline bucket. The trench is kept open with a bentonite slurry which prevents the trench

hTailings

Liner

(a) (b)

hmaxhmax

UnderdrainsTailings

Figure 11.50 Clay liner (a) with and (b) without underdrains.

Upstreaminterceptor drain

Strip foundation

Original surface

Rock facing

Planned future raisingsPossible future raising

All gravelsstripped

2.7521 1

326

322313

314.5

Zone 1 Zone 2

Downstreaminterceptor drain

Rolled earthcut-off

Figure 11.51 Rolled earth cut-off.


walls from caving in. The depth to which the excavation can be made depends on the equip-ment used, but approximate depths of excavation are

1. Excavator – 10 m 2. Dragline – 25 m 3. Clamshell – 25 m

When complete, the trench is backfilled from one end, displacing the bentonite. The back-fill should be well graded but impermeable, the coarse fraction being desirable to limit post-construction settlement.

A typical composition for the backfill may be a bentonite slurry with a Marsh funnel viscosity of greater than 40 s and with water losses not greater than 15 cm3 in 30 min. The Marsh viscosity is obtained by placing 1500 cm3 of slurry in a funnel and recording the time for 946 cm3 (1 U.S. quart) to flow through the nozzle. The percentage of bentonite may range from 5% to 15%. Aggregate added to the mixture should have a continuous grading with particle sizes ranging from 0.02 to 30 mm and if necessary aggregate may have to be imported from elsewhere to adjust the grading (ICOLD 1985).

The backfill mixture can be made with a bulldozer beside the trench or in a concrete type batching plant and the aggregate and bentonite mixture blended until a homogeneous mate-rial with a standard concrete cone slump of 100–200 mm is obtained.

Khancoban Dam, which is part of the Snowy Mountains Scheme in NSW, was constructed using a slurry cut-off trench and the grading of the material used is shown in Figure 11.52.

Care must be taken that the mixture does not segregate as it is placed, so to avoid this, the trench can first be filled at one end using a clamshell bucket and then the backfill material placed at a shallow slope of between 6H:1V and 8H:1V.

Post-construction settlement of material in a trench constructed in this manner would be of the order of 25–150 mm for trenches of 15–20 m deep.

Owaidat and Day (1988) describe a slurry trench that was constructed for a copper mine in Arizona. The soil–bentonite mix had a 100 mil (1/10 inches) thick HDPE membrane placed vertically into the trench to create a sealing wall. The HDPE panels were each 8 feet

0

20

40

60

80

100

75 m

m

150

75 μ

m

300

600

9.5

191.18

mm

2.36

4.75

37.5

U.S. Standard sieve size

Perc

ent p

assin

g

Khancoban DamBackfill material

200

100 50 30 16 8 4

3/8"

3/4"

1½" 3"

Figure 11.52 Grading of material used for a slurry trench – Khancoban Dam.


wide and joined together with interlocking joints that contained a cord that would swell on contact with water thus sealing the joint. The soil that was mixed with the bentonite con-sisted of 5.6% gravel, 39.4% sand, 55% silt, and 3% bentonite was added to this to form the wall.

11.10.10.5 Grout diaphragm walls

These cast-in-place walls are usually of thickness between 0.5 and 1.5 m. A trench is exca-vated under a cementitious bentonite slurry referred to as grout, and when complete the grout in the trench is allowed to harden. The construction sequence is shown in Figure 11.53. Panels are excavated in the sequence A–B–C–D–E–F–G with the secondary panels being excavated before the primary panels have fully hardened. This has the advantage that there are no end stop joints (as are used for concrete walls).

Grouts are produced in batching plants which consist of the cement and bentonite storage bins, and the bentonite slurry storage bins for storing the grout mixture. These plants are highly automated and can produce 20–50 m3 per hour.

A grout may typically contain

1. 80–350 kg cement per m3

2. 30–50 kg of bentonite per m3

3. Cement/water ratios in the range 0.2–0.3 if Portland cement is used or 0.1–0.25 if granulated blast furnace slag is used.

4. If there is a possibility of a groundwater attack on the grout, fly ash is added in propor-tions varying between 10% and 100% of cement by weight. Pure water can dissolve free lime in the cement and selenic water can destroy the grout by forming a salt (tri-calcium sulphoaluminate) which expands in the pores of the hardening mixture.

5. A set retarder can be added. The set retarder allows the grout to remain as a slurry for a longer period of time (about 15 h) before hardening to allow more time to construct the adjacent panel.

A B C D E F

G F D E B C A

Order of constructing panels

Figure 11.53 Construction sequence for A grout diaphragm wall.


This type of wall needs to be flexible so that it can endure the deformations which are caused by the self-weight of the dam and the water loads without cracking or joints open-ing. The modulus of the wall needs to be only 4–5 times that of the surrounding soil. On average, the unconfined compressive strength of the grout is only 100 kPa at 28 days and 150 kPa at 90 days (ICOLD 1985).

An example of a grout diaphragm wall is the 27 m deep cut-off wall that was constructed beneath the Harris Dam, which is 10 km north of Collie in Western Australia (see Bradbury 1990, Potulski 1990). The wall, that was constructed using an 8 tonne clamshell grab, was 800 mm wide and was constructed in panels 2.7 m long that had an overlap of 350 mm at the surface.

The grout consisted of

1. Cementing agent that was 65% blast furnace slag and 35% ordinary Portland cement (225 kg/m3)

2. Bentonite (30 kg/m3) 3. Water (913 L/m3) 4. Retarder 0%–2% of cement weight (Daratard combined plasticiser/retarder)

The properties obtained for the grout were

1. Laboratory permeability (average) 2.5 × 10−8 m/s 2. Elastic modulus (average) 67 MPa 3. Unconfined compressive strength 500–800 kPa 4. Plastic behaviour occurred up to 5% strain (often up to 10% strain) 5. ND1 classification in the pinhole test (see Section 11.10.9)

The mix design had to achieve a modulus in the grout which was equal to that of the sur-rounding soil, so that the wall and soil would settle by the same amount. This was because it was expected that the embankment construction would cause 500 mm of settlement of the soil and if the wall were more compressible than the surrounding soil it would not carry load and shed it to the surrounding soil, thus making the wall vulnerable to hydraulic fracture. If less compressible, the slurry wall would attract load and it could be overstressed and cracked.

11.10.10.6 Grouting

Grouting of soil has been discussed in Section 10.11 in Chapter 19 (“Soil Improvement”) and the same techniques can be used for reducing the permeability of soils to reduce losses of contaminated water from tailings impoundments.

An example of the use of chemical grouts was the sealing of the foundation for the refinery catchment lake dam, which was part of the Worsley Aluminium project in Australia. The foundation consisting of silts, clays, and sand lenses was grouted using a resin forming grout (Geoseal MQ4). This grout seal was used in conjunction with a conventional rolled earth cut-off, which was taken through the more porous surface soils, and downstream monitor-ing bores that could accept a pump if it was necessary to return any seepage to the storage.

If it is necessary to grout the rock below cut-off level because the rock is fractured, holes can be drilled into the rock and cement or chemical grout injected under pressure so that it is forced into the fissures in the rock (Houlsby 1990).

Curtain grouting usually involves drilling a row (or several rows: 3–5) of holes which are drilled and grouted in sequence. This is shown in Figure 11.54. First, the primary holes are


grouted, then the secondary holes which are at intermediate positions and then the tertiary holes and so on, with the distance between holes being halved each time. This allows packer testing of the foundation to determine whether the grouting has been successful before further grouting is carried out. Water/cement ratios (by weight) used for the grout typically vary between 0.5 and 5 depending on the ease of grouting.

One of the problems with cement grouts is that they tend to ‘bleed’. This occurs if the grout is being injected slowly or is motionless so that the cement particles can sediment out from the water. For this reason, the grout must be continuously agitated in a grout mixer before injection. For slow grout takes that may occur in low permeability rocks or toward the end of a grouting programme, the grout hole must be bled off to allow the thin grout and water which rises to the surface to be replaced by thicker well mixed grout.

An example of a grouting operation has been described by Houlsby (in Baker 1985) for the foundation of Copeton dam. The dam foundation consisted of coarse-grained massive biotite granite with an inclusion of a small mass of fine-grained granite. The main joints in the rock were relatively widely spaced and of considerable extent with wide-open surface joints.

The primary grout holes were drilled at 24.4 m (80 ft) apart, then intermediate holes were drilled at spacings of 12.2 m (40 ft), 6.1 m (10 ft), and finally 1.5 m (5 ft). Water cement ratios used varied depending on the water takes indicated in the holes but varied from 0.6:1 up to 3:1 with the thinner mixes being used to start grouting and the thicker mixes to finish (e.g. start at 2:1 and finish at 0.6:1). The permeability of the foundation was reduced from 21 lugeons down to about 5 lugeons at the final hole opening of 1.5 m. Average grout takes dropped from 0.8 L/cm in the primary holes down to 0.1 L/cm in the quinery holes.

A lugeon is defined as a water loss of 1 litre per minute per metre length of a test section at an effective pressure of 1 MPa. The test is performed in a borehole section sealed by inflatable packers. The borehole should be between 46 and 74 mm in diameter. Rock masses with lugeon values which are greater than about 20 may have their permeabilities effectively lowered by grouting. If the lugeon values are less than this, then a great deal of expensive grouting may not make much improvement.

11.10.11 Synthetic liners

Synthetic liners which have been used for tailings impoundments to control seepage include concrete, shotcrete, asphalt, synthetic rubbers such as butyl rubber, and EDPM (ethylene propylene diene monomer), although such materials are not commonly used. More common

(P) Primary grout holes(S) Secondary grout holes(T) Tertiary grout holes

P T T PS

Figure 11.54 Sequence of grouting.


are the thermoplastic membranes. These materials nearly all have good resistance to acids, alkalis, and salts produced by the ore processing. The following thermoplastics can be used in tailings applications, and are similar to plastics used for landfill liners.

1. Polyvinyl chloride: PVC is among the least expensive of liners; however, it is subject to degradation by sunlight (ultra-violet light). It must therefore be kept under water or under cover by soil or tailings. PVC used in tailings impoundment applications is generally from 20 to 30 mil thickness (a mil is 1/1000 inches).

2. High-density polyethylene: HDPE has good chemical and sunlight resistance and is available in thicknesses ranging from 20 to 140 mil. It is easy to heat ‘weld’ the sheets together to form strong joints that do not leak.

3. ‘Hypalon’ – chlorosulphonated polyethylene: This is among the most commonly used liner in tailings applications. It has good ageing characteristics and is resistant to sun-light (UV radiation). It is commonly used in 30–36 mil thicknesses but is about twice the cost of the equivalent PVC sheet. It is not as easy to join sheets together as it is for materials such as HDPE.

4. Chlorinated polyethylene (CPE): This material is similar to Hypalon in most charac-teristics, but has the advantage that it is cheaper.

5. Elasticised polyolefin (EP): This material is resistant to sunlight and also has a high resistance to petroleum derivatives. It costs about the same as Hypalon.

Tests carried out by Abbott and Corless (1993) indicated that PVC, HDPE, and CPE were satisfactory for containing acid wastes. However, there was only limited experience for highly alkaline wastes stored for long periods of time.

Tests performed on PVC liners, which had been in service for 8 years and had been per-manently inundated and seasonally inundated, showed that the PVC performed satisfacto-rily with only a slight loss in elongation properties.

11.10.12 Seepage return systems

Some seepage, which may be small, will always get past artificial barriers of the kinds mentioned previously. In some cases, it may be necessary to collect seepage water because of its toxic nature or it may be cheaper to collect and return seepage water than to invest in expensive cut-off systems. There are several kinds of collection systems that can be used, and these are examined in the following sections.

11.10.12.1 Collector ditches

Ditches can be a useful way of intercepting seepage if the seepage emerges in the region of the ditch. Deeper flows will pass beneath the ditch unless it is made deep enough to go down to an impermeable stratum.

Such drains will also collect water that runs off the face of the dam and groundwater from downstream of the dam, and such flows may exacerbate water management problems if water is returned to the tailings storage.

11.10.12.2 Collector wells

Collector wells can be constructed in a line in order to intercept seepage, which can then be pumped back into the storage. Like collector ditches, wells are most effective if they are


taken down to a depth where there is a less permeable stratum. There are several disadvan-tages of using collector wells:

1. The pumps lower the phreatic surface downstream of the storage and therefore tend to increase seepage rates.

2. The wells have to be pumped (perhaps continuously) and this is costly. 3. The wells draw in water from downstream (of the wells) and this can add to water

management problems. 4. After close down of the storage, the wells will usually not be operated and so other

collection schemes are necessary. 5. The well screens and pumps are liable to corrode or to block, and therefore require

maintenance or replacement.

Hydraulic barriers may be set up in some circumstances. An injection well and an extrac-tion well are used in conjunction in this case. Water is pumped into the outer line of wells, and because the hydraulic head is higher there than at the line of extraction wells, water will always flow towards the extraction wells.

To do this, it is necessary to have a supply of fresh water to inject into the injection wells. This is because water can flow downstream from the injector wells.

A schematic diagram of a collector ditch and collector wells is shown in Figure 11.55 (Vick 1983).

11.10.12.3 Collection and dilution dams

Seepage dams can be used where seepage can be collected in a natural channel. Usually, the water is collected and pumped back into the storage, so the collector dam must be close to the storage so that excess water from runoff is not collected.

Alternatively, the collector dam can be located far enough downstream so that it will col-lect sufficient water to dilute any seepage to acceptable standards.

11.10.13 Acid rock drainage

Acid generation is caused by the exposure of sulphide minerals to oxygen and water. Minerals such as galena (PbS) and sphalerite (ZnS) can be oxidised to form acid, however by far the

Impervious layer

Pervious layer

Well

Drawndownwater surface

(a)

(b)

Figure 11.55 Seepage return systems: (a) collector ditch or sump; (b) collector wells.


most reactive mineral is pyrite (FeS2). Pyrite can be found in the tailings from lead–zinc, nickel, gold, copper, coal, and uranium processing. The oxidation of pyrite is represented by the following equation:

FeS / O / H O Fe(OH) +2SO +4H22

2 2 3 415 4 7 2+ + +( ) ( ) → − (11.74)

Normally, the sulphide bearing minerals lie at depth in rock beneath the water table, so that their contact with oxygen is minimised and very little acid is formed. When these min-erals are processed and exposed to the air, they will oxidise and acid will form much more readily.

11.10.13.1 Factors affecting acid generation

The mining of the rock, or the crushing of the ore, greatly increases the surface area of the sulphate minerals and this will accelerate oxidation. Waste rock dumps are permeable to both air and water, so that the sulphide minerals can be more easily oxidised. Most tailings deposits have low permeability and remain saturated during the life of the storage so it is less likely that oxidation will take place. Figure 11.56 shows how acid generation can occur in waste rock dumps and tailings impoundments.

Potential airflowpathway

Rainfall infiltration

Groundwater inflow

Surface water runoff

Potential acidrock drainage (ARD)

Potential sources ofwater

Rainfall infiltration

Oxidation frontZone of oxidationSurface water

runoff Air

Partially saturatedzone

Saturated zoneGroundwaterseepage

Mixture of residualprocess fluids andARD

Figure 11.56 Acid rock drainage (ARD) formation in a waste rock dump and in a tailings impoundment.


Factors which assist in the chemical oxidation of pyrite are

1. The pH of the media: Chemical oxidation of pyrite is sensitive to the pH of the reaction media. At low pH levels, ferric iron acts as an oxidant of pyrite.

2. Oxygen concentrations: Oxygen is the major factor contributing to the chemical oxi-dation of pyrite as mentioned previously.

3. Temperature: Cooler conditions will reduce the rate of reaction and slow down the rate of acid formation. Temperatures as high as 60–80°C can be generated in waste dumps and this rise in temperature causes a greater flow of air into the dump as the hot air rises, sucking fresh air into the base of the dump.

4. Moisture content: Small amounts of moisture are necessary to assist in the oxidation process. However, if the sulphides are totally submerged in water, the process will be slowed considerably due to lack of oxygen.

5. The presence of micro-organisms: Certain bacteria such as thiobacillus ferro-oxidans can speed up the oxidation process. These organisms are most active at pH 2.0–5.0. The rate of ferrous iron oxidation can increase 10–50 times through the addition of these bacteria to the pyrite system. The optimum conditions for the bacteria are a pH of about 3.2, a moist environment, and the presence of carbon dioxide and oxygen.

6. The presence of acid neutralising chemicals in the tailings: Acid produced from sul-phide oxidation can be totally or partially neutralised by minerals contained in the ore or from chemicals added during processing. Minerals such as calcite CaCO3 and dolomite CaMg(CO3)2 are the most common minerals responsible for neutralisation.

11.10.13.2 Control of acid generation

Several measures can be used to combat acid generation. Many of these methods are ways of counteracting the conditions mentioned above which are conducive to acid generation. These measures are

1. Pre-treatment of tailings: Flotation can be used prior to disposal to remove sulphide minerals, or limestone, cement or bentonite can be added to the tailings.

2. Covers and seals: Clay seals can be placed over the top of the tailings which will reduce the ingress of water. Multi-layer covers can cut the infiltration to 1%–3% of rainfall.

3. Covering tailings with water: If the tailings are covered with water, the oxygen avail-able is greatly reduced. Tailings can be dumped into existing water bodies such as lakes or oceans, although today this would not be common. Alternatively, they can be kept underwater in an old open pit or underground workings or in a constructed impound-ment where water levels would have to be kept high enough to cover the tailings.

4. Addition of chemicals to control pH: Alkaline additives such as lime, limestone, or sodium hydroxide NaOH can be added to neutralise any acid formed.

5. Bactericides: As bacteria can increase the rate at which oxidation of sulphides occurs, killing the bacteria will slow the process. Bactericides such as sodium lauryl sulphate can be used for this purpose.

11.10.13.3 Control of acid migration

Not only can measures be taken to stop the production of acid, but measures can be taken to try to reduce the migration of any acid which may be produced. Measures which can be taken are


1. Covers and seals: As discussed above, clay covers with underlying sand drainage lay-ers can be used to intercept water infiltrating the tailings and transporting the acid produced into the groundwater.

2. Interception and diversion of surface water: Surface water can be intercepted and channelled away from the tailings by surface drains.

3. Collection and treatment of acid drainage: The leachate from the waste rock dump or tailings impoundment can be collected downstream from the waste and monitored to see if acid levels are too high. If so, the leachate can be contained or treated to bring down the acid level, although the second option is more costly and would have to be continued for a long period of time after mining had ceased.

11.10.14 Remediation

After mining operations cease, it is generally required by government bodies that the tailings repositories be made secure for the future. Drainage and leakage from tailings may contain unacceptable levels of toxicity for many years into the future, and this has to be taken into consideration. It is therefore necessary to make sure that

1. The impoundment has long-term mass stability. 2. The waste has long-term erosion stability. 3. Contamination of the environment will not occur in the long term. 4. The area may be returned to productive use.

11.10.14.1 Hydrological stability

The best way to guard against damage due to water inflows is to prevent water entering the waste by providing a cap. The cap needs to be provided with a slope to assist drainage and prevent ponding of water that could infiltrate the tailings.

Placing the capping may not be easy in some cases. The additional weight of the cap can cause soft tailings to consolidate, requiring large quantities of capping material to infill depressions which would cause water to pond if left unfilled.

An example of this is the residue areas in Kwinana, Western Australia, where historically tailings were placed into the impoundments in a wet state (Coreless and Glenister 1990). The tailings sands from perimeter beaches have been used to cover the slimes. This was a time consuming process, because if it was carried out too fast, displacement of the mud at depths of up to 8 m would occur. One technique which was used was to produce ditches using an amphibious screw tractor. The ditches would drain the upper layers of the tailings allowing them to form a crust to a depth of about 500 mm. This was then strong enough to support the cover of sand.

One of the problems with providing a cover is that often large settlements can occur over the soft slimes zones with less settlement over firmer regions. This means that large amounts of fill can be required to infill depressions which can be of the order of 3 m deep. Providing a slope for drainage on the cap can also be a problem as a slope of only 0.5%–1.0% can mean significant thicknesses of cap material are required at the centre of a large area.

An example of the design of a cover for a gold mine in Washington State, has been given by Frechette (1994). Conventional oedometer tests were carried out on the tailings, and Terzaghi’s one-dimensional consolidation theory was used to predict the amount of consoli-dation that would be caused by the cover. The compression index Cc, of the tailings ranged from 0.18 to 0.33.


Ramps were constructed on the tailings initially so that access could be obtained for plac-ing the cover. These ramps consisted of a single layer of geotextile fabric 5.3 m wide, or a double layer of fabric 7.6 m wide in softer areas. Fill of between 0.9 and 2.4 m thickness and 4 m wide at the crest was placed above the fabric.

Fill placed over the tailings from the ramps ranged in thickness, but was up to 6 m thick with an average of 3 m.

The design of the cap for another gold mine in northern California is also described by Frechette (1994). First, a geofabric was laid on the tailings, and an initial layer of rock fill was placed on the geofabric. The fabric and rockfill (which was up to 6 m thick in some areas) were placed to allow initial drainage and consolidation of the tailings. The rockfill was placed in lifts (initial lift 60 cm) and kept 15–30 m back from the edge of the fabric.

The final cover (see Figure 11.57) consisted of (from the bottom up):

• A geofabric• An initial rock layer• An impermeable soil layer (90 cm min)• A soil/bentonite mix (15 cm min)• A topsoil layer (30 cm min)• A gravel cover

Settlements of up to 3 m were predicted for the cover in this case.Experience has shown that for gold tailings reclaimed immediately after deposition,

the tailings will consolidate a further 10% during the initial 2–3 years following closure. Placement of fill over the tailings to re-contour the impoundment so as to shed surface

Foundation rockFinished grade

Initial foundation rock layer

Geofabric

Firs

t lay

er o

f soi

l

Soil

with

ben

toni

te m

ix

Top

laye

r of s

oil

Gra

vel c

over

Impoundment limits

0 50 100

Scale in meters

Figure 11.57 Construction sequence for cover of gold tailings.


water, may require quantities of up to 10% of the tailings thickness. Additional settlement caused by cover fill placement may be of the order of 25% of the cover thickness.

11.10.14.2 Long-term erosion stability

On flat, unbroken surfaces, wind erosion is primarily a factor but on embankment slopes erosion by water is most often a problem. Vick (1983) states that slopes flatter than about 3:1 are usually required for ‘reasonable erosion resistance and establishment of vegetation’. Eurenius (1990) suggests slopes should be flatter than 2:1 or 3:1 for dams with heights of between 15 and 30 m.

11.10.14.3 Vegetation

Resistance to erosion (both wind and water) is best provided by vegetation, and it is neces-sary to provide the correct type of grasses and plants for the climatic conditions in the region. Native plants are an obvious choice, and should be used wherever possible (Pickersgill 1994).

Where tailings have a low pH (acid) or a high pH (alkaline) or high concentrations of heavy metals or salts, establishment of vegetation may be difficult. Adjustment of soil pH can be carried out (e.g. with lime), or drainage layers can be added beneath the final cover to prevent migration upwards of pollutants by capillary action.

It is often necessary therefore, to strip the natural topsoil before constructing the tailings impoundment and storing it for later use for establishing vegetation. Topsoil stored too long in this manner can lose much of its organic matter, nutrients, and bacteria and therefore can be much less suitable for plant growth than it was originally.

Forrest et al. (1990) describe the rehabilitation of a copper–silver tailings impoundment. Soil which was stockpiled was placed to a depth of 30 cm over the top of the tailings and the face of the embankment. Grass was then planted by hydro-mulching and conventional means. Riprap was placed on the crest and upper part of the embankment (which had a slope of 3:1) to prevent erosion gullies forming. The estimated cost of this work was esti-mated at about 19% of the construction cost of the impoundment.

Brett (1990) reports that coal washery discard at the Duncan colliery in Tasmania was rehabilitated by establishing native flora. The site was firstly contoured and then deep ripped to a depth of 500 mm and fertiliser added at the rate of 300 kg per hectare. Both native seedlings and native seeds were planted and these consisted of acacias, casuarinas, banksias, and eucalypts.

11.10.14.4 Riprap

Wind erosion may also be prevented by placing a layer of riprap over the surface of the tail-ings. This can consist of either gravel- or rock-sized particles and smelter slag has also been used successfully.

The disadvantages of riprap are, however, that it can be costly to place and that it prevents natural plant growth.

11.10.14.5 Prevention of environmental contamination

Long-term prevention of environmental contamination is also of great importance. As has been discussed before, in many cases the loss of water from pyrite rich tailings will result in exposure of the tailings to oxygen. One long-term measure against acid drainage is to seal the tailings with a clay cap to prevent the entry of water which can carry acid from the tailings.


Uranium waste is another example. Upon desaturation, the tailings will be able to emit radon gas to the atmosphere (nominally saturated tailings will not emit the gas). Generally, a cover of about 3 m is required to prevent long-term radon gas emission.

An example of a cover for uranium mine tailings is presented by Euranius (1990). A test pile of tailings was covered by two types of covers as shown in Figure 11.58. One consisted of limestone, moraine, and topsoil, while the other consisted of a bentonite–moraine layer, overlain by a moraine layer and a topsoil layer. Overall, each cover was 1.5 m in total thick-ness. Tests on both of these covers have shown that the infiltration is about 1%–4% of pre-cipitation, that the weathering is reduced 98%–99% as compared to uncovered tailings and that radon attenuation has about the same value as the natural background.

≈ 100 m

Top soil Top soil

Limestone Bentonitemoraine

0.2 m

1.0 m

Tailings

0.3 m

MoraineMoraine

Figure 11.58 Cover design for uranium tailings.

471

Chapter 12

Basic rock mechanics

12.1 INTRODUCTION

The behaviour of rock and rock masses is different to that of soils mainly due to the fact that the intact rock generally has a higher strength and deforms less than soil. In addition, rock mass behaviour is often dictated by any discontinuities or planes of weakness such as fractures and fissures in the rock, especially if the fissures are infilled with soft material.

Special methods of testing rock and for analysis of rock slopes and structures constructed on rock (as opposed to soil) have been developed, and some of these methods are included in this chapter.

The study of rock mechanics is a large field, so only basic rock mechanics are covered here. The reader is referred to the many specialist texts and papers that have been referenced if further information on some of the topics covered is required.

12.2 ENGINEERING PROPERTIES OF ROCKS

The properties of rocks may be found from a number of specialist laboratory or field tests. Many of the test procedures are outlined in national or international codes or standards that provide details of how the test should be performed, and where possible these stan-dards are referred to in the following sections.

12.2.1 Point load strength index

This is a simple test that can be performed in the field or in the laboratory, using a hand operated hydraulic jack. Generally, rock core is placed between two cone shaped platens (of included angle 60° and with a 5 mm radius spherical tip, see Figure 12.1) and loaded until the core undergoes a tensile failure, although the test can be performed on blocks or lumps of rock. For rock core, the test may be performed by loading across the diameter or by loading in the direction of the axis of the core.

The uncorrected point load index Is is given by Equation 12.1.

I

PD

s = ×10002

(12.1)

whereP is the load at failure in kilonewtonsD is the platen separation in millimetres


The value is then corrected to a value that would have been measured for core of 50 mm diameter (or dimension) using the following equation:

I I

Ds s( )

.

50

0 45

50=

(12.2)

Details of the test procedure are given in the standards AS 4133.4.1 (2007) or ASTM D5731-08 (2008).

For axial, block, or irregular lump tests, an equivalent distance De is used instead of D in Equations 12.1 and 12.2 where

D

Ae2 4= ×

π (12.3)

and A is the minimum cross-sectional area of the plane through the two platen points. Typical values of the point load index are given in Table 12.1.

Conical platens

Hydraulic pump

Figure 12.1 Point load testing machine.

Table 12.1 Point load index values for various rock types

Type of rock Is(50) (MPa)

Tertiary sandstone 0.05–1Coal 0.2–2.0Limestone 0.25–8Mudstone 0.2–8Volcanic lava 3–15Dolomite 6–11Hawkesbury sandstone (Sydney) 0.3–1.6

Basic rock mechanics 473

12.2.2 Unconfined compression test

The unconfined compressive strength (UCS) of an intact rock sample can be found by com-pressing a piece of core axially until the core fails in compression. Generally, the core needs to have a length to diameter ratio of between 2.5 and 3 so that the end effects are minimised. No lateral pressure is applied during the test, and so the sample is ‘unconfined’.

The UCS qu(max) is then simply calculated from

q

PA

u( )max =0

(12.4)

whereP is the applied load at failure andA0 is the original cross-sectional area of the rock sample.

Details of the test are given in standards AS 4133.4.3.2 (2013), ASTM D7012-14 (2014), or BS EN 1926 (2006).

The relationship between the UCS and the point load index can be established by test-ing any particular rock type (in unconfined compression and point load) but for hard rock Bieniawski (1975) indicates a relationship of

UCS kI Is s= =( ) ( )50 5024 (12.5)

For Sydney sandstones, the relationship of UCS to axial point load index value has been found to have the best fit to the data if UCS = 20 Is(50) for axial point load tests (on saturated specimens) but with a range of k from 15 to 29, and UCS = 24 Is(50) for diametral tests on saturated specimens with a range of k from 14 to 35 (Pells 2004).

12.2.3 Modulus of rock from unconfined compression test

Rock core can be instrumented during the unconfined compression test so as to obtain the deformation characteristics. Changes in length and diameter can be measured with LVDTs, optical devices, or strain gauges. If strain gauges are used the circumferential strain can be measured, and this is equal to the diametral strain.

The elastic modulus is calculated from

E

PA

lla

= =σε 0 ∆

(12.6)

where P is the axial load at any stage of loading, A0 is the original cross-sectional area, and l is the length of the specimen. The Poisson’s ratio ν of the core is the ratio of the diam-etral strain to the axial strain. Details of the test are provided in AS 4133.4.3.1 (2009), AS 4133.4.3.2 (2013), and ASTM D7012-14 (2014).

Typical values of UCS and elastic modulus are presented in Table 12.2 for various rock types.

12.2.4 Confined compressive strength

Rock core may be tested under triaxial conditions much like soil, where the core is placed in a cell and all round hydraulic pressure applied to the sample. The pressures used are gen-erally higher than those required for soil, and so the cell is made from steel with specially


designed membranes that self-seal against the cell (Hoek and Franklin 1968). This cell can apply about 70 MPa pressure to the specimen. The lateral pressure applied to the sample as well as the axial pressure allow the strength envelope of the rock to be obtained, and this is examined later (Section 12.3), as the strength envelopes of rock are found to be different to those of soils.

Design of the cell is such that strain gauges can be attached to the rock sample and the wires passed out of the cell through the base (see Figure 12.2), hence avoiding the need to pass the wires through the membrane. The strain gauges that are attached axially and radi-ally to the rock sample, allow the moduli of the rock and its Poisson’s ratio to be calculated. The procedures for the test may be found in ASTM D7012-14 (2014).

12.2.5 Sonic velocity test

Elastic constants can be found for intact rock by propagation of sonic waves through a rock sample (usually core). Both p and s waves can be passed through the rock and the

Steel cellbody

Rock sample

Inlet for cell oil

Diametralstrain gauges

Rubber sealing sheath

Axial straingauges

Spherical seatfor axial ram

Figure 12.2 Steel triaxial cell for testing of rock samples.

Table 12.2 Typical rock strengths and moduli

Rock type UCS (MPa) Elastic modulus E (GPa)

Granite 150–200 50–70Limestone 100–150 50–100Sandstone 20–50 2–10Hawkesbury sandstone (Sydney) ~20 ~6


compressive vp and shear vs wave velocities measured. Assuming that the rock is homoge-neous, isotropic, and linear elastic, the shear modulus G, modulus of elasticity E, and the Poisson’s ratio ν of the rock can be calculated from Equations 12.7 through 12.9.

Since

G E vs= + =/ /2 1 2( )ν ρ (12.7)

E v

v vv v

sp s

p s

=−−

ρ 2

2

2

3 41

( )( )

//

(12.8)

ν =

−−

12

21

2

2

( )( )v vv v

p s

p s

//

(12.9)

where ρ is the density of the rock.ASTM D2845-08 (2008) gives the details for performing the test.

12.3 FAILURE CRITERION FOR ROCK

It has been found through testing of rock under triaxial conditions, that the failure surface when plotted in σ1–σ3 space (σ1 and σ3 are the maximum and minimum principal stresses) is not a straight line like the Mohr–Coulomb failure surface for soils, but is curved. A typical curved failure surface is shown in Figure 12.3.

Triaxial compressive strength

Uniaxial compressive strengthBrazilian tensilestrength

Uniaxial tensile strength

Biaxial tensile strength

σ3 (MPa)

σci = 100 MPa

σ 1 (M

Pa)

190

160

130

100

70

40

5 15–5

–20

10

–15

s = 1a = 0.5

mi = 10σtU = –9.9 MPaσtB = –10 MPa

Figure 12.3 Curved failure surface for rock.


12.3.1 Hoek–Brown criterion for intact rock

The curved failure surface of Figure 12.3 can be approximated by an empirical equation that was developed by Hoek and Brown (1980a) and is therefore known as the Hoek–Brown failure surface. The equation for the failure envelope is given in Equation 12.10 where the constant m is used to fit the equation to test data.

σ σ σ

1 33

0 5

1 0= + +

q mq

uu

..

(12.10)

Values of m will range from about 5.4 for limestone or marble up to 28 for hard rocks like granite. More values are given in Hoek and Brown (1980b).

12.3.2 Hoek–Brown criterion for rock mass

The failure criterion of Equation 12.10 was developed for intact rock. However, rock masses contain fractures or jointing that means that the equation for intact rock will not apply to rock masses. Hoek and Brown (1980b) proposed the Equation 12.11 for rock masses where they introduced the parameter s. For intact rock, s is equal to 1.0, but for a completely granulated rock mass s = 0.0.

′ = ′ + ′ +

σ σ σ1 3

30 5

q mq

sui iui

.

(12.11)

Later, Hoek and Brown (1997) introduced a more general equation that had the form

′ = ′ + ′ +

σ σ σ1 3

3q mq

sui bui

a

(12.12)

where mb, s, and a are all material constants for the rock mass (qui is the unconfined com-pressive strength of the intact rock) and mb is found from the intact rock value of mi.

m mGSI

D

sGSI

D

a e

b i= −−

= −−

= + −

exp

exp

10028 14

1009 3

12

16

(( ) ( )GSI e/ /15 20 3+( )−

(12.13)

where GSI is the geological strength index which can be found from Table 12.3, and D is the damage index which is a measure of blast and stress relaxation around an excavation. D can range from 0 where there is no damage to 0.7–1.0 where the rock is badly damaged. Hoek et al. (2002) give examples of obtaining values of D, and their descriptions are given in Table 12.4.


Table 12.3 Values of GSI

Source: Adapted from Hoek, E. and Brown, E.T. 1997. International Journal of Rock Mechanics and Mining Sciences & Geomechanics Abstracts, Vol. 34, No. 8, pp. 1165–1186.


12.4 CLASSIFICATION OF ROCKS AND ROCK MASSES

The purpose of rock classification systems is to group together rocks having similar characteristics. This is useful in designing structures in rock where engineers need informa-tion about the intact rock and the rock mass.

Rock mass classifications can be used with empirical schemes based on experience and judgement to carry out engineering design.

12.4.1 Classification on strength

The UCS of the intact rock can be used as a means of classifying rocks. This is not indicative of the rock mass strength or behaviour if the rock mass is fractured, but is a means of clas-sifying the parent rock by its strength. Table 12.5 shows such a classification (after Deere and Miller 1966).

12.4.2 Classification by jointing

As mentioned in the previous section, the rock mass behaviour is generally affected by the jointing in the rock. Hence, rock classification can be carried out for the rock mass on the basis of the joint spacing. Such a classification is shown in Table 12.6.

Table 12.4 Suggested values of the parameter D

Description of rock mass Suggested value of D

Excellent quality controlled blasting or excavation by tunnel boring machine results in minimal disturbance to the confined rock mass surrounding a tunnel.

D = 0

Mechanical or hand excavation in poor quality rock masses (no blasting) results in minimal disturbance to the surrounding rock mass. Where squeezing problems result in significant floor heave, disturbance can be severe unless a temporary invert is placed.

D = 0 or D = 0.5 No invert

Very poor quality blasting in a hard rock tunnel results in severe local damage, extending 2 or 3 m, in the surrounding rock mass.

D = 0.8

Small scale blasting in civil engineering slopes results in modest rock mass damage, particularly if controlled blasting is used. However, stress relief results in some disturbance.

D = 0.7 Good blasting D = 1.0 Poor blasting

Very large open pit mine slopes suffer significant disturbance due to heavy production blasting and also due to stress relief from overburden removal. In some softer rocks excavation can be carried out by ripping and dozing and the degree of damage to the slopes is less.

D = 1.0Production blastingD = 0.7Mechanical excavation

Source: Adapted from Hoek, E., Carranza-Torres, C., and Corkum, B. 2002. Proceedings 5th North American Rock Mechanics Symposium and 17th Tunnelling Association of Canada Conference, University of Toronto, Toronto, Vol. 1, pp. 267–273.

Table 12.5 Classification of rocks by intact rock strength

DescriptionUniaxial compressive

strength (MPa) Examples of rock types

Very low strength 1–25 Chalk, rock saltLow strength 25–50 Coal, siltstone, schistMedium strength 50–100 Sandstone, slate, shaleHigh strength 100–200 Marble, granite, gneissVery high strength >200 Quartzite, dolerite, gabbro, basalt


12.4.3 Rock quality designation

One system for indicating the quality of rock is the rock quality designation (RQD) that was originally introduced by Deere et al. (1967). The RQD is defined as the percent-age of intact core pieces longer than 10 cm (4 inches) in the total length of core. The test was developed for NX size core (54.7 mm or 2.16 inches in diameter) but NQ core (47.5 mm or 1.87 inches) is also commonly used. The core should be drilled with a double or triple-tube core barrel so that good recovery is achieved. A section of rock core is shown in Figure 12.4 showing how the RQD should be calculated. Deere and Deere (1988) recommend that the length of core used be about 1.5 m as the result is influenced by the length of core selected.

The method is outlined in detail in the Standard ASTM D6032 (2008) or British Standard BS 5930. The description of the rock can then be made as shown in Table 12.7.

Table 12.6 Classification of rock mass due to joint spacing

Rock description Spacing of joints (m) Rock mass grading

Very wide >3 SolidWide 1–3 MassiveModerately close 0.3–1 Blocky/seamyClose 0.05–0.3 FracturedVery close <0.05 Crushed and shattered

L = 38 cm

L = 17 cm

L = 0 cm

L = 43 cm

L = 0 cmNo recovery

Mechanical breakcaused by drillingprocess

RQD(rock qualitydesignation)

0%–25%25%–50%50%–75%75%–90%

90%–100%

Description of rock qualityVery poorPoorFairGoodExcellent

RQD =

Length ofcore pieces >10 cm (4″)Total core run length × 100%

RQD =38 + 17 + 20 + 43

200 × 100%

RQD = 59% (FAIR)

Core

run

tota

l len

gth

= 20

0 cm

L = 0 cm as nocentrelinepieces longerthan 10 cm

L = 20 cm

Figure 12.4 Method of assessing RQD value from rock core. (After Deere, D.U. and Deere, D.W. 1988. Rock Classification Systems for Engineering Purposes, Ed. Kirkaldie, L., ASTM Special Publication No. 984, pp. 91–101. Philadelphia: American Society for Testing of Materials. Reprinted with per-mission from ASTM STP 984 Rock Classification Systems for Engineering Purposes, Copyright ASTM International, 100 Barr Harbor Drive, West Conshohocken, PA 19428.)


12.4.4 Classification of individual parameters used in the NGI tunnelling quality index

The Norwegian Geotechnical Institute (NGI) has developed a system of classifying rock based on several factors that include the RQD, jointing set numbers, joint roughness, joint alteration, joint water, and stress reduction. The effect of these factors is quantified by assigning numerical values. The factors are listed in Appendix 12A and are defined below:

RQD is the rock quality designationJn is the joint set numberJr is the joint roughness numberJa is the joint alteration numberJw is the joint water reduction numberSRF is the stress reduction factor

Three parameters that are crude measurements of the block size RQD/Jn, the inter-block shear strength Jr/Jn, and the active stress Jw/SRF can then be calculated and multiplied together to give the Q index for the rock (see Equation 12.24).

The Q index can be used for design of tunnels and this is discussed more fully in Section 12.6.5.1.

12.4.5 Rock mass rating method

The Council for Scientific and Industrial Research CSIR (South Africa) method proposed originally by Bieniawski (1973) provides a general rock mass rating (RMR) with a ‘score’ between 1 and 100 given to the rock mass.

The RMR is based on five rock parameters and one parameter depending on use. These parameters are

1. The strength of the intact rock from UCS or point load test Is

2. The quality of the drill core RQD 3. The groundwater conditions 4. The spacing of joints or fractures 5. The condition of joints 6. Joint orientations relative to the tunnel or foundation or slope

The ratings are given in Appendix 12B from an updated scheme by Bieniawski (1989). Use of the method for tunnel design is given in Section 12.6.5.2.

Table 12.7 Description of rock mass based on RQD

Description of rock mass RQD Ratio Efield/Elab

Very poor 0–25 0.2Poor 25–50 0.2Fair 50–75 0.2–0.5Good 75–90 0.5–0.8Excellent 90–100 0.8–1.0


12.5 PLANES OF WEAKNESS

Planes of weakness include fissures, joints, shear zones formed by interlayer slip during folding, and faults which are major dislocations caused by tectonic forces. Such planes cause the rock mass to be weaker than the intact rock and it may be more deformable and aniso-tropic because of the jointing. In addition, the rock may be highly permeable parallel to the planes of weakness.

It is rare to find planar weaknesses distributed in a truly random pattern, and often planes of weakness are oriented in one or more preferred directions. The orientation of joints can be measured by the dip and the dip direction as shown in Figure 12.5.

12.5.1 Stereographic projections

In order to be able to plot the directions of all planes of weakness in a rock mass, stereographic projection is often used. The idea is to take a sphere and allow the plane of weakness to cut through the sphere (also passing through the centre of the sphere).

There are different types of projections that are then used to obtain a projection of the great circle where the plane intersects the sphere. The trace of the great circle can be plotted or, to make the plot simpler, the pole of the plane can be plotted. The pole is a point that shows where the normal to the plane of weakness would cut the sphere. Hence, the pole is just a single point rather than an arc such as the trace of the great circle.

Shown in Figure 12.6 is a plane of weakness cutting through a sphere. The very top point of the sphere is the zenith, and if lines are drawn from the zenith to the great circle (where the plane cuts the sphere) then the trace of this on the horizontal plane through the centre of the sphere gives the stereographic projection of the great circle. This particular type of projection is called an equal angle projection.

The normal to the plane of weakness through the centre of the sphere will cut the sphere at the pole. The pole can be projected onto the horizontal plane as well by joining the zenith to the pole and plotting where it cuts the horizontal plane.

A stereonet can be produced (as shown in Figure 12.7) and the poles or great circles plot-ted by hand on the net. For example, for the net shown in Figure 12.7, if the dip direction was 120° and the dip was 50°, then the great circle and pole can be plotted as shown in the figure. The dip direction is found by rotating clockwise from the north position as shown in Figure 12.7a. Then the dip angle is found by moving in from the circumference of the circle

Plane of interest

Dip

StrikeN

Dip direction

Figure 12.5 Dip and dip direction of a plane of weakness.


to the given angle of dip, that is, 50° as shown in Figure 12.7b. By using tracing paper, the great circle can be copied and then rotated to the 120° position as shown in Figure 12.7c.

Because the pole is found from a perpendicular to the plane of weakness, the dip direction of the pole will plot at 180° away from the dip direction of the plane of weakness. The dip of the pole will be (90° – dip angle) of the plane of weakness.

Computer programs that can be downloaded from the Internet such as GEOrient (2014) or DIPS (2014) are available to perform the plotting and it is not necessary to plot the jointing data by hand on a stereonet any longer. Figure 12.8 shows the equal angle plots of two sets of joint data. Because the joints are in a similar direction, the traces of the planes can be seen to fall into two groups. The plots of the poles in Figure 12.8 also shows that the poles fall into two regions, and such groupings indicate that there are probably two joint sets in the rock.

Stereonets are very useful for finding intersections between planes. For instance, they can be used to find the dip and dip direction of the line of intersection of two planes of weakness. Such information is useful when examining the stability of rock wedges (see Section 12.7.2).

Pole

Zenith

Stereographicprojection of pole

Stereographicprojection of great circle

Great circlePlane ofweakness

Figure 12.6 Plane of weakness (shaded) cutting sphere and equal angle projection onto a horizontal plane.

EW

S

Pole 90°

Great circle

N(a) (b) (c)

Greatcircle

Pole

α = 120°β = 50° β = 50°

Figure 12.7 Drawing a great circle (and pole) manually on a stereonet where the dip direction is 120° and the dip is 50°. (a) Dip direction; (b) dip angle; (c) rotation of great circle to dip direction.


This can be done manually by using a stereonet and tracing paper, but this is an outdated method, and the intersection data can easily be found using computer programs such as those listed above.

12.5.2 Roughness of joints

The roughness of joints has an effect on the shearing resistance of the joint. Smooth, slick-ensided joints will slip easily, while joints that are rough with a lot of interlocking will be most resistant to shearing.

Barton (1973) suggested a method of finding the shear resistance τ of a joint based on the joint roughness coefficient (JRC). This was expressed in terms of an empirical Equation 12.14.

τσ σ

φ′

=′

+ ′

n nrJRC

JCStan log10

(12.14)

In this equation, JCS is the joint wall compressive strength, ′φr is the residual angle of friction of the joint, and ′σn is the normal stress acting on the joint. The equation is therefore similar to the Mohr–Coulomb criterion where the angle of friction is the residual angle increased by the joint roughness term (the first term in the square brackets of Equation 12.14).

The joint roughness coefficient ranges from 0 for very smooth joints to 20 for very rough joints. A means of estimating what the joint roughness is can be seen in Figure 12.9 where a section of joint 10 cm long is depicted along with the JRC to be assigned to that roughness (Barton and Choubey 1977).

The JCS is equal to the unconfined compressive strength of the rock if the joint is unweathered, but less if it is weathered. The value for weathered joints can be found by using a Schmidt hammer on the joint surface (see Barton and Choubey 1977).

(a) (b)

Figure 12.8 Plots showing (a) projections of planes; (b) plots of the associated poles for two joint sets. (Adapted from GEOrient. 2014. V9.x http://www.holcombe.net.au.)


12.6 UNDERGROUND EXCAVATION

Underground excavations may be carried out for highway, railway, water supply, and sew-age tunnels as well as tunnels for mine access and hydro-power. Chambers may be exca-vated for hydro machine halls, mine plant rooms, drawpoints in mines, or for radioactive waste disposal.

Rock mechanics can be used in planning the dimensions, shapes, and orientations of the openings. It can also be used in the design of the support systems, the planning of blasting and excavation operations, and in monitoring the excavation process.

12.6.1 Support systems

If the stresses in the rock are too large, then there may be failure in the rock surrounding the excavation. This can manifest itself as rock bursts and spalling of the walls and floor. Even if there is no failure in the rock, movements may be too large and may increase with time (creep).

The rock may therefore need to be supported by bolting or by linings. If the rock is allowed to move and stabilise before the support systems are put in place, then this may lead to more economic design as the support systems do not have to deform as much and therefore carry as much load.

Design of support systems is examined in Section 12.6.6.

JRC = 0–2

JRC = 2–4

JRC = 4–6

JRC = 12–14

JRC = 8–10

JRC = 10–12

JRC = 6–8

5 cm 10 0

JRC = 14–16

JRC = 16–18

JRC = 18–20

Figure 12.9 Joint roughness profiles. (After Barton, N.R. and Choubey, V. 1977. Rock Mechanics, Vol. 10, No. 1–2, pp. 1–54.)


12.6.2 Design process

The design process for excavations can involve performing two- or three-dimensional stress analysis, and this commonly involves finite element or finite difference (FLAC) analysis. This can be used as a preliminary analysis to refine the shapes and spacings of tunnels. Joints can be included either as major discontinuities or as ‘ubiquitous joints’ in numerical analyses.

The major driving force for the analysis is the initial stress state used, as it is the gravity stresses and stresses locked into the rock that cause the rock to move when it is unloaded by excavation.

If only loosening of the rock is expected from the initial analyses, then the primary support system can be based on experience, that is, empirical methods based on rock classification schemes. However, if the analysis indicates the strength of the rock being exceeded then the support systems need to be placed rapidly after excavation. Design can still be performed using empirical means in this case, but more sophisticated numerical methods may need to be used.

12.6.3 In situ stresses

The behaviour of excavations and tunnels is largely driven by the stresses being released as the rock is removed. If there are very high lateral stresses, a tunnel will be squeezed inwards (along the horizontal diameter) when the rock is excavated.

In some cases, any high lateral stresses can be of benefit to the engineer when a tunnel is being excavated near the surface. In this case, the lateral stresses tend to compress the rock in the roof of the tunnel, thus holding any blocks of rock in place. For example, the roof of the Opera House car park in Sydney is comprised of 8–9 m of rock spanning over 17.5–19 m (Pells et al. 1991). This is possible because the stress field is such that the lateral stress is 2–5 times as large as the vertical stress, that is, σh = 2–5σv.

The stress in rocks can be measured by using overcoring methods (see Goodman 1989). The process involves drilling a borehole into the rock and then installing a strain gauged device into a smaller pilot hole at the base of the borehole. The smaller hole is then over-cored, thus releasing the stresses in the rock as shown in Figure 12.10. Strain gauges in

Overcoring

Pilot hole

Main hole

Figure 12.10 Overcoring stress measurement.


the device record movements that can be used to calculate the stress released. The CSIRO HI cell is a device used in a pilot hole (Worotnicki and Walton 1976). The South African CSIR doorstopper gauge system (Thompson et al. 1997) is slightly different as the gauge is attached to the base of the drill hole before overcoring.

Another way to measure stresses in rocks is to use hydraulic fracturing. A section of borehole is isolated by packers, and this section is subjected to hydraulic pressure until the rock fractures at the fracture initiation pressure pf. The pressure in the borehole is then reduced and reapplied over several cycles. The fracture re-opening pressure pr is the pressure required to open the fracture on re-pressurising (see Figure 12.11). It is assumed that the minimum horizontal principal stress σh is equal to the re-opening pressure, that is,

σh rp= (12.15)

This is then used (Hubbert and Willis 1957) in calculating the maximum horizontal stress in the ground σH through use of Equation 12.16 (assuming that the tensile stress in the rock is almost zero).

p p p pr h H− = − − −0 0 03( () )σ σ (12.16)

wherepr is the fracture re-opening pressureσh is the minor lateral pressure in the rockσH is the major lateral stress in the rockp0 is the groundwater pressure

Details of the test are given in ASTM standard D4645-08 (2008).The direction of the fracture in the rock can be found by using an impression packer,

which is a partially cured thin rubber sleeve that is pressurised against the side of the borehole to obtain an imprint of the fracture.

TimeFlow

rate

Pres

sure

Cycle 1 Cycle 2 Cycle 3

Shut-inFormation porepressure

Pf = Fracture initiation pressurePs = Shut-in pressure Shut-in

Drillhole

Drillhole

Straddlepacker

ImpressionpackerFracture

interval ≈0.9 m

≈1.1 m

Pressuretransducerhousing

Highpressurehoses

To pump

To pump, flowmeter,pressure transducers To pump

Drill rod

Compass

≈1.2 m

Pr = Fracture re-opening pressure

Ps = Shut-in pressure

Figure 12.11 Hydraulic fracturing in borehole.


12.6.4 Stresses around underground openings

Analytic solutions to the stresses caused by tunnel excavation can be found in the book by Poulos and Davis (1974) and in Hoek and Brown (1980a), and some of these are reproduced here.

12.6.4.1 Circular tunnel

For a circular tunnel in an infinite mass under conditions of plane strain, the stresses can be calculated for a stress field that involves a vertical stress pz and a lateral stress px = Npz The radial σr tangential σθ and shear τrθ stresses as shown in Figure 12.12 are given separately for each of the vertical and lateral stresses, but can be superimposed if both are acting. The stresses due to each of the vertical and lateral stresses can be scaled and added also because the solution is based on linear elasticity theory.

Stresses due to uniform vertical pressure pz

σ θ

σθ

rz z

z

p ar

p ar

ar

p ar

= −

+ + −

= +

21

21

3 42

21

2

2

4

4

2

2

2

2

cos

− +

= − +

p ar

p ar

ar

z

rz

21

32

21

3 2

4

4

4

4

2

2

cos

sin

θ

τ θ 22θ

(12.17)

Stresses due to uniform lateral pressure px

σ θ

σθ

rp a

rp a

rar

p ar

= −

− + −

= +

x x

x

21

21

3 42

21

2

2

4

4

2

2

2

2

cos

+ +

= − +

p ar

p ar

ar

r

x

x

21

32

21

3 2

4

4

4

4

2

2

cos

sin

θ

τ θ 22θ

(12.18)

a θr

x

z

σr

τθrσθτrθ

pz

px = Npz

Figure 12.12 Circular tunnel in an infinite mass.


It may be noted that at the boundary (r = a) the stresses become

σ τσ θ

θ

θ

r r

zp N N

= == + − −

0 0

1 2 1 2

,

( ) ( )cos (12.19)

At the floor and roof (θ = 0° and 180°)

σθ = − p Nz 3 1

(12.20)

and at the sidewalls θ = 90° and 270°

σθ = − p Nz 3

(12.21)

Other observations are that there are zones of tensile stress if N N< / or >1 3 3 and that the stress concentrations reduce rapidly away from the opening as they vary with a2/r2, and at values of r/a ≈ 3 the stress field is not very different to the original stress field.

12.6.4.2 Elliptical tunnel

For an elliptical tunnel under the action of the stress state shown in Figure 12.13, the stresses in the tangential direction at the points A and C are given by

σA zp

ac

N= + −

1 2

(12.22)

σC zp N

ca

= +

−

1 2 1

(12.23)

12.6.5 Support design

Many types of direct support are available including props, arches, rock bolts, and precast concrete segments. Different support systems are used in mining to civil engineering applications, as support does not need to be permanent.

Empirical systems have been developed from years of experience in tunnelling and mining, and the design methods are based on various rock classification schemes.

za a

cc

x

z

x0A

Cα1

α2

π2

β1 β2σα

σβα0

β = 0

β =

pz

px = Npz

Figure 12.13 Elliptical tunnel in an infinite mass.


Table 12.7a Stresses on axes of elliptical tunnel (a/c = 0.5)

z/a or x/a

pz = 1.0 px = 0 pz = 0 px = 1.0 pz = 1.0 px = 0.25

σv σh σv σh σv σh

Stresses along the z-axis2.00 0 −1.000 0 5.000 0 0.2502.03 −0.042 −0.790 0.254 4.423 0.021 0.3162.06 −0.565 −0.636 0.421 3.974 0.049 0.3572.12 −0.039 −0.413 0.630 3.281 0.118 0.4082.35 0.155 −0.102 0.746 2.155 0.341 0.4372.88 0.491 0.017 0.547 1.470 0.628 0.3854.67 0.828 0.019 0.192 1.114 0.876 0.2987.89 0.943 0.008 0.065 1.035 0.959 0.266∞ 1.000 0 0 1.000 1.000 0.250

Stresses along the x-axis1.00 2.000 0 −1.000 0 1.750 01.06 1.925 0.028 −0.900 −0.013 1.700 0.0251.11 1.859 0.069 −0.814 −0.021 1.656 0.0641.23 1.747 0.092 −0.657 −0.026 1.583 0.0851.59 1.476 0.166 −0.311 0.023 1.398 0.1722.30 1.219 0.188 −0.030 0.234 1.211 0.2474.33 1.043 0.099 0.055 0.660 1.057 0.2647.70 1.010 0.038 0.028 0.875 1.017 0.257∞ 1.000 0 0 1.000 1.000 0.250

Table 12.7b Stresses on axes of elliptical tunnel (a/c = 2.0)

z/a or x/a

pz = 1.0 px= 0 Pz = 0 px = 1.0 pz = 1.0 px= 0.25

σv σh σv σh σv σh

Stresses along the z-axis0.50 0 −1.000 0 2.000 0 −0.5000.53 −0.013 −0.900 0.028 1.925 −0.006 −0.4190.56 −0.021 −0.814 0.069 1.859 −0.003 −0.3490.61 −0.026 −0.657 0.092 1.747 −0.003 −0.2210.80 0.023 −0.311 0.166 1.476 0.065 0.0581.15 0.234 −0.030 0.188 1.219 0.281 0.2742.1 0.660 0.055 0.099 1.043 0.685 0.3163.85 0.875 0.028 0.038 1.010 0.885 0.280∞ 1.000 0 0 1.000 1.000 0.250

Stresses along the x-axis1.00 5.000 0 −1.00 0 4.750 01.01 4.423 0.254 −0.790 −0.042 4.226 0.2431.03 3.974 0.421 −0.636 −0.057 3.815 0.4071.06 3.281 0.630 −0.413 −0.039 3.178 0.6211.18 2.155 0.746 −0.102 0.155 2.129 0.7851.44 1.470 0.547 0.017 0.491 1.474 0.6702.33 1.114 0.192 0.019 0.828 1.119 0.3993.94 1.035 0.065 0.008 0.943 1.036 0.301∞ 1.000 0 0 1.000 1.000 0.250


12.6.5.1 Q index method

Several rock properties, that is, the J factors and RQD discussed in Sections 12.4.3 and 12.4.4 need to be estimated to calculate the Q index. Once the factors have been estimated from the tables of Appendix 10A, the rock mass quality Q can be calculated from

Q

RQDJ

JJ

JSRFn

r

a

w= × ×

(12.24)

Another quantity called the excavation support ratio (ESR) can be introduced which is related to the intended use and the importance of a tunnel. Barton et al. (1974) apply the values suggested in Table 12.8.

These values of Q and ESR can be used with a chart such as that from Grimstad and Barton (1993) (Figure 12.14) to make estimates of the bolt spacing and the thickness of shotcrete for the rock support system of a tunnel.

Figure 12.14 also gives rock bolt lengths L for an ESR of 1. However, if the ESR is other than 1, Barton et al. (1980) provide additional information on rock bolt length L as given by Equation 12.25, where B is the excavation width.

L

BESR

= +20 15.

(12.25)

They also give the formula of Equation 12.26 for estimating the unsupported tunnel span

Unsupported span = ⋅ ⋅2 0 4ESR Q .

(12.26)

The estimated permanent roof support pressure Proof can also be found in terms of the NGI Q index factors as shown in Equation 12.27.

P

J Q

Jn

rroof

( / )

=−2

3

1 3

(12.27)

Although these systems provide a practical method of designing tunnel support, they have their limitations and should be used with caution. Palmstrom and Brock (2006) have discussed the limitations of the system and Pells and Bertuzzi (2007) have pointed out that in the Sydney sandstone, the Q method suggests less support than has actually been used on major projects (as designed by more sophisticated methods).

Table 12.8 The various excavation support ratio (ESR) categories

Class Description ESR

A Temporary mine openings 3–5B Permanent mine openings, water tunnels for hydro-power (excluding high pressure penstocks),

pilot tunnels, drifts, and headings for large excavations1.6

C Storage rooms, water treatment plants, minor road and railway tunnels, surge chambers, and access tunnels

1.3

D Power stations, major road and railway tunnels, civil defence chambers, and portal intersections 1.0E Underground nuclear power stations, railway stations, sports and public facilities, and factories 0.8

Source: Adapted from Barton N., Lien R., and Lunde J. 1974. Rock Mechanics, Vol. 6, No. 4, pp. 189–236.


12.6.5.2 RMR method

RMR (Section 12.4.5) can also be used to design tunnel openings in rock. Table 12.9 from Bieniawski (1989) provides a guide to the support that may be needed for tunnels in rock for which the RMR has been calculated. Steel sets and shotcrete and rock bolts that are sug-gested as support systems are discussed in Section 12.6.7.

12.6.6 Support types

Where the rock is not of sufficient quality to support itself or the tunnel spans are wide, linings are used to support the tunnel. Linings can be concrete, made of steel sets or be shotcrete used in conjunction with rock bolts. Rock bolts alone can be used if no shot-crete is deemed necessary to support the rock. Steel sets used as support are shown in Figure 12.15.

In cases where the rock tends to squeeze into the opening, it is of advantage to delay the placing of the lining as this will allow the rock to carry some of the stresses released on excavation and results in less load on the lining. In Austria, innovative linings that contain a gap that can close as the rock squeezes have been used. Metal cylinders that can collapse are placed in the gap to control the gradual closing of the lining (Schubert 2007).

Reinforcement categories(1) Unsupported bolting (5) Fibre reinforced shotcrete, 50–90 mm, and bolting(2) Spot bolting (6) Fibre reinforced shotcrete, 90–120 mm, and bolting(3) Systematic bolting (7) Fibre reinforced shotcrete,120–150 mm, and bolting(4) Systematic bolting with 40–100 mm unreinforced shotcrete

(8) Fibre reinforced shotcrete, >150 mm with reinforced ribs of shotcrete and bolting(9) Cast concrete lining

0.001

G F E D C B ARock classes

Exceptionallypoor

Bolt spacing in shotcreted area

Bolt spacing in unshotcreted area

Extremelypoor

Verypoor Poor Fair Good

Verygood

Extremelygood

Excep.good

0.004 0.01 0.04 0.1 0.4 1 4 10 40 4001001

235

10

Span

or h

eigh

t in

met

res

ESR

20

50

Rock mass quality Q = × ×RQDJn

JrJa

Bolt

leng

th in

m fo

r ESR

= 1

1.5

2.4

3

5711

20

1 m1.2 m

1.3 m

1.6 m2.0 m

3.0 m4.0 m

1.5 m1.7 m 2.1 m 2.3 m 2.5 m

1234569 8 7

1.3 m1.0 m

100

1000

250 mm

150 mm

120 mm

90 mm

50 m

m

40 m

m

JwSRF

Figure 12.14 Estimated support categories based on the tunnelling quality index Q. (After Grimstad, E. and Barton, N. 1993. Proceedings of the International Symposium on Sprayed Concrete – Modern Use of Wet Mix Sprayed Concrete for Underground Support, Fagernes, Eds. Kompen, Opsahl and Berg, Norwegian Concrete Association, Oslo, pp. 46–66.)


Table 12.9 Guidelines for excavation and support of 10 m span rock tunnels in accordance with the RMR system

Rock mass class Excavation

Rock bolts (20 mm diameter,

fully grouted) Shotcrete Steel sets

I – Very good rock RMR: 81–100

Full face 3 m advance Generally no support required except spot bolting

II – Good rock RMR: 61–80

Full face1–1.5 m advance Complete support 20 m from face

Locally, bolts in crown 3 m long, spaced 2.5 m with occasional wire mesh

50 mm in crown where required

None

III – Fair rock RMR: 41–60

Top heading and bench 1.5–3 m advance in top heading

Commence support after each blast

Complete support 10 m from face

Systematic bolts 4 m long, spaced 1.5– 2 m in crown and walls with wire mesh in crown

50–100 mm in crown and 30 mm in sides

None

IV – Poor rock RMR: 21–40

Top heading and bench 1.0–1.5 m advance in top heading. Install support concurrently with excavation, 10 m from face

Systematic bolts 4–5 m long, spaced 1–1.5 m in crown and walls with wire mesh

100–150 mm in crown and 100 mm in sides

Light to medium ribs spaced 1.5 m where required

V – Very poor rock RMR: <20

Multiple drifts 0.5–1.5 m advance in top heading Install support concurrently with excavation. Shotcrete as soon as possible after blasting

Systematic bolts 5–6 m long, spaced 1–1.5 m in crown and walls with wire mesh. Bolt invert

150–200 mm in crown, 150 mm in sides, and 50 mm on face

Medium to heavy ribs spaced 0.75 m with steel lagging and fore-poling if required. Close invert

Source: Adapted from Bieniawski, Z.T. 1989. Engineering Rock Mass Classifications. Wiley, New York.

Figure 12.15 Steel sets used to support a tunnel roof.


Support can be primary or secondary support. Primary support is applied immediately after excavation to make the tunnel safe for subsequent excavation, after which a more permanent secondary support system is constructed.

12.6.7 Rock bolts and shotcrete

The support for tunnels can be through rock bolts or by rock bolts and shotcrete lining. For permanent bolts in civil structures such as railway or road tunnels, the bolts need to be protected from corrosion (if steel) or they can be made from stainless steel. Carbon steel bolts in a plastic sheath and grouted can last for more than 100 years without corrosion of the steel.

The bolts serve to tie any rock wedges back to the rock behind the tunnel walls and roof, and are usually tensioned so as to compress the rock and increase the shear resistance of potential sliding blocks. Various types of bolts are described in the book by Brady and Brown (2006).

Bolts may be made of fibreglass, stainless steel, or black steel that can be epoxy or resin coated. The steel bolts are placed inside a plastic sheath and grout is pumped into the annu-lus between the bolt and the plastic sheath and between the region outside the plastic sheath and the rock. A section through a grouted rock bolt is shown in Figure 12.16, and a drawing of a grouted bolt is shown in Figure 12.17.

Shotcrete is applied by spraying the cement, sand, and fine gravel mixture onto the surface of the rock. A steel mesh can be placed first to act as reinforcement for the shotcrete, or steel or propylene fibres can be incorporated in the shotcrete mixture. Often admixtures are added, such as air entraining agents, set accelerators or retarders, and water reducers, to obtain the required properties of the grout.

The shotcrete is sprayed using compressed air, and this may be done wet or dry. In the dry process, the water is added at the shotcreting nozzle, whereas in the wet process, the grout already has the water added and is pumped wet. The dry process has the advantage that the dry mix can be conveyed over longer distances.

12.7 ROCK SLOPES

Slopes are cut in rock for civil works such as roads, railways, and canals or for spillways, and foundations for dams. Open pit mines also require stable slopes that are cut to access

Figure 12.16 Rock bolt section showing plastic sheathing and surrounding grout.


the ore body. For mining applications, the slopes need to be as steep as possible so that the volume of rock removed is not large while still being safe. Often slopes are steeper than they would be for civil applications, but monitoring allows warning of any impending failures.

The stability of slopes in rock is often dictated by the planes of weakness in the rock. Modes of failure can include

1. Planar sliding 2. Wedge sliding 3. Block toppling 4. Circular slip, in highly fractured or highly weathered rocks

12.7.1 Planar sliding

If a rock slide can be taken as being a two-dimensional planar slide, then the forces acting on the wedge of sliding rock are as shown in Figure 12.18. The weight of the wedge of rock is W and the water forces acting on the wedge along the plane of weakness is U and from any water in a crack behind the wedge is V.

Plastic sheath 36 mmdiameter and min2 mm thick

Extra high strengthdeformed bar of24 mm diameter

150 mm diameterstainless steel plate

Water seal

Steel nut

Hemispherical dome and groutinjection hole

Cement groutinjection

45 mm diameter hole(nominally)

Isolation washer

300

mm

max

mec

hani

cal o

rre

sin gr

out a

ncho

rage

50 mm min cover

Embe

dded

leng

th va

ries

Mechanical anchor

Figure 12.17 Typical rock bolt used in tunnel support.


By resolving forces, we can calculate the force normal to the plane of sliding as

N W U V= − −cos sinα α (12.28)

and the force acting parallel to the plane of sliding

T W V= +sin cosα α (12.29)

The factor of safety F can then be calculated from the force acting parallel to the plane of sliding divided by the force resisting sliding (Equation 12.30)

F

c NT

c W U VW V

= ′ + ′ = ′ + − − ′+

tan ( cos sin ) tansin cos

φ α α φα α

(12.30)

where the angle of shearing resistance on the plane of sliding is ϕ′. As well, the forces in the above equation can be calculated from

= −= −

=

( ) sin

. ( ) sin

.

H z

U z H z

V z

w w

w w

/

/

αγ α

γ

0 5

0 5 2

(12.31)

If the vertical joint is behind the crest of the slope

W HzH

= −

−

0 5 12

. γ α β2 cot cot

(12.32)

Tension crack in slope face

Failure surface

Slopeface

Failure surface

Slopeface

Tension crackbehind slope

U

W

l

l

V

W

UVH

H

α

α

β

β

zw

zw

z

z

Figure 12.18 Planar failure of a rock wedge.


If the vertical crack is on the slope face

W HzH

= −

−

0 5 1 122

. cot (cot tan )γ α α β

(12.33)

where the plane of weakness is at the angle α to the horizontal and the slope face is at an angle β to the horizontal.

12.7.2 Wedge failure

A wedge of rock may slide out of a slope and if the slide occurs on two planes as shown in Figure 12.19, then the factor of safety may be calculated through the use of charts. The angle of shearing resistance on each of the two planes of sliding can be different; in Figure 12.19, these are called ϕA on plane A and ϕB on plane B. The factor of safety for the wedge (considering friction only) can then be calculated from Equation 12.34.

F A BA B= +tan tanφ φ (12.34)

where the coefficients A and B may be found from charts (Hoek and Bray 1981; Wyllie and Mah 2004). The values of the dip direction and the dip of each of the planes need to be known before the charts can be used to find A and B.

The charts are presented for various differences in the dip of the two planes of sliding, and an example for a dip difference of 0° is given in Figure 12.20. Further charts for different dip differences are provided in Appendix 12C.

12.8 FOUNDATIONS ON ROCK

Foundations may be constructed on the surface of rock or they may be socketed into the rock. Drilled or socketed foundations may be used to resist downward or upward forces or large lateral forces, so it is of interest to be able to make estimates of the ultimate load that the sockets can support as well as what the expected deflections of the foundation may be.

Plane B

Plane A

Note: The flatter of the twoplanes is always called plane A

Factor of safety FF = A · tan ϕA + B · tan ϕB

Figure 12.19 Wedge failure in a rock slope.


12.8.1 Surface foundations

For surface foundations on rock, the allowable vertical bearing pressure can be found from

q q Kba u sp=

(12.35)

where qu is the UCS of the rock.The coefficient Ksp can be found from the plot shown in Figure 12.21 (Canadian

Geotechnical Society 2006), where it may be seen that the coefficient depends on the spac-ing of discontinuities c and the aperture of the discontinuities δ. The factor Ksp contains a factor of safety of 3 against the lower bound bearing capacity.

Carter and Kulhawy (1988) have suggested a method for obtaining the ultimate capacity through the use of the Hoek and Brown strength criterion

q s m s s qb uult = + +( )

0 5.

(12.36)

where m and s are the coefficients in the Hoek–Brown failure criterion for the rock mass (see Section 12.3).

12.8.2 Shafts in rock

Shafts constructed in rock gain their bearing resistance from end bearing and from friction on the shaft. In order to get good resistance, the base of the shaft should be clean and free of debris, and the sides of the shaft should not be smooth or caked with drilling mud.

A/B chart – dip difference 0°

0 20 40 60 80 100 120 140 160 1800

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

A o

r B

360 340 320 300 280 260 240 220 200

20

3040

5060 70

80

Difference in dip direction (degrees)

Dip of plane (degrees)

Figure 12.20 Chart for rock wedge stability (dip difference 0°). See Appendix 12C for a complete set of charts.


12.8.2.1 Base resistance

The end resistance can be calculated from an empirical relationship such as that given by Ladanyi and Roy (1971) in Equation 12.37, where the allowable base load qba is calculated by modifying the surface foundation bearing pressure by a factor λ.

q q Kba u sp= λ

(12.37)

where λ = + ≤( . ) .1 0 4 3 4L Bs s/ is a factor to allow for the depth of the foundation, and Ls is the depth and Bs is the diameter or width of the socket. The coefficient Ksp is found from the plot of Figure 12.21 (for footings), and includes a factor of safety of 3. qu is taken as the UCS of concrete in the shaft or the rock, whichever is the lower.

Another formula often used is one such as that in Equation 12.38 that gives the ultimate base load for a socketed pile (Zhang and Einstein 1998).

q K qbu b ult MPa= ( ) ( ).0 5

(12.38)

The constant Kb is determined empirically for any given rock mass, but for the Sydney sandstone it is taken as Kb = 4.8 if qu (the UCS) is in MPa.

12.8.2.2 Shaft resistance

The ultimate side friction of shafts (rock sockets) may be expressed in terms of a similar power law to that for the base bearing capacity, that is,

qp

qp

sf

a

u

a

=

λ0 5.

(12.39)

0.020

Ratio c/B0 0.4 0.8 1.2 1.6 2.0

0.4

0.5

0.3

0.001

0.002

0.005

0.0100.2

0.1

0

Valu

e of K

sp

Ksp =10 1 + 300 δ/c

c = Spacing of discontinuities δ = Aperture of discontinuitiesB = Footing width

Valid for 0.05 < c/B < 2.00 < δ/c < 0.02

δ/c = 0

3 + c/B

Figure 12.21 Bearing pressure coefficient Ksp. (Adapted from Canadian Geotechnical Society. 2006. Canadian Foundation Engineering Manual, 4th ed. 488pp.)


where the expression has been normalised by using the atmospheric pressure pa which can be taken as 100 kPa, and the UCS qu is the lower of the values of the shaft concrete or of the rock.

Values of λ have been given by several authors, and the value depends on whether the shaft has been grooved or whether it is smooth. Shafts can have a groove cut in them by a rotary grooving tool to make the shaft rougher and therefore more resistant to slip.

For conventional drilled shafts with walls that have not been grooved, values of λ have been given as 1.41 (Rowe and Armitage 1984), 0.63–0.94 (Horvath et al. 1983) while Carter and Kulhawy indicate that λ = 0.63 is a conservative value that may be used in design.

For artificially grooved socket walls, Rowe and Armitage (1984) suggest using a value of λ = 1.89 in Equation 12.39.

Williams and Pells (1981) provided a method that took into account the roughness of a shaft and of the discontinuities in the rock. The ultimate shaft friction is found from

q qsf u= αβ

(12.40)

whereα is a factor for strength and includes a roughness effectβ is a factor expressing the effect if discontinuities

The factors α and β may be found from the plots of Figures 12.22 and 12.23.Zhang and Einstein (1998) also give the formula of Equation 12.41 for the ultimate shaft

friction

q K qsf s u= ( ) .0 5(MPa)

(12.41)

where they found for a smooth socket Ks = 0.4 and for a rough socket Ks = 0.8. For the Sydney sandstone, the factor Ks is often taken as being about 0.3 for an unroughened socket. This equation is the same as Equation 12.39, without non-dimensionalisation so units must be specified as MPa.

0.6

0.4

0.2

0

0.8

1.0

1 10 1000.1Unconfined compressive strength (MPa)

Redu

ctio

n fa

ctor

α

×

×

×

××

× × ××××

× ××

Figure 12.22 Factors α for use in Equation 12.40. (After Williams, A.F. and Pells, P.J.N. 1981. Canadian Geotechnical Journal, Vol. 18, No. 4, pp. 502–513.)


12.8.2.3 Lateral capacity

The ultimate lateral resistance of rock sockets has not been as well researched as vertical load capacity, but there are some results that may be used as guidance. Zhang et al. (2000) have suggested that the lateral resistance is a combination of the direct pressure in front of the pier plus a component at the sides of the pier due to shear. This leads to the formula in (Equation 12.42) for the ultimate lateral pressure pult

p pult L= +( )τmax (12.42)

The limit pressure in front of the pile or pier pL can be found from the Hoek–Brown criterion for a rock mass (Equation 12.11), where the minor principal stress is given by the pressure due to the self-weight of the rock, that is, σ3 = γ ′z and z is the depth below the surface. This is shown in Figure 12.24. Hence, we can write the limit pressure (in terms of effective stress) as

p z q m

zq

sL uu

a

= ′ = ′ + ′ +

σ γ γ1

(12.43)

The parameters m, s, and a can be found for the rock mass as outlined in Section 12.3.2.The side shear can be calculated as for the vertical side shear (Equation 12.41) where

Zhang et al. suggest that the factor Ks be taken as 0.2, that is,

τmax = 0 2 0 5. ( ) .qu (12.44)

Carter and Kulhawy (1988) use a similar formula to that of Equation 12.42, but they calculate pL from cavity expansion theory (as the limit pressure of an expanding cylindrical cavity). These authors make the assumption that the lateral resistance increases linearly to a depth of 3 diameters and then remains constant (at greater depth). Hence, they give the following two equations (Equation 12.45) for the maximum lateral force Hult that can be applied to a rock-socketed pier.

0.2

0.4

0.6

0.8

1.0

00 0.2 0.4 0.6 0.8 1.0

Emass/Eintact

Fact

or β

Figure 12.23 Reduction factor β for rock mass stiffness.


Hp L

D L L D

Hp

D p

L

LL

ult max

ult max

for= +

≤

= +

+ +

63

63 2

τ

τ τ( mmax for)( )L D D L D− >3 3

(12.45)

In the above formulae, L is the length of the pier and D is the diameter.

12.8.2.4 Uplift capacity

Shafts in rock may be used as anchors for such things as power transmission cables and structures subjected to wind load or earthquake. The capacity in uplift is different in some aspects to the capacity in compression loading, and so different approaches are taken in design.

In uplift, it is assumed that the base of the pier does not take any load and is able to break away from the base of the hole. In tension sockets also, there is a negative Poisson’s ratio effect, with the shaft of the pier contracting under load (i.e. the diameter becomes smaller). However, this effect is only important for a flexible shaft, and not for rigid shafts. Carter and Kulhawy (1988) have shown that if (Ec/Er)(D/L)2 > 4 (D and L are the diameter and length of the pier and Ec and Er are the elastic moduli of the concrete in the pier and the rock mass, respectively) then the pile can be considered rigid, and the uplift capacity can be calculated by reducing the side resistance by 30% from that of a pier in compression loading.

For uplift in a highly fractured rock mass, it is often assumed that the rock can provide no resistance to the uplift force and that the only resistance is due to the weight of a cone of the rock that is pulled out by the drilled pier foundation.

MH

D

D

zEnlargedrock mass element

Enlargedcross section D

Side shear resistance τmax

Front normalresistance pL

MH

pL = σ1′

γz = σ3′

Figure 12.24 Laterally loaded pier in rock socket.


12.8.2.5 Piles on sandstones and shales of the Sydney region

The design of piles or piers in rock for piles socketed into the shales and sandstone of the Sydney region is based on experience and has been developed for a specific region. It may not be applicable to other locations, but similar design charts have been developed for other cities.

Sockets can be excavated by hand or by drilling and can be constructed dry or under water. The sockets must have clean walls and bases so that there is good contact between the concrete of the pier and the rock. Sockets drilled in moist weathered shale or sandstone can have very smooth sides or crushed rock can be smeared on the walls. The base may also contain debris, so sockets must be inspected and cleaned before concreting.

Inspection can be carried out by descending into the hole in a cage or using a special TV camera. Safety rules apply if personnel descend into the hole in NSW, where breathing apparatus and safety harness must be used.

If sidewalls are not clean, they can be roughened with a roughening tool. The tool can be attached to a drilling rig and can cut grooves in the walls of the socket. Appendix 12D con-tains Tables 12.1 through 12.3 that give the rock class classification and the recommended design values for end bearing pressure and of shaft adhesion for sockets in shale and in sandstone (see Pells et al. 1998). It may be seen that the design pressures depend on defects in the rock as well as rock strength.

12.8.3 Deformation of foundations on rock

The deformation of foundations on rock is often analysed using the theory of elasticity, and for rock, the rock mass stiffness or modulus. The rock mass stiffness depends upon any fractures or fissures in the rock as these will influence the deformation behaviour as they will tend to close upon loading.

Hoek et al. (2002) suggest that the modulus of the rock mass can be estimated from the formula in Equation 12.46.

E

D qm

ui GSI (GPa) = −

−12 100

10 10 40. ([ ] / )

(12.46)

where the GSI and damage factor D have been given in Tables 12.3 and 12.4 previously and qui is the unconfined compressive strength of the intact rock.

12.8.3.1 Vertical deformation

In compression, Carter and Kulhawy (1988) suggest that the compressive deformation yc can be found (for a rigid shaft in contact with the base) from

y

PE D E L

cc

b b m m

=− + +( ( )() )/ / /1 12ν π ζ ν

(12.47)

In the above equation, Pc is the compressive load, L is the socket length, Eb is the modu-lus of the rock below the base of the socketed pier, Em is the rock mass modulus, νb is the Poisson’s ratio below the base of the pier and νm is the Poisson’s ratio of the rock mass. ζ = ln[5(1 − νm)L/D] where D is the diameter or width of the foundation as Em ≈ Eb and ν ≈ νb the equation simplifies to


y

P E DL D

cc m m

m

= −+ −

( ) ( )( )( )( )

11 1

2νπ ζ ν

// /

(12.48)

In uplift, the same authors suggest that the deflection yu of a rigid shaft in a shear socket can be obtained by taking Eb = 0 in Equation 20.47 giving

y

G LPu

mu=

ζπ2

1

(12.49)

where Gm is the shear modulus of the rock mass and Pu is the uplift load.

12.8.3.2 Lateral deformation

Lateral deflection u and rotation θ of flexible piers in rock sockets may be estimated by sim-ple formulae given by Randolph (1981) and used by Carter and Kulhawy (1988) (Equations 12.50 and 12.51).

u

HG D

EG

MG D

EG

e e=

+

− −

0 50 1 081 7

2

3

. .* *

/

* *

//7

(12.50)

θ =

+

− −

1 08 6 402

3 7

3. .* *

/

* *

HG D

EG

MG D

EG

e e55 7/

(12.51)

In the above equations, H is the horizontal load applied, M is the moment applied to the head of the pier, D is the diameter of the pier, and

G Gr

r* = +

134ν

(12.52)

where Gr and νr are the shear modulus and Poisson’s ratio of the rock (Gr = Er/2(1 + νr)) for an isotropic rock mass. The effective Ee Young’s modulus of the concrete shaft is given by

E

EID

ec= ( )

π 4 64/ (12.53)

where I is the second moment of inertia of the pier cross section if the pier is not circular.

12.9 VIBRATION THROUGH ROCK

A common problem encountered when excavating rock is the vibration caused by the equip-ment used for breaking the rock. Measurements of ground vibrations are generally made in terms of peak particle velocities (PPV).


The Australian code for blasting, AS 2187.2-1993, provides a recommended limit for PPV of less than 10 mm/s for residential structures; while the German standard, DIN 4150-3 1999, recommends values of less than 5–20 mm/s depending on the frequency of the vibra-tion (the higher the frequency, the greater the allowable peak particle velocity). Higher PPV values are allowed for industrial buildings; AS2187.2-1993 allows up to 25 mm/s and DIN 4150-3 1999 allows between 20 and 50 mm/s at ground level, again depending on frequency.

Measurements for several different types of machinery have been measured by Wiss (1981) and presented on a plot of PPV to distance using a log–log scale in which the data plotted as a straight line. However, it is not clear which soil or rock types this data was collected for.

Measurements have been taken in Sydney sandstone by Hackney (2002) for different kinds of rock breaking equipment. These were classified into 250–500 kg, 500–1000 kg, 100–1500 kg, and >1500 kg rock hammers (used for breaking rock) as well as rotary rock grinders. The results were also plotted for different classes of sandstone, classes I/II and classes II/III. Sandstone classes have been discussed previously in Section 12.8.2.5 and also in Appendix 12D.

Generally the PPV versus distance from the source of vibration plots as a straight line on a log–log plot as shown in Figure 12.25, but there is a good deal of scatter, so upper and lower bounds to the data are shown as straight lines on the plot.

From the data collected, the plot of Figure 12.26 may be produced showing how far from the source of vibration a PPV of 10 mm/s will be likely to occur for different rock hammer weights. The data is plotted for classes I/II sandstone and classes II/III sandstone (broken line). The data for rotary rock grinders lies at approximately the same location as for 250–500 kg hammers.

12.10 NUMERICAL METHODS

Numerical methods can be used to model the rock mass when designing footings, piers, cut-tings, and tunnels in rock. The models that are used depend on the type of jointing in the rock. When the jointing is spaced widely or there are just a few well-defined joints, finite ele-ment programs can be used where the joints are specifically modelled in a continuum that is used to model the remaining rock mass. Such a model is shown in Figure 12.27 for a tunnel

1

10

100

0.11 10 100

Distance (m)× Hammer > 1500 kg

Peak

par

ticle

velo

city

PPV

(mm

/s)

Figure 12.25 Peak particle velocities versus distance for >1500 kg hammers in class I/II sandstone. (Adapted from Hackney G.A. 2002. Proceedings 5th Australian New Zealand Young Geotechnical Professional Conference, New Zealand.)


in a jointed rock mass where the commercial finite element software Phase2 (Rocscience 2014) has been used.

Joint networks can be generated in finite element analyses to simulate jointing patterns, and software such as Phase2 allows this to be done. Rock bolts and lining systems can also be included in such numerical analyses.

If the rock has several joint sets and the jointing is closely spaced, discrete element modelling can be carried out, where each rock block can be modelled individually. Commercial software such as UDEC (2014) may be used to model the rock behaviour.

When the rock is fairly competent, non-linear finite element analysis may be used, and the rock treated as a continuum. An example of this is shown in Figure 12.28 (Liu et al. 2008) where the finite element software ABACUS (2014) has been used to model the effects of a new tunnel being constructed next to an existing tunnel.

250 to 500 500 to 1000 1000 to1500

>15000

5

10

15

Rock hammer size (kg)

Class I or II sandstoneClass II or III sandstone

Dist

ance

to m

aint

ain

<10

mm

/s P

PV (m

)

Figure 12.26 Distance to cause <10 mm/s peak particle velocity PPV for various rock hammer sizes in dif-ferent classes of sandstone.

0.0015

0.0007

0.0006

0.0006

0.0006

0.0006End of basement excavation – defects, staged tunnel excavation

Defects

0.00060.00

05

0.00

05

0.00

05

0.00

05

–20

5456

5860

–18 –16 –14 –12 –10 –8 –6

0.00

040.0004

0.0013

0.0013

0.0014

0.0014

0.0011

0.0012

0.0000

Totaldisplacementm

Total displacement and deformation vectors (scaled up by 100) at end of basement excavation (stage 8)

0.00020.00040.00060.00080.00100.00130.00150.00170.00190.00210.00230.0025

0.00

13

0.00

14

0.00

15

0.00

15

0.00

15

0.00

Figure 12.27 Specific joints modelled for tunnel excavation by Phase2 finite element software.


Figure 12.28 shows a cutaway view of the two tunnels and the shotcrete lining and rock bolts in the roof and sides of the tunnels. Construction of the new tunnel is modelled by removing elements to simulate excavation, then adding shell elements for the shotcrete and then adding structural elements to model the rock bolts.

APPENDIX 12A

Figure 12.28 Finite element (ABACUS) analysis of two parallel tunnels in sandstone.

Table 12A Classification of individual parameters used in the NGI tunnelling quality index

1 Rock quality designation RQD

A Very poor 0%–25%B Poor 25%–50%C Fair 50%–75%D Good 75%–90%E Excellent 90%–100%

Notes:

1. Where RQD is reported or measured as ≤10, (including 0) a nominal value of 10 is used to evaluate Q.

2. RQD intervals of 5, i.e. 100, 95, 90, etc., are sufficiently accurate.

2 Joint set number JnA Massive, no or few joints 0.5–1.0B One joint set 2C One joint set plus random 3D Two joint sets 4E Two joint sets plus random 6F Three joint sets 9G Three joint sets plus random 12H Four or more joint sets, random,

heavily jointed, ‘sugar-cube’, etc.15

J Crushed rock, earthlike 20

Notes:

1. For intersections use (3.0 × Jn).2. For portals use (2.0 × Jn).

(Continued )


Table 12A (Continued) Classification of individual parameters used in the NGI tunnelling quality index

3 Joint roughness number Jra. Rock wall contact andb. Rock wall contact before 10 cm shear

A Discontinuous joints 4B Rough or irregular, undulating 3C Smooth, undulating 2D Slickensided, undulating 1.5E Rough or irregular, planar 1.5F Smooth, planar 1.0G Slickensided, planar 0.5

c. No rock wall contact when shearedH Zone containing clay minerals thick enough to prevent rock wall contact 1.0J Sandy, gravelly, or crushed zone thick enough to prevent rock wall contact 1.0

Notes:

1. Add 1.0 if the mean spacing of the relevant joint set is greater than 3 m.2. Jr = 0.5 can be used for planar slickensided joints having lineations, provided the lineations are orientated for minimum

strength.

4 Joint alteration number Ja ϕr (approx.)

a. Rock wall contactA Tightly healed, hard, non-softening, impermeable filling, i.e. quartz or epidote 0.75 –B Unaltered joint walls, surface staining only 1.0 (25–35°)C Slightly altered joint walls. Non-softening mineral coatings, sandy

particles, clay-free disintegrated rock, etc.2.0 (25–30°)

D Silty-, or sandy-clay coatings, small clay fraction (non-soft) 3.0 (20–25°)E Softening or low friction clay mineral coatings, i.e. kaolinite or mica. Also

chlorite, talc, gypsum, graphite, etc., and small quantities of swelling clays4.0 (8–16°)

b. Rock wall contact before 10 cm shearF Sandy particles, clay-free disintegrated

rock, etc.4.0 (25–30°)

G Strongly overconsolidated non-softening clay mineral fillings (continuous, but <5 mm thick)

6.0 (16–24°)

H Medium or low overconsolidation softening clay, mineral fillings (continuous but <5 mm thickness)

8.0 (12–16°)

J Swelling-clay fillings, i.e. montmorillonite (continuous, but <5mm thick). Value of Ja depends on percent of swelling clay-size particles, and access to water

8.0–12.0 (6–12°)

c. No rock wall contact when shearedKLM

Zones or bands, of disintegrated or crushed rock and clay (see G, H, J for clay conditions)

6.08.0

8.0–12.0

(6–24°)

N Zones or bands of silty- or sandy clay, small clay fraction (non-softening) 5.0OPR

Thick, continuous zones or bands of clay (see G, H, J for clay conditions) 10.0–13.013.0–20.0

(6–24°)

Notes:

1. Values of ϕr, the residual friction angle, are intended as an approximate guide to the mineralogical properties of the altera-tion products, if present.

(Continued )


Table 12A (Continued) Classification of individual parameters used in the NGI tunnelling quality index

5 Joint water reduction factor JwApprox. water

press. (kgf/cm2)

A Dry excavations or minor inflow, i.e. <5 l/min locally 1.0 <1B Medium inflow or pressure, occasional outwash of joint fillings 0.66 1–2.5C Large inflow or high pressure in competent rock with unfilled joints 0.5 2.5–10D Large inflow or high pressure, considerable outwash of joint fillings 0.33 2.5–10E Exceptionally high inflow or water pressure at blasting, decaying with time 0.2–0.1 >10F Exceptionally high inflow or water pressure continuing without decay 0.1–0.05 >10

Notes:

1. Factors C–F are crude estimates. Increase Jw if drainage measures are installed.2. Special problems caused by ice formation are not considered.

6 Stress reduction factor SRF

a. Weakness zones intersecting excavation, which may cause loosening of rock mass when tunnel is excavatedA Multiple occurrences of weakness zones containing clay or chemically

disintegrated rock, very loose surrounding rock (any depth)10.0

B Single weakness zones containing clay or chemically disintegrated rock (depth of excavation <50 m)

5.0

C Single weakness zones containing clay or chemically disintegrated rock (depth of excavation >50 m)

2.5

D Multiple shear zones in competent rock (clay-free), loose surrounding rock (any depth)

7.5

E Single shear zones in competent rock (clay-free) (depth of excavation < 50 m)

5.0

F Single shear zones in competent rock (clay-free) (depth of excavation >50 m)

2.5

G Loose open joints, heavily jointed or ‘sugar-cube’ (any depth) 5.0

b. Competent rock, rock stress problems

H Low stress, near surface σc/σ1>200

σt/σ1>13

2.5

J Medium stress 200–10 13–0.66 1.0K High stress, very tight structure (usually favourable to stability, may be

unfavourable for wall stability)10–5 0.66–0.33 0.5–2.0

L Mild rock burst (massive rock) 5–2.5 0.33–0.16 5–10M Heavy rock burst (massive rock) <2.5 <0.16 10–20c. Squeezing rock plastic flow of incompetent rock under the influence of high rock pressureN Mild squeezing rock pressure 5–10O Heavy squeezing rock pressure 10–20d. Swelling rock chemical swelling activity depending on the presence of waterP Mild swelling rock pressure 5–10R Heavy swelling rock pressure 10–20

Notes:

1. Reduce these values of SRF by 25%–50% if the relevant shear zones only influence but do not intersect the excavation.2. For strongly anisotropic virgin stress field (if measured): when 5 ≤ σ1/σ3 ≤ 10, reduce σc to 0.8σc and σt to 0.8σt. When σ1/σ3 > 10, reduce σc and σt to 0.6σc and 0.6σt, where, σc = unconfined compression strength, and σt = tensile

strength (point load), and σ1 and σ3 are the major and minor principal stresses.3. Few case records available where depth of crown below surface is less than span width. Suggest SRF increase from 2.5 to

5 for such cases (see H).


ADDITIONAL NOTES ON THE USE OF THESE TABLES

When a making estimates of the rock quality (Q) the following guidelines should be fol-lowed, in addition to the notes listed in the tables.

1. When a borehole core is unavailable, RQD can be estimated from the number of joints per unit volume, in which the number of joints per metre for each joint set are added. A simple relation can be used to convert this number to RQD for the case of clay-free rock masses:

RQD = 115–3.3 Jv (approx.) where Jv = total number of joints per m3 (RQD = 100 for Jv < 4.5).

2. The parameter Jn representing the number of joint sets will often be affected by folia-tion, schistosity, slatey cleavage, or bedding. If strongly developed these parallel ‘joints’ should obviously be counted as a complete joint set. However, if there are few ‘joints’ visible, or only occasional breaks in bore core due to these features, then it will be more appropriate to count them as ‘random joints’ when evaluating Jn.

3. The parameters Jr and Ja (representing shear strength) should be relevant to the weakest significant joint set or clay filled discontinuity in the given zone. However, if the joint set or discontinuity with the minimum value of (Jr/Ja) is favourably oriented for stabil-ity, then a second, less favourably orientated joint set or discontinuity may sometimes be of more significance, and its higher value of (Jr/Ja) should be used when evaluating Q. The value of (Jr/Ja) should in fact relate to the surface most likely to allow failure to initiate.

4. When a rock mass contains clay, the factor SRF appropriate to loosening loads should be evaluated. In such cases, the strength of the intact rock is of little interest. However, when jointing is minimal and clay is completely absent the strength of the intact rock may become the weakest link, and the stability will then depend on the ratio rock-stress/rock strength. A strongly anisotropic stress field is unfavourable for stability and is roughly accounted for as in Note 2 in the table for stress reduction factor evaluation.

5. The compressive and tensile strengths (σc and σt) of the intact rock should be evaluated in the saturated condition if this is appropriate to present or future in situ conditions. A very conservative estimate of strength should be made for those rocks that deterio-rate when exposed to moist or saturated conditions.


APPENDIX 12B

Table 12B Rock mass rating system RMR

A. Classification parameters and their ratings

Parameter Range of values

1 Strength of intact rock material

Point-load strength index

>10 MPa 4–10 MPa 2–4 MPa 1–2 MPa For this low range—Uniaxial compressive test is preferred

Uniaxial comp. strength

>250 MPa 100–250 MPa

50–100MPa 25–50 MPa 5–25 MPa

1–5 MPa

<1 MPa

Rating 15 12 7 4 2 1 02 Drill core quality RQD 90%–100% 75%–90% 50%–75% 25%–50% <25%

Rating 20 17 13 8 33 Spacing of discontinuities >2 m 0.6–2 m 200–

600 mm60–200 mm <60 mm

Rating 20 15 10 8 54 Condition of

discontinuities (see E)Very rough surfaces

Not continuous

No separation

Unweathered wall rock

Slightly rough surfaces

Separation < 1 mm

Slightly weathered walls

Slightly rough surfaces

Separation < 1 mm

Highly weathered walls

Slickensided surfaces or

Gouge < 5 mm thick or Separation 1–5 mm

Continuous

Soft gouge > 5 mm thick or

Separation > 5 mmContinuous

Rating 30 25 20 10 05 Groundwater Inflow per

10 m tunnel length (L/m)

None <10 10–25 25–125 >125

(joint water pressure)/(major principal σ)

0 <0.1 0.1–0.2 0.2–0.5 >0.5

General conditions

Completely dry

Damp Wet Dripping Flowing

Rating 15 10 7 4 0

B. Rating adjustments for discontinuity orientations (see F)

Strike and dip orientations Very favourable Favourable Fair Unfavourable Very unfavourable

Ratings Tunnels and mines 0 −2 −5 −10 −12Foundations 0 −2 −7 −15 −25Slopes 0 −5 −25 −50

C. Rock mass classes determined from total ratings

Rating 100 ← 81 80 ← 61 60 ← 41 40 ← 21 <21Class number I II III IV VDescription Very good rock Good rock Fair rock Poor rock Very poor rock

(Continued )


Table 12B (Continued) Rock mass rating system RMR

D. Meaning of rock classes

Class number I II III IV VAverage stand-up time

20 years for 15 m span

1 year for 10 m span

1 week for 5 m span

10 h for 2.5 m span

30 min for 1 m span

Cohesion of rock mass (kPa)

>400 300–400 200–300 100–200 <100

Friction angle of rock mass (deg)

>45 35–45 25–35 15–25 <15

E. Guidelines for classification of discontinuity conditions

Discontinuity length (persistence)

< 1 m 1–3 m 3–10 m 10–20 m >20 m

Rating 6 4 2 1 0Separation (aperture)

None <0.1 mm 0.1–1.0 mm 1–5 mm >5 mm

Rating 6 5 4 1 0Roughness Very rough Rough Slightly rough Smooth SlickensidedRating 6 5 3 1 0Infilling (gouge) None Hard filling

< 5 mmHard filling > 5 mm

Soft filling < 5 mm

Soft filling > 5 mm

Rating 6 4 2 2 0Weathering Unweathered Slightly

weatheredModerately weathered

Highly weathered

Decomposed

Rating 6 5 3 1 0

F. Effect of discontinuity strike and dip orientation in tunnellinga

Strike perpendicular to tunnel axis Strike parallel to tunnel axis

Drive with dip – Dip 45–90°

Drive with dip – Dip 20–45°

Dip 45–90° Dip 20–45°

Very favourable Favourable Very unfavourable FairDrive against dip – Dip 45–90°

Drive against dip – Dip 20–45°

Dip 0–20° Irrespective of strikeb

Fair Unfavourable Fair

Source: Adapted from Bieniawski, Z.T. 1989. Engineering Rock Mass Classifications. Wiley, New York.a Modified after Wickham et al. (1972).b Some conditions are mutually exclusive. For example, if infilling is present, the roughness of the surface will

be overshadowed by the influence of the gouge. In such cases, use A4 directly.


APPENDIX 12C

Table 12C Rock wedge stability factors A and B

3040

5060

7080

20

2030

4060

70 8090

203040

50

0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

B

00 20 40 60 80 100 120 140 160 180

360 340 320 300 280 260 240 220 200Difference in dip direction (degrees)

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

A

B chart – Dip difference 10°A chart – Dip difference 10°

Dip of plane B (degrees)

Dip of plane A (degrees)

20

30

4050

6070

3040

50

8090

60 70

304050

0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

B

0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

A

B chart – Dip difference 20°A chart – Dip difference 20°


Dip of plane A(degrees)

(Continued)


Table 12C (Continued) Rock wedge stability factors A and B

20

3040

5060 4050

7080 90

60405060

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

A

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

B

0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


A chart – Dip difference 30° B chart – Dip difference 30°



20

3040 50

50 60708090

5060

0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200





0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

A

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

B

(Continued)



20

3040

6090

70

80

0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200



0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

A

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

B



20

30

70 80

90

0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

B

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

A




(Continued)


APPENDIX 12D


8090

20

0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200


0 20 40 60 80 100 120 140 160 180360 340 320 300 280 260 240 220 200





0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

B

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Ratio

B

Table 12D Engineering classification of shales and sandstones in the Sydney region – A summary guide (Adapted from Pells, Douglas, Rodway, Thorne, and McMahon, 1978, as revised by Pells et al. 1998.)

The classification system is based on rock strength, defect spacing and allowable seams as set out below. All three factors must be satisfied.

TABLE I Classification for sandstone

ClassUnconfined compressive

strength qu (MPa) Defect spacing Allowable seams

I >24 >600 mm <1.5%II >12 >600 mm <3%III >7 >200 mm <5%IV >2 >60 mm <10%V >1 N.A. N.A.

Classification for shale

ClassUnconfined compressive

strength qu (MPa) Defect spacing Allowable seams

I >16 >600 mm <2%II >7 >200 mm <4%III >2 >60 mm <8%IV >1 >20 mm <25%V >1 N.A. N.A.


DEFECT SPACING

The terms relate to spacing of natural fractures in NMLC, NO, and HO diamond drill cores and have the following definitions:

Defect spacingmm

Terms used to describe defectspacinga

>2000 Very widely spaced600–2000 Widely spaced200–600 Moderately spaced60–200 Closely spaced20–60 Very closely spaced<20 Extremely closely spaceda After ISO/CD 14689. 1995. Geotechnics in Civil Engineering:

Identification and classification of rock ISRM (International Society for Rock Mechanics). 1978. International Journal of Rock Mechanics and Mineral Science, Vol. 5, pp. 316–368.

ALLOWABLE SEAMS

Seams include clay, fragmented, highly weathered or similar zones, usually sub-parallel to the loaded surface. The limits suggested in the tables relate to a defined zone of influence. For pad footings, the zone of influence is defined as 1.5 times the least footing dimension. For socketed footings, the zone includes the length of the socket plus a further depth equal to the width of the footing. For tunnel or excavation assessment purposes, the defects are assessed over a length of core of similar characteristics.

Table III Design values for vertical loading on sandstone

ClassUltimate end

bearinga (MPa)Serviceability end bearing

pressureb (MPa)Ultimate shaft adhesionc (kPa) Typical Efield (MPa)

I >120 12 3000 >2000II 60–120 0.5qu 1500–3000 900–2000

Max. 12III 20–40 0.5qu 800–1500 350–1200

Max. 6IV 4–1 5 0.5qu 250–800 100–700

Max. 3.5V >3 1.0 150 50–100a Ultimate values occur at large settlements (>5% of minimum footing dimensions).b End bearing pressure to cause settlement of <1% of minimum footing dimension.c Clean socket of roughness category R2 or better.

Table II Design values for foundations on shale

ClassUltimate end

bearinga (MPa)Serviceability end bearing

pressureb (MPa)Ultimate shaft adhesionc (kPa) Typical Efield MPa

I >120 Max. 8 1000 >2000II 30–120 0.5qu 600–1000 700–2000

Max. 6III 6–30 0.5qu 350–600 200–1200

Max. 3.5IV >3 1.0 150 100–500V >3 0.7 50–100 50–300a Ultimate values occur at large settlements (>5% of minimum footing dimensions).b End bearing pressure to cause settlement of <1% of minimum footing dimension.c Clean socket of roughness category R2 or better. Values may have to be reduced because of smear.

517

References

Aas, G., Lacasse, S., Lunne, I., and Hoeg, K. 1986. Use of in situ tests for foundation design in clay, Proceedings, In Situ 86, American Society of Civil Engineers, ASCE Geotechnical Special Publication 6, Blacksburg, VA, pp. 1–30.

ABACUS finite element software V6.14. 2014. Dassault Systèmes.Abbott, T.M. and Corless, B.M. 1993. Use of clay and geomembrane liners as base seals for baux-

ite residue disposal, Geotechnical Management of Waste and Contamination, Eds. Fell, R., Phillips, T., and Gerrard, C., Balkema, Rotterdam, pp. 315–322.

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